CBSE Paper (All India – 2023) Class 12: Mathematics Solution – ICSE / ISC / CBSE Mathematics Portal for K12 Students

MATHEMATICS

$\displaystyle \text{Series EF1GH/4} \hspace{1.0cm} | \hspace{1.0cm} \text{Q. P. Code 65/4/3} \hspace{1.0cm} | \hspace{1.0cm} \text{Set 3 }$

$\displaystyle \text{Time Allowed : 3 hours} \hspace{5.0cm} \text{Maximum Marks : 80 }$

General Instructions :

Read the following instructions very carefully and strictly follow them :

(i) This question paper contains 38 questions. All questions are compulsory.

(ii) This question paper is divided into five Sections A, B, C, D and E.

(iii) In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs) and questions number 19 and 20 are Assertion-Reason based questions of 1 mark each.

(iv) In Section B, Questions no. 21 to 25 are very short answer (VSA) type questions, carrying 2 marks each.

(v) In Section C, Questions no. 26 to 31 are short answer (SA) type questions, carrying 3 marks each.

(vi) In Section D, Questions no. 32 to 35 are long answer (LA) type questions carrying 5 marks each.

(vii) In Section E, Questions no. 36 to 38 are case study based questions carrying 4 marks each.

(viii) There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 3 questions in Section C, 2 questions in Section D and 2 questions in Section E.

(ix) Use of calculators is not allowed.

SECTION A

This section comprises multiple choice questions (MCQs) of 1 mark each

$\displaystyle \text{Question 1: If } A \text{ is a } 3 \times 4 \text{ matrix and } B \text{ is a matrix such that } A'B \text{ and } AB' \\ \text{ are both defined, then the order of the matrix } B \text{ is : }$

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

Answer:

$\fbox{ a }$

$\displaystyle \text{Order of matrix } A \text{ is } 3 \times 4 \text{ and therefor } A' \text{ is } 4 \times 3$

$\displaystyle \text{Let order of matrix } B \text{ is } m \times n \text{ therefore } B' \text{ is } n \times m$

$\displaystyle \text{Therefore, number of columns } of A' \text{ must equal rows of } B, \text{because } \\ A'B \text{ is defined on } m = 3.$

$\displaystyle \text{Also, number of columns } of B \text{ must equal rows of } A', \text{because } \\ AB' \text{ is defined on } n = 4.$

$\displaystyle \text{So the order of matrix } B = 3 \times 4$

$\\$

$\displaystyle \text{Question 2: If the area of a triangle with vertices } (2, 6), (5, 4) \text{ and } (k, 4) \text{ is } 35 \text{ sq units, } \\ \text{ then } k \text{ is }$

$\displaystyle (a) \ {12 } \hspace{1.0cm} (b) \ {-2 } \hspace{1.0cm}(c) \ {-12, -2 } \hspace{1.0cm}(d) \ {12, -2 }$

Answer:

$\fbox{ d }$

$\displaystyle \text{The area of a triangle with vertices } (2, 6), (5, 4), (k, 4) \text{ is }$

$\displaystyle A = \frac{1}{2} \begin{bmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{bmatrix} = 35$

$\displaystyle\Rightarrow \frac{1}{2} |[ 2(4-4) + 5(4-(-6)) + k(-6-4) ]| = 35$

$\displaystyle\Rightarrow \frac{1}{2} | 0+ 50 - 10k| = 35$

$\displaystyle\Rightarrow 25 - 5 k = \pm 35$

$\displaystyle\Rightarrow 5k = -10 \text{ or } 5k = 60$

$\displaystyle\Rightarrow k = -2 \text{ or } k = 12$

$\\$

$\displaystyle \text{Question 3: If } f(x) = 2|x| + 3 |\sin x|+6, \text{ then the right hand derivative of } f(x) \text{ at } x = 0 \text{ is : }$

$\displaystyle (a) \ {6 } \hspace{1.0cm} (b) \ {5 } \hspace{1.0cm}(c) \ {3 } \hspace{1.0cm}(d) \ {2 }$

Answer:

$\fbox{ b }$

$\displaystyle \text{Let }x = 0 + h, \text{ where }h \to 0$

$\displaystyle \Rightarrow f'(x) = \lim \limits_{h \to 0} \frac{f(0+h) - f(0)}{h}$

$\displaystyle \Rightarrow f'(x) = \lim \limits_{h \to 0} \frac{2|h|+ 3 |\sin h| + 6 - 6}{h}$

$\displaystyle \Rightarrow f'(x) = \lim \limits_{h \to 0} \Big( \frac{2h}{h} + \frac{3 \sin h}{h} \Big) = 5$

$\\$

$\displaystyle \text{Question 4: if } x \begin{bmatrix} 1 \\ 2 \end{bmatrix} + y \begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} 4 \\ 9 \end{bmatrix}, \text{ then: }$

$\displaystyle (a) \ { x=1, y=2} \hspace{1.0cm} (b) \ {x=2, y=1 } \hspace{1.0cm}(c) \ { x=1, y=-1} \hspace{1.0cm}(d) \ {x=3, y=2 }$

Answer:

$\fbox{ b }$

$\displaystyle \text{Given } x \begin{bmatrix} 1 \\ 2 \end{bmatrix} + y \begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} 4 \\ 9 \end{bmatrix}$

We can write the equation as,

$\displaystyle x+ 2y = 4 \ldots \ldots (i)$

$\displaystyle 2x+5y = 9 \ldots \ldots (ii)$

$\displaystyle \text{Multiplying equation (i) by 2 and subtracting equation (ii) from equation (i) }$

$\displaystyle \begin{array}{l} 2x+ 4y = 8 \\ 2x+5y = 9 \\ (-) \hspace{0.2cm} (-) \\ \hline \hspace{0.5cm} -y = - 1 \end{array}$

$\displaystyle \text{Putting } y = 1 \text{ in equation (i) }$

$\displaystyle \Rightarrow x + 2 = 4 \hspace{1.2cm} \Rightarrow x = 2$

$\\$

$\displaystyle \text{Question 5: If a matrix } A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}, \text{ then the matrix } AA' \text{ (where } A' \\ \text{ is the transpose of } A \text{ ) is : }$

$\displaystyle (a) \ {14 } \hspace{1.0cm} (b) \ { \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} } \hspace{1.0cm}(c) \ {\begin{bmatrix} 1 &2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{bmatrix} } \hspace{1.0cm}(d) \ {[14] }$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given, } A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$

$\displaystyle \text{The } A' = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

$\displaystyle \Rightarrow AA' = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

$\displaystyle \Rightarrow AA' = \begin{bmatrix} 1+4+9 \end{bmatrix} = 14$

$\\$

$\displaystyle \text{Question 6: The product of } \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & -a \end{bmatrix} \text{ is equal to:}$

$\displaystyle (a) \ {\begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix} } \hspace{1.0cm} (b) \ {\begin{bmatrix} (a+b)^2 & 0 \\ (a+b)^2 & 0 \end{bmatrix} } \hspace{1.0cm}(c) \ { \begin{bmatrix} a^2+b^2 & 0 \\ a^2+b^2 & 0 \end{bmatrix}} \hspace{1.0cm}(d) \ {\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} }$

Answer:

$\fbox{ a }$

Product of given matrix is defined as:

