\displaystyle \textbf{MATHEMATICS} \ \

\displaystyle \textit{Maximum Marks: 80} \ \
\displaystyle \textit{Time Allowed: Three Hours} \ \
\displaystyle \text{(Candidates are allowed additional 15 minutes for only reading the paper.} \ \
\displaystyle \text{They must NOT start writing during this time.)} \ \

\displaystyle \text{This Question Paper consists of three sections A, B and C.} \ \
\displaystyle \text{Candidates are required to attempt all questions from Section A and all questions} \ \
\displaystyle \text{EITHER from Section B OR Section C.} \ \
\displaystyle \text{Section A: Internal choice has been provided in two questions of two marks each, two questions} \ \
\displaystyle \text{of four marks each and two questions of six marks each.} \ \
\displaystyle \text{Section B: Internal choice has been provided in one question of two marks and} \ \
\displaystyle \text{one question of four marks.} \ \
\displaystyle \text{Section C: Internal choice has been provided in one question of two marks and} \ \
\displaystyle \text{one question of four marks.} \ \
\displaystyle \text{All working, including rough work, should be done on the same sheet as, and adjacent to the rest} \ \
\displaystyle \text{of the answer.} \ \
\displaystyle \text{The intended marks for questions or parts of questions are given in brackets [ ].} \ \
\displaystyle \text{Mathematical tables and graph papers are provided.} \ \

\displaystyle \textbf{SECTION A - 65 MARKS} \ \

\displaystyle \textbf{Question 1} \ \
\displaystyle \text{In subparts (i) to (xi) choose the correct options and in subparts (xii) to (xv), answer the} \ \
\displaystyle \text{questions as instructed.} \ \

\displaystyle \textbf{(i) } \text{If }A=\begin{bmatrix}0&a\\0&0\end{bmatrix}\text{, then }A^{16}\text{ is:} \ \
\displaystyle \text{(a) Unit matrix} \ \
\displaystyle \text{(b) Null matrix} \ \
\displaystyle \text{(c) Diagonal matrix} \ \
\displaystyle \text{(d) Skew matrix} \ \
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}0&a\\0&0\end{bmatrix}
\displaystyle A^{2}=\begin{bmatrix}0&a\\0&0\end{bmatrix}\begin{bmatrix}0&a\\0&0\end{bmatrix}
\displaystyle \phantom{A^{2}}=\begin{bmatrix}0\cdot 0+a\cdot 0,\ 0\cdot a+a\cdot 0\\0\cdot 0+0\cdot 0,\ 0\cdot a+0\cdot 0\end{bmatrix}
\displaystyle \phantom{A^{2}}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \therefore\ A^{2}=O
\displaystyle \text{Hence, }A^{16}=A^{2}\cdot A^{2}\cdot A^{2}\cdot A^{2}\cdot A^{2}\cdot A^{2}\cdot A^{2}\cdot A^{2}=O
\displaystyle \therefore\ A^{16}\text{ is a null matrix.}
\displaystyle \text{Hence, the correct option is (b).}
\\
\displaystyle \textbf{(ii) } \text{Which of the following is a homogenous differential equation?} \ \
\displaystyle \text{(a) }(4x^{2}+6y+5)\,dy-(3y^{2}+2x+4)\,dx=0 \ \
\displaystyle \text{(b) }(xy)\,dx-(x^{3}+y^{3})\,dy=0 \ \
\displaystyle \text{(c) }(x^{3}+2y^{2})\,dx+2xy\,dy=0 \ \
\displaystyle \text{(d) }y^{2}\,dx+(x^{2}-xy-y^{2})\,dy=0 \ \
\displaystyle \text{Answer:}
\displaystyle \text{A differential equation }M(x,y)\,dx+N(x,y)\,dy=0\text{ is homogeneous if } \\ M\text{ and }N\text{ are homogeneous functions of the same degree.}
\displaystyle \text{(a) }M=-(3y^{2}+2x+4),\ N=(4x^{2}+6y+5)
\displaystyle \text{Degrees in }M:\ 2,1,0\ \text{(not same)};\ \text{in }N:\ 2,1,0\ \text{(not same)}
\displaystyle \therefore\ \text{Not homogeneous}
\displaystyle \text{(b) }M=xy,\ N=-(x^{3}+y^{3})
\displaystyle \deg(M)=2,\ \deg(N)=3\ \Rightarrow\ \text{Not homogeneous}
\displaystyle \text{(c) }M=x^{3}+2y^{2},\ N=2xy
\displaystyle \deg(M):\ 3,2\ \text{(not same)},\ \deg(N)=2\ \Rightarrow\ \text{Not homogeneous}
\displaystyle \text{(d) }M=y^{2},\ N=x^{2}-xy-y^{2}
\displaystyle \deg(M)=2,\ \deg(N)=2\ \text{(all terms of same degree)}
\displaystyle \therefore\ \text{Homogeneous differential equation}
\displaystyle \text{Hence, the correct option is (d).}
\\
\displaystyle \textbf{(iii) } \text{Consider the graph of the function }f(x)\text{ shown below:} \ \ \displaystyle \textbf{Statement 1: The function }f(x)\text{ is increasing in }\left(\frac{1}{2},2\right). \ \
\displaystyle \textbf{Statement 2: The function }f(x)\text{ is strictly increasing in }\left(\frac{1}{2},1\right). \ \
\displaystyle \text{Which of the following is correct with respect to the above statements?} \ \
\displaystyle \text{(a) Statement 1 is true and Statement 2 is false.} \ \
\displaystyle \text{(b) Statement 2 is true and Statement 1 is false.} \ \
\displaystyle \text{(c) Both the statements are true.} \ \
\displaystyle \text{(d) Both the statements are false.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{From the graph, the function decreases from }x=0\text{ to }x=\frac{1}{2},\text{ and then increases from} \\ x=\frac{1}{2}\text{ to }x=1.
\displaystyle \text{From }x=1\text{ to }x=2,\text{ the function is constant.}
\displaystyle \text{Statement 1: The function is increasing in }\left(\frac{1}{2},2\right).
\displaystyle \text{This is false because in the interval }(1,2),\text{ the function is constant, not increasing.}
\displaystyle \text{Statement 2: The function is strictly increasing in }\left(\frac{1}{2},1\right).
