\displaystyle \Large {\textbf{Series W1XZY \quad Set - 1 \quad Q.P. Code 65/1/1}}

\displaystyle \text{Candidates must write the Q.P. Code on the title page of the answer-book.}
\displaystyle \text{Please check that this question paper contains 23 printed pages.}
\displaystyle \text{Please check that this question paper contains 38 questions.}
\displaystyle \text{Q.P. Code given on the right hand side of the question paper should be written}
\displaystyle \text{on the title page of the answer-book by the candidate.}
\displaystyle \text{Please write down the serial number of the question in the answer-book before}
\displaystyle \text{attempting it.}
\displaystyle \text{15 minute time has been allotted to read this question paper. The question}
\displaystyle \text{paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the}
\displaystyle \text{students will read the question paper only and will not write any answer on}
\displaystyle \text{the answer-book during this period.}

\displaystyle \textbf{MATHEMATICS}
\displaystyle \text{Time allowed : 3 hours \quad Maximum Marks : 80}

\displaystyle \textbf{General Instructions :}
\displaystyle \text{Read the following instructions very carefully and strictly follow them :}
\displaystyle \text{(i) This question paper contains 38 questions. All questions are compulsory.}
\displaystyle \text{(ii) This question paper is divided into five Sections - A, B, C, D and E.}
\displaystyle \text{(iii) In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs)}
\displaystyle \text{and questions number 19 and 20 are Assertion-Reason based questions of}
\displaystyle \text{1 mark each.}
\displaystyle \text{(iv) In Section B, Questions no. 21 to 25 are very short answer (VSA) type}
\displaystyle \text{questions, carrying 2 marks each.}
\displaystyle \text{(v) In Section C, Questions no. 26 to 31 are short answer (SA) type questions,}
\displaystyle \text{carrying 3 marks each.}
\displaystyle \text{(vi) In Section D, Questions no. 32 to 35 are long answer (LA) type questions}
\displaystyle \text{carrying 5 marks each.}
\displaystyle \text{(vii) In Section E, Questions no. 36 to 38 are case study based questions}
\displaystyle \text{carrying 4 marks each.}
\displaystyle \text{(viii) There is no overall choice. However, an internal choice has been provided}
\displaystyle \text{in 2 questions in Section B, 3 questions in Section C, 2 questions in Section D}
\displaystyle \text{and 2 questions in Section E.}
\displaystyle \text{(ix) Use of calculators is not allowed.}

\displaystyle \textbf{SECTION A}
\displaystyle \text{This section comprises multiple choice questions (MCQs) of 1 mark each.}

\displaystyle \textbf{1. } \text{If } A=\begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \text{ then } A^{-1} \text{ is}
\displaystyle \text{(A) } \begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \quad \text{(B) } \begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}
\displaystyle \text{(C) } \begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \quad \text{(D) } \begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle A=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&1\end{pmatrix}
\displaystyle \text{Since }A\text{ is diagonal, }A^{-1}\text{ is obtained by taking reciprocals of diagonal elements}
\displaystyle A^{-1}=\begin{pmatrix}\frac{1}{-1}&0&0\\0&\frac{1}{1}&0\\0&0&\frac{1}{1}\end{pmatrix}
\displaystyle =\begin{pmatrix}-1&0&0\\0&1&0\\0&0&1\end{pmatrix}
\displaystyle \therefore\ A^{-1}=A
\displaystyle \therefore\ \text{Correct option is (D)}

\\
\displaystyle \textbf{2. } \text{If vector } \overrightarrow{a}=3\widehat{i}+2\widehat{j}-\widehat{k} \text{ and vector } \overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}, \text{ then which is correct ?}
\displaystyle \text{(A) } \overrightarrow{a}\parallel \overrightarrow{b} \quad \text{(B) } \overrightarrow{a}\perp \overrightarrow{b}
\displaystyle \text{(C) } |\overrightarrow{b}|>|\overrightarrow{a}| \quad \text{(D) } |\overrightarrow{a}|=|\overrightarrow{b}|
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}=3\widehat{i}+2\widehat{j}-\widehat{k},\quad \overrightarrow{b}=\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=3(1)+2(-1)+(-1)(1)=3-2-1=0
\displaystyle \therefore\ \overrightarrow{a}\perp \overrightarrow{b}
\displaystyle \therefore\ \text{Correct option is (B)}

\\
\displaystyle \textbf{3. } \int_{-1}^{1} \frac{|x|}{x}\,dx,\ x\neq 0 \text{ is equal to}
\displaystyle \text{(A) } -1 \quad \text{(B) } 0 \quad \text{(C) } 1 \quad \text{(D) } 2
\displaystyle \text{Answer:}
\displaystyle \int_{-1}^{1}\frac{|x|}{x}\,dx
\displaystyle \text{For }x>0,\ \frac{|x|}{x}=1;\quad \text{for }x<0,\ \frac{|x|}{x}=-1
\displaystyle \therefore\ \int_{-1}^{1}\frac{|x|}{x}\,dx=\int_{-1}^{0}(-1)\,dx+\int_{0}^{1}1\,dx
\displaystyle =[-x]_{-1}^{0}+[x]_{0}^{1}
\displaystyle =(0-1)+(1-0)
\displaystyle =-1+1
\displaystyle =0
\displaystyle \therefore\ \text{Correct option is (B)}

\\
\displaystyle \textbf{4. } \text{Which of the following is not a homogeneous function of } x \text{ and } y ?
\displaystyle \text{(A) } y^{2}-xy \quad \text{(B) } x-3y
\displaystyle \text{(C) } \sin^{2}\left(\frac{y}{x}\right)+\frac{y}{x} \quad \text{(D) } \tan x-\sec y
\displaystyle \text{Answer:}
\displaystyle \text{A function }f(x,y)\text{ is homogeneous if }f(\lambda x,\lambda y)=\lambda^{n}f(x,y)
\displaystyle \textbf{(A)}\ y^{2}-xy\Rightarrow \lambda^{2}(y^{2}-xy)\ \text{homogeneous of degree }2
\displaystyle \textbf{(B)}\ x-3y\Rightarrow \lambda(x-3y)\ \text{homogeneous of degree }1
\displaystyle \textbf{(C)}\ \sin^{2}\left(\frac{y}{x}\right)+\frac{y}{x}\Rightarrow \text{function of }\frac{y}{x}\ \text{homogeneous of degree }0
\displaystyle \textbf{(D)}\ \tan x-\sec y\Rightarrow \tan(\lambda x)-\sec(\lambda y)\neq \lambda^{n}(\tan x-\sec y)
\displaystyle \therefore\ \text{(D) is not homogeneous}

\\
\displaystyle \textbf{5. } \text{If } f(x)=|x|+|x-1|, \text{ then which of the following is correct ?}
\displaystyle \text{(A) } f(x) \text{ is both continuous and differentiable at } x=0 \text{ and } x=1
\displaystyle \text{(B) } f(x) \text{ is differentiable but not continuous at } x=0 \text{ and } x=1
\displaystyle \text{(C) } f(x) \text{ is continuous but not differentiable at } x=0 \text{ and } x=1
\displaystyle \text{(D) } f(x) \text{ is neither continuous nor differentiable at } x=0 \text{ and } x=1
\displaystyle \text{Answer:}
\displaystyle f(x)=|x|+|x-1|
\displaystyle \text{Absolute value functions are continuous everywhere}
\displaystyle \therefore\ f(x)\text{ is continuous at all }x,\text{ including }x=0,1
\displaystyle \text{Now check differentiability}
\displaystyle \text{For }x<0,\ f(x)=-x-(x-1)=-2x+1
\displaystyle \text{For }0<x<1,\ f(x)=x-(x-1)=1
\displaystyle \text{For }x>1,\ f(x)=x+(x-1)=2x-1
\displaystyle \text{At }x=0,\ \text{LHD}=\lim_{x\to 0^-}f'(x)=-2,\ \text{RHD}=\lim_{x\to 0^+}f'(x)=0
\displaystyle \Rightarrow\ f'(0)\text{ does not exist}
\displaystyle \text{At }x=1,\ \text{LHD}=\lim_{x\to 1^-}f'(x)=0,\ \text{RHD}=\lim_{x\to 1^+}f'(x)=2
\displaystyle \Rightarrow\ f'(1)\text{ does not exist}
\displaystyle \therefore\ f(x)\text{ is continuous but not differentiable at }x=0\text{ and }x=1
\displaystyle \therefore\ \text{Correct option is (C)}