$\displaystyle \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & -a \end{bmatrix} = \begin{bmatrix} a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2 \end{bmatrix} \hspace{0.5cm} \Rightarrow \begin{bmatrix} a^2+b^2 & 0 \\ 0 & b^2+a^2 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 7: Distance of the point } (p, q, r) \text{ from y-axis is : }$

$\displaystyle (a) \ {q } \hspace{1.0cm} (b) \ {|q| } \hspace{1.0cm}(c) \ {|q|+|r| } \hspace{1.0cm}(d) \ {\sqrt{p^2+r^2} }$

Answer:

$\fbox{ d }$

$\displaystyle \text{Let point on y axis is } (0, q, 0) \text{ that is nearest from } (p, q, r)$

$\displaystyle \text{Distance } (d) = \sqrt{p^2 + (q-q)^2+ r^2}= \sqrt{p^2 + r^2}$

$\\$

$\displaystyle \text{Question 8: The solution set of the inequation } 3x + 5y < 7 \text{ is : }$
$\displaystyle \text{(a) whole xy-plane except the points lying on the line } 3x + 5y = 7.$
$\displaystyle \text{(b) whole xy-plane along with the points lying on the line } 3x + 5y = 7.$
$\displaystyle \text{(c) open half plane containing the origin except the points of line } \\ 3x + 5y = 7.$
$\displaystyle \text{(d) open half plane not containing the origin. }$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given inequation: } 3x + 5y < 7$

$\displaystyle \text{Solution defines in open half plane containing the origin except the points of the line } \\ 3x + 5y = 7$

$\\$

$\displaystyle \text{Question 9: if } \int \limits_{0}^{a} 3x^2 \ dx = 8, \text{ then the value of } a \text{ is : }$

$\displaystyle (a) \ {2 } \hspace{1.0cm} (b) \ {4 } \hspace{1.0cm}(c) \ {8 } \hspace{1.0cm}(d) \ {10 }$

Answer:

$\fbox{ a }$

$\displaystyle \text{Given } \int \limits_{0}^{a} 3x^2 \ dx = 8$

Integrating the equation, we get

$\displaystyle \Rightarrow 3 \Big[ \frac{x^3}{3} \Big] = 8$

$\displaystyle \Rightarrow a^3 = 8 \hspace{1.0cm} \Rightarrow a = 2$

$\\$

$\displaystyle \text{Question 10: The sine of the angle between the vectors } \overrightarrow{a} = 3 \hat{i}+ \hat{j}+2\hat{k} \\ \text{ and } \overrightarrow{b} = \hat{i}+ \hat{j}+2\hat{k}$

$\displaystyle (a) \ {\sqrt{\frac{5}{21}} } \hspace{1.0cm} (b) \ {\frac{5}{\sqrt{21}} } \hspace{1.0cm}(c) \ {\sqrt{\frac{3}{21}} } \hspace{1.0cm}(d) \ {\frac{4}{\sqrt{21}} }$

Answer:

$\fbox{ a }$

$\displaystyle \text{Let the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ be } \theta$

$\displaystyle \text{Then, } \cos \theta = \frac{\overrightarrow{a} . \overrightarrow{b}}{|\overrightarrow{a}|| \overrightarrow{b}|} = \frac{(3 \hat{i}+ \hat{j}+2\hat{k})(\hat{i}+ \hat{j}+2\hat{k})}{\sqrt{3^2+1^2+2^2} \sqrt{1^2+1^2+2^2}}$

$\displaystyle \Rightarrow \cos \theta = \frac{3+1+4}{\sqrt{14} \sqrt{6}} = \frac{4}{\sqrt{21}}$

$\displaystyle \Rightarrow \sin \theta = \sqrt{1- \cos^2 \theta} = \sqrt{1 - \frac{16}{21} } = \sqrt{\frac{5}{21} }$

$\\$

$\displaystyle \text{Question 11: The order and degree (if defined) of the differential equation, } \\ \\ \Big( \frac{d^2y}{dx^2} \Big)^2 + \Big( \frac{dy}{x} \Big)^2= x \sin x \Big( \frac{dy}{dx} \Big) \text{ respectively are: }$

$\displaystyle (a) \ {2, 2 } \hspace{1.0cm} (b) \ { 1, 3} \hspace{1.0cm}(c) \ {2, 3 } \hspace{1.0cm}(d) \ {2, \text{ degree not defined } }$

Answer:

$\fbox{ d }$

Given differential equation is not a polynomial form, order 2, but degree is not defined.

$\\$

$\displaystyle \text{Question 12: } \int e^{5 \log x} dx \text{ is equal to }$

$\displaystyle (a) \ { \frac{x^5}{5}+C} \hspace{1.0cm} (b) \ {\frac{x^6}{6}+C } \hspace{1.0cm}(c) \ {5x^4+C } \hspace{1.0cm}(d) \ {6x^5+C }$

Answer:

$\fbox{ b }$

$\displaystyle \text{Given } \int e^{5 \log x} dx$

$\displaystyle \text{Where, } e^{5 \log x} = e^{{\log x}^5} = x^5$

$\displaystyle \text{Now, } \int e^{5 \log x} dx = \int x^5 \ dx$

$\displaystyle \text{By integrating the given equation, we get } \Rightarrow \frac{x^6}{6} +C$

$\\$

$\displaystyle \text{Question 13: A unit vector along the vector} 4 \hat{i} - 3\hat{k} \text{ is }$

$\displaystyle (a) \ {\frac{1}{7} (4 \hat{i} - 3\hat{k}) } \hspace{1.0cm} (b) \ {\frac{1}{5} (4 \hat{i} - 3\hat{k}) } \hspace{1.0cm}(c) \ {\frac{1}{\sqrt{7}} (4 \hat{i} - 3\hat{k}) } \hspace{1.0cm}(d) \ { \frac{1}{\sqrt{5}}(4 \hat{i} - 3\hat{k})}$

Answer:

$\fbox{ b }$

$\displaystyle \text{Given } 4 \hat{i} - 3\hat{k}$

$\displaystyle \text{Unit Vector } = \frac{4 \hat{i} - 3\hat{k}}{\sqrt{4^2+3^2}} = \frac{4 \hat{i}}{5} - \frac{3\hat{k}}{5} = \frac{1}{5} (4 \hat{i} - 3\hat{k})$

$\\$

$\displaystyle \text{Question 14: Which of the following points satisfies both the inequations }\\ 2x + y \leq 10 \text{ and } x + 2y \geq 8 ?$

$\displaystyle (a) \ {(-2, 4) } \hspace{1.0cm} (b) \ {(3, 2) } \hspace{1.0cm}(c) \ { )-5, 6)} \hspace{1.0cm}(d) \ {(4, 2) }$

Answer:

$\fbox{ d }$

$\displaystyle \text{At } (-2, 4)$

$\displaystyle \Rightarrow 2x + y \leq 10 \Rightarrow -4+4 \leq 10 \Rightarrow 0 \leq 10$

$\displaystyle \Rightarrow x + 2y \geq 8 \Rightarrow -2 + 8 \geq 8 \Rightarrow 6 \ngeq 8$

$\displaystyle \text{At } (3, 2)$

$\displaystyle \Rightarrow 2x + y \leq 10 \Rightarrow 6+2 \leq 10 \Rightarrow 8 \leq 10$

$\displaystyle \Rightarrow x + 2y \geq 8 \Rightarrow 3+4 \geq 8 \Rightarrow 7 \ngeq 8$