\displaystyle \text{This is true since the function increases continuously with no flat portion in this interval.}
\displaystyle \therefore\ \text{Statement 2 is true and Statement 1 is false.}
\displaystyle \text{Hence, the correct option is (b).}
\\
\displaystyle \textbf{(iv) }\int_{0}^{1}\frac{x^{4}-1}{x^{2}+1}\,dx\text{ is equal to:} \ \
\displaystyle \text{(a) }\frac{2}{3} \ \
\displaystyle \text{(b) }\frac{1}{3} \ \
\displaystyle \text{(c) }\frac{-2}{3} \ \
\displaystyle \text{(d) }0 \ \
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{1}\frac{x^{4}-1}{x^{2}+1}\,dx
\displaystyle \text{Divide: }\frac{x^{4}-1}{x^{2}+1}=\frac{(x^{2}-1)(x^{2}+1)}{x^{2}+1}=x^{2}-1
\displaystyle \therefore\ I=\int_{0}^{1}(x^{2}-1)\,dx
\displaystyle I=\int_{0}^{1}x^{2}\,dx-\int_{0}^{1}1\,dx
\displaystyle I=\left[\frac{x^{3}}{3}\right]_{0}^{1}-\left[x\right]_{0}^{1}
\displaystyle I=\frac{1}{3}-1
\displaystyle I=-\frac{2}{3}
\displaystyle \text{Hence, the correct option is (c).}
\\
\displaystyle \textbf{(v) } \text{Assertion: }\text{Consider the two events }A\text{ and }B\text{ such that }n(A)=n(B)\text{ and} \ \
\displaystyle \mathrm{P}\left(\frac{A}{B}\right)=\mathrm{P}\left(\frac{B}{A}\right). \ \
\displaystyle \textbf{Reason: }\text{The events }A\text{ and }B\text{ are mutually exclusive.} \ \
\displaystyle \text{(a) Both Assertion and Reason are true and Reason is the correct explanation for} \ \
\displaystyle \text{Assertion.} \ \
\displaystyle \text{(b) Both Assertion and Reason are true but Reason is not the correct explanation} \ \
\displaystyle \text{for Assertion.} \ \
\displaystyle \text{(c) Assertion is true and Reason is false.} \ \
\displaystyle \text{(d) Assertion is false and Reason is true.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given }n(A)=n(B)\text{ and }P\!\left(\frac{A}{B}\right)=P\!\left(\frac{B}{A}\right)
\displaystyle \text{Here }P\!\left(\frac{A}{B}\right)\text{ means }P(A\mid B)\text{ and }P\!\left(\frac{B}{A}\right)\text{ means }P(B\mid A)
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)},\quad P(B\mid A)=\frac{P(A\cap B)}{P(A)}
\displaystyle \text{Since }n(A)=n(B),\text{ we get }P(A)=P(B)\text{ (for equally likely outcomes)}
\displaystyle \therefore\ \frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{P(A)}
\displaystyle \text{Hence }P(A\mid B)=P(B\mid A)
\displaystyle \therefore\ \text{Assertion is true}
\displaystyle \text{Reason says that }A\text{ and }B\text{ are mutually exclusive}
\displaystyle \text{But this is not necessarily true from the given information}
\displaystyle \text{Hence Reason is false}
\displaystyle \therefore\ \text{the correct option is (c)}
\\
\displaystyle \textbf{(vi) } \text{The existence of unique solution of the system of equations }x+y=\lambda\text{ and} \ \
\displaystyle 5x+ky=2\text{ depends on:} \ \
\displaystyle \text{(a) }\lambda\text{ only} \ \
\displaystyle \text{(b) }\frac{\lambda}{k}=1 \ \
\displaystyle \text{(c) both }k\text{ and }\lambda \ \
\displaystyle \text{(d) }k\text{ only} \ \
\displaystyle \text{Answer:}
\displaystyle \text{For the system }x+y=\lambda\text{ and }5x+ky=2,\text{ a unique solution exists if}
\displaystyle \frac{1}{5}\neq\frac{1}{k}
\displaystyle \text{i.e. }k\neq 5
\displaystyle \text{Thus, the existence of unique solution depends only on }k
\displaystyle \therefore\ \text{the correct option is (d)}
\\
\displaystyle \textbf{(vii) } \text{A cylindrical popcorn tub of radius }10\text{ cm is being filled with popcorns at the rate} \ \
\displaystyle \text{of }314\ \mathrm{cm}^{3}\text{ per minute. The level of the popcorns in the tub is increasing at the rate} \ \
\displaystyle \text{of:} \ \
\displaystyle \text{(a) }1\text{ cm/minute} \ \
\displaystyle \text{(b) }0.1\text{ cm/minute} \ \
\displaystyle \text{(c) }1.1\text{ cm/minute} \ \
\displaystyle \text{(d) }0.5\text{ cm/minute} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }h\text{ be the level of popcorns in the cylindrical tub of radius }10\text{ cm}
\displaystyle V=\pi r^{2}h=\pi(10)^{2}h=100\pi h
\displaystyle \frac{dV}{dt}=100\pi\frac{dh}{dt}
\displaystyle \text{Given }\frac{dV}{dt}=314\ \text{cm}^{3}\text{/minute}
\displaystyle 314=100\pi\frac{dh}{dt}
\displaystyle \frac{dh}{dt}=\frac{314}{100\pi}
\displaystyle \text{Using }\pi=3.14,\ \frac{dh}{dt}=\frac{314}{314}=1\ \text{cm/minute}
\displaystyle \therefore\ \text{the correct option is (a)}
\\
\displaystyle \textbf{(viii) } \text{If }f(x)=\left\{\begin{matrix}x+2,&x<0\\-x^{2}-2,&0\leq x<1\\x,&x\geq 1\end{matrix}\right. \ \
\displaystyle \text{then the number of point(s) of discontinuity of }f(x)\text{, is/are:} \ \
\displaystyle \text{(a) }1 \ \
\displaystyle \text{(b) }3 \ \
\displaystyle \text{(c) }2 \ \
\displaystyle \text{(d) }0 \ \
\displaystyle \text{Answer:}
\displaystyle f(x)=\begin{cases}x+2,&x<0\\-x^{2}-2,&0\leq x<1\\x,&x\geq 1\end{cases}
\displaystyle \text{Possible points of discontinuity are }x=0\text{ and }x=1
\displaystyle \text{At }x=0,\ \lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}(x+2)=2
\displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}(-x^{2}-2)=-2
\displaystyle \text{Since }2\neq -2,\ f(x)\text{ is discontinuous at }x=0
\displaystyle \text{At }x=1,\ \lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}(-x^{2}-2)=-3
\displaystyle \lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}x=1
\displaystyle \text{Since }-3\neq 1,\ f(x)\text{ is discontinuous at }x=1
