ISC Paper (2022) Class 12: Mathematics Solution – Semester 1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

MATHEMATICS

(Maximum Marks: 40)

(Time Allowed: One and a half hours)

(Candidates are allowed additional 10 minutes for only reading the paper.

They must NOT start writing during this time)

The Question Paper consists of three sections A, B and C.

Candidates are required to attempt all questions from Section A and all question EITHER from Section B OR Section C

All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.

The intended marks for questions or parts of questions are given in brackets [ ].

Mathematical tables and graphs papers are provided.

SECTION – A [34 Marks]

$\displaystyle \text{Question 1: Let } 'R' \text{ be a relation on } N, \text{ set of all natural given by } R= \{(a, b): a - b = 2 \}. \text{ Then: }$

$\displaystyle (a) \ (2, 4) \in R \hspace{1.0cm} (b) \ (10, 8) \in R \hspace{1.0cm}(c) \ (6, 8) \in R \hspace{1.0cm}(d) \ (8, 7) \in R$

Answer:

$\fbox{ b }$

$\displaystyle \text{The given relation is, } R= \{(a, b): a - b = 2 \}$

$\displaystyle \text{As the relation is on set of all natural numbers therefore, } 'a' \text{ should be greater than } 'b'.$

$\displaystyle \text{In the given options, in option } (b) 10 > 8 \text{ and } 10 - 8 = 2$

$\displaystyle \text{Thus, } (10, 8) \in R.$

$\\$

$\displaystyle \text{Question 2: If } A = \begin{vmatrix} zy & x & yz \\ xz & y & zx \\ yx & z & xy \end{vmatrix}, \text{ then the value of } A \text{ is equal to }$

$\displaystyle (a) \ 0 \hspace{1.0cm} (b) \ xyz \hspace{1.0cm}(c) \ 1 \hspace{1.0cm}(d) \ \frac{1}{xyz}$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given determinant } A = \begin{vmatrix} zy & x & yz \\ xz & y & zx \\ yx & z & xy \end{vmatrix}$

$\displaystyle \text{In } A, C_1 \text{ and } C_3 \text{ are identical. }$

According to the property of determinants, if two rows or columns are identical, then its value is zero.

$\displaystyle \text{Therefore, } A = 0$

$\\$

$\displaystyle \text{Question 3: The function} f(x) = \frac{x^3}{3} - x \text{ is decreasing in the interval }$

$\displaystyle (a) \ (-1, 1) \hspace{1.0cm} (b) \ (-\infty, -1) \hspace{1.0cm}(c) \ (1, \infty ) \hspace{1.0cm}(d) \ (-\infty, - 1) \cup (1, \infty)$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given function } f(x) = \frac{x^3}{3} - x$

$\displaystyle \text{Differentiating both sides w.r.t } x$

$\displaystyle \Rightarrow \hspace{1.0cm} f'(x) = \frac{3x^2}{3} - 1$

$\displaystyle \Rightarrow \hspace{1.0cm} f(x) = x^2 - 1$

$\displaystyle \text{Put } f(x) = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} x^2 - 1 = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} (x-1)(x+1) = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} x = -1, 1$

$\displaystyle \text{In the interval } ( -1, 1) f(x) \text{ is decreasing. }$

$\\$

$\displaystyle \text{Question 4: If } \cot^{-1} \frac{1}{5} + \tan^{-1} x = \frac{\pi}{2}, \text{ than value of } x \text{ is}$

$\displaystyle (a) \ \frac{1}{5} \hspace{1.0cm} (b) \ 1 \hspace{1.0cm}(c) \ 0 \hspace{1.0cm}(d) \ \frac{-1}{5}$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given } \cot^{-1} \frac{1}{5} + \tan^{-1} x = \frac{\pi}{2}$

$\displaystyle \text{We know, }\cot^{-1} x + \tan^{-1} x = \frac{\pi}{2}$

$\displaystyle \text{Therefore }x = \frac{1}{5}$

$\\$

$\displaystyle \text{Question 5: Let the two functions} f(x) \text{ and } g(x) \text{ be defined as } \\ \\ f(x) = x^2 - 1 \text{ and } g(x) = \sqrt{x} \text{ then } (fog)(6) \text{ is } :$

$\displaystyle (a) \ 5 \hspace{1.0cm} (b) \ 7 \hspace{1.0cm}(c) \ 35 \hspace{1.0cm}(d) \ -35$

Answer:

$\fbox{ a }$

$\displaystyle \text{The given function are } f(x) = x^2 - 1 \text{ and } g(x) = \sqrt{x}$

$\displaystyle (fog)(x) = f(g(x)) = f(\sqrt{x}) = (\sqrt{x})^2 - 1 = x - 1$

$\displaystyle (fog)(6) = 6 - 1 = 5$

$\\$

$\displaystyle \text{Question 6: If} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}, \text{ then the matrix } A^2 \text{ is equal to: }$

$\displaystyle (a) \ A^2 = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 16 \end{bmatrix} \hspace{1.0cm} (b) \ A^2 = \begin{bmatrix} 0 & 0 & 4 \\ 0 & 9 & 0 \\ 16 & 0 & 0 \end{bmatrix} \hspace{1.0cm} \\ \\ \\ (c) \ A^2 = \begin{bmatrix} 4 & 0 & 16 \\ 0 & 0 & 0 \\ 0 & 9 & 0 \end{bmatrix} \hspace{1.0cm} (d) \ A^2 = \begin{bmatrix} 0 & 0 & 0 \\ 4 & 9 & 16 \\ 0 & 0 & 0 \end{bmatrix}$

Answer:

$\fbox{ a }$

$\displaystyle A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$

$\displaystyle A^2 = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+9+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+16 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 16 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 7: The value of } \lim \limits_{x \to 0} \frac{1 - \cos x}{x^2} \text{ is eual to: }$

$\displaystyle (a) \ \frac{1}{2} \hspace{1.0cm} (b) \ 1 \hspace{1.0cm}(c) \ -1 \hspace{1.0cm}(d) \ \frac{-1}{2}$

Answer:

