Mathematics

\displaystyle \text{Time Allowed : 3 Hours} \hspace{6cm} \text{Maximum Marks : 100}

\displaystyle \textbf{General Instructions:}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) The question paper consists of 29 questions divided into three Sections A, B and C.} \\ \text{Section A comprises of 10 questions of one mark each, Section B comprises of 12} \\ \text{questions of four marks each and Section C comprises of 07 questions of six marks each.}
\displaystyle \text{(iii) All questions in Section A are to be answered in one word, one sentence or as} \\ \text{per the exact requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, internal choice has been provided in 04} \\ \text{questions of four marks each and 02 questions of six marks each. You have to attempt }  \\ \text{only one of the alternatives in all such questions.}
\displaystyle \text{(v) Use of calculators is not permitted. You may ask for logarithmic tables, if } \\ \text{required.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Question numbers 1 to 10 carry 1 mark each.}

\displaystyle \textbf{1. } \text{Let } * \text{ be a binary operation, on the set of all non-zero real numbers, given by } \\  a*b=\frac{ab}{5} \text{ for all } a,b\in R-\{0\}. \text{ Find the value of } x,\ \text{given that } 2*(x*5)=10.
\displaystyle \text{Answer:}
\displaystyle  a*b=\frac{ab}{5},\ \forall a,b\in R-\{0\}
\displaystyle 2*(x*5)=10 \Rightarrow 2*\frac{5x}{5}=10
\displaystyle \Rightarrow 2*x=10 \Rightarrow \frac{2x}{5}=10
\displaystyle \Rightarrow x=25

\displaystyle \textbf{2. } \text{If } \sin\left(\sin^{-1}\frac{1}{5}+\cos^{-1}x\right)=1,\ \text{then find the value of } x.
\displaystyle \text{Answer:}
\displaystyle \sin\left(\sin^{-1}\frac{1}{5}+\cos^{-1}x\right)=1
\displaystyle \Rightarrow \sin^{-1}\frac{1}{5}+\cos^{-1}x=\sin^{-1}1
\displaystyle \Rightarrow \sin^{-1}\frac{1}{5}+\cos^{-1}x=\frac{\pi}{2}
\displaystyle \Rightarrow \sin^{-1}\frac{1}{5}=\frac{\pi}{2}-\cos^{-1}x
\displaystyle \Rightarrow \sin^{-1}\frac{1}{5}=\sin^{-1}x
\displaystyle \Rightarrow x=\frac{1}{5}

\displaystyle \textbf{3. } \text{If } 2\begin{bmatrix}3&4\\5&x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix},\ \text{find } (x-y).
\displaystyle \text{Answer:}
\displaystyle   2\begin{bmatrix}3&4\\5&x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}6&8\\10&2x\end{bmatrix}+\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}7&8+y\\10&2x+1\end{bmatrix}=\begin{bmatrix}7&0\\10&5\end{bmatrix}
\displaystyle \Rightarrow 8+y=0 \Rightarrow y=-8
\displaystyle \Rightarrow 2x+1=5 \Rightarrow 2x=4 \Rightarrow x=2
\displaystyle \Rightarrow x-y=2-(-8)=10

\displaystyle \textbf{4. } \text{Solve the following matrix equation of } x:  \  [x\ \ 1]\begin{bmatrix}1&0\\-2&0\end{bmatrix}=0.
\displaystyle \text{Answer:}
\displaystyle  [x\ \ 1]\begin{bmatrix}1&0\\-2&0\end{bmatrix}=0
\displaystyle \Rightarrow [x-2\ \ 0]=[0\ \ 0]
\displaystyle \Rightarrow x-2=0 \Rightarrow x=2

\displaystyle \textbf{5. } \text{If } \begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix},\ \text{write the value of } x.
\displaystyle \text{Answer:}
\displaystyle \begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}
\displaystyle \Rightarrow 2x^{2}-40=18+14
\displaystyle \Rightarrow 2x^{2}=72 \Rightarrow x^{2}=36
\displaystyle \Rightarrow x=\pm 6

\displaystyle \textbf{6. } \text{Write the anti derivative of } \left(3\sqrt{x}+\frac{1}{\sqrt{x}}\right).
\displaystyle \text{Answer:}
\displaystyle \int \left(3\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx
\displaystyle =3\frac{x^{3/2}}{3/2}+\frac{x^{1/2}}{1/2}+C
\displaystyle =2x^{3/2}+2x^{1/2}+C=2\sqrt{x}(x+1)+C

\displaystyle \textbf{7. } \text{Evaluate : } \int_{0}^{3}\frac{dx}{9+x^{2}}
\displaystyle \text{Answer:}
\displaystyle \int_{0}^{3}\frac{dx}{9+x^{2}}=\frac{1}{3}\left[\tan^{-1}\frac{x}{3}\right]_{0}^{3}=\frac{1}{3}\left(\frac{\pi}{4}-0\right)=\frac{\pi}{12}

\displaystyle \textbf{8. } \text{Find the projection of the vector } \widehat{i}+3\widehat{j}+7\widehat{k} \text{ on the vector }  2\widehat{i}-3\widehat{j}+6\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{Let } \overrightarrow{a}=\widehat{i}+3\widehat{j}+7\widehat{k},\ \overrightarrow{b}=2\widehat{i}-3\widehat{j}+6\widehat{k}
\displaystyle \text{Projection of } \overrightarrow{a} \text{ on } \overrightarrow{b}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{b}|}
\displaystyle =\frac{2-9+42}{\sqrt{4+9+36}}=\frac{35}{7}=5

\displaystyle \textbf{9. } \text{If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are two unit vectors such that } \overrightarrow{a}+\overrightarrow{b} \text{ is also a unit vector, then find}
\displaystyle \text{ the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{Given } |\overrightarrow{a}|=1,\ |\overrightarrow{b}|=1,\ |\overrightarrow{a}+\overrightarrow{b}|=1
\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=1
\displaystyle \Rightarrow 1+1+2\overrightarrow{a}\cdot\overrightarrow{b}=1
\displaystyle \Rightarrow 2\overrightarrow{a}\cdot\overrightarrow{b}=-1
\displaystyle \Rightarrow 2\cos\theta=-1 \Rightarrow \cos\theta=-\frac{1}{2}
\displaystyle \Rightarrow \theta=120^{\circ}

\displaystyle \textbf{10. } \text{Write the vector equation of the plane, passing through } \text{point } (a,b,c)
\displaystyle \text{ and parallel to the plane } \overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=2.
\displaystyle \text{Answer:}
\displaystyle  \overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=2 \Rightarrow x+y+z=2
\displaystyle \text{Equation of plane parallel to it is } x+y+z=k
\displaystyle \text{Passing through } (a,b,c),\ k=a+b+c
\displaystyle \therefore \overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Question numbers 11 to 22 carry 4 marks each.}

