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} \\ \text{and C. 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 per} \\ \text{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 required.}

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

\displaystyle \textbf{1. } \text{Write the principal value of } \tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}).
\displaystyle \text{Answer:}
\displaystyle \text{Consider } \tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})
\displaystyle =\tan^{-1}\left(\tan\frac{\pi}{3}\right)-\cot^{-1}\left(-\cot\frac{\pi}{6}\right)
\displaystyle =\frac{\pi}{3}-\cot^{-1}\left(\cot\left(\pi-\frac{\pi}{6}\right)\right)
\displaystyle =\frac{\pi}{3}-\frac{5\pi}{6}=-\frac{\pi}{2}

\displaystyle \textbf{2. } \text{Write the value of } \tan^{-1}\left[ 2\sin\left( 2\cos^{-1}\frac{\sqrt{3}}{2} \right) \right].
\displaystyle \text{Answer:}
\displaystyle \text{Consider } \tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]
\displaystyle =\tan^{-1}\left[2\sin\left(2\cos^{-1}\left(\cos\frac{\pi}{6}\right)\right)\right]
\displaystyle =\tan^{-1}\left[2\sin\left(2\times\frac{\pi}{6}\right)\right]=\tan^{-1}\left[2\sin\frac{\pi}{3}\right]
\displaystyle =\tan^{-1}\left[2\cdot\frac{\sqrt{3}}{2}\right]
\displaystyle =\tan^{-1}\sqrt{3}=\tan^{-1}\left(\tan\frac{\pi}{3}\right)=\frac{\pi}{3}

\displaystyle \textbf{3. } \text{For what value of } x,\ \text{is the matrix } A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix}   \text{a skew-symmetric matrix?}
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}0&1&-2\\-1&0&3\\x&-3&0\end{bmatrix},\ A'=\begin{bmatrix}0&-1&x\\1&0&-3\\-2&3&0\end{bmatrix}
\displaystyle \text{We know that } A \text{ is skew-symmetric, if } A'=-A
\displaystyle \Rightarrow \begin{bmatrix}0&-1&x\\1&0&-3\\-2&3&0\end{bmatrix}=\begin{bmatrix}0&-1&2\\1&0&-3\\-x&3&0\end{bmatrix}
\displaystyle \text{On comparing values of elements of matrices}
\displaystyle \Rightarrow x=2 \text{ or } -x=-2
\displaystyle \Rightarrow x=2

\displaystyle \textbf{4. } \text{If matrix } A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \text{ and } A^2 = kA,\ \text{then write the value of } k.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}
\displaystyle \Rightarrow A^2=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}=\begin{bmatrix}2&-2\\-2&2\end{bmatrix}
\displaystyle \text{Given } A^2=kA
\displaystyle \Rightarrow \begin{bmatrix}2&-2\\-2&2\end{bmatrix}=k\begin{bmatrix}1&-1\\-1&1\end{bmatrix}=\begin{bmatrix}k&-k\\-k&k\end{bmatrix}
\displaystyle \Rightarrow k=2

\displaystyle \textbf{5. } \text{Write the differential equation representing the family of curves } y = mx,
\displaystyle \text{where } m \text{ is an arbitrary constant.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider } y=mx \qquad \ldots (i)
\displaystyle \Rightarrow \frac{dy}{dx}=m
\displaystyle \text{From (i), we have } y=\frac{dy}{dx}\,x \Rightarrow x\frac{dy}{dx}-y=0   \text{ which is the required differential equation.}

\displaystyle \textbf{6. } \text{If } A_{ij} \text{ is the cofactor of the element } a_{ij} \text{ of the determinant }   \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}, \\ \text{ then write the value of } a_{32}A_{32}.
\displaystyle \text{Answer:}
\displaystyle \Delta=\begin{vmatrix}2&-3&5\\6&0&4\\1&5&-7\end{vmatrix}
\displaystyle A_{32}=\text{Cofactor of element } a_{32}
\displaystyle =-\begin{vmatrix}2&5\\6&4\end{vmatrix}=-\left(8-30\right)=22
\displaystyle \therefore\ a_{32}A_{32}=5(22)=110 \qquad [\because\ a_{32}=5]

\displaystyle \textbf{7. } \text{P and Q are two points with position vectors } 3\overrightarrow{a} - 2\overrightarrow{b} \text{ and } \overrightarrow{a} + \overrightarrow{b} \text{ respectively. }
\displaystyle \text{Write the position vector of a point R }   \text{which divides the line segment PQ in} \\ \text{the ratio } 2:1 \text{ externally.}
\displaystyle \text{Answer:}
\displaystyle \text{Let P } (3\overrightarrow{a}-2\overrightarrow{b}) \text{ and Q } (\overrightarrow{a}+\overrightarrow{b}) \text{ be the given points} \displaystyle \text{The position vector of point R dividing PQ externally in the}   \text{ratio } 2:1 \text{ is}
\displaystyle \frac{2(\overrightarrow{a}+\overrightarrow{b})-1(3\overrightarrow{a}-2\overrightarrow{b})}{2-1}=-\overrightarrow{a}+4\overrightarrow{b}

\displaystyle \textbf{8. } \text{Find } |\overrightarrow{x}|, \text{ if for a unit vector } \overrightarrow{a},   (\overrightarrow{x} - \overrightarrow{a}) \cdot (\overrightarrow{x} + \overrightarrow{a}) = 15.
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \overrightarrow{a} \text{ is a unit vector } \therefore\ |\overrightarrow{a}|=1
\displaystyle \text{Now, consider } (\overrightarrow{x}-\overrightarrow{a})\cdot(\overrightarrow{x}+\overrightarrow{a})=15
\displaystyle \Rightarrow |\overrightarrow{x}|^2-|\overrightarrow{a}|^2=15
\displaystyle \Rightarrow |\overrightarrow{x}|^2-1=15 \Rightarrow |\overrightarrow{x}|^2=16 \Rightarrow |\overrightarrow{x}|=4

\displaystyle \textbf{9. } \text{Find the length of the perpendicular drawn from the origin to the plane}
\displaystyle 2x - 3y + 6z + 21 = 0.
\displaystyle \text{Answer:}
\displaystyle \text{Required length of the perpendicular from origin } (0,0,0) \text{ to}   \text{the given plane}
\displaystyle =\frac{|2(0)-3(0)+6(0)+21|}{\sqrt{2^2+(-3)^2+(6)^2}}=\frac{21}{7}=3 \text{ units.}

