Mathematics

\displaystyle \text{Time Allowed : 3 Hours} \hfill \text{Maximum Marks : 100}

\displaystyle \textbf{General Instructions:}

\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) Please check that this Question Paper contains 26 questions.}
\displaystyle \text{(iii) Marks for each question are indicated against it.}
\displaystyle \text{(iv) Question 1 to 6 in Section-A are Very Short Answer Type Questions carrying }
\displaystyle \text{one mark each.}
\displaystyle \text{(v) Question 7 to 19 in Section-B are Long Answer Type Questions carrying }
\displaystyle \text{4 marks each.}
\displaystyle \text{(vi) Question 20 to 26 in Section-C are Long Answer II Type Questions carrying }
\displaystyle \text{6 marks each.}
\displaystyle \text{(vii) Please write down the serial number of the question before } \text{attempting it.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Questions numbers 1 to 6 carry 1 Mark each}

\displaystyle \textbf{1. }  \text{Write the value of } \overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{a}).
\displaystyle \text{Answer:}
\displaystyle \text{Let } \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 \therefore \overrightarrow{a}\cdot(\overrightarrow{b}\times \overrightarrow{a})
\displaystyle =(a_{1}\widehat{i}+a_{2}\widehat{j}+a_{3}\widehat{k})\cdot[(b_{1}\widehat{i}+b_{2}\widehat{j}+b_{3}\widehat{k})\times(a_{1}\widehat{i}+a_{2}\widehat{j}+a_{3}\widehat{k})]
\displaystyle =\left|\begin{matrix}a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ a_{1} & a_{2} & a_{3}\end{matrix}\right|
\displaystyle =(b_{2}a_{3}-b_{3}a_{2})a_{1}-(b_{1}a_{3}-b_{3}a_{1})a_{2}+(b_{1}a_{2}-b_{2}a_{1})a_{3}
\displaystyle =a_{1}a_{3}b_{2}-a_{1}a_{2}b_{3}-a_{2}a_{3}b_{1}+a_{1}a_{2}b_{3}+a_{2}a_{3}b_{1}-a_{1}a_{3}b_{2}=0
\displaystyle \text{Alternate Method: } \overrightarrow{b}\times \overrightarrow{a}\ \text{is perpendicular to both } \overrightarrow{a}\ \text{and}\ \overrightarrow{b}
\displaystyle \therefore \overrightarrow{b}\times \overrightarrow{a}\perp \overrightarrow{a}
\displaystyle \Rightarrow \overrightarrow{a}\cdot(\overrightarrow{b}\times \overrightarrow{a})=|\overrightarrow{a}||\overrightarrow{b}\times \overrightarrow{a}|\cos 90^{\circ}=0

\displaystyle \textbf{2. }  \text{If } \overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\ \overrightarrow{b}=2\widehat{i}+\widehat{j}+\widehat{k}\ \text{and}\ \overrightarrow{c}=5\widehat{i}-4\widehat{j}+3\widehat{k}, \text{then find the value of }
\displaystyle (\overrightarrow{a}+\overrightarrow{b})\cdot \overrightarrow{c}.
\displaystyle \text{Answer:}
\displaystyle  (\overrightarrow{a}+\overrightarrow{b})\cdot \overrightarrow{c}
\displaystyle =(\widehat{i}+2\widehat{j}-\widehat{k}+2\widehat{i}+\widehat{j}+\widehat{k})\cdot(5\widehat{i}-4\widehat{j}+3\widehat{k})
\displaystyle =(3\widehat{i}+3\widehat{j})\cdot(5\widehat{i}-4\widehat{j}+3\widehat{k})
\displaystyle =(3\times 5)+(3\times(-4))+(0\times 3)=15-12=3

\displaystyle \textbf{3. }  \text{Write the direction ratios of the following line:}   x=-3,\ \frac{y-4}{3}=\frac{2-z}{1}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the given line can be rewritten as:}
\displaystyle \frac{x+3}{0}=\frac{y-4}{3}=\frac{z-2}{-1}
\displaystyle \text{Thus, the given line has direction ratios proportional to } 0,3,-1.

\displaystyle \textbf{4. }  \text{If } A=\begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}\text{, then write } A^{-1}.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}
\displaystyle |A|=2(-2)-5(3)=-4-15=-19\ne 0
\displaystyle \text{So, A is a non-singular matrix. Therefore, it is invertible.}
\displaystyle \text{Now, } \mathrm{adj}A=\begin{bmatrix}-2 & -3 \\ -5 & 2\end{bmatrix}
\displaystyle A^{-1}=\frac{1}{|A|}\mathrm{adj}A=\frac{1}{-19}\begin{bmatrix}-2 & -3 \\ -5 & 2\end{bmatrix}
\displaystyle =\begin{bmatrix}\frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & -\frac{2}{19}\end{bmatrix}

\displaystyle \textbf{5. }  \text{Find the differential equation representing the curve } y=cx+c^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have } y=cx+c^{2}
\displaystyle \frac{dy}{dx}=c
\displaystyle \text{Substituting } c=\frac{dy}{dx}\ \text{in equation, we get}
\displaystyle y=x\frac{dy}{dx}+\left(\frac{dy}{dx}\right)^{2}
\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)^{2}+x\frac{dy}{dx}-y=0

\displaystyle \textbf{6. }  \text{Write the integrating factor of the following differential equation:}
\displaystyle (1+y^{2})dx-(\tan^{-1}y-x)dy=0.
\displaystyle \text{Answer:}
\displaystyle  (1+y^{2})dx-(\tan^{-1}y-x)dy=0
\displaystyle \Rightarrow (1+y^{2})\frac{dx}{dy}=\tan^{-1}y-x
\displaystyle \Rightarrow (1+y^{2})\frac{dx}{dy}+x=\tan^{-1}y
\displaystyle \Rightarrow \frac{dx}{dy}+\frac{1}{1+y^{2}}x=\frac{\tan^{-1}y}{1+y^{2}}
\displaystyle \text{Integrating factor (IF)}=e^{\int \frac{1}{1+y^{2}}dy}=e^{\tan^{-1}y}

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Questions numbers 7 to 19 carry 4 Mark each}

\displaystyle \textbf{7. }  \text{Using the properties of determinants, prove the following:} \\
\displaystyle \left|\begin{matrix}1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(1-x) & x(x-1)(x-2) & x(x+1)(x-1)\end{matrix}\right|=6x^{2}(1-x^{2}).
\displaystyle \text{Answer:}
\displaystyle  \left|\begin{matrix}1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(1-x) & x(x-1)(x-2) & x(x+1)(x-1)\end{matrix}\right|
\displaystyle =x(1+x)\left|\begin{matrix}1 & 1 & 1 \\ 2x & x-1 & x \\ 3x(1-x) & (x-1)(x-2) & x(x-1)\end{matrix}\right|
\displaystyle =x(1-x)(1+x)\left|\begin{matrix}1 & 1 & 1 \\ 2x & x-1 & x \\ 3x & -(x-2) & -x\end{matrix}\right|
\displaystyle =x(1-x^{2})\left|\begin{matrix}0 & 0 & 1 \\ x+1 & -1 & x \\ 4x-2 & 2 & -x\end{matrix}\right|
\displaystyle =x(1-x^{2})[2(x+1)+4x-2]
\displaystyle =6x^{2}(1-x^{2})
\displaystyle \text{Hence, proved.}

