MATHEMATICS

\displaystyle \text{Series EF1GH/5} \hspace{1.0cm} | \hspace{1.0cm} \text{Q. P. Code 65/5/2} \hspace{1.0cm} | \hspace{1.0cm} \text{Set 2 }    

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

\displaystyle \textbf{Time Allowed: 3 Hours}\hspace{6cm}\textbf{Maximum Marks: 100}
\displaystyle \textbf{General Instructions:}
\displaystyle \text{Read the following instructions very carefully and follow them:}
\displaystyle \text{(i) This question paper contains 38 questions. All questions are compulsory.}
\displaystyle \text{(ii) Question paper is divided into FIVE Sections - Section A, B, C, D and E.}
\displaystyle \text{(iii) In Section A - Question Number 1 to 18 are Multiple Choice Questions (MCQ)}
\displaystyle \text{type and Question Number 19 \& 20 are Assertion-Reason based questions of 1 mark each.}
\displaystyle \text{(iv) In Section B - Question Number 21 to 25 are Very Short Answer (VSA)}
\displaystyle \text{type questions of 2 marks each.}
\displaystyle \text{(v) In Section C - Question Number 26 to 31 are Short Answer (SA) type questions,}
\displaystyle \text{carrying 3 marks each.}
\displaystyle \text{(vi) In Section D - Question Number 32 to 35 are Long Answer (LA) type questions}
\displaystyle \text{carrying 5 marks each.}
\displaystyle \text{(vii) In Section E - Question Number 36 to 38 are case study based questions}
\displaystyle \text{carrying 4 marks each where 2 VSA type questions are of 1 mark each}
\displaystyle \text{and 1 SA type question is of 2 marks. Internal choice is provided in 2 marks}
\displaystyle \text{question in each case-study.}
\displaystyle \text{(viii) There is no overall choice. However, an internal choice has been provided}
\displaystyle \text{in 2 questions in Section B, 3 questions in Section C, 2 questions in Section D}
\displaystyle \text{and 2 questions in Section E.}
\displaystyle \text{(ix) Use of calculators is NOT allowed.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Select the correct option out of the four given options:}

\displaystyle \textbf{1. }\sin\left[\frac{\pi}{3}+\sin^{-1}\left(\frac{1}{2}\right)\right]\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }\frac{1}{2}
\displaystyle \text{(c) }\frac{1}{3}\qquad\text{(d) }\frac{1}{4}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a) }\sin\left[\frac{\pi}{3}+\sin^{-1}\left(\frac{1}{2}\right)\right]=\sin\left[\frac{\pi}{3}+\frac{\pi}{6}\right]=\sin\left(\frac{2\pi+\pi}{6}\right)
\displaystyle =\sin\left(\frac{3\pi}{6}\right)=\sin\left(\frac{\pi}{2}\right)=1\qquad

\displaystyle \textbf{2. }\text{Let }A=\{3,5\}\text{. Then number of reflexive relations on }A\text{ is}
\displaystyle \text{(a) }2\qquad\text{(b) }4
\displaystyle \text{(c) }0\qquad\text{(d) }8
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given set }A=\{3,5\}\Rightarrow n(A)=2
\displaystyle \text{Number of reflexive relations on }A=2^{(n^2-n)}=2^{(2^2-2)}=2^{(4-2)}=2^2=4

\displaystyle \textbf{3. }\text{If }A=\begin{bmatrix}1&0\\2&1\end{bmatrix},\ B=\begin{bmatrix}x&0\\1&1\end{bmatrix}\text{ and }A=B^2\text{, then equals}
\displaystyle \text{(a) }\pm 1\qquad\text{(b) }-1
\displaystyle \text{(c) }1\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }A=\begin{bmatrix}1&0\\2&1\end{bmatrix}\text{ and }B=\begin{bmatrix}x&0\\1&1\end{bmatrix}
\displaystyle \text{Since, given }A=B^2
\displaystyle \Rightarrow \begin{bmatrix}1&0\\2&1\end{bmatrix}=\begin{bmatrix}x\cdot x+0\cdot 1&x\cdot 0+0\cdot 1\\x\cdot 1+1\cdot 1&1\cdot 0+1\cdot 1\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}1&0\\2&1\end{bmatrix}=\begin{bmatrix}x^2&0\\x+1&1\end{bmatrix}
\displaystyle \text{If }x^2=1\Rightarrow x=\pm 1
\displaystyle \text{If }x+1=2\Rightarrow x=1
\displaystyle \text{So }x=1\text{ only}\qquad

\displaystyle \textbf{4. }\text{If }A=[a_{ij}]\text{ is a square matrix of order 2 such that}
\displaystyle a_{ij}=\begin{cases}1,\ \text{when }i\neq j\\0,\ \text{when }i=j\end{cases}\text{ then }A^2\text{ is}
\displaystyle \text{(a) }\begin{bmatrix}1&0\\1&0\end{bmatrix}\qquad\text{(b) }\begin{bmatrix}1&1\\0&0\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}1&1\\1&0\end{bmatrix}\qquad\text{(d) }\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }A=[a_{ij}]_{2\times 2}
\displaystyle \text{Such that }a_{ij}=\begin{cases}1,\ \text{when }i\neq j\\0,\ \text{when }i=j\end{cases}
\displaystyle \text{So, }A=\begin{bmatrix}0&1\\1&0\end{bmatrix}
\displaystyle \text{Now, }A^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\qquad

\displaystyle \textbf{5. }\text{The value of the determinant }\begin{vmatrix}6&0&-1\\2&1&4\\1&1&3\end{vmatrix}\text{ is}
\displaystyle \text{(a) }10\qquad\text{(b) }8
\displaystyle \text{(c) }7\qquad\text{(d) }-7
\displaystyle \text{Answer:}
\displaystyle \text{(d) Since, }\begin{vmatrix}6&0&-1\\2&1&4\\1&1&3\end{vmatrix}
\displaystyle =6(3-4)-0(6-4)-1(2-1)
\displaystyle =6\times(-1)-0-1\times 1=-6-1=-7\qquad

\displaystyle \textbf{6. }\text{The function }f(x)=[x]\text{, where }[x]\text{ denotes the greatest integer less than or equal} \\ \text{to }x\text{, is continuous at}
\displaystyle \text{(a) }x=1\qquad\text{(b) }x=1.5
\displaystyle \text{(c) }x=-2\qquad\text{(d) }x=4
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given the function }f(x)=[x]
\displaystyle \Rightarrow [x]=\begin{cases}-2,\ -2\leq x<-1\\-1,\ -1\leq x<0\\0,\ 0\leq x<1\\1,\ 1\leq x<2\\2,\ 2\leq x<3\end{cases}
\displaystyle \text{We can see easily }f(x)=[x]\text{ is not continuous at all integers.}
\displaystyle \text{So }f(x)=[x]\text{ is continuous at }x=1.5

\displaystyle \textbf{7. }\text{The derivative of }x^{2x}\text{ w.r.t. }x\text{ is}
\displaystyle \text{(a) }x^{2x-1}\qquad\text{(b) }2x^{2x}\log x
\displaystyle \text{(c) }2x^{2x}(1+\log x)\qquad\text{(d) }2x^{2x}(1-\log x)
\displaystyle \text{Answer:}
\displaystyle \text{(c) Let }y=x^{2x}
\displaystyle \text{Since }y=x^{2x}=e^{\ln(x^{2x})}=e^{2x\ln x}
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{d}{dx}\left(e^{2x\ln x}\right)=e^{2x\ln x}\frac{d}{dx}(2x\ln x)
\displaystyle =y\left(2x\frac{d}{dx}(\ln x)+2\ln x\frac{d}{dx}(x)\right)
\displaystyle =y\left(2x\cdot\frac{1}{x}+2\ln x\right)=y(2+2\ln x)
\displaystyle =x^{2x}(2+2\ln x)=2x^{2x}(1+\ln x)

