A relation R on set A = {1, 2, 3} defined as R = {(1, 1), (2, 2), (1, 2)} is
- Reflexive only
- Reflexive and Transitive
- Symmetric and Transitive
- Transitive only
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(D) Transitive only
CBSE Board papers 2022 to 2026, plus CBSE Sample Question Papers. Chapters as per CBSE Curriculum 2026-27. 1325 questions. Click Reveal answer under any question to see the official marking scheme.
A relation R on set A = {1, 2, 3} defined as R = {(1, 1), (2, 2), (1, 2)} is
(D) Transitive only
Assertion (A) : A function f : \mathbb{N} \to \mathbb{N} given by f(x) = x^{3} + 2, \forall\, x \in \mathbb{N} is one-one but not onto.
Reason (R) : Since \forall\, y \in \mathbb{N} (Codomain), there does not exist x = (y - 2)^{1/3} in \mathbb{N} (Domain) such that f(x) = x^{3} + 2 = y.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1]
Assertion (A) : A relation R on the set {1, 2, 3} defined as R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} is an equivalence relation.
Reason (R) : A relation that is reflexive, symmetric and transitive is an equivalence relation.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Assertion (A) : Let R be a relation on set A of all books in a school library given by R = \{(x, y) : x, y \in A, x and y have the same number of pages\} and S is a relation on set A from the same library such that S = \{(p, q) : p and q are Mathematics books\}. Also, R \cap S \ne \phi. Then, R \cap S is an equivalence relation.
Reason (R) : The intersection of two equivalence relations on a set is an equivalence relation on the set.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A) : Let Z be the set of integers. A function f:Z\to Z defined as f(x)=3x-5,\ \forall x\in Z is a bijective.
Reason (R) : A function is a bijective if it is both surjective and injective.
(D) Assertion (A) is false, but Reason (R) is true.
For real x, let f(x)=x^3+5x+1. Then :
(C) f is one-one and onto on R
If f:N\to W is defined as
then f is :
(B) surjective only
Assertion (A) : Let A=\{x\in R:-1\le x\le1\}. If f:A\to A be defined as f(x)=x^2, then f is not an onto function.
Reason (R) : If y=-1\in A, then x=\pm\sqrt{-1}\notin A.
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
Assertion (A) : Let f(x) = e^x and g(x) = \log x. Then (f + g)\, x = e^x + \log x where domain of (f + g) is R.
Reason (R) : \text{Dom}(f + g) = \text{Dom}(f) \cap \text{Dom}(g).
(D) Assertion (A) is false, but Reason (R) is true.
Consider the function f : R \longrightarrow R, defined as f(x) = x^3.
Assertion (A) : f(x) is a one-one function.
Reason (R) : f(x) is a one-one function, if co-domain = range.
(C) Assertion (A) is true, but Reason (R) is false.
A function f : R_+ \to R (where R_+ is the set of all non-negative real numbers) defined by f(x) = 4x + 3 is :
(A) one-one but not onto
Let f : R_+ \to [-5, \infty) be defined as f(x) = 9x^2 + 6x - 5, where R_+ is the set of all non-negative real numbers. Then, f is :
(C) Bijective
Let R_+ denote the set of all non-negative real numbers. Then the function f : R_+ \to R_+ defined as f(x) = x^2 + 1 is :
(A) one-one but not onto
Assertion (A) : The relation R = \{(x, y) : (x + y) \text{ is a prime number and } x, y \in N\} is not a reflexive relation.
Reason (R) : The number ‘2n’ is composite for all natural numbers n.
(C) Assertion (A) is true, but Reason (R) is false.
[1]A function f:\mathbb{R}\to\mathbb{R} defined as f(x)=x^2-4x+5 is :
(D) neither injective nor surjective. [1]
Let A = {3, 5}. Then number of reflexive relations on A is
(B) 4
Check whether f : R - \{3\} \to R defined as f(x) = \frac{x-2}{x-3} is onto or not.
Let y = \frac{x-2}{x-3} i.e. x = \frac{3y-2}{y-1} [1]
Here y \ne 1 \therefore R_f = R - \{1\} \ne \text{Codomain}
\therefore f is not onto. [1]
Check whether f : Z \times Z \to Z \times Z (where Z is the set of integers) defined as f(x, y) = (2y, 3x) is injective or not.
Let (x_1, y_1), (x_2, y_2) \in Z \times Z such that f(x_1, y_1) = f(x_2, y_2)
\Rightarrow (2y_1, 3x_1) = (2y_2, 3x_2)
\Rightarrow y_1 = y_2,\ x_1 = x_2 [1]
So, (x_1, y_1) = (x_2, y_2)
\therefore f is injective. [1]
A relation R on A = \{1, 2, 3\} is defined as R = \{(1, 1)\ (3, 3), (1, 2)\}. Is R a symmetric relation ? Justify. Write the smallest relation set R_1 such that R \cup R_1 becomes an equivalence relation on the set \{1, 2, 3\}.
As (1,2) \in R, but (2,1) \notin R \Rightarrow R is not symmetric relation. [1]
R_1 = \{(2,2), (2,1)\} [1]
Students are trying to fix thumb pins on a graph sheet on the class notice board. A relation R is defined on the set A of thumb pins as
R = {(P, Q) : P, Q \in A and distance between P and Q is less than 3 cm}
Determine if R is an equivalence relation.
Let d(P, Q) denotes distance between any point P and Q
For any point P on set A, the distance from P to itself is zero.
d(P, P) = 0 \lt 3, (P, P) \in R. Therefore, R is reflexive. [½]
Let (P, Q) \in R, then d(P, Q) \lt 3.
As, d(P, Q) = d(Q, P). Thus, d(Q, P) \lt 3, which means (Q, P) \in R.
Therefore, R is symmetric. [½]
Let (P, Q) \in R, so d(P, Q) \lt 3 and (Q, S) \in R so d(Q, S) \lt 3,
then (P, S) \notin R for example
Let P, Q and R be any three different points on the same line such that d(P,Q) = 2 cm and d(Q, R) = 2 cm so d(P, R) = 4 cm >3
Therefore, R is not transitive. Thus, R is not an equivalence relation. [1]
**Note: If a student shows only Transitive part correctly (Without mentioning anything about Reflexive and Symmetric part), then also solution deserve full 2 marks. Alternative example are also acceptable.
Let f:A\to B be defined by f(x)=\frac{x-2}{x-3}, where A=\mathbb{R}-\{3\} and B=\mathbb{R}-\{1\}.
Discuss the bijectivity of the function.
Let x_1,x_2\in A such that f(x_1)=f(x_2)\Rightarrow\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}\Rightarrow x_1=x_2, \therefore ‘f’ is one-one. [1]
For each y\in B, there exists x=\frac{3y-2}{y-1}\in\mathbb{R}-\{3\}, such that f(x)=y, \therefore ‘f’ is onto [½]
\Rightarrow ‘f’ is one-one & onto, or ‘f’ is a bijective function. [½]
A function f:A\to B defined as f(x)=2x is both one-one and onto. If A=\{1,2,3,4\}, then find the set B.
f(1)=2,\ f(2)=4,\ f(3)=6,\ f(4)=8 [1½]
\therefore B=\{2,4,6,8\} [½]
Prove that the greatest integer function f : \mathbb{R} \to \mathbb{R}, given by f(x) = [x], is neither one-one nor onto.
Official marking scheme for this paper is not available.
Let A = \mathbb{R} - \{3\} and B = \mathbb{R} - \{1\}. A function f : A \to B is defined by f(x) = \left(\frac{x - 2}{x - 3}\right). Find whether f is one-one and onto.
Let f(a) = f(b) for some a, b \in \mathbb{R} - \{3\}
so, \left(\frac{a - 2}{a - 3}\right) = \left(\frac{b - 2}{b - 3}\right) \Rightarrow ab - 2b - 3a + 6 = ab - 3b - 2a + 6 [1]
\Rightarrow a = b, Thus f is one-one function. [½]
Now, let us assume for some y \in \mathbb{R} - \{1\},\ y = f(x) = \frac{x - 2}{x - 3}
For getting, x = \frac{3y - 2}{y - 1} [1]
Range = \mathbb{R} - \{1\} = co-domain, Thus f is onto function. [½]
Let n be a fixed positive integer. A relation R is defined in set Z such that R = \{(x, y) : (x - y) \text{ is divisible by n},\ x, y \in Z\}. Determine if R is an equivalence relation.
Let for some a \in \mathbb{Z}, (a, a) \in R
\Rightarrow 0 is divisible by n, which is true
\therefore (a, a) \in R, so R is reflexive. [½]
Let (a, b) \in R \Rightarrow a - b = np, p \in \mathbb{Z}
\Rightarrow b - a = n(-p) = nq, q \in \mathbb{Z}
\Rightarrow b - a is also divisble by n.
\therefore (b, a) \in R, so R is symmetric. [1]
Let (a, b) \in R, (b, c) \in R for some a, b, c \in \mathbb{Z}
\Rightarrow a - b = np and b - c = nr, where p, r \in \mathbb{Z}
\Rightarrow a - b + b - c = n(p + r) \Rightarrow (a - c) is divisble by n.
\therefore (a, c) \in R, so R is transitive. [1]
Thus, R is an equivalence relation. [½]
If R is a relation on set of natural numbers N such that
R = \{(x, y) : x, y \in N, x + 2y = 25\},
then find the Domain and Range of the relation R. Also, determine if it is an equivalence relation.
R = {(23,1), (21,2), (19,3), (17,4), (15,5), (13,6), (11,7), (9,8), (7,9), (5,10), (3,11), (1,12)} [1]
Domain of relation R = {23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1} [½]
Range of relation R = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} [½]
Since, 1 \in N, but (1,1) \notin R, So R is not reflexive. Therefore, R is not an equivalence relation. [1]
Determine whether the function f : \mathbb{R} \to \mathbb{R} defined by f(x) = 2x^3 - 7, \forall\, x \in \mathbb{R} (Real numbers) is bijective or not.
Let f(a) = f(b), for some a, b \in \mathbb{R} \Rightarrow 2a^3 - 7 = 2b^3 - 7
\Rightarrow a^3 = b^3 \Rightarrow a = b [1]
Thus, f is one-one function. [½]
Now, Let y = f(x) = 2x^3 - 7 \Rightarrow x = \left(\frac{y+7}{2}\right)^{\frac{1}{3}}
For any real number y, the expression \left(\frac{y+7}{2}\right)^{\frac{1}{3}} is a real number.
Therefore, Range = \mathbb{R} = Co-domain, so f is onto function. [1]
Since f is both one-one and onto, so it is a bijective function. [½]
Let R be a relation defined over N, where N is set of natural numbers, defined as “mRn if and only if m is a multiple of n, m, n \in N.” Find whether R is reflexive, symmetric and transitive or not.
Let x \in N. Then we know that x is a multiple of itself.
\Rightarrow xRx
Hence, R is reflexive. [1]
We have 2, 8 \in N such that 8 is a multiple of 2
\Rightarrow 8R2
But, 2 is not a multiple of 8. Hence, 2 is not R-related to 8.
Therefore, R is not symmetric. [1]
Let x, y, z \in N such that xRy, yRz
Then x = my, y = nz for some m, n \in N
\Rightarrow x = mnz \Rightarrow x = pz, where p = mn \in N. Hence, xRz
Therefore, R is transitive. [1]
If f:R^+\to R is defined as f(x)=\log_a x (a > 0 and a\ne 1), prove that f is a bijection.
(R^+ is a set of all positive real numbers.)
f(x)=\log_a x\quad(a\gt 0,a\ne 1)
Let x_1,x_2\in R^+ such that f(x_1)=f(x_2)
\Rightarrow\log_a x_1=\log_a x_2
\Rightarrow x_1=x_2\Rightarrow f is one-one. [1½]
Let f(x)=y\Rightarrow\log_a x=y\Rightarrow a^y=x
\therefore for every y\in R, there exists x\in R^+
\therefore f is onto.
f is a bijection. [1½]
Let A = {1, 2, 3} and B = {4, 5, 6}. A relation R from A to B is defined as R = {(x, y) : x + y = 6, x\in A,\ y\in B}.
(i) Write all elements of R.
(ii) Is R a function ? Justify.
(iii) Determine domain and range of R.
(i) R=\{(1,5),(2,4)\} [1]
(ii) R is not a function as 3\in A do not have an image in co-domain. [1]
(iii) Domain of R=\{1,2\}, Range of R=\{4,5\} [1]
A student wants to pair up natural numbers in such a way that they satisfy the equation 2x + y = 41,\ x, y \in N. Find the domain and range of the relation. Check if the relation thus formed is reflexive, symmetric and transitive. Hence, state whether it is an equivalence relation or not.
R = {(1, 39), (2, 37), _ _ _ , (20, 1)}
Domain = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} [½]
Range = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39} [½]
(1, 1) does not belong to R hence not reflexive [½]
(1, 39) belongs to R but (39, 1) does not belong to R hence not symmetric [½]
(11, 19) and (19, 3) belong to R but (11, 3) does not belong to R hence not transitive.
Hence R is not an equivalence relation. [1]
Show that the function f : N \to N, where N is a set of natural numbers, given by f(n) = \begin{cases} n - 1, & \text{if } n \text{ is even} \\ n + 1, & \text{if } n \text{ is odd} \end{cases} is a bijection.
(a) Let f(x) = f(y)
Let x and y are both odd or both even
Then either x + 1 = y + 1 or x - 1 = y - 1 gives x = y [½]
x odd and y even is rejected as x + 1 = y - 1 gives x - y = -2 not possible as odd number and even number cannot differ by 2 [½]
Hence f is one-one [½]
For onto: Let f(x) = y gives x = y + 1 or x = y - 1
If y is odd, x is even and if y is even, x is odd.
Range = N = co-domain, hence onto [1]
As f is both one-one and onto hence bijective [½]
Show that the function f:\mathbb{R}\to\mathbb{R} defined by f(x)=4x^3-5,\ \forall\ x\in\mathbb{R} is one-one and onto.
One-One: Let x_1,x_2\in\mathbb{R} such that f(x_1)=f(x_2)\Rightarrow4x_1^3-5=4x_2^3-5\Rightarrow x_1^3=x_2^3\Rightarrow x_1=x_2, \therefore ‘f’ is one-one [1½]
Onto: x\in\mathbb{R}\ (D_f)\Rightarrow x^3\in\mathbb{R}\Rightarrow4x^3-5\in\mathbb{R}\Rightarrow f(x)\in\mathbb{R}, \therefore R_f=\text{Co-domain}(f) [1½]
\therefore ‘f’ is an onto function
\Rightarrow ‘f’ is one-one & onto both
Let R be a relation defined on a set N of natural numbers such that R=\{(x,y):xy\text{ is a square of a natural number, }x,y\in\mathbb{N}\}. Determine if the relation R is an equivalence relation.
Reflexive: For any x\in\mathbb{N}, x\cdot x=x^2, which is square of the natural number 'x'.
\Rightarrow(x,x)\in R
\therefore ‘R’ is a Reflexive relation. [1]
Symmetric: Let (x,y)\in R\Rightarrow xy is a square of a natural number
\Rightarrow yx is a square of a natural number, \because xy=yx.
\Rightarrow(y,x)\in R
\therefore ‘R’ is a Symmetric relation. [1]
Transitive: Let (x,y),(y,z)\in R\Rightarrow xy=a^2,\ yz=b^2 for some a,b\in\mathbb{N},
\therefore\frac{a^2}{y}=x,\ \frac{b^2}{y}=z\in\mathbb{N}
\Rightarrow xz=\frac{a^2}{y}\cdot\frac{b^2}{y}=\left(\frac{ab}{y}\right)^2,\ \frac{ab}{y}\in\mathbb{N} [½]
\Rightarrow(x,z)\in R
\therefore ‘R’ is a Transitive relation.
Hence, R is an Equivalence relation [½]
A relation R on set A = {1, 2, 3, 4, 5} is defined as R = \{(x, y) : |x^2 - y^2| \lt 8\}. Check whether the relation R is reflexive, symmetric and transitive.
(a) Reflexive:
\because |x^2 - x^2| \lt 8\ \forall\, x \in A \Rightarrow (x, x) \in R\ \therefore R is reflexive. [½]
(b) Symmetric:
Let (x, y) \in R for some x, y \in A
\therefore |x^2 - y^2| \lt 8 \Rightarrow |y^2 - x^2| \lt 8 \Rightarrow (y, x) \in R
Hence R is symmetric. [1]
(c) Transitive:
(1, 2), (2, 3) \in R as |1^2 - 2^2| \lt 8,\ |2^2 - 3^2| \lt 8 respectively
But |1^2 - 3^2| \not\lt 8 \Rightarrow (1, 3) \notin R
Hence R is not transitive. [1½]
A function f is defined from R \to R as f(x) = ax + b, such that f(1) = 1 and f(2) = 3. Find function f(x). Hence, check whether function f(x) is one-one and onto or not.
f(x) = ax + b
Solving a + b = 1 and 2a + b = 3 to get a = 2,\ b = -1 [1]
f(x) = 2x - 1
Let f(x_1) = f(x_2) for some x_1, x_2 \in R
2x_1 - 1 = 2x_2 - 1 \Rightarrow x_1 = x_2
Hence f is one – one. [1]
Let y = 2x - 1,\ y \in R (Codomain)
\Rightarrow x = \frac{y+1}{2} \in R (domain)
Also, f(x) = f\left(\frac{y+1}{2}\right) = y
\therefore f is onto. [1]
Case Study -1
A city’s traffic management department is planning to optimize traffic flow by analyzing the connectivity between various traffic signals. The city has five major spots labelled A,B,C,D, and E.
The department has collected the following data regarding one-way traffic flow between spots:
1. Traffic flows from A to B, A to C, and A to D.
2. Traffic flows from B to C and B to E.
3. Traffic flows from C to E.
4. Traffic flows from D to E and D to C.
The department wants to represent and analyze this data using relations and functions. Use the given data to answer the following questions:
I. Is the traffic flow reflexive? Justify. [1]
II. Is the traffic flow transitive? Justify. [1]
III A. Represent the relation describing the traffic flow as a set of ordered pairs. Also state the domain and range of the relation.
OR
III B. Does the traffic flow represent a function? Justify your answer. [2]
I. Traffic flow is not reflexive as (A,A)\notin R (or no major spot is connected with itself) [1]
II. Traffic flow is not transitive as (A,B)\in R\ and\ (B,E)\in R,but\ (A,E)\notin R [1]
III A. R=\{(A,B),(A,C),(A,D),(B,C),(B,E),(C,E),(D,E),(D,C)\} [1]
\text{Domain}=\{A,B,C,D\}
\text{Range}=\{B,C,D,E\} [½+½]
OR
III B. No, the traffic flow doesn’t represent a function as A has three images. [1+1]
Case Study – 1
A school wants the students of class XII to do a project on ‘Sustainability’ keeping the world environment in mind. They select the student participants on the basis of an essay writing competition.
7 students out of 80 are selected for the project and are categorized into two sets such that :
Girl students belong to Set A = \{G_1, G_2, G_3, G_4\},
Boy students belong to Set B = \{B_1, B_2, B_3\}.
Based on the above information, answer the following questions :
(i) How many relations are possible from Set A \to Set B ? [1]
(ii) Let R be a relation from A \to B such that R = \{(G_1, B_1), (G_2, B_2), (G_3, B_2), (G_4, B_3), (G_1, B_2)\}. Is R an injective function ? Justify your answer. [1]
(iii) (a) Let the relation R from A \to A be such that R = {(x, y), x, y \in A, x and y are students from the same colony in the city}. Verify if R is an equivalence relation. [2]
OR
(iii) (b) Verify if any function f : B \to A is bijective. Give reason to support your answer. [2]
(i) Number of relations from A to B is 2^{12} or 4096 [1]
(ii) No R is not injective [½]
It is not a function as G_1 has two images B_1 and B_2 [½]
(iii) (a) R is reflexive as (x, x) \in R\ \forall \in A because x and x are students from same colony [½]
Let (x, y) \in R so x and y are students from same colony
Hence y and x are students from same colony
Thus, (y, x) \in R
\therefore R is symmetric [½]
Let (x, y) \in R and (y, z) \in R so x and y are students from same colony, y and z are students from same colony
Thus, x and z are students from same colony
\therefore (x, z) \in R
Thus, R is transitive [½]
As R is reflexive, symmetric and transitive, hence R is an equivalence relation [½]
OR
(iii) (b) Any function f from B to A will not be bijective [1]
As f from B to A cannot be surjective because n(B) < n(A)
or
A and B are finite sets and n(A) \ne n(B) [1]
A class-room teacher is keen to assess the learning of her students the concept of “relations” taught to them. She writes the following five relations each defined on the set A = {1, 2, 3} :
R_1=\{(2,3),(3,2)\}
R_2=\{(1,2),(1,3),(3,2)\}
R_3=\{(1,2),(2,1),(1,1)\}
R_4=\{(1,1),(1,2),(3,3),(2,2)\}
R_5=\{(1,1),(1,2),(3,3),(2,2),(2,1),(2,3),(3,2)\}
The students are asked to answer the following questions about the above relations :
(i) Identify the relation which is reflexive, transitive but not symmetric.
(ii) Identify the relation which is reflexive and symmetric but not transitive.
(iii) (a) Identify the relations which are symmetric but neither reflexive nor transitive.
OR
(iii) (b) What pairs should be added to the relation R_2 to make it an equivalence relation ?
(i) R_4 [1]
(ii) R_5 [1]
(iii)(a) R_1 and R_3 [1+1]
OR
(iii)(b) Required pairs to be added to make the relation R_2 as an equivalence relation are:
(1,1),(2,2),(3,3),(2,1),(3,1) and (2,3) [2]
A school is organizing a debate competition with participants as speakers S = \{S_1, S_2, S_3, S_4\} and these are judged by judges J = \{J_1, J_2, J_3\}. Each speaker can be assigned one judge. Let R be a relation from set S to J defined as R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y,\ x \in S,\ y \in J\}.
Based on the above, answer the following :
(i) How many relations can be there from S to J ? [1]
(ii) A student identifies a function from S to J as f = \{(S_1, J_1), (S_2, J_2), (S_3, J_2), (S_4, J_3)\} Check if it is bijective. [1]
(iii) (a) How many one-one functions can be there from set S to set J ? [2]
OR
(iii) (b) Another student considers a relation R_1 = \{(S_1, S_2), (S_2, S_4)\} in set S. Write minimum ordered pairs to be included in R_1 so that R_1 is reflexive but not symmetric. [2]
(i) The number of relations = 2^{4\times 3} = 2^{12} [1]
(ii) Since, S_2 and S_3 have been assigned the same judge J_2, the function is not one-one.
Hence, it is not bijective. [1]
(iii) (a) There cannot exist any one-one function from S to J as n(S) > n(J). Hence, the number of one-one functions from S to J is 0. [2]
OR
(iii) (b) To make R_1 reflexive and not symmetric we need to add the following ordered pairs:
(S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4) [2]
Case Study – 2
Let A be the set of 30 students of class XII in a school. Let f:A\to N, N is a set of natural numbers such that function f(x) = Roll Number of student x.
On the basis of the given information, answer the following :
(i) Is f a bijective function ? [1]
(ii) Give reasons to support your answer to (i). [1]
(iii) (a) Let R be a relation defined by the teacher to plan the seating arrangement of students in pairs, where
R=\{(x,y):x,y are Roll Numbers of students such that y=3x\}.
List the elements of R. Is the relation R reflexive, symmetric and transitive ? Justify your answer. [2]
OR
(iii) (b) Let R be a relation defined by
R=\{(x,y):x,y are Roll Numbers of students such that y=x^3\}.
List the elements of R. Is R a function ? Justify your answer. [2]
(i) No, f is not bijective function [1]
(ii) Range =\{1,2,3,4,\ldots\ldots,30\} and codomain =N [½]
Since, Range \ne codomain \Rightarrow f is not onto and hence f is not bijective. [½]
(iii) (a) R=\{(1,3),(2,6),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)\} [1]
Since (1,1)\notin R\Rightarrow R is not reflexive.
(1,3)\in R but (3,1)\notin R\Rightarrow R is not symmetric
(1,3)\in R,(3,9)\in R but (1,9)\notin R\Rightarrow R is not transitive. [1]
OR
(iii) (b) R=\{(1,1),(2,8),(3,27)\} [1]
\because elements 4, 5, 6 … 30 do not have an image. Hence the above relation is not a function. [1]
Case Study – 2
During the festival season, there was a mela organized by the Resident Welfare Association at a park, near the society. The main attraction of the mela was a huge swing installed at one corner of the park. The swing is traced to follow the path of a parabola given by x^2 = y.
Based on the above information, answer the following questions :
(i) Let f : N \longrightarrow R is defined by f(x) = x^2. What will be the range ? [1]
(ii) Let f : N \longrightarrow N is defined by f(x) = x^2. Check if the function is injective or not. [1]
(iii) (a) Let f : \{1, 2, 3, 4 \ldots\} \longrightarrow \{1, 4, 9, 16 \ldots\} be defined by f(x) = x^2. Prove that the function is bijective. [2]
OR
(iii) (b) Let f : \mathbb{R} \longrightarrow \mathbb{R} is defined by f(x) = x^2. Show that f is neither injective nor surjective. [2]
(i) R_f = \{1, 4, 9, 16, \ldots\} i.e. set of perfect squares of natural numbers. [1]
(ii) Let x_1, x_2 \in N (domain)
Let f(x_1) = f(x_2)
\Rightarrow x_1^2 = x_2^2
\Rightarrow x_1 = \pm x_2
\Rightarrow x_1 = x_2 as x_1, x_2 \in N
\therefore f is injective. [1]
(iii)(a) f(x) = x^2
Let x_1, x_2 \in \{1, 2, 3, 4, \ldots\}
Let f(x_1) = f(x_2)
\Rightarrow x_1^2 = x_2^2
\Rightarrow x_1 = x_2
\therefore f is one-one. [1]
As Co-domain = Range = \{1, 4, 9, 16, \ldots\}
\therefore f is onto. [1]
Since, f is one-one and onto, so f is bijective.
OR
(iii)(b) f : R \to R, f(x) = x^2
-1, 1 \in R (domain)
As f(-1) = f(1) = 1 but -1 \ne 1
\therefore f is not injective. [1]
Co-domain = R, but Range = [0, \infty)
Since Co-domain \ne Range, f is not surjective. [1]
Students of a school are taken to a railway museum to learn about railways heritage and its history.
An exhibit in the museum depicted many rail lines on the track near the railway station. Let L be the set of all rail lines on the railway track and R be the relation on L defined by
R=\{(l_1,l_2):l_1\text{ is parallel to }l_2\}
On the basis of the above information, answer the following questions :
(i) Find whether the relation R is symmetric or not.
(ii) Find whether the relation R is transitive or not.
(iii) If one of the rail lines on the railway track is represented by the equation y=3x+2, then find the set of rail lines in R related to it.
(i) Let (l_1,l_2)\in R\Rightarrow l_1\parallel l_2\Rightarrow l_2\parallel l_1\Rightarrow(l_2,l_1)\in R, \therefore R is a symmetric relation [1]
(ii) Let (l_1,l_2),(l_2,l_3)\in R\Rightarrow l_1\parallel l_2,\ l_2\parallel l_3\Rightarrow l_1\parallel l_3\Rightarrow(l_1,l_3)\in R, \therefore R is a transitive relation [1]
(iii) The set is \{l:\ l\text{ is a line of type }y=3x+c,c\in R\} [2]
** Due to printing error Part (a) or Part(b), both parts be taken as independent questions of 4 marks each
Students of a school are taken to a railway museum to learn about railways heritage and its history.
An exhibit in the museum depicted many rail lines on the track near the railway station. Let L be the set of all rail lines on the railway track and R be the relation on L defined by
R=\{(l_1,l_2):l_1\text{ is parallel to }l_2\}
Let S be the relation defined by S=\{(l_1,l_2):l_1\text{ is perpendicular to }l_2\} check whether the relation S is symmetric and transitive.
Let (l_1,l_2)\in R\Rightarrow l_1\perp l_2\Rightarrow l_2\perp l_1\Rightarrow(l_2,l_1)\in R, \therefore R is a symmetric relation [2]
Let (l_1,l_2),(l_2,l_3)\in R\Rightarrow l_1\perp l_2,l_2\perp l_3\Rightarrow l_1\parallel l_3\Rightarrow(l_1,l_3)\notin R, \therefore R is not a transitive relation [2]
** Due to printing error Part (a) or Part(b), both parts be taken as independent questions of 4 marks each
Case Study-I
An organization conducted bike race under two different categories – Boys and Girls. There were 28 participants in all. Among all of them, finally three from category 1 and two from category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his college project.
Let B=\{b_1,b_2,b_3\} and G=\{g_1,g_2\}, where B represents the set of Boys selected and G the set of Girls selected for the final race.
Based on the above information, answer the following questions :
(I) How many relations are possible from B to G ? [1]
(II) Among all the possible relations from B to G, how many functions can be formed from B to G ? [1]
(III) Let R:B\to B be defined by R=\{(x,y):x\text{ and }y\text{ are students of the same sex}\}. Check if R is an equivalence relation. [2]
OR
(III) A function f:B\to G be defined by f=\{(b_1,g_1),(b_2,g_2),(b_3,g_1)\}. Check if f is bijective. Justify your answer. [2]
(I) Number of relations =2^6=64 [1]
(II) Number of possible functions =2^3=8 [1]
(III) R is an equivalence relation as it is reflexive, symmetric and transitive [2]
OR
Since f is not one-one function [1]
\therefore f is not bijective [1]
Let A=N\times N, where N is the set of natural numbers and a relation R is defined on A such that (a,b)R(c,d)\ if\ and\ only\ if\ a^2+d^2=b^2+c^2.
Check whether R is an equivalence relation or not?
Given relation R is defined on A=N\times N such that
(a,b)R(c,d)\ if\ and\ only\ if\ a^2+d^2=b^2+c^2
Reflexive:
Let (p,q)\in A=N\times N
Then as p,q\in N, so p^2+q^2=q^2+p^2
\Rightarrow(p,q)R(p,q)\ \forall(p,q)\in N\times N
\Rightarrow R is a reflexive relation. [1½]
Symmetric:
Let (p,q)\ and\ (r,s)\in A=N\times N such that (p,q)\ R\ (r,s)
\Rightarrow p^2+s^2=q^2+r^2
\Rightarrow q^2+r^2=p^2+s^2
\Rightarrow r^2+q^2=s^2+p^2
\Rightarrow(r,s)\ R\ (p,q)
Thus, R is a symmetric relation. [1½]
Transitive:
Let (p,q),(r,s),(u,v)\in A such that (p,q)\ R\ (r,s)\ and\ (r,s)R(u,v)
\Rightarrow p^2+s^2=q^2+r^2\ \ldots(i) and r^2+v^2=s^2+u^2\ \ldots(ii)
Adding (i) and (ii), p^2+s^2+r^2+v^2=q^2+r^2+s^2+u^2
\Rightarrow p^2+v^2=q^2+u^2
\Rightarrow(p,q)\ R\ (u,v)
Thus, R is a transitive relation [1½]
So, R is a reflexive, symmetric and transitive relation.
Hence R is an equivalence relation [½]
A relation R:(-\infty,\infty)\to[0,\infty) is defined as R=\{(x,y):y=x^2+5\}
Draw its graph and check whether it is a function or not.
If the given relation is a function, then examine whether the function is injective, and whether it is surjective or not.
For Visually Impaired: A relation R:(-\infty,\infty)\to[0,\infty) is defined as R=\{(x,y):y=x^2+7\}.
Check whether it is a function or not. If the given relation is a function, then examine whether the function is injective, surjective, or both, and conclude whether it is bijective or not.
[Graph of y=x^2+5: upward parabola with vertex (0,5)] [1 for correct graph]
Yes, R is a function as the graph shows every element of the set (-\infty,\infty) has a unique image in the set [0,\infty). [1]
Now, let us denote this function by f:(-\infty,\infty)\to[0,\infty) as f(x)=x^2+5
Injective Function:
Here f(-1)=(-1)^2+5=6 and f(1)=1^2+5=6
So f(-1)=f(1)=6
Thus f is a many-one function, so not an injective function. [1½]
Surjective Function:
Range of f(x)=[5,\infty)\ne co-domain=[0,\infty)
So, it is not surjective. [1½]
Alternative method
Correct graph as above [1]
Yes, R is a function as the graph shows every element of the set (-\infty,\infty) has a unique image in the set [0,\infty). [1]
Now, let us denote this function by f:(-\infty,\infty)\to[0,\infty) as f(x)=x^2+5
Injective Function:
Let a,b\in(-\infty,\infty), such that f(a)=f(b)
\Rightarrow a^2+5=b^2+5\Rightarrow a^2=b^2\Rightarrow a=\pm b
So f is not injective. [1½]
Surjective Function:
For y=1 there is no pre-image as if 1=x^2+5\Rightarrow x^2=-4 not possible.
So, it is not surjective. [1½]
For Visually Impaired:
For every element a of the set (-\infty,\infty), a^2\in[0,\infty) and therefore a^2+7\in[7,\infty)
Hence every element of the set (-\infty,\infty) has a unique image in the set [0,\infty).
So, f:(-\infty,\infty)\to[0,\infty), f(x)=x^2+7 is a function. [1]
Now, let us denote this function by f:(-\infty,\infty)\to[0,\infty) as f(x)=x^2+7
Injective Function:
Here f(-1)=(-1)^2+7=8 and f(1)=1^2+7=8
So f(-1)=f(1)=8
Thus f is a many-one function, so not an injective function. [1½]
Surjective Function:
Range of f(x)=[7,\infty)\ne co-domain=[0,\infty)
So, it is not surjective. [1½]
Thus f(x) is neither injective nor surjective and hence not bijective [1]
A relation R is defined on Z, the set of integers, as
R = \{(x, y) : |x - y| \text{ is divisible by a prime number 'p'},\ x, y \in Z\}
check whether R is an equivalence relation or not.
Let x \in Z. Then |x - x| = 0, which is divisible by a prime number p.
Hence, (x, x) \in R. Thus, R is reflexive. [1½]
Let x, y \in Z such that (x, y) \in R.
Then |x - y| is divisible by the prime number p.
\Rightarrow |y - x| is divisible by the prime number p as |x - y| = |y - x|
Hence, (y, x) \in R. Thus, R is symmetric. [1½]
Let x, y, z \in Z such that (x, y), (y, z) \in R.
Then |x - y| is divisible by the prime number p and |y - z| is divisible by the prime number p.
\Rightarrow x - y is divisible by the prime number p and y - z is divisible by the prime number p.
\Rightarrow x - y + y - z is divisible by the prime number p
\Rightarrow x - z is divisible by the prime number p
\Rightarrow |x - z| is divisible by the prime number p
Hence, (x, z) \in R. Thus, R is transitive. [1½]
Since, R is reflexive, symmetric and transitive, therefore, R is an equivalence relation. [½]
A function f : R - \left\{\frac{3}{5}\right\} \longrightarrow R - \left\{\frac{3}{5}\right\} is defined as f(x) = \frac{3x + 2}{5x - 3}. Show that f is one-one and onto.
Let x_1, x_2 \in R - \left\{\frac{3}{5}\right\} such that f(x_1) = f(x_2)
\Rightarrow \frac{3x_1 + 2}{5x_1 - 3} = \frac{3x_2 + 2}{5x_2 - 3}
\Rightarrow 15x_1x_2 - 9x_1 + 10x_2 - 6 = 15x_1x_2 + 10x_1 - 9x_2 - 6
\Rightarrow x_1 = x_2
Hence, f is one-one. [2½]
Let y \in R - \left\{\frac{3}{5}\right\} (Codomain).
Then f(x) = y
or, \frac{3x+2}{5x-3} = y \Rightarrow x = \frac{3y+2}{5y-3} \in Domain.
Hence, for every y \in R - \left\{\frac{3}{5}\right\}\ (codomain),
there exists x = \frac{3y+2}{5y-3} \in R - \left\{\frac{3}{5}\right\}\ (domain) such that f\left(\frac{3y+2}{5y-3}\right) = y
Thus, f is onto. [2½]
Show that f : \mathbb{R} \to \mathbb{R} defined as f(x) = \frac{x}{\sqrt{1 + x^2}} is one-one but not onto.
One one: For x_1, x_2 \in \mathbb{R}; let f(x_1) = f(x_2) \Rightarrow \frac{x_1}{\sqrt{1 + x_1^2}} = \frac{x_2}{\sqrt{1 + x_2^2}} [1]
[\Rightarrow x_1 and x_2 must be of same sign.]
\Rightarrow x_1^2(1 + x_2^2) = x_2^2(1 + x_1^2)
\Rightarrow x_1^2 = x_2^2 \Rightarrow x_1 = x_2 (Rejecting x_1 = -x_2)
Hence, f is one − one. [1½]
Onto: y = \frac{x}{\sqrt{1 + x^2}} \Rightarrow x = \pm\sqrt{\frac{y^2}{1 - y^2}} [1]
x is defined \forall\, y \in (-1, 1) \Rightarrow \text{Range} = (-1, 1) [1]
\because Range \ne codomain \Rightarrow f is not onto. [½]
Show that a function f : R \to R defined by f(x) = \frac{2x}{1+x^2} is neither one-one nor onto. Further, find set A so that the given function f : R \to A becomes an onto function.
Let f(x_1) = f(x_2)\ \text{for some } x_1, x_2 \in R
Then \frac{2x_1}{1+{x_1}^2} = \frac{2x_2}{1+{x_2}^2}
\Rightarrow x_1 + x_1{x_2}^2 = x_2 + {x_1}^2x_2
\Rightarrow (x_1 - x_2) - x_1x_2(x_1 - x_2) = 0
\Rightarrow (x_1 - x_2)(1 - x_1x_2) = 0
\Rightarrow x_1 - x_2 = 0 or 1 - x_1x_2 = 0
\Rightarrow x_1 = x_2 or x_1x_2 = 1, so if x_1x_2 = 1, x_1 \ne x_2
Hence f is not one -one [2]
Let y = f(x) where x \in R
Then y = \frac{2x}{1+x^2}. Here, for x = 0, y = 0
If y \ne 0, then y = \frac{2x}{1+x^2}
\Rightarrow yx^2 - 2x + y = 0
\Rightarrow x = \frac{2 \pm \sqrt{4-4y^2}}{2y}
For x to be real, 4 - 4y^2 \ge 0
\Rightarrow y^2 \le 1
\Rightarrow -1 \le y \le 1
Hence, range = [-1, 1] \ne codomain
Hence, f is not onto. [2]
For the given function to become onto, A = [-1, 1] [1]
A relation R is defined on N \times N (where N is the set of natural numbers) as :
(a, b)\ R\ (c, d) \Leftrightarrow a - c = b - d
Show that R is an equivalence relation.
Let (a, b) \in N \times N
We have
a - a = b - b
This implies that (a, b) R (a, b) \forall (a, b) \in N \times N
Hence R is reflexive [1½]
Let (a, b) R (c, d) for some (a, b), (c, d) \in N \times N
Then a - c = b - d
\Rightarrow c - a = d - b
\Rightarrow(c, d) R (a, b)
Hence, R is symmetric. [1½]
Let (a, b) R (c, d), (c, d) R (e, f) for some (a, b), (c, d), (e, f) \in N \times N
Then a - c = b - d, c - e = d - f
\Rightarrow a - c + c - e = b - d + d - f
\Rightarrow a - e = b - f
\Rightarrow(a, b) R (e, f)
Hence, R is transitive [2]
Thus, R is an equivalence relation.
A relation R on set A = {– 4, – 3, – 2, – 1, 0, 1, 2, 3, 4} be defined as R = {(x, y) : x + y is an integer divisible by 2}. Show that R is an equivalence relation. Also, write the equivalence class [2].
For reflexive: clearly x + x i.e. 2x is integer divisible by 2.
\Rightarrow (x, x) \in R \Rightarrow R is reflexive. [1]
For symmetric : (x, y) \in R \Rightarrow x + y is integer divisible by 2.
\Rightarrow y + x is integer divisible by 2 \Rightarrow (y, x) \in R [1]
For transitive : (x, y) \in R \Rightarrow x + y is integer divisible by 2.
and (y, z) \in R \Rightarrow y + z is integer divisible by 2.
so, (x + z) + 2y is integer divisible by 2.
\Rightarrow x + z is integer divisible by 2 \Rightarrow (x, z) \in R [2]
Equivalence class [2] = \{-4, -2, 0, 2, 4\} [1]
Let A = R – {5} and B = R – {1}. Consider the function f : A \to B, defined by f(x) = \frac{x-3}{x-5}. Show that f is one-one and onto.
Let f(x_1) = f(x_2), for some x_1, x_2 \in A
\Rightarrow \frac{x_1-3}{x_1-5} = \frac{x_2-3}{x_2-5}
\Rightarrow (x_1-3)(x_2-5) = (x_2-3)(x_1-5)
\Rightarrow x_1 = x_2, So f is one – one Function. [2½]
Let y = f(x) = \frac{x-3}{x-5} \Rightarrow y(x-5) = x-3
\Rightarrow yx - 5y = x - 3
\Rightarrow x = \frac{5y-3}{y-1}, We observe that x is defined for all values of y except y = 1,
So, Range = R – {1} and Co-domain is Given R – {1} [As, f : A \to B]
Since, Range = Co-domain, f is onto Function.
Thus, f is one-one & onto function. [2½]
Check whether the relation S in the set of real numbers R defined by S = {(a, b) : where a - b + \sqrt{2} is an irrational number} is reflexive, symmetric or transitive.
Reflexive: For a \in S
\Rightarrow a - a + \sqrt{2} is irrational number
\Rightarrow \sqrt{2} is irrational number
\Rightarrow (a, a) \in S
Thus, S is Reflexive Relation. [1½]
Symmetric : Let (a, b) \in S \Rightarrow a - b + \sqrt{2} is irrational number
but b - a + \sqrt{2} may not be irrational number
For example, (\sqrt{2}, 1) \in S \Rightarrow \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1 is irrational number
(1, \sqrt{2}) \notin S as 1 - \sqrt{2} + \sqrt{2} = 1 is not irrational number
\therefore (b, a) \notin S, So S is NOT Symmetric Relation. [1½]
Transitive : Let (a, b) \in S \Rightarrow a - b + \sqrt{2} is irrational number
& (b, c) \in S \Rightarrow b - c + \sqrt{2} is irrational number
but a - c + \sqrt{2} may not be irrational number
For example, (1, \sqrt{3}) \in S \Rightarrow 1 - \sqrt{3} + \sqrt{2} is irrational number
(\sqrt{3}, \sqrt{2}) \in S \Rightarrow \sqrt{3} - \sqrt{2} + \sqrt{2} = \sqrt{3} is irrational number
But (1, \sqrt{2}) \notin S as 1 - \sqrt{2} + \sqrt{2} = 1 is not irrational number
\therefore (a, c) \notin S, So S is NOT Transitive Relation.
Thus, S is Reflexive But Neither Symmetric nor Transitive Relation. [2]
Check whether the relation S in the set of all real numbers (\mathbb{R}) defined by S=\{(a,b):a\le b^{3}\} is reflexive, symmetric or transitive.
Reflexive: Let \frac{1}{2}\not\le\left(\frac{1}{2}\right)^{3}, \therefore S is not reflexive. [1]
Symmetric: 1\le2^{3}\Rightarrow(1,2)\in S but 2\not\le1^{3}\Rightarrow(2,1)\notin S. \therefore S is not symmetric. [2]
Transitive: 10\le6^{3}\Rightarrow(10,6)\in S & 6\le2^{3}\Rightarrow(6,2)\in S but 10\not\le2^{3}\Rightarrow(10,2)\notin S
\therefore S is not Transitive. [2]
A relation R is defined on a set of real numbers \mathbb{R} as
R=\{(x,y):x\cdot y\text{ is an irrational number}\}.
Check whether R is reflexive, symmetric and transitive or not.
For reflexive
(1,1)\notin R as 1^2 is rational (or any other counter example)
R is not reflexive [1½]
For symmetric
Let (x,y)\in R\ \therefore\ x.y is an irrational number
\therefore\ (y.x) is an irrational number
\therefore (y,x)\in R
\therefore R is symmetric [1½]
For Transitive
(1,\sqrt2)\in R,\ (\sqrt2,2)\in R (or any other counter example)
but (1,2)\notin R
\therefore R is not transitive [2]
A function f: [-4, 4] \to [0, 4] is given by f(x) = \sqrt{16 - x^2}. Show that f is an onto function but not a one-one function. Further, find all possible values of ‘a’ for which f(a) = \sqrt{7}.
Onto: Let y = \sqrt{16 - x^2} \Rightarrow y \ge 0 [½]
Squaring we get, x^2 = 16 - y^2 \Rightarrow x = \pm\sqrt{16 - y^2} [½]
For each y \in [-4, 4], ‘x’ is a real number, \therefore 0 \le y \le 4 \Rightarrow R_f = [0, 4] = \text{Co-domain}
\therefore ‘f’ is an onto function. [1½]
One-One: f(-1) = f(1) = \sqrt{15} but -1 \ne 1,\ \therefore ‘f’ is not a one-one function. [1½]
f(a) = \sqrt{7} \Rightarrow \sqrt{16 - a^2} = \sqrt{7} \Rightarrow a = \pm 3 [1]
Show that a function f : \mathbb{R} \to \mathbb{R} defined as f(x) = \frac{5x - 3}{4} is both one-one and onto.
Let x_1, x_2 \in R (Domain) such that f(x_1) = f(x_2)
\Rightarrow \frac{5x_1 - 3}{4} = \frac{5x_2 - 3}{4}
\Rightarrow 5x_1 = 5x_2 \Rightarrow x_1 = x_2
\therefore f is one-one. [2½]
Let y \in R (co-domain). Then f(x) = y for some x.
if, y = \frac{5x - 3}{4}, i.e., if, x = \frac{4y + 3}{5}, which \in R (Domain)
Thus, for every y \in R (co-domain), there exists \frac{4y + 3}{5} \in R (domain) such that f\left(\frac{4y + 3}{5}\right) = y
\therefore Range of f = R = codomain of f. Hence, f is onto. [2½]
If N denotes the set of all natural numbers and R is the relation on N\times N defined by (a,b)\,R\,(c,d), if ad(b+c)=bc(a+d). Show that R is an equivalence relation.
Reflexive : Here, (a,b)\,R\,(a,b)\ \forall\ (a,b)\in N\times N since ab(b+a)=ba(a+b) is always true. [1]
Symmetric: Let (a,b)\,R\,(c,d)\ \forall\ (a,b),(c,d)\in N\times N. Then,
ad(b+c)=bc(a+d)
\Rightarrow bc(a+d)=ad(b+c)
\Rightarrow(c,d)\,R\,(a,b) [1½ (For Symmetric)]
Transitive: Let (a,b)\,R\,(c,d) and (c,d)\,R\,(e,f)\ \forall\ (a,b),(c,d),(e,f)\in N\times N. Then
ad(b+c)=bc(a+d) and cf(d+e)=de(c+f)
\Rightarrow\frac{b+c}{bc}=\frac{a+d}{ad} and \frac{d+e}{de}=\frac{c+f}{cf}
\frac{1}{c}+\frac{1}{b}=\frac{1}{d}+\frac{1}{a}\Rightarrow\frac{1}{e}+\frac{1}{d}=\frac{1}{f}+\frac{1}{c}
Adding, we get
\frac{1}{c}+\frac{1}{b}+\frac{1}{e}+\frac{1}{d}=\frac{1}{d}+\frac{1}{a}+\frac{1}{f}+\frac{1}{c}
\Rightarrow\frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}
\Rightarrow\frac{e+b}{be}=\frac{f+a}{af}
\Rightarrow af(b+e)=be(a+f)
\Rightarrow(a,b)\,R\,(e,f) [2 For Transitive]
Hence, R is equivalence relation. [½]
Let f:\mathbb{R}-\left\{-\frac{4}{3}\right\}\to\mathbb{R} be a function defined as f(x)=\frac{4x}{3x+4}. Show that f is a one-one function. Also, check whether f is an onto function or not.
For one-one
Let f(x_1)=f(x_2) for some x_1,x_2\in R-\left\{-\frac{4}{3}\right\}
\Rightarrow\frac{4x_1}{3x_1+4}=\frac{4x_2}{3x_2+4}
\Rightarrow 12x_1x_2+16x_1=12x_1x_2+16x_2 [1]
\Rightarrow 16x_1=16x_2
\Rightarrow x_1=x_2 [1]
\Rightarrow f is one-one
For onto let y\in R, and for some x.
Let y=\frac{4x}{3x+4}
\Rightarrow 3xy+4y=4x
\Rightarrow x(3y-4)=-4y
\Rightarrow x=-\frac{4y}{3y-4} or x=\frac{4y}{4-3y} [1]
x is real if y\ne\frac{4}{3}. So R_f=R-\left\{\frac{4}{3}\right\}\ne Codomain\ (f) [1]
So, f is not onto. [1]
Show that the relation S in set \mathbb{R} of real numbers defined by
is neither reflexive, nor symmetric, nor transitive.
Official marking scheme for this paper is not available.
Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by
R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Hence, find the elements of equivalence class [1].
Official marking scheme for this paper is not available.
Identify the function whose graph is given below:
For Visually Impaired: Find the range of \frac{\pi}{2}+\sec^{-1}\left(\frac{x}{2}\right)
(A) [0,\pi] (B) [0,\pi]-\left\{\frac{\pi}{2}\right\} (C) \left[\frac{\pi}{2},\frac{3\pi}{2}\right] (D) \left[\frac{\pi}{2},\frac{3\pi}{2}\right]-\{\pi\}
(D) \pi+\cos^{-1}x
For Visually Impaired: (D) \left[\frac{\pi}{2},\frac{3\pi}{2}\right]-\{\pi\}
Domain of the function: \sec^{-1}(2x) is
(D) \left(-\infty,-\frac12\right]\cup\left[\frac12,\infty\right)
Value of \cot^{-1}\left(2\cos\left(\frac12\sin^{-1}\left(\frac{\sqrt3}{2}\right)\right)\right)
(A) \frac{\pi}{6}
\cot^{-1}\left(2\cos\left(\frac12\sin^{-1}\left(\frac{\sqrt3}{2}\right)\right)\right)=\cot^{-1}\left(2\cos\left(\frac12\times\frac{\pi}{3}\right)\right)
=\cot^{-1}\left(2\cos\left(\frac{\pi}{6}\right)\right)=\cot^{-1}\left(2\times\frac{\sqrt3}{2}\right)=\cot^{-1}(\sqrt3)=\frac{\pi}{6}
Identify the function shown in the graph
For Visually Impaired: Inverse Trigonometric Function, whose domain is \left[-\frac{1}{3},\frac{1}{3}\right], is ...
(A) \cos^{-1}x (B) \cos^{-1}\left(\frac{x}{3}\right) (C) \cos^{-1}(3x) (D) 3\cos^{-1}x
Clearly from the graph Domain is \left[-\frac{1}{2},\frac{1}{2}\right]
So graph is of the function \sin^{-1}(2x)
(B) \sin^{-1}(2x) [1]
For Visually Impaired:
Domain is \left[-\frac{1}{3},\frac{1}{3}\right]
So the function is \cos^{-1}(3x)
(C) \cos^{-1}(3x) [1]
Assertion (A): Value of the expression \sin^{-1}\left(\frac{\sqrt3}{2}\right)+\tan^{-1}1-\sec^{-1}(\sqrt2) is \frac{\pi}{4}.
Reason (R): Principal value branch of \sin^{-1}x is \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and that of \sec^{-1}x is [0,\pi]-\left\{\frac{\pi}{2}\right\}.
\sin^{-1}\left(\frac{\sqrt3}{2}\right)+\tan^{-1}1-\sec^{-1}(\sqrt2)=\frac{\pi}{3}+\frac{\pi}{4}-\frac{\pi}{4}=\frac{\pi}{3}\ne\frac{\pi}{4}
So, A is false.
Principal Value branch of \sin^{-1}x is \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and that of \sec^{-1}x is [0,\pi]-\left\{\frac{\pi}{2}\right\}.
So, R is true
(D) Assertion is false, but Reason is true [1]
If 2\cos^{-1}x = y, then
(C) 0 \le y \le 2\pi
The domain of f(x) = \cos^{-1}(2x - 5) is :
(D) [2, 3]
The following graph represents :
(D) y = \text{cosec}^{-1} x [1]
The principal value of \sec^{-1}\left(\sqrt{2}\right) + 2\,\text{cosec}^{-1}\left(-\sqrt{2}\right) is :
(B) \frac{-\pi}{4} [1]
For the inverse trigonometric functions, which of the following Principal Value Branch is not correctly defined ?
(D) \text{cosec}^{-1} : R - (-1, 1) \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
The value of \sin^{-1}\left(\sin\frac{3\pi}{5}\right) is :
(D) \frac{2\pi}{5}
[\tan^{-1}\sqrt{3} - \sec^{-1}(-2)] is equal to :
(B) -\frac{\pi}{3}
The graph of a trigonometric function is as shown. Which of the following will represent graph of its inverse ?
(C) Graph (C)
If y = \sin^{-1}x,\ -1 \le x \le 0, then the range of y is
(B) \left[-\frac{\pi}{2}, 0\right]
The principal value of \sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right) is :
(C) \frac{\pi}{3}
The principal value of \cot^{-1}\left(-\frac{1}{\sqrt3}\right) is :
(D) \frac{2\pi}{3}
The following graph is a combination of :
(C) y = \sin^{-1} x and y = \sin x
\left[\sec^{-1}(-\sqrt{2}) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\right] is equal to :
(D) \frac{7\pi}{12}
The given graph illustrates :
(A) y=\tan^{-1}x [1]
Domain of f(x)=\cos^{-1}x+\sin x is :
(C) [-1,1] [1]
Assertion (A) : Set of values of \sec^{-1}\left(\frac{\sqrt3}{2}\right) is a null set.
Reason (R) : \sec^{-1}x is defined for x\in\mathbb{R}-(-1,1).
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1]
The domain of f(x) = \cos^{-1}(2x) is :
(D) \left[-\frac{1}{2}, \frac{1}{2}\right]
Assertion (A) : Domain of y=\cos^{-1}(x) is [-1,1].
Reason (R) : The range of the principal value branch of y=\cos^{-1}(x) is [0,\pi]-\left\{\frac{\pi}{2}\right\}.
(C) Assertion (A) is true, but Reason (R) is false. [1]
Assertion (A) : \cos^{-1}\left(\cos\frac{13\pi}{6}\right) is equal to \frac{\pi}{6}.
Reason (R) : The range of the principal value branch of the function y=\cos^{-1}x is [0,\pi].
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
[1]Assertion (A) : The range of the function f(x)=2\sin^{-1}x+\frac{3\pi}{2}, where x\in[-1,1], is \left[\frac{\pi}{2},\frac{5\pi}{2}\right].
Reason (R) : The range of the principal value branch of \sin^{-1}(x) is [0,\pi].
(C) Assertion is True, Reason is False
Assertion (A) : Maximum value of (\cos^{-1} x)^2 is \pi^2.
Reason (R) : Range of the principal value branch of \cos^{-1}x is \left[\frac{-\pi}{2}, \frac{\pi}{2}\right].
(C) (A) is true, but (R) is false.
Assertion (A) : Range of [\sin^{-1} x + 2\cos^{-1} x] is [0, \pi].
Reason (R) : Principal value branch of \sin^{-1} x has range \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
(D) Assertion (A) is false and Reason (R) is true.
Assertion (A) : All trigonometric functions have their inverses over their respective domains.
Reason (R) : The inverse of \tan^{-1}x exists for some x\in\mathbb{R}.
(d) Assertion (A) is false and reason (R) is true.
\sin\left[\frac{\pi}{3}+\sin^{-1}\left(\frac{1}{2}\right)\right] is equal to
(A) 1
Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b), (c) and (d) as given below.
Assertion (A) : The principal value of \cot^{-1}(\sqrt{3}) is \frac{\pi}{6}.
Reason (R) : Domain of \cot^{-1} x is \mathbb{R} - \{-1, 1\}.
Official marking scheme for this paper is not available.
Evaluate \tan\left(\tan^{-1}(-1)+\frac{\pi}{3}\right)
\tan\left(\tan^{-1}(-1)+\frac{\pi}{3}\right)=\tan\left(-\frac{\pi}{4}+\frac{\pi}{3}\right) [½]
=\frac{\tan\frac{\pi}{3}-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\tan\frac{\pi}{4}} [1]
=\frac{\sqrt3-1}{1+\sqrt3}\ or\ 2-\sqrt3 [½]
Find the domain of \cos^{-1}(3x-2)
For domain, -1\le3x-2\le1 [½]
\Rightarrow 1\le3x\le3 [½]
\Rightarrow \frac13\le x\le1 [½]
So, domain of \cos^{-1}(3x-2) is \left[\frac13,1\right] [½]
Simplify : \tan^{-1}\left(\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right),\ 0 \lt x \lt \frac{\pi}{4}.
Given expression = \tan^{-1}\left(\frac{1 - \tan 2x}{1 + \tan 2x}\right) [1]
= \tan^{-1}\left[\tan\left(\frac{\pi}{4} - 2x\right)\right] = \frac{\pi}{4} - 2x [1]
Evaluate : \tan\left(\sin^{-1}1 - \cos^{-1}\left(-\frac{1}{2}\right)\right)
Given expression = \tan\left(\frac{\pi}{2} - \frac{2\pi}{3}\right) [1]
= \tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} [1]
Find the value of \sin[\cot^{-1}\sqrt{2}\,(\cos(\tan^{-1} 1))].
\sin\left[\cot^{-1}\sqrt{2}(\cos(\tan^{-1}1))\right] = \sin\left[\cot^{-1}\sqrt{2}\left(\cos\frac{\pi}{4}\right)\right] = \sin(\cot^{-1}1) [1]
= \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} [1]
Evaluate :
\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left(\sin\left(\frac{-\pi}{2}\right)\right) + \tan^{-1}\left(\tan\frac{2\pi}{3}\right)
= -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4} - \frac{\pi}{3} [1½] (For any 3 correct result)
= \frac{-5\pi}{12} [½]
Evaluate \sin\left[\tan^{-1}\tan\left(\frac{3\pi}{4}\right)\right].
Given expression = \sin[\tan^{-1}(-1)] or \sin\left[\tan^{-1}\tan\left(\pi - \frac{\pi}{4}\right)\right] [1]
= \sin\left(-\frac{\pi}{4}\right) [½]
= -\frac{1}{\sqrt{2}} [½]
Evaluate : \tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]
\tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]
=\tan^{-1}\left[2\sin\left(2\times\frac{\pi}{6}\right)\right]=\tan^{-1}\left[2\sin\frac{\pi}{3}\right] [1]
=\tan^{-1}\left[2\times\frac{\sqrt{3}}{2}\right]=\tan^{-1}\sqrt{3}=\frac{\pi}{3} [1]
Simplify \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right).
Put x=\tan\theta\Rightarrow\theta=\tan^{-1}x [½]
Now \sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)
=\sin^{-1}\left(\frac{\tan\theta}{\sec\theta}\right)=\sin^{-1}(\sin\theta) [1]
=\theta=\tan^{-1}x [½]
Find domain of \sin^{-1}\sqrt{x-1}.
Here -1\le\sqrt{x-1}\le 1 [1]
\Rightarrow 0\le x-1\le 1\Rightarrow 1\le x\le 2
Hence, domain is x\in[1,2] [1]
Find the domain of the function f(x)=\cos^{-1}(x^2-4).
Domain of \cos^{-1}x is [-1,1] [1]
\Rightarrow -1\le x^2-4\le1\Rightarrow 3\le x^2\le5 [½]
\Rightarrow x\in[-\sqrt5,-\sqrt3]\cup[\sqrt3,\sqrt5] [½]
Find the domain of f(x) = \sin^{-1}(-x^2).
-1 \le -x^2 \le 1 \Rightarrow -1 \le -x^2 \le 0 [1]
\Rightarrow 0 \le x^2 \le 1 \Rightarrow -1 \le x \le 1 [1]
Find the principal value of \cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right).
\cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)
= \left(\pi - \frac{\pi}{3}\right) + 2\left(\frac{\pi}{6}\right) [1 + ½]
= \pi [½]
Prove that :
Put x = \tan^2\theta \Rightarrow \theta = \tan^{-1}\sqrt{x} [½]
RHS = \frac{1}{2}\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)
= \frac{1}{2}\cos^{-1}(\cos 2\theta) [1]
= \frac{1}{2}(2\theta)
= \theta = \tan^{-1}\sqrt{x} = LHS [½]
Find the value of \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right].
The given expression = \frac{-\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4} [1½]
= \frac{-\pi}{12} [½]
Find the domain of the function f(x) = \sin^{-1}(x^2 - 4). Also, find its range.
-1 \le x^2 - 4 \le 1 [½]
\Rightarrow 3 \le x^2 \le 5
Domain = [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}] [½]
Range = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] [1]
Evaluate :
\sec^2\left(\tan^{-1}\frac{1}{2}\right) + \operatorname{cosec}^2\left(\cot^{-1}\frac{1}{3}\right)
= \left[1 + \tan^2\left(\tan^{-1}\frac{1}{2}\right)\right] + \left[1 + \cot^2\left(\cot^{-1}\frac{1}{3}\right)\right] [1]
= \left[1 + \left(\frac{1}{2}\right)^2\right] + \left[1 + \left(\frac{1}{3}\right)^2\right] [½]
= \frac{85}{36} [½]
Express \tan^{-1}\left(\frac{\cos x}{1-\sin x}\right), where \frac{-\pi}{2} \lt x \lt \frac{\pi}{2} in the simplest form.
y = \tan^{-1}\left[\frac{\cos x}{1-\sin x}\right] = \tan^{-1}\left[\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{(\cos\frac{x}{2}-\sin\frac{x}{2})^2}\right] [½]
y = \tan^{-1}\left[\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right] = \tan^{-1}\left[\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right] = \left(\frac{\pi}{4}+\frac{x}{2}\right) [1½]
Find the principal value of \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right).
\tan^{-1}(1) + \left[\pi - \cos^{-1}\left(\frac{1}{2}\right)\right] - \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} + \left(\pi - \frac{\pi}{3}\right) - \frac{\pi}{4} [1½]
= \frac{2\pi}{3} [½]
Find value of k if
k\tan\left(2\cos^{-1}\frac{\sqrt3}{2}\right)=\sin\frac{\pi}{3} [½]
\Rightarrow k\tan\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2} [½+½]
\Rightarrow k\sqrt3=\frac{\sqrt3}{2}\quad\therefore k=\frac12 [½]
Find the value of \cos^{-1}\left(\frac{1}{2}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\text{cosec}^{-1}(-2).
\cos^{-1}\left(\frac{1}{2}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\text{cosec}^{-1}(-2)
=\frac{\pi}{3}+\frac{\pi}{6}-\frac{\pi}{6} [1½]
=\frac{\pi}{3} [½]
Evaluate :
Required value =\frac{\pi}{4}+\frac{3\pi}{4}+\frac{\pi}{4} [1½]
=\frac{5\pi}{4} [½]
Evaluate \sin^{-1}\left(\sin\frac{3\pi}{4}\right) + \cos^{-1}(\cos\pi) + \tan^{-1}(1).
\sin^{-1}\left(\sin\frac{3\pi}{4}\right) + \cos^{-1}(\cos\pi) + \tan^{-1}(1) = \frac{\pi}{4} + \pi + \frac{\pi}{4} [1½]
= \frac{3\pi}{2} [½]
Draw the graph of \cos^{-1} x, where x \in [-1, 0]. Also, write its range.
Correct Graph [1]
Range: \left[\frac{\pi}{2}, \pi\right] [1]
Evaluate : 3\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) + 2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \cos^{-1}(0)
Given expression = \frac{3\pi}{4} + \frac{2\pi}{6} + \frac{\pi}{2} [1½]
= \frac{19\pi}{12} [½]
Draw the graph of f(x) = \sin^{-1} x, x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]. Also, write range of f(x).
Correct graph [1]
Here, the points A, B, C and D are respectively \left(0, \frac{\pi}{4}\right), \left(0, -\frac{\pi}{4}\right), \left(\frac{1}{\sqrt{2}}, 0\right), \left(-\frac{1}{\sqrt{2}}, 0\right).
Range = \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] [1]
Find the domain of y=\sin^{-1}(x^2-4).
Domain of \sin^{-1}x is -1\le x\le 1
\therefore -1\le x^2-4\le 1 [½]
or x^2\ge 3,\ x^2\le 5
\Rightarrow x\ge\sqrt{3} or x\le-\sqrt{3},\ x\le\sqrt{5} or x\ge-\sqrt{5} [1]
\therefore Domain is [-\sqrt{5},-\sqrt{3}]\cup[\sqrt{3},\sqrt{5}] [½]
Evaluate :
\cos^{-1}\left[\cos\left(-\frac{7\pi}{3}\right)\right]=\cos^{-1}\left[\cos\left(\frac{7\pi}{3}\right)\right] [1]
=\cos^{-1}\left[\cos\left(2\pi+\frac{\pi}{3}\right)\right] [½]
=\cos^{-1}\left[\cos\frac{\pi}{3}\right]=\frac{\pi}{3}. [½]
Write the domain and range (principle value branch) of the following functions :
f(x)=\tan^{-1}x
Domain = R ; Range = \left(-\frac{\pi}{2},\frac{\pi}{2}\right) [1+1]
Simplify :
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Case Study – 3
If a function f : X \to Y defined as f(x) = y is one-one and onto, then we can define a unique function g : Y \to X such that g(y) = x, where x \in X and y = f(x),\ y \in Y. Function g is called the inverse of function f.
The domain of sine function is R and function sine : R \to R is neither one-one nor onto. The following graph shows the sine function.
Let sine function be defined from set A to [-1, 1] such that inverse of sine function exists, i.e., \sin^{-1}x is defined from [-1, 1] to A.
On the basis of the above information, answer the following questions :
(i) If A is the interval other than principal value branch, give an example of one such interval. [1]
(ii) If \sin^{-1}(x) is defined from [-1, 1] to its principal value branch, find the value of \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1). [1]
(iii) (a) Draw the graph of \sin^{-1}x from [-1, 1] to its principal value branch. [2]
OR
(iii) (b) Find the domain and range of f(x) = 2\sin^{-1}(1 - x). [2]
(i) \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] or any other interval corresponding to the domain [-1,1] [1]
(ii) \sin^{-1}\left(\frac{-1}{2}\right) - \sin^{-1}(1)
= \frac{-\pi}{6} - \frac{\pi}{2}
= \frac{-4\pi}{6}\ \text{or}\ \frac{-2\pi}{3} [1]
(iii) (a) [Graph of y = \sin^{-1}x for -1 \le x \le 1, rising from \left(-1, \frac{-\pi}{2}\right) through O to \left(1, \frac{\pi}{2}\right)] Correct graph - [2]
OR
(b) f(x) = 2\sin^{-1}(1 - x)
-1 \le 1 - x \le 1 [½]
\Rightarrow -2 \le -x \le 0
\Rightarrow 0 \le x \le 2
Domain = [0, 2] [½]
\frac{-\pi}{2} \le \sin^{-1}(1 - x) \le \frac{\pi}{2} [½]
-\pi \le 2\sin^{-1}(1 - x) \le \pi
So range = [-\pi, \pi] [½]
Find the domain of g(x) = \cos^{-1}(x^2 - 1). Hence, find the value of x for which g(x) = \frac{\pi}{3}.
Also, write the range of \cos^{-1}x other than its principal branch.
Since domain of \cos^{-1}x is [-1, 1]. \therefore -1 \le x^2 - 1 \le 1 [1]
\Rightarrow 0 \le x^2 \le 2 \Rightarrow -\sqrt{2} \le x \le \sqrt{2}
\therefore Domain of g(x) = [-\sqrt{2}, \sqrt{2}] [1]
When g(x) = \frac{\pi}{3}, we have \cos^{-1}(x^2 - 1) = \frac{\pi}{3}
\Rightarrow x^2 - 1 = \cos\frac{\pi}{3} = \frac{1}{2} [1]
\Rightarrow x^2 = \frac{3}{2} \Rightarrow x = \sqrt{\frac{3}{2}},\ -\sqrt{\frac{3}{2}} [1]
Range of \cos^{-1}x other than its principal branch is [-\pi, 0] or [\pi, 2\pi]. [1]
Note: Any valid branch like [2\pi, 3\pi], [-2\pi, -\pi], ...etc can be considered.
If A=\begin{bmatrix}2&1&-2\\3a&0&4\\b&2c&1\end{bmatrix} is a symmetric matrix, then value of \sqrt{6a+2b+3c} is:
(B) 2
3a=1,\ b=-2,\ 2c=4\Rightarrow\sqrt{6a+2b+3c}=\sqrt{2-4+6}=\sqrt4=2
If \begin{bmatrix}1&x\end{bmatrix}\begin{bmatrix}1&0\\0&2\end{bmatrix}\begin{bmatrix}4\\-1\end{bmatrix}=O, then value of x is
(D) 2
\begin{bmatrix}1&x\end{bmatrix}\begin{bmatrix}1&0\\0&2\end{bmatrix}\begin{bmatrix}4\\-1\end{bmatrix}=O\Rightarrow\begin{bmatrix}1&2x\end{bmatrix}\begin{bmatrix}4\\-1\end{bmatrix}=O\Rightarrow4-2x=0\Rightarrow x=2
If for three matrices A=[a_{ij}]_{m\times 4}, B=[b_{ij}]_{n\times 3} and C=[c_{ij}]_{p\times q} products AB and AC both are defined and are square matrices of same order, then value of m,n,p and q are:
AB is defined so n=4
AC is defined so p=4
AB and AC are square matrices of same order
so m\times 3=m\times q\Rightarrow q=3=m
(A) m=q=3\ \text{and}\ n=p=4 [1]
If the matrix A=\begin{bmatrix}0&r&-2\\3&p&t\\q&-4&0\end{bmatrix} is skew-symmetric, then value of \frac{q+t}{p+r} is....
As A is skew symmetric
So p=0,q=2,r=-3,t=4
So \frac{q+t}{p+r}=\frac{6}{-3}=-2
(A) -2 [1]
Which of the following cannot be the order of a row-matrix ?
(A) 2 \times 1
Which of the following properties is/are true for two matrices of suitable orders ?
(i) (A + B)' = A' + B' (ii) (A - B)' = B' - A'
(iii) (AB)' = A'B' (iv) (kAB)' = kB'A' (k is a scalar)
(D) (i) and (iv)
If A and B are skew symmetric matrices of same order, then which of the following matrices is also skew symmetric ?
(D) A – B
If A and B are square matrices of same order, then which of the following statements is/are always true ?
(i) (A + B)(A - B) = A^2 - B^2
(ii) AB = BA
(iii) (A + B)^2 = A^2 + AB + BA + B^2
(iv) AB = 0 \Rightarrow A = 0 \text{ or } B = 0
(C) Only (iii)
If A = \begin{bmatrix} 1 & a & b \\ -1 & 2 & c \\ 0 & 5 & 3 \end{bmatrix} is a symmetric matrix, then the value of 3a + b + c is
(A) 2
If A = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} and A + A' = I, then the value of x \in \left[0, \frac{\pi}{2}\right] is
(C) \frac{\pi}{3}
If A^2 = 4A + 3I and A^{-1} = xA + yI, then the value of (x + y) is :
(A) -1
If A and B are skew-symmetric matrices of same order, then AB' + BA' is a/an :
(A) symmetric matrix
If \begin{bmatrix}4\\1\\3\end{bmatrix}A = \begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}, then order of A must be :
(B) 1 \times 3
If a square matrix A is such that A^2 = A and (I - A)^3 = xA + I, then value of x must be :
(D) -1
If A = [a_{ij}] is a 2 \times 2 matrix whose elements are given by a_{ij} = \frac{|i - 3j|}{2}, then A' is :
(B) \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{5}{2} & 2 \end{bmatrix} [1]
Assertion (A) : If A and B are two square matrices such that AB and BA are defined, then it is not necessary that AB = BA.
Reason (R) : Product of two diagonal matrices of same order is commutative.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1]
If matrix A = \begin{bmatrix}-p & q\\ r & p\end{bmatrix} is such that A^2 = I, then :
(B) 1 - p^2 - qr = 0
If A is a square matrix such that A^2 = A, then (A - I)^3 - A is equal to :
(B) -I
Let A = \begin{bmatrix}0 & -3 & 4\\ 1 & 0 & 2\end{bmatrix} and B = \begin{bmatrix}-3 & 0 & 1\\ 2 & 4 & 0\end{bmatrix}. If A + B + C = O, then matrix C is :
(C) \begin{bmatrix}3 & 3 & -5\\ -3 & -4 & -2\end{bmatrix}
If A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} satisfies A' + A = I, then the value of \theta is :
(B) \frac{\pi}{3}
If A is a square matrix, then which of the following is not true ?
(C) A + A' is a skew symmetric matrix
If A = \begin{bmatrix} 1 & -2 & 3 \\ 3 & 2 & -1 \end{bmatrix}, B = \begin{bmatrix} 2 & 3 \\ -1 & 2 \\ 4 & -5 \end{bmatrix} and C = [C_{ij}] = AB, then C_{12} is :
(B) -16
If A=\begin{bmatrix}-1&0&0\\0&1&0\\0&0&1\end{bmatrix}, then A^{-1} is
(D) \begin{bmatrix}-1&0&0\\0&1&0\\0&0&1\end{bmatrix}
Let A=\begin{bmatrix}1&-2&-1\\0&4&-1\\-3&2&1\end{bmatrix}, B=\begin{bmatrix}-2\\-5\\-7\end{bmatrix}, C=\begin{bmatrix}9&8&7\end{bmatrix}, which of the following is defined ?
(A) Only AB
If A=\begin{bmatrix}7&0&x\\0&7&0\\0&0&y\end{bmatrix} is a scalar matrix, then y^x is equal to
(B) 1
If A = \begin{bmatrix}1 & 12 & 4y\\ 6x & 5 & 2x\\ 8x & 4 & 6\end{bmatrix} is a symmetric matrix, then (2x + y) is
(D) 8
Which of the following can be both a symmetric and skew-symmetric matrix ?
(C) Null Matrix
Four friends Abhay, Bina, Chhaya and Devesh were asked to simplify 4AB + 3(AB + BA) - 4BA, where A and B are both matrices of order 2 \times 2. It is known that A \ne B \ne I and A^{-1} \ne B.
Their answers are given as :
Abhay : 6 AB
Bina : 7 AB – BA
Chhaya : 8 AB
Devesh : 7 BA – AB
Who answered it correctly ?
(B) Bina
If A and B are square matrices of order m such that A^2 - B^2 = (A - B)(A + B), then which of the following is always correct ?
(B) AB = BA
Assertion (A) : A = diag [3 5 2] is a scalar matrix of order 3 \times 3.
Reason (R) : If a diagonal matrix has all non-zero elements equal, it is known as a scalar matrix.
(D) Assertion (A) is false and Reason (R) is true.
If A and B are square matrices of same order such that AB = A and BA = B, then A^2+B^2 is equal to :
(A) A + B
The matrix \begin{bmatrix}0&1&-2\\-1&0&-7\\2&7&0\end{bmatrix} is a :
(C) skew symmetric matrix
If A=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}, then A^3 is :
(B) \begin{bmatrix}125&0&0\\0&125&0\\0&0&125\end{bmatrix}
If A=\begin{bmatrix}1&2&3\\-4&3&7\end{bmatrix} and B=\begin{bmatrix}4&3\\-1&2\\0&5\end{bmatrix}, then the correct statement is :
(C) AB and BA, both are defined.
Let both AB' and B'A be defined for matrices A and B. If order of A is n \times m, then the order of B is :
(B) n \times m
If A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}, then A is a/an :
(C) symmetric matrix
Sum of two skew-symmetric matrices of same order is always a/an :
(A) skew-symmetric matrix
What is the total number of possible matrices of order 3\times3 with each entry as \sqrt2 or \sqrt3 ?
(B) 512 [1]
The matrix A=\begin{bmatrix}\sqrt3&0&0\\0&\sqrt2&0\\0&0&\sqrt5\end{bmatrix} is a/an :
(D) symmetric matrix [1]
If \begin{bmatrix}2x-1&3x\\0&y^2-1\end{bmatrix}=\begin{bmatrix}x+3&12\\0&35\end{bmatrix}, then the value of (x-y) is :
(B) -2 or 10 [1]
If the matrix A = [a_{ij}]_{2\times 2} is such that a_{ij} = \begin{cases}1, & \text{if } i \ne j\\ 0, & \text{if } i = j\end{cases}, then A + A^2 is equal to :
(C) \begin{bmatrix}1&1\\1&1\end{bmatrix}
For a non-singular matrix X, if X^2 = I, then X^{-1} is equal to :
(A) X
If a matrix has 36 elements, the number of possible orders it can have, is :
(D) 9
If \begin{bmatrix} x+y & 2 \\ 5 & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}, then the value of \left(\frac{24}{x} + \frac{24}{y}\right) is :
(D) 18
If A and B are two non-zero square matrices of same order such that (A + B)^2 = A^2 + B^2, then :
(B) AB = -BA
If the sum of all the elements of a 3 \times 3 scalar matrix is 9, then the product of all its elements is :
(A) 0
If A = [a_{ij}] be a 3 \times 3 matrix, where a_{ij} = i - 3j, then which of the following is false ?
(C) a_{13} \gt a_{31}
If F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} and [F(x)]^2 = F(kx), then the value of k is :
(B) 2
Assertion (A) : For any symmetric matrix A, B′AB is a skew-symmetric matrix.
Reason (R) : A square matrix P is skew-symmetric if P′ = – P.
(D) Assertion (A) is false, but Reason (R) is true
If A = [a_{ij}] is an identity matrix, then which of the following is true ?
(D) a_{ij} = \begin{cases}0, & \text{if } i \ne j\\ 1, & \text{if } i = j\end{cases}
If A and B are two skew symmetric matrices, then (AB + BA) is :
(B) a symmetric matrix
Assertion (A) : Every scalar matrix is a diagonal matrix.
Reason (R) : In a diagonal matrix, all the diagonal elements are 0.
(C) Assertion (A) is true, but Reason (R) is false.
If \begin{bmatrix} a & c & 0 \\ b & d & 0 \\ 0 & 0 & 5 \end{bmatrix} is a scalar matrix, then the value of a + 2b + 3c + 4d is :
(D) 25
[1]Given that A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}, matrix A is :
(B) \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}
[1]If A = \begin{bmatrix} 2 & 1 \\ -4 & -2 \end{bmatrix}, then the value of I - A + A^2 - A^3 + \ldots is :
(A) \begin{bmatrix} -1 & -1 \\ 4 & 3 \end{bmatrix}
[1]Given that [1 \;\; x]\begin{bmatrix} 4 & 0 \\ -2 & 0 \end{bmatrix} = 0, the value of x is :
(C) 2
[1]If A=\begin{bmatrix}a & c & -1\\ b & 0 & 5\\ 1 & -5 & 0\end{bmatrix} is a skew-symmetric matrix, then the value of 2a-(b+c) is :
(A) 0 [1]
If \begin{bmatrix}x & 2 & 0\end{bmatrix}\begin{bmatrix}5\\ -1\\ x\end{bmatrix}=\begin{bmatrix}3 & 1\end{bmatrix}\begin{bmatrix}-2\\ x\end{bmatrix}, then value of x is :
(A) -1 [1]
Find the matrix A^2, where A=[a_{ij}] is a 2\times2 matrix whose elements are given by a_{ij}=\text{maximum}(i,j)-\text{minimum}(i,j) :
(C) \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} [1]
For two matrices A and B, given that A^{-1}=\frac{1}{4}B, then inverse of (4A) is :
(D) \frac{1}{16}B
[1]If X, Y and XY are matrices of order 2\times3, m\times n and 2\times5 respectively, then number of elements in matrix Y is :
(C) 15
[1]If for a square matrix A, A^2-3A+I=O and A^{-1}=xA+yI, then the value of x+y is :
(B) 2
If A=\begin{bmatrix}3&4\\5&2\end{bmatrix} and 2A+B is a null matrix, then B is equal to :
(B) \begin{bmatrix}-6&-8\\-10&-4\end{bmatrix}
If A=\begin{bmatrix}0&1\\-1&0\end{bmatrix} and (3I+4A)(3I-4A)=x^2I, then the value(s) x is/are :
(C) \pm5
If A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, then A^{2023} is equal to
(C) \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
If \begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix} = P + Q, where P is a symmetric and Q is a skew symmetric matrix, then Q is equal to
(B) \begin{bmatrix} 0 & -5/2 \\ 5/2 & 0 \end{bmatrix}
If A = \begin{bmatrix} 1 & 4 & x\\ z & 2 & y\\ -3 & -1 & 3\end{bmatrix} is a symmetric matrix, then the value of x + y + z is :
(D) 0
A and B are skew-symmetric matrices of same order. AB is symmetric, if :
(C) AB = BA
For what value of x \in \left[0, \frac{\pi}{2}\right], is A + A' = \sqrt{3}\, I, where A = \begin{bmatrix} \cos x & \sin x\\ -\sin x & \cos x\end{bmatrix} ?
(B) \frac{\pi}{6}
If x\begin{bmatrix}1\\2\end{bmatrix}+y\begin{bmatrix}2\\5\end{bmatrix}=\begin{bmatrix}4\\9\end{bmatrix}, then :
(b) x=2,\ y=1
The product \begin{bmatrix}a & b\\-b & a\end{bmatrix}\begin{bmatrix}a & -b\\b & a\end{bmatrix} is equal to :
(a) \begin{bmatrix}a^2+b^2 & 0\\0 & a^2+b^2\end{bmatrix}
If A is a square matrix and A^2=A, then (I+A)^2-3A is equal to :
(a) I
If a matrix A=[1\ \ 2\ \ 3], then the matrix AA' (where A' is the transpose of A) is :
(d) [14]
If for a square matrix A, A^2-A+I=O, then A^{-1} equals
(C) I – A
If A=\begin{bmatrix}1&0\\2&1\end{bmatrix}, B=\begin{bmatrix}x&0\\1&1\end{bmatrix} and A=B^2, then x equals
(C) 1
If matrix A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} and A^2 = kA, then the value of k is :
Official marking scheme for this paper is not available.
It is given that X\begin{bmatrix} 3 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}. Then matrix X is :
Official marking scheme for this paper is not available.
Solve the matrix equation for x and y.
x^2 - 3x = -2 \Rightarrow x^2 - 3x + 2 = 0 \Rightarrow x = 1, 2 [1]
y^2 - 9y = -20 \Rightarrow y^2 - 9y + 20 = 0 \Rightarrow y = 4, 5 [1]
If A=\begin{bmatrix}2&3\\-1&2\end{bmatrix}, then show that A^2-4A+7I=0.
A^2=\begin{bmatrix}2&3\\-1&2\end{bmatrix}\begin{bmatrix}2&3\\-1&2\end{bmatrix}=\begin{bmatrix}1&12\\-4&1\end{bmatrix}. [1]
L.H.S.=A^2-4A+7I=\begin{bmatrix}1&12\\-4&1\end{bmatrix}-\begin{bmatrix}8&12\\-4&8\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O=R.H.S. [1]
For two matrices, P=\begin{bmatrix}2&0&0\\0&1&2\end{bmatrix} and Q=\begin{bmatrix}1&0\\0&2\\1&0\end{bmatrix}, find PQ and QP, whichever is defined. Also, find the inverse of the product matrix, if it exists.
Let A=PQ and B=QP. Then
A=PQ=\begin{bmatrix}2&0&0\\0&1&2\end{bmatrix}\begin{bmatrix}1&0\\0&2\\1&0\end{bmatrix}=\begin{bmatrix}2&0\\2&2\end{bmatrix} [1]
B=QP=\begin{bmatrix}1&0\\0&2\\1&0\end{bmatrix}\begin{bmatrix}2&0&0\\0&1&2\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&4\\2&0&0\end{bmatrix} [1]
Here |A|=4\ne0
So, A is a non-singular matrix, hence its inverse exists.
Now A^{-1}=\frac{1}{|A|}adj\,A=\frac14\begin{bmatrix}2&0\\-2&2\end{bmatrix}=\begin{bmatrix}\frac12&0\\-\frac12&\frac12\end{bmatrix} [½]
Here |B|=0
So, B is a singular matrix, hence its inverse does not exist. [½]
If A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix}, then compute A^{2} - 7A + 10I.
A^{2} = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix}\begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} [1½] (½ for any three correct elements)
A^{2} - 7A + 10I = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} - \begin{bmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{bmatrix} + \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix} [1]
A^{2} - 7A + 10I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} [½]
Express the matrix A = \begin{bmatrix} 2 & 4 \\ 5 & 6 \end{bmatrix} as the sum of a symmetric matrix and a skew symmetric matrix.
As, A = \begin{bmatrix} 2 & 4 \\ 5 & 6 \end{bmatrix} \Rightarrow A' = \begin{bmatrix} 2 & 5 \\ 4 & 6 \end{bmatrix} [½]
So, A + A' = \begin{bmatrix} 4 & 9 \\ 9 & 12 \end{bmatrix} and A - A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} [1]
Thus, A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 6 \end{bmatrix} + \begin{bmatrix} 0 & \frac{-1}{2} \\ \frac{1}{2} & 0 \end{bmatrix} [1½]
Let A = \begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix} and C = \begin{bmatrix} 3 & 4 & 2 \\ 12 & 16 & 8 \\ -6 & -8 & -4 \end{bmatrix} be two matrices. Then, find the matrix B if AB = C.
Let B = \begin{bmatrix} x & y & z \end{bmatrix} [½]
AB = C \Rightarrow \begin{bmatrix} x & y & z \\ 4x & 4y & 4z \\ -2x & -2y & -2z \end{bmatrix} = \begin{bmatrix} 3 & 4 & 2 \\ 12 & 16 & 8 \\ -6 & -8 & -4 \end{bmatrix} which gives x = 3,\ y = 4 and z = 2 [2]
B = \begin{bmatrix} 3 & 4 & 2 \end{bmatrix} [½]
A shopkeeper sells 50 Chemistry, 60 Physics and 35 Maths books on day I and sells 40 Chemistry, 45 Physics and 50 Maths books on day II. If the selling price for each such subject book is ₹ 150 (Chemistry), ₹ 175 (Physics) and ₹ 180 (Maths), then find his total sale in two days, using matrix method. If cost price of all the books together is ₹ 35,000, what profit did he earn after the sale of two days ?
Let A=\begin{bmatrix}50&60&35\\40&45&50\end{bmatrix}\begin{matrix}\text{Day I}\\\text{Day II}\end{matrix}, B=\begin{bmatrix}150\\175\\180\end{bmatrix} be the day wise sale and the selling price per subject, matrices respectively. [1]
Total sales day wise =\begin{bmatrix}50&60&35\\40&45&50\end{bmatrix}\begin{bmatrix}150\\175\\180\end{bmatrix}=\begin{bmatrix}24{,}300\\22{,}875\end{bmatrix}\begin{matrix}\text{Day I}\\\text{Day II}\end{matrix} [1]
Total sales in two days = ₹ 24,300 + ₹ 22,875 = ₹ 47,175 [½]
Profit = ₹ 47,175 – ₹ 35,000 = ₹ 12,175. [½]
Find a matrix A such that
Also, find A^{-1}.
A=\begin{bmatrix}17&10\\0&-16\end{bmatrix}\begin{bmatrix}4&0\\-1&-2\end{bmatrix}^{-1} [½]
=-\frac{1}{8}\begin{bmatrix}17&10\\0&-16\end{bmatrix}\begin{bmatrix}-2&0\\1&4\end{bmatrix} [1]
=\begin{bmatrix}3&-5\\2&8\end{bmatrix} [½]
Also, A^{-1}=\frac{1}{34}\begin{bmatrix}8&5\\-2&3\end{bmatrix} [1]
Given a square matrix A of order 3 such that A^{2}=\begin{bmatrix}3&-4&4\\0&-1&0\\-2&2&-3\end{bmatrix}, show that A^{3}=A^{-1}.
A^{3}=A^{-1}\Leftrightarrow A^{4}=I [1½]
A^{4}=A^{2}A^{2}=\begin{bmatrix}3&-4&4\\0&-1&0\\-2&2&-3\end{bmatrix}\begin{bmatrix}3&-4&4\\0&-1&0\\-2&2&-3\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I [1½]
If A=\begin{bmatrix}1&2&3\\3&-2&1\\4&2&1\end{bmatrix}, then show that A^3-23A-40I=O.
Getting A^2=\begin{bmatrix}19&4&8\\1&12&8\\14&6&15\end{bmatrix} [1]
Getting A^3=\begin{bmatrix}63&46&69\\69&-6&23\\92&46&63\end{bmatrix} [1]
A^3-23A-40I=
\begin{bmatrix}63&46&69\\69&-6&23\\92&46&63\end{bmatrix}-\begin{bmatrix}23&46&69\\69&-46&23\\92&46&23\end{bmatrix}-\begin{bmatrix}40&0&0\\0&40&0\\0&0&40\end{bmatrix} [½]
=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O [½]
Given that P = \begin{bmatrix}2 & -1\\3 & 4\end{bmatrix}, Q = \begin{bmatrix}5 & 2\\7 & 4\end{bmatrix} and R = \begin{bmatrix}2 & 5\\3 & 8\end{bmatrix}, find a matrix S such that PQ - RS is a null matrix.