$\displaystyle \text{At } (-5, 6)$

$\displaystyle \Rightarrow 2x + y \leq 10 \Rightarrow -10+6 \leq 10 \Rightarrow 4 \leq 10$

$\displaystyle \Rightarrow x + 2y \geq 8 \Rightarrow -5+12 \geq 8 \Rightarrow 7 \ngeq 8$

$\displaystyle \text{At } (4, 2)$

$\displaystyle \Rightarrow 2x + y \leq 10 \Rightarrow 8+2 \leq 10 \Rightarrow 10 \leq 10$

$\displaystyle \Rightarrow x + 2y \geq 8 \Rightarrow 4+4 \geq 8 \Rightarrow 8 \geq 8$

$\\$

$\displaystyle \text{Question 15: If } y = \sin^2 (x^3), \text{ then } \frac{dy}{dx} \text{ is equal to : }$

$\displaystyle (a) \ {2 \sin x^3 \cos x^3 } \hspace{1.0cm} (b) \ {3x^3 \sin x^3 \cos x^3 } \hspace{1.0cm}(c) \ { 6x^2 \sin x^3 \cos x^3} \hspace{1.0cm}(d) \ { 2x^2 \sin^2 (x^3)}$

Answer:

$\fbox{ c }$

$\displaystyle \text{Given } y = \sin^2 (x^3)$

By Chain Rule differentiation the above equation

$\displaystyle \Rightarrow \frac{dy}{dx} = 2 \sin ( x^3) \frac{d}{dx} \Big(\sin (x^3) \Big)$

$\displaystyle \Rightarrow \frac{dy}{dx} = 2 \sin ( x^3) \cos (x^3) . (3x^2)$

$\displaystyle \Rightarrow \frac{dy}{dx} = 6x^2 \sin x^3 . \cos x^3$

$\\$

$\displaystyle \text{Question 16: The point } (x, y, 0) \text{ on the xy-plane divides the line segment joining the points } \\ \\ (1, 2, 3) \text{ and } (3, 2, 1) \text{ in the ratio : }$

$\displaystyle (a) \ {1 : 2 \text{ internally} } \hspace{1.0cm} (b) \ {2 : 1 \text{ internally} } \hspace{1.0cm}(c) \ {3 : 1 \text{ internally} } \hspace{1.0cm}(d) \ {3 : 1 \text{ externally} }$

Answer:

$\fbox{ c }$

$\displaystyle \text{Let } (x, y, 0) \text{ divides the point in } k:1 \text{ ratio }$

$\displaystyle \text{The, } x = \frac{3k-1}{k+1} , y = \frac{2k-2}{k+1}, z = \frac{k-3}{k+1}$

$\displaystyle \text{In } (x, y, 0), z=0$

$\displaystyle \Rightarrow \frac{k-3}{k+1} = 0 \Rightarrow k = 3$

$\displaystyle \text{Therefore } (x, y, 0) \text{ divides the line segment in the ratio } 3:1 \text{ internally. }$

$\\$

$\displaystyle \text{Question 17: The events E and F are independent. If } P(E) = 0.3 \text{ and } \\ P(E \cup F) = 0.5, \text{ then } P(E/F) - P(F/E) \text{ equals : }$

$\displaystyle (a) \ {\frac{1}{7} } \hspace{1.0cm} (b) \ { \frac{2}{7}} \hspace{1.0cm}(c) \ {\frac{3}{35} } \hspace{1.0cm}(d) \ {\frac{1}{70} }$

Answer:

$\fbox{ d }$

$\displaystyle \text{The events } E \text{ and } F \text{ are independent so, }$

$\displaystyle P(E \cap F) = P(E).P(F) = (0.3) (P(F))$

$\displaystyle \text{We know, } P(E \cup F) = P(E) + P(F) - P(E \cap F)$

$\displaystyle \Rightarrow 0.5 = 0.3 + P(F) - (0.3 )P(F)$

$\displaystyle \Rightarrow (0.7) P(F) = ).2 \Rightarrow P(F) = \frac{2}{7}$

$\displaystyle \text{Now } P \Big( \frac{E}{F} \Big) = P(E) \text{ and } P \Big( \frac{F}{E} \Big) = P(F) \text{ because } E \text{ and } F \text{are independent. }$

$\displaystyle \Rightarrow P(E)-P(F) = 0.3 - \frac{2}{7} = \frac{1}{70}$

$\\$

$\displaystyle \text{Question 18: The integrating factor for solving the differential equation } \\ \\ x \frac{dy}{dx} - y = 2x^2 \text{ is }$

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

Answer:

$\fbox{ d }$

$\displaystyle \text{Given } x \frac{dy}{dx} - y = 2x^2 \ \ \ \text{ or } \ \ \ \frac{dy}{dx} - \frac{y}{x} = 2x$

$\displaystyle \text{Integrating factor } = e^{- \int \frac{1}{x} \ dx} \\ \\ \Rightarrow e^{-(\log x)} = e^{\log \frac{1}{x}} = \frac{1}{x}$

$\\$

Questions number 19 and 20 are Assertion and Reason based questions carrying 1 mark each. 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.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(d) Assertion (A) is false and Reason (R) is true.

$\displaystyle \text{Question 19: } \\ \text{Assertion (A): The lines } \overrightarrow{r} = \overrightarrow{a_1}+ \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2}+ \mu \overrightarrow{b_2} \text{ are perpendicular when } \\ \overrightarrow{b_1} . \overrightarrow{b_2} = 0. \\ \\ \text{Reason (R) : The angle } \theta \text{ between the lines }\overrightarrow{r} = \overrightarrow{a_1}+ \lambda \overrightarrow{b_1} \text{ and } \\ \overrightarrow{r} = \overrightarrow{a_2}+ \mu \overrightarrow{b_2} \text{ is given by } \cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{|\overrightarrow{b_1} || \overrightarrow{b_2}|} .$

Answer:

$\fbox{ a }$

Given Assertion is true and Reason is correct explanation of Assertion.

$\displaystyle \Rightarrow \cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{|\overrightarrow{b_1} || \overrightarrow{b_2}|}$

$\displaystyle \text{If } \overrightarrow{b_1} \text{ and } \overrightarrow{b_1} \text{ are perpendicular then }$

$\displaystyle \Rightarrow \cos 90^{\circ} = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{|\overrightarrow{b_1} || \overrightarrow{b_2}|} = 0$

$\displaystyle \Rightarrow \overrightarrow{b_1} . \overrightarrow{b_2} = 0$

$\\$

$\displaystyle \text{Question 20: } \\ \text{Assertion (A) : All trigonometric functions have their inverses over their respective} \\ \text{domains. }$

$\displaystyle \text{ Reason (R) : The inverse of } \tan^{-1} x \text{ exists for some } x$

Answer:

$\fbox{ d }$

Assertion is not true for all trigonometric function.

Domain of $\sin x$ is $R$ but $\sin^{-1} x$ is not defined on $R.$

This implies Reason is true $\tan^{-1} x$ inverse exist for some $x \in R.$

$\\$

SECTION B

This section comprises very short answer (VSA) type questions of 2 marks each.