\displaystyle \therefore\ \text{the number of points of discontinuity is }2
\displaystyle \therefore\ \text{the correct option is (c)}
\\
\displaystyle \textbf{(ix) } \text{Assertion: If Set }A\text{ has }m\text{ elements, Set }B\text{ has }n\text{ elements and }n<m\text{, then the number} \ \
\displaystyle \text{of one-one function(s) from }A\rightarrow B\text{ is zero.} \ \
\displaystyle \text{Reason: A function }f:A\rightarrow B\text{ is defined only if all elements in Set }A\text{ have an image} \ \
\displaystyle \text{in Set }B. \ \
\displaystyle \text{(a) Both Assertion and Reason are true and Reason is the correct explanation for} \ \
\displaystyle \text{Assertion.} \ \
\displaystyle \text{(b) Both Assertion and Reason are true but Reason is not the correct explanation} \ \
\displaystyle \text{for Assertion.} \ \
\displaystyle \text{(c) Assertion is true and Reason is false.} \ \
\displaystyle \text{(d) Assertion is false and Reason is true.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given }|A|=m,\ |B|=n\text{ and }n<m
\displaystyle \text{For a one-one (injective) function }f:A\to B,\ \text{we must have }|A|\leq |B|
\displaystyle \text{But here }m>n,\ \text{so such a function is not possible}
\displaystyle \therefore\ \text{the number of one-one functions from }A\text{ to }B\text{ is }0
\displaystyle \therefore\ \text{Assertion is true}
\displaystyle \text{Reason: A function }f:A\to B\text{ is defined only if every element of }A\text{ has an image in }B
\displaystyle \text{This statement is true, but it does not explain why one-one functions are not possible}
\displaystyle \therefore\ \text{Reason is true but not the correct explanation}
\displaystyle \therefore\ \text{the correct option is (b)}
\\
\displaystyle \textbf{(x) } \text{Let }X\text{ be a discrete random variable. The probability distribution of }X\text{ is given} \ \
\displaystyle \text{below:} \ \
\displaystyle \begin{array}{|c|c|c|c|}\hline X&30&10&-10\\\hline \mathrm{P}(X)&\frac{1}{5}&\frac{3}{10}&\frac{1}{2}\\\hline \end{array} \ \
\displaystyle \text{Then }E(X)\text{ will be:} \ \
\displaystyle \text{(a) }1 \ \
\displaystyle \text{(b) }4 \ \
\displaystyle \text{(c) }2 \ \
\displaystyle \text{(d) }30 \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given }X\text{ takes values }30,10,-10\text{ with probabilities }\frac{1}{5},\frac{3}{10},\frac{1}{2}
\displaystyle \text{Check: }\frac{1}{5}+\frac{3}{10}+\frac{1}{2}=\frac{2}{10}+\frac{3}{10}+\frac{5}{10}=1
\displaystyle E(X)=\sum xP(x)=30\cdot\frac{1}{5}+10\cdot\frac{3}{10}+(-10)\cdot\frac{1}{2}
\displaystyle E(X)=6+3-5
\displaystyle E(X)=4
\displaystyle \therefore\ \text{the correct option is (b)}
\\
\displaystyle \textbf{(xi) } \text{Statement 1: If }A\text{ is an invertible matrix, then }(A^{2})^{-1}=(A^{-1})^{2} \ \
\displaystyle \text{Statement 2: If }A\text{ is an invertible matrix, then }|A^{-1}|=|A|^{-1} \ \
\displaystyle \text{(a) Statement 1 is true and Statement 2 is false.} \ \
\displaystyle \text{(b) Statement 2 is true and Statement 1 is false.} \ \
\displaystyle \text{(c) Both the statements are true.} \ \
\displaystyle \text{(d) Both the statements are false.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Statement 1: If }A\text{ is invertible, then }(A^{2})^{-1}=(A^{-1})^{2}
\displaystyle (A^{2})^{-1}=(A\cdot A)^{-1}=A^{-1}\cdot A^{-1}=(A^{-1})^{2}
\displaystyle \therefore\ \text{Statement 1 is true}
\displaystyle \text{Statement 2: If }A\text{ is invertible, then }|A^{-1}|=|A|^{-1}
\displaystyle \text{Using property }|A^{-1}|=\frac{1}{|A|}
\displaystyle \therefore\ |A^{-1}|=|A|^{-1}
\displaystyle \therefore\ \text{Statement 2 is true}
\displaystyle \therefore\ \text{both the statements are true}
\displaystyle \therefore\ \text{the correct option is (c)}
\\
\displaystyle \textbf{(xii) } \text{Write the smallest equivalence relation from the set }A\text{ to }A\text{, where }A=\{1,2,3\}. \ \
\displaystyle \text{Answer:}
\displaystyle \text{The smallest equivalence relation on }A=\{1,2,3\}\text{ is the identity relation}
\displaystyle R=\{(1,1),(2,2),(3,3)\}
\displaystyle \text{This relation is reflexive, symmetric and transitive}
\displaystyle \therefore\ \text{the smallest equivalence relation on }A\text{ is }\{(1,1),(2,2),(3,3)\}
\\
\displaystyle \textbf{(xiii) } \text{For what value of }x\text{, is }A=\begin{bmatrix}0&1&-2\\-1&0&3\\x&-3&0\end{bmatrix}\text{ a skew symmetric matrix?} \ \
\displaystyle \text{Answer:}
\displaystyle \text{For a skew symmetric matrix, }A^{T}=-A
\displaystyle A=\begin{bmatrix}0&1&-2\\-1&0&3\\x&-3&0\end{bmatrix}
\displaystyle \text{Comparing corresponding entries, }a_{31}=-a_{13}
\displaystyle x=-(-2)=2
\displaystyle \therefore\ x=2
\\
\displaystyle \textbf{(xiv) } \text{ Three critics review a book. Odds in favour of the book are } \\ 5:2,\ 4:3\text{ and }3:4 \ \
\displaystyle \text{respectively for the three critics. Find the probability that all critics are in favour of} \ \
\displaystyle \text{the book.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Odds in favour }5:2\Rightarrow P(E_{1})=\frac{5}{7}
\displaystyle \text{Odds in favour }4:3\Rightarrow P(E_{2})=\frac{4}{7}
\displaystyle \text{Odds in favour }3:4\Rightarrow P(E_{3})=\frac{3}{7}
\displaystyle \text{Assuming independence, probability that all three critics are in favour}
\displaystyle =P(E_{1}\cap E_{2}\cap E_{3})=\frac{5}{7}\cdot\frac{4}{7}\cdot\frac{3}{7}
\displaystyle =\frac{60}{343}
\displaystyle \therefore\ \text{the required probability is }\frac{60}{343}