Let S = \begin{bmatrix}a & b\\c & d\end{bmatrix} [½]
PQ = \begin{bmatrix}3 & 0\\43 & 22\end{bmatrix} [1]
RS = \begin{bmatrix}2a + 5c & 2b + 5d\\3a + 8c & 3b + 8d\end{bmatrix} [1]
Given that PQ - RS = null matrix \Rightarrow PQ = RS
On comparing corresponding elements,
2a + 5c = 3,\quad 2b + 5d = 0,\quad 3a + 8c = 43,\quad 3b + 8d = 22 [½]
By solving,
a = -191,\ b = -110,\ c = 77,\ d = 44 [1½]
Hence, S = \begin{bmatrix}-191 & -110\\77 & 44\end{bmatrix} [½]
Alternative method
Given that PQ - RS = null matrix \Rightarrow RS = PQ [1]
\begin{bmatrix}2 & 5\\3 & 8\end{bmatrix}S = \begin{bmatrix}3 & 0\\43 & 22\end{bmatrix} [1]
\Rightarrow S = \begin{bmatrix}2 & 5\\3 & 8\end{bmatrix}^{-1}\begin{bmatrix}3 & 0\\43 & 22\end{bmatrix} [1]
\Rightarrow S = \begin{bmatrix}8 & -5\\-3 & 2\end{bmatrix}\begin{bmatrix}3 & 0\\43 & 22\end{bmatrix} [1½]
\Rightarrow S = \begin{bmatrix}-191 & -110\\77 & 44\end{bmatrix} [½]
If f(\alpha) = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}, prove that f(\alpha)\cdot f(-\beta) = f(\alpha - \beta)
LHS = f(\alpha)f(-\beta) = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos\beta & \sin\beta & 0 \\ -\sin\beta & \cos\beta & 0 \\ 0 & 0 & 1 \end{bmatrix} [1]
= \begin{bmatrix} \cos\alpha\cos\beta + \sin\alpha\sin\beta & \cos\alpha\sin\beta - \sin\alpha\cos\beta & 0 \\ \sin\alpha\cos\beta - \cos\alpha\sin\beta & \sin\alpha\sin\beta + \cos\alpha\cos\beta & 0 \\ 0 & 0 & 1 \end{bmatrix} [2]
= \begin{bmatrix} \cos(\alpha - \beta) & -\sin(\alpha - \beta) & 0 \\ \sin(\alpha - \beta) & \cos(\alpha - \beta) & 0 \\ 0 & 0 & 1 \end{bmatrix} [1½]
= f(\alpha - \beta) = RHS [½]
If A = \begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3\end{bmatrix}, then show that A^3 - 6A^2 + 7A + 2I = O.
Getting, A^2 = \begin{bmatrix} 5 & 0 & 8\\ 2 & 4 & 5\\ 8 & 0 & 13\end{bmatrix} [1½]
Getting, A^3 = \begin{bmatrix} 21 & 0 & 34\\ 12 & 8 & 23\\ 34 & 0 & 55\end{bmatrix} [1½]
\therefore A^3 - 6A^2 + 7A + 2I =
\begin{bmatrix} 21 & 0 & 34\\ 12 & 8 & 23\\ 34 & 0 & 55\end{bmatrix} - \begin{bmatrix} 30 & 0 & 48\\ 12 & 24 & 30\\ 48 & 0 & 78\end{bmatrix} + \begin{bmatrix} 7 & 0 & 14\\ 0 & 14 & 7\\ 14 & 0 & 21\end{bmatrix} + \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{bmatrix} [1]
= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix} = O [1]
If P and Q are matrices of same order such that P=\begin{bmatrix}2&1\\0&1\end{bmatrix}, |Q|=3, then find |adj(PQ)|
(C) 6
|adj(PQ)|=|PQ|=|P||Q|=2\times3=6 (\because order of P and Q=2\times2)
If A is a square matrix of order 4 and |adj\,A|=27, then A\,(adj\,A) is equal to
|adj\,A|=27\Rightarrow |A|^3=27=3^3\Rightarrow |A|=3
A\,(adj\,A)=|A|\,I=3\,I
(C) 3\,I [1]
The inverse of the matrix \begin{bmatrix}3&0&0\\0&2&0\\0&0&5\end{bmatrix} is...
Inverse of the matrix \begin{bmatrix}3&0&0\\0&2&0\\0&0&5\end{bmatrix}=\begin{bmatrix}\frac13&0&0\\0&\frac12&0\\0&0&\frac15\end{bmatrix}
(B) [1]
Value of the determinant \begin{vmatrix}\cos 67^o&\sin 67^o\\\sin 23^o&\cos 23^o\end{vmatrix} is
\begin{vmatrix}\cos 67^o&\sin 67^o\\\sin 23^o&\cos 23^o\end{vmatrix}=\cos 67^o\cos 23^o-\sin 67^o\sin 23^o=\cos(67^o+23^o)=\cos 90^o=0
(A) 0 [1]
If \Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix} and \Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix}, then
(B) \Delta_2 = -2\Delta_1
One of the values of x for which \begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = 1 is
(B) \frac{\pi}{4}
For a square matrix A, (3A)^{-1} =
(C) \frac{1}{3}A^{-1}
If \begin{vmatrix} -1 & -2 & 5 \\ -2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = -86, then the sum of all possible values of a is
(C) –4
If B(\text{adj } B) = \begin{bmatrix}\frac{1}{3} & 0 & 0\\0 & \frac{1}{3} & 0\\0 & 0 & \frac{1}{3}\end{bmatrix}, then the value of \det(B^{-1}) =
(C) 3
If points (2, 3), (0, 4) and (p, 2) are collinear, then the value of p is :
(C) 4 [1]
If A is a non-singular matrix, then which of the following is not true ?
(A) adj A is singular
If the area of \Delta ABC with vertices A(3, 1), B(-2, 1) and C(0, k) is 5 sq. units, then values of k are :
(B) -1, 3
If \begin{vmatrix} 3x & x+2 \\ 3(x-1) & x-2 \end{vmatrix} = \begin{vmatrix} 1 & 5 \\ 3 & 3 \end{vmatrix}, then the value of x is :
(A) 2
If A and B are square matrices of order 3 such that |A| = 4 and |B| = 7, then |2AB| is :
(D) 224
If A is a square matrix of order 2 such that \det(A)=4, then \det(4\,\text{adj}\,A) is equal to :
(B) 64
If A and B are invertible matrices, then which of the following is not correct ?
(A) (A+B)^{-1}=B^{-1}+A^{-1}
If M and N are square matrices of order 3 such that det (M) = m and MN = mI, then det (N) is equal to :
(D) m^2
If \begin{vmatrix}2x&5\\12&x\end{vmatrix}=\begin{vmatrix}6&-5\\4&3\end{vmatrix}, then the value of x is :
(C) \pm7
If A=[a_{ij}] is a 3\times3 diagonal matrix such that a_{11}=1, a_{22}=5 and a_{33}=-2, then |A| is :
(B) -10
If \begin{bmatrix}4+x&x-1\\-2&3\end{bmatrix} is a singular matrix, then the value of x is :
(C) -2
If A and B are two square matrices each of order 3 with |A|=3 and |B|=5, then |2AB| is :
(B) 120 [1]
Let A be a square matrix of order 3. If |A|=5, then |\text{adj}\,A| is :
(C) 25 [1]
The value of the determinant \begin{vmatrix}\cos 75^\circ & \sin 75^\circ\\ \sin 15^\circ & \cos 15^\circ\end{vmatrix} is :
(B) zero
The cofactor of the element a_{32} in the determinant \Delta = \begin{vmatrix}1&2&3\\2&3&1\\3&1&2\end{vmatrix} is :
(C) 5
If A is an identity matrix of order n, then A (Adj A) is a/an :
(A) identity matrix
\begin{vmatrix} x+1 & x-1 \\ x^2+x+1 & x^2-x+1 \end{vmatrix} is equal to :
(B) 2
Assertion (A) : For matrix A = \begin{bmatrix} 1 & \cos\theta & 1 \\ -\cos\theta & 1 & \cos\theta \\ -1 & -\cos\theta & 1 \end{bmatrix}, where \theta \in [0, 2\pi], |A| \in [2, 4].
Reason (R) : \cos\theta \in [-1, 1],\ \forall\, \theta \in [0, 2\pi].
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of Assertion (A).
If \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc, then the value of k is :
(D) 4
Let A = \begin{bmatrix} a & b\\ c & d\end{bmatrix} be a square matrix such that adj A = A. Then, (a + b + c + d) is equal to :
(A) 2a
If \begin{vmatrix} 1 & 3 & 1\\ k & 0 & 1\\ 0 & 0 & 1\end{vmatrix} = \pm 6, then the value of k is :
(D) \mp 2
If A = \begin{bmatrix} 2 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 5\end{bmatrix}, then A^{-1} is :
(A) \begin{bmatrix} \frac12 & 0 & 0\\ 0 & \frac13 & 0\\ 0 & 0 & \frac15\end{bmatrix}
If A = \begin{bmatrix} -2 & 0 & 0 \\ 1 & 2 & 3 \\ 5 & 1 & -1 \end{bmatrix}, then the value of |A(\text{adj. } A)| is :
(D) 1000
[1]If A is a square matrix of order 3 such that the value of |\text{adj}\cdot A|=8, then the value of |A^T| is :
(D) 2\sqrt{2} [1]
If inverse of matrix \begin{bmatrix}7 & -3 & -3\\ -1 & 1 & 0\\ -1 & 0 & 1\end{bmatrix} is the matrix \begin{bmatrix}1 & 3 & 3\\ 1 & \lambda & 3\\ 1 & 3 & 4\end{bmatrix}, then value of \lambda is :
(D) 4 [1]
If A=\begin{bmatrix}0&1&-1\\1&2&1\\0&3&-2\end{bmatrix}, then the value of |A\,\text{adj}(A)| is :
(A) -1
[1]If |A|=2, where A is a 2\times2 matrix, then |4A^{-1}| equals :
(C) 8
Let A be a 3\times3 matrix such that |\text{adj }A|=64. Then |A| is equal to :
(D) 8 or -8
If \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & a & 1 \end{bmatrix} is non-singular matrix and a \in A, then the set A is
(D) \mathbb{R} - \{4\}
If |A| = |kA|, where A is a square matrix of order 2, then sum of all possible values of k is
(D) 0
If (a, b), (c, d) and (e, f) are the vertices of \Delta ABC and \Delta denotes the area of \Delta ABC, then \begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}^2 is equal to
(B) 4\Delta^2
If A \cdot (\text{adj } A) = \begin{bmatrix} 3 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 3\end{bmatrix}, then the value of |A| + |\text{adj } A| is equal to :
(A) 12
Let A be the area of a triangle having vertices (x_1, y_1), (x_2, y_2) and (x_3, y_3). Which of the following is correct ?
(B) \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix} = \pm 2A
The value of \begin{vmatrix}x+y & y+z & z+x\\z & x & y\\1 & 1 & 1\end{vmatrix} is
(a) 0
If \begin{vmatrix}\alpha&3&4\\1&2&1\\1&4&1\end{vmatrix}=0, then the value of \alpha is
(D) 4
If A is a square matrix of order 3 and |A| = 6, then the value of |\text{adj } A| is :
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For which value of x, are the determinants \begin{vmatrix} 2x & -3 \\ 5 & x \end{vmatrix} and \begin{vmatrix} 10 & 1 \\ -3 & 2 \end{vmatrix} equal ?
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The value of the cofactor of the element of second row and third column in the matrix \begin{bmatrix} 4 & 3 & 2 \\ 2 & -1 & 0 \\ 1 & 2 & 3 \end{bmatrix} is :
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Let A and B be two square matrices of order 3 such that det (A) = 3 and det (B) = − 4. Find the value of det (− 6AB).
|-6AB|=(-6)^3|A||B| [1]
=-216\times 3\times(-4)=2592 [1]
Two lines are represented by equations 5x - 7y - 2 = 0 and 7x - 5y - 3 = 0 in the xy-plane. Using matrix method, find their point of intersection.
Given equations can be written in matrix form as
\underset{A}{\begin{bmatrix} 5 & -7 \\ 7 & -5 \end{bmatrix}} \underset{X}{\begin{bmatrix} x \\ y \end{bmatrix}} = \underset{B}{\begin{bmatrix} 2 \\ 3 \end{bmatrix}} \Rightarrow AX = B \Rightarrow X = A^{-1}B (|A| = 24 \ne 0, \therefore A^{-1} exists ) [½+ ½]
Now, A^{-1} = \frac{1}{24}\begin{bmatrix} -5 & 7 \\ -7 & 5 \end{bmatrix} [1]
Thus, X = \frac{1}{24}\begin{bmatrix} -5 & 7 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{11}{24} \\ \frac{1}{24} \end{bmatrix} \Rightarrow x = \frac{11}{24}, y = \frac{1}{24} [1]
\therefore The point of intersection of the two lines is \left(\frac{11}{24}, \frac{1}{24}\right).
Let 2x+5y-1=0 and 3x+2y-7=0 represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.
The system of equations in matrices is:
AX=B, where A=\begin{bmatrix}2&5\\3&2\end{bmatrix},X=\begin{bmatrix}x\\y\end{bmatrix},B=\begin{bmatrix}1\\7\end{bmatrix} [1]
The solution is given by X=A^{-1}B\Rightarrow\begin{bmatrix}x\\y\end{bmatrix}=\frac{-1}{11}\begin{bmatrix}2&-5\\-3&2\end{bmatrix}\begin{bmatrix}1\\7\end{bmatrix}=\begin{bmatrix}3\\-1\end{bmatrix} [1½]
Point common to paths of the ants is (3,-1). [½]
To celebrate their child's birthday, a family decided to distribute fruits among children in three orphanages. The fruit vendor packed three different types of baskets, according to the quantity required at each orphanage, details are given as follows:
Basket A contains 5 kg Apples, 10 kg Oranges and 5 kg Pomegranate
Basket B contains 10 kg Apples, 10 kg Oranges and 5kg Pomegranate
Basket C contains 8 kg Apples, 15 kg Oranges and 7kg Pomegranate
Cost of basket A, B and C are respectively ₹2750, ₹3500, and ₹4100.
Based on the given information, answer the following questions:
(i) Write the matrix form of the equations involved in this situation and find the adjoint of the coefficient matrix involved. [2]
(ii) Using inverse of the matrix, calculate the price of each fruit per kg. [2]
(i) Taking price of each kg of apples, oranges and pomegranate as ₹x, ₹y, ₹z respectively.
Matrix form is AX=B,
where A=\begin{pmatrix}5&10&5\\10&10&5\\8&15&7\end{pmatrix}, X=\begin{pmatrix}x\\y\\z\end{pmatrix}, B=\begin{pmatrix}2750\\3500\\4100\end{pmatrix} [1]
Here |A|=25\ne0
So, the system is consistent and has unique solution
Adjoint of the matrix A = adjA = \begin{pmatrix}-5&5&0\\-30&-5&25\\70&5&-50\end{pmatrix} [1]
(ii) Unique solution of the system is given by
X=A^{-1}B=\frac{1}{25}\begin{pmatrix}-5&5&0\\-30&-5&25\\70&5&-50\end{pmatrix}\begin{pmatrix}2750\\3500\\4100\end{pmatrix} [1]
\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}150\\100\\200\end{pmatrix}\Rightarrow x=150, y=100, z=200
So the price of Apples, Oranges and Pomegranate are respectively ₹150 per kg, ₹100 per kg, ₹200 per kg. [1]
Case Study – 2
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.
On the basis of the above information, answer the following questions :
(i) Write the equations representing the various dimensions and express them as the matrix equation AX = B. [1]
(ii) Find if A^{-1} exists. Justify your answer. [1]
(iii) (a) Find A^{-1}. [2]
OR
(iii) (b) Find A^{2} + 7I. [2]
(i) \underset{A}{\begin{pmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ -3 & 1 & 7 \end{pmatrix}}\underset{X}{\begin{pmatrix} l \\ b \\ h \end{pmatrix}} = \underset{B}{\begin{pmatrix} 3 \\ 10 \\ -1 \end{pmatrix}} [1]
(ii) |A| = -8 \ne 0, so A^{-1} exists. [1]
(iii) (a) A^{-1} = \frac{1}{|A|}(\text{adj}A) = \frac{1}{-8}\begin{pmatrix} 20 & -8 & 4 \\ -17 & 4 & -3 \\ 11 & -4 & 1 \end{pmatrix} [2] (1 Mark for any 4 correct entries)
OR
(iii) (b) A^{2} = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ -3 & 1 & 7 \end{pmatrix}\begin{pmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ -3 & 1 & 7 \end{pmatrix} = \begin{pmatrix} 6 & 3 & -7 \\ 5 & 12 & 8 \\ -22 & 7 & 53 \end{pmatrix} [1½]
Thus, A^{2} + 7I = \begin{pmatrix} 13 & 3 & -7 \\ 5 & 19 & 8 \\ -22 & 7 & 60 \end{pmatrix} [½]
Alternative method
Alternative : If A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 3 & -1 & -7 \end{pmatrix}, then |A| = 8 & A^{-1} = \frac{1}{8}\begin{pmatrix} -20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1 \end{pmatrix} [1 + 2]
A^{2} = \begin{pmatrix} 0 & 5 & 7 \\ 11 & 10 & -6 \\ -20 & 7 & 45 \end{pmatrix} [1], A^{2} + 7I = \begin{pmatrix} 7 & 5 & 7 \\ 11 & 17 & -6 \\ -20 & 7 & 52 \end{pmatrix} [1]
Note: In this question, students may take different A by rearranging of terms so check the correctness of the solution accordingly.
Case Study – 1
Three students, Neha, Rani and Sam go to a market to purchase stationery items. Neha buys 4 pens, 3 notepads and 2 erasers and pays ₹ 60. Rani buys 2 pens, 4 notepads and 6 erasers for ₹ 90. Sam pays ₹ 70 for 6 pens, 2 notepads and 3 erasers.
Based upon the above information, answer the following questions :
(i) Form the equations required to solve the problem of finding the price of each item, and express it in the matrix form AX = B. [1]
(ii) Find |A| and confirm if it is possible to find A^{-1}. [1]
(iii) (a) Find A^{-1}, if possible, and write the formula to find X. [2]
OR
(iii) (b) Find A^2 - 8I, where I is an identity matrix. [2]
(i) Let the price of each pen, notepad, eraser be ₹x, ₹y and ₹z respectively
Given system in the form AX = B is \begin{pmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 60 \\ 90 \\ 70 \end{pmatrix} [1]
(ii) |A| = 50 \ne 0, hence A^{-1} exists [1]
(iii) (a) A^{-1} = \frac{\text{adj}A}{|A|} = \frac{1}{50}\begin{pmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{pmatrix} [1½]
X = A^{-1}B [½]
OR
(iii) (b) A^2 = \begin{pmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{pmatrix}\begin{pmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 34 & 28 & 32 \\ 52 & 34 & 46 \\ 46 & 32 & 33 \end{pmatrix} [1½]
A^2 - 8I = \begin{pmatrix} 26 & 28 & 32 \\ 52 & 26 & 46 \\ 46 & 32 & 25 \end{pmatrix} [½]
Case Study – 3
A scholarship is a sum of money provided to a student to help him or her pay for education. Some students are granted scholarships based on their academic achievements, while others are rewarded based on their financial needs.
Every year a school offers scholarships to girl children and meritorious achievers based on certain criteria. In the session 2022 – 23, the school offered monthly scholarship of ₹ 3,000 each to some girl students and ₹ 4,000 each to meritorious achievers in academics as well as sports.
In all, 50 students were given the scholarships and monthly expenditure incurred by the school on scholarships was ₹ 1,80,000.
Based on the above information, answer the following questions :
(i) Express the given information algebraically using matrices. [1]
(ii) Check whether the system of matrix equations so obtained is consistent or not. [1]
(iii) (a) Find the number of scholarships of each kind given by the school, using matrices. [2]
OR
(iii) (b) Had the amount of scholarship given to each girl child and meritorious student been interchanged, what would be the monthly expenditure incurred by the school ? [2]
(i) Let No. of girl child scholarships = x
No. of meritorious achievers = y
x + y = 50
3000x + 4000y = 180000 or 3x + 4y = 180
\begin{bmatrix}1 & 1\\ 3 & 4\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix} = \begin{bmatrix}50\\ 180\end{bmatrix} [1]
(ii) \begin{vmatrix}1 & 1\\ 3 & 4\end{vmatrix} = 1 \ne 0
\therefore system is consistent. [1]
(iii)(a) Let A = \begin{bmatrix}1 & 1\\ 3 & 4\end{bmatrix},\ X = \begin{bmatrix}x\\ y\end{bmatrix},\ B = \begin{bmatrix}50\\ 180\end{bmatrix}
AX = B \Rightarrow X = A^{-1}B [½]
X = \begin{bmatrix}4 & -1\\ -3 & 1\end{bmatrix}\begin{bmatrix}50\\ 180\end{bmatrix} = \begin{bmatrix}20\\ 30\end{bmatrix} [1]
\Rightarrow x = 20, y = 30 [½]
OR
(iii)(b) Required expenditure = ₹ [30(3000) + 20(4000)] [1]
= ₹ 1,70,000 [1]
Case Study – 3
An architect is developing a plot of land for a commercial complex. When asked about the dimensions of the plot, he said that if the length is decreased by 25 m and the breadth is increased by 25 m, then its area increases by 625 m2. If the length is decreased by 20 m and the breadth is increased by 10 m, then its area decreases by 200 m2.
On the basis of the above information, answer the following questions :
(i) Formulate the linear equations in x and y to represent the given information.
(ii) Find the dimensions of the plot of land by matrix method.
(i) (x-25)(y+25)=xy+625\Rightarrow x-y=50 [1]
(x-20)(y+10)=xy-200\Rightarrow x-2y=0 [1]
(ii) The system of linear equations can be written in matrix form as
\begin{bmatrix}1&-1\\1&-2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}50\\0\end{bmatrix} [½]
\Rightarrow\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&-1\\1&-2\end{bmatrix}^{-1}\begin{bmatrix}50\\0\end{bmatrix}=-\begin{bmatrix}-2&1\\-1&1\end{bmatrix}\begin{bmatrix}50\\0\end{bmatrix}=\begin{bmatrix}100\\50\end{bmatrix} [1]
x=100\text{m},\ y=50\text{m} [½]
Case Study-II
Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of ₹ 160. From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of ₹ 190. Also Ankur buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of ₹ 250.
Based on the above information, answer the following questions :
(I) Convert the given above situation into a matrix equation of the form AX = B. [1]
(II) Find |A|. [1]
(III) Find A^{-1}. [2]
OR
(III) Determine P=A^2-5A. [2]
(I) Matrix equation is AX = B, where
where x is the number of pens bought, y the number of bags and z the number of instrument boxes.
(II) |A|=5(4-6)-3(8-3)+1(4-1)=-22 [1]
(III) \text{adj}(A)=\begin{bmatrix}-2&-5&3\\-10&19&-7\\8&-13&-1\end{bmatrix}'=\begin{bmatrix}-2&-10&8\\-5&19&-13\\3&-7&-1\end{bmatrix} [1]
\Rightarrow A^{-1}=\frac{1}{(-22)}\begin{bmatrix}-2&-10&8\\-5&19&-13\\3&-7&-1\end{bmatrix} [1]
OR
P=A^2-5A=\begin{bmatrix}32&20&18\\15&13&17\\13&13&23\end{bmatrix}-\begin{bmatrix}25&15&5\\10&5&15\\5&10&20\end{bmatrix} [1+½]
=\begin{bmatrix}7&5&13\\5&8&2\\8&3&3\end{bmatrix} [½]
For two matrices A=\begin{bmatrix}3&-6&-1\\2&-5&-1\\-2&4&1\end{bmatrix} and B=\begin{bmatrix}1&-2&-1\\0&-1&-1\\2&0&3\end{bmatrix}, find the product AB and hence solve the system of equations:
3x-6y-z=3
2x-5y-z+2=0
-2x+4y+z=5
AB=\begin{bmatrix}3&-6&-1\\2&-5&-1\\-2&4&1\end{bmatrix}\begin{bmatrix}1&-2&-1\\0&-1&-1\\2&0&3\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I [1]
So, A^{-1}=B\ and\ B^{-1}=A [½]
Given system of equations is
3x-6y-z=3,\ 2x-5y-z+2=0,\ -2x+4y+z=5
In matrix form it can be written as: AX=C, [½]
where X=\begin{bmatrix}x\\y\\z\end{bmatrix}\ and\ C=\begin{bmatrix}3\\-2\\5\end{bmatrix} [½]
Here |A|=-3-0+2=-1\ne0 [½]
So, the system is consistent and has unique solution given by the expression X=A^{-1}C=BC [½]
\Rightarrow X=\begin{bmatrix}1&-2&-1\\0&-1&-1\\2&0&3\end{bmatrix}\begin{bmatrix}3\\-2\\5\end{bmatrix}\Rightarrow\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-3\\21\end{bmatrix} [1]
Thus x=2,y=-3,z=21 [½]
If A = \begin{bmatrix} 0 & 2 & 1 \\ -2 & -1 & -2 \\ 1 & -1 & 0 \end{bmatrix}, find A^{-1} and use it to solve the following system of equations :
-2y + z = 7,\ 2x - y - z = 8,\ x - 2y = 10
|A| = -2(2) + 1(3) = -1 \ne 0 [1]
adj A = \begin{bmatrix} -2 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix} [1½]
A^{-1} = \frac{1}{|A|}adj A = \begin{bmatrix} 2 & 1 & 3 \\ 2 & 1 & 2 \\ -3 & -2 & -4 \end{bmatrix} [½]
Given system of equations is equivalent to A^T X = B, where X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix} [½]
X = (A^T)^{-1}B = (A^{-1})^T B
\Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 & 2 & -3 \\ 1 & 1 & -2 \\ 3 & 2 & -4 \end{bmatrix}\begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix} [1½]
\therefore x = 0, y = -5, z = -3
If \begin{bmatrix} 3 & -1 & \sin 3x \\ -7 & 4 & \cos 2x \\ -11 & 7 & 2 \end{bmatrix} is a singular matrix, then find all values of x where x \in \left[0, \frac{\pi}{2}\right].
Given matrix is singular, hence, its determinant = 0,
i.e., \begin{vmatrix} 3 & -1 & \sin 3x \\ -7 & 4 & \cos 2x \\ -11 & 7 & 2 \end{vmatrix} = 0 [1]
\Rightarrow 2\cos 2x + \sin 3x = 2 [1]
\Rightarrow 4\sin^3 x + 4\sin^2 x - 3\sin x = 0 [1]
\Rightarrow \sin x(4\sin^2 x + 4\sin x - 3) = 0
\Rightarrow \sin x(2\sin x - 1)(2\sin x + 3) = 0 [1]
Solving to get x = 0, \frac{\pi}{6} [1]
If P = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} and Q = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}, find (QP) and hence solve the following system of equations using matrices :
x - y = 3,\ 2x + 3y + 4z = 17,\ y + 2z = 7
QP = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} i.e. QP = 6I [2]
\Rightarrow P^{-1} = \frac{1}{6}Q [½]
Given system of equations is given by PX = R
where, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ R = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} [½]
PX = R \Rightarrow X = P^{-1}R
\Rightarrow X = \frac{1}{6}Q.R
= \frac{1}{6}\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}\begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} [½]
\Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{6}\begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} [1½]
\Rightarrow x = 2,\ y = -1,\ z = 4
Obtain the value of \Delta = \begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix} in terms of x, y and z.
Further, if \Delta = 0 and x, y, z are non-zero real numbers, prove that x^{-1} + y^{-1} + z^{-1} = -1.
\Delta = \begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix}
= (1 + x)[(1 + y)(1 + z) - 1] - 1[(1 + z) - 1] + 1[1 - (1 + y)] [2]
= (1 + x)(1 + z + y + yz - 1) - z - y
= xy + yz + zx + xyz [1]
\Delta = 0 \Rightarrow xy + yz + zx + xyz = 0
\Rightarrow yz + zx + xy = -xyz [1]
Dividing both sides by xyz,
x^{-1} + y^{-1} + z^{-1} = -1 [1]
Note: Proportionate marks may be allotted if a student uses properties of determinants to solve this question.
On the inauguration day of a new showroom, a lucky draw was organized and some vouchers of ₹ 1,000 and ₹ 500 were given to the lucky draw winners.
A total of 60 vouchers were given on the day. The number of ₹ 1,000 vouchers added to 3 times the number of ₹ 500 vouchers, gives 100. Express the given information as a system of linear equations in two variables. Hence, find the number of vouchers of each type by matrix method.
Let, number of ₹ 1000 vouchers = x
and number of ₹ 500 vouchers = y
Hence, x + y = 60 and x + 3y = 100 [1]
In matrix form, \begin{bmatrix}1 & 1\\1 & 3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}60\\100\end{bmatrix}
\Rightarrow AX = B where A = \begin{bmatrix}1 & 1\\1 & 3\end{bmatrix}, B = \begin{bmatrix}60\\100\end{bmatrix} and X = \begin{bmatrix}x\\y\end{bmatrix}
\Rightarrow X = A^{-1}B \quad \ldots(1) [1]
|A| = 2 \quad \because |A| \ne 0 \Rightarrow A is an invertible matrix. [½]
\text{adj } A = \begin{bmatrix}3 & -1\\-1 & 1\end{bmatrix} and A^{-1} = \frac{1}{2}\begin{bmatrix}3 & -1\\-1 & 1\end{bmatrix} [1½]
By eq.(1), \begin{bmatrix}x\\y\end{bmatrix} = \frac{1}{2}\begin{bmatrix}3 & -1\\-1 & 1\end{bmatrix}\begin{bmatrix}60\\100\end{bmatrix} = \begin{bmatrix}40\\20\end{bmatrix}
On comparing, x = 40 and y = 20 [1]
Hence, number of ₹ 1000 vouchers = 40
and number of ₹ 500 vouchers = 20
A man goes to buy fruits from the market. The shopkeeper informs him that 4 apples, 3 oranges and 2 bananas cost ₹ 60; 2 apples, 4 oranges and 6 bananas cost ₹ 90; whereas 6 apples, 2 oranges and 3 bananas cost ₹ 70. Using matrix method, find the cost of one fruit of each kind.
Let cost of one apple, one orange, one banana be ₹ x, ₹ y and ₹ z respectively
According to the question,
[1]
Let A = \begin{bmatrix}4 & 3 & 2\\ 2 & 4 & 6\\ 6 & 2 & 3\end{bmatrix}, X = \begin{bmatrix}x\\ y\\ z\end{bmatrix}, B = \begin{bmatrix}60\\ 90\\ 70\end{bmatrix}
|A| = 50 \ne 0 \Rightarrow A^{-1} exists
System becomes AX = B. So, X = A^{-1}B [1]
\text{adj}A = \begin{bmatrix}0 & -5 & 10\\ 30 & 0 & -20\\ -20 & 10 & 10\end{bmatrix} [1½]
A^{-1} = \frac{1}{50}\begin{bmatrix}0 & -5 & 10\\ 30 & 0 & -20\\ -20 & 10 & 10\end{bmatrix} [½]
X = \frac{1}{50}\begin{bmatrix}0 & -5 & 10\\ 30 & 0 & -20\\ -20 & 10 & 10\end{bmatrix}\begin{bmatrix}60\\ 90\\ 70\end{bmatrix}
X = \begin{bmatrix}5\\ 8\\ 8\end{bmatrix} [1]
\therefore cost of one apple, one orange, one banana is ₹ 5, ₹ 8 and ₹ 8 respectively
A school wants to allocate students into three clubs : Sports, Music and Drama, under following conditions :
Find the number of students allocated to different clubs, using matrix method.
Let x, y and z be the no. of students allocated to Sports, Music and Drama clubs respectively.
Here, x=y+z,\ y=\frac{x}{2}+20,\ x+y+z=180
\Rightarrow x-y-z=0,\ x-2y=-40,\ x+y+z=180 [1½]
Given equations can be written as AX=B
where, A=\begin{bmatrix}1&-1&-1\\1&-2&0\\1&1&1\end{bmatrix},\ B=\begin{bmatrix}0\\-40\\180\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix} [½]
|A|=-4\ne 0\Rightarrow A^{-1} exists. [½]
adjA=\begin{bmatrix}-2&0&-2\\-1&2&-1\\3&-2&-1\end{bmatrix} [1]
A^{-1}=\frac{1}{|A|}\times adjA=\frac{1}{4}\begin{bmatrix}2&0&2\\1&-2&1\\-3&2&1\end{bmatrix} [½]
X=A^{-1}B
=\frac{1}{4}\begin{bmatrix}2&0&2\\1&-2&1\\-3&2&1\end{bmatrix}\begin{bmatrix}0\\-40\\180\end{bmatrix}=\begin{bmatrix}90\\65\\25\end{bmatrix} [1]
\therefore x=90,\ y=65,\ z=25
Number of students allocated in sports, music and drama are 90 , 65 and 25 respectively .
Given A = \begin{bmatrix}-4 & 4 & 4\\ -7 & 1 & 3\\ 5 & -3 & -1\end{bmatrix} and B = \begin{bmatrix}1 & -1 & 1\\ 1 & -2 & -2\\ 2 & 1 & 3\end{bmatrix}, find AB. Hence, solve the system of linear equations :
x - y + z = 4
x - 2y - 2z = 9
2x + y + 3z = 1
AB = \begin{bmatrix}8 & 0 & 0\\ 0 & 8 & 0\\ 0 & 0 & 8\end{bmatrix} = 8I [2]
The system of equations is equivalent to the matrix equation:
BX = C, where C = \begin{bmatrix}4\\ 9\\ 1\end{bmatrix}, X = \begin{bmatrix}x\\ y\\ z\end{bmatrix} [½]
\Rightarrow X = B^{-1}C
AB = 8I
\Rightarrow B^{-1} = \frac{1}{8}A [1]
X = \frac{1}{8}\begin{bmatrix}-4 & 4 & 4\\ -7 & 1 & 3\\ 5 & -3 & -1\end{bmatrix}\begin{bmatrix}4\\ 9\\ 1\end{bmatrix} = \frac{1}{8}\begin{bmatrix}24\\ -16\\ -8\end{bmatrix} = \begin{bmatrix}3\\ -2\\ -1\end{bmatrix}
\therefore x = 3, y = -2, z = -1 [1½]
If A = \begin{bmatrix}1 & 2 & 0\\ -2 & -1 & -2\\ 0 & -1 & 1\end{bmatrix}, then find A^{-1}.
Hence, solve the system of linear equations :
x - 2y = 10
2x - y - z = 8
-2y + z = 7
|A| = 1 \ne 0 \Rightarrow A^{-1} exists. [1]
adjA = \begin{bmatrix}-3 & -2 & -4\\ 2 & 1 & 2\\ 2 & 1 & 3\end{bmatrix} [1½]
A^{-1} = \frac{1}{|A|}adjA = \begin{bmatrix}-3 & -2 & -4\\ 2 & 1 & 2\\ 2 & 1 & 3\end{bmatrix}
The given system of equations is equivalent to the matrix equation
A^TX = B, where B = \begin{bmatrix}10\\ 8\\ 7\end{bmatrix}, X = \begin{bmatrix}x\\ y\\ z\end{bmatrix} [½]
\Rightarrow X = (A^T)^{-1}B
\Rightarrow X = (A^{-1})^TB [½]
\Rightarrow X = \begin{bmatrix}-3 & 2 & 2\\ -2 & 1 & 1\\ -4 & 2 & 3\end{bmatrix}\begin{bmatrix}10\\ 8\\ 7\end{bmatrix} = \begin{bmatrix}0\\ -5\\ -3\end{bmatrix}
\therefore x = 0, y = -5, z = -3 [1½]
If A is a 3\times 3 invertible matrix, show that for any scalar k\ne 0, (kA)^{-1}=\frac{1}{k}A^{-1}. Hence calculate (3A)^{-1}, where
Consider (kA)\left(\frac{1}{k}A^{-1}\right)=k\cdot\frac{1}{k}(A.A^{-1})=I
\Rightarrow kA and \frac{1}{k}A^{-1} are inverse of each other.
\therefore(kA)^{-1}=\frac{1}{k}A^{-1} [1]
\therefore(3A)^{-1}=\frac{1}{3}A^{-1}
Here, |A|=4\ne 0\ \therefore A^{-1} exists. [1]
adjA=\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix} [2]
\therefore A^{-1}=\frac{1}{|A|}.adjA=\frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix} [½]
\therefore(3A)^{-1}=\frac{1}{12}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix} [½]
A furniture workshop produces three types of furniture – chairs, tables and beds each day. On a particular day the total number of furniture pieces produced is 45. It was also found that production of beds exceeds that of chairs by 8, while the total production of beds and chairs together is twice the production of tables. Determine the units produced of each type of furniture, using matrix method.
Let the numbers of chairs, tables and beds produced be x, y and z respectively.
\therefore x+y+z=45;\quad -x+0.y+z=8;\quad x-2y+z=0 [1½]
Let A=\begin{bmatrix}1&1&1\\-1&0&1\\1&-2&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}45\\8\\0\end{bmatrix}
|A|=1(0+2)-1(-1-1)+1(2-0)=6\ne0
\therefore A^{-1} exists [½]
AX=B\Rightarrow X=A^{-1}B [½]
\text{adj}(A)=\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix}
A^{-1}=\frac16\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix} [1½]
\therefore\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac16\begin{bmatrix}2&-3&1\\2&0&-2\\2&3&1\end{bmatrix}\begin{bmatrix}45\\8\\0\end{bmatrix}=\begin{bmatrix}11\\15\\19\end{bmatrix} [1]
So, x=11,y=15,z=19
Hence the numbers of chairs, tables and beds produced are 11, 15 and 19 respectively.
If A = \begin{bmatrix}3&2&1\\4&-1&2\\7&3&-3\end{bmatrix}, find A^{-1}. Using A^{-1}, solve the given system of equations 3x + 4y + 7z = 14;\ 2x - y + 3z = 4;\ x + 2y - 3z = 0.
|A| = 3(-3) - 2(-26) + 1(19) = 62 \ne 0 \Rightarrow A^{-1} exists. [½]
cofactor Matrix = \begin{bmatrix}-3&26&19\\9&-16&5\\5&-2&-11\end{bmatrix}
adj A = \begin{bmatrix}-3&9&5\\26&-16&-2\\19&5&-11\end{bmatrix} [2]
A^{-1} = \frac{1}{62}\begin{bmatrix}-3&9&5\\26&-16&-2\\19&5&-11\end{bmatrix} [½]
Given system of equations can be written as A'.X = B
where X = \begin{bmatrix}x\\y\\z\end{bmatrix}, B = \begin{bmatrix}14\\4\\0\end{bmatrix}
Now, A'.X = B \Rightarrow X = (A')^{-1}.B [½]
\Rightarrow X = \left(A^{-1}\right)'.B = \frac{1}{62}\begin{bmatrix}-3&26&19\\9&-16&5\\5&-2&-11\end{bmatrix}\begin{bmatrix}14\\4\\0\end{bmatrix} [½]
= \frac{1}{62}\begin{bmatrix}62\\62\\62\end{bmatrix} = \begin{bmatrix}1\\1\\1\end{bmatrix} [1]
\Rightarrow x = 1, y = 1, z = 1
If A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}, find A^{-1} and use it to solve the following system of equations :
x - 2y = 10,\ 2x - y - z = 8,\ -2y + z = 7
|A| = 1 \ne 0 hence A^{-1} exists. [1]
Adj\,A = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix} [2]
A^{-1} = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix} [½]
AX = B \Rightarrow \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}
X = A^{-1}B \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix}\begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix}
\Rightarrow x = 0,\ y = -5,\ z = -3 [1½]
If A = \begin{bmatrix} -1 & a & 2 \\ 1 & 2 & x \\ 3 & 1 & 1 \end{bmatrix} and A^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ b & y & 3 \end{bmatrix}, find the value of (a + x) - (b + y).
AA^{-1} = I [1]
\begin{bmatrix} -1 & a & 2 \\ 1 & 2 & x \\ 3 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ b & y & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} -1 - 8a + 2b & 1 + 7a + 2y & 5 - 5a \\ -15 + bx & 13 + xy & 3x - 9 \\ -5 + b & 4 + y & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} [1½]
-5 + b = 0 \Rightarrow b = 5,\quad 5 - 5a = 0 \Rightarrow a = 1 [1]
4 + y = 0 \Rightarrow y = -4,\quad 3x - 9 = 0 \Rightarrow x = 3 [1]
\therefore (a + x) - (b + y) = (1 + 3) - (5 - 4) = 3 [½]
Solve the following system of equations, using matrices :
\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4,\ \ \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1,\ \ \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2
where x, y, z \ne 0
Given system of linear equations is equivalent to AX = B, where
A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}, X = \begin{bmatrix} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{bmatrix}, B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} [½]
|A| = 1200 \ne 0 [½]
Cofactors of the elements of A are
A_{11} = 75, A_{12} = 110, A_{13} = 72
A_{21} = 150, A_{22} = -100, A_{23} = 0
A_{31} = 75, A_{32} = 30, A_{33} = -24
adjA = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} [2]
A^{-1} = \frac{adjA}{|A|} = \frac{1}{1200}\begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} [½]
X = A^{-1}B = \frac{1}{1200}\begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}
= \frac{1}{1200}\begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{bmatrix} [1]
\thereforex = 2, y = 3, z = 5 [½]
If A = \begin{bmatrix} 1 & \cot x \\ -\cot x & 1 \end{bmatrix}, show that A'A^{-1} = \begin{bmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix}.
|A| = 1 + \cot^2 x = \text{cosec}^2 x [½]
adjA = \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix} [1½]
A^{-1} = \frac{adjA}{|A|} = \frac{1}{\text{cosec}^2 x}\begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix} [1]
A' = \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix} [½]
A'A^{-1} = \frac{1}{\text{cosec}^2 x}\begin{bmatrix} 1-\cot^2 x & -2\cot x \\ 2\cot x & 1-\cot^2 x \end{bmatrix}
= \begin{bmatrix} \sin^2 x - \cos^2 x & -2\sin x\cos x \\ 2\sin x\cos x & \sin^2 x - \cos^2 x \end{bmatrix}
= \begin{bmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix} [1½]
If A = \begin{bmatrix} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{bmatrix}, find A^{-1} and hence solve the following system of equations :
2x + y - 3z = 13
3x + 2y + z = 4
x + 2y - z = 8
For Matrix A = \begin{pmatrix} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{pmatrix}, |A| = -16 \ne 0 so, A^{-1} exists. [1]
\text{adj}A = \begin{pmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{pmatrix}, [1½]
Thus, A^{-1} = \frac{-1}{16}\begin{pmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{pmatrix} [½]
so, Given equation can be written into a matrix equation as
\underset{A}{\begin{pmatrix} 2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1 \end{pmatrix}}\underset{X}{\begin{pmatrix} x \\ y \\ z \end{pmatrix}} = \underset{B}{\begin{pmatrix} 13 \\ 4 \\ 8 \end{pmatrix}} \Rightarrow X = A^{-1}.B [½]
\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{-1}{16}\begin{pmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{pmatrix}\begin{pmatrix} 13 \\ 4 \\ 8 \end{pmatrix} = \frac{-1}{16}\begin{pmatrix} -16 \\ -32 \\ 48 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} \Rightarrow x = 1, y = 2, z = -3 [1½]
If A=\begin{bmatrix}1 & 2 & -3\\ 2 & 0 & -3\\ 1 & 2 & 0\end{bmatrix}, then find A^{-1} and hence solve the following system of equations :
x+2y-3z=1
2x-3z=2
x+2y=3
|A|=1(6)-2(3)-3(4)=-12\ne0, \therefore A^{-1} exist [1]
\text{adj}A=\begin{bmatrix}6 & -6 & -6\\ -3 & 3 & -3\\ 4 & 0 & -4\end{bmatrix} [2]
\therefore A^{-1}=-\frac{1}{12}\begin{bmatrix}6 & -6 & -6\\ -3 & 3 & -3\\ 4 & 0 & -4\end{bmatrix} [½]
The given system of equations can be written as AX=B, X=\begin{bmatrix}x\\ y\\ z\end{bmatrix},B=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}
X=A^{-1}B\Rightarrow\begin{bmatrix}x\\ y\\ z\end{bmatrix}=-\frac{1}{12}\begin{bmatrix}6 & -6 & -6\\ -3 & 3 & -3\\ 4 & 0 & -4\end{bmatrix}\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}=\begin{bmatrix}2\\ \frac12\\ \frac23\end{bmatrix} [1½]
\therefore The solution of the given system of equations is: x=2,y=\frac12,z=\frac23
Find the product of the matrices \begin{bmatrix}1 & 2 & -3\\ 2 & 3 & 2\\ 3 & -3 & -4\end{bmatrix}\begin{bmatrix}-6 & 17 & 13\\ 14 & 5 & -8\\ -15 & 9 & -1\end{bmatrix} and hence solve the system of linear equations :
x+2y-3z=-4
2x+3y+2z=2
3x-3y-4z=11
\begin{bmatrix}1 & 2 & -3\\ 2 & 3 & 2\\ 3 & -3 & -4\end{bmatrix}\begin{bmatrix}-6 & 17 & 13\\ 14 & 5 & -8\\ -15 & 9 & -1\end{bmatrix}=\begin{bmatrix}67 & 0 & 0\\ 0 & 67 & 0\\ 0 & 0 & 67\end{bmatrix} [2½]
\Rightarrow\begin{bmatrix}1 & 2 & -3\\ 2 & 3 & 2\\ 3 & -3 & -4\end{bmatrix}^{-1}=\frac{1}{67}\begin{bmatrix}-6 & 17 & 13\\ 14 & 5 & -8\\ -15 & 9 & -1\end{bmatrix} [1]
Solution of the system of equations is given by:
\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}1 & 2 & -3\\ 2 & 3 & 2\\ 3 & -3 & -4\end{bmatrix}^{-1}\begin{bmatrix}-4\\ 2\\ 11\end{bmatrix}=\frac{1}{67}\begin{bmatrix}-6 & 17 & 13\\ 14 & 5 & -8\\ -15 & 9 & -1\end{bmatrix}\begin{bmatrix}-4\\ 2\\ 11\end{bmatrix}=\begin{bmatrix}3\\ -2\\ 1\end{bmatrix},
\therefore x=3,y=-2,z=1 [1½]
If A=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix} and B^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}, find (AB)^{-1}.
A=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix},\ B^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
(AB)^{-1}=B^{-1}A^{-1} [½]
|A|=1(3)-2(-1)-2(2)=3+2-4=1\ne0 [1]
\text{adj}(A)=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix} [2]
A^{-1}=\frac11\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix} [½]
\therefore B^{-1}A^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}
=\begin{bmatrix}10&7&21\\-49&-34&-103\\17&12&36\end{bmatrix} [1]
Solve the following system of equations by matrix method :
Given system is
\begin{bmatrix}1&2&3\\2&-1&1\\3&2&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\2\\3\end{bmatrix}
A\cdot X=B\Rightarrow X=A^{-1}B
|A|=35\ne0 [½]
A_{11}=0\quad A_{12}=7\quad A_{13}=7 [1]
A_{21}=10\quad A_{22}=-11\quad A_{23}=4
A_{31}=5\quad A_{32}=5\quad A_{33}=-5 [1½]
\therefore A^{-1}=\frac{1}{35}\begin{bmatrix}0&10&5\\7&-11&5\\7&4&-5\end{bmatrix} [1]
\Rightarrow X=\frac{1}{35}\begin{bmatrix}0&10&5\\7&-11&5\\7&4&-5\end{bmatrix}\begin{bmatrix}6\\2\\3\end{bmatrix}=\frac{1}{35}\begin{bmatrix}35\\35\\35\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix} [1]
\therefore x=1\quad y=1\quad z=1
If A = \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}, B = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}, then find AB and use it to solve the following system of equations :
x - 2y = 3
2x - y - z = 2
-2y + z = 3
AB = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} [1]
\Rightarrow B^{-1} = A [1]
The given system of equations can be written as:
B^T\cdot X = C, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, C = \begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix} [½]
X = (B^T)^{-1}\cdot C = (B^{-1})^T\cdot C = A^T\cdot C [1]
\Rightarrow X = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix}\begin{bmatrix} 3 \\ 2 \\ 3 \end{bmatrix} [½]
\Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \therefore x = 1, y = -1, z = 1 [1]
If A = \begin{bmatrix} 3 & 2\\ 5 & -7\end{bmatrix}, then find A^{-1} and use it to solve the following system of equations :
3x + 5y = 11,\ 2x - 7y = -3.
adj A = \begin{bmatrix} -7 & -2\\ -5 & 3\end{bmatrix} [1]
|A| = -31 [1]
A^{-1} = \frac{-1}{31}\begin{bmatrix} -7 & -2\\ -5 & 3\end{bmatrix} [½]
Given system of equations is
\begin{bmatrix} 3 & 5\\ 2 & -7\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix} = \begin{bmatrix} 11\\ -3\end{bmatrix}
which is A'X = B, where X = \begin{bmatrix} x\\ y\end{bmatrix}, B = \begin{bmatrix} 11\\ -3\end{bmatrix} [½]
\Rightarrow X = (A')^{-1}B [½]
\Rightarrow X = (A^{-1})'B [½]
= \frac{-1}{31}\begin{bmatrix} -7 & -5\\ -2 & 3\end{bmatrix}\begin{bmatrix} 11\\ -3\end{bmatrix} = \begin{bmatrix} 2\\ 1\end{bmatrix} [1]
\therefore x = 2,\ y = 1
Find the inverse of the matrix A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}. Using the inverse, A^{-1}, solve the system of linear equations
x-y+2z=1;\ 2y-3z=1;\ 3x-2y+4z=3.
A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}
|A|=1(8-6)+1(0+9)+2(0-6)=-1\ne 0 [1]
\therefore A is invertible.
\text{adj }A=\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix} [1½]
\Rightarrow A^{-1}=\frac{1}{|A|}\text{adj }A=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix} [½]
The given system of equation can be written as AX = B, when
A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}1\\1\\3\end{bmatrix}
\Rightarrow X=A^{-1}B [1]
\Rightarrow\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\begin{bmatrix}1\\1\\3\end{bmatrix}=\begin{bmatrix}1\\2\\1\end{bmatrix}
\Rightarrow x=1,\ y=2,\ z=1 [1]
If matrix A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix}, find A^{-1} and hence solve the following system of linear equations :
Official marking scheme for this paper is not available.
The derivative of \cos^{-1}(2x^2-1) with respect to \cos^{-1}(x) is
(A) 2
Let y=\cos^{-1}(2x^2-1) and u=\cos^{-1}(x)
\Rightarrow y=\cos^{-1}(2\cos^2u-1) as \cos u=x
\Rightarrow y=\cos^{-1}(\cos(2u))=2u
\Rightarrow\frac{dy}{du}=2
If a,b,c are arbitrary constants such that the function defined by
f(x)=\begin{cases}ax^2+b,\ b\ne0,\ x\le1\\ bx^2+ax+c,\ x\gt1\end{cases}
is differentiable at x=1, then which of the following could be a triplet for (a,b,c)?
(D) \left(3,\frac32,0\right)
f(x) is differentiable \Rightarrow f(x) is continuous
\Rightarrow c=0 and a=2b
\Rightarrow (a,b,c) becomes (2b,b,0) which is satisfied by option (D) \left(3,\frac32,0\right) only
If a function defined by f(x)=\begin{cases}kx+1, & x\le\pi\\ \cos x, & x\gt\pi\end{cases} is continuous at x=\pi, then the value of k is
f(x) is continuous at x=\pi
\Rightarrow \lim_{x\to\pi^-}(kx+1)=\lim_{x\to\pi^+}\cos x=f(\pi)
\Rightarrow \lim_{h\to0}[k(\pi-h)+1]=\lim_{h\to0}\cos(\pi+h)=k\pi+1
\Rightarrow k\pi+1=-1 \quad\Rightarrow k=\frac{-2}{\pi}
(D) \frac{-2}{\pi} [1]
If f(x)=x\tan^{-1}x, then f'(1) is equal to
f(x)=x\tan^{-1}x
f'(x)=1.\tan^{-1}x+x.\frac{1}{1+x^2}
f'(1)=1.\tan^{-1}1+\frac{1}{1+1}=\frac{\pi}{4}+\frac12
(B) \frac{\pi}{4}+\frac12 [1]
If e^{-x} + e^{-y} = 2, then \frac{dy}{dx} is
(D) -e^{y-x}
The value of k for which the function f(x) = \begin{cases} x^2 \sin\frac{1}{x}, & x \ne 0\\ k(x+1), & x = 0\end{cases} is a continuous function, is :
(D) 0
If \sin^{-1} x = y, then \frac{dy}{dx} is :
(D) \sec y
Differential of e^{e^{x}} with respect to x is :
(C) e^{x} . e^{e^{x}} [1]
If f(x) = \begin{cases} \frac{\sin x}{x} + \cos x, & x \ne 0 \\ k, & x = 0 \end{cases} is continuous at x = 0, then the value of k is :
(D) 2 [1]
The greatest integer function, f(x) = [x], 0 \lt x \lt 3 is not differentiable at how many points ?
(B) At only two points [1]
If f(x) = \begin{cases}\dfrac{x^2 - 4x - 5}{x + 1}, & x \ne -1\\ k, & x = -1\end{cases}
is continuous at x = -1, then the value of k is :
(D) -6
Derivative of \cos^{-1}\left(\frac{\sin x + \cos x}{\sqrt{2}}\right), -\frac{\pi}{4} \lt x \lt \frac{\pi}{4} with respect to x is :
(A) -1
If f(x) = \begin{cases} \frac{|x|}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}, then which of the following statements is not correct ?
(B) f is continuous for every x \in \mathbb{R}.
Differential of \cos^{-1}(e^x) with respect to x is :
(C) \frac{-e^x}{\sqrt{1-e^{2x}}}
If y = \log\sqrt{\tan x}, then the value of \frac{dy}{dx} at x = \frac{\pi}{4} is :
(C) 1
If f(x)=|x|+|x-1|, then which of the following is correct ?
(C) f(x) is continuous but not differentiable, at x=0 and x=1.
Assertion (A) : f(x)=\begin{cases}3x-8, & x\le 5\\ 2k, & x\gt 5\end{cases} is continuous at x=5 for k=\frac{5}{2}.
Reason (R) : For a function f to be continuous at x=a, \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a).
(D) Assertion (A) is false, but Reason (R) is true.
If A denotes the set of continuous functions and B denotes set of differentiable functions, then which of the following depicts the correct relation between set A and B ?
(B) (B drawn inside A)
If y=\sin^{-1}x, then (1-x^2)\frac{d^2y}{dx^2} is equal to :
(A) x\frac{dy}{dx}
If f(x)=\begin{cases}3x-2, & 0\lt x\le 1\\ 2x^2+ax, & 1\lt x\lt 2\end{cases} is continuous for x\in(0,2), then a is equal to :
(D) -1
The function f defined by
is not continuous at :
(B) x = 1
If f(x)=\begin{cases}\dfrac{\sin^2 ax}{x^2}, & x\ne0\\ 1, & x=0\end{cases} is continuous at x=0, then the value of a is :
(C) \pm1
If f(x)=\{[x],\ x\in R\} is the greatest integer function, then the correct statement is :
(B) f is neither continuous nor differentiable at x=2.
If f(x) = \begin{cases} \dfrac{\log(1+ax) + \log(1-bx)}{x}, & \text{for } x \ne 0 \\ k, & \text{for } x = 0 \end{cases}
is continuous at x = 0, then the value of k is :
(C) a - b
If \tan^{-1}(x^2 - y^2) = a, where ‘a’ is a constant, then \frac{dy}{dx} is :
(A) \frac{x}{y}
If y = a\cos(\log x) + b\sin(\log x), then x^2y_2 + xy_1 is :
(C) -y
Let f(x) = |x|,\ x \in R. Then, which of the following statements is incorrect ?
(D) f is differentiable at x = 0.
If f(x)=\begin{cases}1, & \text{if } x\le3\\ ax+b, & \text{if } 3\lt x\lt5\\ 7, & \text{if } 5\le x\end{cases} is continuous in R, then the values of a and b are :
(A) a=3,\ b=-8 [1]
If f(x)=-2x^8, then the correct statement is :
(B) f'\left(\frac12\right)=-f'\left(-\frac12\right) [1]
Assertion (A) : f(x)=\begin{cases}x\sin\frac1x, & x\ne0\\ 0, & x=0\end{cases} is continuous at x=0.
Reason (R) : When x\to0, \sin\frac1x is a finite value between -1 and 1.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1]
If x = t^3 and y = t^2, then \frac{d^2y}{dx^2} at t = 1 is :
(B) -\frac{2}{9}
Assertion (A) : f(x) = [x], x \in \mathbb{R}, the greatest integer function is not differentiable at x = 2.
Reason (R) : The greatest integer function is not continuous at any integral value.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
Which of the following statements is true for the function f(x) = \begin{cases} x^2 + 3, & x \ne 0 \\ 1, & x = 0 \end{cases} ?
(C) f(x) is continuous and differentiable \forall\, x \in R - \{0\}
The derivative of \sin(x^2) w.r.t. x, at x = \sqrt{\pi} is :
(C) -2\sqrt{\pi}
The number of points of discontinuity of f(x) = \begin{cases} |x|+3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 \lt x \lt 3 \\ 6x+2, & \text{if } x \ge 3 \end{cases} is :
(B) 1
The derivative of \tan^{-1}(x^2) w.r.t. x is :
(B) \frac{2x}{1+x^4}
A function f(x) = |1 - x + |x|| is :
(D) continuous everywhere
The derivative of 2^x w.r.t. 3^x is :
(C) \left(\frac{2}{3}\right)^x \frac{\log 2}{\log 3}
Derivative of e^{2x} with respect to e^{x}, is :
(B) 2e^{x}
[1]For what value of k, the function given below is continuous at x = 0 ?
(B) \frac{1}{4}
[1]If xe^y=1, then the value of \frac{dy}{dx} at x=1 is :
(A) -1 [1]
Derivative of e^{\sin^2 x} with respect to \cos x is :
(C) -2\cos x\, e^{\sin^2 x} [1]
The number of discontinuities of the function f given by
is :
(C) 2
[1]Let y=f\left(\frac{1}{x}\right) and f'(x)=x^{3}. What is the value of \frac{dy}{dx} at x=\frac{1}{2} ?
(C) -32
[1]If y=\log\sqrt{\sec\sqrt{x}}, then the value of \frac{dy}{dx} at x=\frac{\pi^{2}}{16} is :
(A) \frac{1}{\pi}
[1]If x=3\cos\theta and y=5\sin\theta, then \frac{dy}{dx} is equal to :
(B) -\frac{5}{3}\cot\theta
[1]The greatest integer function defined by f(x)=[x], 1\lt x\lt 3 is not differentiable at x=
(C) 2
[1]The value of k for which f(x)=\begin{cases}3x+5, & x\ge2\\ kx^2, & x\lt2\end{cases} is a continuous function, is :
(D) \frac{11}{4}
The function f(x) = x|x| is
(A) continuous and differentiable at x = 0.
If \tan\left(\frac{x+y}{x-y}\right) = k, then \frac{dy}{dx} is equal to
(B) \frac{y}{x}
The value of k for which function f(x) = \begin{cases} x^2, & x \ge 0\\ kx, & x \lt 0\end{cases} is differentiable at x = 0 is :
(D) 0
If y = \frac{\cos x - \sin x}{\cos x + \sin x}, then \frac{dy}{dx} is :
(A) -\sec^2\left(\frac{\pi}{4} - x\right)
The function f(x)=|x| is
(c) continuous everywhere, but differentiable everywhere except at x=0
If y=\sin^2(x^3), then \frac{dy}{dx} is equal to :
(c) 6x^2\sin(x^3)\cos(x^3)
The derivative of x^{2x} w.r.t. x is
(C) 2x^{2x}(1+\log x)
The function f(x)=[x], where [x] denotes the greatest integer less than or equal to x, is continuous at
(B) x=1\cdot5
If x=A\cos 4t+B\sin 4t, then \frac{d^2x}{dt^2} is equal to
(D) -16x
If (\cos x)^y = (\cos y)^x, then \frac{dy}{dx} is equal to :
Official marking scheme for this paper is not available.
If x = a\cos\theta + b\sin\theta, y = a\sin\theta - b\cos\theta, then which one of the following is true ?
Official marking scheme for this paper is not available.
If y=\sin(x+y) and \frac{dy}{dx}=0, then prove that y=\pm1.
y=\sin(x+y)\Rightarrow\sin^{-1}y=x+y
\Rightarrow\frac{1}{\sqrt{1-y^2}}=\frac{dx}{dy}+1\Rightarrow\frac{dx}{dy}=\frac{1-\sqrt{1-y^2}}{\sqrt{1-y^2}} [1]
\Rightarrow\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{1-\sqrt{1-y^2}}
So, \frac{dy}{dx}=0\Rightarrow y=\pm1 [1]
Alternative method
y=\sin(x+y)\Rightarrow\frac{dy}{dx}=\cos(x+y)\left(1+\frac{dy}{dx}\right)
\Rightarrow\frac{dy}{dx}(1-\cos(x+y))=\cos(x+y) [1]
\Rightarrow\cos(x+y)=0 as \frac{dy}{dx}=0
\Rightarrow\sin(x+y)=\pm1\Rightarrow y=\pm1 [1]
If y=\log\tan\left(\frac{\pi}{4}+\frac{x}{2}\right), then prove that \frac{dy}{dx}-\sec x=0
y=\log\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)
Differentiating with respect to x
\frac{dy}{dx}=\frac{1}{\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)}.\sec^2\left(\frac{\pi}{4}+\frac{x}{2}\right).\frac12 [½]
=\frac{\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)}.\frac{1}{\cos^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}.\frac12
=\frac{1}{2\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)}=\frac{1}{\sin\left(\frac{\pi}{2}+x\right)}=\frac{1}{\cos x} [1]
\Rightarrow\frac{dy}{dx}-\sec x=0 [½]
If f(x+y)=f(x)f(y) for all x,y\in R and f(5)=2, f'(0)=3, then using the definition of derivatives, find f'(5).
Given f(x+y)=f(x)f(y)
f(0+5)=f(0)f(5)
\Rightarrow f(0)=1 [½]
f'(5)=\lim_{h\to0}\frac{f(5+h)-f(5)}{h}=\lim_{h\to0}\frac{f(5)f(h)-f(5)}{h} [\because f(x+y)=f(x)f(y)]
=\lim_{h\to0}\frac{2f(h)-2}{h} [\because f(5)=2] [1]
=2\lim_{h\to0}\frac{f(h)-1}{h}=2\lim_{h\to0}\frac{f(h)-f(0)}{h}=2f'(0)
=2(3) [\because f'(0)=3]
=6 [½]
Check whether function f(x) defined as
is continuous at x = 3 or not ?
f(x) = \begin{cases} \frac{-(x-3)}{2(x-3)}, & x \lt 3 \\ \frac{x-6}{6}, & x \ge 3 \end{cases} \Rightarrow f(x) = \begin{cases} \frac{-1}{2}, & x \lt 3 \\ \frac{x-6}{6}, & x \ge 3 \end{cases}
f(3) = \frac{-1}{2}
LHL = \lim_{x \to 3^-} f(x) = \lim_{x\to 3^-} \frac{-1}{2} = \frac{-1}{2} [1]
RHL = \lim_{x \to 3^+} f(x) = \lim_{x\to 3^+} \frac{x-6}{6} = \frac{-1}{2} [½]
LHL = RHL = f(3), hence the function f(x) is continuous at x = 3. [½]
If \sqrt{3}(x^2 + y^2) = 4xy, then find \frac{dy}{dx} at \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).
Differentiating both sides w.r.to x, we get \sqrt{3}\left(2x + 2y\frac{dy}{dx}\right) = 4\left(x\frac{dy}{dx} + y\right) [1]
\Rightarrow \frac{dy}{dx} = \frac{2y - \sqrt{3}x}{\sqrt{3}y - 2x} [½]
\Rightarrow \left(\frac{dy}{dx}\right)_{\left(\frac12, \frac{\sqrt{3}}{2}\right)} = \sqrt{3} [½]
If x = a\sin^3 t,\ y = b\cos^3 t, then find \frac{dy}{dx} at t = \frac{\pi}{4}.
\frac{dx}{dt} = 3a\sin^2 t\cos t,\ \frac{dy}{dt} = -3b\cos^2 t\sin t [1]
\frac{dy}{dx} = -\frac{b}{a}\cot t [½]
\left.\frac{dy}{dx}\right]_{t = \frac{\pi}{4}} = -\frac{b}{a} [½]
If x = t + \frac{1}{t} and y = t - \frac{1}{t}, find \frac{dy}{dx} at t = 2.
x = t + \frac{1}{t} \Rightarrow \frac{dx}{dt} = 1 - \frac{1}{t^2} [½]
y = t - \frac{1}{t} \Rightarrow \frac{dy}{dt} = 1 + \frac{1}{t^2} [½]
\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1} [½]
\left(\frac{dy}{dx}\right)_{(t=2)} = \frac{5}{3} [½]
Show that the function f(x) = \begin{cases} \frac{\cos x}{-x + \frac{\pi}{2}}, & x \ne \frac{\pi}{2} \\ 1, & x = \frac{\pi}{2} \end{cases} is continuous at x = \frac{\pi}{2}.
For showing: \lim_{x \to \frac{\pi}{2}} f(x) = 1 [1]
For showing: f\left(\frac{\pi}{2}\right) = 1 [½]
\therefore f(x) is continuous at x = \frac{\pi}{2}. [½]
Find whether the function f(x) = \begin{cases} x - 1, & x \lt 2 \\ 2x - 3, & x \ge 2 \end{cases} at x = 2 is differentiable or not.
Getting LHD = 1 [1]
Getting RHD = 2 [½]
\therefore LHD \ne RHD, so f(x) is not differentiable at x = 2. [½]
Differentiate x^x with respect to x log x.
Let u = x^x and v = x \log x [½]
\frac{du}{dx} = x^x(1 + \log x) [½]
\frac{dv}{dx} = 1 + \log x [½]
\therefore \frac{du}{dv} = x^x [½]
If y = P cos ux + Q sin ux, show that \frac{d^2y}{dx^2} + u^2y = 0.
\frac{dy}{dx} = -P u \sin ux + Q u \cos ux [½]
\frac{d^2y}{dx^2} = -P u^2 \cos ux - Q u^2 \sin ux [½]
\frac{d^2y}{dx^2} = -u^2(P \cos ux + Q \sin ux) [½]
\therefore \frac{d^2y}{dx^2} + u^2y = 0 [½]
Determine a and b if f(x) is a continuous function, where
For getting 3a + b = 2 [½]
For getting 5a + b = 5 [½]
Thus, a = \frac{3}{2} and b = \frac{-5}{2} [½ + ½]
If 2y^3 + y = 2x, then show \frac{d^2y}{dx^2} = \frac{-48y}{(6y^2+1)^3}.
As, 2y^3 + y = 2x \Rightarrow (6y^2 + 1)\frac{dy}{dx} = 2 [½]
Thus, \frac{dy}{dx} = \frac{2}{6y^2+1} [½]
On again Differentiating w.r.t. x, we get
\frac{d^2y}{dx^2} = \frac{-24y}{(6y^2+1)^2} \times \frac{dy}{dx} [½]
\frac{d^2y}{dx^2} = \frac{-24y}{(6y^2+1)^2} \times \frac{2}{6y^2+1} = \frac{-48y}{(6y^2+1)^3} [½]
Differentiate 2^{\cos^2 x} w.r.t \cos^2 x.
Let u=2^{\cos^2 x}\Rightarrow\frac{du}{dx}=2^{\cos^2 x}(-2\cos x\sin x)\log 2 [1]
Let v=\cos^2 x\Rightarrow\frac{dv}{dx}=-2\cos x\sin x [½]
Now \frac{du}{dv}=\frac{\left(\frac{du}{dx}\right)}{\left(\frac{dv}{dx}\right)}=2^{\cos^2 x}\log 2 [½]
If \tan^{-1}(x^2+y^2)=a^2, then find \frac{dy}{dx}.
\tan^{-1}(x^2+y^2)=a^2\Rightarrow x^2+y^2=\tan a^2 [½]
Differentiate both sides wrt x,
2x+2y\frac{dy}{dx}=0 [1]
\Rightarrow\frac{dy}{dx}=-\frac{x}{y} [½]
If x = e^{\frac{x}{y}}, then prove that \frac{dy}{dx} = \frac{x - y}{x\log x}.
x = e^{\frac{x}{y}}
\Rightarrow \log x = \frac{x}{y}
\Rightarrow y\log x = x [½]
Differentiating both sides w.r.to x, we get
\frac{y}{x} + \log x\frac{dy}{dx} = 1 [1]
\Rightarrow \frac{dy}{dx} = \frac{x - y}{x\log x} [½]
If f(x) = \begin{cases} 2x - 3, & -3 \le x \le -2\\ x + 1, & -2 \lt x \le 0\end{cases}
Check the differentiability of f(x) at x = -2.
Lf'(-2) = \lim_{h\to 0}\frac{f(-2-h) - f(-2)}{-h}\quad (h \gt 0)
= \lim_{h\to 0}\frac{2(-2-h) - 3 - (-7)}{-h}
= \lim_{h\to 0} 2 = 2 [1]
Rf'(-2) = \lim_{h\to 0}\frac{f(-2+h) - f(-2)}{h}\quad (h \gt 0)
= \lim_{h\to 0}\frac{-2 + h + 1 - (-7)}{h}
= \lim_{h\to 0}\frac{6 + h}{h}, which does not exist, i.e., RHD does not exist.
Therefore, the function is not differentiable at -2. [1]
Note: (1) If a student finds only RHD and concludes the result, full marks may be awarded.
(2) If a student proves that the function is discontinuous at -2 and hence not differentiable at -2, full marks may be awarded.
Differentiate \frac{\sin x}{\sqrt{\cos x}} with respect to x.
Let y=\frac{\sin x}{\sqrt{\cos x}}
\frac{dy}{dx}=\frac{\sqrt{\cos x}\cdot\cos x-\sin x\cdot\left(\frac{-\sin x}{2\sqrt{\cos x}}\right)}{\cos x} [1½]
\Rightarrow\frac{dy}{dx}=\frac{2\cos^2x+\sin^2x}{2(\cos x)^{3/2}} or \frac{1+\cos^2x}{2(\cos x)^{3/2}} [½]
If y=5\cos x-3\sin x, prove that \frac{d^2y}{dx^2}+y=0.
y=5\cos x-3\sin x, then \frac{dy}{dx}=-5\cdot\sin x-3\cdot\cos x [1]
\Rightarrow\frac{d^2y}{dx^2}=-5\cdot\cos x+3\cdot\sin x=-y [½]
\Rightarrow\frac{d^2y}{dx^2}+y=0 [½]
Differentiate \sqrt{e^{\sqrt{2x}}} with respect to e^{\sqrt{2x}} for x \gt 0.
Let u = \sqrt{e^{\sqrt{2x}}} and v = e^{\sqrt{2x}} [½]
Derivative of \sqrt{v} w.r.t. v = \frac{1}{2\sqrt{v}} [1]
Required derivative = \frac{1}{2\sqrt{e^{\sqrt{2x}}}} [½]
If (x)^y = (y)^x, then find \frac{dy}{dx}.
Taking log on both sides, we get y \log x = x \log y [½]
Differentiating both sides w.r.t. x, we get
\frac{y}{x} + \log x \frac{dy}{dx} = \frac{x}{y}\frac{dy}{dx} + \log y [1]
\Rightarrow \frac{dy}{dx} = \frac{y(x\log y - y)}{x(y\log x - x)} [½]
Differentiate \left(\frac{5^x}{x^5}\right) with respect to x.
Let, y=\frac{5^x}{x^5}=5^x\cdot x^{-5}\Rightarrow\frac{dy}{dx}=\left(5^x\right)'\cdot x^{-5}+5^x\cdot\left(x^{-5}\right)' [1]
=\frac{5^x}{x^5}\log5-\frac{5^{x+1}}{x^6} [1]
If -2x^2-5xy+y^3=76, then find \frac{dy}{dx}.
Differentiating -2x^2-5xy+y^3=76, with respect to ‘x’
-4x-5y-5x\frac{dy}{dx}+3y^2\frac{dy}{dx}=0 [1½]
\Rightarrow\frac{dy}{dx}=\frac{4x+5y}{3y^2-5x} [½]
If e^y(x + 1) = 1, prove that \frac{dy}{dx} = -e^y.
e^y(x+1) = 1 \Rightarrow e^y = \frac{1}{x+1}
\Rightarrow y = -\log(x+1) [½]
\Rightarrow \frac{dy}{dx} = -\frac{1}{x+1} [1]
= -e^y \quad \left[\because \frac{1}{x+1} = e^y\right] [½]
Check whether the function f(x) = x^2\,|x| is differentiable at x = 0 or not.
f(x) = \begin{cases} x^3, & x \ge 0 \\ -x^3, & x \le 0 \end{cases} [½]
RHD = \lim_{h\to 0}\frac{f(0+h) - f(0)}{h} = \lim_{h\to 0} h^2 = 0 [½]
LHD = \lim_{h\to 0}\frac{f(0-h) - f(0)}{-h} = \lim_{h\to 0}(-h^2) = 0 [½]
\because RHD = LHD = 0, So f(x) is differentiable at x = 0 [½]
If y = \sqrt{\tan\sqrt{x}}, prove that \sqrt{x}\,\frac{dy}{dx} = \frac{1 + y^4}{4y}.
y = \sqrt{\tan\sqrt{x}}
\frac{dy}{dx} = \frac{\sec^2\sqrt{x}}{2\sqrt{\tan\sqrt{x}}} \times \frac{1}{2\sqrt{x}}
\sqrt{x}\,\frac{dy}{dx} = \frac{\sec^2\sqrt{x}}{4\sqrt{\tan\sqrt{x}}} [1]
= \frac{1 + (\tan\sqrt{x})^2}{4\sqrt{\tan\sqrt{x}}} = \frac{1 + y^4}{4y} [1]
If f(x) = |\tan 2x|, then find the value of f'(x) at x = \frac{\pi}{3}.
f(x) = -\tan 2x,\ \frac{\pi}{4} \lt x \lt \frac{\pi}{2} [1]
f'(x) = -2\sec^2 2x,\ \frac{\pi}{4} \lt x \lt \frac{\pi}{2} [½]
f'\left(\frac{\pi}{3}\right) = -2(-2)^2 = -8 [½]
If y = \text{cosec}(\cot^{-1} x), then prove that \sqrt{1 + x^2}\,\frac{dy}{dx} - x = 0.
y = \sqrt{1 + \cot^2(\cot^{-1}x)} = \sqrt{1 + x^2} [½]
\Rightarrow \frac{dy}{dx} = \frac{x}{\sqrt{1+x^2}} [1]
\Rightarrow \sqrt{1+x^2}\,\frac{dy}{dx} - x = 0 [½]
If x = e^{x/y}, prove that \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}
x = e^{\frac{x}{y}} \Rightarrow \log x = \frac{x}{y} \Rightarrow y = \frac{x}{\log x} [1]
\Rightarrow \frac{dy}{dx} = \frac{(\log x)(1) - x\left(\frac{1}{x}\right)}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} [1]
Check the differentiability of f(x) = \begin{cases} x^2 + 1, & 0 \le x \lt 1\\ 3 - x, & 1 \le x \le 2\end{cases} at x = 1.
LHD at x = 1
= \lim_{h\to 0}\frac{f(1-h) - f(1)}{-h} = \lim_{h\to 0}\frac{\left[(1-h)^2 + 1\right] - 2}{-h} = 2 [1]
RHD at x = 1
= \lim_{h\to 0}\frac{f(1+h) - f(1)}{h} = \lim_{h\to 0}\frac{\left[3 - (1+h)\right] - 2}{h} = -1 [½]
as LHD \ne RHD, so f(x) is not differentiable at x = 1 [½]
If y = \cos^3(\sec^2 2t), find \frac{dy}{dt}.
y = \cos^3(\sec^2 2t)
\Rightarrow \frac{dy}{dt} = 3\cos^2(\sec^2 2t)[-\sin(\sec^2 2t)] \times \frac{d(\sec^2 2t)}{dt} [½]
\Rightarrow \frac{dy}{dt} = -3\cos^2(\sec^2 2t).\sin(\sec^2 2t) \times 2\sec 2t.\sec 2t\tan 2t.2 [1]
\therefore \frac{dy}{dt} = -12\cos^2(\sec^2 2t) \times \sin(\sec^2 2t) \times \sec^2 2t \times \tan 2t. [½]
If x^y = e^{x-y}, prove that \frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}.
As, x^y = e^{x-y} \Rightarrow \log(x^y) = \log(e^{x-y})
\Rightarrow y\log x = (x-y) \Rightarrow y = \frac{x}{1+\log x} [1]
Now, Differentiating both the sides wrt x
\frac{dy}{dx} = \frac{(\log x+1).1 - x\left(\frac{1}{x}\right)}{(\log x+1)^2} = \frac{\log x}{(1+\log x)^2} [1]
Verify whether the function f defined by
is continuous at x=0 or not.
\lim_{x\to0}f(x)=\lim_{x\to0}x\cdot\sin\frac{1}{x}=0\times\text{Finite value in }[-1,1]=0=f(0) [1½]
\therefore f is a continuous function. [½]
Check for differentiability of the function f defined by f(x)=|x-5|, at the point x=5.
\text{LHD}=\lim_{x\to5^-}\frac{|x-5|-0}{x-5}=\lim_{x\to5^-}\frac{-(x-5)}{x-5}=-1 [1]
\text{RHD}=\lim_{x\to5^+}\frac{|x-5|-0}{x-5}=\lim_{x\to5^+}\frac{(x-5)}{x-5}=1 [½]
\text{LHD}\ne\text{RHD}, \therefore f is not differentiable at x=5 [½]
If y=(\sin^{-1}x)^{2}, then find (1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}.
\frac{dy}{dx}=\frac{2\sin^{-1}x}{\sqrt{1-x^{2}}}\Rightarrow(1-x^{2})\left(\frac{dy}{dx}\right)^{2}=4y [1]
Differentiating with respect to ‘x’, we get
(1-x^{2})2\frac{dy}{dx}\frac{d^{2}y}{dx^{2}}-2x\left(\frac{dy}{dx}\right)^{2}=4\frac{dy}{dx} [½]
\Rightarrow(1-x^{2})\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=2 [½]
If y^{x}=x^{y}, then find \frac{dy}{dx}.
y^{x}=x^{y}\Rightarrow x\log y=y\log x [½]
Differentiating with respect to ‘x’,
\frac{x}{y}\frac{dy}{dx}+\log y=\frac{y}{x}+\frac{dy}{dx}\log x [1]
\Rightarrow\frac{dy}{dx}=\frac{y}{x}\left(\frac{y-x\log y}{x-y\log x}\right) [½]
If y=\left(x+\sqrt{x^2-1}\right)^2, then show that (x^2-1)\left(\frac{dy}{dx}\right)^2=4y^2.
\frac{dy}{dx}=2\left(x+\sqrt{x^2-1}\right)\left(1+\frac{x}{\sqrt{x^2-1}}\right)=\frac{2\left(x+\sqrt{x^2-1}\right)^2}{\sqrt{x^2-1}} [1½]
\sqrt{x^2-1}\,\frac{dy}{dx}=2y [½]
(x^2-1)\left(\frac{dy}{dx}\right)^2=4y^2
If y = \sqrt{ax + b}, prove that y\left(\frac{d^2y}{dx^2}\right) + \left(\frac{dy}{dx}\right)^2 = 0.
y = \sqrt{ax + b} \Rightarrow y^2 = ax + b
Differentiate with respect to ‘x’, 2y\frac{dy}{dx} = a [1]
Differentiate with respect to ‘x’, 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 = 0 \Rightarrow y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 [1]
If f(x) = \begin{cases} ax + b & ; 0 \lt x \le 1 \\ 2x^2 - x & ; 1 \lt x \lt 2 \end{cases} is a differentiable function in (0, 2), then find the values of a and b.
f(x) is differentiable in (0, 2) \Rightarrow f(x) is continuous on (0, 2) [½]
\Rightarrow f(x) is continuous at x = 1\ \therefore \lim_{x\to 1^-}(ax + b) = \lim_{x\to 1^+}(2x^2 - x) \Rightarrow a + b = 1 [½]
Also, f(x) is differentiable at x = 1,\ \therefore L.H.D.(x = 1) = R.H.D.(x = 1)
\Rightarrow a = 4(1) - 1 \therefore a = 3 & b = 1 - a = -2 [½+½]
If y = x^{\frac{1}{x}}, then find \frac{dy}{dx} at x = 1.
y = x^{1/x}
\Rightarrow \log y = \frac{1}{x}\log x [½]
\Rightarrow \frac{1}{y}\frac{dy}{dx} = -\frac{\log x}{x^2} + \frac{1}{x^2} \Rightarrow \frac{dy}{dx} = x^{\frac{1}{x}}\,\frac{(1 - \log x)}{x^2} [1]
\Rightarrow \left(\frac{dy}{dx}\right)_{x=1} = 1 [½]
If x = a\sin 2t, y = a(\cos 2t + \log \tan t), then find \frac{dy}{dx}.
\frac{dx}{dt} = 2a\cos 2t [½]
\frac{dy}{dt} = 2a\left(-\sin 2t + \frac{\sec^2 t}{2\tan t}\right) = 2a\,\frac{\cos^2 2t}{\sin 2t} [1]
\frac{dy}{dx} = \cot 2t [½]
If (x^2+y^2)^2=xy, then find \frac{dy}{dx}.
(x^2+y^2)^2=xy gives
2(x^2+y^2)\left[2x+2y\frac{dy}{dx}\right]=x\frac{dy}{dx}+y [1]
\Rightarrow[4y(x^2+y^2)-x]\frac{dy}{dx}=y-4x(x^2+y^2) [½]
\Rightarrow\frac{dy}{dx}=\frac{y-4x(x^2+y^2)}{4y(x^2+y^2)-x}. [½]
If f(x)=\begin{cases}x^2, & \text{if } x\ge1\\ x, & \text{if } x\lt1\end{cases}, then show that f is not differentiable at x=1.
Here
\text{RHD}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=2
\text{LHD}=\lim_{h\to0}\left[\frac{f(1-h)-f(1)}{-h}\right]=1 [1½]
Since RHD \ne LHD
\therefore f is not differentiable at x = 1. [½]
Find the value(s) of ‘\lambda’, if the function
is continuous at x=0.
\lim_{x\to0}f(x)=\lim_{x\to0}\left(\frac{\sin^2\lambda x}{x^2}\right)=\lim_{x\to0}\left[\frac{\sin^2\lambda x}{(\lambda x)^2}\cdot\lambda^2\right]=\lambda^2 [1]
Since f(x) is continuous at x = 0
\lim_{x\to0}f(x)=f(0) [½]
\Rightarrow\lambda^2=1\Rightarrow\lambda=\pm1 [½]
Function f is defined as
Find the value of k for which the function f is continuous at x = 2.
Official marking scheme for this paper is not available.
If x=\sin^3t, y=\cos^3t, Evaluate \frac{d^2y}{dx^2} at \left(\frac{1}{2\sqrt2},\frac{1}{2\sqrt2}\right).
\frac{dx}{dt}=3\sin^2t\cos t and \frac{dy}{dt}=-3\cos^2t\sin t
\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=-\cot t [1]
\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}=cosec^2t\left(\frac{1}{3\sin^2t\cos t}\right)=\frac13\left(\sec t\cdot cosec^4t\right) [1]
At the point \left(\frac{1}{2\sqrt2},\frac{1}{2\sqrt2}\right), \sin^3t=\frac{1}{2\sqrt2}\Rightarrow t=\frac{\pi}{4}
\left.\frac{d^2y}{dx^2}\right]_{t=\frac{\pi}{4}}=\frac{4\sqrt2}{3} [1]
If y=x^{\log x}+(\log x)^x; x\gt1, then evaluate \frac{dy}{dx}
Let y=x^{\log x}+(\log x)^x=u+v where u=x^{\log x} and v=(\log x)^x
\log u=(\log x)^2 and \log v=x\cdot\log(\log x)
\frac1u\frac{du}{dx}=2(\log x)\cdot\frac1x and \frac1v\frac{dv}{dx}=\log(\log x)+\frac{1}{\log x} [1]
\frac{du}{dx}=x^{\log x}\left(2(\log x)\cdot\frac1x\right) and \frac{dv}{dx}=(\log x)^x\left(\log(\log x)+\frac{1}{\log x}\right) [1]
\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
\Rightarrow\frac{dy}{dx}=\frac2x\,x^{\log x}\log x+(\log x)^x\left(\log(\log x)+\frac{1}{\log x}\right) [1]
Alternative method
Since x=e^{\log x}
y=x^{\log x}+(\log x)^x=e^{\log(x^{\log x})}+e^{\log((\log x)^x)}
=e^{(\log x)^2}+e^{x\log(\log x)} [1]
\frac{dy}{dx}=e^{(\log x)^2}\left[\frac{2\log x}{x}\right]+e^{x\log(\log x)}\left[\log(\log x)+x\left(\frac{1}{\log x}\right)\left(\frac1x\right)\right]
OR x^{\log x}\left[\frac{2\log x}{x}\right]+(\log x)^x\left[\log(\log x)+\left(\frac{1}{\log x}\right)\right] [1+1]
If x^y=e^{x-y} prove that \frac{dy}{dx}=\frac{\log x}{(\log(xe))^2} and hence find its value at x=e.
x^y=e^{x-y}
Taking log of both sides
y\log x=(x-y)\log e
y\log x+y=x (since \log e=1)
\Rightarrow y=\frac{x}{1+\log x} [1]
Differentiating with respect to x
\frac{dy}{dx}=\frac{(1+\log x).1-x.\frac1x}{(1+\log x)^2}
=\frac{\log x}{(\log e+\log x)^2}
=\frac{\log x}{(\log(xe))^2} [1]
Now \left.\frac{dy}{dx}\right]_{x=e}=\frac{\log e}{(\log e^2)^2}=\frac{1}{(2\log e)^2}=\frac{1}{2^2}=\frac14 (as \log e=1) [1]
Alternative method
x^y=e^{x-y}
Taking log of both sides
y\log x=(x-y)\log e
y\log x+y=x (since \log e=1)
Differentiating both sides w.r.t. x
\log x\frac{dy}{dx}+\frac{y}{x}+\frac{dy}{dx}=1
\Rightarrow\frac{dy}{dx}(1+\log x)=1-\frac{y}{x}
\Rightarrow\frac{dy}{dx}=\frac{x-y}{x(1+\log x)}=\frac{x-\frac{x}{1+\log x}}{x(1+\log x)}=\frac{x(1+\log x)-x}{x(1+\log x)^2}=\frac{x(1+\log x-1)}{x(\log e+\log x)^2}=\frac{\log x}{(\log(xe))^2}
Now \left.\frac{dy}{dx}\right]_{x=e}=\frac{\log e}{(\log e^2)^2}=\frac{1}{(2\log e)^2}=\frac{1}{2^2}=\frac14 (as \log e=1)
If x=a(\theta-\sin\theta), y=a(1-\cos\theta) find \frac{d^2y}{dx^2}.
\frac{dx}{d\theta}=a(1-\cos\theta), \frac{dy}{d\theta}=a(0+\sin\theta), [1]
\Rightarrow\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{a\sin\theta}{a(1-\cos\theta)}
=\frac{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}{2\sin^2\left(\frac{\theta}{2}\right)}=\cot\frac{\theta}{2} [1]
\Rightarrow\frac{d^2y}{dx^2}=-\frac12 cosec^2\left(\frac{\theta}{2}\right)\frac{d\theta}{dx}
=-\frac{1}{2a}cosec^2\left(\frac{\theta}{2}\right)\frac{1}{2\sin^2\left(\frac{\theta}{2}\right)}
=-\frac{1}{4a}cosec^4\left(\frac{\theta}{2}\right) [1]
If (\sin x)^y = y^{\cos x}, then find \frac{dy}{dx}.
(\sin x)^y = y^{\cos x} \Rightarrow y\log\sin x = \cos x\log y [1]
On diff. wrt x both sides, we get
\frac{y\cos x}{\sin x} + \log\sin x\frac{dy}{dx} = \frac{\cos x}{y}\frac{dy}{dx} - \log y \cdot \sin x [1½]
\Rightarrow \frac{dy}{dx} = \frac{y(y\cos x + \log y \cdot \sin^2 x)}{\sin x(\cos x - y\log\sin x)} or \frac{(y\cot x + \sin x \cdot \log y)y}{\cos x - y\log\sin x} [½]
If xy = e^{x - y}, then find \frac{dy}{dx}.
xy = e^{x-y} \Rightarrow x\frac{dy}{dx} + y = e^{x-y}\left(1 - \frac{dy}{dx}\right) [1½]
\Rightarrow \left(x + e^{x-y}\right)\frac{dy}{dx} = e^{x-y} - y \Rightarrow \frac{dy}{dx} = \frac{e^{x-y} - y}{x + e^{x-y}} or \frac{y(x - 1)}{x(y + 1)} [1½]
Alternative method
xy = e^{x-y} \Rightarrow \log x + \log y = (x - y) [1]
\Rightarrow \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx} [1]
\Rightarrow \frac{dy}{dx} = \frac{1 - \frac{1}{x}}{\frac{1}{y} + 1} = \frac{y(x - 1)}{x(y + 1)} [1]
Differentiate \tan^{-1}\left(\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}\right) with respect to \cos^{-1} x^{2}.
Let y = \tan^{-1}\left(\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}\right) & put x^{2} = \cos\theta
y = \tan^{-1}\left(\frac{\sqrt{1 + \cos\theta} - \sqrt{1 - \cos\theta}}{\sqrt{1 + \cos\theta} + \sqrt{1 - \cos\theta}}\right) [½]
y = \tan^{-1}\left(\frac{\cos\left(\frac{\theta}{2}\right) - \sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right) + \sin\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right)\right) [1½]
Thus, y = \left(\frac{\pi}{4} - \frac{\theta}{2}\right) \Rightarrow \frac{dy}{d\theta} = \frac{-1}{2} [1]
Differentiate y = e^{\cos^{-1}\sqrt{1-x^2}} with respect to x and hence show that \frac{dy}{dx} = \frac{y}{\sqrt{1-x^2}}.
y = e^{\cos^{-1}\sqrt{1-x^2}}
\frac{dy}{dx} = e^{\cos^{-1}\sqrt{1-x^2}} \times \frac{-1}{\sqrt{1 - \left(\sqrt{1-x^2}\right)^2}} \times \frac{1}{2\sqrt{1-x^2}} \times (-2x) [1½]
\Rightarrow \frac{dy}{dx} = e^{\cos^{-1}\sqrt{1-x^2}} \times \frac{1}{\sqrt{1-x^2}} [1]
\Rightarrow \frac{dy}{dx} = \frac{y}{\sqrt{1-x^2}} [½]
If y = \log\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2, then show that x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2.