\displaystyle \textbf{10. } \text{The money to be spent for the welfare of the employees of a firm is proportional}
\displaystyle \text{to the rate of change of its total revenue (marginal revenue). If the total revenue}
\displaystyle \text{(in rupees)  received } \text{from the sale of } x \text{ units of a product is given by }
\displaystyle  R(x) = 3x^2 + 36x + 5,   \text{find the marginal revenue, when } x = 5, \text{ and write which value}
\displaystyle \text{does the question indicate.}
\displaystyle \text{Answer:}
\displaystyle \text{Here, total revenue (in } \text{Rs}\text{) received from sale of } x \text{ units of a}  \text{product is given by}
\displaystyle R(x)=3x^2+36x+5
\displaystyle \text{Differentiate w.r.t. } x
\displaystyle \Rightarrow \frac{d}{dx}R(x)=6x+36
\displaystyle \Rightarrow \left.\frac{d}{dx}R(x)\right|_{x=5}=6\times 5+36=\text{Rs }66
\displaystyle \text{Welfare of the employees indicate social and moral values.}

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

\displaystyle \textbf{11. } \text{Consider } f : R^{+} \rightarrow [4,\infty) \text{ given by } f(x) = x^2 + 4. \text{Show that } f \text{ is invertible with }
\displaystyle \text{the inverse } f^{-1} \text{ of } f \text{ given by }   f^{-1}(y) = \sqrt{y - 4}, \text{ where } R^{+} \text{ is the set of all non-negative } \\ \text{real numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Here, } f:R^{+}\rightarrow[4,\infty) \text{ is given by } f(x)=x^2+4
\displaystyle \text{If } f \text{ is one-one and onto then it is also invertible}
\displaystyle \text{To show one-one}
\displaystyle \text{Let } x_1,x_2\in R^{+}, \text{ such that}
\displaystyle f(x_1)=f(x_2)
\displaystyle \Rightarrow x_1^2+4=x_2^2+4
\displaystyle \Rightarrow x_1^2=x_2^2 \Rightarrow x_1=x_2 \qquad [\because\ x_1,x_2\geq 0]
\displaystyle \text{Thus, } f \text{ is one-one.}
\displaystyle \text{For onto}
\displaystyle \text{Let } y\in R_f,\ \text{then } y=f(x), \text{ for all } x\in D_f=R^{+}
\displaystyle \Rightarrow y=x^2+4
\displaystyle \Rightarrow x=\sqrt{y-4} \qquad [\because\ x\geq 0]
\displaystyle \text{As } x \text{ is real, } y-4\geq 0 \Rightarrow R_f=[4,\infty)
\displaystyle R_f=\text{Co-domain} \Rightarrow f \text{ is onto.}
\displaystyle \text{Now, } f \text{ is one-one and onto}
\displaystyle \therefore\ f^{-1} \text{ exists.}
\displaystyle \text{For } f^{-1}, \text{ we have}
\displaystyle f\circ f^{-1}(x)=x,\ \text{for all } x\in R_f
\displaystyle \Rightarrow f(f^{-1}(x))=x
\displaystyle \Rightarrow (f^{-1}(x))^2+4=x \qquad (\text{By definition of } f(x))
\displaystyle \Rightarrow (f^{-1}(x))^2=x-4
\displaystyle \Rightarrow f^{-1}(x)=\sqrt{x-4}

\displaystyle \textbf{12. } \text{Show that } \tan\left( \frac{1}{2} \sin^{-1} \frac{3}{4} \right) = \frac{4 - \sqrt{7}}{3}
\displaystyle \text{OR}
\displaystyle \text{Solve the following equation: } \cos(\tan^{-1} x) = \sin\left( \cot^{-1} \frac{3}{4} \right).
\displaystyle \text{Answer:}
\displaystyle \text{Let } \sin^{-1}\frac{3}{4}=\theta \Rightarrow \sin\theta=\frac{3}{4}
\displaystyle \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\left(\frac{3}{4}\right)^2}=\frac{\sqrt{7}}{4}
\displaystyle \text{Now, L.H.S : } \tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right)=\tan\frac{\theta}{2}
\displaystyle \text{As } \tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}
\displaystyle \therefore\ \tan\frac{\theta}{2}=\sqrt{\frac{1-\frac{\sqrt{7}}{4}}{1+\frac{\sqrt{7}}{4}}}
\displaystyle =\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}\times\frac{4-\sqrt{7}}{4-\sqrt{7}}}
\displaystyle =\frac{4-\sqrt{7}}{\sqrt{16-7}}=\frac{4-\sqrt{7}}{3}=\text{R.H.S.}
\displaystyle \text{OR}
\displaystyle \text{Let } \cot^{-1}\frac{3}{4}=\theta \Rightarrow \cot\theta=\frac{3}{4}
\displaystyle \Rightarrow \tan\theta=\frac{4}{3} \Rightarrow \sin\theta=\frac{4}{5}
\displaystyle \Rightarrow \theta=\sin^{-1}\frac{4}{5},\ \therefore\ \cot^{-1}\frac{3}{4}=\sin^{-1}\frac{4}{5} \qquad \ldots (i)
\displaystyle \text{and let } \tan^{-1}x=\phi
\displaystyle \Rightarrow \tan\phi=x \Rightarrow \sec\phi=\sqrt{1+\tan^2\phi}
\displaystyle \Rightarrow \sec\phi=\sqrt{1+x^2}
\displaystyle \Rightarrow \cos\phi=\frac{1}{\sqrt{1+x^2}}
\displaystyle \Rightarrow \phi=\cos^{-1}\frac{1}{\sqrt{1+x^2}}
\displaystyle \therefore\ \tan^{-1}x=\cos^{-1}\frac{1}{\sqrt{1+x^2}} \qquad \ldots (ii)
\displaystyle \text{Now, consider } \cos(\tan^{-1}x)=\sin\left(\cot^{-1}\frac{3}{4}\right)
\displaystyle \Rightarrow \cos\left(\cos^{-1}\frac{1}{\sqrt{1+x^2}}\right)=\sin\left(\sin^{-1}\frac{4}{5}\right)
\displaystyle \text{[From equations (i) and (ii)]}
\displaystyle \Rightarrow \frac{1}{\sqrt{1+x^2}}=\frac{4}{5} \Rightarrow x=\pm\frac{3}{4}
\displaystyle \text{Since } \tan^{-1}x \text{ is acute here, } x=\frac{3}{4}

