CBSE Paper (All India – 2022) Term I Class 12: Mathematics Solution – ICSE / ISC / CBSE Mathematics Portal for K12 Students

MATHEMATICS

$\displaystyle \text{Series SSJ/2} \hspace{1.0cm} | \hspace{1.0cm} \text{Q. P. Code 065/2/4 } \hspace{1.0cm} | \hspace{1.0cm} \text{Set 4 }$

$\displaystyle \text{Time Allowed : 2 hours} \hspace{5.0cm} \text{Maximum Marks : 40 }$

General Instructions :

This question paper comprises 50 questions out of which 40 questions are to be attempted as per instructions. All questions call equal marks.
This question paper contains three Sections – Section A, B and C.
Section A contains 20 questions. Attempt any 16 questions from Q. No. 1 to 20.
Section B also contains 20 questions. Attempt any 16 questions from Q. No. 21 to 40.
Section C contains 10 questions including one Case Study. Attempt any 8 from Q. No. 41 to 50.
There is only one correct option for every Multiple Choice Question (MCQ). Marks will not be awarded for answering more than one option.
There is no negative marking.

SECTION A

In this section, attempt any 16 questions out of questions 1 – 20. Each question is of one mark.

$\displaystyle \textbf{1: } \text{Differential of } \log [\log ( \log x^5) ] \text{ w.r.t. } x \text{ is }$

$\displaystyle (a) \ \frac{5}{x \log ( x^5) \log ( \log x^5)} \hspace{1.0cm} (b) \ \frac{5}{x \log ( \log x^5) } \hspace{1.0cm} \\ \\ (c) \ \frac{5x^4}{\log (x^5) \log ( \log x^5)} \hspace{1.4cm}(d) \ \frac{5x^4}{\log x^5 \log (\log x^5) }$

Answer:

$\fbox{ a }$

$\displaystyle \text{Let } y = \log [\log ( \log x^5) ]$

$\displaystyle \text{Differentiate w.r.t. } x$

$\displaystyle \frac{dy}{dx} = \frac{1}{\log ( \log x^5)} \times \frac{dy}{dx} \log ( \log x^5)$

$\displaystyle = \frac{1}{\log ( \log x^5)} \times \frac{1}{\log x^5} \times \frac{dy}{dx} \log x^5$

$\displaystyle = \frac{1}{\log ( \log x^5)} \times \frac{1}{\log x^5} \times 5 \frac{dy}{dx} \log x$

$\displaystyle = \frac{1}{\log ( \log x^5)} \times \frac{1}{\log x^5} \times \frac{5}{x}$

$\displaystyle = \frac{5}{x \log x^5 \log ( \log x^5)}$

$\\$

$\displaystyle \textbf{2: } \text{The number of all possible matrices of order } 2 \times 3 \text{ with each entry } 1 \text{ or } 2 \text{ is }$

$\displaystyle (a) \ 16 \hspace{1.0cm} (b) \ 6 \hspace{1.0cm}(c) \ 64 \hspace{1.0cm}(d) \ 24$

Answer:

$\fbox{ c }$

$\displaystyle \text{Let the matrix be } \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}_{2 \times 3}$

$\displaystyle \text{The given matrix of order } 2 \times 3 \text{ has 6 elements and each of these elements can either } \\ \text{be 1 or 2. }$

$\displaystyle \text{Then the 6 elements can be filled in two possible ways. Therefore by multiplication} \\ \text{principle }$

$\displaystyle \text{No. of possible matrices } 2^{2 \times 3} = 2^6 = 64$

$\\$

$\displaystyle \textbf{3: } \text{The function } f: R \rightarrow R \text{ is defined as } f(x) = x^3 + 1 . \text{ Then the function has }$

$\displaystyle (a) \ \text{no minimum value} \hspace{1.0cm} \\ (b) \ \text{no maximum value} \hspace{1.0cm} \\ (c) \ \text{both maximum and minimum value } \hspace{1.0cm} \\ (d) \ \text{neither maximum value nor minimum value }$

Answer:

$\fbox{ d }$

$\displaystyle \text{Given that } f(x) = x^3 + 1$

$\displaystyle \text{Then } f'(x) = 3x^2$

$\displaystyle \text{For critical point we have }$

$\displaystyle 3x^2 = 0 \Rightarrow x = 0$

$\displaystyle \text{Now } II^{nd} \text{ derivative test at point } x = 0, \text{ we have }$

$\displaystyle f''(x) = 6x$

$\displaystyle f''(x)\Big|_{x=0 }= 6 \times 0 = 0$

$\displaystyle \text{Zero is neither positive not negative. } \\ \text{Therefore function has neither maximum value not minimum value. }$

$\\$

$\displaystyle \textbf{4: } \text{If } \sin y = x \cos ( a+y) , \text{ then } \frac{dy}{dx} \text{ is }$

$\displaystyle (a) \ \frac{\cos a}{\cos^2 (a+y)} \hspace{1.0cm} (b) \ \frac{-\cos a}{\cos^2 (a+y)} \hspace{1.0cm}(c) \ \frac{\cos a}{\sin^2 y} \hspace{1.0cm}(d) \ \frac{-\cos a}{\sin^2 y}$

Answer:

$\fbox{ a }$

$\displaystyle \sin y = x \cos ( a+y)$

$\displaystyle \Rightarrow x = \frac{\sin y}{\cos (a+y)}$

$\displaystyle \text{Differentiate both sides w.r.t. } y$

$\displaystyle \frac{dy}{dx}= \frac{\cos (a+y) \frac{d}{dy} \sin y - \sin y \frac{d}{dy} \cos (a+y)}{[ \cos (a+y) ]^2}$

$\displaystyle = \frac{\cos (a+y) \cos y - \sin y [-\sin (a+y)]}{[ \cos (a+y) ]^2}$

$\displaystyle = \frac{\cos (a+y) \cos y + \sin y \sin (a+y)}{\cos^2 (a+y) }$

$\displaystyle = \frac{\cos( a+y-y)}{\cos^2 (a+y) }$

$\displaystyle = \frac{\cos a}{\cos^2 (a+y) }$

$\\$

$\displaystyle \textbf{5: } \text{The points on the curve } \frac{x^2}{9}+\frac{y^2}{25} = 1 , \text{ where tangent is parallel to x-axis are }$

$\displaystyle (a) \ (\pm 5, 0) \hspace{1.0cm} (b) \ (0, \pm 5) \hspace{1.0cm}(c) \ (0, \pm, 3) \hspace{1.0cm}(d) \ (\pm 3, 0)$

Answer:

$\fbox{ b }$

$\displaystyle \text{Given curve }$

$\displaystyle \frac{x^2}{9}+\frac{y^2}{25} = 1 \ldots \ldots (i)$

$\displaystyle \Rightarrow \frac{y^2}{25} = 1 - \frac{x^2}{9}$

$\displaystyle \text{Differentiate both sides, we have }$

$\displaystyle \frac{1}{25} . 2y \frac{dy}{dx} = - \frac{2x}{9}$

$\displaystyle \Rightarrow \frac{dy}{dx} = - \frac{2x}{9} \times \frac{25}{2y} = -\frac{25x}{9y}$

$\displaystyle \text{Since tangent is parallel to x-axis }$

$\displaystyle \frac{dy}{dx} = 0 \Rightarrow - \frac{25}{9y} = 0 \Rightarrow x = 0$

$\displaystyle \text{Putting } x = 0 \text{ in equation (i) we get }$

$\displaystyle \frac{y^2}{25} = 1 \Rightarrow y^2 = 25 \Rightarrow y = \pm 5$

$\displaystyle \text{Therefore the required points are (0, 5) and (0, -5) }$

$\\$

$\displaystyle \textbf{6: } \text{Three points } P(2x, x+3), Q(0, x) \text{ and } R(x+3, x+6) \text{ are collinear, then } x \text{ is equal to }$

$\displaystyle (a) \ 0 \hspace{1.0cm} (b) \ 2 \hspace{1.0cm}(c) \ 3 \hspace{1.0cm}(d) \ 1$

Answer:

$\fbox{ c }$

$\displaystyle \text{If } P(2x, x+3), Q(0, x) \text{ and } R(x+3, x+6) \text{ are collinear then the area of the } \triangle \text{ will be zero. }$

$\displaystyle \text{Area of }\triangle = \frac{1}{2} [ x_1 ( y_2-y_3) + x_2(y_3 - y_1) + x_3( y_1-y_2) ]$