$\displaystyle \text{ Question 21: If } xy = e^{x-y} , \text{ then show that } \frac{dy}{dx} = \frac{y(x-1_)}{x(y+1)}$

Answer:

$\displaystyle \text{Given, } xy = e^{x-y} = e^x . e^y$

Differentiate above function w.r.t $x$

$\displaystyle \Rightarrow \hspace{1.0cm} x \frac{dy}{dx} + y = e^x \frac{d}{dx} e^{-y} + e^{-y}\frac{d}{dx} e^x$

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

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

$\displaystyle \Rightarrow \hspace{1.0cm} (x+xy) \frac{dy}{dx} = -y +xy$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}$

$\\$

$\displaystyle \text{Question 22: } \\ \text{(a) Find the domain of } y = \sin^{-1`} (x^2-4)$

$\displaystyle \text{(b) Evaluate: } \cos^{-1} \Big[ \cos \Big( -\frac{7 \pi}{3} \Big) \Big]$

Answer:

$\displaystyle \text{(a) Domain of } \sin^{-1} x \text{ is } [-1, 1]$

$\displaystyle \text{Therefore, domain of } \sin^{-1} (x^2 - 4) \text{ is }$

$\displaystyle \Rightarrow \hspace{1.0cm} -1 \leq x^2 - 4 \leq 1$

$\displaystyle \Rightarrow \hspace{1.0cm} 3 \leq x^2 \leq 5$

$\displaystyle \Rightarrow \hspace{1.0cm} \sqrt{3} \leq |x| \leq \sqrt{5}$

(b)

$\displaystyle \cos^{-1} \Big[ \cos \Big( -\frac{7 \pi}{3} \Big) \Big] = \cos^{-1} \Big[ \cos \Big( -2\pi -\frac{ \pi}{3} \Big) \Big] = \cos^{-1} \cos \Big( \frac{\pi}{3} \Big) = \frac{\pi}{3}$

$\\$

$\displaystyle \text{Question 23: If the projection of the vector } \hat{i} + \hat{j} + \hat{k} \text{ on the vector } p\hat{i} + \hat{j} -2 \hat{k} \\ \text{ is } \frac{1}{3}, \text{then find the value (s) of } p.$

Answer:

$\displaystyle \text{Let } \overrightarrow{a} = \hat{i} + \hat{j} + \hat{k} \text{ and } \overrightarrow{b} = p\hat{i} + \hat{j} -2 \hat{k}$

$\displaystyle \text{Then projection of } \overrightarrow{a} \text{ on } \overrightarrow{b} \text{ is } \frac{\overrightarrow{a}. \overrightarrow{b}}{|\overrightarrow{b}|}$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{\overrightarrow{a}. \overrightarrow{b}}{|\overrightarrow{b}|}= \frac{(\hat{i} + \hat{j} + \hat{k} ).(p\hat{i} + \hat{j} -2 \hat{k})}{\sqrt{P^2+1^2+2^2}} = \frac{1}{3}$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{P-1+2}{\sqrt{P^2+5}} = \frac{1}{3}$

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

$\\$

$\displaystyle \text{Question 24: Find the point on the curve } y^2 = 8x \text{ for which the abscissa and } \\ \text{ordinate change at the same rate. }$

Answer:

$\displaystyle \text{Given, } y^2 = 8x$

Derivative of given equation w.r.t $x$

$\displaystyle \Rightarrow \hspace{1.0cm} 2y \frac{dy}{dx} = 8$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{dy}{dx} = \frac{8}{2y} = 1$

$\displaystyle \Rightarrow \hspace{1.0cm} y = 4$

Derivative of given equation w.r.t $y$

$\displaystyle \Rightarrow \hspace{1.0cm} 2y = 8 \frac{dy}{dx}$

$\displaystyle \Rightarrow \hspace{1.0cm} 2y = 8$

$\displaystyle \Rightarrow \hspace{1.0cm} y = 4$

$\displaystyle \text{At } y = 4, x = \frac{16}{8} = 2 \Rightarrow x = 2$

$\\$

$\displaystyle \text{ Question 25:}$

$\displaystyle \text{(a) Find the vector equation of the line passing through the point } (2, 1, 3) \\ \text{ and perpendicular to both the lines}$

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

$\displaystyle \text{(b) The equations of a line are } 5x-3 = 15y + 7 = 3 - 10z. \text{ Write the direction } \\ \text{cosines of the line and find the coordinates of a point through which it passes. }$

Answer:

(a)

Let the cartesian equation of the line passing through $(2, 1, 3)$ be

$\displaystyle \frac{x-2}{a}= \frac{y-2}{b} = \frac{z-3}{c} \ldots \ldots (i)$

Since, line (i) is perpendicular to given lines

$\displaystyle \frac{x-1}{1}= \frac{y-2}{2} = \frac{z-3}{3} \ldots \ldots (ii)$

$\displaystyle \text{and } \frac{x}{-3} = \frac{y}{2} = \frac{z}{5} \ldots \ldots (iii)$

$\displaystyle \therefore a + 2b+3c=0 \ldots \ldots (iv)$

$\displaystyle -3a+2b+5c=0 \ldots \ldots (v)$

From equation (iv) and (v).

$\displaystyle \frac{a}{10-6}= \frac{b}{-9-5} = \frac{c}{2+6}$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{a}{4}= \frac{b}{-14} = \frac{c}{8} = \lambda$

$\displaystyle \Rightarrow \hspace{1.0cm} a= 4\lambda, b = - 14 \lambda, c = 8\lambda$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{x-2}{2}= \frac{y-1}{-7} = \frac{z-3}{4}$

which is the cartesian form The vector form is

$\displaystyle \overrightarrow{r} = (2\hat{i}+\hat{j}+3\hat{j} ) + \lambda(2\hat{i}-7\hat{j}+4\hat{j} )$

(b)

Given equations of a line is

$\displaystyle 5x-3 = 15y+3 = 3-10z \ldots \ldots (i)$

Let us first convert the equation in standard form

$\displaystyle \frac{x-x_1}{a}= \frac{y-y_1}{b}= \frac{z-z_1}{c} \ldots \ldots (ii)$

$\displaystyle \text{Let us divide equation (i) by LCM (coefficient of } x, y, \text{ and } z \text{ ) i.e. LCM of } (5, 15, 10) = 30$

Now equation (i) becomes

$\displaystyle \frac{5x-3}{30}= \frac{15y+7}{30}= \frac{3-10z}{30}$

$\displaystyle \text{or } \frac{5\Big( x - \frac{3}{5} \Big)}{30}= \frac{15\Big( y + \frac{7}{15} \Big)}{30}= \frac{-10\Big( z - \frac{3}{10} \Big)}{30}$

$\displaystyle \text{or } \frac{\Big( x - \frac{3}{5} \Big)}{6}= \frac{\Big( y + \frac{7}{15} \Big)}{2}= \frac{\Big( z - \frac{3}{10} \Big)}{-3}$

Comparing the above equation with Equation (ii) we get (6, 2, -3) are the direction ratios of the given line. Now the direction cosine of the given line are

$\displaystyle \frac{6}{\sqrt{6^2+2^2+(-3)^2}}, \frac{2}{\sqrt{6^2+2^2+(-3)^2}}, \text{ and } \frac{-3}{\sqrt{6^2+2^2+(-3)^2}}$

$\displaystyle \text{i.e. } \Big( \frac{6}{7}, \frac{2}{7}, \frac{-3}{7} \Big)$

$\\$

SECTION C

This section comprises short answer (SA) type questions of 3 marks each.