\displaystyle \textbf{13. } \text{Using properties of determinants, prove the following:}
\displaystyle \begin{vmatrix} x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x \end{vmatrix} = 9y^2(x + y).
\displaystyle \text{Answer:}
\displaystyle \text{Let } \Delta=\begin{vmatrix} x & x+y & x+2y \\ x+2y & x & x+y \\ x+y & x+2y & x \end{vmatrix}
\displaystyle \text{Operating } C_1 \rightarrow C_1+C_2+C_3
\displaystyle \Delta=\begin{vmatrix} 3(x+y) & x+y & x+2y \\ 3(x+y) & x & x+y \\ 3(x+y) & x+2y & x \end{vmatrix}
\displaystyle \text{Taking } 3(x+y) \text{ common from } C_1
\displaystyle \Delta=3(x+y)\begin{vmatrix} 1 & x+y & x+2y \\ 1 & x & x+y \\ 1 & x+2y & x \end{vmatrix}
\displaystyle \text{Operating } R_2 \rightarrow R_2-R_1 \text{ and } R_3 \rightarrow R_3-R_1
\displaystyle \Delta=3(x+y)\begin{vmatrix} 1 & x+y & x+2y \\ 0 & -y & -y \\ 0 & y & -2y \end{vmatrix}
\displaystyle \text{Taking } y \text{ common from } R_2 \text{ and } R_3
\displaystyle \Delta=3y^2(x+y)\begin{vmatrix} 1 & x+y & x+2y \\ 0 & -1 & -1 \\ 0 & 1 & -2 \end{vmatrix}
\displaystyle \text{Expanding by } C_1,\ \text{we get}
\displaystyle \Delta=3y^2(x+y)\{1(2+1)\}=9y^2(x+y)
\displaystyle \text{Hence Proved.}

\displaystyle \textbf{14. } \text{If } y^x = e^{y - x}, \text{ prove that } \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y}.
\displaystyle \text{Answer:}
\displaystyle \text{ Given } y^x=e^{y-x}
\displaystyle \text{Taking log on both sides, we get}
\displaystyle x\log y=(y-x)\log e
\displaystyle \Rightarrow x\log y=y-x \Rightarrow x(1+\log y)=y
\displaystyle \Rightarrow x=\frac{y}{1+\log y}
\displaystyle \text{Differentiate both sides w.r.t. } y,\ \text{we get}
\displaystyle \frac{dx}{dy}=\frac{(1+\log y)\cdot 1-y\left(0+\frac{1}{y}\right)}{(1+\log y)^2}
\displaystyle \Rightarrow \frac{dx}{dy}=\frac{1+\log y-1}{(1+\log y)^2}=\frac{\log y}{(1+\log y)^2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{(1+\log y)^2}{\log y}

\displaystyle \textbf{15. } \text{Differentiate the following with respect to } x: \  \sin^{-1} \left( \frac{2^{x+1} \cdot 3^x}{1 + 36^x} \right).
\displaystyle \text{Answer:}
\displaystyle \text{Let } y=\sin^{-1}\left(\frac{2^{x+1}\cdot 3^x}{1+(36)^x}\right)
\displaystyle =\sin^{-1}\left(\frac{2\cdot 2^x\cdot 3^x}{1+(6)^{2x}}\right)=\sin^{-1}\left(\frac{2\cdot 6^x}{1+(6)^{2x}}\right)
\displaystyle \text{Put } 6^x=\tan\theta
\displaystyle \Rightarrow 1+(6)^{2x}=1+\tan^2\theta=\sec^2\theta
\displaystyle \text{Now, } y=\sin^{-1}\left(\frac{2\tan\theta}{\sec^2\theta}\right)
\displaystyle =\sin^{-1}\left(2\frac{\sin\theta}{\cos\theta}\cdot \cos^2\theta\right)=\sin^{-1}(\sin 2\theta)
\displaystyle \Rightarrow y=2\theta=2\tan^{-1}(6^x)
\displaystyle \text{Differentiate both sides w.r.t. } x,\ \text{we have}
\displaystyle \frac{dy}{dx}=2\cdot \frac{1}{1+(6^x)^2}\cdot \frac{d}{dx}(6^x)
\displaystyle =2\cdot \frac{1}{1+(6^x)^2}\cdot 6^x\log 6
\displaystyle =\frac{2\cdot 6^x\log 6}{1+(6)^{2x}}

\displaystyle \textbf{16. } \text{Find the value of } k,\ \text{for which }   f(x)=\begin{cases}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & -1\le x<0 \\ \frac{2x+1}{x-1}, & 0\le x<1 \end{cases}
\displaystyle \text{is continuous at } x=0.
\displaystyle \text{OR}
\displaystyle \text{If } x=a\cos^3\theta \text{ and } y=a\sin^3\theta,\ \text{then find the value of } \frac{d^2y}{dx^2}   \text{ at } \theta=\frac{\pi}{6}.
\displaystyle \text{Answer:}
\displaystyle  \lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}\left(\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\right)
\displaystyle =\lim_{x\to 0^-}\left(\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\times\frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}}\right)
\displaystyle =\lim_{x\to 0^-}\left(\frac{1+kx-1+kx}{x\left(\sqrt{1+kx}+\sqrt{1-kx}\right)}\right)
\displaystyle =\lim_{x\to 0^-}\left(\frac{2kx}{x\left(\sqrt{1+kx}+\sqrt{1-kx}\right)}\right)
\displaystyle =\lim_{x\to 0^-}\left(\frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}}\right)=\frac{2k}{2}=k
\displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\frac{2x+1}{x-1}=\frac{0+1}{0-1}=-1
\displaystyle \text{Since } f(x) \text{ is continuous at } x=0
\displaystyle \Rightarrow \lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)\Rightarrow k=-1
\displaystyle \text{OR}
\displaystyle \text{Let } x=a\cos^3\theta
\displaystyle \text{Diff. w.r.t. } \theta
\displaystyle \Rightarrow \frac{dx}{d\theta}=-3a\cos^2\theta\sin\theta \qquad \ldots (i)
\displaystyle \text{and } y=a\sin^3\theta
\displaystyle \text{Diff. w.r.t. } \theta
\displaystyle \Rightarrow \frac{dy}{d\theta}=3a\sin^2\theta\cos\theta \qquad \ldots (ii)
\displaystyle \text{From (i) and (ii), we have}
\displaystyle \frac{dy}{dx}=\frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}
\displaystyle \Rightarrow \frac{dy}{dx}=-\tan\theta
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=-\sec^2\theta\frac{d\theta}{dx}
\displaystyle \Rightarrow \frac{d^2y}{dx^2}=-\sec^2\theta\cdot\frac{1}{-3a\cos^2\theta\sin\theta}
\displaystyle =\frac{1}{3a}\sec^4\theta\cdot\mathrm{cosec}\,\theta
\displaystyle \text{Now, } \left.\frac{d^2y}{dx^2}\right|_{\theta=\frac{\pi}{6}}=\frac{1}{3a}\cdot\sec^4\frac{\pi}{6}\cdot\mathrm{cosec}\,\frac{\pi}{6}
\displaystyle =\frac{1}{3a}\left(\frac{2}{\sqrt{3}}\right)^4\cdot 2=\frac{32}{27a}