$\fbox{ a }$

$\displaystyle \lim \limits_{x \to 0} \frac{1 - \cos x}{x^2}$

$\displaystyle \text{Using : } 1 - \cos x = 2 \sin^2 \frac{x}{2}$

$\displaystyle = \lim \limits_{x \to 0} \frac{2 \sin^2 \frac{x}{2}}{x^2}$

$\displaystyle = \lim \limits_{x \to 0} 2 \Bigg( \frac{\sin \frac{x}{2}}{x} \Bigg)^2$

$\displaystyle = \lim \limits_{x \to 0} 2 \Bigg( \frac{\sin \frac{x}{2}}{\frac{2x}{2}} \Bigg)^2$

$\displaystyle \text{Using : } \lim \limits_{x \to 0} \frac{\sin x}{x} = 1$

$\displaystyle = 2 \times \frac{1}{4}$

$\\$

$\displaystyle \text{Question 8: The expression of } \cos (\tan^{-1}) \text{ is equal to: }$

$\displaystyle (a) \ \frac{1}{\sqrt{1-x^2}} \hspace{1.0cm} (b) \ \frac{1}{\sqrt{1+x^2}} \hspace{1.0cm}(c) \ \frac{\sqrt{1-x^2}}{2} \hspace{1.0cm}(d) \ \sqrt{1-x^2}$

Answer:

$\fbox{ b }$

$\displaystyle \text{Let } \theta = \tan^{-1} x$

$\displaystyle \text{Also, } \cos \theta = \frac{1}{\sec \theta } = \frac{1}{\sqrt{1 + \tan^2 \theta}} = \frac{1}{\sqrt{1+x^2}}$

$\displaystyle \therefore \cos \Big( \tan^{-1} x \Big) = \frac{1}{\sqrt{1+x^2}}$

$\\$

$\displaystyle \text{Question 9: If } 2 \begin{pmatrix} a & 9 \\ 6 & d \end{pmatrix} + 3 \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix} = 3 \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix} \text{ then the values of } a \text { and } d \text{ respectively are: }$

$\displaystyle (a) \ 3, 6 \hspace{1.0cm} (b) \ 3, 5 \hspace{1.0cm}(c) \ 3, 9 \hspace{1.0cm}(d) \ 3, 7$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given, } \ \ \ 2 \begin{pmatrix} a & 9 \\ 6 & d \end{pmatrix} + 3 \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix} = 3 \begin{pmatrix} 3 & 5 \\ 4 & 6 \end{pmatrix}$

$\displaystyle \Rightarrow \hspace{1.0cm} \begin{pmatrix} 2a & 18 \\ 12 & 2d \end{pmatrix} + \begin{pmatrix} 3 & -3 \\ 0 & 6 \end{pmatrix} = \begin{pmatrix} 9 & 15 \\ 12 & 18 \end{pmatrix}$

$\displaystyle \Rightarrow \hspace{1.0cm} \begin{pmatrix} 2a+3 & 15 \\ 12 & 2d+6 \end{pmatrix} = \begin{pmatrix} 9 & 15 \\ 12 & 18 \end{pmatrix}$

$\displaystyle \text{On comparing the corresponding elements, we get }$

$\displaystyle 2a+3 = 9 \text{ and } 2d+6= 18$

$\displaystyle \Rightarrow \hspace{1.0cm}2a = 6 \text{ and } 2d = 12$

$\displaystyle \Rightarrow \hspace{1.0cm}a = 3 \text{ and } d = 6$

$\\$

$\displaystyle \text{Question 10: Differentiation of } \log (1+x^2) \text{ with respect to } \tan^{-1} x \text{ is: }$

$\displaystyle (a) \ \frac{1}{1+x^2} \hspace{1.0cm} (b) \ 2x \hspace{1.0cm}(c) \ \frac{-1}{1+x^2} \hspace{1.0cm}(d) \ -x$

Answer:

$\fbox{ b }$

$\displaystyle \text{Let } u = \log (1+x^2) \text{ and } v = \tan^{-1} x$

$\displaystyle \frac{du}{dv} = \frac{d}{dx} \Bigg[ \log(1+x^2) \Bigg] = \frac{1}{1+x^2} (2x) = \frac{2x}{1+x^2}$

$\displaystyle \frac{du}{dx} = \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}$

$\displaystyle \frac{du}{dv} = \frac{du}{dx} \times \frac{dx}{dv} = \frac{2x}{1+x^2} \times \frac{1+x^2}{1} = 2x$

$\\$

$\displaystyle \text{Question 11: The relation } R = \{ (a, a), (b, b), (c, c) \} \text{ on the set } \{ a, b, c\} \text{ is: }$

$\displaystyle (a) \ \text{symmetric only } \hspace{1.0cm} (b) \ \text{reflexive only } \hspace{1.0cm} \\ \\ (c) \ \text{transitive only } \hspace{1.0cm}(d) \ \text{an euivalence relation }$

Answer:

$\fbox{ d }$

$\displaystyle \text{The given relation } R = \{ (a, a), (b, b), (c, c) \} \text{ is an identity relation. }$

We know that, identity relation is always an equivalence relation, therefore the given relation is an equivalence relation.

$\\$

$\displaystyle \text{Question 12: If the function } f(x) = \Bigg\{ \begin{array}{ll} 3x-1 & x < 2 \\ k, & x = 2 \text{ is continuous at } \\ 2x+1, & x > 2 \end{array} \\ x= 2, \text{ then the value of } 'K' \text{ is: }$

$\displaystyle (a) \ k=2 \hspace{1.0cm} (b) \ k=3 \hspace{1.0cm}(c) \ k=5 \hspace{1.0cm}(d) \ k=1$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given the function } f(x) \text{ is continuous at } x = 2.$

$\displaystyle \text{Then, } \lim \limits_{x \to 2^{-}} f(x) = \lim \limits_{x \to 2^{+}} f(x) = f(2)$

$\displaystyle \text{Now, } \lim \limits_{x \to 2^{-}} (3x-1) = 3(2) -1 = 5$

$\displaystyle \text{and } \lim \limits_{x \to 2^{+}} (2x+1) = 2(2) + 1 = 5$

$\displaystyle \text{Hence, } k = 5$

$\\$

$\displaystyle \text{Question 13: If } x^2 + y^2 = 42 , \text{ then } \frac{dy}{dx} \text{ is: }$