$\displaystyle \text{ Question 26: Find } \int \frac{2}{(1-x)(1+x^2)} \ dx$

Answer:

$\displaystyle \text{Let } I = \int \frac{2}{(1-x)(1+x^2)} \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{2}{(1-x)(1+x^2)} = \frac{A}{(x-1)}+\frac{Bx+C}{1+x^2}$

$\displaystyle \Rightarrow \hspace{1.0cm} -2 = A(1+x^2) + (Bx+C)(x-1)$

On comparing, we get

$\displaystyle A+B = 0 \ldots \ldots (i)$

$\displaystyle -B+C = 0 \ldots \ldots (ii)$

$\displaystyle A-C = 2 \ldots \ldots (iii)$

On solving Equation (i), (ii) and (iii) we get

$\displaystyle A = -1 , B = 1, C = 1$

$\displaystyle \therefore \int \frac{2}{(1-x)(1+x^2)} \ dx = \int \frac{-1}{x-1} \ dx + \int \frac{x+1}{x^2+1} \ dx$

$\displaystyle = - \log |x-1| + \frac{1}{2} \int \frac{2x \ dx}{x^2+1} + \int \frac{dx}{x^2+1}$

$\displaystyle I = - \log |x-1| + \frac{1}{2} \log (x^2+1) + \tan^{-1} x + C$

$\\$

$\displaystyle \text{Question 27: }$

$\displaystyle \text{(a) Evaluate: } \int \limits_{\frac{1}{3}}^{1} \frac{(x-x^3)^{1/3}}{x^4} \ dx$

$\displaystyle \text{(b) Evaluate: } \int \limits_{1}^{3} \Bigg\{ \Big| (x-1) \Big| +\Big|(x-2) \Big| \Bigg\} \ dx$

Answer:

(a)

$\displaystyle \text{Let } I = \int \limits_{\frac{1}{3}}^{1} \frac{(x-x^3)^{1/3}}{x^4} \ dx$

$\displaystyle = \int \limits_{\frac{1}{3}}^{1} \frac{x \Big(\frac{1}{x^2} -1 \Big)^{1/3} }{x^3} \ dx$

$\displaystyle =\int \limits_{\frac{1}{3}}^{1} \frac{\Big(\frac{1}{x^2} -1 \Big)^{1/3} }{x^2} \ dx$

$\displaystyle \text{Let } \frac{1}{x^2} -1 = t, \text{ then } -\frac{2}{x^3} \ dx = dt$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int -t^{1/3} \ dt$

$\displaystyle = \frac{-3t^{4/3}}{4} \times \frac{1}{2}$

$\displaystyle = \frac{-3}{8} \Bigg[ \Big( \frac{1}{x^2} -1 \Big)^{4/3} \Bigg]_{1/3}^{1}$

$\displaystyle = \frac{-3}{8} [ 0 - (8)^{4/3}]$

$\displaystyle = \frac{3}{8} ( 2^3)^{4/3}$

$\displaystyle = \frac{3}{8} \times 16$

$\displaystyle = 2$

(b)

$\displaystyle I = \int \limits_{1}^{3} \Bigg\{ \Big| (x-1) \Big| +\Big|(x-2) \Big| \Bigg\} \ dx$

$\displaystyle \text{In given limit } |x-1| = (x-x)$

$\displaystyle \text{and } |x-2| = \Bigg\{ \begin{array}{rl} -(x-2), & x< 2 \\ x-2, & x> 2 \end{array}$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int \limits_{1}^{2} [(x-1) - (x-2) ] \ dx + I = \int \limits_{2}^{3} [(x-1)+(x-2) ] \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int \limits_{1}^{2} \ dx + I = \int \limits_{2}^{3} (2x-3) \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = [x]_{1}^{2} + \Big[x^2 - 3x \Big]_{2}^{3} \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = (2-1) + [0-(4-6)] \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = 1+2 = 3$

$\\$

$\displaystyle \text{Question 28: Solve the following linear programming problem graphically :} \\ \text{Maximise } z = 5x + 3y \text{ subject to the constraints } 3x + 5y \leq 15, 5x + 2y \leq 10, \\ x, y \geq 0.$

Answer:

$\displaystyle \text{Our problem is to maximize } Z = 5x + 3y \ldots \ldots (i)$

Subject to constraints

$\displaystyle 3x + 5y \leq 15 \ldots \ldots (ii)$

$\displaystyle 5x + 2y \leq 10 \ldots \ldots (iii)$

$\displaystyle x \geq 0, y \geq 0 \ldots \ldots (iv)$

$\displaystyle \text{Firstly, draw the graph of the line } 3x + 5y = 15$

$\displaystyle \text{Secondly, draw the graph of the line } 5x + 2y = 10$

$\displaystyle \text{On solving given equations } 3x + 5y = 15 \text{ and } 5x + 2y = 10, \text{ we get }\\ \\ x = \frac{20}{19} , y = \frac{45}{19 }$

$\displaystyle \text{Feasible region is OABCO (see the below figure). }$

$\displaystyle \text{The corner points of the feasible region are } O(0, 0), A(2, 0), B\Bigg( \frac{20}{19} , \frac{45}{19 } \Bigg) \\ \text{ and } C(0, 3). \text{The values of } Z \text{ at these points are as follows: }$

$\displaystyle \text{Therefore, the maximum value of } Z \text{ is } 235/19 \text{ at the point } B \Bigg( \frac{20}{19} , \frac{45}{19 } \Bigg).$

$\\$

$\displaystyle \text{Question 29: From a lot of 30 bulbs which include 6 defective bulbs, a sample } \\ \text{of 2 bulbs is drawn at random one by one with replacement. Find the probability } \\ \text{distribution of the number of defective bulbs and hence find the mean number} \\ \text{of defective bulbs. }$

Answer:

Non defective bulbs $= 30 - 6 = 24$

Let $X$ be the random variable that denotes the number of defective bulbs.

$\displaystyle P(X=0) = P( \text{ 2 non defective and 0 defective} ) = ^2C_0 \times \frac{24}{30} .\times \frac{24}{30} = \frac{16}{25}$

$\displaystyle P(X=1) = P(\text{ 1 non defective and 1 defective} ) = ^2C_1 \times .\frac{24}{30}\times \frac{6}{30} = \frac{8}{25}$

$\displaystyle P(X=2) = P(\text{ 0 non defective and 2 defective} ) = ^2C_2 \times \frac{6}{30} \times \frac{6}{30} = \frac{1}{25}$

Required probability distribution is.