\displaystyle \textbf{17. } \text{Evaluate : } \int \frac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}\,dx
\displaystyle \text{OR}
\displaystyle \text{Evaluate: } \int \frac{x+2}{\sqrt{x^2+2x+3}}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx
\displaystyle =\int\frac{(2\cos^2x-1)-(2\cos^2\alpha-1)}{\cos x-\cos\alpha}\,dx
\displaystyle =\int\frac{2(\cos^2x-\cos^2\alpha)}{\cos x-\cos\alpha}\,dx
\displaystyle =2\int(\cos x+\cos\alpha)\,dx
\displaystyle =2(\sin x+x\cos\alpha)+C
\displaystyle \text{OR}
\displaystyle \text{Let } I=\int\frac{x+2}{\sqrt{x^2+2x+3}}\,dx
\displaystyle =\int\frac{\frac{1}{2}(2x+2)+1}{\sqrt{x^2+2x+3}}\,dx
\displaystyle =\frac{1}{2}\int\frac{2x+2}{\sqrt{x^2+2x+3}}\,dx+\int\frac{1}{\sqrt{x^2+2x+3}}\,dx
\displaystyle \text{Let } x^2+2x+3=t
\displaystyle (2x+2)\,dx=dt
\displaystyle \frac{1}{2}\int\frac{dt}{\sqrt{t}}=\frac{1}{2}\cdot\frac{t^{1/2}}{1/2}=\sqrt{x^2+2x+3}
\displaystyle \therefore\ \int\frac{1}{\sqrt{x^2+2x+3}}\,dx=\int\frac{1}{\sqrt{x^2+2x+3+1^2-1^2}}\,dx
\displaystyle =\int\frac{1}{\sqrt{(x+1)^2+(\sqrt{2})^2}}\,dx
\displaystyle =\log\left|x+1+\sqrt{x^2+2x+3}\right|+C
\displaystyle \therefore\ I=\sqrt{x^2+2x+3}+\log\left|x+1+\sqrt{x^2+2x+3}\right|+C

\displaystyle \textbf{18. } \text{Evaluate : } \int \frac{dx}{x(x^5+3)}
\displaystyle \text{Answer:}
\displaystyle \text{Here, Let } I=\int\frac{dx}{x(x^5+3)}
\displaystyle \text{Put } x^5=t \Rightarrow 5x^4\,dx=dt \Rightarrow dx=\frac{dt}{5x^4}
\displaystyle \text{Now, } I=\int\frac{1}{x(t+3)}\cdot\frac{dt}{5x^4}=\int\frac{dt}{5x^5(t+3)}=\int\frac{dt}{5t(t+3)}
\displaystyle =\frac{1}{5}\int\left(\frac{1}{3}\cdot\frac{1}{t}-\frac{1}{3}\cdot\frac{1}{t+3}\right)dt \qquad (\text{By using partial fraction})
\displaystyle =\frac{1}{5}\left[\frac{1}{3}\int\frac{1}{t}\,dt-\frac{1}{3}\int\frac{1}{t+3}\,dt\right]
\displaystyle =\frac{1}{15}\log|t|-\frac{1}{15}\log|t+3|+C
\displaystyle =\frac{1}{15}\log x^5-\frac{1}{15}\log|x^5+3|+C=\frac{1}{15}\log\left|\frac{x^5}{x^5+3}\right|+C

\displaystyle \textbf{19. } \text{Evaluate: } \int_{0}^{2\pi} \frac{1}{1+e^{\sin x}}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{2\pi}\frac{1}{1+e^{\sin x}}\,dx \qquad \ldots (i)
\displaystyle \text{Also, } I=\int_{0}^{2\pi}\frac{1}{1+e^{\sin(2\pi-x)}}\,dx \qquad \left[\because\ \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\right]
\displaystyle =\int_{0}^{2\pi}\frac{1}{1+e^{-\sin x}}\,dx
\displaystyle =\int_{0}^{2\pi}\frac{e^{\sin x}}{e^{\sin x}+1}\,dx \qquad \ldots (ii)
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{2\pi}\frac{1+e^{\sin x}}{e^{\sin x}+1}\,dx=\int_{0}^{2\pi}1\,dx \Rightarrow 2I=\left[x\right]_{0}^{2\pi}=2\pi
\displaystyle \Rightarrow I=\pi

\displaystyle \textbf{20. } \text{If } \overrightarrow{a}=\widehat{i}-\widehat{j}+7\widehat{k} \text{ and } \overrightarrow{b}=5\widehat{i}-\widehat{j}+\lambda\widehat{k},   \text{then find the value of } \lambda,\ \text{so that } \\  \overrightarrow{a}+\overrightarrow{b} \text{ and } \overrightarrow{a}-\overrightarrow{b}   \text{are perpendicular vectors.}
\displaystyle \text{Answer:}
\displaystyle  \overrightarrow{a}+\overrightarrow{b}=\widehat{i}-\widehat{j}+7\widehat{k}+5\widehat{i}-\widehat{j}+\lambda\widehat{k}
\displaystyle =6\widehat{i}-2\widehat{j}+(7+\lambda)\widehat{k}
\displaystyle \overrightarrow{a}-\overrightarrow{b}=\widehat{i}-\widehat{j}+7\widehat{k}-(5\widehat{i}-\widehat{j}+\lambda\widehat{k})
\displaystyle =\widehat{i}-\widehat{j}+7\widehat{k}-5\widehat{i}+\widehat{j}-\lambda\widehat{k}
\displaystyle =-4\widehat{i}+(7-\lambda)\widehat{k}
\displaystyle \text{Since } (\overrightarrow{a}+\overrightarrow{b}) \text{ is perpendicular to } (\overrightarrow{a}-\overrightarrow{b})
\displaystyle \Rightarrow (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}-\overrightarrow{b})=0
\displaystyle \Rightarrow (6\widehat{i}-2\widehat{j}+(7+\lambda)\widehat{k})\cdot(-4\widehat{i}+(7-\lambda)\widehat{k})=0
\displaystyle \Rightarrow -24-2(0)+(7+\lambda)(7-\lambda)=0
\displaystyle \Rightarrow -24+49-\lambda^2=0
\displaystyle \Rightarrow \lambda^2=25 \Rightarrow \lambda=\pm 5