\\
\displaystyle \textbf{(xv) } \text{Evaluate: }\int\frac{5}{\sqrt{2x+7}}\,dx \ \
\displaystyle \text{Answer:}
\displaystyle  I=\int \frac{5}{\sqrt{2x+7}}\,dx
\displaystyle \text{Let }2x+7=t
\displaystyle \therefore\ \frac{dt}{dx}=2\Rightarrow dx=\frac{dt}{2}
\displaystyle I=\int \frac{5}{\sqrt{t}}\cdot\frac{dt}{2}
\displaystyle I=\frac{5}{2}\int t^{-\frac{1}{2}}\,dt
\displaystyle I=\frac{5}{2}\cdot\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C
\displaystyle I=5\sqrt{t}+C
\displaystyle I=5\sqrt{2x+7}+C
\\
\displaystyle \textbf{Question 2} \ \
\displaystyle \text{Find the point on the curve }y=2x^{2}-6x-4\text{ at which the tangent is parallel to the }x\text{-axis.} \ \
\displaystyle \text{Answer:}
\displaystyle y=2x^{2}-6x-4
\displaystyle \frac{dy}{dx}=4x-6
\displaystyle \text{For the tangent to be parallel to the }x\text{-axis, }\frac{dy}{dx}=0
\displaystyle 4x-6=0
\displaystyle 4x=6
\displaystyle x=\frac{3}{2}
\displaystyle y=2\left(\frac{3}{2}\right)^{2}-6\left(\frac{3}{2}\right)-4
\displaystyle y=2\cdot\frac{9}{4}-9-4
\displaystyle y=\frac{9}{2}-13
\displaystyle y=\frac{9}{2}-\frac{26}{2}
\displaystyle y=-\frac{17}{2}
\displaystyle \therefore\ \text{the required point is }\left(\frac{3}{2},-\frac{17}{2}\right)
\\
\displaystyle \textbf{Question 3} \ \
\displaystyle \text{Find the value of }\tan^{-1}x-\cot^{-1}x\text{, if }\left(\tan^{-1}x\right)^{2}-\left(\cot^{-1}x\right)^{2}=\frac{5\pi}{8} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }\tan^{-1}x=A,\ \cot^{-1}x=B
\displaystyle \text{Then }A^{2}-B^{2}=\frac{5\pi}{8}
\displaystyle A^{2}-B^{2}=(A-B)(A+B)
\displaystyle \text{Also, }\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}
\displaystyle \therefore\ A+B=\frac{\pi}{2}
\displaystyle (A-B)\cdot\frac{\pi}{2}=\frac{5\pi^{2}}{8}
\displaystyle A-B=\frac{5\pi^{2}}{8}\cdot\frac{2}{\pi}
\displaystyle A-B=\frac{5\pi}{4}
\displaystyle \therefore\ \tan^{-1}x-\cot^{-1}x=\frac{5\pi}{4}
\\
\displaystyle \textbf{Question 4} \ \
\displaystyle \textbf{(i) } \text{If }x^{y}=e^{x-y}\text{, prove that }\frac{dy}{dx}=\frac{\log x}{(1+\log x)^{2}} \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{If }f(x)=\log(1+x)+\frac{1}{1+x}\text{, show that }f(x)\text{ attains its minimum value at }x=0. \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{Given }x^{y}=e^{x-y}
\displaystyle \text{Taking log on both sides,}
\displaystyle y\log x=x-y
\displaystyle y\log x+y=x
\displaystyle y(1+\log x)=x
\displaystyle y=\frac{x}{1+\log x}
\displaystyle \text{Differentiating with respect to }x,\ \frac{dy}{dx}=\frac{(1+\log x)\cdot 1-x\cdot\frac{1}{x}}{(1+\log x)^{2}}
\displaystyle \frac{dy}{dx}=\frac{1+\log x-1}{(1+\log x)^{2}}
\displaystyle \frac{dy}{dx}=\frac{\log x}{(1+\log x)^{2}}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{\log x}{(1+\log x)^{2}}
\displaystyle \textbf{OR} 
\displaystyle \textbf{(ii)}
\displaystyle f(x)=\log(1+x)+\frac{1}{1+x}
\displaystyle \text{Differentiating,}
\displaystyle f'(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}
\displaystyle f'(x)=\frac{1+x-1}{(1+x)^{2}}
\displaystyle f'(x)=\frac{x}{(1+x)^{2}}
\displaystyle \text{For critical points, }f'(x)=0
\displaystyle \frac{x}{(1+x)^{2}}=0
\displaystyle \therefore\ x=0
\displaystyle \text{Now, for }-1<x<0,\ f'(x)<0
\displaystyle \text{and for }x>0,\ f'(x)>0
\displaystyle \text{So }f(x)\text{ decreases on }(-1,0)\text{ and increases on }(0,\infty)
\displaystyle \therefore\ f(x)\text{ attains its minimum value at }x=0
\displaystyle \text{Also, }f(0)=\log 1+\frac{1}{1}=0+1=1
\displaystyle \therefore\ \text{the minimum value is }1\text{ at }x=0
\\
\displaystyle \textbf{Question 5} \ \
\displaystyle \text{Three shopkeepers Gaurav, Rizwan and Jacob use carry bags made of polythene,} \ \
\displaystyle \text{handmade paper and newspaper. The number of polythene bags, handmade bags and} \ \
\displaystyle \text{newspaper bags used by Gaurav, Rizwan and Jacob are }(20,30,40),\ (30,40,20)\text{ and} \ \
\displaystyle (40,20,30)\text{ respectively. One polythene bag costs Rs }1\text{, one handmade bag is for Rs }5\text{ and} \ \
\displaystyle \text{one newspaper bag costs Rs }2\text{. Gaurav, Rizwan and Jacob spend Rs }A\text{, Rs }B\text{ and Rs }C\text{ respectively} \ \
\displaystyle \text{on these carry bags.} \ \
\displaystyle \text{Using the concepts of matrices and determinants, answer the following questions:} \ \
\displaystyle \textbf{(i) } \text{Represent the above information in Matrix form.} \ \
\displaystyle \textbf{(ii) } \text{Find the values of Rs }A\text{, Rs }B\text{ and Rs }C. \ \
\displaystyle \text{Answer:}
\displaystyle \text{The numbers of polythene, handmade paper and newspaper bags used by}
\displaystyle \text{Gaurav, Rizwan and Jacob are respectively }(20,30,40),\ (30,40,20)\text{ and }
\displaystyle (40,20,30)
\displaystyle \text{Cost of one polythene bag = Rs }1,\ \text{cost of one handmade bag = Rs }5,
\displaystyle \text{cost of one newspaper bag = Rs }2
\displaystyle \textbf{(i) } \text{Matrix form:}
\displaystyle \begin{bmatrix}20&30&40\\30&40&20\\40&20&30\end{bmatrix} \begin{bmatrix}1\\5\\2\end{bmatrix}=\begin{bmatrix}A\\B\\C\end{bmatrix}
\displaystyle \textbf{(ii) } \text{Now,}
\displaystyle A=20\cdot 1+30\cdot 5+40\cdot 2
\displaystyle A=20+150+80=250
\displaystyle \therefore\ A=\text{Rs }250
\displaystyle B=30\cdot 1+40\cdot 5+20\cdot 2
\displaystyle B=30+200+40=270
\displaystyle \therefore\ B=\text{Rs }270
