\displaystyle \textbf{Time Allowed : 3 Hours} \qquad \textbf{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.}
\displaystyle \text{Section A comprises of 10 questions of one mark each, Section B comprises of 12 questions}
\displaystyle \text{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 }
\displaystyle \text{per the exact requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, internal choice has been provided }
\displaystyle \text{in 04 questions of four marks each and 02 questions of six marks each. You have to }
\displaystyle \text{attempt only one of the alternatives in all such questions.}
\displaystyle \text{(v) Use of calculator is not permitted. You may ask for logarithmic tables, if required.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Question number 1 to 10 carry 1 Mark each}

\displaystyle \textbf{1. } \text{If } R=\{(x,y):x+2y=8\} \text{ is a relation on N, write the range of R.}
\displaystyle \text{Answer:}
\displaystyle R=\{(x,y):x+2y=8\}
\displaystyle x=1,\ y=\frac{7}{2}\notin N;\ x=5,\ y=\frac{3}{2}\notin N
\displaystyle x=2,\ y=3\in N;\ x=6,\ y=1\in N
\displaystyle x=3,\ y=\frac{5}{2}\notin N;\ x=7,\ y=\frac{1}{2}\notin N
\displaystyle x=4,\ y=2\in N;\ x=8,\ y=0\notin N
\displaystyle \therefore\ \text{Range}(R)=\{1,2,3\}

\displaystyle \textbf{2. } \text{If } \tan^{-1}x+\tan^{-1}y=\frac{\pi}{4},\ xy<1,\ \text{then write the value of } x+y+xy.
\displaystyle \text{Answer:}
\displaystyle  \tan^{-1}x+\tan^{-1}y=\frac{\pi}{4},\ xy<1
\displaystyle \Rightarrow \tan^{-1}\left(\frac{x+y}{1-xy}\right)=\frac{\pi}{4}
\displaystyle \Rightarrow \frac{x+y}{1-xy}=\tan\frac{\pi}{4}=1
\displaystyle \Rightarrow x+y=1-xy
\displaystyle \Rightarrow x+y+xy=1

\displaystyle \textbf{3. } \text{If A is a square matrix such that } A^{2}=A,\ \text{then write the value of }
\displaystyle 7A-(I+A)^{3},  \text{where I is an identity matrix.}
\displaystyle \text{Answer:}
\displaystyle   A^{2}=A
\displaystyle 7A-(I+A)^{3}
\displaystyle =7A-(I^{3}+A^{3}+3I^{2}A+3IA^{2})
\displaystyle =7A-(I+A^{3}+3IA+3IA)
\displaystyle =7A-(I+A^{2}+3A+3A)
\displaystyle =7A-(I+A+6A)
\displaystyle =7A-I-7A
\displaystyle \therefore\ 7A-(I+A)^{3}=-I

\displaystyle \textbf{4. } \text{If } \begin{bmatrix} x-y & z \\ 2x-y & w \end{bmatrix}=\begin{bmatrix} -1 & 4 \\ 0 & 5 \end{bmatrix},\ \text{find the value of } x+y.
\displaystyle \text{Answer:}
\displaystyle \text{We have}
\displaystyle \begin{bmatrix}x-y&z\\2x-y&w\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix}
\displaystyle \text{By the equality of matrices, we have}
\displaystyle x-y=-1,\ 2x-y=0
\displaystyle \text{On subtracting, we get}
\displaystyle 2x-y=0
\displaystyle x-y=-1
\displaystyle \therefore\ x=1
\displaystyle y=2
\displaystyle \therefore\ x+y=3

\displaystyle \textbf{5. } \text{If } \begin{vmatrix} 3x & 7 \\ -2 & 4 \end{vmatrix}=\begin{vmatrix} 8 & 7 \\ 6 & 4 \end{vmatrix},\ \text{find the value of } x.
\displaystyle \text{Answer:}
\displaystyle  \begin{vmatrix}3x&7\\-2&4\end{vmatrix}=\begin{vmatrix}8&7\\6&4\end{vmatrix}
\displaystyle \Rightarrow 12x+14=32-42
\displaystyle \Rightarrow 12x=-10-14
\displaystyle \Rightarrow 12x=-24
\displaystyle \Rightarrow x=-2

\displaystyle \textbf{6. } \text{If } f(x)=\int_{0}^{x}t \sin t\,dt,\ \text{then write the value of } f'(x).
\displaystyle \text{Answer:}
\displaystyle \text{Let } f(x)=\int_{0}^{x}t\sin t\,dt
\displaystyle f(x)=-t\cos t\Big|_{0}^{x}+\int_{0}^{x}\cos t\,dt
\displaystyle f(x)=-x\cos x+\sin t\Big|_{0}^{x}
\displaystyle f(x)=-x\cos x+\sin x
\displaystyle f'(x)=-x(-\sin x)-\cos x+\cos x
\displaystyle =x\sin x

\displaystyle \textbf{7. } \text{Evaluate: } \int_{2}^{4}\frac{x}{x^{2}+1}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{2}^{4}\frac{x}{x^{2}+1}\,dx
\displaystyle \text{Put } x^{2}+1=t
\displaystyle 2x\,dx=dt
\displaystyle \therefore\ x\,dx=\frac{dt}{2}
\displaystyle \text{When } x=2,\ t=5
\displaystyle \text{When } x=4,\ t=17
\displaystyle I=\int_{5}^{17}\frac{1}{2t}\,dt=\frac{1}{2}\log t\Big|_{5}^{17}
\displaystyle I=\frac{1}{2}\log\left(\frac{17}{5}\right)

\displaystyle \textbf{8. } \text{Find the value of 'p' for which the vectors } 3\widehat{i}+2\widehat{j}+9\widehat{k} \text{and } \widehat{i}-2p\widehat{j}+3\widehat{k} 
\displaystyle \text{ are parallel.}
\displaystyle \text{Answer:}
\displaystyle \text{Given } 3\widehat{i}+2\widehat{j}+9\widehat{k} \text{ and } \widehat{i}-2p\widehat{j}+3\widehat{k} \text{ are parallel.}
\displaystyle \frac{3}{1}=\frac{2}{-2p}=\frac{9}{3}
\displaystyle \Rightarrow \frac{3}{1}=\frac{-1}{p}\Rightarrow p=-\frac{1}{3}