\\
\displaystyle \textbf{6. } \text{If } A \text{ is a square matrix of order 2 such that } \det(A)=4, \text{ then } \det(4\,\mathrm{adj}\,A)
\displaystyle \text{is equal to :}
\displaystyle \text{(A) } 16 \quad \text{(B) } 64 \quad \text{(C) } 256 \quad \text{(D) } 512
\displaystyle \text{Answer:}
\displaystyle \text{Given }|A|=4,\ \text{order of }A=2
\displaystyle \text{We know }|kB|=k^{n}|B|\ \text{for an }n\times n\text{ matrix}
\displaystyle \therefore\ |4\,\mathrm{adj}(A)|=4^{2}\,|\mathrm{adj}(A)|=16\,|\mathrm{adj}(A)|
\displaystyle \text{Also, }|\mathrm{adj}(A)|=|A|^{n-1}=4^{1}=4
\displaystyle \therefore\ |4\,\mathrm{adj}(A)|=16\cdot 4=64
\displaystyle \therefore\ \text{Correct option is (B)}

\\
\displaystyle \textbf{7. } \text{If E and F are two independent events such that } P(E)=\frac{2}{3}, P(F)=\frac{3}{7}, \text{ then}
\displaystyle P(E|\overline{F}) \text{ is equal to :}
\displaystyle \text{(A) } \frac{1}{6} \quad \text{(B) } \frac{1}{2} \quad \text{(C) } \frac{2}{3} \quad \text{(D) } \frac{7}{9}
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(E)=\frac{2}{3},\ P(F)=\frac{3}{7}
\displaystyle \text{Since }E\text{ and }F\text{ are independent, }E\text{ and }\overline{F}\text{ are also independent}
\displaystyle \therefore\ P(E\mid \overline{F})=P(E)=\frac{2}{3}
\displaystyle \therefore\ \text{Correct option is (C)}

\\
\displaystyle \textbf{8. } \text{The absolute maximum value of function } f(x)=x^{3}-3x+2 \text{ in } [0,2] \text{ is :}
\displaystyle \text{(A) } 0 \quad \text{(B) } 2 \quad \text{(C) } 4 \quad \text{(D) } 5
\displaystyle \text{Answer:}
\displaystyle f(x)=x^{3}-3x+2
\displaystyle f'(x)=3x^{2}-3=3(x^{2}-1)
\displaystyle \text{Critical points: }x=\pm 1
\displaystyle \text{In }[0,2],\ x=1\text{ is relevant}
\displaystyle f(0)=2,\quad f(1)=1-3+2=0,\quad f(2)=8-6+2=4
\displaystyle \therefore\ \text{Absolute maximum value is }4
\displaystyle \therefore\ \text{Correct option is (C)}

\\
\displaystyle \textbf{9. } \text{Let } A=\begin{bmatrix} 1 & -2 & -1 \\ 0 & 4 & -1 \\ -3 & 2 & 1 \end{bmatrix},
\displaystyle B=\begin{bmatrix} -2 \\ -5 \\ -7 \end{bmatrix}, \ C=[9\ 8\ 7], \text{ which of the following is}
\displaystyle \text{defined ?}
\displaystyle \text{(A) Only AB \quad (B) Only AC \quad (C) Only BA \quad (D) All AB, AC and BA}
\displaystyle \text{Answer:}
\displaystyle A=\begin{pmatrix}1&-2&-1\\0&4&-1\\-3&2&1\end{pmatrix}\text{ is of order }3\times 3
\displaystyle B=\begin{pmatrix}-2\\-5\\-7\end{pmatrix}\text{ is of order }3\times 1
\displaystyle C=\begin{pmatrix}9&8&7\end{pmatrix}\text{ is of order }1\times 3
\displaystyle \text{Now, }AB\text{ is defined since }(3\times 3)(3\times 1)\text{ is possible}
\displaystyle \text{Also, }AC\text{ is defined since }(3\times 3)(1\times 3)\text{ is not possible? No, inner orders differ}
\displaystyle \therefore\ AC\text{ is not defined}
\displaystyle \text{Further, }BA\text{ is defined since }(3\times 1)(3\times 3)\text{ is not possible? No, inner orders differ}
\displaystyle \therefore\ BA\text{ is not defined}
\displaystyle \therefore\ \text{Only }AB\text{ is defined}
\displaystyle \therefore\ \text{Correct option is (A)}

\\
\displaystyle \textbf{10. } \text{If } \int \frac{2^{\frac{1}{x}}}{x^{2}}\,dx=k\cdot 2^{\frac{1}{x}}+C, \text{ then } k \text{ is equal to}
\displaystyle \text{(A) } -\frac{1}{\log 2} \quad \text{(B) } -\log 2 \quad \text{(C) } -1 \quad \text{(D) } \frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \int \frac{2^{\frac{1}{x}}}{x^{2}}\,dx
\displaystyle \text{Let }t=\frac{1}{x}\Rightarrow dt=-\frac{1}{x^{2}}dx
\displaystyle \Rightarrow\ \frac{1}{x^{2}}dx=-dt
\displaystyle \therefore\ \int \frac{2^{\frac{1}{x}}}{x^{2}}\,dx=\int 2^{t}(-dt)
\displaystyle =-\int 2^{t}\,dt
\displaystyle =-\frac{2^{t}}{\log 2}+C
\displaystyle =-\frac{2^{\frac{1}{x}}}{\log 2}+C
\displaystyle \therefore\ k=-\frac{1}{\log 2}
\displaystyle \therefore\ \text{Correct option is (A)}

\\
\displaystyle \textbf{11. } \text{If } \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}, \ |\overrightarrow{a}|=\sqrt{37},
\displaystyle |\overrightarrow{b}|=3 \text{ and } |\overrightarrow{c}|=4, \text{ then angle between }
\displaystyle \overrightarrow{b} \text{ and } \overrightarrow{c} \text{ is}
\displaystyle \text{(A) } \frac{\pi}{6} \quad \text{(B) } \frac{\pi}{4} \quad \text{(C) } \frac{\pi}{3} \quad \text{(D) } \frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\Rightarrow \overrightarrow{a}=-(\overrightarrow{b}+\overrightarrow{c})
\displaystyle |\overrightarrow{a}|^{2}=|\overrightarrow{b}+\overrightarrow{c}|^{2}
\displaystyle \Rightarrow\ 37=|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}+2\overrightarrow{b}\cdot\overrightarrow{c}
\displaystyle \Rightarrow\ 37=3^{2}+4^{2}+2\overrightarrow{b}\cdot\overrightarrow{c}
\displaystyle \Rightarrow\ 37=9+16+2\overrightarrow{b}\cdot\overrightarrow{c}
\displaystyle \Rightarrow\ 37=25+2\overrightarrow{b}\cdot\overrightarrow{c}
\displaystyle \Rightarrow\ 2\overrightarrow{b}\cdot\overrightarrow{c}=12
\displaystyle \Rightarrow\ \overrightarrow{b}\cdot\overrightarrow{c}=6
\displaystyle \text{Now } \overrightarrow{b}\cdot\overrightarrow{c}=|\overrightarrow{b}||\overrightarrow{c}|\cos\theta
\displaystyle 6=3\cdot 4\cos\theta
\displaystyle \Rightarrow\ 6=12\cos\theta
\displaystyle \Rightarrow\ \cos\theta=\frac{1}{2}
\displaystyle \Rightarrow\ \theta=\frac{\pi}{3}
\displaystyle \therefore\ \text{Correct option is (C)}

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\displaystyle \textbf{12. } \text{The integrating factor of differential equation } (x+2y^{3})\frac{dy}{dx}=2y \text{ is}
\displaystyle \text{(A) } e^{\frac{y^{2}}{2}} \quad \text{(B) } \frac{1}{\sqrt{y}} \quad \text{(C) } \frac{1}{y^{2}} \quad \text{(D) } e^{-\frac{1}{y^{2}}}
\displaystyle \text{Answer:}
\displaystyle (x+2y^{3})\frac{dy}{dx}=2y
\displaystyle \Rightarrow\ \frac{dx}{dy}=\frac{x+2y^{3}}{2y}
\displaystyle \Rightarrow\ \frac{dx}{dy}-\frac{x}{2y}=y^{2}
\displaystyle \text{This is a linear differential equation in }x
\displaystyle \text{Integrating factor }=\exp\left(\int -\frac{1}{2y}\,dy\right)
\displaystyle =\exp\left(-\frac{1}{2}\log y\right)
\displaystyle =y^{-\frac{1}{2}}=\frac{1}{\sqrt{y}}
\displaystyle \therefore\ \text{Correct option is (B)}

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\displaystyle \textbf{13. } \text{If } A=\begin{bmatrix} 7 & 0 & x \\ 0 & 7 & 0 \\ 0 & 0 & y \end{bmatrix} \text{ is a scalar matrix, then } y^{x} \text{ is equal to}
\displaystyle \text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 7 \quad \text{(D) } \pm 7
\displaystyle \text{Answer:}
\displaystyle A=\begin{pmatrix}7&0&x\\0&7&0\\0&0&y\end{pmatrix}
\displaystyle \text{For }A\text{ to be a scalar matrix, it must be }kI
\displaystyle \Rightarrow\ \text{all diagonal elements equal and all non-diagonal elements }=0
\displaystyle \Rightarrow\ x=0,\ y=7
\displaystyle \therefore\ y^{x}=7^{0}=1
\displaystyle \therefore\ \text{Correct option is (B)}