\displaystyle \textbf{8. } \text{If } x=\alpha\sin 2t(1+\cot 2t)\ \text{and}\ y=\beta\cos 2t(1-\cos 2t),  \text{show that } \\ \frac{dy}{dx}=\frac{\beta}{\alpha}\tan t.
\displaystyle \text{Answer:}
\displaystyle  x=\alpha\sin 2t(1+\cos 2t)
\displaystyle \Rightarrow x=\alpha\sin 2t+\frac{\alpha}{2}\times 2\sin 2t\cos 2t
\displaystyle \Rightarrow x=\alpha\sin 2t+\frac{\alpha}{2}\sin 4t
\displaystyle \text{Differentiating both sides w.r.t. } t,\ \text{we get}
\displaystyle \frac{dx}{dt}=\alpha\cos 2t\times 2+\frac{\alpha}{2}\cos 4t\times 4
\displaystyle \Rightarrow \frac{dx}{dt}=2\alpha(\cos 2t+\cos 4t)
\displaystyle \Rightarrow \frac{dx}{dt}=2\alpha(\cos 2t+2\cos^{2}2t-1)
\displaystyle \Rightarrow \frac{dx}{dt}=2\alpha(2\cos^{2}2t+\cos 2t-1)
\displaystyle =2\alpha(2\cos^{2}2t+2\cos 2t-\cos 2t-1)
\displaystyle \Rightarrow \frac{dx}{dt}=2\alpha(\cos 2t+1)(2\cos 2t-1)
\displaystyle \text{Now, } y=\beta\cos 2t(1-\cos 2t)
\displaystyle \Rightarrow y=\beta\cos 2t-\beta\cos^{2}2t
\displaystyle \text{Differentiating both sides w.r.t. } t,\ \text{we get}
\displaystyle \frac{dy}{dt}=-\beta\sin 2t\times 2+\beta\times 2\cos 2t\times \sin 2t\times 2
\displaystyle \Rightarrow \frac{dy}{dt}=-2\beta\sin 2t+4\beta\cos 2t\sin 2t
\displaystyle \Rightarrow \frac{dy}{dt}=2\beta\sin 2t(2\cos 2t-1)
\displaystyle \text{We know } \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2\beta\sin 2t(2\cos 2t-1)}{2\alpha(\cos 2t+1)(2\cos 2t-1)}
\displaystyle =\frac{\beta\sin 2t}{\alpha(\cos 2t+1)}=\frac{\beta\times 2\sin t\cos t}{\alpha\times 2\cos^{2}t}=\frac{\beta}{\alpha}\tan t

\displaystyle \textbf{9. }  \text{Find } \frac{d}{dz}\cos^{-1}\left(\frac{z-z^{-1}}{z+z^{-1}}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let } y=\cos^{-1}\left(\frac{z-z^{-1}}{z+z^{-1}}\right)=\cos^{-1}\left(\frac{z^{2}-1}{z^{2}+1}\right)
\displaystyle \text{Let } z=\tan \theta
\displaystyle \therefore y=\cos^{-1}\left(\frac{\tan^{2}\theta-1}{\tan^{2}\theta+1}\right)=\cos^{-1}\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)
\displaystyle =\pi-\cos^{-1}(\cos 2\theta)\qquad [\because \cos^{-1}(-x)=\pi-\cos^{-1}x]
\displaystyle =\pi-2\theta=\pi-2\tan^{-1}z
\displaystyle \text{Differentiating both sides w.r.t. } z,\ \text{we have}
\displaystyle \frac{dy}{dz}=0-2\times \frac{1}{1+z^{2}}
\displaystyle \therefore \frac{d}{dz}\cos^{-1}\left(\frac{z-z^{-1}}{z+z^{-1}}\right)=\frac{-2}{1+z^{2}}

\displaystyle \textbf{10. }  \text{Find the derivative of the following function}
\displaystyle f(x)=\cos^{-1}\left[\sin\sqrt{\frac{1+x}{2}}\right]+x^{x}\ \text{w.r.t.}\ x,\ \text{at}\ x=1.
\displaystyle \text{Answer:}
\displaystyle  f(x)=\cos^{-1}\left[\sin\sqrt{\frac{1+x}{2}}\right]+x^{x}
\displaystyle \text{Let } f_{1}(x)=\cos^{-1}\left[\sin\sqrt{\frac{1+x}{2}}\right]\ \text{and}\ f_{2}(x)=x^{x}
\displaystyle \text{Now, } f_{1}(x)=\cos^{-1}\left[\sin\sqrt{\frac{1+x}{2}}\right]
\displaystyle =\cos^{-1}\left[\cos\left(\frac{\pi}{2}-\sqrt{\frac{1+x}{2}}\right)\right]\qquad [\because \cos\left(\frac{\pi}{2}-x\right)=\sin x]
\displaystyle =\frac{\pi}{2}-\sqrt{\frac{1+x}{2}}
\displaystyle \Rightarrow f_{1}'(x)=-\frac{1}{2\sqrt{2(1+x)}}
\displaystyle \text{and } f_{2}(x)=x^{x}
\displaystyle \text{Taking log on both sides, we get } \log f_{2}(x)=x\log x
\displaystyle \Rightarrow \frac{1}{f_{2}(x)}f_{2}'(x)=\log x+x\cdot \frac{1}{x}
\displaystyle \Rightarrow \frac{1}{f_{2}(x)}f_{2}'(x)=\log x+1
\displaystyle \Rightarrow f_{2}'(x)=f_{2}(x)(\log x+1)
\displaystyle \Rightarrow f_{2}'(x)=x^{x}(\log x+1)
\displaystyle \therefore f(x)=f_{1}(x)+f_{2}(x)
\displaystyle \therefore f'(x)=f_{1}'(x)+f_{2}'(x)
\displaystyle =-\frac{1}{2\sqrt{2(1+x)}}+x^{x}(\log x+1)
\displaystyle \text{At } x=1
\displaystyle f'(1)=-\frac{1}{2\sqrt{2(1+1)}}+1^{1}(\log 1+1)
\displaystyle =-\frac{1}{4}+1=\frac{3}{4}