$\displaystyle (a) \ \frac{dy}{dx} = \frac{-3y^2}{2x} \hspace{1.0cm} (b) \ \frac{dy}{dx} = \frac{3y^2}{2x} \hspace{1.0cm}(c) \ \frac{dy}{dx} = \frac{2x}{3y^2} \hspace{1.0cm}(d) \ \frac{dy}{dx} = \frac{-2x}{3y^2}$

Answer:

$\fbox{ d }$

$\displaystyle \text{The given equation is } x^2 + y^2 = 42$

$\displaystyle \text{Differentiating both sides } w.r.t. x, \text{ we get }$

$\displaystyle \frac{d}{dx} (x^2) + \frac{d}{dx} (y^3) = \frac{d}{dx} (42)$

$\displaystyle \Rightarrow \hspace{1.0cm} 2x + 3y^2 \frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{dy}{dx} = \frac{-2x}{3y^2}$

$\\$

$\displaystyle \text{Question 14: If the matrix } A = \begin{pmatrix} 1 & x & -1 \\ -1 & 3 & 2 \\ 2 & 1 & 1 \end{pmatrix} \text{ is singular, then the value of } 'x' \text{ is: }$

$\displaystyle (a) \ x = \frac{8}{5} \hspace{1.0cm} (b) \ x = \frac{-8}{5} \hspace{1.0cm}(c) \ x = \frac{5}{8} \hspace{1.0cm}(d) \ x=1$

Answer:

$\fbox{ b }$

$\displaystyle \text{When } det(A) \text{ is zero, then } A \text{ is a singular matrix. Therefore, } |A| = 0$

$\displaystyle A = \begin{vmatrix} 1 & x & -1 \\ -1 & 3 & 2 \\ 2 & 1 & 1 \end{vmatrix} = 0$

$\displaystyle 1 \begin{vmatrix} 3& 2 \\ 1 & 1 \end{vmatrix} - x \begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} + (-1) \begin{vmatrix} -1& 3 \\ 2 & 1 \end{vmatrix} = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} 1+5x + 7 = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} 5x = - 8$

$\displaystyle \Rightarrow \hspace{1.0cm} x = \frac{-8}{5}$

$\\$

$\displaystyle \text{Question 15: The value of } \tan^{-1} 1 + \cos^{-1} \frac{1}{2} \text{ is: }$

$\displaystyle (a) \ \frac{5\pi}{12} \hspace{1.0cm} (b) \ \frac{\pi}{4} \hspace{1.0cm}(c) \ \frac{\pi}{2} \hspace{1.0cm}(d) \ \frac{7\pi}{12}$

Answer:

$\fbox{ d }$

$\displaystyle \tan^{-1} 1 = \frac{\pi}{4} \text{ and } \cos^{-1} \frac{1}{2} = \frac{\pi}{3}$

$\displaystyle \Rightarrow \hspace{1.0cm} \tan^{-1} 1 + \cos^{-1} \frac{1}{2} = \frac{\pi}{4} + \frac{\pi}{3} = \frac{7\pi}{12}$

$\\$

$\displaystyle \text{Question 16: The slope of the tangent to the curve } \sqrt{x} +\sqrt{y} = a \text{ at } \Big( \frac{a^2}{4} , \frac{a^2}{4} \Big) \text{ is: }$

$\displaystyle (a) \ 1 \hspace{1.0cm} (b) \ -1 \hspace{1.0cm}(c) \ \frac{a}{4} \hspace{1.0cm}(d) \ \frac{a}{2}$

Answer:

$\fbox{ b }$

$\displaystyle \text{The given equation is, } \sqrt{x} +\sqrt{y} = a$

$\displaystyle \text{Differentiating both sides } w.r.t. x, \text{ we get }$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{d}{dx} (\sqrt{x} )+ \frac{d}{dx} (\sqrt{y}) = \frac{d}{dx} (a)$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{dy}{dx} = - \sqrt{\frac{y}{x}}$

$\displaystyle \text{At } \Big( \frac{a^2}{4} , \frac{a^2}{4} \Big), \ \ \ \ \ \ \frac{dy}{dx} = -1$

$\\$

$\displaystyle \text{Question 17: If } A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \text{ then the matrix } A + A^T \text{ is: }$

$\displaystyle (a) \ \text{Symmetric matrix } \hspace{1.0cm} (b) \ \text{Skew-symmetric matrix } \hspace{1.0cm}\\ \\ (c) \ \text{Diagonal matrix } \hspace{1.0cm}(d) \ \text{Identity matrix }$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given } A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$

$\displaystyle A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$

$\displaystyle \text{And, } A + A^T = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} + \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 5 & 8 \end{pmatrix}$

$\displaystyle \text{Now, } (A + A^T)^T = \begin{pmatrix} 2 & 5 \\ 5 & 8 \end{pmatrix}$

From equation (i) and (ii),

$\displaystyle A + A^T = (A + A^T)^T$

$\displaystyle \text{Therefore, } A + A^T \text{ is a symmetric matrix. }$

$\\$

$\displaystyle \text{Question 18: If} x = a \cos \theta, y = a \sin \theta , \text{ then } \frac{dy}{dx} \text{ at } \theta = \frac{\pi}{2} \text{ will be: }$

$\displaystyle (a) \ \frac{dy}{dx} = -1 \hspace{1.0cm} (b) \ \frac{dy}{dx} = 1 \hspace{1.0cm}(c) \ \frac{dy}{dx} = 0 \hspace{1.0cm}(d) \ \frac{dy}{dx} = 2$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given } x = a \cos \theta, \text{ and } y = a \sin \theta$

$\displaystyle \therefore \frac{dx} {d \theta} = \frac{d}{d \theta} (a \cos \theta) = - a \sin \theta$

$\displaystyle \text{and } \hspace{1.0cm} \frac{dy}{d \theta} = \frac{d}{d \theta} (a \sin \theta) = a \cos \theta$