$\displaystyle = \frac{1}{2} [ 2x(x-x-3) + 0(x+6-x-3) + (x+3)(x+3 - x) ]$

$\displaystyle = \frac{1}{2} [ -6x+ 3x+9]$

$\displaystyle = \frac{1}{2} [9 - 3x]$

$\displaystyle \text{Since the points are collinear we have } 9-3x=0 \Rightarrow x = 3$

$\\$

$\displaystyle \textbf{7: } \text{The principal value of } \cos^{-1} \Big( \frac{1}{2} \Big) + \sin^{-1} \Big( - \frac{1}{\sqrt{2}} \Big) \text{ is }$

$\displaystyle (a) \ \frac{\pi}{12} \hspace{1.0cm} (b) \ \pi \hspace{1.0cm}(c) \ \frac{\pi}{3} \hspace{1.0cm}(d) \ \frac{\pi}{6}$

Answer:

$\fbox{ a }$

$\displaystyle \cos^{-1} \Big( \frac{1}{2} \Big) + \sin^{-1} \Big( - \frac{1}{\sqrt{2}} \Big)$

$\displaystyle = \cos^{-1} \Big( \cos \frac{\pi}{3} \Big) - \sin^{-1} \Big( \frac{1}{\sqrt{2}} \Big)$

$\displaystyle = \cos^{-1} \Big( \cos \frac{\pi}{3} \Big) - \sin^{-1} \Big(\sin \frac{\pi}{4} \Big)$

$\displaystyle = \frac{\pi}{3} - \frac{\pi}{4}$

$\displaystyle = \frac{\pi}{12}$

$\\$

$\displaystyle \textbf{8: } \text{If } (x^2+y^2)^2 = xy, \text{ then } \frac{dy}{dx} \text{ is }$

$\displaystyle (a) \ \frac{y + 4x ( x^2 + y^2)}{4y(x^2+y^2) - x} \hspace{1.0cm} (b) \ \frac{y - 4x ( x^2 + y^2)}{x + 4 ( x^2 + y^2)} \hspace{1.0cm} \\ \\ \\ (c) \ \frac{y - 4x ( x^2 + y^2)}{4y(x^2+y^2) - x} \hspace{1.0cm}(d) \ \frac{4y(x^2+y^2)-x}{y - 4x ( x^2 + y^2)}$

Answer:

$\fbox{ c }$

$\displaystyle (x^2+y^2)^2 = xy$

$\displaystyle \Rightarrow x^4 + y^4 + 2x^2 y^2 = xy$

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

$\displaystyle 4x^3 + 4y^3 \frac{dy}{dx} + 2 \Bigg[ x^2 . 2y \frac{dy}{dx} + y^2.2x \Bigg] = x \frac{dy}{dx} + y$

$\displaystyle \Rightarrow 4x^3 + 4y^3 \frac{dy}{dx} + 4x^2 . y \frac{dy}{dx} + 4xy^2 = x \frac{dy}{dx} + y$

$\displaystyle \Rightarrow \frac{dy}{dx} ( 4y^3 + 4x^2 y - x) = y - 4x^3 - 4xy^2$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y - 4x^3 - 4xy^2}{4y^3 + 4x^2 y - x}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{y-4x(x^2+y^2)}{4y(x^2+y^2) - x}$

$\\$

$\displaystyle \textbf{9: } \text{If a matrix A is both symmetric and skew symmetric, then A is necessarily a}$

$\displaystyle (a) \ \text{Diagonal matrix} \hspace{1.0cm} (b) \ \text{Zero square matrix} \hspace{1.0cm} \\ (c) \ \text{Square matrix} \hspace{1.4cm}(d) \ \text{Identity matrix}$

Answer:

$\fbox{ b }$

$\displaystyle \text{Given that matrix A is both symmetric and skew symmetric }$

$\displaystyle \text{i.e. } A' = A \text{ and } A' = - 1$

$\displaystyle \text{From above we get } \\ \\ A = - A \Rightarrow 2A = 0 \Rightarrow A = 0$

$\displaystyle \text{Therefore A is a zero matrix. }$

$\\$

$\displaystyle \textbf{10: } \text{Let set } X = \{ 1, 2, 3 \} \text{ and a relation } R \text{ is defined in } X \text{ as: } R = \{ (1, 3), (2, 2), (3, 2) \}, \\ \text{ then minimum ordered pairs which should be added in relation } R \text{ to make it reflexive} \\ \text{and symmetric are }$

$\displaystyle (a) \ \{ (1, 1), (2, 3), (1, 2) \} \hspace{2.0cm} (b) \ \{ (3, 3), (3, 1), (1, 2) \} \hspace{1.0cm}\\ \\ (c) \ \{ (1, 1), (3, 3), (3, 1), (2, 3) \} \hspace{1.0cm}(d) \ \{ (1, 1), (3, 3), (3, 1), (1, 2) \}$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given that set } X = \{ 1, 2, 3 \}$

$\displaystyle R \text{ is reflexive since } (1, 1), (2, 2) \text{ and } ( 3, 3) \text{ lie in } R.$

$\displaystyle \text{Also, } R \text{ is symmetric if }$

$\displaystyle (1, 3) \in X \Rightarrow (3, 1) \in X \Rightarrow (3, 2) \in X \Rightarrow (2, 3) \in X$

$\displaystyle \text{Therefore } R = \{ (1,1), (2,2), (3,3), (1, 3), (3, 1), ( 3,2 ), (2, 3) \}$

$\\$

$\displaystyle \textbf{11: } \text{A linear programing problem is as follows:} \\ \text{Minimize} \hspace{3.5cm} Z = 2x+ y \\ \text{Subject to the constraints } \ \ \ x \geq 3, \ \ x \leq 9, \ \ y \geq 0, x-y \geq 0 , \ \ x+y \leq 14$

The feasible region has

$\displaystyle (a) \ \text{5 corner points including } (0, 0 ) \text{ and } (9, 5) \\ (b) \ \text{5 corner points including } (7, 7 ) \text{ and } (3, 3) \\ (c) \ \text{5 corner points including } (14, 0 ) \text{ and } (9, 0) \\ (d) \ \text{5 corner points including } (3, 0 ) \text{ and } (9, 5)$

Answer:

$\fbox{ b }$

$\displaystyle \text{Objective function: } Z = 2x+ y$

$\displaystyle \text{Constraints: } x \geq 3, \ \ x \leq 9, \ \ y \geq 0, x-y \geq 0 , \ \ x+y \leq 14$

Consider the line

$\displaystyle x-y = 0 \ldots \ldots (i)$

$\displaystyle x+y = 14 \ldots \ldots (ii)$

$\displaystyle \begin{array}{ |c|c|c|c|} \hline x & 0 & 1 & 2 \\ \hline y & 0 & 1 & 2 \\ \hline \end{array} \ \ \ \begin{array}{ |c|c|c|} \hline x & 0 & 14 \\ \hline y & 14 & 0 \\ \hline \end{array}$

$\displaystyle \text{From equation (i) and equation (ii) we get } x = 7 \text{ and } y = 7$

$\displaystyle \text{Therefore feasible region has five corners, } \text{ (3, 0), (9, 0), (9, 5), (7, 7) } \text{ and } (3, 3) \\ \text{respectively. }$

$\\$

$\displaystyle \textbf{12: } \text{The function } f(x) = \Bigg\{ \begin{array}{cc} \frac{e^{3x} -e^{-5x}}{x}, & \text{if } x \neq 0 \\ & \\ k, & \text{if } x= 0 \end{array} \ \ \ \text{is continuous at } x=0 \\ \text{ for the value of } k, \text{ is }$

$\displaystyle (a) \ 3 \hspace{1.0cm} (b) \ 5 \hspace{1.0cm}(c) \ 2 \hspace{1.0cm}(d) \ 8$

Answer:

$\fbox{ d }$

$\displaystyle f(x) = \Bigg\{ \begin{array}{cc} \frac{e^{3x} -e^{-5x}}{x}, & \text{if } x \neq 0 \\ & \\ k, & \text{if } x= 0 \end{array}$

$\displaystyle \text{Given that } f(0) = k$

$\displaystyle \text{Also given that function is continuous at } x=0$

$\displaystyle \text{Therefore, RHL } = f(0)^+ \Rightarrow \lim \limits_{x\to 0^+} \frac{e^{3x} -e^{-5x}}{x} \Rightarrow \frac{1-1}{0} = \frac{0}{0}$