\displaystyle \textbf{9. } \text{Find } \overrightarrow{a}\cdot(\overrightarrow{b}\times \overrightarrow{c}),\ \text{if } \overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k},\ \overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k}\ \text{ and } \overrightarrow{c}=3\widehat{i}+\widehat{j}+2\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle  \overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k},\ \overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k},\ \overrightarrow{c}=3\widehat{i}+\widehat{j}+2\widehat{k}
\displaystyle \text{We have,}
\displaystyle \overrightarrow{b}\times \overrightarrow{c}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&2&1\\3&1&2\end{vmatrix}
\displaystyle =(4-1)\widehat{i}-(-2-3)\widehat{j}+(-1-6)\widehat{k}
\displaystyle =3\widehat{i}+5\widehat{j}-7\widehat{k}
\displaystyle \text{Now, } \overrightarrow{a}\cdot(\overrightarrow{b}\times \overrightarrow{c})=(2\widehat{i}+\widehat{j}+3\widehat{k})\cdot(3\widehat{i}+5\widehat{j}-7\widehat{k})
\displaystyle =6+5-21=-10

\displaystyle \textbf{10. } \text{If the cartesian equations of a line are } \frac{3-x}{5}=\frac{y+4}{7}=\frac{2z-6}{4},
\displaystyle \text{write the vector equation for the line.}
\displaystyle \text{Answer:}
\displaystyle \text{Given equations of line}
\displaystyle \frac{x-3}{-5}=\frac{y+4}{7}=\frac{z-3}{2}
\displaystyle \text{Comparing with } \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c},\ \text{we get}
\displaystyle x_{1}=3,\ y_{1}=-4,\ z_{1}=3
\displaystyle a=-5,\ b=7,\ c=2
\displaystyle \text{Let } \overrightarrow{r} \text{ be the position vector of any point on the line,}
\displaystyle \text{then the vector equation is}
\displaystyle \overrightarrow{r}=3\widehat{i}-4\widehat{j}+3\widehat{k}+\lambda(-5\widehat{i}+7\widehat{j}+2\widehat{k})

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Question number 11 to 22 carry 4 Mark each}

\displaystyle \textbf{11. } \text{If the function } f:R\rightarrow R \text{ be given by } f(x)=x^{2}+2 \text{ and } \text{g:R}\rightarrow R \text{ be given by } 
\displaystyle g(x)=\frac{x}{x-1},\ x\neq 1,\ \text{find fog}   \text{and gof and hence find fog(2) and gof(-3).}
\displaystyle \text{Answer:}
\displaystyle \text{Let } f(x)=x^{2}+2,\ g(x)=\frac{x}{x-1},\ x\neq 1
\displaystyle \text{Consider } fog(x)=f[g(x)]=f\left(\frac{x}{x-1}\right)=\left(\frac{x}{x-1}\right)^{2}+2
\displaystyle =\frac{x^{2}+2x^{2}+2-4x}{x^{2}+1-2x}=\frac{3x^{2}-4x+2}{x^{2}-2x+1}
\displaystyle gof(x)=g[f(x)]=g[x^{2}+2]=\frac{x^{2}+2}{x^{2}+2-1}=\frac{x^{2}+2}{x^{2}+1}
\displaystyle \text{Now, } gof(-3)=\frac{(-3)^{2}+2}{(-3)^{2}+1}=\frac{9+2}{9+1}=\frac{11}{10}
\displaystyle \text{and } fog(2)=\frac{3(2)^{2}-4(2)+2}{(2)^{2}-2(2)+1}=\frac{12-8+2}{4-4+1}=6

\displaystyle \textbf{12. } \text{Prove that: }   \tan^{-1}\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x,\ \frac{-1}{\sqrt{2}}\leq x\leq 1
\displaystyle \text{OR}
\displaystyle \text{If } \tan^{-1}\left(\frac{x-2}{x-4}\right)+\tan^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\pi}{4},\ \text{find the value of } x.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}
\displaystyle \tan^{-1}\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]
\displaystyle \text{Put } x=\cos 2\theta\Rightarrow \theta=\frac{1}{2}\cos^{-1}x
\displaystyle =\tan^{-1}\left[\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\right]
\displaystyle =\tan^{-1}\left[\frac{\sqrt{2\cos^{2}\theta}-\sqrt{2\sin^{2}\theta}}{\sqrt{2\cos^{2}\theta}+\sqrt{2\sin^{2}\theta}}\right]
\displaystyle =\tan^{-1}\left[\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\right]
\displaystyle =\tan^{-1}\left[\frac{1-\tan\theta}{1+\tan\theta}\right]=\tan^{-1}[\tan(\frac{\pi}{4}-\theta)]
\displaystyle =\frac{\pi}{4}-\theta=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x,\ -\frac{1}{\sqrt{2}}\leq x\leq 1
\displaystyle \text{RHS. Hence proved.}
\displaystyle \text{OR}
\displaystyle \text{Let } \tan^{-1}\left[\frac{x-2}{x-4}\right]+\tan^{-1}\left[\frac{x+2}{x+4}\right]=\frac{\pi}{4}
\displaystyle \Rightarrow \tan^{-1}\left(\frac{\frac{x-2}{x-4}+\frac{x+2}{x+4}}{1-\frac{x-2}{x-4}\times \frac{x+2}{x+4}}\right)=\frac{\pi}{4}
\displaystyle \Rightarrow \frac{(x-2)(x+4)+(x+2)(x-4)}{(x-4)(x+4)-(x-2)(x+2)}=\tan\frac{\pi}{4}
\displaystyle \Rightarrow (x-2)(x+4)+(x+2)(x-4)=(x^{2}-16)-(x^{2}-4)
\displaystyle \Rightarrow x^{2}+2x-8+x^{2}-2x-8=-12
\displaystyle \Rightarrow 2x^{2}-16+12=0
\displaystyle \Rightarrow 2x^{2}=4\Rightarrow x^{2}=2
\displaystyle \Rightarrow x=\pm \sqrt{2}