\displaystyle \textbf{8. }\text{The interval in which the function }f(x)=2x^3+9x^2+12x-1\text{ is decreasing, is}
\displaystyle \text{(a) }(-1,\infty)\qquad\text{(b) }(-2,-1)
\displaystyle \text{(c) }(-\infty,-2)\qquad\text{(d) }[-1,1]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given the function }f(x)=2x^3+9x^2+12x-1
\displaystyle \text{Now, }f'(x)=\frac{d}{dx}(2x^3+9x^2+12x-1)
\displaystyle \Rightarrow f'(x)=6x^2+18x+12
\displaystyle \text{Since, the interval for which the given function is decreasing }f'(x)<0
\displaystyle \Rightarrow 6x^2+18x+12<0
\displaystyle \Rightarrow (x+2)(x+1)<0
\displaystyle \text{So }x\in(-2,-1)

\displaystyle \textbf{9. }\text{The function }f(x)=x|x|,\ x\in R\text{ is differentiable}
\displaystyle \text{(a) only at }x=0\qquad\text{(b) only at }x=1
\displaystyle \text{(c) in }R\qquad\text{(d) in }R-\{0\}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given the function }f(x)=x|x|,\ x\in R
\displaystyle f(x)=\begin{cases}-x^2,\ \text{if }x<0\\x^2,\ \text{if }x\geq 0\end{cases}\Rightarrow f'(x)=\begin{cases}-2x,\ \text{if }x<0\\2x,\ \text{if }x\geq 0\end{cases}
\displaystyle \text{at }x=0
\displaystyle f'(0)=0\Rightarrow f(x)\text{ is differentiable at }x=0
\displaystyle \text{So }f(x)\text{ is differentiable }\forall x\in R

\displaystyle \textbf{10. }\int \frac{\sec x}{\sec x-\tan x}\,dx\text{ equals}
\displaystyle \text{(a) }\sec x-\tan x+c\qquad\text{(b) }\sec x+\tan x+c
\displaystyle \text{(c) }\tan x-\sec x+c\qquad\text{(d) }-(\sec x+\tan x)+c
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let }I=\int \frac{\sec x}{\sec x-\tan x}\,dx
\displaystyle \Rightarrow I=\int \frac{(\sec^2 x-\tan^2 x)\sec x}{\sec x-\tan x}\,dx
\displaystyle =\int \frac{(\sec x-\tan x)(\sec x+\tan x)\sec x}{\sec x-\tan x}\,dx
\displaystyle =\int (\sec x+\tan x)\sec x\,dx=\int \sec^2 x\,dx+\int \sec x\tan x\,dx
\displaystyle =\tan x+\sec x+c

\displaystyle \textbf{11. }\text{The value of }\int_{0}^{\frac{\pi}{4}} \sin 2x\,dx\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1
\displaystyle \text{(c) }\frac{1}{2}\qquad\text{(d) }-\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Let }I=\int_{0}^{\frac{\pi}{4}}\sin(2x)\,dx
\displaystyle =\left(\frac{-\cos 2x}{2}\right)_{0}^{\frac{\pi}{4}}=-\frac{1}{2}\left\{\cos\left(2\times \frac{\pi}{4}\right)-\cos(2\times 0)\right\}
\displaystyle =-\frac{1}{2}\left\{\cos\left(\frac{\pi}{2}\right)-\cos(0)\right\}=-\frac{1}{2}(0-1)=\frac{1}{2}

\displaystyle \textbf{12. }\text{The sum of the order and the degree of the differential } \text{equation }\frac{d}{dx}\left(\left(\frac{dy}{dx}\right)^3\right)\text{ is}
\displaystyle \text{(a) }2\qquad\text{(b) }3
\displaystyle \text{(c) }5\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{(b) Since, }\frac{d}{dx}\left(\left(\frac{dy}{dx}\right)^3\right)
\displaystyle =3\left(\frac{dy}{dx}\right)^2\cdot \frac{d}{dx}\left(\frac{dy}{dx}\right)=3\left(\frac{dy}{dx}\right)^2\cdot \frac{d^2y}{dx^2}
\displaystyle \text{Now, order }=2,\ \text{degree }=1
\displaystyle \text{So, order + degree }=2+1=3

\displaystyle \textbf{13. }\text{Two vectors }\overrightarrow{a}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}\text{ and }\overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}\text{ are }  \text{collinear if}
\displaystyle \text{(a) }a_1b_1+a_2b_2+a_3b_3=0
\displaystyle \text{(b) }\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}
\displaystyle \text{(c) }a_1=b_1,\ a_2=b_2,\ a_3=b_3
\displaystyle \text{(d) }a_1+a_2+a_3=b_1+b_2+b_3
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given the vectors are}
\displaystyle \overrightarrow{a}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k},\ \overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}
\displaystyle \text{For collinear vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}=\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}

\displaystyle \textbf{14. }\text{A unit vector }\overrightarrow{a}\text{ makes equal but acute angles on the}  \text{co-ordinate axes. The} \\ \text{projection of the vector }\overrightarrow{a}\text{ on the}   \text{vector }\overrightarrow{b}=5\widehat{i}+7\widehat{j}-\widehat{k}\text{ is}
\displaystyle \text{(a) }\frac{11}{15}\qquad\text{(b) }\frac{11}{5\sqrt{3}}
\displaystyle \text{(c) }\frac{4}{5}\qquad\text{(d) }\frac{3}{5\sqrt{3}}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given }|\overrightarrow{a}|=1
\displaystyle \text{Since, }\ell=\cos\alpha=m=n
\displaystyle \text{Now, }\ell^2+m^2+n^2
\displaystyle \Rightarrow \cos^2\alpha+\cos^2\alpha+\cos^2\alpha=1\Rightarrow 3\cos^2\alpha=1
\displaystyle \Rightarrow \cos\alpha=\frac{1}{\sqrt{3}}\text{ So, }\overrightarrow{a}=\frac{1}{\sqrt{3}}\widehat{i}+\frac{1}{\sqrt{3}}\widehat{j}+\frac{1}{\sqrt{3}}\widehat{k}
\displaystyle \text{Now, projection of the vector }\overrightarrow{a}\text{ on the vector }\overrightarrow{b}=5\widehat{i}+7\widehat{j}-\widehat{k}\text{ is}
\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|}=\frac{\left(\frac{1}{\sqrt{3}}\widehat{i}+\frac{1}{\sqrt{3}}\widehat{j}+\frac{1}{\sqrt{3}}\widehat{k}\right).(5\widehat{i}+7\widehat{j}-\widehat{k})}{\sqrt{5^2+7^2+(-1)^2}}
\displaystyle =\frac{\frac{5}{\sqrt{3}}+\frac{7}{\sqrt{3}}-\frac{1}{\sqrt{3}}}{\sqrt{25+49+1}}=\frac{5+7-1}{\sqrt{75}\cdot \sqrt{3}}=\frac{11}{5\sqrt{3}\cdot \sqrt{3}}=\frac{11}{15}