\displaystyle \textbf{11. }  \text{Evaluate:}
\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}}dx
\displaystyle \text{OR}
\displaystyle \int_{0}^{\frac{3}{2}}\left|x\cdot \cos(\pi x)\right|dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}}dx\qquad ...(1)
\displaystyle \text{Then, } I=\int_{0}^{\frac{\pi}{2}}\frac{2^{\sin\left(\frac{\pi}{2}-x\right)}}{2^{\sin\left(\frac{\pi}{2}-x\right)}+2^{\cos\left(\frac{\pi}{2}-x\right)}}dx
\displaystyle \qquad [\because \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx]
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{2^{\cos x}}{2^{\cos x}+2^{\sin x}}dx\qquad ...(2)
\displaystyle \text{Adding equations } (1)\ \text{and } (2),\ \text{we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}}dx+\int_{0}^{\frac{\pi}{2}}\frac{2^{\cos x}}{2^{\cos x}+2^{\sin x}}dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}
\displaystyle I=\frac{\pi}{4}
\displaystyle \therefore \int_{0}^{\frac{\pi}{2}}\frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}}dx=\frac{\pi}{4}
\displaystyle \text{OR}
\displaystyle f(x)=|x\cdot \cos(\pi x)|=\begin{cases}x\cdot \cos(\pi x), & \text{for } 0<x\le \frac{1}{2} \\ -x\cdot \cos(\pi x), & \text{for } \frac{1}{2}<x\le \frac{3}{2}\end{cases}
\displaystyle \therefore \int_{0}^{\frac{3}{2}}|x\cdot \cos(\pi x)|dx=\int_{0}^{\frac{1}{2}}x\cdot \cos(\pi x)dx+\int_{\frac{1}{2}}^{\frac{3}{2}}-x\cdot \cos(\pi x)dx
\displaystyle \text{Integrating both integrals on right hand side, we get}
\displaystyle \int_{0}^{\frac{3}{2}}|x\cdot \cos(\pi x)|dx=\left[\frac{x\sin \pi x}{\pi}+\frac{\cos \pi x}{\pi^{2}}\right]_{0}^{\frac{1}{2}}-\left[\frac{x\sin \pi x}{\pi}+\frac{\cos \pi x}{\pi^{2}}\right]_{\frac{1}{2}}^{\frac{3}{2}}
\displaystyle =\left(\frac{1}{2\pi}+0\right)-\left(0+\frac{1}{\pi^{2}}\right)-\left[\left(\frac{3}{2}\times \frac{-1}{\pi}+0\right)-\left(\frac{1}{2}\times \frac{1}{\pi}+0\right)\right]
\displaystyle =\frac{1}{2\pi}-\frac{1}{\pi^{2}}+\frac{3}{2\pi}+\frac{1}{2\pi}
\displaystyle =\frac{5}{2\pi}-\frac{1}{\pi^{2}}

\displaystyle \textbf{12. }  \text{To raise money for an orphanage, students of three schools A, B and C organised }
\displaystyle \text{an exhibition in their locality, where they sold paper bags, scrap-books and pastel }
\displaystyle \text{sheets made by them using recycled paper, at the rate of Rs 20, Rs 15 and Rs 5 per unit }
\displaystyle \text{respectively. School A sold 25 paper bags, 12 scrap-books and 34 pastel sheets. School B sold }
\displaystyle \text{22 paper bags, 15 scrap-books and 28 pastel sheets while school C sold 26 paper bags, }
\displaystyle \text{18 scrap-books and 36 pastel sheets. Using matrix, find the total amount raised }
\displaystyle \text{by each school. By such exhibition, which values are generated in the students?}
\displaystyle \text{Answer:}
\displaystyle \text{School}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Article} & A & B & C \\ \hline \text{Paper Bags} & 25 & 22 & 26 \\ \hline \text{Scrapbooks} & 12 & 15 & 18 \\ \hline \text{Pastel Sheets} & 34 & 28 & 36 \\ \hline \end{array}
\displaystyle \text{The number of articles sold by each school can be written in the matrix form as follows:}
\displaystyle X=\begin{bmatrix}25 & 22 & 26 \\ 12 & 15 & 18 \\ 34 & 28 & 36\end{bmatrix}
\displaystyle \text{The rate of each article can be written in the matrix form as follows:}
\displaystyle Y=\begin{bmatrix}20 & 15 & 5\end{bmatrix}
\displaystyle \text{The amount collected by each school is given by}
\displaystyle YX=\begin{bmatrix}20 & 15 & 5\end{bmatrix}\begin{bmatrix}25 & 22 & 26 \\ 12 & 15 & 18 \\ 34 & 28 & 36\end{bmatrix}
\displaystyle YX=\begin{bmatrix}850 & 805 & 970\end{bmatrix}
\displaystyle \text{Thus, the amount collected by schools A, B and C are Rs } 850\text{, Rs } 805\text{ and Rs } 970\text{, respectively.}
\displaystyle \therefore \text{Total amount collected }=\text{Rs }(850+805+970)
\displaystyle =\text{Rs }2625
\displaystyle \text{The situation highlights the helpful nature of the students.}

\displaystyle \textbf{13. }  \text{Prove that: } 2\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right)=\cos^{-1}\left(\frac{a\cos x+b}{a+b\cos x}\right).
\displaystyle \text{OR} \\ \text{Solve the following for } x:
\displaystyle \tan^{-1}\left(\frac{x-2}{x-3}\right)+\tan^{-1}\left(\frac{x+2}{x+3}\right)=\frac{\pi}{4},\ |x|<1.
\displaystyle \text{Answer:}
\displaystyle 2\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right)
\displaystyle =\cos^{-1}\left(\frac{1-\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right)^{2}}{1+\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right)^{2}}\right)
\displaystyle \qquad [\because 2\tan^{-1}(x)=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)]
\displaystyle =\cos^{-1}\left(\frac{1-\frac{a-b}{a+b}\tan^{2}\frac{x}{2}}{1+\frac{a-b}{a+b}\tan^{2}\frac{x}{2}}\right)
\displaystyle =\cos^{-1}\left(\frac{a+b-(a-b)\tan^{2}\frac{x}{2}}{a+b+(a-b)\tan^{2}\frac{x}{2}}\right)
\displaystyle =\cos^{-1}\left(\frac{a+b-a\tan^{2}\frac{x}{2}+b\tan^{2}\frac{x}{2}}{a+b+a\tan^{2}\frac{x}{2}-b\tan^{2}\frac{x}{2}}\right)
\displaystyle =\cos^{-1}\left(\frac{a\left(1-\tan^{2}\frac{x}{2}\right)+b\left(1+\tan^{2}\frac{x}{2}\right)}{a\left(1+\tan^{2}\frac{x}{2}\right)+b\left(1-\tan^{2}\frac{x}{2}\right)}\right)
\displaystyle \text{Dividing the numerator and the denominator by } 1+\tan^{2}\frac{x}{2}
\displaystyle =\cos^{-1}\left(\frac{a\left(\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)+b\left(\frac{1+\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)}{a\left(\frac{1+\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)+b\left(\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)}\right)
\displaystyle =\cos^{-1}\left(\frac{a\left(\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)+b}{a+b\left(\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)}\right)
\displaystyle =\cos^{-1}\left(\frac{a\cos x+b}{a+b\cos x}\right)
\displaystyle \text{OR}
\displaystyle \tan^{-1}\left(\frac{x-2}{x-3}\right)+\tan^{-1}\left(\frac{x+2}{x+3}\right)=\frac{\pi}{4}
\displaystyle \Rightarrow \tan^{-1}\left(\frac{x-2}{x-3}\right)+\tan^{-1}\left(\frac{x+2}{x+3}\right)=\tan^{-1}1
\displaystyle \Rightarrow \tan^{-1}\left(\frac{\frac{x-2}{x-3}+\frac{x+2}{x+3}}{1-\left(\frac{x-2}{x-3}\right)\left(\frac{x+2}{x+3}\right)}\right)=\tan^{-1}1
\displaystyle \Rightarrow \tan^{-1}\left(\frac{x^{2}+x-6+x^{2}-x-6}{x^{2}-9-x^{2}+4}\right)=\tan^{-1}1
\displaystyle \Rightarrow \tan^{-1}\left(\frac{2x^{2}-12}{-5}\right)=\tan^{-1}1
\displaystyle \Rightarrow 2x^{2}-12=-5
\displaystyle \Rightarrow 2x^{2}=7
\displaystyle \Rightarrow x=\pm \sqrt{\frac{7}{2}}