\displaystyle \textbf{13. } \text{Using properties of determinants, prove that}   \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix}=x^{3}
\displaystyle \text{Answer:}
\displaystyle \text{Let } \Delta=\begin{vmatrix}x+y&x&x\\5x+4y&4x&2x\\10x+8y&8x&3x\end{vmatrix}
\displaystyle \text{Take } x \text{ common from } C_{2} \text{ and } C_{3}
\displaystyle \Delta=x^{2}\begin{vmatrix}x+y&1&1\\5x+4y&4&2\\10x+8y&8&3\end{vmatrix}
\displaystyle \text{Apply } R_{3}\to R_{3}-R_{2}
\displaystyle \Delta=x^{2}\begin{vmatrix}x+y&1&1\\5x+4y&4&2\\5x+4y&4&1\end{vmatrix}
\displaystyle \text{Apply } R_{2}\to R_{2}-R_{3}
\displaystyle \Delta=x^{2}\begin{vmatrix}x+y&1&1\\0&0&1\\5x+4y&4&1\end{vmatrix}
\displaystyle =x^{2}[(x+y)(0-4)-1(5x+4y)+1(0)]
\displaystyle =x^{2}[-4x-4y+5x+4y]=x^{2}(x)=x^{3}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{14. } \text{Find the value of } \frac{dy}{dx} \text{ at } \theta=\frac{\pi}{4},\ \text{if } x=ae^{\theta}(\sin\theta-\cos\theta) \text{ and } 
\displaystyle y=ae^{\theta}(\sin\theta+\cos\theta).
\displaystyle \text{Answer:}
\displaystyle \text{Let } x=ae^{\theta}(\sin\theta-\cos\theta)
\displaystyle \frac{dx}{d\theta}=ae^{\theta}(\cos\theta+\sin\theta)+(\sin\theta-\cos\theta)ae^{\theta}
\displaystyle =2ae^{\theta}\sin\theta
\displaystyle y=ae^{\theta}(\sin\theta+\cos\theta)
\displaystyle \frac{dy}{d\theta}=ae^{\theta}(\cos\theta-\sin\theta)+[\sin\theta+\cos\theta]ae^{\theta}
\displaystyle =2ae^{\theta}\cos\theta
\displaystyle \text{Now, } \frac{dy}{dx}=\frac{dy}{d\theta}\times \frac{d\theta}{dx}=\frac{2ae^{\theta}\cos\theta}{2ae^{\theta}\sin\theta}=\cot\theta
\displaystyle \text{Now, } \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{4}}=\cot\frac{\pi}{4}=1

\displaystyle \textbf{15. } \text{If } y=Pe^{ax}+Qe^{bx},\ \text{show that } \frac{d^{2}y}{dx^{2}}-(a+b)\frac{dy}{dx}+aby=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let } y=Pe^{ax}+Qe^{bx}
\displaystyle \frac{dy}{dx}=Pae^{ax}+Qbe^{bx}\qquad \ldots (1)
\displaystyle \frac{d^{2}y}{dx^{2}}=Pa^{2}e^{ax}+Qb^{2}e^{bx}\qquad \ldots (2)
\displaystyle aby=ab[Pe^{ax}+Qe^{bx}]
\displaystyle aby=Pabe^{ax}+Qabe^{bx}\qquad \ldots (3)
\displaystyle \text{Consider}
\displaystyle \frac{d^{2}y}{dx^{2}}-(a+b)\frac{dy}{dx}+aby
\displaystyle =(Pa^{2}e^{ax}+Qb^{2}e^{bx})-(a+b)(Pae^{ax}+Qbe^{bx})+Pabe^{ax}+Qabe^{bx}\qquad \text{(from 1, 2 and 3)}
\displaystyle =Pa^{2}e^{ax}+Qb^{2}e^{bx}-Pa^{2}e^{ax}-Pabe^{ax}-Qabe^{bx}-Qb^{2}e^{bx}+Pabe^{ax}+Qabe^{bx}=0
\displaystyle \text{Hence proved.}

\displaystyle \textbf{16. } \text{Find the value(s) of x for which } y=[x(x-2)]^{2} \text{ is an increasing function.}
\displaystyle \text{OR}
\displaystyle \text{Find the equations of the tangent and normal to the curve } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
\displaystyle \text{at the point } (\sqrt{2}a,b).
\displaystyle \text{Answer:}
\displaystyle \text{Let } f(x)=[x(x-2)]^{2}
\displaystyle f'(x)=2[x(x-2)][x+x-2]
\displaystyle =2[x^{2}-2x][2x-2]
\displaystyle =4x(x-2)(x-1)
\displaystyle \text{For } f(x) \text{ to be increasing } f'(x)\geq 0
\displaystyle \Rightarrow 4x(x-2)(x-1)\geq 0
\displaystyle x\in [0,1]\cup [2,\infty)
\displaystyle \text{OR}
\displaystyle \text{We have } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
\displaystyle \text{Differentiating w.r.t. } x,\ \text{we get}
\displaystyle \frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0
\displaystyle \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}
\displaystyle \text{Now}
\displaystyle \left(\frac{dy}{dx}\right)_{(\sqrt{2}a,b)}=\frac{b^{2}\sqrt{2}a}{a^{2}b}
\displaystyle \text{slope of the tangent }=\frac{\sqrt{2}b}{a}
\displaystyle \text{Since } (x_{0},y_{0}) \text{ lies on } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
\displaystyle \therefore\ \frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}}=1\qquad \ldots (i)
\displaystyle \text{Equation of the tangent is}
\displaystyle y-y_{0}=\frac{\sqrt{2}b}{a}(x-x_{0})\Rightarrow y-b=\frac{\sqrt{2}b}{a}(x-\sqrt{2}a)
\displaystyle b\sqrt{2}x-ay-ab=0
\displaystyle \text{Now, slope of the normal is } -\frac{1}{\left(\frac{dy}{dx}\right)_{(\sqrt{2}a,b)}}=-\frac{a}{\sqrt{2}b}
\displaystyle \text{Equation of the normal is}
\displaystyle y-y_{0}=-\frac{a}{\sqrt{2}b}(x-x_{0})\Rightarrow \frac{y-y_{0}}{a^{2}y_{0}}+\frac{x-x_{0}}{b^{2}x_{0}}=0
\displaystyle \Rightarrow \frac{y-b}{a^{2}b}+\frac{x-\sqrt{2}a}{\sqrt{2}ab^{2}}=0
\displaystyle \Rightarrow (y-b)\sqrt{2}b+a(x-\sqrt{2}a)=0
\displaystyle \Rightarrow b\sqrt{2}y-\sqrt{2}b^{2}+ax-\sqrt{2}a^{2}=0
\displaystyle \Rightarrow ax+b\sqrt{2}y-\sqrt{2}(a^{2}+b^{2})=0