\displaystyle \textbf{11. } \text{Let } A=\{1,2,3,\ldots,9\} \text{ and } R \text{ be the relation in } A\times A \text{ defined by } (a,b)\ R\ (c,d)
\displaystyle \text{if } a+d=b+c \text{ for } (a,b),(c,d)\in A\times A. \text{ Prove that } R \text{ is an equivalence relation.}
\displaystyle \text{Also obtain the equivalence class } [(2,5)].
\displaystyle \text{Answer:}
\displaystyle \text{Here } A=\{1,2,3,\ldots,9\} \text{ and } R \text{ is a relation on } A\times A \text{ defined by}
\displaystyle (a,b)\ R\ (c,d) \Leftrightarrow a+d=b+c \ \forall a,b,c,d\in A
\displaystyle \text{(i) Let } (a,b)\in A\times A
\displaystyle a+b=b+a
\displaystyle (a,b)\ R\ (a,b)\ \forall (a,b)\in A\times A
\displaystyle \Rightarrow R \text{ is reflexive on } A
\displaystyle \text{(ii) Let } (a,b)\ R\ (c,d)
\displaystyle a+d=b+c
\displaystyle b+c=d+a
\displaystyle (c,d)\ R\ (a,b)\ \forall (a,b),(c,d)\in A\times A
\displaystyle \Rightarrow R \text{ is symmetric on } A
\displaystyle \text{(iii) Let } (a,b)\ R\ (c,d) \text{ and } (c,d)\ R\ (e,f)
\displaystyle a+d=b+c \text{ and } c+f=d+e
\displaystyle \text{On adding } (a+d)+(c+f)=(b+c)+(d+e)
\displaystyle \Rightarrow a+f=b+e
\displaystyle (a,b)\ R\ (e,f)
\displaystyle \Rightarrow R \text{ is transitive on } A
\displaystyle \text{Hence } R \text{ is an equivalence relation on } A
\displaystyle \text{Also equivalence class } [(2,5)]
\displaystyle =\{(a,b)\in A\times A \mid (2,5)\ R\ (a,b)\}
\displaystyle =\{(a,b)\in A\times A \mid 2+b=5+a\}
\displaystyle =\{(a,b)\in A\times A \mid b=a+3\}
\displaystyle =\{(a,a+3)\mid a\in A\}
\displaystyle =\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}

\displaystyle \textbf{12. } \text{Prove that } \cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2};\ x\in \left(0,\frac{\pi}{4}\right).
\displaystyle \text{OR}
\displaystyle \text{Prove that } 2\tan^{-1}\left(\frac{1}{5}\right)+\sec^{-1}\left(\frac{5\sqrt{2}}{7}\right)+2\tan^{-1}\left(\frac{1}{8}\right)=\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S. }=\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)
\displaystyle \Rightarrow \cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\cdot\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right)
\displaystyle =\cot^{-1}\left(\frac{1+\sin x+1-\sin x+2\sqrt{(1+\sin x)(1-\sin x)}}{(1+\sin x)-(1-\sin x)}\right)
\displaystyle =\cot^{-1}\left(\frac{2+2\sqrt{1-\sin^{2}x}}{2\sin x}\right)
\displaystyle =\cot^{-1}\left(\frac{1+\cos x}{\sin x}\right)
\displaystyle =\cot^{-1}\left(\frac{2\cos^{2}\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\right)
\displaystyle =\cot^{-1}\left(\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\right)
\displaystyle =\cot^{-1}\left(\cot\frac{x}{2}\right)=\frac{x}{2}=\text{R.H.S.}
\displaystyle \text{Hence proved.}
\displaystyle \text{OR}
\displaystyle \text{L.H.S. }=2\tan^{-1}\left(\frac{1}{5}\right)+\sec^{-1}\left(\frac{5\sqrt{2}}{7}\right)+2\tan^{-1}\left(\frac{1}{8}\right)
\displaystyle =2\left(\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{8}\right)+\tan^{-1}\sqrt{\left(\frac{5\sqrt{2}}{7}\right)^{2}-1}
\displaystyle =2\tan^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\cdot\frac{1}{8}}\right)+\tan^{-1}\left(\frac{1}{7}\right)
\displaystyle =2\tan^{-1}\left(\frac{13/40}{39/40}\right)+\tan^{-1}\left(\frac{1}{7}\right)
\displaystyle =2\tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{7}\right)
\displaystyle =\tan^{-1}\left(\frac{2\cdot\frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)+\tan^{-1}\left(\frac{1}{7}\right)
\displaystyle =\tan^{-1}\left(\frac{3}{4}\right)+\tan^{-1}\left(\frac{1}{7}\right)
\displaystyle =\tan^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\cdot\frac{1}{7}}\right)
\displaystyle =\tan^{-1}(1)=\frac{\pi}{4}=\text{R.H.S.}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{13. } \text{Using properties of determinants, prove that}
\displaystyle \begin{vmatrix}2y&y-z-x&2y\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}=(x+y+z)^{3}
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S. Let } \Delta=\begin{vmatrix}2y&y-z-x&2y\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}
\displaystyle \text{Applying } R_{1}\to R_{1}+R_{2}+R_{3}
\displaystyle \Delta=\begin{vmatrix}x+y+z&x+y+z&x+y+z\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}
\displaystyle =(x+y+z)\begin{vmatrix}1&1&1\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}
\displaystyle \text{Applying } C_{2}\to C_{2}-C_{1} \text{ and } C_{3}\to C_{3}-C_{1}
\displaystyle \Delta=(x+y+z)\begin{vmatrix}1&0&0\\2z&0&-(x+y+z)\\x-y-z&x+y+z&x+y+z\end{vmatrix}
\displaystyle \text{Expanding along } R_{1}
\displaystyle \Delta=(x+y+z)\cdot 1\cdot (x+y+z)^{2}
\displaystyle \Delta=(x+y+z)^{3}=\text{R.H.S.}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{14. } \text{Differentiate } \tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right) \text{ with respect to } \cos^{-1}(2x\sqrt{1-x^{2}}),\ \\ \text{when } x\neq 0.
\displaystyle \text{Answer:}
\displaystyle \text{Let } y=\tan^{-1}\left[\frac{\sqrt{1-x^{2}}}{x}\right] \text{ and } t=\cos^{-1}(2x\sqrt{1-x^{2}})
\displaystyle y=\tan^{-1}\left[\frac{\sqrt{1-x^{2}}}{x}\right]
\displaystyle \text{Let } x=\cos\theta
\displaystyle y=\tan^{-1}\left[\frac{\sqrt{1-\cos^{2}\theta}}{\cos\theta}\right]
\displaystyle =\tan^{-1}\left[\frac{\sin\theta}{\cos\theta}\right]
\displaystyle =\tan^{-1}\tan\theta=\theta
\displaystyle \text{Differentiating w.r.t. } \theta
\displaystyle \frac{dy}{d\theta}=1\qquad \ldots (1)
\displaystyle t=\cos^{-1}(2x\sqrt{1-x^{2}})\qquad (x\neq 0)
\displaystyle \text{Putting } x=\cos\theta
\displaystyle t=\cos^{-1}(2\cos\theta\sqrt{1-\cos^{2}\theta})
\displaystyle t=\cos^{-1}(2\cos\theta\sin\theta)
\displaystyle t=\cos^{-1}(\sin 2\theta)
\displaystyle t=\cos^{-1}\cos\left(\frac{\pi}{2}-2\theta\right)
\displaystyle t=\frac{\pi}{2}-2\theta
\displaystyle \text{Differentiating w.r.t. } \theta
\displaystyle \frac{dt}{d\theta}=0-2=-2\qquad \ldots (2)
\displaystyle \text{Dividing (1), (2)}
\displaystyle \frac{dy}{dt}=\frac{\frac{dy}{d\theta}}{\frac{dt}{d\theta}}=-\frac{1}{2}