The given function can be written as
y = 2\log(x + 1) - \log x
\Rightarrow y_1 = \frac{2}{x+1} - \frac{1}{x} = \frac{x - 1}{x(x + 1)} [1]
\Rightarrow (x + 1)y_1 = \frac{x - 1}{x} = 1 - \frac{1}{x}
\Rightarrow (x + 1)y_2 + y_1 = \frac{1}{x^2} [1]
\Rightarrow x(x + 1)^2y_2 + x(x + 1)y_1 = 1 + \frac{1}{x}
\Rightarrow x(x + 1)^2y_2 + x(x + 1)y_1 = 1 + 1 - (x + 1)y_1
\Rightarrow x(x + 1)^2y_2 + (x + 1)^2y_1 = 2 [1]
If x\sqrt{1 + y} + y\sqrt{1 + x} = 0,\ -1 \lt x \lt 1,\ x \ne y, then prove that \frac{dy}{dx} = \frac{-1}{(1 + x)^2}.
x\sqrt{1 + y} + y\sqrt{1 + x} = 0
\Rightarrow x\sqrt{1 + y} = -y\sqrt{1 + x}
\Rightarrow x^2(1 + y) = y^2(1 + x) [½]
\Rightarrow (x - y)(x + y) + xy(x - y) = 0
\Rightarrow (x - y)(x + y + xy) = 0 [1]
x \ne y \Rightarrow x + y + xy = 0
\Rightarrow y = \frac{-x}{1 + x} [½]
\Rightarrow \frac{dy}{dx} = \frac{-1}{(1 + x)^2} [1]
Find k so that
is continuous at x = − 1.
\lim_{x\to -1}\frac{x^2-2x-3}{x+1}=\lim_{x\to -1}\frac{(x-3)(x+1)}{x+1}=\lim_{x\to -1}(x-3)=-4 [2]
Also, f(-1)=k [½]
as f is continuous, k=-4 [½]
Check the differentiability of function f(x)=x|x| at x = 0.
f(x)=x|x|=\begin{cases}-x^2, & x\le 0\\ x^2, & x\gt 0\end{cases} [1]
\text{LHD}=\lim_{h\to 0}\frac{f(0-h)-f(0)}{-h}=\lim_{h\to 0}\frac{-h^2-0}{-h}=0 [1]
\text{RHD}=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{h^2-0}{h}=0 [½]
Since LHD = RHD, f is differentiable at x=0 [½]
Differentiate y = \sin^{-1}(3x - 4x^3) w.r.t. x, if x \in \left[-\frac{1}{2}, \frac{1}{2}\right].
x = \sin t gives y = \sin^{-1}(\sin 3t) = 3t = 3\sin^{-1}x [½+1+½]
\frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}} [1]
Alternative method
Aliter: \frac{dy}{dx} = \frac{3 - 12x^2}{\sqrt{1 - (3x - 4x^3)^2}} [3]
Differentiate y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) with respect to x, when x \in (0, 1).
x = \tan t gives y = \cos^{-1}(\cos 2t) = 2t = 2\tan^{-1}x [½+1+½]
\frac{dy}{dx} = \frac{2}{1+x^2} [1]
Alternative method
Aliter: \frac{dy}{dx} = \frac{-1}{\sqrt{1 - \left(\frac{1-x^2}{1+x^2}\right)^2}} \cdot \frac{-4x}{(1+x^2)^2} [3]
Differentiate y=\sqrt{\log\left\{\sin\left(\frac{x^3}{3}-1\right)\right\}} with respect to x.
\frac{dy}{dx}=\frac{1}{2\sqrt{\log\left\{\sin\left(\frac{x^3}{3}-1\right)\right\}}}\cdot\frac{1}{\sin\left(\frac{x^3}{3}-1\right)}\cdot\cos\left(\frac{x^3}{3}-1\right)\cdot\frac{3x^2}{3} [1+1+½]
=\frac{x^2\cot\left(\frac{x^3}{3}-1\right)}{2\sqrt{\log\left\{\sin\left(\frac{x^3}{3}-1\right)\right\}}} [½]
If \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y), prove that \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}.
\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)
Put x = \sin\theta,\ y = \sin\phi
\Rightarrow \cos\theta + \cos\phi = a(\sin\theta - \sin\phi) [½]
\Rightarrow 2\cos\left(\frac{\theta + \phi}{2}\right)\cos\left(\frac{\theta - \phi}{2}\right) = 2a\sin\left(\frac{\theta - \phi}{2}\right)\cos\left(\frac{\theta + \phi}{2}\right) [½]
\Rightarrow \cot\left(\frac{\theta - \phi}{2}\right) = a
\Rightarrow \theta - \phi = 2\cot^{-1}a
\Rightarrow \sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a [½]
\Rightarrow \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}}\frac{dy}{dx} = 0 [1]
\Rightarrow \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} [½]
If y = (\tan x)^x, then find \frac{dy}{dx}.
y = (\tan x)^x
\log y = x\log(\tan x) [½]
\frac{1}{y}\frac{dy}{dx} = x\left(\frac{\sec^2 x}{\tan x}\right) + \log(\tan x) [2]
\frac{dy}{dx} = (\tan x)^x\left[\left(\frac{x\sec^2 x}{\tan x}\right) + \log(\tan x)\right] [½]
If x = e^{\cos 3t} and y = e^{\sin 3t}, prove that \frac{dy}{dx} = -\frac{y \log x}{x \log y}.
\frac{dx}{dt} = e^{\cos 3t} \times (-\sin 3t) \times 3 [1]
\frac{dy}{dt} = e^{\sin 3t} \times (\cos 3t) \times 3 [1]
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{e^{\sin 3t} \times (\cos 3t)}{-e^{\cos 3t} \times (\sin 3t)}
x = e^{\cos 3t} \Rightarrow \cos 3t = \log x
y = e^{\sin 3t} \Rightarrow \sin 3t = \log y
\therefore \frac{dy}{dx} = \frac{-y\log x}{x\log y} [1]
Show that :
\frac{d(|x|)}{dx} = \frac{d(\sqrt{x^2})}{dx},\ x \ne 0 [1]
= \frac{1}{2}(x^2)^{-\frac{1}{2}} \times \frac{d(x^2)}{dx} [1½]
= \frac{1}{2\sqrt{x^2}}\,2x = \frac{x}{|x|} [½]
If x\cos(p + y) + \cos p \sin(p + y) = 0, prove that \cos p\,\frac{dy}{dx} = -\cos^2(p + y), where p is a constant.
x\cos(p + y) + \cos p\sin(p + y) = 0
\Rightarrow x = \frac{-\cos p\sin(p + y)}{\cos(p + y)} \Rightarrow x = -\cos p.\tan(p + y) [1]
\Rightarrow \frac{dx}{dy} = -\cos p.\sec^2(p + y) [1]
\Rightarrow \frac{dy}{dx} = \frac{-1}{\cos p.\sec^2(p + y)} [½]
\Rightarrow \cos p\frac{dy}{dx} = -\cos^2(p + y) [½]
Find the value of a and b so that function f defined as :
is a continuous function.
\lim_{x\to 2^-} f(x) = -1 + a,\ \lim_{x\to 2^+} f(x) = 1 + b and f(2) = a + b [1]
as f is continous at x = 2\ \therefore -1 + a = 1 + b = a + b [1]
\Rightarrow a = 1, b = -1 [½+½]
Given that y = (\sin x)^x \cdot x^{\sin x} + a^x, find \frac{dy}{dx}.
As, y = (\sin x)^x.x^{\sin x} + a^x = u + a^x \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{d(a^x)}{dx} [½]
where u = (\sin x)^x.x^{\sin x} \Rightarrow \log u = x\log(\sin x) + \sin x.\log x
on differentiating both sides with respect to x, we get
\Rightarrow \frac{du}{dx} = (\sin x)^x.x^{\sin x}\left[\log(\sin x) + x\cot x + \frac{\sin x}{x} + \log x.\cos x\right] [1½]
Thus, \frac{dy}{dx} = (\sin x)^x.x^{\sin x}\left[\log(\sin x) + x\cot x + \frac{\sin x}{x} + \log x.\cos x\right] + a^x\log a [1]
Find \frac{dy}{dx}, if (\cos x)^y=(\cos y)^x.
Taking ‘log’ on both sides of (\cos x)^y=(\cos y)^x, we get
y\log\cos x=x\log\cos y [1]
\Rightarrow\frac{dy}{dx}\log\cos x+y(-\tan x)=\log\cos y+x(-\tan y)\frac{dy}{dx} [1½]
\Rightarrow\frac{dy}{dx}=\frac{\log\cos y+y\tan x}{\log\cos x+x\tan y} [½]
If \sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y), prove that \frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}.
Let x=\sin A,\ y=\sin B\ \therefore A=\sin^{-1}x,\ B=\sin^{-1}y [½]
\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)
\Rightarrow\cos A+\cos B=a(\sin A-\sin B)
\Rightarrow2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)=2a\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) [1]
\Rightarrow\cot\left(\frac{A-B}{2}\right)=a\Rightarrow A-B=2\cot^{-1}a, \therefore\sin^{-1}x-\sin^{-1}y=2\cot^{-1}a, differentiating with respect to ‘x’. [½]
\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0\Rightarrow\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}} [1]
If x=a\sin^3\theta,\ y=b\cos^3\theta, then find \frac{d^2y}{dx^2} at \theta=\frac{\pi}{4}.
\frac{dx}{d\theta}=3a\sin^2\theta\cos\theta,\ \frac{dy}{d\theta}=-3b\cos^2\theta\sin\theta [1]
\Rightarrow\frac{dy}{dx}=\frac{-3b\cos^2\theta\sin\theta}{3a\sin^2\theta\cos\theta}=-\frac{b}{a}\cot\theta [½]
\Rightarrow\frac{d^2y}{dx^2}=\frac{b}{a}\text{cosec}^2\theta\frac{d\theta}{dx}=\frac{b}{a}\text{cosec}^2\theta\cdot\frac{1}{3a\sin^2\theta\cos\theta}=\frac{b}{3a^2}\sec\theta\,\text{cosec}^4\theta [1]
\Rightarrow\left.\frac{d^2y}{dx^2}\right]_{\theta=\frac{\pi}{4}}=\frac{4\sqrt2 b}{3a^2} [½]
If x\sin(a+y)-\sin y=0,
prove that \frac{dy}{dx}=\frac{\sin^{2}(a+y)}{\sin a}
x=\frac{\sin y}{\sin(a+y)}\Rightarrow\frac{dx}{dy}=\frac{\sin(a+y)\cos y-\sin y\cos(a+y)}{\sin^{2}(a+y)} [2]
\Rightarrow\frac{dx}{dy}=\frac{\sin a}{\sin^{2}(a+y)},\ \therefore\frac{dy}{dx}=\frac{\sin^{2}(a+y)}{\sin a} [1]
Find \frac{dy}{dx}, if y=(\cos x)^{x}+\cos^{-1}\sqrt{x}.
Let y=u+v\Rightarrow\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
u=(\cos x)^{x}\Rightarrow\log u=x\log\cos x [½]
\Rightarrow\frac{du}{dx}=(\cos x)^{x}[\log\cos x-x\tan x] [1]
v=\cos^{-1}\sqrt{x}\Rightarrow\frac{dv}{dx}=-\frac{1}{2\sqrt{x}\sqrt{1-x}} [1]
\therefore\frac{dy}{dx}=(\cos x)^{x}[\log\cos x-x\tan x]-\frac{1}{2\sqrt{x}\sqrt{1-x}} [½]
Differentiate \sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right) w.r.t. \sin^{-1}\left(2x\sqrt{1-x^2}\right).
Let x=\sin\theta. Then
U=\sec^{-1}\left(\frac{1}{\sqrt{1-\sin^2\theta}}\right)=\sec^{-1}\left(\frac{1}{\cos\theta}\right)
=\sec^{-1}(\sec\theta)=\theta=\sin^{-1}x [1]
\Rightarrow\frac{dU}{dx}=\frac{1}{\sqrt{1-x^2}}
and V=\sin^{-1}\{2\sin\theta\sqrt{1-\sin^2\theta}\}
=\sin^{-1}[2\sin\theta\cos\theta]=2\theta=2\sin^{-1}x
\Rightarrow\frac{dV}{dx}=\frac{2}{\sqrt{1-x^2}} [1]
\Rightarrow\frac{dU}{dV}=\frac{dU/dx}{dV/dx}=\frac{1}{2} [1]
Note: If the substitution is made as x=\cos\theta, answer will be -\frac{1}{2}
If y=\tan x+\sec x, then prove that \frac{d^2y}{dx^2}=\frac{\cos x}{(1-\sin x)^2}.
y=\tan x+\sec x=\frac{\sin x+1}{\cos x}
\Rightarrow\frac{dy}{dx}=\frac{\cos x(\cos x)+(\sin x+1)\sin x}{\cos^2x}
=\frac{\cos^2x+\sin^2x+\sin x}{\cos^2x}=\frac{1+\sin x}{1-\sin^2x}=\frac{1}{1-\sin x} [1½]
\Rightarrow\frac{d^2y}{dx^2}=\frac{(1-\sin x)\cdot0-1(0-\cos x)}{(1-\sin x)^2}=\frac{\cos x}{(1-\sin x)^2} [1½]
Sports car racing is a form of motorsport which uses sports car prototypes. The competition is held on special tracks designed in various shapes.
The equation of one such track is given as follows :
Based on given information, answer the following questions :
(i) Find f'(x) for 0 \lt x \lt 3. [1]
(ii) Find f'(4) [1]
(iii) (a) Test for continuity of f(x) at x = 3. [2]
OR
(iii) (b) Test for differentiability of f(x) at x = 3. [2]
(i) For 0 \lt x \lt 3,\ f(x) = x^4 - 4x^2 + 4
\therefore f'(x) = 4x^3 - 8x [1]
(ii) For x \ge 3,\ f(x) = x^2 + 40
\therefore f'(x) = 2x
Hence, f'(4) = 2 \times 4 = 8 [1]
(iii)(a) Here f(3) = 49
LHL : \lim_{x \to 3^-}(x^4 - 4x^2 + 4) = 49
RHL : \lim_{x \to 3^+}(x^2 + 40) = 49 [1½]
Since \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3), so f(x) is continuous at x = 3. [½]
OR
(iii)(b) Here f(3) = 49
getting LHD = 84
getting RHD = 6 [1½]
Since LHD \ne RHD at x = 3, so f(x) is not differentiable at x = 3. [½]
Case Study – 3
The equation of the path traced by a roller-coaster is given by the polynomial f(x)=a(x+9)(x+1)(x-3). If the roller-coaster crosses y-axis at a point (0, –1), answer the following :
(i) Find the value of ‘a’. [2]
(ii) Find f''(x) at x=1. [2]
(i) -1=a(-27)\Rightarrow a=\frac{1}{27} [1 + 1]
(ii) f(x)=\frac{1}{27}(x+9)(x+1)(x-3) [½]
=\frac{1}{27}(x^3+7x^2-21x-27)
f'(x)=\frac{1}{27}(3x^2+14x-21) [½]
f''(x)=\frac{6x+14}{27} [½]
f''(1)=\frac{20}{27} [½]
Let f(x) be a real valued function. Then its
• Left Hand Derivative (L.H.D.) : Lf'(a)=\lim_{h\to 0}\frac{f(a-h)-f(a)}{-h}
• Right Hand Derivative (R.H.D.) : Rf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}
Also, a function f(x) is said to be differentiable at x=a if its L.H.D. and R.H.D. at x=a exist and both are equal.
For the function f(x)=\begin{cases}|x-3|, & x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}, & x\lt 1\end{cases}
answer the following questions :
(i) What is R.H.D. of f(x) at x=1 ? [1]
(ii) What is L.H.D. of f(x) at x=1 ? [1]
(iii) (a) Check if the function f(x) is differentiable at x=1. [2]
OR
(iii) (b) Find f'(2) and f'(-1). [2]
(i) R.H.D. of f(x) at x=1=\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}
=\lim_{h\to 0}\frac{|1+h-3|-|-2|}{h}=\lim_{h\to 0}\frac{2-h-2}{h}=-1 [1]
(ii) L.H.D. of f(x) at x=1=\lim_{h\to 0}\frac{f(1-h)-f(1)}{-h}
=\lim_{h\to 0}\frac{\left[\frac{(1-h)^2}{4}-\frac{3(1-h)}{2}+\frac{13}{4}-2\right]}{-h}
=\lim_{h\to 0}\left[\frac{h^2-2h+1-6+6h+13-8}{-4h}\right]
=\lim_{h\to 0}\left[\frac{h^2+4h}{-4h}\right]=-1. [1]
(iii) (a) Since L.H.D. of f(x) at x=1 is same as R.H.D. of f(x) at x=1, f(x) is differentiable at x=1. [2]
OR
(iii) (b) f(x)=\begin{cases}x-3, & x\ge 3\\ 3-x, & 1\le x\lt 3\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}, & x\lt 1\end{cases}
[f'(x)]_{x=2}=0-1=-1 [1]
[f'(x)]_{x=-1}=\frac{2(-1)}{4}-\frac{3}{2}=-2 [1]
If x = \cos t, y = \cos mt, prove that (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + m^2y = 0.
\frac{dx}{dt} = -\sin t, \frac{dy}{dt} = -m \times \sin(mt) [1]
\frac{dy}{dx} = \frac{m \times \sin(mt)}{\sin t} or \sin t\frac{dy}{dx} = m \times \sin(mt) [1]
\Rightarrow \sin t\frac{d^2y}{dx^2} + \frac{dy}{dx}\cos t \times \frac{dt}{dx} = m^2\cos(mt) \times \frac{dt}{dx} [2]
\Rightarrow (1 - \cos^2 t)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = -m^2\cos(mt)
\Rightarrow (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + m^2y = 0 [1]
If y\sqrt{x^2 + 1} = \log \sqrt{x^2 + 1} - x, show that
Differentiating both sides with respect to x, we get
\sqrt{x^2+1}\,\frac{dy}{dx} + y.\frac{x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1}}.\frac{x}{\sqrt{x^2+1}} - 1 [2 + 2]
(x^2 + 1)\frac{dy}{dx} + y.x = \frac{x}{\sqrt{x^2+1}} - \sqrt{x^2+1} [1]
Note: In case student, considers \sqrt{x^2+1} - x as part of log and writes following solution
Differentiating both sides with respect to x, we get
\sqrt{x^2+1}\,\frac{dy}{dx} + y.\frac{x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1} - x}.\left(\frac{x}{\sqrt{x^2+1}} - 1\right) [2]
\sqrt{x^2+1}\,\frac{dy}{dx} + y.\frac{x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1} - x}.\left(\frac{x - \sqrt{x^2+1}}{\sqrt{x^2+1}}\right) [1]
\sqrt{x^2+1}\,\frac{dy}{dx} + y.\frac{x}{\sqrt{x^2+1}} = \frac{-1}{\sqrt{x^2+1}} [1]
gives
(x^2 + 1)\frac{dy}{dx} + y.x = -1 or (x^2 + 1)\frac{dy}{dx} + x.y + 1 = 0 [1]
Find the differential of x^{\cot x} + \frac{2x^2 - 3}{2x^2 - x + 2} with respect to x.
Let u = x^{\cot x} and v = \frac{2x^2 - 3}{2x^2 - x + 2} [½]
\log u = \cot x.\log x [½]
\frac{1}{u}\frac{du}{dx} = \frac{\cot x}{x} + \log x(-\text{cosec}^2x) [1]
\frac{du}{dx} = x^{\cot x}\left(\frac{\cot x}{x} - \text{cosec}^2x.\log x\right) [½]
\frac{dv}{dx} = \frac{(2x^2 - x + 2)(4x) - (2x^2 - 3)(4x - 1)}{(2x^2 - x + 2)^2} [1]
= \frac{-2x^2 + 20x - 3}{(2x^2 - x + 2)^2} [1]
Required derivative = x^{\cot x}\left(\frac{\cot x}{x} - \text{cosec}^2x.\log x\right) + \frac{-2x^2 + 20x - 3}{(2x^2 - x + 2)^2} [½]
If \sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y), then prove that \frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}.
Let x=\sin A,\ y=\sin B\Rightarrow A=\sin^{-1}x,\ B=\sin^{-1}y [1]
\therefore\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)
\Rightarrow\cos A+\cos B=a(\sin A-\sin B)
\Rightarrow 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)=2a\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) [1]
\Rightarrow\cot\left(\frac{A-B}{2}\right)=a\Rightarrow A-B=2\cot^{-1}a [1]
\Rightarrow\sin^{-1}x-\sin^{-1}y=2\cot^{-1}a [½]
differentiate both sides wrt x,
\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0 [1½]
\Rightarrow\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}
If x=a\left(\cos\theta+\log\tan\frac{\theta}{2}\right) and y=\sin\theta, then find \frac{d^2y}{dx^2} at \theta=\frac{\pi}{4}.
x=a\left(\cos\theta+\log\tan\frac{\theta}{2}\right)
\Rightarrow\frac{dx}{d\theta}=a\left(-\sin\theta+\frac{1}{\tan\frac{\theta}{2}}\times\sec^2\frac{\theta}{2}\times\frac{1}{2}\right) [½]
=a\left(-\sin\theta+\frac{1}{\sin\theta}\right)=a\left(\frac{1-\sin^2\theta}{\sin\theta}\right) [½]
\frac{dx}{d\theta}=a\cot\theta\cos\theta [½]
Also, y=\sin\theta\Rightarrow\frac{dy}{d\theta}=\cos\theta [½]
\therefore\frac{dy}{dx}=\frac{\tan\theta}{a} [1]
Differetiating wrt x,
\frac{d^2y}{dx^2}=\frac{\sec^2\theta}{a}\times\frac{d\theta}{dx}
=\frac{\sec^3\theta\tan\theta}{a^2} [1]
\left.\frac{d^2y}{dx^2}\right]_{\text{at }\theta=\frac{\pi}{4}}=\frac{2\sqrt{2}}{a^2} [1]
For a positive constant ‘a’, differentiate a^{t+\frac1t} with respect to \left(t+\frac1t\right)^a, where t is a non-zero real number.
Let u=a^{t+\frac1t}\Rightarrow\frac{du}{dt}=a^{t+\frac1t}\cdot\log a\cdot\left(1-\frac{1}{t^2}\right) [2]
v=\left(t+\frac1t\right)^a\Rightarrow\frac{dv}{dt}=a\left(t+\frac1t\right)^{a-1}\cdot\left(1-\frac{1}{t^2}\right) [2]
\frac{du}{dv}=\frac{du/dt}{dv/dt}=\frac{a^{t+\frac1t}\cdot\log a}{a\left(t+\frac1t\right)^{a-1}} [1]
Find \frac{dy}{dx} if y^x+x^y+x^x=a^b, where a and b are constants.
Let u=y^x,\quad v=x^y and w=x^x
\Rightarrow\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0 ...............(i) [1]
u=y^x\Rightarrow\log u=x\cdot\log y\Rightarrow\frac1u\cdot\frac{du}{dx}=\frac{x}{y}\cdot\frac{dy}{dx}+\log y
\Rightarrow\frac{du}{dx}=y^x\left(\frac{x}{y}\cdot\frac{dy}{dx}+\log y\right)=xy^{x-1}\frac{dy}{dx}+y^x\log y [1]
v=x^y\Rightarrow\log v=y\cdot\log x\Rightarrow\frac1v\cdot\frac{dv}{dx}=\frac{y}{x}+\log x\cdot\frac{dy}{dx}
\Rightarrow\frac{dv}{dx}=x^y\left(\frac{y}{x}+\log x\cdot\frac{dy}{dx}\right)=yx^{y-1}+x^y\log x\,\frac{dy}{dx} [1]
w=x^x\Rightarrow\log w=x\cdot\log x\Rightarrow\frac1w\cdot\frac{dw}{dx}=1+\log x
\Rightarrow\frac{dw}{dx}=x^x\cdot(1+\log x) [1]
\therefore From (i), we get
xy^{x-1}\cdot\frac{dy}{dx}+y^x\cdot\log y+yx^{y-1}+x^y\cdot\log x\cdot\frac{dy}{dx}+x^x\cdot(1+\log x)=0
\Rightarrow\frac{dy}{dx}=-\frac{x^x\cdot(1+\log x)+y^x\cdot\log y+yx^{y-1}}{x\cdot y^{x-1}+x^y\cdot\log x} [1]
If y = \cos x^2 + \cos^2 x + \cos^2(x^2) + \cos(x^x), find \frac{dy}{dx}.
y = \cos x^2 + \cos^2 x + \cos^2(x^2) + \cos(x^x)
\frac{d}{dx}(\cos x^2) = -2x\sin x^2 [1]
\frac{d}{dx}(\cos^2 x) = 2\cos x(-\sin x) = -2\sin x\cos x [1]
\frac{d}{dx}(\cos^2(x^2)) = 2\cos(x^2)(-\sin(x^2))(2x) = -4x\sin x^2\cos x^2 [1]
\frac{d}{dx}(\cos(x^x)) = -\sin(x^x)\,[x^x(1 + \log x)] [1½]
\frac{dy}{dx} = -2x\sin x^2 - 2\sin x\cos x - 4x\sin x^2\cos x^2 - \sin(x^x)\,[x^x(1 + \log x)] [½]
If y=a\left\{x+2a\sin\left(\frac{x}{a}\right)\right\}, a\gt0, x\in[0,\pi a]; is a decreasing function, then the interval in which x lies is
(D) \left(\frac{2\pi a}{3},\pi a\right]
y is decreasing \Rightarrow\frac{dy}{dx}\lt0\Rightarrow a\left\{1+2\cos\left(\frac{x}{a}\right)\right\}\lt0\Rightarrow\cos\left(\frac{x}{a}\right)\lt-\frac12
\Rightarrow\frac{x}{a}\in\left(\frac{2\pi}{3},\pi\right]\Rightarrow x\in\left(\frac{2\pi a}{3},\pi a\right]
The values of x for which the derivative of the function y=x^4 is greater than the value of the function are
(A) 0\lt x\lt4
\frac{dy}{dx}\gt y\Rightarrow4x^3\gt x^4\Rightarrow x^3(4-x)\gt0\Rightarrow x(x-4)\lt0\Rightarrow0\lt x\lt4
A function f(x)=10-x-2x^2 is increasing on the interval
f(x)=10-x-2x^2
\Rightarrow f'(x)=-1-4x
For increasing function f'(x)\ge0
\Rightarrow -(1+4x)\ge0
\Rightarrow (1+4x)\le0
\Rightarrow x\le-\frac14
\Rightarrow x\in\left(-\infty,-\frac14\right]
(A) \left(-\infty,-\frac14\right] [1]
The least value of f(x) = x^3 - 12x,\ x \in [0, 3] is
(A) -16
For f(x) = x + \frac{1}{x}\ (x \ne 0)
(C) local maximum value is –2
The rate of change of volume of a sphere with respect to its diameter, when its radius is 5 cm, is :
(C) 50\pi \text{ cm}^3/\text{cm}
The surface area of a sphere when its volume changes at the same rate as its radius is :
(B) 1 sq. units [1]
Absolute minimum value of f(x) = (x - 2)^2 + 5 in the interval [-3, 2] is :
(C) 5
For f(x) = -|x+1| + 5, which of the following statements is true ?
(B) Max f(x) = 5 at x = -1.
The values of x for which the rate of increase of f(x) = x^3 - 5x^2 + 6x - 7 is thrice the rate of increase of x, are :
(C) 3, \frac{1}{3}
The absolute maximum value of function f(x)=x^3-3x+2 in [0,2] is :
(C) 4
The function f(x) = x^2 - 4x + 6 is increasing in the interval
(D) [2, \infty)
A cylindrical tank of radius 10 cm is being filled with sugar at the rate of 100\pi\ \text{cm}^3/\text{s}. The rate, at which the height of the sugar inside the tank is increasing, is :
(C) 1 cm/s
The values of \lambda so that f(x)=\sin x-\cos x-\lambda x+C decreases for all real values of x are :
(C) \lambda\ge\sqrt{2}
If f(x)=2x+\cos x, then f(x) :
(C) is an increasing function
The slope of the curve y=-x^3+3x^2+8x-20 is maximum at :
(A) (1, -10)
A spherical ball has a variable diameter \frac52(3x+1). The rate of change of its volume w.r.t. x, when x=1, is :
(C) 375\pi [1]
If f:\mathbb{R}\to\mathbb{R} is defined as f(x)=2x-\sin x, then f is :
(B) an increasing function [1]
If the rate of change of volume of a sphere is twice the rate of change of its radius, then the surface area of the sphere is :
(B) 2 sq units
Let f(x) be a continuous function on [a, b] and differentiable on (a, b). Then, this function f(x) is strictly increasing in (a, b) if
(B) f'(x) \gt 0,\ \forall\, x \in (a, b)
The function f(x) = x^3 - 3x^2 + 12x - 18 is :
(B) Strictly increasing on R
If the sides of a square are decreasing at the rate of 1·5 cm/s, the rate of decrease of its perimeter is :
(B) 6 cm/s
The function f(x) = kx - \sin x is strictly increasing for
(A) k \gt 1
[1]The function f(x)=\frac{x}{2}+\frac{2}{x} has a local minima at x equal to :
(A) 2 [1]
Given a curve y=7x-x^3 and x increases at the rate of 2 units per second. The rate at which the slope of the curve is changing, when x=5 is :
(A) -60 units/sec [1]
For the function f(x)=x^{3}, x=0 is a point of :
(D) inflexion
[1]If the radius of a circle is increasing at the rate of 0.5 cm/s, then the rate of increase of its circumference is :
(B) \pi cm/s
[1]If f(x)=a(x-\cos x) is strictly decreasing in \mathbb{R}, then ‘a’ belongs to
(C) (-\infty,0)
The interval in which the function f(x)=2x^3+9x^2+12x-1 is decreasing, is
(B) (-2,-1)
A room freshner bottle in the shape of an inverted cone sprays the perfume at regular intervals such that volume of the perfume in the bottle decreases at the steady rate of 1 mm3/min. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is 10 mm, if the semi-vertical angle of conical bottle is \frac{\pi}{6}.
Let v, r, h, respectively, be the volume, radius and the height of perfume at any time t.
\frac{dv}{dt} = -1\ mm^3/min
\frac{r}{h} = \tan\frac{\pi}{6} \Rightarrow r = \frac{h}{\sqrt{3}} [½]
v = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\frac{h^3}{3} [½]
\Rightarrow \frac{dv}{dt} = \frac{\pi}{9}3h^2\frac{dh}{dt} [½]
\Rightarrow \frac{dh}{dt} = -\frac{3}{\pi h^2}
\Rightarrow \left(\frac{dh}{dt}\right)_{h=10} = -\frac{3}{100\pi} [½]
The level of perfume is dropping at the rate of \frac{3}{100\pi}\ mm/min
Find the absolute maximum value of f(x) = \cos x + \sin^2 x,\ x \in [0, \pi]
f(x) = \cos x + \sin^2 x,\ x \in [0, \pi]
f'(x) = -\sin x + 2\sin x\cos x [1]
f'(x) = 0 \Rightarrow \sin x(2\cos x - 1) = 0
\Rightarrow \sin x = 0 \text{ or } \cos x = \frac{1}{2}
\therefore x = \frac{\pi}{3} [½]
Now, f(0) = 1,\ f\left(\frac{\pi}{3}\right) = \frac{5}{4},\ f(\pi) = -1
\therefore Absolute maximum value of f(x) is \frac{5}{4}. [½]
If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.
V = \frac{2}{3}\pi r^3 \Rightarrow \frac{dV}{dt} = 2\pi r^2\frac{dr}{dt} = C (say) [1]
\Rightarrow \frac{dr}{dt} = \frac{C}{2\pi r^2}
Now, S = 3\pi r^2
\Rightarrow \frac{dS}{dt} = 6\pi r\frac{dr}{dt} = 6\pi r \times \frac{C}{2\pi r^2} = \frac{3C}{r} [1]
\therefore S varies inversely as its radius.
Find the sub-interval(s) of \left(0, \frac{\pi}{2}\right) in which f(x) = \tan x - 4x is increasing.
f'(x) = \sec^2 x - 4 [1]
Put f'(x) = 0 \Rightarrow x = \frac{\pi}{3} [½]
(Sign of f'(x): - on \left(0, \frac{\pi}{3}\right), + on \left(\frac{\pi}{3}, \frac{\pi}{2}\right))
Hence f is increasing in \left(\frac{\pi}{3}, \frac{\pi}{2}\right). [½]
Note: Interval [\pi/3, \pi/2) can be considered.
Find the values of x for which f(x) = x^{x}, x \gt 0 is increasing.
f'(x) = x^{x}(1 + \log x) [1]
Put, f'(x) = 0 \Rightarrow x^{x}(1 + \log x) = 0
As x^{x} \gt 0 for x \gt 0, (1 + \log x) = 0 \Rightarrow x = \frac{1}{e} [½]
\therefore f(x) is increasing when x \gt \frac{1}{e} or x \in \left(\frac{1}{e}, \infty\right) or x \in \left[\frac{1}{e}, \infty\right) [½]
Determine the values of x for which f(x) = \frac{x - 3}{x + 1}, x \ne -1 is an increasing function.
f^{\prime}(x) = \frac{(x+1).1 - (x-3).1}{(x+1)^2} [1]
= \frac{4}{(x+1)^2} \gt 0 [½]
\therefore f(x) is an increasing function for all x \in R - \{-1\} [½]
Find the intervals in which function f(x)=5x^{\frac{3}{2}}-3x^{\frac{5}{2}} is (i) increasing (ii) decreasing.
f(x)=5x^{3/2}-3x^{5/2}\Rightarrow f'(x)=\frac{15}{2}\sqrt{x}(1-x) [1]
For increasing / decreasing, put f'(x)=0
\Rightarrow x=0,1
(i) When x\in[0,1], f'(x)\ge 0. So, f is increasing when x\in[0,1] [½]
(The intervals (0,1), [0,1) or (0,1] can also be considered.)
(ii) When x\in[1,\infty), f'(x)\le 0. So, f is decreasing when x\in[1,\infty) [½]
(The interval (1,\infty) can also be considered.)
Find the values of ‘a’ for which f(x) = \sin x - ax + b is increasing on R.
f'(x) = \cos x - a
For f(x) to be increasing, f'(x) \ge 0
i.e., \cos x \ge a [1]
Since, -1 \le \cos x \le 1
\Rightarrow a \le -1
Hence, a \in (-\infty, -1]. (Also, accept a \in (-\infty, -1)) [1]
Find the least value of ‘a’ so that f(x)=2x^2-ax+3 is an increasing function on [2, 4].
f(x)=2x^2-ax+3\Rightarrow f'(x)=4x-a [½]
Now 2\le x\le 4\Rightarrow 8-a\le 4x-a\le 16-a [1]
For f to be an increasing function, f'(x)\ge 0
\Rightarrow 8-a\ge 0\Rightarrow a\le 8 [½]
\therefore Least value of a does not exist.
If f(x)=x+\frac{1}{x},\ x\ge 1, show that f is an increasing function.
f(x)=x+\frac{1}{x}\Rightarrow f'(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2} [1]
Now \frac{x^2-1}{x^2}\ge 0 for all x\ge 1 [½]
\Rightarrow f'(x)\ge 0\Rightarrow f is an increasing function. [½]
For the curve y=5x-2x^3, if x increases at the rate of 2 units/s, then how fast is the slope of the curve changing when x = 2 ?
y=5x-2x^3
Given \frac{dx}{dt}=2 units/s [½]
slope of the curve =\frac{dy}{dx}=5-6x^2=m [½]
\frac{dm}{dt}=-12x\frac{dx}{dt}=-12x(2)=-24x [½]
at x=2,\ \frac{dm}{dt}=-24(2)=-48 [½]
Hence, slope of curve is decreasing at the rate of 48
Surface area of a balloon (spherical), when air is blown into it, increases at a rate of 5 mm2/s. When the radius of the balloon is 8 mm, find the rate at which the volume of the balloon is increasing.
\frac{dS}{dt}=5\text{ mm}^2/\text{s},\quad \left(\frac{dV}{dt}\right)_{r=8}=?
S=4\pi r^2\Rightarrow\frac{dS}{dt}=8\pi r\cdot\frac{dr}{dt}\Rightarrow\frac{dr}{dt}=\frac{5}{8\pi r} [½]
V=\frac43\pi r^3\Rightarrow\frac{dV}{dt}=4\pi r^2\cdot\frac{dr}{dt}\Rightarrow\frac{dV}{dt}=\frac52 r [1]
\Rightarrow\left(\frac{dV}{dt}\right)_{r=8}=20\text{ mm}^3/\text{s} [½]
Determine the values of x for which f(x) = \frac{x-4}{x+1},\ x \ne -1 is an increasing or a decreasing function.
f'(x) = \frac{x+1-x+4}{(x+1)^2} = \frac{5}{(x+1)^2} \gt 0 [1½]
Hence f is increasing in its domain. [½]
A ladder 13 m long is leaning against the wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 2 m/s. How fast is the height on the wall decreasing when the foot of the ladder is 12 m away from the wall ?
x^2 + y^2 = 169 [½]
Differentiate both sides w.r.t. t
2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 [1]
\Rightarrow 12(2) + 5\left(\frac{dy}{dt}\right) = 0\ [\because \text{when } x = 12m, y = 5m]
\Rightarrow \frac{dy}{dt} = -\frac{24}{5} [½]
Hence, the height decreases at the rate of \frac{24}{5} m/s
Show that the function f(x) = 4x^3 - 18x^2 + 27x - 7 has neither maxima nor minima.
f'(x) = 12x^2 - 36x + 27 [½]
= 3(2x - 3)^2 \ge 0 for all x \in R [½]
\therefore f is increasing on R. [½]
Hence f(x) does not have maxima or minima. [½]
If M and m denote the local maximum and local minimum values of the function f(x) = x + \frac{1}{x}\ (x \ne 0) respectively, find the value of (M – m).
f'(x) = 1 - \frac{1}{x^2} = \frac{x^2-1}{x^2} = \frac{(x+1)(x-1)}{x^2}
f'(x) = 0 \Rightarrow x = -1, 1 [½]
f''(x) = \frac{2}{x^3} \Rightarrow f''(-1) = -2 \lt 0
\therefore -1 is a point of local maximum
The local maximum value = f(-1) = -2 = M [½]
f''(1) = 2 \gt 0
\therefore 1 is point of local minimum
The local minimum value = f(1) = 2 = m [½]
M - m = -4 [½]
Show that f(x) = e^x - e^{-x} + x - \tan^{-1} x is strictly increasing in its domain.
f'(x) = e^x + e^{-x} + 1 - \frac{1}{1+x^2} [1]
= e^x + \frac{1}{e^x} + \frac{x^2}{1+x^2} \gt 0\ \text{for all } x \in R
\therefore f is strictly increasing over its domain R [1]
Find the interval in which the function f(x) = x^4 - 4x^3 + 10 is strictly decreasing.
f'(x) = 4x^3 - 12x^2 = 4x^2(x-3) [1]
\Rightarrow 4x^2(x-3) \lt 0 for x \lt 3, x \ne 0
\Rightarrow f'(x) \lt 0 for x \lt 3, x \ne 0 [½]
Thus, f(x) = x^4 - 4x^3 + 10 is strictly decreasing on (-\infty, 0) \cup (0, 3).
OR
Thus, f(x) = x^4 - 4x^3 + 10 is strictly decreasing on (-\infty, 0] \cup [0, 3] [½]
The volume of a cube is increasing at the rate of 6 cm3/s. How fast is the surface area of cube increasing, when the length of an edge is 8 cm ?
Given, \frac{dV}{dt} = 6 \text{ cm}^3/\text{sec}. Since, V = x^3
\frac{dV}{dt} = 3x^2\frac{dx}{dt} \Rightarrow 6 = 3x^2\frac{dx}{dt} \Rightarrow \frac{dx}{dt} = \frac{2}{x^2} \text{ cm/sec} [1]
Now, Surface Area = S = 6x^2 \Rightarrow \frac{dS}{dt} = 12x\frac{dx}{dt} = 3 \text{ cm}^2/\text{sec} [1]
The area of the circle is increasing at a uniform rate of 2 cm2/sec. How fast is the circumference of the circle increasing when the radius r = 5 cm ?
Let C=2\pi r, be the circumference of the circle,
\frac{d(\pi r^2)}{dt}=2\Rightarrow\frac{dr}{dt}=\frac{1}{\pi r}\text{ cm/sec} [1]
\frac{dC}{dt}=2\pi\frac{dr}{dt}=\frac{2}{r}=\frac{2}{5}\text{ cm/sec at }r=5\text{ cm} [1]
Given that f(x)=\frac{\log x}{x}, find the point of local maximum of f(x).
f'(x)=\frac{1-\log x}{x^{2}},\ \therefore f'(x)=0\Rightarrow\log x=1\Rightarrow x=e [1]
f''(x)=\frac{2x\log x-3x}{x^{4}}\Rightarrow f''(e)=-\frac{1}{e^{3}}\lt 0 i.e. x=e is a point of local maximum. [1]
Show that the function f(x)=\frac{16\sin x}{4+\cos x}-x, is strictly decreasing in \left(\frac{\pi}{2},\pi\right).
f'(x)=\frac{16[4+\cos x]\cos x+16\sin^2x}{(4+\cos x)^2}-1 [1]
=\frac{\cos x\,(56-\cos x)}{(4+\cos x)^2} [½]
in \left(\frac{\pi}{2},\pi\right), \cos x\lt0\Rightarrow f'(x)\lt0 [½]
\therefore f(x) in strictly decreasing in \left(\frac{\pi}{2},\pi\right)
A particle moves along the curve 3y = ax^3 + 1 such that at a point with x-coordinate 1, y-coordinate is changing twice as fast at x-coordinate. Find the value of a.
Differentiating equation 3y = ax^3 + 1 with respect to ‘x’, 3\frac{dy}{dx} = 3ax^2 [1]
Taking x = 1, \frac{dy}{dx} = 2,\ 3(2) = 3a(1)^2 \Rightarrow a = 2 [1]
If f(x) = a(\tan x - \cot x), where a \gt 0, then find whether f(x) is increasing or decreasing function in its domain.
f'(x) = a(\sec^2 x + \text{cosec}^2 x) [1]
As a \gt 0 and \sec^2 x, \text{cosec}^2 x are squares, f'(x) \gt 0 [½]
\therefore f(x) is an increasing function in its domain. [½]
Find the maximum and minimum values of the function given by f(x)=5+\sin 2x.
We know that -1\le\sin 2x\le 1 [1]
\Rightarrow 4\le\sin 2x+5\le 1+5
\Rightarrow 4\le\sin 2x+5\le 6 [½]
So, maximum value is 6 and minimum value is 4 [½]
Find the intervals in which the function f(x) = x^4 - 4x^3 + 4x^2 + 15, is strictly increasing.
Official marking scheme for this paper is not available.
The circular area of vision on the ground by a surveillance drone fitted with camera increases at the rate of 44\ m^2\ \text{per metre} as it gains vertical height. The radius of the circle on the ground increases at the rate of 6 m/s. Find the rate at which the height of the drone increases with time, at the instant when the radius of the circular area on the ground is 14 m. \left(Use\ \pi=\frac{22}{7}\right)
Let A be the area of the circular region covered by the drone and let r be its radius. Let h be the height of the drone at any time t.
Given \frac{dA}{dh}=44\ m^2\ per\ metre and \frac{dr}{dt}=6\ m/s To find \frac{dh}{dt} when r = 14 m. [½]
A=\pi r^2\quad\Rightarrow\frac{dA}{dt}=2\pi r\frac{dr}{dt}\quad\Rightarrow\frac{dA}{dh}\cdot\frac{dh}{dt}=2\pi r\frac{dr}{dt} [1½]
44\frac{dh}{dt}=2\left(\frac{22}{7}\right)6r\quad\Rightarrow\left.\frac{dh}{dt}\right]_{r=14m}=12 m/s [1]
The height of the drone increases at the rate of 12 m/s at that moment.
A spherical ball of ice melts in such a way that the rate at which its volume decreases at any instant is directly proportional to its surface area. Prove that the radius of the ice ball decreases at a constant rate.
Let r be the radius of ice ball at time t .
V=\frac43\pi r^3 ............ (1)
S=4\pi r^2 ..............(2) [½]
Given \frac{dV}{dt}\propto S
\Rightarrow\frac{dV}{dt}=-kS (where k is some positive constant) .........(3) [½]
Differentiating (1) w.r.t. t, we get
\frac{dV}{dt}=\frac43\pi.(3r^2)\frac{dr}{dt}
\frac{dV}{dt}=4\pi r^2\frac{dr}{dt} .......... (4) [1]
\Rightarrow -kS=4\pi r^2\frac{dr}{dt} (from (3) and (4)) [½]
\Rightarrow -kS=S\frac{dr}{dt} (using (2))
\Rightarrow\frac{dr}{dt}=-k [½]
\Rightarrow radius of the ice-ball decreases at a constant rate
A spherical balloon loses its volume due to escape of air from it in such a way that decrease of volume at any instant is proportional to its surface area. Show that the radius is decreasing at a constant rate.
Let r be the radius, V = \frac{4}{3}\pi r^3 be the volume
\frac{dV}{dt} = -k(4\pi r^2) where k \gt 0 [1]
4\pi r^2\frac{dr}{dt} = -4k\pi r^2 [1]
\therefore \frac{dr}{dt} = -k which is a constant
Thus, radius is decreasing at a constant rate [1]
Find the values of ‘a’ and ‘b’, if y = \frac{ax - b}{(x-1)(x-4)} has a critical point (3, 1).
y = \frac{ax - b}{(x-1)(x-4)}
Since (3,1) lies on the given curve,
\Rightarrow 1 = \frac{3a - b}{(3-1)(3-4)}
\Rightarrow 3a - b = -2 \quad \ldots(i) [1]
Now \frac{dy}{dx} = \frac{(x-1)(x-4)(a) - (ax-b)(2x-5)}{(x-1)^2(x-4)^2} [1]
Since (3,1) is a critical point, \frac{dy}{dx} = 0
\Rightarrow b = 5a \quad \ldots(ii) [½]
solving (i) and (ii)
a = 1 and b = 5 [½]
The side of an equilateral triangle is increasing at the rate of 3 cm/s. At what rate its area increasing when the side of the triangle is 15 cm ?
Let 'a' be the side of the triangle, so \frac{da}{dt}=3 cm/s [½]
Now area of an equilateral triangle, A=\frac{\sqrt{3}}{4}a^2
\Rightarrow\frac{dA}{dt}=\frac{\sqrt{3}a}{2}\times\frac{da}{dt} [1½]
\therefore\left.\frac{dA}{dt}\right]_{a=15\text{ cm}}=\frac{\sqrt{3}\times 15}{2}\times 3=\frac{45\sqrt{3}}{2}\ \text{cm}^2/\text{s} [1]
Find the value of ‘a’ for which f(x)=\sqrt3\sin x-\cos x-2ax+6 is decreasing in R.
Since f(x) is a decreasing function \Rightarrow f'(x)\le0
\Rightarrow\sqrt3\cdot\cos x+\sin x-2a\le0 [1]
\Rightarrow2\left(\frac{\sqrt3}{2}\cdot\cos x+\frac12\sin x\right)-2a\le0
\Rightarrow\cos\left(x-\frac{\pi}{6}\right)\le a [1]
Since, -1\le\cos\left(x-\frac{\pi}{6}\right)\le1\Rightarrow a\ge1 i.e. a\in[1,\infty) or (1,\infty) [1]
Amongst all pairs of positive integers with product as 289, find which of the two numbers add up to the least.
Let numbers be ‘x’ and ‘y’ such that xy=289\Rightarrow y=\frac{289}{x}, ‘S’ be their sum, then
S=x+y=x+\frac{289}{x} [1]
\frac{dS}{dx}=1-\frac{289}{x^2},\ \frac{dS}{dx}=0\Rightarrow x=17, a positive integer [1½]
\left.\frac{d^2S}{dx^2}\right]_{x=17}=289\left.\left(\frac{2}{x^3}\right)\right]_{x=17}\gt0, \therefore S is minimum when x=17,y=17 [½]
Find the maximum slope of the curve y = -x^3 + 3x^2 + 9x - 30.
y = -x^3 + 3x^2 + 9x - 30
Slope of the curve, m = \frac{dy}{dx} = -3x^2 + 6x + 9 [1]
\Rightarrow \frac{dm}{dx} = -6x + 6 [½]
For maximum/ minimum slope, put \frac{dm}{dx} = 0
\Rightarrow x = 1 [½]
As \frac{d^2m}{dx^2} = -6 \lt 0 \therefore m is maximum at x = 1 [½]
Maximum slope = -3(1)^2 + 6(1) + 9 = 12 [½]
Find the intervals in which the function f(x) = \frac{\log x}{x} is strictly increasing or strictly decreasing.
f(x) = \frac{\log x}{x} \Rightarrow f'(x) = \frac{1 - \log x}{x^2};\ x \gt 0 [1]
for strictly increasing/decreasing, put f'(x) = 0 \Rightarrow x = e [1]
for strictly increasing, x \in (0, e) and for strictly decreasing x \in (e, \infty) [½+½]
Find the absolute maximum and absolute minimum values of the function f given by f(x) = \frac{x}{2} + \frac{2}{x}, on the interval [1, 2].
f(x) = \frac{x}{2} + \frac{2}{x};\ x \in [1, 2]
\Rightarrow f'(x) = \frac{1}{2} - \frac{2}{x^2} [1]
for absolute maximum / minimum, put f'(x) = 0
\Rightarrow x^2 = 4 \Rightarrow x = 2 [½]
Now, f(1) = \frac{5}{2} and f(2) = 2 [½+½]
\therefore absolute maximum value = \frac{5}{2} and absolute minimum value = 2 [½]
Water is essential for life, and rivers are vital water sources across the world. The Yamuna Action Plan (YAP), one of India's largest river restoration projects, was launched to clean the Yamuna, but its effectiveness (WQI) the Water Quality Index must be measured through cost-benefit, economic and environmental analysis. Therefore, comprehensive pollutant treatment, prevention of further contamination, and community awareness are crucial for sustainable river management.
Implementing the same concept to the river Yamuna, if an initial base line WQI is 35, a quadratic expression is formed to measure the impact of investment of ₹x crore on the WQI i.e., W(x) as
W(x)=35+1.6x-0.02x^2, where x is in Rupees in crores
Use the given W(x) to answer the following questions:
(i) Find the value of x at the stationary point for the function W(x). [1]
(ii) Find the maximum value of the WQI using the second derivative test? [1]
(iii) (a) If the investment is ₹ 35 crores, find by how much the WQI is short of its maximum achievable value. [2]
Or
(b) Evaluate the WQI when the investment is ₹60 crore. Determine whether this indicates an enhancement or a diminishing return in our objective when compared with the optimum WQI. [2]
(i) W(x)=35+1.6x-0.02x^2
W'(x)=1.6-0.04x=0\Rightarrow x=40
Therefore, the value of x is 40 crores at the stationary point of W(x). [1]
(ii) W''(x)=-0.04\lt0\Rightarrow W(x) is maximum at x=40 by the second derivative test. [½]
Thus, the WQI is maximum when ₹ 40 crores are spent on the revival.
Max. WQI = W(40)=35+64-32=67 [½]
(iii) (a) If the investment is ₹35 crores, then the WQI = W(35)
W(35)=35+(1.6)(35)-(0.02)(35)^2
=(2.6)(35)-(0.02)(1225)
=91-24.5=66.5 [1]
Also, 67-66.5 = 0.5
So, the WQI is short by 0.5 units of its maximum level. [1]
OR
(b) W(60)=35+96-72=59 [1]
59 is less than the maximum value 67 of WQI, even when the investment has been increased from 40 crores to 60 crores. This shows that the returns are diminishing. [1]
Case Study -2
LED bulbs are energy-efficient because they use significantly less electricity than traditional bulbs while producing the same amount of light. They convert more energy into light rather than heat, reducing waste. Additionally, their long lifespan means fewer replacements, saving resources and money over time.
A company manufactures a new type of energy-efficient LED bulb. The cost of production and the revenue generated by selling x bulbs (in an hour) are modelled as
C(x)=0.5x^2-10x+150 and R(x)=-0.3x^2+20x respectively, where C(x) and R(x) are both in ₹.
To maximize the profit, the company needs to analyze these functions using calculus. Use the given models to answer the following questions:
I. Derive the profit function P(x) [1]
II. Find the critical points of P(x). [1]
III A. Determine whether the critical points correspond to a maximum or a minimum profit by using the second derivative test.
OR
III B. Identify the possible practical value of x (i.e., the number of bulbs that can realistically be produced and sold) that can maximize the profit, if the resources available and the expenditure on machines allows to produce minimum 10 but not more than 18 bulbs per hour. Also calculate the maximum profit. [2]
I. P(x)=R(x)-C(x)=-0.3x^2+20x-(0.5x^2-10x+150)
=-0.8x^2+30x-150 [1]
II. For critical points P'(x)=0\Rightarrow-1.6x+30=0
\Rightarrow x=\frac{30}{1.6}=\frac{300}{16}=18.75 [1]
III A. Now P''(x)=-1.6
In particular P''(18.75)=-1.6\lt0 [1]
So, critical value x=18.75 corresponds to a maximum profit. [1]
OR
III B. As x is the number of bulbs, so practically 18 bulbs correspond to a maximum profit. [1]
Maximum profit is P(18)=-0.8\times18^2+30\times18-150
=-259.2+540-150
=540-409.2=₹130.80 [1]
An online delivery company in a city has 5000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries.
The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹ 1, ten subscribers will discontinue. Assume that the company increased the annual fee by ₹ x.
Based on the given information, answer the following questions :
(i) How many subscribers will discontinue after an increase of ₹ x in annual fee ? [1]
(ii) If R(x) denotes the total revenue collected after the increase of ₹ x in subscription fee, express R(x) as a function of x. [1]
(iii) (a) Find the value of x for which R(x) is maximum. [2]
OR
(iii) (b) Find the sub-intervals of (0, 5000) in which R(x) is increasing and decreasing. [2]
(i) 10x [1]
(ii) R(x) = (5000 - 10x)(300 + x) [1]
(iii) (a) R'(x) = 2000 - 20x [½]
R'(x) = 0 \Rightarrow x = 100 [½]
R''(x) = -20 \lt 0 at x = 100 [½]
At x = 100, R(x) is maximum [½]
OR
(iii) (b) R'(x) = 2000 - 20x [1/2]
R'(x) = 0 \Rightarrow x = 100 [1/2]
In (0, 100), R'(x) \gt 0. Therefore, R(x) is increasing in (0, 100) or (0, 100] [1/2]
In (100, 5000), R'(x) \lt 0. Therefore, R(x) is decreasing in (100, 5000) or [100, 5000) [1/2]
Case Study – 1
Two vertical light poles of height 22 m and 16 m stand on the opposite sides of a 20 m wide road as shown below in the figure.
Two ladders of length l_1 and l_2 are placed from a common point R on the road at a distance of x m from the smaller pole.
Based on the above information, answer the following questions :
(i) Express p(x) = l_1 + l_2 in terms of x. [1]
(ii) Find p'(x). [1]
(iii) (a) Find the value of x for which l_1^2 + l_2^2 is minimum. [2]
OR
(iii) (b) If the 22 m long pole is also replaced by a 16 m long pole, at what distance from either pole should the ladders be kept so that the sum of squares of lengths of ladders needed to reach the top of the pole is minimum ? [2]
(i) p(x) = \sqrt{22^2 + (20 - x)^2} + \sqrt{16^2 + x^2} or
p(x) = \sqrt{x^2 - 40x + 884} + \sqrt{x^2 + 256} [1]
(ii) p'(x) = \frac{x - 20}{\sqrt{x^2 - 40x + 884}} + \frac{x}{\sqrt{x^2 + 256}} [1]
(iii) (a) Let q(x) = {l_1}^2 + {l_2}^2 = 2x^2 - 40x + 1140 [½]
q'(x) = 4x - 40 [½]
Put q'(x) = 0 \Rightarrow x = 10 [½]
q''(x = 10) = 4 \gt 0 \Rightarrow q(x) is minimum at x = 10. [½]
OR
(iii) (b) Let, ladder is kept at y m distance from one pole.
Suppose, r(y) = sum of squares of lengths of ladders
\Rightarrow r(y) = (16^2 + y^2) + [16^2 + (20 - y)^2]
\Rightarrow r(y) = 2y^2 - 40y + 912 [½]
On diff. wrt y, we get r'(y) = 4y - 40 [½]
Put r'(y) = 0 \Rightarrow y = 10 [½]
r''(y) = 4 \gt 0 \Rightarrow r(y) is minimum at y = 10. [½]
Hence, the required distance of the ladder from each pole = 10 m
Case Study – 1
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.
Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :
(i) Establish a relation between the height h of the juice in the cup and radius r of the surface of the juice in the cup, if the semi-vertical angle of the cone is \alpha. [1]
(ii) At what rate is the juice level in the cup rising when the juice is 6 cm deep ? [1]
(iii) (a) When the juice is 6 cm deep, then find at what rate is the upper surface area of juice increasing ? [2]
OR
(iii) (b) When the juice is 6 cm deep, then find the rate at which the wetted surface area of the cup is increasing. [2]
(i) \tan\alpha = \frac{r}{h} = \frac{5}{15} \Rightarrow r = \frac{h}{3} [1]
(ii) Volume, V = \frac{1}{27}\pi h^{3} \Rightarrow \frac{dV}{dt} = \frac{1}{9}\pi h^{2}\frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{1}{40\pi} cm/sec [1]
(iii) (a) Upper surface area of juice, A = \pi r^{2} = \frac{\pi h^{2}}{9} [1]
\frac{dA}{dt} = \frac{2\pi h}{9}\frac{dh}{dt} = \frac{1}{30} cm2/sec [1]
OR
(iii) (b) Wetted surface area, S = \pi r l = \frac{\sqrt{10}\pi h^{2}}{9} [1]
\frac{dS}{dt} = \frac{2\sqrt{10}\pi h}{9}\frac{dh}{dt} = \frac{\sqrt{10}}{30} cm2/sec [1]
Case Study – 3
A company produces cylindrical tumblers, open from the top. Since they want uniformity in the product, they fix the surface area of the tumblers produced.
Based on the above information, answer the following questions :
If for a tumbler, V is its volume, h the height and r the radius of the circular base, then :
(i) Differentiate its volume with respect to radius of the base, where the surface area is constant. [2]
(ii) If the company wants to maximize the volume of each tumbler, then establish a relation between its height and the radius of the base. [2]
(i) Surface area S = 2\pi rh + \pi r^2 [½]
V = \pi r^2 h = \frac{\pi r^2(S - \pi r^2)}{2\pi r} = \frac{(Sr - \pi r^3)}{2} [1]
\frac{dV}{dr} = \frac{(S - 3\pi r^2)}{2} [½]
(ii) \frac{dV}{dr} = 0 gives S = 3\pi r^2 [1]
\frac{d^2V}{dr^2} = -3\pi r \lt 0 [½]
V is maximum when S = 3\pi r^2
Thus, h = \frac{3\pi r^2 - \pi r^2}{2\pi r} = r [½]
Case Study – 2
A window is in the form of a rectangle of length x and breadth y (in metres), surmounted by an equilateral triangle on its length.
Based on the above information, answer the following questions :
(i) If the perimeter of the window is 18 m, write the relation between x and y. [1]
(ii) From (i), write an expression for the area of the window as a function of x only. [1]
(iii) (a) Find the dimensions of the rectangle that will allow the maximum light through the window. [2]
OR
(iii) (b) If it is given that the area of the window is 100 m2, then find the expression for its perimeter in terms of x. [2]
(i) Perimeter = 18 = 3x + 2y [1]
(ii) Area, A = xy + \frac{\sqrt{3}x^2}{4} = \left(9x - \frac{3x^2}{2}\right) + \frac{\sqrt{3}x^2}{4} [1]
(iii) (a) \frac{dA}{dx} = 9 - 3x + \frac{\sqrt{3}x}{2} = 0 \Rightarrow 9 = 3x - \frac{\sqrt{3}x}{2} [1]
So, x = \frac{18}{6 - \sqrt{3}} m [½]
Now, \frac{d^2A}{dx^2} = -3 + \frac{\sqrt{3}}{2} \lt 0
Thus, area is maximum when x = \frac{18}{6-\sqrt{3}} m, y = \frac{27 - 9\sqrt{3}}{6 - \sqrt{3}} m [½]
OR
(iii) (b) xy + \frac{\sqrt{3}x^2}{4} = 100 \Rightarrow y = \frac{100}{x} - \frac{\sqrt{3}x}{4} [1]
So, Perimeter = 3x + 2y = \left(3x - \frac{\sqrt{3}x}{2} + \frac{200}{x}\right) m [1]
A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.
Let the length of the side perpendicular to the partition be x metres and with parallel to the partition be y metres.
Based on this information, answer the following questions :
(i) Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of x and y. [1]
(ii) Write the area of the solar panel as a function of x. [1]
(iii) (a) Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area. [2]
OR
(iii) (b) Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition. [2]
(i) 2x+3y=300 [1]
(ii) A=xy=\frac{x}{3}(300-2x) [1]
(iii)(a) A=\frac{x}{3}(300-2x)=\frac{1}{3}(300x-2x^2)
\Rightarrow\frac{dA}{dx}=\frac{1}{3}(300-4x) [½]
For critical points, put \frac{dA}{dx}=0\Rightarrow x=75 [½]
Also, \frac{d^2A}{dx^2}=-\frac{4}{3}\lt 0. So, A is maximum at x=75 [½]
Also, maximum area is A=\frac{75}{3}(300-150)=3750\ \text{m}^2 [½]
OR
(iii)(b) A=\frac{x}{3}(300-2x)=\frac{1}{3}(300x-2x^2)
\Rightarrow\frac{dA}{dx}=\frac{1}{3}(300-4x) [½]
For critical points, put \frac{dA}{dx}=0\Rightarrow x=75 [½]
As \frac{dA}{dx} changes its sign from positive to negative as x passes through x=75 from left to right, which means x=75 is the point of maximum. [½]
Also, maximum area is A=\frac{75}{3}(300-150)=3750\ \text{m}^2 [½]
Note : Full credit to be given if the student takes equation as 2x+2y=300 or 2x+4y=300 or 4x+4y=300 or 4x+3y=300. The solutions of sub-parts will differ and marks may be given accordingly.
A small town is analyzing the pattern of a new street light installation. The lights are set up in such a way that the intensity of light at any point x metres from the start of the street can be modelled by f(x) = e^x \sin x, where x is in metres.
Based on the above, answer the following :
(i) Find the intervals on which the f(x) is increasing or decreasing, x \in [0, \pi]. [2]
(ii) Verify, whether each critical point when x \in [0, \pi] is a point of local maximum or local minimum or a point of inflexion. [2]
(i) f'(x) = e^x(\cos x + \sin x)
For critical points, f'(x) = 0
\Rightarrow \cos x + \sin x = 0
\Rightarrow \cos x = -\sin x [½]
For x to be a critical point x \in (0, \pi), hence, x = \frac{3\pi}{4} [½]
For all x \in \left[0, \frac{3\pi}{4}\right], f'(x) \ge 0
Hence, f is increasing in \left[0, \frac{3\pi}{4}\right] [½]
Note: If a student concludes the answer in any of the following intervals, full marks may be awarded:
\left(0, \frac{3\pi}{4}\right) or \left[0, \frac{3\pi}{4}\right) or \left(0, \frac{3\pi}{4}\right]
For all x \in \left[\frac{3\pi}{4}, \pi\right], f'(x) \le 0
Hence, f is decreasing in \left[\frac{3\pi}{4}, \pi\right] [½]
Note: If a student concludes the answer in any of the following intervals, full marks may be awarded:
\left(\frac{3\pi}{4}, \pi\right) or \left(\frac{3\pi}{4}, \pi\right] or \left[\frac{3\pi}{4}, \pi\right)
(ii) x = \frac{3\pi}{4} is a critical point
f''(x) = e^x(\cos x - \sin x) + e^x(\cos x + \sin x) [1]
= 2e^x\cos x
f''\left(\frac{3\pi}{4}\right) = -ve [½]
Hence, \frac{3\pi}{4} is a point of local maximum. [½]
Case Study – 1
A carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed volume. Since he is short of the paint required to paint the box on completion, he wants the surface area to be minimum.
On the basis of the above information, answer the following questions :
(i) Taking length = breadth = x m and height = y m, express the surface area (S) of the box in terms of x and its volume (V), which is constant. [1]
(ii) Find \frac{dS}{dx}. [1]
(iii) (a) Find a relation between x and y such that the surface area (S) is minimum. [2]
OR
(iii) (b) If surface area (S) is constant, the volume (V)=\frac14(Sx-2x^3), x being the edge of base. Show that volume (V) is maximum for x=\sqrt{\frac{S}{6}}. [2]
(i) V=x^2y\Rightarrow y=\frac{V}{x^2} ...............(i)
Hence, S=2x^2+4xy=2x^2+\frac{4V}{x} [1]
(ii) \frac{dS}{dx}=4\left(x-\frac{V}{x^2}\right) [1]
(iii) (a) \frac{dS}{dx}=0\Rightarrow V=x^3\Rightarrow x^2y=x^3\Rightarrow y=x [1]
\frac{d^2S}{dx^2}=4\left(1+\frac{2V}{x^3}\right)\gt0\Rightarrow S is minimum if y=x. [1]
OR
(iii) (b) V=\frac14(Sx-2x^3)\Rightarrow\frac{dV}{dx}=\frac14(S-6x^2) [1]
Put \frac{dV}{dx}=0\Rightarrow x=\sqrt{\frac{S}{6}} [½]
\left(\frac{d^2V}{dx^2}\right)_{x=\sqrt{\frac{S}{6}}}=-3\sqrt{\frac{S}{6}}\lt0\Rightarrow Volume is maximum for x=\sqrt{\frac{S}{6}}. [½]
Case Study – 2
A ladder of fixed length ‘h’ is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions :
(i) Express the distance (y) between the wall and foot of the ladder in terms of ‘h’ and height (x) on the wall at a certain instant. Also, write an expression in terms of h and x for the area (A) of the right triangle, as seen from the side by an observer. [1]
(ii) Find the derivative of the area (A) with respect to the height on the wall (x), and find its critical point. [1]
(iii) (a) Show that the area (A) of the right triangle is maximum at the critical point. [2]
OR
(iii) (b) If the foot of the ladder whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance (y) is 2 m/s, then at what rate is the height on the wall (x) increasing, when the foot of the ladder is 3 m away from the wall ? [2]
(i) y^2 = h^2 - x^2 [½]
A = \frac{1}{2}xy = \frac{1}{2}x\sqrt{h^2 - x^2} [½]
(ii) \frac{dA}{dx} = \frac{1}{2}\sqrt{h^2 - x^2} + \frac{1}{2}x\frac{-x}{\sqrt{h^2 - x^2}} [½]
\frac{dA}{dx} = 0 gives x = \frac{h}{\sqrt{2}} [½]
(iii) (a) A'' = \frac{1}{2}\,\frac{-4x\sqrt{h^2 - x^2} - (h^2 - 2x^2)\frac{-x}{\sqrt{h^2 - x^2}}}{h^2 - x^2} is \lt 0 at x = \frac{h}{\sqrt{2}} [1½]
Hence A is maximum at critical point [½]
OR
(iii) (b) y^2 = 25 - x^2 hence y = 3 gives x = 4 [½]
2y\frac{dy}{dt} = -2x\frac{dx}{dt} [1]
\frac{dx}{dt} = 1.5\ \text{m/s} [½]
Case Study – 1
A window is in the form of a rectangle surmounted by an equilateral triangle on its length. Let the rectangular part have length and breadth x and y metres respectively.
Based on the given information, answer the following questions :
(i) If the perimeter of the window is 12 m, find the relation between x and y. [1]
(ii) Using the expression obtained in (i), write an expression for the area of the window as a function of x only. [1]
(iii) (a) Find the dimensions of the rectangle that will allow maximum light through the window. (use expression obtained in (ii)) [2]
OR
(iii) (b) If it is given that the area of the window is 50 m2, find an expression for its perimeter in terms of x. [2]
(i) Perimeter (P) = 3x + 2y = 12 [1]
(ii) Area (A) = xy + \frac{\sqrt{3}}{4}x^2
= x\left(\frac{12 - 3x}{2}\right) + \frac{\sqrt{3}}{4}x^2
= 6x - \frac{3}{2}x^2 + \frac{\sqrt{3}}{4}x^2 [1]
(iii)(a) \frac{dA}{dx} = 6 - 3x + \frac{\sqrt{3}}{2}x [½]
For maximum light, \frac{dA}{dx} = 0
\Rightarrow 6 - 3x + \frac{\sqrt{3}}{2}x = 0 \Rightarrow x = \frac{12}{6 - \sqrt{3}}\,m [½]
Also, \frac{d^2A}{dx^2} = -3 + \frac{\sqrt{3}}{2} \lt 0 \therefore A is maximum when x = \frac{12}{6 - \sqrt{3}}\,m [½]
Now, y = \frac{12 - 3x}{2} = 6 - \frac{3}{2}\left(\frac{12}{6 - \sqrt{3}}\right) = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}\,m [½]
OR
(iii)(b) xy + \frac{\sqrt{3}}{4}x^2 = 50 [½]
\Rightarrow y = \frac{50}{x} - \frac{\sqrt{3}}{4}x [1]
Now, P = 3x + 2y
= 3x + 2\left(\frac{50}{x} - \frac{\sqrt{3}}{4}x\right)m [½]
Case Study – 1
The traffic police has installed Over Speed Violation Detection (OSVD) system at various locations in a city. These cameras can capture a speeding vehicle from a distance of 300 m and even function in the dark.
A camera is installed on a pole at the height of 5 m. It detects a car travelling away from the pole at the speed of 20 m/s. At any point, x m away from the base of the pole, the angle of elevation of the speed camera from the car C is \theta.
On the basis of the above information, answer the following questions :
(i) Express \theta in terms of height of the camera installed on the pole and x. [1]
(ii) Find \frac{d\theta}{dx}. [1]
(iii) (a) Find the rate of change of angle of elevation with respect to time at an instant when the car is 50 m away from the pole. [2]
OR
(iii) (b) If the rate of change of angle of elevation with respect to time of another car at a distance of 50 m from the base of the pole is \frac{3}{101} rad/s, then find the speed of the car. [2]
(i) \tan\theta = \frac{5}{x} \Rightarrow \theta = \tan^{-1}\left(\frac{5}{x}\right) [1]
(ii) \frac{d\theta}{dx} = \frac{-5}{5^2 + x^2} [1]
(iii) (a) \frac{d\theta}{dt} = \frac{d\theta}{dx} \times \frac{dx}{dt} = \left.\frac{-5}{5^2 + x^2} \times 20\right]_{x = 50} [1½]
= \frac{-100}{2525}\ \text{or}\ \frac{-4}{101}\ rad/s [½]
OR
(b) \frac{d\theta}{dt} = \frac{d\theta}{dx} \times \frac{dx}{dt} \Rightarrow \frac{3}{101} = \left.\frac{-5}{5^2 + x^2}\right]_{x = 50} \times \frac{dx}{dt} [1½]
\Rightarrow \frac{3}{101} = \frac{-5}{2525} \times \frac{dx}{dt} \Rightarrow \frac{dx}{dt} = -15\ m/s [½]
Hence the speed is 15 m/s
Overspeeding increases fuel consumption and decreases fuel economy as a result of tyre rolling friction and air resistance. While vehicles reach optimal fuel economy at different speeds, fuel mileage usually decreases rapidly at speeds above 80 km/h.
The relation between fuel consumption F (l/100 km) and speed V (km/h) under some constraints is given as F = \frac{V^2}{500} - \frac{V}{4} + 14.
On the basis of the above information, answer the following questions :
(i) Find F, when V = 40 km/h. [1]
(ii) Find \frac{dF}{dV}. [1]
(iii) (a) Find the speed V for which fuel consumption F is minimum. [2]
OR
(iii) (b) Find the quantity of fuel required to travel 600 km at the speed V at which \frac{dF}{dV} = -0{\cdot}01. [2]
(i) When V = 40 km/h, F = 36/5 ℓ/100km [1]
(ii) \frac{dF}{dV} = \frac{V}{250} - \frac{1}{4} [1]
(iii)(a) \frac{dF}{dV} = 0
\RightarrowV = 62.5 km/h [1]
\frac{d^2F}{dV^2} = \frac{1}{250} \gt 0 at V = 62.5 km/h [½]
Hence, F is minimum when V = 62.5 km/h [½]
OR
(iii) (b) \frac{dF}{dV} = -0.01
\Rightarrow \frac{V}{250} - \frac{1}{4} = \frac{-1}{100}
\Rightarrow V = 60 km/h [1]
F = \frac{60^2}{500} - \frac{60}{4} + 14 = 6.2\ \ell/100km [½]
Quantity of fuel required for 600 km
= 6.2 \times 6 = 37.2\ \ell [½]
A store has been selling calculators at ₹ 350 each. A market survey indicates that a reduction in price (p) of calculator increases the number of units (x) sold. The relation between the price and quantity sold is given by the demand function p = 450 - \frac{1}{2}x.
Based on the above information, answer the following questions :
(i) Determine the number of units (x) that should be sold to maximise the revenue R(x) = xp(x). Also, verify the result.
(ii) What rebate in price of calculator should the store give to maximise the revenue ?
(i) Revenue by selling x items = R(x) = x.p(x) = 450x - \frac{x^2}{2} [½]
\frac{dR}{dx} = 450 - x
For Maxima or Minima, \frac{dR}{dx} = 0 \Rightarrow x = 450 [1]
\frac{d^2R}{dx^2} = -1 \lt 0
(Revenue is maximum when x = 450 units are sold) [½]
(ii) At x = 450, p = 450 - \frac{450}{2} = 225 [1]
So, Rebate = 350 – 225 = Rs.125 per calculator [1]
A rectangular visiting card is to contain 24 sq.cm. of printed matter. The margins at the top and bottom of the card are to be 1 cm and the margins on the left and right are to be 1½ cm as shown below :
On the basis of the above information, answer the following questions :
(i) Write the expression for the area of the visiting card in terms of x.
(ii) Obtain the dimensions of the card of minimum area.
(i) Let A(x) be the area of the visiting card then,
As xy=24, A(x)=(x+3)(y+2)=2x+3y+xy+6=2x+\frac{72}{x}+30 [2]
(ii) A'(x)=2-\frac{72}{x^2} and A''(x)=\frac{144}{x^3}, [1]
solving A'(x)=0\Rightarrow x=6 is the critical point. [½]
A''(6)=\frac{144}{6}\gt0, \therefore Area of the card is minimum at x=6,y=4
The dimension of the card with minimum area is Length = 9 cm, Breadth = 6 cm [½]
Case Study – 1
A sandbag is dropped from a balloon at a height of 60 metres.
When the angle of elevation of the sun is 30^{\circ}, the position of the sandbag is given by the equation y=60-4.9t^{2}, where y is the height of the sandbag above the ground and t is the time in seconds.
On the basis of the above information, answer the following questions :
(i) Find the relation between x and y, where x is the distance of the shadow at P from the point Q and y is the height of the sandbag above the ground.
(ii) After how much time will the sandbag be 35 metres above the ground ?
(iii) (a) Find the rate at which the shadow of the sandbag is travelling along the ground when the sandbag is at a height of 35 metres.
OR
(iii) (b) How fast is the height of the sandbag decreasing when 2 seconds have elapsed ?
(i) \frac{y}{x}=\tan30^{\circ}\Rightarrow y=\frac{x}{\sqrt{3}} or x=\sqrt{3}y [1]
(ii) Using y=35\text{m}\Rightarrow60-4.9t^{2}=35\Rightarrow4.9t^{2}=25\Rightarrow t=\frac{5\sqrt{10}}{7} seconds. [1]
(iii) (a) x=\sqrt{3}y\Rightarrow x=60\sqrt{3}-4.9\sqrt{3}t^{2}\Rightarrow\left.\frac{dx}{dt}\right]_{t=\frac{5\sqrt{10}}{7}}=\left.-4.9\sqrt{3}(2t)\right]_{t=\frac{5\sqrt{10}}{7}}=-7\sqrt{30} m/s [2]
Or
(iii) (b) \frac{dy}{dt}=-9.8t=-9.8\times2=-19.6 m/s
Height of the sandbag is decreasing at the rate of 19.6 m/s [2]
Case Study – 2
A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drainage as compared to a flat bottomed tank.
A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of 2 cm3/s. The semi-vertical angle of the conical tank is 45°.
On the basis of given information, answer the following questions :
(i) Find the volume of water in the tank in terms of its radius r. [1]
(ii) Find rate of change of radius at an instant when r=2\sqrt2 cm. [1]
(iii) (a) Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius r=2\sqrt2 cm. [2]
OR
(iii) (b) Find the rate of change of height ‘h’ at an instant when slant height is 4 cm. [2]
(i) v=\frac13\pi r^2h=\frac13\pi r^3 [ as \theta=45^\circ gives r = h] [1]
(ii) \frac{dv}{dt}=\pi r^2\frac{dr}{dt} [½]
\Rightarrow\left(\frac{dr}{dt}\right)_{r=2\sqrt2}=-\frac{1}{4\pi} cm/sec [½]
(iii)(a) C=\pi rl=\pi r\sqrt2\,r=\sqrt2\,\pi r^2 [1]
\frac{dC}{dt}=\sqrt2\,\pi\,2r\,\frac{dr}{dt} [½]
\left(\frac{dC}{dt}\right)_{r=2\sqrt2}=-2 cm2/sec [½]
OR
(iii)(b) l^2=h^2+r^2
l=4\Rightarrow r=h=2\sqrt2 [1]
h=r\Rightarrow\frac{dh}{dt}=\frac{dr}{dt}=-\frac{1}{4\pi} cm/sec [1]
Engine displacement is the measure of the cylinder volume swept by all the pistons of a piston engine. The piston moves inside the cylinder bore
The cylinder bore in the form of circular cylinder open at the top is to be made from a metal sheet of area 75\pi cm2.
Based on the above information, answer the following questions :
(i) If the radius of cylinder is r cm and height is h cm, then write the volume V of cylinder in terms of radius r. [1]
(ii) Find \frac{dV}{dr}. [1]
(iii) (a) Find the radius of cylinder when its volume is maximum. [2]
OR
(iii) (b) For maximum volume, h > r. State true or false and justify. [2]
(i) \pi r^2 + 2\pi rh = 75\pi \Rightarrow h = \frac{75 - r^2}{2r},\ \therefore V = \pi r^2 h = \frac{\pi}{2}(75r - r^3) [1]
(ii) \frac{dV}{dr} = \frac{\pi}{2}(75 - 3r^2) [1]
(iii) \frac{dV}{dr} = 0 \Rightarrow r = 5, \frac{d^2V}{dr^2}\Big]_{r=5} = \frac{\pi}{2}(-6r) \lt 0 \therefore volume is maximum when r = 5 [2]
Or
False, [½]
\frac{dV}{dr} = 0 \Rightarrow r = 5, \frac{d^2V}{dr^2}\Big]_{r=5} = \frac{\pi}{2}(-6r) \lt 0 \therefore volume is maximum when r = 5 [1]
As volume is maximum at r = 5 \Rightarrow h = \frac{75 - 5^2}{2(5)} = 5 \Rightarrow h = r [½]
The use of electric vehicles will curb air pollution in the long run.
The use of electric vehicles is increasing every year and estimated electric vehicles in use at any time t is given by the function V :
where t represents the time and t = 1, 2, 3.... corresponds to year 2001, 2002, 2003, ....... respectively.
Based on the above information, answer the following questions :
(i) Can the above function be used to estimate number of vehicles in the year 2000 ? Justify. [2]
(ii) Prove that the function V(t) is an increasing function. [2]
(i) For the year 2000, t = 0 & V(0) = -2 and the number of vehicles cannot be negative \therefore the given function V(t) cannot be used. [2]
(ii) V^{\prime}(t) = \frac{3}{5}t^2 - 5t + 25 = \frac{3}{5}\left[\left(t - \frac{25}{6}\right)^2 + \frac{875}{36}\right] \gt 0,\ \therefore V(t) is an increasing function. [2]
Case Study – 2
In order to set up a rain water harvesting system, a tank to collect rain water is to be dug. The tank should have a square base and a capacity of 250 m3. The cost of land is ₹ 5,000 per square metre and cost of digging increases with depth and for the whole tank, it is ₹ 40,000 h2, where h is the depth of the tank in metres. x is the side of the square base of the tank in metres.
ELEMENTS OF A TYPICAL RAIN WATER HARVESTING SYSTEM
Based on the above information, answer the following questions :
(i) Find the total cost C of digging the tank in terms of x. [1]
(ii) Find \frac{dC}{dx}. [1]
(iii) (a) Find the value of x for which cost C is minimum. [2]
OR
(iii) (b) Check whether the cost function C(x) expressed in terms of x is increasing or not, where x \gt 0. [2]
(i) C = 40000h^2 + 5000x^2
as x^2h = 250 [½]
\Rightarrow C = \frac{40000\,(250)^2}{x^4} + 5000x^2 [½]
(ii) \frac{dC}{dx} = \frac{-160000\,(250)^2}{x^5} + 10000x [1]
(iii)(a) For minimum cost \frac{dC}{dx} = 0 [½]
\Rightarrow 10000x^6 = 250 \times 250 \times 160000
\Rightarrow x = 10 [1]
showing \frac{d^2C}{dx^2} \gt 0 at x = 10 [½]
\therefore cost is minimum when x = 10
OR
(iii)(b) \frac{dC}{dx} = \frac{-160000\,(250)^2}{x^4} + 10000x [½]
\frac{dC}{dx} = 0 gives x = 10 [1]
\frac{dC}{dx} \gt 0 in (10, \infty) and \frac{dC}{dx} \lt 0 in (0, 10).
Hence, cost function is neither increasing nor decreasing for x \gt 0 [½]
Case Study – 3
A volleyball player serves the ball which takes a parabolic path given by the equation h(t) = -\frac{7}{2}t^2 + \frac{13}{2}t + 1, where h(t) is the height of ball at any time t (in seconds), (t \ge 0).
Based on the above information, answer the following questions :
(i) Is h(t) a continuous function ? Justify. [2]
(ii) Find the time at which the height of the ball is maximum. [2]
(i) h(t) = -\frac{7}{2}t^2 + \frac{13}{2}t + 1
Clearly h(t) is a polynomial function, hence continuous.
Hence h(t) is a continuous function. [2]
(ii) For maximum height,
\frac{dh}{dt} = 0 \Rightarrow -7t + \frac{13}{2} = 0 [1]
\Rightarrow t = \frac{13}{14} [½]
\frac{d^2h}{dt^2} = -7 \lt 0 \therefore height is maximum at t = \frac{13}{14} [½]
Sooraj’s father wants to construct a rectangular garden using a brick wall on one side of the garden and wire fencing for the other three sides as shown in the figure. He has 200 metres of fencing wire.
Based on the above information, answer the following questions :
(i) Let ‘x’ metres denote the length of the side of the garden perpendicular to the brick wall and ‘y’ metres denote the length of the side parallel to the brick wall. Determine the relation representing the total length of fencing wire and also write A(x), the area of the garden. [2]
(ii) Determine the maximum value of A(x). [2]
(i) (a) 2x+y=200 [1]
(b) A(x)=xy=x(200-x) [1]
(ii) From (a) and (b) of (i) we have
A(x)=x(200-2x)
=200x-2x^2
From max./min of A(x)
\frac{dA}{dx}=0 i.e., 200-4x=0
\Rightarrow x=50.
\frac{d^2A}{dx^2}=-4
\left(\frac{d^2A}{dx^2}\right)_{x=50}\lt 0
Hence, A(x) is maximum at x=50 [1]
Thus, Max A(x)=200(50)-2(50)^2
=10000-5000
=5000 sqm. [1]
Case Study – 2
A housing society wants to commission a swimming pool for its residents. For this, they have to purchase a square piece of land and dig this to such a depth that its capacity is 250 cubic metres. Cost of land is ₹ 500 per square metre. The cost of digging increases with the depth and cost for the whole pool is ₹ 4000 (\text{depth})^2.
Suppose the side of the square plot is x metres and depth is h metres.
On the basis of the above information, answer the following questions :
(i) Write cost C(h) as a function in terms of h. [1]
(ii) Find critical point. [1]
(iii) (a) Use second derivative test to find the value of h for which cost of constructing the pool is minimum. What is the minimum cost of construction of the pool ? [2]
OR
(iii) (b) Use first derivative test to find the depth of the pool so that cost of construction is minimum. Also, find relation between x and h for minimum cost. [2]
Official marking scheme for this paper is not available.
Case Study – 3
In an agricultural institute, scientists do experiments with varieties of seeds to grow them in different environments to produce healthy plants and get more yield.
A scientist observed that a particular seed grew very fast after germination. He had recorded growth of plant since germination and he said that its growth can be defined by the function
where x is the number of days the plant is exposed to sunlight.
On the basis of the above information, answer the following questions :
(i) What are the critical points of the function f(x) ? [2]
(ii) Using second derivative test, find the minimum value of the function. [2]
Official marking scheme for this paper is not available.
Find the sub intervals in which f(x) = \cot^{-1}(\sin x + \cos x),\ x \in (0, \pi) is increasing and decreasing.
f(x) = \cot^{-1}(\sin x + \cos x),\ x \in (0, \pi)
f'(x) = \frac{-(\cos x - \sin x)}{1 + (\sin x + \cos x)^2} = \frac{\sin x - \cos x}{1 + (\sin x + \cos x)^2} [2]
For critical points of f(x), put f'(x) = 0
\Rightarrow \sin x = \cos x \Rightarrow x = \frac{\pi}{4} [1]
f'(x) \gt 0 when x \in \left(\frac{\pi}{4}, \pi\right) \Rightarrow f(x) is increasing in \left(\frac{\pi}{4}, \pi\right). [1]
Note: Alternatively f(x) is also increasing in the interval \left[\frac{\pi}{4}, \pi\right).
f'(x) \lt 0 when x \in \left(0, \frac{\pi}{4}\right) \Rightarrow f(x) is decreasing in \left(0, \frac{\pi}{4}\right). [1]
Note: Alternatively f(x) is also decreasing in the interval \left(0, \frac{\pi}{4}\right].
A rectangle of perimeter 36 cm is revolved around one of its sides to sweep out a cylinder of maximum volume.
Find the dimensions of the rectangle.
Let the lengths of the sides of the rectangle be x and (18 – x).
Let it be revolved around the side of length (18 – x), so that x becomes the radius of the cylinder. [1]
Volume of cylinder, V = \pi x^2(18 - x) = \pi(18x^2 - x^3) [1]
\frac{dV}{dx} = \pi(36x - 3x^2) [½]
For critical points, put \frac{dV}{dx} = 0
\Rightarrow x = 12\text{ cm}\ (\because x \ne 0) [1]
Now \frac{d^2V}{dx^2} = \pi(36 - 6x)\ ;\ \left.\frac{d^2V}{dx^2}\right]_{x = 12\text{ cm}} = -36\pi \lt 0 [1]
So, volume is maximum when x = 12 cm.
Hence the dimensions of rectangle are 12 cm and 6 cm. [½]
Find the absolute maximum and absolute minimum of function f(x)=2x^3-15x^2+36x+1 on [1,5].
f(x)=2x^3-15x^2+36x+1
\Rightarrow f'(x)=6(x^2-5x+6)=6(x-2)(x-3) [1]
f'(x)=0\Rightarrow x=2,3\in[1,5] [1]
Now f(1)=24,\ f(2)=29,\ f(3)=28,\ f(5)=56 [2]
Hence, the absolute maximum value is 56 and the absolute minimum value is 24. [1]
The relation between the height of the plant (y cm) with respect to exposure to sunlight is governed by the equation y=4x-\frac{1}{2}x^2, where x is the number of days exposed to sunlight.
(i) Find the rate of growth of the plant with respect to sunlight. [2]
(ii) In how many days will the plant attain its maximum height ? What is the maximum height ? [3]
(i) y=4x-\frac{1}{2}x^2\Rightarrow\frac{dy}{dx}=(4-x) cm/day [2]
(ii) For maximum height, \frac{dy}{dx}=0\Rightarrow x=4 days [1]
as \frac{d^2y}{dx^2}\lt 0, number of days =4 [1]
Now, Maximum height =y(4)=16-\frac{1}{2}(16)=8 cm [1]
Find the intervals in which the function given by
f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11 is :
(i) strictly increasing.
(ii) strictly decreasing.
f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11
\Rightarrow f'(x) = \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5} = \frac{6}{5}\left(x^3 - 2x^2 - 5x + 6\right) [2]
= \frac{6}{5}(x-1)(x+2)(x-3) [1]
For strictly inc/dec, put f'(x) = 0
\Rightarrow x = 1, -2, 3 [1]
(i) f is strictly increasing when x \in (-2, 1) \cup (3, \infty)
(ii) f is strictly decreasing when x \in (-\infty, -2) \cup (1, 3) [1]
Note: Closed intervals are also acceptable.
It is given that function f(x) = x^4 - 62x^2 + ax + 9 attains local maximum value at x = 1. Find the value of ‘a’, hence obtain all other points where the given function f(x) attains local maximum or local minimum values.
f(x) = x^4 - 62x^2 + ax + 9 \Rightarrow f'(x) = 4x^3 - 124x + a [½]
as at x = 1, f attains local maximum value, f'(1) = 0 \Rightarrow a = 120 [1]
now, f'(x) = 4x^3 - 124x + 120 = 4(x - 1)(x^2 + x - 30) = 4(x - 1)(x - 5)(x + 6) [1]
Critical points are x = -6, 1, 5 [1]
f''(x) = 12x^2 - 124
f''(-6) \gt 0,\ f''(1) \lt 0,\ f''(5) \gt 0 [½]
so f attains local maximum value at x = 1 and local minimum value at x = -6, 5 [1]
The perimeter of a rectangular metallic sheet is 300 cm. It is rolled along one of its sides to form a cylinder. Find the dimensions of the rectangular sheet so that volume of cylinder so formed is maximum.
Let length of rectangle be x cm and breadth be (150 - x) cm.