\displaystyle \textbf{17. } \text{Evaluate: } \int_{0}^{\pi}\frac{4x\sin x}{1+\cos^{2}x}\,dx
\displaystyle \text{OR}
\displaystyle \text{Evaluate: } \int \frac{x+2}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{\pi}\frac{4x\sin x}{1+\cos^{2}x}\,dx\qquad \ldots (1)
\displaystyle I=\int_{0}^{\pi}\frac{4(\pi-x)\sin(\pi-x)}{1+\cos^{2}(\pi-x)}\,dx
\displaystyle \qquad \left[\because\ \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\right]
\displaystyle I=\int_{0}^{\pi}\frac{(4\pi-4x)\sin x}{1+\cos^{2}x}\,dx\qquad \ldots (2)
\displaystyle \text{On adding (1) and (2), we get}
\displaystyle 2I=\int_{0}^{\pi}\frac{4\pi\sin x}{1+\cos^{2}x}\,dx
\displaystyle =-4\pi\int_{1}^{-1}\frac{dt}{1+t^{2}}\qquad \left[\because\ \cos x=t,\ \sin x\,dx=-dt\right]
\displaystyle \text{when } t=\cos x
\displaystyle \text{when } x=0,\ t=1
\displaystyle x=\pi,\ t=-1
\displaystyle =-4\pi\int_{-1}^{1}\frac{dt}{1+t^{2}}=4\pi(\tan^{-1}t)\Big|_{-1}^{1}
\displaystyle \qquad \left[\because\ \int_{a}^{b}f(x)\,dx=-\int_{b}^{a}f(x)\,dx\right]
\displaystyle =-4\pi\left[\tan^{-1}(1)-\tan^{-1}(-1)\right]
\displaystyle =-4\pi\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]=-2\pi^{2}
\displaystyle \therefore\ I=\pi^{2}
\displaystyle \text{OR}
\displaystyle \text{Let } I=\int \frac{x+2}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle =\frac{1}{2}\int \frac{2x+4}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle =\frac{1}{2}\int \frac{2x+5-1}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle =\frac{1}{2}\int \frac{(2x+5)\,dx}{\sqrt{x^{2}+5x+6}}-\frac{1}{2}\int \frac{dx}{\sqrt{x^{2}+5x+6}}
\displaystyle \text{Consider } I_{1}=\frac{1}{2}\int \frac{(2x+5)\,dx}{\sqrt{x^{2}+5x+6}}
\displaystyle \text{Let } x^{2}+5x+6=t^{2}
\displaystyle (2x+5)\,dx=2t\,dt
\displaystyle I_{1}=\frac{1}{2}\int \frac{2t\,dt}{t}=\sqrt{x^{2}+5x+6}
\displaystyle I_{2}=\int \frac{dx}{\sqrt{x^{2}+5x+6}}=\int \frac{dx}{\sqrt{\left(x+\frac{5}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}
\displaystyle =\log\left|x+\frac{5}{2}+\sqrt{\left(x+\frac{5}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}\right|+C
\displaystyle =\log\left|x+\frac{5}{2}+\sqrt{x^{2}+5x+6}\right|+C
\displaystyle =\log\left|\frac{2x+5}{2}+\sqrt{x^{2}+5x+6}\right|+C
\displaystyle I=\sqrt{x^{2}+5x+6}+\frac{1}{2}\left[\log\left|\frac{2x+5}{2}+\sqrt{x^{2}+5x+6}\right|\right]+C

\displaystyle \textbf{18. } \text{Find the particular solution of the differential equation}
\displaystyle \frac{dy}{dx}=1+x+y+xy,\ \text{given that } y=0 \text{ when } x=1.
\displaystyle \text{Answer:}
\displaystyle \text{Consider}
\displaystyle \frac{dy}{dx}=1+x+y+xy
\displaystyle =1+x+y(1+x)
\displaystyle =(1+x)(1+y)
\displaystyle \Rightarrow \frac{dy}{1+y}=(1+x)\,dx\Rightarrow \int \frac{dy}{1+y}=\int (1+x)\,dx
\displaystyle \Rightarrow \log(1+y)=x+\frac{x^{2}}{2}+C
\displaystyle \text{Putting } y=0 \text{ and } x=1,\ \text{we get}
\displaystyle \log 1=1+\frac{1}{2}+C\Rightarrow C=-\frac{3}{2}
\displaystyle \therefore\ \text{Particular solution is}
\displaystyle \log(1+y)=x+\frac{x^{2}}{2}-\frac{3}{2}

\displaystyle \textbf{19. } \text{Solve the differential equation }   (1+x^{2})\frac{dy}{dx}+y=e^{\tan^{-1}x}
\displaystyle \text{Answer:}
\displaystyle \text{Consider } (1+x^{2})\frac{dy}{dx}+y=e^{\tan^{-1}x}
\displaystyle \text{Dividing both the sides by } (1+x^{2}),\ \text{we get}
\displaystyle \frac{dy}{dx}+\frac{y}{1+x^{2}}=\frac{e^{\tan^{-1}x}}{1+x^{2}}\qquad \text{(Linear form)}
\displaystyle \text{I.F.}=e^{\int \frac{1}{1+x^{2}}\,dx}=e^{\tan^{-1}x}
\displaystyle y(\text{I.F.})=\int Q\cdot \text{I.F.}\,dx+C
\displaystyle y\cdot e^{\tan^{-1}x}=\int \frac{(e^{\tan^{-1}x})e^{\tan^{-1}x}}{1+x^{2}}\,dx+C
\displaystyle y\cdot e^{\tan^{-1}x}=\int \frac{e^{2\tan^{-1}x}}{1+x^{2}}\,dx+C
\displaystyle \text{Let } t=\tan^{-1}x,\ dt=\frac{1}{1+x^{2}}\,dx
\displaystyle y\cdot e^{\tan^{-1}x}=\int e^{2t}\,dt+C
\displaystyle \Rightarrow y\cdot e^{\tan^{-1}x}=\frac{e^{2\tan^{-1}x}}{2}+C