$\displaystyle \text{Now, } \hspace{1.0cm} \frac{dy}{dx} = \frac{dy} {d \theta} \times \frac{d \theta}{dx} = \frac{a \cos \theta}{-a \sin \theta} = - \cot \theta$

$\displaystyle \text{At } \hspace{1.0cm} \theta = \frac{\pi}{2}, \frac{dy}{dx}= \cot \frac{\pi}{2}=0$

$\\$

$\displaystyle \text{Question 19: If } y = \tan^{-1} x \text{ then }$

$\displaystyle (a) \ (1+x^2) \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} = 0 \hspace{1.0cm} (b) \ \sqrt{(1-x^2)}\frac{d^2y}{dx^2} + 2x \frac{dy}{dx} = 0 \hspace{1.0cm}\\ \\ \\ (c) \ (1-x^2) \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} = 0 \hspace{1.0cm}(d) \ \sqrt{(1+x^2)}\frac{d^2y}{dx^2} + 2x \frac{dy}{dx} = 0$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given } y = \tan^{-1} x$

$\displaystyle \text{Differentiating both sides } w.r.t. x, \text{ we get }$

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}$

$\displaystyle (1+x^2) \frac{dy}{dx} = 1$

$\displaystyle \text{Again differentiating both sides } w.r.t. x, \text{ we get }$

$\displaystyle (1+x^2) \frac{d^2y}{dx^2} + 2x \frac{dy}{dx} = 0$

$\\$

$\displaystyle \text{Question 20: If } A = \begin{pmatrix} -1 & 1 \\ 2 & 3 \end{pmatrix}, \text{ then } A(adj \ A) \text{ is equal to: }$

$\displaystyle (a) \ \begin{pmatrix} 4 & 0 \\ 10 & 24 \end{pmatrix} \hspace{1.0cm} (b) \ \begin{pmatrix} -5 & 0 \\ 0 & 5 \end{pmatrix} \hspace{1.0cm}(c) \ \begin{pmatrix} 4 & 5 \\ 5 & 4 \end{pmatrix} \hspace{1.0cm}(d) \ \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}$

Answer:

$\fbox{ b }$

$\displaystyle \text{Given } A = \begin{pmatrix} -1 & 1 \\ 2 & 3 \end{pmatrix}$

$\displaystyle \text{Minors of } A \text{ are: }$

$\displaystyle M_{11} = 3, \ \ \ M_{12} = 2, \ \ \ M_{21} =1, \ \ \ M_{22} = - 1$

$\displaystyle \text{Cofactors of } A \text{ are: }$

$\displaystyle C_{11} = 3, \ \ \ C_{12} = -2, \ \ \ C_{21} = -1, \ \ \ C_{22}= -1$

Therefore,

$\displaystyle Adj( A) = \begin{pmatrix} 3 & -2 \\ -1 & -1 \end{pmatrix}^T = \begin{pmatrix} 3 & -1 \\ -2 & -1 \end{pmatrix}$

Now,

$\displaystyle A(adj \ A) = \begin{pmatrix} -1 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 3 & -1 \\ -2 & -1 \end{pmatrix} = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix}$

$\\$

$\displaystyle \text{Question 21: Consider the two functions} f:R \to R \text{ given by } \\ \\ f(x) = x-2 \text{ and } g : R \to R \text{ given by } g(x) = x^2$

$\displaystyle \text{(i) The function } f(x) \text{ is }$

$\displaystyle (a) \ \text{One to one but not onto } \hspace{1.0cm} (b) \ \text{Onto but not one to one } \hspace{1.0cm}\\ \\ (c) \ \text{Neither one to nor onto } \hspace{1.0cm}(d) \ \text{Bijective }$

$\displaystyle \text{(ii) The value of } f(1) + g(1)\text{ is }$

$\displaystyle (a) \ \frac{1}{2} \hspace{1.0cm} (b) \ 0 \hspace{1.0cm}(c) \ 1 \hspace{1.0cm}(d) \ \frac{-1}{2}$

$\displaystyle \text{(iii) The expression for } (gof)(x) \text{ is }$

$\displaystyle (a) \ x-2 \hspace{1.0cm} (b) \ (x-2)^2 \hspace{1.0cm}(c) \ x^2 - 2 \hspace{1.0cm}(d) \ x^2 - 3$

$\displaystyle \text{(iv) If } (gof)(x) = 0, \text{ then the value of } x \text{ will be:}$

$\displaystyle (a) \ x = \pm 2 \hspace{1.0cm} (b) \ x=2 \hspace{1.0cm}(c) \ x = \pm \sqrt{2} \hspace{1.0cm}(d) \ x=3$

Answer:

$\text{(i) } \ \ \ \fbox{ d }$

$\displaystyle \text{Let } x_1, x_2 \in R$

$\displaystyle \text{Now, } f(x_1) = f(x_2)$

$\displaystyle \Rightarrow \hspace{1.0cm} x_1 - 2 = x_2 - 2$

$\displaystyle \Rightarrow \hspace{1.0cm} x_1 = x_2$

$\displaystyle \Rightarrow \hspace{1.0cm} f(x) \text{ is one-one }$

$\displaystyle \text{Let } y \in R$

$\displaystyle \text{Let } y = f(x_0)$

$\displaystyle \text{Then, } x_0 - 2 = y$

$\displaystyle \Rightarrow \hspace{1.0cm} x_0 = y + 2$

$\displaystyle \text{Now, } y \in R$

$\displaystyle \Rightarrow \hspace{1.0cm} y+2 \in R$

$\displaystyle \Rightarrow \hspace{1.0cm} x_0 \in R$

$\displaystyle f(x_0) = x_0 + 2 = 7$

$\displaystyle \text{Therefore, for for each } y \in R, \text{ there exists }x_0 \in R \text{ such that } f(x_o) = y$