\displaystyle \textbf{14. }  \text{If } A=\begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{bmatrix}\text{, find } A^{2}-5A+16I.
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{bmatrix}
\displaystyle \therefore A^{2}=A\cdot A=\begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{bmatrix}\begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{bmatrix}
\displaystyle \Rightarrow A^{2}=\begin{bmatrix}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{bmatrix}
\displaystyle \text{Thus, } A^{2}-5A+16I=\begin{bmatrix}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{bmatrix}-5\begin{bmatrix}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{bmatrix}+16\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
\displaystyle =\begin{bmatrix}5-10+16 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+16 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+16\end{bmatrix}
\displaystyle =\begin{bmatrix}11 & -1 & -3 \\ -1 & 9 & -10 \\ -5 & 4 & 14\end{bmatrix}

\displaystyle \textbf{15. }  \text{Show that four points A, B, C and D whose position vectors are }
\displaystyle 4\widehat{i}+5\widehat{j}+\widehat{k},\ -\widehat{j}-\widehat{k},\ 3\widehat{i}+9\widehat{j}+4\widehat{k}\ \text{and}
\displaystyle 4(-\widehat{i}+\widehat{j}+\widehat{k})\ \text{respectively are coplanar.}
\displaystyle \text{Answer:}
\displaystyle \text{The position vectors of the points A, B, C and D are }
\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})
\displaystyle \text{respectively.}
\displaystyle \overrightarrow{BA}=(4\widehat{i}+5\widehat{j}+\widehat{k})-(0\widehat{i}-\widehat{j}-\widehat{k})=4\widehat{i}+6\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{BC}=(3\widehat{i}+9\widehat{j}+4\widehat{k})-(0\widehat{i}-\widehat{j}-\widehat{k})=3\widehat{i}+10\widehat{j}+5\widehat{k}
\displaystyle \overrightarrow{BD}=(-4\widehat{i}+4\widehat{j}+4\widehat{k})-(0\widehat{i}-\widehat{j}-\widehat{k})=-4\widehat{i}+5\widehat{j}+5\widehat{k}
\displaystyle \text{The given points are coplanar if } \overrightarrow{BA},\ \overrightarrow{BC}\ \text{and } \overrightarrow{BD}\ \text{are coplanar.}
\displaystyle \overrightarrow{BA}\cdot(\overrightarrow{BC}\times \overrightarrow{BD})=\left|\begin{matrix}4 & 6 & 2 \\ 3 & 10 & 5 \\ -4 & 5 & 5\end{matrix}\right|
\displaystyle =4(50-25)-6(15+20)+2(15+40)
\displaystyle =100-210+110=0
\displaystyle \text{Hence, the four points A, B, C and D are coplanar.}

\displaystyle \textbf{16. }  \text{Show that the following two lines are coplanar:}
\displaystyle \frac{x-a+d}{a-d}=\frac{y-a}{a}=\frac{z-a-d}{a+d}\ \text{and}  \frac{x-b+c}{\beta-\tau}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\tau}.
\displaystyle \text{OR } \\ \text{Find the acute angle between the plane } 5x-4y+7z-13=0\ \text{and the } y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the lines } \frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}}\ \text{and}
\displaystyle \frac{x-x_{2}}{l_{2}}=\frac{y-y_{2}}{m_{2}}=\frac{z-z_{2}}{n_{2}}\ \text{are coplanar if}
\displaystyle \left|\begin{matrix}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2}\end{matrix}\right|=0
\displaystyle \text{Now, the equations of the given lines are}
\displaystyle \frac{x-a+d}{a-d}=\frac{y-a}{a}=\frac{z-a-d}{a+d}\ \text{and}\ \frac{x-b+c}{\beta-\tau}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\tau}
\displaystyle \text{These can be rewritten as}
\displaystyle \frac{x-(a-d)}{a-d}=\frac{y-a}{a}=\frac{z-(a+d)}{a+d}\ \text{and}\ \frac{x-(b-c)}{\beta-\tau}=\frac{y-b}{\beta}=\frac{z-(b+c)}{\beta+\tau}
\displaystyle \Rightarrow \left|\begin{matrix}(b-c)-(a-d) & b-a & (b+c)-(a+d) \\ a-d & a & a+d \\ \beta-\tau & \beta & \beta+\tau\end{matrix}\right|
\displaystyle \text{Applying } C_{1}\rightarrow C_{1}+C_{3}
\displaystyle =\left|\begin{matrix}2(b-a) & b-a & (b+c)-(a+d) \\ 2a & a & a+d \\ 2\beta & \beta & \beta+\tau\end{matrix}\right|
\displaystyle =2\left|\begin{matrix}b-a & b-a & (b+c)-(a+d) \\ a & a & a+d \\ \beta & \beta & \beta+\tau\end{matrix}\right|
\displaystyle =0\ \text{(since } C_{1}\ \text{and } C_{2}\ \text{are identical)}
\displaystyle \text{Hence, the given lines are coplanar.}
\displaystyle \text{OR The equation of the } y\text{-axis is } \frac{x-0}{0}=\frac{y-0}{1}=\frac{z-0}{0}
\displaystyle \text{The direction ratios of the } y\text{-axis are } 0,1,0
\displaystyle \text{The equation of the given plane is } 5x-4y+7z-13=0
\displaystyle \text{So, the direction ratios of the normal to the plane are } 5,-4,7
\displaystyle \text{Let } \theta\ \text{be the acute angle between the given plane and the } y\text{-axis}
\displaystyle \sin\theta=\frac{|0\times 5+1\times(-4)+0\times 7|}{\sqrt{0^{2}+1^{2}+0^{2}}\sqrt{5^{2}+(-4)^{2}+7^{2}}}
\displaystyle =\frac{4}{\sqrt{1}\sqrt{25+16+49}}=\frac{4}{3\sqrt{10}}
\displaystyle \Rightarrow \theta=\sin^{-1}\left(\frac{4}{3\sqrt{10}}\right)=\sin^{-1}\left(\frac{2}{15}\sqrt{10}\right)
\displaystyle \text{Hence, the acute angle between the given plane and the } y\text{-axis is } \sin^{-1}\left(\frac{2}{15}\sqrt{10}\right)