$\displaystyle \begin{array}{ |l|l|l|l|} \hline \text{Variable } X & 0 \hspace{1.0cm} & 1 \hspace{1.0cm} & 2 \hspace{1.0cm} \\ \hline \text{Mean } P(x) & \frac{16}{25} & \frac{8}{25} & \frac{1}{25} \\ \hline \end{array}$

$\\$

$\displaystyle \text{ Question 30: }$

$\displaystyle \text{ (a) Find the particular solution of the differential equation }$

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

$\displaystyle \text{ (b) Find the general solution of the differential equation }$

$\displaystyle e^x \tan y \ dx + ( 1 - e^x) \sec^2 y \ dy = 0$

Answer:

(a)

$\displaystyle \text{Given } \frac{dy}{dx} = \frac{x+y}{x}, y(1) = 0$

$\displaystyle Let y = vx$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{dy}{dx} = v + x \frac{dv}{dx}$

$\displaystyle \Rightarrow \hspace{1.0cm} v + x \frac{dv}{dx} = \frac{x+vx}{x}$

$\displaystyle \Rightarrow \hspace{1.0cm} v + x \frac{dv}{dx} = 1 + v$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{dv}{dx} = = \frac{1}{x}$

$\displaystyle \Rightarrow \hspace{1.0cm} \int dv = \int \frac{dv}{x}$

$\displaystyle \Rightarrow \hspace{1.0cm} v = \log x + c$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{y}{x} = \log c + c$

$\displaystyle \text{At } x = 1, y = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} c=0$

$\displaystyle \text{Hence, } y = x \log x$

(b)

$\displaystyle \text{Given } e^x \tan y \ dx + ( 1 - e^x) \sec^2 y \ dy = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{e^x}{1 - e^x} dx + \frac{\sec^2 y}{\tan y} dy = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} \int \frac{\sec^2 y}{\tan y} dy = \int \frac{e^x}{e^x - 1} dx$

$\displaystyle \Rightarrow \hspace{1.0cm} \log |\tan y | = \log |e^x - 1| + c$

$\\$

$\displaystyle \text{ Question 31: }$

$\displaystyle \text{(a) Evaluate: } \int \limits_{\pi/4}^{\pi/2} e^{2x} \Big( \frac{1- \sin 2x}{1 - \cos 2x} \Big) \ dx$

$\displaystyle \text{(b) Evaluate: } \int \limits_{-2}^{2} \frac{x^2}{1+5^x} \ dx$

Answer:

(a)

$\displaystyle Let I = \int \limits_{\pi/4}^{\pi/2} e^{2x} \Big( \frac{1- \sin 2x}{1 - \cos 2x} \Big) \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int \limits_{\pi/4}^{\pi/2} e^{2x} \Big( \frac{1}{1- \cos 2x} -\frac{\sin 2x}{1 - \cos 2x} \Big) \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int \limits_{\pi/4}^{\pi/2} e^{2x} \Big( \frac{1}{2 \sin^2 x} -\frac{2 \sin x \cos x}{2 \sin^2 x} \Big) \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int \limits_{\pi/4}^{\pi/2} e^{2x} \Big( \frac{\mathrm{cosec}^2 x}{2} - \cot x \Big) \ dx$

$\displaystyle \text{Put } 2x = t$

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

$\displaystyle \Rightarrow \hspace{1.0cm} dx = \frac{dt}{2}$

$\displaystyle \Rightarrow \hspace{1.0cm} I = - \int e^t \Big( \cot \frac{t}{2} - \frac{1}{2} \mathrm{cosec}^2 \frac{t}{2} \Big) \frac{dt}{2}$

$\displaystyle \Rightarrow \hspace{1.0cm} I = - \frac{1}{2} \Big( e^t \cot \frac{t}{2} \Big)$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \Bigg[ - \frac{1}{2} e^{2x} \cot x \Bigg]_{\pi/4}^{\pi/2}$

$\displaystyle \Rightarrow \hspace{1.0cm} I = - \Bigg[ 0 - \frac{1}{2} e^{\pi/2} \Bigg] = \frac{e^{\pi/2}}{2}$

(b)

$\displaystyle \text{Let } I = \int \limits_{-2}^{2} \frac{x^2}{1+5^x} \ dx \ldots \ldots (i)$

$\displaystyle \text{Using the property, } \int \limits_{a}^{b} f(x) \ dx = \int \limits_{a}^{b} f(a+b-x) \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int \limits_{-2}^{2} \frac{(-2+2-x)^2}{1+5^{-2+2-x}} \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int \limits_{-2}^{2} \frac{x^2}{1+\frac{1}{5^x}} \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \int \limits_{-2}^{2} \frac{5^x x^2}{1+5^x} \ dx \ldots \ldots (ii)$

Adding equation (i) and (ii) we get

$\displaystyle \Rightarrow \hspace{1.0cm} 2I = \int \limits_{-2}^{2} \Bigg[ \frac{x^2}{1+5^x} + \frac{5^x x^2}{1+5^x} \Bigg] \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} 2I = \int \limits_{-2}^{2} \Bigg[ \frac{x^2(1+5^x)}{1+5^x} \Bigg] \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} 2I = \int \limits_{-2}^{2} x^2 \ dx = 2 \int \limits_{0}^{2} x^2 \ dx$

$\displaystyle \Rightarrow \hspace{1.0cm} 2I = 2 \Bigg[ \frac{x^3}{3} \Bigg]_{0}^{2} dx$

$\displaystyle \Rightarrow \hspace{1.0cm} I = \frac{1}{2} \times 2 \times \frac{8}{3} = \frac{8}{3}$

$\\$

SECTION D

This section comprises long answer (LA) type questions of 5 marks each.

$\displaystyle \text{ Question 32: }$

$\displaystyle \text{(a) Find the image of the point (2, -1, 5) on line}$

$\displaystyle \frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11}$

$\displaystyle \text{(b) Vertices B and C of } \triangle ABC \text{ lie on the line } \frac{x+2}{2} = \frac{y-1}{1} = \frac{z}{4}. \\ \text{ Find the area of the } \triangle ABC \text{ given that the point } A \text{ has coordinates } (1, -1, 2) \\ \text{ and the line segment } BC \text{ has length of 5 units. }$

Answer:

(a)

$\displaystyle \text{The given line } \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11}$

$\displaystyle \text{Let } N \text{ be the foot of the perpendicular drawn from the point } P(2, -1, 5).$

$\displaystyle \text{ point on line (1) is } N (11 + 10t, -2 - 4t, -8 - 11t).$

$\displaystyle \text{Now, direction ratio of NP is: } ( 9 + 10t,-1 - 4t, -13 -11t ) \text{ and }$

$\displaystyle \text{Direction ratio of line (1) is } ( 10, -4,-11 )$

$\displaystyle 10(9 + 10t) + 4(1 + 4t) + 11(13 + 11t) = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} 90 + 100t + 4 + 16t+ 143+ 121t = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} 237t + 237 = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} t = -1 N = (1, 2, 3)$

$\displaystyle \text{Now, image of point } P (2, -1, 5) \text{ is } \Big( \frac{2+x}{2}, \frac{-1+y}{2}, \frac{5+z}{2} \Big)$

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

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{-1+y}{2} =2 \Rightarrow \hspace{1.0cm} y = 5$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{5+z}{2} =3 \Rightarrow \hspace{1.0cm} z = 1$

Hence, image is $(0, 5, 1)$

(b)

$\displaystyle BC \text{ lie on } \frac{x+2}{2} = \frac{y-1}{1} = \frac{z}{4}$

$\displaystyle \text{Coordinates of } D \text{ can be expressed as } ( 2 \lambda -2, \lambda + 1 , 4 \lambda)$

$\displaystyle \text{Direction ratio of } AD \text{ are } (2 \lambda - 3, \lambda + 2, 4 \lambda - 2)$

$\displaystyle AD \text{ is perpendicular to } BC$

$\displaystyle \text{So, } 2 ( 2 \lambda - 3) + ( \lambda + 2) + 4 ( 4 \lambda - 2) = 0$

$\displaystyle \Rightarrow \hspace{1.0cm} \lambda = \frac{12}{21} \Rightarrow z = 1$

$\displaystyle |AD| = \sqrt{ ( 2 \lambda - 3)^2 + ( \lambda + 2)^2 + 4 ( 4 \lambda - 2)^2 }$