$\displaystyle \lim \limits_{x\to 0^+} \frac{e^{3x} -e^{-5x}}{x} = \lim \limits_{x\to 0^+} \Bigg( 3 \Big\{ \frac{e^{3x} -1}{3x} \Big\} - 5 \Big\{ \frac{e^{-5x} -1}{(-5x)} \Big\}\Bigg) = 3 \times 1 + 5 \times 1 = 8$

$\displaystyle \text{Therefore } k = 8$

$\\$

$\displaystyle \textbf{13: } \text{If } C_y \text{ denotes the cofactor of the element } p_{ij} \text{ of the matrix } p = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & 2 & 4 \end{bmatrix} , \\ \text{ then the value of } C_{31} . C_{23} \text{ is }$

$\displaystyle (a) \ 5 \hspace{1.0cm} (b) \ 24 \hspace{1.0cm}(c) \ -24 \hspace{1.0cm}(d) \ 5$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given matrix, } p = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & 2 & 4 \end{bmatrix}$

$\displaystyle \text{Now, cofactor, } C_{31} = (-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} = 1 ( 3-4) = -1$

$\displaystyle \text{and, cofactor, } C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = -1 ( 2+3) = -5$

$\displaystyle \text{Therefore } C_{31} \times C_{23} = (-1) \times (-5) = 5$

$\\$

$\displaystyle \textbf{14: } \text{The function } y=x^2 e^{-x} \text{ is decreasing in the interval }$

$\displaystyle (a) \ (0, 2) \hspace{1.0cm} (b) \ (2, \infty ) \hspace{1.0cm}(c) \ (- \infty , 0) \hspace{1.0cm}(d) \ (- \infty , 0) \cup (2, \infty)$

Answer:

$\fbox{ d }$

$\displaystyle y = x^2 e^{-x}$

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

$\displaystyle \frac{dy}{dx} = e^{-x} x (2 - x)$

$\displaystyle \text{For decreasing}$

$\displaystyle \frac{dy}{dx} < 0$

$\displaystyle e^{-x} x (2 - x) < 0$

$\displaystyle e^{-x} \text{ is always positive for } \forall x \in R$

$\displaystyle \text{For } x(2-x) = 0 \Rightarrow x = 0 \text{ or } x = 2$

$\displaystyle \text{Therefore, the point } x = 0 \text{ and } x = 2 \text{ divide the real line into three disjoint intervals. }$

$\displaystyle \text{For } -\infty < x < 0, \frac{dy}{dx} = (-)(+) = (-) [\text{ negative }]$

$\displaystyle \text{For } 0 < x < 2, \frac{dy}{dx} = (+)(+) = (+) [\text{ positive }]$

$\displaystyle \text{For } 2 < x < \infty, \frac{dy}{dx} = (+)(-) = (-) [\text{ negative }]$

$\displaystyle \text{Therefore function is decreasing in the interval } (-\infty , 0) \cup (2, \infty)$

$\\$

$\displaystyle \textbf{15: } \text{If } R = \{ (x, y): x, y \in Z, \ \ x^2+y^2 \leq 4 \} \text{ is a relation in set } Z, \text{ then domain of } R \text{ is }$

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

Answer:

$\fbox{ b }$

$\displaystyle \text{We have } R = \{ (x, y): x, y \in Z, x^2+y^2 \leq 4 \}$

$\displaystyle \text{When } x = 0$

$\displaystyle x^2 + y^2 \leq 4 \Rightarrow y^2 \leq 4 \Rightarrow y = 0, \pm 1, \pm 2$

$\displaystyle \text{When } x = \pm 1$

$\displaystyle x^2 + y^2 \leq 4 \Rightarrow 1 + y^2 \leq 4 \Rightarrow y^2 \leq 3 \Rightarrow y = 0, \pm 1$

$\displaystyle \text{When } x = \pm 2$

$\displaystyle x^2 + y^2 \leq 4 \Rightarrow 4+ y^2 \leq 4 \Rightarrow y^2 \leq 0 \Rightarrow y = 0$

$\displaystyle \text{Therefore } \\ R = \{ (0, 0), (0, -1), (0, 1), (0, -2), (0, 2), (-1, 0), (1, 0), (1, 1), (-1, 1), (-1, -1), (2, 0), (-2, 0) \}$

$\displaystyle \text{Hence, Domain of } R= \{ x : (x, y) \in R \} = \{ 0, -1, 1, -2, 2 \}$

$\\$

$\displaystyle \textbf{16: } \text{The system of linear equations } 5x+ky = 5, 3x+3y=5 \text{ will be consistent }$

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

Answer:

$\fbox{ d }$

Given,

$\displaystyle 5x+ky = 5 \ldots \ldots (i)$

$\displaystyle 3x+3y=5 \ldots \ldots (ii)$

$\displaystyle \text{By rule of consistency } \frac{5}{3}\neq \frac{k}{3}$

$\displaystyle 3k \neq 15 \Rightarrow k \neq 5$

$\\$

$\displaystyle \textbf{17: } \text{The equation of the tangent to the curve } y ( 1 + x^2) = 2-x, \text{ where it cross the} \\ \text{ x-axis is }$

$\displaystyle (a) \ x-5y=2 \hspace{1.0cm} (b) \ 5x-y=2 \hspace{1.0cm}(c) \ x+5y=2 \hspace{1.0cm}(d) \ 5x+y=2$

Answer:

$\fbox{ c }$

$\displaystyle \text{Equation of the curve: } y ( 1 + x^2) = 2-x$

$\displaystyle \text{Curve crosses the x axis which means y coordinate is 0. }$

$\displaystyle \text{Hence, } (1+x^2) \times 0 = 2- x \Rightarrow 0 = 2 - x \Rightarrow x = 2$

$\displaystyle \text{Therefore, the curve passes through the point } (2, 0).$

$\displaystyle \text{Equation of the curve, }$

$\displaystyle y(1+x^2) = 2 - x$

$\displaystyle y + y x^2 = 2 - x$

$\displaystyle \frac{dy}{dx} + x^2 \frac{dy}{dx} + y . 2x = - 1$

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

$\displaystyle \text{Slope at point } (2, 0)$

$\displaystyle \frac{dy}{dx} \Bigg]_{(2, 0)} = \frac{-(1+ 2 \times 2 \times 0)}{1 + (2)^2} \Rightarrow -\frac{1}{1+4}$

$\displaystyle \text{or } \frac{dy}{dx} \Bigg]_{(2, 0)} = - \frac{1}{5}$

$\displaystyle \text{Equation of tangent }$

$\displaystyle y - y_1 = \frac{dy}{dx} \Bigg]_{(2, 0)} (x - x_1) \Rightarrow y - 0 = - \frac{1}{5} (x-2)$

$\displaystyle 5y = - x + 2$

$\displaystyle \text{or } x + 5y = 2$

$\\$

$\displaystyle \textbf{18: } \text{If } \begin{bmatrix} 3c+6 & a-d \\ a+d & 2-3b \end{bmatrix} = \begin{bmatrix} 12 & 2 \\ -8 &-4 \end{bmatrix} \text{ are equal then the value of } ab - cd \text{ is }$

$\displaystyle (a) \ 4 \hspace{1.0cm} (b) \ 16 \hspace{1.0cm}(c) \ -4 \hspace{1.0cm}(d) \ -16$

Answer:

$\fbox{ a }$

$\displaystyle \begin{bmatrix} 3c+6 & a-d \\ a+d & 2-3b \end{bmatrix} = \begin{bmatrix} 12 & 2 \\ -8 &-4 \end{bmatrix}$

By equality of property of matrix

$\displaystyle 3c+6 = 12 \ldots \ldots (i)$

$\displaystyle a-d=2 \ldots \ldots (ii)$

$\displaystyle a+d=-8 \ldots \ldots (iii)$

$\displaystyle 2-3b=-4 \ldots \ldots (iv)$

$\displaystyle \text{From equation (i) } c = 2$

$\displaystyle \text{From equation (ii) and equation (iii) } a = - 3 \text{ and } d = - 5$

$\displaystyle \text{From equation (iv) } b = 2$

$\displaystyle \text{Therefore, } ab - cd = ( -3)(2) - (2)(- 5) = - 6 + 0 = 4$

$\\$

$\displaystyle \textbf{19: } \text{The principal value of } \tan^{-1} \Big( \tan \frac{9\pi}{8} \Big) \text{ is }$

$\displaystyle (a) \ \frac{\pi}{8} \hspace{1.0cm} (b) \ \frac{3\pi}{8} \hspace{1.0cm}(c) \ -\frac{\pi}{8} \hspace{1.0cm}(d) \ -\frac{3\pi}{8}$

Answer:

$\fbox{ a }$

$\displaystyle \tan^{-1} \Big( \tan \frac{9\pi}{8} \Big) = \tan^{-1} \Bigg[ \tan \Big( \pi + \frac{\pi}{8} \Big) \Bigg] = \tan^{-1} \Bigg[ \tan \frac{\pi}{8} \Bigg] = \frac{\pi}{8}$

$\\$

$\displaystyle \textbf{20: } \text{For two matrices } P = \begin{bmatrix} 2 & 3 \\ -3 & 0 \\ 0 & 1 \end{bmatrix} \text{ and } Q^T = \begin{bmatrix} -1 & 2 & 1 \\ 1 &2 & 3 \end{bmatrix} \\ P-Q \text{ is }$

$\displaystyle (a) \ \begin{bmatrix} 2 & 3 \\ -3 & 0 \\ 0 & -3 \end{bmatrix} \hspace{1.0cm} (b) \ \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix} \hspace{1.0cm}(c) \ \begin{bmatrix} 4 & 3 \\ 0 & -3 \\ -1 & -2 \end{bmatrix} \hspace{1.0cm}(d) \ \begin{bmatrix} 2 & 3 \\ 0 & -3 \\ 0 & -3 \end{bmatrix}$

Answer:

$\fbox{ b }$

$\displaystyle \text{Transpose of } (Q^T) = \begin{bmatrix} -1 & 2 & 1 \\ 1 &2 & 3 \end{bmatrix}^T$

$\displaystyle \therefore Q = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$

$\displaystyle \text{Therefore } P - Q = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \Rightarrow \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$

$\\$

SECTION B

In this section, attempt any 16 questions out of questions 1 – 20. Each question is of one mark.

$\displaystyle \textbf{21: } \text{The function } f(x) = 2x^3- 15x^2 + 36x + 6 \text{ is increasing in the interval }$

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

Answer:

$\fbox{ c }$

$\displaystyle f(x) = 2x^3 - 15 x^2 + 36 x + 6$

$\displaystyle f'(x) = 6x^2 - 30x+ 36$

$\displaystyle \text{For increasing } f'(x) \geq 0$

$\displaystyle 6x^2 - 30x+ 36 \geq 0$

$\displaystyle \Rightarrow 6 [ x^2 - 5x + 6] \geq 0$

$\displaystyle \Rightarrow 6 [ x^2 - 3x - 2x + 6] \geq 0$

$\displaystyle \Rightarrow 6 [ (x-2)(x-3)] \geq 0$

$\displaystyle \text{When } x \leq 2, f'(x) \text{ is +ve }$

$\displaystyle \text{When } x \geq 3, f'(x) \text{ is +ve }$

$\displaystyle \text{Therefore, for increasing. Interval is } (-\infty, 2] \cup [3, \infty )$

$\\$

$\displaystyle \textbf{22: } \text{If } x = 2 \cos \theta - \cos 2\theta \text{ and } y = 2 \sin \theta - \sin 2\theta, \text{ then } \frac{dy}{dx} \text{ is }$

$\displaystyle (a) \ \frac{\cos \theta + \cos 2\theta}{\sin \theta - \sin 2\theta} \hspace{1.0cm} (b) \ \frac{\cos \theta - \cos 2\theta}{\sin 2\theta - \sin \theta} \hspace{1.0cm} \\ \\ (c) \ \frac{\cos \theta - \cos 2\theta}{\sin \theta - \sin 2\theta} \frac{}{} \hspace{1.0cm}(d) \ \frac{\cos 2\theta - \cos \theta}{\sin 2\theta + \sin \theta} \frac{}{}$

Answer:

$\fbox{ b }$

$\displaystyle x = 2 \cos \theta - \cos 2\theta$

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

$\displaystyle \frac{dx}{d \theta} = 2(- \sin \theta) - ( - \sin 2 \theta . 2) = - 2 \sin \theta + 2 \sin 2 \theta = 2 \sin 2 \theta - 2 \sin \theta$

$\displaystyle \text{Now, } y = 2 \sin \theta - \sin 2\theta$

$\displaystyle \frac{dy}{d \theta} = 2 \cos \theta - \cos 2 \theta . 2 = 2 \cos \theta - 2 \cos 2 \theta$

$\displaystyle \text{Therefore } \frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}} = \frac{2 \cos \theta - 2 \cos 2 \theta}{2 \sin 2 \theta - 2 \sin \theta}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{\cos \theta - \cos 2 \theta}{\sin 2 \theta - \sin \theta}$

$\\$

$\displaystyle \textbf{23: } \text{What is the domain of the function} \cos^{-1} (2x-3) ?$

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

Answer:

$\fbox{ d }$

$\displaystyle \text{We know that domain of } \cos^{-1} x \text{ is } [-1, 1]$

$\displaystyle \text{For domain of } \cos^{-1} (2x-3) \text{ is the set of all values of x which satisfy }$

$\displaystyle -1 \leq 2x-3 \leq 1$

$\displaystyle -1 + 3 \leq 2x-3 + 3 \leq 1 + 3$

$\displaystyle \text{add 3 in equality }$

$\displaystyle 2 \leq 2x \leq 4$

$\displaystyle \text{or } 1 \leq x \leq 2$

$\displaystyle \text{Hence the domain of } \cos^{-1} (2x-3) \text{ is } [1, 2]$

$\\$

$\displaystyle \textbf{24: } \text{A matrix } A = [a_{ij}]_{3 \times 3} \text{ is defined by } a_{ij} = \Bigg\{ \begin{array}{cc} 2i+3j, & i < j \\5, & i = j \\3i-2j, & i > j \end{array} \\ \text{The number of elements in A which are more than 5, is }$

$\displaystyle (a) \ 3 \hspace{1.0cm} (b) \ 4 \hspace{1.0cm}(c) \ 5 \hspace{1.0cm}(d) \ 6$

Answer:

$\fbox{ b }$

$\displaystyle A = [a_{ij}]_{3 \times 3} \Rightarrow \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$

$\displaystyle \text{Given, } a_{ij} = \Bigg\{ \begin{array}{cc} 2i+3j, & i < j \\5, & i = j \\3i-2j, & i > j \end{array}$

From the defined condition for elements, we have

$\displaystyle a_{11} \ \ \ (i=j) \hspace{0.5cm} \Rightarrow a_{11}=5$

$\displaystyle a_{12} \ \ \ (i < j) \hspace{0.5cm} \Rightarrow a_{12} = 2 \times 1 + 3 \times 2 =8$

$\displaystyle a_{13} \ \ \ (i < j) \hspace{0.5cm} \Rightarrow a_{13} = 2 \times 1 + 3 \times 3 =11$

$\displaystyle a_{21} \ \ \ (i > j) \hspace{0.5cm} \Rightarrow a_{21} = 3 \times 2 - 2 \times 1 =4$

$\displaystyle a_{22} \ \ \ (i=j) \hspace{0.5cm} \Rightarrow a_{22}=5$

$\displaystyle a_{23} \ \ \ (i < j) \hspace{0.5cm} \Rightarrow a_{23} = 2 \times 2 + 3 \times 3 =13$

$\displaystyle a_{31} \ \ \ (i > j) \hspace{0.5cm} \Rightarrow a_{31} = 3 \times 3 - 2 \times 1 =7$

$\displaystyle a_{32} \ \ \ (i > j) \hspace{0.5cm} \Rightarrow a_{32} = 3 \times 3 - 2 \times 2 =5$

$\displaystyle a_{33} \ \ \ (i=j) \hspace{0.5cm} \Rightarrow a_{33}=5$

$\displaystyle \text{Therefore } A = \begin{bmatrix} 5 & 8 & 11 \\ 4 & 5 & 13 \\ 7 & 5 & 5 \end{bmatrix}$

$\displaystyle \text{Hence, 4 elements are in A which are more than 5. }$

$\\$

$\displaystyle \textbf{25: } \text{If a function} f \text{ is defined by } f(x) = \Bigg\{ \begin{array}{cc} \frac{k \cos x}{\pi - 2x}, & \text{ if } x \neq \frac{\pi}{2} \\ & \\3, & \text{ if } x = \frac{\pi}{2} \end{array} \\ \text{ is continuous at } x = \frac{\pi}{2}, \text{then the value of } k \text{ is }$

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

Answer:

$\fbox{ c }$

$\displaystyle f(x) = \Bigg\{ \begin{array}{cc} \frac{k \cos x}{\pi - 2x}, & \text{ if } x \neq \frac{\pi}{2} \\ & \\3, & \text{ if } x = \frac{\pi}{2} \end{array}$