Let r be the radius of cylinder \Rightarrow 2\pi r = x \Rightarrow r = \frac{x}{2\pi} [1]
V = \pi r^2 h = \pi\left(\frac{x^2}{4\pi^2}\right)(150 - x) = \frac{75x^2}{2\pi} - \frac{x^3}{4\pi} [1]
\frac{dV}{dx} = \frac{150x}{2\pi} - \frac{3x^2}{4\pi} [1]
\frac{dV}{dx} = 0 \Rightarrow x = 100 cm [1]
\left.\frac{d^2V}{dx^2}\right|_{x = 100\text{ cm}} = -\frac{75}{\pi} \lt 0 \Rightarrow V is maximum when x = 100 cm. [½]
Length of rectangle is 100 cm and breadth of rectangle is 50 cm. [½]
The median of an equilateral triangle is increasing at the rate of 2\sqrt3 cm/s. Find the rate at which its side is increasing.
In an equilateral triangle, median is same as altitude. Let ‘h’ denote the length of the median (or altitude) and ‘x’ be the side of \Delta ABC. [1]
[Figure: equilateral triangle ABC with altitude from A]
Then, h=\frac{\sqrt3}{2}x or x=\frac{2h}{\sqrt3} ________ (i) [2]
It is given that \frac{dh}{dt}=2\sqrt3 So, by (i) we have [1]
\frac{dx}{dt}=\frac{2}{\sqrt3}\frac{dh}{dt}\Rightarrow\frac{dx}{dt}=4 [1]
Thus, the side of \Delta ABC is increasing at the rate of 4 cm/sec.
Sum of two numbers is 5. If the sum of the cubes of these numbers is least, then find the sum of the squares of these numbers.
Let the two numbers be x and y. Then, x + y = 5 or y = 5 – x [½]
Let S denote the sum of the cubes of these numbers. Then
S=x^3+y^3=x^3+(5-x)^3 [1]
\frac{dS}{dx}=3x^2-3(5-x)^2=15(2x-5) [1]
Now \frac{dS}{dx}=0, gives x=\frac{5}{2} [½]
Showing S is minimum at x=\frac{5}{2} [1]
So, the two numbers are \frac{5}{2} and \frac{5}{2}
\Rightarrow x^2+y^2=\frac{25}{4}+\frac{25}{4}=\frac{25}{2} [1]
If \int\frac{1}{1+e^{2x}}dx=x-P\log(1+e^{2x})+C, then ‘P’ is equal to
(D) \frac12
\int\frac{1}{1+e^{2x}}dx=\int\left(\frac{1+e^{2x}}{1+e^{2x}}-\frac{e^{2x}}{1+e^{2x}}\right)dx=x-\frac12\log|1+e^{2x}|+C So P=\frac12
Assertion (A): The value of the integral \int_{-1}^{1}|x-2|\,dx is 4
Reason (R): If f(-x)=f(x) then \int_{-a}^{a}f(x)dx=2\int_0^af(x)dx
(B) Both assertion and reason are true, but reason is not the correct explanation of the assertion
\int_{-1}^{1}|x-2|\,dx=\int_{-1}^{1}(-x+2)\,dx=4
If f(a+b-x)=f(x), then \int_a^b x\,f(x)dx is equal to
I=\int_a^b x\,f(x)dx=\int_a^b(a+b-x)f(a+b-x)dx
\Rightarrow I=\int_a^b(a+b-x)f(x)dx (given f(a+b-x)=f(x))
\Rightarrow I=\int_a^b(a+b)f(x)dx-\int_a^b x\,f(x)dx
\Rightarrow 2I=(a+b)\int_a^b f(x)dx
\Rightarrow I=\frac12(a+b)\int_a^b f(x)dx
(D) \frac{a+b}{2}\int_a^b f(x)dx [1]
If \int x^3\sin^4(x^4)\cos(x^4)\,dx=a\sin^5(x^4)+C, then a is equal to
Let I=\int x^3\sin^4(x^4)\cos(x^4)\,dx
Let \sin(x^4)=t\Rightarrow 4x^3\cos(x^4)\,dx=dt\Rightarrow x^3\cos(x^4)=\frac14dt
Thus I=\int t^4\left(\frac14dt\right)=\frac{1}{20}t^5+C=\frac{1}{20}\sin^5(x^4)+C
\Rightarrow I=\frac{1}{20}\sin^5(x^4)+C=a\sin^5(x^4)+C
So, a=\frac{1}{20}
(B) \frac{1}{20} [1]
If \int \frac{3ax}{b^2 + c^2x^2}\,dx = A \log\left|b^2 + c^2x^2\right| + K, then the value of A is
(D) \frac{3a}{2c^2}
The value of \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1}\,dx is
(A) 0
If \int_{0}^{2a} \frac{1}{1 + 4x^2}\,dx = \frac{\pi}{6}, then the value of a is
(A) \frac{\sqrt{3}}{4}
\int \frac{dx}{2^x + 2^{-x}} is equal to :
(C) \frac{\tan^{-1}(2^x)}{\log 2} + C
\int_{-1}^{1} (1 - |x|)\, dx is equal to :
(B) 2\int_{-1}^{0}(1 + x)\, dx
\int \frac{dx}{\sqrt{25 - 16x^{2}}} is equal to :
(C) \frac{1}{4}\sin^{-1}\left(\frac{4x}{5}\right) + c [1]
If \int_{0}^{1} \frac{dx}{e^{x} + e^{-x}} = \tan^{-1} e + k, then the value of k is :
(D) \frac{-\pi}{4} [1]
\int \frac{1}{\sqrt{1 + \cos 2x}}\,dx is equal to :
(B) \frac{1}{\sqrt{2}} \log |\sec x + \tan x| + C
The value of \int_{-5}^{-1} \frac{1}{x}\,dx is equal to :
(A) -\log 5
If \int_{0}^{1} (9x^2 - 4x - 2k)\,dx = 0, then the value of ‘k’ is :
(C) \frac{1}{2}
\int \left[\frac{1}{\log x} - \frac{1}{(\log x)^2}\right] dx is equal to :
(A) \frac{x}{\log x} + C
If \int_{0}^{4} \frac{1}{x + \sqrt{x}}\,dx = k\log 3, then the value of k is :
(D) 2
\int_{-1}^{1}\frac{|x|}{x}\,dx,\ x\ne 0 is equal to
(B) 0
If \int\frac{2^{\frac{1}{x}}}{x^2}\,dx=k\cdot 2^{\frac{1}{x}}+C, then k is equal to
(A) \frac{-1}{\log 2}
If f(2a - x) = f(x), then \int_{0}^{2a} f(x)\,dx is
(D) 2\int_{0}^{a} f(x)\,dx
\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx is equal to :
(A) 2(\sin x+x\cos\alpha)+C
The value of \int_{0}^{1}\frac{dx}{e^x+e^{-x}} is :
(C) \tan^{-1}e-\frac{\pi}{4}
\int\sqrt{1+\sin x}\,dx is equal to :
(B) 2\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)+C
\int_{0}^{\pi/2}\cos x\cdot e^{\sin x}\,dx is equal to :
(C) e-1
Let f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5,\ f(1) = 0. Then, f(x) is :
(B) x^3 + 3x^2 + \frac{2}{x^2} + 5x - 11
\int \frac{x+5}{(x+6)^2}\, e^x\, dx is equal to :
(C) \frac{e^x}{x+6} + C
\displaystyle\int\frac{e^{9\log x}-e^{8\log x}}{e^{6\log x}-e^{5\log x}}\,dx is equal to :
(C) \frac{x^4}{4}+C [1]
For a function f(x), which of the following holds true ?
(A) \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx [1]
\displaystyle\int\frac{e^x}{\sqrt{4-e^{2x}}}\,dx is equal to :
(D) \sin^{-1}\left(\frac{e^x}{2}\right)+C [1]
\int \frac{3\cos\sqrt{x}}{\sqrt{x}}\,dx is equal to :
(D) 6\sin\sqrt{x} + C
If \frac{d}{dx}f(x) = 3x^2 - \frac{3}{x^4} such that f(1) = 0, then f(x) is :
(C) x^3 + \frac{1}{x^3} - 2
\int_{a}^{b} f(x)\, dx is equal to :
(B) \int_{a}^{b} f(a + b - x)\, dx
\displaystyle\int_{0}^{\pi/2} \frac{\sin x - \cos x}{1+\sin x \cos x}\,dx is equal to :
(B) Zero (0)
\int_{-a}^{a} f(x)\,dx = 0, if :
(B) f(-x) = -f(x)
The value of \displaystyle\int_{0}^{3} \frac{dx}{\sqrt{9-x^2}} is :
(C) \frac{\pi}{2}
[1]\int\frac{1}{x(\log x)^2}\,dx is equal to :
(B) -\frac{1}{\log x}+c [1]
The value of \int_{-1}^{1}x|x|\,dx is :
(D) 0 [1]
\int_{-\pi/4}^{\pi/4} x^{3}\cos^{2}x\,dx is equal to :
(A) 0
[1]\int\frac{x-3}{(x-1)^{3}}e^{x}\,dx is equal to :
(D) \frac{e^{x}}{(x-1)^{2}}+C
[1]If \frac{d}{dx}(f(x))=\log x, then f(x) equals :
(B) x(\log x-1)+C
\int_{0}^{\frac{\pi}{6}}\sec^2\left(x-\frac{\pi}{6}\right)dx is equal to :
(A) \frac{1}{\sqrt3}
If \frac{d}{dx}[f(x)] = ax + b and f(0) = 0, then f(x) is equal to
(B) \frac{ax^2}{2} + bx
Anti-derivative of \frac{\tan x - 1}{\tan x + 1} with respect to x is
(C) \log\left|\sec\left(\frac{\pi}{4} - x\right)\right| + c
\int 2^{x+2}\, dx is equal to :
(C) \frac{2^{x+2}}{\log 2} + C
\int \frac{2\cos 2x - 1}{1 + 2\sin x}\, dx is equal to :
(B) x + 2\cos x + C
\int e^{5\log x}\,dx is equal to :
(b) \frac{x^6}{6}+C
If \int_{0}^{a}3x^2\,dx=8, then the value of ‘a’ is :
(a) a=2
\int\frac{\sec x}{\sec x-\tan x}\,dx equals
(B) \sec x+\tan x+c
\int_{-1}^{1}\frac{|x-2|}{x-2}\,dx,\ x\ne2 is equal to
(D) -2
Assertion (A) : \int_{2}^{8}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx=3
Reason (R) : \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx
(A) Both (A) and (R) are true and (R) is the correct explanation of (A)
The value of \displaystyle\int_{0}^{\pi/6} \sin 3x \, dx is :
Official marking scheme for this paper is not available.
\displaystyle\int \frac{\cos 2x}{\sin^2 x \cdot \cos^2 x}\,dx is equal to
Official marking scheme for this paper is not available.
Evaluate \int\frac{1}{x\sqrt{x^6-1}}dx
I=\int\frac{1}{x\sqrt{x^6-1}}dx=\frac16\int\frac{6x^5}{x^6\sqrt{x^6-1}}dx
Let x^6-1=t^2\quad\Rightarrow6x^5dx=2t\,dt [½]
I=\frac13\int\frac{1}{(t^2+1)}dt=\frac13\tan^{-1}\left(\sqrt{x^6-1}\right)+C [1+½]
Evaluate \int\frac{\cos x-\sin x}{\sqrt{\sin(2x)}}\,dx
Let I=\int\frac{\cos x-\sin x}{\sqrt{\sin(2x)}}dx=\int\frac{\cos x-\sin x}{\sqrt{1+\sin(2x)-1}}dx=\int\frac{\cos x-\sin x}{\sqrt{(\sin x+\cos x)^2-1}}dx [½]
Let \sin x+\cos x=t\quad\Rightarrow(\cos x-\sin x)dx=dt [½]
I=\int\frac{1}{\sqrt{t^2-1}}dt=\log\left|t+\sqrt{t^2-1}\right|+C [½]
=\log\left|\sin x+\cos x+\sqrt{(\sin x+\cos x)^2-1}\right|+C
=\log\left|\sin x+\cos x+\sqrt{\sin(2x)}\right|+C [½]
Find: \int\frac{(x-3)}{(x-1)^3}e^x\,dx
\int\frac{(x-3)e^x}{(x-1)^3}dx=\int\frac{(x-1-2)e^x}{(x-1)^3}dx
=\int\left(\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\right)e^x\,dx=\int\left(\frac{1}{(x-1)^2}+\frac{d}{dx}\left(\frac{1}{(x-1)^2}\right)\right)e^x\,dx [1]
=\frac{e^x}{(x-1)^2}+c (as \int(f(x)+f'(x))e^x\,dx=e^x f(x)+c) [1]
Evaluate : \int_{0}^{\frac{\pi}{4}} \sqrt{1 + \sin 2x}\,dx
Given definite integral = \int_{0}^{\frac{\pi}{4}} \sqrt{(\sin x + \cos x)^2}\,dx [1]
= \int_{0}^{\frac{\pi}{4}} (\sin x + \cos x)\,dx
= [-\cos x + \sin x]_{0}^{\frac{\pi}{4}}
= 1 [1]
Find :
1 + 2x = t^2
2\,dx = 2t\,dt [½]
\frac{1}{2}\int (t^4 - t^2)\, dt = \frac{1}{2}\left[\frac{t^5}{5} - \frac{t^3}{3}\right] + C [1]
= \frac{(1+2x)^{\frac{5}{2}}}{10} - \frac{(1+2x)^{\frac{3}{2}}}{6} + C [½]
Evaluate :
\int_{0}^{\frac{\pi^2}{4}} \frac{\sin\sqrt{x}}{\sqrt{x}}\, dx Put \sqrt{x} = t \Rightarrow dx = 2t\,dt [½]
2\int_{0}^{\frac{\pi}{2}} \sin t\, dt = 2\left[-\cos t\right]_{0}^{\frac{\pi}{2}} [1]
= 2 [½]
Find :
Given integral = \int \frac{e^{4x}-1}{e^{4x}+1}\,dx
= \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx [½]
= \frac{1}{2}\int \frac{2(e^{2x}-e^{-2x})}{e^{2x}+e^{-2x}}\,dx [½]
= \frac{1}{2}\log|e^{2x}+e^{-2x}| + C [1]
Evaluate :
I = \int_{0}^{\frac{\pi}{2}} \sin 2x \cos 3x\, dx
= \frac{1}{2}\int_{0}^{\frac{\pi}{2}} (\sin 5x - \sin x)\, dx [1]
= \frac{1}{2}\left[-\frac{1}{5}\cos 5x + \cos x\right]_{0}^{\frac{\pi}{2}} [½]
= -\frac{2}{5} [½]
Given \frac{d}{dx}F(x) = \frac{1}{\sqrt{2x - x^2}} and F(1) = 0, find F(x).
F(x) = \int \frac{1}{\sqrt{2x - x^2}}\, dx
= \int \frac{1}{\sqrt{1 - (x-1)^2}}\, dx [½]
= \sin^{-1}(x - 1) + c [½]
when x = 1, y = 0 gives c = 0 [½]
\therefore F(x) = \sin^{-1}(x - 1) [½]
Find : \displaystyle\int \frac{1}{x(x^2-1)}\,dx.
I = \int \frac{dx}{x(x^2-1)} = \int \frac{dx}{x^3\left(1-\frac{1}{x^2}\right)} = \frac{1}{2}\int \frac{\left(\frac{2}{x^3}\right)dx}{\left(1-\frac{1}{x^2}\right)} [1]
Put \left(1-\frac{1}{x^2}\right) = t \Rightarrow \left(\frac{2}{x^3}\right)dx = dt
I = \frac{1}{2}\int \frac{dt}{t} = \frac{1}{2}\log\left|\left(1-\frac{1}{x^2}\right)\right| + c \;\text{ OR }\; \frac{1}{2}\log\left|\left(\frac{x^2-1}{x^2}\right)\right| + c [1]
Find : \int\cos^3 x\, e^{\log\sin x}\,dx
\int\cos^3x\,e^{\log\sin x}dx=\int\cos^3x\cdot\sin x\,dx, Assuming \cos x=t and \sin x\,dx=-dt [1]
=-\int t^3dt [½]
=-\frac{t^4}{4}+C=-\frac{\cos^4x}{4}+C [½]
Find : \int\frac{1}{5+4x-x^2}\,dx
\int\frac{1}{5+4x-x^2}dx=\int\frac{1}{3^2-(x-2)^2}dx [1]
=\frac16\log\left|\frac{1+x}{5-x}\right|+C [1]
Find :
\int\frac{x^{3}-1}{x^{3}-x}dx=\int\left(1+\frac{1}{x}-\frac{1}{x+1}\right)dx [1]
=x+\log|x|-\log|x+1|+c [1]
Evaluate :
\int_{-4}^{0}|x+2|dx=\int_{-4}^{-2}-(x+2)dx+\int_{-2}^{0}(x+2)dx [1]
=-\left[\frac{(x+2)^{2}}{2}\right]_{-4}^{-2}+\left[\frac{(x+2)^{2}}{2}\right]_{-2}^{0}=2+2=4 [1]
If \frac{d}{dx}[F(x)]=\frac{\sec^4 x}{\text{cosec}^4 x} and F\left(\frac{\pi}{4}\right)=\frac{\pi}{4} , then find F(x).
F(x)=\int \tan^4 x\,dx=\int \tan^2 x.(\sec^2 x-1)\,dx [½]
=\int(\tan^2 x\sec^2 x-\sec^2 x+1)\,dx
F(x)=\frac{\tan^3 x}{3}-\tan x+x+C [1]
x=\frac{\pi}{4},\ F(x)=\frac{\pi}{4} gives C=\frac{2}{3}
F(x)=\frac{\tan^3 x}{3}-\tan x+x+\frac{2}{3} [½]
Find : \int \frac{\log x-3}{(\log x)^4}\,dx .
Put x=e^t or \log x=t dx=e^t dt [½]
\therefore \int\frac{\log x-3}{(\log x)^4}dx=\int\frac{t-3}{t^4}e^t\,dt
=\int\left(\frac{1}{t^3}-\frac{3}{t^4}\right)e^t\,dt [½]
=\frac{e^t}{t^3}+C [½]
=\frac{x}{(\log x)^3}+C [½]
Find :
\int\frac{\sin 3x}{\sin x}dx=\int\frac{3\sin x-4\sin^3 x}{\sin x}dx [½]
=\int\left[3-4\frac{(1-\cos 2x)}{2}\right]dx [½]
=\int(1+2\cos 2x)\,dx
=x+\sin 2x+C [1]
Evaluate :
Let e^x=t,\ e^x dx=dt [½]
\therefore \int_0^{\frac12\log 3}\frac{e^x}{e^{2x}+1}dx=\int_1^{\sqrt3}\frac{dt}{t^2+1} [½]
=\tan^{-1}t\Big|_1^{\sqrt3} [½]
=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12} [½]
Find : \int \frac{dx}{x^2-6x+13}
\int \frac{dx}{x^2-6x+13}
= \int \frac{dx}{(x-3)^2+(2)^2} [1]
= \frac12 \tan^{-1}\frac{(x-3)}{2}+C [1]
Evaluate :
Find : \int \frac{dx}{\sqrt{4x-x^2}}
\int \frac{dx}{\sqrt{4x-x^2}}
= \int \frac{dx}{\sqrt{2^2-(x-2)^2}} [1]
= \sin^{-1}\left(\frac{x-2}{2}\right)+C [1]
Evaluate :
I = \int_{0}^{\pi/2} \frac{1}{1+\cot^{5/2}x}dx = \int_{0}^{\pi/2} \frac{\sin^{5/2}x}{\sin^{5/2}x+\cos^{5/2}x}dx ...(i)
\Rightarrow I = \int_{0}^{\pi/2} \frac{\cos^{5/2}x}{\sin^{5/2}x+\cos^{5/2}x}dx ...(ii) [1]
adding (i) and (ii)
2I = \int_{0}^{\pi/2} 1\cdot dx \Rightarrow 2I = x\big]_{0}^{\pi/2} = \frac{\pi}{2} [½]
\Rightarrow I = \frac{\pi}{4} [½]
Evaluate : \int_{0}^{1} x\tan^{-1}x\,dx.
\int_0^1 x\tan^{-1}x\,dx = \left[\tan^{-1}x \times \frac{x^2}{2}\right]_0^1 - \int_0^1 \frac{1}{1+x^2} \times \frac{x^2}{2}dx [1]
= \frac{\pi}{8} - \int_0^1 \frac{1 + x^2 - 1}{1 + x^2} \times \frac{1}{2}dx
= \frac{\pi}{8} - \frac{1}{2}\left[\int_0^1 dx - \int_0^1 \frac{1}{1+x^2}dx\right]
= \frac{\pi}{8} - \frac{1}{2}\left[x - \tan^{-1}x\right]_0^1 [1]
= \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{\pi}{4} - \frac{1}{2} [1]
Find \int \sqrt{\frac{x+2}{x-2}}\,dx
I = \int \sqrt{\frac{x+2}{x-2}}dx = \int \sqrt{\frac{x+2}{x-2} \times \frac{x+2}{x+2}}dx = \int \frac{x+2}{\sqrt{x^2-4}}dx [1]
= \frac{1}{2}\int \frac{2x}{\sqrt{x^2-4}}dx + 2\int \frac{1}{\sqrt{x^2-4}}dx [1]
I_1 = \frac{1}{2}\int \frac{2x}{\sqrt{x^2-4}}dx = \frac{1}{2}\int \frac{dt}{\sqrt{t}}\quad [(x^2 - 4) = t \Rightarrow 2x\,dx = dt]
= \sqrt{t} = \sqrt{x^2 - 4} [1/2]
I_2 = 2\int \frac{1}{\sqrt{x^2-4}}dx = 2\log\left|x + \sqrt{x^2-4}\right| [1/2]
I = I_1 + I_2 = \sqrt{x^2-4} + 2\log\left|x + \sqrt{x^2-4}\right| + C
Find : \int \frac{x^2}{(x^2 + 9)(x^2 + 16)}\,dx
Put x^2 = t to get \frac{x^2}{(x^2+9)(x^2+16)} = \frac{t}{(t+9)(t+16)} = \frac{A}{t+9} + \frac{B}{t+16} [½]
Getting A = \frac{-9}{7}, B = \frac{16}{7} [1]
Given integral = \frac{-9}{7}\int \frac{1}{x^2+9}dx + \frac{16}{7}\int \frac{1}{x^2+16}dx [½]
= \frac{-3}{7}\tan^{-1}\left(\frac{x}{3}\right) + \frac{4}{7}\tan^{-1}\left(\frac{x}{4}\right) + C [1]
If I_1 = \int_{-\pi/4}^{\pi/4} \frac{dx}{1 + \cos 2x} and I_2 = \int_{-1/2}^{1/2} |x|\,dx, then show that I_1 - 4I_2 = 0.
I_1 = 2\int_0^{\frac{\pi}{4}} \frac{dx}{1 + \cos 2x}, \left(\frac{1}{1 + \cos 2x}\text{ is an even function}\right)
= 2\int_0^{\frac{\pi}{4}} \frac{dx}{2\cos^2 x} = \int_0^{\frac{\pi}{4}} \sec^2 x\,dx
= [\tan x]_0^{\frac{\pi}{4}} = 1 [1½]
Note: Students attempting I_1 without the application of any property should be marked proportionately.
I_2 = 2\int_0^{\frac12} |x|\,dx, (|x| is an even function) [1/2]
= 2\int_0^{\frac12} x\,dx
= [x^2]_0^{\frac12} = \frac{1}{4} [½]
I_1 - 4I_2 = 0 [½]
Find : \int \frac{x + 2}{\sqrt{9x - x^2}}\,dx
I = \int \frac{x + 2}{\sqrt{9x - x^2}}\,dx
Let x + 2 = A(9 - 2x) + B
\Rightarrow A = -\frac{1}{2},\ B = \frac{13}{2} [1]
I = -\frac{1}{2}\int \frac{9 - 2x}{\sqrt{9x - x^2}}\,dx + \frac{13}{2}\int \frac{1}{\sqrt{9x - x^2}}\,dx
= -\frac{1}{2}\left(2\sqrt{9x - x^2}\right) + \frac{13}{2}\int \frac{1}{\sqrt{\left(\frac{9}{2}\right)^2 - \left(x - \frac{9}{2}\right)^2}}\,dx [1]
= -\sqrt{9x - x^2} + \frac{13}{2}\sin^{-1}\left(\frac{2x - 9}{9}\right) + C [1]
Evaluate : \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{dx}{1 + \sqrt{\cot x}}
I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{dx}{1 + \sqrt{\cot x}} = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \quad \text{(i)} [½]
Applying \int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx, we get
I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx \quad \text{(ii)} [1]
Adding (i) and (ii)
2I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} 1.dx [1]
= [x]_{\pi/12}^{5\pi/12} = \frac{\pi}{3}
\Rightarrow I = \frac{\pi}{6} [½]
Evaluate : \int_{\frac{-\pi}{6}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)\,dx
I = \int_{\frac{-\pi}{6}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)\,dx
= \int_{\frac{-\pi}{6}}^{0} (-\sin x + \cos x)\,dx + \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x)\,dx [1]
= [\cos x + \sin x]_{-\pi/6}^{0} + [-\cos x + \sin x]_{0}^{\pi/2} [1]
= \left(1 - \frac{\sqrt{3}}{2} + \frac{1}{2}\right) + (1 + 1)
= \frac{1}{2}\left(7 - \sqrt{3}\right) [1]
If \frac{d}{dx}(F(x)) = \frac{1}{e^x + 1}, then find F(x) given that F(0) = \log\frac{1}{2}.
\frac{d}{dx}(F(x)) = \frac{1}{e^x + 1}
\Rightarrow F(x) = \int \frac{1}{e^x + 1}\,dx [1]
= \int \frac{e^{-x}}{1 + e^{-x}}\,dx
= -\log(1 + e^{-x}) + C [1]
Now, F(0) = -\log 2 + C \Rightarrow \log\frac{1}{2} + \log 2 = C \Rightarrow C = 0 [½]
\therefore F(x) = -\log(1 + e^{-x}) [½]
Note: The above expression is equivalent to F(x) = x - \log(e^x + 1)
Find :
Put x = t^6 \Rightarrow dx = 6t^5 dt
\int \frac{dx}{x^{\frac12} + x^{\frac13}} = 6\int \frac{t^5}{t^3 + t^2}dt = 6\int \frac{t^3}{t + 1}dt [1]
= 6\int\left(t^2 - t + 1 - \frac{1}{t+1}\right)dt
= 6\left(\frac{t^3}{3} - \frac{t^2}{2} + t - \log|t + 1|\right) + c [1]
= 2x^{\frac12} - 3x^{\frac13} + 6x^{\frac16} - 6\log\left|x^{\frac16} + 1\right| + c [1]
Find :
\int \tan^{-1}\left(\frac{1-x}{1+x}\right)dx
Getting \int\left(\frac{\pi}{4} - \tan^{-1}x\right)dx [1]
= \frac{\pi x}{4} - \left[x\tan^{-1}x - \int\frac{x}{1+x^2}dx\right] [1½]
= \frac{\pi x}{4} - x\tan^{-1}x + \frac{1}{2}\log(1 + x^2) + c [½]
Evaluate :
Let I = \int_{0}^{\pi}\frac{\sin^{2026}x}{\sin^{2026}x + \cos^{2026}x}dx = 2\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2026}x}{\sin^{2026}x + \cos^{2026}x}dx [1]
Getting I = 2\int_{0}^{\frac{\pi}{2}}\frac{\cos^{2026}x}{\sin^{2026}x + \cos^{2026}x}dx [1]
On adding, we get 2I = 2\int_{0}^{\frac{\pi}{2}} 1\, dx \Rightarrow I = \frac{\pi}{2} [1]
Find :
Put \sin x = t \Rightarrow \cos x\, dx = dt [1]
\int\frac{\cos x}{(2+\sin x)(4+\sin x)}dx = \int\frac{1}{(2+t)(4+t)}dt
= \frac{1}{2}\int\left(\frac{1}{2+t} - \frac{1}{4+t}\right)dt = \frac{1}{2}\log\left|\frac{2+t}{4+t}\right| + c [1½]
= \frac{1}{2}\log\left|\frac{2 + \sin x}{4 + \sin x}\right| + c [½]
Find :
\int\frac{x+3}{x^2+4x+5}dx = \frac{1}{2}\int\frac{(2x+4)+2}{x^2+4x+5}dx
= \frac{1}{2}\int\frac{2x+4}{x^2+4x+5}dx + \int\frac{1}{x^2+4x+5}dx [1]
= \frac{1}{2}\log|x^2+4x+5| + \int\frac{1}{(x+2)^2+1}dx [1]
= \frac{1}{2}\log|x^2+4x+5| + \tan^{-1}(x+2) + c [1]
Find :
Let I = \int \frac{x - \sin x}{1 - \cos x}\,dx
= \int \frac{x}{1-\cos x}\,dx - \int \frac{\sin x}{1-\cos x}\,dx [½]
= \int \frac{x}{2\sin^2\frac{x}{2}}\,dx - \int \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}\,dx [1]
= -\frac{x}{2}.2\cot\frac{x}{2} - \frac{1}{2}\int -2\cot\frac{x}{2}\,dx - \int \cot\frac{x}{2}\,dx [1]
= -x\cot\frac{x}{2} + C [½]
Evaluate :
I = \int_{0}^{2} \frac{1}{\sqrt{x^2 + 2x + 3}}\,dx
= \int_{0}^{2} \frac{1}{\sqrt{(x+1)^2 + 2}}\,dx [1]
= \left[\log\left(x + 1 + \sqrt{x^2 + 2x + 3}\right)\right]_0^2 [1]
= \log\left(3 + \sqrt{11}\right) - \log\left(1 + \sqrt{3}\right) or \log\left(\frac{3 + \sqrt{11}}{1 + \sqrt{3}}\right) [1]
Find : \int\frac{x+\sin x}{1+\cos x}\,dx
\int\frac{x+\sin x}{1+\cos x}dx
=\int\frac{x+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}dx [1]
=\int x\left(\frac{1}{2}\sec^2\frac{x}{2}\right)dx+\int\tan\frac{x}{2}dx [½]
=x\tan\frac{x}{2}-\int\tan\frac{x}{2}dx+\int\tan\frac{x}{2}dx [1]
=x\tan\frac{x}{2}+C [½]
Evaluate : \int_{0}^{\frac{\pi}{4}}\frac{dx}{\cos^3 x\sqrt{2\sin 2x}}
\int_{0}^{\pi/4}\frac{dx}{\cos^3 x\sqrt{2\sin 2x}}
=\frac{1}{2}\int_{0}^{\pi/4}\frac{dx}{\cos^4 x\sqrt{\tan x}} [½]
=\frac{1}{2}\int_{0}^{\pi/4}\frac{(1+\tan^2 x)\sec^2 x}{\sqrt{\tan x}}dx
Put \tan x=t\Rightarrow\sec^2 x\,dx=dt [½]
\therefore I=\frac{1}{2}\int_{0}^{1}\frac{1+t^2}{\sqrt{t}}dt [½]
=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{\sqrt{t}}+t^{3/2}\right)dt
=\frac{1}{2}\left[2\sqrt{t}+\frac{2}{5}t^{5/2}\right]_{0}^{1} [1]
=\frac{6}{5} [½]
Find : \int \frac{1}{x}\sqrt{\frac{x + a}{x - a}}\,dx.
I = \int \frac{1}{x}\frac{x + a}{\sqrt{x^2 - a^2}}dx = \int\frac{1}{\sqrt{x^2 - a^2}}dx + a\int\frac{1}{x\sqrt{x^2 - a^2}}dx [1]
= \log\left|x + \sqrt{x^2 - a^2}\right| + \sec^{-1}\left(\frac{x}{a}\right) + C [1+1]
Evaluate :
I=\int_{\pi/2}^{\pi}e^x\left(\frac{1-\sin x}{1-\cos x}\right)dx
=\int_{\pi/2}^{\pi}e^x\left(\frac{1-2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}\right)dx [1]
=\int_{\pi/2}^{\pi}e^x\left(\frac{1}{2}\text{cosec}^2\frac{x}{2}-\cot\frac{x}{2}\right)dx
\therefore I=-\left[e^x\cot\frac{x}{2}\right]_{\pi/2}^{\pi} [1]
=-\left[e^{\pi}\cot\frac{\pi}{2}-e^{\frac{\pi}{2}}\cot\frac{\pi}{4}\right]
=e^{\frac{\pi}{2}} [1]
Find :
Let I=\int\frac{2x}{(x^2+3)(x^2-5)}dx
Put x^2=t\Rightarrow2x\cdot dx=dt [½]
\Rightarrow I=\int\frac{dt}{(t+3)(t-5)}
=\int\left(-\frac{1}{8(t+3)}+\frac{1}{8(t-5)}\right)dt [1]
=\frac18[\log|t-5|-\log|t+3|]+c [1]
=\frac18\log\left|\frac{x^2-5}{x^2+3}\right|+c [½]
Evaluate :
\int_1^4(|x-2|+|x-4|)\,dx
=\int_1^2(2-x)\,dx+\int_2^4(x-2)\,dx-\int_1^4(x-4)\,dx [1½]
=\left[\frac{(2-x)^2}{-2}\right]_1^2+\left[\frac{(x-2)^2}{2}\right]_2^4-\left[\frac{(x-4)^2}{2}\right]_1^4 [1]
=\frac12+2+\frac92=7 [½]
Find :
I = \int \sqrt{4x^2 - 4x + 10}\,dx
= \int \sqrt{(2x-1)^2 + (3)^2}\,dx [1]
= \frac{1}{2}\left[\left(\frac{2x-1}{2}\right)\sqrt{4x^2 - 4x + 10} + \frac{9}{2}\log\left|(2x-1) + \sqrt{4x^2 - 4x + 10}\right|\right] + C [1+1]
Evaluate :
I = \int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx \quad \ldots(i)
= \int_{0}^{\pi} \frac{(\pi - x)\sin(\pi - x)}{1+\cos^2(\pi - x)}\,dx
I = \int_{0}^{\pi} \frac{(\pi - x)\sin x}{1+\cos^2 x}\,dx \quad \ldots(ii) [½]
Adding (i) and (ii)
2I = \int_{0}^{\pi} \frac{\pi\sin x}{1+\cos^2 x}\,dx [½]
\Rightarrow I = \frac{\pi}{2}\int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx
Put \cos x = t \Rightarrow -\sin x\,dx = dt [½]
I = -\frac{\pi}{2}\int_{1}^{-1} \frac{dt}{1+t^2} [½]
= \pi\left[\tan^{-1} t\right]_0^1 = \frac{\pi^2}{4} [1]
Find :
Let I = \int \frac{x^2}{(x^2 + 4)(x^2 + 9)}\, dx
Put x^2 = t [½]
\frac{t}{(t + 4)(t + 9)} = \frac{A}{t + 4} + \frac{B}{t + 9} \Rightarrow A = \frac{-4}{5},\ B = \frac{9}{5}
I = \frac{-4}{5}\int \frac{1}{2^2 + x^2}\, dx + \frac{9}{5}\int \frac{1}{3^2 + x^2}\, dx [1½]
= \frac{-2}{5}\tan^{-1}\left(\frac{x}{2}\right) + \frac{3}{5}\tan^{-1}\left(\frac{x}{3}\right) + C [1]
Evaluate :
\int_{1}^{3} (|x - 1| + |x - 2| + |x - 3|)\, dx
= \int_{1}^{3} (x - 1)\, dx + \int_{1}^{2} -(x - 2)\, dx + \int_{2}^{3} (x - 2)\, dx - \int_{1}^{3} (x - 3)\, dx [1½]
= \int_{1}^{3} 2\, dx + \int_{1}^{2} (2 - x)\, dx + \int_{2}^{3} (x - 2)\, dx
= [2x]_{1}^{3} + \left[\frac{(2 - x)^2}{-2}\right]_{1}^{2} + \left[\frac{(x - 2)^2}{2}\right]_{2}^{3}
= 4 + \frac{1}{2} + \frac{1}{2} = 5 [1½]
Evaluate :
\int_{-2}^{2} \sqrt{\frac{2-x}{2+x}}\,dx
= \int_{-2}^{2} \frac{2-x}{\sqrt{4-x^2}}\,dx [½]
= \int_{-2}^{2} \frac{2}{\sqrt{4-x^2}}\,dx - \int_{-2}^{2} \frac{x}{\sqrt{4-x^2}}\,dx [½]
= 2\int_{0}^{2} \frac{2}{\sqrt{4-x^2}}\,dx - 0 [\frac{2}{\sqrt{4-x^2}} is even, \frac{x}{\sqrt{4-x^2}} is odd] [1]
= 4\int_{0}^{2} \frac{1}{\sqrt{4-x^2}}\,dx
= 4\sin^{-1}\frac{x}{2}\Big|_{0}^{2} [½]
= 2\pi [½]
Find :
Let \log x = t \Rightarrow \frac{1}{x}dx = dt [½]
The given integral becomes = \int \frac{1}{t^2-3t-4}\,dt [½]
= \int \frac{1}{\left(t-\frac{3}{2}\right)^2 - \left(\frac{5}{2}\right)^2}\,dx [1]
= \frac{1}{5}\log\left|\frac{t-4}{t+1}\right| + C [½]
= \frac{1}{5}\log\left|\frac{\log x-4}{\log x+1}\right| + C [½]
Find :
Let x^{\frac{3}{2}} = t
\Rightarrow \frac{3}{2}x^{\frac{1}{2}}dx = dt [½]
The given integral becomes \frac{2}{3}\int t\sin^{-1}t\,dt [½]
= \frac{2}{3}\left[\sin^{-1}t \times \frac{t^2}{2} - \int \frac{1}{\sqrt{1-t^2}} \times \frac{t^2}{2}\,dt\right] [1]
= \frac{1}{3}\left[\sin^{-1}t \times t^2 + \int \frac{1-t^2-1}{\sqrt{1-t^2}}\,dt\right]
= \frac{1}{3}\left[\sin^{-1}t \times t^2 + \int \sqrt{1-t^2}\,dt - \int \frac{1}{\sqrt{1-t^2}}\,dt\right]
= \frac{1}{3}\left[t^2\sin^{-1}t + \frac{t}{2}\sqrt{1-t^2} + \frac{1}{2}\sin^{-1}t - \sin^{-1}t\right] + C
= \frac{1}{3}\left[t^2\sin^{-1}t + \frac{t}{2}\sqrt{1-t^2} - \frac{1}{2}\sin^{-1}t\right] + C
= \frac{1}{3}\left[x^3\sin^{-1}\left(x^{\frac{3}{2}}\right) + \frac{x^{\frac{3}{2}}}{2}\sqrt{1-x^3} - \frac{1}{2}\sin^{-1}\left(x^{\frac{3}{2}}\right)\right] + C [1]
Find :
I = \int \frac{x^2 + 1}{(x^2 + 2)(x^2 + 4)}\, dx
Let x^2 = y, then \frac{x^2 + 1}{(x^2 + 2)(x^2 + 4)} = \frac{y + 1}{(y + 2)(y + 4)} [1]
Let \frac{y + 1}{(y + 2)(y + 4)} = \frac{A}{y + 2} + \frac{B}{y + 4}
this gives A = -\frac{1}{2}, B = \frac{3}{2}
\therefore I = -\frac{1}{2}\int \frac{1}{x^2 + 2}\, dx + \frac{3}{2}\int \frac{1}{x^2 + 4}\, dx [1]
\Rightarrow I = -\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{3}{4}\tan^{-1}\left(\frac{x}{2}\right) + c [1]
Find :
I = \int \frac{2 + \sin 2x}{1 + \cos 2x} e^x\, dx
= \int \frac{2 + 2\sin x\cos x}{2\cos^2 x} e^x\, dx [1]
= \int (\sec^2 x + \tan x)\, e^x\, dx [1]
= e^x.\tan x + c [1]
Evaluate :
I = \int_{0}^{\frac{\pi}{4}} \frac{1}{\sin x + \cos x}\, dx
= \frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x}\, dx [1]
= \frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin\left(x + \frac{\pi}{4}\right)}\, dx = = \frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{4}} \operatorname{cosec}\left(x + \frac{\pi}{4}\right) dx
= \frac{1}{\sqrt{2}}\left[\log\left|\operatorname{cosec}\left(x + \frac{\pi}{4}\right) - \cot\left(x + \frac{\pi}{4}\right)\right|\right]_{0}^{\frac{\pi}{4}} [1]
= \frac{1}{\sqrt{2}}\log\left(\sqrt{2} + 1\right) or -\frac{1}{\sqrt{2}}\log\left(\sqrt{2} - 1\right) [1]
Evaluate : \displaystyle\int_{0}^{\frac{\pi}{4}} \frac{x\,dx}{1+\cos 2x+\sin 2x}
I = \int_{0}^{\frac{\pi}{4}} \frac{x}{1+\cos 2x+\sin 2x}dx \quad \ldots(1)
On applying \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx,
we get I = \int_{0}^{\frac{\pi}{4}} \frac{\frac{\pi}{4}-x}{1+\cos 2x+\sin 2x}dx \quad \ldots(2) [½]
On adding Eq. (1) and (2), we get 2I = \frac{\pi}{4}\int_{0}^{\frac{\pi}{4}} \frac{1}{1+\cos 2x+\sin 2x}dx [½]
I = \frac{\pi}{16}\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos^2 x+\sin x\cos x}dx = \frac{\pi}{16}\int_{0}^{\frac{\pi}{4}} \frac{\sec^2 x\,dx}{1+\tan x} [1]
I = \frac{\pi}{16}\left(\log|1+\tan x|\right)_{0}^{\frac{\pi}{4}} [½]
I = \frac{\pi}{16}\log 2 [½]
Find : \displaystyle\int e^x\left[\frac{1}{(1+x^2)^{\frac{3}{2}}} + \frac{x}{\sqrt{1+x^2}}\right]dx
I = \int e^x\left(\frac{x}{\sqrt{1+x^2}} + \frac{1}{(1+x^2)^{\frac{3}{2}}}\right)dx
Let f(x) = \frac{x}{\sqrt{1+x^2}}, f'(x) = \frac{\sqrt{1+x^2} - x\frac{x}{\sqrt{1+x^2}}}{1+x^2} = \frac{1}{(1+x^2)^{\frac{3}{2}}} [½ + 1½]
On applying \int e^x(f(x)+f'(x))dx = e^x f(x) + c,
I = e^x\frac{x}{\sqrt{1+x^2}} + c [1]
Find : \displaystyle\int \frac{3x+5}{\sqrt{x^2+2x+4}}\,dx
I = \int \frac{3x+5}{\sqrt{x^2+2x+4}}dx = \frac{3}{2}\int \frac{2x+2}{\sqrt{x^2+2x+4}}dx + 2\int \frac{1}{\sqrt{x^2+2x+4}}dx [1]
I = \frac{3}{2}\left(2\sqrt{x^2+2x+4}\right) + 2\int \frac{1}{\sqrt{(x+1)^2+\sqrt{3}^2}}dx [1 + ½]
I = 3\sqrt{x^2+2x+4} + 2\log\left|(x+1)+\sqrt{x^2+2x+4}\right| + c [½]
Evaluate : \int_0^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx
Let I=\int_0^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}dx ---------- (i)
\Rightarrow I=\int_0^{\pi}\frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}}dx=\int_0^{\pi}\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}dx ---------- (ii) [1½]
Adding (i) and (ii), we get
2I=\int_0^{\pi}dx=x\Big]_0^{\pi}=\pi,\ \therefore I=\frac{\pi}{2} [1½]
Find : \int\frac{2x+1}{(x+1)^2(x-1)}\,dx
\int\frac{2x+1}{(x+1)^2(x-1)}dx=-\frac34\int\frac{1}{x+1}dx+\frac12\int\frac{1}{(x+1)^2}dx+\frac34\int\frac{1}{x-1}dx [1½]
=-\frac34\log|x+1|-\frac{1}{2(x+1)}+\frac34\log|x-1|+C [1½]
or, =\frac34\log\left|\frac{x-1}{x+1}\right|-\frac{1}{2(x+1)}+C
Find :
\int\frac{dx}{\cos x\sqrt{\cos 2x}}=\int\frac{dx}{\cos x\sqrt{\cos^{2}x-\sin^{2}x}} [½]
=\int\frac{\sec^{2}x}{\sqrt{1-\tan^{2}x}}dx [1]
=\int\frac{1}{\sqrt{1-t^{2}}}dt, using \tan x=t,\ \sec^{2}x\cdot dx=dt [½]
=\sin^{-1}t+c=\sin^{-1}(\tan x)+c [1]
Find :
\int\frac{5x-3}{\sqrt{1+4x-2x^{2}}}dx=-\frac{5}{4}\int\frac{-4x+4}{\sqrt{1+4x-2x^{2}}}dx+\sqrt{2}\int\frac{1}{\sqrt{\left(\sqrt{\frac{3}{2}}\right)^{2}-(x-1)^{2}}}dx [1½]
=-\frac{5}{2}\sqrt{1+4x-2x^{2}}+\sqrt{2}\sin^{-1}\left(\sqrt{\frac{2}{3}}(x-1)\right)+c [1½]
Evaluate :
Let I=\int_{0}^{\pi/2}[\log\sin x-\log(2\cos x)]dx=\int_{0}^{\pi/2}\log\left(\frac{\tan x}{2}\right)dx [½]
Using property \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx
We get, I=\int_{0}^{\pi/2}\log\left(\frac{\cot x}{2}\right)dx [1]
\therefore 2I=\int_{0}^{\pi/2}\log\left(\frac{\tan x}{2}\times\frac{\cot x}{2}\right)dx=\int_{0}^{\pi/2}\log\left(\frac14\right)dx [½]
2I=\log\left(\frac14\right)x\Big|_{0}^{\pi/2}=\frac{\pi}{2}\log\frac14 [½]
I=\frac{\pi}{4}\log\frac14 OR -\frac{\pi}{2}\log 2 [½]
Find :
Let I=\int\frac{dx}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}+2)}
Let \sqrt{x}=t,\ \frac{1}{2\sqrt{x}}dx=dt [½]
\therefore I=2\int\frac{dt}{(t+1)(t+2)}
=2\int\left(\frac{1}{t+1}-\frac{1}{t+2}\right)dt [1]
=2[\log|t+1|-\log|t+2|]+C [1]
=2[\log(\sqrt{x}+1)-\log(\sqrt{x}+2)]+C or 2\log\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}\right)+C [½]
Evaluate :
Let I=\int e^x\sin x\,dx
=e^x\sin x-\int\cos x\,e^x\,dx [1]
=e^x\sin x-\cos x\,e^x-I [½]
\therefore I=\frac12e^x(\sin x-\cos x) [½]
\therefore\int_{0}^{\pi/2}e^x\sin x\,dx=\frac12e^{\pi/2}+\frac12 or \frac12\left(e^{\pi/2}+1\right) [1]
Find :
Let I=\int\frac{1}{\cos(x-a)\cos(x-b)}dx
=\frac{1}{\sin(a-b)}\int\frac{\sin[(x-b)-(x-a)]}{\cos(x-a)\cos(x-b)}dx [1]
=\frac{1}{\sin(a-b)}\left[\int\frac{\sin(x-b)\cos(x-a)}{\cos(x-a)\cos(x-b)}-\frac{\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}\right]dx [½]
=\frac{1}{\sin(a-b)}\left[\int[\tan(x-b)-\tan(x-a)]dx\right] [½]
=\frac{1}{\sin(a-b)}[\log|\sec(x-b)|-\log|\sec(x-a)|]+C [1]
Evaluate \int_{0}^{\pi/4} \log(1 + \tan x)\,dx.
Let I = \int_{0}^{\pi/4} \log(1 + \tan x)\,dx ---------- (i)
\Rightarrow I = \int_{0}^{\pi/4} \log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right)dx = \int_{0}^{\pi/4} \log\left(\frac{2}{1 + \tan x}\right)dx ---------- (ii) [1]
Add (i) and (ii), 2I = \int_{0}^{\pi/4} \log 2\,dx = \log 2 \cdot x\Big]_{0}^{\pi/4} = \frac{\pi}{4}\log 2 [1½]
\Rightarrow I = \frac{\pi}{8}\log 2 [½]
Find \int \frac{dx}{\sqrt{\sin^3 x\cos(x - \alpha)}}.
Let I = \int \frac{1}{\sqrt{\sin^3 x\cos(x - \alpha)}}dx = \int \frac{\text{cosec}^2 x}{\sqrt{\sin\alpha + \cos\alpha\cot x}}dx [1]
Put \sin\alpha + \cos\alpha\cot x = t \Rightarrow \text{cosec}^2 x\,dx = -\frac{1}{\cos\alpha}dt [½]
\therefore I = -\int \frac{1}{\cos\alpha\sqrt{t}}dt = -\frac{2\sqrt{t}}{\cos\alpha} + c [1]
\Rightarrow I = -\frac{2}{\cos\alpha}\sqrt{\sin\alpha + \cos\alpha\cot x} + c [½]
Find \int e^{\cot^{-1} x}\left(\frac{1 - x + x^2}{1 + x^2}\right)dx.
Put \cot^{-1} x = t \therefore x = \cot t and \frac{1}{1 + x^2}dx = -\,dt [1]
\therefore \int e^{\cot^{-1} x}\left(\frac{1 - x + x^2}{1 + x^2}\right)dx = -\int e^t(1 - \cot t + \cot^2 t)\,dt = \int e^t(\cot t - \text{cosec}^2 t)\,dt [1]
= e^t\cot t + c [½]
= xe^{\cot^{-1} x} + c [½]
Evaluate \int_{\log\sqrt{2}}^{\log\sqrt{3}} \frac{1}{(e^x + e^{-x})(e^x - e^{-x})}dx
Let I = \int_{\log\sqrt{2}}^{\log\sqrt{3}} \frac{1}{(e^x + e^{-x})(e^x - e^{-x})}dx = \int_{\log\sqrt{2}}^{\log\sqrt{3}} \frac{e^{2x}}{(e^{2x})^2 - 1}dx [1]
Put e^{2x} = t \Rightarrow e^{2x}dx = \frac{1}{2}dt, Upper limit = 3, Lower Limit = 2 [½]
\therefore I = \frac{1}{2}\int_{2}^{3} \frac{1}{t^2 - 1}dt = \frac{1}{4}\log\left|\frac{t-1}{t+1}\right|\Big]_{2}^{3} [1]
\Rightarrow I = \frac{1}{4}\left[\log\frac{2}{4} - \log\frac{1}{3}\right] = \frac{1}{4}\log\frac{3}{2} [½]
Evaluate :
Let I = \int_{1}^{3} \frac{\sqrt{4 - x}}{\sqrt{x} + \sqrt{4 - x}}dx
I = \int_{1}^{3} \frac{\sqrt{4 - (4 - x)}}{\sqrt{4 - x} + \sqrt{4 - (4 - x)}}dx [1]
2I = \int_{1}^{3} \frac{\sqrt{4 - x} + \sqrt{x}}{\sqrt{4 - x} + \sqrt{x}}dx = \int_{1}^{3} 1\,dx [1]
2I = x\Big|_{1}^{3} = 2 [½]
\Rightarrow I = 1 [½]
Evaluate :
Let I = \int_{1}^{e} \frac{1}{\sqrt{4x^2 - (x\log x)^2}}dx
= \int_{1}^{e} \frac{1}{x\sqrt{4 - (\log x)^2}}dx [Let \log x = t \Rightarrow \frac{1}{x}dx = dt] [1½]
= \int_{0}^{1} \frac{dt}{\sqrt{4 - t^2}} [½]
= \sin^{-1}\frac{t}{2}\Big|_{0}^{1} = \frac{\pi}{6} [1]
Find :
I = \int \frac{\cos x}{3\sin x - 4\sin^3 x}dx
Let \sin x = t \Rightarrow \cos x\, dx = dt [½]
I = \int \frac{dt}{3t - 4t^3} [½]
= \int \frac{1}{t^3\left(\frac{3}{t^2} - 4\right)}dt [½]
Let \frac{3}{t^2} - 4 = z \Rightarrow -\frac{6}{t^3}dt = dz [½]
I = -\frac{1}{6}\int \frac{dz}{z}
= -\frac{1}{6}\log|z| + C [½]
= -\frac{1}{6}\log|3\,\text{cosec}^2 x - 4| + C [½]
Find :
Let I = \int x^2 \log(x^2 + 1)dx
= \log(x^2 + 1) \cdot \frac{x^3}{3} - \int \frac{2x}{x^2 + 1} \cdot \frac{x^3}{3}dx [1]
= \frac{x^3}{3}\log(x^2 + 1) - \frac{2}{3}\int \frac{x^4}{x^2 + 1}dx [½]
= \frac{x^3}{3}\log(x^2 + 1) - \frac{2}{3}\int\left(x^2 - 1 + \frac{1}{x^2 + 1}\right)dx [½]
= \frac{x^3}{3}\log(x^2 + 1) - \frac{2}{3}\left[\frac{x^3}{3} - x + \tan^{-1} x\right] + C [1]
Find :
Let I=\int\frac{x^2+x+1}{(x+1)^2(x+2)}\,dx
Here \frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{x+2}
\Rightarrow x^2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^2 [½]
On comparing, we get
A=-2,\ B=1 and C=3. [1½]
\therefore I=\int\frac{-2\,dx}{x+1}+\int\frac{dx}{(x+1)^2}+3\int\frac{dx}{x+2}
=-2\log|x+1|-\frac{1}{x+1}+3\log|x+2|+C [1]
Evaluate :
I=\int_{\pi/4}^{\pi/2}e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx
Put 2x=t so that 2\,dx=dt [½]
When x=\frac{\pi}{2},\ t=\pi\ ;\ x=\frac{\pi}{4},\ t=\frac{\pi}{2}
Thus
I=\int_{\pi/2}^{\pi}e^{t}\left(\frac{1-\sin t}{1-\cos t}\right)\frac{dt}{2}
=\int_{\pi/2}^{\pi}e^{t}\left(\frac{1-2\sin t/2\cos t/2}{2\sin^2 t/2}\right)\frac{dt}{2} [1]
=\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}e^{t}\left(\frac{1}{2}\text{cosec}^2\frac{t}{2}-\cot\frac{t}{2}\right)dt [½]
=-\frac{1}{2}\left|e^{t}\cot\frac{t}{2}\right|_{\pi/2}^{\pi} [½]
=\frac{1}{2}e^{\pi/2}. [½]
Evaluate :
I=\int_{-2}^{2}\frac{x^2}{1+5^x}\,dx -----------(1)
Replacing x by -2+2-x
=\int_{-2}^{2}\frac{(-2+2-x)^2}{1+5^{(-2+2-x)}}\,dx [1]
I=\int_{-2}^{2}\frac{x^2 5^x}{1+5^x}\,dx -------------(2) [½]
Adding (1) and (2)
2I=\int_{-2}^{2}\frac{x^2(1+5^x)}{1+5^x}\,dx=\int_{-2}^{2}x^2\,dx=\left|\frac{x^3}{3}\right|_{-2}^{2}=\frac{8}{3}+\frac{8}{3}=\frac{16}{3} [1]
\Rightarrow I=\frac{16}{6} or \frac{8}{3} [½]
Find :
Let e^x=t, so that e^x\,dx=dt. Then, [½]
I=\int\frac{dt}{\sqrt{5-4t-t^2}}=\int\frac{dt}{\sqrt{3^2-(t+2)^2}} [1]
=\sin^{-1}\left(\frac{t+2}{3}\right)+C [1]
=\sin^{-1}\left(\frac{e^x+2}{3}\right)+C [½]
Evaluate :
I=\int_{0}^{\pi/2}\sqrt{\sin x}\,.\,(1-\sin^2 x)^2\cos x\,dx
Put \sin x=t^2 [½]
=\int_{0}^{1}\sqrt{t}\,(1-t^2)^2\,dt=\int_{0}^{1}\left(\sqrt{t}+t^{\frac{9}{2}}-2t^{\frac{5}{2}}\right)dt [1]
=\left.\frac{2t^{3/2}}{3}+\frac{2t^{11/2}}{11}-\frac{4t^{7/2}}{7}\right]_{0}^{1}=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}=\frac{64}{231} [1+½]
Evaluate : \int_{0}^{2\pi}\frac{1}{1+e^{\sin x}}\,dx
Let I=\int_{0}^{2\pi}\frac{1}{1+e^{\sin x}}dx=\int_{0}^{2\pi}\frac{1}{1+e^{\sin(2\pi-x)}}dx [1]
=\int_{0}^{2\pi}\frac{1}{1+e^{-\sin x}}dx=\int_{0}^{2\pi}\frac{e^{\sin x}}{e^{\sin x}+1}dx [1]
\Rightarrow2I=\int_{0}^{2\pi}\frac{e^{\sin x}+1}{e^{\sin x}+1}dx=\int_{0}^{2\pi}1\cdot dx=2\pi [½]
\Rightarrow I=\pi [½]
Find : \int\frac{x^4}{(x-1)(x^2+1)}\,dx
I=\int\frac{x^4}{(x-1)(x^2+1)}dx=\int\left[x+1+\frac{1}{(x-1)(x^2+1)}\right]dx [1]
=\frac{x^2}{2}+x+\int\left[\frac{1}{2(x-1)}-\frac{1}{2}\frac{(x+1)}{(x^2+1)}\right]dx [½+1]
(Using partial fractions)
=\frac{x^2}{2}+x+\frac{1}{2}\log|x-1|-\frac{1}{4}\log|x^2+1|-\frac{1}{2}\tan^{-1}x+C [½]
Evaluate :
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Evaluate :
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Find :
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Find :
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Find : \int\frac{dx}{\sqrt{x}+\sqrt[3]{x}} .
I=\int\frac{dx}{\sqrt{x}+\sqrt[3]{x}}
=\int\frac{6y^5\,dy}{y^3+y^2} x=y^6 so that dx=6y^5dy [1]
=6\int\frac{y^3}{y+1}dy
=6\int\left[(y^2-y+1)-\frac{1}{y+1}\right]dy [1]
=6\left[\frac{y^3}{3}-\frac{y^2}{2}+y-\log|y+1|\right]+C [½]
=2\sqrt{x}-3\sqrt[3]{x}+6x^{1/6}-6\log(x^{1/6}+1)+C [½]
Evaluate : \int_{0}^{\pi/2}\frac{\cos x}{(1+\sin x)(4+\sin x)}\,dx .
Let \sin x=t , then \cos x\,dx=dt [½]
\therefore \int_0^{\pi/2}\frac{\cos x}{(1+\sin x)(4+\sin x)}dx=\int_0^1\frac{dt}{(1+t)(4+t)} [½]
=\frac{1}{3}\left[\int_0^1\frac{1}{1+t}dt-\int_0^1\frac{1}{4+t}dt\right] [1]
=\frac{1}{3}\left[\log(1+t)\Big|_0^1-\log(4+t)\Big|_0^1\right] [½]
=\frac{1}{3}[\log 2-\log 5+\log 4] [½]
or =\frac{1}{3}\log\frac{8}{5}
Evaluate :
Consider \int(\tan^{-1}x)\,dx=\tan^{-1}x\cdot x-\int\frac{1}{1+x^2}\cdot x\,dx [1]
=x\tan^{-1}x-\frac12\log(1+x^2) [1]
\int_0^1(\tan^{-1}x)\,dx=\left.x\tan^{-1}x-\frac12\log(1+x^2)\right]_0^1
=\frac{\pi}{4}-\frac12\log 2 [1]
Find :
I=\int\frac{2x\cdot dx}{x^2+3x+2}=\int\frac{2x}{(x+1)(x+2)}dx [½]
=\int\left(\frac{-2}{x+1}+\frac{4}{x+2}\right)dx using partial fraction [1½]
=-2\log|x+1|+4\log|x+2|+C [1]
Evaluate : \int_{0}^{2\pi}\frac{dx}{1+e^{\sin x}}
I=\int_{0}^{2\pi}\frac{dx}{1+e^{\sin x}} ... (1)
I=\int_{0}^{2\pi}\frac{dx}{1+e^{\sin(2\pi-x)}} [½]
I=\int_{0}^{2\pi}\frac{dx}{1+e^{-\sin x}} [½]
I=\int_{0}^{2\pi}\frac{e^{\sin x}}{1+e^{\sin x}}dx ... (2) [½]
Adding (1) and (2)
2I=\int_{0}^{2\pi}1\,dx\Rightarrow 2I=2\pi [1]
I=\pi [½]
Find :
(a) Let, e^x=t,\ e^xdx=dt [½]
Evaluate :
=6+4=10 [½]
Evaluate :
I=\int_{0}^{\pi/4}\frac{dx}{1+\tan x}
I=\int_{0}^{\pi/4}\frac{dx}{1+\tan\left(\frac{\pi}{4}-x\right)} (using property) [½]
I=\int_{0}^{\pi/4}\frac{dx}{1+\frac{1-\tan x}{1+\tan x}} [½]
I=\frac{1}{2}\int_{0}^{\pi/4}(1+\tan x)\,dx [½]
I=\frac{1}{2}\left[x+\log\sec x\right]_{0}^{\pi/4} [1]
=\frac{1}{2}\left[\frac{\pi}{4}+\log\sqrt{2}\right]=\frac{1}{2}\left(\frac{\pi}{4}+\frac{1}{2}\log 2\right)
=\frac{\pi}{8}+\frac{1}{4}\log 2 [½]
Find :
I=\int \underset{\text{II}}{e^x}\,\underset{\text{I}}{\sin 2x}\,dx
I=\sin 2x\,e^x-\int \underset{\text{I}}{2\cos 2x}\,\underset{\text{II}}{e^x}\,dx [1]
=e^x\sin 2x-2[\cos 2x\,e^x-\int(-2\sin 2x)\,e^x dx] [½]
I=e^x\sin 2x-2\cos 2x\,e^x-4I [½]
5I=e^x\sin 2x-2\cos 2x\,e^x [½]
\therefore I=\frac{1}{5}e^x[(\sin 2x-2\cos 2x)]+C [½]
Find :
\int\frac{2x}{(x^2+1)(x^2+2)}\,dx
Let x^2=t,\ 2x\,dx=dt [½]
\therefore \int\frac{2x}{(x^2+1)(x^2+2)}\,dx=\int\frac{dt}{(t+1)(t+2)}
Getting \frac{1}{(t+1)(t+2)}=\frac{1}{t+1}+\frac{-1}{t+2} (by Partial Fraction) [1]
\int\frac{dt}{(t+1)(t+2)}=\int\frac{dt}{t+1}-\int\frac{dt}{t+2}
=\log|t+1|-\log|t+2|+C [1]
=\log\left(x^2+1\right)-\log\left(x^2+2\right)+C or \log\left(\frac{x^2+1}{x^2+2}\right)+C [½]
Find :
I = \int \frac{\sin x}{\sin(x-2a)}dx
= \int \frac{\sin[(x-2a)+2a]}{\sin(x-2a)}dx [½]
= \int \frac{\sin(x-2a)\cos 2a+\cos(x-2a)\sin 2a}{\sin(x-2a)}dx [1]
= \int \cos 2a\,dx+\sin 2a\int \cot(x-2a)\,dx [½]
= x\cos 2a+\sin 2a\cdot\log|\sin(x-2a)|+C [1]
Evaluate : \int_{0}^{\pi}\frac{x}{1+\sin x}\,dx .
I=\int_0^\pi\frac{x}{1+\sin x}dx=\int_0^\pi\frac{\pi-x}{1+\sin(\pi-x)}dx=\pi\int_0^\pi\frac{1}{1+\sin x}dx-\int_0^\pi\frac{x}{1+\sin x}dx [1]
\therefore 2I=\pi\int_0^\pi\frac{dx}{1+\sin x} [½]
\therefore I=\frac{\pi}{2}\int_0^\pi\frac{dx}{1+\sin x}=\frac{\pi}{2}\int_0^\pi\frac{1}{1+\cos\left(\frac{\pi}{2}-x\right)}dx [½]
=\frac{\pi}{2}\int_0^\pi\frac{1}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}dx
=\frac{\pi}{4}\left[-2\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right]_0^\pi [1]
=\frac{\pi}{4}[2-(-2)]=\pi [1]
Evaluate :
I=\int_0^1 x(1-x)^n\,dx
=\int_0^1(1-x)[1-(1-x)]^n\,dx [using property] [1]
=\int_0^1 x^n(1-x)\,dx
=\int_0^1 x^n dx-\int_0^1 x^{n+1}dx [1]
=\left[\frac{x^{n+1}}{n+1}\right]_0^1-\left[\frac{x^{n+2}}{n+2}\right]_0^1 [1]
=\frac{1}{n+1}-\frac{1}{n+2} Or \frac{1}{(n+1)(n+2)} [1]
Find : \int\frac{x^2}{(x^2+1)(3x^2+4)}dx
\int\frac{x^2dx}{(x^2+1)(3x^2+4)}
Let x^2=t [½]
\frac{t}{(t+1)(3t+4)}=\frac{-1}{t+1}+\frac{4}{3t+4} (by Partial fraction) [1]
\int\frac{x^2}{(x^2+1)(3x^2+4)}dx=\int-\frac{1}{x^2+1}dx+\int\frac{4}{3x^2+4}dx [½]
=-\tan^{-1}x+\frac43\times\frac{\sqrt3}{2}\tan^{-1}\frac{\sqrt3x}{2}+C [1½]
=-\tan^{-1}x+\frac{2}{\sqrt3}\tan^{-1}\frac{\sqrt3x}{2}+C [½]
Evaluate : \int_{-2}^{1}\sqrt{5-4x-x^2}\,dx
\int_{-2}^{1}\sqrt{5-4x-x^2}\,dx
=\int_{-2}^{1}\sqrt{9-(x+2)^2}\,dx [1½]
=\left.\frac{x+2}{2}\sqrt{9-(x+2)^2}+\frac92\sin^{-1}\frac{x+2}{3}\right]_{-2}^{1} [1½]
=0+\frac92\cdot\frac{\pi}{2}-0
=\frac{9\pi}{4} [1]
Evaluate :
Adding (1) and (2)
Find :
I = \int \cos x\cdot\tan^{-1}(\sin x)\,dx
Put \sin x = t \Rightarrow \cos x\,dx = dt [1]
\therefore I = \int \tan^{-1}t\cdot 1\,dt
= \tan^{-1}t\cdot t-\frac{1}{2}\int \frac{2t}{1+t^2}dt [1½]
= t\cdot\tan^{-1}t-\frac{1}{2}\log|1+t^2|+C [1]
= \sin x\cdot\tan^{-1}(\sin x)-\frac{1}{2}\log|1+\sin^2 x|+C [½]
Find :
I = \int \frac{e^{x}}{(e^{x}+1)(e^{x}+3)}dx
Put e^{x} = t \Rightarrow e^{x}dx = dt [1]
I = \int \frac{dt}{(t+1)(t+3)}
= \frac{1}{2}\int\left(\frac{1}{t+1}-\frac{1}{t+3}\right)dt [1½]
= \frac{1}{2}[\log|t+1|-\log|t+3|]+C [1]
= \frac{1}{2}[\log|e^{x}+1|-\log|e^{x}+3|]+C [½]
or \frac{1}{2}\log\left|\frac{e^{x}+1}{e^{x}+3}\right|+C
Evaluate \int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx
Let I=\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx
I=\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}\times\frac{a-x}{a-x}}\,dx [½]
=\int_{-a}^{a}\frac{a-x}{\sqrt{a^2-x^2}}dx=\int_{-a}^{a}\frac{a}{\sqrt{a^2-x^2}}dx-\int_{-a}^{a}\frac{x}{\sqrt{a^2-x^2}}dx=I_1-I_2 [1]
I_1=a\int_{-a}^{a}\frac{1}{\sqrt{a^2-x^2}}dx
=2a\int_0^a\frac{1}{\sqrt{a^2-x^2}}dx using \int_{-a}^{a}f(x)\,dx=2\int_0^af(x)\,dx if f(-x)=f(x) [1]
I_1=2a\left[\sin^{-1}\left(\frac{x}{a}\right)\right]_0^a=2a\left(\frac{\pi}{2}\right)=\pi a [1]
and
I_2=\int_{-a}^{a}\frac{x}{\sqrt{a^2-x^2}}dx=0 using \int_{-a}^{a}f(x)\,dx=0 if f(-x)=-f(x) [1]
I=I_1-I_2=\pi a [½]
Alternative method
Let x=a\cos2\theta\quad\Rightarrow dx=-2a\sin(2\theta)\,d\theta [½]
When x=-a,\theta=\frac{\pi}{2} and x=a,\theta=0 [½]
I=\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx=-2a\int_{\frac{\pi}{2}}^{0}\tan\theta\sin(2\theta)\,d\theta
I=2a\int_0^{\frac{\pi}{2}}\tan\theta\sin(2\theta)\,d\theta ………(i) using \int_a^bf(x)dx=-\int_b^af(x)dx [1]
I=2a\int_0^{\frac{\pi}{2}}\tan\left(\frac{\pi}{2}-\theta\right)\sin(\pi-2\theta)d\theta using \int_0^af(x)dx=\int_0^af(a-x)dx [1]
I=2a\int_0^{\frac{\pi}{2}}\cot(\theta)\sin(2\theta)\,d\theta ………(ii)
Adding equations (i) and (ii)
2I=2a\int_0^{\frac{\pi}{2}}(\tan\theta+\cot\theta)\sin(2\theta)\,d\theta [1]
\Rightarrow I=2a\int_0^{\frac{\pi}{2}}(\sin^2\theta+\cos^2\theta)\,d\theta
\Rightarrow I=2a\int_0^{\frac{\pi}{2}}1\,d\theta=2a\left(\frac{\pi}{2}\right)=\pi a [1]
Evaluate: \int_0^1\frac{\log(1+x)}{1+x^2}dx
Let x=\tan\theta\Rightarrow dx=\sec^2\theta\,d\theta [½]
I=\int_0^{\frac{\pi}{4}}\frac{\log(1+\tan\theta)}{1+\tan^2\theta}.\sec^2\theta\,d\theta
I=\int_0^{\frac{\pi}{4}}\log(1+\tan\theta)d\theta=\int_0^{\frac{\pi}{4}}\log\left[1+\tan\left(\frac{\pi}{4}-\theta\right)\right]d\theta [1]
=\int_0^{\frac{\pi}{4}}\log\left[1+\frac{1-\tan\theta}{1+\tan\theta}\right]d\theta [1]
=\int_0^{\frac{\pi}{4}}\log\left[\frac{1+\tan\theta+1-\tan\theta}{1+\tan\theta}\right]d\theta
=\int_0^{\frac{\pi}{4}}\log\left[\frac{2}{1+\tan\theta}\right]d\theta
=\int_0^{\frac{\pi}{4}}\log2\,d\theta-\int_0^{\frac{\pi}{4}}\log[1+\tan\theta]\,d\theta [1]
=\log2\times x\Big]_0^{\frac{\pi}{4}}-I
\Rightarrow 2I=\frac{\pi}{4}\log2 [1]
\Rightarrow I=\frac{\pi}{8}\log2 [½]
Find \int\frac{(3\sin\theta-2)\cos\theta}{5-\cos^2\theta-4\sin\theta}d\theta
I=\int\frac{(3\sin\theta-2)\cos\theta}{5-\cos^2\theta-4\sin\theta}d\theta=\int\frac{(3\sin\theta-2)\cos\theta}{5-(1-\sin^2\theta)-4\sin\theta}d\theta [½]
Let \sin\theta=t\Rightarrow\cos\theta\,d\theta=dt
I=\int\frac{(3t-2)}{5-(1-t^2)-4t}dt
=\int\frac{(3t-2)}{t^2-4t+4}dt=\int\frac{3t-2}{(t-2)^2}dt [1]
Let \frac{3t-2}{(t-2)^2}=\frac{A}{(t-2)}+\frac{B}{(t-2)^2}
3t-2=A(t-2)+B
Comparing the coefficients of t and constant terms on both sides
A=3,-2A+B=-2,B=4 [½+½]
\int\frac{(3\sin\theta-2)\cos\theta}{5-\cos^2\theta-4\sin\theta}d\theta=\int\frac{3}{t-2}dt+\int\frac{4}{(t-2)^2}dt
=3\log|t-2|-\frac{4}{t-2}+C [1+1]
=3\log|\sin\theta-2|-\frac{4}{\sin\theta-2}+C [½]
Find :
Let \frac{x}{(x - 1)(x^{2} + 4)} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + 4} [1]
on solving we get, A = \frac{1}{5}, B = \frac{-1}{5}, C = \frac{4}{5} [1½]
so, \int \frac{x}{(x - 1)(x^{2} + 4)}dx = \frac{1}{5}\int \frac{dx}{x - 1} - \frac{1}{10}\int \frac{2x\,dx}{x^{2} + 4} + \frac{4}{5}\int \frac{dx}{x^{2} + 4}
\int \frac{x}{(x - 1)(x^{2} + 4)}dx = \frac{1}{5}\log|x - 1| - \frac{1}{10}\log\left(x^{2} + 4\right) + \frac{2}{5}\tan^{-1}\left(\frac{x}{2}\right) + c [2½]
Evaluate :
Let x = \tan t \Rightarrow dx = \sec^{2} t\,dt [1]
I(\text{Let}) = \int_{0}^{1} \frac{x\tan^{-1} x}{\left(1 + x^{2}\right)^{\frac{3}{2}}}dx = \int_{0}^{\frac{\pi}{4}} \frac{t\tan t}{\sec^{3} t}\sec^{2} t\,dt = \int_{0}^{\frac{\pi}{4}} t\sin t\,dt [2]
I = \left(-t\cos t + \sin t\right)_{0}^{\frac{\pi}{4}} [1]
I = \frac{-\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}} or \frac{-\pi + 4}{4\sqrt{2}} [1]
Evaluate :
Let I = \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}}\,dx
Put \sin x - \cos = t \Rightarrow (\sin x + \cos x)\,dx = dt [1]
when x = \frac{\pi}{6}, t = \frac{1-\sqrt{3}}{2} and when x = \frac{\pi}{3}, t = \frac{\sqrt{3}-1}{2} [1]
Now \cos x - \sin x = t on squaring gives \sin 2x = 1 - t^2
\therefore I = \int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^2}} [1]
= \left[\sin^{-1} t\right]_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} [1]
= 2\sin^{-1}\left(\frac{\sqrt{3}-1}{2}\right) [1]
Find : \int \frac{x^2 + x + 1}{(x + 2)(x^2 + 1)}\,dx.
\frac{x^2 + x + 1}{(x + 2)(x^2 + 1)} = \frac{A}{(x + 2)} + \frac{Bx + c}{x^2 + 1} [2]
Getting A = \frac{3}{5}, B = \frac{2}{5}, C = \frac{1}{5} [1½]
Given integral = \frac{3}{5}\int\frac{1}{x+2}dx + \frac{1}{5}\int\frac{2x}{x^2+1}dx + \frac{1}{5}\int\frac{1}{x^2+1}dx
= \frac{3}{5}\log|x + 2| + \frac{1}{5}\log(x^2 + 1) + \frac{1}{5}\tan^{-1}x + C [1½]
Find :
I=\int\frac{\cos x}{(4+\sin^2x)(5-4\cos^2x)}dx
=\int\frac{\cos x}{(4+\sin^2x)(1+4\sin^2x)}dx [1]
\sin x=t gives [½]
I=\int\frac{dt}{(4+t^2)(1+4t^2)}
=-\frac{1}{15}\int\frac{dt}{4+t^2}+\frac{4}{15}\int\frac{dt}{1+4t^2} (\because using Partial Fraction) [2]
=-\frac{1}{30}\tan^{-1}\left(\frac{t}{2}\right)+\frac{2}{15}\tan^{-1}(2t)+C [1]
=-\frac{1}{30}\tan^{-1}\left(\frac{\sin x}{2}\right)+\frac{2}{15}\tan^{-1}(2\sin x)+C [½]
Evaluate :
I=\int_{0}^{\pi}\frac{dx}{a^2\cos^2x+b^2\sin^2x}
=2\int_{0}^{\pi/2}\frac{dx}{a^2\cos^2x+b^2\sin^2x} [½]
=2\int_{0}^{\pi/2}\frac{\sec^2x}{a^2+b^2\tan^2x}dx [1]
\tan x=t gives [½]
I=2\int_{0}^{\infty}\frac{dt}{a^2+b^2t^2} [½]
=\frac{2}{b^2}.\frac{b}{a}\tan^{-1}\left(\frac{bt}{a}\right)\Big]_0^{\infty} [1½]
=\frac{\pi}{ab} [1]
Find :
\frac{x^2+1}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3} = \frac{3/8}{x-1} + \frac{1/2}{(x-1)^2} + \frac{5/8}{x+3} [1 + 2]
I = \frac{3}{8}\log|x - 1| - \frac{1}{2(x-1)} + \frac{5}{8}\log|x + 3| + C [2]
Evaluate :
Let I = \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x}\, dx
I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} - x}{\cos x + \sin x}\, dx using property [1]
2I = \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2}}{\sin x + \cos x}\, dx [1]
= \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin\left(\frac{\pi}{4} + x\right)}\, dx [1]
2I = \frac{\pi}{2\sqrt{2}} \log\left|\operatorname{cosec}\left(\frac{\pi}{4} + x\right) - \cot\left(\frac{\pi}{4} + x\right)\right|_{0}^{\frac{\pi}{2}} [1]
I = \frac{\pi}{4\sqrt{2}} \log\frac{\sqrt{2}+1}{\sqrt{2}-1} [1]
Find :
\int\frac{x^2+1}{\left(x^2+2\right)\left(2x^2+1\right)}dx=\frac13\int\frac{1}{x^2+2}dx+\frac13\int\frac{1}{2x^2+1}dx (Using Partial Fractions) [2½]
=\frac13\cdot\frac{1}{\sqrt2}\tan^{-1}\left(\frac{x}{\sqrt2}\right)+\frac13\cdot\frac{1}{\sqrt2}\tan^{-1}\left(\sqrt2x\right)+C [2½]
or =\frac{1}{3\sqrt2}\left(\tan^{-1}\frac{x}{\sqrt2}+\tan^{-1}\sqrt2x\right)+C
Evaluate :
Let I=\int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}dx -- (i)
\Rightarrow I=\int_{0}^{\pi}\frac{(\pi-x)\tan x}{\sec x+\tan x}dx -- (ii) [1½]
Adding (i) & (ii), we get
2I=\pi\int_{0}^{\pi}\frac{\tan x}{\sec x+\tan x}dx\Rightarrow2I=\pi\int_{0}^{\pi}\frac{\tan x(\sec x-\tan x)}{\sec^2x-\tan^2x}dx [1]
=\pi\int_{0}^{\pi}\left(\sec x\tan x-\sec^2x+1\right)dx [1]
=\pi\left.\left(\sec x-\tan x+x\right)\right]_0^{\pi} [1]
=\pi(-1+\pi-1)=\pi(\pi-2)
\therefore I=\frac{\pi}{2}(\pi-2) or \pi\left(\frac{\pi}{2}-1\right) [½]
Evaluate :
Let I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 16\sin 2x}\, dx
Put \sin x - \cos x = t, so that (\cos x + \sin x)\, dx = dt [1]
\sin^2 x + \cos^2 x - \sin 2x = t^2 \Rightarrow \sin 2x = 1 - t^2 [½]
I = \int_{-1}^{0} \frac{dt}{25 - 16t^2} [1]
= \frac{1}{16}\int_{-1}^{0} \frac{dt}{\left(\frac{5}{4}\right)^2 - t^2}
= \frac{1}{40}\left[\log\left|\frac{5 + 4t}{5 - 4t}\right|\right]_{-1}^{0} [1½]
= \frac{1}{40}\left[\log 1 - \log\left(\frac{1}{9}\right)\right] = \frac{1}{40}\log 9\ \text{or}\ \frac{1}{20}\log 3 [1]
Evaluate :
Let I = \int_{0}^{\frac{\pi}{2}} \sin 2x\,\tan^{-1}(\sin x)\, dx
Put \sin x = t so that \cos x\, dx = dt [1]
I = 2\int_{0}^{1} t\,\tan^{-1}t\, dt [1]
= 2\left[\tan^{-1}t\left(\frac{t^2}{2}\right) - \frac{1}{2}\int \frac{t^2}{1 + t^2}\, dt\right]_{0}^{1} [1½]
= 2\left[\left(\frac{t^2}{2}\right)\tan^{-1}t - \frac{1}{2}t + \frac{1}{2}\tan^{-1}t\right]_{0}^{1} [1]
= 2\left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{\pi}{2} - 1 [½]
Evaluate : \int_{0}^{\frac{\pi}{2}}\sin 2x\tan^{-1}(\sin x)\,dx
Let I=\int_{0}^{\pi/2}\sin 2x\tan^{-1}(\sin x)\,dx.
=\int_{0}^{\pi/2}2\sin x\cos x\tan^{-1}(\sin x)\,dx [½]
Put \sin x=t so that \cos x\,dx=dt [½]
Thus, I=2\int_{0}^{1}t\tan^{-1}t\,dt [½]
=2\left[\left|\frac{t^2}{2}\tan^{-1}t\right|_{0}^{1}-\int_{0}^{1}\frac{1}{1+t^2}\cdot\frac{t^2}{2}\,dt\right] [1]
=2\cdot\frac{1}{2}\cdot\frac{\pi}{4}-\int_{0}^{1}\frac{t^2}{1+t^2}\,dt
=\frac{\pi}{4}-\int_{0}^{1}\left[1-\frac{1}{1+t^2}\right]dt [1]
=\frac{\pi}{4}-|t|_{0}^{1}+\left|\tan^{-1}t\right|_{0}^{1} [1]
=\frac{\pi}{4}-1+\frac{\pi}{4}
=\frac{\pi}{2}-1 [½]
The area bounded by the curve y = x|x|, x-axis and the ordinates x = -1 and x = 1 is given by
(C) \frac{2}{3}
Which of the following expressions will give the area of region bounded by the curve y = x^2 and line y = 16 ?
(D) 2\int_{0}^{16} \sqrt{y}\,dy
The area of the shaded region of the circle given below is equal to :
(B) 2\int_{1}^{3}\sqrt{9 - y^2}\, dy
The area of the region bounded by the curve y = x and x-axis, between x = 0 and x = 2 is :
(A) 2 sq. units [1]
An ant is observed crawling on a sheet of paper along a straight line given by equation y = 2x − 4. Area of the surface covered by the ant bounded by y-axis, x-axis and x = 1 is :
(B) 3 sq. units
The area of the shaded region bounded by the curves y^2=x, x=4 and the x-axis is given by
(D) \int_{0}^{4}\sqrt{x}\,dx
The area of the shaded region (figure) represented by the curves y = x^2,\ 0 \le x \le 2 and y-axis is given by
(D) \int_{0}^{4} \sqrt{y}\,dy
The area of the region enclosed by the curve y=\sqrt{x} and the lines x = 0 and x = 4 and x-axis is :
(C) \frac{16}{3} sq. units
The area of the region enclosed between the curve y=x|x|, x-axis, x=-2 and x=2 is :
(B) \frac{16}{3}
The area of the region bounded by the curve y^2 = x between x = 0 and x = 1 is :
(D) \frac{4}{3} sq units
The area bounded by the parabola x^2 = y and the line y = 1 is :
(C) \frac{4}{3} sq units
Area of the region bounded by curve y^2=4x and the X-axis between x=0 and x=1 is :
(D) \frac{4}{3} [1]
The area (in sq. units) of the region bounded by the curve y=x, x-axis, x=0 and x=2 is :
(C) 2
[1]Find out the area of shaded region in the enclosed figure.
For Visually Impaired: Find out the area of the region enclosed by the curve y^2=x, x=3 and x-axis in the first quadrant.
A=\int_0^4 x\,dy=\int_0^4\sqrt{y}\,dy [1]
=\frac23\times y^{3/2}\Big]_{y=0}^{y=4}=\frac{16}{3} sq. units [1]
For Visually Impaired:
A=\int_0^3 y\,dx=\int_0^3\sqrt{x}\,dx [1]
=\frac23\times x^{3/2}\Big]_{x=0}^{x=3}=2\sqrt3 sq. units [1]
Calculate the area of the region bounded by the curve \frac{x^2}{9}+\frac{y^2}{4}=1 and the x-axis using integration.
(½ for correct figure) [½]
A=2\times\frac{2}{3}\int_{0}^{3}\sqrt{9-x^2}\,dx [½]
=\frac{4}{3}\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)\right]_0^3 [½]
=\frac{4}{3}\left[\left(0+\frac{9}{2}\sin^{-1}1\right)-0\right]
=3\pi [½]
Sketch the region bounded by the lines 2x+y=8, y = 2, y = 4 and the y-axis. Hence, obtain its area using integration.
[Figure: region bounded by 2x+y=8, y = 2, y = 4 and the y-axis] [½ for correct figure]
Required area =\int_{2}^{4}\frac{1}{2}(8-y)\,dy [½]
=\frac{1}{2}\left|8y-\frac{y^2}{2}\right|_{2}^{4} [½]
=5 [½]
Sketch the graph y=|x+1|. Evaluate \int_{-4}^{2}|x+1|dx. What does the value of this integral represent on the graph?
For Visually Impaired: Define the function y=|x+1|. Evaluate \int_{-4}^{2}|x+1|dx. What does the value of this integral represent?
[Graph of y=|x+1| with shaded region between x=-4 and x=2 — see figure] [1]
\int_{-4}^{2}|x+1|dx=\int_{-4}^{-1}(-x-1)\,dx+\int_{-1}^{2}(x+1)\,dx [½]
=-\frac{(x+1)^2}{2}\Big]_{-4}^{-1}+\frac{(x+1)^2}{2}\Big]_{-1}^{2} [½]
=-\left(0-\frac92\right)+\left(\frac92-0\right)=9 [½]
It represent the area of shaded region bounded by the curve y=|x+1|, x-axis and the lines x=-4\ and\ x=2 [½]
For Visually Impaired:
y=|x+1|=f(x)=\begin{cases}-x-1, & x\lt-1\\ x+1, & x\ge-1\end{cases} [1]
\int_{-4}^{2}|x+1|dx=\int_{-4}^{-1}(-x-1)\,dx+\int_{-1}^{2}(x+1)\,dx
=-\frac{(x+1)^2}{2}\Big]_{-4}^{-1}+\frac{(x+1)^2}{2}\Big]_{-1}^{2} [1]
=-\left(0-\frac92\right)+\left(\frac92-0\right)=9
It represent the area of shaded region bounded by the curve y=|x+1|, x-axis and the lines x=-4\ and\ x=2 [1]
Using integration find the area of the region \{(x,y):x^2-4y\le0,y-x\le0\}
For Visually Impaired: Using integration find the area enclosed within the curve: 25x^2+16y^2=400
[Graph of the region — see figure] [1]
Required Area =\int_0^4 x\,dx-\int_0^4\frac{x^2}{4}dx [1]
=\frac{x^2}{2}\Big]_0^4-\frac{1}{12}[x^3]_0^4 [½]
=\frac12(16-0)-\frac{1}{12}(64-0)=8-\frac{16}{3}=\frac83 sq. units [½]
For Visually Impaired:
25x^2+16y^2=400\Rightarrow\frac{x^2}{16}+\frac{y^2}{25}=1\Rightarrow\frac{x^2}{4^2}+\frac{y^2}{5^2}=1\Rightarrow y=\frac54\sqrt{4^2-x^2}
Required Area =4\int_0^4\frac54\sqrt{4^2-x^2}\,dx [1]
=5\left[\frac{x\sqrt{4^2-x^2}}{2}+\frac{4^2}{2}\sin^{-1}\left(\frac{x}{4}\right)\right]_0^4 [1]
=5[0+8\sin^{-1}(1)-0]
=40\times\frac{\pi}{2}=20\pi sq. units [1]
Sketch the graph of y=|x+3| and find the area of the region enclosed by the curve, x-axis, between x=-6 and x=0, using integration.
[Graph] [1] (for correct graph)
Required Area =\int_{-6}^{0}y\,dx [½]
=2\int_{-3}^{0}(x+3)dx [½]
=2\left[\frac{(x+3)^2}{2}\right]_{-3}^{0} [½]
=9 [½]
Find the area of the following region using integration :
\{(x,y):y^2\le2x\ \text{and}\ y\ge x-4\}
[Figure: region between y^2=2x and y=x-4, meeting at (2, –2) and (8, 4)] [1 mark for correct figure]
Solving y^2=2x and y=x-4, we get
y = 4 or – 2 [½]
Required area =\int_{-2}^{4}\left[(y+4)-\frac{y^2}{2}\right]dy [1]
=\left|\frac{y^2}{2}+4y-\frac{1}{6}y^3\right|_{-2}^{4}
= 18 [½]
Find the area of the region \{(x,y):x^2\le y\le x+2\}, using integration.
(Figure: region between x^2=y and y=x+2) (for correct figure) [1]
x-coordinates of points of intersection are -1,\ 2. [½]
Required area =\int_{-1}^{2}[(x+2)-x^2]dx [½]
Find the area of the region \{(x, y) : x^2+y^2 \le 9,\ x+y \ge 3\} , using integration.
Point of intersection (3, 0) and (0, 3)
Correct figure [½]
Required Area
= \int_{0}^{3}\sqrt{9-x^2}\,dx-\int_{0}^{3}(3-x)dx [1]
= \left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)\right]_{0}^{3}-\left[\frac{(3-x)^2}{-2}\right]_{0}^{3} [1]
= \frac{9}{2}\sin^{-1}1-\frac{9}{2} = \frac{9}{2}\left(\frac{\pi}{2}-1\right) [½]
Using integration, find the area of the region bounded by the parabola y^2 = 4x , the lines x = 0 and x = 3 and the x-axis.
Required Area
Correct Figure [1]
= \int_{0}^{3} 2\sqrt{x}\,dx [1]
= 2\times\frac{2}{3}[x^{3/2}]_{0}^{3} [½]
= \frac{4}{3}\times 3^{3/2} = 4\sqrt{3} [½]
Roundabouts are often made on busy roads to ease the traffic and avoid red lights.
One such round-about is made such that equation representing its boundary is given by C_1 ; x^2 + y^2 = 64.
There is a circular pond with a fountain in the middle of the roundabout whose equation is given by C_2 : x^2 + y^2 = 4.