\displaystyle \textbf{15. }\text{The angle between the lines }2x=3y=-z\text{ and }6x=-y=-4z\text{ is}
\displaystyle \text{(a) }0^\circ\qquad\text{(b) }30^\circ
\displaystyle \text{(c) }45^\circ\qquad\text{(d) }90^\circ
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given the lines }2x=3y=-z\text{ and }6x=-y=-4z
\displaystyle \frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{-1}\text{ and }\frac{x}{\frac{1}{6}}=\frac{y}{-1}=\frac{z}{-\frac{1}{4}}
\displaystyle \text{Now, }a_1a_2+b_1b_2+c_1c_2=\frac{1}{2}\times \frac{1}{6}+\frac{1}{3}\times (-1)+(-1)\times \left(-\frac{1}{4}\right)
\displaystyle =\frac{1}{12}-\frac{1}{3}+\frac{1}{4}=\frac{1}{12}-\frac{1}{12}=0
\displaystyle \text{So, angle between the given lines is }90^\circ

\displaystyle \textbf{16. }\text{If a line makes angles of }90^\circ,\ 135^\circ\text{ and }45^\circ\text{ with the }x,\ y  \text{and }z\text{ axes respectively,} \\ \text{then its direction cosines are}
\displaystyle \text{(a) }0,\ -\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}\qquad\text{(b) }-\frac{1}{\sqrt{2}},\ 0,\ \frac{1}{\sqrt{2}}
\displaystyle \text{(c) }\frac{1}{\sqrt{2}},\ 0,\ -\frac{1}{\sqrt{2}}\qquad\text{(d) }0,\ \frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }\alpha=90^\circ,\ \beta=135^\circ,\ \gamma=45^\circ
\displaystyle \text{Now, }\cos\alpha=\cos(90^\circ)=0
\displaystyle \cos\beta=\cos(135^\circ)=\cos(90^\circ+45^\circ)=-\sin(45^\circ)=-\frac{1}{\sqrt{2}}\Rightarrow \cos\gamma=\cos(45^\circ)=\frac{1}{\sqrt{2}}
\displaystyle \text{So, direction cosines are }0,\ -\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}

\displaystyle \textbf{17. }\text{If for any two events }A\text{ and }B,\ P(A)=\frac{4}{5}\text{ and }   P(A\cap B)=\frac{7}{10}\text{, then }P(B/A)\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{10}\qquad\text{(b) }\frac{1}{8}
\displaystyle \text{(c) }\frac{7}{8}\qquad\text{(d) }\frac{17}{20}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given }P(A)=\frac{4}{5},\ P(A\cap B)=\frac{7}{10}
\displaystyle \text{Now, }P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{7}{10}}{\frac{4}{5}}=\frac{7\times 5}{10\times 4}=\frac{7}{8}

\displaystyle \textbf{18. }\text{If }A\text{ and }B\text{ are two independent events such that }  P(A)=\frac{1}{3}\text{ and} \\ P(B)=\frac{1}{4}\text{, then }P(B'/A)\text{ is}
\displaystyle \text{(a) }\frac{1}{4}\qquad\text{(b) }\frac{1}{8}
\displaystyle \text{(c) }\frac{3}{4}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given }P(A)=\frac{1}{3},\ P(B)=\frac{1}{4}\text{ and }A\text{ and }B\text{ are independent events.}
\displaystyle \text{Now, }P\left(\frac{B'}{A}\right)=1-P\left(\frac{B}{A}\right)=1-\frac{P(A\cap B)}{P(A)}
\displaystyle =1-\frac{P(A)\cdot P(B)}{P(A)}=1-P(B)=1-\frac{1}{4}=\frac{3}{4}

\displaystyle \textbf{Assertion - Reason Based Questions}
\displaystyle \text{In the following questions 19 and 20, a statement of Assertion (A) is followed by a}
\displaystyle \text{statement of Reason (R). Choose the correct answer out of the following choices:}
\displaystyle \text{(a) Both (A) and (R) are true and (R) is the correct explanation of (A).}
\displaystyle \text{(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).}
\displaystyle \text{(c) (A) is true and (R) is false.}
\displaystyle \text{(d) (A) is false, but (R) is true.}

\displaystyle \textbf{19. }\text{Assertion (A): }\int_{2}^{8}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx=3
\displaystyle \text{Reason (R): }\int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }I=\int_{2}^{8}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx\qquad \ldots (i)
\displaystyle \Rightarrow I=\int_{2}^{8}\frac{\sqrt{10-10+x}}{\sqrt{10-x}+\sqrt{10-10+x}}\,dx
\displaystyle \Rightarrow I=\int_{2}^{8}\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\,dx\qquad \ldots (ii)
\displaystyle \text{By adding (i) and (ii), we get}
\displaystyle \Rightarrow 2I=\int_{2}^{8}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx+\int_{2}^{8}\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\,dx
\displaystyle \Rightarrow 2I=\int_{2}^{8} dx\Rightarrow 2I=(x)_{2}^{8}=8-2\Rightarrow 2I=6\Rightarrow I=3
\displaystyle \text{and }\int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle \text{So, both A and R are correct and R is correct explanation of A.}

\displaystyle \textbf{20. }\text{Assertion (A): Two coins are tossed simultaneously. The probability of getting}
\displaystyle \text{two heads, if it is known that at least one head comes up, is }\frac{1}{3}\text{.}
\displaystyle \text{Reason (R): Let }E\text{ and }F\text{ be two events with a random experiment, then}
\displaystyle P(F/E)=\frac{P(E\cap F)}{P(E)}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }E_1=\text{getting two head}
\displaystyle E_2=\text{at least one head}
\displaystyle \text{and sample space }=\{(HH),(HT),(TH),(TT)\}
\displaystyle \text{So, }P(E_1)=\frac{1}{4},\ P(E_2)=\frac{3}{4},\ P(E_1\cap E_2)=\frac{1}{4}
\displaystyle \text{Now, }P\left(\frac{E_1}{E_2}\right)=\frac{P(E_1\cap E_2)}{P(E_2)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}
\displaystyle \text{So, Both A and R is correct and R is correct explanation of A.}

\displaystyle \textbf{SECTION - B}
\displaystyle \text{This section comprises of Very Short Answer (VSA) type}
\displaystyle \text{questions of 2 marks each.}

\displaystyle \textbf{21. }\text{Draw the graph of the principal branch of the function}   f(x)=\cos^{-1}x\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Given the function }f(x)=\cos^{-1}x\displaystyle \text{Since, }-1\leq x\leq 1\text{ and }0\leq f(x)\leq \pi
\displaystyle \text{So, graph of the principal branch is a decreasing curve from }(-1,\pi)\text{ to }(1,0)

\displaystyle \textbf{22. (a) }\text{If the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are such that }   |\overrightarrow{a}|=3,\ |\overrightarrow{b}|=\frac{2}{3}\text{ and } \\ \overrightarrow{a}\times\overrightarrow{b}\text{ is a unit vector,}   \text{then find the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{.}
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{Find the area of a parallelogram whose adjacent sides}   \text{are determined} \\ \text{by the vectors}   \overrightarrow{a}=\widehat{i}-\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-7\widehat{j}+\widehat{k}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }|\overrightarrow{a}|=3,\ |\overrightarrow{b}|=\frac{2}{3}\text{ and }|\overrightarrow{a}\times \overrightarrow{b}|=1
\displaystyle \text{Since, }|\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{a}|\times |\overrightarrow{b}|\sin\theta=3\times \frac{2}{3}\sin\theta
\displaystyle \Rightarrow 1=2\sin\theta\Rightarrow \sin\theta=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{6}
\displaystyle \text{So, angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }\frac{\pi}{6}
\displaystyle \text{OR}
\displaystyle \text{(b) Given the adjacent sides of parallelogram are}
\displaystyle \overrightarrow{a}=\widehat{i}-\widehat{j}+3\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}-7\widehat{j}+\widehat{k}
\displaystyle \text{Since, we know that Area of parallelogram }=|\overrightarrow{a}\times \overrightarrow{b}|
\displaystyle \text{Now, }\overrightarrow{a}\times \overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&-1&3\\2&-7&1\end{vmatrix}=20\widehat{i}+5\widehat{j}-5\widehat{k}
\displaystyle \text{So Area of parallelogram }=|20\widehat{i}+5\widehat{j}-5\widehat{k}|
\displaystyle =\sqrt{20^2+5^2+(-5)^2}=\sqrt{400+25+25}=\sqrt{450}=15\sqrt{2}\text{ sq unit.}