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\displaystyle \textbf{14. } \text{The corner points of the feasible region in graphical representation of a}
\displaystyle \text{L.P.P. are } (2,72), (15,20) \text{ and } (40,15). \text{ If } Z=18x+9y \text{ be the objective}
\displaystyle \text{function, then}
\displaystyle \text{(A) } Z \text{ is maximum at } (2,72), \text{ minimum at } (15,20)
\displaystyle \text{(B) } Z \text{ is maximum at } (15,20) \text{ minimum at } (40,15)
\displaystyle \text{(C) } Z \text{ is maximum at } (40,15), \text{ minimum at } (15,20)
\displaystyle \text{(D) } Z \text{ is maximum at } (40,15), \text{ minimum at } (2,72)
\displaystyle \text{Answer:}
\displaystyle Z=18x+9y
\displaystyle \text{At }(2,72),\ Z=18(2)+9(72)=36+648=684
\displaystyle \text{At }(15,20),\ Z=18(15)+9(20)=270+180=450
\displaystyle \text{At }(40,15),\ Z=18(40)+9(15)=720+135=855
\displaystyle \text{Hence, }Z\text{ is maximum at }(40,15)\text{ and minimum at }(15,20)
\displaystyle \therefore\ \text{Correct option is (C)}

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\displaystyle \textbf{15. } \text{If A and B are invertible matrices, then which of the following is not correct ?}
\displaystyle \text{(A) } (A+B)^{-1}=B^{-1}+A^{-1} \quad \text{(B) } (AB)^{-1}=B^{-1}A^{-1}
\displaystyle \text{(C) } \mathrm{adj}(A)=|A|A^{-1} \quad \text{(D) } |A^{-1}|=|A^{-1}|
\displaystyle \text{Answer:}
\displaystyle \textbf{(A)}\ (A+B)^{-1}=B^{-1}+A^{-1}\ \text{is not generally true}
\displaystyle \textbf{(B)}\ (AB)^{-1}=B^{-1}A^{-1}\ \text{is true}
\displaystyle \textbf{(C)}\ \mathrm{adj}(A)=|A|A^{-1}\ \text{is true for invertible matrices}
\displaystyle \textbf{(D)}\ |A|^{-1}=|A^{-1}|\ \text{is true}
\displaystyle \therefore\ \text{the incorrect statement is (A)}

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\displaystyle \textbf{16. } \text{If the feasible region of a linear programming problem with objective}
\displaystyle \text{function } Z=ax+by, \text{ is bounded, then which of the following is correct ?}
\displaystyle \text{(A) It will only have a maximum value.}
\displaystyle \text{(B) It will only have a minimum value.}
\displaystyle \text{(C) It will have both maximum and minimum values.}
\displaystyle \text{(D) It will have neither maximum nor minimum value.}
\displaystyle \text{Answer:}
\displaystyle \text{If the feasible region of an L.P.P. is bounded, then the objective function }Z=ax+by
\displaystyle \text{attains both its maximum and minimum values at some corner points of the region}
\displaystyle \therefore\ \text{Correct option is (C)}

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\displaystyle \textbf{17. } \text{The area of the shaded region bounded by the curves } y^{2}=x, \ x=4 \text{ and the}
\displaystyle x\text{-axis is given by} \displaystyle \text{(A) } \int_{0}^{4} x\,dx \quad \text{(B) } \int_{0}^{2} y^{2}\,dy
\displaystyle \text{(C) } 2\int_{0}^{4} \sqrt{x}\,dx \quad \text{(D) } \int_{0}^{4} \sqrt{x}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Given }y^{2}=x\Rightarrow y=\sqrt{x}\text{ in the upper half-plane}
\displaystyle \text{The shaded region is bounded above by }y=\sqrt{x},\text{ below by the }x\text{-axis, and on the right by }x=4
\displaystyle \text{So the area is } \int_{0}^{4}(\sqrt{x}-0)\,dx
\displaystyle =\int_{0}^{4}\sqrt{x}\,dx
\displaystyle \therefore\ \text{Correct option is (D)}

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\displaystyle \textbf{18. } \text{The graph of a trigonometric function is as shown. Which of the following}
\displaystyle \text{will represent graph of its inverse ?} \displaystyle \text{Answer:}
\displaystyle \text{The given graph corresponds to }y=\cos x\text{ restricted to }[0,\pi]\text{ (one-one part)}
\displaystyle \text{Hence its inverse is }y=\cos^{-1}x
\displaystyle \text{Graph of inverse is reflection about }y=x
\displaystyle \text{Domain of }\cos^{-1}x=[-1,1],\ \text{range }[0,\pi]
\displaystyle \text{It is decreasing from }( -1,\pi )\text{ to }(1,0)
\displaystyle \therefore\ \text{Correct option is (C)}

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\displaystyle \textbf{Assertion - Reason Based Questions}
\displaystyle \text{Direction : Question numbers 19 and 20 are Assertion (A) and Reason (R)}
\displaystyle \text{based questions carrying 1 mark each. Two statements are given, one}
\displaystyle \text{labelled Assertion (A) and other labelled Reason (R). Select the correct}
\displaystyle \text{answer from the options (A), (B), (C) and (D) as given below.}
\displaystyle \text{(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the}
\displaystyle \text{correct explanation of the Assertion (A).}
\displaystyle \text{(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the}
\displaystyle \text{correct explanation of the Assertion (A).}
\displaystyle \text{(C) Assertion (A) is true, but Reason (R) is false.}
\displaystyle \text{(D) Assertion (A) is false, but Reason (R) is true.}

\displaystyle \textbf{19. } \text{Assertion (A) : Let } Z \text{ be the set of integers. A function } f:Z\rightarrow Z
\displaystyle \text{defined as } f(x)=3x-5, \forall x\in Z \text{ is a bijective.}
\displaystyle \text{Reason (R) : A function is a bijective if it is both surjective and injective.}
\displaystyle \text{Answer:}
\displaystyle \text{Assertion (A): }f:Z\to Z,\ f(x)=3x-5
\displaystyle \text{For }f(x)\text{ to be onto }Z,\text{ every }y\in Z\text{ must be of the form }y=3x-5
\displaystyle \Rightarrow\ x=\frac{y+5}{3}
\displaystyle \text{But for every integer }y,\ \frac{y+5}{3}\text{ need not be an integer}
\displaystyle \text{Hence }f\text{ is not onto }Z,\text{ so it is not bijective}
\displaystyle \therefore\ \text{Assertion (A) is false}
\displaystyle \text{Reason (R): A function is bijective if it is both injective and surjective}
\displaystyle \text{This is true}
\displaystyle \therefore\ \text{Correct option is (D)}

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\displaystyle \textbf{20. } \text{Assertion (A) : } f(x)=\begin{cases} 3x-8, & x\leq 5 \\ 2k, & x>5 \end{cases}
\displaystyle \text{is continuous at } x=5 \text{ for } k=\frac{5}{2}.
\displaystyle \text{Reason (R) : For a function } f \text{ to be continuous at } x=a,
\displaystyle \lim_{x\rightarrow a^{-}} f(x)=\lim_{x\rightarrow a^{+}} f(x)=f(a).
\displaystyle \text{Answer:}
\displaystyle f(x)=\begin{cases}3x-8,&x\leq 5\\2k,&x>5\end{cases}
\displaystyle f(5)=3(5)-8=7
\displaystyle \text{For continuity at }x=5,\ \lim_{x\to 5^-}f(x)=\lim_{x\to 5^+}f(x)=f(5)
\displaystyle \lim_{x\to 5^-}f(x)=7,\quad \lim_{x\to 5^+}f(x)=2k
\displaystyle \Rightarrow\ 2k=7
\displaystyle \Rightarrow\ k=\frac{7}{2}
\displaystyle \text{So the assertion that continuity occurs for }k=\frac{5}{2}\text{ is false}
\displaystyle \therefore\ \text{Assertion (A) is false}
\displaystyle \text{Reason (R) is true, since it states the condition for continuity at }x=a
\displaystyle \therefore\ \text{Correct option is (D)}

\displaystyle \textbf{SECTION - B} \quad 5 \times 2 = 10
\displaystyle \text{This section comprises of 5 Very Short Answer (VSA) type questions of}
\displaystyle \text{2 marks each.}