\displaystyle C=40\cdot 1+20\cdot 5+30\cdot 2
\displaystyle C=40+100+60=200
\displaystyle \therefore\ C=\text{Rs }200
\\
\displaystyle \textbf{Question 6} \ \
\displaystyle \textbf{(i) } \text{Differentiate }\sin^{-1}\left(\frac{2^{x+1}3^{x}}{1+(36)^{x}}\right)\text{ with respect to }x. \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{Show that }\tan^{-1}x+\tan^{-1}y=C\text{ is the general solution of the differential equation} \ \
\displaystyle (1+x^{2})\,dy+(1+y^{2})\,dx=0 \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle y=\sin^{-1}\!\left(\frac{2^{x+1}3^{x}}{1+36^{x}}\right)
\displaystyle y=\sin^{-1}\!\left(\frac{2\cdot 6^{x}}{1+36^{x}}\right)
\displaystyle \text{Let }u=\frac{2\cdot 6^{x}}{1+36^{x}}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{1}{\sqrt{1-u^{2}}}\cdot\frac{du}{dx}
\displaystyle \frac{du}{dx}=\frac{(1+36^{x})(2\cdot 6^{x}\log 6)-(2\cdot 6^{x})(36^{x}\log 36)}{(1+36^{x})^{2}}
\displaystyle \text{Since }\log 36=2\log 6,\ \frac{du}{dx}=\frac{2\cdot 6^{x}\log 6\,(1+36^{x}-2\cdot 36^{x})}{(1+36^{x})^{2}}
\displaystyle \therefore\ \frac{du}{dx}=\frac{2\cdot 6^{x}\log 6\,(1-36^{x})}{(1+36^{x})^{2}}
\displaystyle 1-u^{2}=1-\frac{4\cdot 36^{x}}{(1+36^{x})^{2}}
\displaystyle 1-u^{2}=\frac{(1+36^{x})^{2}-4\cdot 36^{x}}{(1+36^{x})^{2}}
\displaystyle 1-u^{2}=\frac{1-2\cdot 36^{x}+36^{2x}}{(1+36^{x})^{2}}=\frac{(1-36^{x})^{2}}{(1+36^{x})^{2}}
\displaystyle \therefore\ \sqrt{1-u^{2}}=\frac{|1-36^{x}|}{1+36^{x}}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{2\cdot 6^{x}\log 6\,(1-36^{x})}{(1+36^{x})^{2}}\cdot \frac{1+36^{x}}{|1-36^{x}|}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{2\cdot 6^{x}\log 6}{1+36^{x}}\cdot \frac{1-36^{x}}{|1-36^{x}|}
\displaystyle \text{Hence, } \frac{dy}{dx}=\frac{2\cdot 6^{x}\log 6}{1+36^{x}}\ \text{for }x<0, \ \text{and }\frac{dy}{dx}=-\frac{2\cdot 6^{x}\log 6}{1+36^{x}}\ \text{for }x>0
\displaystyle \text{Also, the derivative is not defined at }x=0
\displaystyle \textbf{OR} 
\displaystyle \textbf{(ii)}
\displaystyle (1+x^{2})\,dy+(1+y^{2})\,dx=0
\displaystyle \therefore\ (1+x^{2})\frac{dy}{dx}+(1+y^{2})=0
\displaystyle \therefore\ \frac{dy}{dx}=-\frac{1+y^{2}}{1+x^{2}}
\displaystyle \therefore\ \frac{dy}{1+y^{2}}=-\frac{dx}{1+x^{2}}
\displaystyle \text{Integrating both sides,}
\displaystyle \int \frac{dy}{1+y^{2}}=-\int \frac{dx}{1+x^{2}}
\displaystyle \tan^{-1}y=-\tan^{-1}x+C
\displaystyle \therefore\ \tan^{-1}x+\tan^{-1}y=C
\displaystyle \text{Hence, }\tan^{-1}x+\tan^{-1}y=C\text{ is the general solution}
\\
\displaystyle \textbf{Question 7} \ \
\displaystyle \text{If }x+y+z=0\text{ then show that }\left|\begin{matrix}1&1&1\\x&y&z\\x^{3}&y^{3}&z^{3}\end{matrix}\right|=0\text{, using properties of determinant.} \ \
\displaystyle \text{Answer:}
\displaystyle \Delta=\begin{vmatrix}1&1&1\\x&y&z\\x^{3}&y^{3}&z^{3}\end{vmatrix}
\displaystyle \text{Apply }C_{1}\to C_{1}-C_{2}\text{ and }C_{2}\to C_{2}-C_{3}
\displaystyle \Delta=\begin{vmatrix}0&0&1\\x-y&y-z&z\\x^{3}-y^{3}&y^{3}-z^{3}&z^{3}\end{vmatrix}
\displaystyle \Delta=1\cdot\begin{vmatrix}x-y&y-z\\x^{3}-y^{3}&y^{3}-z^{3}\end{vmatrix}
\displaystyle \Delta=(x-y)(y^{3}-z^{3})-(y-z)(x^{3}-y^{3})
\displaystyle \Delta=(x-y)(y-z)(y^{2}+yz+z^{2})-(y-z)(x-y)(x^{2}+xy+y^{2})
\displaystyle \Delta=(x-y)(y-z)\left[(y^{2}+yz+z^{2})-(x^{2}+xy+y^{2})\right]
\displaystyle \Delta=(x-y)(y-z)(z^{2}+yz-x^{2}-xy)
\displaystyle \text{Given }x+y+z=0,\ \text{so }z=-(x+y)
\displaystyle z^{2}+yz-x^{2}-xy=(x+y)^{2}-y(x+y)-x^{2}-xy
\displaystyle =x^{2}+2xy+y^{2}-xy-y^{2}-x^{2}-xy
\displaystyle =0
\displaystyle \therefore\ \Delta=(x-y)(y-z)\cdot 0=0
\displaystyle \therefore\ \begin{vmatrix}1&1&1\\x&y&z\\x^{3}&y^{3}&z^{3}\end{vmatrix}=0
\\
\displaystyle \textbf{Question 8} \ \
\displaystyle \textbf{(i) } \text{Evaluate: }\int\frac{\cos x}{3\cos x-5}\,dx \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{Evaluate: }\int(\log x)^{2}\,dx \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i) }I=\int \frac{\cos x}{3\cos x-5}\,dx
\displaystyle \text{Write }\cos x=\frac{1}{3}(3\cos x-5)+\frac{5}{3}
\displaystyle I=\int \frac{\frac{1}{3}(3\cos x-5)+\frac{5}{3}}{3\cos x-5}\,dx
\displaystyle I=\frac{1}{3}\int dx+\frac{5}{3}\int \frac{dx}{3\cos x-5}
\displaystyle I=\frac{x}{3}+\frac{5}{3}\int \frac{dx}{3\cos x-5}
\displaystyle \text{Now use standard result }\int \frac{dx}{a+b\cos x}=\frac{2}{\sqrt{a^{2}-b^{2}}} \tan^{-1}\!\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right)
\displaystyle \text{Here }a=-5,\ b=3
\displaystyle \int \frac{dx}{3\cos x-5}=\frac{2}{\sqrt{25-9}} \tan^{-1}\!\left(\sqrt{\frac{-5-3}{-5+3}}\tan\frac{x}{2}\right)
\displaystyle =\frac{2}{4}\tan^{-1}\!\left(\sqrt{\frac{-8}{-2}}\tan\frac{x}{2}\right)
\displaystyle =\frac{1}{2}\tan^{-1}\!\left(2\tan\frac{x}{2}\right)
\displaystyle \therefore\ I=\frac{x}{3}+\frac{5}{6}\tan^{-1}\!\left(2\tan\frac{x}{2}\right)+C
\displaystyle \textbf{OR} 
\displaystyle \textbf{(ii) }I=\int (\log x)^{2}\,dx
\displaystyle \text{Use integration by parts: let }u=(\log x)^{2},\ dv=dx
\displaystyle \therefore\ du=\frac{2\log x}{x}\,dx,\ v=x
\displaystyle I=x(\log x)^{2}-\int x\cdot\frac{2\log x}{x}\,dx
\displaystyle I=x(\log x)^{2}-2\int \log x\,dx
\displaystyle \int \log x\,dx=x\log x-x
\displaystyle \therefore\ I=x(\log x)^{2}-2(x\log x-x)
\displaystyle I=x(\log x)^{2}-2x\log x+2x+C
\\
\displaystyle \textbf{Question 9} \ \