\displaystyle \textbf{20. } \text{Show that the four points A, B, C and D with position } \text{vectors }
\displaystyle 4\widehat{i}+5\widehat{j}+\widehat{k},\ -\widehat{j}-\widehat{k},\ 3\widehat{i}+9\widehat{j}+4\widehat{k} \text{ and }   4(-\widehat{i}+\widehat{j}+\widehat{k}) \text{ respectively are coplanar.}
\displaystyle \text{OR}
\displaystyle \text{The scalar product of the vector } \overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k} \text{ with a unit vector along the sum}
\displaystyle \text{of vectors } \overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k} \text{ and } \overrightarrow{c}=\lambda\widehat{i}+2\widehat{j}+3\widehat{k} \text{ is equal to one. Find the value of }
\displaystyle \lambda \text{ and }   \text{hence find the unit vector along } \overrightarrow{b}+\overrightarrow{c}.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=4\widehat{i}+5\widehat{j}+\widehat{k}
\displaystyle B=-\widehat{j}-\widehat{k}
\displaystyle C=3\widehat{i}+9\widehat{j}+4\widehat{k}
\displaystyle D=4(-\widehat{i}+\widehat{j}+\widehat{k})
\displaystyle \text{Now, } \overrightarrow{AB}=-4\widehat{i}-6\widehat{j}-2\widehat{k}
\displaystyle \overrightarrow{AC}=-\widehat{i}+4\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{AD}=-8\widehat{i}-\widehat{j}+3\widehat{k}
\displaystyle \text{These points are coplanar if the vectors } \overrightarrow{AB},\ \overrightarrow{AC} \text{ and } \overrightarrow{AD} \text{ are coplanar.}
\displaystyle [\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]=\begin{vmatrix}-4&-6&-2\\-1&4&3\\-8&-1&3\end{vmatrix}
\displaystyle =-4(12+3)+6(-3+24)-2(1+32)
\displaystyle =-60+126-66=0
\displaystyle \therefore\ \text{Points } A,\ B,\ C,\ D \text{ are coplanar.}
\displaystyle \text{OR}
\displaystyle \text{Let } \overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\ \overrightarrow{b}=2\widehat{i}+4\widehat{j}-5\widehat{k}
\displaystyle \overrightarrow{c}=\lambda\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle \overrightarrow{b}+\overrightarrow{c}=(2+\lambda)\widehat{i}+6\widehat{j}-2\widehat{k}
\displaystyle \text{Let } \overrightarrow{r} \text{ be the unit vector along } \overrightarrow{b}+\overrightarrow{c}
\displaystyle \therefore\ \overrightarrow{r}=\frac{\overrightarrow{b}+\overrightarrow{c}}{|\overrightarrow{b}+\overrightarrow{c}|}=\frac{(2+\lambda)\widehat{i}+6\widehat{j}-2\widehat{k}}{\sqrt{(2+\lambda)^{2}+40}}
\displaystyle \text{Now, } (\widehat{i}+\widehat{j}+\widehat{k})\cdot \overrightarrow{r}=1
\displaystyle \Rightarrow (\widehat{i}+\widehat{j}+\widehat{k})\cdot \left\{(2+\lambda)\widehat{i}+6\widehat{j}-2\widehat{k}\right\}=\sqrt{(2+\lambda)^{2}+40}
\displaystyle \Rightarrow (\lambda+6)^{2}=(2+\lambda)^{2}+40
\displaystyle \Rightarrow 8\lambda=8\Rightarrow \lambda=1

\displaystyle \textbf{21. } \text{A line passes through } (2,-1,3) \text{ and is perpendicular to the lines}
\displaystyle \overrightarrow{r}=(\widehat{i}+\widehat{j}-\widehat{k})+\lambda(2\widehat{i}-2\widehat{j}+\widehat{k}) \text{ and }   \overrightarrow{r}=(2\widehat{i}-\widehat{j}-3\widehat{k})+\mu(\widehat{i}+2\widehat{j}+2\widehat{k}).
\displaystyle \text{Obtain its equation in vector and cartesian form.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } \overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b} \text{ be the vector form of the required equation,}
\displaystyle \text{where } \overrightarrow{a} \text{ is the position vector and vector equation parallel to vector } \overrightarrow{b}. \text{ Since, } a\widehat{i}+b\widehat{j}+c\widehat{k}
\displaystyle \text{is perpendicular to the given lines}
\displaystyle \therefore\ 2a-2b+c=0\qquad \ldots (1)
\displaystyle a+2b+2c=0\qquad \ldots (2)
\displaystyle \text{On solving (1) and (2), we get}
\displaystyle a=2b,\ c=-2b
\displaystyle \therefore\ a:b:c=(2:1:-2)
\displaystyle \text{Also, required vector equation passes through } (2,-1,3).
\displaystyle \text{Hence, required vector equation is } \overrightarrow{r}=(2\widehat{i}-\widehat{j}+3\widehat{k})+\lambda(2\widehat{i}+\widehat{j}-2\widehat{k})
\displaystyle \text{and Cartesian form of equation is}
\displaystyle \frac{x-2}{2}=\frac{y+1}{1}=\frac{z-3}{-2}

\displaystyle \textbf{22. } \text{An experiment succeeds thrice as often as it fails. Find the probability that }
\displaystyle \text{in the next  five trials, there will be at least 3 }   \text{successes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } p \text{ denote the probability of succeeding in the experiment.}
\displaystyle \text{Given: } p=3(1-p)\Rightarrow p=\frac{3}{4}
\displaystyle \text{Let } X=\text{number of successes in 5 trials.}
\displaystyle \Rightarrow X \text{ follows binomial distribution with}
\displaystyle n=5,\ p=\frac{3}{4},\ q=\frac{1}{4}
\displaystyle P(X=r)={}^{5}C_{r}\left(\frac{3}{4}\right)^{r}\left(\frac{1}{4}\right)^{5-r},\ r=0,1,2,\ldots,5
\displaystyle \text{Required probability}=P(X\geq 3)
\displaystyle =P(X=3)+P(X=4)+P(X=5)
\displaystyle ={}^{5}C_{3}\left(\frac{3}{4}\right)^{3}\left(\frac{1}{4}\right)^{2}+{}^{5}C_{4}\left(\frac{3}{4}\right)^{4}\left(\frac{1}{4}\right)+{}^{5}C_{5}\left(\frac{3}{4}\right)^{5}
\displaystyle =10\left(\frac{3}{4}\right)^{3}\left(\frac{1}{4}\right)^{2}+5\left(\frac{3}{4}\right)^{4}\left(\frac{1}{4}\right)+\left(\frac{3}{4}\right)^{5}
\displaystyle =\left(\frac{3}{4}\right)^{3}\left[\frac{10}{16}+5\times \frac{3}{4}\times \frac{1}{4}+\frac{3\times 3}{4\times 4}\right]
\displaystyle =\frac{27}{64}\left[\frac{10}{16}+\frac{15}{16}+\frac{9}{16}\right]
\displaystyle =\frac{27}{64}\times \frac{34}{16}=\frac{27}{32}\times \frac{17}{16}=\frac{459}{512}

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question number 22 to 29 carry 6 Mark each}