Based on the given information, answer the following questions :
(i) Represent the given equations C_1 and C_2 with the help of a diagram. [1]
(ii) Express y as a function of x, (y = f(x)), for both C_1 an C_2. [1]
(iii) (a) Using integration find the area of region covered by the roundabout. [2]
OR
(iii) (b) Using integration, find the area of region covered by circular pond. [2]
(i) [Diagram showing circles C_1 (radius 8) and C_2 (radius 2) centred at the origin O] [1]
(ii) Circle C_1 : y = \sqrt{64 - x^2} or y = -\sqrt{64 - x^2} [½]
Circle C_2 : y = \sqrt{4 - x^2} or y = -\sqrt{4 - x^2} [½]
(iii) (a) Required area = 4 \times \int_0^8 \sqrt{64 - x^2}\,dx [1]
= 2\left[x\sqrt{64 - x^2} + 64\sin^{-1}\frac{x}{8}\right]_0^8 = 64\pi [1]
OR
(iii) (b) Required area = 4 \times \int_0^2 \sqrt{4 - x^2}\,dx [1]
= 2\left[x\sqrt{4 - x^2} + 4\sin^{-1}\frac{x}{2}\right]_0^2 = 4\pi [1]
A racing track is build around an elliptical ground whose equation is given by 9x^2 + 16y^2 = 144. The width of the track is 3 m as shown below :
Based on given information, answer the following questions :
(i) Express y as a function of x from the given equation of ellipse. [1]
(ii) Integrate the function obtained in (i) with respect to x. [1]
(iii) (a) Find the area of the region enclosed within the elliptical ground excluding the track using integration. [2]
OR
(iii) (b) Write the co-ordinates of the points P and Q where the outer edge of the track cuts x axis and y axis in first quadrant and find the area of the triangle formed by points P, O, Q using integration. [2]
(i) 9x^2 + 16y^2 = 144 \Rightarrow y = \frac{3}{4}\sqrt{16 - x^2} [1]
(ii) \int y\,dx = \int \frac{3}{4}\sqrt{16 - x^2}\,dx
= \frac{3}{4}\left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right] + C [1]
(iii)(a) Required Area = 4 \times \int_0^4 \frac{3}{4}\sqrt{16 - x^2}\,dx [1]
= 3\left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right]_0^4
= 3(8\sin^{-1}1 - 0) = 24 \times \frac{\pi}{2} = 12\pi [1]
OR
(iii)(b) The track has a width of 3 m.
Clearly P(7, 0) and Q(0, 6) [1]
The equation of line passing through P(7, 0) and Q(0, 6) is \frac{x}{7} + \frac{y}{6} = 1
Here, y = \frac{6}{7}(7 - x)
Required area = \int_0^7 y\,dx = \int_0^7 \frac{6}{7}(7 - x)\,dx [½]
= \frac{6}{7}\left[\frac{(7 - x)^2}{-2}\right]_0^7
= 21 [½]
Case Study – 3
There is a triangular park in the society. The park is divided into two sections as shown in the figure.
In the region OAC, children are allowed to play games like cricket, football, while in the region AOB, activities which involve running are not allowed.
The vertices of the triangular park ABC are A(0, 4), B(– 2, 0) and C(3, 0).
Based on the above information, answer the following questions :
(i) Write the equation of the boundary line AB of the park. [1]
(ii) Write the equation of the boundary line AC of the park. [1]
(iii) (a) Using integration, find the area of region OAC, in which children are allowed to play cricket, football. [2]
OR
(iii) (b) Using integration, find the area of region AOB. [2]
(i) Equation of the boundary line AB is 2x - y + 4 = 0 [1]
(ii) Equation of the boundary line AC is 4x + 3y = 12 [1]
(iii) (a) Area of region OAC = \int_{0}^{3}\frac{12 - 4x}{3}dx [1]
= \frac{1}{3}\left[12x - 2x^2\right]_0^3 = 6 [1]
OR
(iii) (b) Area of region AOB = \int_{-2}^{0}(2x + 4)\, dx [1]
= \left[x^2 + 4x\right]_{-2}^{0} = 4 [1]
Using integration, find the area of the region \{(x, y) : 4x^2+9y^2\le 36,\ 2x+3y\ge 6\} .
Correct figure [1]
Clearly points of intersection are (3, 0) and (0, 2) [½]
Required area =\frac{2}{3}\int_0^3\sqrt{9-x^2}\,dx-\frac{2}{3}\int_0^3(3-x)\,dx [1]
=\frac{2}{3}\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\frac{x}{3}+\frac{(3-x)^2}{2}\right]_0^3 [1]
=\frac{2}{3}\left[\frac{9\pi}{4}-\frac{9}{2}\right] or \frac{3\pi}{2}-3 [½]
Using integration, find the area of the region bounded by lines x-y+1=0 , x=-2 , x=3 and x-axis.
Correct figure [1]
Required area =-\int_{-2}^{-1}(x+1)\,dx+\int_{-1}^{3}(x+1)\,dx [1]
=-\frac{(x+1)^2}{2}\Big|_{-2}^{-1}+\frac{(x+1)^2}{2}\Big|_{-1}^{3} [1]
=\frac{1}{2}+8
=\frac{17}{2} [1]
Using integration, find the area of the smaller region enclosed by the curve 4x^2+4y^2=9 and the line 2x+2y=3.
Clearly point of intersection are
\left(\frac32,0\right)\ \&\ \left(0,\frac32\right) [½]
Correct Figure [1]
Required area =\int_0^{3/2}\sqrt{\frac94-x^2}\,dx-\int_0^{3/2}\left(\frac32-x\right)dx [1]
=\left.\frac{x}{2}\sqrt{\frac94-x^2}+\frac98\sin^{-1}\frac{2x}{3}\right|_0^{3/2}+\left.\frac{\left(\frac32-x\right)^2}{2}\right|_0^{3/2} [1]
=\frac{9\pi}{16}-\frac98 [½]
If the area of the region bounded by the curve y^2=4ax and the line x=4a is \frac{256}{3} sq. units, then using integration, find the value of a, where a > 0.
Given area =\frac{256}{3}
Correct Figure [1]
Area of Shaded region =2\int_0^{4a}\sqrt{4ax}\,dx [1]
=8\sqrt{a}\left.\frac{x^{3/2}}{3}\right|_0^{4a} [1]
=\frac{64a^2}{3} [½]
\frac{64a^2}{3}=\frac{256}{3}
\Rightarrow a^2=4 gives a=2 (as a\gt 0) [½]
Find the area of the region bounded by curve 4x^2=y and the line y=8x+12, using integration.
[Figure] Correct Figure [1]
Point of intersection
4x^2=8x+12
x^2-2x-3=0
(x-3)(x+1)=0
x=3,\ -1 [½]
Area =\int_{-1}^{3}[(8x+12)-4x^2]dx [1]
=\left.4x^2+12x-\frac43x^3\right]_{-1}^{3} [1]
=36+36-36-\left(4-12+\frac43\right)
=44-\frac43=\frac{128}{3} [½]
Find the area bounded by the curves y=|x-1| and y=1, using integration.
(Figure: region between y=1-x, y=x-1 and y=1, with vertices (0,1), (1,0), (2,1)) (for correct figure) [1]
Area of the bounded region is
\int_{0}^{1}\left[1-(1-x)\right]dx+\int_{1}^{2}\left[1-(x-1)\right]dx [1]
\int_{0}^{1}x\,dx+\int_{1}^{2}(2-x)\,dx
=\left[\frac{x^2}{2}\right]_{0}^{1}+\left[2x-\frac{x^2}{2}\right]_{1}^{2} [1]
=\frac{1}{2}+\left[4-\frac{4}{2}-2+\frac{1}{2}\right]
=-1+2=1 [1]
Using Integration, find the area of the region \left\{(x,y):0\le y\le\sqrt{4-x^2}, x\ge2-y\right\}.
For Visually Impaired: Using Integration, find the area of the region enclosed within 25x^2+16y^2=400
[Figure: semicircle y=\sqrt{4-x^2} and line x+y=2; region between them shaded] [1] (1 mark for correct figure)
y=\sqrt{4-x^2} is a semi-circle with radius 2 units and centre (0,0)
x=2-y which is x + y = 2 is a line with intercepts 2 on each axis.
Curves meet at (0,2) and (2,0) [1]
Area = \int_0^2\left(\sqrt{4-x^2}-(2-x)\right)dx [1]
=\left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)-2x+\frac{x^2}{2}\right]_0^2 [1]
=\pi-4+2=(\pi-2) sq. units [1]
For Visually Impaired:
25x^2+16y^2=400 is an Ellipse with the standard equation as \frac{x^2}{16}+\frac{y^2}{25}=1 [½]
\Rightarrow y=\frac54\sqrt{4^2-x^2} [1]
Area = 4\int_0^4\left(\frac54\sqrt{4^2-x^2}\right)dx
=5\int_0^4\left(\sqrt{4^2-x^2}\right)dx [1]
=5\left[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)\right]_0^4 [1½]
=5[4(\pi)]=20\pi sq. units [1]
Using Integration, find the area of the region bounded by the curve y^2=2x and the line x-2y=0.
For Visually Impaired: Using Integration find the area of the region in the first quadrant bounded by the curve y^2=8x, and the lines x=2, x=5 and x-axis.
[Figure: parabola y^2=2x and line x=2y meeting at (0,0) and (8,4); region between them shaded] [1] (1 mark for correct figure)
y^2=2x represents a parabola, whose points of intersection with the given line x=2y are (0,0), (8,4) [1]
Area = \int_0^8\left(\sqrt{2x}-\frac{x}{2}\right)dx [1]
=\left[\sqrt2\,\frac23x^{\frac32}-\frac{x^2}{4}\right]_0^8 [1]
=\frac{64}{3}-16=\frac{16}{3} sq. units [1]
For Visually Impaired:
y^2=8x represents a parabola, and the area of the region above the x-axis, below the parabola, between the lines x=2 and x=5 is given by [½]
Area = \int_2^5\left(\sqrt{8x}\right)dx [2]
=\left[2\sqrt2\,\frac23x^{\frac32}\right]_2^5 [1½]
=\sqrt2\left(\frac43\right)\left[5\sqrt5-2\sqrt2\right]=\frac43\left[5\sqrt{10}-4\right] sq. units [1]
Using integration, find the area of the region enclosed by the curve y = |x - 6|, the x-axis, and between x = 4 and x = 8.
[Graph of y = 6 - x and y = x - 6 with the region between x = 4 and x = 8 shaded] [1] (For correct graph)
Required Area = \int_{4}^{8} y\,dx = \int_{4}^{8} |x - 6|\,dx [1]
Area = -\int_{4}^{6}(x - 6)\,dx + \int_{6}^{8}(x - 6)\,dx [1]
Area = -\left(\frac{(x - 6)^{2}}{2}\right)_{4}^{6} + \left(\frac{(x - 6)^{2}}{2}\right)_{6}^{8} = 2 + 2 = 4 or 4 sq. units [2]
Sketch the curve \{(x, y) : 100x^2 + 25y^2 = 2500\} and find the area of the region enclosed by it, using integration.
[Sketch: ellipse \frac{x^2}{25} + \frac{y^2}{100} = 1 through (\pm 5, 0) and (0, \pm 10)] [1½]
Required area = 4\int_{0}^{5} \frac{10}{5}\sqrt{25 - x^2}\,dx [1½]
= 8\left[\frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\frac{x}{5}\right]_0^5 [1]
= 50\pi [1]
Using integration, find the area bounded by the curve y = x^2 and the line y = 2.
[Graph: parabola y = x^2 and line y = 2 meeting at A and B; region between them shaded] [1 (For correct graph)]
Required Area = 2\int_0^2 x\,dy = 2\int_0^2 \sqrt{y}\,dy [2]
= 2\left(\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right)_0^2 [1]
= \frac{4}{3}\left(2\sqrt{2}\right) = \frac{8\sqrt{2}}{3} [1]
Using integration, find the area bounded by the curve 9x^2 + 16y^2 = 144 and the lines x = 2, x = -2.
[Graph: ellipse 9x^2 + 16y^2 = 144 with lines x = -2 and x = 2; region between the lines inside the ellipse shaded] [1 (For correct graph)]
Required Area = 4\int_0^2 y\,dx = 4 \times \frac{3}{4}\int_0^2 \sqrt{4^2 - x^2}\,dx [2]
= 3 \times \left(\frac{x\sqrt{4^2 - x^2}}{2} + 8\sin^{-1}\frac{x}{4}\right)_0^2 [1]
= 3\left(2\sqrt{3} + \frac{8 \times \pi}{6}\right) = \left(6\sqrt{3} + 4\pi\right) [1]
Using integration, find the area of the region bounded by the line y = 5x + 2, the x – axis and the ordinates x = -2 and x = 2.
[Graph of y = 5x + 2 with the region between x = −2 and x = 2 shaded]
Correct sketch and shading [2]
The required area
= \left|\int_{-2}^{-\frac{2}{5}}(5x + 2)dx\right| + \int_{-\frac{2}{5}}^{2}(5x + 2)dx [1]
= \left|\left[\frac{(5x + 2)^2}{10}\right]_{-2}^{-\frac{2}{5}}\right| + \left[\frac{(5x + 2)^2}{10}\right]_{-\frac{2}{5}}^{2} [1]
= \frac{64}{10} + \frac{144}{10} = \frac{104}{5} [1]
Sketch a graph of y=x^2. Using integration, find the area of the region bounded by y=9, x=0 and y=x^2.
[Graph of y=x^2 and y=9, with the region between the curve, the y-axis and y=9 in the first quadrant shaded] [1½ for correct figure and shading]
Required area =\int_0^9\sqrt{y}\,dy [1½]
=\frac23\left[y^{3/2}\right]_0^9 [1]
=18 [1]
Note: If area is found in second quadrant, may be considered.
Draw a rough sketch for the curve y = 2 + |x + 1|. Using integration, find the area of the region bounded by the curve y = 2 + |x + 1|, x = -4, x = 3 and y = 0.
Correct Graph: [1]
Required area = \int_{-4}^{-1} (2 + |x + 1|)\, dx + \int_{-1}^{3} (2 + |x + 1|)\, dx [1]
= \int_{-4}^{-1} (1 - x)\, dx + \int_{-1}^{3} (3 + x)\, dx [1]
= -\frac{(1-x)^2}{2}\Big|_{-4}^{-1} + \frac{(3+x)^2}{2}\Big|_{-1}^{3} [1]
= \frac{21}{2} + 16 = \frac{53}{2} [1]
A woman discovered a scratch along a straight line on a circular table top of radius 8 cm. She divided the table top into 4 equal quadrants and discovered the scratch passing through the origin inclined at an angle \frac{\pi}{4} anticlockwise along the positive direction of x-axis. Find the area of the region enclosed by the x-axis, the scratch and the circular table top in the first quadrant, using integration.
Correct graph [1]
Equation of the circular tabletop: x^2+y^2=64 [½]
Equation of line (scratch): x=y [½]
The line and circle intersect at x=4\sqrt2 [½]
Area of the shaded region
=\int_{0}^{4\sqrt2}x\,dx+\int_{4\sqrt2}^{8}\sqrt{64-x^2}\,dx [1]
=\left.\frac{x^2}{2}\right]_0^{4\sqrt2}+\left[\frac{x}{2}\sqrt{64-x^2}+32\sin^{-1}\frac{x}{8}\right]_{4\sqrt2}^{8} [1]
=\frac{32}{2}+32\sin^{-1}1-2\sqrt2\cdot4\sqrt2-32\sin^{-1}\frac{1}{\sqrt2}
=16+16\pi-16-8\pi=8\pi\text{ cm}^2 [½]
Using integration, find the area of the region
[Figure: region under y = x^2 from 0 to 1 and under y = x from 1 to 3, shaded] [1] (1 mark for correct figure)
Required Area
= \int_0^1 x^2\,dx + \int_1^3 x\,dx [1+1]
= \left.\frac{x^3}{3}\right]_0^1 + \left.\frac{x^2}{2}\right]_1^3 [1]
= \frac{1}{3} + 4 = \frac{13}{3} [1]
Using integration, find the area of the ellipse \frac{x^2}{16} + \frac{y^2}{4} = 1, included between the lines x = -2 and x = 2.
[Figure: ellipse with the region between x = -2 and x = 2 shaded] Correct graph - [1]
Area = 4\int_{0}^{2} y\, dx
= 4\left[\frac{1}{2}\int_{0}^{2}\sqrt{4^2 - x^2}\, dx\right] [1]
= 2\left[\frac{x}{2}\sqrt{4^2 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right]_{0}^{2} [2]
= 2\left[\sqrt{12} + \frac{8\pi}{6}\right] = 4\sqrt{3} + \frac{8\pi}{3} [1]
If A_1 denotes the area of region bounded by y^2 = 4x, x = 1 and x-axis in the first quadrant and A_2 denotes the area of region bounded by y^2 = 4x, x = 4, find A_1 : A_2.
[Figure] [2]
A_1 = \text{Area (region OABO)} = \int_{0}^{1} 2\sqrt{x}\,dx = 2\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1} = \frac{4}{3} [1]
A_2 = \text{Area (region ODEO)} = 2\int_{0}^{4} 2\sqrt{x}\,dx = 4 \times \frac{2}{3}[2^3] = \frac{64}{3} [1]
A_1 : A_2 = \frac{4}{3} : \frac{64}{3} = 1 : 16 [1]
Using integration, find the area of the region enclosed between the circle x^2 + y^2 = 16 and the lines x = -2 and x = 2.
[Figure: circle x^2 + y^2 = 16 with the region between the lines x = -2 and x = 2 shaded]
For correct figure [1]
Required area = 4\int_{0}^{2}\sqrt{16 - x^2}\, dx [1]
= 4\left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right]_{0}^{2} [2]
= 8\sqrt{3} + \frac{16\pi}{3} [1]
Sketch the graph of y = x|x| and hence find the area bounded by this curve, X-axis and the ordinates x = -2 and x = 2, using integration.
[Figure] [For correct figure 1½]
As, y = x|x| = \begin{cases} -x^2, & x \lt 0 \\ x^2, & x \ge 0 \end{cases}
Area of the shaded region = \int_{-2}^{2} y\,dx = 2\int_{0}^{2} y\,dx = 2\int_{0}^{2} x^2\,dx [1½]
= 2\left(\frac{x^3}{3}\right)_{0}^{2} [1]
= 2\left(\frac{8}{3}\right) = \frac{16}{3} [1]
Using integration, find the area bounded by the ellipse 9x^2 + 25y^2 = 225, the lines x = -2, x = 2, and the X-axis.
[Figure] [For correct figure 1½]
As, 9x^2 + 25y^2 = 225 \Rightarrow \boxed{y = \pm\frac{3}{5}\sqrt{5^2 - x^2}}
Required Area = \int_{-2}^{2} \frac{3}{5}\sqrt{5^2 - x^2}\,dx = \frac{6}{5}\int_{0}^{2} \sqrt{5^2 - x^2}\,dx [1½]
= \frac{6}{5}\left(\frac{x\sqrt{5^2 - x^2}}{2} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right)\right)_{0}^{2} [1]
= \frac{6}{5}\left(\frac{2\sqrt{21}}{2} + \frac{25}{2}\sin^{-1}\left(\frac{2}{5}\right)\right)
= \left(\frac{6\sqrt{21}}{5} + 15\sin^{-1}\left(\frac{2}{5}\right)\right) [1]
Find the area of the region bounded by the curve 4x^2+y^2=36 using integration.
The given equation can be written as: \frac{x^2}{9}+\frac{y^2}{36}=1, which is an ellipse.
Correct Graph [1]
Area of the region bounded by the curve
=4\times\frac63\int_0^3\sqrt{9-x^2}\,dx [1½]
=8\left[\frac{x}{2}\sqrt{9-x^2}+\frac92\sin^{-1}\frac{x}{3}\right]_0^3 [1½]
=18\pi [1]
Using integration, find the area of the region bounded by the curve y=\sqrt{4-x^{2}}, the lines x=-\sqrt{2} and x=\sqrt{3} and the x-axis.
Correct Graph [1]
Area of the region bounded by the curve
=\int_{-\sqrt{2}}^{\sqrt{3}}\sqrt{4-x^{2}}\,dx [1½]
=\left[\frac{x}{2}\sqrt{4-x^{2}}+2\sin^{-1}\frac{x}{2}\right]_{-\sqrt{2}}^{\sqrt{3}} [1½]
=\frac{\sqrt{3}}{2}+2\cdot\frac{\pi}{3}+1+2\cdot\frac{\pi}{4}=\frac{\sqrt{3}}{2}+1+\frac{7\pi}{6} [1]
Using integration, evaluate the area of the region bounded by the curve y=x^{2}, the lines y=1 and y=3 and the y-axis.
Correct Graph [1]
Area of the region bounded by the curve
=\int_{1}^{3}\sqrt{y}\,dy [2]
=\left[\frac{2y^{3/2}}{3}\right]_{1}^{3}=\frac{2}{3}(3\sqrt{3}-1) [2]
Find the area of the region bounded by the curves x^2=y,\ y=x+2 and x-axis, using integration.
Correct Graph : [1½]
x coordinates of point of intersection are – 1, 2 [½]
Required area =\int_{-2}^{-1}(x+2)\,dx+\int_{-1}^{0}x^2\,dx [1½]
=\frac{(x+2)^2}{2}\Big|_{-2}^{-1}+\frac{x^3}{3}\Big|_{-1}^{0} [1]
=\frac12+\frac13=\frac56 [½]
Using integration, find the area of region bounded by line y = \sqrt{3}x, the curve y = \sqrt{4 - x^2} and y-axis in first quadrant.
Correct Figure [1]
Shaded region [½]
Point of intersection at x = 1 [½]
ar(OAB) = \int_{0}^{1}\sqrt{4 - x^2}\,dx - \int_{0}^{1}\sqrt{3}x\,dx [1½]
= \left(\frac{x\sqrt{4 - x^2}}{2} + 2\sin^{-1}\left(\frac{x}{2}\right)\right)\Big]_{0}^{1} - \frac{\sqrt{3}}{2}x^2\Big]_{0}^{1} [1]
= \frac{\sqrt{3}}{2} + 2\times\frac{\pi}{6} - \frac{\sqrt{3}}{2} = \frac{\pi}{3} [½]
The area of the region bounded by the line y = mx (m \gt 0), the curve x^2 + y^2 = 4 and the x-axis in the first quadrant is \frac{\pi}{2} units. Using integration, find the value of m.
Correct figure [1]
x^2 + y^2 = 4 and y = mx
\Rightarrow x^2 + m^2x^2 = 4 \Rightarrow x = \frac{2}{\sqrt{1 + m^2}}
x- coordinate of the required point of intersection is \frac{2}{\sqrt{1 + m^2}}. [1]
According to question,
\int_{0}^{\frac{2}{\sqrt{1 + m^2}}} mx\, dx + \int_{\frac{2}{\sqrt{1 + m^2}}}^{2} \sqrt{4 - x^2}\, dx = \frac{\pi}{2} [1+1]
\Rightarrow m\frac{x^2}{2}\Big|_{0}^{\frac{2}{\sqrt{1 + m^2}}} + \left.\frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\frac{x}{2}\right|_{\frac{2}{\sqrt{1 + m^2}}}^{2} = \frac{\pi}{2} [½]
\Rightarrow \frac{2m}{1 + m^2} + \pi - \frac{2m}{1 + m^2} - 2\sin^{-1}\frac{1}{\sqrt{1 + m^2}} = \frac{\pi}{2}
\Rightarrow \frac{\pi}{4} = \sin^{-1}\frac{1}{\sqrt{1 + m^2}}
\Rightarrow \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{1 + m^2}} \Rightarrow m^2 + 1 = 2
\Rightarrow m = 1 (as m \gt 0) [½]
Using integration, find the area of the region bounded by the parabola y^2=4ax and its latus rectum.
[Figure: parabola y^2=4ax with latus rectum x=a; the region between them is shaded] [1]
Required area =2\int_{0}^{a}2\sqrt{ax}\,dx [1½]
=4\sqrt{a}\int_{0}^{a}\sqrt{x}\,dx [½]
=4\sqrt{a}\left|\frac{2x^{3/2}}{3}\right|_{0}^{a} [1]
=\frac{8}{3}a\sqrt{a}\sqrt{a}
=\frac{8}{3}a^2 [1]
Find the area of the triangle ABC bounded by the lines represented by the equations 5x - 2y - 10 = 0, x - y - 9 = 0 and 3x - 4y - 6 = 0, using integration method.
Official marking scheme for this paper is not available.
The degree and order of the differential equation y-x\left(\frac{dy}{dx}\right)^3=a\left(y+\frac{d^2y}{dx^2}\right)^2 are respectively
(B) 2, 2
The order is 2 and the term containing \frac{d^2y}{dx^2} has the maximum power 2 in the equation, so the degree is 2
Assertion (A): The general solution of a differential equation represents a family of curves.
Reason (R): The derivative of an arbitrary constant is zero.
(A) Both assertion and reason are true, and reason is the correct explanation of the assertion.
The solution of the differential equation xdx+ydy=0 represents a family of
xdx+ydy=0
\Rightarrow \int xdx=-\int ydy
\Rightarrow \frac{x^2}{2}=-\frac{y^2}{2}+k
\Rightarrow x^2+y^2=2k
Solution is x^2+y^2=2k, k being an arbitrary constant.
(C) Circles [1]
The integrating factor of differential equation R\frac{dx}{dy} + Px = Q where P, Q, R are functions of y is
(C) e^{\int \frac{P}{R}dy}
The order and degree of the differential equation \frac{d}{dx}(e^y) = 0 respectively are
(B) 1, 1
The general solution of the differential equation \frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}} is
(C) \sqrt{y} - \sqrt{x} = C
The integrating factor of the differential equation 2x\frac{dy}{dx} - y = 3 is
(B) \frac{1}{\sqrt{x}}
\frac{dy}{dx} = F(x, y) will be a homogeneous differential equation for which of the following functions ?
(i) F(x, y) = 3x + 2y (ii) F(x, y) = \sin\frac{y}{x} + \log y - \log x
(iii) F(x, y) = e^{y/x} + 1 (iv) F(x, y) = \sqrt{x^2 + y^2} - y
(D) (ii) and (iii)
Assertion (A) : One of the particular solutions of the differential equation \frac{dy}{dx} = e^{x+y} can be e^x + e^{-y} = -2.
Reason (R) : e^x + e^{-y} = C is the general solution of the differential equation \frac{dy}{dx} = e^{x+y}.
(D) Assertion (A) is false, but Reason (R) is true.
Product of the order and degree of differential equation 1 + \left(\frac{dy}{dx}\right)^{3} = \lambda \left(\frac{d^{3}y}{dx^{3}}\right)^{2} is :
(B) 6 [1]
Which of the following is not a Linear Differential Equation ?
(C) x\left(1 + y^{2}\right)dx - y\left(1 + x^{2}\right)dy = 0 [1]
The order and degree of the differential equation
is :
(A) Order = 3, Degree = 3
The general solution for the differential equation \frac{dy}{dx} = e^{3x - y} is :
(A) 3e^y = e^{3x} + C
Which of the following is not a homogeneous function of x and y ?
(D) \tan x-\sec y
The integrating factor of differential equation (x+2y^3)\frac{dy}{dx}=2y is
(B) \frac{1}{\sqrt{y}}
If p and q are respectively the order and degree of the differential equation \frac{d}{dx}\left(\frac{dy}{dx}\right)^3 = 0, then (p - q) is
(B) 1
The order and degree of the differential equation \left(\frac{d^2y}{dx^2}\right)^2+\left(\frac{dy}{dx}\right)^2=x\sin\left(\frac{dy}{dx}\right) are :
(C) order 2, degree not defined
The integrating factor of the differential equation \left(\frac{e^{-2\sqrt x}}{\sqrt x}-\frac{y}{\sqrt x}\right)\frac{dx}{dy}=1 is :
(C) e^{2\sqrt x}
The sum of the order and degree of the differential equation \left[1+\left(\frac{dy}{dx}\right)^2\right]^3=\frac{d^2y}{dx^2} is :
(C) 3
The order and degree of the following differential equation are, respectively :
(B) 4, not defined
The solution for the differential equation \log\left(\frac{dy}{dx}\right) = 3x + 4y is :
(D) 3e^{-4y} + 4e^{3x} + 12C = 0
The general solution of the differential equation \frac{dy}{dx} = 2x \cdot e^{x^2 + y} is :
(B) e^{x^2} + e^{-y} = C
If ‘m’ and ‘n’ are the degree and order respectively of the differential equation 1 + \left(\frac{dy}{dx}\right)^3 = \frac{d^2y}{dx^2}, then the value of (m + n) is :
(B) 3
The integrating factor of the differential equation (1 - x^2)\frac{dy}{dx} + xy = ax,\ -1 \lt x \lt 1, is :
(D) \frac{1}{\sqrt{1 - x^2}}
The order and degree of the differential equation \left[1 + \left(\frac{dy}{dx}\right)^2\right]^3 = \frac{d^2y}{dx^2} respectively are :
(C) 2, 1
The differential equation \frac{dy}{dx} = F(x, y) will not be a homogeneous differential equation, if F(x, y) is :
(A) \cos x - \sin\left(\frac{y}{x}\right)
The degree of the differential equation (y'')^2 + (y')^3 = x \sin(y') is :
(D) not defined
x \log x \frac{dy}{dx} + y = 2 \log x is an example of a :
(C) first order linear differential equation.
The general solution of the differential equation x\,dy + y\,dx = 0 is :
(A) xy = c
[1]The integrating factor of the differential equation (x + 2y^2)\frac{dy}{dx} = y \; (y \gt 0) is :
(D) \frac{1}{y}
[1]The order of the differential equation \frac{d^4y}{dx^4}-\sin\left(\frac{d^2y}{dx^2}\right)=5 is :
(A) 4 [1]
The number of arbitrary constants in the general solution of the differential equation \frac{dy}{dx}+y=0 is :
(B) 1
[1]The sum of the order and the degree of the differential equation \frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^3=\sin y is :
(C) 3
The general solution of the differential equation x\,dy-(1+x^2)\,dx=dx is :
(D) y=2\log x+\frac{x^2}{2}+C
Degree of the differential equation \sin x + \cos\left(\frac{dy}{dx}\right) = y^2 is
(C) not defined
The integrating factor of the differential equation
is
(D) \frac{1}{\sqrt{1 - y^2}}
The solution of the differential equation \frac{dx}{x} + \frac{dy}{y} = 0 is :
(C) xy = C
What is the product of the order and degree of the differential equation \frac{d^2y}{dx^2} \sin y + \left(\frac{dy}{dx}\right)^3 \cos y = \sqrt{y} ?
(B) 2
The integrating factor for solving the differential equation x\frac{dy}{dx}-y=2x^2 is :
(d) \frac{1}{x}
The order and degree (if defined) of the differential equation, \left(\frac{d^2y}{dx^2}\right)^2+\left(\frac{dy}{dx}\right)^3=x\sin\left(\frac{dy}{dx}\right) respectively are :
(d) 2, degree not defined
The sum of the order and the degree of the differential equation \frac{d}{dx}\left(\left(\frac{dy}{dx}\right)^3\right) is
Due to error in the question, 1 mark should be awarded to each student who attempted the question [1]
The difference of the order and the degree of the differential equation \left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^3 + x^4 = 0 is :
Official marking scheme for this paper is not available.
The integrating factor of the differential equation (3x^2 + y)\frac{dx}{dy} = x is
Official marking scheme for this paper is not available.
Find the sum of the order and the degree of the differential equation :
Given differential equation can be written as
i.e. x^2+2x\frac{dy}{dx}=1 ; Order = 1, degree = 1 [½ + ½]
Sum of order and degree = 1 + 1 = 2 [1]
Find the product of the order and the degree of the differential equation \left[\frac{d}{dx}(xy^2)\right]\cdot\frac{dy}{dx}+y=0.
Given differential equation can be written as
2xy\left(\frac{dy}{dx}\right)^2+y^2\frac{dy}{dx}+y=0 Order = 1, Degree = 2 [1]
Order \times degree = 1 × 2 = 2 [1]
Find the general solution of the differential equation : e^{dy/dx}=x^2.
e^{dy/dx}=x^2
\frac{dy}{dx}=\log x^2=2\log x [½]
\int dy=2\int \log x\,dx [½]
y=2\left[\log x\times x-\int 1\,dx\right] [½]
y=2[x\log x-x]+C [½]
Find the general solution of the differential equation
Given differential equation can be written as \frac{\sec^2 x}{\tan x}dx+\frac{\sec^2 y}{\tan y}dy=0 [1]
Integrating, \log|\log x|+\log|\tan y|=\log C [½]
\tan x\cdot\tan y=C [½]
Find the general solution of the following differential equation :
\int \frac{dy}{e^{-y}}=\int (x^2+e^x)\,dx [1]
e^y=\frac{x^3}{3}+e^x+C [1]
Find the general solution of the differential equation
x\cos y\,dy = (x\log x+1)e^{x}dx
\Rightarrow \int \cos y\,dy = \int \left(\log x+\frac{1}{x}\right)\cdot e^{x}dx [½]
\Rightarrow \sin y = \log x.e^{x}+C \quad \left(\because \int [f(x)+f'(x)]\cdot e^{x}dx = f(x)\cdot e^{x}+C\right) [1+½]
Find the value of (2a − 3b), if a and b represent respectively the order and the degree of the differential equation
order = 2, degree = 3 [1+½]
\therefore 2a-3b = 4-9 = -5 [½]
Find the particular solution of the differential equation xdy+(y-x^2e^x)dx=0, given that y (1) =0
xdy+(y-x^2e^x)dx=0 is a linear differential equation given by
\frac{dy}{dx}+\frac1x\,y=xe^x [½]
I.F = e^{\int\frac1xdx}=e^{\log x}=x [½]
Solution is y\cdot x=\int x^2e^x\,dx
using by parts
yx=x^2e^x-\int2xe^x\,dx [½]
using by parts again
yx=x^2e^x-2\{xe^x-e^x\}+C [½]
y (1) = 0 \Rightarrow C=-e [½]
Particular solution is
yx=x^2e^x-2\{xe^x-e^x\}-e=e^x\{x^2-2x+2\}-e [½]
Find the particular solution of the differential equation (x^2+y^2)dy=xydx, given that y(0)=1
\frac{dy}{dx}=\frac{xy}{x^2+y^2} is a homogeneous differential equation.
Let y=vx\quad\Rightarrow\frac{dy}{dx}=v+x\frac{dv}{dx}
So, equation reduces to
v+x\frac{dv}{dx}=\frac{v}{1+v^2}\quad\Rightarrow x\frac{dv}{dx}=\frac{v-v-v^3}{1+v^2} [1]
\int\frac{1+v^2}{v^3}dv=-\int\frac{dx}{x}
\int\left(\frac{1}{v^3}+\frac1v\right)dv=-\log|x|+\log C\Rightarrow\frac{-1}{2v^2}+\log|v|=-\log|x|+\log C
\log\left|\frac{y}{C}\right|=\frac{x^2}{2y^2} or y=C\,e^{\frac{x^2}{2y^2}} [1]
C=1 when x=0 and y=1.
The particular solution is y=e^{\frac{x^2}{2y^2}} [1]
Find the general solution of the following differential equation :
Given differential equation is \frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 [1/2]
Put y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} [1/2]
The given differential equation becomes x\frac{dv}{dx} = 1 + v^2
\Rightarrow \frac{dv}{1 + v^2} = \frac{dx}{x} [½]
\Rightarrow \int \frac{dv}{1+v^2} = \int \frac{dx}{x}
\Rightarrow \tan^{-1}v = \log|x| + C [1]
\Rightarrow \tan^{-1}\left(\frac{y}{x}\right) = \log|x| + C [½]
Find the particular solution of the differential equation xy\frac{dy}{dx} = (x + 2)(y + 2), given that y(1) = -1.
Given differential equation is \frac{y}{y+2}dy = \frac{(x+2)}{x}dx [1]
\Rightarrow \int \left(1 - \frac{2}{y+2}\right)dy = \int \left(1 + \frac{2}{x}\right)dx
\Rightarrow y - 2\log|y + 2| = x + 2\log|x| + C or y = x + 2\log|x(y+2)| + C.
When x = 1, y = -1 \Rightarrow C = -2 [1½]
The required particular solution is
y - 2\log|y + 2| = x + 2\log|x| - 2 or y = x + 2\log|x(y+2)| - 2 [½]
Solve the following differential equation :
x\frac{dy}{dx} = y - x\sin^2\left(\frac{y}{x}\right), given that y(1) = \frac{\pi}{6}
Here, \frac{dy}{dx} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right) [½]
Put \frac{y}{x} = v \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} [½]
Differential equation reduces to
v + x\frac{dv}{dx} = v - \sin^2 v
i.e. x\frac{dv}{dx} = -\sin^2 v
\Rightarrow -\int \operatorname{cosec}^2 v\,dv = \int \frac{dx}{x} [½]
\Rightarrow \cot v = \log|x| + C
\Rightarrow \cot\left(\frac{y}{x}\right) = \log|x| + C [1]
Now y(1) = \frac{\pi}{6} gives C = \sqrt{3}
Required solution is : \cot\left(\frac{y}{x}\right) = \log|x| + \sqrt{3} [½]
Find the general solution of the differential equation : y\log y\frac{dx}{dy} + x = \frac{2}{y}.
Given differential equation can be written as
\frac{dx}{dy} + \frac{1}{y\log y}\cdot x = \frac{2}{y^2\log y} [1]
\text{I.F.} = e^{\int \frac{1}{y\log y}dy} = e^{\log(\log y)} = \log y [1]
Solution is given by
x \times \log y = \int \frac{2}{y^2\log y} \times \log y\,dy = \int \frac{2}{y^2}\,dy [½]
\Rightarrow x\log y = -\frac{2}{y} + C \text{ or } xy\log y = -2 + Cy [½]
Find the general solution of the differential equation 2x^2\frac{dy}{dx} = y^2 + 2xy.
2x^2\frac{dy}{dx} = y^2 + 2xy \Rightarrow \frac{dy}{dx} = \frac{1}{2}\left(\frac{y}{x}\right)^2 + \frac{y}{x} [½]
Put y = vx and \frac{dy}{dx} = v + x\frac{dv}{dx} [½]
Hence, v + x\frac{dv}{dx} = \frac{1}{2}v^2 + v
\Rightarrow \int\frac{1}{v^2}dv = \int\frac{1}{2x}dx [1]
\Rightarrow -\frac{1}{v} = \frac{1}{2}\log x + c
\Rightarrow -\frac{x}{y} = \frac{1}{2}\log x + c [1]
Find a particular solution of the differential equation (x + 1)\frac{dy}{dx} = 2e^{-y} - 1, given that y = 0 when x = 0.
(x+1)\frac{dy}{dx} = 2e^{-y} - 1 \Rightarrow \int\frac{e^y}{2 - e^y}dy = \int\frac{1}{x+1}dx [1]
Put 2 - e^y = t \Rightarrow e^y dy = -dt [½]
Hence, -\int\frac{dt}{t} = \log|x+1|
\Rightarrow -\log|t| = \log|x+1| - \log|c|
\Rightarrow (x+1)(2 - e^y) = c [1]
Put x = y = 0; c = 1
Hence, particular solution is given by
(x+1)(2 - e^y) = 1 or \log|x+1| + \log|2 - e^{-y}| = 0 [½]
Solve the differential equation (x + 2y^3)\,dy = y\,dx.
Given differential equation can be written as
\frac{dx}{dy} - \frac{x}{y} = 2y^2 [½]
Integrating factor = e^{\int -\frac{1}{y}dy} = e^{-\log y} = \frac{1}{y} [1]
Solution is x.\frac{1}{y} = \int 2y\,dy + C [1]
i.e., x.\frac{1}{y} = y^2 + C or x = y^3 + Cy [½]
Solve the differential equation 2(y + 3) - xy\frac{dy}{dx} = 0; given y(1) = -2.
Given differential equation can be written as
\frac{y}{y+3}dy = \frac{2}{x}dx
\Rightarrow \int\left(1 - \frac{3}{y+3}\right)dy = 2\int\frac{1}{x}dx [1]
\Rightarrow y - 3\log|y + 3| = 2\log|x| + C [1½]
y = -2, when x = 1 \Rightarrow C = -2 [½]
Hence, the required particular solution is
\Rightarrow y - 3\log|y + 3| = 2\log|x| - 2
Solve the following differential equation :
Given differential equation can be written as
\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2}, which is linear in y.
I.F. = e^{\int\frac{2x}{1+x^2}dx} = e^{\log(1+x^2)} = 1 + x^2 [1]
The solution is given by
y(1 + x^2) = \int 4x^2\,dx [1]
\Rightarrow y(1 + x^2) = \frac{4}{3}x^3 + C [1]
or y = \frac{4x^3}{3(1+x^2)} + C\frac{1}{(1+x^2)}, which is the required general solution
Find the particular solution of the differential equation
\left[x\cdot\sin^2\frac{y}{x}-y\right]\cdot dx+x\cdot dy=0
\Rightarrow\frac{dy}{dx}=\frac{y}{x}-\sin^2\frac{y}{x}
Put y=vx\Rightarrow\frac{dy}{dx}=x\frac{dv}{dx}+v [1]
\therefore x\frac{dv}{dx}+v=v-\sin^2v
\Rightarrow-\int\text{cosec}^2v\,dv=\int\frac{dx}{x}
\Rightarrow\cot v=\log|x|+c [1]
\Rightarrow\cot\frac{y}{x}=\log|x|+c [½]
x=1,y=\frac{\pi}{4}\Rightarrow c=1
\Rightarrow\cot\frac{y}{x}=\log|x|+1 [½]
Find the general solution of the differential equation
(2x^2 + y)\,dx = x\,dy.
(2x^2 + y)\,dx = x\,dy
\Rightarrow \frac{dy}{dx} - \frac{1}{x}\cdot y = 2x [1]
I.F. = e^{-\int \frac{1}{x}dx} = \frac{1}{x} [1]
Solution is given by,
y\cdot\left(\frac{1}{x}\right) = \int 2x\cdot\frac{1}{x}\,dx [½]
\Rightarrow \frac{y}{x} = 2x + C or y = 2x^2 + Cx [½]
For the differential equation \frac{dy}{dx} - \frac{y}{x} + \operatorname{cosec}\left(\frac{y}{x}\right) = 0, find the particular solution, given that y = 0 when x = 1.
\frac{dy}{dx} = \frac{y}{x} - \operatorname{cosec}\left(\frac{y}{x}\right)
Put \frac{y}{x} = v i.e. y = vx
\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} [½]
The differential equation reduces to
v + x\frac{dv}{dx} = v - \operatorname{cosec} v [½]
\sin v\,dv = -\frac{1}{x}dx \Rightarrow \int \sin v\,dv = -\int \frac{dx}{x}
\Rightarrow -\cos v = -\log|x| - C [1]
or, \cos v = \log|x| + C'
\Rightarrow \cos\left(\frac{y}{x}\right) = \log|x| + C'
y = 0, x = 1 gives C' = 1 [½]
\therefore Solution is given by \cos\left(\frac{y}{x}\right) = \log|x| + 1 [½]
Find the particular solution of the differential equation given by x^2\frac{dy}{dx} - xy = x^2\cos^2\left(\frac{y}{2x}\right), given that when x = 1,\ y = \frac{\pi}{2}.
\frac{dy}{dx} = \frac{y}{x} + \cos^2\left(\frac{y}{2x}\right) [½]
Put y = vx so that \frac{dy}{dx} = v + x\frac{dv}{dx} [½]
\Rightarrow v + x\frac{dv}{dx} = v + \cos^2\left(\frac{v}{2}\right)
\Rightarrow \int \sec^2\left(\frac{v}{2}\right) dv = \int \frac{1}{x}\, dx [½]
\Rightarrow 2\tan\left(\frac{v}{2}\right) = \log|x| + C [1]
\Rightarrow 2\tan\left(\frac{y}{2x}\right) = \log|x| + C
2\tan\frac{\pi}{4} = \log 1 + C \Rightarrow C = 2 \Rightarrow 2\tan\left(\frac{y}{2x}\right) = \log|x| + 2 [½]
Find the particular solution of the differential equation given by 2xy + y^2 - 2x^2\frac{dy}{dx} = 0; y = 2, when x = 1.
Given differential equation can be written as
\frac{dy}{dx} = \frac{2xy+y^2}{2x^2} = \frac{y}{x} + \frac{y^2}{2x^2} [½]
Let y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} [½]
The equation becomes
x\frac{dv}{dx} = \frac{1}{2}v^2
\Rightarrow \frac{dv}{v^2} = \frac{1}{2} \times \frac{dx}{x} [½]
Integrating both sides, we get
\frac{-1}{v} = \frac{1}{2}\log|x| + C
\Rightarrow -\frac{x}{y} = \frac{1}{2}\log|x| + C [1]
y = 2, x = 1 gives C = -\frac{1}{2}
The particular solution is
-\frac{x}{y} = \frac{1}{2}\log|x| - \frac{1}{2}\ \text{or}, y = \frac{2x}{1-\log|x|} [½]
Find the general solution of the differential equation :
Given differential equation can be written as
\frac{dx}{dy} - \frac{x}{y} = 2y [1]
Integrating Factor = e^{\int \frac{-1}{y}dy} = \frac{1}{y} [1]
Solution is x\frac{1}{y} = \int 2dy [½]
\Rightarrow \frac{x}{y} = 2y + C
\Rightarrow x = 2y^2 + Cy [½]
Find the particular solution of the differential equation \frac{dy}{dx} = y\cot 2x, given that y\left(\frac{\pi}{4}\right) = 2.
\frac{dy}{dx} = y\cot 2x \Rightarrow \int\frac{dy}{y} = \int\cot 2x\,dx [1]
\Rightarrow \log|y| = \frac{1}{2}\log|\sin 2x| + \log c [1]
y = c.\sqrt{\sin 2x}
when y\left(\frac{\pi}{4}\right) = 2, gives c = 2 [½]
\therefore y = 2\sqrt{\sin 2x} is the required Particular solution of given D.E. [½]
Find the particular solution of the differential equation
given that y = 1 when x = 1.
\frac{dy}{dx} = e^{\frac{y}{x}} + \frac{y}{x} = f\left(\frac{y}{x}\right) so, its a homogeneous differential equation
Let y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} [1]
Now, v + x\frac{dv}{dx} = e^v + v
\Rightarrow \int e^{-v}dv = \int\frac{1}{x}dx [½]
\Rightarrow -e^{-v} = \log|x| + c \Rightarrow -e^{\frac{-y}{x}} = \log|x| + c \quad \ldots(1) [1]
Now, x = 1, y = 1, gives c = -e^{-1}
Thus, \boxed{\log|x| + e^{\frac{-y}{x}} = e^{-1}} [½]
Find the particular solution of the differential equation
Given differential equation is a linear order differential equation with:
P=-2x,\ Q=3x^2e^{x^2} [½]
Integrating Factor =e^{\int-2x\,dx}=e^{-x^2} [1]
The general solution is: y\cdot e^{-x^2}=\int e^{-x^2}\cdot3x^2e^{x^2}dx+C\Rightarrow y\cdot e^{-x^2}=x^3+C [1]
Putting x=0,y=5, we get, C=5
\therefore The Particular solution is: y\cdot e^{-x^2}=x^3+5 or y=(x^3+5)e^{x^2} [½]
Solve the following differential equation :
x^2\,dy+y(x+y)\,dx=0
x^2dy+y(x+y)dx=0\Rightarrow\frac{dy}{dx}=-\frac{y}{x}-\left(\frac{y}{x}\right)^2 [½]
Putting \frac{y}{x}=v\Rightarrow y=vx,\ \frac{dy}{dx}=v+x\frac{dv}{dx} [½]
v+x\frac{dv}{dx}=-v-v^2,
separating the variable and integrating
\int\frac{1}{v^2+2v}dv=-\int\frac1x dx [½]
\Rightarrow\int\frac{1}{(v+1)^2-1}dv=-\int\frac1x dx [½]
\Rightarrow\frac12\log\left|\frac{v}{v+2}\right|=\log\left|\frac{C}{x}\right| [½]
The solution of the differential equation is, \left|\frac{y}{y+2x}\right|=\frac{C^2}{x^2} or x^2y=k(y+2x) [½]
Find the general solution of the differential equation
The given differential equation can be written as, \frac{dy}{dx}-\frac{1}{x}y=\log x [½]
\therefore P=-\frac{1}{x},\ Q=\log x [½]
\text{I.F.}=e^{\int-\frac{1}{x}dx}=e^{-\log x}=e^{\log\frac{1}{x}}=\frac{1}{x} [½]
The solution of the differential equation is:
y\cdot\frac{1}{x}=\int\log x\cdot\frac{1}{x}dx [1]
\Rightarrow\frac{y}{x}=\frac{(\log x)^{2}}{2}+c, or y=\frac{1}{2}x\cdot(\log x)^{2}+cx [½]
Find the particular solution of the differential equation \frac{dy}{dx}+\sec^2x\cdot y=\tan x\cdot\sec^2x, given that y(0)=0.
Integrating factor =e^{\int\sec^2x\,dx}=e^{\tan x} [½]
Solution is ye^{\tan x}=\int\tan x\sec^2x\,e^{\tan x}dx+C [½]
Let \tan x=t\quad\sec^2x\,dx=dt [½]
\therefore\int e^{\tan x}\tan x\sec^2x\,dx=\int e^t\,t\,dt=e^t(t-1) [½]
\therefore ye^{\tan x}=e^{\tan x}(\tan x-1)+C [½]
y(0)=0 gives C=1
Particular solution is ye^{\tan x}=e^{\tan x}(\tan x-1)+1 or y=\tan x-1+e^{-\tan x} [½]
Solve the differential equation given by x\,dy-y\,dx-\sqrt{x^2+y^2}\,dx=0.
Given differential equation can be written as
\frac{dy}{dx}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2} --------- (i) [½]
Let y=vx\Rightarrow\frac{dy}{dx}=v+x\frac{dv}{dx} substituting in (i) [½]
We get v+x\frac{dv}{dx}=v+\sqrt{1+v^2}
\Rightarrow\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x} [½]
Integrating both sides, we get
\log\left|\sqrt{1+v^2}+v\right|=\log|x|+\log C [1]
y+\sqrt{x^2+y^2}=Cx^2 [½]
Find the general solution of the differential equation :
Given differential equation can be written as
Put y = ux, \frac{dy}{dx} = u + x\frac{du}{dx} [½]
\therefore u + x\frac{du}{dx} = \frac{u^2}{u - 1} \Rightarrow x\frac{du}{dx} = \frac{u}{u - 1} [½]
Separating the variables and integrating
\int\left(1 - \frac{1}{u}\right)du = \int\frac{dx}{x} \Rightarrow u - \log u = \log|x| + c \Rightarrow \frac{y}{x} - \log\frac{y}{x} = \log|x| + c [1+½]
Find the general solution of the differential equation :
Rewriting the given differential equation as:
Integrating factor = e^{\int\frac{2x}{1+x^2}dx} = e^{\log(1+x^2)} = 1 + x^2 [1]
\therefore solution of the differential equation is
y(1 + x^2) = \int\frac{\sqrt{x^2 + 4}}{1 + x^2}\cdot(1 + x^2)\,dx + c [½]
= \int\sqrt{x^2 + 4}\,dx + c
\therefore y(1 + x^2) = \frac{x\sqrt{x^2 + 4}}{2} + 2\log\left|x + \sqrt{x^2 + 4}\right| + c [1]
Find the general solution of the differential equation :
Given differential equation is
2xy\frac{dy}{dx} + y^2 = 2y(1 + x^2)
\Rightarrow \frac{dy}{dx} + \frac{y}{2x} = \frac{1}{x} + x [½]
Integrating factor = e^{\int \frac{1}{2x}dx} = e^{\log\sqrt{x}} = \sqrt{x} [1]
Solution is given by y\sqrt{x} = \int\left(\frac{1}{\sqrt{x}} + x^{\frac{3}{2}}\right)dx [1]
\Rightarrow y\sqrt{x} = 2\sqrt{x} + \frac{2x^{\frac{5}{2}}}{5} + C, or y = 2 + \frac{2x^2}{5} + \frac{C}{\sqrt{x}} [½]
Solve the following differential equation :
Given differential equation is \frac{dy}{dx} = \frac{y}{x} - e^{\frac{y}{x}} [½]
Let y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} [½]
The given equation becomes v + x\frac{dv}{dx} = v - e^v
\Rightarrow -e^{-v}dv = \frac{dx}{x} [½]
Integrating both sides, we get
e^{-v} = \log|x| + C [1]
\Rightarrow e^{-\frac{y}{x}} = \log|x| + C [½]
Find the particular solution of the differential equation \frac{dy}{dx}=\frac{x+y}{x},\ y(1)=0.
\frac{dy}{dx}=\frac{x+y}{x}\Rightarrow\frac{dy}{dx}=1+\frac{y}{x}
Let \frac{y}{x}=v. Then x\frac{dv}{dx}+v=\frac{dy}{dx}
So, Differential equation becomes.
x\frac{dv}{dx}+v=1+v [1]
\Rightarrow dv=\frac{dx}{x}
\Rightarrow v=\log|x|+c
\Rightarrow y=x\log|x|+cx [1]
\Rightarrow x=1,\ y=0\Rightarrow c=0,\ y=x\log|x| [1]
[can also be solved, taking first order linear diff. eqn]
Find the general solution of the differential equation e^x\tan y\,dx+(1-e^x)\sec^2 y\,dy=0.
The given D.E. is
\frac{\sec^2 y}{\tan y}dy=-\frac{e^x}{1-e^x}dx [1]
Integrating
\Rightarrow\log|\tan y|=\log|1-e^x|+\log C [1½]
\Rightarrow\tan y=C(1-e^x) [½]
Find the particular solution of the differential equation \frac{dy}{dx} = \frac{xy}{x^2+y^2}, given that y = 1 when x = 0.
Official marking scheme for this paper is not available.
Find the particular solution of the differential equation (1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}, given that y = 0 when x = 1.
Official marking scheme for this paper is not available.
Find the particular solution of the differential equation x\frac{dy}{dx}+x\cos^2\left(\frac{y}{x}\right)=y ; given that when x=1 , y=\frac{\pi}{4} .
Given differential equation can be written as
\frac{dy}{dx}=\frac{y}{x}-\cos^2\frac{y}{x} ... (1) [½]
Let y=vx\ \therefore \frac{dy}{dx}=v+x\frac{dv}{dx} [½]
(1) Becomes v+x\frac{dv}{dx}=v-\cos^2 v
\sec^2 v\,dv=-\frac{dx}{x}
Integrating both sides we get
\tan v=-\log|x|+c [½]
\tan\frac{y}{x}=-\log|x|+c
x=1,\ y=\frac{\pi}{4}\Rightarrow c=1 [1]
\therefore Particular solution is \tan\frac{y}{x}=-\log|x|+1 [½]
Find the particular solution of the differential equation (y+3x^2)\frac{dx}{dy}=x, given that y = 1, when x = 1.
Given differential equation can be written as
x\frac{dy}{dx}-y=3x^2 or \frac{dy}{dx}-\frac{1}{x}y=3x [½]
I.F =e^{\int-\frac{1}{x}dx}=e^{-\log x}=-x^{-1}=\frac{1}{x} [1]
Solution is y\cdot\frac{1}{x}=\int 3x\frac{1}{x}dx+C [½]
\frac{y}{x}=3x+C [½]
x=1,\ y=1 gives C=-2
Particular solution is \frac{y}{x}=3x-2 or y=3x^2-2x [½]
Find the particular solution of the differential equation x\frac{dy}{dx}-y=x^2\cdot e^x, given y(1)=0.
\frac{dy}{dx}-\frac{y}{x}=\left(\frac{x^2}{x}\right)e^x=xe^x [½]
I\cdot F=e^{\int-\frac{1}{x}dx}=e^{-\log x}=e^{\log x^{-1}}=\frac1x [1]
Solution is y\times\frac1x=\int e^x dx
\frac{y}{x}=e^x+C [½]
y=xe^x+Cx
C=-e [½]
Particular solution is y=xe^x-ex [½]
Find the general solution of the differential equation x\frac{dy}{dx}=y(\log y-\log x+1).
x\frac{dy}{dx}=y[\log y-\log x+1]
\frac{dy}{dx}=\frac{y}{x}\left[\log\frac{y}{x}+1\right] [½]
y=vx
\frac{dy}{dx}=v+x\frac{dv}{dx} [½]
v+x\frac{dv}{dx}=v[\log v+1]
x\frac{dv}{dx}=v\log v [½]
\int\frac{dv}{v\log v}=\int\frac1x dx [½]
\log|\log v|=\log|x|+\log C [½]
\log\left(\frac{y}{x}\right)=Cx [½]
During a heavy gaming session, the temperature of a student’s laptop processor increases significantly. After the session, the processor begins to cool down, and the rate of cooling is proportional to the difference between the processor’s temperature and the room temperature (25°C). Initially the processor’s temperature is 85°C. The rate of cooling is defined by the equation \frac{d}{dt}(T(t))=-k(T(t)-25),
where T(t) represents the temperature of the processor at time t (in minutes) and k is a constant.
Based on the above information, answer the following questions :
(i) Find the expression for temperature of processor, T(t) given that T(0) = 85°C. [2]
(ii) How long will it take for the processor’s temperature to reach 40°C ? Given that k = 0·03, \log_e 4=1\cdot 3863. [2]
(i) \frac{dT}{dt}=-k(T-25)
\Rightarrow\frac{dT}{T-25}=-k\,dt [½]
\Rightarrow\int\frac{dT}{T-25}=-k\int dt
\Rightarrow\log|T-25|=-kt+C\quad\ldots(a) [1]
When t=0,T=85
\Rightarrow\log 60=C
Using in equation (a), \log|T-25|=-kt+\log 60\quad\ldots(b) [½]
(ii) When k=0.03, \log|T-25|=-0.03t+\log 60
\Rightarrow\log\left|\frac{T-25}{60}\right|=-0.03t
\Rightarrow T-25=60.e^{-0.03t} [1]
When T=40,t=t_1
\Rightarrow\frac{15}{60}=e^{-0.03t_1} [½]
\Rightarrow e^{-0.03t_1}=\frac{1}{4}\Rightarrow -0.03t_1=-\log 4
\Rightarrow t_1=\frac{\log 4}{0.03}=\frac{1.3863}{0.03}=46.21\text{m} [½]
Case Study – 2
Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation, if left in the open at room temperature.
(Cylindrical-shaped Camphor tablets)
A cylindrical camphor tablet whose height is equal to its radius (r) evaporates when exposed to air such that the rate of reduction of its volume is proportional to its total surface area. Thus, \frac{dV}{dt}=kS is the differential equation, where V is the volume, S is the surface area and t is the time in hours.
Based upon the above information, answer the following questions :
(i) Write the order and degree of the given differential equation. [1]
(ii) Substituting V=\pi r^3 and S=2\pi r^2, we get the differential equation \frac{dr}{dt}=\frac23k. Solve it, given that r(0) = 5 mm. [1]
(iii) (a) If it is given that r = 3 mm when t = 1 hour, find the value of k. Hence, find t for r = 0 mm. [2]
OR
(iii) (b) If it is given that r = 1 mm when t = 1 hour, find the value of k. Hence, find t for r = 0 mm. [2]
(i) Order = 1, Degree = 1 [½+½]
(ii) Separating the variable and integrating, \int dr=\frac{2k}{3}\int dt\Rightarrow r=\frac23kt+C [½]
Putting t=0,r=5, we get C=5
r=\frac23kt+5 [½]
(iii) (a) Putting r=3,t=1, 3=\frac23k(1)+5\Rightarrow k=-3 [1]
r=-2t+5, For r=0, t=\frac52 hours or 2.5 hours [1]
OR
(iii) (b) Putting r=1,t=1, 1=\frac23k+5\Rightarrow k=-6 [1]
\therefore r=-4t+5, For r=0, t=\frac54 hours or 1.25 hours [1]
Case Study – 2
A bacteria sample of certain number of bacteria is observed to grow exponentially in a given amount of time. Using exponential growth model, the rate of growth of this sample of bacteria is calculated.
The differential equation representing the growth of bacteria is given as :
\frac{dP}{dt} = kP, where P is the population of bacteria at any time ‘t’.
Based on the above information, answer the following questions :
(i) Obtain the general solution of the given differential equation and express it as an exponential function of ‘t’. [2]
(ii) If population of bacteria is 1000 at t = 0, and 2000 at t = 1, find the value of k. [2]
(i) \frac{dP}{dt} = kP \Rightarrow \int\frac{dP}{P} = \int k\, dt [1]
\Rightarrow \log P = kt + C or P = e^{kt + C} [1]
(ii) \log P = kt + C
when t = 0, P = 1000 \Rightarrow C = \log 1000 [½]
when t = 1, P = 2000 \Rightarrow \log 2000 = k + \log 1000 [½]
\Rightarrow k = \log 2 [1]
Case Study-III
An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form \frac{dy}{dx}=F(x,y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogenous function of degree n if F(\lambda x,\lambda y)=\lambda^nF(x,y). To solve a homogeneous differential equation of the type \frac{dy}{dx}=F(x,y)=g\left(\frac{y}{x}\right), we make the substitution y = vx and then separate the variables.
Based on the above, answer the following questions :
(I) Show that (x^2-y^2)\,dx+2xy\,dy=0 is a differential equation of the type \frac{dy}{dx}=g\left(\frac{y}{x}\right). [2]
(II) Solve the above equation to find its general solution. [2]
(I) (x^2-y^2)dx+2xydy=0
\Rightarrow\frac{dy}{dx}=\frac{y^2-x^2}{2xy} [1]
=\frac{\left(\frac{y}{x}\right)^2-1}{2\left(\frac{y}{x}\right)}
=g\left(\frac{y}{x}\right) [1]
(II) y=vx\Rightarrow\frac{dy}{dx}=v+x\frac{dy}{dx} [½]
v+x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{-1-v^2}{2v}
\Rightarrow\int\frac{2v}{1+v^2}dv=-\int\frac{dx}{x} [1]
\Rightarrow\log|1+v^2|+\log|x|=\log C [½]
or x\left(1+\frac{y^2}{x^2}\right)=C
or x^2+y^2=Cx
Find the particular solution of the differential equation x\frac{dy}{dx}+y+\frac{1}{1+x^2}=0, given that y(1)=0.
(a) Given differential equation can be written as \frac{dy}{dx}+\frac{1}{x}y=\frac{-1}{x(1+x^2)} [½]
I.F. =e^{\int\frac{1}{x}dx}=e^{\log x}=x [1]
Solution is y\cdot x=\int\frac{-1}{1+x^2}dx+C [1]
\Rightarrow xy=-\tan^{-1}x+C [1]
Now y(1)=0\Rightarrow C=\frac{\pi}{4}
\therefore Particular solution is xy=\frac{\pi}{4}-\tan^{-1}x [½]
Find the general solution of the differential equation x(y^3+x^3)\,dy=(2y^4+5x^3y)\,dx.
(b) Given differential equation can be written as
Let, y=vx\Rightarrow\frac{dy}{dx}=v+x\frac{dv}{dx} [½]
Equation (1) becomes v+x\frac{dv}{dx}=\frac{2v^4+5v}{v^3+1}
\Rightarrow x\frac{dv}{dx}=\frac{v^4+4v}{v^3+1}
\frac{v^3+1}{v^4+4v}dv=\frac{dx}{x} [1]
\int\frac{4v^3+4}{v^4+4v}dv=4\int\frac{dx}{x} [½]
\log|v^4+4v|=\log(x)^4+\log C [1]
\log\left|\frac{y^4+4yx^3}{x^4}\right|=\log Cx^4
y^4+4yx^3=Cx^8 [½]
Solve the following differential equation :
(y-\sin^2 x)\,dx+\tan x\,dy=0
\frac{dy}{dx}+\frac{y}{\tan x}=\frac{\sin^2 x}{\tan x}
\frac{dy}{dx}+(\cot x)\,y=\sin x\cos x [1]
I.F. =e^{\int\cot x\,dx}=e^{\log(\sin x)}=\sin x [1]
Solution is given by
y\sin x=\int\sin^2 x\cos x\,dx [½]
y\sin x=\int t^2dt \quad \because \sin x=t,\ \cos x\,dx=dt [½]
y\sin x=\frac{t^3}{3}+C [½]
y\sin x=\frac{\sin^3 x}{3}+C [½]
Find the general solution of the differential equation :
\frac{dy}{dx}=\frac{x^2y}{x^3+y^3} [½]
Put y=vx,\ \frac{dy}{dx}=v+x\frac{dv}{dx} [½]
v+x\frac{dv}{dx}=\frac{v}{1+v^3}\Rightarrow x\frac{dv}{dx}=\frac{-v^4}{1+v^3}
\int\frac{1+v^3}{v^4}\,dv=-\int\frac{dx}{x} [1]
\int\frac{1}{v^4}\,dv+\int\frac{1}{v}\,dv=-\log|x|+C
\frac{1}{-3v^3}+\log|v|=-\log|x|+C [1½]
\frac{-x^3}{3y^3}+\log\left|\frac{y}{x}\right|=-\log|x|+C or \frac{-x^3}{3y^3}+\log|y|=C [½]
Find the particular solution of the differential equation
x\frac{dy}{dx}+2y = x^2\log x
\Rightarrow \frac{dy}{dx}+\frac{2}{x}\cdot y = x\log x [½]
I.F. = e^{\int\frac{2}{x}dx} = e^{2\log x} = x^2 [½]
Solution is given by
y\cdot x^2 = \int x^2(x\log x)dx [1]
\Rightarrow yx^2 = \int x^3\log x\,dx
= \log x\cdot\frac{x^4}{4}-\int\frac{1}{x}\cdot\frac{x^4}{4}dx
= \frac{x^4}{4}\log x-\frac{1}{16}\cdot x^4+C [1]
\Rightarrow y = \frac{x^2}{4}\log x-\frac{1}{16}x^2+\frac{C}{x^2}
when x = 1, y = 1, we get C = \frac{17}{16} [½]
Required particular solution is
y = \frac{x^2}{4}\log x-\frac{1}{16}x^2+\frac{17}{16x^2} [½]
Solve the differential equation: y+\frac{d}{dx}(xy)=x(\sin x+x)
y+\frac{d}{dx}(xy)=x(\sin x+x)
\Rightarrow y+\left(x\frac{dy}{dx}+y\right)=x(\sin x+x)
\Rightarrow 2y+x\frac{dy}{dx}=x(\sin x+x)
\Rightarrow\frac{dy}{dx}+\frac{2y}{x}=(\sin x+x) [1]
This a linear differential equation of the form \frac{dy}{dx}+Py=Q
P=\frac2x, Q=(\sin x+x)
I.F=e^{\int\frac2x dx}=e^{2\log x}=e^{\log x^2}=x^2 [1]
Solution will be y.I.F=\int Q.IF\,dx
yx^2=\int(\sin x+x)x^2\,dx
yx^2=\int\sin x.x^2\,dx+\int x^3\,dx [1]
\Rightarrow yx^2=-x^2\cos x+2\int x\cos x\,dx+\frac{x^4}{4}+C [1]
\Rightarrow yx^2=-x^2\cos x+2(x\sin x+\cos x)+\frac{x^4}{4}+C
Which is the required solution [1]
Find the particular solution of the differential equation:
2ye^{x/y}dx+(y-2xe^{x/y})\,dy=0 given that y(0)=1
2ye^{x/y}dx+(y-2xe^{x/y})\,dy=0
\Rightarrow\frac{dx}{dy}=\frac{2xe^{x/y}-y}{2ye^{x/y}}=\frac{2\frac{x}{y}e^{\frac{x}{y}}-1}{2e^{\frac{x}{y}}}
It is a homogeneous differential equation. [1]
Let x=vy\Rightarrow\frac{dx}{dy}=v+y\frac{dv}{dy} [1]
v+y\frac{dv}{dy}=\frac{2ve^v-1}{2e^v}
\Rightarrow y\frac{dv}{dy}=\frac{2ve^v-1}{2e^v}-v=\frac{2ve^v-1-2ve^v}{2e^v}
\Rightarrow y\frac{dv}{dy}=\frac{-1}{2e^v}
\Rightarrow 2e^v\,dv=-\frac{dy}{y} [1]
\int2e^v\,dv=-\int\frac{dy}{y}
\Rightarrow 2e^v=-\log|y|+C
\Rightarrow 2e^{\frac{x}{y}}+\log|y|=C [1]
When x=0,y=1, C = 2
Required solution 2e^{\frac{x}{y}}+\log|y|=2 [1]
Solve the differential equation y\,e^{y}\,dx = (y^{3} + 2x\,e^{y})\,dy, when y(0) = 1.
Given differential eq. can be written as \frac{dx}{dy} - \frac{2}{y}x = y^{2}e^{-y} [1]
Integrating factor = e^{\int \frac{-2}{y}dy} = \frac{1}{y^{2}} [1]
General solution: x\left(\frac{1}{y^{2}}\right) = \int \frac{1}{y^{2}}.y^{2}e^{-y}dy [1]
x\left(\frac{1}{y^{2}}\right) = -e^{-y} + c or x = y^{2}\left(-e^{-y} + c\right) [1]
Now, when x = 0, y = 1 : 0 = -e^{-1} + c \Rightarrow c = e^{-1} [½]
Required particular solution is, x = y^{2}\left(e^{-1} - e^{-y}\right) [½]
Find the general solution of the differential equation
\frac{dy}{dx} = \frac{x^{3} - 3xy^{2}}{y^{3} - 3x^{2}y} [1]
Put y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}, so [1]
v + x\frac{dv}{dx} = \frac{1 - 3v^{2}}{v^{3} - 3v} \Rightarrow x\frac{dv}{dx} = \frac{1 - 3v^{2}}{v^{3} - 3v} - v = \frac{1 - v^{4}}{v^{3} - 3v}
\Rightarrow \int \frac{3v - v^{3}}{v^{4} - 1}dv = \int \frac{1}{x}dx [1]
\Rightarrow \frac{3}{4}\log\left|\frac{v^{2} - 1}{v^{2} + 1}\right| - \frac{1}{4}\log\left|v^{4} - 1\right| = \log x + c [1½]
\therefore \frac{3}{4}\log\left|\frac{y^{2} - x^{2}}{y^{2} + x^{2}}\right| - \frac{1}{4}\log\left|\frac{y^{4} - x^{4}}{x^{4}}\right| = \log x + c [½]
Solve the differential equation x^2\,dy + y(x + y)\,dx = 0, given that y = 1 when x = 1.
x^2 dy + y(x+y)dx = 0
\Rightarrow \frac{dy}{dx} = -\frac{(xy - x^2)}{x^2} = -\frac{y}{x} - \left(\frac{y}{x}\right)^2 \quad \ldots(i) [1]
Let \frac{y}{x} = v i.e. y = vx \quad \ldots(ii)
\Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \quad \ldots(iii) [1]
Using (ii) and (iii) in equation (i),
v + x\frac{dv}{dx} = -v - v^2
\Rightarrow \frac{dv}{v(v+2)} = -\frac{dx}{x} [½]
\Rightarrow \frac{1}{2}\int\left(\frac{1}{v} - \frac{1}{v+2}\right)dv = -\int\frac{dx}{x} [½]
\Rightarrow \frac{1}{2}\log\left|\frac{v}{v+2}\right| = -\log|x| + \log c' [1]
\Rightarrow \log\left|\frac{y}{y+2x}\right| = \log\left|\frac{c}{x^2}\right|
\Rightarrow yx^2 = c(y + 2x)
x = 1, y = 1 gives c = \frac{1}{3} [½]
Solution is given by 3yx^2 = y + 2x or y = \frac{2x}{3x^2 - 1} [½]
Solve the differential equation : x^2y\, dx - (x^3 + y^3)\, dy = 0.
Given differential equation can be written as \frac{dy}{dx} = \frac{yx^2}{x^3 + y^3} [½]
Put y = vx, so \frac{dy}{dx} = v + x\frac{dv}{dx} [1]
Therefore, v + x\frac{dv}{dx} = \frac{vx^3}{x^3 + v^3x^3} = \frac{v}{1 + v^3}
x\frac{dv}{dx} = \frac{-v^4}{1 + v^3} [1]
\left(\frac{1}{v^4} + \frac{1}{v}\right) dv = \frac{-dx}{x} [1]
Integrating we get
\frac{-1}{3v^3} + \log|v| = -\log|x| + C [1]
\frac{-x^3}{3y^3} + \log|y| = C [½]
Solve the differential equation (1 + x^2)\frac{dy}{dx} + 2xy - 4x^2 = 0 subject to initial condition y(0) = 0.
Given D.E. is \frac{dy}{dx} + \frac{2x}{1+x^2}\, y = \frac{4x^2}{1+x^2} [1]
Integrating factor is e^{\int \frac{2x}{1+x^2} dx} = e^{\log(1+x^2)} = (1 + x^2) [1]
Solution is y(1 + x^2) = \int 4x^2\, dx + C [1]
y(1 + x^2) = \frac{4x^3}{3} + C [1]
y(0) = 0 gives C = 0, hence solution is y(1 + x^2) = \frac{4x^3}{3} [1]
Solve the differential equation \frac{dy}{dx}=\cos x-2y.
The given differential equation can be written as:
\frac{dy}{dx}+2y=\cos x, Taking P=2,Q=\cos x [½]
Integrating factor is given by, I=e^{\int2dx}=e^{2x} [1]
\therefore The solution is, y\cdot e^{2x}=\int e^{2x}\cos x\,dx [1½]
Let, I_1=\int\cos x\cdot e^{2x}dx=\cos x\frac{e^{2x}}{2}-\int(-\sin x)\frac{e^{2x}}{2}dx
=\frac{e^{2x}\cos x}{2}+\frac12\left[\sin x\cdot\frac{e^{2x}}{2}-\int\cos x\cdot\frac{e^{2x}}{2}dx\right]
\Rightarrow I_1=\frac{e^{2x}\cos x}{2}+\frac{e^{2x}\sin x}{4}-\frac14I_1\Rightarrow I_1=\frac{e^{2x}}{5}(2\cos x+\sin x) [1½]
\therefore The solution of the differential equation is
y\cdot e^{2x}=\frac{e^{2x}}{5}(2\cos x+\sin x)+C\Rightarrow y=\frac15(2\cos x+\sin x)+Ce^{-2x} [½]
If vector 2\hat{i}-u\hat{j}+3\hat{k} has magnitude 4, then value of ‘u’ is
(C) \pm\sqrt3
Magnitude of 2\hat{i}-u\hat{j}+3\hat{k}=4\Rightarrow\sqrt{4+u^2+9}=4
\Rightarrow u^2=16-13=3\Rightarrow u=\pm\sqrt3
Angle made by the vector \hat{i}+\hat{j}+\sqrt2\,\hat{k} with positive direction of z-axis is
(B) \frac{\pi}{4}
Direction cosines of the vector \hat{i}+\hat{j}+\sqrt2\,\hat{k} are \frac12,\frac12,\frac{1}{\sqrt2}
Let angle = \gamma. Then \cos\gamma=\frac{1}{\sqrt2}\Rightarrow\gamma=\frac{\pi}{4}
A vector perpendicular to the vectors \hat{i}+\hat{j} and \hat{i}-\hat{j} is
(B) 3\hat{k}
Both the vectors \hat{i}+\hat{j} and \hat{i}-\hat{j} are in XY-plane
So a vector perpendicular to both is 3\hat{k}
A bird flies through a distance in a straight line given by the vector \hat{i}+2\hat{j}+\hat{k}. A man standing beside a straight metro rail track given by \vec{r}=(3+\lambda)\hat{i}+(2\lambda-1)\hat{j}+3\lambda\hat{k} is observing the bird. The projected length of its flight on the metro track is
The projection of the vector \hat{i}+2\hat{j}+\hat{k} on the line \vec{r}=(3\hat{i}-\hat{j})+\lambda(\hat{i}+2\hat{j}+3\hat{k}) is \frac{1\times1+2\times2+1\times3}{\sqrt{1^2+2^2+3^2}}=\frac{8}{\sqrt{14}} units
(C) \frac{8}{\sqrt{14}} units [1]
If \vec{a}=3\hat{i}+2\hat{j}+4\hat{k}, \vec{b}=\hat{i}+\hat{j}-3\hat{k} and \vec{c}=6\hat{i}-\hat{j}+2\hat{k} are three given vectors, then (2\vec{a}.\hat{i})\hat{i}-(\vec{b}.\hat{j})\hat{j}+(\vec{c}.\hat{k})\hat{k} is same as the vector
(2\vec{a}.\hat{i})\hat{i}-(\vec{b}.\hat{j})\hat{j}+(\vec{c}.\hat{k})\hat{k}=(2\times3)\hat{i}-(1)\hat{j}+(2)\hat{k}
=6\hat{i}-\hat{j}+2\hat{k}=\vec{c}
(D) \vec{c} [1]
Assertion(A): Given two non-zero vectors \vec{a} and \vec{b}. If \vec{r} is another non-zero vector such that \vec{r}\times(\vec{a}+\vec{b})=\vec{0}. Then \vec{r} is perpendicular to \vec{a}\times\vec{b}.
Reason (R): The vector (\vec{a}+\vec{b}) is perpendicular to the plane of \vec{a} and \vec{b}
\vec{r}\times(\vec{a}+\vec{b})=\vec{0}\Rightarrow\vec{r} is parallel to (\vec{a}+\vec{b}) and (\vec{a}+\vec{b}) lies on the plane of \vec{a} and \vec{b}.
So, \vec{r} is parallel to the plane of \vec{a} and \vec{b}\Rightarrow\vec{r} is perpendicular to (\vec{a}\times\vec{b}).
So, Assertion is true
But (\vec{a}+\vec{b}) lies on the plane of \vec{a} and \vec{b}, so (\vec{a}+\vec{b}) is not perpendicular to the plane of \vec{a} and \vec{b}
Therefore, Reason is false.
(C) Assertion is true, but Reason is false [1]
The value of p for which vectors \hat{i} + 2\hat{j} + 3\hat{k} and 2\hat{i} - p\hat{j} + \hat{k} are perpendicular to each other is
(C) \frac{5}{2}
The value of m for which the points with position vectors -\hat{i} - \hat{j} + 2\hat{k}, 2\hat{i} + m\hat{j} + 5\hat{k} and 3\hat{i} + 11\hat{j} + 6\hat{k} are collinear, is
(A) 8
If |\vec{a}| = 8, |\vec{b}| = 3 and |\vec{a} \times \vec{b}| = 12, then the value of |\vec{a} \cdot \vec{b}|
(C) 12\sqrt{3}
If |\vec{a}| = 5 and -2 \le \lambda \le 1, then the sum of greatest and the smallest value of |\lambda\vec{a}| is
(C) 10
Vector of magnitude 3 making equal angles with x and y axes and perpendicular to z axis is
(C) \frac{3\sqrt{2}}{2}\hat{i} + \frac{3\sqrt{2}}{2}\hat{j}
For two vectors \vec{a} and \vec{b}
Assertion (A) : |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2\,|\vec{b}|^2
Reason (R) : |\vec{a} \times \vec{b}| = (\vec{a} \cdot \vec{b})\tan\theta,\ \left(\theta \ne \frac{\pi}{2}\right)
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
For any two vectors \vec{a} and \vec{b}, which of the following statements is always true ?
(A) \vec{a} \cdot \vec{b} \le |\vec{a}|\,|\vec{b}|
If (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 198 and |\vec{a}| = 10|\vec{b}|, then :
(B) |\vec{b}| = \sqrt{2}
Assertion (A) : The vectors \vec{a} and (-2\vec{a}), where \vec{a} \ne \vec{0} are collinear vectors.
Reason (R) : \vec{a} \cdot (-2\vec{a}) = 0.
(C) Assertion (A) is true, but Reason (R) is false.
If (3\hat{i} - 2\hat{j} + 5\hat{k}) \times (4\hat{i} + p\hat{j} + q\hat{k}) = \vec{0}, then the values of p and q are :
(B) p = \frac{-8}{3}, q = \frac{20}{3} [1]
Three points A(0, 1, 1), B(2, 0, -1) and C(1, 0, 3) form \DeltaABC. The ar (\Delta ABC) is :
(A) \frac{\sqrt{53}}{2} sq. units [1]
If position vector \vec{p} of a point (24, n) is such that |\vec{p}| = 25, then the value of n is :
(D) \pm 7
If vectors \vec{a} = 3\hat{i} + 2\hat{j} + \lambda\hat{k} and \vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}, represent the two strips of the Red Cross sign placed outside a doctor’s clinic, then the value of \lambda is :
(C) \frac{2}{5}
If vectors \vec{a} = 3\hat{i} - 2\hat{j} + 4\hat{k} and \vec{b} = \lambda\hat{i} + \hat{j} are perpendicular to each other, then the value of \lambda is :
(C) \frac{2}{3}
Assertion (A) : Points A(6, – 7, 0), B(16, –19, – 4) and C(1, – 1, 2) are collinear.
Reason (R) : Three points A, B and C are collinear, when point C is the mid-point of line segment joining points A and B.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
If vector \vec{a}=3\hat{i}+2\hat{j}-\hat{k} and vector \vec{b}=\hat{i}-\hat{j}+\hat{k}, then which of the following is correct ?
(B) \vec{a}\perp\vec{b}
If \vec{a}+\vec{b}+\vec{c}=\vec{0}, |\vec{a}|=\sqrt{37}, |\vec{b}|=3 and |\vec{c}|=4, then angle between \vec{b} and \vec{c} is
(C) \frac{\pi}{3}
The projection vector of vector \vec{a} on vector \vec{b} is
(A) \left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}
Let \vec{p} and \vec{q} be two unit vectors and \alpha be the angle between them. Then (\vec{p} + \vec{q}) will be a unit vector for what value of \alpha ?
(D) \frac{2\pi}{3}
If the sides AB and AC of \Delta ABC are represented by vectors \hat{j}+\hat{k} and 3\hat{i}-\hat{j}+4\hat{k} respectively, then the length of the median through A on BC is :
(C) \frac{\sqrt{34}}{2} units
Let \vec{a} be a position vector whose tip is the point (2,-3). If \overrightarrow{AB}=\vec{a}, where coordinates of A are (-4,5), then the coordinates of B are :
(C) (-2,2)
The respective values of |\vec{a}| and |\vec{b}|, if given (\vec{a}-\vec{b})\cdot(\vec{a}+\vec{b})=512 and |\vec{a}|=3|\vec{b}|, are :
(C) 24 and 8
Let |\vec{a}| = 5 and -2 \le \lambda \le 1. Then, the range of |\lambda \vec{a}| is :
1 mark for any attempt as correct answer is not given in any option
Assertion (A) : If |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 256 and |\vec{b}| = 8, then |\vec{a}| = 2.
Reason (R) : \sin^2\theta + \cos^2\theta = 1 and |\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}| \sin\theta and \vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}| \cos\theta.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
A student tries to tie ropes, parallel to each other from one end of the wall to the other. If one rope is along the vector 3\hat{i}+15\hat{j}+6\hat{k} and the other is along the vector 2\hat{i}+10\hat{j}+\lambda\hat{k}, then the value of \lambda is :
(D) 4 [1]
If |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| for any two vectors, then vectors \vec{a} and \vec{b} are :
(A) orthogonal vectors [1]
If |\vec{a}| = 1, |\vec{b}| = 2 and \vec{a}\cdot\vec{b} = 2, then the value of |\vec{a} + \vec{b}| is :
(B) 3
Two vectors \vec{a} and \vec{b} are such that |\vec{a}\times\vec{b}| = \vec{a}\cdot\vec{b}. The angle between the two vectors is :
(C) 45^\circ
Let \theta be the angle between two unit vectors \hat{a} and \hat{b} such that \sin\theta = \frac{3}{5}. Then, \hat{a} \cdot \hat{b} is equal to :
(C) \pm \frac{4}{5}
The vector with terminal point A (2, -3, 5) and initial point B (3, -4, 7) is :
(D) -\hat{i} + \hat{j} - 2\hat{k}
For any two vectors \vec{a} and \vec{b}, which of the following statements is always true ?
(C) \vec{a} \cdot \vec{b} \le |\vec{a}||\vec{b}|
The unit vector perpendicular to both vectors \hat{i} + \hat{k} and \hat{i} - \hat{k} is :
(B) \hat{j}
Assertion (A) : For two non-zero vectors \vec{a} and \vec{b}, \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}.
Reason (R) : For two non-zero vectors \vec{a} and \vec{b}, \vec{a} \times \vec{b} = \vec{b} \times \vec{a}.
(C) Assertion (A) is true, but Reason (R) is false.
If \vec{a} = 2\hat{i} - \hat{j} + \hat{k} and \vec{b} = \hat{i} + \hat{j} - \hat{k}, then \vec{a} and \vec{b} are :
(C) perpendicular vectors
If |\vec{a}| = 2 and -3 \le k \le 2, then |k\vec{a}| \in :
(D) [0, 6]
Assertion (A) : Projection of \vec{a} on \vec{b} is same as projection of \vec{b} on \vec{a}.
Reason (R) : Angle between \vec{a} and \vec{b} is same as angle between \vec{b} and \vec{a} numerically.
(D) Assertion (A) is false, but Reason (R) is true.
If \vec{a} and \vec{b} are two vectors such that |\vec{a}| = 1, |\vec{b}| = 2 and \vec{a}\cdot\vec{b} = \sqrt{3}, then the angle between 2\vec{a} and -\vec{b} is :
(C) \frac{5\pi}{6}
[1]The vectors \vec{a} = 2\hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} - 3\hat{j} - 5\hat{k} and \vec{c} = -3\hat{i} + 4\hat{j} + 4\hat{k} represents the sides of
(D) a right-angled triangle
[1]Let \vec{a} be any vector such that |\vec{a}| = a. The value of
is :
(B) 2a^2
[1]The position vectors of points P and Q are \vec{p} and \vec{q} respectively. The point R divides line segment PQ in the ratio 3 : 1 and S is the mid-point of line segment PR. The position vector of S is :
(D) \frac{5\vec{p}+3\vec{q}}{8} [1]
Assertion (A) : The vectors
\vec{a}=6\hat{i}+2\hat{j}-8\hat{k}
\vec{b}=10\hat{i}-2\hat{j}-6\hat{k}
\vec{c}=4\hat{i}-4\hat{j}+2\hat{k}
represent the sides of a right angled triangle.
Reason (R) : Three non-zero vectors of which none of two are collinear forms a triangle if their resultant is zero vector or sum of any two vectors is equal to the third.
(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of the Assertion (A). [1]
What is the value of \dfrac{\text{projection of }\vec{a}\text{ on }\vec{b}}{\text{projection of }\vec{b}\text{ on }\vec{a}} for vectors \vec{a}=2\hat{i}-3\hat{j}-6\hat{k} and \vec{b}=2\hat{i}-2\hat{j}+\hat{k} ?
(B) \frac{7}{3}
[1]If \vec{a} and \vec{b} are two vectors such that \vec{a}\cdot\vec{b}\gt 0 and |\vec{a}\cdot\vec{b}|=|\vec{a}\times\vec{b}|, then the angle between \vec{a} and \vec{b} is :
(A) \frac{\pi}{4}
[1]The value of p for which the vectors 2\hat{i}+p\hat{j}+\hat{k} and -4\hat{i}-6\hat{j}+26\hat{k} are perpendicular to each other, is :
(A) 3
The value of (\hat{i}\times\hat{j})\cdot\hat{j}+(\hat{j}\times\hat{i})\cdot\hat{k} is :
(D) -1
If \vec{a}+\vec{b}=\hat{i} and \vec{a}=2\hat{i}-2\hat{j}+2\hat{k}, then |\vec{b}| equals :
(B) 3
Unit vector along \overrightarrow{PQ}, where coordinates of P and Q respectively are (2, 1, –1) and (4, 4, –7), is
(D) \frac{2\hat{i}}{7} + \frac{3\hat{j}}{7} - \frac{6\hat{k}}{7}
Position vector of the mid-point of line segment AB is 3\hat{i} + 2\hat{j} - 3\hat{k}. If position vector of the point A is 2\hat{i} + 3\hat{j} - 4\hat{k}, then position vector of the point B is
(B) 4\hat{i} + \hat{j} - 2\hat{k}
Projection of vector 2\hat{i} + 3\hat{j} on the vector 3\hat{i} - 2\hat{j} is
(A) 0
If a vector makes an angle of \frac{\pi}{4} with the positive directions of both x-axis and y-axis, then the angle which it makes with positive z-axis is :
(C) \frac{\pi}{2}
\vec{a} and \vec{b} are two non-zero vectors such that the projection of \vec{a} on \vec{b} is 0. The angle between \vec{a} and \vec{b} is :
(A) \frac{\pi}{2}
In \Delta ABC, \overrightarrow{AB} = \hat{i} + \hat{j} + 2\hat{k} and \overrightarrow{AC} = 3\hat{i} - \hat{j} + 4\hat{k}. If D is mid-point of BC, then vector \overrightarrow{AD} is equal to :
(D) 2\hat{i} + 3\hat{k}
A unit vector along the vector 4\hat{i}-3\hat{k} is :
(b) \frac{1}{5}(4\hat{i}-3\hat{k})
If \theta is the angle between two vectors \vec{a} and \vec{b}, then \vec{a}\cdot\vec{b}\ge 0 only when :
(b) 0\le\theta\le\frac{\pi}{2}
Two vectors \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} and \vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} are collinear if
(B) \frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}
The magnitude of the vector 6\hat{i}-2\hat{j}+3\hat{k} is
(C) 7
If \vec{a}, \vec{b} and (\vec{a}+\vec{b}) are all unit vectors and \theta is the angle between \vec{a} and \vec{b}, then the value of \theta is :
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The projection of vector \hat{i} on the vector \hat{i}+\hat{j}+2\hat{k} is :
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If ABCD is a parallelogram and AC and BD are its diagonals, then \overrightarrow{AC} + \overrightarrow{BD} is :
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A rectangle ABCD is such that \overrightarrow{AC}=\vec{u}, and \overrightarrow{BD}=\vec{v}, then find the vectors representing four sides of the rectangle (in a cyclic order starting from \overrightarrow{AB}).