\displaystyle \textbf{23. (a) }\text{If }f(x)=\begin{cases}x^2,\ \text{if }x\geq 1\\x,\ \text{if }x<1\end{cases}\text{, then show that }f\text{ is not }   \text{differentiable at } \\ x=1\text{.}
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{Find the value(s) of }\lambda\text{, if the function}
\displaystyle f(x)=\begin{cases}\frac{\sin^2 \lambda x}{x^2},\ \text{if }x\neq 0\\1,\ \text{if }x=0\end{cases}\text{ is continuous at }x=0\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given the function }f(x)=\begin{cases}x^2,\ \text{if }x\geq 1\\x,\ \text{if }x<1\end{cases}
\displaystyle \text{at }x=1,\ \text{Left hand derivative (L.H.D)}
\displaystyle f'_-(1)=\lim_{h\to 0}\frac{f(1-h)-f(1)}{-h}
\displaystyle =\lim_{h\to 0}\frac{(1-h)-(1)^2}{-h}=\lim_{h\to 0}\frac{-h}{-h}=1
\displaystyle \text{at }x=1,\ \text{Right hand derivative (R.H.D)}
\displaystyle f'_+(1)=\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim_{h\to 0}\frac{(1+h)^2-1}{h}=\lim_{h\to 0}\frac{1+h^2+2h-1}{h}
\displaystyle =\lim_{h\to 0}\frac{h(h+2)}{h}=\lim_{h\to 0}(h+2)=2
\displaystyle \text{Since, L.H.D }\neq \text{ R.H.D., so }f(x)\text{ is not differentiable at }x=1

\displaystyle \text{(b) Given the function }f(x)=\begin{cases}\frac{\sin^2(\lambda x)}{x^2},\ \text{if }x\neq 0\\1,\ \text{if }x=0\end{cases}
\displaystyle \text{at }x=0,\ \text{Left hand limit (L.H.L)}
\displaystyle \lim_{x\to 0^-} f(x)=\lim_{x\to 0^-}\frac{\sin^2(\lambda x)}{x^2}
\displaystyle =\lim_{x\to 0^-}\left(\frac{\sin(\lambda x)}{\lambda x}\right)^2\lambda^2=\lambda^2
\displaystyle \text{at }x=0,\ \text{Right hand limit (R.H.L)}
\displaystyle \lim_{x\to 0^+} f(x)=\lim_{x\to 0^+}\frac{\sin^2(\lambda x)}{x^2}
\displaystyle =\lim_{x\to 0^+}\left(\frac{\sin(\lambda x)}{\lambda x}\right)^2\lambda^2=\lambda^2
\displaystyle \text{Since, }f(x)\text{ is continuous at }x=0
\displaystyle \Rightarrow \lim_{x\to 0} f(x)=f(0)
\displaystyle \Rightarrow \lambda^2=1\Rightarrow \lambda=\pm 1

\displaystyle \textbf{24. }\text{Sketch the region bounded by the lines }2x+y=8,\ y=2,\ y=4\text{ and the }y\text{-axis.}
\displaystyle \text{Hence, obtain its area using integration.}
\displaystyle \text{Answer:}
\displaystyle \text{Given the lines }2x+y=8,\ y=2,\ y=4\displaystyle \text{Since, }2x+y=8\Rightarrow 2x=8-y\Rightarrow x=4-\frac{y}{2}
\displaystyle \text{So, required area }=\int_{2}^{4} x\,dy=\int_{2}^{4}\left(4-\frac{y}{2}\right)dy
\displaystyle =\left(4y-\frac{y^2}{4}\right)_{2}^{4}=\left(16-\frac{16}{4}\right)-\left(8-\frac{4}{4}\right)
\displaystyle =(16-4)-(8-1)=12-7=5

\displaystyle \textbf{25. }\text{Find the angle between the following two lines:}
\displaystyle \overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda(3\widehat{i}+2\widehat{j}+6\widehat{k});
\displaystyle \overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu(\widehat{i}+2\widehat{j}+2\widehat{k})
\displaystyle \text{Answer:}
\displaystyle \text{Given the lines }\overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda(3\widehat{i}+2\widehat{j}+6\widehat{k})
\displaystyle \text{and }\overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu(\widehat{i}+2\widehat{j}+2\widehat{k})
\displaystyle \text{Let }\overrightarrow{b_1}=(3,2,6),\ \overrightarrow{b_2}=(1,2,2)
\displaystyle \text{Now, angle between these two lines is given by}
\displaystyle \cos\theta=\frac{\overrightarrow{b_1}\cdot \overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}
\displaystyle =\frac{(3,2,6)\cdot(1,2,2)}{\sqrt{3^2+2^2+6^2}\cdot \sqrt{1^2+2^2+2^2}}
\displaystyle =\frac{3+4+12}{\sqrt{9+4+36}\cdot \sqrt{1+4+4}}=\frac{19}{\sqrt{49}\cdot \sqrt{9}}=\frac{19}{7\times 3}=\frac{19}{21}
\displaystyle \Rightarrow \theta=\cos^{-1}\left(\frac{19}{21}\right)

\displaystyle \textbf{SECTION - C}
\displaystyle \text{This section comprises of Short Answer (SA) type questions of}
\displaystyle \text{3 marks each.}

\displaystyle \textbf{26. }\text{Using determinants, find the area of }\Delta PQR\text{ with vertices }   P(3,1),\ Q(9,3)\text{ and }R(5,7). \\ \text{Also, find the equation of line }   PQ\text{ using determinants.}
\displaystyle \text{Answer:}
\displaystyle \text{Given the vertices of the }\Delta PQR
\displaystyle P(3,1),\ Q(9,3)\text{ and }R(5,7)
\displaystyle \text{Now, area of }\Delta PQR=\frac{1}{2}\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|=\frac{1}{2}\left|\begin{matrix}3&1&1\\9&3&1\\5&7&1\end{matrix}\right|
\displaystyle =\frac{1}{2}\left\{3(3-7)-1(9-5)+1(63-15)\right\}
\displaystyle =\frac{1}{2}(-12-4+48)=\frac{1}{2}\times 32=16
\displaystyle \Rightarrow \text{Area of }\Delta PQR=16\text{ sq. units}
\displaystyle \text{Now, equation of line }PQ\text{ is given by }\frac{1}{2}\left|\begin{matrix}3&1&1\\9&3&1\\x&y&1\end{matrix}\right|=0
\displaystyle \Rightarrow \left|\begin{matrix}3&1&1\\9&3&1\\x&y&1\end{matrix}\right|=0\Rightarrow 3(3-y)-1(9-x)+(9y-3x)=0
\displaystyle \Rightarrow -2x+6y=0\Rightarrow x-3y=0
\displaystyle \text{It is equation of line }PQ.