\displaystyle \textbf{(i) } \text{If }x=\tan\left(\frac{1}{a}\log y\right)\text{ then show that }(1+x^{2})\frac{d^{2}y}{dx^{2}}+(2x-a)\frac{dy}{dx}=0 \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{The graph of }f(x)=-x^{3}+27x-2\text{ is given below:} \ \ \displaystyle \text{(a) Find the slope of the above graph.} \ \
\displaystyle \text{(b) Find the co-ordinates of turning points, }A\text{ and }B. \ \
\displaystyle \text{(c) Evaluate }f''(-2),\ f(0)\text{ and }f'(3)\text{ and arrange them in ascending order.} \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle x=\tan\left(\frac{1}{a}\log y\right)
\displaystyle \therefore\ \tan^{-1}x=\frac{1}{a}\log y
\displaystyle \therefore\ \log y=a\tan^{-1}x
\displaystyle \therefore\ y=e^{a\tan^{-1}x}
\displaystyle \text{Differentiating with respect to }x,
\displaystyle \frac{dy}{dx}=e^{a\tan^{-1}x}\cdot a\cdot\frac{1}{1+x^{2}}
\displaystyle \therefore\ \frac{dy}{dx}=\frac{ay}{1+x^{2}}
\displaystyle \text{Differentiating again,}
\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{a(1+x^{2})\frac{dy}{dx}-2axy}{(1+x^{2})^{2}}
\displaystyle \text{Using }\frac{dy}{dx}=\frac{ay}{1+x^{2}},\ \frac{d^{2}y}{dx^{2}}= \frac{a(1+x^{2})\cdot\frac{ay}{1+x^{2}}-2axy}{(1+x^{2})^{2}}
\displaystyle \therefore\ \frac{d^{2}y}{dx^{2}}=\frac{a^{2}y-2axy}{(1+x^{2})^{2}}= \frac{ay(a-2x)}{(1+x^{2})^{2}}
\displaystyle \text{Now, }(1+x^{2})\frac{d^{2}y}{dx^{2}}+(2x-a)\frac{dy}{dx}
\displaystyle =(1+x^{2})\cdot\frac{ay(a-2x)}{(1+x^{2})^{2}}+(2x-a)\cdot\frac{ay}{1+x^{2}}
\displaystyle =\frac{ay(a-2x)}{1+x^{2}}+\frac{ay(2x-a)}{1+x^{2}}
\displaystyle =0
\displaystyle \therefore\ (1+x^{2})\frac{d^{2}y}{dx^{2}}+(2x-a)\frac{dy}{dx}=0
\displaystyle \textbf{OR} 
\displaystyle \textbf{(ii)}
\displaystyle f(x)=-x^{3}+27x-2
\displaystyle \text{(a) Slope of the graph }=f'(x)
\displaystyle f'(x)=-3x^{2}+27
\displaystyle \therefore\ \text{the slope function is }-3x^{2}+27
\displaystyle \text{(b) Turning points occur where }f'(x)=0
\displaystyle -3x^{2}+27=0
\displaystyle x^{2}=9
\displaystyle x=\pm 3
\displaystyle f(-3)=-(-3)^{3}+27(-3)-2=27-81-2=-56
\displaystyle f(3)=-(3)^{3}+27(3)-2=-27+81-2=52
\displaystyle \therefore\ A(-3,-56)\text{ and }B(3,52)
\displaystyle \text{(c) }f''(x)=-6x
\displaystyle f''(-2)=-6(-2)=12
\displaystyle f(0)=-(0)^{3}+27(0)-2=-2
\displaystyle f'(3)=-3(3)^{2}+27=-27+27=0
\displaystyle \text{Ascending order: }f(0)<f'(3)<f''(-2)
\displaystyle \text{i.e. }-2<0<12
\\
\displaystyle \textbf{Question 10} \ \
\displaystyle \text{Pia, Sia and Dia displayed their paintings in an art exhibition. The three artists displayed} \ \
\displaystyle 15,\ 5\text{ and }10\text{ of their paintings respectively. A person bought three paintings from the} \ \
\displaystyle \text{exhibition.} \ \
\displaystyle \textbf{(i) } \text{Find the probability that he bought one painting from each of them.} \ \
\displaystyle \textbf{(ii) } \text{Find the probability that he bought all the three paintings from the same person.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Total paintings }=15+5+10=30
\displaystyle \text{Total ways of selecting 3 paintings }={}^{30}C_{3}=\frac{30\cdot 29\cdot 28}{6}=4060
\displaystyle \textbf{(i) } \text{One from each artist:}
\displaystyle \text{Ways }=15\cdot 5\cdot 10=750
\displaystyle \text{Probability }=\frac{750}{4060}=\frac{75}{406}
\displaystyle \textbf{(ii) } \text{All three from the same artist:}
\displaystyle \text{Ways }={}^{15}C_{3}+{}^{5}C_{3}+{}^{10}C_{3}
\displaystyle {}^{15}C_{3}=\frac{15\cdot 14\cdot 13}{6}=455,\quad {}^{5}C_{3}=10,\quad {}^{10}C_{3}=120
\displaystyle \text{Total favourable ways }=455+10+120=585
\displaystyle \text{Probability }=\frac{585}{4060}=\frac{117}{812}
\\
\displaystyle \textbf{Question 11} \ \
\displaystyle \textbf{(i) } \text{Prove: }\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{x\,dx}{1+\sin x}=(\sqrt{2}-1)\pi \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{Evaluate: }\int\frac{x^{2}}{(x-1)^{2}(x^{2}+1)}\,dx \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle I=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{x\,dx}{1+\sin x}
\displaystyle \text{Using the substitution }x=\pi-t,\ dx=-dt
\displaystyle I=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{(\pi-x)\,dx}{1+\sin x}
\displaystyle \text{Adding the two expressions for }I,
\displaystyle 2I=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\pi\,dx}{1+\sin x}
\displaystyle 2I=\pi\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{dx}{1+\sin x}
\displaystyle \frac{1}{1+\sin x}=\frac{1-\sin x}{1-\sin^{2}x}=\frac{1-\sin x}{\cos^{2}x}
\displaystyle \therefore\ \frac{1}{1+\sin x}=\sec^{2}x-\sec x\tan x
\displaystyle 2I=\pi\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}(\sec^{2}x-\sec x\tan x)\,dx
\displaystyle 2I=\pi\left[\tan x-\sec x\right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}
\displaystyle 2I=\pi\left[\left(-1-(-\sqrt{2})\right)-\left(1-\sqrt{2}\right)\right]
\displaystyle 2I=\pi\left[(\sqrt{2}-1)-(1-\sqrt{2})\right]
\displaystyle 2I=\pi(2\sqrt{2}-2)
\displaystyle I=(\sqrt{2}-1)\pi
\displaystyle \therefore\ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{x\,dx}{1+\sin x}=(\sqrt{2}-1)\pi
\displaystyle \textbf{OR} 
\displaystyle \textbf{(ii)}
\displaystyle I=\int \frac{x^{2}}{(x-1)^{2}(x^{2}+1)}\,dx
\displaystyle \text{Resolve into partial fractions:}
\displaystyle \frac{x^{2}}{(x-1)^{2}(x^{2}+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{Cx+D}{x^{2}+1}
\displaystyle x^{2}=A(x-1)(x^{2}+1)+B(x^{2}+1)+(Cx+D)(x-1)^{2}