$\displaystyle \text{So, } f(x) \text{ is onto. }$

$\displaystyle \text{Thus, } f(x) \text{ is one-one and onto or bijective. }$

$\text{(ii) } \ \ \ \fbox{ d }$

$\displaystyle f(1) = 1-2 = - 1 \text{ and } g(1) = (1)^2= 1$

$\displaystyle \text{Now, } f(1) +g(1) = - 1 + 1 = 0$

$\text{(iii) } \ \ \ \fbox{ d }$

$\displaystyle gof(x) = g(f(x)) = g(x-2) = (x-2)^2$

$\text{(iv) } \ \ \ \fbox{ d }$

$\displaystyle gof(x) = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} (x-2)^2 = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} x-2 = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} x = 2$

$\\$

$\displaystyle \text{Question 22: A school wants to award its students for their achievement in sports, } \\ \text{Music and Literature with a total cash prize of Rs. 6000.}$

$\displaystyle \text{Three times the prize money for Literature added to the prize money given for } \\ \text{sports is equal to Rs. 11000 }$

$\displaystyle \text{The prize money given for Sports and Literature together is equal to two times} \\ \text{of the prize money given for Music. }$

$\displaystyle \text{If x, y and z represent the prize money given for Sports, Music and Literature respectively} \\ \text{then:. }$

$\displaystyle \text{(i) The set of linear equations representing the above information will be: }$

$\displaystyle (a) \ x+y+z = 6000, \ \ x+3z=11000 \text{ and } \ x - 2y + z = 0 \hspace{1.0cm} \\ (b) \ x+y+z = 6000, \ \ x+3z=11000 \text{ and } \ x +y + 2z = 0 \hspace{1.0cm}\\ (c) \ x+y+z = 6000, \ \ 3x+z=11000 \text{ and } \ x +y -2z = 0 \hspace{1.0cm}\\ (d) \ x+y+z = 6000, \ \ x+3z=11000 \text{ and } \ 2x + 2y - z = 0$

$\displaystyle \text{(ii) Consider } A= \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{pmatrix} \text{ Then the value of } |A| \text{ is: }$

$\displaystyle (a) \ 3 \hspace{1.0cm} (b) \ 6 \hspace{1.0cm}(c) \ 4 \hspace{1.0cm}(d) \ -1$

$\displaystyle \text{(iii) Consider } adj(A) = \begin{pmatrix} 6 & 9 & 3 \\ q & 0 & r \\ -2 & 3 & -1 \end{pmatrix} \text{ the values of } p, q, \text{ and } r \text{ respectively will be }$

$\displaystyle (a) \ 3, 2, -2 \hspace{1.0cm} (b) \ 3, -2, -2 \hspace{1.0cm}(c) \ -3, 2, -2 \hspace{1.0cm}(d) \ 3, 2, 2$

$\displaystyle \text{(iv) Using } |A| \text{ and } adj(A) \text{ calculate the prize money for Sports} (x)$

$\displaystyle (a) \ x=1500 \hspace{1.0cm} (b) \ x=500 \hspace{1.0cm}(c) \ x=1000 \hspace{1.0cm}(d) \ x=2000$

Answer:

$\text{(i) } \ \ \ \fbox{ a }$

Given,

$\displaystyle \text{Cash prize for achievement in sports is } x.$

$\displaystyle \text{Cash prize for achievement in Music is } y.$

$\displaystyle \text{Cash prize for achievement in Literature is } z.$

$\displaystyle \text{Three times the prize money for literature is added to the money given } \\ \text{for sports is equal to Rs. 11000.}$

$\displaystyle \Rightarrow \hspace{1.0cm} x + 3z = 11000$

$\displaystyle \text{The prize money given for sports and literature is equal to two times of } \\ \text{ prize money for music. }$

$\displaystyle \Rightarrow \hspace{1.0cm} x + z = 2y \text{ or } x - 2y + z = 0$

$\displaystyle \text{Total cash prize for all the three subjects is Rs. 6000.}$

$\displaystyle x+y+z = 6000$

$\text{(ii) } \ \ \ \fbox{ b }$

$\displaystyle |A| = 1 \begin{vmatrix} 0 & 3 \\ -2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 0 \\ 1 & -2 \end{vmatrix}$

$\displaystyle = 1[0+6]-1[1-3]+1[-2-0] = 6+2-2 = 6$

$\text{(iii) } \ \ \ \fbox{ c }$

$\displaystyle \text{The above system of equations can be written in matrix form as } AX = B$

$\displaystyle \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6000 \\ 11000 \\ 0 \end{pmatrix}$

$\displaystyle \text{Here, } |A| = 6 \text{ which means } A \text{is non-singular. Hence, it is invertible. }$

$\displaystyle \text{Now, co-factors of } A \text{ are: }$

$\displaystyle A_{11} = (-1)^{1+1} \begin{vmatrix} 0 & 3 \\ -2 & 1 \end{vmatrix} = 0+6= 6$

$\displaystyle A_{12} = (-1)^{1+2} \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} = -(1-3) = 2$

$\displaystyle A_{13} = (-1)^{1+3} \begin{vmatrix} 1 & 0 \\ 1 & -2 \end{vmatrix} = -2-0= -2$

$\displaystyle A_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 1 \\ -2 & 1 \end{vmatrix} = -(1+2)= -3$

$\displaystyle A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 1-1= 0$

$\displaystyle A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = -(-2-1)= 3$

$\displaystyle A_{31} = (-1)^{3+1} \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = 3-0= 3$

$\displaystyle A_{32} = (-1)^{3+2} \begin{vmatrix} 1 &1 \\ 1 & 3\end{vmatrix} =-(3-1)= -2$

$\displaystyle A_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 0-1= 1$

$\displaystyle \therefore \hspace{1.0cm} adj(A) = \begin{pmatrix} 6 & 2 & -2 \\ -3 & 0 & 3 \\ 3 & -2 & 1 \end{pmatrix}^T$

$\displaystyle \therefore \hspace{1.0cm} A = \begin{pmatrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{pmatrix}$

$\displaystyle \text{On comparing, we get } p= -3, q = 2, \text{ and } r = - 2$

$\text{(iv) } \ \ \ \fbox{ b }$

$\displaystyle X = A^{-1} B$

$\displaystyle X = \frac{adj(A)}{|A|} B$

$\displaystyle \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1 \end{pmatrix} \begin{pmatrix} 6000 \\ 11000 \\ 0 \end{pmatrix}$