\displaystyle \textbf{27. (a) }\text{Differentiate }\sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\text{ w.r.t.}   \sin^{-1}(2x\sqrt{1-x^2})\text{.}
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{If }y=\tan x+\sec x\text{, then prove that }\frac{d^2y}{dx^2}=\frac{\cos x}{(1-\sin x)^2}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }y=\sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)
\displaystyle z=\sin^{-1}(2x\sqrt{1-x^2})\text{ and }x=\sin\theta
\displaystyle \text{So, }y=\sec^{-1}\left(\frac{1}{\sqrt{1-\sin^2\theta}}\right)=\sec^{-1}\left(\frac{1}{\sqrt{\cos^2\theta}}\right)
\displaystyle \Rightarrow \sec^{-1}\left(\frac{1}{\cos\theta}\right)=\sec^{-1}(\sec\theta)=\theta\Rightarrow y=\sin^{-1}x
\displaystyle \text{and }z=\sin^{-1}\left(2\sin\theta\sqrt{1-\sin^2\theta}\right)
\displaystyle =\sin^{-1}\left(2\sin\theta\sqrt{\cos^2\theta}\right)=\sin^{-1}(2\sin\theta\cos\theta)
\displaystyle =\sin^{-1}(\sin 2\theta)=2\theta=2\sin^{-1}x
\displaystyle \text{Now, }\frac{dy}{dx}=\frac{d(\sin^{-1}x)}{dx}=\frac{1}{\sqrt{1-x^2}}
\displaystyle \text{and }\frac{dz}{dx}=\frac{d(2\sin^{-1}x)}{dx}=\frac{2}{\sqrt{1-x^2}}
\displaystyle \text{Since, }\frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{\frac{1}{\sqrt{1-x^2}}}{\frac{2}{\sqrt{1-x^2}}}=\frac{1}{2}
\displaystyle \textbf{Note: }\sin 2\theta=2\sin\theta\cos\theta\text{ and }\sin^2\theta+\cos^2\theta=1
\displaystyle \text{OR}
\displaystyle \text{(b) Given }y=\tan x+\sec x
\displaystyle \text{Now }\frac{dy}{dx}=\frac{d}{dx}(\tan x+\sec x)=\frac{d(\tan x)}{dx}+\frac{d(\sec x)}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\sec^2 x+\sec x\tan x=\frac{1}{\cos^2 x}+\frac{1}{\cos x}\times \frac{\sin x}{\cos x}
\displaystyle =\frac{1}{\cos^2 x}+\frac{\sin x}{\cos^2 x}=\frac{1+\sin x}{\cos^2 x}
\displaystyle =\frac{1+\sin x}{1-\sin^2 x}=\frac{1+\sin x}{(1+\sin x)(1-\sin x)}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{1-\sin x}
\displaystyle \text{Since, }\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{1}{1-\sin x}\right)
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=\frac{\cos x}{(1-\sin x)^2}

\displaystyle \textbf{28. (a) }\text{Evaluate: }\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos 2x}{1+\cos 2x}\,dx
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{Find: }\int e^{x^2}(x^5+2x^3)\,dx
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos 2x}{1+\cos 2x}\,dx\Rightarrow I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{2\cos^2 x-1}{2\cos^2 x}\,dx
\displaystyle I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(1-\frac{1}{2\cos^2 x}\right)dx
\displaystyle =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(1-\frac{\sec^2 x}{2}\right)dx=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}1\,dx-\frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2 x\,dx
\displaystyle =\left(x\right)_{-\frac{\pi}{4}}^{\frac{\pi}{4}}-\frac{1}{2}\left(\tan x\right)_{-\frac{\pi}{4}}^{\frac{\pi}{4}}
\displaystyle =\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)-\frac{1}{2}\left\{\tan\left(\frac{\pi}{4}\right)-\tan\left(-\frac{\pi}{4}\right)\right\}
\displaystyle =\frac{\pi}{4}+\frac{\pi}{4}-\frac{1}{2}\left\{1-(-1)\right\}=\frac{2\pi}{4}-\frac{1+1}{2}=\frac{\pi}{2}-1
\displaystyle =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos 2x}{1+\cos 2x}\,dx=\frac{\pi}{2}-1
\displaystyle \text{OR}
\displaystyle \text{(b) Let }I=\int e^{x^2}(x^5+2x^3)\,dx=\int e^{x^2}(x^4+2x^2)x\,dx
\displaystyle =\int e^{x^2}\left((x^2)^2+2x^2\right)x\,dx
\displaystyle \text{Let }x^2=t\Rightarrow 2x\,dx=dt\Rightarrow x\,dx=\frac{1}{2}dt
\displaystyle \text{Now }I=\frac{1}{2}\int e^t(t^2+2t)\,dt
\displaystyle =\frac{1}{2}e^t t^2+c\qquad\text{(Since } \int e^x(f(x)+f'(x))\,dx=e^x f(x)+c\text{)}
\displaystyle I=\frac{1}{2}e^{x^2}x^4+c=\frac{x^4 e^{x^2}}{2}+c

\displaystyle \textbf{29. }\text{Find the area of the minor segment of the circle }x^2+y^2=4   \text{cut off by the line} \\ x=1\text{, using integration.}
\displaystyle \text{Answer:}
\displaystyle \text{Given equation of circle }x^2+y^2=4\text{ and line }x=1\displaystyle \text{Now, }x^2+y^2=4\Rightarrow y^2=4-x^2\Rightarrow y=\sqrt{4-x^2}
\displaystyle \text{So, required area }=2\int_{1}^{2} y\,dx=2\int_{1}^{2}\sqrt{4-x^2}\,dx
\displaystyle =2\int_{1}^{2}\sqrt{2^2-x^2}\,dx=2\left\{\frac{x}{2}\sqrt{2^2-x^2}+\frac{2^2}{2}\sin^{-1}\left(\frac{x}{2}\right)\right\}_{1}^{2}
\displaystyle =\left\{x\sqrt{4-x^2}+4\sin^{-1}\left(\frac{x}{2}\right)\right\}_{1}^{2}
\displaystyle =\left[2\sqrt{4-4}+4\sin^{-1}\left(\frac{2}{2}\right)\right]-\left[1\sqrt{4-1}+4\sin^{-1}\left(\frac{1}{2}\right)\right]
\displaystyle =4\times \frac{\pi}{2}-\sqrt{3}-4\times \frac{\pi}{6}=2\pi-\frac{2\pi}{3}-\sqrt{3}=\frac{4\pi}{3}-\sqrt{3}
\displaystyle \text{Required area }=\frac{4\pi}{3}-\sqrt{3}\text{ sq. unit}

\displaystyle \textbf{30. }\text{Find the distance between the lines:}
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k});
\displaystyle \overrightarrow{r}=(3\widehat{i}+3\widehat{j}-5\widehat{k})+\mu(4\widehat{i}+6\widehat{j}+12\widehat{k})
\displaystyle \text{Answer:}
\displaystyle \text{Given the lines }\overrightarrow{r}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \overrightarrow{r}=(3\widehat{i}+3\widehat{j}-5\widehat{k})+\mu(4\widehat{i}+6\widehat{j}+12\widehat{k})=(3\widehat{i}+3\widehat{j}-5\widehat{k})+2\mu(2\widehat{i}+3\widehat{j}+6\widehat{k})
\displaystyle \text{Since the given both lines are parallel.}
\displaystyle \text{Let }\overrightarrow{a_1}=(\widehat{i}+2\widehat{j}-4\widehat{k}),\ \overrightarrow{a_2}=3\widehat{i}+3\widehat{j}-5\widehat{k}\text{ and }\overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}
\displaystyle \text{Now, distance between these two parallel lines is given by}
\displaystyle d=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|}{|\overrightarrow{b}|}
\displaystyle \text{Now }\overrightarrow{a_2}-\overrightarrow{a_1}=(3\widehat{i}+3\widehat{j}-5\widehat{k})-(\widehat{i}+2\widehat{j}-4\widehat{k})=2\widehat{i}+\widehat{j}-\widehat{k}
\displaystyle \text{and, }(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}=\left|\begin{matrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&1&-1\\2&3&6\end{matrix}\right|=(6+3)\widehat{i}-(12+2)\widehat{j}+(6-2)\widehat{k}
\displaystyle \Rightarrow (\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}=9\widehat{i}-14\widehat{j}+4\widehat{k}
\displaystyle \text{So, required distance }=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|}{|\overrightarrow{b}|}=\frac{|9\widehat{i}-14\widehat{j}+4\widehat{k}|}{|2\widehat{i}+3\widehat{j}+6\widehat{k}|}
\displaystyle =\frac{\sqrt{9^2+(-14)^2+4^2}}{\sqrt{2^2+3^2+6^2}}=\frac{\sqrt{81+196+16}}{\sqrt{4+9+36}}=\frac{\sqrt{293}}{7}\text{ units}