\displaystyle \textbf{21. (a) } \text{Differentiate } 2^{\cos^{2}x} \text{ w.r.t. } \cos^{2}x.
\displaystyle \textbf{OR}
\displaystyle \text{(b) If } \tan^{-1}(x^{2}+y^{2})=a^{2}, \text{ then find } \frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{Let }u=\cos^{2}x
\displaystyle \text{Then the function becomes }2^{u}
\displaystyle \frac{d}{du}(2^{u})=2^{u}\log 2
\displaystyle \therefore\ \frac{d}{d(\cos^{2}x)}\left(2^{\cos^{2}x}\right)=2^{\cos^{2}x}\log 2
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle \tan^{-1}(x^{2}+y^{2})=a^{2}
\displaystyle \text{Differentiate w.r.t. }x
\displaystyle \frac{1}{1+(x^{2}+y^{2})^{2}}\cdot(2x+2y\frac{dy}{dx})=0
\displaystyle \Rightarrow\ 2x+2y\frac{dy}{dx}=0
\displaystyle \Rightarrow\ x+y\frac{dy}{dx}=0
\displaystyle \Rightarrow\ \frac{dy}{dx}=-\frac{x}{y}

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\displaystyle \textbf{22. } \text{Evaluate : } \tan^{-1}\left[2\sin\left(2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)\right]
\displaystyle \text{Answer:}
\displaystyle \tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]
\displaystyle \text{Let }\theta=\cos^{-1}\frac{\sqrt{3}}{2}\Rightarrow \theta=\frac{\pi}{6}
\displaystyle \Rightarrow\ 2\theta=\frac{\pi}{3}
\displaystyle \Rightarrow\ \sin\left(2\theta\right)=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}
\displaystyle \Rightarrow\ 2\sin\left(2\theta\right)=2\cdot\frac{\sqrt{3}}{2}=\sqrt{3}
\displaystyle \therefore\ \tan^{-1}(\sqrt{3})=\frac{\pi}{3}
\displaystyle \therefore\ \text{Value is }\frac{\pi}{3}

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\displaystyle \textbf{23. } \text{The diagonals of a parallelogram are given by } \overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}
\displaystyle \text{and } \overrightarrow{b}=\widehat{i}+3\widehat{j}-\widehat{k}. \text{ Find the area of the parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Given diagonals }\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{b}=\widehat{i}+3\widehat{j}-\widehat{k}
\displaystyle \text{Area of parallelogram }=\frac{1}{2}\left|\overrightarrow{a}\times\overrightarrow{b}\right|
\displaystyle \overrightarrow{a}\times\overrightarrow{b}= \begin{vmatrix} \widehat{i}&\widehat{j}&\widehat{k}\\ 2&-1&1\\ 1&3&-1 \end{vmatrix}
\displaystyle =\widehat{i}\left((-1)(-1)-1\cdot 3\right)-\widehat{j}\left(2(-1)-1\cdot 1\right)+\widehat{k}\left(2\cdot 3-(-1)\cdot 1\right)
\displaystyle =\widehat{i}(1-3)-\widehat{j}(-2-1)+\widehat{k}(6+1)
\displaystyle =-2\widehat{i}+3\widehat{j}+7\widehat{k}
\displaystyle \left|\overrightarrow{a}\times\overrightarrow{b}\right|=\sqrt{(-2)^{2}+3^{2}+7^{2}}
\displaystyle =\sqrt{4+9+49}=\sqrt{62}
\displaystyle \therefore\ \text{Area}=\frac{1}{2}\sqrt{62}

\\
\displaystyle \textbf{24. } \text{Find the intervals in which function } f(x)=5x^{\frac{3}{2}}-3x^{\frac{5}{2}} \text{ is}
\displaystyle \text{(i) increasing \quad (ii) decreasing.}
\displaystyle \text{Answer:}
\displaystyle f(x)=5x^{\frac{3}{2}}-3x^{\frac{5}{2}}
\displaystyle f'(x)=5\cdot\frac{3}{2}x^{\frac{1}{2}}-3\cdot\frac{5}{2}x^{\frac{3}{2}}
\displaystyle =\frac{15}{2}x^{\frac{1}{2}}-\frac{15}{2}x^{\frac{3}{2}}
\displaystyle =\frac{15}{2}x^{\frac{1}{2}}(1-x)
\displaystyle \text{Critical points: }x=0,\ x=1
\displaystyle \text{Domain: }x\geq 0
\displaystyle \text{For }0<x<1,\ x^{\frac{1}{2}}>0,\ (1-x)>0\Rightarrow f'(x)>0
\displaystyle \Rightarrow\ f(x)\text{ is increasing on }(0,1)
\displaystyle \text{For }x>1,\ x^{\frac{1}{2}}>0,\ (1-x)<0\Rightarrow f'(x)<0
\displaystyle \Rightarrow\ f(x)\text{ is decreasing on }(1,\infty)
\displaystyle \therefore\ \text{Increasing on }(0,1),\ \text{Decreasing on }(1,\infty)

\\
\displaystyle \textbf{25. (a) } \text{Two friends while flying kites from different locations, find the}
\displaystyle \text{strings of their kites crossing each other. The strings can be represented by}
\displaystyle \overrightarrow{a}=3\widehat{i}+\widehat{j}+2\widehat{k} \text{ and } \overrightarrow{b}=2\widehat{i}-2\widehat{j}+4\widehat{k}.
\displaystyle \text{Determine the angle formed between the kite strings. Assume there is no}
\displaystyle \text{slack in the strings.}
\displaystyle \textbf{OR}
\displaystyle \text{(b) Find a vector of magnitude 21 units in the direction opposite to that of}
\displaystyle \overrightarrow{AB} \text{ where A and B are the points } A(2,1,3) \text{ and } B(8,-1,0).
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \overrightarrow{a}=3\widehat{i}+\widehat{j}+2\widehat{k},\quad \overrightarrow{b}=2\widehat{i}-2\widehat{j}+4\widehat{k}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=3(2)+1(-2)+2(4)=6-2+8=12
\displaystyle |\overrightarrow{a}|=\sqrt{9+1+4}=\sqrt{14},\quad |\overrightarrow{b}|=\sqrt{4+4+16}=\sqrt{24}=2\sqrt{6}
\displaystyle \cos\theta=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{12}{\sqrt{14}\cdot 2\sqrt{6}}=\frac{12}{2\sqrt{84}}=\frac{6}{\sqrt{84}}
\displaystyle =\frac{6}{2\sqrt{21}}=\frac{3}{\sqrt{21}}
\displaystyle \therefore\ \theta=\cos^{-1}\left(\frac{3}{\sqrt{21}}\right)
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle A(2,1,3),\ B(8,-1,0)
\displaystyle \overrightarrow{AB}=(8-2)\widehat{i}+(-1-1)\widehat{j}+(0-3)\widehat{k}=6\widehat{i}-2\widehat{j}-3\widehat{k}
\displaystyle \text{Opposite direction vector }=-\overrightarrow{AB}=-6\widehat{i}+2\widehat{j}+3\widehat{k}
\displaystyle |\overrightarrow{AB}|=\sqrt{36+4+9}=\sqrt{49}=7
\displaystyle \text{Unit vector in opposite direction }=\frac{-6\widehat{i}+2\widehat{j}+3\widehat{k}}{7}
\displaystyle \text{Required vector of magnitude }21=21\times \text{unit vector}
\displaystyle =3(-6\widehat{i}+2\widehat{j}+3\widehat{k})
\displaystyle =-18\widehat{i}+6\widehat{j}+9\widehat{k}

\displaystyle \textbf{SECTION - C } \quad 6 \times 3 = 18
\displaystyle \text{This section comprises of 6 Short Answer (SA) type questions of 3 marks}
\displaystyle \text{each.}

\displaystyle \textbf{26. } \text{The side of an equilateral triangle is increasing at the rate of } 3 \text{ cm/s. At}
\displaystyle \text{what rate is its area increasing when the side of the triangle is } 15 \text{ cm ?}
\displaystyle \text{Answer:}
\displaystyle \text{Let side }=a,\ \frac{da}{dt}=3\ \text{cm/s}
\displaystyle \text{Area of equilateral triangle }A=\frac{\sqrt{3}}{4}a^{2}
\displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{4}\cdot 2a\cdot \frac{da}{dt}
\displaystyle =\frac{\sqrt{3}}{2}a\cdot \frac{da}{dt}
\displaystyle \text{At }a=15,\ \frac{dA}{dt}=\frac{\sqrt{3}}{2}\cdot 15\cdot 3
\displaystyle =\frac{45\sqrt{3}}{2}
\displaystyle \therefore\ \text{Rate of increase of area }=\frac{45\sqrt{3}}{2}\ \text{cm}^{2}\text{/s}