\displaystyle \text{Comparing coefficients, }A=\frac{1}{2},\ B=\frac{1}{2},\ C=-\frac{1}{2},\ D=0
\displaystyle \therefore\ \frac{x^{2}}{(x-1)^{2}(x^{2}+1)}=\frac{1}{2(x-1)}+\frac{1}{2(x-1)^{2}}-\frac{x}{2(x^{2}+1)}
\displaystyle I=\frac{1}{2}\int \frac{dx}{x-1}+\frac{1}{2}\int \frac{dx}{(x-1)^{2}}-\frac{1}{2}\int \frac{x\,dx}{x^{2}+1}
\displaystyle I=\frac{1}{2}\log|x-1|-\frac{1}{2(x-1)}-\frac{1}{4}\log(x^{2}+1)+C
\\
\displaystyle \textbf{Question 12} \ \
\displaystyle \textbf{(i) } \text{Solve the differential equation:} \ \
\displaystyle (x+5y^{2})\frac{dy}{dx}=y\text{ when }x=2\text{ and }y=1 \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{Find the particular solution of the differential equation:} \ \
\displaystyle (x^{2}-2y^{2})\,dx+2xy\,dy=0,\text{ when }x=1\text{ and }y=1 \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle (x+5y^{2})\frac{dy}{dx}=y
\displaystyle \therefore\ \frac{dx}{dy}=\frac{x+5y^{2}}{y}
\displaystyle \therefore\ \frac{dx}{dy}-\frac{x}{y}=5y
\displaystyle \text{This is a linear differential equation in }x
\displaystyle \text{Integrating factor }=e^{\int -\frac{1}{y}\,dy}=e^{-\log y}=\frac{1}{y}
\displaystyle \therefore\ \frac{1}{y}\frac{dx}{dy}-\frac{x}{y^{2}}=5
\displaystyle \therefore\ \frac{d}{dy}\left(\frac{x}{y}\right)=5
\displaystyle \text{Integrating,}
\displaystyle \frac{x}{y}=5y+C
\displaystyle \therefore\ x=5y^{2}+Cy
\displaystyle \text{Using }x=2,\ y=1,
\displaystyle 2=5+C
\displaystyle \therefore\ C=-3
\displaystyle \therefore\ x=5y^{2}-3y
\displaystyle \textbf{(ii)}
\displaystyle (x^{2}-2y^{2})\,dx+2xy\,dy=0
\displaystyle \therefore\ 2xy\frac{dy}{dx}=-(x^{2}-2y^{2})
\displaystyle \therefore\ \frac{dy}{dx}=-\frac{x^{2}-2y^{2}}{2xy}
\displaystyle \text{This is a homogeneous differential equation}
\displaystyle \text{Put }y=vx,\ \text{so that }\frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=-\frac{x^{2}-2v^{2}x^{2}}{2x\cdot vx}
\displaystyle v+x\frac{dv}{dx}=-\frac{1-2v^{2}}{2v}
\displaystyle v+x\frac{dv}{dx}=v-\frac{1}{2v}
\displaystyle \therefore\ x\frac{dv}{dx}=-\frac{1}{2v}
\displaystyle \therefore\ 2v\,dv=-\frac{dx}{x}
\displaystyle \text{Integrating,}
\displaystyle v^{2}=-\log x+C
\displaystyle \therefore\ \left(\frac{y}{x}\right)^{2}=-\log x+C
\displaystyle \therefore\ \frac{y^{2}}{x^{2}}+\log x=C
\displaystyle \text{Using }x=1,\ y=1,
\displaystyle \frac{1^{2}}{1^{2}}+\log 1=C
\displaystyle \therefore\ C=1
\displaystyle \therefore\ \frac{y^{2}}{x^{2}}+\log x=1
\\
\displaystyle \textbf{Question 13} \ \
\displaystyle \text{Observe the two graphs, Graph 1 and Graph 2 given below and answer the questions that} \ \
\displaystyle \text{follow.} \ \ \displaystyle \textbf{(i) } \text{Which one of the graphs represents }y=\sin^{-1}x\ ? \ \
\displaystyle \textbf{(ii) } \text{Write the domain and range of }y=\sin^{-1}x. \ \
\displaystyle \textbf{(iii) } \text{Prove that }\sin^{-1}\frac{1}{\sqrt{5}}+\sin^{-1}\frac{2}{\sqrt{5}}=\frac{\pi}{2} \ \
\displaystyle \text{(iv) Find the value of }\tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right] \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i) } \text{The graph of }y=\sin^{-1}x\text{ has domain }[-1,1]\text{ and range } \left[-\frac{\pi}{2},\frac{\pi}{2}\right]
\displaystyle \text{Also, it passes through }(0,0)\text{ and is increasing throughout its domain}
\displaystyle \text{Hence, Graph 1 represents }y=\sin^{-1}x
\displaystyle \textbf{(ii) } \text{Domain of }y=\sin^{-1}x\text{ is }[-1,1]
\displaystyle \text{Range of }y=\sin^{-1}x\text{ is }\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
\displaystyle \textbf{(iii) } \text{Let }\sin^{-1}\frac{1}{\sqrt{5}}=A,\ \sin^{-1}\frac{2}{\sqrt{5}}=B
\displaystyle \text{Then }\sin A=\frac{1}{\sqrt{5}},\ \sin B=\frac{2}{\sqrt{5}}
\displaystyle \text{Since }A,B\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right],\ \cos A=\sqrt{1-\frac{1}{5}}=\frac{2}{\sqrt{5}}
\displaystyle \text{and }\cos B=\sqrt{1-\frac{4}{5}}=\frac{1}{\sqrt{5}}
\displaystyle \sin(A+B)=\sin A\cos B+\cos A\sin B
\displaystyle =\frac{1}{\sqrt{5}}\cdot\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}}\cdot\frac{2}{\sqrt{5}}
\displaystyle =\frac{1}{5}+\frac{4}{5}=1
\displaystyle \text{Now }A,B\in \left(0,\frac{\pi}{2}\right)\Rightarrow A+B\in (0,\pi)
\displaystyle \text{and }\sin(A+B)=1\Rightarrow A+B=\frac{\pi}{2}
\displaystyle \therefore\ \sin^{-1}\frac{1}{\sqrt{5}}+\sin^{-1}\frac{2}{\sqrt{5}}=\frac{\pi}{2}
\displaystyle \text{(iv) } \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}
\displaystyle \therefore\ \cos\theta=\frac{\sqrt{3}}{2}\Rightarrow \theta=\frac{\pi}{6}
\displaystyle \therefore\ 2\sin(2\theta)=2\sin\left(\frac{\pi}{3}\right)=2\cdot\frac{\sqrt{3}}{2}=\sqrt{3}
\displaystyle \therefore\ \tan^{-1}\!\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]=\tan^{-1}(\sqrt{3})
\displaystyle \therefore\ \tan^{-1}\!\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]=\frac{\pi}{3}
\\
\displaystyle \textbf{Question 14} \ \
\displaystyle \text{An international conference takes place in a metropolitan city. International leaders,} \ \
\displaystyle \text{scientists and industrialists participate in it.} \ \
\displaystyle \text{The organisers of the conference appoint three agencies namely }X,\ Y\text{ and }Z\text{ for the security} \ \
\displaystyle \text{of the participants. The track record of the success of }X,\ Y\text{ and }Z\text{ in providing security} \ \
\displaystyle \text{services is }99\%,\ 98.5\%\text{ and }98\%\text{ respectively. The organisers assign the responsibility} \ \