\displaystyle \textbf{17. }  \text{A and B throw a die alternatively till one of them gets a number greater than four}
\displaystyle \text{and wins the game. If A starts the game, what is the probability of B winning?}
\displaystyle \text{OR } \\ \text{A die is thrown three times. Events A and B are defined as below:}
\displaystyle \text{A: 5 on the first and 6 on the second throw.}   \text{B: 3 or 4 on the third throw. Find the }
\displaystyle \text{probability of B, given that A has already occurred.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } S\ \text{denote the success, i.e. getting a number greater than four and } F\ \text{denote the failure, i.e. getting a number less than four.}
\displaystyle \therefore P(S)=\frac{2}{6}=\frac{1}{3},\ P(F)=1-\frac{1}{3}=\frac{2}{3}
\displaystyle \text{Now, B gets the second throw, if A fails in the first throw.}
\displaystyle \therefore P(\text{B wins in the second throw})=P(FS)=P(F)P(S)
\displaystyle =\frac{2}{3}\times \frac{1}{3}
\displaystyle \text{Similarly, } P(\text{B wins in the fourth throw})=P(FFFS)
\displaystyle =P(F)P(F)P(F)P(S)=\left(\frac{2}{3}\right)^{3}\times \frac{1}{3}
\displaystyle P(\text{B wins in the sixth throw})=P(FFFFFS)
\displaystyle =P(F)P(F)P(F)P(F)P(F)P(S)=\left(\frac{2}{3}\right)^{5}\times \frac{1}{3}\ \text{and so on.}
\displaystyle \text{Hence,}
\displaystyle P(\text{B wins})=\frac{2}{3}\times \frac{1}{3}+\left(\frac{2}{3}\right)^{3}\times \frac{1}{3}+\left(\frac{2}{3}\right)^{5}\times \frac{1}{3}+\cdots
\displaystyle =\frac{2}{3}\times \frac{1}{3}\left[1+\left(\frac{2}{3}\right)^{2}+\left(\frac{2}{3}\right)^{4}+\cdots\right]
\displaystyle =\frac{2}{3}\times \frac{1}{3}\left(\frac{1}{1-\frac{4}{9}}\right)
\displaystyle =\frac{2}{9}\times \frac{9}{5}=\frac{2}{5}
\displaystyle \text{Thus, the probability that B wins is } \frac{2}{5}.
\displaystyle \text{OR}
\displaystyle \text{A is an event getting 5 on the first throw and 6 on the second throw.}
\displaystyle \text{Then } A=\{(5,6,1),(5,6,2),(5,6,3),(5,6,4),(5,6,5),(5,6,6)\}
\displaystyle \text{Also B is an event of getting 3 or 4 on the third throw.}
\displaystyle \therefore A\cap B=\{(5,6,3),(5,6,4)\}
\displaystyle \text{Required probability, } P(B|A)=\frac{n(A\cap B)}{n(A)}=\frac{2}{6}=\frac{1}{3}
\displaystyle \text{Thus, the probability of B, given that A has already occurred, is } \frac{1}{3}.

\displaystyle \textbf{18. }  \text{Evaluate: } \int (\sqrt{\cot x}+\sqrt{\tan x})dx.
\displaystyle \text{Answer:}
\displaystyle I=\int (\sqrt{\cot x}+\sqrt{\tan x})dx
\displaystyle =\int \sqrt{\tan x}(1+\cot x)\,dx
\displaystyle \text{Let } \tan x=t^{2}
\displaystyle \text{Differentiating both sides w.r.t. } x,\ \text{we get}
\displaystyle \sec^{2}x\,dx=2t\,dt
\displaystyle (1+\tan^{2}x)\,dx=2t\,dt
\displaystyle (1+t^{4})\,dx=2t\,dt
\displaystyle \Rightarrow dx=\frac{2t\,dt}{1+t^{4}}
\displaystyle \therefore I=\int t\left(1+\frac{1}{t^{2}}\right)\times \frac{2t}{1+t^{4}}\,dt
\displaystyle =2\int \frac{t^{2}+1}{t^{4}+1}\,dt=2\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}\,dt
\displaystyle =2\int \frac{\left(1+\frac{1}{t^{2}}\right)dt}{\left(t-\frac{1}{t}\right)^{2}+2}
\displaystyle \text{Let } t-\frac{1}{t}=y
\displaystyle \Rightarrow \left(1+\frac{1}{t^{2}}\right)dt=dy
\displaystyle \therefore I=2\int \frac{dy}{y^{2}+(\sqrt{2})^{2}}
\displaystyle =2\times \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{y}{\sqrt{2}}\right)+C
\displaystyle =\sqrt{2}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+C
\displaystyle =\sqrt{2}\tan^{-1}\left(\frac{t^{2}-1}{\sqrt{2}\,t}\right)+C
\displaystyle =\sqrt{2}\tan^{-1}\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+C

\displaystyle \textbf{19. }  \text{Find: } \int \frac{x^{3}-1}{x^{3}+x}dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let}
\displaystyle I=\int \frac{x^{3}-1}{x^{3}+x}dx=\int \frac{x^{3}+x-x-1}{x^{3}+x}dx
\displaystyle =\int \left[\frac{x^{3}+x}{x^{3}+x}-\frac{x+1}{x^{3}+x}\right]dx
\displaystyle =\int \left[1-\frac{x+1}{x^{3}+x}\right]dx
\displaystyle =\int 1\,dx-\int \frac{x+1}{x^{3}+x}dx
\displaystyle =x+C_{1}-\int \frac{x+1}{x^{3}+x}dx
\displaystyle \text{Let } I_{1}=\int \frac{x+1}{x^{3}+x}dx
\displaystyle \text{Then } I=x+C_{1}-I_{1}\qquad ...(i)
\displaystyle \text{Now, } I_{1}=\int \frac{x+1}{x^{3}+x}dx
\displaystyle \text{Let } \frac{x+1}{x(x^{2}+1)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}
\displaystyle \Rightarrow \frac{x+1}{x(x^{2}+1)}=\frac{(A+B)x^{2}+Cx+A}{x(x^{2}+1)}
\displaystyle \text{Comparing the coefficients of numerator, we get } A=1,\ B=-1\ \text{and } C=1
\displaystyle \text{So, } I_{1}=\int \frac{x+1}{x(x^{2}+1)}dx=\int \frac{1}{x}dx+\int \frac{-x+1}{x^{2}+1}dx
\displaystyle \Rightarrow I_{1}=\log|x|+\int \frac{-x+1}{x^{2}+1}dx
\displaystyle \Rightarrow I_{1}=\log|x|-\frac{1}{2}\int \frac{2x}{x^{2}+1}dx+\int \frac{1}{x^{2}+1}dx
\displaystyle \Rightarrow I_{1}=\log|x|-\frac{1}{2}\log|x^{2}+1|+\tan^{-1}x+C_{2}\qquad ...(ii)
\displaystyle \text{From } (i)\ \text{and } (ii),\ \text{we get}
\displaystyle I=x-\log|x|+\frac{1}{2}\log|x^{2}+1|-\tan^{-1}x+C