[Figure: rectangle ABCD with sides \vec{a},\vec{b},\vec{c},\vec{d} and diagonals \vec{u},\vec{v}]
Let vectors representing the four sides of the rectangle are as follows
\overrightarrow{AB}=\vec{a},\overrightarrow{BC}=\vec{b},\overrightarrow{CD}=\vec{c},and\ \overrightarrow{DA}=\vec{d}
Then \overrightarrow{AC}=\vec{u}=\vec{a}+\vec{b} …(i) and \overrightarrow{BD}=\vec{v}=\vec{b}+\vec{c}=\vec{b}-\vec{a} …(ii) [½]
Solving (i) and (ii),
\vec{a}=\frac12(\vec{u}-\vec{v})\ and\ \vec{b}=\frac12(\vec{u}+\vec{v}) [1]
Also, \vec{c}=-\vec{a}=\frac12(\vec{v}-\vec{u}) and \vec{d}=-\vec{b}=-\frac12(\vec{u}+\vec{v}) [½]
For two non-zero vectors \vec{a} and \vec{b}, |\vec{a}|=\sqrt6, \vec{a}\cdot\vec{b}=1 and \vec{b}\times\vec{a}=3\hat{i}+4\hat{j}-2\hat{k}. Find \left|\vec{b}\right|.
Given |\vec{a}|=\sqrt6, \vec{a}\cdot\vec{b}=1 and \vec{b}\times\vec{a}=3\hat{i}+4\hat{j}-2\hat{k}
So \left|\vec{a}\times\vec{b}\right|=\left|\vec{b}\times\vec{a}\right|=\sqrt{29} [½]
We know, \left|\vec{a}\times\vec{b}\right|^2+\left|\vec{a}\cdot\vec{b}\right|^2=|\vec{a}|^2\left|\vec{b}\right|^2 [½]
\Rightarrow29+1=6\left|\vec{b}\right|^2\Rightarrow\left|\vec{b}\right|^2=\frac{30}{6}=5\Rightarrow\left|\vec{b}\right|=\sqrt5 [1]
The two vectors \hat{i}+\hat{j}+\hat{k} and 3\hat{i}-\hat{j}+3\hat{k} represent the two sides OA and OB, respectively of a \Delta OAB, where O is the origin. The point P lies on AB such that OP is a median. Find the area of the parallelogram formed by the two adjacent sides as OA and OP.
The vector \overrightarrow{OP}=\frac12(4\hat{i}+4\hat{k})=2\hat{i}+2\hat{k} [½]
Area of the parallelogram formed by the two adjacent sides as OA and OP
=\left|\overrightarrow{OA}\times\overrightarrow{OP}\right|=\left\|\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\2&0&2\end{vmatrix}\right\| [½]
=|2\hat{i}-2\hat{k}| [½]
=2\sqrt2 square units. [½]
Find the vector of magnitude 14 in the direction of \overrightarrow{QP}, where P and Q are the points (1, 3, 2) and (–1, 0, 8) respectively.
\overrightarrow{QP} = 2\hat{i} + 3\hat{j} - 6\hat{k} [½]
|\overrightarrow{QP}| = \sqrt{4 + 9 + 36} = 7 [½]
The unit vector in the direction of \overrightarrow{QP} = \frac{\overrightarrow{QP}}{|\overrightarrow{QP}|} = \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}
The required vector = 14\left(\frac{\overrightarrow{QP}}{|\overrightarrow{QP}|}\right) = 4\hat{i} + 6\hat{j} - 12\hat{k} [1]
Vectors \vec{a} = 3\hat{i} - 2\hat{j} + 2\hat{k} and \vec{b} = \hat{i} + 2\hat{k} represent the two adjacent sides of a parallelogram. Find the vectors representing its diagonals and hence find their lengths.
\vec{a} + \vec{b} = 4\hat{i} - 2\hat{j} + 4\hat{k},\ \vec{a} - \vec{b} = 2\hat{i} - 2\hat{j}
The vector representing one of its diagonals is 4\hat{i} - 2\hat{j} + 4\hat{k} or -(4\hat{i} - 2\hat{j} + 4\hat{k}) [½]
The vector representing the other diagonal is 2\hat{i} - 2\hat{j} or -(2\hat{i} - 2\hat{j}) [½]
The lengths of the two diagonals are \sqrt{16 + 4 + 16} = 6 and \sqrt{4 + 4} = 2\sqrt{2} [1]
If \overrightarrow{AB} = \hat{j} + \hat{k} and \overrightarrow{AC} = 3\hat{i} - \hat{j} + 4\hat{k} represent the two vectors along the sides AB and AC of \Delta ABC, prove that the median \overrightarrow{AD} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2}, where D is midpoint of BC.
Hence, find the length of median AD.
Considering A as the origin, \overrightarrow{AB} and \overrightarrow{AC} are p.v. of B and C respectively.
Hence p.v. of the mid point D is \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2} [1]
So, \overrightarrow{AD} = \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2}
\overrightarrow{AD} = \frac{3\hat{i} + 5\hat{k}}{2} [½]
|\overrightarrow{AD}| = \frac{1}{2}\sqrt{3^2 + 5^2} = \frac{\sqrt{34}}{2} [½]
If for two unit vectors \vec{a} and \vec{b}, |\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}|, then find the angle between \vec{a} and \vec{b}.
Given that |\vec{a}| = |\vec{b}| = 1 [½]
|\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}| \Rightarrow |\vec{a} + 2\vec{b}|^2 = |2\vec{a} - \vec{b}|^2 [½]
\Rightarrow |\vec{a}|^2 + 4|\vec{b}|^2 + 4\vec{a}\cdot\vec{b} = 4|\vec{a}|^2 + |\vec{b}|^2 - 4\vec{a}\cdot\vec{b} \Rightarrow \vec{a}\cdot\vec{b} = 0 [½]
\Rightarrow Angle between \vec{a} and \vec{b} = \frac{\pi}{2} [½]
If the position vectors of three points A, B and C are 3\hat{i} + \hat{j}, 5\hat{i} + 6\hat{j} - 3\hat{k} and 4\hat{j} respectively, then show that they form an isosceles triangle.
\overrightarrow{AB} = 2\hat{i} + 5\hat{j} - 3\hat{k}, \overrightarrow{BC} = -5\hat{i} - 2\hat{j} + 3\hat{k}, \overrightarrow{CA} = 3\hat{i} - 3\hat{j} [1]
Now, \left|\overrightarrow{AB}\right| = \left|\overrightarrow{BC}\right| = \sqrt{38} units, \left|\overrightarrow{CA}\right| = \sqrt{18} units [½]
As, \left|\overrightarrow{AB}\right| = \left|\overrightarrow{BC}\right| & sum of two sides is greater than the third so, given points form an isosceles triangle. [½]
Let two rods placed on the ground be represented by vectors 4\hat{i} - \hat{j} + 3\hat{k} and -2\hat{i} + \hat{j} - 2\hat{k}. Find a vector representing a flag-post of height 5 m that has to be erected perpendicular to both the rods.
Let \vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}, \vec{b} = -2\hat{i} + \hat{j} - 2\hat{k}
\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} = -\hat{i} + 2\hat{j} + 2\hat{k} [1]
Now, Required vector = 5\left(\frac{\vec{a} \times \vec{b}}{\left|\vec{a} \times \vec{b}\right|}\right) = \begin{cases} \frac{-5}{3}\hat{i} + \frac{10}{3}\hat{j} + \frac{10}{3}\hat{k} \\ or \\ \frac{5}{3}\hat{i} - \frac{10}{3}\hat{j} - \frac{10}{3}\hat{k} \end{cases} [1] (For any one result)
Alternative method
Alternative approach: Let \vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}, \vec{b} = -2\hat{i} + \hat{j} - 2\hat{k}
Let required vector be \vec{p} = x\hat{i} + y\hat{j} + z\hat{k}
so, 25 = x^{2} + y^{2} + z^{2}, 4x - y + 3z = 0, -2x + y - 2z = 0 [1]
on solving we get, z = -2x,\ y = z \Rightarrow \frac{100}{9} = z^{2} \Rightarrow z = \pm\frac{10}{3}
\therefore Required vector, \vec{p} = \begin{cases} \frac{-5}{3}\hat{i} + \frac{10}{3}\hat{j} + \frac{10}{3}\hat{k} \\ or \\ \frac{5}{3}\hat{i} - \frac{10}{3}\hat{j} - \frac{10}{3}\hat{k} \end{cases} [1] (For any one result)
A unit vector \vec{a} is such that it makes an angle \frac{\pi}{4} with x-axis, \frac{\pi}{3} with y-axis and an acute angle \theta with z-axis. Find \theta and the components of \vec{a}.
\cos^{2}\left(\frac{\pi}{4}\right) + \cos^{2}\left(\frac{\pi}{3}\right) + \cos^{2}\theta = 1 \Rightarrow \cos\theta = \frac{1}{2} (As, \theta is acute) [1]
\Rightarrow \theta = \frac{\pi}{3} & Components of \vec{a} = \frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2} [1]
Three honey bees were found flying along the vectors \vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}, \vec{b} = 4\hat{j} - 2\hat{k} and \vec{c} = 3\hat{i} + 2\hat{k} respectively. Find the value of \lambda such that the path for \vec{a} + \lambda\vec{b} is perpendicular to \vec{c}.
As \vec{a} + \lambda\vec{b} is perpendicular to \vec{c}
\therefore (\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0 [½]
i.e. \vec{a}\cdot\vec{c} + \lambda\vec{b}\cdot\vec{c} = 0
which gives 8 + \lambda(-4) = 0 [1]
\lambda = 2 [½]
If A, B and C be three non-collinear points such that \overrightarrow{AB} = \hat{i} + 2\hat{j} - \hat{k} and \overrightarrow{AC} = 2\hat{i} - 3\hat{j}, then find the area of \Delta ABC.
Area of \Delta ABC = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|
= \frac{1}{2}|-3\hat{i} - 2\hat{j} - 7\hat{k}| [1]
= \frac{\sqrt{62}}{2} [1]
The diagonals of a parallelogram are given by \vec{a}=2\hat{i}-\hat{j}+\hat{k} and \vec{b}=\hat{i}+3\hat{j}-\hat{k}. Find the area of the parallelogram.
\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-1&1\\1&3&-1\end{vmatrix}=-2\hat{i}+3\hat{j}+7\hat{k} [1]
Area of parallelogram =\frac{1}{2}|\vec{a}\times\vec{b}|
=\frac{1}{2}\sqrt{(-2)^2+3^2+7^2}=\frac{\sqrt{62}}{2} [1]
Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors \vec{a}=3\hat{i}+\hat{j}+2\hat{k} and \vec{b}=2\hat{i}-2\hat{j}+4\hat{k}. Determine the angle formed between the kite strings. Assume there is no slack in the strings.
Let the required angle between the kite strings be \theta.
Then, \cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}
\Rightarrow\cos\theta=\frac{(3\hat{i}+\hat{j}+2\hat{k})\cdot(2\hat{i}-2\hat{j}+4\hat{k})}{\sqrt{9+1+4}\sqrt{4+4+16}}=\frac{12}{\sqrt{336}}=\frac{3}{\sqrt{21}} [1½]
\Rightarrow\theta=\cos^{-1}\left(\frac{12}{\sqrt{336}}\right) or \cos^{-1}\left(\frac{3}{\sqrt{21}}\right) [½]
Find a vector of magnitude 21 units in the direction opposite to that of \overrightarrow{AB} where A and B are the points A(2, 1, 3) and B(8, –1, 0) respectively.
\overrightarrow{BA}=-6\hat{i}+2\hat{j}+3\hat{k} [1]
Required unit vector of magnitude 21
=21\times\left(\frac{-6\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{36+4+9}}\right) [½]
=3(-6\hat{i}+2\hat{j}+3\hat{k}) or -18\hat{i}+6\hat{j}+9\hat{k} [½]
A vector \vec{a} makes equal angles with all the three axes. If the magnitude of the vector is 5\sqrt{3} units, then find \vec{a}.
Let \alpha be the angle which the vector \vec{a} makes with all the three axes.
Then 3\cos^2\alpha = 1
\Rightarrow \cos\alpha = \frac{1}{\sqrt{3}} [1]
The unit vector along the vector \vec{a} = \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) [½]
\vec{a} = 5(\hat{i} + \hat{j} + \hat{k}) [½]
If \vec{\alpha} and \vec{\beta} are position vectors of two points P and Q respectively, then find the position vector of a point R in QP produced such that QR = \frac{3}{2}QP.
[Figure: points R(\vec{x}), P(\vec{\alpha}), Q(\vec{\beta}) on a line]
\frac{QR}{QP} = \frac{3}{2}
Hence, R divides PQ, externally, in the ratio 1:3. [1]
The Position vector of R = \vec{x} = \frac{\vec{\beta} - 3\vec{\alpha}}{1 - 3} = \frac{3\vec{\alpha} - \vec{\beta}}{2} [1]
If \vec{a} and \vec{b} are two non-collinear vectors, then find x, such that \vec{\alpha} = (x - 2)\vec{a} + \vec{b} and \vec{\beta} = (3 + 2x)\vec{a} - 2\vec{b} are collinear.
\vec{\alpha} and \vec{\beta} are collinear
\Rightarrow \frac{x - 2}{3 + 2x} = \frac{1}{-2} [1½]
\Rightarrow x = \frac{1}{4} [½]
Find a vector of magnitude 5 which is perpendicular to both the vectors 3\hat{i}-2\hat{j}+\hat{k} and 4\hat{i}+3\hat{j}-2\hat{k}.
Let \vec{a}=3\hat{i}-2\hat{j}+\hat{k},\quad\vec{b}=4\hat{i}+3\hat{j}-2\hat{k}
\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-2&1\\4&3&-2\end{vmatrix}=\hat{i}+10\hat{j}+17\hat{k} [½]
|\vec{a}\times\vec{b}|=\sqrt{1^2+10^2+17^2}=\sqrt{390} [½]
Unit vector \hat{n}=\frac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}=\frac{1}{\sqrt{390}}(\hat{i}+10\hat{j}+17\hat{k}) [½]
\therefore Required vector =\frac{5}{\sqrt{390}}(\hat{i}+10\hat{j}+17\hat{k}) [½]
Let \vec{a}, \vec{b} and \vec{c} be three vectors such that \vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c} and \vec{a}\times\vec{b}=\vec{a}\times\vec{c}, \vec{a}\ne0. Show that \vec{b}=\vec{c}.
\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}\Rightarrow\vec{a}\cdot(\vec{b}-\vec{c})=0
\Rightarrow either \vec{b}=\vec{c} or \vec{a}\perp(\vec{b}-\vec{c}), since \vec{a}\ne0 [1]
Also, \vec{a}\times\vec{b}=\vec{a}\times\vec{c}\Rightarrow\vec{a}\times(\vec{b}-\vec{c})=0
\Rightarrow either \vec{b}=\vec{c} or \vec{a}\parallel(\vec{b}-\vec{c}), since \vec{a}\ne0 [½]
Since vectors \vec{a} and (\vec{b}-\vec{c}) cannot be \parallel and \perp simultaneously
Hence \vec{b}=\vec{c} [½]
A man needs to hang two lanterns on a straight wire whose end points have coordinates A (4,1,-2) and B (6,2,-3). Find the coordinates of the points where he hangs the lanterns such that these points trisect the wire AB.
Let P and Q trisect the wire AB.
P divides AB in the ratio 1:2 then, coordinate of point P =\left(\frac{14}{3},\frac43,-\frac73\right) [1]
Q divides AB in the ratio 2:1 then, coordinate of point Q =\left(\frac{16}{3},\frac53,-\frac83\right) [1]
If \vec{a} and \vec{b} are position vectors of point A and point B respectively, find the position vector of point C on BA produced such that BC = 3BA.
C divides BA in the ratio 3 : 2 externally [1]
Required vector = \vec{c} = \frac{3\vec{a} - 2\vec{b}}{3-2} = 3\vec{a} - 2\vec{b} [1]
Vector \vec{r} is inclined at equal angles to the three axes x, y and z. If magnitude of \vec{r} is 5\sqrt{3} units, then find \vec{r}.
Unit vector equally inclined along coordinate axes is \frac{\hat{i}}{\sqrt{3}} + \frac{\hat{j}}{\sqrt{3}} + \frac{\hat{k}}{\sqrt{3}} [1]
\vec{r} = 5\sqrt{3}\left(\frac{\hat{i}}{\sqrt{3}} + \frac{\hat{j}}{\sqrt{3}} + \frac{\hat{k}}{\sqrt{3}}\right) = 5\hat{i} + 5\hat{j} + 5\hat{k} (or -5\hat{i} - 5\hat{j} - 5\hat{k}) [1]
Find the value of \lambda, if the points (-1, -1, 2), (2, 8, \lambda) and (3, 11, 6) are collinear.
A(-1,-1,2), B(2,8,\lambda), C(3,11,6)
\overrightarrow{AB} = 3\hat{i} + 9\hat{j} + (\lambda - 2)\hat{k} and \overrightarrow{BC} = \hat{i} + 3\hat{j} + (6 - \lambda)\hat{k} [1]
Since A, B and C are collinear, \frac{3}{1} = \frac{9}{3} = \frac{\lambda - 2}{6 - \lambda} [½]
\Rightarrow \lambda = 5 [½]
\vec{a} and \vec{b} are two co-initial vectors forming the adjacent sides of a parallelogram such that |\vec{a}| = 10, |\vec{b}| = 2 and \vec{a}\cdot\vec{b} = 12. Find the area of the parallelogram.
Let \theta is the angle between \vec{a} and \vec{b}.
\vec{a}\cdot\vec{b} = 12 \Rightarrow |\vec{a}||\vec{b}|\cos\theta = 12 [½]
\Rightarrow (10)(2)\cos\theta = 12 \Rightarrow \cos\theta = \frac{3}{5} [½]
\therefore \sin\theta = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} [½]
Now, area of parallelogram = |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta
= (10)(2)\left(\frac{4}{5}\right) = 16 [½]
\therefore area of parallelogram = 16
If \vec{a} and \vec{b} are two non-zero vectors such that (\vec{a} + \vec{b}) \perp \vec{a} and (2\vec{a} + \vec{b}) \perp \vec{b}, then prove that |\vec{b}| = \sqrt{2}\,|\vec{a}|.
(\vec{a} + \vec{b})\cdot\vec{a} = 0 \Rightarrow |\vec{a}|^2 + \vec{b}\cdot\vec{a} = 0 ----------(1) [½]
(2\vec{a} + \vec{b})\cdot\vec{b} = 0 \Rightarrow 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 0 ----------(2) [½]
2(-|\vec{a}|^2) + |\vec{b}|^2 = 0 {Using (1) and (2)} [½]
|\vec{b}|^2 = 2|\vec{a}|^2 \Rightarrow |\vec{b}| = \sqrt{2}|\vec{a}| [½]
In the given figure, ABCD is a parallelogram. If \overrightarrow{AB} = 2\hat{i} - 4\hat{j} + 5\hat{k} and \overrightarrow{DB} = 3\hat{i} - 6\hat{j} + 2\hat{k}, then find \overrightarrow{AD} and hence find the area of parallelogram ABCD.
\overrightarrow{AD} + \overrightarrow{DB} = \overrightarrow{AB}
\overrightarrow{AD} = (2\hat{i} - 4\hat{j} + 5\hat{k}) - (3\hat{i} - 6\hat{j} + 2\hat{k})
= -\hat{i} + 2\hat{j} + 3\hat{k} [½]
\overrightarrow{AD} \times \overrightarrow{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 2 & -4 & 5 \end{vmatrix} = 22\hat{i} + 11\hat{j} [1]
Area = |\overrightarrow{AD} \times \overrightarrow{AB}| = |22\hat{i} + 11\hat{j}|
= \sqrt{605}\ \text{or}\ 11\sqrt{5} [½]
Find the position vector of point C which divides the line segment joining points A and B having position vectors \hat{i} + 2\hat{j} - \hat{k} and -\hat{i} + \hat{j} + \hat{k} respectively in the ratio 4 : 1 externally. Further, find |\overrightarrow{AB}| : |\overrightarrow{BC}|.
Position vector of C = \vec{r} = \frac{4\vec{b} - \vec{a}}{3}
i.e. \vec{r} = \frac{1}{3}\left(-5\hat{i} + 2\hat{j} + 5\hat{k}\right) [1]
Now, \overrightarrow{AB} = -2\hat{i} - \hat{j} + 2\hat{k} \Rightarrow |\overrightarrow{AB}| = 3
\overrightarrow{BC} = -\frac{1}{3}\left(2\hat{i} + \hat{j} - 2\hat{k}\right) \Rightarrow |\overrightarrow{BC}| = 1
|\overrightarrow{AB}| : |\overrightarrow{BC}| = 3 : 1 [1]
Let \vec{a} and \vec{b} be two non-zero vectors.
Prove that |\vec{a} \times \vec{b}| \le |\vec{a}|\,|\vec{b}|.
State the condition under which equality holds, i.e., |\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|.
|\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\,|\sin\theta| [½]
As, 0 \le |\sin\theta| \le 1 [½]
\Rightarrow |\vec{a}|\,|\vec{b}|\,|\sin\theta| \le |\vec{a}|\,|\vec{b}| [½]
\Rightarrow |\vec{a} \times \vec{b}| \le |\vec{a}|\,|\vec{b}|
For equality, \sin\theta = 1 \Rightarrow \theta = \frac{\pi}{2} \Rightarrow \vec{a} is perpendicular to \vec{b}. [½]
Find all the vectors of magnitude 3\sqrt3 which are collinear to vector \hat{i}+\hat{j}+\hat{k}.
Unit vector along \hat{i}+\hat{j}+\hat{k} is \frac{\hat{i}}{\sqrt3}+\frac{\hat{j}}{\sqrt3}+\frac{\hat{k}}{\sqrt3} [1]
Required vectors are 3\hat{i}+3\hat{j}+3\hat{k} and -3\hat{i}-3\hat{j}-3\hat{k} [½ + ½]
Position vectors of the points A, B and C as shown in the figure below are \vec{a}, \vec{b} and \vec{c} respectively.
If \overrightarrow{AC}=\frac{5}{4}\overrightarrow{AB}, express \vec{c} in terms of \vec{a} and \vec{b}.
According to question, \vec{c}-\vec{a}=\frac{5}{4}(\vec{b}-\vec{a}) [1½]
\therefore\vec{c}=\frac{5\vec{b}}{4}-\frac{\vec{a}}{4} [½]
If \vec{a}, \vec{b}, \vec{c} are three non-zero unequal vectors such that \vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}, then find the angle between \vec{a} and \vec{b} - \vec{c}.
\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c} \Rightarrow \vec{a}\cdot(\vec{b} - \vec{c}) = 0 \Rightarrow \vec{a} = \vec{0}\ ;\ \vec{b} = \vec{c} \text{ or } \vec{a} \perp (\vec{b} - \vec{c}) [1½]
As, \vec{a} \ne \vec{0}\ ;\ \vec{b} \ne \vec{c}\ \therefore the angle between \vec{a} and \vec{b} - \vec{c} is \frac{\pi}{2} [½]
If \vec{r} = 3\hat{i} - 2\hat{j} + 6\hat{k}, find the value of (\vec{r} \times \hat{j}) \cdot (\vec{r} \times \hat{k}) - 12.
(\vec{r} \times \hat{j}) \cdot (\vec{r} \times \hat{k}) - 12 = (3\hat{k} - 6\hat{i}) \cdot (-3\hat{j} - 2\hat{i}) - 12 [1½]
= 12 - 12 = 0 [½]
If the projection of the vector \hat{i}+\hat{j}+\hat{k} on the vector p\hat{i}+\hat{j}-2\hat{k} is \frac{1}{3}, then find the value(s) of p.
Here, \left[\frac{(\hat{i}+\hat{j}+\hat{k})\cdot(p\hat{i}+\hat{j}-2\hat{k})}{\sqrt{p^2+1+4}}\right]=\frac{1}{3} [½]
\Rightarrow\frac{p-1}{\sqrt{p^2+5}}=\frac{1}{3} [½]
\Rightarrow 8p^2-18p+4=0
4p^2-9p+2=0 [½]
4p^2-8p-p+2=0
(4p-1)(p-2)=0
\Rightarrow p=2 or p=\frac{1}{4} [½]
If the vectors \vec{a} and \vec{b} are such that |\vec{a}|=3, |\vec{b}|=\frac{2}{3} and \vec{a}\times\vec{b} is a unit vector, then find the angle between \vec{a} and \vec{b}.
Let \theta be the angle between \vec{a} and \vec{b}
Since \vec{a}\times\vec{b} is a unit vector, we have |\vec{a}\times\vec{b}|=1
\Rightarrow|\vec{a}|\,|\vec{b}|\sin\theta=1 [1]
\Rightarrow\sin\theta=\frac{1}{2}, or \theta=30^\circ (or \frac{\pi}{6}) [1]
Find the area of a parallelogram whose adjacent sides are determined by the vectors \vec{a}=\hat{i}-\hat{j}+3\hat{k} and \vec{b}=2\hat{i}-7\hat{j}+\hat{k}.
Here
\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&-7&1\end{vmatrix}=20\hat{i}+5\hat{j}-5\hat{k} [1½]
\Rightarrow|\vec{a}\times\vec{b}|=\sqrt{400+25+25}=\sqrt{450}
Area of parallelogram =|\vec{a}\times\vec{b}|=\sqrt{450}=15\sqrt2 [½]
If three non-zero vectors are \vec{a}, \vec{b} and \vec{c} such that \vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c} and \vec{a}\times\vec{b} = \vec{a}\times\vec{c}, then show that \vec{b} = \vec{c}.
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If \vec{a}, \vec{b} and \vec{c} are three vectors such that |\vec{a}| = 7, |\vec{b}| = 24, |\vec{c}| = 25 and \vec{a}+\vec{b}+\vec{c} = \vec{0}, then find the value of \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}.
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In a parallelogram PQRS, \overrightarrow{PQ}=3\hat{i}-2\hat{j}+2\hat{k} and \overrightarrow{PS}=-\hat{i}-2\hat{k} . Find |\overrightarrow{PR}| and |\overrightarrow{QS}| .
\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{PS} \quad \because \overrightarrow{PS}=\overrightarrow{QR}
=2\hat{i}-2\hat{j}
|\overrightarrow{PR}|=2\sqrt{2} [1]
\overrightarrow{QS}=\overrightarrow{QP}+\overrightarrow{PS}=\overrightarrow{PS}-\overrightarrow{PQ}
=-4\hat{i}+2\hat{j}-4\hat{k}
|\overrightarrow{QS}|=6 [1]
\vec{a} and \vec{b} are two unit vectors such that |2\vec{a}+3\vec{b}|=|3\vec{a}-2\vec{b}|. Find the angle between \vec{a} and \vec{b}.
|2\vec{a}+3\vec{b}|=|3\vec{a}-2\vec{b}|
\Rightarrow|2\vec{a}+3\vec{b}|^2=|3\vec{a}-2\vec{b}|^2
\Rightarrow 4|\vec{a}|^2+12\,\vec{a}\cdot\vec{b}+9|\vec{b}|^2=9|\vec{a}|^2-12\,\vec{a}\cdot\vec{b}+4|\vec{b}|^2 [1]
As |\vec{a}|=|\vec{b}|=1
\therefore 24\,\vec{a}\cdot\vec{b}=5|\vec{a}|^2-5|\vec{b}|^2=0\Rightarrow\vec{a}\cdot\vec{b}=0 [½]
So, \vec{a}\perp\vec{b} or Angle between them is \frac{\pi}{2} [½]
Write the projection of the vector (\vec{b}+\vec{c}) on the vector \vec{a}, where \vec{a}=2\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-2\hat{k} and \vec{c}=2\hat{i}-\hat{j}+4\hat{k}.
\vec{b}+\vec{c}=3\hat{i}+\hat{j}+2\hat{k} [½]
Projection of \vec{b}+\vec{c} on \vec{a}=\frac{(\vec{b}+\vec{c})\cdot\vec{a}}{|\vec{a}|}
=\frac{(3\hat{i}+\hat{j}+2\hat{k})\cdot(2\hat{i}-2\hat{j}+\hat{k})}{\sqrt{9}} [1]
=\frac{6-2+2}{3}=\frac{6}{3}=2 [½]
If |\vec{a}\times\vec{b}|^2+|\vec{a}\cdot\vec{b}|^2=400 and |\vec{b}|=5, then find the value of |\vec{a}|.
|\vec{a}|^2|\vec{b}|^2\sin^2\theta+|\vec{a}|^2|\vec{b}|^2\cos^2\theta=400 [1]
|\vec{a}|^2\cdot 25(1)=400 [½]
|\vec{a}|^2=16
|\vec{a}|=4 [½]
Find all the possible vectors of magnitude 5\sqrt{3} which are equally inclined to the coordinate axes.
Let the required vector be x\hat{i}+x\hat{j}+x\hat{k} [½]
\sqrt{3x^2}=5\sqrt{3} [½]
x^2=25\Rightarrow x=\pm 5 [½]
Required vectors are 5\hat{i}+5\hat{j}+5\hat{k} or -5\hat{i}-5\hat{j}-5\hat{k}. [½]
If \vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{a}\cdot\vec{b}=1 and \vec{a}\times\vec{b}=\hat{j}-\hat{k}, then find \left|\vec{b}\right|
Let \vec{b}=x\hat{i}+y\hat{j}+z\hat{k}
\vec{a}\cdot\vec{b}=1 \Rightarrow x+y+z=1 \quad \ldots (1) [½]
\vec{a}\times\vec{b}=\hat{j}-\hat{k}
\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\x&y&z\end{vmatrix}=\hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x)=\hat{j}-\hat{k} [½]
z-y=0 \Rightarrow y=z
\Rightarrow x-z=1 \quad \ldots (2)
x-y=1 \quad \ldots (3)
Solving (1), (2), (3)
x=1,\ y=0,\ z=0 [½]
\vec{b}=\hat{i}, so |\vec{b}|=1 [½]
If \vec{a} = \hat{i}+\hat{j}-2\hat{k},\ \vec{b} = -\hat{i}+2\hat{j}+2\hat{k} and \vec{c} = -\hat{i}+2\hat{j}-\hat{k} are three vectors, then find a vector perpendicular to both the vectors (\vec{a}+\vec{b}) and (\vec{b}-\vec{c}) .
\vec{a}+\vec{b} = 3\hat{j},\ \vec{b}-\vec{c} = 3\hat{k} [½+½]
Vector perpendicular to \vec{a}+\vec{b} and \vec{b}-\vec{c}
= (3\hat{j})\times(3\hat{k}) = 9\hat{i} [1]
If \vec{a}, \vec{b} and \vec{c} are unit vectors such that \vec{a}+\vec{b}+\vec{c} = \vec{0} , then find the value of \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} .
|\vec{a}| = |\vec{b}| = |\vec{c}| = 1
\vec{a}+\vec{b}+\vec{c} = 0 \Rightarrow (\vec{a}+\vec{b}+\vec{c})(\vec{a}+\vec{b}+\vec{c}) = 0
\Rightarrow |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0 [1]
\Rightarrow 3+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0 [½]
\Rightarrow \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = \frac{-3}{2} [½]
Let three toys A, B and C be placed in the same straight line. If the position vectors of A, B and C are 55\hat{i} - 2\hat{j}, 5\hat{i} + 8\hat{j} and a\hat{i} - 52\hat{j} respectively, find the value of ‘a’.
\overrightarrow{AB} = -50\hat{i} + 10\hat{j} and \overrightarrow{BC} = (a - 5)\hat{i} - 60\hat{j} [1]
As \overrightarrow{AB} and \overrightarrow{BC} are collinear vectors
\therefore \frac{-50}{a-5} = \frac{10}{-60} [1]
a = 305 [1]
If \vec{a}, \vec{b} and \vec{c} are unit vectors, then prove that
|\vec{a}| = |\vec{b}| = |\vec{c}| = 1 [½]
Now, consider
|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2
= 2|\vec{a}|^2 + 2|\vec{b}|^2 + 2|\vec{c}|^2 - 2\vec{a}.\vec{b} - 2\vec{b}.\vec{c} - 2\vec{c}.\vec{a} [1½]
= 3|\vec{a}|^2 + 3|\vec{b}|^2 + 3|\vec{c}|^2 - (|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{a}.\vec{b} + 2\vec{b}.\vec{c} + 2\vec{c}.\vec{a}) [½]
= 9 - |\vec{a} + \vec{b} + \vec{c}|^2 \le 9
|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \le 9 [½]
During a cricket match, the position of the bowler, the wicket keeper and the leg slip fielder are in a line given by \vec{B}=2\hat{i}+8\hat{j}, \vec{W}=6\hat{i}+12\hat{j} and \vec{F}=12\hat{i}+18\hat{j} respectively. Calculate the ratio in which the wicketkeeper divides the line segment joining the bowler and the leg slip fielder.
Let the wicket keeper divides the line segment in ratio k:1
\therefore\vec{W}=\frac{k\vec{F}+1.\vec{B}}{k+1} [1]
\Rightarrow 6\hat{i}+12\hat{j}=\left(\frac{12k+2}{k+1}\right)\hat{i}+\left(\frac{18k+8}{k+1}\right)\hat{j} [1]
\Rightarrow k=\frac{2}{3}
Hence, the required ratio is 2:3 [1]
If \vec{a}+\vec{b}+\vec{c}=\vec{0} such that |\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7, then find the angle between \vec{a} and \vec{b}.
Given \vec{a}+\vec{b}+\vec{c}=\vec{0}\Rightarrow|\vec{a}+\vec{b}|=|-\vec{c}| [1]
\Rightarrow|\vec{a}+\vec{b}|^2=|\vec{c}|^2\Rightarrow|\vec{a}|^2+|\vec{b}|^2+2\,\vec{a}\cdot\vec{b}=|\vec{c}|^2
\Rightarrow9+25+2\,\vec{a}\cdot\vec{b}=49 [1]
\Rightarrow2|\vec{a}||\vec{b}|\cos\theta=15
\Rightarrow\cos\theta=\frac12\ \therefore\ \theta=\frac{\pi}{3} [1]
If \vec{a} and \vec{b} are unit vectors inclined with each other at an angle \theta, then prove that \frac12|\vec{a}-\vec{b}|=\sin\frac{\theta}{2}.
|\vec{a}|=|\vec{b}|=1 [½]
|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b} [1]
=1+1-2|\vec{a}||\vec{b}|\cos\theta [½]
=2-2\cos\theta
=2\left(2\sin^2\frac{\theta}{2}\right)=4\sin^2\frac{\theta}{2} [½]
\Rightarrow\sin\frac{\theta}{2}=\frac12|\vec{a}-\vec{b}| [½]
The scalar product of the vector \vec{a}=\hat{i}-\hat{j}+2\hat{k} with a unit vector along sum of vectors \vec{b}=2\hat{i}-4\hat{j}+5\hat{k} and \vec{c}=\lambda\hat{i}-2\hat{j}-3\hat{k} is equal to 1. Find the value of \lambda.
Let \vec{d}=\vec{b}+\vec{c}=(2+\lambda)\hat{i}-6\hat{j}+2\hat{k} [1]
\hat{d}=\frac{(2+\lambda)\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{(2+\lambda)^2+40}} [½]
\vec{a}\cdot\hat{d}=\left(\hat{i}-\hat{j}+2\hat{k}\right)\cdot\frac{(2+\lambda)\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{(2+\lambda)^2+40}}=1 [1]
\Rightarrow(2+\lambda)+6+4=\sqrt{(2+\lambda)^2+40}\Rightarrow\lambda=-5 [½]
If \hat{a}, \hat{b} and \hat{c} are unit vectors such that \hat{a}\cdot\hat{b} = \hat{a}\cdot\hat{c} = 0 and the angle between \hat{b} and \hat{c} is \frac{\pi}{6}, then prove that \hat{a} = \pm 2(\hat{b}\times\hat{c}).
\hat{a}\cdot\hat{b} = 0 \Rightarrow \hat{a}\perp\hat{b},\ \hat{a}\cdot\hat{c} = 0 \Rightarrow \hat{a}\perp\hat{c} [1]
\Rightarrow \hat{a} is perpendicular to both \hat{b} and \hat{c} \Rightarrow \hat{a} \parallel \left(\hat{b}\times\hat{c}\right)
Let \hat{a} = \lambda\left(\hat{b}\times\hat{c}\right) [½]
\Rightarrow |\hat{a}| = |\lambda|\left|\left(\hat{b}\times\hat{c}\right)\right| \Rightarrow |\hat{a}| = |\lambda||\hat{b}||\hat{c}|\sin\frac{\pi}{6} [1]
\Rightarrow |\lambda| = 2 \Rightarrow \lambda = \pm 2 [½]
\therefore \hat{a} = \pm 2\left(\hat{b}\times\hat{c}\right)
The position vectors of vertices of \Delta ABC are A(2\hat{i} - \hat{j} + \hat{k}), B(\hat{i} - 3\hat{j} - 5\hat{k}) and C(3\hat{i} - 4\hat{j} - 4\hat{k}). Find all the angles of \Delta ABC.
\cos A = \frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{(-\hat{i}-2\hat{j}-6\hat{k})\cdot(\hat{i}-3\hat{j}-5\hat{k})}{\sqrt{41}\sqrt{35}} = \frac{35}{\sqrt{41}\sqrt{35}} = \frac{\sqrt{35}}{\sqrt{41}}
A = \cos^{-1}\left(\frac{\sqrt{35}}{\sqrt{41}}\right) [1]
\cos B = \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} = \frac{(\hat{i}+2\hat{j}+6\hat{k})\cdot(2\hat{i}-\hat{j}+\hat{k})}{\sqrt{41}\sqrt{6}} = \frac{6}{\sqrt{41}\sqrt{6}} = \frac{\sqrt{6}}{\sqrt{41}}
B = \cos^{-1}\left(\frac{\sqrt{6}}{\sqrt{41}}\right) [1]
\cos C = \frac{\overrightarrow{CB}\cdot\overrightarrow{CA}}{|\overrightarrow{CB}||\overrightarrow{CA}|} = \frac{(-2\hat{i}+\hat{j}-\hat{k})\cdot(-\hat{i}+3\hat{j}+5\hat{k})}{|\overrightarrow{CB}||\overrightarrow{CA}|} = 0
cos C = 0 \Rightarrow C = \frac{\pi}{2} [1]
Find a vector of magnitude 4 units perpendicular to each of the vectors 2\hat{i}-\hat{j}+\hat{k} and \hat{i}+\hat{j}-\hat{k} and hence verify your answer.
Let \vec{a}=2\hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-\hat{k} and \vec{c} be the vector perpendicular to both \vec{a}\ \&\ \vec{b}
then, \vec{c}=\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 2 & -1 & 1\\ 1 & 1 & -1\end{vmatrix}=3\hat{j}+3\hat{k} [1]
Let \vec{d} the vector perpendicular to both the vectors \vec{a}\ \&\ \vec{b} and having magnitude 4,
\vec{d}=4\hat{c}=2\sqrt2\hat{j}+2\sqrt2\hat{k} (or -2\sqrt2\hat{j}-2\sqrt2\hat{k}) [1]
Verification: |\vec{d}|=\sqrt{8+8}=4,\ \vec{d}\cdot\vec{a}=\vec{d}\cdot\vec{b}=0\Rightarrow\vec{d}\perp\vec{a} and \vec{d}\perp\vec{b} [1]
If the vectors \vec{a}, \vec{b} and \vec{c} represent the three sides of a triangle, then show that \vec{a}\times\vec{b}=\vec{b}\times\vec{c}=\vec{c}\times\vec{a}.
\vec{a}+\vec{b}+\vec{c}=\vec{O} [1]
\Rightarrow\vec{a}\times(\vec{a}+\vec{b}+\vec{c})=\vec{O} [½]
\Rightarrow\vec{a}\times\vec{a}+\vec{a}\times\vec{b}+\vec{a}\times\vec{c}=\vec{O}
\Rightarrow\vec{a}\times\vec{b}=-\vec{a}\times\vec{c},\ \therefore\vec{a}\times\vec{b}=\vec{c}\times\vec{a} [1]
Similarly, \vec{b}\times\vec{c}=\vec{c}\times\vec{a}, \therefore\vec{a}\times\vec{b}=\vec{b}\times\vec{c}=\vec{c}\times\vec{a} [½]
Three vectors \vec{a}, \vec{b} and \vec{c} satisfy the condition \vec{a}+\vec{b}+\vec{c}=\vec{0}. Evaluate the quantity \mu=\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}, if |\vec{a}|=3, |\vec{b}|=4 and |\vec{c}|=2.
(\vec{a}+\vec{b}+\vec{c})^2=0 [1]
\Rightarrow\vec{a}^2+\vec{b}^2+\vec{c}^2+2(\mu)=0 [1]
\Rightarrow\mu=-\frac{29}{2} [1]
If \vec{a},\vec{b},\vec{c} and \vec{d} are four non-zero vectors such that \vec{a}\times\vec{b}=\vec{c}\times\vec{d} and \vec{a}\times\vec{c}=4\vec{b}\times\vec{d} , then show that (\vec{a}-2\vec{d}) is parallel to (2\vec{b}-\vec{c}) where \vec{a}\ne 2\vec{d},\ \vec{c}\ne 2\vec{b} .
Consider (\vec{a}-2\vec{d})\times(2\vec{b}-\vec{c}) [½]
=\vec{a}\times 2\vec{b}-\vec{a}\times\vec{c}-4\vec{d}\times\vec{b}+2\vec{d}\times\vec{c} [1]
=0 [1]
\therefore (\vec{a}-2\vec{d})\parallel(2\vec{b}-\vec{c}) [½]
The two adjacent sides of a parallelogram are represented by 2\hat{i}-4\hat{j}-5\hat{k} and 2\hat{i}+2\hat{j}+3\hat{k} . Find the unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram also.
Let ABCD be a parallelogram with
\overrightarrow{AB}=\overrightarrow{DC}=2\hat{i}-4\hat{j}-5\hat{k} and \overrightarrow{BC}=\overrightarrow{AD}=2\hat{i}+2\hat{j}+3\hat{k}
\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=4\hat{i}-2\hat{j}-2\hat{k} and \overrightarrow{BD}=6\hat{j}+8\hat{k} [1]
\therefore |\overrightarrow{AC}|=2\sqrt{6} and |\overrightarrow{BD}|=10
\therefore Required unit vectors \hat{d}_1 and \hat{d}_2 are
\hat{d}_1=\frac{2}{\sqrt{6}}\hat{i}-\frac{1}{\sqrt{6}}\hat{j}-\frac{\hat{k}}{\sqrt{6}} and \hat{d}_2=\frac{3}{5}\hat{j}+\frac{4}{5}\hat{k} [1]
Now,
=\frac{1}{2}|-4\hat{i}-32\hat{j}+24\hat{k}|
=\frac{1}{2}\sqrt{1616}=2\sqrt{101} [1]
ABCD is a parallelogram such that \overrightarrow{AC}=\hat{i}+\hat{j} and \overrightarrow{BD}=2\hat{i}+\hat{j}+\hat{k}. Find \overrightarrow{AB} and \overrightarrow{AD}. Also, find the area of the parallelogram ABCD.
Let \overrightarrow{AB}=\vec{a} and \overrightarrow{AD}=\vec{b}
\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\vec{a}+\vec{b}=\hat{i}+\hat{j}
\overrightarrow{BD}=\overrightarrow{BC}+\overrightarrow{CD}=\vec{b}-\vec{a}=2\hat{i}+\hat{j}+\hat{k}
Adding we get, 2\overrightarrow{AD}=\overrightarrow{AC}+\overrightarrow{BD}=3\hat{i}+2\hat{j}+\hat{k}
\Rightarrow\overrightarrow{AD}=\frac{3}{2}\hat{i}+\hat{j}+\frac{1}{2}\hat{k} [1]
Subtracting, we get
2\overrightarrow{AB}=\overrightarrow{AC}-\overrightarrow{BD}=-\hat{i}-\hat{k}\Rightarrow\overrightarrow{AB}=-\frac{1}{2}\hat{i}-\frac{1}{2}\hat{k} [1]
\left|\overrightarrow{AC}\times\overrightarrow{BD}\right|=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&0\\2&1&1\end{vmatrix}=\hat{i}-\hat{j}-\hat{k}
Area =\frac{1}{2}\left|\overrightarrow{AC}\times\overrightarrow{BD}\right|
=\frac{\sqrt3}{2} [1]
The two adjacent sides of a parallelogram are represented by vectors 2\hat{i}-4\hat{j}+5\hat{k} and \hat{i}-2\hat{j}-3\hat{k}. Find the unit vector parallel to one of its diagonals. Also, find the area of the parallelogram.
One diagonal of the parallelogram
=(2\hat{i}-4\hat{j}+5\hat{k})+(\hat{i}-2\hat{j}-3\hat{k})
=3\hat{i}-6\hat{j}+2\hat{k} [1]
Unit vector parallel to the diagonal
=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{9+36+4}}
=\frac37\hat{i}-\frac67\hat{j}+\frac27\hat{k} [1]
Vector area of parallelogram =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-4&5\\1&-2&-3\end{vmatrix}
=\hat{i}(22)-\hat{j}(-11)+\hat{k}(0)
=22\hat{i}+11\hat{j}
Area =\sqrt{484+121}=\sqrt{605}=11\sqrt5 [1]
If \vec{a}=2\hat{i}+2\hat{j}+3\hat{k}, \vec{b}=-\hat{i}+2\hat{j}+\hat{k} and \vec{c}=3\hat{i}+\hat{j} are such that the vector (\vec{a}+\lambda\vec{b}) is perpendicular to vector \vec{c}, then find the value of \lambda.
(\vec{a}+\lambda\vec{b})\cdot\vec{c}=0\Rightarrow\left[(2-\lambda)\hat{i}+(2+2\lambda)\hat{j}+(3+\lambda)\hat{k}\right]\cdot(3\hat{i}+\hat{j})=0 [½+1]
\Rightarrow 3(2-\lambda)+(2+2\lambda)\cdot1=0 [1]
\Rightarrow -3\lambda+2\lambda+6+2=0
\Rightarrow\lambda=8 [½]
If \vec{a},\ \vec{b} and \vec{c} are mutually perpendicular vectors of equal magnitude, then prove that the vector (2\vec{a}+\vec{b}+2\vec{c}) is equally inclined to both \vec{a} and \vec{c}. Also, find the angle between \vec{a} and (2\vec{a}+\vec{b}+2\vec{c}).
Let, |\vec{a}|=|\vec{b}|=|\vec{c}|=m [½]
|2\vec{a}+\vec{b}+2\vec{c}|=3m
Let \theta be angle between 2\vec{a}+\vec{b}+2\vec{c} and \vec{a}.
Let, \alpha be angle between 2\vec{a}+\vec{b}+2\vec{c} and \vec{c}
Similarly, \cos\alpha=\frac{2m^2}{3m\cdot m}=\frac{2}{3}
\therefore\cos\theta=\cos\alpha\Rightarrow\theta=\alpha [½]
\therefore Required angle =\cos^{-1}\left(\frac{2}{3}\right) [½]
If \vec{a} and \vec{b} are two vectors such that \left|\vec{a}+\vec{b}\right|=\left|\vec{b}\right|, then prove that \left(\vec{a}+2\vec{b}\right) is perpendicular to \vec{a}.
|\vec{a}+\vec{b}|=|\vec{b}|
\left(\vec{a}+\vec{b}\right)^2=\left(\vec{b}\right)^2 [1]
\vec{a}^2+\vec{b}^2+2\vec{a}.\vec{b}=\vec{b}^2 [½]
\vec{a}^2+2\vec{a}.\vec{b}=0
(\vec{a}+2\vec{b}).\vec{a}=0 [1]
\therefore (\vec{a}+2\vec{b})\perp\vec{a} [½]
If \vec{a} and \vec{b} are unit vectors and \theta is the angle between them, then prove that
Consider
|\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b}).(\vec{a}-\vec{b})=|\vec{a}|^2-2\vec{a}.\vec{b}+|\vec{b}|^2 [1]
=1-2|\vec{a}||\vec{b}|\cos\theta+1
=2-2\cos\theta [1]
=2\left(2\sin^2\frac{\theta}{2}\right)
\therefore \sin\frac{\theta}{2}=\frac{1}{2}|\vec{a}-\vec{b}| [1]
Case Study – 1
Three friends A, B and C move out from the same location O at the same time in three different directions to reach their destinations. They move out on straight paths and decide that A and B after reaching their destinations will meet up with C at his predecided destination, following straight paths from A to C and B to C in such a way that \overrightarrow{OA}=\vec{a}, \overrightarrow{OB}=\vec{b} and \overrightarrow{OC}=5\vec{a}-2\vec{b} respectively.
Based upon the above information, answer the following questions :
(i) Complete the given figure to explain their entire movement plan along the respective vectors. [1]
(ii) Find vectors \overrightarrow{AC} and \overrightarrow{BC}. [1]
(iii) (a) If \vec{a}\cdot\vec{b}=1, distance of O to A is 1 km and that from O to B is 2 km, then find the angle between \overrightarrow{OA} and \overrightarrow{OB}. Also, find |\vec{a}\times\vec{b}|. [2]
OR
(iii) (b) If \vec{a}=2\hat{i}-\hat{j}+4\hat{k} and \vec{b}=\hat{j}-\hat{k}, then find a unit vector perpendicular to (\vec{a}+\vec{b}) and (\vec{a}-\vec{b}). [2]
(i) The Complete figure of their entire movement plan is: [figure: triangle paths O→A, O→B, O→C with A→C and B→C] [1]
(ii) \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=4\vec{a}-2\vec{b}, \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=5\vec{a}-3\vec{b} [½+½]
(iii) (a) we are given: |\vec{a}|=1,|\vec{b}|=2, assuming ‘\theta’ as the angle between \overrightarrow{OA} and \overrightarrow{OB}.
\theta=\cos^{-1}\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\right)=\cos^{-1}\frac{1}{1\times2}=\cos^{-1}\frac12=\frac{\pi}{3} [1]
|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta=1(2)\frac{\sqrt3}{2}=\sqrt3 [1]
OR
(iii) (b) \vec{a}+\vec{b}=2\hat{i}+3\hat{k}, \vec{a}-\vec{b}=2\hat{i}-2\hat{j}+5\hat{k}, let \vec{c} be \perp to both (\vec{a}+\vec{b}) and (\vec{a}-\vec{b}) [½+½]
Then, \vec{c}=(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&0&3\\2&-2&5\end{vmatrix}=6\hat{i}-4\hat{j}-4k and |\vec{c}|=\sqrt{68} [½]
The required unit vector is, \hat{c}=\frac{1}{2\sqrt{17}}\left(6\hat{i}-4\hat{j}-4\hat{k}\right)=\frac{1}{\sqrt{17}}\left(3\hat{i}-2\hat{j}-2\hat{k}\right) [½]
An instructor at the astronomical centre shows three among the brightest stars in a particular constellation. Assume that the telescope is located at O(0, 0, 0) and the three stars have their locations at the points D, A and V having position vectors 2\hat{i} + 3\hat{j} + 4\hat{k}, 7\hat{i} + 5\hat{j} + 8\hat{k} and -3\hat{i} + 7\hat{j} + 11\hat{k} respectively.
Based on the above information, answer the following questions :
(i) How far is the star V from star A ? [1]
(ii) Find a unit vector in the direction of \overrightarrow{DA}. [1]
(iii) Find the measure of \angle VDA. [2]
OR
(iii) What is the projection of vector \overrightarrow{DV} on vector \overrightarrow{DA} ? [2]
(i) \overrightarrow{AV} = Position Vector of V – Position Vector of A
= -3\hat{i} + 7\hat{j} + 11\hat{k} - 7\hat{i} - 5\hat{j} - 8\hat{k} = -10\hat{i} + 2\hat{j} + 3\hat{k} [½]
Thus, |\overrightarrow{AV}| = \sqrt{100+4+9} = \sqrt{113} units [½]
(ii) \overrightarrow{DA} = Position Vector of A – Position Vector of D
= 7\hat{i} + 5\hat{j} + 8\hat{k} - 2\hat{i} - 3\hat{j} - 4\hat{k} = 5\hat{i} + 2\hat{j} + 4\hat{k} [½]
Unit vector in the direction of \overrightarrow{DA} = \frac{5\hat{i} + 2\hat{j} + 4\hat{k}}{3\sqrt{5}} [½]
(iii) \overrightarrow{DV} = -5\hat{i} + 4\hat{j} + 7\hat{k} [½]
\angle VDA = \cos^{-1}\left(\frac{\overrightarrow{DV}.\overrightarrow{DA}}{|\overrightarrow{DV}||\overrightarrow{DA}|}\right) = \cos^{-1}\left(\frac{11\sqrt{2}}{90}\right) [1½]
OR
(iii) \overrightarrow{DV} = -5\hat{i} + 4\hat{j} + 7\hat{k} [½]
Projection of \overrightarrow{DV} on \overrightarrow{DA} = \left(\frac{\overrightarrow{DV}.\overrightarrow{DA}}{|\overrightarrow{DA}|}\right) = \frac{11\sqrt{5}}{15} [1½]
Let \vec{\alpha}, \vec{\beta}, \vec{\gamma} be three non-zero vectors, such that \vec{\alpha} \times \vec{\beta} = \vec{\gamma} and \vec{\beta} \times \vec{\gamma} = \vec{\alpha}. Show that \alpha, \beta, \gamma are mutually perpendicular vectors such that |\vec{\beta}| = 1 and |\vec{\alpha}| = |\vec{\gamma}|.
Find projection of vector \vec{\alpha} on \vec{\beta}. Give reasons to support your answer.
As, \vec{\alpha} \times \vec{\beta} = \vec{\gamma} \Rightarrow \vec{\gamma} \perp \vec{\alpha}\ and\ \vec{\gamma} \perp \vec{\beta} [1]
Similarly, \vec{\beta} \times \vec{\gamma} = \vec{\alpha} \Rightarrow \vec{\alpha} \perp \vec{\gamma}\ and\ \vec{\alpha} \perp \vec{\beta}
Thus, \vec{\alpha}, \vec{\beta}\ and\ \vec{\gamma} are mutually perpendicular vectors. [1]
Now, |\vec{\alpha} \times \vec{\beta}| = |\vec{\gamma}| \Rightarrow |\vec{\alpha}||\vec{\beta}| = |\vec{\gamma}| and
|\vec{\beta} \times \vec{\gamma}| = |\vec{\alpha}| \Rightarrow |\vec{\beta}||\vec{\gamma}| = |\vec{\alpha}| [1]
Thus, |\vec{\beta}||\vec{\alpha}||\vec{\beta}| = |\vec{\alpha}| \Rightarrow |\vec{\beta}| = 1 (since, \vec{\alpha} \ne 0) [1]
On putting |\vec{\beta}| = 1 in the |\vec{\beta}||\vec{\gamma}| = |\vec{\alpha}|, we get |\vec{\gamma}| = |\vec{\alpha}| [½]
Projection of \vec{\alpha}\ on\ \vec{\beta} = 0, as \vec{\alpha} \perp \vec{\beta}. [½]
If points A(3, –1, 2), B(6, – 3, 6), C(4, – 3, 1) and D(1, –1, – 3) are vertices of a quadrilateral ABCD, write the position vectors of A, B, C and D and show that ABCD represents a parallelogram. Also, find the area of the parallelogram ABCD.
Position vector of A = 3\hat{i} - \hat{j} + 2\hat{k}, Position vector of B = 6\hat{i} - 3\hat{j} + 6\hat{k} [½]
Position vector of C = 4\hat{i} - 3\hat{j} + \hat{k}, Position vector of D = \hat{i} - \hat{j} - 3\hat{k} [½]
Now, \overrightarrow{AB} = 3\hat{i} - 2\hat{j} + 4\hat{k}, \overrightarrow{CD} = -3\hat{i} + 2\hat{j} - 4\hat{k} = -\overrightarrow{AB} [1]
Also, |\overrightarrow{CD}| = |\overrightarrow{AB}|
As, AB and CD are parallel and equal so ABCD represents a parallelogram. [1]
Since, \overrightarrow{AB} = 3\hat{i} - 2\hat{j} + 4\hat{k}, \overrightarrow{BC} = -2\hat{i} - 5\hat{k}
\overrightarrow{AB} \times \overrightarrow{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 4 \\ -2 & 0 & -5 \end{vmatrix} = 10\hat{i} + 7\hat{j} - 4\hat{k} [1]
Area of Parallelogram = \sqrt{100 + 49 + 16} = \sqrt{165} [1]
Show that the area of a parallelogram whose diagonals are represented by \vec{a} and \vec{b} is given by \frac{1}{2}|\vec{a}\times\vec{b}|. Also find the area of a parallelogram whose diagonals are 2\hat{i}-\hat{j}+\hat{k} and \hat{i}+3\hat{j}-\hat{k}.
Let ABCD be the parallelogram with diagonals \overrightarrow{AB}=\vec{a} and \overrightarrow{BD}=\vec{b}.
\therefore\overrightarrow{AB}=\frac{1}{2}(\vec{a}-\vec{b}) and \overrightarrow{AD}=\frac{1}{2}(\vec{a}+\vec{b}) [½]
Area of ABCD
=|\overrightarrow{AB}\times\overrightarrow{AD}|
=\left|\frac{1}{2}(\vec{a}-\vec{b})\times\frac{1}{2}(\vec{a}+\vec{b})\right| [½]
=\frac{1}{4}|\vec{a}\times\vec{a}+\vec{a}\times\vec{b}-\vec{b}\times\vec{a}-\vec{b}\times\vec{b}| [½]
=\frac{1}{4}|\vec{a}\times\vec{b}+\vec{a}\times\vec{b}|\quad(\because\vec{a}\times\vec{a}=\vec{0})
=\frac{1}{4}|2(\vec{a}\times\vec{b})|
=\frac{1}{2}|\vec{a}\times\vec{b}| [½]
Given \vec{a}=2\hat{i}-\hat{j}+\hat{k},\ \vec{b}=\hat{i}+3\hat{j}-\hat{k}
Area of parallelogram =\frac{1}{2}|\vec{a}\times\vec{b}|
Now \vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-1&1\\1&3&-1\end{vmatrix}=-2\hat{i}+3\hat{j}+7\hat{k} [2]
|\vec{a}\times\vec{b}|=\sqrt{62} [½]
Area of parallelogram =\frac{1}{2}\sqrt{62} [½]
The distance of the point with position vector 3\hat{i}+4\hat{j}+5\hat{k} from the y-axis is
The distance of the point (a, b, c) from the y-axis is \sqrt{a^2+c^2}
So, the distance is \sqrt{3^2+5^2}=\sqrt{34} units.
(B) \sqrt{34} units [1]
The length of perpendicular drawn from point (2, 5, 7) on line \frac{x}{1} = \frac{y}{0} = \frac{z}{0} is
(C) \sqrt{74}
Assertion (A) : Lines given by x = py + q, z = ry + s and x = p'y + q', z = r'y + s' are perpendicular to each other when pp' + rr' = 1.
Reason (R) : Two lines \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 and \vec{r} = \vec{a}_2 + \mu\vec{b}_2 are perpendicular to each other if \vec{b}_1 \cdot \vec{b}_2 = 0.
(D) Assertion (A) is false and Reason (R) is true.
Direction cosines of the line given by equations : \frac{2x-1}{4} = \frac{1-y}{3} = \frac{-z}{6} are
(B) \frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}
Assertion (A) : A line can have direction cosines < 1, 1, 1 >
Reason (R) : \cos\theta = 1 is possible for \theta = 0.
(D) Assertion (A) is false, but Reason (R) is true.
If l_1, m_1, n_1 and l_2, m_2, n_2 are direction cosines of lines L_1 and L_2 respectively and \theta is the acute angle between them, then :
(D) \cos\theta = |l_1l_2 + m_1m_2 + n_1n_2|
Direction ratios of lines l_1 and l_2 are \lt 12, -3, 9 \gt and \lt 4, q, -p \gt respectively. The values of p and q for which l_1 and l_2 are parallel are respectively :
(C) -3, -1
If a line makes angles of \frac{3\pi}{4}, \frac{\pi}{3} and \theta with the positive directions of x, y and z-axis respectively, then \theta is
No option is correct. Full marks may be awarded for attempting the question. [1]
The equation of a line parallel to the vector 3\hat{i} + \hat{j} + 2\hat{k} and passing through the point (4, -3, 7) is :
(C) x = 3t + 4,\ y = t - 3,\ z = 2t + 7
The line x = 1 + 5\mu,\ y = -5 + \mu,\ z = -6 - 3\mu passes through which of the following point ?
(C) (1, -5, -6)
If P is a point on the line segment joining (3, 6, −1) and (6, 2, −2) and y-coordinate of P is 4, then its z-coordinate is :
(A) -\frac{3}{2}
If the direction cosines of a line are \sqrt{3}k,\ \sqrt{3}k,\ \sqrt{3}k, then the value of k is :
(D) \pm \frac{1}{3}
The distance of point P(a, b, c) from y-axis is :
(C) \sqrt{a^2 + c^2}
Assertion (A) : A line in space cannot be drawn perpendicular to x, y and z axes simultaneously.
Reason (R) : For any line making angles, \alpha, \beta, \gamma with the positive directions of x, y and z axes respectively, \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.
(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of Assertion (A).
The coordinates of the foot of the perpendicular drawn from the point (0, 1, 2) on the x-axis are given by :
(D) (0, 0, 0)
Direction ratios of a vector parallel to line \frac{x-1}{2} = -y = \frac{2z+1}{6} are :
(D) 2, -1, 3
If a line makes an angle of 30° with the positive direction of x-axis, 120° with the positive direction of y-axis, then the angle which it makes with the positive direction of z-axis is :
(A) 90^{0}
If \alpha, \beta and \gamma are the angles which a line makes with positive directions of x, y and z axes respectively, then which of the following is not true ?
(D) \cos\alpha + \cos\beta + \cos\gamma = 1
If a line makes an angle of \frac{\pi}{4} with the positive directions of both x-axis and z-axis, then the angle which it makes with the positive direction of y-axis is :
(C) \frac{\pi}{2}
The vector equation of a line passing through the point (1, –1, 0) and parallel to Y-axis is :
(B) \vec{r} = \hat{i} - \hat{j} + \lambda\hat{j}
[1]The lines \frac{1-x}{2} = \frac{y-1}{3} = \frac{z}{1} and \frac{2x-3}{2p} = \frac{y}{-1} = \frac{z-4}{7} are perpendicular to each other for p equal to :
(C) 2
[1]The angle which the line \frac{x}{1}=\frac{y}{-1}=\frac{z}{0} makes with the positive direction of Y-axis is :
(B) \frac{3\pi}{4} [1]
The Cartesian equation of the line passing through the point (1,-3,2) and parallel to the line :
\vec{r}=(2+\lambda)\hat{i}+\lambda\hat{j}+(2\lambda-1)\hat{k} is
(D) \frac{x-1}{1}=\frac{y+3}{1}=\frac{z-2}{2} [1]
The direction ratios of the line \frac{x-1}{3}=\frac{2-y}{1}=\frac{3z}{2} are :
(C) 9, – 3, 2
[1]The Cartesian equation of the line passing through the point (1,-3,2) and parallel to the line \vec{r}=2\hat{i}-\hat{k}+\lambda(\hat{i}+\hat{j}+2\hat{k}) is :
(D) \frac{x-1}{1}=\frac{y+3}{1}=\frac{2-z}{-2}
[1]Direction cosines of the line \frac{x-1}{2}=\frac{1-y}{3}=\frac{2z-1}{12} are :
(D) \frac{2}{7},-\frac{3}{7},\frac{6}{7}
Assertion (A) : Equation of a line passing through the points (1, 2, 3) and (3, – 1, 3) is \frac{x-3}{2}=\frac{y+1}{3}=\frac{z-3}{0}.
Reason (R) : Equation of a line passing through points (x_1,y_1,z_1), (x_2,y_2,z_2) is given by \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}.
(D) Assertion is False, Reason is True
Equation of a line passing through point (1, 1, 1) and parallel to z-axis is
(D) \frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{1}
Assertion (A) : If a line makes angles \alpha, \beta, \gamma with positive direction of the coordinate axes, then \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2.
Reason (R) : The sum of squares of the direction cosines of a line is 1.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
The value of \lambda for which the angle between the lines \vec{r} = \hat{i} + \hat{j} + \hat{k} + p(2\hat{i} + \hat{j} + 2\hat{k}) and \vec{r} = (1 + q)\hat{i} + (1 + q\lambda)\hat{j} + (1 + q)\hat{k} is \frac{\pi}{2} is :
(A) -4
Assertion (A) : A line through the points (4, 7, 8) and (2, 3, 4) is parallel to a line through the points (– 1, – 2, 1) and (1, 2, 5).
Reason (R): Lines \vec{r} = \vec{a_1} + \lambda \vec{b_1} and \vec{r} = \vec{a_2} + \mu \vec{b_2} are parallel if \vec{b_1} \cdot \vec{b_2} = 0.
(C) Assertion (A) is true and Reason (R) is false.
Distance of the point (p, q, r) from y-axis is :
(d) \sqrt{p^2+r^2}
If the direction cosines of a line are \left(\frac{1}{a},\frac{1}{a},\frac{1}{a}\right), then :
(d) a=\pm\sqrt{3}
Assertion (A) : The lines \vec{r}=\vec{a}_1+\lambda\vec{b}_1 and \vec{r}=\vec{a}_2+\mu\vec{b}_2 are perpendicular, when \vec{b}_1\cdot\vec{b}_2=0.
Reason (R) : The angle \theta between the lines \vec{r}=\vec{a}_1+\lambda\vec{b}_1 and \vec{r}=\vec{a}_2+\mu\vec{b}_2 is given by \cos\theta=\frac{\vec{b}_1\cdot\vec{b}_2}{|\vec{b}_1||\vec{b}_2|}
(a) Both assertion(A) and reason(R) are true and reason (R) is the correct explanation of assertion (A)
If a line makes angles of 90°, 135° and 45° with the x, y and z axes respectively, then its direction cosines are
(A) 0,-\frac{1}{\sqrt2},\frac{1}{\sqrt2}
The angle between the lines 2x=3y=-z and 6x=-y=-4z is
(D) 90°
The vector equation of a line which passes through the point (2, -4, 5) and is parallel to the line \frac{x+3}{3} = \frac{4-y}{2} = \frac{z+8}{6} is :
Official marking scheme for this paper is not available.
Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b), (c) and (d) as given below.
Assertion (A) : Quadrilateral formed by vertices A(0, 0, 0), B(3, 4, 5), C(8, 8, 8) and D(5, 4, 3) is a rhombus.
Reason (R) : ABCD is a rhombus if AB = BC = CD = DA, AC \ne BD.
Official marking scheme for this paper is not available.
A line passing through the points (2,6,5), and (3,p,7) is parallel to the line joining the points (-4,3,4) and (-1,0,q). Find the value of \frac{q}{p}.
Given points are A(2,6,5), B(3,p,7), C(-4,3,4) and D(-1,0,q)
Direction ratios of the line \overleftrightarrow{AB} are \langle1,\ p-6,\ 2\rangle and
Direction ratios of the line \overleftrightarrow{CD} are \langle3,-3,q-4\rangle
As the lines are parallel, so we have
\frac13=\frac{p-6}{-3}=\frac{2}{q-4}
\Rightarrow-1=p-6\ and\ q-4=3\times2 [1]
\Rightarrow p=-1+6=5,\ q=6+4=10 [½]
Thus, \frac{q}{p}=\frac{10}{5}=2 [½]
Find the co-ordinates of the point on the line \vec{r} = -\hat{j} + 3\hat{k} + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) such that the sum of co-ordinates is 3.
\vec{r} = -\hat{j} + 3\hat{k} + \lambda(2\hat{i} - 2\hat{j} + \hat{k})
i.e. \vec{r} = 2\lambda\hat{i} + (-2\lambda - 1)\hat{j} + (\lambda + 3)\hat{k}
According to question,
2\lambda + (-2\lambda - 1) + (\lambda + 3) = 3 \Rightarrow \lambda = 1 [1]
\therefore \vec{r} = 2\hat{i} - 3\hat{j} + 4\hat{k}
Co-ordinates of the required point is (2, -3, 4). [1]
If the lines \frac{x - 3}{1} = \frac{1 - y}{1} = \frac{z + 2}{p} and \frac{2 - x}{3} = \frac{y + 1}{5} = \frac{z + 56}{2p} are perpendicular to each other, then find the value(s) of p.
Direction Ratios of line \frac{x-3}{1} = \frac{1-y}{1} = \frac{z+2}{p} are \lt 1, -1, p \gt [½]
Direction Ratios of line \frac{2-x}{3} = \frac{y+1}{5} = \frac{z+56}{2p} are \lt -3, 5, 2p \gt [½]
\because Given both lines are perpendicular to each other.
\therefore 1 \times (-3) + (-1) \times 5 + p \times 2p = 0 [½]
\Rightarrow p = \pm 2 [½]
Find the vector equation of a line passing through the origin and perpendicular to both the lines \vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k}) and \vec{r} = \mu(\hat{i} - \hat{j} + \hat{k}).
The vector parallel to required line is given by
\vec{b} = (3\hat{i} + 4\hat{j} + 2\hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\3 & 4 & 2\\1 & -1 & 1\end{vmatrix} = 6\hat{i} - \hat{j} - 7\hat{k} [1½]
Equation of required line is given by
\vec{r} = \delta(6\hat{i} - \hat{j} - 7\hat{k}) [½]
Find the angle between the following pair of lines :
Drs of given lines are <3, 2, 6> and <1, 2, 2> [½]
Let \theta be angle between given lines
\therefore \cos\theta = \frac{3\times1 + 2\times2 + 6\times2}{7\times3} [1]
Thus, \theta = \cos^{-1}\left(\frac{19}{21}\right) [½]
Determine if the lines \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) and \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) intersect with each other.
\vec{b}_1 = 3\hat{i} - \hat{j},\ \vec{b}_2 = 2\hat{i} + 3\hat{k},\ \vec{a}_2 = 4\hat{i} - \hat{k},\ \vec{a}_1 = \hat{i} + \hat{j} - \hat{k}
(\vec{a}_2 - \vec{a}_1) = 3\hat{i} - \hat{j} [½]
\vec{b}_1 \times \vec{b}_2 = -3\hat{i} - 9\hat{j} + 2\hat{k} [1]
(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (3\hat{i} - \hat{j}) \cdot (-3\hat{i} - 9\hat{j} + 2\hat{k}) = 0, hence lines intersect [½]
Find the angle between the lines
Given lines are: \vec{r} = \left(3\hat{i} - 2\hat{j} + 6\hat{k}\right) + \lambda\left(2\hat{i} + 2\hat{j} + 2\hat{k}\right) [½]
and \vec{r} = \left(2\hat{j} - 5\hat{k}\right) + \mu\left(6\hat{i} + 3\hat{j} + 2\hat{k}\right)
Let \theta be the angle between these two lines.
\cos\theta = \frac{2(6) + 2(3) + 2(2)}{\sqrt{4+4+4}\sqrt{36+9+4}} = \frac{22}{2\sqrt{3}\times 7} [1]
\Rightarrow \cos\theta = \frac{11}{21}\sqrt{3} \Rightarrow \theta = \cos^{-1}\left(\frac{11}{21}\sqrt{3}\right) [½]
Find the vector equation of the line passing through the point (2,3,-5) and making equal angles with the co-ordinate axes.
\cos\alpha=\cos\beta=\cos\gamma=l\Rightarrow l^2+l^2+l^2=1\Rightarrow3l^2=1,\ \therefore l=\frac{1}{\sqrt3} [1]
Direction cosines of the line are \frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3}\Rightarrow the direction ratios are 1,1,1 [½]
\therefore Vector equation of the line is: \vec{r}=2\hat{i}+3\hat{j}-5\hat{k}+\lambda(\hat{i}+\hat{j}+\hat{k}) [½]
Find the angle between the lines
The direction ratios of the two lines are: 7,-5,1 and 1,2,3 respectively. [½]
\therefore The angle between the two lines is given by:
\theta=\cos^{-1}\left(\frac{7(1)-5(2)+1(3)}{\sqrt{49+25+1}\sqrt{1+4+9}}\right) [1]
\Rightarrow\theta=\cos^{-1}0=\frac{\pi}{2} [½]
Check whether the lines given by equations x=2\lambda+2,\ y=7\lambda+1,\ z=-3\lambda-3 and x=-\mu-2,\ y=2\mu+8,\ z=4\mu+5 are perpendicular to each other or not.
D.r.s. of lines are < 2, 7, – 3 > and < – 1, 2, 4 > [1]
Now 2\cdot(-1)+7\cdot2+(-3)\cdot4=0 [1]
\therefore given lines are perpendicular
Find the coordinates of points on line \frac{x}{1} = \frac{y-1}{2} = \frac{z+1}{2} which are at a distance of \sqrt{11} units from origin.
General point on the curve is P(k, 2k+1, 2k-1),\ k \in R [½]
OP = \sqrt{11} \Rightarrow OP^2 = 11
\therefore k^2 + (2k+1)^2 + (2k-1)^2 = 11 \Rightarrow k = \pm 1 [1]
\therefore Coordinates of points are (1, 3, 1) & (-1, -1, -3) [½]
If the angle between the lines \frac{x - 5}{\alpha} = \frac{y + 2}{-5} = \frac{z + \frac{24}{5}}{\beta} and \frac{x}{1} = \frac{y}{0} = \frac{z}{1} is \frac{\pi}{4}, find the relation between \alpha and \beta.
\cos\frac{\pi}{4} = \frac{|\alpha \cdot 1 + 0 + \beta|}{\sqrt{\alpha^2 + \beta^2 + 25}\,\sqrt{2}} \Rightarrow \frac{1}{\sqrt{2}} = \frac{|\alpha + \beta|}{\sqrt{\alpha^2 + \beta^2 + 25}\,\sqrt{2}} [1]
Squaring both sides, we get
\alpha^2 + \beta^2 + 2\alpha\beta = \alpha^2 + \beta^2 + 25 [½]
\Rightarrow \alpha\beta = \frac{25}{2} [½]
Find the vector equation of the line passing through the point (2, 1, 3) and perpendicular to both the lines
Vector equation of the line passing through (2, 1, 3) is
\vec{r}=(2\hat{i}+\hat{j}+3\hat{k})+\lambda(a\hat{i}+b\hat{j}+c\hat{k}) [½]
Line \vec{r} is perpendicular to the given lines then
a+2b+3c=0;\ \ -3a+2b+5c=0 [1]
\Rightarrow\frac{a}{4}=\frac{b}{-14}=\frac{c}{8}=k'(say)
\Rightarrow a=2k,\ b=-7k and c=4k.
Thus, the required vector equation is
\vec{r}=(2\hat{i}+\hat{j}+3\hat{k})+\lambda(2\hat{i}-7\hat{j}+4\hat{k}) [½]
The equations of a line are 5x-3=15y+7=3-10z. Write the direction cosines of the line and find the coordinates of a point through which it passes.
The equation of the given line is
\frac{x-3/5}{1/5}=\frac{y+7/15}{1/15}=\frac{z-3/10}{-1/10} [1]
Its direction ratios are.
\left\langle\frac{1}{5},\frac{1}{15},-\frac{1}{10}\right\rangle or \langle 6,2,-3\rangle
Direction cosines are \left\langle\pm\frac{6}{7},\pm\frac{2}{7},\mp\frac{3}{7}\right\rangle [½]
The point through which it passes is \left(\frac{3}{5},\frac{-7}{15},\frac{3}{10}\right) [½]
Find the vector and the cartesian equations of a line that passes through the point A(1, 2, –1) and parallel to the line 5x-25=14-7y=35z.
The given line is
So, the required vector equation of the line passing through (1,2,-1) is
\vec{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(7\hat{i}-5\hat{j}+\hat{k}) [½]
Cartesian equation of the line is
\frac{x-1}{7}=\frac{y-2}{-5}=\frac{z+1}{1} [½]
If a line makes angles \alpha, \beta and \gamma with x-axis, y-axis and z-axis respectively, then prove that \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2.
Official marking scheme for this paper is not available.
Find the values of \lambda , for which the distance of point (2, 1, \lambda) from plane 3x+5y+4z=11 is 2\sqrt{2} units.
\frac{|3\cdot 2+5\cdot 1+4\lambda-11|}{\sqrt{9+25+16}}=2\sqrt{2} [1]
|4\lambda|=20
\lambda=\pm 5 [1]
Write the cartesian equation of the line PQ passing through points P(2, 2, 1) and Q(5, 1, – 2). Hence, find the y-coordinate of the point on the line PQ whose z-coordinate is – 2.
Required equation of line is given by
putting z=-2, we get \frac{y-2}{-1}=\frac{-3}{-3}=1 [½]
y-2=-1\Rightarrow y=1 [½]
If the distance of the point (1, 1, 1) from the plane x-y+z+\lambda=0 is \frac{5}{\sqrt{3}}, find the value(s) of \lambda.
Distance of point (1, 1, 1) from x-y+z+\lambda=0 is \frac{|1+\lambda|}{\sqrt{3}} [½]
\frac{|1+\lambda|}{\sqrt{3}}=\frac{5}{\sqrt{3}}\Rightarrow 1+\lambda=\pm5 [1]
\lambda=-6 or +4 [½]
The foot of a perpendicular drawn from the point (-2,-1,-3) on a plane is (1,-3,3). Find the equation of the plane.
D.R.'s of normal vector are \langle 3,-2,6\rangle [1]
Equation of plane is :
\Rightarrow 3x-2y+6z=27 [1]
If a line makes an angle \alpha, \beta, \gamma with the coordinate axes, then find the value of \cos 2\alpha+\cos 2\beta+\cos 2\gamma.
l=\cos\alpha,\ m=\cos\beta,\ n=\cos\gamma \Rightarrow \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 [1]
\cos 2\alpha+\cos 2\beta+\cos 2\gamma
=2\cos^2\alpha-1+2\cos^2\beta-1+2\cos^2\gamma-1
=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3
=2-3
=-1 [1]
Find the foot of the perpendicular drawn from the point (2,1,4) on the line x-2=\frac{2y+3}{4}=2-z. Also find the equation of the line passing through this point and the point (2,1,4).
A point on the line x-2=\frac{2y+3}{4}=2-z=k\ is\ P\left(k+2,\frac{4k-3}{2},2-k\right) [½]
And direction ratios of the given line are \langle1,2,-1\rangle.
Direction Ratios of the line joining the points A(2,1,4) and this point P are
\left\langle k+2-2,\frac{4k-3}{2}-1,2-k-4\right\rangle=\left\langle k,\frac{4k-5}{2},-k-2\right\rangle [½]
As \overrightarrow{AP} is perpendicular to the given line, So
1\times k+2\times\frac{4k-5}{2}+(-1)\times(-k-2)=0
\Rightarrow k+4k-5+k+2=0\Rightarrow6k-3=0\Rightarrow k=\frac12 [½]
So, foot of the perpendicular is
P\left(k+2,\frac{4k-3}{2},2-k\right)=\left(\frac12+2,\frac{2-3}{2},2-\frac12\right)=\left(\frac52,\frac{-1}{2},\frac32\right) [½]
Direction ratios of the line \overleftrightarrow{AP} are \left\langle k,\frac{4k-5}{2},-k-2\right\rangle=\left\langle\frac12,-\frac32,-\frac52\right\rangle or \langle1,-3,-5\rangle [½]
And equation of \overleftrightarrow{AP} is
\frac{x-2}{1}=\frac{y-1}{-3}=\frac{z-4}{-5} OR x-2=\frac{1-y}{3}=\frac{4-z}{5} [½]
Find the angle between the lines: \frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1} and \frac{x-5}{2}=y-\frac12=\frac{3-z}{2}. Also find the cartesian equation of a line passing through the point (1,1,1) and perpendicular to both lines.
Given lines are \frac{x-1}{2}=\frac{y-1}{2}=\frac{z-2}{1} and
\frac{x-5}{2}=y-\frac12=\frac{3-z}{2}\Rightarrow\frac{x-5}{2}=\frac{y-\frac12}{1}=\frac{z-3}{-2} [½]
Let the angle between the lines be \theta, then
\cos\theta=\frac{2\times2+2\times1+1\times(-2)}{\sqrt9\times\sqrt9}=\frac49\Rightarrow\theta=\cos^{-1}\left(\frac49\right) [1]
Now required line will be in the direction of the vector
\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&2&1\\2&1&-2\end{vmatrix}=-5\hat{i}+6\hat{j}-2\hat{k} [1]
So required line is \frac{x-1}{-5}=\frac{y-1}{6}=\frac{z-1}{-2} OR \frac{x-1}{5}=\frac{y-1}{-6}=\frac{z-1}{2} [½]
Find the distance of the point (2,-1,3) from the line \vec{r}=(2\hat{i}-\hat{j}+2\hat{k})+\mu(3\hat{i}+6\hat{j}+2\hat{k}) measured parallel to the z-axis.
The line through (2,-1,3) parallel to the z-axis is given by
\vec{r}=(2\hat{i}-\hat{j}+3\hat{k})+\lambda(\hat{k}) [1]
Any point on this line is P(2,-1,3+\lambda) [½]
Any point on the given line \vec{r}=(2\hat{i}-\hat{j}+2\hat{k})+\mu(3\hat{i}+6\hat{j}+2\hat{k}) is
Q(2+3\mu,-1+6\mu,2+2\mu) [½]
For the intersection point
Q(2+3\mu,-1+6\mu,2+2\mu)=P(2,-1,3+\lambda)\Rightarrow 2=2+3\mu\Rightarrow\mu=0 [½]
The point of intersection is (2,-1,2) [½]
The distance from (2,-1,3) to (2,-1,2) is clearly 1 unit.
Alternative method
Any point on the line through (2,-1,3) parallel to the z-axis is (2,-1,\lambda) [1]
Any point on the given line is (2+3\mu,-1+6\mu,2+2\mu)
Therefore, 2=2+3\mu\Rightarrow\mu=0 [1]
The point of intersection is (2,-1,2) [½]
The distance from (2,-1,3) to (2,-1,2) is clearly 1 unit. [½]
Find the point of intersection of the line \vec{r}=(3\hat{i}+\hat{k})+\mu(\hat{i}+\hat{j}+\hat{k}) and the line through (2,-1,1) parallel to the z-axis. How far is this point from the z-axis?
The line through (2,-1,1) parallel to the z-axis is \vec{r}=(2\hat{i}-\hat{j}+\hat{k})+\lambda(\hat{k}) [1]
Any point on this line is P(2,-1,1+\lambda)
Any point on the given line is A(3+\mu,\mu,1+\mu)
A(3+\mu,\mu,1+\mu)=P(2,-1,1+\lambda)\Rightarrow\mu=-1 [1]
The point of intersection is (2,-1,0) [½]
The distance of (2,-1,0) from the z-axis is \sqrt{2^2+(-1)^2}=\sqrt5 units. [½]
Find a point on the line \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} at a distance of \sqrt{2} units from the point (1, 2, 3).
Let \frac{x - 2}{3} = \frac{y - 1}{-2} = \frac{z - 3}{2} = k
A general point P on the line be (3k + 2, -2k + 1, 2k + 3) [1]
It is given that PQ = \sqrt{2} where Q(1,2,3), so
PQ = \sqrt{(3k + 1)^{2} + (-2k - 1)^{2} + (2k)^{2}} = \sqrt{2}
\Rightarrow 17k^{2} + 10k = 0 [1]
\therefore k = 0 or \frac{-10}{17} [½]
Thus, required points are (2, 1, 3) & \left(\frac{4}{17}, \frac{37}{17}, \frac{31}{17}\right) [½]
Find the shortest distance between the lines
Here, \vec{a_1} = 4\hat{i} - \hat{j}, \vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}, \vec{a_2} = \hat{i} - \hat{j} + 2\hat{k}, \vec{b_2} = 2\hat{i} + 4\hat{j} - 5\hat{k} [1]
Now,
\vec{a_2} - \vec{a_1} = -3\hat{i} + 2\hat{k},\ \vec{b_1} \times \vec{b_2} = 2\hat{i} - \hat{j} and \left|\vec{b_1} \times \vec{b_2}\right| = \sqrt{5} [1½]
Required distance = \left|\frac{\left(-3\hat{i} + 2\hat{k}\right) \cdot \left(2\hat{i} - \hat{j}\right)}{\sqrt{5}}\right| = \frac{6}{\sqrt{5}} or \frac{6\sqrt{5}}{5} [½]
Find the equation of a line passing through the point (1, –1, 2) and parallel to x-axis. Also, find the shortest distance between this line and the line \vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k}).
Required Eq. of line: \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i}) [1]
Now, \vec{a_1} = \hat{i} - \hat{j} + 2\hat{k}, \vec{b_1} = \hat{i}
\vec{a_2} = 2\hat{i} + \hat{j} - \hat{k}, \vec{b_2} = 3\hat{i} - 5\hat{j} + 2\hat{k}
So, \vec{a_2} - \vec{a_1} = \hat{i} + 2\hat{j} - 3\hat{k}, \vec{b_1} \times \vec{b_2} = -2\hat{j} - 5\hat{k}, |\vec{b_1} \times \vec{b_2}| = \sqrt{29} [1½]
Thus, Shortest distance = \left|\frac{-4+15}{\sqrt{29}}\right| = \frac{11}{\sqrt{29}} [½]
Find the equations of the lines AC and BD, if coordinates of points A, B, C and D respectively, are (3, – 4, 2), (–1, – 2, 0), (4, 1, –1) and (4, 0, 5). Show that the lines AC and BD are perpendicular to each other.
Required eq. of line AC: \frac{x-3}{1} = \frac{y+4}{5} = \frac{z-2}{-3} OR \frac{x-4}{1} = \frac{y-1}{5} = \frac{z+1}{-3} [1]
Required eq. of line BD: \frac{x+1}{5} = \frac{y+2}{2} = \frac{z}{5} OR \frac{x-4}{5} = \frac{y}{2} = \frac{z-5}{5} [1]
Since, 1 \times 5 + 5 \times 2 + (-3) \times 5 = 5 + 10 - 15 = 0
Therefore, the lines AC and BD are perpendicular to each other. [1]
Verify that lines given by \vec{r}=(1-\lambda)\hat{i}+(\lambda-2)\hat{j}+(3-2\lambda)\hat{k} and \vec{r}=(\mu+1)\hat{i}+(2\mu-1)\hat{j}-(2\mu+1)\hat{k} are skew lines. Hence, find shortest distance between the lines.
Rewriting the lines, we get
\vec{r}=(\hat{i}-2\hat{j}+3\hat{k})+\lambda(-\hat{i}+\hat{j}-2\hat{k}) and \vec{r}=(\hat{i}-\hat{j}-\hat{k})+\mu(\hat{i}+2\hat{j}-2\hat{k}) [½]
Let \vec{a}_1=\hat{i}-2\hat{j}+3\hat{k},\ \vec{a}_2=\hat{i}-\hat{j}-\hat{k},\ \vec{b}_1=-\hat{i}+\hat{j}-2\hat{k},\ \vec{b}_2=\hat{i}+2\hat{j}-2\hat{k}
Note that the dr's of given lines are not proportional so, they are not parallel lines.
The lines will be skew if they do not intersect each other also.
Here \vec{a}_2-\vec{a}_1=\hat{j}-4\hat{k}, \vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&1&-2\\1&2&-2\end{vmatrix}=2\hat{i}-4\hat{j}-3\hat{k} [½ + ½]
Consider (\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)
=(\hat{j}-4\hat{k})\cdot(2\hat{i}-4\hat{j}-3\hat{k})=8\ne 0
Hence lines will not intersect. So the lines are skew. [½]
Shortest Distance =\frac{|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)|}{|\vec{b}_1\times\vec{b}_2|}
=\frac{8}{\sqrt{4+16+9}}=\frac{8}{\sqrt{29}} [1]
Find the distance of the point (−1, −5, −10) from the point of intersection of the lines \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} and \frac{x-4}{5}=\frac{y-1}{2}=z.
l_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda
Any point on l_1 is (2\lambda+1,3\lambda+2,4\lambda+3)
l_2:\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}=\mu
Any point on l_2 is (5\mu+4,2\mu+1,\mu) [1]
For point of intersection,
2\lambda+1=5\mu+4,\ 3\lambda+2=2\mu+1
Solving, \lambda=\mu=-1 [1]
Since, \lambda=\mu=-1 satisfy 4\lambda+3=\mu
\therefore Point of intersection is (-1,-1,-1) [½]
Now distance of (-1,-5,-10) from (-1,-1,-1) is:
\sqrt{(-1+1)^2+(-1+5)^2+(-1+10)^2}=\sqrt{97} units [½]
Find the distance of the point P(2, 4, -1) from the line \frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9}.
Let \vec{a}_2 = 2\hat{i} + 4\hat{j} - \hat{k},\ \vec{a}_1 = -5\hat{i} - 3\hat{j} + 6\hat{k} and \vec{b} = \hat{i} + 4\hat{j} - 9\hat{k}
Distance between point and line is given by d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}
Here (\vec{a}_2 - \vec{a}_1) = 7\hat{i} + 7\hat{j} - 7\hat{k} [½]
(\vec{a}_2 - \vec{a}_1) \times \vec{b} = -35\hat{i} + 56\hat{j} + 21\hat{k} [1½]
d = \frac{49\sqrt{2}}{7\sqrt{2}} = 7 [1]
Let the position vectors of the points A, B and C be 3\hat{i} - \hat{j} - 2\hat{k}, \hat{i} + 2\hat{j} - \hat{k} and \hat{i} + 5\hat{j} + 3\hat{k} respectively. Find the vector and cartesian equations of the line passing through A and parallel to line BC.
Vector parallel to given line = 3\hat{j} + 4\hat{k} [1]
Vector equation is \vec{r} = 3\hat{i} - \hat{j} - 2\hat{k} + \mu(3\hat{j} + 4\hat{k}) [1]
Cartesian equation is \frac{x-3}{0} = \frac{y+1}{3} = \frac{z+2}{4} [1]
Find the shortest distance between the lines :
\vec{r}=(2\hat{i}-\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})
\vec{r}=(\hat{i}+4\hat{k})+\mu(3\hat{i}-6\hat{j}+9\hat{k}).
The two given lines are parallel with,
\vec{a}_1=2\hat{i}-\hat{j}+3\hat{k},\ \vec{a}_2=\hat{i}+4\hat{k}
Then \vec{a}_2-\vec{a}_1=-\hat{i}+\hat{j}+\hat{k} and the parallel vector is \vec{b}=\hat{i}-2\hat{j}+3\hat{k} [½]
\vec{b}\times(\vec{a}_2-\vec{a}_1)=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-2&3\\-1&1&1\end{vmatrix}=-5\hat{i}-4\hat{j}-\hat{k} [1½]
Shortest Distance =\frac{\left|\vec{b}\times(\vec{a}_2-\vec{a}_1)\right|}{\left|\vec{b}\right|}=\frac{\sqrt{42}}{\sqrt{14}}=\sqrt3 [1]
Find the coordinates of the foot of the perpendicular drawn from the point P(0, 2, 3) to the line \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}.