$\displaystyle \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 36000-33000+0 \\ 12000+0-0 \\ -12000+33000-0 \end{pmatrix}$

$\displaystyle \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{3000}{6} \\ \frac{12000}{6} \\ \frac{21000}{6} \end{pmatrix}$

$\displaystyle \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 500 \\ 2000 \\ 3500 \end{pmatrix}$

$\displaystyle \text{Therefore the przie money for sports } (x) \text{ is Rs. 500 }$

$\\$

$\displaystyle \text{Question 23: A person has manufactured a water tank in the shape of a closed right } \\ \text{circulr cylinder. The volume of the cylinder is } \frac{539}{2} \text{ cubic units. If the height and} \\ \text{the raduis of the cylinder be } h \text{ and } r, \text{ then: }$

$\displaystyle \text{(i) The height } h \text{ in terms of radius } r \text{ and given volume will be }$

$\displaystyle (a) \ h = \frac{539}{\pi r^2} \hspace{1.0cm} (b) \ h = \frac{539}{2\pi r^2} \hspace{1.0cm}(c) \ h = \frac{539}{2 \pi r} \hspace{1.0cm}(d) \ h = \frac{539}{\pi r}$

$\displaystyle \text{(ii) Let the total surface area of the closed cylinderical tank be } \\ S, \text{ given by } S = \frac{539}{r} + 2 \pi r^2. \text{ If the total surface area of the tank is minimum, } \\ \text{then the value of } r \text{ will be: }$

$\displaystyle (a) \ r = 7 \text{ cm } \hspace{1.0cm} (b) \ r = 14 \text{ cm } \hspace{1.0cm}(c) \ r = 49 \text{ cm } \hspace{1.0cm}(d) \ r = \frac{7}{2} \text{ cm }$

$\displaystyle \text{(iii) The height of the tank } h \text{ is equal to:}$

$\displaystyle (a) \ h = 7 \text{ cm } \hspace{1.0cm} (b) \ h = 14 \text{ cm } \hspace{1.0cm}(c) \ h = 28 \text{ cm } \hspace{1.0cm}(d) \ h = 2 \text{ cm }$

$\displaystyle \text{(iv) The minimum total surface area of the tank } S \text{ will be:}$

$\displaystyle (a) \ 231 \text{ sq. cm } \hspace{1.0cm} (b) \ 321 \text{ sq. cm } \hspace{1.0cm}(c) \ 230 \text{ sq. cm } \hspace{1.0cm}(d) \ 221 \text{sq. cm }$

Answer:

$\text{(i) } \ \ \ \fbox{ b }$

$\displaystyle \text{Given radius }= r \text{ and height } = h$

$\displaystyle \text{Given, Volume of the cylinder }= \frac{539}{2} \text{ cubic units}$

$\displaystyle \Rightarrow \hspace{1.0cm} \pi r^2 h = \frac{539}{2}$

$\displaystyle \Rightarrow \hspace{1.0cm} h = \frac{539}{2 \pi r^2}$

$\text{(ii) } \ \ \ \fbox{ d }$

Total surface area,

$\displaystyle S = 2 \pi r^2 h + 2 \pi r^2$

$\displaystyle = 2 \pi r \frac{539}{2 \pi r^2} + 2 \pi r^2$

$\displaystyle = \frac{539}{r} + 2 \pi r^2$

$\displaystyle \text{Now, } \frac{ds}{dr} = \frac{539}{r^2} + 4 \pi r \text{ and } \frac{d^2s}{dr^2} = \frac{1078}{r^3} + 4 \pi$

$\displaystyle \text{Put } \frac{ds}{dr} = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} r^3 = \frac{539 \times 7}{4 \times 22}$

$\displaystyle \Rightarrow \hspace{1.0cm} r^3 = \frac{7 \times 7 \times 7}{2 \times 2 \times 2}$

$\displaystyle \Rightarrow \hspace{1.0cm} r = \frac{7}{2}$

$\displaystyle \text{Here, at } r = \frac{7}{2}, \frac{d^2s}{dr^2} > 0. \text{Hence, minimum. }$

$\text{(iii) } \ \ \ \fbox{ a }$

$\displaystyle \text{As, } h = \frac{539}{2 \pi r^2} = \frac{539}{2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} } = 7 \text{ cm}$

$\text{(iv) } \ \ \ \fbox{ a }$

$\displaystyle \text{Surface area is minimum at } r = \frac{7}{2} \text{ cm }$

$\displaystyle S = \frac{539}{r} + 2 \pi r^2 = \frac{539}{\frac{7}{2}} + 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 154 + 77 = 231 \text{ sq. cm }$

$\\$

SECTION – B [16 Marks]

(Answer all Questions)

$\displaystyle \text{Question 24: If } \overrightarrow{a} = 2 \hat{i} + 4 \hat{j} - \hat{k} \text{ and } \overrightarrow{b} = 3 \hat{i} -2 \hat{j} - \lambda \hat{k} \\ \\ \text{ such that they are perpendicular to each other, then the value of } \lambda , \text{ will be: }$

$\displaystyle (a) \ 2 \hspace{1.0cm} (b) \ -2 \hspace{1.0cm}(c) \ 3 \hspace{1.0cm}(d) \ -3$

Answer:

$\fbox{ a }$

$\displaystyle \text{When two vectors } \overrightarrow{a} , \text{ and } \overrightarrow{b} \text{ are perpendicular, then }$

$\displaystyle \overrightarrow{a} . \overrightarrow{b} = 0$

$\displaystyle \text{Therefore, } (2 \hat{i} + 4 \hat{j} - \hat{k}).(3 \hat{i} -2 \hat{j} - \lambda \hat{k}) = 0$

$\displaystyle \Rightarrow \hspace{1.0cm}6 - 8 + \lambda = 0$

$\displaystyle \Rightarrow \hspace{1.0cm}\lambda = 2$

$\\$

$\displaystyle \text{Question 25: The equation of the line passing through the points } (0, 1, 2) \text{ and } (1, 3, 5) \text{ is: }$