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Questions numbers 20 to 26 carry 6 Mark each}

\displaystyle \textbf{20. }  \text{Using integration, find the area of the region bounded by the lines }
\displaystyle y=2+x,\ y=2-x,\ x=2.
\displaystyle \text{Answer:}
\displaystyle \text{The equations of the given lines are}\displaystyle y=2+x\qquad ...(1)
\displaystyle y=2-x\qquad ...(2)
\displaystyle x=2\qquad ...(3)
\displaystyle \text{Solving } (1),\ (2)\ \text{and } (3),\ \text{we obtain the coordinates of the point of intersection}
\displaystyle A(0,2),\ B(2,4)\ \text{and } C(2,0).
\displaystyle \text{Here, the shaded region represents the area bounded by the given lines.}
\displaystyle \therefore \text{Required area = Area of the region } ABCA
\displaystyle =\text{Area of region } OABCO-\text{Area of region } OACO
\displaystyle =\int_{0}^{2}x_{AB}\,dx-\int_{0}^{2}x_{AC}\,dx
\displaystyle =\int_{0}^{2}(2+x)\,dx-\int_{0}^{2}(2-x)\,dx
\displaystyle =\left(2x+\frac{x^{2}}{2}\right)_{0}^{2}-\left(2x-\frac{x^{2}}{2}\right)_{0}^{2}
\displaystyle =[(4+2)-(0-0)]-(4-2)
\displaystyle =6-2
\displaystyle =4\ \text{square units}
\displaystyle \text{Thus, the area of the region bounded by the given lines is } 4\ \text{square units.}

\displaystyle \textbf{21. }  \text{Find the differential equation for all the straight lines, which are at a unit}
\displaystyle \text{distance from the origin.}
\displaystyle \text{OR } \\ \text{Show that the differential equation } 2xy\frac{dy}{dx}=x^{2}+3y^{2}\ \text{is homogeneous and solve it.}
\displaystyle \text{Answer:}
\displaystyle \text{The general equation of a line that is at unit distance from the origin is given by}
\displaystyle x\cos \alpha+y\sin \alpha=1\qquad ...(i)
\displaystyle \text{Differentiating } (i)\ \text{w.r.t. } x,\ \text{we get}
\displaystyle \cos \alpha+\frac{dy}{dx}\sin \alpha=0
\displaystyle \Rightarrow \cot \alpha=-\frac{dy}{dx}\qquad ...(ii)
\displaystyle \text{Dividing } (i)\ \text{by } \sin \alpha,\ \text{we get}
\displaystyle x\frac{\cos \alpha}{\sin \alpha}+y\frac{\sin \alpha}{\sin \alpha}=\frac{1}{\sin \alpha}
\displaystyle \Rightarrow x\cot \alpha+y=\mathrm{cosec}\,\alpha
\displaystyle \Rightarrow x\cot \alpha+y=\sqrt{1+\cot^{2}\alpha}\qquad ...(iii)
\displaystyle \text{Putting the value of } (ii)\ \text{in } (iii),\ \text{we get}
\displaystyle x\left(-\frac{dy}{dx}\right)+y=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\qquad ...(iv)
\displaystyle \text{Squaring } (iv),\ \text{we get}
\displaystyle \left(-x\frac{dy}{dx}+y\right)^{2}=\left(\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\right)^{2}
\displaystyle \Rightarrow (x^{2}-1)\left(\frac{dy}{dx}\right)^{2}-2xy\frac{dy}{dx}+y^{2}-1=0
\displaystyle \text{OR}
\displaystyle \text{The given differential equation can be expressed as}
\displaystyle \frac{dy}{dx}=\frac{x^{2}+3y^{2}}{2xy}\qquad ...(i)
\displaystyle \text{Let } F(x,y)=\frac{x^{2}+3y^{2}}{2xy}
\displaystyle \text{Now, } F(\lambda x,\lambda y)=\frac{(\lambda x)^{2}+3(\lambda y)^{2}}{2(\lambda x)(\lambda y)}
\displaystyle =\frac{\lambda^{2}(x^{2}+3y^{2})}{\lambda^{2}(2xy)}
\displaystyle =F(x,y)
\displaystyle \text{Therefore, } F(x,y)\ \text{is a homogeneous function of degree zero. So, the given}
\displaystyle \text{differential equation is a homogeneous differential equation.}
\displaystyle \text{Let } y=vx\qquad ...(ii)
\displaystyle \text{Differentiating } (ii)\ \text{w.r.t. } x,\ \text{we get}
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{Substituting the value of } y\ \text{and}\ \frac{dy}{dx}\ \text{in } (i),\ \text{we get}
\displaystyle v+x\frac{dv}{dx}=\frac{1+3v^{2}}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+3v^{2}}{2v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+3v^{2}-2v^{2}}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{2v}
\displaystyle \Rightarrow \frac{2v}{1+v^{2}}dv=\frac{dx}{x}\qquad ...(iii)
\displaystyle \text{Integrating both sides of } (iii),\ \text{we get}
\displaystyle \int \frac{2v}{1+v^{2}}dv=\int \frac{dx}{x}
\displaystyle \Rightarrow \log|1+v^{2}|=\log|x|+\log|C_{1}|
\displaystyle \Rightarrow \log\left|\frac{1+v^{2}}{x}\right|=\log|C_{1}|
\displaystyle \Rightarrow \frac{1+v^{2}}{x}=\pm C_{1}
\displaystyle \Rightarrow \frac{1+\frac{y^{2}}{x^{2}}}{x}=\pm C_{1}
\displaystyle \Rightarrow \frac{x^{2}+y^{2}}{x^{3}}=\pm C_{1}
\displaystyle \text{or } x^{2}+y^{2}=Cx^{3}