\displaystyle \textbf{21. } \text{Show that the lines } \overrightarrow{r}=3\widehat{i}+2\widehat{j}-4\widehat{k}+\lambda(\widehat{i}+2\widehat{j}+2\widehat{k});
\displaystyle \overrightarrow{r}=5\widehat{i}-2\widehat{j}+\mu(3\widehat{i}+2\widehat{j}+6\widehat{k}) \text{ are intersecting. Hence, find}   \text{ their point of intersection.}
\displaystyle \text{OR}
\displaystyle \text{Find the vector equation of the plane through the points } (2,1,-1) \text{ and } (-1,3,4)
\displaystyle \text{and perpendicular to the plane } x-2y+4z=10.
\displaystyle \text{Answer:}
\displaystyle \text{Position vector of arbitrary points on the given lines are given by}
\displaystyle (3+\lambda)\widehat{i}+(2+2\lambda)\widehat{j}+(-4+2\lambda)\widehat{k} \qquad \ldots (i)
\displaystyle (5+3\mu)\widehat{i}+(-2+2\mu)\widehat{j}+6\mu\widehat{k} \qquad \ldots (ii)
\displaystyle \text{If the given lines intersect, then given lines have one common point}
\displaystyle \therefore\ \text{For some values of } \lambda \text{ and } \mu,\ \text{we have}
\displaystyle (3+\lambda)\widehat{i}+(2+2\lambda)\widehat{j}+(-4+2\lambda)\widehat{k}=(5+3\mu)\widehat{i}
\displaystyle +(-2+2\mu)\widehat{j}+6\mu\widehat{k}
\displaystyle \text{Equating coefficients of } \widehat{i},\widehat{j} \text{ and } \widehat{k},\ \text{we have } 3+\lambda=5+3\mu;
\displaystyle 2+2\lambda=-2+2\mu \text{ and } -4+2\lambda=6\mu
\displaystyle \text{which implies } \lambda=2+3\mu
\displaystyle \lambda=-2+\mu
\displaystyle \Rightarrow 2+3\mu=-2+\mu \Rightarrow 2\mu=-4
\displaystyle \Rightarrow \mu=-2 \text{ and } \lambda=3(-2)+2=-6+2=-4
\displaystyle \text{Thus, } \lambda=-4 \text{ and } \mu=-2
\displaystyle \text{Now, } \lambda=3\mu+2
\displaystyle \Rightarrow -4=3(-2)+2
\displaystyle \Rightarrow -4=-4
\displaystyle \text{which is true.}
\displaystyle \text{Thus, given lines intersect.}
\displaystyle \text{From (i), by putting } \lambda=-4,\ \text{we have}
\displaystyle \text{Position vector of point of intersection is given by}
\displaystyle \overrightarrow{r}=(3-4)\widehat{i}+(2+2(-4))\widehat{j}+(-4+2(-4))\widehat{k}
\displaystyle =-\widehat{i}-6\widehat{j}-12\widehat{k}
\displaystyle \text{OR}
\displaystyle \text{Let equation of plane through } (2,1,-1) \text{ be}
\displaystyle a(x-2)+b(y-1)+c(z+1)=0 \qquad \ldots (i)
\displaystyle \text{It passes through } (-1,3,4)
\displaystyle \Rightarrow a(-1-2)+b(3-1)+c(4+1)=0
\displaystyle \Rightarrow -3a+2b+5c=0 \qquad \ldots (ii)
\displaystyle \text{Also, (i) is perpendicular to}
\displaystyle x-2y+4z=10
\displaystyle \Rightarrow a-2b+4c=0 \qquad \ldots (iii)
\displaystyle \text{From equations (ii) and (iii)}
\displaystyle \text{Now, } \frac{a}{\begin{vmatrix}2&5\\-2&4\end{vmatrix}}=\frac{b}{\begin{vmatrix}5&-3\\4&1\end{vmatrix}}=\frac{c}{\begin{vmatrix}-3&2\\1&-2\end{vmatrix}}
\displaystyle \Rightarrow \frac{a}{8+10}=\frac{b}{5+12}=\frac{c}{6-2}=k \text{(say)}
\displaystyle \Rightarrow a=18k,\ b=17k \text{ and } c=4k
\displaystyle \text{From (i), we have}
\displaystyle 18k(x-2)+17k(y-1)+4k(z+1)=0
\displaystyle \Rightarrow 18x-36+17y-17+4z+4=0
\displaystyle \Rightarrow 18x+17y+4z-49=0

\displaystyle \textbf{22. } \text{The probabilities of two students A and B coming to the } \text{school in time are }
\displaystyle \frac{3}{7} \text{ and } \frac{5}{7} \text{ respectively. Assuming that the events, 'A coming in time' and 'B }
\displaystyle \text{coming in time' are independent, find the probability of only one of them coming}
\displaystyle \text{to the school in time. Write at least one advantage of coming to school in time.}
\displaystyle \text{Answer:}
\displaystyle  P(A)=\text{Probability of student A coming to school in time}=\frac{3}{7}
\displaystyle P(B)=\text{Probability of student B coming to school in time}=\frac{5}{7}
\displaystyle \text{Given, events A and B are independent.}
\displaystyle \therefore\ \text{Probability that only one of them will come in time}
\displaystyle =P(A)\cdot P(\overline{B})+P(B)\cdot P(\overline{A})
\displaystyle =P(A)(1-P(B))+P(B)(1-P(A))
\displaystyle =\frac{3}{7}\left(1-\frac{5}{7}\right)+\frac{5}{7}\left(1-\frac{3}{7}\right)=\frac{3}{7}\cdot\frac{2}{7}+\frac{5}{7}\cdot\frac{4}{7}
\displaystyle =\frac{6}{49}+\frac{20}{49}=\frac{26}{49}
\displaystyle \text{Advantages of coming to school in time are:}
\displaystyle \text{(i) developing habit of being punctual in everything.}
\displaystyle \text{(ii) never misses any class.}