\\
\displaystyle \textbf{27. } \text{Solve the following linear programming problem graphically :}
\displaystyle \text{Maximise } Z=x+2y
\displaystyle \text{Subject to the constraints :}
\displaystyle x-y\geq 0
\displaystyle x-2y\geq -2
\displaystyle x\geq 0,\ y\geq 0
\displaystyle \text{Answer:} \displaystyle \text{Given objective function }Z=x+2y
\displaystyle \text{Subject to the constraints }x-y\geq 0,\ x-2y\geq -2,\ x\geq 0,\ y\geq 0
\displaystyle \text{First convert the inequalities into boundary lines:}
\displaystyle x-y=0\Rightarrow y=x
\displaystyle x-2y=-2\Rightarrow y=\frac{x+2}{2}
\displaystyle x=0,\quad y=0
\displaystyle \text{Now determine the feasible side of each line.}
\displaystyle x-y\geq 0\Rightarrow y\leq x
\displaystyle x-2y\geq -2\Rightarrow x+2\geq 2y\Rightarrow y\leq \frac{x+2}{2}
\displaystyle x\geq 0,\ y\geq 0\Rightarrow \text{feasible region lies in the first quadrant}
\displaystyle \text{Hence the feasible region is the common region in the first quadrant below both lines }
\displaystyle y=x\text{ and }y=\frac{x+2}{2}
\displaystyle \text{Find the corner points of the feasible region.}
\displaystyle \text{Intersection of }y=x\text{ and }y=\frac{x+2}{2}:
\displaystyle x=\frac{x+2}{2}
\displaystyle \Rightarrow 2x=x+2
\displaystyle \Rightarrow x=2
\displaystyle \Rightarrow y=2
\displaystyle \therefore\ \text{point of intersection is }(2,2)
\displaystyle \text{Intersection of }y=x\text{ with }y=0:
\displaystyle y=0\Rightarrow x=0
\displaystyle \therefore\ (0,0)
\displaystyle \text{Intersection of }y=\frac{x+2}{2}\text{ with }y=0:
\displaystyle 0=\frac{x+2}{2}\Rightarrow x=-2
\displaystyle \text{But }x\geq 0,\text{ so this point is not in the feasible region}
\displaystyle \text{On the }x\text{-axis, the feasible region starts from }(0,0)\text{ and extends to the right.}
\displaystyle \text{Thus the feasible region is unbounded.}
\displaystyle \text{Now evaluate }Z=x+2y\text{ at the corner points.}
\displaystyle Z(0,0)=0+2(0)=0
\displaystyle Z(2,2)=2+2(2)=6
\displaystyle \text{But since the feasible region is unbounded, check whether }Z\text{ can increase further.}
\displaystyle \text{Take any point on the ray }y=x\text{ with }x\geq 2
\displaystyle \text{Then }Z=x+2x=3x
\displaystyle \text{As }x\to \infty,\ Z\to \infty
\displaystyle \therefore\ Z\text{ has no maximum value}
\displaystyle \text{Also, from the corner point values, the least value is at }(0,0)
\displaystyle \therefore\ \text{Minimum value of }Z=0\text{ at }(0,0)
\displaystyle \therefore\ \text{the L.P.P. has no maximum value, since the feasible region} \\ \text{is unbounded in the direction of increase of }Z

\\
\displaystyle \textbf{28. (a) } \text{Find : } \int \frac{x+\sin x}{1+\cos x}\,dx
\displaystyle \textbf{OR}
\displaystyle \text{(b) Evaluate : } \int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^{3}x\sqrt{2\sin 2x}}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle I=\int \frac{x+\sin x}{1+\cos x}\,dx
\displaystyle =\int \frac{x}{1+\cos x}\,dx+\int \frac{\sin x}{1+\cos x}\,dx
\displaystyle =\int \frac{x}{2\cos^{2}\frac{x}{2}}\,dx+\int \frac{\sin x}{1+\cos x}\,dx
\displaystyle =\frac{1}{2}\int x\sec^{2}\frac{x}{2}\,dx+\int \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^{2}\frac{x}{2}}\,dx
\displaystyle =\frac{1}{2}\int x\sec^{2}\frac{x}{2}\,dx+\int \tan\frac{x}{2}\,dx
\displaystyle \text{For }\frac{1}{2}\int x\sec^{2}\frac{x}{2}\,dx,\ \text{use integration by parts}
\displaystyle u=x,\ dv=\frac{1}{2}\sec^{2}\frac{x}{2}\,dx
\displaystyle \Rightarrow\ du=dx,\ v=\tan\frac{x}{2}
\displaystyle \therefore\ \frac{1}{2}\int x\sec^{2}\frac{x}{2}\,dx=x\tan\frac{x}{2}-\int \tan\frac{x}{2}\,dx
\displaystyle \text{Hence }I=x\tan\frac{x}{2}-\int \tan\frac{x}{2}\,dx+\int \tan\frac{x}{2}\,dx
\displaystyle \therefore\ I=x\tan\frac{x}{2}+C
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{dx}{\cos^{3}x\sqrt{2\sin 2x}}
\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{dx}{\cos^{3}x\sqrt{4\sin x\cos x}}
\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{dx}{2\cos^{3}x\sqrt{\sin x\cos x}}
\displaystyle =\frac{1}{2}\int_{0}^{\frac{\pi}{4}}\frac{dx}{\cos^{2}x\sqrt{\sin x}\sqrt{\cos x}}
\displaystyle =\frac{1}{2}\int_{0}^{\frac{\pi}{4}}\frac{\sec^{2}x}{\sqrt{\tan x}}\,dx
\displaystyle \text{Let }t=\tan x\Rightarrow dt=\sec^{2}x\,dx
\displaystyle \text{When }x=0,\ t=0;\quad \text{when }x=\frac{\pi}{4},\ t=1
\displaystyle \therefore\ I=\frac{1}{2}\int_{0}^{1}t^{-\frac{1}{2}}\,dt
\displaystyle =\frac{1}{2}\left[2t^{\frac{1}{2}}\right]_{0}^{1}
\displaystyle =\left[\sqrt{t}\right]_{0}^{1}
\displaystyle =1

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\displaystyle \textbf{29. (a) } \text{Verify that lines given by } \overrightarrow{r}=(1-\lambda)\widehat{i}+(\lambda-2)\widehat{j}+(3-2\lambda)\widehat{k} \text{ and}
\displaystyle \overrightarrow{r}=(\mu+1)\widehat{i}+(2\mu-1)\widehat{j}-(2\mu+1)\widehat{k} \text{ are skew lines. Hence, find}
\displaystyle \text{shortest distance between the lines.}
\displaystyle \textbf{OR}
\displaystyle \text{(b) During a cricket match, the position of the bowler, the wicket keeper}
\displaystyle \text{and the leg slip fielder are in a line given by } \overrightarrow{B}=2\widehat{i}+8\widehat{j},
\displaystyle \overrightarrow{W}=6\widehat{i}+12\widehat{j} \text{ and } \overrightarrow{F}=12\widehat{i}+18\widehat{j} \text{ respectively. Calculate the ratio}
\displaystyle \text{in which the wicketkeeper divides the line segment joining the bowler and the}
\displaystyle \text{leg slip fielder.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{Line }L_{1}:\ \overrightarrow{r}=(1-\lambda)\widehat{i}+(\lambda-2)\widehat{j}+(3-2\lambda)\widehat{k}
\displaystyle \text{A point on }L_{1}\text{ is }A(1,-2,3)\text{ and its direction ratios are }(-1,1,-2)
\displaystyle \text{Line }L_{2}:\ \overrightarrow{r}=(\mu+1)\widehat{i}+(2\mu-1)\widehat{j}-(2\mu+1)\widehat{k}
\displaystyle \text{A point on }L_{2}\text{ is }B(1,-1,-1)\text{ and its direction ratios are }(1,2,-2)
\displaystyle \text{Since }(-1,1,-2)\text{ and }(1,2,-2)\text{ are not proportional, the lines are not parallel}
\displaystyle \text{Now check whether they intersect}
\displaystyle 1-\lambda=\mu+1,\quad \lambda-2=2\mu-1,\quad 3-2\lambda=-(2\mu+1)
\displaystyle \Rightarrow\ -\lambda=\mu,\quad \lambda-2=2\mu-1,\quad 3-2\lambda=-2\mu-1
\displaystyle \text{From }\mu=-\lambda,\ \lambda-2=-2\lambda-1
\displaystyle \Rightarrow\ 3\lambda=1\Rightarrow \lambda=\frac{1}{3},\ \mu=-\frac{1}{3}
\displaystyle \text{Now in third equation, }3-2\cdot\frac{1}{3}=\frac{7}{3}\text{ but }-2\left(-\frac{1}{3}\right)-1=-\frac{1}{3}
\displaystyle \text{which is not equal}
\displaystyle \therefore\ \text{the lines do not intersect}
\displaystyle \text{Hence, the given lines are skew lines}
\displaystyle \text{Shortest distance }d=\frac{|(\overrightarrow{B}-\overrightarrow{A})\cdot(\overrightarrow{d_{1}}\times \overrightarrow{d_{2}})|}{|\overrightarrow{d_{1}}\times \overrightarrow{d_{2}}|}
\displaystyle \overrightarrow{d_{1}}=-\widehat{i}+\widehat{j}-2\widehat{k},\quad \overrightarrow{d_{2}}=\widehat{i}+2\widehat{j}-2\widehat{k},\quad \overrightarrow{B}-\overrightarrow{A}=0\widehat{i}+\widehat{j}-4\widehat{k}
\displaystyle \overrightarrow{d_{1}}\times \overrightarrow{d_{2}}= \begin{vmatrix} \widehat{i}&\widehat{j}&\widehat{k}\\ -1&1&-2\\ 1&2&-2 \end{vmatrix}
\displaystyle =\widehat{i}\left(1(-2)-(-2)(2)\right)-\widehat{j}\left((-1)(-2)-(-2)(1)\right)+\widehat{k}\left((-1)(2)-1(1)\right)
\displaystyle =2\widehat{i}-4\widehat{j}-3\widehat{k}
\displaystyle |\overrightarrow{d_{1}}\times \overrightarrow{d_{2}}|=\sqrt{2^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{29}
\displaystyle (\overrightarrow{B}-\overrightarrow{A})\cdot(\overrightarrow{d_{1}}\times \overrightarrow{d_{2}})=(0,1,-4)\cdot(2,-4,-3)
\displaystyle =0-4+12=8
\displaystyle \therefore\ d=\frac{8}{\sqrt{29}}
\displaystyle \therefore\ \text{Shortest distance between the lines is }\frac{8}{\sqrt{29}}
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle \overrightarrow{B}=2\widehat{i}+8\widehat{j},\quad \overrightarrow{W}=6\widehat{i}+12\widehat{j},\quad \overrightarrow{F}=12\widehat{i}+18\widehat{j}
\displaystyle \text{Let wicketkeeper }W\text{ divide }BF\text{ in the ratio }m:n
\displaystyle \text{Then by section formula, }\overrightarrow{W}=\frac{m\overrightarrow{F}+n\overrightarrow{B}}{m+n}
\displaystyle \Rightarrow\ (6,12)=\left(\frac{12m+2n}{m+n},\frac{18m+8n}{m+n}\right)
\displaystyle \text{From }x\text{-coordinates, }6m+6n=12m+2n
\displaystyle \Rightarrow\ 4n=6m\Rightarrow 2n=3m
\displaystyle \Rightarrow\ m:n=2:3
\displaystyle \therefore\ \text{the wicketkeeper divides the line segment joining the bowler and leg slip fielder in the ratio }2:3