\displaystyle \textbf{22. }  \text{Find the direction ratios of the normal to the plane, which passes through}
\displaystyle \text{the points } (1,0,0)\ \text{and}\ (0,1,0)\ \text{and makes angle } \frac{\pi}{4}\ \text{with the plane }x+y=3.
\displaystyle \ \text{Also find the equation of the plane.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the plane passing through } (1,0,0)\ \text{be}
\displaystyle a(x-1)+b(y-0)+c(z-0)=0\qquad ...(1)
\displaystyle \text{Here, } a,\ b\ \text{and } c\ \text{are the direction ratios of the normal to the plane.}
\displaystyle \text{If the plane passes through } (0,1,0),\ \text{then}
\displaystyle a(0-1)+b(1-0)+c(0-0)=0
\displaystyle \Rightarrow -a+b=0
\displaystyle \Rightarrow a=b\qquad ...(2)
\displaystyle \text{It is given that plane } (1)\ \text{makes an angle of } \frac{\pi}{4}\ \text{with the plane } x+y=3.
\displaystyle \therefore \cos \frac{\pi}{4}=\frac{(a\times 1)+(b\times 1)+(c\times 0)}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{1^{2}+1^{2}+0^{2}}}
\displaystyle \Rightarrow \frac{1}{\sqrt{2}}=\frac{a+b}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{2}}
\displaystyle \Rightarrow \sqrt{a^{2}+b^{2}+c^{2}}=a+b
\displaystyle \text{Squaring on both sides, we get}
\displaystyle a^{2}+b^{2}+c^{2}=a^{2}+b^{2}+2ab
\displaystyle \Rightarrow c^{2}=2ab\qquad ...(3)
\displaystyle \text{From equations } (2)\ \text{and } (3),\ \text{we get}
\displaystyle c=\pm \sqrt{2}\,a
\displaystyle \text{So, the equation of the plane is}
\displaystyle a(x-1)+a(y-0)\pm \sqrt{2}\,a(z-0)=0
\displaystyle \Rightarrow x-1+y\pm \sqrt{2}\,z=0
\displaystyle \Rightarrow x+y\pm \sqrt{2}\,z=1
\displaystyle \text{Hence, the direction ratios of the normal to the plane are proportional to } 1,1,\pm \sqrt{2}.

\displaystyle \textbf{23. }  \text{If the function } f:R\to R\ \text{be defined by } f(x)=2x-3\ \text{and }
\displaystyle  g:R\to R\ \text{by }g(x)=x^{3}+5\text{, then find the value of } (fog)^{-1}(x).
\displaystyle \text{OR } \\ \text{Let } A=Q\times Q,\ \text{where } Q\ \text{is the set of all rational numbers, } \text{and } \times \text{ be a binary }
\displaystyle \text{operation defined on } A\ \text{by} (a,b)\times(c,d)=(ac,b+ad),\ \text{for all } (a,b),(c,d)\in A.
\displaystyle \text{Find (i) the identity element in } A\ \text{(ii) the invertible element of } A.
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle f(x)=2x-3
\displaystyle g(x)=x^{3}+5
\displaystyle (f\circ g)(x)=f[g(x)]=f(x^{3}+5)=2(x^{3}+5)-3
\displaystyle =2x^{3}+10-3=2x^{3}+7
\displaystyle \text{Let } (f\circ g)(x)=y
\displaystyle \Rightarrow 2x^{3}+7=y
\displaystyle \Rightarrow x=\left(\frac{y-7}{2}\right)^{\frac{1}{3}}
\displaystyle \Rightarrow (f\circ g)^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}}
\displaystyle \text{Thus, } (f\circ g)^{-1}:R\rightarrow R\ \text{be defined by}
\displaystyle (f\circ g)^{-1}(x)=\left(\frac{x-7}{2}\right)^{\frac{1}{3}}
\displaystyle \text{OR}
\displaystyle \text{Let } X=(a,b)\ \text{and } Y=(c,d)\in A,\ \forall a,b,c,d\in Q
\displaystyle \text{Let } E=(x,y)\ \text{be the identity element in } A\ \text{with respect to } *,
\displaystyle \forall X,Y\in Q\ \text{such that}
\displaystyle X*E=X=E*X,\ \forall X\in A
\displaystyle \Rightarrow X*E=X\ \text{and } E*X=X
\displaystyle \Rightarrow (ax,b+ay)=(a,b)\ \text{and } (xa,y+xb)=(a,b)
\displaystyle \text{Considering } (ax,b+ay)=(a,b)
\displaystyle \Rightarrow ax=a
\displaystyle \Rightarrow x=1
\displaystyle \text{and } b+ay=b
\displaystyle \Rightarrow y=0\ \ [\because x=1]
\displaystyle \text{Also, considering } (xa,y+xb)=(a,b)
\displaystyle \Rightarrow xa=a
\displaystyle \Rightarrow x=1
\displaystyle \text{and } y+xb=b
\displaystyle \Rightarrow y=0\ \ [\because x=1]
\displaystyle \therefore (1,0)\ \text{is the identity element in } A\ \text{with respect to } *
\displaystyle \text{(ii) Let } F=(m,n)\ \text{be the inverse in } A,\ \forall m,n\in Q
\displaystyle X*F=E\ \text{and } F*X=E
\displaystyle \Rightarrow (am,b+an)=(1,0)\ \text{and } (ma,n+mb)=(1,0)
\displaystyle \text{Considering } (am,b+an)=(1,0)
\displaystyle \Rightarrow am=1
\displaystyle \Rightarrow m=\frac{1}{a}
\displaystyle \text{and } b+an=0
\displaystyle \Rightarrow n=-\frac{b}{a}
\displaystyle \text{Also, considering } (ma,n+mb)=(1,0)
\displaystyle \Rightarrow ma=1
\displaystyle \Rightarrow m=\frac{1}{a}
\displaystyle \text{and } n+mb=0
\displaystyle \Rightarrow n=-\frac{b}{a}\ \ [\because m=\frac{1}{a}]
\displaystyle \therefore \text{The inverse of } (a,b)\in A\ \text{with respect to } *\ \text{is } \left(\frac{1}{a},-\frac{b}{a}\right)