\displaystyle \textbf{15. } \text{If } y=x^{x},\ \text{prove that } \frac{d^{2}y}{dx^{2}}-\frac{1}{y}\left(\frac{dy}{dx}\right)^{2}-\frac{y}{x}=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } y=x^{x}
\displaystyle \text{Taking log on both sides}
\displaystyle \log y=x\log x
\displaystyle \text{Differentiating both sides w.r.t. } x
\displaystyle \frac{1}{y}\frac{dy}{dx}=x\cdot \frac{1}{x}+\log x
\displaystyle \frac{dy}{dx}=y(1+\log x)\qquad \ldots (1)
\displaystyle \text{Again differentiating w.r.t. } x
\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}(1+\log x)+y\cdot \frac{1}{x}
\displaystyle =\frac{dy}{dx}\left(\frac{1}{y}\frac{dy}{dx}\right)+\frac{y}{x}\qquad \text{(from (1))}
\displaystyle \frac{d^{2}y}{dx^{2}}-\frac{1}{y}\left(\frac{dy}{dx}\right)^{2}-\frac{y}{x}=0
\displaystyle \text{Hence proved.}

\displaystyle \textbf{16. } \text{Find the intervals in which the function }   f(x)=3x^{4}-4x^{3}-12x^{2}+5 \text{ is}
\displaystyle \text{(a) strictly increasing}                \displaystyle \text{(b) strictly decreasing}
\displaystyle \text{OR}
\displaystyle \text{Find the equations of the tangent and normal to the curve } x=a\sin^{3}\theta \text{ and } 
\displaystyle y=a\cos^{3}\theta \text{ at } \theta=\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle  f(x)=3x^{4}-4x^{3}-12x^{2}+5
\displaystyle \text{Differentiating w.r.t. } x
\displaystyle f'(x)=12x^{3}-12x^{2}-24x
\displaystyle =12x(x^{2}-x-2)
\displaystyle =12x(x^{2}-2x+x-2)
\displaystyle =12x(x-2)(x+1)
\displaystyle \text{Critical values for } f \text{ are } 0,\ 2,\ -1
\displaystyle \text{For } -\infty<x<-1,\ f'(x)=(-)(-)(-)<0
\displaystyle \therefore\ \text{Function is strictly decreasing}
\displaystyle \text{For } -1<x<0,\ f'(x)=(-)(+)(-) >0
\displaystyle \therefore\ \text{Function is strictly increasing}
\displaystyle \text{For } 0<x<2,\ f'(x)=(+)(+)(-)<0
\displaystyle \therefore\ \text{Function is strictly decreasing}
\displaystyle \text{For } 2<x<\infty,\ f'(x)=(+)(+)(+)>0
\displaystyle \therefore\ \text{Function is strictly increasing}
\displaystyle \text{Hence, } f(x) \text{ is strictly increasing in } (-1,0)\cup (2,\infty)
\displaystyle \text{and strictly decreasing in } (-\infty,-1)\cup (0,2)
\displaystyle \text{OR}
\displaystyle x=a\sin^{3}\theta
\displaystyle \Rightarrow \frac{dx}{d\theta}=3a\sin^{2}\theta\cos\theta\qquad \ldots (1)
\displaystyle y=a\cos^{3}\theta
\displaystyle \Rightarrow \frac{dy}{d\theta}=-3a\cos^{2}\theta\sin\theta\qquad \ldots (2)
\displaystyle \text{Dividing (2) by (1)}
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{-3a\cos^{2}\theta\sin\theta}{3a\sin^{2}\theta\cos\theta}=-\cot\theta
\displaystyle \Rightarrow \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{4}}=-\cot\frac{\pi}{4}=-1
\displaystyle \text{Slope of tangent at } \left(\theta=\frac{\pi}{4}\right)=-1
\displaystyle \text{Slope of normal}=\frac{-1}{\text{Slope of tangent}}
\displaystyle \text{Slope of normal}=1
\displaystyle x=a\sin^{3}\theta,\ \text{at } \theta=\frac{\pi}{4}
\displaystyle x=a\sin^{3}\frac{\pi}{4}=a\left(\frac{1}{\sqrt{2}}\right)^{3}=\frac{a}{2\sqrt{2}}
\displaystyle y=a\cos^{3}\theta,\ \text{at } \theta=\frac{\pi}{4}\Rightarrow y=\frac{a}{2\sqrt{2}}
\displaystyle \text{Point } P\left(\frac{a}{2\sqrt{2}},\frac{a}{2\sqrt{2}}\right)
\displaystyle \text{Equation of the tangent at } P
\displaystyle y-\frac{a}{2\sqrt{2}}=-1\left(x-\frac{a}{2\sqrt{2}}\right)
\displaystyle \Rightarrow x+y=\frac{a}{\sqrt{2}}
\displaystyle \text{Equation of the normal at } P
\displaystyle y-\frac{a}{2\sqrt{2}}=1\left(x-\frac{a}{2\sqrt{2}}\right)
\displaystyle \Rightarrow y=x