\displaystyle \textbf{31. (a) }\text{Find the coordinates of the foot of the perpendicular }  \text{drawn from the point} \\ P(0,2,3)\text{ to the line }   \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\text{,}
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{Three vectors }\overrightarrow{a},\ \overrightarrow{b}\text{ and }\overrightarrow{c}\text{ satisfy the condition }  \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}. \\ \text{Evaluate the quantity }\mu=\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\text{, if}   |\overrightarrow{a}|=3,\ |\overrightarrow{b}|=4 \\ \text{ and }|\overrightarrow{c}|=2\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given the line }\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}
\displaystyle \text{Let }\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=t
\displaystyle \Rightarrow x=5t-3,\ y=2t+1,\ z=3t-4
\displaystyle \text{So the point }(5t-3,\ 2t+1,\ 3t-4)\text{ lies on the given line.}
\displaystyle \text{Now, the point }(5t-3,\ 2t+1,\ 3t-4)\text{ is foot of perpendicular from }P(0,\ 2,\ 3)
\displaystyle \text{So }(5t-3,\ 2t-1,\ 3t-7)\text{ is perpendicular to }(5,\ 2,\ 3)
\displaystyle \text{So }(5t-3,\ 2t-1,\ 3t-7).(5,\ 2,\ 3)=0
\displaystyle \Rightarrow 5(5t-3)+2(2t-1)+3(3t-7)=0
\displaystyle \Rightarrow 25t-15+4t-2+9t-21=0
\displaystyle \Rightarrow 38t-38=0\Rightarrow t=1
\displaystyle \text{So foot of perpendicular is }(5-3,\ 2+1,\ 3-4)=(2,\ 3,\ -1)
\displaystyle \text{OR}
\displaystyle \text{(b) Given the vectors }\overrightarrow{a},\ \overrightarrow{b}\text{ and }\overrightarrow{c}\text{ such that }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\text{ and}
\displaystyle |\overrightarrow{a}|=3,\ |\overrightarrow{b}|=4,\ |\overrightarrow{c}|=2
\displaystyle \text{Since, given }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=0
\displaystyle (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=0
\displaystyle (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})=0
\displaystyle \Rightarrow |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0
\displaystyle \Rightarrow 3^2+4^2+2^2+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0
\displaystyle \Rightarrow 9+16+4+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0
\displaystyle \Rightarrow 29+2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})=0
\displaystyle \Rightarrow \overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a}=-\frac{29}{2}

\displaystyle \textbf{SECTION - D}
\displaystyle \text{This section comprises of Long Answer (LA) type questions}
\displaystyle \text{of 5 marks each.}

\displaystyle \textbf{32. }\text{Evaluate: }\int_{0}^{1}\frac{x}{1+\sin x}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi}\frac{x}{1+\sin x}\,dx\qquad\text{(i)}
\displaystyle \text{Using } \int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx
\displaystyle I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin(\pi-x)}\,dx=\int_{0}^{\pi}\frac{\pi-x}{1+\sin x}\,dx\qquad\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2I=\int_{0}^{\pi}\frac{x}{1+\sin x}\,dx+\int_{0}^{\pi}\frac{\pi-x}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi}\frac{\pi}{1+\sin x}\,dx
\displaystyle =\pi\int_{0}^{\pi}\frac{1}{1+\sin x}\,dx
\displaystyle =\pi\int_{0}^{\pi}\frac{1-\sin x}{1-\sin^2 x}\,dx=\pi\int_{0}^{\pi}\frac{1-\sin x}{\cos^2 x}\,dx
\displaystyle =\pi\int_{0}^{\pi}\left(\sec^2 x-\sec x\tan x\right)dx
\displaystyle =\pi\left[\tan x-\sec x\right]_{0}^{\pi}
\displaystyle =\pi\left[(\tan\pi-\tan0)-(\sec\pi-\sec0)\right]
\displaystyle =\pi\left[(0-0)-(-1-1)\right]=\pi(2)=2\pi
\displaystyle \Rightarrow 2I=2\pi\Rightarrow I=\pi

\displaystyle \textbf{33. (a) }\text{The median of an equilateral triangle is increasing at } \text{the rate of }2\sqrt{3}
\displaystyle \text{cm/s. Find the rate at which its side }  \text{is increasing.}
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{Sum of two numbers is 5. If the sum of the cubes of these numbers is least,}
\displaystyle \text{then find the sum of the squares}  \text{of these numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }ABC\text{ be an equilateral triangle in which }AD\text{ is a median}
\displaystyle \text{Let side of triangle be }a \displaystyle \text{Since, }\angle ADC=90^\circ\displaystyle \text{In }\triangle ADC,\ AD^2=AC^2-CD^2
\displaystyle \Rightarrow AD^2=a^2-\left(\frac{a}{2}\right)^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}
\displaystyle \Rightarrow AD=\frac{\sqrt{3}a}{2}
\displaystyle \text{Given }\frac{d}{dt}(AD)=2\sqrt{3}\text{ cm/s}
\displaystyle \Rightarrow \frac{d}{dt}\left(\frac{\sqrt{3}}{2}a\right)=2\sqrt{3}
\displaystyle \Rightarrow \frac{\sqrt{3}}{2}\frac{da}{dt}=2\sqrt{3}
\displaystyle \Rightarrow \frac{da}{dt}=2\sqrt{3}\times \frac{2}{\sqrt{3}}=4\text{ cm/s}
\displaystyle \text{OR}
\displaystyle \text{(b) Let first number be }x,\ \text{second number be }5-x
\displaystyle y=x^3+(5-x)^3
\displaystyle \frac{dy}{dx}=\frac{d(x^3)}{dx}+\frac{d(5-x)^3}{dx}=3x^2-3(5-x)^2
\displaystyle \text{For least }y,\ \frac{dy}{dx}=0\Rightarrow 3x^2-3(5-x)^2=0
\displaystyle \Rightarrow x^2-(5-x)^2=0
\displaystyle \Rightarrow (x-(5-x))(x+(5-x))=0\Rightarrow (2x-5)(5)=0
\displaystyle \Rightarrow x=\frac{5}{2}
\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(3x^2-3(5-x)^2\right)=6x+6(5-x)
\displaystyle =6x+30-6x=30
\displaystyle \text{At }x=\frac{5}{2},\ \frac{d^2y}{dx^2}=30>0
\displaystyle \Rightarrow y\text{ is minimum at }x=\frac{5}{2}
\displaystyle \text{So, first number }=\frac{5}{2},\ \text{second number }=\frac{5}{2}
\displaystyle \text{Required number }=\left(\frac{5}{2}\right)^2+\left(\frac{5}{2}\right)^2=\frac{25}{4}+\frac{25}{4}=\frac{50}{4}=\frac{25}{2}