$\displaystyle \text{Given: } f \Big(\frac{\pi}{2} \Big) = 3$

$\displaystyle \text{Function is continuous at } x = \frac{\pi}{2}, \text{ so }$

$\displaystyle \text{LHL = RHL = } f \Big(\frac{\pi}{2} \Big)$

$\displaystyle \text{Now, LHL } \Rightarrow \lim \limits_{x \to \frac{\pi^-}{2}} \frac{k \cos x}{\pi - 2x}$

$\displaystyle \text{Put } x = \Big( \frac{\pi}{2} -h\Big)$

$\displaystyle \lim \limits_{h \to 0} \frac{k \cos \Big( \frac{\pi}{2} -h\Big)}{\pi - 2\Big( \frac{\pi}{2} -h\Big)}$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{k \sin h}{\pi - \pi + 2h }$

$\displaystyle = \lim \limits_{h \to 0} \frac{k}{2} \ \frac{ \sin h}{h }$

$\displaystyle = \frac{k}{2}$

$\displaystyle \text{Now, RHL } \Rightarrow \lim \limits_{x \to \frac{\pi^+}{2}} \frac{k \cos x}{\pi - 2x}$

$\displaystyle \text{Put } x = \Big( \frac{\pi}{2} +h\Big)$

$\displaystyle \lim \limits_{h \to 0} \frac{k \cos \Big( \frac{\pi}{2} +h\Big)}{\pi - 2\Big( \frac{\pi}{2} +h\Big)}$

$\displaystyle \Rightarrow \lim \limits_{h \to 0} \frac{k \sin h}{-2h }$

$\displaystyle = \frac{k}{2}$

$\displaystyle \text{Hence } k = 3$

$\\$

$\displaystyle \textbf{26: } \text{For the matrix } X = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} , (X^2 - X) \text{ is }$

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

Answer:

$\fbox{ a }$

$\displaystyle X = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix}$

$\displaystyle X^2 = X.X = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} . \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}$

$\displaystyle Now, X^2 - X = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} = 2\begin{bmatrix} 1 & 0 & 0 \\ 0& 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 2I$

$\\$

$\displaystyle \textbf{27: } \text{Let } X = \{ x^2 : x \in N \} \text{ and the function } f: N \rightarrow X \text{ is defined by } \\ f(x) = x^2 , x \in N. \text{ Then this function is }$

$\displaystyle (a) \ \text{ injective only } \hspace{1.0cm} (b) \ \text{ not bijective } \hspace{1.0cm} \\ (c) \ \text{ surjective only } \hspace{1.0cm}(d) \ \text{ bijective }$

Answer:

$\fbox{ d }$

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

$\displaystyle \Rightarrow {x_1}^2= {x_2}^2$

$\displaystyle \Rightarrow x_1= x_2$

$\displaystyle \therefore f(x) \text{ is invective }$

$\displaystyle \because f:N \rightarrow X$

$\displaystyle \text{Angle range of }$

$\displaystyle f(x) = x^2, x \in N \text{ is } X$

$\displaystyle \therefore \text{ Range }= \text{ codomain }$

$\displaystyle \text{Hence, it is surjective also. }$

$\displaystyle \text{Hence, it is bijective. }$

$\\$

$\displaystyle \textbf{28: } \text{The corner points of the feasible region for a linear programming problem are} \\ P(0, 5), Q (1, 5), R(4, 2) \text{ and } S(12, 0). \text{ The minimum value of the objective function } \\ Z = 2x + 5y \text{ is at the point }$

$\displaystyle (a) \ P \hspace{1.0cm} (b) \ Q \hspace{1.0cm}(c) \ R \hspace{1.0cm}(d) \ S$

Answer:

$\fbox{ c }$

$\displaystyle \begin{array}{ |c|c|} \hline \text{Coner Points} & \text{Objective Function } Z = 2x + 5y\\ \hline P(0, 5) & Z=2 \times 0+5 \times 5 = 0 + 25 = 25 \\ \hline Q(1, 5) & Z=2 \times 1+5 \times 5 = 1 + 25 = 27 \\ \hline R(4, 2)& Z=2 \times 4+5 \times 2 = 8 + 10 = 18\\ \hline S(12, 0) & Z=2 \times 12+5 \times 0 = 24 + 0 = 24 \\ \hline \end{array}$

Therefore, minimum value will be at point R.

$\\$

$\displaystyle \textbf{29: } \text{The equation of the normal to the curve } ay^2 = x^3 \text{ at the point } (am^2, am^3) \text{ is }$

$\displaystyle (a) \ 2y-3mx+am^3=0 \hspace{2.0cm} (b) \ 2x+3my-3am^4-am^2 = 0 \hspace{1.0cm} \\ \\ (c) \ 2x+ 3my + 3am^4 - 2am^2 \hspace{1.0cm}(d) \ 2x+ 3my- 3am^4- 2am^2$

Answer:

$\fbox{ d }$

$\displaystyle \text{Equation of curve: } ay^2 = x^3$

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

$\displaystyle a \cdot 2y \frac{dy}{dx} = 3x^2 \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ay}$

$\displaystyle \text{Slope of tangent at point } (am^2, am^3)$

$\displaystyle \frac{dy}{dx} \Bigg]_{(am^2, \ am^3)} = \frac{3(am^2)^2}{2a(am^3)} = \frac{3m}{2}$

$\displaystyle \text{Slope of normal} = -\frac{1}{\text{Slope of tangent}} = - \frac{1}{\frac{3m}{2}} = -\frac{2}{3m}$

$\displaystyle \text{Therefore, equation of normal is }$

$\displaystyle (y - am^3) = - \frac{2}{3m} (x-am^2)$

$\displaystyle \Rightarrow 3my - 3am^4 = - 2x + 2am^2$

$\displaystyle \Rightarrow 2x + 3my - 3am^4 - 2am^2 = 0$

$\\$

$\displaystyle \textbf{30: } \text{If} A \text{ is a square matrix of order 3 and } |A|= - 5 \text{ then } |adj A | \text{ is }$

$\displaystyle (a) \ 125 \hspace{1.0cm} (b) \ -25 \hspace{1.0cm}(c) \ 25 \hspace{1.0cm}(d) \ \pm 25$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given, } n = 3 , |A| = - 5$

$\displaystyle \text{We know that } |Adj \ A| = |A|^{n-1}, \text{ where n is the order of the matrix } \\ \\ = (-5)^{3-1} = (-5)^2 = 25$

$\displaystyle \text{Therefore, } |Adj \ A|= 25$

$\\$

$\displaystyle \textbf{31: } \text{The simplest form of } \tan^{-1} \Bigg[ \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x } + \sqrt{1-x}} \Bigg] \text{ is }$

$\displaystyle (a) \ \frac{\pi}{4} - \frac{x}{2} \hspace{1.0cm} (b) \ \frac{\pi}{4} + \frac{x}{2} \hspace{1.0cm}(c) \ \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x \hspace{1.0cm}(d) \ \frac{\pi}{4} + \frac{1}{2} \cos^{-1}x$

Answer:

$\fbox{ c }$

$\displaystyle \tan^{-1} \Bigg[ \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x } + \sqrt{1-x}} \Bigg]$

$\displaystyle \text{Let } x = \cos 2 \theta \Rightarrow 2 \theta = \cos^{-1} x$

$\displaystyle \theta = \frac{1}{2} \cos^{-1} x$

$\displaystyle \tan^{-1} \Bigg[ \frac{\sqrt{1+\cos 2 \theta} - \sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta } + \sqrt{1-\cos 2 \theta}} \Bigg]$

$\displaystyle = \tan^{-1} \Bigg[ \frac{\sqrt{1+2 \cos^2 \theta - 1} - \sqrt{1 - 1 + 2 \sin^2 \theta}}{\sqrt{1 + 2 \cos^2 \theta - 1 } + \sqrt{1-1 + 2 \sin^2 \theta}} \Bigg]$

$\displaystyle = \tan^{-1} \Bigg[ \frac{\sqrt{2 \cos^2 \theta} - \sqrt{2 \sin^2 \theta}}{\sqrt{2 \cos^2 \theta } + \sqrt{2 \sin^2 \theta}} \Bigg]$

$\displaystyle = \tan^{-1} \Bigg[ \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \Bigg]$

$\displaystyle = \tan^{-1} \Bigg[ \frac{\cos (1 - \tan \theta)}{\cos (1 + \tan \theta)} \Bigg]$