\displaystyle \textbf{17. } \text{Evaluate : } \int \frac{\sin^{6}x+\cos^{6}x}{\sin^{2}x.\cos^{2}x}\,dx
\displaystyle \text{OR}
\displaystyle \text{Evaluate : } \int (x-3)\sqrt{x^{2}+3x-18}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\sin^{6}x+\cos^{6}x}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle I=\int \frac{(\sin^{2}x)^{3}+(\cos^{2}x)^{3}}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle I=\int \frac{(\sin^{2}x+\cos^{2}x)(\sin^{4}x+\cos^{4}x-\sin^{2}x\cos^{2}x)}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle I=\int \frac{\sin^{4}x+\cos^{4}x-\sin^{2}x\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle =\int \frac{(\sin^{2}x)^{2}+(\cos^{2}x)^{2}+2\sin^{2}x\cos^{2}x-2\sin^{2}x\cos^{2}x-\sin^{2}x\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle I=\int \frac{(\sin^{2}x+\cos^{2}x)^{2}-3\sin^{2}x\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle I=\int \frac{1-3\sin^{2}x\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle I=\int \frac{1}{\sin^{2}x\cos^{2}x}\,dx-3\int dx+c
\displaystyle \text{Multiplying Num. \& Den. by } \cos^{2}x
\displaystyle I=\int \frac{\cos^{2}x}{\sin^{2}x\cos^{4}x}\,dx-3x+c=\int \frac{\sec^{4}x}{\tan^{2}x}\,dx-3x+c
\displaystyle =\int \frac{\sec^{2}x\sec^{2}x}{\tan^{2}x}\,dx-3x+c
\displaystyle =\int \frac{(1+\tan^{2}x)\sec^{2}x}{\tan^{2}x}\,dx-3x+c
\displaystyle \text{Let, } \tan x=t,\ \sec^{2}x\,dx=dt
\displaystyle I=\int \frac{1+t^{2}}{t^{2}}\,dt-3x+c
\displaystyle I=\int \left(\frac{1}{t^{2}}+1\right)\,dt-3x+c
\displaystyle I=-\frac{1}{t}+t-3x+c
\displaystyle I=\tan x-\frac{1}{\tan x}-3x+c
\displaystyle \text{OR}
\displaystyle \text{Let } I=\int (x-3)\sqrt{x^{2}+3x-18}\,dx
\displaystyle I=\int \left[\frac{1}{2}(2x+3)-\frac{9}{2}\right]\sqrt{x^{2}+3x-18}\,dx
\displaystyle I=\frac{1}{2}\int (2x+3)\sqrt{x^{2}+3x-18}\,dx-\frac{9}{2}\int \sqrt{x^{2}+3x-18}\,dx
\displaystyle \text{Let } x^{2}+3x-18=t
\displaystyle (2x+3)\,dx=dt
\displaystyle I=\frac{1}{2}\int \sqrt{t}\,dt-\frac{9}{2}\int \sqrt{x^{2}+3x-18+\left(\frac{3}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}}\,dx
\displaystyle I=\frac{1}{2}\cdot \frac{t^{3/2}}{3/2}-\frac{9}{2}\int \sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}}\,dx
\displaystyle I=\frac{1}{3}(x^{2}+3x-18)^{3/2}-\frac{9}{2}\left[\frac{1}{2}\left(x+\frac{3}{2}\right)\sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}}\right.
\displaystyle \left.-\frac{1}{2}\left(\frac{9}{2}\right)^{2}\log\left|x+\frac{3}{2}+\sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}}\right|\right]+C
\displaystyle \qquad \left[\because\ \int \sqrt{x^{2}-a^{2}}\,dx=\frac{1}{2}x\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}\log|x+\sqrt{x^{2}-a^{2}}|+C\right]
\displaystyle I=\frac{1}{3}(x^{2}+3x-18)^{3/2}-\frac{9}{4}\left(x+\frac{3}{2}\right)\sqrt{x^{2}+3x-18}+\frac{81}{8}\log\left|x+\frac{3}{2}+\sqrt{x^{2}+3x-18}\right|+C

\displaystyle \textbf{18. } \text{Find the particular solution of the differential equation }
\displaystyle e^{x}\sqrt{1-y^{2}}\,dx+\frac{y}{x}\,dy=0,\ \text{given that } y=1 \text{ when } x=0.
\displaystyle \text{Answer:}
\displaystyle e^{x}\sqrt{1-y^{2}}\,dx+\frac{y}{x}\,dy=0
\displaystyle xe^{x}\,dx+\frac{y}{\sqrt{1-y^{2}}}\,dy=0
\displaystyle \text{On integrating}
\displaystyle \int xe^{x}\,dx+\int \frac{y}{\sqrt{1-y^{2}}}\,dy=c
\displaystyle \text{Let } 1-y^{2}=t
\displaystyle -2y\,dy=dt\Rightarrow y\,dy=-\frac{1}{2}dt
\displaystyle x\int e^{x}\,dx-\int 1\cdot e^{x}\,dx-\frac{1}{2}\int \frac{dt}{\sqrt{t}}=c
\displaystyle \qquad \left[\because\ \int uvdx=u\int vdx-\int \left(\frac{du}{dx}\int vdx\right)dx\right]
\displaystyle x e^{x}-e^{x}-\frac{1}{2}\cdot \frac{t^{1/2}}{1/2}=c
\displaystyle \qquad \left[\because\ \int x^{n}\,dx=\frac{x^{n+1}}{n+1}\right]
\displaystyle e^{x}(x-1)-\sqrt{t}=c\Rightarrow e^{x}(x-1)-\sqrt{1-y^{2}}=c
\displaystyle \text{Putting } y=1,\ x=0
\displaystyle e^{0}(-1)-0=c
\displaystyle c=-1
\displaystyle e^{x}(x-1)-\sqrt{1-y^{2}}=-1

\displaystyle \textbf{19. } \text{Solve the following differential equation: }   (x^{2}-1)\frac{dy}{dx}+2xy=\frac{2}{x^{2}-1}.
\displaystyle \text{Answer:}
\displaystyle (x^{2}-1)\frac{dy}{dx}+2xy=\frac{2}{x^{2}-1}
\displaystyle \frac{dy}{dx}+\frac{2x}{x^{2}-1}y=\frac{2}{(x^{2}-1)^{2}}
\displaystyle \text{This is a linear differential equation of the form}
\displaystyle \frac{dy}{dx}+py=Q
\displaystyle \text{where } P=\frac{2x}{x^{2}-1},\ Q=\frac{2}{(x^{2}-1)^{2}}
\displaystyle \text{I.F.}=e^{\int p\,dx}=e^{\int \frac{2x}{x^{2}-1}\,dx}
\displaystyle \text{Let } x^{2}-1=t\Rightarrow 2x\,dx=dt
\displaystyle \text{I.F.}=e^{\int \frac{1}{t}\,dt}=e^{\log t}=e^{\log(x^{2}-1)}=x^{2}-1
\displaystyle y\cdot \text{I.F.}=\int Q(\text{I.F.})\,dx+c
\displaystyle y(x^{2}-1)=\int \frac{2}{(x^{2}-1)^{2}}(x^{2}-1)\,dx+c
\displaystyle y(x^{2}-1)=2\int \frac{dx}{x^{2}-1}+c
\displaystyle \Rightarrow y(x^{2}-1)=2\cdot \frac{1}{2}\log\left|\frac{x-1}{x+1}\right|+c
\displaystyle \qquad \left[\because\ \int \frac{1}{x^{2}-a^{2}}\,dx=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+c\right]
\displaystyle \Rightarrow y(x^{2}-1)=\log\left|\frac{x-1}{x+1}\right|+c