\displaystyle \textbf{34. (a) }\text{In answering a question on a multiple choice test, a student either knows the}
\displaystyle \text{answer or guesses. Let }\frac{3}{5}\text{ be} \text{the probability that he knows the answer and } \frac{2}{5}\text{ be the}
\displaystyle \text{probability that he guesses. Assuming that a} \text{student who guesses at the answer will be}
\displaystyle \text{correct with probability }\frac{1}{3}\text{. } \text{What is the probability that the student knows the answer,}
\displaystyle \text{given that he answered it correctly?}
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{A box contains 10 tickets, 2 of which carry a prize of} \text{Rs }8\text{ each, 5 of which carry}
\displaystyle \text{a prize of Rs }4\text{ each, and }  \text{remaining 3 carry a prize of Rs }2\text{ each. If one ticket is drawn}
\displaystyle \text{at random, find the mean value of the prize.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }E_1=\text{Student knows the answer, }E_2=\text{Student guesses the answer}
\displaystyle \text{Since, }P(E_1)=\frac{3}{5},\ P(E_2)=\frac{2}{5}
\displaystyle \text{Let }A=\text{event that answer is correct}
\displaystyle \text{Now, }P\left(\frac{A}{E_1}\right)=1,\ P\left(\frac{A}{E_2}\right)=\frac{1}{3}
\displaystyle \text{So, }P\left(\frac{E_1}{A}\right)=\frac{P(E_1)P\left(\frac{A}{E_1}\right)}{P(E_1)P\left(\frac{A}{E_1}\right)+P(E_2)P\left(\frac{A}{E_2}\right)}
\displaystyle =\frac{\frac{3}{5}\times 1}{\frac{3}{5}\times 1+\frac{2}{5}\times \frac{1}{3}}=\frac{\frac{3}{5}}{\frac{3}{5}+\frac{2}{15}}=\frac{\frac{9}{15}}{\frac{9}{15}+\frac{2}{15}}
\displaystyle =\frac{9}{11}
\displaystyle \text{OR}
\displaystyle \text{(b) Since, 2 tickets carry prize of Rs. 8 each, 5 tickets carry prize of Rs. 4 each,} \\ \text{3 tickets carry prize of Rs. 2 each}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline X & 2 & 4 & 8 & \text{Total} \\ \hline P(X) & \frac{3}{10} & \frac{5}{10} & \frac{2}{10} & 1 \\ \hline XP(X) & \frac{6}{10} & \frac{20}{10} & \frac{16}{10} & \frac{42}{10} \\ \hline \end{array}
\displaystyle \text{Mean }=\frac{42}{10}=4.2

\displaystyle \textbf{35. }\text{Solve the following Linear Programming Problem}   \text{graphically:}
\displaystyle \text{Maximize: }P=70x+40y
\displaystyle \text{Subject to: }3x+2y\leq 9\text{,}
\displaystyle 3x+y\leq 9\text{,}
\displaystyle x\geq 0,\ y\geq 0
\displaystyle \text{Answer:}
\displaystyle \text{Given }P=70x+40y
\displaystyle \text{Subject to }3x+2y\leq 9,\ 3x+y\leq 9,\ x\geq 0,\ y\geq 0
\displaystyle \text{Now graph of the given constraints}
\displaystyle \text{Corner points are }(0,0),\ (3,0),\ \left(0,\frac{9}{2}\right)
\displaystyle \text{For maximizing }P:
\displaystyle P(0,0)=0
\displaystyle P(3,0)=70\times 3+40\times 0=210
\displaystyle P\left(0,\frac{9}{2}\right)=70\times 0+40\times \frac{9}{2}=20\times 9=180
\displaystyle \Rightarrow P_{\max}=210\text{ at }(3,0)

\displaystyle \textbf{SECTION - E}
\displaystyle \text{This section comprises of 3 Case Study/Passage-Based}
\displaystyle \text{questions of 4 marks each with two sub-parts. First two case}
\displaystyle \text{study questions have three sub-parts (I), (II), (III) of marks 1,}
\displaystyle \text{1, 2 respectively. The third case study question has two sub-}
\displaystyle \text{parts (I) and (II) of marks 2 each.}

\displaystyle \textbf{Case Study - I}
\displaystyle \textbf{36. }\text{Gautam buys 5 pens, 3 bags and 1 instrument box and pays } \text{a sum of Rs }160.
\displaystyle \text{From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a}
\displaystyle \text{sum of Rs }190\text{. Also Ankur buys} \text{1 pen, 2 bags and 4 instrument boxes and pays a}
\displaystyle \text{sum of Rs }250\text{. }\text{Based on the above information, answer the following questions:}
\displaystyle \text{(I) Convert the given above situation into a matrix equation of the form } \\ AX=B\text{.}
\displaystyle \text{(II) Find }|A|\text{.}
\displaystyle \text{(III) Find }A^{-1}\text{.}
\displaystyle \text{OR}
\displaystyle \text{Determine }P=A^2-5A\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let cost of a pen, a bag and an instrument box be Rs }x,\ y\text{ and }z\text{ respectively.}
\displaystyle \text{Given, }5x+3y+z=160
\displaystyle 2x+y+3z=190
\displaystyle x+2y+4z=250
\displaystyle \text{(I) Now, above system of equation can be represented into matrix form}
\displaystyle \begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}160\\190\\250\end{bmatrix}
\displaystyle \text{Where }A=\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}160\\190\\250\end{bmatrix}
\displaystyle \text{(II) }|A|=\left|\begin{matrix}5&3&1\\2&1&3\\1&2&4\end{matrix}\right|
\displaystyle =5(4-6)-3(8-3)+1(4-1)
\displaystyle =5\times (-2)-3\times 5+1\times 3=-10-15+3
\displaystyle \Rightarrow |A|=-22
\displaystyle \text{(III) Since }|A|=-22\neq 0
\displaystyle \text{So }A^{-1}\text{ exists}
\displaystyle \text{Now }C_{11}=(-1)^{1+1}(4-6)=-2
\displaystyle C_{12}=(-1)^{1+2}(8-3)=-5
\displaystyle C_{13}=(-1)^{1+3}(4-1)=3
\displaystyle C_{21}=(-1)^{2+1}(12-2)=-10
\displaystyle C_{22}=(-1)^{2+2}(20-1)=19
\displaystyle C_{23}=(-1)^{2+3}(10-3)=-7
\displaystyle C_{31}=(-1)^{3+1}(9-1)=8
\displaystyle C_{32}=(-1)^{3+2}(15-2)=-13
\displaystyle C_{33}=(-1)^{3+3}(5-6)=-1
\displaystyle \text{So adj}(A)=\begin{bmatrix}-2&-10&8\\-5&19&-13\\3&-7&-1\end{bmatrix}
\displaystyle \text{Now, }A^{-1}=\frac{\text{adj}(A)}{|A|}=-\frac{1}{22}\begin{bmatrix}-2&-10&8\\-5&19&-13\\3&-7&-1\end{bmatrix}
\displaystyle \Rightarrow A^{-1}=\frac{1}{22}\begin{bmatrix}2&10&-8\\5&-19&13\\-3&7&1\end{bmatrix}
\displaystyle \text{OR}
\displaystyle \text{Now, }P=A^2-5A
\displaystyle =\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix}\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix}-5\begin{bmatrix}5&3&1\\2&1&3\\1&2&4\end{bmatrix}
\displaystyle =\begin{bmatrix}25+6+1&15+3+2&5+9+4\\10+2+3&6+1+6&2+3+12\\5+4+4&3+2+8&1+6+16\end{bmatrix}-\begin{bmatrix}25&15&5\\10&5&15\\5&10&20\end{bmatrix}
\displaystyle =\begin{bmatrix}32&20&18\\15&13&17\\13&13&23\end{bmatrix}-\begin{bmatrix}25&15&5\\10&5&15\\5&10&20\end{bmatrix}
\displaystyle =\begin{bmatrix}7&5&13\\5&8&2\\8&3&3\end{bmatrix}