\displaystyle \textbf{23. } \text{Two schools A and B want to award their selected students on the values of }
\displaystyle \text{sincerity, truthfulness and helpfulness. } \text{The school A wants to award Rs } x
\displaystyle \text{ each, Rs } y \text{ each and Rs } z \text{ each for the three respective values  to 3, 2 and 1 students }
\displaystyle \text{respectively with a total award money of Rs } 1600. \text{ School B } \text{wants to spend Rs } 2300
\displaystyle \text{ to award its 4, 1 and 3 students on the respective values (by giving the same }
\displaystyle \text{ award money to the three values as before). If the total amount of award for one}
\displaystyle \text{prize on each value is Rs } 900,\ \text{using matrices, find the award money for each value.}
\displaystyle \text{Apart from these three values, suggest one more value which should be considered}
\displaystyle \text{for award.}
\displaystyle \text{Answer:}
\displaystyle \text{ The given information can be written as}
\displaystyle 3x+2y+z=1600\qquad \ldots (1)
\displaystyle 4x+y+3z=2300\qquad \ldots (2)
\displaystyle x+y+z=900\qquad \ldots (3)
\displaystyle \text{where,}
\displaystyle A=\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix},\ B=\begin{bmatrix}1600\\2300\\900\end{bmatrix}\ \text{and}\ X=\begin{bmatrix}x\\y\\z\end{bmatrix}
\displaystyle \Rightarrow AX=B
\displaystyle \Rightarrow X=A^{-1}B
\displaystyle \text{Now, } |A|=[3(1-3)-2(4-3)+1(4-1)]
\displaystyle =[3(-2)-2\times 1+3]
\displaystyle |A|=-5\neq 0
\displaystyle \text{So, } A \text{ is invertible.}
\displaystyle \text{Let } C_{ij} \text{ be the cofactor of } a_{ij} \text{ in } A=[a_{ij}]
\displaystyle \text{We have}
\displaystyle C_{11}=(-1)^{2}\begin{vmatrix}1&3\\1&1\end{vmatrix}=-2
\displaystyle C_{12}=(-1)^{3}\begin{vmatrix}4&3\\1&1\end{vmatrix}=1
\displaystyle C_{13}=(-1)^{4}\begin{vmatrix}4&1\\1&1\end{vmatrix}=3
\displaystyle C_{21}=(-1)^{3}\begin{vmatrix}2&1\\1&1\end{vmatrix}=-1
\displaystyle C_{22}=(-1)^{4}\begin{vmatrix}3&1\\1&1\end{vmatrix}=2
\displaystyle C_{23}=(-1)^{5}\begin{vmatrix}3&2\\1&1\end{vmatrix}=-1
\displaystyle C_{31}=(-1)^{4}\begin{vmatrix}2&1\\1&3\end{vmatrix}=5
\displaystyle C_{32}=(-1)^{5}\begin{vmatrix}3&1\\4&3\end{vmatrix}=-5
\displaystyle C_{33}=(-1)^{6}\begin{vmatrix}3&2\\4&1\end{vmatrix}=-5
\displaystyle \therefore\  {adj \ } A=\begin{bmatrix}-2&1&3\\-1&2&-1\\5&-5&-5\end{bmatrix}^{T}
\displaystyle =\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 \text{Thus, the solution of the system of equations is given by}
\displaystyle X=A^{-1}B
\displaystyle X=\frac{1}{-5}\begin{bmatrix}-2&-1&5\\1&2&-5\\3&-1&-5\end{bmatrix}\begin{bmatrix}1600\\2300\\900\end{bmatrix}
\displaystyle =\frac{1}{-5}\begin{bmatrix}-3200-2300+4500\\1600+4600-4500\\4800-2300-4500\end{bmatrix}
\displaystyle =\frac{1}{-5}\begin{bmatrix}-1000\\1700\\-2000\end{bmatrix}=\begin{bmatrix}200\\-340\\400\end{bmatrix}
\displaystyle \text{Hence, the money awarded for sincerity, truthfulness and helpfulness are} \\ \text{Rs } 200,\ \text{Rs } 300 \text{ and Rs } 400 \text{ respectively.}
\displaystyle \therefore\ x,\ y \text{ and } z \text{ will have unique solution: } x=\text{Rs } 200,\ y=\text{Rs } 300,\ z=\text{Rs } 400.

\displaystyle \textbf{24. } \text{Show that the altitude of the right circular cone of maximum volume that can be }
\displaystyle \text{inscribed in a sphere of radius } r \text{ is } \frac{4r}{3} \text{. Also show that the maximum volume of  the}
\displaystyle \text{cone is } \frac{8}{27} \text{ of }   \text{the volume of the sphere.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } R \text{ and } h \text{ be the radius and height of the cone and } r \text{ be the radius of sphere.}\displaystyle \text{To show: } h=\frac{4r}{3} \text{ and Maximum volume of cone }=\frac{8}{27}(\text{Volume of sphere})
\displaystyle \text{Proof:}
\displaystyle \text{In } \triangle ABC
\displaystyle AC=h-r
\displaystyle r^{2}=(h-r)^{2}+R^{2}\ (\text{By Pythagoras theorem})
\displaystyle R^{2}=r^{2}-(h-r)^{2}
\displaystyle \text{Volume of cone}=\frac{1}{3}\pi R^{2}h
\displaystyle V=\frac{1}{3}\pi [r^{2}-(h-r)^{2}]h\qquad (1)
\displaystyle =\frac{1}{3}\pi [2h^{2}r-h^{3}]
\displaystyle \text{For maxima or minima, } \frac{dV}{dh}=0
\displaystyle \Rightarrow \frac{1}{3}\pi [4hr-3h^{2}]=0
\displaystyle \Rightarrow h=\frac{4r}{3}
\displaystyle \frac{d^{2}V}{dh^{2}}=\frac{1}{3}\pi [4r-6h]
\displaystyle \text{Putting } h=\frac{4r}{3}
\displaystyle \frac{d^{2}V}{dh^{2}}=\frac{1}{3}\pi \left[4r-\frac{24r}{3}\right]=-\frac{4}{3}\pi r<0
\displaystyle \therefore\ \text{Volume is maximum at } h=\frac{4r}{3}
\displaystyle V=\frac{1}{3}\pi [r^{2}-(h-r)^{2}]h\ \text{[from (1)]}
\displaystyle =\frac{1}{3}\pi \left[r^{2}-\left(\frac{4r}{3}-r\right)^{2}\right]\frac{4r}{3}
\displaystyle =\frac{1}{3}\pi \left[\frac{8r^{2}}{9}\right]\frac{4r}{3}
\displaystyle =\frac{8}{27}\left(\frac{4}{3}\pi r^{3}\right)=\frac{8}{27}(\text{Volume of sphere})
\displaystyle \text{Hence, proved.}

\displaystyle \textbf{25. } \text{Evaluate: } \int \frac{1}{\cos^{4}x+\sin^{4}x}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{1}{\cos^{4}x+\sin^{4}x}\,dx
\displaystyle =\int \frac{\sec^{4}x\,dx}{1+\tan^{4}x}
\displaystyle =\int \frac{\sec^{2}x\sec^{2}x\,dx}{1+\tan^{4}x}
\displaystyle \text{Putting } \tan x=t\Rightarrow \sec^{2}x\,dx=dt
\displaystyle =\int \frac{(1+\tan^{2}x)\,dt}{1+t^{4}}=\int \frac{1+t^{2}}{1+t^{4}}\,dt
\displaystyle =\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}\,dt
\displaystyle =\int \frac{1+\frac{1}{t^{2}}}{\left(t-\frac{1}{t}\right)^{2}+2}\,dt
\displaystyle \text{Putting } t-\frac{1}{t}=z\Rightarrow 1+\frac{1}{t^{2}}\,dt=dz
\displaystyle =\int \frac{dz}{z^{2}+2}=\int \frac{dz}{z^{2}+(\sqrt{2})^{2}}
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{z}{\sqrt{2}}\right)+C
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x-\frac{1}{\tan x}}{\sqrt{2}}\right)+C
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)+C