\displaystyle \textbf{20. } \text{Prove that, for any three vectors } \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}
\displaystyle [\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a}]=2[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]
\displaystyle \text{OR}
\displaystyle \text{Vectors } \overrightarrow{a},\overrightarrow{b} \text{ and } \overrightarrow{c} \text{ are such that } \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0 \text{ and } |\overrightarrow{a}|=3,\   
\displaystyle |\overrightarrow{b}|=5 \text{ and } |\overrightarrow{c}|=7. \text{ Find the angle between } \overrightarrow{a} \text{ and } \overrightarrow{b}.
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S. } [\overrightarrow{a}+\overrightarrow{b},\ \overrightarrow{b}+\overrightarrow{c},\ \overrightarrow{c}+\overrightarrow{a}]
\displaystyle =(\overrightarrow{a}+\overrightarrow{b})\cdot[(\overrightarrow{b}+\overrightarrow{c})\times(\overrightarrow{c}+\overrightarrow{a})]
\displaystyle =(\overrightarrow{a}+\overrightarrow{b})\cdot[(\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{b}\times\overrightarrow{a}+\overrightarrow{c}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a})]
\displaystyle =(\overrightarrow{a}+\overrightarrow{b})\cdot[(\overrightarrow{b}\times\overrightarrow{c}-\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{0}+\overrightarrow{c}\times\overrightarrow{a})]\qquad [\because\ \overrightarrow{c}\times\overrightarrow{c}=0]
\displaystyle =\overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c})-\overrightarrow{a}\cdot(\overrightarrow{a}\times\overrightarrow{b})+\overrightarrow{a}\cdot(\overrightarrow{c}\times\overrightarrow{a})+\overrightarrow{b}\cdot(\overrightarrow{b}\times\overrightarrow{c})
\displaystyle \quad +\overrightarrow{b}\cdot(\overrightarrow{c}\times\overrightarrow{a})-\overrightarrow{b}\cdot(\overrightarrow{a}\times\overrightarrow{b})
\displaystyle =\overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c})+\overrightarrow{b}\cdot(\overrightarrow{c}\times\overrightarrow{a})\qquad [\because\ \overrightarrow{a}\cdot(\overrightarrow{a}\times\overrightarrow{b})=0,\ \overrightarrow{b}\cdot(\overrightarrow{a}\times\overrightarrow{b})=0]
\displaystyle =[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]+[\overrightarrow{b}\ \overrightarrow{c}\ \overrightarrow{a}]
\displaystyle =[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]+[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]
\displaystyle =2[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]=\text{R.H.S.}
\displaystyle \text{OR}
\displaystyle \text{Given: } |\overrightarrow{a}|=3,\ |\overrightarrow{b}|=5,\ |\overrightarrow{c}|=7
\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\Rightarrow \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}
\displaystyle \text{Squaring on both sides}
\displaystyle |\overrightarrow{a}+\overrightarrow{b}|^{2}=|-\overrightarrow{c}|^{2}
\displaystyle (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=7^{2}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}=49
\displaystyle |\overrightarrow{a}|^{2}+2\overrightarrow{a}\cdot\overrightarrow{b}+|\overrightarrow{b}|^{2}=49
\displaystyle 3^{2}+2|\overrightarrow{a}||\overrightarrow{b}|\cos\theta+5^{2}=49
\displaystyle 9+2(3)(5)\cos\theta+25=49
\displaystyle 30\cos\theta=15
\displaystyle \cos\theta=\frac{1}{2}
\displaystyle \theta=\cos^{-1}\left(\frac{1}{2}\right)=60^{\circ}

\displaystyle \textbf{21. } \text{Show that the lines } \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \text{ and } \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}
\displaystyle \text{intersect. Also find their point of intersection.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=p\qquad \ldots (1)
\displaystyle \text{Any point on this line is } (3p-1,\ 5p-3,\ 7p-5)
\displaystyle \text{Let } \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=k\qquad \ldots (2)
\displaystyle \text{Any point on this line is } (k+2,\ 3k+4,\ 5k+6)
\displaystyle \text{For lines (1) and (2) to intersect}
\displaystyle 3p-1=k+2\Rightarrow 3p-k=3
\displaystyle 5p-3=3k+4\Rightarrow 5p-3k=7
\displaystyle 7p-5=5k+6\Rightarrow 7p-5k=11
\displaystyle \text{Solving equations } k=3p-3;\ 5p-3k=7
\displaystyle \Rightarrow 5p-3(3p-3)=7
\displaystyle \Rightarrow p=\frac{1}{2}
\displaystyle k=3p-3=3\left(\frac{1}{2}\right)-3\Rightarrow k=-\frac{3}{2}
\displaystyle \text{Putting } p=\frac{1}{2},\ k=-\frac{3}{2} \text{ in } 7p-5k=11
\displaystyle 7\left(\frac{1}{2}\right)-5\left(-\frac{3}{2}\right)=11\Rightarrow \frac{7}{2}+\frac{15}{2}=11\Rightarrow 11=11
\displaystyle \text{Since, it satisfied third equation.}
\displaystyle \text{So, both lines intersect each other.}
\displaystyle \therefore\ \text{Lines (1) and (2) intersect at point}
\displaystyle (3p-1,\ 5p-3,\ 7p-5)=\left(\frac{1}{2},-\frac{1}{2},-\frac{3}{2}\right)

\displaystyle \textbf{22. } \text{Assume that each born child is equally likely to be a boy or a girl. If a family }
\displaystyle \text{has two children, what is the conditional probability that both are girls? Given that}
\displaystyle \text{probability that both are girls? Given that}
\displaystyle \text{(a) the youngest is a girl.}
\displaystyle \text{(b) atleast one is a girl.}
\displaystyle \text{Answer:}
\displaystyle \text{Sample space of having two children in a family,}
\displaystyle S=[B_{1}B_{2},\ B_{1}G_{2},\ G_{1}G_{2},\ G_{1}B_{2}]
\displaystyle n(S)=4
\displaystyle \text{Let } A \text{ be the event that both children are girls}
\displaystyle A=[G_{1},G_{2}],\ n(A)=1
\displaystyle P(A)=\frac{1}{4}
\displaystyle \text{Let } B \text{ be the event that the youngest child is a girl}
\displaystyle B=[G_{1}G_{2},\ B_{1}G_{2}],\ n(B)=2
\displaystyle P(B)=\frac{2}{4}
\displaystyle \text{Let } C \text{ be the event that at least one of the children is girl}
\displaystyle C=[B_{1}G_{2},\ G_{1}G_{2},\ G_{1}B_{2}]
\displaystyle n(C)=3,\ P(C)=\frac{3}{4}
\displaystyle \text{(i) } P(A/B)=\frac{P(A\cap B)}{P(B)}
\displaystyle A\cap B=\{G_{1},G_{2}\}
\displaystyle P(A\cap B)=\frac{1}{4}
\displaystyle P(A/B)=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}
\displaystyle \text{(ii) } P(A/C)=\frac{P(A\cap C)}{P(C)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}
\displaystyle \qquad [\because\ A\cap C=\{G_{1},G_{2}\}]