General point on the given line is M(5\lambda-3,\ 2\lambda+1,\ 3\lambda-4)
Direction ratios of PM are 5\lambda-3,\ 2\lambda-1,\ 3\lambda-7 [1]
If this point is the foot of the perpendicular from the point P (0, 2, 3), then PM is perpendicular to the line. Thus,
(5\lambda-3)\cdot5+(2\lambda-1)\cdot2+(3\lambda-7)\cdot3=0 [1]
\Rightarrow\lambda=1
Hence co-ordinates of M are (2, 3, – 1) [1]
Find the distance between the lines :
\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda(2\hat{i}+3\hat{j}+6\hat{k});
\vec{r}=(3\hat{i}+3\hat{j}-5\hat{k})+\mu(4\hat{i}+6\hat{j}+12\hat{k})
Here
\vec{a_1}=\hat{i}+2\hat{j}-4\hat{k},\ \vec{b_1}=2\hat{i}+3\hat{j}+6\hat{k}
\vec{a_2}=3\hat{i}+3\hat{j}-5\hat{k},\ \vec{b_2}=4\hat{i}+6\hat{j}+12\hat{k}
Here, \vec{b_1} and \vec{b_2} are parallel vectors. [½]
\vec{a_2}-\vec{a_1}=2\hat{i}+\hat{j}-\hat{k},\quad\vec{b}=2\hat{i}+3\hat{j}+6\hat{k} [½]
Thus, (\vec{a_2}-\vec{a_1})\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\2&3&6\end{vmatrix}=9\hat{i}-14\hat{j}+4\hat{k} [1]
Distance between the lines =\left|\frac{(\vec{a_2}-\vec{a_1})\times\vec{b}}{|\vec{b}|}\right|
=\frac{\sqrt{81+196+16}}{\sqrt{4+9+36}}
=\frac{\sqrt{293}}{7} units. [1]
Find the vector equation of the plane passing through the intersection of the planes \vec{r}\cdot(2\hat{i}+2\hat{j}-3\hat{k})=7 and \vec{r}\cdot(2\hat{i}+5\hat{j}+3\hat{k})=9 and through the point (2, 1, 3).
Equation of the plane is \vec{r}\cdot\left(2\hat{i}+2\hat{j}-3\hat{k}\right)-7+\lambda\left[\vec{r}\cdot\left(2\hat{i}+5\hat{j}+3\hat{k}\right)-9\right]=0 [1]
or \vec{r}\cdot\left[(2+2\lambda)\hat{i}+(2+5\lambda)\hat{j}+(-3+3\lambda)\hat{k}\right]-7-9\lambda=0
(2, 1, 3) lies on it
\therefore 2(2+2\lambda)+1(2+5\lambda)+3(-3+3\lambda)-7-9\lambda=0
9\lambda=10\Rightarrow\lambda=10/9 [1½]
Equation is \vec{r}\cdot(38\hat{i}+68\hat{j}+3\hat{k})=153 [½]
Find the equation of the plane passing through points (2, 1, 0), (3, – 2, – 2) and (1, 1, – 7). Also, obtain its distance from the origin.
Equation of plane is given by
21(x-2)+9(y-1)-3z=0
i.e., 7x+3y-z=17 [1]
Distance of plane from origin is
d=\frac{|0+0-0-17|}{\sqrt{59}}=\frac{17}{\sqrt{59}} [½]
Find the distance between the lines x=\frac{y-1}{2}=\frac{z-2}{3} and x+1=\frac{y+2}{2}=\frac{z-1}{3}.
For lines \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} and \frac{x+1}{1}=\frac{y+2}{2}=\frac{z-1}{3}
Let \vec{a_1}=\hat{j}+2\hat{k},\ \vec{b}=\hat{i}+2\hat{j}+3\hat{k}
\vec{a_2}=-\hat{i}-2\hat{j}+\hat{k},\ \vec{b}=\hat{i}+2\hat{j}+3\hat{k}
Clearly lines are parallel [½]
Hence, Shortest distance or distance is given by
\vec{a_2}-\vec{a_1}=-\hat{i}-3\hat{j}-\hat{k} [½]
(\vec{a_2}-\vec{a_1})\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&-3&-1\\1&2&3\end{vmatrix}
=-7\hat{i}+2\hat{j}+\hat{k} [1]
Required distance =\frac{\sqrt{49+4+1}}{\sqrt{1+4+9}}=\frac{\sqrt{27}}{\sqrt7} or \frac{3\sqrt{21}}{7} [1]
Show that the lines : \frac{1-x}{2}=\frac{y-3}{4}=\frac{z}{-1} and \frac{x-4}{3}=\frac{2y-2}{-4}=z-1 are coplanar.
Equations of given lines in standard form are
\frac{x-1}{-2}=\frac{y-3}{4}=\frac{z}{-1} and [½]
\frac{x-4}{3}=\frac{y-1}{-2}=\frac{z-1}{1} [½]
Lines are coplanar if
\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}=0 [½]
LHS = \begin{vmatrix}3&-2&1\\-2&4&-1\\3&-2&1\end{vmatrix}
=3(4-2)+2(-2+3)+1(4-12) [½]
=6+2-8
=0
\Rightarrow Coplanar. [1]
If a line makes 60^\circ and 45^\circ angles with the positive directions of x-axis and z-axis respectively, then find the angle that it makes with the positive direction of y-axis. Hence, write the direction cosines of the line.
(a) l=\cos 60^\circ=\frac{1}{2},\ n=\cos 45^\circ=\frac{1}{\sqrt2} [½+½]
Now, l^2+m^2+n^2=1
\Rightarrow\frac{1}{4}+m^2+\frac{1}{2}=1 [½]
\Rightarrow m^2=\frac{1}{4}\Rightarrow m=\pm\frac{1}{2}
\theta=60^\circ [½]
Required direction cosines are \left\langle\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt2}\right\rangle [1]
Check whether the lines \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} and \frac{x-4}{5}=\frac{y-1}{2}=z are skew or not.
(b) Let, \vec{a}_1=\hat{i}+2\hat{j}+3\hat{k},\ \vec{a}_2=4\hat{i}+\hat{j}
\vec{b}_1=2\hat{i}+3\hat{j}+4\hat{k},\ \vec{b}_2=5\hat{i}+2\hat{j}+\hat{k}
\vec{a}_2-\vec{a}_1=3\hat{i}-\hat{j}-3\hat{k} [½]
\vec{b}_1\times\vec{b}_2=-5\hat{i}+18\hat{j}-11\hat{k} [1½]
Here, (\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)=-15-18+33=0 [½]
Hence given lines are not skew lines. [½]
Find the equation of the plane passing through the line of intersection of the planes \vec{r}\cdot\left(\hat{i}+\hat{j}+\hat{k}\right)=10 and \vec{r}\cdot\left(2\hat{i}+3\hat{j}-\hat{k}\right)+4=0 and passing through the point (–2, 3, 1).
Equation of plane through the intersection of given two planes is
\vec{r}\cdot[(\hat{i}+\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k})]-10+4\lambda=0 [1]
\vec{r}\cdot[(1+2\lambda)\hat{i}+(1+3\lambda)\hat{j}+(1-\lambda)\hat{k}]-10+4\lambda=0
Point (–2, 3, 1) lies on it.
\therefore -2-4\lambda+3+9\lambda+1-\lambda-10+4\lambda=0
8\lambda=8\Rightarrow\lambda=1 [1]
\therefore Equation of plane is
\vec{r}\cdot(3\hat{i}+4\hat{j})-6=0 [1]
Find the equation of the plane passing through three points whose position vectors are -\hat{j},\ 3\hat{i}+3\hat{j} and \hat{i}+\hat{j}+\hat{k} .
Given points are (0, −1, 0), (3, 3, 0) and (1, 1, 1)
Equation of required plane is
\begin{vmatrix} x & y+1 & z \\ 3 & 4 & 0 \\ 1 & 2 & 1 \end{vmatrix} = 0 [1½]
\Rightarrow 4x-3y+2z = 3 [1½]
Find the distance between the following parallel lines :
\vec{a}_2-\vec{a}_2 = -\hat{i}-3\hat{j}+2\hat{k},\ \vec{b} = \hat{i}+\hat{j}-\hat{k} [½]
(\vec{a}_2-\vec{a}_1)\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -3 & 2 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}+\hat{j}+2\hat{k} [1]
Required distance = \frac{|(\vec{a}_2-\vec{a}_1)\times\vec{b}|}{|\vec{b}|}
= \frac{\sqrt{1+1+4}}{\sqrt{1+1+1}} = \sqrt{\frac{6}{3}} = \sqrt{2} [1½]
Find the coordinates of the point where the line through the points (−1, 1, −8) and (5, −2, 10) crosses the ZX-plane.
Equation of line through (−1, 1, −8) and (5, −2, 10)
is \frac{x+1}{5-(-1)} = \frac{y-1}{-2-1} = \frac{z+8}{10-(-8)}
i.e. \frac{x+1}{6} = \frac{y-1}{-3} = \frac{z+8}{18} = \lambda [1]
Any point on this line is (6\lambda-1, -3\lambda+1, 18\lambda-8) [½]
Line crosses ZX-plane i.e. y = 0
\Rightarrow -3\lambda+1 = 0 \Rightarrow \lambda = \frac{1}{3} [1]
Required point is (1, 0, −2) [½]
Two security sensors in a multi-storey mall emit laser beams for movement detection. Three of these beams travel along the following lines:
l_1:\vec{r}=(\lambda+1)\hat{i}+(3\lambda+2)\hat{j}+(3-\lambda)\hat{k} and
l_2:\vec{r}=2(\mu+1)\hat{i}+(3\mu-1)\hat{j}+(\mu+1)\hat{k}
l_3:\vec{r}=(\gamma+3)\hat{i}+(3\gamma-1)\hat{j}+(1-\gamma)\hat{k}
Based on the given information, answer the following questions:
(i) Show that the lines l_1 and l_2 are not parallel to each other. [1]
(ii) Show that the lines l_1 and l_3 are parallel to each other. [1]
(iii) (a) Find the shortest distance between the lines l_1 and l_2. [2]
OR
(b) Find the shortest distance between the lines l_1 and l_3. [2]
(i) Given lines are
l_1:\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(\hat{i}+3\hat{j}-\hat{k})\equiv\vec{a}+\lambda\vec{b}
and
l_2:\vec{r}=2\hat{i}-\hat{j}+\hat{k}+\mu(2\hat{i}+3\hat{j}+\hat{k})\equiv\vec{c}+\mu\vec{d}
Here \vec{a}=\hat{i}+2\hat{j}+3\hat{k},\vec{b}=\hat{i}+3\hat{j}-\hat{k},\ \vec{c}=2\hat{i}-\hat{j}+\hat{k},\ \vec{d}=2\hat{i}+3\hat{j}+\hat{k}
For direction vectors \vec{b} and \vec{d}, \vec{b}\ne\lambda\vec{d}, for any \lambda\in R
\Rightarrow\vec{b} is not parallel to \vec{d}
So, lines are not parallel. [1]
(ii) Given lines are
l_1:\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(\hat{i}+3\hat{j}-\hat{k})\equiv\vec{a}+\lambda\vec{b}
and
l_3:\vec{r}=3\hat{i}-\hat{j}+\hat{k}+\gamma\left(\hat{i}+3\hat{j}-\hat{k}\right)\equiv\vec{e}+\gamma\vec{f}
Since \vec{b}=\vec{f}, the lines l_1 and l_3 are parallel. [1]
(iii) (a) \vec{c}-\vec{a}=2\hat{i}-\hat{j}+\hat{k}-\left(\hat{i}+2\hat{j}+3\hat{k}\right)=\hat{i}-3\hat{j}-2\hat{k}
\vec{b}\times\vec{d}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&3&-1\\2&3&1\end{vmatrix}=6\hat{i}-3\hat{j}-3\hat{k}=3(2\hat{i}-\hat{j}-\hat{k}) [1]
(\vec{c}-\vec{a})\cdot(\vec{b}\times\vec{d})=(\hat{i}-3\hat{j}-2\hat{k})\cdot3(2\hat{i}-\hat{j}-\hat{k})=3(2+3+2)=21\ne0
So, the lines are skew lines.
\left|\vec{b}\times\vec{d}\right|=3\sqrt{2^2+(-1)^2+(-1)^2}=3\sqrt6
Thus, the shortest distance between the given lines is
D=\left|\frac{(\vec{c}-\vec{a})\cdot(\vec{b}\times\vec{d})}{|\vec{b}\times\vec{d}|}\right|=\left|\frac{21}{3\sqrt6}\right|=\frac{7}{\sqrt6}=\frac{7\sqrt6}{6}\ units [1]
OR
(b) l_1:\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(\hat{i}+3\hat{j}-\hat{k})\equiv\vec{a}+\lambda\vec{b}
and
l_3:\vec{r}=3\hat{i}-\hat{j}+\hat{k}+\gamma\left(\hat{i}+3\hat{j}-\hat{k}\right)\equiv\vec{e}+\gamma\vec{b} [1]
Since the lines are parallel, the shortest distance = \frac{\left|\vec{b}\times(\vec{e}-\vec{a})\right|}{|\vec{b}|}
=\frac{1}{\sqrt{11}}\left|\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&3&-1\\2&-3&-2\end{vmatrix}\right|=\frac{|-9\hat{i}-9\hat{k}|}{\sqrt{11}}=\frac{9\sqrt{22}}{11}units [1]
Case Study – 1
In a school, the swimming pool is shaped as a cuboid, and the breadth of the pool is represented by line \frac{-x+2}{-5} = y = z - 3. Two students A and B are swimming along the length of the pool.
Based on the above information, answer the following questions :
(i) If the path of student A is represented by line \frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-3p}, find the value of p. [1]
(ii) Write the vector equation of the path of student B swimming parallel to the path of student A, if at a certain instant he passes through the point with coordinates (2, 4, –1). [1]
(iii) (a) Find the foot of the perpendicular drawn from (2, 4, –1) to the path represented by student A. [2]
OR
(iii) (b) If a bird flies across the top of the swimming pool with its path represented by \frac{x+4}{2} = \frac{y+3}{8} = \frac{z-5}{-18}, then find the shortest distance between its path and the path of student A. [2]
(i) 5 \times 1 + 1 \times 4 + 1 \times (-3p) = 0 \Rightarrow p = 3 [1]
(ii) Vector equation of student B’s path:
\vec{r} = (2\hat{i} + 4\hat{j} - \hat{k}) + \lambda(\hat{i} + 4\hat{j} - 9\hat{k}) [1]
(iii) (a) Let the point be Q(2, 4, -1)
Let P(\lambda - 5, 4\lambda - 3, -9\lambda + 6) be the foot of the perpendicular [½]
So, drs of PQ = \lt \lambda - 7, 4\lambda - 7, -9\lambda + 7 \gt [½]
Since PQ is perpendicular to path of student A, so
\lambda - 7 + 16\lambda - 28 + 81\lambda - 63 = 0 \Rightarrow 98\lambda = 98 \Rightarrow \lambda = 1 [½]
Thus, required foot of the perpendicular from Q to path of A is (-4, 1, -3). [½]
OR
(iii) (b) As, drs of bird path and drs of student A’s path are proportional
So, both paths (line) are parallel. [½]
Here, \vec{a_1} = -5\hat{i} - 3\hat{j} + 6\hat{k}, \vec{a_2} = -4\hat{i} - 3\hat{j} + 5\hat{k}, \vec{b} = \hat{i} + 4\hat{j} - 9\hat{k}
Now, (\vec{a_2} - \vec{a_1}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 1 & 4 & -9 \end{vmatrix} = 4\hat{i} + 8\hat{j} + 4\hat{k} [1]
Thus, required shortest distance = \frac{\sqrt{96}}{\sqrt{98}} or \frac{4\sqrt{3}}{7} [½]
An engineer is designing a new metro rail network in a city.
Initially, two metro lines, Line A and Line B, each consisting of multiple stations are designed. The track for Line A is represented by l_1:\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-3}{4}, while the track for Line B is represented by l_2:\frac{x-1}{2}=\frac{y-3}{1}=\frac{z+2}{-3}.
Based on the above information, answer the following questions :
(i) Find whether the two metro tracks are parallel. [1]
(ii) Solar panels are to be installed on the rooftop of the metro stations. Determine the equation of the line representing the placement of solar panels on the rooftop of Line A’s stations, given that panels are to be positioned parallel to Line A’s track (l_1) and pass through the point (1, − 2, − 3). [1]
(iii) (a) To connect the stations, a pedestrian pathway perpendicular to the two metro lines is to be constructed which passes through point (3, 2, 1). Determine the equation of the pedestrian walkway. [2]
OR
(iii) (b) Find the shortest distance between Line A and Line B. [2]
l_1:\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-3}{4}\ ;\ l_2:\frac{x-1}{2}=\frac{y-3}{1}=\frac{z+2}{-3}
(i) Drs of l_1 are \lt 3,-2,4\gt, drs of l_2 are \lt 2,1,-3\gt
as drs are not proportional, hence l_1 is not parallel to l_2. [1]
(ii) Equations of line parallel to l_1 and passing through (1,-2,-3) is
\frac{x-1}{3}=\frac{y+2}{-2}=\frac{z+3}{4} or \vec{r}=(\hat{i}-2\hat{j}-3\hat{k})+\lambda(3\hat{i}-2\hat{j}+4\hat{k}) [1]
(iii)(a) The pathway is perpendicular to l_1 and l_2. \therefore It is parallel to \vec{b}_1\times\vec{b}_2
\vec{b}=\vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-2&4\\2&1&-3\end{vmatrix}=2\hat{i}+17\hat{j}+7\hat{k} [1]
\therefore Equation of pathway is \vec{r}=(3\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+17\hat{j}+7\hat{k}) [1]
OR
(iii)(b) \vec{a}_1=2\hat{i}-\hat{j}+3\hat{k},\ \vec{a}_2=\hat{i}+3\hat{j}-2\hat{k}
\vec{b}_1=3\hat{i}-2\hat{j}+4\hat{k},\ \vec{b}_2=2\hat{i}+\hat{j}-3\hat{k}
d=\left|\frac{(\vec{a}_2-\vec{a}_1).(\vec{b}_1\times\vec{b}_2)}{|\vec{b}_1\times\vec{b}_2|}\right|
=\left|\frac{(-\hat{i}+4\hat{j}-5\hat{k}).(2\hat{i}+17\hat{j}+7\hat{k})}{\sqrt{4+289+49}}\right| [1]
=\frac{31}{\sqrt{342}} [1]
Electrical transmission wires which are laid down in winters are stretched tightly to accommodate expansion in summers.
Two such wires lie along the following lines :
Based on the given information, answer the following questions :
(i) Are the lines l_1 and l_2 coplanar ? Justify your answer. [2]
(ii) Find the point of intersection of the lines l_1 and l_2 . [2]
(i) Consider \begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\A_1&B_1&C_1\\A_2&B_2&C_2\end{vmatrix}
=\begin{vmatrix}+1&4&-5\\3&-2&-1\\-1&3&-2\end{vmatrix} [1]
=1(+7)-4(-7)-5(7)
=0
\therefore lines l_1 and l_2 are coplanar [1]
(ii) Any point on l_1 : \left(3\lambda-1,-2\lambda+3,-\lambda-2\right) [1]
Substituting in equation of l_2 ,
\frac{3\lambda-1}{-1}=\frac{-2\lambda+3-7}{3}
\Rightarrow 9\lambda-3=2\lambda+4
\Rightarrow 7\lambda=7\Rightarrow\lambda=1 [½]
Point is (2,1,-3) [½]
Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12} and the plane x – y + z = 5.
General point of the line is
(3\lambda+2,\ 4\lambda-1,\ 12\lambda+2) [1]
This will be the point of intersection with the plane, if
(3\lambda+2)-(4\lambda-1)+(12\lambda+2)=5
\Rightarrow 11\lambda+5=5,\ \Rightarrow\lambda=0 [1]
Thus, the point of intersection is (2, –1, 2) [1]
Distance =\sqrt{(2+1)^2+(-1+5)^2+(2+10)^2}
=\sqrt{9+16+144}
=\sqrt{169}=13 [1]
Find the distance of the point (1, –2, 9) from the point of intersection of the line \vec{r}=4\hat{i}+2\hat{j}+7\hat{k}+\lambda(3\hat{i}+4\hat{j}+2\hat{k}) and the plane \vec{r}\cdot(\hat{i}-\hat{j}+\hat{k})=10.
Given line is \vec{r}=(4\hat{i}+2\hat{j}+7\hat{k})+\lambda(3\hat{i}+4\hat{j}+2\hat{k})
Any point on the line is (4+3\lambda,\ 2+4\lambda,\ 7+2\lambda) [1]
This point lies on the plane \vec{r}\cdot(\hat{i}-\hat{j}+\hat{k})=10
\Rightarrow 4+3\lambda-2-4\lambda+7+2\lambda=10
\Rightarrow\lambda+9=10\Rightarrow\lambda=1 [1]
\therefore Point of intersection is (7, 6, 9) [1]
Distance of (7, 6, 9) from (1, –2, 9) is
d=\sqrt{(7-1)^2+(6+2)^2+0}
=\sqrt{36+64}=10 [1]
Find the equations of the planes passing through the line of intersection of the planes \vec{r}\cdot(\hat{i}+3\hat{j})=6 and \vec{r}\cdot(3\hat{i}-\hat{j}-4\hat{k})=0, which are at a distance of 1 unit from the origin.
Let the equations of required planes are: \vec{r}\cdot(\hat{i}+3\hat{j})-6+\lambda\left[\vec{r}\cdot(3\hat{i}-\hat{j}-4\hat{k})\right]=0 [1]
or \vec{r}\cdot[(1+3\lambda)\hat{i}+(3-\lambda)\hat{j}-4\lambda\hat{k}]=6 for some \lambda
According to given condition
\Rightarrow 26\lambda^2=26\Rightarrow\lambda=\pm1 [1]
Equations of planes are: \vec{r}\cdot(4\hat{i}+2\hat{j}-4\hat{k})=6,\ \vec{r}\cdot(-2\hat{i}+4\hat{j}+4\hat{k})=6 [½+½]
Case Study Based Question
Two motorcycles A and B are running at the speed more than the allowed speed on the roads represented by the lines \vec{r}=\lambda\left(\hat{i}+2\hat{j}-\hat{k}\right) and \vec{r}=\left(3\hat{i}+3\hat{j}\right)+\mu\left(2\hat{i}+\hat{j}+\hat{k}\right) respectively.
Based on the above information, answer the following questions :
(a) Find the shortest distance between the given lines. [2]
(b) Find the point at which the motorcycles may collide. [2]
(a) \vec{a}_1=0\hat{i}+0\hat{j}+0\hat{k},\ \vec{a}_2=3\hat{i}+3\hat{j}
\vec{a}_2-\vec{a}_1=3\hat{i}+3\hat{j} [½]
\vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&-1\\2&1&1\end{vmatrix}
=3\hat{i}-3\hat{j}-3\hat{k} [1]
SD =\frac{\left|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)\right|}{|\vec{b}_1\times\vec{b}_2|}
Now, (\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)=(3\hat{i}+3\hat{j})(3\hat{i}-3\hat{j}-3\hat{k})
=9-9=0
Shortest distance between two lines = 0 [½]
(b) Any point on the line \vec{r}=\lambda(\hat{i}+2\hat{j}-\hat{k}) is \lambda\hat{i}+2\lambda\hat{j}-\lambda\hat{k}
Any point on the line \vec{r}=3\hat{i}+3\hat{j}+\mu(2\hat{i}+\hat{j}+\hat{k}) is (2\mu+3)\hat{i}+(\mu+3)\hat{j}+\mu\hat{k} [½]
As the lines are intersecting,
\lambda=2\mu+3,\ 2\lambda=\mu+3 [½]
On solving \mu=-1,\ \lambda=1 [½]
Point of intersection is \hat{i}+2\hat{j}-\hat{k} or (1, 2, –1) [½]
Find the equation of the plane passing through the intersection of the planes \vec{r}\cdot(2\hat{i}+2\hat{j}-3\hat{k}) = 7 and \vec{r}\cdot(2\hat{i}+5\hat{j}+3\hat{k}) = 9 and through the point (2, 1, 3).
Equation of required plane is
\vec{r}\cdot[(2\hat{i}+2\hat{j}-3\hat{k})+\lambda(2\hat{i}+5\hat{j}+3\hat{k})] = 7+9\lambda [1]
or \vec{r}\cdot[(2+2\lambda)\hat{i}+(2+5\lambda)\hat{j}+(-3+3\lambda)\hat{k}] = 7+9\lambda
As the plane passes through (2, 1, 3), we have
(2\hat{i}+\hat{j}+3\hat{k})\cdot[(2+2\lambda)\hat{i}+(2+5\lambda)\hat{j}+(-3+3\lambda)\hat{k}] = 7+9\lambda [1]
\Rightarrow 2(2+2\lambda)+1(2+5\lambda)+3(-3+3\lambda) = 7+9\lambda
\Rightarrow 9\lambda = 10 \Rightarrow \lambda = \frac{10}{9} [1]
Required plane is \vec{r}\cdot\left(\frac{38}{9}\hat{i}+\frac{68}{9}\hat{j}+\frac{3}{9}\hat{k}\right) = 17 [1]
or \vec{r}\cdot(38\hat{i}+68\hat{j}+3\hat{k}) = 153
The two lines \frac{x-1}{3}=-y, z+1=0 and \frac{-x}{2}=\frac{y+1}{2}=z+2 intersect at a point whose y-coordinate is 1. Find the co-ordinates of their point of intersection. Find the vector equation of a line perpendicular to both the given lines and passing through this point of intersection.
Let \frac{x-1}{3}=\frac{y-0}{-1}=\frac{z+1}{0}=\lambda\Rightarrow Any point on it is (3\lambda+1,-\lambda,-1) [½]
For the point where y=1\Rightarrow\lambda=-1 [1]
\Rightarrow The point is (-2,1,-1) [½]
The directions of the two lines are \vec{m}=3\hat{i}-\hat{j} [1]
and \vec{n}=-2\hat{i}+2\hat{j}+\hat{k} [½]
\vec{m}\times\vec{n}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-1&0\\-2&2&1\end{vmatrix}=-\hat{i}-3\hat{j}+4\hat{k} [1]
The equation of the required line is
\vec{r}=(-2\hat{i}+\hat{j}-\hat{k})+\mu(-\hat{i}-3\hat{j}+4\hat{k}) [½]
Alternative method
Let \frac{x-1}{3}=\frac{y-0}{-1}=\frac{z+1}{0}=\lambda\Rightarrow Any point on it is (3\lambda+1,-\lambda,-1) [½]
For the point where y=1\Rightarrow\lambda=-1 [1]
\Rightarrow The point is (-2,1,-1) [½]
Let the direction ratios of the required line be a, b, c
Then 3a-b=0 [1]
And -2a+2b+c=0 [1]
Solving we get \frac{a}{-1}=\frac{-b}{3}=\frac{c}{4}\Rightarrow\frac{a}{-1}=\frac{b}{-3}=\frac{c}{4}
The required line is \frac{x+2}{-1}=\frac{y-1}{-3}=\frac{z+1}{4}=\mu [½]
In vector form \vec{r}=(-2\hat{i}+\hat{j}-\hat{k})+\mu(-\hat{i}-3\hat{j}+4\hat{k}) [½]
Check whether the lines given by \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} and \frac{x-4}{5} = \frac{y-1}{2} = z are parallel or not. If parallel, find the distance between them, otherwise find their point of intersection, if the lines are intersecting.
Drs of first line are < 2, 3, 4 >
Drs of second line are < 5, 2, 1 > [1]
Drs of the two lines are not proportional. Hence, the lines are not parallel. [1]
Any point on the first line is (2\lambda + 1, 3\lambda + 2, 4\lambda + 3)
Any point on the second line is (5\mu + 4, 2\mu + 1, \mu) [1]
For the lines to intersect, we must have some \lambda and \mu, for which these coordinates must coincide, i.e., we must have 2\lambda + 1 = 5\mu + 4, 3\lambda + 2 = 2\mu + 1, 4\lambda + 3 = \mu, for some \lambda and \mu
The first two equations, when solved, give us \lambda = -1 and \mu = -1. These values satisfy the third equation. Hence, the lines intersect. [1]
The point of intersection is (-1, -1, -1) [1]
A line passing through the points A(1, 2, 3) and B(5, 8, 11) intersects the line \vec{r} = 4\hat{i} + \hat{j} + \lambda(5\hat{i} + 2\hat{j} + \hat{k}). Find the co-ordinates of the point of intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines.
The line passing through the points A(1, 2, 3) and B(5, 8, 11) is given by
l_1 : \frac{x - 1}{5 - 1} = \frac{y - 2}{8 - 2} = \frac{z - 3}{11 - 3} i.e. \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = \mu [1]
Also the given line in its cartesian form is, l_2 : \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1} = \lambda
Any point on l_1 is P(2\mu + 1, 3\mu + 2, 4\mu + 3)
Any point on l_2 is Q(5\lambda + 4, 2\lambda + 1, \lambda) [½]
When the lines intersect, the points P and Q must coincide.
\therefore 2\mu + 1 = 5\lambda + 4,\ 3\mu + 2 = 2\lambda + 1 \text{ and } 4\mu + 3 = \lambda
\Rightarrow 2\mu - 5\lambda = 3 \ldots\text{(i)},\ 3\mu - 2\lambda = -1 \ldots\text{(ii)} \text{ and } 4\mu - \lambda = -3 \ldots\text{(iii)}
solving any two above equations, we get \mu = \lambda = -1 [1]
The point of intersection is (-1, -1, -1). [½]
Let the required line (l) passing through (-1, -1, -1) be \frac{x + 1}{a} = \frac{y + 1}{b} = \frac{z + 1}{c}
Since l \perp l_1 and l \perp l_2,
2a + 3b + 4c = 0,\ 5a + 2b + c = 0
\Rightarrow \frac{a}{3 - 8} = \frac{b}{20 - 2} = \frac{c}{4 - 15} i.e. \frac{a}{-5} = \frac{b}{18} = \frac{c}{-11} [1]
Hence, the required line is \frac{x + 1}{-5} = \frac{y + 1}{18} = \frac{z + 1}{-11} or \frac{x + 1}{5} = \frac{y + 1}{-18} = \frac{z + 1}{11} [1]
Note: The required line may also be written in the vector form.
i.e. \vec{r} = (-\hat{i} - \hat{j} - \hat{k}) + \lambda(5\hat{i} - 18\hat{j} + 11\hat{k})
Note: Student can alternatively use point B to write the equation of line l_1.
Represent the equations of lines l_1 and l_2 in vector form and check whether they are intersecting or not.
l_1 : \frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}
l_2 : \frac{x + 1}{-1} = \frac{2 - y}{-2} = \frac{z - 5}{5}
l_1 : \vec{r} = (-3\hat{i} + \hat{j} + 5\hat{k}) + \lambda(-3\hat{i} + \hat{j} + 5\hat{k}) [1]
l_2 : \vec{r} = (-\hat{i} + 2\hat{j} + 5\hat{k}) + \mu(-\hat{i} + 2\hat{j} + 5\hat{k}) [1]
\vec{a_1} = -3\hat{i} + \hat{j} + 5\hat{k};\quad \vec{b_1} = -3\hat{i} + \hat{j} + 5\hat{k}
\vec{a_2} = -\hat{i} + 2\hat{j} + 5\hat{k};\quad \vec{b_2} = -\hat{i} + 2\hat{j} + 5\hat{k}
\vec{a_2} - \vec{a_1} = 2\hat{i} + \hat{j} [1]
\vec{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\-3 & 1 & 5\\-1 & 2 & 5\end{vmatrix} = -5\hat{i} + 10\hat{j} - 5\hat{k} [1]
\because (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0 [½]
\therefore Both lines are intersecting. [½]
Opposite sides of a square are along the lines :
\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})
\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})
Find the area of the square if direction ratios of other pair of opposite sides of the square are given by \lt -3, 6, p \gt. Also, find the value of p.
\vec{a_1} = \hat{i} + 2\hat{j} - 4\hat{k};\quad \vec{a_2} = 3\hat{i} + 3\hat{j} - 5\hat{k};\quad \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}
\vec{a_2} - \vec{a_1} = 2\hat{i} + \hat{j} - \hat{k} [1]
(\vec{a_2} - \vec{a_1}) \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\2 & 1 & -1\\2 & 3 & 6\end{vmatrix} = 9\hat{i} - 14\hat{j} + 4\hat{k} [1]
|(\vec{a_2} - \vec{a_1}) \times \vec{b}| = \sqrt{293} [½]
Distance between given parallel lines = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|} = \frac{\sqrt{293}}{7} [½]
\therefore Length of side of square = distance between parallel lines
= \frac{\sqrt{293}}{7} or \frac{\sqrt{293}}{7} units
Area of square = \frac{293}{49} or \frac{293}{49} square units [1]
\because Adjacent sides of square are perpendicular to each other.
\therefore -3 \times 2 + 6 \times 3 + 6p = 0 \Rightarrow p = -2 [1]
Find the equation of a line (in vector and cartesian form) that passes through the point of intersection of lines \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 4}{5} = \frac{y - 1}{2} = z and is parallel to the vector 3\hat{i} + 2\hat{j} - 8\hat{k}.
Let \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = a & \frac{x - 4}{5} = \frac{y - 1}{2} = z = b
So any general point on these are P(2a + 1, 3a + 2, 4a + 3) and Q(5b + 4, 2b + 1, b) respectively. [1]
For intersection points: 2a + 1 = 5b + 4, 4a + 3 = b
on solving we get, a = b = -1 [1]
Thus, required point of intesection is (-1, -1, -1). [1]
Required equation of line (cartesian form): \frac{x + 1}{3} = \frac{y + 1}{2} = \frac{z + 1}{-8} [1]
Required equation of line (Vector form):
\vec{r} = \left(-\hat{i} - \hat{j} - \hat{k}\right) + \lambda\left(3\hat{i} + 2\hat{j} - 8\hat{k}\right) [1]
Find the foot of the perpendicular from the point (0, 2, 3) on the line \frac{-x-3}{-5} = \frac{1-y}{-2} = \frac{3z+12}{9} and hence find the length of the perpendicular.
Given line is \frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = \lambda [1]
Any general point on line l is (5\lambda - 3, 2\lambda + 1, 3\lambda - 4) [½]
Drs of given line are <5, 2, 3>
Drs of perpendicular line are \lt 5\lambda - 3, 2\lambda - 1, 3\lambda - 7 \gt [½]
As lines are perpendicular
\therefore 5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0 gives, \lambda = 1 [1]
\therefore coordinates of foot of perpendicular are (2, 3, -1) [1]
length of perpendicular = \sqrt{21} [1]
Find the value of p if the shortest distance between the lines
\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and
\vec{r} = (p\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})
is \frac{3}{\sqrt{2}} units.
Let \vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}, \vec{a}_2 = p\hat{i} - \hat{j} - \hat{k}, and
\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}, \vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}
Here (\vec{a}_2 - \vec{a}_1) = (p - 1)\hat{i} - 3\hat{j} - 2\hat{k} [1]
\vec{b}_1 \times \vec{b}_2 = -3\hat{i} + 3\hat{k} [1½]
S.D. = \left|\frac{(\vec{a}_2 - \vec{a}_1).(\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|}\right|
\frac{-3p - 3}{3\sqrt{2}} = \pm\frac{3}{\sqrt{2}} [1½]
gives p = -4, 2 [1]
Find the image A' of the point A(1, 6, 3) in the line \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}. Also, find the equation of the line joining A and A'.
The equation of given line is \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda
Any arbitrary point on the line is M(\lambda,2\lambda+1,3\lambda+2) [1]
dr's of AM are \lt\lambda-1,2\lambda-5,3\lambda-1\gt
Here 1(\lambda-1)+2(2\lambda-5)+3(3\lambda-1)=0 [1]
\Rightarrow\lambda=1 [½]
\therefore M(1,3,5) is the foot perpendicular of the point A to the given line.
Let image of point A in the line be A'(\alpha,\beta,\gamma)
Since M is the mid-point of AA', so M\left(\frac{1+\alpha}{2},\frac{6+\beta}{2},\frac{3+\gamma}{2}\right)=M(1,3,5) [½]
\Rightarrow A'(1,0,7) is the image of A. [1]
Also, Equation of AA' is \frac{x-1}{0}=\frac{y-6}{-3}=\frac{z-3}{2} [1]
Find a point P on the line \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} such that its distance from point Q(2, 4, –1) is 7 units. Also, find the equation of line joining P and Q.
The given line is \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda and Q(2,4,-1)
Any random point on the line will be given by P(\lambda-5,4\lambda-3,-9\lambda+6) [1]
Since PQ=7\Rightarrow\sqrt{(\lambda-7)^2+(4\lambda-7)^2+(-9\lambda+7)^2}=7 [1]
\Rightarrow 98(\lambda^2-2\lambda+1)=0\Rightarrow\lambda=1 [1]
Hence, the required point is P(-4,1,-3) [1]
The equation of line PQ is \frac{x+4}{6}=\frac{y-1}{3}=\frac{z+3}{2} or \frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2} [1]
Find the shortest distance between the lines :
The vector equations of the lines are
\vec{r} = -\hat{i} + \hat{j} + 9\hat{k} + \lambda(2\hat{i} + \hat{j} - 3\hat{k})
\vec{r} = 3\hat{i} - 15\hat{j} + 9\hat{k} + \mu(2\hat{i} - 7\hat{j} + 5\hat{k})
\vec{a_1} = -\hat{i} + \hat{j} + 9\hat{k},\ \vec{a_2} = 3\hat{i} - 15\hat{j} + 9\hat{k}
\vec{b_1} = 2\hat{i} + \hat{j} - 3\hat{k},\ \vec{b_2} = 2\hat{i} - 7\hat{j} + 5\hat{k} [1]
\vec{a_2} - \vec{a_1} = 4\hat{i} - 16\hat{j} [1]
\vec{b_1}\times\vec{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 2 & 1 & -3\\ 2 & -7 & 5\end{vmatrix} = -16\hat{i} - 16\hat{j} - 16\hat{k} [2]
S.D. = \frac{|(\vec{a_2} - \vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} = \frac{12}{\sqrt{3}} = 4\sqrt{3} [1]
Find the image A′ of the point A(2, 1, 2) in the line l : \vec{r} = 4\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - \hat{j} - \hat{k}). Also, find the equation of line joining AA′. Find the foot of perpendicular from point A on the line l.
[Figure: A(2, 1, 2), foot P on line l, image A′(α, β, γ)]
Let the image of A in the line be A'(\alpha, \beta, \gamma)
The point P, which is the point of intersection of the lines l and AA', will have coordinates (\lambda + 4, -\lambda + 2, -\lambda + 2) for some λ. [½]
Drs of AP are \lt \lambda + 2, -\lambda + 1, -\lambda \gt [½]
AP \perp l
(\lambda + 2) - (-\lambda + 1) - (-\lambda) = 0
\Rightarrow \lambda = -\frac{1}{3} [1]
Therefore, the coordinates of P are \left(\frac{11}{3}, \frac{7}{3}, \frac{7}{3}\right) [½]
P is the mid-point of AA'
\Rightarrow \frac{2 + \alpha}{2} = \frac{11}{3}, \frac{1 + \beta}{2} = \frac{7}{3}, \frac{2 + \gamma}{2} = \frac{7}{3}
\Rightarrow \alpha = \frac{16}{3}, \beta = \frac{11}{3}, \gamma = \frac{8}{3} [1½]
The coordinates of the image are \left(\frac{16}{3}, \frac{11}{3}, \frac{8}{3}\right)
The equation of AA' is
\frac{x - 2}{\frac{10}{3}} = \frac{y - 1}{\frac{8}{3}} = \frac{z - 2}{\frac{2}{3}} [1]
or, \frac{3(x - 2)}{5} = \frac{3(y - 1)}{4} = \frac{3(z - 2)}{1}
Find the equation of a line in vector and cartesian form which passes through the point (1, 2, − 4) and is perpendicular to the lines \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}, and
\vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu(3\hat{i}+8\hat{j}-5\hat{k}).
Given lines are \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}
and \vec{r}=(15\hat{i}+29\hat{j}+5\hat{k})+\mu(3\hat{i}+8\hat{j}-5\hat{k})
The first line in vector form is \vec{r}=(8\hat{i}-19\hat{j}+10\hat{k})+\lambda(3\hat{i}-16\hat{j}+7\hat{k}) [1]
\vec{a}_1=8\hat{i}-19\hat{j}+10\hat{k},\ \vec{a}_2=15\hat{i}+29\hat{j}+5\hat{k}
\vec{b}_1=3\hat{i}-16\hat{j}+7\hat{k},\ \vec{b}_2=3\hat{i}+8\hat{j}-5\hat{k} [1]
\vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-16&7\\3&8&-5\end{vmatrix}=24\hat{i}+36\hat{j}+72\hat{k} [1]
\therefore Equation of line passing through (1,2,-4) and parallel to \vec{b} is
\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+t(24\hat{i}+36\hat{j}+72\hat{k}) or \vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+t'(2\hat{i}+3\hat{j}+6\hat{k}) [1]
Cartesian form of line is \frac{x-1}{24}=\frac{y-2}{36}=\frac{z+4}{72} or \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} [1]
Find the foot of the perpendicular drawn from the point (1,1,4) on the line \frac{x+2}{5}=\frac{y+1}{2}=\frac{-z+4}{-3}.
Let \frac{x+2}{5}=\frac{y+1}{2}=\frac{z-4}{3}=\lambda
Coordinate of general point on the given line are M(5\lambda-2,2\lambda-1,3\lambda+4) [1]
Direction Ratios of PM vector are \lt5\lambda-3,2\lambda-2,3\lambda\gt [1]
Since, PM\perp l
\Rightarrow5(5\lambda-3)+2(2\lambda-2)+3(3\lambda)=0 [1]
\Rightarrow\lambda=\frac12 [1]
Hence, coordinates of M are \left(\frac12,0,\frac{11}{2}\right) [1]
Find the point on the line \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-4}{3} at a distance of 2\sqrt2 units from the point (-1,-1,2).
Equation of given line be \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-4}{3}=\lambda (say)
Coordinate of any general point on the line are P(3\lambda+1,2\lambda-1,3\lambda+4). [1]
Let distance of point P from (-1,-1,2) is 2\sqrt2.
\Rightarrow\sqrt{(3\lambda+2)^2+(2\lambda)^2+(3\lambda+2)^2}=2\sqrt2 [1½]
\Rightarrow22\lambda^2+24\lambda=0
\Rightarrow\lambda=0 or \lambda=-\frac{12}{11} [1]
Hence, coordinates of point P are (1,-1,4) or \left(-\frac{25}{11},-\frac{35}{11},\frac{8}{11}\right) [1½]
Let the polished side of the mirror be along the line \frac{x}{1} = \frac{1-y}{-2} = \frac{2z-4}{6}.
A point P(1, 6, 3), some distance away from the mirror, has its image formed behind the mirror. Find the coordinates of the image point and the distance between the point P and its image.
Equation of given line is \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} [1]
Let coordinates of point on the line be (\lambda,\ 2\lambda + 1,\ 3\lambda + 2) for some \lambda [½]
Drs of line perpendicular to line along mirror are \langle \lambda - 1,\ 2\lambda - 5,\ 3\lambda - 1 \rangle [½]
(\lambda - 1) \cdot 1 + (2\lambda - 5) \cdot 2 + (3\lambda - 1) \cdot 3 = 0 gives \lambda = 1 [1]
Coordinates of foot of perpendicular are (1, 3, 5) [1]
For image
\frac{x+1}{2} = 1,\ \frac{y+6}{2} = 3,\ \frac{z+3}{2} = 5 gives image as (1, 0, 7) [½]
Required distance = \sqrt{0 + 36 + 16} = 2\sqrt{13} [½]
Find the point Q on the line \frac{2x+4}{6}=\frac{y+1}{2}=\frac{-2z+6}{-4} at a distance of 3\sqrt2 from the point P(1, 2, 3).
The general point on the line (3\lambda-2,2\lambda-1,2\lambda+3) is Q, from some \lambda\in\mathbb{R} [2]
PQ=3\sqrt2\Rightarrow(PQ)^2=18\Rightarrow(3\lambda-3)^2+(2\lambda-3)^2+(2\lambda)^2=18 [1]
17\lambda^2-30\lambda=0\Rightarrow\lambda=0 or \lambda=\frac{30}{17} [1]
Thus, the point is Q(-2,-1,3) or Q\left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right) [1]
Find the image of the point (-1,5,2) in the line \frac{2x-4}{2}=\frac{y}{2}=\frac{2-z}{3}. Find the length of the line segment joining the points (given point and the image point).
Let A'(a,b,c) be the image of the point A(-1,5,2) in the given line, also assume ‘M’ as the point of intersection of AA' with the given line, then ‘M’ is the mid-point of the line segment AA'
The Line in the standard form is: \frac{x-2}{1}=\frac{y}{2}=\frac{z-2}{-3}, then [1]
M is the point (\lambda+2,2\lambda,-3\lambda+2), for some \lambda\in\mathbb{R} [½]
Direction Ratios of AM are \lambda+3,2\lambda-5,-3\lambda [½]
AM\perp\text{Line}, \therefore1(\lambda+3)+2(2\lambda-5)-3(-3\lambda)=0\Rightarrow\lambda=\frac12 [1]
M\left(\frac52,1,\frac12\right)=M\left(\frac{a-1}{2},\frac{b+5}{2},\frac{c+2}{2}\right)\Rightarrow a=6,b=-3,c=-1 [1]
\therefore The Image of A in the line is A'(6,-3,-1) [½]
And, AA'=\sqrt{49+64+9}=\sqrt{122} [½]
Find the shortest distance between the lines l_1 and l_2 given by :
l_1 : \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(4\hat{i} + 6\hat{j} + 12\hat{k})
and l_2 : \vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(6\hat{i} + 9\hat{j} + 18\hat{k})
Given lines are : \vec{r} = \left(\hat{i} + 2\hat{j} - 4\hat{k}\right) + 2\lambda\left(2\hat{i} + 3\hat{j} + 6\hat{k}\right)
and \vec{r} = \left(\hat{i} + 2\hat{j} - 4\hat{k}\right) + 3\mu\left(2\hat{i} + 3\hat{j} + 6\hat{k}\right)
Clearly, the given lines are parallel.
Here, \vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}, \vec{a}_2 = \hat{i} + 2\hat{j} - 4\hat{k} and \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} [1]
\vec{a}_2 - \vec{a}_1 = 2\hat{i} + \hat{j} - \hat{k} [½]
(\vec{a}_2 - \vec{a}_1)\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\2&3&6\end{vmatrix} = 9\hat{i} - 14\hat{j} + 4\hat{k} [1 ½]
\therefore \left|(\vec{a}_2 - \vec{a}_1)\times\vec{b}\right| = \sqrt{81 + 196 + 16} = \sqrt{293} [1]
Also, |\vec{b}| = \sqrt{4 + 9 + 36} = 7 [½]
S.D. = \frac{\left|(\vec{a}_2 - \vec{a}_1)\times\vec{b}\right|}{|\vec{b}|}
= \frac{\sqrt{293}}{7} [½]
Show that the lines \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} and \frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1} intersect. Also, find their point of intersection.
Let the given lines be
l_1 : \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda and l_2 : \frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1} = \mu
Any point on the line l_1 is (2\lambda + 1, 3\lambda + 2, 4\lambda + 3) [1]
Any point on the line l_2 is (5\mu + 4, 2\mu + 1, \mu) [1]
For the given lines to intersect, there must be a common point.
\therefore 2\lambda + 1 = 5\mu + 4 \Rightarrow 2\lambda - 5\mu = 3 \quad \ldots(i)
3\lambda + 2 = 2\mu + 1 \Rightarrow 3\lambda - 2\mu = -1 \quad \ldots(ii)
4\lambda + 3 = \mu \Rightarrow 4\lambda - \mu = -3 \quad \ldots(iii) [1]
Solving (i) and (ii) gives, \lambda = \mu = -1 [1]
We notice that \lambda = \mu = -1 also satisfies equation (iii)
\therefore The given lines intersect.
Point of intersection is (2(-1) + 1, 3(-1) + 2, 4(-1) + 3) i.e. (-1, -1, -1) [1]
The image of point P(x, y, z) with respect to line \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} is P′ (1, 0, 7). Find the coordinates of point P.
Let foot of the perpendicular on the given line from point P be M (\lambda, 2\lambda + 1, 3\lambda + 2) [1]
D. ratios of PP′ are \lambda - 1,\ 2\lambda + 1,\ 3\lambda - 5 [1]
1(\lambda - 1) + 2(2\lambda + 1) + 3(3\lambda - 5) = 0
\Rightarrow \lambda = 1 [1]
Coordinates of M(1, 3, 5) [1]
\frac{x + 1}{2} = 1,\ \frac{y + 0}{2} = 3,\ \frac{z + 7}{2} = 5
\Rightarrow x = 1,\ y = 6,\ z = 3 \Rightarrow P(1, 6, 3) [1]
Find the equation of the line which bisects the line segment joining points A(2, 3, 4) and B(4, 5, 8) and is perpendicular to the lines \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} and \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}.
Let direction ratios of the required line be a, b, c.
\because the required line is perpendicular to both the given lines
\therefore 3a - 16b + 7c = 0 [1]
and 3a + 8b - 5c = 0 [½]
\Rightarrow \frac{a}{24} = \frac{b}{36} = \frac{c}{72}
\Rightarrow \frac{a}{2} = \frac{b}{3} = \frac{c}{6} [1½]
The mid-point of the line-segment AB is (3, 4, 6) [1]
Hence, the required equation of the line is
\frac{x-3}{2} = \frac{y-4}{3} = \frac{z-6}{6} [1]
Find the equation of the line passing through the point of intersection of the lines \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} and \frac{x - 1}{0} = \frac{y}{-3} = \frac{z - 7}{2} and perpendicular to these given lines.
l_1 : \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda\ ;\ l_2 : \frac{x - 1}{0} = \frac{y}{-3} = \frac{z - 7}{2} = \mu
any point on l_1 is (\lambda, 2\lambda + 1, 3\lambda + 2) & any point on l_2 is (1, -3\mu, 2\mu + 7) [1]
If l_1 and l_2 intersect,
\lambda = 1, 2\lambda + 1 = -3\mu and 3\lambda + 2 = 2\mu + 7 \Rightarrow \lambda = 1 and \mu = -1 [1]
Point of intersection of l_1 and l_2 is (1, 3, 5). [1]
Let d.r.'s of required line be \lt a, b, c \gt. Then,
a + 2b + 3c = 0 and -3b + 2c = 0 \Rightarrow \frac{a}{13} = \frac{b}{-2} = \frac{c}{-3} [1]
Required equation of line is \frac{x - 1}{13} = \frac{y - 3}{-2} = \frac{z - 5}{-3} [1]
Two vertices of the parallelogram ABCD are given as A(–1, 2, 1) and B(1, –2, 5). If the equation of the line passing through C and D is \frac{x - 4}{1} = \frac{y + 7}{-2} = \frac{z - 8}{2}, then find the distance between sides AB and CD. Hence, find the area of parallelogram ABCD.
[Figure: parallelogram ABCD with A(–1, 2, 1), B(1, –2, 5)]
d.r’s of CD are \lt 1, -2, 2 \gt
\therefore d.r’s of AB are \lt 1, -2, 2 \gt [½]
\therefore Equation of AB is \frac{x + 1}{1} = \frac{y - 2}{-2} = \frac{z - 1}{2} [½]
\therefore Equation of CD is \frac{x - 4}{1} = \frac{y + 7}{-2} = \frac{z - 8}{2}
Let \vec{a}_1 = -\hat{i} + 2\hat{j} + \hat{k},\ \vec{a}_2 = 4\hat{i} - 7\hat{j} + 8\hat{k} & \vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}
Now, \vec{a}_2 - \vec{a}_1 = 5\hat{i} - 9\hat{j} + 7\hat{k} [1]
(\vec{a}_2 - \vec{a}_1)\times\vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 5 & -9 & 7\\ 1 & -2 & 2\end{vmatrix} = -4\hat{i} - 3\hat{j} - \hat{k} [1]
Distance between AB and CD is given by d = \frac{\left|(\vec{a}_2 - \vec{a}_1)\times\vec{b}\right|}{|\vec{b}|}
d = \frac{\sqrt{16 + 9 + 1}}{\sqrt{1 + 4 + 4}} = \frac{\sqrt{26}}{3} [½]
CD = \sqrt{2^2 + (-4)^2 + (4)^2} = 6 [½]
Area of parallelogram ABCD = b \times h = 6 \times \frac{\sqrt{26}}{3} = 2\sqrt{26} [1]
Find the distance between the line \frac{x}{2} = \frac{2y-6}{4} = \frac{1-z}{-1} and another line parallel to it passing through the point (4, 0, –5).
Equation of the given line in standard form is
L_1 : \frac{x}{2} = \frac{y-3}{2} = \frac{z-1}{1} [½]
Equation of the line parallel to L_1 & passing through (4, 0, –5) is
L_2 : \frac{x-4}{2} = \frac{y}{2} = \frac{z+5}{1} [1]
Vector Equation of Lines are L_1 : \vec{r} = (0\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 2\hat{j} + \hat{k})
L_2 : \vec{r} = (4\hat{i} + 0\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})
Now, \vec{a_2} - \vec{a_1} = (4\hat{i} + 0\hat{j} - 5\hat{k}) - (0\hat{i} + 3\hat{j} + \hat{k}) = (4\hat{i} - 3\hat{j} - 6\hat{k}) [½]
\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}
(\vec{a_2} - \vec{a_1}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -3 & -6 \\ 2 & 2 & 1 \end{vmatrix} = 9\hat{i} - 16\hat{j} + 14\hat{k} [1]
|\vec{b}| = \sqrt{4+4+1} = 3 [½]
Thus, distance between the lines is
\text{S.D.} = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|} = \frac{\sqrt{81+256+196}}{3} = \frac{\sqrt{533}}{3} \text{ units} [1½]
If the lines \frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2} and \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-7} are perpendicular to each other, find the value of k and hence write the vector equation of a line perpendicular to these two lines and passing through the point (3, –4, 7).
L_1 : \frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2} \Rightarrow direction ratio's of L_1 = \langle -3, 2k, 2 \rangle [½]
L_2 : \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-7} \Rightarrow direction ratio's of L_2 = \langle 3k, 1, -7 \rangle [½]
Since L_1 \perp L_2,
-9k + 2k - 14 = 0 \Rightarrow k = -2 [1]
Thus, d.r.'s of L_1 = \langle -3, -4, 2 \rangle, d.r.'s of L_2 = \langle -6, 1, -7 \rangle
Now the vector perpendicular to both L_1 & L_2 is given by
\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -4 & 2 \\ -6 & 1 & -7 \end{vmatrix} = 26\hat{i} - 33\hat{j} - 27\hat{k} [2]
Thus, Equation of the required line is
\vec{r} = (3\hat{i} - 4\hat{j} + 7\hat{k}) + \lambda(26\hat{i} - 33\hat{j} - 27\hat{k}) [1]
Find the co-ordinates of the foot of the perpendicular drawn from the point (2,3,-8) to the line \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.
Also, find the perpendicular distance of the given point from the line.
The standard form of the equation of the line is \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3} [½]
Let foot of the perpendicular from the point A(2,3,-8) to the given line be B(-2\lambda+4,6\lambda,-3\lambda+1) [½]
D-ratios of AB is: -2\lambda+2,\ 6\lambda-3,\ -3\lambda+9 [1]
As AB is perpendicular to the given line: -2(-2\lambda+2)+6(6\lambda-3)-3(-3\lambda+9)=0
\Rightarrow\lambda=1 [1]
\therefore Foot of the perpendicular is: B(2,6,-2) [1]
Perpendicular distance =AB=3\sqrt5 [1]
Find the shortest distance between the lines L_1 & L_2 given below :
L_1 : The line passing through (2,-1,1) and parallel to \frac{x}{1}=\frac{y}{1}=\frac{z}{3}
L_2:\ \vec{r}=\hat{i}+(2\mu+1)\hat{j}-(\mu+2)\hat{k}.
Equation of L_1:\ \vec{r}=2\hat{i}-\hat{j}+\hat{k}+\lambda(\hat{i}+\hat{j}+3\hat{k}) [1]
Equation of L_2:\ \vec{r}=(\hat{i}+\hat{j}-2\hat{k})+\mu(2\hat{j}-\hat{k}) [½]
Taking
\vec{a}_1=2\hat{i}-\hat{j}+\hat{k},\ \vec{b}_1=\hat{i}+\hat{j}+3\hat{k}
\vec{a}_2=\hat{i}+\hat{j}-2\hat{k},\ \vec{b}_2=2\hat{j}-\hat{k} [½]
\vec{a}_2-\vec{a}_1=-\hat{i}+2\hat{j}-3\hat{k},\ \vec{b}_1\times\vec{b}_2=-7\hat{i}+\hat{j}+2\hat{k} [½+1]
Shortest Distance =\frac{|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)|}{|\vec{b}_1\times\vec{b}_2|}=\frac{1}{\sqrt6} [1½]
Find the shortest distance between the lines \frac{x-8}{3}=\frac{y+9}{-16}=\frac{z-10}{7} and \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}.
\vec{r}=(8\hat{i}-9\hat{j}+10\hat{k})+\lambda(3\hat{i}-16\hat{j}+7\hat{k})
\vec{r}=(15\hat{i}+29\hat{j}+5\hat{k})+\mu(3\hat{i}+8\hat{j}-5\hat{k})
We have, \vec{a}_1=8\hat{i}-9\hat{j}+10\hat{k},\ \vec{b}_1=3\hat{i}-16\hat{j}+7\hat{k},\ \vec{a}_2=15\hat{i}+29\hat{j}+5\hat{k},\ \vec{b}_2=3\hat{i}+8\hat{j}-5\hat{k} [2]
\vec{a}_2-\vec{a}_1=7\hat{i}+38\hat{j}-5\hat{k},\ \vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-16&7\\3&8&-5\end{vmatrix}=24\hat{i}+36\hat{j}+72\hat{k} [½+1½]
\text{Shortest Distance}=\left|\frac{(\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1)}{|\vec{b}_1\times\vec{b}_2|}\right|=\frac{98}{7}=14 [1]
Find the point of intersection of the lines
\vec{r}=\hat{i}-\hat{j}+6\hat{k}+\lambda(3\hat{i}-\hat{k}), and
\vec{r}=-3\hat{j}+3\hat{k}+\mu(\hat{i}+2\hat{j}-\hat{k}).
Also, find the vector equation of the line passing through the point of intersection of the given lines and perpendicular to both the lines.
If the given lines intersect then,
\hat{i}-\hat{j}+6\hat{k}+\lambda(3\hat{i}-\hat{k})=-3\hat{j}+3\hat{k}+\mu(\hat{i}+2\hat{j}-\hat{k}) [1]
Solving the equations, 1+3\lambda=\mu,\ -1=2\mu-3, we get \lambda=0,\mu=1 [2]
which do not satisfy the equation 6-\lambda=3-\mu. [1]
\therefore The lines do not intersect, hence no point of intersection [1]
Find the vector and the Cartesian equations of a line passing through the point (1, 2, – 4) and parallel to the line joining the points A(3, 3, – 5) and B(1, 0, – 11). Hence, find the distance between the two lines.
Vector equation of required line through (1, 2, – 4) is
\vec{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda(2\hat{i}+3\hat{j}+6\hat{k}) [1]
and cartesian equation: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} [1]
Equation of line through A(3, 3, – 5) and B(1, 0, – 11) is
\vec{r}=3\hat{i}+3\hat{j}-5\hat{k}+\mu(2\hat{i}+3\hat{j}+6\hat{k}) [½]
Distance between parallel lines is given by d=\frac{|(\vec{a}_2-\vec{a}_1)\times\vec{b}|}{|\vec{b}|}
Here \vec{b}=2\hat{i}+3\hat{j}+6\hat{k},\ \vec{a}_1=\hat{i}+2\hat{j}-4\hat{k},\ \vec{a}_2=3\hat{i}+3\hat{j}-5\hat{k}
(\vec{a}_2-\vec{a}_1)=2\hat{i}+\hat{j}-\hat{k} [½]
(\vec{a}_2-\vec{a}_1)\times\vec{b}=9\hat{i}-14\hat{j}+4\hat{k} [1]
\therefore d=\frac{\sqrt{293}}{7} [1]
Find the equations of the line passing through the points A(1, 2, 3) and B(3, 5, 9). Hence, find the coordinates of the points on this line which are at a distance of 14 units from point B.
Equation of line AB is \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{6} [1]
Let coordinates of required point on AB be (2\lambda+1,3\lambda+2,6\lambda+3) for some \lambda [1]
According to Question
(2\lambda-2)^2+(3\lambda-3)^2+(6\lambda-6)^2=14^2 gives \lambda^2-2\lambda-3=0 [1]
Solving we get \lambda=3 and -1 [1]
\therefore required points are (7, 11, 21) and (– 1, – 1, – 3) [1]
Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, –6), Q(5, –3, 1), R(12, 4, 5) and S(11, 9, –2). Use these equations to find the point of intersection of diagonals.
Equation of diagonal PR: \frac{x-4}{8} = \frac{y-2}{2} = \frac{z+6}{11} [1]
Equation of diagonal QS: \frac{x-5}{6} = \frac{y+3}{12} = \frac{z-1}{-3} [1]
General points on PR & QS are (8k+4, 2k+2, 11k-6) and (6t+5, 12t-3, -3t+1) for real numbers ‘k’ and ‘t’ respectively. [1]
For point of intersection of PR and QS: 8k + 4 = 6t + 5,\ 2k + 2 = 12t - 3 [½]
Solving, we get k = \frac{1}{2}, t = \frac{1}{2}\ \therefore The point of intersection is \left(8, 3, -\frac{1}{2}\right) [1½]
A line l passes through point (–1, 3, –2) and is perpendicular to both the lines \frac{x}{1} = \frac{y}{2} = \frac{z}{3} and \frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}. Find the vector equation of the line l. Hence, obtain its distance from origin.
Let direction ratios of the required line be a, b, c
Since it is perpendicular to the two given lines, a + 2b + 3c = 0\ ;\ -3a + 2b + 5c = 0 [1]
Solving together, a = 4k,\ b = -14k,\ c = 8k [1]
\therefore Equation of line is: \frac{x+1}{4k} = \frac{y-3}{-14k} = \frac{z+2}{8k} \Rightarrow \frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4} [1]
Vector equation: \vec{r} = -\hat{i} + 3\hat{j} - 2\hat{k} + \lambda(2\hat{i} - 7\hat{j} + 4\hat{k}) [1]
Distance from origin = \frac{\left|(-\hat{i} + 3\hat{j} - 2\hat{k})\times(2\hat{i} - 7\hat{j} + 4\hat{k})\right|}{\left|2\hat{i} - 7\hat{j} + 4\hat{k}\right|} = \frac{\left|-2\hat{i} + \hat{k}\right|}{\left|2\hat{i} - 7\hat{j} + 4\hat{k}\right|} = \frac{\sqrt{5}}{\sqrt{69}} \text{ or } \sqrt{\frac{5}{69}} [1]
Find the value of b so that the lines \frac{x - 1}{2} = \frac{y - b}{3} = \frac{z - 3}{4} and \frac{x - 4}{5} = \frac{y - 1}{2} = z are intersecting lines. Also, find the point of intersection of these given lines.
As lines are intersecting, (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0
\Rightarrow \begin{vmatrix} 3 & 1 - b & -3\\ 2 & 3 & 4\\ 5 & 2 & 1\end{vmatrix} = 0 [1]
\Rightarrow b = 2 [1]
Any point on the line \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} is
(2\lambda + 1, 3\lambda + 2, 4\lambda + 3),\ \lambda \in R [1]
For the point of intersection, this point must lie on the line \frac{x - 4}{5} = \frac{y - 1}{2} = z
\Rightarrow \frac{2\lambda + 1 - 4}{5} = \frac{3\lambda + 2 - 1}{2} = 4\lambda + 3 [1]
\Rightarrow \lambda = -1 [½]
\therefore point of intersection is (– 1, – 1, – 1) [½]
Find the equations of all the sides of the parallelogram ABCD whose vertices are A(4, 7, 8), B(2, 3, 4), C(– 1, – 2, 1) and D(1, 2, 5). Also, find the coordinates of the foot of the perpendicular from A to CD.
Equation of the line AB : \frac{x - 4}{2} = \frac{y - 7}{4} = \frac{z - 8}{4} [½]
Equation of the line BC : \frac{x - 2}{3} = \frac{y - 3}{5} = \frac{z - 4}{3} [½]
Equation of the line CD : \frac{x + 1}{1} = \frac{y + 2}{2} = \frac{z - 1}{2} [½]
Equation of the line DA : \frac{x - 4}{3} = \frac{y - 7}{5} = \frac{z - 8}{3} [½]
Let P be foot of perpendicular from A to CD.
\therefore Coordinates of P are (\lambda - 1, 2\lambda - 2, 2\lambda + 1) for some \lambda [1]
d.r.'s of AP are (\lambda - 5, 2\lambda - 9, 2\lambda - 7) [½]
since AP \perp CD
\Rightarrow 1(\lambda - 5) + 2(2\lambda - 9) + 2(2\lambda - 7) = 0 [½]
\Rightarrow 9\lambda = 37 \Rightarrow \lambda = \frac{37}{9} [½]
\therefore Coordinates of P are \left(\frac{28}{9}, \frac{56}{9}, \frac{83}{9}\right) [½]
Show that the following lines do not intersect each other :
Shortest Distance =\frac{\begin{vmatrix}-3&2&-2\\3&2&5\\4&3&-2\end{vmatrix}}{\sqrt{(2\times-2-5\times3)^2+(3\times-2-4\times5)^2+(3\times3-2\times4)^2}} [2]
=\frac{-3(-19)-2(-26)-2(1)}{\sqrt{361+676+1}} [1]
=\frac{57+52-2}{\sqrt{1038}} [1]
=\frac{107}{\sqrt{1038}}\ne 0 [1]
So, the line will not intersect each other.
Find the angle between the lines
2x=3y=-z and 6x=-y=-4z.
The given lines are
\frac{x-0}{1/2}=\frac{y-0}{1/3}=\frac{z-0}{-1} and \frac{x-0}{1/6}=\frac{y-0}{-1}=\frac{z-0}{-1/4} [1+1]
Or \frac{x}{3}=\frac{y}{2}=\frac{z}{-6} and \frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}
Let \theta be the angle between the two lines, then.
\cos\theta=\left|\frac{(3\times2)+(2\times-12)+(-6)(-3)}{\sqrt{9+4+36}\sqrt{4+144+9}}\right| [2]
=\left|\frac{6-24+18}{7\times\sqrt{157}}\right|
=0
\Rightarrow\theta=90^\circ [1]
Show that the lines \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7} and \frac{x-2}{1} = \frac{y-4}{3} = \frac{z-6}{5} intersect. Also find their point of intersection.
Official marking scheme for this paper is not available.
Find the shortest distance between the pair of lines \frac{x-1}{2} = \frac{y+1}{3} = z and \frac{x+1}{5} = \frac{y-2}{1};\ z = 2.
Official marking scheme for this paper is not available.
A student of class XII studying Mathematics comes across an incomplete question in a book.
Maximise Z=3x+2y+1
Subject to the constraints x\ge0,y\ge0,3x+4y\le12,
He/ She notices the below shown graph for the said LPP problem, and finds that a constraint is missing in it:
Help him/her choose the required constraint from the graph.
The missing constraint is
For Visually Impaired: If Z=ax+by+c, where a,b,c\gt0, attains its maximum value at two of its corner points (4,0) and (0,3) of the feasible region determined by the system of linear inequalities, then
(A) 4a=3b (B) 3a=4b (C) 4a+c=3b (D) 3a+c=4b
The points (1,0) and (0,2) satisfy the equation 2x+y=2
And shaded region shows that (0,0) doesn’t lie in the feasible solution region
So, the inequality is 2x+y\ge2
(B) 2x+y\ge2 [1]
For Visually Impaired:
(4,0) and (0,3) gives maximum value so
Z_{(4,0)}=Z_{(0,3)}\Rightarrow 4a+c=3b+c\Rightarrow 4a=3b
(A) 4a=3b [1]
The feasible region of a linear programming problem is bounded but the objective function attains its minimum value at more than one point. One of the points is (5,0).
Then one of the other possible points at which the objective function attains its minimum value is
For Visually Impaired: The graph of the inequality 3x+5y\lt10 is the
(A) Entire XY-plane
(B) Open Half plane that doesn’t contain origin
(C) Open Half plane that contains origin, but not the points of the line 3x+5y=10
(D) Half plane that contains origin and the points of the line 3x+5y=10
The student may read the point (2,9) from the line on the graph.
The student may find the equation 3x+y=15 joining (5,0) and (0,15) and then verify the point (2,9) satisfies it.
(A) (2,9) [1]
For Visually Impaired: (C) Open Half plane that contains origin, but not the points of the line 3x+5y=10 [1]
The feasible region of a linear programming problem with objective function Z = 5x + 7y is shown below :
The maximum value of Z – minimum value of Z is
(D) 43
The degree of an objective function of a linear programming problem is
(B) 1
In a linear programming problem, the linear function which has to be maximized or minimized is called
(B) an objective function
For the feasible region shown below, the non-trivial constraints of the linear programming problem are
(C) x + y \ge 5,\ x + 3y \le 9
The region represented by the system of inequations 3x + y \ge 3, 2x - y \ge -5, x, y \ge 0 is :
(A) unbounded in 1st Quadrant [1]
In the graph, the feasible region representing the Linear Programming Problem for maximising objective function Z = px + qy, p, q > 0 is shaded. If all points on segment AB give max (Z), then which of the following is true ?
(C) q = 3p [1]
The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40) (60, 20) and (60, 0). If the objective function of an LPP is Z = 4x + 3y, then the maximum value is :
(B) 300
Assertion (A) : Consider a Linear Programming Problem with minimise Z = x + 2y subject to constraints 2x + y \ge 3,\ x + 2y \ge 6,\ x, y \ge 0 which gives minimum Z at infinitely many points. The corner points of feasible region are (0, 3) and (6, 0).
Reason (R) : If two corner points produce the same minimum value of the objective function, then every point on the line segment joining the points will give the same minimum value.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
The corner points of the feasible region in graphical representation of a L.P.P. are (2, 72), (15, 20) and (40, 15). If Z=18x+9y be the objective function, then
(C) Z is maximum at (40, 15), minimum at (15, 20)
If the feasible region of a linear programming problem with objective function Z=ax+by, is bounded, then which of the following is correct ?
(C) It will have both maximum and minimum values.
A factory produces two products X and Y. The profit earned by selling X and Y is represented by the objective function Z = 5x + 7y, where x and y are the number of units of X and Y respectively sold. Which of the following statement is correct ?
(C) The objective function maximizes the combined profit earned from selling X and Y
Assertion (A) : Every point of the feasible region of a Linear Programming Problem is an optimal solution.
Reason (R) : The optimal solution for a Linear Programming Problem exists only at one or more corner point(s) of the feasible region.
(D) Assertion (A) is false and Reason (R) is true.
The corner points of the feasible region of a Linear Programming Problem are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). If Z = ax + by; (a, b > 0) be the objective function, and maximum value of Z is obtained at (0, 2) and (3, 0), then the relation between a and b is :
(D) 3a = 2b
Assertion (A) : In a Linear Programming Problem, if the feasible region is empty, then the Linear Programming Problem has no solution.
Reason (R) : A feasible region is defined as the region that satisfies all the constraints.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
For a Linear Programming Problem (LPP), the given objective function Z=3x+2y is subject to constraints :
x+2y\le10
3x+y\le15
x,y\ge0
The correct feasible region is :
(B) AOEC
Assertion (A) : The shaded portion of the graph represents the feasible region for the given Linear Programming Problem (LPP).
Min Z=50x+70y
subject to constraints
2x+y\ge8,\ x+2y\ge10,\ x,y\ge0
Z=50x+70y has a minimum value =380 at B(2,4).
Reason (R) : The region representing 50x+70y\lt380 does not have any point common with the feasible region.
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
For a Linear Programming Problem (LPP), the given objective function is Z = x + 2y. The feasible region PQRS determined by the set of constraints is shown as a shaded region in the graph.
(Note : The figure is not to scale)
Which of the following statements is correct ?
(A) Z is minimum at S\left(\frac{18}{7}, \frac{2}{7}\right)
In a Linear Programming Problem (LPP), the objective function Z = 2x + 5y is to be maximised under the following constraints :
Study the graph and select the correct option.
(Note : The figure is not to scale)
The solution of the given LPP :
(C) does not exist.
In an LPP, corner points of the feasible region determined by the system of linear constraints are (1, 1), (3, 0) and (0, 3). If Z = ax + by, where a, b \gt 0 is to be minimized, the condition on a and b, so that the minimum of Z occurs at (3, 0) and (1, 1), will be :
(B) a = \frac{b}{2}
The maximum value of Z = 3x + 4y subject to the constraints x + y \le 1, x, y \ge 0 is :
(B) 4
A linear programming problem deals with the optimization of a/an :
(B) linear function
The number of corner points of the feasible region determined by constraints x \ge 0,\ y \ge 0,\ x + y \ge 4 is :
(C) 2
The common region determined by all the constraints of a linear programming problem is called :
(D) a feasible region
The restrictions imposed on decision variables involved in an objective function of a linear programming problem are called :
(B) constraints
Of the following, which group of constraints represents the feasible region given below ?
(C) x + 2y \ge 76,\ 2x + y \le 104,\ x, y \ge 0
The maximum value of Z = 4x + y for a L.P.P. whose feasible region is given below is :
(C) 120
[1]Assertion (A) : The corner points of the bounded feasible region of a L.P.P. are shown below. The maximum value of Z = x + 2y occurs at infinite points.
Reason (R) : The optimal solution of a LPP having bounded feasible region must occur at corner points.
(B) Both A and R are true but R is not the correct explanation of A.
[1]The corner points of the feasible region in the graphical representation of a linear programming problem are (2, 72), (15, 20) and (40, 15). If z=18x+9y be the objective function, then :
(C) z is maximum at (40, 15), minimum at (15, 20)
The number of corner points of the feasible region determined by the constraints x-y\ge0,\ 2y\le x+2,\ x\ge0,\ y\ge0 is :
(A) 2
The objective function Z = ax + by of an LPP has maximum value 42 at (4, 6) and minimum value 19 at (3, 2). Which of the following is true ?
(C) a = 3, b = 5
The corner points of the feasible region of a linear programming problem are (0, 4), (8, 0) and \left(\frac{20}{3}, \frac{4}{3}\right). If Z = 30x + 24y is the objective function, then (maximum value of Z – minimum value of Z) is equal to
Give 1 Mark to those who have attempted as the correct option is not given [1]
The number of feasible solutions of the linear programming problem given as
Maximize z = 15x + 30y subject to constraints :
3x + y \le 12,\ x + 2y \le 10,\ x \ge 0,\ y \ge 0 is
(D) infinite
The feasible region of a linear programming problem is shown in the figure below :
Which of the following are the possible constraints ?
(C) x + 2y \ge 4,\ x + y \ge 3,\ x \ge 0,\ y \ge 0
The solution set of the inequation 3x+5y\lt 7 is :
(c) open half plane containing the origin except the points of line 3x+5y=7
Which of the following points satisfies both the inequations 2x+y\le 10 and x+2y\ge 8 ?
(d) (4, 2)
The point which lies in the half-plane 2x + y - 4 \le 0 is :
Official marking scheme for this paper is not available.
The corner points of the bounded feasible region of an LPP are O(0, 0), A(250, 0), B(200, 50) and C(0, 175). If the maximum value of the objective function Z = 2ax + by occurs at the points A(250, 0) and B(200, 50), then the relation between a and b is :
Official marking scheme for this paper is not available.
The feasible region of a linear programming problem is shown in the given figure:
Write all the constraints on the decision variables involved
For Visually Impaired: For a linear programming problem, the maximum value of an objective function Z=3x+4y exists at the points (p,q) and \left(2p,\frac{q}{2}\right). Find the relation between p and q.
Constraints are
3x+2y\ge10, x+2y\le6, [1½]
y\ge0 [½]
For Visually Impaired:
As maximum value exists at (p,q) and \left(2p,\frac{q}{2}\right)
So Z\ at\ (p,q)=Z\ at\ \left(2p,\frac{q}{2}\right)
\Rightarrow3p+4q=3\times2p+4\times\frac{q}{2} [1]
\Rightarrow3p+4q=6p+2q
\Rightarrow6p-3p=4q-2q
\Rightarrow3p=2q [1]
For an LPP with constraints
shade the feasible region on the graph and write the coordinates of its corner points.
[Graph: Feasible region ABCO, with A(0, 40), B(30, 10), C(35, 0), O(0, 0); lines meet axes at (0, 70), (0, 40), (35, 0), (40, 0)] [1]
Corner Points are (0,0), (35,0), (30,10), (0,40) [1]
In a Linear Programming Problem, the objective function Z=5x+4y needs to be maximised under constraints 3x+y\le6, x\le1, x,y\ge0. Express the LPP on the graph and shade the feasible region and mark the corner points.
[Graph: lines 3x+y=6 and x=1; feasible region OABC shaded with corner points O(0, 0), A(0, 6), B(1, 3), C(1, 0)]
Correct Plotting of the two lines [1½]
Correct shading of the feasible region and marking the corner points [½]
Solve the linear programming problem graphically:
Maximize Z=40x+30y
Subject to constraints: x+y\le6, 2x\ge y, y\ge2
For Visually Impaired: For a linear programming problem, vertices of the feasible region are (20,0),(40,0),(30,10),(0,20) and (0,10). Minimize and maximize the objective function Z=50x-20y
[Graph: feasible region bounded by x+y=6, y=2x and y=2, with vertices A(1,2), B(4,2), C(2,4)] [1½]
Feasible region is bounded with three vertices A(1,2),B(4,2),C(2,4)
Now Z_A=100,Z_B=220,Z_C=200 [1]
So, Maximum Value of Z is 220 at the point B(4,2). [½]
For Visually Impaired:
The vertices of the feasible region are A(20,0),B(40,0),C(30,10), D(0,20) and E(0,10).
Now Z_A=50\times20-20\times0=1000
Z_B=50\times40-20\times0=2000
Z_C=50\times30-20\times10=1300
Z_D=50\times0-20\times20=-400
Z_E=50\times0-20\times10=-200 [2]
Minimum value of Z is -400 at D(0,20)
Maximum Value of Z is 2000 at B(40,0) [1]
Solve graphically:
Maximise Z=2x+y subject to
x+y\le1200
x+y\ge600
y\le\frac{x}{2}
x\ge0,y\ge0.
For Visually Impaired: The objective function Z=3x+2y of a linear programming problem under some constraints is to be maximized and minimized. The corner points of the feasible region are A(600,0),B(1200,0),C(800,400) and D(400,200). Find the point at which Z is maximum and the point at which Z is minimum. Also, find the corresponding maximum and minimum values of Z.)
Sketching the graph [1½]
Corner points A(600,0), B(1200,0), C(800,400), D(400,200) [½]
Values of Z: Z_A=1200,Z_B=2400,Z_C=2000,Z_D=1000 [½]
Maximum Z=2400 when x=1200 and y=0 [½]
For Visually Impaired:
At Corner points A(600,0), B(1200,0), C(800,400), D(400,200)
Values of Z are Z_A=1800,Z_B=3600,Z_C=3200,Z_D=1600 [1]
Maximum Value of Z = 3600 at B(1200,0) [1]
And Minimum Value of Z= 1600 at D(400,200) [1]
Solve the following linear programming problem graphically :
Minimize Z = 13x - 15y
Subject to constraints
[Graph showing the feasible region with corner points (0, 0), (7, 0), (3, 4), (0, 2)] [1½] (Correct graph and shading)
| Corner point | Value of Z = 13x - 15y |
| (0, 0) | 0 |
| (7, 0) | 91 |
| (3, 4) | -21 |
| (0, 2) | -30 |
[1]
The minimum value of Z is -30 at x = 0, y = 2 [½]
Solve the following linear programming problem graphically :
Maximize Z = 10500x + 9000y
Subject to constraints
x + y \le 50
2x + y \le 80
x, y \ge 0
(Graph showing the lines 2x + y = 80 and x + y = 50; shaded feasible region OABC with corner points A(0, 50), B(30, 20), C(40, 0).)
For correct graph and shading [1½]
| Corner Point | Value of Z = 10500x + 9000y |
| O(0, 0) | 0 |
| A(0, 50) | 4,50,000 |
| B(30, 20) | 4,95,000 |
| C(40, 0) | 4,20,000 |
For correct table [1]
Z_{\max} = 4,95,000 when x = 30,\ y = 20 [½]
Solve the following Linear Programming Problem graphically :
Maximise Z = 200x + 120y
subject to the constraints
[Graph of the feasible region with corner points A, B, C, D, E] [1½] (For correct graph & shading)
| Corner point | Value of Z = 200x + 120y |
| A (100, 200) | 44000 |
| B (150, 150) | 48000 |
| C (0, 100) | 12000 |
| D (0, 0) | 0 |
| E (200, 0) | 40000 |
[1]
Maximum value of Z is 48000 at x = 150 & y = 150. [½]
Solve the following Linear Programming Problem graphically :
Maximize Z = \frac{2x}{5} + \frac{3y}{10}
subject to constraints
[Graph: feasible region OBCD with A = (0, 0), B = (500, 0), C = (200, 600), D = (0, 800)]
Correct Lines and Shading [1½]
| Corner Points | Value of Z |
| (0, 0) | 0 |
| (500, 0) | 200 |
| (200, 600) | 260 |
| (0, 800) | 240 |
[1]
Z_{\max} = 260 at x = 200, y = 600 [½]
Solve the following Linear Programming Problem graphically :
Minimize Z = 4x + y
subject to
[Graph: unbounded feasible region with corner points A(0, 3), B(2, 1), C(4, 0); lines x + y = 3, x + 2y = 4 and dashed line 4x + y = 3] [For correct graph and shading 1½]
| Corner Point | Value of Z = 4x + y |
| A(0,3) | 3 |
| B(2,1) | 9 |
| C(4,0) | 16 |
[For correct table 1]
Since half plane 4x + y \lt 3 has no common point with the feasible region, so Z_{\min} = 3 at (0,3) [½]
Solve the following linear programming problem graphically :
Maximise Z=x+2y
Subject to the constraints :
x-y\ge 0
x-2y\ge -2
x\ge 0,\ y\ge 0
[Graph of the feasible region] [1½] (for correct graph and shading)
| Corner Point | Value of Z=x+2y |
| O(0,0) | 0 |
| A(2,2) | 6 |
For correct table [1]
Since feasible region is unbounded. Plot x+2y\gt 6 which has common region with feasible region, thus Z has no maximum value. [½]
Solve the following linear programming problem graphically :
Minimise Z = x - 5y
subject to the constraints :
x - y \ge 0
-x + 2y \ge 2
x \ge 3,\ y \le 4,\ y \ge 0
[Graph: feasible region ABCD bounded by x = 3, y = 4, x − y = 0 and −x + 2y = 2]
Correct graph and shading [1½]
| Corner point | Value of Z = x - 5y |
| A (3, 2.5) | -9.5 |
| B (3, 3) | -12 |
| C (4, 4) | -16 |
| D (6, 4) | -14 |
[1]
The minimum value of Z is -16, which is attained at x = 4, y = 4. [½]
Solve the following Linear Programming Problem using graphical method :
Maximise Z = 100x + 50y
subject to the constraints
3x+y\le 600
x+y\le 300
y\le x+200
x\ge 0,\ y\ge 0
For correct graph and shading [1½]
| Corner Point | Value of Z=100x+50y |
| O(0,0) | 0 |
| A(0,200) | 10000 |
| B(50,250) | 17500 |
| C(150,150) | 22500 |
| D(200,0) | 20000 |
For correct table [1]
Z_{\max}=22500 when x=150,\ y=150 [½]
In the Linear Programming Problem (LPP), find the point/points giving maximum value for Z=5x+10y
subject to constraints
x+2y\le120
x+y\ge60
x-2y\ge0
x,y\ge0
[Graph showing the feasible region with corner points A(60, 30), B(120, 0), C(60, 0), D(40, 20)] [1½ for correct graph and correct feasible region]
| Corner Points | Value of Z |
| A (60, 30) | 600 |
| B (120, 0) | 600 |
| C (60, 0) | 300 |
| D (40, 20) | 400 |
[1]
Since Z is maximum on points A and B
Hence all points lying on segment AB give maximum Z. [½]
Consider the Linear Programming Problem, where the objective function Z = (x + 4y) needs to be minimized subject to constraints
Draw a neat graph of the feasible region and find the minimum value of Z.
Correct Graph and shading: [1½]
| Corner points | Value of Z |
| (800, 0) | 800 |
| (400, 200) | 1200 |
| (0, 1000) | 4000 |
[1]
x + 4y \lt 800 has no region common with feasible region, hence 800 is minimum [½]
In the Linear Programming Problem for objective function Z=18x+10y subject to constraints
4x+y\ge20
2x+3y\ge30
x,y\ge0
find the minimum value of Z.
[Graph: lines 4x+y=20, 2x+3y=30; unbounded feasible region with corner points A(0, 20), B(3, 8), C(15, 0); line 18x+10y=134 shown]
Correct Fig. [1½]
| Corner points | Value of Z = 18x + 10y |
| A (0, 20) | 200 |
| B (3, 8) | 134 |
| C (15, 0) | 270 |
[1]
Also, Z\lt134, does not have any common point with the feasible region,
\therefore\text{Min}(Z)=134 at B (3, 8) [½]
Solve the following LPP graphically :
Maximize Z = 2x + 3y
subject to the constraints x + 4y \le 8
2x + 3y \le 12
3x + y \le 9
x \ge 0, y \ge 0.
[Graph: feasible region OABC shaded, bounded by x + 4y = 8, 3x + y = 9 and the axes; see figure] [2] (For correct graph and shading)
| Corner Point | Value of Z = 2x + 3y |
| O(0,0) | 0 |
| A(0,2) | 6 |
| B\left(\frac{28}{11}, \frac{15}{11}\right) | \frac{101}{11} Maximum |
| C(3,0) | 6 |
Z_{\max} = \frac{101}{11} when x = \frac{28}{11}, y = \frac{15}{11} [1] (For correct table)
Solve the following linear programming problem graphically :
Maximise z = 500x + 300y,
subject to constraints
x + 2y \le 12
2x + y \le 12
4x + 5y \ge 20
x \ge 0,\ y \ge 0
Max z = 500x + 300y
[Graph: feasible region bounded by x + 2y = 12, 2x + y = 12, 4x + 5y = 20 and the axes, with corner points A = (0, 4), B = (0, 6), C = (4, 4), D = (6, 0), E = (5, 0)] Correct Graph - [1½]
| Corner Point | Z |
| A (0,4) | 1200 |
| B (0,6) | 1800 |
| C (4,4) | 3200 |
| D (6,0) | 3000 |
| E (5,0) | 2500 |
Correct Table - [1]
Max z = 3200 at x = 4,\ y = 4 [½]
Solve the following linear programming problem graphically :
Maximise z = 4x + 3y,
subject to the constraints
x + y \le 800
2x + y \le 1000
x \le 400
x, y \ge 0.
[Graph of the feasible region OABCD with corner points O(0, 0), A(400, 0), B(400, 200), C(200, 600), D(0, 800); lines x + y = 800, 2x + y = 1000, x = 400]
For correct graph [1½]
| Corner Point | Value of z = 4x + 3y |
| O(0, 0) | 0 |
| A(400, 0) | 1600 |
| B(400, 200) | 2200 |
| C(200, 600) | 2600 |
| D(0, 800) | 2400 |
For correct table [1]
z_{\max} = 2600 when x = 200, y = 600 [½]
Solve the following linear programming problem graphically :
Maximise Z = 2x + 3y
subject to the constraints :
x + y \le 6
x \ge 2
y \le 3
x, y \ge 0
On plotting the graph of x + y \le 6, x \ge 2, y \le 3, \& x \ge 0, y \ge 0 we get the following graph and common shaded region is the region ABCD.
[Graph] [For correct Graph 1½]
Now, Corner points of the common shaded region are A(2, 3), B(3, 3), C(6, 0) & D(2, 0). Thus,
| Corner points | Value of Z = 2x + 3y |
| A(2, 3) | 13 |
| B(3, 3) | 15 |
| C(6, 0) | 12 |
| D(2, 0) | 4 |
[For correct Table 1]
So, Maximum Value of Z is 15 at x = 3, y = 3. [½]
Solve graphically the following linear programming problem :
Maximise z=6x+3y,
subject to the constraints
Correct Graph [2]
| Corner points | Value of Z |
| (2, 72) | (12 + 216 = 228) |
| (15, 20) | (90 + 60 = 150) |
| (40, 15) | (240 + 45 = 285) Maximum |
[1]
Solve the following linear programming problem graphically :
Minimize : Z = 5x + 10y
subject to constraints : x + 2y \le 120,\ x + y \ge 60,\ x - 2y \ge 0,\ x \ge 0,\ y \ge 0
Correct graph [2]
| Corner points | Value of Z |
| A (40,20) | 400 |
| B (60,30) | 600 |
| C (120,0) | 600 |
| D (60,0) | 300 (Min) |
Min(Z) = 300 at x = 60; y = 0 [1]
Determine graphically the minimum value of the following objective function :
z = 500x + 400y
subject to constraints
x + y \le 200,
x \ge 20,
y \ge 4x,
y \ge 0.