\displaystyle \textbf{26. } \text{Using integration, find the area of the region bounded by the triangle }
\displaystyle \text{whose vertices are } (-1,2),\ (1,5) \text{ and } (3,4).
\displaystyle \text{Answer:} \displaystyle \text{The points } A(-1,2),\ B(1,5) \text{ and } C(3,4) \text{ are plotted and joined.}
\displaystyle \text{Equation of AB:}
\displaystyle \frac{y-2}{x+1}=\frac{3}{2}\Rightarrow y=\frac{3x+7}{2}
\displaystyle \text{Equation of BC:}
\displaystyle \frac{y-5}{x-1}=-\frac{1}{2}\Rightarrow 2y-10=-x+1
\displaystyle \Rightarrow y=\frac{-x+11}{2}
\displaystyle \text{Equation of CA:}
\displaystyle \frac{y-4}{x-3}=\frac{1}{2}\Rightarrow 2y-8=x-3
\displaystyle \Rightarrow y=\frac{x+5}{2}
\displaystyle \text{Area of } \triangle ABC=(\text{Area of trapezium ABML}+\text{Area of trapezium BCMN})-(\text{Area of trapezium ACNL})
\displaystyle =\int_{-1}^{1}\frac{3x+7}{2}\,dx+\int_{1}^{3}\frac{-x+11}{2}\,dx-\int_{-1}^{3}\frac{x+5}{2}\,dx
\displaystyle \text{Required Area}=\left[\frac{3x^{2}}{4}+\frac{7x}{2}\right]_{-1}^{1}+\left[\frac{-x^{2}}{4}+\frac{11x}{2}\right]_{1}^{3}-\left[\frac{x^{2}}{4}+\frac{5x}{2}\right]_{-1}^{3}
\displaystyle =\left(\frac{3}{4}+\frac{7}{2}\right)-\left(\frac{3}{4}-\frac{7}{2}\right)+\left[\left(-\frac{9}{4}+\frac{33}{2}\right)-\left(-\frac{1}{4}+\frac{11}{2}\right)\right]-\left[\left(\frac{9}{4}+\frac{15}{2}\right)-\left(\frac{1}{4}-\frac{5}{2}\right)\right]
\displaystyle =\left(\frac{17}{4}+\frac{11}{4}\right)+\frac{36}{4}-\frac{48}{4}
\displaystyle =\frac{16}{4}=4\ \text{sq. unit}

\displaystyle \textbf{27. } \text{Find the equation of the plane through the line of intersection of the planes}
\displaystyle x+y+z=1 \text{ and } 2x+3y+4z=5 \text{which is perpendicular to the plane }
\displaystyle x-y+z=0. \text{ Also find the distance of the plane obtained above, from the origin.}
\displaystyle \text{OR}
\displaystyle \text{Find the distance of the point } (2,12,5) \text{ from the point of intersection of the line }
\displaystyle \overrightarrow{r}=2\widehat{i}-4\widehat{j}+2\widehat{k}+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k})   \text{and the plane } \overrightarrow{r}\cdot(\widehat{i}-2\widehat{j}+\widehat{k})=0
\displaystyle \text{Answer:}
\displaystyle \text{Given planes are}
\displaystyle x+y+z-1=0\qquad \ldots (i)
\displaystyle 2x+3y+4z-5=0\qquad \ldots (ii)
\displaystyle x-y+z=0\qquad \ldots (iii)
\displaystyle \text{Any plane through the intersection of (i) and (ii) is}
\displaystyle (1+2\lambda)x+(1+3\lambda)y+(1+4\lambda)z-1-5\lambda=0\qquad \ldots (iv)
\displaystyle \text{Direction ratios of normal of (iii) are } 1,-1,1
\displaystyle \text{Also direction ratios of normal of (iv) are } 1+2\lambda,\ 1+3\lambda,\ 1+4\lambda
\displaystyle \text{Two planes are perpendicular if their normals are perpendicular.}
\displaystyle \Rightarrow (1+2\lambda)- (1+3\lambda)+ (1+4\lambda)=0
\displaystyle \Rightarrow 1+2\lambda-1-3\lambda+1+4\lambda=0
\displaystyle \Rightarrow 1+3\lambda=0\Rightarrow \lambda=-\frac{1}{3}
\displaystyle \text{Now equation of the required plane is}
\displaystyle x-z+2=0
\displaystyle \text{Distance } d=\left|\frac{ax_{1}+by_{1}+cz_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|
\displaystyle =\left|\frac{1(0)+0(0)+(-1)(0)+2}{\sqrt{1^{2}+0^{2}+(-1)^{2}}}\right|
\displaystyle =\frac{2}{\sqrt{2}}=\sqrt{2}
\displaystyle \text{OR}
\displaystyle \text{The line and plane are}
\displaystyle \overrightarrow{r}=(2\widehat{i}-4\widehat{j}+2\widehat{k})+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k})\qquad (1)
\displaystyle \overrightarrow{r}\cdot(\widehat{i}-2\widehat{j}+\widehat{k})=0\qquad (2)
\displaystyle \text{On solving (1) and (2)}
\displaystyle [(2\widehat{i}-4\widehat{j}+2\widehat{k})+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k})]\cdot(\widehat{i}-2\widehat{j}+\widehat{k})=0
\displaystyle \Rightarrow (2+8+2)+\lambda(3-8+2)=0
\displaystyle \Rightarrow 12+\lambda(-3)=0\Rightarrow \lambda=4
\displaystyle \therefore\ \text{Point of intersection is } 14\widehat{i}+12\widehat{j}+10\widehat{k}
\displaystyle \text{The other point is } (2,12,5)=2\widehat{i}+12\widehat{j}+5\widehat{k}
\displaystyle \text{Distance between these points}
\displaystyle =\sqrt{(14-2)^{2}+(12-12)^{2}+(10-5)^{2}}
\displaystyle =\sqrt{144+25}=\sqrt{169}=13