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question numbers 23 to 29 carry 6 marks each.}

\displaystyle \textbf{23. } \text{Two schools P and Q want to award their selected students on the values of }
\displaystyle \text{Discipline, Politeness and Punctuality. The } \text{school P wants to award Rs } x \text{ each, Rs } y
\displaystyle \text{ each and Rs } z \text{ each for } \text{the three respective values to its } 3,\ 2 \text{ and } 1 \text{ students with}
\displaystyle \text{a total award money of Rs } 1000. \text{ School Q wants to spend } \text{Rs } 1500 \text{ to award its } 4,\ 1 \text{ and }
\displaystyle 3 \text{ students on the respective }   \text{values (by giving the same award money for the three values}
\displaystyle \text{as before). If the total amount of awards for one prize on } \text{each value is Rs } 600, 
\displaystyle \text{using matrices, find the award money for each value. Apart from the above three values, }
\displaystyle \text{suggest one more value for awards.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the awards for Discipline, Politeness and Punctuality be Rs } x,\ \text{Rs } y \text{ and Rs } z  \\ \text{ respectively.}
\displaystyle 3x+2y+z=1000
\displaystyle 4x+y+3z=1500
\displaystyle x+y+z=600
\displaystyle \text{Matrix form of given equations is}
\displaystyle AX=B
\displaystyle A=\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\\z\end{bmatrix},\ B=\begin{bmatrix}10\\15\\6\end{bmatrix}
\displaystyle \begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\15\\6\end{bmatrix}
\displaystyle |A|=\begin{vmatrix}3&2&1\\4&1&3\\1&1&1\end{vmatrix}=3(1-3)-2(4-3)+1(4-1)
\displaystyle =-6-2+3=-5\neq 0
\displaystyle \therefore\ A^{-1}\ \text{exists}
\displaystyle A_{11}=(-1)^{2}(1-3)=-2,\ A_{12}=(-1)^{3}(4-3)=-1
\displaystyle A_{13}=(-1)^{4}(4-1)=3,\ A_{21}=(-1)^{3}(2-1)=-1
\displaystyle A_{22}=(-1)^{4}(3-1)=2,\ A_{23}=(-1)^{5}(3-2)=-1
\displaystyle A_{31}=(-1)^{4}(6-1)=5,\ A_{32}=(-1)^{5}(9-4)=-5
\displaystyle A_{33}=(-1)^{6}(3-8)=-5
\displaystyle \therefore {adj \ } A=\begin{bmatrix}-2&-1&3\\-1&2&-1\\5&-5&-5\end{bmatrix}^{T}=\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|} {adj \ }A=-\frac{1}{5}\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}
\displaystyle AX=B
\displaystyle X=A^{-1}B
\displaystyle \begin{bmatrix}x\\y\\z\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}\begin{bmatrix}10\\15\\6\end{bmatrix}
\displaystyle =-\frac{1}{5}\begin{bmatrix}-2000-1500+3000\\-1000+3000-3000\\3000-1500-3000\end{bmatrix}=\frac{1}{5}\begin{bmatrix}-5\\-10\\-15\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}
\displaystyle \therefore\ x=\text{Rs }100,\ y=\text{Rs }200,\ z=\text{Rs }300
\displaystyle \text{Apart from the three values, Discipline, Politeness and Punctuality, another value} \\ \text{for award should be hard work.}

\displaystyle \textbf{24. } \text{Show that the semi-vertical angle of the cone of the maximum volume and of }
\displaystyle \text{given slant height is } \cos^{-1}\frac{1}{\sqrt{3}}.
\displaystyle \text{Answer:}
\displaystyle \text{24. Let } \theta \text{ be the semi vertical angle, } h \text{ is height, } r \text{ base radius and } \\  l \text{ be slant height of a cone}
\displaystyle \text{In } \triangle OAB
\displaystyle \sin\theta=\frac{r}{l}
\displaystyle r=l\sin\theta
\displaystyle \cos\theta=\frac{h}{l}
\displaystyle h=l\cos\theta
\displaystyle \text{Let } V \text{ be the volume of cone}
\displaystyle V=\frac{1}{3}\pi r^{2}h
\displaystyle V=\frac{1}{3}\pi (l\sin\theta)^{2}(l\cos\theta)
\displaystyle V=\frac{1}{3}\pi l^{3}\sin^{2}\theta\cdot\cos\theta
\displaystyle \text{Differentiating w.r.t. } \theta
\displaystyle \frac{dV}{d\theta}=\frac{1}{3}\pi l^{3}(2\sin\theta\cdot\cos\theta\cdot\cos\theta-\sin^{2}\theta\sin\theta)
\displaystyle =\frac{1}{3}\pi l^{3}\sin\theta(2\cos^{2}\theta-\sin^{2}\theta)
\displaystyle \text{For maxima or minima, } \frac{dV}{d\theta}=0
\displaystyle \sin\theta(2\cos^{2}\theta-\sin^{2}\theta)=0
\displaystyle \Rightarrow \sin\theta=0 \text{ or } 2\cos^{2}\theta-\sin^{2}\theta=0
\displaystyle \Rightarrow \sin\theta\neq 0 \qquad [\because\ \theta \text{ cannot be } 0]
\displaystyle \Rightarrow 2\cos^{2}\theta-\sin^{2}\theta=0
\displaystyle \Rightarrow 2\cos^{2}\theta-(1-\cos^{2}\theta)=0
\displaystyle \Rightarrow 3\cos^{2}\theta=1\Rightarrow \cos^{2}\theta=\frac{1}{3}
\displaystyle \theta=\cos^{-1}\frac{1}{\sqrt{3}}
\displaystyle \text{Also, } \cos\theta=\frac{1}{\sqrt{3}}\Rightarrow \sin\theta=\frac{\sqrt{2}}{\sqrt{3}}
\displaystyle \text{Again differentiating } \frac{dV}{d\theta}
\displaystyle \frac{d^{2}V}{d\theta^{2}}=\frac{1}{3}\pi l^{3}\left[\cos\theta(2\cos^{2}\theta-\sin^{2}\theta)+\sin\theta(-4\cos\theta\sin\theta-2\sin\theta\cos\theta)\right]
\displaystyle =\frac{1}{3}\pi l^{3}\left[\frac{1}{\sqrt{3}}\left(2\cdot \frac{1}{3}-\frac{2}{3}\right)-6\cdot \frac{2}{3}\cdot \frac{1}{\sqrt{3}}\right]
\displaystyle =\frac{1}{3}\pi l^{3}\left(-\frac{4}{\sqrt{3}}\right)<0
\displaystyle \therefore\ V \text{ is maximum for } \theta=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)
\displaystyle \text{Hence proved.}