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\displaystyle \textbf{30. (a) } \text{The probability distribution for the number of students being absent}
\displaystyle \text{in a class on a Saturday is as follows :}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline X & 0 & 2 & 4 & 5 \\\hline P(X) & p & 2p & 3p & p \\\hline \end{array}
\displaystyle \text{Where } X \text{ is the number of students absent.}
\displaystyle \text{(i) Calculate } p.
\displaystyle \text{(ii) Calculate the mean of the number of absent students on Saturday.}
\displaystyle \textbf{OR}
\displaystyle \text{(b) For the vacancy advertised in the newspaper, } 3000 \text{ candidates}
\displaystyle \text{submitted their applications. From the data it was revealed that two}
\displaystyle \text{third of the total applicants were females and other were males. The}
\displaystyle \text{selection for the job was done through a written test. The performance of}
\displaystyle \text{the applicants indicates that the probability of a male getting a distinction}
\displaystyle \text{in written test is } 0.4 \text{ and that a female getting a distinction is } 0.35. \text{ Find the}
\displaystyle \text{probability that the candidate chosen at random will have a distinction in}
\displaystyle \text{the written test.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle X:0,2,4,5;\quad P(X):p,2p,3p,p
\displaystyle \text{(i) }p+2p+3p+p=1
\displaystyle 7p=1\Rightarrow p=\frac{1}{7}
\displaystyle \text{(ii) Mean }E(X)=\sum xP(x)
\displaystyle =0\cdot p+2\cdot 2p+4\cdot 3p+5\cdot p
\displaystyle =4p+12p+5p=21p
\displaystyle =21\cdot\frac{1}{7}=3
\displaystyle \therefore\ \text{Mean }=3
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle P(F)=\frac{2}{3},\quad P(M)=\frac{1}{3}
\displaystyle P(D\mid M)=0.4,\quad P(D\mid F)=0.35
\displaystyle \text{Required }P(D)=P(M)P(D\mid M)+P(F)P(D\mid F)
\displaystyle =\frac{1}{3}\cdot 0.4+\frac{2}{3}\cdot 0.35
\displaystyle =\frac{0.4}{3}+\frac{0.7}{3}=\frac{1.1}{3}
\displaystyle =\frac{11}{30}
\displaystyle \therefore\ \text{Probability }=\frac{11}{30}

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\displaystyle \textbf{31. } \text{Sketch the graph of } y=|x+3| \text{ and find the area of the region enclosed}
\displaystyle \text{by the curve, } x\text{-axis, \ between } x=-6 \text{ and } x=0, \text{ using integration.}
\displaystyle \text{Answer:}
\displaystyle y=|x+3|=\begin{cases}-(x+3),&x<-3\\x+3,&x\geq -3\end{cases}\displaystyle \text{The graph is V-shaped with vertex at }(-3,0)
\displaystyle \text{Required area }=\int_{-6}^{-3}|x+3|\,dx+\int_{-3}^{0}|x+3|\,dx
\displaystyle =\int_{-6}^{-3}-(x+3)\,dx+\int_{-3}^{0}(x+3)\,dx
\displaystyle =\left[-\frac{x^{2}}{2}-3x\right]_{-6}^{-3}+\left[\frac{x^{2}}{2}+3x\right]_{-3}^{0}
\displaystyle =\left(\frac{9}{2}\right)+\left(\frac{9}{2}\right)
\displaystyle =9
\displaystyle \therefore\ \text{The required area is }9\text{ square units}

\displaystyle \textbf{SECTION - D} \quad 4 \times 5 = 20
\displaystyle \text{This section comprises of 4 Long Answer (LA) type questions of 5 marks}
\displaystyle \text{each.}

\displaystyle \textbf{32. (a) } \text{If } \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y), \text{ then prove that } \frac{dy}{dx}=
\displaystyle \sqrt{\frac{1-y^{2}}{1-x^{2}}}.
\displaystyle \textbf{OR}
\displaystyle \text{(b) If } x=a\left(\cos\theta+\log\tan\frac{\theta}{2}\right) \text{ and } y=\sin\theta, \text{ then find } \frac{d^{2}y}{dx^{2}} \text{ at } \theta=\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)
\displaystyle \text{Differentiating both sides w.r.t. }x,
\displaystyle \frac{-x}{\sqrt{1-x^{2}}}+\frac{-y}{\sqrt{1-y^{2}}}\frac{dy}{dx}=a\left(1-\frac{dy}{dx}\right)
\displaystyle \Rightarrow\ a+\frac{x}{\sqrt{1-x^{2}}}=\left(a-\frac{y}{\sqrt{1-y^{2}}}\right)\frac{dy}{dx}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{a+\frac{x}{\sqrt{1-x^{2}}}}{a-\frac{y}{\sqrt{1-y^{2}}}}
\displaystyle \text{From the given equation, }a=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}
\displaystyle \text{Substituting this in the numerator,}
\displaystyle a+\frac{x}{\sqrt{1-x^{2}}}=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}+\frac{x}{\sqrt{1-x^{2}}}
\displaystyle =\frac{(1-x^{2})+\sqrt{(1-x^{2})(1-y^{2})}+x(x-y)}{(x-y)\sqrt{1-x^{2}}}
\displaystyle =\frac{1-xy+\sqrt{(1-x^{2})(1-y^{2})}}{(x-y)\sqrt{1-x^{2}}}
\displaystyle \text{Similarly,}
\displaystyle a-\frac{y}{\sqrt{1-y^{2}}}=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}-\frac{y}{\sqrt{1-y^{2}}}
\displaystyle =\frac{\sqrt{(1-x^{2})(1-y^{2})}+(1-y^{2})-y(x-y)}{(x-y)\sqrt{1-y^{2}}}
\displaystyle =\frac{1-xy+\sqrt{(1-x^{2})(1-y^{2})}}{(x-y)\sqrt{1-y^{2}}}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{\frac{1-xy+\sqrt{(1-x^{2})(1-y^{2})}}{(x-y)\sqrt{1-x^{2}}}}{\frac{1-xy+\sqrt{(1-x^{2})(1-y^{2})}}{(x-y)\sqrt{1-y^{2}}}}
\displaystyle =\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}
\displaystyle =\sqrt{\frac{1-y^{2}}{1-x^{2}}}
\displaystyle \therefore\ \frac{dy}{dx}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle x=a\left(\cos\theta+\log\tan\frac{\theta}{2}\right),\quad y=\sin\theta
\displaystyle \frac{dx}{d\theta}=a\left(-\sin\theta+\frac{d}{d\theta}\log\tan\frac{\theta}{2}\right)
\displaystyle \text{Now, }\frac{d}{d\theta}\log\tan\frac{\theta}{2}=\frac{1}{\sin\theta}
\displaystyle \therefore\ \frac{dx}{d\theta}=a\left(-\sin\theta+\frac{1}{\sin\theta}\right)=a\frac{1-\sin^{2}\theta}{\sin\theta}=a\frac{\cos^{2}\theta}{\sin\theta}
\displaystyle \frac{dy}{d\theta}=\cos\theta
\displaystyle \therefore\ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\cos\theta}{a\cos^{2}\theta/\sin\theta}=\frac{\sin\theta}{a\cos\theta}=\frac{\tan\theta}{a}
\displaystyle \frac{d}{d\theta}\left(\frac{dy}{dx}\right)=\frac{1}{a}\sec^{2}\theta
\displaystyle \therefore\ \frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}=\frac{\frac{1}{a}\sec^{2}\theta}{a\cos^{2}\theta/\sin\theta}
\displaystyle =\frac{1}{a^{2}}\cdot\frac{1}{\cos^{2}\theta}\cdot\frac{\sin\theta}{\cos^{2}\theta}=\frac{\sin\theta}{a^{2}\cos^{4}\theta}
\displaystyle \text{At }\theta=\frac{\pi}{4},\ \sin\theta=\frac{1}{\sqrt{2}},\ \cos\theta=\frac{1}{\sqrt{2}}
\displaystyle \therefore\ \left.\frac{d^{2}y}{dx^{2}}\right|_{\theta=\frac{\pi}{4}}=\frac{\frac{1}{\sqrt{2}}}{a^{2}\left(\frac{1}{\sqrt{2}}\right)^{4}}
\displaystyle =\frac{\frac{1}{\sqrt{2}}}{a^{2}\cdot\frac{1}{4}}
\displaystyle =\frac{4}{a^{2}\sqrt{2}}=\frac{2\sqrt{2}}{a^{2}}
\displaystyle \therefore\ \left.\frac{d^{2}y}{dx^{2}}\right|_{\theta=\frac{\pi}{4}}=\frac{2\sqrt{2}}{a^{2}}