\displaystyle \textbf{Case Study - II}
\displaystyle \textbf{37. }\text{An organization conducted bike race under two different categories - Boys and Girls. }
\displaystyle \text{There were 28 participants in all. Among all of them, finally three from category 1 and }
\displaystyle \text{two from category 2 were selected} \text{for the final race. Ravi forms two sets }B\text{ and }G\text{ with}
\displaystyle \text{these participants for his college project. } \text{Let }B=\{b_1,b_2,b_3\}\text{ and } G=\{g_1,g_2\},
\displaystyle \text{where }B\text{ represents}  \text{the set of Boys selected and }G\text{ the set of Girls selected for the final}


\displaystyle \text{race. Based on the above information, answer the following question:}
\displaystyle \text{(I) How many relations are possible from }B\text{ to }G\text{?}
\displaystyle \text{(II) Among all the possible relations from }B\text{ to }G\text{, how many} \text{functions can be} \\ \text{formed from }B\text{ to }G\text{? }
\displaystyle \text{(III) Let }R:B\to B\text{ be defined by }R=\{(x,y):x\text{ and }y\text{ are students } \text{of the same sex.}\}
\displaystyle \text{Check if }R\text{ is an equivalence relation.}
\displaystyle \text{OR}
\displaystyle \text{A function }f:B\to G\text{ is defined by }f=\{(b_1,g_1),(b_2,g_2),(b_3,g_1)\}\text{.}
\displaystyle \text{Check if it is bijective. Justify your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }B=\{b_1,b_2,b_3\}\text{ and }G=\{g_1,g_2\}

\displaystyle \text{(I) }n(B)=3\text{ and }n(G)=2
\displaystyle \therefore n(B\times G)=3\times 2=6
\displaystyle \text{Now, number of relations from }B\text{ to }G=2^6=64

\displaystyle \text{(II) We know that, a relation }f\text{ is said to be function, if every element of a non-empty}
\displaystyle \text{set }B\text{ has only one image in set }G
\displaystyle \therefore \text{Number of choice of }b_1=2
\displaystyle \text{Number of choice of }b_2=2
\displaystyle \text{Number of choice of }b_3=2
\displaystyle \text{So, number of functions }=2\times 2\times 2=8

\displaystyle \text{(III) Let }R:B\to B
\displaystyle R=\{(x,y):x\text{ and }y\text{ are students of the same sex}\}
\displaystyle \text{Let }x,\ y,\ z\in B
\displaystyle \text{For reflexive}
\displaystyle \therefore x\text{ and }x\text{ are students of the same sex}
\displaystyle \therefore (x,x)\in R
\displaystyle \text{So, }R\text{ is reflexive.}
\displaystyle \text{For symmetric}
\displaystyle \text{Let }(x,y)\in R
\displaystyle \Rightarrow x\text{ and }y\text{ are students of the same sex.}
\displaystyle \Rightarrow y\text{ and }x\text{ are students of the same sex.}
\displaystyle \Rightarrow (y,x)\in R
\displaystyle \text{So, }R\text{ is symmetric.}
\displaystyle \text{For transitive}
\displaystyle \text{Let }(x,y)\in R
\displaystyle \Rightarrow x\text{ and }y\text{ are students of the same sex}\qquad \ldots (i)
\displaystyle \text{and }(y,z)\in R
\displaystyle \Rightarrow y\text{ and }z\text{ are students of the same sex}\qquad \ldots (ii)
\displaystyle \text{From (i) and (ii), }x\text{ and }z\text{ are students of the same sex}
\displaystyle \Rightarrow (x,z)\in R
\displaystyle \text{So, }R\text{ is transitive.}
\displaystyle \text{Hence, }R\text{ is an equivalence relation.}
\displaystyle \text{OR}
\displaystyle f:B\to G
\displaystyle f=\{(b_1,g_1),(b_2,g_2),(b_3,g_1)\}
\displaystyle \text{Since }f(b_1)=g_1\text{ and }f(b_3)=g_1
\displaystyle \text{So }f\text{ is not one-one.}
\displaystyle \text{Range of }f=\{g_1,g_2\}=G=\text{codomain}
\displaystyle \text{So }f\text{ is onto function.}
\displaystyle \text{Hence }f\text{ is not bijective.}
\\

\displaystyle \textbf{Case Study - III}
\displaystyle \textbf{38. }\text{An equation involving derivatives of the dependent variable with respect to the}
\displaystyle \text{independent variables is called a differential equation.} \text{A differential equation of the form }
\displaystyle \frac{dy}{dx}=F(x,y)\text{ is said to be}  \text{homogeneous if }F(x,y)\text{ is a homogeneous function of degree}
\displaystyle \text{zero, whereas a function }F(x,y)\text{ is a homogeneous function of degree }n\text{ if}
\displaystyle F(\lambda x,\lambda y)=\lambda^n F(x,y)\text{. To solve a homogeneous } \text{differential equation of the type }
\displaystyle \frac{dy}{dx}=F(x,y)=g\left(\frac{y}{x}\right)\text{, we}  \text{make the substitution }y=vx\text{ and then separate the variables.}
\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(I) Show that }(x^2-y^2)dx+2xy\,dy=0\text{ is a differential equation of }  \text{the type } \\ \frac{dy}{dx}=g\left(\frac{y}{x}\right)\text{.}
\displaystyle \text{(II) Solve the above equation to find its general solution.}
\displaystyle \text{Answer:}
\displaystyle \text{(I) Given differential equation is}
\displaystyle (x^2-y^2)\,dx+2xy\,dy=0
\displaystyle \Rightarrow 2xy\,dy=-(x^2-y^2)\,dx
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}=\frac{1}{2}\left(\frac{y}{x}-\frac{1}{\frac{y}{x}}\right)=g\left(\frac{y}{x}\right)
\displaystyle \text{Put }x=\lambda x,\ y=\lambda y
\displaystyle f(\lambda x,\lambda y)=\frac{1}{2}\left(\frac{\lambda y}{\lambda x}-\frac{1}{\frac{\lambda y}{\lambda x}}\right)
\displaystyle =\frac{1}{2}\left(\frac{y}{x}-\frac{1}{\frac{y}{x}}\right)=f(x,y)
\displaystyle \text{So, given differential equation has type }\frac{dy}{dx}=g\left(\frac{y}{x}\right)

\displaystyle \text{(II) }\frac{dy}{dx}=\frac{1}{2}\left(\frac{y}{x}-\frac{1}{\frac{y}{x}}\right)
\displaystyle \text{Let }\frac{y}{x}=v\text{ i.e. }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=\frac{v^2-1}{2v}\Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}
\displaystyle x\frac{dv}{dx}=-\frac{1+v^2}{2v}
\displaystyle \int \frac{2v}{1+v^2}\,dv=-\int \frac{1}{x}\,dx\Rightarrow \log_e(1+v^2)=\log_e\left(\frac{c}{x}\right)
\displaystyle \Rightarrow 1+v^2=\frac{c}{x}\Rightarrow 1+\frac{y^2}{x^2}=\frac{c}{x}\Rightarrow \frac{x^2+y^2}{x^2}=\frac{c}{x}
\displaystyle \Rightarrow x^2+y^2=cx

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