\displaystyle \textbf{25. } \text{Evaluate : } \int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\cot x}}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\cot x}}
\displaystyle I=\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\frac{\cos x}{\sin x}}}\qquad \left[\because\ \cot x=\frac{\cos x}{\sin x}\right]
\displaystyle I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\qquad \ldots (1)
\displaystyle I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}+\sqrt{\cos\left(\frac{\pi}{2}-x\right)}}\,dx
\displaystyle \qquad \left[\because\ \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx,\ \text{where } a+b=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}\right]
\displaystyle I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx\qquad \ldots (2)
\displaystyle \text{Adding (1) and (2)}
\displaystyle 2I=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx\Rightarrow 2I=\int_{\pi/6}^{\pi/3}1\,dx
\displaystyle 2I=(x)_{\pi/6}^{\pi/3}\Rightarrow 2I=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}
\displaystyle I=\frac{\pi}{12}

\displaystyle \textbf{26. } \text{Find the area of the region in the first quadrant}   \text{enclosed by the } x\text{-axis, the line} \\  y=x \text{ and the circle}   x^{2}+y^{2}=32.
\displaystyle \text{Answer:}
\displaystyle \text{Equation of line: } y=x\qquad \ldots (1)
\displaystyle \text{Equation of circle: } x^{2}+y^{2}=32\qquad \ldots (2)
\displaystyle y=\sqrt{32-x^{2}}
\displaystyle y^{2}+y^{2}=32\qquad \text{[from (1) and (2)]}
\displaystyle \Rightarrow 2y^{2}=32
\displaystyle \Rightarrow y=\pm 4,\ x=\pm 4
\displaystyle \therefore\ \text{Point of intersection of circle and line is } (4,4) \text{ and } (-4,-4)
\displaystyle x^{2}+y^{2}=32
\displaystyle x^{2}+y^{2}=(4\sqrt{2})^{2}
\displaystyle \text{Circle meets } x\text{-axis at } p(4\sqrt{2},0) \text{ and } p'(-4\sqrt{2},0)\displaystyle \text{Required area = Area } OPQ
\displaystyle =\int_{0}^{4}y_{1}\,dx+\int_{4}^{4\sqrt{2}}y_{2}\,dx=\int_{0}^{4}x\,dx+\int_{4}^{4\sqrt{2}}\sqrt{32-x^{2}}\,dx
\displaystyle \text{Required area}=\left[\frac{x^{2}}{2}\right]_{0}^{4}+\left[\frac{x\sqrt{32-x^{2}}}{2}+\frac{32}{2}\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right)\right]_{4}^{4\sqrt{2}}
\displaystyle \qquad \left[\because\ \int \sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}+c\right]
\displaystyle =8+\left[0+16\sin^{-1}(1)-\frac{4\cdot 4}{2}-16\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\right]
\displaystyle =16\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=16\times \frac{\pi}{4}=4\pi\ \text{sq. units}

\displaystyle \textbf{27.  } \text{Find the distance between the point } (7,2,4) \text{ and the plane determined by the points}
\displaystyle A(2,5,-3), B(-2,-3,5) \text{ and } C(5,3,-3).
\displaystyle \text{OR}
\displaystyle \text{Find the distance of the point } (-1,-5,-10) \text{ from the point of intersection of the line}
\displaystyle \overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k})   \text{and the plane } \overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+\widehat{k})=5.
\displaystyle \text{Answer:}
\displaystyle \text{The plane passing through } A(2,5,-3) \text{ is}
\displaystyle a(x-2)+b(y-5)+c(z+3)=0\qquad \ldots (1)
\displaystyle \text{It also passes through } B(-2,-3,5) \text{ and } C(5,3,-3)
\displaystyle -4a-8b+8c=0\qquad \ldots (2)
\displaystyle 3a-2b+0c=0\qquad \ldots (3)
\displaystyle \text{On solving equations (2) and (3)}
\displaystyle \frac{a}{0+16}=\frac{b}{24-0}=\frac{c}{8+24}
\displaystyle \frac{a}{2}=\frac{b}{3}=\frac{c}{4}\qquad \ldots (4)
\displaystyle \text{From (1) \& (4) equation of required plane is}
\displaystyle 2(x-2)+3(y-5)+4(z+3)=0
\displaystyle 2x-4+3y-15+4z+12=0
\displaystyle 2x+3y+4z-7=0
\displaystyle \text{Distance of the point from plane is}
\displaystyle d=\left|\frac{ax_{1}+by_{1}+cz_{1}-d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|=\left|\frac{2\times 7+3\times 2+4\times 4-7}{\sqrt{2^{2}+3^{2}+4^{2}}}\right|
\displaystyle d=\left|\frac{14+6+16-7}{\sqrt{4+9+16}}\right|=\frac{29}{\sqrt{29}}=\sqrt{29}
\displaystyle \text{OR}
\displaystyle \text{Given plane: } \overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+\widehat{k})=5
\displaystyle \text{Cartesian form } x-y+z=5\qquad \ldots (1)
\displaystyle \text{The given line is}
\displaystyle \overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k})
\displaystyle \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda
\displaystyle \text{Point on the given line is } (3\lambda+2,\ 4\lambda-1,\ 2\lambda+2)
\displaystyle \text{As both the plane and line are intersecting, point will also lie on plane}
\displaystyle 3\lambda+2-4\lambda+1+2\lambda+2=5
\displaystyle \lambda=0
\displaystyle \therefore\ \text{Point of intersection is } (2,-1,2)
\displaystyle \text{Distance between } (2,-1,2) \text{ and } (-1,-5,-10) \text{ is given as}
\displaystyle d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}
\displaystyle d=\sqrt{9+16+144}=\sqrt{169}=13
\displaystyle \text{Distance }=13\ \text{units.}