\\
\displaystyle \textbf{33. } \text{Find the absolute maximum and absolute minimum of function}
\displaystyle f(x)=2x^{3}-15x^{2}+36x+1 \text{ on } [1,5].
\displaystyle \text{Answer:}
\displaystyle f(x)=2x^{3}-15x^{2}+36x+1
\displaystyle f'(x)=6x^{2}-30x+36
\displaystyle =6(x^{2}-5x+6)
\displaystyle =6(x-2)(x-3)
\displaystyle \text{Critical points in }[1,5]\text{ are }x=2,\ 3
\displaystyle f(1)=2-15+36+1=24
\displaystyle f(2)=16-60+72+1=29
\displaystyle f(3)=54-135+108+1=28
\displaystyle f(5)=250-375+180+1=56
\displaystyle \text{Comparing these values, the absolute maximum value is }56\text{ at }x=5
\displaystyle \text{and the absolute minimum value is }24\text{ at }x=1
\displaystyle \therefore\ \text{Absolute maximum }=56,\ \text{Absolute minimum }=24

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\displaystyle \textbf{34. (a) } \text{Find the image } A' \text{ of the point } A(1,6,3) \text{ in the line } \frac{x}{1}=
\displaystyle \frac{y-1}{2}=\frac{z-2}{3}. \text{ Also, find the equation of the line joining } A \text{ and } A'.
\displaystyle \textbf{OR}
\displaystyle \text{(b) Find a point } P \text{ on the line } \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} \text{ such that its distance}
\displaystyle \text{from point } Q(2,4,-1) \text{ is } 7 \text{ units. Also, find the equation of line joining}
\displaystyle P \text{ and } Q.
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{Given line: }\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=t
\displaystyle \Rightarrow x=t,\ y=2t+1,\ z=3t+2
\displaystyle \text{Let foot of perpendicular from }A(1,6,3)\text{ to the line be }P(t,2t+1,3t+2)
\displaystyle \overrightarrow{AP}\cdot \overrightarrow{d}=0,\ \text{where }\overrightarrow{d}=(1,2,3)
\displaystyle (t-1,2t+1-6,3t+2-3)\cdot (1,2,3)=0
\displaystyle (t-1)+(2t-5)\cdot 2+(3t-1)\cdot 3=0
\displaystyle t-1+4t-10+9t-3=0
\displaystyle 14t-14=0\Rightarrow t=1
\displaystyle \therefore P(1,3,5)
\displaystyle \text{Image }A'\text{ satisfies }P\text{ is midpoint of }AA'
\displaystyle \Rightarrow A'=2P-A=(2,6,10)-(1,6,3)=(1,0,7)
\displaystyle \text{Equation of line }AA':\ \overrightarrow{r}=(1,6,3)+\lambda(0,-6,4)
\displaystyle \textbf{OR}
\displaystyle \textbf{(b)}
\displaystyle \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=t
\displaystyle \Rightarrow P(t-5,4t-3,-9t+6)
\displaystyle \text{Distance }PQ=7,\ Q(2,4,-1)
\displaystyle (t-5-2)^{2}+(4t-3-4)^{2}+(-9t+6+1)^{2}=49
\displaystyle (t-7)^{2}+(4t-7)^{2}+(-9t+7)^{2}=49
\displaystyle t^{2}-14t+49+16t^{2}-56t+49+81t^{2}-126t+49=49
\displaystyle 98t^{2}-196t+147=49
\displaystyle 98t^{2}-196t+98=0
\displaystyle t^{2}-2t+1=0\Rightarrow (t-1)^{2}=0\Rightarrow t=1
\displaystyle \therefore P(-4,1,-3)
\displaystyle \text{Line }PQ:\ \overrightarrow{r}=(2,4,-1)+\lambda(-6,-3,-2)

\\
\displaystyle \textbf{35. } \text{A school wants to allocate students into three clubs : Sports, Music and}
\displaystyle \text{Drama, under following conditions :}
\displaystyle \bullet \ \text{The number of students in Sports club should be equal to the sum of}
\displaystyle \text{the number of students in Music and Drama club.}
\displaystyle \bullet \ \text{The number of students in Music club should be } 20 \text{ more than half}
\displaystyle \text{the number of students in Sports club.}
\displaystyle \bullet \ \text{The total number of students to be allocated in all three clubs are}
\displaystyle 180.
\displaystyle \text{Find the number of students allocated to different clubs, using matrix}
\displaystyle \text{method.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of students in Sports, Music and Drama clubs be }S,\ M,\ D\text{ respectively}
\displaystyle \text{According to the conditions,}
\displaystyle S=M+D
\displaystyle M=\frac{S}{2}+20
\displaystyle S+M+D=180
\displaystyle \text{Rewriting,}
\displaystyle S-M-D=0
\displaystyle -S+2M=40
\displaystyle S+M+D=180
\displaystyle \text{Matrix form is}
\displaystyle \begin{pmatrix}1&-1&-1\\-1&2&0\\1&1&1\end{pmatrix}\begin{pmatrix}S\\M\\D\end{pmatrix}=\begin{pmatrix}0\\40\\180\end{pmatrix}
\displaystyle \text{Now, from }S=M+D\text{ and }S+M+D=180
\displaystyle \Rightarrow\ S+S=180
\displaystyle \Rightarrow\ 2S=180
\displaystyle \Rightarrow\ S=90
\displaystyle \text{Then }M=\frac{90}{2}+20=45+20=65
\displaystyle \text{Also }D=S-M=90-65=25
\displaystyle \therefore\ \text{Number of students in Sports club }=90
\displaystyle \therefore\ \text{Number of students in Music club }=65
\displaystyle \therefore\ \text{Number of students in Drama club }=25

\displaystyle \textbf{SECTION - E} \quad 3 \times 4 = 12
\displaystyle \text{This section comprises of 3 case study based questions of 4 marks each.}