\displaystyle \textbf{24. }  \text{If the function } f(x)=2x^{3}-9mx^{2}+12m^{2}x+1,\ \text{where } m>0 \text{ attains its}
\displaystyle \text{maximum and minimum at } p\ \text{and}\ q\ \text{respectively such that} p^{2}=q,\ \text{then find the value}
\displaystyle \text{of } m.
\displaystyle \text{Answer:}
\displaystyle \text{We have}
\displaystyle f(x)=2x^{3}-9mx^{2}+12m^{2}x+1
\displaystyle \Rightarrow f'(x)=6x^{2}-18mx+12m^{2}
\displaystyle \text{Also, } f''(x)=12x-18m
\displaystyle \text{Since, } f(x)\ \text{attains its maximum and minimum values at } x=p\ \text{and } x=q,
\displaystyle \text{respectively, so } f'(p)=0\ \text{and } f'(q)=0
\displaystyle f'(p)=0
\displaystyle \Rightarrow 6p^{2}-18mp+12m^{2}=0
\displaystyle \Rightarrow p^{2}-3mp+2m^{2}=0
\displaystyle \Rightarrow (p-2m)(p-m)=0
\displaystyle \Rightarrow p=2m\ \text{or } p=m
\displaystyle \text{Similarly, } f'(q)=0
\displaystyle \Rightarrow q=2m\ \text{or } q=m
\displaystyle \text{Now, consider the following cases:}
\displaystyle \text{Case I: If } p=2m\ \text{and } q=2m,\ \text{then}
\displaystyle p^{2}=q
\displaystyle \Rightarrow 4m^{2}=2m
\displaystyle \Rightarrow 2m^{2}-m=0
\displaystyle \Rightarrow m(2m-1)=0
\displaystyle \Rightarrow m=\frac{1}{2}\ \ [\because m>0]
\displaystyle \text{But, this gives } p=1\ \text{as the point of minima, which is not true.}
\displaystyle \text{Case II: If } p=2m\ \text{and } q=m,\ \text{then}
\displaystyle p^{2}=q
\displaystyle \Rightarrow 4m^{2}=m
\displaystyle \Rightarrow 4m^{2}-m=0
\displaystyle \Rightarrow m(4m-1)=0
\displaystyle \Rightarrow m=\frac{1}{4}\ \ [\because m>0]
\displaystyle \text{But, this gives } p=\frac{1}{2}\ \text{as the point of minima, which is not true.}
\displaystyle \text{Case III: If } p=m\ \text{and } q=2m,\ \text{then}
\displaystyle p^{2}=q
\displaystyle \Rightarrow m^{2}=2m
\displaystyle \Rightarrow m(m-2)=0
\displaystyle \Rightarrow m=2\ \ [\because m>0]
\displaystyle \text{For this case, } p=2\ \text{and } q=4\ \text{are the points of maxima and minima, respectively.}
\displaystyle \text{Case IV: If } p=m\ \text{and } q=m,\ \text{then}
\displaystyle p^{2}=q
\displaystyle \Rightarrow m^{2}=m
\displaystyle \Rightarrow m(m-1)=0
\displaystyle \Rightarrow m=1\ \ [\because m>0]
\displaystyle \text{But, this gives } q=1\ \text{as the point of maxima, which is not true.}
\displaystyle \text{Hence, the value of } m\ \text{is } 2.

\displaystyle \textbf{25. } \text{The postmaster of a local post office wishes to hire extra helpers during the}
\displaystyle \text{Deepawali season because of large increase in the volume of mail handling and delivery.}
\displaystyle \text{Because of the limited office space and the budgetary conditions, the number of temporary}
\displaystyle \text{helpers must not exceed 10. According to past experience, a man can handle 300 letters}
\displaystyle \text{and 80 packages per day, on the average, and a woman can handle 400 letters and }
\displaystyle \text{50 packets per day. The postmaster believes that the daily volume of extra mail and }
\displaystyle \text{packages will be no less than 3400 and 680 respectively. A man receives Rs 225 a day }
\displaystyle \text{and a woman receives Rs 200 a day. How many men and women helpers should be }
\displaystyle \text{hired to keep the pay-roll at a minimum? Formulate an LPP and solve it graphically.}
\displaystyle \text{Let the postmaster hire } x\ \text{men and } y\ \text{women.}
\displaystyle \text{Clearly, } x\ge 0,\ y\ge 0
\displaystyle \text{Also, it is given that the number of temporary helpers must not exceed } 10.
\displaystyle \therefore x+y\le 10
\displaystyle \text{The given information can be represented in the tabular form as}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{} & \text{Men }(x) & \text{Women }(y) & \text{Minimum Volume of Mails} \\ \hline \text{Letters} & 300 & 400 & 3400 \\ \hline \text{Packages} & 80 & 50 & 680 \\ \hline \text{Payroll (Rs)} & 225 & 200 & \text{} \\ \hline \end{array}
\displaystyle \text{Thus, the given LPP can be stated mathematically as follows:}
\displaystyle \text{Minimise } Z=225x+200y
\displaystyle \text{Subject to the constraints:}
\displaystyle x+y\le 10\qquad ...(1)
\displaystyle 300x+400y\ge 3400\qquad \text{(constraint on letters)}
\displaystyle \Rightarrow 3x+4y\ge 34\qquad ...(2)
\displaystyle 80x+50y\ge 680\qquad \text{(constraint on packages)}
\displaystyle \Rightarrow 8x+5y\ge 68\qquad ...(3)
\displaystyle x,\ y\ge 0\qquad ...(4)
\displaystyle \text{Converting the inequalities into equations, we obtain the lines } x+y=10,
\displaystyle 3x+4y=34,\ 8x+5y=68,\ x=0\ \text{and}\ y=0.
\displaystyle \text{These lines are drawn and the feasible region of the LPP is shaded. It is observed that}
\displaystyle \text{the feasible region is a point } (6,4).
\displaystyle \text{The value of the objective function at this point is given as}\displaystyle Z=225(6)+200(4)=1350+800=2150
\displaystyle \text{So, the minimum value of } Z\ \text{is Rs } 2150\ \text{at the point } (6,4).
\displaystyle \text{Hence, the postmaster should hire } 6\ \text{men and } 4\ \text{women to keep pay-roll}
\displaystyle \text{at a minimum of Rs } 2150.

\displaystyle \textbf{26. }  \text{40\% students of a college reside in hostel and the remaining reside outside.}
\displaystyle \text{At the end of the year, 50\% of the hostellers got A grade while from outside students,}
\displaystyle \text{only 30\% got A grade in the examination. At the end of the year, a student }
\displaystyle \text{of the college was chosen at random and was found to have gotten A grade. What is the }
\displaystyle \text{probability that the selected student was a hosteller?}
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_{1}\ \text{and } E_{2}\ \text{be the events that the student is a hosteller and}
\displaystyle \text{an outside student, respectively and } A\ \text{be the event that the chosen student gets A grade.}
\displaystyle \therefore P(E_{1})=40\%=0.4
\displaystyle P(E_{2})=(100-40)\%=60\%=0.6
\displaystyle P(A|E_{1})=P(\text{student getting A grade is a hosteller})=50\%=0.5
\displaystyle P(A|E_{2})=P(\text{student getting A grade is an outside student})=30\%=0.3
\displaystyle \text{The probability that a randomly chosen student is a hosteller, given that he got}
\displaystyle \text{A grade, is given by } P(E_{1}|A).
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle P(E_{1}|A)=\frac{P(E_{1})\cdot P(A|E_{1})}{P(E_{1})\cdot P(A|E_{1})+P(E_{2})\cdot P(A|E_{2})}
\displaystyle =\frac{0.4\times 0.5}{0.4\times 0.5+0.6\times 0.3}=\frac{0.20}{0.38}=\frac{10}{19}
\displaystyle \text{Hence, the required probability is } \frac{10}{19}.

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