$\displaystyle (a) \ \frac{x-1}{0} = \frac{y-2}{1} = \frac{z-3}{2} \hspace{1.0cm} (b) \ \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \hspace{1.0cm}\\ \\ \\ (c) \ \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} \hspace{1.7cm}(d) \ \frac{x-1}{1} = \frac{y-3}{3} = \frac{z-5}{5}$

Answer:

$\fbox{ c }$

$\displaystyle \text{The equation of a line passing through two points } (x_1, y_1, z_1) \text{ and } (x_2, y_2, z_2) \text{ is given by }$

$\displaystyle \frac{x-x_1}{x_2 - x_1}=\frac{y - y_1}{y_2 - y_1}=\frac{z-z_1}{z_2 - z_1}$

$\displaystyle \text{Therefore the line passing through the two points } (0, 1, 2) \text{ and } (1, 3, 5) \text{ is }$

$\displaystyle \frac{x-0}{1-0} = \frac{y-1}{3-1} = \frac{z-2}{5-2}$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{x}{1}= \frac{y-1}{2} = \frac{z-2}{3}$

$\\$

$\displaystyle \text{Question 26: The direction cosine of a line parallel to } \frac{x-1}{2} = \frac{y+3}{3} = \frac{z-6}{-6} \text{ are: }$

$\displaystyle (a) \ \Big( \frac{-2}{7} , \frac{-2}{7} , \frac{-6}{7} \Big) \hspace{1.0cm} (b) \ \Big( \frac{2}{7} , \frac{-3}{7} , \frac{6}{7} \Big) \hspace{1.0cm}(c) \ \Big( \frac{2}{7} , \frac{3}{7} , \frac{6}{7} \Big) \hspace{1.0cm}(d) \ \Big( \frac{2}{7} , \frac{3}{7} , \frac{-6}{7} \Big)$

Answer:

$\fbox{ d }$

$\displaystyle \text{Given equation: } \frac{x-1}{2} = \frac{y+3}{3} = \frac{z-6}{-6}$

$\displaystyle \text{Direction ratio of this line is } <2, 3, -6>$

$\displaystyle \text{Direction ratio of the line parallel to the given line are also same } < 2, 3, -6>$

$\displaystyle \text{Let the direction cosines of the second line are } < l, m, n >$

$\displaystyle l = \frac{2}{\sqrt{2^2 + 3^2 + (-6)^2}} = \frac{2}{\sqrt{4 + 9 + 36}} = \frac{2}{\sqrt{49}} = \frac{2}{7}$

$\displaystyle \text{Similarly, } m = \frac{3}{7} \text{ and } n = \frac{-6}{7}$

$\\$

$\displaystyle \text{Question 27: The angle between the pair of the lines } \\ \\ \overrightarrow{r} = 3 \hat{i}+ 2 \hat{j} - 4 \hat{k} + \lambda( \hat{i}+ 2 \hat{j} +2 \hat{k}) \text{ and } \overrightarrow{r} = 5 \hat{i} - 2 \hat{k} + \mu( 3\hat{i}+ 2 \hat{j} +6 \hat{k}) \text{ is: }$

$\displaystyle (a) \ \theta = \sin^{-1} \frac{19}{21} \hspace{1.0cm} (b) \ \theta = \cos^{-1} \frac{22}{21} \hspace{1.0cm} \\ \\ \\ (c) \ \theta = \cos^{-1} \frac{19}{20} \hspace{1.0cm}(d) \ \theta = \cos^{-1} \frac{19}{21}$

Answer:

$\fbox{ d }$

$\displaystyle \text{The angle between two lines } \\ \\ \overrightarrow{r_1} = \overrightarrow{a_1} + \gamma \overrightarrow{b_1} \text{ and } \overrightarrow{r_2} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \text{ is } \cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}$

$\displaystyle \therefore \cos \theta = \frac{( \hat{i}+ 2 \hat{j} +2 \hat{k}). ( 3\hat{i}+ 2 \hat{j} +6 \hat{k})}{\sqrt{1^2+2^2+2^2} \sqrt{3^2+2^2+6^2} } = \frac{3+4+12}{\sqrt{9} \times \sqrt{49} } = \frac{19}{21}$

$\displaystyle \theta = \cos^{-1} \frac{19}{21}$

$\\$

$\displaystyle \text{Question 28: Consider the two vectors } \overrightarrow{a} = 3 \hat{i}+ 2 \hat{j} + 4 \hat{k} \text{ and } \overrightarrow{b} = \hat{i} -3 \hat{j} + \hat{k}$

$\displaystyle \text{(i) The vector perpendicular to both } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ will be: }$

$\displaystyle (a) \ 14 \hat{i}+ \hat{j} -12 \hat{k} \hspace{1.0cm} (b) \ 14 \hat{i} - \hat{j} +11 \hat{k} \hspace{1.0cm}(c) \ 14 \hat{i}+ \hat{j} -11 \hat{k} \hspace{1.0cm}(d) \ 14 \hat{i}+ \hat{j} +11 \hat{k}$

$\displaystyle \text{(ii) The unit vector perpendicular to both } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ is: }$

$\displaystyle (a) \ \frac{14 \hat{i}+ \hat{j} -12 \hat{k}}{\sqrt{308}} \hspace{1.0cm} (b) \ \frac{14 \hat{i} - \hat{j} +11 \hat{k}}{\sqrt{318}} \hspace{1.0cm} \\ \\ \\ (c) \ \frac{14 \hat{i}+ \hat{j} -11 \hat{k}}{\sqrt{318}} \hspace{1.0cm}(d) \ \frac{14 \hat{i}+ \hat{j} +11 \hat{k}}{\sqrt{318}}$

$\displaystyle \text{(iii) The value of } |2 \overrightarrow{a}+\overrightarrow{b} | \text{ will be:}$

$\displaystyle (a) \ \sqrt{130} \hspace{1.0cm} (b) \ \sqrt{131} \hspace{1.0cm}(c) \ \sqrt{141} \hspace{1.0cm}(d) \ \sqrt{140}$