$\displaystyle = \tan^{-1} \Bigg[ \tan \Big( \frac{\pi}{4} - \theta \Big) \Bigg]$

$\displaystyle =\frac{\pi}{4} - \theta$

$\displaystyle =\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$

$\\$

$\displaystyle \textbf{32: } \text{If for the matrix } A = \begin{bmatrix} \alpha & -2 \\ -2 & \alpha \end{bmatrix} , |A^3| = 125, \text{ then the value of } \alpha \text{ is }$

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

Answer:

$\fbox{ a }$

$\displaystyle \text{Given } |A^3| = 125$

$\displaystyle |A|^3 = 125 \ \ \ \ (\because |A^m| = |A|^m )$

$\displaystyle \text{ or } |A| = 5I$

$\displaystyle \begin{vmatrix} \alpha & -2 \\ -2 & \alpha \end{vmatrix} = 5 \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$

$\displaystyle \alpha^2 - 4 = 5(1)$

$\displaystyle \alpha^2 - 9 = 0$

$\displaystyle \alpha = \pm 3$

$\\$

$\displaystyle \textbf{33: } \text{If } y = \sin ( m \sin^{-1} x), \text{ then which of the following equations is true }$

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

Answer:

$\fbox{ b }$

$\displaystyle y = \sin (m \sin^{-1} x)$

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

$\displaystyle \frac{dy}{dx} = \cos ( m \sin^{-1} x) \times \frac{d}{dx} m \sin^{-1} x$

$\displaystyle \frac{dy}{dx} = \cos ( m \sin^{-1} x) \times \frac{m}{\sqrt{1 - x^2}}$

$\displaystyle \sqrt{1 - x^2} \frac{dy}{dx} = m \cdot \cos ( m \sin^{-1} x)$

Again differentiating both sides

$\displaystyle \sqrt{1 - x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \Bigg[ \frac{1}{2\sqrt{1-x^2}} \times (-2x) \Bigg] = - m \sin \ ( m \sin^{-1} x) \times \frac{m}{ \sqrt{1-x^2}}$

$\displaystyle \sqrt{1 - x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \Bigg[ \frac{-2x}{2\sqrt{1-x^2}} \Bigg] = - \frac{m^2 \sin \ ( m \sin^{-1} x)}{ \sqrt{1-x^2}}$

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

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

$\\$

$\displaystyle \textbf{34: } \text{The principal value of } [\tan^{-1} \sqrt{3} - \cot^{-1} (-\sqrt{3})] \text{ is }$

$\displaystyle (a) \ \pi \hspace{1.0cm} (b) \ -\frac{\pi}{2} \hspace{1.0cm}(c) \ 0 \hspace{1.0cm}(d) \ 2\sqrt{3}$

Answer:

$\fbox{ b }$

$\displaystyle \tan^{-1} \sqrt{3} - \cot^{-1} (-\sqrt{3})$

$\displaystyle = \tan^{-1} \Big( \tan \frac{\pi}{3} \Big) - \Big( \pi - \cot^{-1} (\sqrt{3}) \Big)$

$\displaystyle = \frac{\pi}{3} - \Big( \pi - \cot^{-1} \Big( \cot \frac{\pi}{6} \Big) \Big)$

$\displaystyle = \frac{\pi}{3} - \pi + \frac{\pi}{6}$

$\displaystyle = \frac{\pi}{2}$

$\\$

$\displaystyle \textbf{35: } \text{The maximum value of } \Big( \frac{1}{x} \Big)^x \text{ is }$

$\displaystyle (a) \ e^{1/e} \hspace{1.0cm} (b) \ e \hspace{1.0cm}(c) \ \Big( \frac{1}{e} \Big)^{1/e} \hspace{1.0cm}(d) \ e^e$

Answer:

$\fbox{ a }$

$\displaystyle y = \Big( \frac{1}{x} \Big)^x$

$\displaystyle \Rightarrow \log y = \log \Big( \frac{1}{x} \Big)^x$

$\displaystyle \Rightarrow \log y = x \cdot \log \Big( \frac{1}{x} \Big)$

$\displaystyle \Rightarrow \log y = x \cdot \log x^{-1}$

$\displaystyle \Rightarrow \log y = - x \log x$

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

$\displaystyle \Rightarrow \frac{1}{y} \times \frac{dy}{dx} = - \Big( x \cdot \frac{1}{x} + \log x \Big)$

$\displaystyle \Rightarrow \frac{dy}{dx} = - y ( 1 + \log x )$

$\displaystyle \Rightarrow \frac{dy}{dx} = - \Big( \frac{1}{x} \Big)^x ( 1 + \log x)$

$\displaystyle \text{For critical point } \frac{dy}{dx} = 0$

$\displaystyle \Big( \frac{1}{x} \Big)^x ( 1 + \log x ) = 0$

$\displaystyle \Rightarrow 1 + \log x = 0$

$\displaystyle \Rightarrow \log x = - 1$

$\displaystyle x = e^{-1} = \frac{1}{e}$

$\displaystyle \frac{d}{dx} \Big( \frac{dy}{dx} \Big) = - \frac{d}{dx} ( y + y \log x )$

$\displaystyle \frac{d^2y}{dx^2} = - \Big( \frac{dy}{dx} + y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx} \Big) = - \Big( \frac{dy}{dx} ( 1+ \log x ) + \frac{y}{x} \Big) < 0$

$\displaystyle \text{Therefore } x = \frac{1}{e} \text{ is the point of maxima and maximum value is }$

$\displaystyle y = \Big(\frac{1}{x} \Big)^x = \Big(\frac{1}{\frac{1}{e}} \Big)^\frac{1}{e} = e^{\frac{1}{e}}$

$\\$

$\displaystyle \textbf{36: } \text{Let matrix } X = [x_{ij}] \text{ is given by } X = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}. \text{ Then the matrix } \\ Y = [m_{ij}], \text{ where } m_{ij} = \text{ Minor of } x_{ij}, \text{is }$

$\displaystyle (a) \ \begin{bmatrix} 7 & -5 & -3 \\ 19 & 1 & -11 \\ -11 & -1 & 3 \end{bmatrix} \hspace{1.0cm} (b) \ \begin{bmatrix} 7 & -19 & -11 \\ 5 & -1 & -1 \\ 3 & 11 & 7 \end{bmatrix} \hspace{1.0cm}\\ \\ \\ (c) \ \begin{bmatrix} 7 & 19 & -11 \\ -3 & 11 & 7 \\ -5 & -1 & -1 \end{bmatrix} \hspace{1.3cm}(d) \ \begin{bmatrix} 7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7 \end{bmatrix}$

Answer:

$\fbox{ d }$

$\displaystyle \text{Given } X = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}$

$\displaystyle \text{Given that minor of element } x _{ij} \text{ is denoted by } m_{ij}, \text{ so }$

$\displaystyle m_{11} = \begin{vmatrix} 4 & -5 \\ -1 & 3 \end{vmatrix} \Rightarrow 12 - 5 = 7$

$\displaystyle m_{12} = \begin{vmatrix} 3 & -5 \\ 2 & 3 \end{vmatrix} \Rightarrow 9+10 = 19$

$\displaystyle m_{13} = \begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix} \Rightarrow -3 - 8 = -11$

$\displaystyle m_{21} = \begin{vmatrix} -1 & 2 \\ -1 & 3 \end{vmatrix} \Rightarrow -3 +2 = -1$

$\displaystyle m_{22} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \Rightarrow 3-4 = -1$

$\displaystyle m_{23} = \begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} \Rightarrow -1+2=1$

$\displaystyle m_{31} = \begin{vmatrix} -1 & 2 \\ 4 & -5 \end{vmatrix} \Rightarrow 5-8 = -3$

$\displaystyle m_{32} = \begin{vmatrix} 1 & 2 \\ 3 & -5 \end{vmatrix} \Rightarrow -5-6 = -11$

$\displaystyle m_{33} = \begin{vmatrix} 1 & -1 \\ 3 & 4 \end{vmatrix} \Rightarrow 4+3 = 7$

Therefore,

$\displaystyle Y = \begin{bmatrix} m_{11} &m_{12} & m_{13} \\ m_{21} &m_{22} & m_{23} \\ m_{31} &m_{32} & m_{33} \end{bmatrix} = \begin{bmatrix} 7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7 \end{bmatrix}$

$\\$

$\displaystyle \textbf{37: } \text{A function } f : R \rightarrow \text{ defined by } f(x) = 2 + x^2 \text{ is }$