\displaystyle \textbf{28. } \text{A dealer in rural area wishes to purchase a number of sewing machines. He has only }
\displaystyle \text{Rs } 5760 \text{ to invest and has space for }\text{at most } 20 \text{ items for storage. An electronic }
\displaystyle \text{sewing machine cost him Rs } 360 \text{ and a manually operated sewing machine }\text{Rs } 240.
\displaystyle \text{ He can sell an electronic sewing machine at a profit } \text{of Rs } 22 \text{ and a manually operated }
\displaystyle \text{sewing machine at a profit of Rs } 18. \text{ Assuming that he can sell all the items that he can}
\displaystyle \text{buy, how should he invest his money in order to maximize his profit? Make it as an LPP}
\displaystyle \text{and solve it graphically.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the electronic and manually operated sewing machine bought by dealer be} \\  x \text{ and } y \text{ respectively}
\displaystyle \text{The LPP is}
\displaystyle \text{Maximize } z=22x+18y
\displaystyle \text{Subject to constraints}
\displaystyle x+y\leq 20
\displaystyle 360x+240y\leq 5760
\displaystyle 3x+2y\leq 48
\displaystyle x\geq 0,\ y\geq 0
\displaystyle x+y=20\displaystyle \begin{array}{|c|c|c|c|}\hline x&20&0&8\\\hline y&0&20&12\\\hline \end{array}
\displaystyle 3x+2y=48
\displaystyle \begin{array}{|c|c|c|c|}\hline x&16&0&8\\\hline y&0&24&12\\\hline \end{array}
\displaystyle \text{Point of intersection of both the lines is } (8,12)
\displaystyle \therefore\ \text{The feasible region is } OCPAO \text{ which is shaded in the figure}
\displaystyle \text{The vertices of feasible region is } O(0,0),\ C(16,0),\ A(0,20),\ P(8,12)
\displaystyle \begin{array}{|c|c|}\hline \text{Corner Points}&Z=22x+18y\\\hline O(0,0)&Z=0\\\hline C(16,0)&Z=352\\\hline P(8,12)&Z=392\ \text{(max)}\\\hline A(0,20)&Z=360\\\hline \end{array}
\displaystyle \therefore\ \text{Maximum profit is Rs } 392 \text{ when } 8 \text{ electronic and } 12 \text{ manually operated machines are purchased}

\displaystyle \textbf{29.  } \text{A card from a pack of } 52 \text{ playing cards is lost. From the remaining cards of the }
\displaystyle \text{pack three cards are drawn at random (without replacement) and are found to be all}
\displaystyle \text{spades. Find the probability of the lost card being a spade.}
\displaystyle \text{OR}
\displaystyle \text{From a lot of } 15 \text{ bulbs which includes } 5 \text{ defectives, a sample } \text{of } 4 \text{ bulbs is drawn one}
\displaystyle  \text{by one with replacement. Find the probability distribution of number of defective}
\displaystyle \text{bulbs. Hence find the mean of the distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the events be}
\displaystyle E_{1}=\text{the missing card is a heart card}
\displaystyle E_{2}=\text{the missing card is a spade card}
\displaystyle E_{3}=\text{the missing card is a club card}
\displaystyle E_{4}=\text{the missing card is a diamond card}
\displaystyle A=\text{drawing three spades card from the remaining card.}
\displaystyle P(E_{1})=\frac{13}{52}=\frac{1}{4}\Rightarrow P(E_{2})=\frac{13}{52}=\frac{1}{4}
\displaystyle P(E_{3})=\frac{13}{52}=\frac{1}{4}
\displaystyle P(E_{4})=\frac{13}{52}=\frac{1}{4}
\displaystyle P(A/E_{1})=\frac{{}^{13}C_{3}}{{}^{51}C_{3}},\ P(A/E_{2})=\frac{{}^{12}C_{3}}{{}^{51}C_{3}}
\displaystyle P(A/E_{3})=\frac{{}^{13}C_{3}}{{}^{51}C_{3}},\ P(A/E_{4})=\frac{{}^{13}C_{3}}{{}^{51}C_{3}}
\displaystyle \text{Required probability }=P(E_{2}/A)
\displaystyle P(E_{2}/A)=\frac{P(E_{2})P(A/E_{2})}{P(A/E_{1})P(E_{1})+P(A/E_{2})P(E_{2})+P(A/E_{3})P(E_{3})+P(A/E_{4})P(E_{4})}
\displaystyle =\frac{\frac{1}{4}\times \frac{{}^{12}C_{3}}{{}^{51}C_{3}}}{\frac{1}{4}\times \frac{{}^{13}C_{3}}{{}^{51}C_{3}}+\frac{1}{4}\times \frac{{}^{12}C_{3}}{{}^{51}C_{3}}+\frac{1}{4}\times \frac{{}^{13}C_{3}}{{}^{51}C_{3}}+\frac{1}{4}\times \frac{{}^{13}C_{3}}{{}^{51}C_{3}}}
\displaystyle =\frac{{}^{12}C_{3}}{{}^{12}C_{3}+3\times {}^{13}C_{3}}
\displaystyle {}^{12}C_{3}=\frac{12!}{(12-3)!3!}=\frac{12\times 11\times 10\times 9!}{9!\times 3\times 2\times 1}=220
\displaystyle {}^{13}C_{3}=\frac{13!}{3!10!}=\frac{13\times 12\times 11\times 10!}{3\times 2\times 1\times 10!}=286
\displaystyle \therefore\ P(E_{2}/A)=\frac{220}{220+3\times 286}
\displaystyle P(E_{2}/A)=\frac{110}{539}
\displaystyle \text{OR}
\displaystyle \text{Let } D \text{ be the event of drawing a defective bulb and } X \text{ denote the variable} \\ \text{showing the number of defective bulbs in } 4 \text{ draws.}
\displaystyle P(D)=\frac{5}{15}=\frac{1}{3}
\displaystyle P(\overline{D})=1-\frac{1}{3}=\frac{2}{3}
\displaystyle \text{The drawn bulb is replaced.}
\displaystyle \therefore\ X \text{ can take values } 0,1,2,3 \text{ and } 4
\displaystyle P(X=0)=P(\text{Getting no defective bulb})={}^{4}C_{0}\left(\frac{1}{3}\right)^{0}\left(\frac{2}{3}\right)^{4}=\left(\frac{2}{3}\right)^{4}=\frac{16}{81}
\displaystyle \qquad [\because\ P(X)={}^{n}C_{r}p^{r}q^{n-r}]
\displaystyle P(X=1)={}^{4}C_{1}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{3}=\frac{32}{81}
\displaystyle P(X=2)={}^{4}C_{2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{2}=\frac{24}{81}
\displaystyle P(X=3)={}^{4}C_{3}\left(\frac{1}{3}\right)^{3}\left(\frac{2}{3}\right)=\frac{8}{81}
\displaystyle P(X=4)={}^{4}C_{4}\left(\frac{1}{3}\right)^{4}=\frac{1}{81}
\displaystyle \therefore\ \text{The probability distribution is}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline X&0&1&2&3&4&\text{Total}\\ \hline P(X=x) &{16}/{81}& {32}/{81}& {24}/{81}& {8}/{81}& {1}/{81}&1 \\ \hline \end{array}
\displaystyle \text{Mean of the distribution }=\sum P_{i}X_{i}
\displaystyle =\frac{16}{81}\times 0+\frac{32}{81}\times 1+\frac{24}{81}\times 2+\frac{8}{81}\times 3+\frac{1}{81}\times 4
\displaystyle =\frac{1}{81}(0+32+48+24+4)=\frac{108}{81}=\frac{4}{3}

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