$\displaystyle = \sqrt{ \Big( \frac{39}{21} \Big)^2 + \Big( \frac{54}{21} \Big)^2 + \Big( \frac{6}{21} \Big)^2}$

$\displaystyle = \sqrt{ \frac{4473}{441} }$

$\displaystyle \text{Area } = \frac{1}{2} \times 5 \times \sqrt{ \frac{4473}{441} } = \sqrt{ \frac{1775}{28} }$

$\\$

$\displaystyle \text{Question 33: Find the inverse of the matrix } A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} . \\ \\ \text{ Using the inverse } A^{-1}, \text{ solve the system of linear equations } \\ \\ x- y + 2z = 1; \hspace{0.5cm} 2y- 3z = 1; \hspace{0.5cm} 3x- 2y + 4z = 3.$

Answer:

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

$\displaystyle \Rightarrow \hspace{1.0cm} A = 1(8-6) + (0+9) -2 (0-6) = -1$

On finding adjoin of A

$\displaystyle F_{11} = \begin{bmatrix} 2 & -3 \\ -2 & 4 \end{bmatrix} = 2 \hspace{1.0cm} F_{12} = \begin{bmatrix} 0 & -3 \\ 3 & 4 \end{bmatrix} = -9 \hspace{1.0cm} F_{13} = \begin{bmatrix} 0 & 2 \\ 3 & -2 \end{bmatrix} = -6$

$\displaystyle F_{21} = \begin{bmatrix} -1 & 2 \\ -2 & 4 \end{bmatrix} = 0 \hspace{1.0cm} F_{22} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = -2 \hspace{1.0cm} F_{23} = -\begin{bmatrix} 1 & -1 \\ 3 & -2 \end{bmatrix} = 1$

$\displaystyle F_{31} = \begin{bmatrix} -1 & 2 \\ 2 & -3 \end{bmatrix} = -1 \hspace{1.0cm} F_{32} = -\begin{bmatrix} 1 & 2 \\ 0 & -3 \end{bmatrix} = 3 \hspace{1.0cm} F_{33} = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 2$

Matrix formed by adjoint of $A$ is

$\displaystyle \text{adj } A = B^{T} = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & -3 \\ -6 & -1 & 2 \end{bmatrix}$

$\displaystyle \text{Then } A^{-1} = \begin{bmatrix} 2 & -9 & -6 \\ 0 & -2 & -1 \\ -1 & 3 & 2 \end{bmatrix} = \frac{1}{-1} \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & 1 & 2 \end{bmatrix}$

$\displaystyle \text{Now, } AX= B$

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

$\displaystyle \Rightarrow \hspace{1.0cm} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{-1} \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & 1 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}= \frac{1}{-1} \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix}$

$\displaystyle \Rightarrow \hspace{1.0cm} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2\\ 1 \end{bmatrix}$

$\displaystyle \Rightarrow \hspace{1.0cm} x=1, y = 2, x = 1$

$\\$

$\displaystyle \text{ Question 34: Using integration, find the area of the region bounded by the parabola } \\ y^2 = 4ax \text{ and its latus rectum. }$

Answer:

$\displaystyle \text{Given } y^2 = 4ax$

$\displaystyle \text{Then the equation for latus rectum is } x = a$

$\displaystyle \text{Required area }= 2 ( \text{ area of AOL } )$

$\displaystyle = 2 \int \limits_{0}^{a} y \ dx = 2 \int \limits_{0}^{a} 2 \sqrt{a} \sqrt{x} \ dx$

$\displaystyle = 4 \sqrt{a} \int \limits_{0}^{a}\sqrt{x} \ dx$

$\displaystyle = 4 \sqrt{a} \Bigg[ \frac{2x^{3/2}}{3} \Bigg]_{0}^{a}$

$\displaystyle = 4 \sqrt{a} \times \frac{2}{3} [ a^{3/2} - 0 ]$

$\displaystyle = \frac{8}{3} a^2 \text{ sq. units }$

$\\$

$\displaystyle \text{Question 35:}$

$\displaystyle \text{(a) If } N \text{denotes the set of all natural numbers and } R \text{ is the relation on } \\ N \times N \text{ defined by } (a, b) R (c, d), \text{ if } ad(b + c) = bc(a + d). \text{ Show that } R \text{ is an equivalence } \\ \text{ relation. }$

$\displaystyle \text{(b) Let } f : R - \Big\{ -\frac{4}{3} \Big\} \rightarrow R \text{ be a function defined as } f(x) = \frac{4x}{3x+4}. \\ \text{ Show that } f \text{ is a one-one function. Also, check whether } f \text{ is an onto function or not. }$

Answer:

(a)

$\displaystyle \text{Given } (a, b) \ R \ (c, d), \text{ as } ad(b + c) = bc(a + d)$

$\displaystyle \therefore \forall a, b \in N$

$\displaystyle \text{or } ab(b + a) = ba(a + b) \ldots \ldots (i)$

$\displaystyle \text{or } (a, b) \ R \ (c, d)$

$\displaystyle \therefore R \text{ is reflexive. }$

$\displaystyle \text{Let } (a, b) \ R \ (c, d) \text{ for } (a, b) , (c, d) \in N \times N$

$\displaystyle \therefore ad(b + c) = bc(a + d) \ldots \ldots (ii)$

$\displaystyle \text{Also, } (c, d) \ R \ (a, b)$

$\displaystyle \because cb(d+a) = da(c+b) \text{ [ By communication of addition and multiplication on N] }$

$\displaystyle \therefore R \text{ is symmetric. } \ldots \ldots (iii)$

$\displaystyle \text{Let } (a, b) \ R \ (c, d) \text{ and } (c, d) \ R \ ( e, f) \text{ for } a, b, c, d, e, f \in N$

$\displaystyle \therefore ad(b+c) = bc( a+d) \ldots \ldots (iv)$

$\displaystyle \text{and } cf(d+e) = de(c+f) \ldots \ldots (v)$

$\displaystyle \text{Dividing equation (iv) by } abcd \text{ and equation (v) by } cdef \text{ we get }$

$\displaystyle \frac{1}{c} + \frac{1}{b} = \frac{1}{d}+ \frac{1}{a}$

$\displaystyle \text{and } \frac{1}{e} + \frac{1}{d} = \frac{1}{f}+ \frac{1}{c}$

On adding, we get

$\displaystyle \frac{1}{c} + \frac{1}{b} + \frac{1}{e} + \frac{1}{d} = \frac{1}{d}+ \frac{1}{a} + \frac{1}{f}+ \frac{1}{c}$

$\displaystyle \text{or } af(b+e) = be(a+f)$

$\displaystyle \text{Hence } (a,b) \ R \ (e,f)$

$\displaystyle R \text{ is transitive. }$

$\displaystyle \text{From equations (i), (iii) and (iv), } R \text{ is an equivalence relation. }$

(b)

$\displaystyle \text{Given } f(x) = \frac{4x}{3x+4}$

$\displaystyle \Rightarrow \hspace{1.0cm} f(x) = f(y)$

$\displaystyle \Rightarrow \hspace{1.0cm} \frac{4x}{3x+4} = \frac{4y}{3y+4}$

$\displaystyle \Rightarrow \hspace{1.0cm} 12xy + 16 x = 12xy + 16 y$

$\displaystyle \Rightarrow \hspace{1.0cm} 16 x = 16 y$

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

$\displaystyle \text{Therefore } f \text{ is one-one }$

$\displaystyle \Rightarrow \hspace{1.0cm} f(y) = \frac{4x}{3x+4}$

$\displaystyle \Rightarrow \hspace{1.0cm} 4x = 3xy + 4y$

$\displaystyle \Rightarrow \hspace{1.0cm} x = \frac{4x}{4-3y}$

$\displaystyle \text{Hence, }y \in R - \Big\{ \frac{4}{3} \Big\}$

$\displaystyle \text{Hence, every element in } R - \Big\{ \frac{4}{3} \Big\} \text{ has a pre-image in } R - \Big\{ -\frac{4}{3} \Big\}$

$\displaystyle \text{Hence,e } f \text{ is onto }$

$\\$

SECTION E

This section comprises 3 case study based questions of 4 marks each.