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

\displaystyle \textbf{23. } \text{Find the area of greatest rectangle that can be inscribed in an ellipse}
\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
\displaystyle \text{OR}
\displaystyle \text{Find the equation of tangent to the curve } 3x^2-y^2=8,\ \text{which pass through the point}
\displaystyle \left(\frac{4}{3},0\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let the coordinates of the vertices of rectangle ABCD be } A
\displaystyle (a\cos\theta,\ b\sin\theta),\ B(-a\cos\theta,\ b\sin\theta),   C(-a\cos\theta,\ -b\sin\theta) \text{ and } D(a\cos\theta,\ -b\sin\theta)
\displaystyle \text{Length of rectangle, } AB=2a\cos\theta
\displaystyle \text{Breadth of rectangle, } AD=2b\sin\theta
\displaystyle \text{Area of rectangle ABCD }=AB\times AD  =2a\cos\theta\times 2b\sin\theta \displaystyle \text{Area of rectangle } (A)=2ab\sin 2\theta
\displaystyle \text{Differentiate w.r.t. } \theta
\displaystyle \frac{dA}{d\theta}=2ab\cos 2\theta\cdot 2
\displaystyle \text{For maximum or minimum, put } \frac{dA}{d\theta}=0 \Rightarrow 4ab\cos 2\theta=0
\displaystyle \Rightarrow \cos 2\theta=0
\displaystyle \Rightarrow 2\theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}
\displaystyle \text{Also, } \frac{d^2A}{d\theta^2}=-8ab\sin 2\theta<0
\displaystyle \therefore\ \text{Area is maximum at } \theta=\frac{\pi}{4}
\displaystyle \text{Hence, maximum area of rectangle ABCD}
\displaystyle =2ab\sin\left(2\cdot\frac{\pi}{4}\right)=2ab\sin\frac{\pi}{2}
\displaystyle =2ab \text{ sq. units}
\displaystyle \text{OR}
\displaystyle \text{The equation of the curve is } 3x^2-y^2=8
\displaystyle \text{On differentiating}
\displaystyle 6x-2y\frac{dy}{dx}=0
\displaystyle \frac{dy}{dx}=\frac{3x}{y}
\displaystyle \text{Let } (x_1,y_1) \text{ be the point on the curve at which tangent passes through the point}
\displaystyle \left(\frac{4}{3},0\right)
\displaystyle \therefore\ 3x_1^2-y_1^2=8 \qquad \ldots (i)
\displaystyle \text{Slope of tangent}
\displaystyle \left(\frac{dy}{dx}\right)_{(x_1,y_1)}=\frac{3x_1}{y_1}
\displaystyle \text{Equation of the tangent passing through the point } (x_1,y_1)   \text{with slope } \frac{3x_1}{y_1} \text{ is}
\displaystyle y-y_1=\frac{3x_1}{y_1}(x-x_1)
\displaystyle \text{Point } \left(\frac{4}{3},0\right) \text{ lies on the above tangent.}
\displaystyle \therefore\ 0-y_1=\frac{3x_1}{y_1}\left(\frac{4}{3}-x_1\right)\Rightarrow y_1^2-3x_1^2+4x_1=0
\displaystyle \Rightarrow y_1^2=3x_1^2-4x_1 \qquad \ldots (ii)
\displaystyle \text{From (i) and (ii)}
\displaystyle 3x_1^2-8=3x_1^2-4x_1 \Rightarrow 4x_1=8
\displaystyle \Rightarrow x_1=2
\displaystyle \text{From (ii)}
\displaystyle y_1^2=3(2)^2-4(2) \Rightarrow y_1^2=4
\displaystyle y_1=\pm 2
\displaystyle \text{Equation of tangent at } (2,2)
\displaystyle y-2=\frac{3\times 2}{2}(x-2)
\displaystyle \Rightarrow y-3x+4=0
\displaystyle \text{Equation of tangent at } (2,-2)
\displaystyle y+2=\frac{3\times 2}{-2}(x-2)
\displaystyle \Rightarrow y+3x-4=0

\displaystyle \textbf{24. } \text{Find the area of the region bounded by the parabola } y=x^2   \text{ and } \\ y=|x|.
\displaystyle \text{Answer:}
\displaystyle \text{Given curves are}
\displaystyle y=x^2 \text{ and } y=|x|
\displaystyle \Rightarrow x^2-|x|=0
\displaystyle \Rightarrow |x|(|x|-1)=0
\displaystyle \Rightarrow |x|=0 \text{ or } |x|-1=0
\displaystyle \Rightarrow x=0 \text{ or } x=\pm 1
\displaystyle \text{Thus, points of intersection are } (0,0) \text{ and } (\pm 1,1)
\displaystyle \text{Since } y=x^2 \text{ is an upward parabola with vertex } (0,0) \text{ and }   \text{symmetric about y-axis.} \displaystyle \therefore\ \text{Required area}=2\int_{0}^{1}(|x|-x^2)\,dx
\displaystyle =2\int_{0}^{1}(x-x^2)\,dx
\displaystyle =2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{1}=2\left(\frac{1}{2}-\frac{1}{3}\right)-0
\displaystyle =\frac{1}{3} \text{ sq. units.}

\displaystyle \textbf{25. } \text{Find the particular solution of the differential equation}
\displaystyle (\tan^{-1}y-x)\,dy=(1+y^2)\,dx,\ \text{given that when } x=0,\ y=0.
\displaystyle \text{Answer:}
\displaystyle \text{Given differential equation is}
\displaystyle (\tan^{-1}y-x)\,dy=(1+y^2)\,dx
\displaystyle \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{\tan^{-1}y}{1+y^2} \qquad \ldots (i)
\displaystyle \text{Which is of the form } \frac{dx}{dy}+Px=Q
\displaystyle \text{where } P=\frac{1}{1+y^2} \text{ and } Q=\frac{\tan^{-1}y}{1+y^2}
\displaystyle \text{I.F.}=e^{\int\frac{1}{1+y^2}\,dy}=e^{\tan^{-1}y}
\displaystyle \text{Now, required solution is}
\displaystyle xe^{\tan^{-1}y}=\int\frac{\tan^{-1}y}{1+y^2}\cdot e^{\tan^{-1}y}\,dy+C \qquad \ldots (ii)
\displaystyle \text{Put } \tan^{-1}y=t \Rightarrow \frac{1}{1+y^2}\,dy=dt
\displaystyle \therefore\ xe^t=\int te^t\,dt+C=t\cdot e^t-\int 1\cdot e^t\,dt+C
\displaystyle \because\ \int u\,dv=uv-\int v\,du
\displaystyle =te^t-e^t+C
\displaystyle xe^t=e^t(t-1)+C
\displaystyle \therefore\ xe^{\tan^{-1}y}=e^{\tan^{-1}y}(\tan^{-1}y-1)+C
\displaystyle x=\tan^{-1}y-1+Ce^{-\tan^{-1}y}
\displaystyle \text{when } x=0,\ y=0
\displaystyle \Rightarrow 0=\tan^{-1}0-1+Ce^{-\tan^{-1}0}
\displaystyle \Rightarrow 0=0-1+C \Rightarrow C=1
\displaystyle \text{Hence, the required solution is}
\displaystyle x=\tan^{-1}y-1+e^{-\tan^{-1}y}