\displaystyle \textbf{28. } \text{A manufacturing company makes two types of teaching aids A and B of }
\displaystyle \text{Mathematics for class XII. Each type of A requires 9 labour hours for fabricating}
\displaystyle \text{and 1 labour hour for finishing. Each type of B requires 12 labour hours for}
\displaystyle \text{fabricating and 3 labour hours for finishing. For fabricating and finishing, the }
\displaystyle \text{maximum labour hours available per week are 180 and 30 respectively. The }
\displaystyle \text{company makes a profit of } \text{Rs } 80 \text{ on each piece of type A and Rs } 120 \text{ on each piece}
\displaystyle \text{of type B. How many pieces of type A and type B should be manufactured per week}
\displaystyle \text{to get a maximum profit? Make it as an LPP and solve graphically. What is the}
\displaystyle \text{maximum profit per week?}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose } x \text{ is the number of teaching aids A and } y \text{ is the number of teaching aids B.}
\displaystyle \text{Then,}
\displaystyle \text{Total profit (in Rs)}=80x+120y
\displaystyle Z=80x+120y
\displaystyle \text{Let, we now have the following mathematical model for the given problem}
\displaystyle \text{Maximize } Z=80x+120y\qquad \ldots (1)
\displaystyle \text{Subject to the constraints}
\displaystyle 9x+12y\leq 180\qquad \text{(fabricating constraint)}
\displaystyle \text{i.e. } 3x+4y\leq 60\qquad \ldots (2)
\displaystyle x+3y\leq 30\qquad \text{(finishing constraint)}\qquad \ldots (3)
\displaystyle x\geq 0,\ y\geq 0\qquad \text{(non-negative constraint)}\qquad \ldots (4)
\displaystyle \text{The feasible region (shaded) OABC determined by the linear inequalities (2) to (4) is shown in the graph.}
\displaystyle \text{The feasible region is bounded.}\displaystyle \text{The objective function } Z \text{ at each corner point is shown below:}
\displaystyle \begin{array}{|c|c|}\hline \text{Corner Point}&Z=80x+120y\\\hline O(0,0)&0\\\hline A(20,0)&1600\\\hline B(12,6)&1680\ \text{(Maximum)}\\\hline C(0,10)&1200\\\hline \end{array}
\displaystyle \text{The maximum value of } Z \text{ is } 1680 \text{ at point } B(12,6). \text{ Hence the company should}
\displaystyle \text{produce 12 pieces of teaching aid A and } 6 \text{ pieces of teaching aid B to realise}
\displaystyle \text{maximum profit and maximum profit will be Rs } 1680.

\displaystyle \textbf{29. } \text{There are three coins. One is a two-headed coin (having head on both faces),}
\displaystyle \text{ another is a biased coin that comes up} \text{heads } 75\% \text{ of the times and third is also a }
\displaystyle  \text{biased coin that } \text{comes up tails } 40\% \text{ of the times. One of the three coins is chosen}
\displaystyle \text{chosen at random and tossed, and it shows heads. What is the probability that it}
\displaystyle \text{was the two headed coin?}
\displaystyle \text{OR}
\displaystyle \text{Two numbers are selected at random (without replacement) from the first six positive }
\displaystyle \text{integers.  Let X denote the larger of the two numbers obtained. Find the probability}
\displaystyle \text{distribution of the random variable X, and hence find the mean of the distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_{1},E_{2},E_{3}, \text{ and } A \text{ denote the following events:}
\displaystyle E_{1}=\text{a two headed coin, } E_{2}=\text{a biased coin that comes up head } 75\% \text{ of the times}
\displaystyle E_{3}=\text{a biased coin that comes up tail } 40\% \text{ of the times}
\displaystyle A=\text{a head is shown}
\displaystyle \text{Now,}
\displaystyle P(A/E_{1})=1,\ P(A/E_{2})=\frac{75}{100}=\frac{3}{4},
\displaystyle P(A/E_{3})=\frac{60}{100}=\frac{3}{5}
\displaystyle \text{By Baye's theorem,}
\displaystyle P(E_{1}/A)=\frac{P(E_{1})P(A/E_{1})}{P(E_{1})P(A/E_{1})+P(E_{2})P(A/E_{2})+P(E_{3})P(A/E_{3})}
\displaystyle P(E_{1}/A)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{3}{5}}
\displaystyle =\frac{\frac{1}{3}}{\frac{1}{3}\left(1+\frac{3}{4}+\frac{3}{5}\right)}=\frac{1}{\frac{20+15+12}{20}}
\displaystyle P(E_{1}/A)=\frac{20}{47}
\displaystyle \text{OR}
\displaystyle \text{There are six numbers } 1,2,3,4,5,6 \text{ and one of them is selected in } 6 \text{ ways.}
\displaystyle \text{When one of the numbers has been selected, } 5 \text{ numbers are left. One number out of } 5 \text{ may be selected in } 5 \text{ ways.}
\displaystyle \text{No. of ways of selecting two numbers without replacement out of } 6 \text{ positive integers }=6\times 5=30
\displaystyle \text{Let } X \text{ denotes larger of two number}
\displaystyle \text{When } X=2 \text{ then the two numbers are } (1,2),\ (2,1)
\displaystyle \therefore\ P(X=2)=\frac{2}{30}
\displaystyle \text{When } X=3 \text{ then the two numbers are } (1,3),\ (2,3),\ (3,1),\ (3,2)
\displaystyle \therefore\ P(X=3)=\frac{4}{30}
\displaystyle \text{When } X=4 \text{ then the two numbers are } (2,4),\ (3,4),\ (4,1),\ (4,2),\ (4,3),\ (1,4)
\displaystyle \therefore\ P(X=4)=\frac{6}{30}
\displaystyle \text{When } X=5 \text{ then the two numbers are } (1,5),\ (2,5),\ (3,5),
\displaystyle (4,5),\ (5,1),\ (5,2),\ (5,3),\ (5,4)
\displaystyle \therefore\ P(X=5)=\frac{8}{30}
\displaystyle \text{When } X=6 \text{ then the two numbers are } (1,6),\ (2,6),\ (3,6),\ (4,6),\ (5,6),
\displaystyle (6,1),\ (6,2),\ (6,3),\ (6,4),\ (6,5)
\displaystyle P(X=6)=\frac{10}{30}
\displaystyle \text{The probability distribution is}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline X&2&3&4&5&6\\\hline P(X)&\frac{2}{30}&\frac{4}{30}&\frac{6}{30}&\frac{8}{30}&\frac{10}{30}\\\hline \end{array}
\displaystyle \text{The mean of the distribution is}
\displaystyle E(X)=\sum XP(X)
\displaystyle =2\times \frac{2}{30}+3\times \frac{4}{30}+4\times \frac{6}{30}+5\times \frac{8}{30}+6\times \frac{10}{30}
\displaystyle =\frac{1}{30}[4+12+24+40+60]
\displaystyle E(X)=\frac{14}{3}

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.