Question 36: 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 ?

(ii) What is the probability that construction will be completed on time ?

(iii)

(a) What is the probability that many workers are not present given that the construction work is completed on time ?

(b) What is the probability that all workers were present given that the construction job was completed on time ?

Answer:

(a)

Given that:

$E_1 :$ 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.

$\displaystyle \therefore P(E_1) = 0.65, \hspace{1.0cm} P(E/E_1) = 0.35 , \hspace{1.0cm} P(E/E_2) = 0.80$

$\displaystyle (i) P ( \text{when all workers were present for the job }) = P(E_2) = 1 - 0.65 = 0.35$

$\displaystyle (ii) P ( \text{construction will be completed on time } )$

$\displaystyle = P(E) = P(E_1) . P(E/E_1) + P(E_2) . P(E/E_2)$

$\displaystyle = 0.65 \times 0.35 + 0.35 \times 0.80$

$\displaystyle = \frac{13}{20} \times \frac{7}{20} + \frac{7}{20} \times \frac{16}{20} = \frac{91}{400} \times \frac{112}{400}$

$\displaystyle = \frac{203}{400} = 0.51$

(iii)

$\displaystyle \text{(a) P(Many workers are not present given that construction work is completed on time) }$

$\displaystyle = P(E_1/E) = \frac{P(E/E_1) . P(E_1)}{P(E/E_1) . P(E_1)+P(E/E_2) . P(E_2)}$

$\displaystyle = \frac{\frac{13}{20} \times \frac{7}{20} }{\frac{203}{400}} = \frac{\frac{91}{400}}{\frac{203}{400}} =\frac{91}{203} = 0.45$

$\displaystyle \text{(b) P(All workers are present given that construction work is completed on time) }$

$\displaystyle = P(E_2/E) = \frac{P(E/E_2) . P(E_2)}{P(E/E_1) . P(E_1)+P(E/E_2) . P(E_2)}$

$\displaystyle = \frac{\frac{16}{20} \times \frac{7}{20} }{\frac{203}{400}} = \frac{\frac{112}{400}}{\frac{203}{400}} =\frac{112}{203} = 0.55$

$\\$

$\displaystyle \text{Question 37: }$

$\displaystyle \text{Let } f(x) \text{ be a real valued function. Then its}$

$\displaystyle \text{Left Hand Derivative (L.H.D.) } : Lf'(a) = \lim \limits_{h \to 0} \frac{f(a-h)-f(a)}{-h}$

$\displaystyle \text{Right Hand Derivative (R.H.D.) } : Lf'(a) = \lim \limits_{h \to 0} \frac{f(a+b)-f(a)}{h}$

$\displaystyle \text{Also, a function } f(x) \text{ is said to be differentiable at }x = a \text{ if its L.H.D. and R.H.D. at } \\ x = a \text{ exist and both are equal. }$

$\displaystyle \text{For the function }= \Bigg\{ \begin{array}{ll} |x-3|, & x \geq 1 \\ & \\ \frac{x^2}{4}-\frac{3x}{2} + \frac{13}{4}, & x < 1 \end{array}$

answer the following questions:

$\displaystyle \text{(i) What is R.H.D. of } f(x) \text{ at } x = 1 ?$

$\displaystyle \text{(ii) What is L.H.D. of } f(x) \text{ at } x = 1 ?$

$\displaystyle \text{(iii) (a) Check if the function } f(x) \text{ is differentiable at } x = 1$

$\displaystyle \text{(iii) (b) Find the } f'(2) \text{ and } f'( 1).$

Answer:

$\displaystyle \text{(i) Given } \Bigg\{ \begin{array}{ll} |x-3|, & x \geq 1 \\ & \\ \frac{x^2}{4}-\frac{3x}{2} + \frac{13}{4}, & x < 1 \end{array}$

$\displaystyle \text{RHD of } f(x) \text{ at } x = 1 \text{ is defined as } \lim \limits_{x \to 1^+} f(1^+)$

$\displaystyle |x-3| = \Bigg\{ \begin{array}{ll} -(x-3), & x < 3\\ & \\ (x-3), & x > 3 \end{array}$

$\displaystyle \Rightarrow \hspace{1.0cm} \lim \limits_{x \to 1^+} |x-3|= \lim \limits_{x \to 1^+} (3-x) = 3-1 = 2$

(ii) $\displaystyle \text{LHD of } f(x) \text{ at } x = 1 \text{ is defined as } \lim \limits_{x \to 1^-} f(1^-)$

$\displaystyle \text{At } x \to 1^{-1}, f(x) = \frac{x^2}{4}-\frac{3x}{2} + \frac{13}{4}$

$\displaystyle \Rightarrow \hspace{1.0cm} \lim \limits_{x \to 1^-} f(1^-) = \frac{1}{4} - \frac{3}{2} + \frac{13}{4} = \frac{1}{4} -\frac{6}{4} + \frac{13}{4} = \frac{8}{4} = 2$

(iii)

$\displaystyle \text{(a) } f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

LHS

$\displaystyle \text{At } x \to 1+ h, \text{where } h \to 0$

$\displaystyle f'(x) = \lim \limits_{h \to 0} \frac{f(1+h) - f(1)}{h} = \frac{-(1+h-3)-2}{h} = \frac{-h+2-2}{h} = -1$

RHS

$\displaystyle \text{At } x \to 1- h, \text{where } h \to 0$

$\displaystyle f'(x) = \lim \limits_{h \to 0} \frac{f(1-h) - f(1)}{h}$

$\displaystyle = \frac{ \frac{(1-h)^2}{4}-\frac{3(1-h)}{2} + \frac{13}{4} - \Big( \frac{1}{4}-\frac{6}{2} + \frac{13}{4} \Big) }{h}$

$\displaystyle = \frac{h^2-2h+6h}{4h}$

$\displaystyle = \frac{h-2+6}{4}$

$\displaystyle = 1$

$\displaystyle \text{Hence, the given function is not differentiable at } x=1 \text{ as RHS } \neq \text{ LHS }$

$\displaystyle \text{(b) At }x = 2, f(x) = -(x-3)$

$\displaystyle \text{Then, } f'(x) = - 1$

$\displaystyle \text{Hence, } f'(2) = - 1$

$\displaystyle \text{And } x = - 1, f(x) = \frac{x^2}{4}-\frac{3x}{2} + \frac{13}{4}$

$\displaystyle \Rightarrow \hspace{1.0cm}f'(x) = \frac{2x}{4} - \frac{3}{2}$

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

$\\$

Question 38:

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 meters of fencing wire.

Based on the above information, answer the following questions :

(i) Let $x'$ meters denote the length of the side of the garden perpendicular to the brick wall and $y'$ meters denotes 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.