\displaystyle \textbf{36. } \text{A technical company is designing a rectangular solar panel installation}
\displaystyle \text{on a roof using } 300 \text{ metres of boundary material. The design includes a}
\displaystyle \text{partition running parallel to one of the sides dividing the area (roof) into}
\displaystyle \text{two sections.}
\displaystyle \text{Let the length of the side perpendicular to the partition be } x \text{ metres and}
\displaystyle \text{with parallel to the partition be } y \text{ metres.} \displaystyle \text{Based on this information, answer the following questions :}
\displaystyle \text{(i) Write the equation for the total boundary material used in the}
\displaystyle \text{boundary and parallel to the partition in terms of } x \text{ and } y.
\displaystyle \text{(ii) Write the area of the solar panel as a function of } x.
\displaystyle \text{(iii) (a) Find the critical points of the area function. Use second}
\displaystyle \text{derivative test to determine critical points at the maximum area. Also,}
\displaystyle \text{find the maximum area.}
\displaystyle \textbf{OR}
\displaystyle \text{(iii) (b) Using first derivative test, calculate the maximum area the}
\displaystyle \text{company can enclose with the } 300 \text{ metres of boundary material,}
\displaystyle \text{considering the parallel partition.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{Since the partition is parallel to the side of length }y,\text{ its length is also }y
\displaystyle \text{Hence total boundary material }=2x+2y+y
\displaystyle \therefore\ 2x+3y=300
\displaystyle \textbf{(ii)}
\displaystyle \text{Area of the solar panel }A=xy
\displaystyle \text{From }2x+3y=300,\ \ y=\frac{300-2x}{3}
\displaystyle \therefore\ A(x)=x\left(\frac{300-2x}{3}\right)
\displaystyle \therefore\ A(x)=100x-\frac{2}{3}x^{2}
\displaystyle \textbf{(iii)(a)}
\displaystyle A(x)=100x-\frac{2}{3}x^{2}
\displaystyle A'(x)=100-\frac{4}{3}x
\displaystyle \text{For critical points, }A'(x)=0
\displaystyle \Rightarrow\ 100-\frac{4}{3}x=0
\displaystyle \Rightarrow\ \frac{4}{3}x=100
\displaystyle \Rightarrow\ x=75
\displaystyle \therefore\ \text{Critical point is }x=75
\displaystyle A''(x)=-\frac{4}{3}
\displaystyle \text{Since }A''(75)<0,\text{ area is maximum at }x=75
\displaystyle \text{Now }y=\frac{300-2(75)}{3}=\frac{150}{3}=50
\displaystyle \therefore\ \text{Maximum area }=xy=75\times 50=3750\ \text{m}^{2}
\displaystyle \textbf{OR}
\displaystyle \textbf{(iii)(b)}
\displaystyle A'(x)=100-\frac{4}{3}x
\displaystyle \text{If }x<75,\text{ then }A'(x)>0\Rightarrow A(x)\text{ is increasing}
\displaystyle \text{If }x>75,\text{ then }A'(x)<0\Rightarrow A(x)\text{ is decreasing}
\displaystyle \therefore\ A(x)\text{ attains maximum at }x=75
\displaystyle \text{Then }y=\frac{300-2(75)}{3}=50
\displaystyle \therefore\ \text{Maximum area }=75\times 50=3750\ \text{m}^{2}

\\
\displaystyle \textbf{37. } \text{A class-room teacher is keen to assess the learning of her students the}
\displaystyle \text{concept of ``relations'' taught to them. She writes the following five}
\displaystyle \text{relations each defined on the set } A=\{1,2,3\} :
\displaystyle R_{1}=\{(2,3),(3,2)\}
\displaystyle R_{2}=\{(1,2),(1,3),(3,2)\}
\displaystyle R_{3}=\{(1,2),(2,1),(1,1)\}
\displaystyle R_{4}=\{(1,1),(1,2),(3,3),(2,2)\}
\displaystyle R_{5}=\{(1,1),(1,2),(3,3),(2,2),(2,1),(2,3),(3,2)\}
\displaystyle \text{The students are asked to answer the following questions about the above}
\displaystyle \text{relations :}
\displaystyle \text{(i) Identify the relation which is reflexive, transitive but not symmetric.}
\displaystyle \text{(ii) Identify the relation which is reflexive and symmetric but not}
\displaystyle \text{transitive.}
\displaystyle \text{(iii) (a) Identify the relations which are symmetric but neither reflexive}
\displaystyle \text{nor transitive.}
\displaystyle \textbf{OR}
\displaystyle \text{(iii) (b) What pairs should be added to the relation } R_{2} \text{ to make it an}
\displaystyle \text{equivalence relation ?}
\displaystyle \text{Answer:}
\displaystyle A=\{1,2,3\}
\displaystyle R_{1}=\{(2,3),(3,2)\}
\displaystyle R_{2}=\{(1,2),(1,3),(3,2)\}
\displaystyle R_{3}=\{(1,2),(2,1),(1,1)\}
\displaystyle R_{4}=\{(1,1),(1,2),(3,3),(2,2)\}
\displaystyle R_{5}=\{(1,1),(1,2),(3,3),(2,2),(2,1),(2,3),(3,2)\}
\displaystyle \textbf{(i) } \text{Relation which is reflexive, transitive but not symmetric:}
\displaystyle R_{4}\text{ is reflexive since }(1,1),(2,2),(3,3)\in R_{4}
\displaystyle R_{4}\text{ is not symmetric since }(1,2)\in R_{4}\text{ but }(2,1)\notin R_{4}
\displaystyle R_{4}\text{ is transitive}
\displaystyle \therefore\ \text{Required relation is }R_{4}
\displaystyle \textbf{(ii) } \text{Relation which is reflexive and symmetric but not transitive:}
\displaystyle R_{5}\text{ is reflexive since }(1,1),(2,2),(3,3)\in R_{5}
\displaystyle R_{5}\text{ is symmetric since }(1,2),(2,1)\in R_{5}\text{ and }(2,3),(3,2)\in R_{5}
\displaystyle R_{5}\text{ is not transitive since }(1,2)\in R_{5}\text{ and }(2,3)\in R_{5}\text{ but }(1,3)\notin R_{5}
\displaystyle \therefore\ \text{Required relation is }R_{5}
\displaystyle \textbf{(iii)(a) } \text{Relations which are symmetric but neither reflexive nor transitive:}
\displaystyle R_{1}\text{ is symmetric, but not reflexive since }(1,1),(2,2),(3,3)\notin R_{1}
\displaystyle R_{1}\text{ is not transitive since }(2,3)\in R_{1}\text{ and }(3,2)\in R_{1}\text{ but }(2,2)\notin R_{1}
\displaystyle R_{3}\text{ is symmetric, but not reflexive since }(2,2)\notin R_{3}\text{ and }(3,3)\notin R_{3}
\displaystyle R_{3}\text{ is not transitive since }(2,1)\in R_{3}\text{ and }(1,2)\in R_{3}\text{ but }(2,2)\notin R_{3}
\displaystyle \therefore\ \text{The required relations are }R_{1}\text{ and }R_{3}
\displaystyle \textbf{OR}
\displaystyle \textbf{(iii)(b) } \text{Pairs to be added to }R_{2}\text{ to make it an equivalence relation:}
\displaystyle R_{2}=\{(1,2),(1,3),(3,2)\}
\displaystyle \text{For reflexivity, add }(1,1),(2,2),(3,3)
\displaystyle \text{For symmetry, add }(2,1),(3,1),(2,3)
\displaystyle \text{Then the relation becomes }A\times A,\text{ which is an equivalence relation}
\displaystyle \therefore\ \text{Pairs to be added are }\{(1,1),(2,2),(3,3),(2,1),(3,1),(2,3)\}

\\
\displaystyle \textbf{38. } \text{A bank offers loan to its customers on different types of interest namely,}
\displaystyle \text{fixed rate, floating rate and variable rate. From the past data with the}
\displaystyle \text{bank, it is known that a customer avails loan on fixed rate, floating rate}
\displaystyle \text{or variable rate with probabilities } 10\%, 20\% \text{ and } 70\% \text{ respectively. A}
\displaystyle \text{customer after availing loan can pay the loan or default on loan}
\displaystyle \text{repayment. The bank data suggests that the probability that a person}
\displaystyle \text{defaults on loan after availing it at fixed rate, floating rate and variable}
\displaystyle \text{rate is } 5\%, 3\% \text{ and } 1\% \text{ respectively.} \displaystyle \text{Based on the above information, answer the following :}
\displaystyle \text{(i) What is the probability that a customer after availing the loan will}
\displaystyle \text{default on the loan repayment ?}
\displaystyle \text{(ii) A customer after availing the loan, defaults on loan repayment.}
\displaystyle \text{What is the probability that he availed the loan at a variable rate of}
\displaystyle \text{interest ?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }F,\ L,\ V\text{ denote fixed, floating and variable rate respectively}
\displaystyle P(F)=0.1,\quad P(L)=0.2,\quad P(V)=0.7
\displaystyle P(D\mid F)=0.05,\quad P(D\mid L)=0.03,\quad P(D\mid V)=0.01
\displaystyle \textbf{(i)}
\displaystyle P(D)=P(F)P(D\mid F)+P(L)P(D\mid L)+P(V)P(D\mid V)
\displaystyle =0.1\times 0.05+0.2\times 0.03+0.7\times 0.01
\displaystyle =0.005+0.006+0.007
\displaystyle =0.018
\displaystyle \therefore\ \text{Probability of default }=0.018
\displaystyle \textbf{(ii)}
\displaystyle P(V\mid D)=\frac{P(V)P(D\mid V)}{P(D)}
\displaystyle =\frac{0.7\times 0.01}{0.018}
\displaystyle =\frac{0.007}{0.018}=\frac{7}{18}
\displaystyle \therefore\ \text{Required probability }=\frac{7}{18}

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