\displaystyle \textbf{26. } \text{Find the equation of the plane passing through the line of intersection of the }
\displaystyle \text{planes } \overrightarrow{r}\cdot(\widehat{i}+3\widehat{j})-6=0 \text{ and } \overrightarrow{r}\cdot(3\widehat{i}-\widehat{j}-4\widehat{k})=0, \ \text{whose perpendicular distance from }
\displaystyle \text{origin is unity.}
\displaystyle \text{OR}
\displaystyle \text{Find the vector equation of the line passing through the point } (1,2,3) \text{ and parallel to }
\displaystyle \text{the planes } \overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+2\widehat{k})=5 \text{ and}   \overrightarrow{r}\cdot(3\widehat{i}+\widehat{j}+\widehat{k})=6.
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the plane through the intersection of given planes be}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}+3\widehat{j})-6+\lambda\,[\overrightarrow{r}\cdot(3\widehat{i}-\widehat{j}-4\widehat{k})]=0
\displaystyle =\overrightarrow{r}\cdot[(1+3\lambda)\widehat{i}+(3-\lambda)\widehat{j}-4\lambda\widehat{k}]-6=0 \qquad \ldots (i)
\displaystyle \text{Since plane (i) is at a unit distance from origin}
\displaystyle \frac{| -6 |}{\sqrt{(1+3\lambda)^2+(3-\lambda)^2+(-4\lambda)^2}}=1
\displaystyle \Rightarrow 36=(1+3\lambda)^2+(3-\lambda)^2+16\lambda^2
\displaystyle \Rightarrow 36=1+9\lambda^2+6\lambda+9+\lambda^2-6\lambda+16\lambda^2
\displaystyle \Rightarrow 36=10+26\lambda^2 \Rightarrow \lambda^2=1 \Rightarrow \lambda=\pm 1
\displaystyle \text{Now, from (i), we have}
\displaystyle \overrightarrow{r}\cdot[(1\pm 3)\widehat{i}+(3\mp 1)\widehat{j}\mp 4\widehat{k}]-6=0
\displaystyle \Rightarrow \overrightarrow{r}\cdot(-2\widehat{i}+4\widehat{j}+4\widehat{k})-6=0
\displaystyle \text{Hence, the equation of the plane is}
\displaystyle \overrightarrow{r}\cdot(4\widehat{i}+2\widehat{j}-4\widehat{k})=6 \text{ or } \overrightarrow{r}\cdot(-2\widehat{i}+4\widehat{j}+4\widehat{k})=6
\displaystyle \text{OR}
\displaystyle \text{Let the vector equation of the line passing through the point } (1,2,3) \text{and having }
\displaystyle  \langle a,b,c\rangle \text{ as direction ratios be}
\displaystyle \overrightarrow{r}=\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda(a\widehat{i}+b\widehat{j}+c\widehat{k})
\displaystyle = (1+a\lambda)\widehat{i}+(2+b\lambda)\widehat{j}+(3+c\lambda)\widehat{k} \qquad \ldots (i)
\displaystyle \text{Line (i) is parallel to plane } \overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+2\widehat{k})=5
\displaystyle \Rightarrow a-b+2c=0
\displaystyle \text{Also, line (i) is parallel to plane } \overrightarrow{r}\cdot(3\widehat{i}+\widehat{j}+\widehat{k})=6
\displaystyle \Rightarrow 3a+b+c=0
\displaystyle \Rightarrow \frac{a}{\begin{vmatrix}-1&2\\1&1\end{vmatrix}}=\frac{b}{\begin{vmatrix}2&1\\1&3\end{vmatrix}}=\frac{c}{\begin{vmatrix}1&-1\\3&1\end{vmatrix}}
\displaystyle \Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{4}=s
\displaystyle \Rightarrow a=-3s,\ b=5s,\ c=4s
\displaystyle \text{Now, from (i), we have}
\displaystyle \overrightarrow{r}=\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda s(-3\widehat{i}+5\widehat{j}+4\widehat{k})
\displaystyle =\widehat{i}+2\widehat{j}+3\widehat{k}+\mu(-3\widehat{i}+5\widehat{j}+4\widehat{k}),\ \text{where } \mu=\lambda s

\displaystyle \textbf{27. } \text{In a hockey match, both teams A and B scored same number of goals up }
\displaystyle \text{to the end of the game, so to decide the winner, the referee asked both the captains to }
\displaystyle \text{throw a die alternatively and decided that the team, whose captain gets a six first, will}
\displaystyle \text{be declared the winner. If the captain of team A was asked to start, find their }
\displaystyle \text{respective  probabilities of winning the match and state whether the decision of the referee }
\displaystyle \text{was fair or not. and state whether the decision of the referee was fair or not.}
\displaystyle \text{Answer:}
\displaystyle \text{Let S denotes the success (getting 6) and F denotes the }   \text{failure (not getting 6).}
\displaystyle \text{Thus, } P(S)=\frac{1}{6} \text{ and } P(F)=\frac{5}{6}
\displaystyle P(\text{A wins in first throw})=P(S)=\frac{1}{6}
\displaystyle \text{A gets the third throw when first throw by A and second }   \text{throw by B results into failures.}
\displaystyle \therefore\ P(\text{A wins in 3rd throw})=P(FFS)=P(F)\cdot P(F)\cdot P(S)
\displaystyle =\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}=\left(\frac{5}{6}\right)^2\frac{1}{6}
\displaystyle \text{Similarly, } P(\text{A wins in 5th throw})=\left(\frac{5}{6}\right)^4\frac{1}{6}
\displaystyle \text{Thus, } P(\text{A wins})=\frac{1}{6}+\left(\frac{5}{6}\right)^2\frac{1}{6}+\left(\frac{5}{6}\right)^4\frac{1}{6}+\cdots
\displaystyle =\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^2+\left(\frac{5}{6}\right)^4+\cdots\right]
\displaystyle =\frac{1}{6}\cdot\frac{1}{1-\left(\frac{25}{36}\right)}=\frac{1}{6}\cdot\frac{36}{11}=\frac{6}{11}
\displaystyle P(\text{B wins})=1-P(\text{A wins})=1-\frac{6}{11}=\frac{5}{11}
\displaystyle \text{Yes, the decision of the referee was fair.}