$\displaystyle \text{(iv) The area of parallelogram formed by } \overrightarrow{a} \text{ and }\overrightarrow{b} \text{ as its diagonal will be:}$

$\displaystyle (a) \ \frac{1}{2} \sqrt{318} \hspace{1.0cm} (b) \ 2\sqrt{318} \hspace{1.0cm}(c) \ \frac{1}{2} \sqrt{308} \hspace{1.0cm}(d) \ 2\sqrt{308}$

Answer:

$\text{(i) } \ \ \ \fbox{ c }$

$\displaystyle \text{The vector which is perpendicular to both }\overrightarrow{a} \text{ and } \overrightarrow{b} \text{ is } \overrightarrow{a} \times \overrightarrow{b}.$

$\displaystyle \overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 4 \\ 1 & -3 & 1 \end{vmatrix} = \hat{i} (2+12)- \hat{j}(3-4) + \hat{k} (-9-2)$

$\displaystyle \Rightarrow \hspace{1.0cm} \overrightarrow{a} \times \overrightarrow{b} = 14 \hat{i} + \hat{j} -11 \hat{k}$

$\text{(ii) } \ \ \ \fbox{ c }$

$\displaystyle \text{Let the unit vector perpendicular to both } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ is } \widehat{C}$

$\displaystyle \widehat{C} = \frac{\overrightarrow{a} \times \overrightarrow{b}}{|\overrightarrow{a} \times \overrightarrow{b}|} = \frac{14 \hat{i} + \hat{j} -11 \hat{k} }{\sqrt{ 14^2+1^2+11^2 }} = \frac{14 \hat{i} + \hat{j} -11 \hat{k} }{\sqrt{ 318 }}$

$\text{(iii) } \ \ \ \fbox{ b }$

$\displaystyle | 2\overrightarrow{a} + \overrightarrow{b}| = |2(3 \hat{i}+ 2 \hat{j} + 4 \hat{k}) + (\hat{i} -3 \hat{j} + \hat{k}) | \\ \\ = | 6\hat{i}+ 4 \hat{j} + 8 \hat{k} + \hat{i}-3 \hat{j} + \hat{k} | \\ \\ = | 7\hat{i}+ \hat{j} + 9 \hat{k} | \\ \\ = \sqrt{7^2+1^2+9^2} \\ \\ = \sqrt{49+1+81} = \sqrt{131}$

$\text{(iv) } \ \ \ \fbox{ a }$

$\displaystyle \text{Area of parallelogram } = \frac{1}{2} \Big| \overrightarrow{a} \times \overrightarrow{b} \Big| = \frac{1}{2} \sqrt{318} \text{ sq. units }$

$\\$

SECTION – B [16 Marks]

(Answer all Questions)

$\displaystyle \text{Question 29: The cost function of the firm is given by } \\ \\ C(x) = 1500 + 25x + \frac{x^2}{10}. \text{Then the marginal cost of the firm } MC \text{ will be: }$

$\displaystyle (a) \ 1500+ \frac{x}{5} \hspace{1.0cm} (b) \frac{-1500}{x^2}+ \frac{1}{10} \hspace{1.0cm}(c) \ 25-\frac{x}{5} \hspace{1.0cm}(d) \ 25 + \frac{x}{5}$

Answer:

$\fbox{ d }$

$\displaystyle \text{Given, } C(x) = 1500 + 25x + \frac{x^2}{10}$

$\displaystyle \text{Therefore, } MC(x) = \frac{d}{dx} \Bigg( 1500 + 25x + \frac{x^2}{10} \Bigg) = 25 + \frac{2x}{10} = 25 + \frac{x}{5}$

$\\$

$\displaystyle \text{Question 30: } \\ \\ \text{The revenue of a monopolist is given by } R(x) = 120x^2 + 300 - x. \text{ Then, the average } \\ \\ \text{revenue function } AR(x) \text{ at } x=10 \text{ will be: }$

$\displaystyle (a) \ 1229 \hspace{1.0cm} (b) 1500 \hspace{1.0cm}(c) \ 1210 \hspace{1.0cm}(d) \ 12310$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given, } R(x) = 120x^2 + 300 - x$

$\displaystyle \text{Therefore, } AR(x) = \frac{R(x)}{x} = \frac{120x^2 + 300 - x}{x}$

$\displaystyle \text{At, } x = 10,$

$\displaystyle AR(10) = \frac{120 \times 10^2 + 300 - 10}{10} = \frac{12000+300-10}{10} = \frac{12290}{10} = 1229$

$\\$

$\displaystyle \text{Question 31: A company sells its products at the rate of Rs. 10 per unit. The variable } \\ \\ \text{costs are estimated to be 25\% of the total revenue received. If the fixed costs for the } \\ \\ \text{product are Rs. 4500, then the cost function will be: }$

$\displaystyle (a) \ \frac{15}{2} - 4500x \hspace{1.0cm} (b) \frac{15}{x} - 4500\hspace{1.0cm}(c) \ \frac{5x}{2} + 4500 \hspace{1.0cm}(d) \ \frac{25x}{2} - 4500$

Answer:

$\fbox{ c }$

$\displaystyle \text{Let } x \text{ be the number of units sold. }$

$\displaystyle \text{Price of 1 unit is Rs. 10 }$

$\displaystyle \text{Total revenue = Rs. } 10x$

$\displaystyle \text{Cost Function , } C(x) = 4500 + 25\% \text{ of Rs. } 10x$

$\displaystyle = 4500 + \frac{25}{100} \times 10 x$

$\displaystyle = 4500 + \frac{5}{2} x$

$\\$

$\displaystyle \text{Question 32: Let the total cost function be } C(x) = 5x + 350 \\ \\ \text{ and the total revenue function be } R(x) = 50x - x^2 \text{ for a company. Then, } \\ \\ \text{the break-even points will be: }$

$\displaystyle (a) \ -35 \text{ and } 10 \hspace{1.0cm} (b) \ 35 \text{ and } 10 \hspace{1.0cm} \\ \\ (c) \ 35 \text{ and } -10 \hspace{1.0cm}(d) \ -35 \text{ and } -10$