$\displaystyle (a) \ \text{not one-one } \hspace{0.5cm} (b) \ \text{one-one } \hspace{0.5cm}(c) \ \text{ not onto} \hspace{0.5cm}(d) \ \text{neither one-one nor onto }$

Answer:

$\fbox{ d }$

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

$\displaystyle \Rightarrow 2 + {x_1}^2 = 2 + {x_2}^2$

$\displaystyle \Rightarrow x_1 = \pm x_2$

$\displaystyle \therefore f(x) s \text{ not one-one }$

$\displaystyle \because x^2 \geq 0 \Rightarrow 2 + x^2 \geq 2$

$\displaystyle \Rightarrow f(x) \geq 2$

$\displaystyle \text{Therefore Range } = [2, \infty ) \neq \text{ co-domain }$

$\displaystyle \text{Hence, } f(x) \text{ is not onto. }$

$\\$

$\displaystyle \textbf{38: } \text{A linear programming problem is as follows: maximize / minimize objective } \\ \text{function } Z = 2x- y + 5 \text{ subject to the constraints } 3x+ 4y \leq 60, x+3y \leq 30, \\ x \geq 0, y \geq 0 \\ \\ \text{If the corner points of the feasible region are } A(0, 10), B(12, 6), C(20, 0) \text{ and } O(0, 0 ), \\ \text{then which of the following is true?}$

$\displaystyle (a) \ \text{ Maximum value of } Z \text{ is } 40 \hspace{1.0cm} \\ (b) \ \text{ Minimum value of } Z \text{ is } -5 \hspace{1.0cm} \\ (c) \ \text{Difference of maximum and minimum values of } Z \text{ is } 35\hspace{1.0cm} \\ (d) \ \text{At two corner points, value of } Z \text{ are equal }$

Answer:

$\fbox{ b }$

$\displaystyle \begin{array}{ |c|c|} \hline \text{Coner Points} & \text{Objective Function } Z = 2x -y + 5 \\ \hline A(0, 10) & Z=2 \times 0-10 + 5 = -10 + 5 = - 5 \\ \hline B(12, 6) & Z=2 \times 12-6 + 5 = 24-6+5 = 23 \\ \hline C(20, 0)& Z=2 \times 20-0 + 5 = 40 + 5 = 45 \\ \hline O(0, 0) & Z=2 \times 0-0 + 5 = 5 \\ \hline \end{array}$

$\displaystyle \text{Therefore, minimum value of } Z = - 5$

$\\$

$\displaystyle \textbf{39: } \text{If } x = - 4 \text{ is a root of } \begin{vmatrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \end{vmatrix}= 0 \text{ then the sum of the other two roots is}$

$\displaystyle (a) \ 4 \hspace{1.0cm} (b) \ -3 \hspace{1.0cm}(c) \ 2 \hspace{1.0cm}(d) \ 5$

Answer:

$\fbox{ a }$

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

$\displaystyle \text{Expanding along } R_1 \text{ we have }$

$\displaystyle x(x^2 - 2) - 2 ( x-3) + 3 ( 2 - 3x) = 0$

$\displaystyle \Rightarrow x^3 - 2x - 2x + 6 + 6 - 9x = 0$

$\displaystyle \Rightarrow x^2 - 13 x + 12 = 0 \ldots \ldots (i)$

$\displaystyle \text{If } x = - 4 \text{is one root then it satisfies equation (i) }$

$\displaystyle x(-4) = (-4)^3 - 13 ( - 4) + 12 = -64 + 52 + 12 = 0$

$\displaystyle \text{Similarly, } x = 3 \text{ and } x = 1 \text{ satisfy equation (i). }$

$\displaystyle x(3) = (3)^3 - 13(3) + 12 = 27 - 39 + 12 = 39 - 39 =0$

$\displaystyle x(1) = (1)^3 - 13(1) + 12 = 1 - 13 + 12 = 13 - 13 = 0$

$\displaystyle \text{Both } x = 3 \text{ and } x = 1 \text{ satisfy the equation, so } x = 3 \text{ and } x = 1 \text{ are two other roots. }$

$\displaystyle \text{Therefore, sum of the other two roots is } = 3 + 1 = 4$

$\\$

$\displaystyle \textbf{40: } \text{The absolute maximum value of the function } f(x) = 4x - \frac{1}{2} x^2 \text{ in the interval } \\ \Big[ -2, \frac{9}{2} \Big] \text{ is }$

$\displaystyle (a) \ 8 \hspace{1.0cm} (b) \ 9 \hspace{1.0cm}(c) \ 6 \hspace{1.0cm}(d) \ 10$

Answer:

$\fbox{ a }$

$\displaystyle f(x) = 4x - \frac{1}{2} x^2$

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

$\displaystyle f'(x) = 4 - x$

$\displaystyle \text{For critical point } f'(x) = 0$

$\displaystyle 4 - x = 0$

$\displaystyle \Rightarrow x = 4$

$\displaystyle \text{Now, find the value of } f(x) \text{ at } x = -2, 4, \frac{9}{2}$

$\displaystyle f(-2) = 4 ( -2) - \frac{1}{2} (-2)^2 = -8-2=-10$

$\displaystyle f(4) = 4(4) - \frac{1}{2} (4)^2 = 16 - 8 = 8$

$\displaystyle f \Big( \frac{9}{2} \Big) = 4 \Big(\frac{9}{2} \Big) - \frac{1}{2}\times \frac{81}{4} = 18 - \frac{81}{8} = \frac{63}{8} = 7.875$

$\displaystyle \text{Therefore absolute minimum value is 8. }$

$\\$

SECTION C

Attempt any 8 questions out of the questions 41-50. Each question is of one mark.

$\displaystyle \textbf{41: } \text{In a sphere of radius } r \text{ a right circular cone of height } h, \text{ having maximum curved } \\ \text{surface area is inscribed. } \text{The expression for the square of the curved surface of the} \\ \text{cone is }$

$\displaystyle (a) \ 2 \pi^2rh (2rh+ h^2) \hspace{1.0cm} (b) \ \pi^2hr (2rh+ h^2) \hspace{1.0cm}\\ \\ (c) \ 2\pi^2r(2rh^2-h^3) \hspace{1.0cm}(d) \ 2\pi^2r^2(2rh-h^2)$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given that: }$

$\displaystyle \text{Radius of the sphere } = r$

$\displaystyle \text{Height of the cone } = h$

$\displaystyle \text{Let radius of cone } = R$

$\displaystyle \text{Let slant height of cone } = l$

$\displaystyle \text{Now, in } \triangle OPQ$

$\displaystyle OP^2 = PQ^2 + QO^2$

$\displaystyle r^2 = R^2 + ( h-r)^2$

$\displaystyle r^2 = R^2 + h^2 + r^2 - 2hr$

$\displaystyle R^2 = r^2 - h^2 - r^2 + 2hr$

$\displaystyle R^2 = 2hr - h^2$

$\displaystyle R = \sqrt{2hr - h^2}$

$\displaystyle \text{In } \triangle MPQ,$

$\displaystyle l^2 = R^2 + h^2$

$\displaystyle l^2= ( 2hr - h^2) + h^2$

$\displaystyle l = \sqrt{2hr}$

$\displaystyle \text{Curved surface area } = \pi R l$

$\displaystyle \text{C.S.A } = \pi ( \sqrt{2hr - h^2})(\sqrt{2hr})$

$\displaystyle \text{Squaring both sides we get }$

$\displaystyle (CSA)^2 = \pi^2 ( 2hr - h^2) ( 2hr) = 2\pi^2 hr ( 2hr - h^2)= 2\pi^2 r ( 2h^2r - h^3)$

$\\$

$\displaystyle \textbf{42: } \text{The corner points of the feasible region determined by a set of constraints } \\ \text{(liner inequalities) are } P(0, 5), Q(3, 5), R(5, 0) \text{ and } S(4, 1) \text{ and the objective function} \\ \text{is } Z = ax + 2by \text{ where } a, b > 0. \text{ The condition on } a \text{ and } b \text{ such that the maximum } Z \\ \text{occurs at } Q \text{ and } S \text{ is }$

$\displaystyle (a) \ a-5b=0 \hspace{1.0cm} (b) \ a-3b=0 \hspace{1.0cm}(c) \ a-2b=0 \hspace{1.0cm}(d) \ a-8b=0$

Answer:

$\fbox{ d }$

$\displaystyle \text{Objective function: } Z = ax + 2by, \text{ where } a, b > 0$