\displaystyle \textbf{28. } \text{A manufacturer considers that men and women workers are equally efficient and }
\displaystyle \text{so he pays them at the same rate. He has } 30 \text{ and } 17  \text{ units of workers (male and female) }
\displaystyle \text{and capital respectively, which he uses to produce two types of goods A and B.}
\displaystyle \text{To produce one unit of A, } 2 \text{ workers and } 3 \text{ units of capital are required while 3 workers}
\displaystyle  \text{and } 1 \text{ unit of capital is required to produce one unit of B. If A and B are priced at}
\displaystyle \text{Rs } 100 \text{ and Rs } 120 \text{ per unit respectively, how should he use his resources to maximise}
\displaystyle \text{the total revenue? Form the above as a LPP and solve graphically. Do you agree with this }
\displaystyle \text{view of the manufacturer that men and women workers are equally efficient and so should}
\displaystyle \text{be paid at the same rate?}
\displaystyle \text{Answer:}
\displaystyle \text{Let } x \text{ and } y \text{ be the number of goods of type A and of type B }  \text{respectively.}
\displaystyle \therefore\ \text{No. of units of labour }=2x+3y
\displaystyle \text{As } 30 \text{ units of labour are available.}
\displaystyle \therefore\ 2x+3y\leq 30
\displaystyle \text{Similarly, constraint for capital is}
\displaystyle 3x+y\leq 17
\displaystyle \text{and non-zero constraints are}
\displaystyle x\geq 0,\ y\geq 0
\displaystyle \text{Objective function}
\displaystyle Z=100x+120y
\displaystyle \text{Consider}
\displaystyle 2x+3y=30
\displaystyle \text{When } x=0,\ \text{then } y=10
\displaystyle \text{When } x=15,\ \text{then } y=0
\displaystyle \therefore\ 2x+3y=30
\displaystyle \text{passes through } A(0,10) \text{ and } B(15,0)
\displaystyle \text{Consider}
\displaystyle 3x+y=17
\displaystyle \text{When } x=0,\ \text{then } y=17
\displaystyle \text{When } y=0,\ \text{then } x=\frac{17}{3}=5.7
\displaystyle \therefore\ 3x+y=17 \text{ passes through } C(0,17) \text{ and } D\left(\frac{17}{3},0\right)
\displaystyle \text{Further above two equations intersect at } E(3,8),\ \text{vertices of the feasible region are }
\displaystyle A(0,10),\ O(0,0),\ D\left(\frac{17}{3},0\right) \text{ and } E(3,8). \displaystyle \text{At } A(0,10),\ Z=100(0)+120(10)=\text{Rs }1200
\displaystyle \text{At } O(0,0),\ Z=100(0)+120(0)=\text{Rs }0
\displaystyle \text{At } D\left(\frac{17}{3},0\right),\ Z=100\left(\frac{17}{3}\right)+120(0)=\text{Rs }566.67
\displaystyle \text{At } E(3,8),\ Z=100(3)+120(8)=\text{Rs }1260
\displaystyle \text{Thus, maximum value of } Z=\text{Rs }1260 \text{ at } x=3 \text{ and } y=8
\displaystyle \text{Yes, the view of manufacturer that men and women workers are equally efficient is correct }
\displaystyle \text{and so they should be paid at the same rate.}

\displaystyle \textbf{29. } \text{The management committee of a residential colony decided to award some }
\displaystyle \text{of its members (say } x) \text{ for honesty, some (say } y) \text{ for helping others and some others}
\displaystyle \text{ (say } z) \text{ for supervising the workers to keep the colony neat and clean. The sum }
\displaystyle \text{of all the awardees is 12. Three times the sum of awardees for cooperation and}
\displaystyle \text{supervision added to two times the number of awardees for honesty is 33. If the sum }
\displaystyle \text{of the number of awardees for honesty and supervision is twice the number of awardees }
\displaystyle \text{for helping others, using matrix method, find the number of awardees of each category. }
\displaystyle \text{Apart from these values, namely, honesty, cooperation and supervision, suggest one more }
\displaystyle \text{value which the management of the colony must include for awards.}
\displaystyle \text{Answer:}
\displaystyle \text{Let award for honesty be denoted by } x,\ \text{for helping others by}    y \text{ and for supervising} \\ \text{the workers by } z,\ \text{we have}
\displaystyle x+y+z=12
\displaystyle 2x+3y+3z=33
\displaystyle x+z=2y
\displaystyle \Rightarrow x-2y+z=0
\displaystyle \text{Its matrix form is}
\displaystyle \begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}12\\33\\0\end{bmatrix}
\displaystyle \text{i.e., } AX=B
\displaystyle \text{Now, } |A|=\begin{vmatrix}1&1&1\\2&3&3\\1&-2&1\end{vmatrix}
\displaystyle =1(3+6)-1(2-3)+1(-4-3)
\displaystyle =9+1-7=3\neq 0
\displaystyle \therefore\ A^{-1} \text{ exists}
\displaystyle A_{11}=9,\qquad A_{12}=1,\qquad A_{13}=-7
\displaystyle A_{21}=-3,\qquad A_{22}=0,\qquad A_{23}=3
\displaystyle A_{31}=0,\qquad A_{32}=-1,\qquad A_{33}=1
\displaystyle \text{adj}A=\begin{bmatrix}9&1&-7\\-3&0&3\\0&-1&1\end{bmatrix}^{\prime}=\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}
\displaystyle A^{-1}=\frac{\text{adj}A}{|A|}=\frac{1}{3}\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}
\displaystyle \text{Now, } AX=B
\displaystyle \Rightarrow X=A^{-1}B
\displaystyle =\frac{1}{3}\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}\begin{bmatrix}12\\33\\0\end{bmatrix}
\displaystyle =\frac{1}{3}\begin{bmatrix}108-99+0\\12+0+0\\-84+99+0\end{bmatrix}=\frac{1}{3}\begin{bmatrix}9\\12\\15\end{bmatrix}=\begin{bmatrix}3\\4\\5\end{bmatrix}
\displaystyle \text{Hence, } x=3,\ y=4 \text{ and } z=5.
\displaystyle \text{The management of the colony must include the award for }   \text{those who help} \\ \text{the committee financially.}

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