Correct graph [2]
| Corner points | Value of Z = 500x + 400y |
| (20, 180) | 82000 |
| (40, 160) | 84000 |
| (20, 80) | 42000 \to Minimum |
[1]
Solve the following linear programming problem graphically :
Minimise : z=-3x+4y
subject to the constraints
[Graph of the feasible region OBED with corner points O(0, 0), B(0, 4), E(2, 3), D(4, 0); lines x+2y=8 and 3x+2y=12] [2 for correct graph]
| Corner Points | Value of Z |
| O(0,0) | 0 |
| B(0,4) | 16 |
| D(4,0) | -12 \to Min |
| E(2,3) | 6 |
[1]
Solve the following Linear Programming Problem graphically :
Minimise z = 3x + 8y
subject to the constraints
Official marking scheme for this paper is not available.
The month of September is celebrated as the Rashtriya Poshan Maah across the country. Following a healthy and well-balanced diet is crucial in order to supply the body with the proper nutrients it needs. A balanced diet also keeps us mentally fit and promotes improved level of energy.
A dietician wishes to minimize the cost of a diet involving two types of foods, food X (x kg) and food Y (y kg) which are available at the rate of ₹ 16/kg and ₹ 20/kg respectively. The feasible region satisfying the constraints is shown in Figure-2.
On the basis of the above information, answer the following questions :
(i) Identify and write all the constraints which determine the given feasible region in Figure-2. [2]
(ii) If the objective is to minimize cost Z = 16x + 20y, find the values of x and y at which cost is minimum. Also, find minimum cost assuming that minimum cost is possible for the given unbounded region. [2]
(i) Constraints are x + 2y \ge 10
x + y \ge 6
3x + y \ge 8 [1½]
x \ge 0
y \ge 0 [½]
(ii)
| Corner points | Value of Z = 16x + 20y |
| A (10, 0) | 160 |
| B (2, 4) | 112 |
| C (1, 5) | 116 |
| D (0, 8) | 160 |
[1½]
The minimum cost is ₹112 [½]
Solve the following Linear Programming Problem graphically :
Maximise Z = 600x + 400y
subject to the constraints
(Graph showing the feasible region with corner points A, B, C, D, E.) For correct graph and correct shaded region [3]
| Corner Points | Value of Z = 600x + 400y |
| A (4,4) | 4000 |
| B (6,0) | 3600 |
| C (5,0) | 3000 |
| D (0,4) | 1600 |
| E (0,6) | 2400 |
[1½]
Hence, Z_{max.} = 4000 at point (4,4). [½]
Solve the following L.P.P. graphically :
Maximise Z=60x+40y
Subject to x+2y\le12
2x+y\le12
4x+5y\ge20
x,y\ge0
Correct Graph [3½]
| Corner Points | Value of Z=60x+40y |
| A(0,4) | Z=160 |
| B(0,6) | Z=240 |
| C(4,4) | Z=400 |
| D(6,0) | Z=360 |
| E(5,0) | Z=300 |
[1]
\text{Max}(Z)=400 at x=4,y=4 [½]
Solve the following linear programming problem graphically :
Minimise Z=6x+7y
subject to constraints
Correct Graph [3½]
| Corner Points | Value of Z=6x+7y |
| A(20,140) | Z=1100 |
| B(40,100) | Z=940 |
| C(140,50) | Z=1190 |
[1]
\text{Min}(Z)=940 at x=40,y=100 [½]
Solve the following Linear Programming Problem graphically :
Maximize : P=70x+40y
subject to : 3x+2y\le9,
3x+y\le9,
x\ge0,\ y\ge0
[Figure: feasible region OBC bounded by 3x+2y=9, 3x+y=9 and the axes] [3 for correct figure and shading]
| Corner Points | Value of P |
| O (0,0) | 0 |
| B (0,4.5) | 180 |
| C (3,0) | 210 \to Max Value |
[1½]
Maximum value of P =210 at x=3 and y=0 [½]
The probability that it rains on a particular day is \frac15. Assuming weather conditions of any two days are independent, find the probability that it rains on three consecutive days.
(D) \frac{1}{125}
The probability that it rains on a particular day is \frac15.
Probability that it rains on three consecutive days = \frac15\times\frac15\times\frac15=\frac{1}{125}
A die is thrown twice. What is the probability that at least one number is 4, given the sum of the numbers appearing is a multiple of 5?
(D) \frac47
Let the events be
A: sum of the numbers appeared is a multiple of 5
B: at least one number is 4
A=\{(1,4),(2,3),(3,2),(4,1),(4,6),(5,5),(6,4)\}
A\cap B=\{(1,4),(4,1),(4,6),(6,4)\}
So required probability = P(B|A)=\frac47
A and B are two independent events such that P(A)=\frac16, P(B)=k. If the probability of occurrence of exactly one of them is \frac13, then find the value of k.
(A) \frac14
Probability of occurrence of any one of them is \frac13
\Rightarrow P(A'\cap B)+P(A\cap B')=\frac13\Rightarrow P(A')P(B)+P(A)P(B')=\frac13
\Rightarrow\left(1-\frac16\right)k+\frac16(1-k)=\frac13\Rightarrow k\left(\frac56-\frac16\right)=\frac13-\frac16\Rightarrow k=\frac16\times\frac64=\frac14
A person observed the first 4 digits of your 6-digit PIN. What is the probability that the person can guess your PIN?
(B) \frac{1}{100} [1]
The person knows the first 4 digits. So the person has to guess the remaining two digits.
P (guessing the PIN )=1\times1\times1\times1\times\frac{1}{10}\times\frac{1}{10}=\frac{1}{100}
Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \frac{2}{3}.
Reason (R) : For any two events A and B, P(A|B) = \frac{P(A \cup B)}{P(B)}
(C) Assertion (A) is true and Reason (R) is false.
For two events A and B such that P(A) \ne 0 and P(B) \ne 1, P(A'/B') =
(D) \frac{1 - P(A \cup B)}{P(B')}
If E and F are two independent events such that P(E) = \frac{3}{10}, P(E \cup F) = \frac{1}{2}, then P(E|F) - P(F|E) is equal to :
(C) \frac{1}{70}
A box contains 4 red, 5 blue and 1 green marble. A child randomly takes out a marble from the box, notes down the colour and puts it back in the box. If the activity is repeated 3 times, what is the probability that at least one marble is red ?
(D) \frac{98}{125} [1]
If 3P(A) = P(B) = \frac{3}{5} and P(A|B) = \frac{2}{3}, then P(A \cup B) is :
(D) \frac{2}{5}
If two events A and B are independent, then which of the following statements is not correct ?
(C) P(\overline{A} \cap \overline{B}) = [1 - P(\overline{A})]\,[1 - P(\overline{B})].
If A and B are independent events such that P(A) = 0{\cdot}45 and P(A \cup B) = 0{\cdot}5, then P(B) is equal to :
(D) \frac{1}{11}
If E and F are two independent events such that P(E)=\frac{2}{3}, P(F)=\frac{3}{7}, then P(E/\overline{F}) is equal to :
(C) \frac{2}{3}
If E and F are two events such that P(E) \gt 0 and P(F) \ne 1, then P(\overline{E}/\overline{F}) is
(D) \frac{1 - P(E \cup F)}{P(\overline{F})}
Assertion (A) : If A and B are two events such that P(A\cap B)=0, then A and B are independent events.
Reason (R) : Two events are independent if the occurrence of one does not effect the occurrence of the other.
(D) Assertion (A) is false, but Reason (R) is true.
If P(A\cup B)=0\cdot9 and P(A\cap B)=0\cdot4, then P(\overline{A})+P(\overline{B}) is :
(D) 0.7
A box has 4 green, 8 blue and 3 red pens. A student picks up a pen at random, checks its colour and replaces it in the box. He repeats this process 3 times. The probability that at least one pen picked was red is :
(C) \frac{61}{125}
If P(A)=\frac17, P(B)=\frac57 and P(A\cap B)=\frac47, then P(\overline{A}\,|\,B) is :
(D) \frac15 [1]
A coin is tossed and a card is selected at random from a well shuffled pack of 52 playing cards. The probability of getting head on the coin and a face card from the pack is :
(B) \frac{3}{26} [1]
A coin is tossed three times. The probability of getting at least two heads is :
(A) \frac{1}{2}
If P(A|B) = P(A'|B), then which of the following statements is true ?
(C) P(A \cap B) = \frac{1}{2}P(B)
Let E be an event of a sample space S of an experiment, then P(S | E) =
(C) 1
Let E and F be two events such that P(E) = 0{\cdot}1, P(F) = 0{\cdot}3, P(E \cup F) = 0{\cdot}4, then P(F|E) is :
(D) 0
The probability distribution of a random variable X is :
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | k | 2k | k | 0.1 |
where k is some unknown constant.
The probability that the random variable X takes the value 2 is :
(B) \frac{2}{5}
[1]If A and B are events such that P(A/B)=P(B/A)\ne0, then :
(D) P(A)=P(B) [1]
Assertion (A) : If R and S are two events such that P(R\,|\,S)=1 and P(S)\gt 0, then S\subset R.
Reason (R) : If two events A and B are such that P(A\cap B)=P(B), then A\subset B.
(C) Assertion (A) is true, but Reason (R) is false.
[1]If P\left(\frac{A}{B}\right)=0\cdot3, P(A)=0\cdot4 and P(B)=0\cdot8, then P\left(\frac{B}{A}\right) is equal to :
(A) 0·6
If the sum of numbers obtained on throwing a pair of dice is 9, then the probability that number obtained on one of the dice is 4, is :
(D) \frac{1}{2}
If P(A \cap B) = \frac{1}{8} and P(\overline{A}) = \frac{3}{4}, then P\left(\frac{B}{A}\right) is equal to :
(A) \frac{1}{2}
The probability that A speaks the truth is \frac{4}{5} and that of B speaking the truth is \frac{3}{4}. The probability that they contradict each other in stating the same fact is :
(a) \frac{7}{20}
If for any two events A and B, P(A)=\frac{4}{5} and P(A\cap B)=\frac{7}{10}, then P(B/A) is equal to
(C) \frac{7}{8}
Five fair coins are tossed simultaneously. The probability of the events that atleast one head comes up is
(C) \frac{31}{32}
Assertion (A) : Two coins are tossed simultaneously. The probability of getting two heads, if it is known that at least one head comes up, is \frac{1}{3}.
Reason (R) : Let E and F be two events with a random experiment, then P(F/E)=\frac{P(E\cap F)}{P(E)}.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A)
A family has 2 children and the elder child is a girl. The probability that both children are girls is :
Official marking scheme for this paper is not available.
A box contains cards numbered 1 to 22. Two cards are drawn simultaneously without replacement and their numbers are noted. What is the probability that both numbers are even ?
As, box contain 11 even numbers and 11 odd numbers. [½]
P (Both even numbers) = \frac{11}{22} \times \frac{10}{21} or \frac{{}^{11}C_2}{{}^{22}C_2} [1]
Thus, P (Both even numbers) = \frac{5}{21} [½]
If P(A) = \frac{3}{8}, P(B) = \frac{1}{2} and P(A \cap B) = \frac{1}{4}, find P(\overline{A}/\overline{B}).
As, P(A \cup B) = \frac{3+4-2}{8} = \frac{5}{8} [1]
So, P(\bar{A}|\bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} = \frac{1 - P(A \cup B)}{1 - P(B)} [½]
Thus, P(\bar{A}|\bar{B}) = \frac{3/8}{1/2} = \frac{3}{4} [½]
10 identical blocks are marked with ‘0’ on two of them, ‘1’ on three of them, ‘2’ on four of them and ‘3’ on one of them and put in a box. If X denotes the number written on the block, then write the probability distribution of X and calculate its mean.
Probability distribution table is:
| X | 0 | 1 | 2 | 3 |
| P(X) | \frac{2}{10} | \frac{3}{10} | \frac{4}{10} | \frac{1}{10} |
[½] [1]
Mean =E(X)=\sum p_ix_i=0\cdot\frac{2}{10}+1\cdot\frac{3}{10}+2\cdot\frac{4}{10}+3\cdot\frac{1}{10}=\frac{14}{10}=\frac75 (or 1.4) [½]
In a village of 8000 people, 3000 go out of the village to work and 4000 are women. It is noted that 30% of women go out of the village to work. What is the probability that a randomly chosen individual is either a woman or a person working outside the village ?
A = A randomly chosen person is a woman
B = A randomly chosen person works outside village.
P(A)=\frac{4000}{8000}=\frac12, P(B)=\frac{3000}{8000}=\frac38, P(A\cap B)=\frac{1200}{8000}=\frac{3}{20} [1½]
Required probability =P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac12+\frac38-\frac{3}{20}=\frac{29}{40} [½]
Let A and B be two events such that P(A)=\frac{5}{8} , P(B)=\frac{1}{2} and P(A/B)=\frac{3}{4} . Find the value of P(B/A).
P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{3}{4}\Rightarrow P(A\cap B)=\frac{3}{8} [1]
P(B/A)=\frac{P(B\cap A)}{P(A)}=\frac{\frac{3}{8}}{\frac{5}{8}}=\frac{3}{5} [1]
Two balls are drawn at random from a bag containing 2 red balls and 3 blue balls, without replacement. Let the variable X denotes the number of red balls. Find the probability distribution of X.
| X | 0 | 1 | 2 |
| P(X) | \frac{3}{5}\times\frac{2}{4}=\frac{3}{10} | 2\left(\frac{2}{5}\times\frac{3}{4}\right)=\frac{6}{10} | \frac{2}{5}\times\frac{1}{4}=\frac{1}{10} |
[½] (values of X) [1½] (probabilities)
A pair of dice is thrown. It is given that the sum of numbers appearing on both dice is an even number. Find the probability that the number appearing on at least one die is 3.
A: number appearing on at least one die is 3
B: sum of numbers appearing on both dice is even
Clearly, A\cap B=\{(3,1),(3,5),(1,3),(5,3),(3,3)\} [½]
P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}=\frac{5/36}{18/36} [1]
=\frac{5}{18} [½]
Probabilities of A and B solving a specific problem are \frac{2}{3} and \frac{3}{5}, respectively. If both of them try independently to solve the problem, then find the probability that the problem is solved.
P (Problem is solved) = 1 – P (Problem not solved) [1]
=1-P(\bar{A})P(\bar{B})
=1-\frac{1}{3}\cdot\frac{2}{5}
=\frac{13}{15} [1]
Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spade cards.
Let X denotes the number of spades
p=\frac{13}{52}=\frac14,\ q=\frac34 [½]
| X | 0 | 1 | 2 |
| P(X) | \frac34\cdot\frac34=\frac{9}{16} | 2\cdot\frac34\cdot\frac14=\frac{6}{16} | \frac14\cdot\frac14=\frac{1}{16} |
[1½]
A pair of dice is thrown and the sum of the numbers appearing on the dice is observed to be 7. Find the probability that the number 5 has appeared on atleast one die.
A: sum is 7
B: 5 has appeared at least on one die [½]
A\cap B=\{(2,5),(5,2)\} [½]
P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{2/36}{6/36}=\frac13 [1]
The probability that A hits the target is \frac13 and the probability that B hits it, is \frac25. If both try to hit the target independently, find the probability that the target is hit.
E_1=A hits the target
E_2=B hits the target
P(target is hit) = 1 – P(target is not hit) [1]
=1-\left(\frac23\cdot\frac35\right)=\frac35 [1]
A bag contains 3 red and 4 white balls. Three balls are drawn at random, one-by-one without replacement from the bag. If the first ball drawn is red in colour, then find the probability that the remaining two balls drawn are also red in colour.
A : First ball drawn is red
B : Remaining two balls are red [½]
A coin is tossed twice. The following table shows the probability distribution of number of tails :
| X | 0 | 1 | 2 |
| P(X) | K | 6K | 9K |
(a) Find the value of K.
(b) Is the coin tossed biased or unbiased ? Justify your answer.
16K=1 Let P(T)=p and P(H)=q
(a) K=\frac{1}{16} [1]
(b) P(0)=\frac{1}{16},\ P(1)=\frac{6}{16},\ P(2)=\frac{9}{16}
q=\frac{1}{4}\Rightarrow p=\frac{3}{4} [½]
Since P(H)\ne P(T), the coin, so tossed, is biased. [½]
Let X be a random variable which assumes values x_1, x_2, x_3, x_4 such that 2P(X=x_1)=3P(X=x_2)=P(X=x_3)=5P(X=x_4).
Find the probability distribution of X.
Let P(X=x_3)=k
P(X=x_1)=\frac{k}{2},\ P(X=x_2)=\frac{k}{3},\ P(X=x_4)=\frac{k}{5} [½]
\frac{k}{2}+\frac{k}{3}+k+\frac{k}{5}=1
k=\frac{30}{61} [½]
Probability Distribution of X is:
| X | x_1 | x_2 | x_3 | x_4 |
| P(X) | 15/61 | 10/61 | 30/61 | 6/61 |
[1]
Events A and B are such that
P(A)=\frac{1}{2},\ P(B)=\frac{7}{12} and P(\overline{A}\cup\overline{B})=\frac{1}{4}
Find whether the events A and B are independent or not.
P(A)=\frac{1}{2},\ P(B)=\frac{7}{12},\ P(\overline{A}\cup\overline{B})=\frac{1}{4}
P(\overline{A}\cup\overline{B})=P(\overline{A\cap B})=1-P(A\cap B) [½]
P(A\cap B)=1-\frac{1}{4}=\frac{3}{4} [½]
P(A)\times P(B)=\frac{1}{2}\times\frac{7}{12}=\frac{7}{24} [½]
P(A)\times P(B)\ne P(A\cap B)
\therefore A and B are not independent [½]
A box B_1 contains 1 white ball and 3 red balls. Another box B_2 contains 2 white balls and 3 red balls. If one ball is drawn at random from each of the boxes B_1 and B_2, then find the probability that the two balls drawn are of the same colour.
P (both balls drawn are of same colour)
= P (both white) + P (both red) [1]
=\frac{1}{4}\times\frac{2}{5}+\frac{3}{4}\times\frac{3}{5}=\frac{11}{20} [1]
A bag contains cards numbered 1 to 25. Two cards are drawn at random, one after the other, without replacement. Find the probability that the number on each card is a multiple of 7.
Multiples of 7 from 1 to 2 are 7, 14, 21 [½]
P (number on each card is a multiple of 7)
= \frac{3}{25}\times\frac{2}{24} = \frac{1}{100} [1½]
One bag contains 4 white and 5 black balls. Another bag contains 6 white and 7 black balls. A ball, drawn at random, is transferred from the first bag to the second bag and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.
Case I: White ball is transferred from bag I to bag II
P(white ball from bag II) = \frac{4}{9}\times\frac{7}{14} [½]
Case II: Black ball is transferred from bag I to bag II
P(white ball from bag II) = \frac{5}{9}\times\frac{6}{14} [½]
Total Probability = \frac{4}{9}\times\frac{7}{14}+\frac{5}{9}\times\frac{6}{14} [½]
= \frac{29}{63} [½]
Two students Mehul and Rashi are seeking admission in a college. The probability that Mehul is selected is 0.4 and the probability of selection of exactly one of the them is 0.5. Chances of selection of them is independent of each other. Find the chances of selection of Rashi. Also find the probability of selection of at least one of them.
Let the events be:
A: Mehul is selected
B: Rashi is selected
Then according to the question,
A and B are independent events and
P(A)=0.4,\ P(A\cap\bar{B})+P(B\cap\bar{A})=0.5 [1]
Let P(B)=x
Then P(A\cap\bar{B})+P(B\cap\bar{A})=0.5
\Rightarrow P(A)P(\bar{B})+P(B)P(\bar{A})=0.5
\Rightarrow 0.4(1-x)+x(1-0.4)=0.5
\Rightarrow 0.4-0.4x+0.6x=0.5
\Rightarrow 0.2x=0.5-0.4=0.1
\Rightarrow x=\frac{0.1}{0.2}=\frac12=0.5 [1]
So, probability of selection of Rashi = 0.5
Probability of selection of at least one of them =1-P(\bar{A}\cap\bar{B})
=1-P(\bar{A})P(\bar{B})
=1-0.6\times0.5
=1-0.3=0.7 [1]
Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball.
Let E_1: The number appearing on the die < 3
E_2: The number appearing on the die \ge 3
A: Red ball is drawn [1/2]
P(E_1) = \frac{2}{6}, P(A/E_1) = \frac{3}{7}, P(E_2) = \frac{4}{6}, P(A/E_2) = \frac{8}{14} [1]
The required probability = P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)
= \frac{2}{6} \times \frac{3}{7} + \frac{4}{6} \times \frac{8}{14} [1]
= \frac{11}{21} [½]
The probability of simultaneous occurrence of atleast one of the two events X and Y is a. If the probability that exactly one of the events X, Y occurs is b, prove that P(X') + P(Y') = 2 - 2a + b.
P(X \cup Y) = a [½]
P(X \cup Y) - P(X \cap Y) = b
Hence, P(X \cap Y) = a - b [1]
P(X') + P(Y') = [1 - P(X)] + [1 - P(Y)] [1]
= 2 - [P(X) + P(Y)]
= 2 - [P(X \cup Y) + P(X \cap Y)]
= 2 - [a + (a - b)]
= 2 - 2a + b [1/2]
The probability of hitting the target by a trained sniper is three times the probability of not hitting the target on a stormy day due to high wind speed.
The sniper fired two shots on the target on a stormy day when wind speed was very high. Find the probability that
(i) target is hit
(ii) atleast one shot misses the target.
(i) Let probability of not hitting = p
\therefore Probability of hitting = 3p
p + 3p = 1 \Rightarrow p = \frac{1}{4}
So, Probability of hitting, P(H) = \frac{3}{4}, Probability of not hitting, P(\overline{H}) = \frac{1}{4} [1]
Now, P(\text{Target is hit}) = 1 - P(\overline{H}\,\overline{H}) = 1 - \frac{1}{4} \times \frac{1}{4} = \frac{15}{16} [1]
(ii) P(\text{at least one shot misses the target})
= 1 - P(HH) = 1 - \frac{3}{4} \times \frac{3}{4} = \frac{7}{16} [1]
Mother, Father and Son line up at random for a family picture. Let events E : Son on one end and F : Father in the middle. Find P(E/F).
There are 3 people - Mother (M), Father (F), Son (S).
Sample Space = \{MFS, MSF, FMS, FSM, SMF, SFM\}
Total possible arrangements = 6 [1]
Given E: Son on one end, F: Father in the middle
P(E \cap F) = P(\text{Son on one end and Father in the middle}) = \frac{2}{6},\ P(F) = \frac{2}{6} [1]
\therefore P(E|F) = \frac{P(E \cap F)}{P(F)} = 1 [1]
Alternative method
For P(E|F), as F has already occured which means that the father is in the middle. So, son has to be on one end. [1]
\therefore P(E|F) = 1 (sure event) [2]
A survey was conducted on the patients who have undergone knee replacement surgeries.
It was found that, Robotic Knee replacement surgeries have 90% success rate.
On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that :
(i) exactly one surgery is successful,
(ii) at most two surgeries are successful.
P(\text{success}) = \frac{9}{10} and P(\text{not success}) = \frac{1}{10} [½]
(i) Probability that exactly one surgery is successful
= \frac{9}{10} \times \frac{1}{10} \times \frac{1}{10} \times 3 = \frac{27}{1000} [1]
(ii) Probability that all three surgeries are successful
= \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} = \frac{729}{1000} [1]
Probability that at most two surgeries are successful
= 1 - \frac{729}{1000} = \frac{271}{1000} [½]
In a school, the probability of holding a debate competition is \frac{1}{3} and that of a quiz competition is \frac{2}{3}. In the two participating teams, A has 4 girls and 6 boys and B has 7 girls and 3 boys. If a debate competition is held, the students are selected from team A and for the quiz competition they are selected from team B. If only two students are to be chosen from the teams, then find the probability that one will be a girl and the other a boy.
Let D be the event a debate competition is held,
Q be the event a quiz competition is held and
E be the event that one girl & one boy are chosen.
P(E|D) = 2 \times \frac{4}{10} \times \frac{6}{9} = \frac{8}{15}, P(E|Q) = 2 \times \frac{7}{10} \times \frac{3}{9} = \frac{7}{15} [1+1]
Now, P(E) = \frac{1}{3} \times \frac{8}{15} + \frac{2}{3} \times \frac{7}{15} = \frac{22}{45} [1]
A die is rolled. Consider events :
A = {1, 2, 5}, B = {3, 5}, C = {2, 3, 4, 5}
and hence find :
(i) P(A|C) and P(C|A)
(ii) P(A \cap B|C) and P(A \cup B|C)
(i) P(A|C) = \frac{n(A\cap C)}{n(C)} = \frac{2}{4} or \frac{1}{2} [1]
and P(C|A) = \frac{n(C\cap A)}{n(A)} = \frac{2}{3} [½]
(ii) P(A \cap B|C) = \frac{n(A\cap B\cap C)}{n(C)} = \frac{1}{4} [1]
and P(A \cup B|C) = \frac{n[(A\cup B)\cap C]}{n(C)} = \frac{3}{4} [½]
A box contains 6 cards numbered 1 to 6. A student is asked to pick up two cards, one by one after replacement and note down the numbers on the cards. Let A be the event of getting sum of the numbers on two cards as 10, and B, the event of a number other than 4 on the first card selected.
Find P(A and B) and find whether the events A and B are independent events or not.
A = {(4, 6), (6, 4), (5, 5)} or n(A) = 3 [½]
n(B) = 30
n(A \cap B) = 2 [½]
P(A \cap B) = \frac{2}{36} [½]
P(A) = \frac{3}{36} = \frac{1}{12} [½]
P(B) = \frac{30}{36} = \frac{5}{6} [½]
P(A).P(B) = \frac{5}{72} \ne P(A \cap B)
Hence A and B are not independent [½]
The probability distribution for the number of students being absent in a class on a Saturday is as follows :
| X | 0 | 2 | 4 | 5 |
| P(X) | p | 2p | 3p | p |
Where X is the number of students absent.
(i) Calculate p. [1]
(ii) Calculate the mean of the number of absent students on Saturday. [2]
(i) Since \sum P(X)=1\Rightarrow p+2p+3p+p=1 [½]
\Rightarrow p=\frac{1}{7} [½]
(ii) Mean =\sum X.P(X)=0(p)+2(2p)+4(3p)+5(p) [1]
=21p=21\left(\frac{1}{7}\right)=3 [1]
For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.
Let E_1 : The applicant is a male
E_2 : The applicant is a female [½]
A : The candidate chosen will have distinction in the written test.
P(E_1)=\frac{1}{3},\ P(E_2)=\frac{2}{3},\ P(A|E_1)=0.4,\ P(A|E_2)=0.35 [1]
\therefore P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)
=\frac{1}{3}\times 0.4+\frac{2}{3}\times 0.35 [1]
=\frac{11}{30} [½]
A die with number 1 to 6 is biased such that P(2) = \frac{3}{10} and probability of other numbers is equal. Find the mean of the number of times number 2 appears on the dice, if the dice is thrown twice.
P(2) = \frac{3}{10},\ P(\text{any other number}) = 1 - \frac{3}{10} = \frac{7}{10} [½]
Let X represent the Random Variable “the number of 2’s”.
Then X = 0, 1, 2 [½]
The probability distribution is
| X | P(X) | XP(X) |
| 0 | \frac{7}{10}\times\frac{7}{10} = \frac{49}{100} | 0 |
| 1 | \frac{3}{10}\times\frac{7}{10}\times 2 = \frac{42}{100} | \frac{42}{100} |
| 2 | \frac{3}{10}\times\frac{3}{10} = \frac{9}{100} | \frac{18}{100} |
[1½]
Mean = \sum XP(X) = \frac{60}{100} = 0.6 [½]
Two dice are thrown. Defined are the following two events A and B :
A = \{(x, y) : x + y = 9\},\ B = \{(x, y) : x \ne 3\}, where (x, y) denote a point in the sample space.
Check if events A and B are independent or mutually exclusive.
A = \{(3,6), (4,5), (5,4), (6,3)\}
P(A) = \frac{4}{36} = \frac{1}{9},\ P(B) = \frac{30}{36} = \frac{5}{6} [1]
P(A \cap B) = \frac{3}{36} = \frac{1}{12} [½]
P(A) \times P(B) = \frac{5}{54} \ne P(A \cap B) [1]
Therefore, A and B are not independent.
A and B are not mutually exclusive as A \cap B \ne \emptyset [½]
Find the probability distribution of the number of boys in families having three children, assuming equal probability for a boy and a girl.
Let X denote the random variable which counts the number of boys.
X=0,1,2,3 [½]
P(\text{Boy})=P(\text{Girl})=\frac{1}{2} [½]
Required Probability Distribution
| X | 0 | 1 | 2 | 3 |
| P(X) | \left(\frac12\right)^3=\frac18 | 3\left(\frac12\right)\left(\frac12\right)^2=\frac38 | 3\left(\frac12\right)^2\left(\frac12\right)=\frac38 | \left(\frac12\right)^3=\frac18 |
[2]
A coin is tossed twice. Let X be a random variable defined as number of heads minus number of tails. Obtain the probability distribution of X and also find its mean.
Possible values of X are -2,0,2 [½]
| X | -2 | 0 | 2 |
| P(X) | \frac14 | \frac24=\frac12 | \frac14 |
[1½]
\text{Mean}=\sum XP(X)=-2\left(\frac14\right)+0\left(\frac12\right)+2\left(\frac14\right)=0 [1]
The probability that a student buys a colouring book is 0\cdot7 and that she buys a box of colours is 0\cdot2. The probability that she buys a colouring book, given that she buys a box of colours, is 0\cdot3. Find the probability that the student :
(i) Buys both the colouring book and the box of colours.
(ii) Buys a box of colours given that she buys the colouring book.
Let A be the event of buying colouring book and B be the event of buying coloured box. [½]
P(A)=0.7,\quad P(B)=0.2,\quad P(A/B)=0.3 [½]
(i) P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}\Rightarrow0.3=\frac{P(A\cap B)}{0.2}
\Rightarrow P(A\cap B)=0.06 or \frac{3}{50} [1]
(ii) P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}
=\frac{0.06}{0.7}=\frac{3}{35} or 0.086 [1]
A person has a fruit box that contains 6 apples and 4 oranges. He picks out a fruit three times, one after the other, after replacing the previous one in the box. Find :
(i) The probability distribution of the number of oranges he draws.
(ii) The expectation of the random variable (number of oranges).
Let X be random variable for number of oranges.
X = 0, 1, 2, 3 [½]
Let A be the event that orange is drawn.
P(A)=\frac{4}{10}=\frac25,\quad P(\overline{A})=1-\frac25=\frac35 [½]
(i)
| X | 0 | 1 | 2 | 3 |
| P(X) | \frac{27}{125} | \frac{54}{125} | \frac{36}{125} | \frac{8}{125} |
[1]
(ii) E(X)=\sum p_ix_i=0\times\frac{27}{125}+1\times\frac{54}{125}+2\times\frac{36}{125}+3\times\frac{8}{125} [½]
=\frac{150}{125} or \frac65 [½]
A person is Head of two independent selection committees I and II. If the probability of making a wrong selection in committee I is 0·03 and that in committee II is 0·01, then find the probability that the person makes the correct decision of selection :
(i) in both committees
(ii) in only one committee
(i) P(correct decision in both committees) = (1 - 0.03) \cdot (1 - 0.01) = 0.9603 [1 + ½]
(ii) P(correct decision in one committee) = 0.03 \cdot (1 - 0.01) + (1 - 0.03) \cdot 0.01 [1]
= 0.0394 [½]
Four students of class XII are given a problem to solve independently. Their chances of solving the problem respectively are \frac{1}{2}, \frac{1}{3}, \frac{2}{3} and \frac{1}{5}. Find the probability that at most one of them will solve the problem.
P(at most one of them will solve the problem)
= P(none of them solves the problem) + P(only one of them solves the problem)
= \left(\frac{1}{2}\times\frac{2}{3}\times\frac{1}{3}\times\frac{4}{5}\right) + \left(\frac{1}{2}\times\frac{2}{3}\times\frac{1}{3}\times\frac{4}{5} + \frac{1}{2}\times\frac{1}{3}\times\frac{1}{3}\times\frac{4}{5} + \frac{1}{2}\times\frac{2}{3}\times\frac{2}{3}\times\frac{4}{5} + \frac{1}{2}\times\frac{2}{3}\times\frac{1}{3}\times\frac{1}{5}\right) [2½]
= \frac{19}{45} [½]
The probability distribution of a random variable X is given below :
| X | 1 | 2 | 4 | 2k | 3k | 5k |
| P(X) | \frac{1}{2} | \frac{1}{5} | \frac{3}{25} | \frac{1}{10} | \frac{1}{25} | \frac{1}{25} |
Find k, if E(X) = 2·94 and also find P(X \le 4).
E(X) = 2.94
\Rightarrow 1\left(\frac{1}{2}\right) + 2\left(\frac{1}{5}\right) + 4\left(\frac{3}{25}\right) + 2k\left(\frac{1}{10}\right) + 3k\left(\frac{1}{25}\right) + 5k\left(\frac{1}{25}\right) = 2.94 [1]
\Rightarrow k = \frac{1.56}{0.52} \Rightarrow k = 3 [1]
Now, P(X \le 4) = P(X = 1) + P(X = 2) + P(X = 4)
= \frac{1}{2} + \frac{1}{5} + \frac{3}{25} [½]
= \frac{41}{50} [½]
E and F are two independent events such that P(\overline{E}) = 0{\cdot}6 and P(E \cup F) = 0{\cdot}6. Find P(F) and P(\overline{E} \cup \overline{F}).
P(\overline{E}) = 0.6 \Rightarrow P(E) = 0.4 [½]
P(E \cup F) = P(E) + P(F) - P(E \cap F) [½]
\Rightarrow 0.6 = 0.4 + P(F) - 0.4\,P(F) \Rightarrow P(F) = \frac{1}{3} [1]
P(\overline{E} \cup \overline{F}) = 1 - P(E \cap F) [½]
= 1 - 0.4 \times \frac{1}{3} = \frac{13}{15} [½]
A pair of dice is thrown simultaneously. If X denotes the absolute difference of the numbers appearing on top of the dice, then find the probability distribution of X.
| X | 0 | 1 | 2 | 3 | 4 | 5 |
| P(X) | \frac{6}{36} = \frac{1}{6} | \frac{10}{36} = \frac{5}{18} | \frac{8}{36} = \frac{2}{9} | \frac{6}{36} = \frac{1}{6} | \frac{4}{36} = \frac{1}{9} | \frac{2}{36} = \frac{1}{18} |
[½ × 6 = 3]
The chances of P, Q and R getting selected as CEO of a company are in the ratio 4 : 1 : 2 respectively. The probabilities for the company to increase its profits from the previous year under the new CEO, P, Q or R are 0·3, 0·8 and 0·5 respectively. If the company increased the profits from the previous year, find the probability that it is due to the appointment of R as CEO.
Let E_1 : P is appointed as CEO,
E_2 : Q is appointed as CEO,
E_3 : R is appointed as CEO
A : company increase profits from previous year [½]
here, P(E_1) = \frac{4}{7}, P(E_3) = \frac{1}{7}, P(E_1) = \frac{2}{7}
P(A|E_1) = 0.3, P(A|E_2) = 0.8, P(A|E_3) = 0.5 [1]
P(E_3|A) = \frac{P(E_3)P(A|E_3)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}
= \frac{\frac{2}{7}\times 0.5}{\frac{4}{7}\times 0.3 + \frac{1}{7}\times 0.8 + \frac{2}{7}\times 0.5} [1]
= \frac{1}{3} [½]
A card from a well shuffled deck of 52 playing cards is lost. From the remaining cards of the pack, a card is drawn at random and is found to be a King. Find the probability of the lost card being a King.
Let E_1 be the event of lost card is King,
E_2 be the event of lost card not a King and
A be the event of drawing a King from remaining 51 cards. [½]
so, P(E_1) = \frac{1}{13}, P(E_2) = \frac{12}{13}, P(A|E_1) = \frac{3}{51}, P(A|E_2) = \frac{4}{51} [1½]
Now, Required probability is P(E_1|A),
P(E_1|A) = \frac{P(A|E_1)\times P(E_1)}{P(A|E_1)\times P(E_1) + P(A|E_2)\times P(E_2)} = \frac{\frac{1}{13}\times\frac{3}{51}}{\frac{1}{13}\times\frac{3}{51} + \frac{12}{13}\times\frac{4}{51}} = \frac{1}{17} [1]
A biased die is twice as likely to show an even number as an odd number. If such a die is thrown twice, find the probability distribution of the number of sixes. Also, find the mean of the distribution.
Let P(1) = P(3) = P(5) = p, so P(2) = P(4) = P(6) = 2p
As, P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \Rightarrow 9p = 1 \Rightarrow p = \frac{1}{9}
P(Getting 6) = \frac{2}{9}, P(Not getting six) = \frac{7}{9} [½]
Let X represents the Number of sixes
Possible values of X are 0, 1 or 2 [½]
Now, P(X=0) = \frac{7}{9}\times\frac{7}{9} = \frac{49}{81}, P(X=1) = 2\times\frac{7}{9}\times\frac{2}{9} = \frac{28}{81}, P(X=2) = \frac{2}{9}\times\frac{2}{9} = \frac{4}{81}
Required probability distribution of number of sixes is
| X | 0 | 1 | 2 |
| P(X) | \frac{49}{81} | \frac{28}{81} | \frac{4}{81} |
[1½]
Mean of X = \sum_{i=1}^{3} X_i P(X_i) = 0 + \frac{28}{81} + \frac{8}{81} = \frac{4}{9} [½]
The random variable X has the following probability distribution where a and b are some constants :
| X | 1 | 2 | 3 | 4 | 5 |
| P(X) | 0.2 | a | a | 0.2 | b |
If the mean E(X) = 3, then find values of a and b and hence determine P(X\ge3).
E(X)=0.2+2a+3a+0.8+5b=5a+5b+1 [½]
\sum p_i=1\Rightarrow2a+b=0.6 [½]
E(X)=3\Rightarrow5a+5b=2 and,
solving the two equations, we get, a=\frac15,b=\frac15 [1]
P(X\ge3)=1-[P(X=1)+P(X=2)]=1-[0.2+a]=1-\frac25=\frac35 [1]
It is known that 20% of the students in a school have above 90% attendance and 80% of the students are irregular. Past year results show that 80% of students who have above 90% attendance and 20% of irregular students get ‘A’ grade in their annual examination. At the end of a year, a student is chosen at random from the school and is found to have an ‘A’ grade. What is the probability that the student is irregular ?
Let B : Student getting ‘A’ grade
E_1 : Student having above 90% attendance
E_2 : Student being irregular [½]
P(E_1)=\frac{20}{100};\ P(E_2)=\frac{80}{100};\ P(B|E_1)=\frac{80}{100};\ P(B|E_2)=\frac{20}{100} [1½]
P(E_2|B)=\frac{P(E_2)P(B|E_2)}{P(E_1)P(B|E_1)+P(E_2)P(B|E_2)}
=\frac{0.8\times0.2}{0.2\times0.8+0.8\times0.2}=\frac{1}{2} [½+½]
The probability distribution of a random variable X is given below :
| X | 1 | 2 | 3 |
| P(X) | \frac{k}{2} | \frac{k}{3} | \frac{k}{6} |
(i) Find the value of k.
(ii) Find P(1\le X\lt3).
(iii) Find E(X), the mean of X.
(i) \frac{k}{2}+\frac{k}{3}+\frac{k}{6}=1 [1]
Gives k=1 [½]
(ii) P(1\le X\lt3)=\frac{5k}{6}=\frac56 [½]
(iii) E(X)=\sum p_ix_i=\frac{k}{2}+\frac{2k}{3}+\frac{k}{2}=\frac{5k}{3} [½]
E(X)=\frac53 [½]
A and B are independent events such that P(A\cap\overline{B})=\frac14 and P(\overline{A}\cap B)=\frac16. Find P(A) and P(B).
P(A)\,P(\overline{B})=\frac14\quad P(\overline{A})\,P(B)=\frac16 [½]
Let P(A)=x\quad P(B)=y
x(1-y)=\frac14,\quad(1-x)y=\frac16\Rightarrow x-y=\frac{1}{12} [1]
eliminating y, we get 12x^2-13x+3=0 [1]
gives x=\frac13,\ \frac34
P(A)=\frac13\Rightarrow P(B)=\frac14
P(A)=\frac34\Rightarrow P(B)=\frac23 [½]
Two balls are drawn at random one by one with replacement from an urn containing equal number of red balls and green balls. Find the probability distribution of number of red balls. Also, find the mean of the random variable.
X: Number of red balls out of the two balls drawn [½]
| X | 0 | 1 | 2 |
| P(X) | \frac{1}{4} | \frac{1}{2} | \frac{1}{4} |
[1½]
\therefore \text{Mean} = 0\cdot\frac{1}{4} + 1\cdot\frac{1}{2} + 2\cdot\frac{1}{4} = 1 [1]
A and B throw a die alternately till one of them gets a ‘6’ and wins the game. Find their respective probabilities of wining, if A starts the game first.
P(\text{getting a six}) = \frac{1}{6};\ P(\text{not getting a six}) = \frac{5}{6} [½+½]
P(\text{A wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^2\cdot\frac{1}{6} + \left(\frac{5}{6}\right)^4\cdot\frac{1}{6} + \ldots = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{6}{11} [1½]
P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{6}{11} = \frac{5}{11} [½]
A pair of dice is thrown simultaneously. If X denotes the absolute difference of numbers obtained on the pair of dice, then find the probability distribution of X.
| X | 0 | 1 | 2 | 3 | 4 | 5 |
| P(X) | \frac{6}{36} | \frac{10}{36} | \frac{8}{36} | \frac{6}{36} | \frac{4}{36} | \frac{2}{36} |
[1½] [1½]
There are two coins. One of them is a biased coin such that P (head) : P (tail) is 1 : 3 and the other coin is a fair coin. A coin is selected at random and tossed once. If the coin showed head, then find the probability that it is a biased coin.
E_1 = Biased coin is selected \Rightarrow P(E_1) = \frac{1}{2}
E_2 = Fair coin is selected \Rightarrow P(E_2) = \frac{1}{2} [½]
A = Head appeared on tossing a selected coin.
P\left(\frac{A}{E_1}\right) = \frac{1}{4},\ P\left(\frac{A}{E_2}\right) = \frac{1}{2} [1]
By Bayes' Theorem P\left(\frac{E_1}{A}\right) = \frac{P(E_1)\,P\left(\frac{A}{E_1}\right)}{P(E_1)\,P\left(\frac{A}{E_1}\right) + P(E_2)\,P\left(\frac{A}{E_2}\right)}
= \frac{\frac{1}{2} \cdot \frac{1}{4}}{\frac{1}{2} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{2}} [1]
= \frac{1}{3} [½]
From a lot of 30 bulbs which include 6 defective bulbs, a sample of 2 bulbs is drawn at random one by one with replacement. Find the probability distribution of the number of defective bulbs and hence find the mean number of defective bulbs.
Let X be the random variable which denotes the number of defective bulbs drawn from a sample of 2 bulbs. Here X may take values 0, 1 or 2. [½]
Let A and B be the event of drawing a defective bulb and non defective bulb respectively
P(A)=\frac{6}{30}=\frac{1}{5};\ P(B)=1-\frac{1}{5}=\frac{4}{5} [½]
Now P(X=0)=\frac{4}{5}\times\frac{4}{5}=\frac{16}{25}
P(X=1)=2\left[\frac{4}{5}\times\frac{1}{5}\right]=\frac{8}{25}
P(X=2)=\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}
Thus P.D. of X is
| X | 0 | 1 | 2 |
| P(X) | \frac{16}{25} | \frac{8}{25} | \frac{1}{25} |
| XP(X) | 0 | \frac{8}{25} | \frac{2}{25} |
[1½]
Mean =\sum X\,P(X)=\frac{10}{25}=\frac{2}{5} [½]
Out of two bags, bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag B.
Official marking scheme for this paper is not available.
Out of a group of 50 people, 20 always speak the truth. Two persons are selected at random from the group (without replacement). Find the probability distribution of number of selected persons who always speak the truth.
Official marking scheme for this paper is not available.
Case Study -3
Excessive use of screens can result in vision problems, obesity, sleep disorders, anxiety, low retention problems and can impede social and emotional comprehension and expression. It is essential to be mindful of the amount of time we spend on screens and to reduce our screen-time by taking regular breaks, setting time limits, and engaging in non-screen-based activities.
In a class of students of the age group 14 to 17, the students were categorised into three groups according to a feedback form filled by them. The first group constituted of the students who spent more than 4 hours per day on the mobile screen or the gaming screens, while the second group spent 2 to 4 hours /day on the same activities. The third group spent less than 2 hours /day on the same. The first group with the high screen time is 60% of all the students, whereas the second group with moderate screen time is 30% and the third group with low screen time is only 10% of the total number of students. It was observed that 80% students of first group faced severe anxiety and low retention issues, with 70% of second group, and 30% of third group having the same symptoms.
I. What is the total percentage of students who suffer from anxiety and low retention issues in the class? [2]
II. A student is selected at random, and he is found to suffer from anxiety and low retention issues. What is the probability that he/she spends screen time more than 4 hours per day? [2]
Let the events be
E_1: the student is in the first group (time spent on screen is more than 4 hours)
E_2: the student is in the second group (time spent on screen is 2 to 4 hours)
E_3: the student is in third group (time spent on screen is less than 2 hours)
A: the event of the student showing symptoms of anxiety and low retention
P(E_1)=\frac{60}{100}, P(E_2)=\frac{30}{100} and P(E_3)=\frac{10}{100}
P(A|E_1)=\frac{80}{100}, P(A|E_2)=\frac{70}{100} and P(A|E_3)=\frac{30}{100}
I. P(A)=P(E_1)\times P(A|E_1)+P(E_2)\times P(A|E_2)+P(E_3)\times P(A|E_3)
=\frac{60}{100}\times\frac{80}{100}+\frac{30}{100}\times\frac{70}{100}+\frac{10}{100}\times\frac{30}{100}=\frac{72}{100}=72\% [2]
II. P(E_1|A)=\frac{P(E_1\cap A)}{P(A)}
=\frac{\left(\frac{60}{100}\times\frac{80}{100}\right)}{\left(\frac{72}{100}\right)}=\frac{48}{72}=\frac23 [2]
In an online jackpot, there is one first prize of ₹ 3,00,000, two second prizes of ₹ 2,00,000 each and three third prizes of ₹ 50,000 each.
A total of 1,00,000 jackpot tickets each costing ₹ 100 were sold there by raising a fund of ₹ 1,00,00,000.
Rohan bought one ticket.
Based on given information, answer the following questions :
(i) What are the possible amounts, the person can win ? [1]
(ii) (a) What is the probability that the person wins atleast ₹ 2,00,000 ? [2]
OR
(ii) (b) What is the probability that the person does not win any amount ? [2]
(iii) In another jackpot, Rohan also bought a ticket having a prize money of ₹ 5,00,000. The chances of winning the jackpot are 1 in 1,00,000. Find the probability that on exactly one of tickets he wins the jackpot. [1]
(i) ₹ 300000, ₹ 200000, ₹ 50000 [1]
(ii) (a) Required probability = \frac{3}{100000} [2]
OR
(ii) (b) Required probability = 1 - \frac{6}{100000} = \frac{99994}{100000} or \frac{49997}{50000} [2]
(iii) P (Exactly on one of the tickets Rohan wins the jackpot) = P [(He wins first and he loses second) or (He loses first and wins second)]
= \frac{6}{100000} \times \frac{99999}{100000} + \frac{99994}{100000} \times \frac{1}{100000} = \frac{699988}{10^{10}} or \frac{174997}{2500000000} [1]
Smoking increases the risk of lung problems.
A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males.
A person is selected at random from these 50 people and tested for lung related problems.
Based on the given information, answer the following questions :
(i) What is the probability that selected person is a female ? [1]
(ii) If a male person is selected, what is the probability that he will not be suffering from lung problems ? [1]
(iii) (a) A person selected at random is detected with lung complications. Find the probability that selected person is a female. [2]
OR
(iii) (b) A person selected at random is not having lung problems, find the probability that the person is a male. [2]
Total number of smokers in the colony = 50,
Number of male smokers = 30, Number of female smokers = 20
Let L represent the event that the person is suffering from the lung complications.
(i) P(F) = \frac{20}{50} = \frac{2}{5} [1]
(ii) P(L'|M) = 1 - P(L|M) = 1 - \frac{170}{1000} = \frac{83}{100} [1]
(iii)(a) P(L|M) = \frac{170}{1000} = \frac{17}{100} and P(L|F) = \frac{120}{1000} = \frac{12}{100}
Using Baye's Theorem,
P(F|L) = \frac{P(F).P(L|F)}{P(M).P(L|M) + P(F).P(L|F)}
= \frac{\frac{2}{5}.\frac{12}{100}}{\frac{3}{5}.\frac{17}{100} + \frac{2}{5}.\frac{12}{100}} [1½]
= \frac{24}{75} \text{ or } \frac{8}{25} [½]
OR
(iii)(b) Here P(L'|M) = \frac{83}{100},\ P(L'|F) = \frac{88}{100}
Using Baye's Theorem,
P(M|L') = \frac{P(M).P(L'|M)}{P(M).P(L'|M) + P(F).P(L'|F)}
= \frac{\frac{3}{5}.\frac{83}{100}}{\frac{3}{5}.\frac{83}{100} + \frac{2}{5}.\frac{88}{100}} [1½]
= \frac{249}{425} [½]
Case Study – 2
A survey was conducted to find out the success rate of students who qualified the entrance examination by dropping a year after class XII.
As per the data collected, 40% students appearing in the examination were dropouts and the remaining students were regular students of class XII.
Of the dropouts, 5% qualify the examination while 10% of the regular students qualify the examination.
Based on the above information, answer the following questions.
(i) Find the probability that a student selected at random is a regular student. [1]
(ii) A student is selected at random from a group of dropout students. What is the probability that the student will not qualify the examination ? [1]
(iii) (a) A student selected at random qualified the examination. Find the probability that student is not a dropout. [2]
OR
(iii) (b) A student selected at random did not qualify the examination. Find the probability that the student was a regular student. [2]
(i) Probability that a student selected at random is a regular student = \frac{3}{5} [1]
(ii) Probability that the selected student from the group of dropout will not qualify the examination = \frac{95}{100} or \frac{19}{20} [1]
(iii) (a) Let A: The student has qualified the examination.
E_1: The student appearing in the examination is dropout.
E_2: The student is regular.
Using Bayes' Theorem, [1½]
P\left(\frac{E_1'}{A}\right) = P\left(\frac{E_2}{A}\right) = \frac{\frac{60}{100} \times \frac{10}{100}}{\frac{40}{100} \times \frac{5}{100} + \frac{60}{100} \times \frac{10}{100}}
= \frac{3}{4} [½]
OR
(iii) (b) P\left(\frac{E_2}{A'}\right) = \frac{\frac{60}{100} \times \frac{90}{100}}{\frac{40}{100} \times \frac{95}{100} + \frac{60}{100} \times \frac{90}{100}} [1½]
= \frac{27}{46} [½]
Case Study – 3
An NGO organises a charity event in which they decide to distribute woollen caps to protect children from winter. The caps to be distributed are in three separate boxes, Box I has 30 red caps, Box II has 20 red and 10 green caps, and Box III has 30 green caps. The probability that a Box i is selected and a cap picked out is \frac{i}{6}, where i = 1, 2, 3.
Based on the above information, answer the following questions :
A person selects a cap.
(i) What is the probability that he selects a red cap ? [2]
(ii) If he selects a green cap, what is the probability that the cap has come from Box II ? [2]
(i) Let B_1, B_2 & B_3 be the event that box I, box II and box III selected respectively.
P(B_1) = \frac{1}{6}, P(B_2) = \frac{2}{6}, P(B_3) = \frac{3}{6} [1]
Now, P(selecting a red cap) = \frac{1}{6} \times 1 + \frac{2}{6} \times \frac{2}{3} + \frac{3}{6} \times 0 = \frac{7}{18} [1]
(ii) P(selecting a green cap) = 1 - \frac{7}{18} = \frac{11}{18} [1]
Now, P\left(\frac{B_2}{\text{Green cap}}\right) = \frac{\frac{2}{6} \times \frac{1}{3}}{\frac{11}{18}} = \frac{2}{11} [1]
Case Study – 2
There are three types of vaccines A_1, A_2, A_3, available in the market to protect the population of the country from spread of certain infection. According to a survey conducted, it was found that 25% of the population was given Vaccine A_1, 35% of the population was given Vaccine A_2 and 40% of the population was given Vaccine A_3. The survey also stated that the probabilities that Vaccines A_1, A_2 and A_3 would protect against the infection were 60%, 55% and 50% respectively.
Based on the above information, answer the following questions :
Find the probability that :
(i) The person taking vaccine A_2 will get infected. [1]
(ii) If a person is chosen randomly, he/she will be protected from the infection. [1]
(iii) (a) The person was given Vaccine A_1, given that the randomly chosen person is infected. [2]
OR
(iii) (b) The person was given Vaccine A_3, given that the randomly chosen person is not infected. [2]
(i) P (Person taking vaccine A_2 will be infected) = 45% or 0.45 [1]
(ii) P (Person is protected from infection)
= \frac{25}{100}.\frac{60}{100} + \frac{35}{100}.\frac{55}{100} + \frac{40}{100}.\frac{50}{100} [½]
= \frac{5425}{10000} or 0.5425 [½]
(iii) (a) P (A_1/person is infected) = \dfrac{\frac{25}{100}.\frac{40}{100}}{\frac{25}{100}.\frac{40}{100} + \frac{35}{100}.\frac{45}{100} + \frac{40}{100}.\frac{50}{100}} [1½]
= \frac{1000}{4575} or \frac{40}{183} [½]
OR
(iii) (b) P(A_3/person is not infected) = \dfrac{\frac{40}{100}.\frac{50}{100}}{\frac{25}{100}.\frac{60}{100} + \frac{35}{100}.\frac{55}{100} + \frac{40}{100}.\frac{50}{100}} [1½]
= \frac{2000}{5425} or \frac{80}{217} [½]
Case Study – 3
An organisation needs to select a building construction company to make houses in cities. Three companies X, Y and Z are short-listed with the probability of their selection being \frac{1}{3}, \frac{2}{9} and \frac{4}{9} respectively. It was found that there were 20%, 40% and 30% chances respectively, of delivering poor construction quality.
Based on the above information, answer the following questions :
(i) If the building constructed was found to be strong, then what is the probability that the construction contract was given to the company X ? [2]
(ii) If a part of the newly constructed building falls off, then what is the probability that it was due to the selection of company Y or Z ? [2]
(i) P(strong construction quality) = \frac{1}{3} \times \frac{80}{100} + \frac{2}{9} \times \frac{60}{100} + \frac{4}{9} \times \frac{70}{100} = \frac{640}{900} [1]
So, P\left(\frac{\text{Construction Contract to X}}{\text{strong construction quality}}\right) = \frac{\frac{240}{900}}{\frac{640}{900}} = \frac{3}{8} [1]
(ii) P(Defective construction quality) = \frac{1}{3} \times \frac{20}{100} + \frac{2}{9} \times \frac{40}{100} + \frac{4}{9} \times \frac{30}{100} = \frac{260}{900} [1]
So, P\left(\frac{\text{Construction Contract to Y or Z}}{\text{strong construction quality}}\right) = 1 - \frac{\frac{60}{900}}{\frac{260}{900}} = \frac{10}{13} [1]
A bank offers loan to its customers on different types of interest namely, fixed rate, floating rate and variable rate. From the past data with the bank, it is known that a customer avails loan on fixed rate, floating rate or variable rate with probabilities 10%, 20% and 70% respectively. A customer after availing loan can pay the loan or default on loan repayment. The bank data suggests that the probability that a person defaults on loan after availing it at fixed rate, floating rate and variable rate is 5%, 3% and 1% respectively.
Based on the above information, answer the following :
(i) What is the probability that a customer after availing the loan will default on the loan repayment ? [2]
(ii) A customer after availing the loan, defaults on loan repayment. What is the probability that he availed the loan at a variable rate of interest ? [2]
E_1 : customer avails loan on fixed rate
E_2 : customer avails loan on floating rate
E_3 : customer avails loan on variable rate
A : the person defaults on the loan
P(E_1)=\frac{1}{10},\ P(E_2)=\frac{2}{10},\ P(E_3)=\frac{7}{10}
P(A|E_1)=\frac{5}{100},\ P(A|E_2)=\frac{3}{100},\ P(A|E_3)=\frac{1}{100}
(i) P(A)=P(E_1).P(A|E_1)+P(E_2).P(A|E_2)+P(E_3).P(A|E_3)
=\frac{1}{10}\times\frac{5}{100}+\frac{2}{10}\times\frac{3}{100}+\frac{7}{10}\times\frac{1}{100} [1]
=\frac{18}{1000} or \frac{9}{500} [1]
(ii) P(E_3|A)=\frac{P(E_3).P(A|E_3)}{P(E_1).P(A|E_1)+P(E_2).P(A|E_2)+P(E_3).P(A|E_3)}
=\frac{\frac{7}{10}\times\frac{1}{100}}{\frac{18}{1000}} [1]
=\frac{7}{18} [1]
Three persons viz. Amber, Bonzi and Comet are manufacturing cars which run on petrol and on battery as well. Their production share in the market is 60%, 30% and 10% respectively. Of their respective production capacities, 20%, 10% and 5% cars respectively are electric (or battery operated).
Based on the above, answer the following :
(i) (a) What is the probability that a randomly selected car is an electric car ? [2]
OR
(i) (b) What is the probability that a randomly selected car is a petrol car ? [2]
(ii) A car is selected at random and is found to be electric. What is the probability that it was manufactured by Comet ? [1]
(iii) A car is selected at random and is found to be electric. What is the probability that it was manufactured by Amber or Bonzi ? [1]
(i) (a) Let A = Amber manufactures the car
B = Bonzi manufactures the car
C = Comet manufactures the car
E = The selected car is electric
P(A) = \frac{60}{100}, P(B) = \frac{30}{100}, P(C) = \frac{10}{100} [½]
P(E) = P(A)\times P\left(\frac{E}{A}\right) + P(B)\times P\left(\frac{E}{B}\right) + P(C)\times P\left(\frac{E}{C}\right)
= \frac{60}{100}\times\frac{20}{100} + \frac{30}{100}\times\frac{10}{100} + \frac{10}{100}\times\frac{5}{100} [1]
= \frac{155}{1000}\ \text{or}\ \frac{31}{200} [½]
OR
(i) (b) Let A = Amber manufactures the car
B = Bonzi manufactures the car
C = Comet manufactures the car
E = The selected car is a petrol car
P(A) = \frac{60}{100}, P(B) = \frac{30}{100}, P(C) = \frac{10}{100} [½]
P(E) = P(A)\times P\left(\frac{E}{A}\right) + P(B)\times P\left(\frac{E}{B}\right) + P(C)\times P\left(\frac{E}{C}\right)
= \frac{60}{100}\times\frac{80}{100} + \frac{30}{100}\times\frac{90}{100} + \frac{10}{100}\times\frac{95}{100} [1]
= \frac{845}{1000}\ \text{or}\ \frac{169}{200} [½]
(ii) P\left(\frac{C}{E}\right) = \frac{P(C)\times P\left(\frac{E}{C}\right)}{P(E)}
= \frac{\frac{10}{100}\times\frac{5}{100}}{\frac{60}{100}\times\frac{20}{100} + \frac{30}{100}\times\frac{10}{100} + \frac{10}{100}\times\frac{5}{100}}
= \frac{\frac{50}{10000}}{\frac{1550}{10000}} = \frac{1}{31} [1]
(iii) P\left(\frac{A \text{ or } B}{E}\right) = 1 - P\left(\frac{C}{E}\right) = 1 - \frac{1}{31} = \frac{30}{31} [1]
Some students are having a misconception while comparing decimals. For example, a student may mention that 78·56 > 78·9 as 7856 > 789. In order to assess this concept, a decimal comparison test was administered to the students of class VI through the following question : In the recently held Sports Day in the school, 5 students participated in a javelin throw competition. The distances to which they have thrown the javelin are shown below in the table :
| Name of student | Distance of javelin (in meters) |
| Ajay | 47·7 |
| Bijoy | 47·07 |
| Kartik | 43·09 |
| Dinesh | 43·9 |
| Devesh | 45·2 |
The students were asked to identify who has thrown the javelin the farthest.
Based on the test attempted by the students, the teacher concludes that 40% of the students have the misconception in the concept of decimal comparison and the rest do not have the misconception. 80% of the students having misconception answered Bijoy as the correct answer in the paper. 90% of the students who are identified with not having misconception, did not answer Bijoy as their answer.
On the basis of the above information, answer the following questions :
(i) What is the probability of a student not having misconception but still answers Bijoy in the test ? [1]
(ii) What is the probability that a randomly selected student answers Bijoy as his answer in the test ? [1]
(iii) (a) What is the probability that a student who answered as Bijoy is having misconception ? [2]
OR
(iii) (b) What is the probability that a student who answered as Bijoy is amongst students who do not have the misconception ? [2]
Let E_1 : Student has a misconception
E_2 : Student does not have misconception
A: Student answered Bijoy as correct
\therefore P(E_1)=\frac{40}{100},\ P(E_2)=\frac{60}{100}
P(A|E_1)=\frac{80}{100},\ P(A|E_2)=\frac{10}{100}
P(\bar{A}|E_1)=\frac{20}{100},\ P(\bar{A}|E_2)=\frac{90}{100}
(i) P(A|E_2)=\frac{10}{100} or \frac{1}{10} [1]
(ii) P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)
=\frac{40}{100}\times\frac{80}{100}+\frac{60}{100}\times\frac{10}{100}
=\frac{38}{100} or \frac{19}{50} [1]
(iii)(a) P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(A)}
=\frac{\frac{40}{100}\times\frac{80}{100}}{\frac{38}{100}}=\frac{16}{19} [2]
OR
(iii)(b) P(E_2|A)=\frac{P(E_2)P(A|E_2)}{P(A)}
=\frac{\frac{60}{100}\times\frac{10}{100}}{\frac{38}{100}}=\frac{3}{19} [2]
Case Study – 3
A gardener wanted to plant vegetables in his garden. Hence he bought 10 seeds of brinjal plant, 12 seeds of cabbage plant and 8 seeds of radish plant. The shopkeeper assured him of germination probabilities of brinjal, cabbage and radish to be 25%, 35% and 40% respectively. But before he could plant the seeds, they got mixed up in the bag and he had to sow them randomly.
Based upon the above information, answer the following questions :
(i) Calculate the probability of a randomly chosen seed to germinate. [2]
(ii) What is the probability that it is a cabbage seed, given that the chosen seed germinates ? [2]
Let A: Event that chosen seed germinates.
B: Event that Brinjal seed is chosen.
C: Event that Cabbage seed is chosen.
R: Event that Radish seed is chosen.
P(B)=\frac{10}{30};\ P(C)=\frac{12}{30};\ P(R)=\frac{8}{30};
P\left(\frac{A}{B}\right)=\frac{25}{100};\ P\left(\frac{A}{C}\right)=\frac{35}{100};\ P\left(\frac{A}{R}\right)=\frac{40}{100} [1]
(i) P(A)=P(B)\cdot P\left(\frac{A}{B}\right)+P(C)\cdot P\left(\frac{A}{C}\right)+P(R)\cdot P\left(\frac{A}{R}\right)
=\frac{10}{30}\times\frac{25}{100}+\frac{12}{30}\times\frac{35}{100}+\frac{8}{30}\times\frac{40}{100}
=\frac{990}{3000} or \frac{33}{100} [1]
(ii) (a) P\left(\frac{C}{A}\right)=\frac{P(C)\cdot P\left(\frac{A}{C}\right)}{P(B)\cdot P\left(\frac{A}{B}\right)+P(C)\cdot P\left(\frac{A}{C}\right)+P(R)\cdot P\left(\frac{A}{R}\right)}
=\frac{\frac{12}{30}\times\frac{35}{100}}{\frac{990}{3000}} [1]
=\frac{42}{99} or \frac{14}{33} [1]
Case Study – 3
A shop selling electronic items sells smartphones of only three reputed companies A, B and C because chances of their manufacturing a defective smartphone are only 5%, 4% and 2% respectively. In his inventory he has 25% smartphones from company A, 35% smartphones from company B and 40% smartphones from company C.
A person buys a smartphone from this shop.
(i) Find the probability that it was defective. [2]
(ii) What is the probability that this defective smartphone was manufactured by company B ? [2]
(i) P(defective smartphone) = 0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02 [1½]
= 0.0345 [½]
(ii) P(B/ Defective) = \frac{0.35 \times 0.04}{0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02} [1½]
= \frac{140}{345} or \frac{28}{69} [½]
Case Study – 3
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let A_1 : People with good health,
A_2 : People with average health,
and A_3 : People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category A_1, A_2 and A_3 are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions :
(i) A person was tested randomly. What is the probability that he/she has contracted the disease ? [2]
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category A_2 ? [2]
(i) Let A: Person contracted the disease
P(A)=P(A_1)\cdot P(A|A_1)+P(A_2)\cdot P(A|A_2)+P(A_3)\cdot P(A|A_3)
=\frac{7}{10}\left(\frac{25}{100}\right)+\frac{2}{10}\left(\frac{35}{100}\right)+\frac{1}{10}\left(\frac{50}{100}\right) [1½]
=\frac{295}{1000}=0.295 or \left(\frac{59}{200}\right) [½]
(ii) P(A_2|\overline{A})=\frac{P(A_2)\cdot P(\overline{A}/A_2)}{P(A_1)\cdot P(\overline{A}/A_1)+P(A_2)\cdot P(\overline{A}/A_2)}
=\frac{\frac{2}{10}\times\frac{65}{100}}{\frac{7}{10}\times\frac{75}{100}+\frac{2}{10}\times\frac{65}{100}+\frac{1}{10}\times\frac{50}{100}} [1½]
=\frac{2\times13}{7\times15+2\times13+1\times10}=\frac{26}{141} [½]
Case Study – 3
Two persons are competing for a position on the Managing Committee of an organisation. The probabilities that the first and the second person will be appointed are 0·5 and 0·6 respectively. Also, if the first person gets appointed, then the probability of introducing waste treatment plant is 0·7 and the corresponding probability is 0·4, if the second person gets appointed.
Based on the above information, answer the following questions :
(i) What is the probability that the waste treatment plant is introduced ? [2]
(ii) After the selection, if the waste treatment plant is introduced, what is the probability that the first person had introduced it ? [2]
E_1 : Event that the first person is appointed.
E_2 : Event that the second person is appointed.
A : Event that the waste treatment plant is introducted.
Here, P(E_1) = 0.5, P(E_2) = 0.6
P(A|E_1) = 0.7, P(A|E_2) = 0.4
(i) P(waste treatment plant is introducted)
= P(E_1)P(A|E_1) + P(E_2)P(A|E_2)
= (0.5)(0.7) + (0.6)(0.4) [1]
= 0.35 + 0.24 = 0.59 [1]
(ii) P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}
= \frac{(0.5)(0.7)}{(0.5)(0.7) + (0.6)(0.4)} [1 ½]
= \frac{0.35}{0.59} = \frac{35}{59} [½]
Note: Full marks to be awarded, in case a student writes “Sum of probabilities of selecting first person and second person should not be greater than 1”.
Case Study – 2
According to recent research, air turbulence has increased in various regions around the world due to climate change. Turbulence makes flights bumpy and often delays the flights.
Assume that, an airplane observes severe turbulence, moderate turbulence or light turbulence with equal probabilities. Further, the chance of an airplane reaching late to the destination are 55%, 37% and 17% due to severe, moderate and light turbulence respectively.
On the basis of the above information, answer the following questions :
(i) Find the probability that an airplane reached its destination late. [2]
(ii) If the airplane reached its destination late, find the probability that it was due to moderate turbulence. [2]
(i) Let A denote the event of airplane reaching its destination late
E_1 = severe turbulence
E_2 = moderate turbulence
E_3 = light turbulence [½]
P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)
= \frac{1}{3} \times \frac{55}{100} + \frac{1}{3} \times \frac{37}{100} + \frac{1}{3} \times \frac{17}{100} [1]
= \frac{1}{3}\left(\frac{109}{100}\right) = \frac{109}{300} [½]
(ii) P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(A)}
= \frac{\frac{1}{3} \times \frac{37}{100}}{\frac{109}{300}} [1½]
= \frac{37}{109} [½]
Airplanes are by far the safest mode of transportation when the number of transported passengers are measured against personal injuries and fatality totals.
Previous records state that the probability of an airplane crash is 0·00001%. Further, there are 95% chances that there will be survivors after a plane crash. Assume that in case of no crash, all travellers survive.
Let E_1 be the event that there is a plane crash and E_2 be the event that there is no crash. Let A be the event that passengers survive after the journey.
On the basis of the above information, answer the following questions :
(i) Find the probability that the airplane will not crash. [1]
(ii) Find P(A \mid E_1) + P(A \mid E_2). [1]
(iii) (a) Find P(A). [2]
OR
(iii) (b) Find P(E_2 \mid A). [2]
(i) P(E_2) = 1 - 0.0000001
= 0.9999999 [1]
(ii) P(A/E_1) + P(A/E_2) = \frac{95}{100} + 1 = \frac{195}{100} [1]
(iii)(a) P(A) = P(E_1) \times P(A/E_1) + P(E_2) \times P(A/E_2) [1]
= \frac{1}{10000000} \times \frac{95}{100} + \frac{9999999}{10000000} \times 1
= \frac{95+999999900}{1000000000} = \frac{999999995}{1000000000} [1]
OR
(iii)(b) P(E_2/A) = \frac{P(E_2) \times P(A/E_2)}{P(E_1) \times P(A/E_1) + P(E_2) \times P(A/E_2)} [1]
= \frac{\frac{9999999}{10000000}}{\frac{999999995}{1000000000}} = \frac{999999900}{999999995} [1]
Case Study – 1
Self-study helps students to build confidence in learning. It boosts the self-esteem of the learners. Recent surveys suggested that close to 50% learners were self-taught using internet resources and upskilled themselves.
A student may spend 1 hour to 6 hours in a day in upskilling self. The probability distribution of the number of hours spent by a student is given below :
where x denotes the number of hours.
Based on the above information, answer the following questions :
(i) Express the probability distribution given above in the form of a probability distribution table. [1]
(ii) Find the value of k. [1]
(iii) (a) Find the mean number of hours spent by the student. [2]
OR
(iii) (b) Find P(1 \lt X \lt 6). [2]
(i)
| X | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X) | k | 4k | 9k | 8k | 10k | 12k |
[1]
(ii) k + 4k + 9k + 8k + 10k + 12k = 1
\Rightarrow k = \frac{1}{44} [1]
(iii)(a) Mean = \sum x_i p_i = k + 8k + 27k + 32k + 50k + 72k [1]
= 190k
= \frac{190}{44} or \frac{95}{22} [1]
OR
(iii)(b) P(1 \lt X \lt 6) = 4k + 9k + 8k + 10k [1]
= 31k
= \frac{31}{44} [1]
Rohit, Jaspreet and Alia appeared for an interview for three vacancies in the same post. The probability of Rohit’s selection is \frac{1}{5}, Jaspreet’s selection is \frac{1}{3} and Alia’s selection is \frac{1}{4}. The event of selection is independent of each other.
Based on the above information, answer the following questions :
(i) What is the probability that at least one of them is selected ? [1]
(ii) Find P(G|\overline{H}) where G is the event of Jaspreet’s selection and \overline{H} denotes the event that Rohit is not selected. [1]
(iii) Find the probability that exactly one of them is selected. [2]
OR
(iii) Find the probability that exactly two of them are selected. [2]
Given P(Rohit) = \frac{1}{5}, P(Jaspreet) = \frac{1}{3}, P(Alia) = \frac{1}{4}
(i) P(atleast one of them is selected) = 1 – P(no one is selected)
= 1 - \left(\frac{4}{5}\times\frac{2}{3}\times\frac{3}{4}\right) = \frac{3}{5} [1]
(ii) P(G|\overline{H}) = \frac{P(G \cap \overline{H})}{P(\overline{H})} = \frac{1}{3} [1]
(iii) P(exactly one of them selected)
= P(R)\times P(\overline{J})\times P(\overline{A}) + P(\overline{R})\times P(J)\times P(\overline{A}) + P(\overline{R})\times P(\overline{J})\times P(A) [1]
= \frac{6+12+8}{60} = \frac{13}{30} [1]
OR
(iii) P(exactly two of them selected)
= P(R)\times P(J)\times P(\overline{A}) + P(R)\times P(\overline{J})\times P(A) + P(\overline{R})\times P(J)\times P(A) [1]
= \frac{3+2+4}{60} = \frac{3}{20} [1]
A departmental store sends bills to charge its customers once a month. Past experience shows that 70% of its customers pay their first month bill in time. The store also found that the customer who pays the bill in time has the probability of 0.8 of paying in time next month and the customer who doesn’t pay in time has the probability of 0.4 of paying in time the next month.
Based on the above information, answer the following questions :
(i) Let E_1 and E_2 respectively denote the event of customer paying or not paying the first month bill in time. Find P(E_1), P(E_2).
(ii) Let A denotes the event of customer paying second month’s bill in time, then find P(A|E_1) and P(A|E_2).
(iii) Find the probability of customer paying second month’s bill in time.
OR
(iii) Find the probability of customer paying first month’s bill in time if it is found that customer has paid the second month’s bill in time.
(i) P(E_1)=\frac{7}{10}=0.7,\ P(E_2)=\frac{3}{10}=0.3 [1]
(ii) P(A|E_1)=0.8,\ P(A|E_2)=0.4 [1]
(iii) P(A)=P(E_1)\cdot P(A|E_1)+P(E_2)\cdot P(A|E_2)=0.7\times0.8+0.3\times0.4=0.68 or \frac{17}{25} [2]
Or
(iii) P(A)=\frac{P(E_1)\cdot P(A|E_1)}{P(E_1)\cdot P(A|E_1)+P(E_2)\cdot P(A|E_2)}=\frac{14}{17} [2]
Case Study – 2
A salesman receives a commission for each sale he makes together with a fixed daily income. The number of sales he makes in a day along with their probabilities are given in the table below :
| X : | 0 | 1 | 2 | 3 | 4 | 5 |
| P(X) : | 0.42 | 3k | 0.3 | 0.05 | 2k | 0.03 |
His daily income Y (in ₹) is given by :
Y=800X+50
On the basis of the above information, answer the following questions :
(i) Find the value of k.
(ii) Evaluate P(X\ge3).
(iii) (a) Calculate the expected weekly income of the salesman assuming he works five days per week.
OR
(iii) (b) Calculate the expected weekly income of the salesman assuming he works only for three days of the week.
(i) 0.42+3k+0.3+0.05+2k+0.03=1\Rightarrow5k=0.2\Rightarrow k=\frac{1}{25}=0.04 [1]
(ii) P(X\ge3)=0.05+2k+0.03=2k+0.08=0.08+0.08=0.16 [1]
(iii) (a) Expected sales per day =\sum x\cdot P(x)
=3k+0.6+0.15+8k+0.15=11k+0.9=1.34 [1]
Expected Daily income =Y=800(1.34)+50= ₹ 1122 [½]
Expectedly sales income for five days per week = ₹ 5610 [½]
Or
(iii) (b) Expected sales per day =\sum x\cdot P(x)
=3k+0.6+0.15+8k+0.15=11k+0.9=1.34 [1]
Expected Daily income =Y=800(1.34)+50= ₹ 1122 [½]
Expectedly sales income for three days of the week = ₹ 3366 [½]
Case Study – 1
There are different types of Yoga which involve the usage of different poses of Yoga Asanas, Meditation and Pranayam as shown in the figure below :
The Venn diagram below represents the probabilities of three different types of Yoga, A, B and C performed by the people of a society. Further, it is given that probability of a member performing type C Yoga is 0·44.
On the basis of the above information, answer the following questions :
(i) Find the value of x. [1]
(ii) Find the value of y. [1]
(iii) (a) Find P\left(\frac{C}{B}\right). [2]
OR
(iii) (b) Find the probability that a randomly selected person of the society does Yoga of type A or B but not C. [2]
(i) x+0\cdot21=0\cdot44\Rightarrow x=0\cdot23 [1]
(ii) 0\cdot41+y+0\cdot44+0.11=1\Rightarrow y=0\cdot04 [1]
(iii) (a) P\left(\frac{C}{B}\right)=\frac{P(C\cap B)}{P(B)}
P(B)=0\cdot09+0\cdot04+0\cdot23=0\cdot36 [1]
P\left(\frac{C}{B}\right)=\frac{0\cdot23}{0\cdot36}=\frac{23}{36} [1]
OR
(iii) (b) P(A or B but not C)
=0\cdot32+0\cdot09+0.04 [1½]
=0\cdot45 [½]
Recent studies suggest that roughly 12% of the world population is left handed.
Depending upon the parents, the chances of having a left handed child are as follows :
A : When both father and mother are left handed :
Chances of left handed child is 24%.
B : When father is right handed and mother is left handed :
Chances of left handed child is 22%.
C : When father is left handed and mother is right handed :
Chances of left handed child is 17%.
D : When both father and mother are right handed :
Chances of left handed child is 9%.
Assuming that P(A) = P(B) = P(C) = P(D) = \frac{1}{4} and L denotes the event that child is left handed.
Based on the above information, answer the following questions :
(i) Find P(L/C) [1]
(ii) Find P(\overline{L}/A) [1]
(iii) (a) Find P(A/L) [2]
OR
(iii) (b) Find the probability that a randomly selected child is left handed given that exactly one of the parents is left handed. [2]
(i) P(L|C) = \frac{17}{100} [1]
(ii) P(\overline{L}|A) = 1 - P(L|A) = 1 - \frac{24}{100} = \frac{76}{100} \text{ or } \frac{19}{25} [1]
(iii) P(A|L) = \frac{\frac{1}{4}\times\frac{24}{100}}{\frac{1}{4}\times\frac{24}{100} + \frac{1}{4}\times\frac{22}{100} + \frac{1}{4}\times\frac{17}{100} + \frac{1}{4}\times\frac{9}{100}} = \frac{24}{72} = \frac{1}{3} [2]
Or
Probability that a randomly selected child is left-handed given that exactly one of the parents is left-handed.
= P(L|B\cup C) = \frac{22}{100} + \frac{17}{100} = \frac{39}{100} [2]
Case Study – 1
An octagonal prism is a three-dimensional polyhedron bounded by two octagonal bases and eight rectangular side faces. It has 24 edges and 16 vertices.
The prism is rolled along the rectangular faces and number on the bottom face (touching the ground) is noted. Let X denote the number obtained on the bottom face and the following table give the probability distribution of X.
| X : | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X) : | p | 2p | 2p | p | 2p | p^2 | 2p^2 | 7p^2 + p |
Based on the above information, answer the following questions :
(i) Find the value of p. [1]
(ii) Find P(X \gt 6). [1]
(iii) (a) Find P(X = 3m), where m is a natural number. [2]
OR
(iii) (b) Find the mean E(X). [2]
(i) 10p^2 + 9p = 1 [½]
\Rightarrow p = \frac{1}{10} [½]
(ii) P(X \gt 6) = 9p^2 + p [½]
= \frac{9}{100} + \frac{1}{10}
= \frac{19}{100} [½]
(iii)(a) P(X = 3m) = P(3) + P(6) [1]
\Rightarrow 2p + p^2 = \frac{21}{100} [1]
OR
(iii)(b) E(X) = \sum XP(X) = p + 4p + 6p + 4p + 10p + 6p^2 + 14p^2 + 56p^2 + 8p [1]
= 33p + 76p^2 [½]
= \frac{406}{100} or \frac{203}{50} [½]
A building contractor undertakes a job to construct 4 flats on a plot along with parking area. Due to strike the probability of many construction workers not being present for the job is 0·65. The probability that many are not present and still the work gets completed on time is 0·35. The probability that work will be completed on time when all workers are present is 0·80.
Let : E_1 : represent the event when many workers were not present for the job;
E_2 : represent the event when all workers were present; and
E : represent completing the construction work on time.
Based on the above information, answer the following questions :
(i) What is the probability that all the workers are present for the job ? [1]
(ii) What is the probability that construction will be completed on time ? [1]
(iii) (a) What is the probability that many workers are not present given that the construction work is completed on time ? [2]
OR
(iii) (b) What is the probability that all workers were present given that the construction job was completed on time ? [2]
(i) P(E_2)=1-P(E_1)=1-0{\cdot}65=0{\cdot}35 [1]
(ii) P(E)=P(E_1)\,.\,P\left(\frac{E}{E_1}\right)+P(E_2)\,.\,P\left(\frac{E}{E_2}\right)
=0{\cdot}65\times0{\cdot}35+0{\cdot}35\times0{\cdot}8
=0{\cdot}35\times1{\cdot}45
=0{\cdot}51 [1]
(iii) (a) P\left(\frac{E_1}{E}\right)=\frac{P(E_1)\,.\,P(E/E_1)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)}=\frac{0{\cdot}65\times0{\cdot}35}{0{\cdot}51}=0{\cdot}45 [2]
OR
(iii) (b) P\left(\frac{E_2}{E}\right)=\frac{P(E_2)\,.\,P(E/E_2)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)}=\frac{0{\cdot}35\times0{\cdot}8}{0{\cdot}51}=0{\cdot}55 [2]
Case Study – 1
In a group activity class, there are 10 students whose ages are 16, 17, 15, 14, 19, 17, 16, 19, 16 and 15 years. One student is selected at random such that each has equal chance of being chosen and age of the student is recorded.
On the basis of the above information, answer the following questions :
(i) Find the probability that the age of the selected student is a composite number. [1]
(ii) Let X be the age of the selected student. What can be the value of X ? [1]
(iii) (a) Find the probability distribution of random variable X and hence find the mean age. [2]
OR
(iii) (b) A student was selected at random and his age was found to be greater than 15 years. Find the probability that his age is a prime number. [2]
Official marking scheme for this paper is not available.
A card from a pack of 52 playing cards is lost. From the remaining cards, 2 cards are drawn at random without replacement, and are found to be both aces. Find the probability that lost card being an ace.
E_1 : Lost card is an ace
E_2 : Lost card is not an ace
A: 2 ace cards are drawn [½]
P(E_1)=\frac{1}{13} P(E_2)=\frac{12}{13} [1]
P(A/E_1)=\frac{{}^3C_2}{{}^{51}C_2} P(A/E_2)=\frac{{}^4C_2}{{}^{51}C_2} [1]
[1 + ½]
Case-Study Based Question
At the start of a cricket match, a coin is tossed and the team winning the toss has the opportunity to choose to bat or bowl. Such a coin is unbiased with equal probabilities of getting head and tail.
Based on the above information, answer the following questions :
(a) If such a coin is tossed 2 times, then find the probability distribution of number of tails. [2]
(b) Find the probability of getting at least one head in three tosses of such a coin. [2]
(a) Let X denote the number of tails: [½]
| X | 0 | 1 | 2 |
| P(X) | \frac14 | \frac12 | \frac14 |
[1½]
(b) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
P (at least one head) = 1 – P (no head) [1]
=1-\frac18
=\frac78 [1]
Case Study Problem :
A shopkeeper sells three types of flower seeds A1, A2, A3. They are sold in the form of a mixture, where the proportions of these seeds are 4 : 4 : 2, respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively.
Based on the above information :
(a) Calculate the probability that a randomly chosen seed will germinate; [2]
(b) Calculate the probability that the seed is of type A2, given that a randomly chosen seed germinates. [2]
E_1 : Seed is of type A_1
E_2 : Seed is of type A_2
E_3 : Seed is of type A_3
A: Seed germinates [½]
P(E_1)=\frac{4}{10},P(E_2)=\frac{4}{10},P(E_3)=\frac{2}{10} [½]
P(A|E_1)=\frac{45}{100},P(A|E_2)=\frac{60}{100},P(A|E_3)=\frac{35}{100} [½]
(a) P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)
=\frac{4}{10}\times\frac{45}{100}+\frac{4}{10}\times\frac{60}{100}+\frac{2}{10}\times\frac{35}{100} [1]
=\frac{180+240+70}{1000}
=\frac{490}{1000}=\frac{49}{100} [½]
(b) P\left(\frac{E_2}{A}\right)=\frac{P(E_2)P(A|E_2)}{P(A)}
=\frac{\frac{4}{10}\times\frac{60}{100}}{\frac{490}{1000}}=\frac{24}{49} [½+½]
In a game of Archery, each ring of the Archery target is valued. The centremost ring is worth 10 points and rest of the rings are allotted points 9 to 1 in sequential order moving outwards.
Archer A is likely to earn 10 points with a probability of 0·8 and Archer B is likely the earn 10 points with a probability of 0·9.
Based on the above information, answer the following questions :
If both of them hit the Archery target, then find the probability that
(a) exactly one of them earns 10 points. [2]
(b) both of them earn 10 points. [2]
P(A)=0\cdot8,\ P(B)=0\cdot9
(a) P (exactly one of them earns 10 points) =P(A)P(\overline{B})+P(\overline{A})P(B)
=0\cdot8\times0\cdot1+0\cdot2\times0\cdot9 [1½]
=0\cdot26 [½]
(b) P(A\cap B)=P(A)P(B)
=0\cdot8\times0\cdot9 [1]
=0\cdot72 [1]
Three persons A, B and C apply for a job of manager in a private company. Chances of their selection are in the ratio 1 : 2 : 4. The probability that A, B and C can introduce changes to increase the profits of a company are 0.8, 0.5 and 0.3 respectively. If increase in the profit does not take place, find the probability that it is due to the appointment of A.
E_1 : A is selected
E_2 : B is selected
E_3 : C is selected
F : increase in profit does not take place [½]
P(E_1)=1/7,\ P(E_2)=2/7,\ P(E_3)=4/7 [1]
P(F|E_1)=0\cdot 2,\ P(F|E_2)=0\cdot 5,\ P(F|E_3)=0\cdot 7 [1]
P(E_1|F)=\frac{P(E_1)\,P(F|E_1)}{P(E_1)\,P(F|E_1)+P(E_2)\,P(F|E_2)+P(E_3)\,P(F|E_3)}
=\frac{\frac{1}{7}\times\frac{2}{10}}{\frac{1}{7}\times\frac{2}{10}+\frac{2}{7}\times\frac{5}{10}+\frac{4}{7}\times\frac{7}{10}} [1]
=\frac{2}{40}=\frac{1}{20} [½]
A biased die is tossed and respective probabilities for various faces to turn up are the following :
| Face | 1 | 2 | 3 | 4 | 5 | 6 |
| Probability | 0·1 | 0·24 | 0·19 | 0·18 | 0·15 | K |
Based on the above information, answer the following questions :
(a) What is the value of K ? [2]
(b) If a face showing an even number has turned up, then what is the probability that it is the face with 2 or 4 ? [2]
(a) P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 [1]
\Rightarrow 0.1+0.24+0.19+0.18+0.15+K = 1
\Rightarrow K = 0.14 [1]
(b) A: face shows 2 or 4
B: even face have turned up
Required probability = P(A/B) = \frac{P(A\cap B)}{P(B)}
here P(A\cap B) = P(2)+P(4)
= 0.24+0.18 = 0.42 [1]
P(B) = P(2)+P(4)+P(6) = 0.56 [½]
\therefore P(A/B) = \frac{0.42}{0.56} = \frac{3}{4} [½]
A fitness app syncs with four types of smartwatches: Alpha, Beta, Gamma and Delta. 45% of the users use Alpha, 30% use Beta, 15% use Gamma and 10% use Delta.
Due to device errors, the probabilities of getting a false heart-rate alert from the Alpha, Beta, Gamma and Delta smartwatches are respectively 0.01, 0.02, 0.05, and 0.01.
Based on the given information, answer the following questions:
(i) Find the probability that a randomly selected user gets a false alert.
(ii) If a false alert occurs, what is the probability that it came from a Beta smartwatch?
Let the events be
E_1: User uses Alpha smartwatch
E_2: User uses Beta smartwatch
E_3: User uses Gamma smartwatch
E_4: User uses Delta smartwatch
A: User gets a false heart-rate alert
Clearly \{E_1,E_2,E_3,E_4\} is a partition of the sample space.
Then P(E_1)=0.45, P(E_2)=0.3, P(E_3)=0.15, P(E_4)=0.1 [1]
Now P(A|E_1)=0.01, P(A|E_2)=0.02, P(A|E_3)=0.05, P(A|E_4)=0.01 [1]
(i) By Theorem of Total Probability,
The total probability that a randomly selected user gets a false alert
=P(A)
=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)+P(E_4)P(A|E_4)
=0.45\times0.01+0.3\times0.02+0.15\times0.05+0.1\times0.01
=0.0045+0.006+0.0075+0.001=0.019 [1½]
(ii) By Bayes Theorem,
The probability that false alert came from a Beta smartwatch
=P(E_2|A)
=\frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)+P(E_4)P(A|E_4)}
=\frac{0.3\times0.02}{0.019}=\frac{0.006}{0.019}=\frac{6}{19} [1½]
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let \frac{3}{5} be the probability that he knows the answer and \frac{2}{5} be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \frac{1}{3}. What is the probability that the student knows the answer, given that he answered it correctly ?
Let events A, B and E be defined as:
A : Student knows the answer
B : Student guesses the answer
E : student answered correctly [1]
P(A)=\frac{3}{5},\ P(B)=\frac{2}{5} [½]
Here, P\left(\frac{E}{A}\right)=1 and P\left(\frac{E}{B}\right)=\frac{1}{3} [1]
By Bayes’ Theorem
=\frac{9}{11} [1]
A box contains 10 tickets, 2 of which carry a prize of ₹ 8 each, 5 of which carry a prize of ₹ 4 each, and remaining 3 carry a prize of ₹ 2 each. If one ticket is drawn at random, find the mean value of the prize.
Let X denote the prize value.
Here X can take values of 8, 4 and 2. [1]
P(X=8)=\frac{2}{10}, or \frac{1}{5}
P(X=4)=\frac{5}{10}, or \frac{1}{2}
P(X=2)=\frac{3}{10}
| X | 8 | 4 | 2 |
| P(X) | \frac{1}{5} | \frac{1}{2} | \frac{3}{10} |
| XP(X) | \frac{8}{5} | \frac{4}{2} | \frac{6}{10} |
[3]
Hence, Mean value of X =\sum X\,P(X)=\frac{8}{5}+2+\frac{6}{10}
=\frac{42}{10} or ₹ 4.20 [1]