$\displaystyle \text{Given: maximum Z occurs at } (3, 5) \text{ and } S(4, 1)$

$\displaystyle \text{Therefore, max } Z ( 3, 5) = \text{ max } Z (4, 1)$

$\displaystyle 3a + 10 b = 4a + 2b$

$\displaystyle 4a - 3a + 2b - 10 b = 0$

$\displaystyle a - 8 b = 0$

$\\$

$\displaystyle \textbf{43: } \text{If curves } y^2=4x \text{ and } xy=c \text{ cut at right angles, then the value of } c \text{ is }$

$\displaystyle (a) \ 4\sqrt{2} \hspace{1.0cm} (b) \ 8 \hspace{1.0cm}(c) \ 2\sqrt{2} \hspace{1.0cm}(d) \ -4 \sqrt{2}$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given curves are } y^2 = 4x \ldots (i) \text{ and } xy = c \ldots (ii)$

$\displaystyle \text{From equation (i) } x = \frac{y^2}{4}$

$\displaystyle \Big( \frac{y^2}{4} \Big) y = c \Rightarrow \frac{y^3}{4} = c$

$\displaystyle y^3 = 4c \text{ or } y = (4c)^{\frac{1}{3}}$

$\displaystyle \text{Substituting } y = (4c)^{\frac{1}{3}} \text{ in equation (i) }$

$\displaystyle [(4c)^{\frac{1}{3}}]^2 = 4x \Rightarrow 4x = (4c)^{\frac{2}{3}} \Rightarrow x = \frac{(4c)^{\frac{2}{3}}}{4}$

$\displaystyle \text{Hence, both curves intersect at point } \Bigg[ (4c)^{\frac{2}{3}}, (4c)^{\frac{1}{3}} \Bigg]$

Now, differentiating equation (i) w.r.t. x

$\displaystyle 2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{4}{2y}$

$\displaystyle \frac{dy}{dx} \Bigg]_{ (4c)^{\frac{2}{3}}, (4c)^{\frac{1}{3}} } = \frac{4}{2(4c)^{\frac{1}{3}}} = \frac{2}{(4c)^{\frac{1}{3}}}$

Differentiating equation (ii) w.r.t. x

$\displaystyle x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = \frac{-y}{x}$

$\displaystyle \frac{dy}{dx} \Bigg]_{ (4c)^{\frac{2}{3}}, (4c)^{\frac{1}{3}} } = \frac{-(4c)^{\frac{1}{3}}}{\frac{(4c)^{\frac{2}{3}}}{4}} = \frac{-4(4c)^{\frac{1}{3}}}{(4c)^{\frac{2}{3}}} = - 4(4c)^{\frac{1}{3} - \frac{2}{3}} = -4(4c)^{-\frac{1}{3}}$

Curves (i) and (ii) cut at right angles. Therefore,

$\displaystyle \frac{2}{(4c)^{\frac{1}{3}}} \times \frac{-4}{(4c)^{\frac{1}{3}}} = -1 \Rightarrow \frac{-8}{(4c)^{\frac{2}{3}}} = -1$

$\displaystyle \Rightarrow 8 = ( 4c)^{\frac{2}{3}} \Rightarrow (4c)^{\frac{2}{3}} = 8 \Rightarrow 4c = 8^{\frac{3}{2}} \\ \\ \Rightarrow 4c = 8\sqrt{8} \Rightarrow c = 2\sqrt{8} \Rightarrow c = 4\sqrt{2}$

$\\$

$\displaystyle \textbf{44: } \text{The inverse of the matrix } X = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} \text{ is }$

$\displaystyle (a) \ 24 \begin{bmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{bmatrix} \hspace{1.0cm} (b) \ \frac{1}{24} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \hspace{1.0cm}\\ \\ \\ (c) \ \frac{1}{24} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} \hspace{2.3cm}(d) \ \begin{bmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{bmatrix}$

Answer:

$\fbox{ d }$

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

$\displaystyle |X| = 2 ( 12-0) + 0 + 0 = 24 \neq 0$

$\displaystyle \text{This is singular matrix. So, } X^{-1} \text{ exists. }$

$\displaystyle \text{Now cofactor, }C = \begin{bmatrix} 12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

$\displaystyle C^T = \begin{bmatrix} 12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

$\displaystyle \text{Adj } X \Rightarrow C^T = \begin{bmatrix} 12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Therefore,

$\displaystyle X^{-1} = \frac{1}{|X|} \cdot \text{Adj } X = \frac{1}{24} \begin{bmatrix} 12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

$\displaystyle \Rightarrow X^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix}$

$\\$

$\displaystyle \textbf{45: } \text{For an L.P.P the objective function is } Z = 4x+ 3y \text{ and the feasible region} \\ \text{determined by a set of constraints (linear inequations) is shown in the graph } \\ \text{Which one of the following statement is true? }$

$\displaystyle (a) \ \text{Maximum value of } Z \text{ is at } R \hspace{1.0cm} \\ (b) \ \text{Maximum value of } Z \text{ is at } Q \hspace{1.0cm} \\ (c) \ \text{Value of } Z \text{ at } R \text{ is less than the value at } P \hspace{1.0cm} \\ (d) \ \text{Value of } Z \text{ at } Q \text{ is less than the value at } P$

Answer:

$\fbox{ b }$

$\displaystyle \begin{array}{ |c|c|} \hline \text{Coner Points} & \text{Objective Function } Z = 4x +3y \\ \hline O(0, 0) & Z=4 \times 0 +3 \times 0 = 0 \\ \hline P(0, 40) & Z=4 \times 0 + 3 \times 40 = 0 + 120 = 120 \\ \hline Q(30, 20)& Z=4 \times 30+ 3 \times 20 = 120 + 60 = 180 \\ \hline R(40, 0) & Z=4 \times 40 + 3 \times 0 = 160 + 0 = 160 \\ \hline \end{array}$

$\displaystyle \text{Therefore, for maximum value of } Z = 10 \text{ at point } Q(30, 20).$

$\\$

$\displaystyle \text{In a residential society comprising of 100 houses, there were 60 children between } \\ \text{the ages of 10-15 years. They were inspired by their teachers to start composing } \\ \text{to ensure that biodegradable waste is recycled. For this purpose, instead of each child } \\ \text{doing it for only his her house, children convinced the Residents welfare association } \\ \text{to do it as a society initiative. For this they identified a square area in the local park. } \\ \text{Local authorities charged amount of Rs. 50 per square meter for space so that there } \\ \text{is no misuse of the space and Resident welfare association takes it seriously. } \\ \text{Association hired a laborer for digging out }250 m^3 \text{ and he charged Rs. } 400 \times (depth)^2. \text{Association will like to have minimum cost. } \\ \\ \text{Based on this information, answer the any 4 of the following questions: }$

$\displaystyle \textbf{46: } \text{Let side of the square plot is } x \text{m and its depth is } h \text{ meters, then cost } c \\ \text{ for the pit is}$

$\displaystyle (a) \ \frac{50}{h} + 400 h^2 \hspace{1.0cm} (b) \ \frac{12500}{h}+400h^2 \hspace{1.0cm}(c) \ \frac{250}{h} + h^2 \hspace{1.0cm}(d) \ \frac{250}{h} + 400 h^2$

Answer:

$\fbox{ b }$

$\displaystyle \text{Total cost = Cost for space + Cost of digging }$

$\displaystyle C = Area \times Rate + 400 \times (Depth)^2$

$\displaystyle C = \frac{250}{h} \times 50 + 400 \times h^2$

$\displaystyle C = \frac{12500}{h} + 400 h^2$

$\\$

$\displaystyle \textbf{47: } \text{Value of } h \text{ (in m) for which } \frac{dc}{dh} = 0 \text{ is }$

$\displaystyle (a) \ 1.5 \hspace{1.0cm} (b) \ 2 \hspace{1.0cm}(c) \ 2.5 \hspace{1.0cm}(d) \ 3$

Answer:

$\fbox{ c }$

$\displaystyle \frac{dc}{dh} = \frac{-12500}{h^2} + 800 h = 0$

$\displaystyle \Rightarrow h^3 = \frac{125}{8} \Rightarrow h =\frac{5}{2} = 2.5 \text{ m }$

$\\$

$\displaystyle \textbf{48: } \frac{d^2c}{dh^2} \text{ is given by }$

$\displaystyle (a) \ \frac{25000}{h^3} + 800 \hspace{1.0cm} (b) \ \frac{500}{h^2}+ 800 \hspace{1.0cm}(c) \ \frac{100}{h^2}+ 800 \hspace{1.0cm}(d) \ \frac{500}{h^2}+ 2$