\displaystyle \text{of ensuring the security of }1000\text{ people to agency }X,\ 2000\text{ people to agency }Y\text{ and }3000\text{ people} \ \
\displaystyle \text{to agency }Z. \ \
\displaystyle \text{At the end of the conference, one participant goes missing from the conference room.} \ \
\displaystyle \text{What is the probability that the missing participant was placed under the responsibility of} \ \
\displaystyle \text{the security agency }X\text{?} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }E_{1},E_{2},E_{3}\text{ be the events that a participant is under agencies } X,Y,Z\text{ respectively}
\displaystyle \text{Let }M\text{ be the event that the participant goes missing}
\displaystyle \text{Given }P(E_{1})=\frac{1000}{6000}=\frac{1}{6},\quad P(E_{2})=\frac{2000}{6000}=\frac{1}{3},\quad P(E_{3})=\frac{3000}{6000}=\frac{1}{2}
\displaystyle \text{Success rates are }99\%,\ 98.5\%,\ 98\%\text{ respectively}
\displaystyle \text{Therefore failure rates are }P(M\mid E_{1})=1\%=0.01,\ P(M\mid E_{2})=1.5\%=0.015,\ P(M\mid E_{3})=2\%=0.02
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(E_{1}\mid M)=\frac{P(E_{1})P(M\mid E_{1})} {P(E_{1})P(M\mid E_{1})+P(E_{2})P(M\mid E_{2})+P(E_{3})P(M\mid E_{3})}
\displaystyle P(E_{1}\mid M)=\frac{\frac{1}{6}\cdot 0.01} {\frac{1}{6}\cdot 0.01+\frac{1}{3}\cdot 0.015+\frac{1}{2}\cdot 0.02}
\displaystyle P(E_{1}\mid M)=\frac{\frac{1}{600}} {\frac{1}{600}+\frac{1}{200}+\frac{1}{100}}
\displaystyle P(E_{1}\mid M)=\frac{\frac{1}{600}}{\frac{1+3+6}{600}}
\displaystyle P(E_{1}\mid M)=\frac{\frac{1}{600}}{\frac{10}{600}}
\displaystyle P(E_{1}\mid M)=\frac{1}{10}
\displaystyle \therefore\ \text{the required probability is }\frac{1}{10}

\displaystyle \textbf{SECTION B - 15 MARKS} \ \

\displaystyle \textbf{Question 15} \ \
\displaystyle \text{In subparts (i) and (ii) choose the correct options and in subparts (iii) and (iv), answer the} \ \
\displaystyle \text{questions as instructed.} \ \
\displaystyle \textbf{(i) } \text{Assertion: }\left(\overrightarrow{a}+\overrightarrow{b}\right)^{2}+\left(\overrightarrow{b}-\overrightarrow{a}\right)^{2}=2\left(a^{2}+b^{2}\right) \ \
\displaystyle \text{Reason: Dot product of any two vectors is commutative.} \ \
\displaystyle \text{(a) Both Assertion and Reason are true and Reason is the correct explanation for} \ \
\displaystyle \text{Assertion.} \ \
\displaystyle \text{(b) Both Assertion and Reason are true but Reason is not the correct explanation} \ \
\displaystyle \text{for Assertion.} \ \
\displaystyle \text{(c) Assertion is true and Reason is false.} \ \
\displaystyle \text{(d) Assertion is false and Reason is true.} \ \
\displaystyle \textbf{(ii) } \text{The angle between the two planes }x+y+2z=9\text{ and }2x-y+z=15\text{ is:} \ \
\displaystyle \text{(a) }\frac{\pi}{2} \ \
\displaystyle \text{(b) }\frac{\pi}{3} \ \
\displaystyle \text{(c) }\pi \ \
\displaystyle \text{(d) }\frac{3\pi}{4} \ \
\displaystyle \textbf{(iii) } \text{Show that points }P(-2,3,5),\ Q(1,2,3)\text{ and }R(7,0,-1)\text{ are collinear.} \ \
\displaystyle \text{(iv) Two honeybees are flying parallel to each other in the garden to collect the nectar.} \ \
\displaystyle \text{The path traced by the bees is given in the form of a straight line. The equation of} \ \
\displaystyle \text{the path traced by one honeybee is }\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k}). \ \ \displaystyle \text{(a) Write the above-mentioned equation in cartesian form.} \ \
\displaystyle \text{(b) Find the equation of the path traced by the other honeybee passing through the} \ \
\displaystyle \text{point }(2,4,5). \ \
\displaystyle \text{Answer:}
\displaystyle (i) (\overrightarrow{a}+\overrightarrow{b})^{2}+ (\overrightarrow{b}-\overrightarrow{a})^{2}
\displaystyle =(\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}+\overrightarrow{b})+ (\overrightarrow{b}-\overrightarrow{a})\cdot (\overrightarrow{b}-\overrightarrow{a})
\displaystyle =(\overrightarrow{a}\cdot\overrightarrow{a}+ 2\overrightarrow{a}\cdot\overrightarrow{b}+ \overrightarrow{b}\cdot\overrightarrow{b})+ (\overrightarrow{b}\cdot\overrightarrow{b}- 2\overrightarrow{a}\cdot\overrightarrow{b}+ \overrightarrow{a}\cdot\overrightarrow{a})
\displaystyle =2(\overrightarrow{a}\cdot\overrightarrow{a}+ \overrightarrow{b}\cdot\overrightarrow{b})
\displaystyle =2(a^{2}+b^{2})
\displaystyle \therefore\ \text{Assertion is true}
\displaystyle \text{Reason: }\overrightarrow{a}\cdot\overrightarrow{b}= \overrightarrow{b}\cdot\overrightarrow{a}\ \text{(commutative)}
\displaystyle \text{This property is used in simplifying cross terms}
\displaystyle \therefore\ \text{Reason is true and explains the Assertion}
\displaystyle \therefore\ \text{the correct option is (a)}
\displaystyle \textbf{(ii) } \text{The angle between two planes is the angle between their normal vectors}
\displaystyle \text{For }x+y+2z=9,\ \text{a normal vector is }\overrightarrow{n_{1}}=(1,1,2)
\displaystyle \text{For }2x-y+z=15,\ \text{a normal vector is }\overrightarrow{n_{2}}=(2,-1,1)
\displaystyle \overrightarrow{n_{1}}\cdot\overrightarrow{n_{2}}=1\cdot 2+1\cdot(-1)+2\cdot 1=3
\displaystyle |\overrightarrow{n_{1}}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{6}
\displaystyle |\overrightarrow{n_{2}}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{6}
\displaystyle \cos\theta=\frac{\overrightarrow{n_{1}}\cdot\overrightarrow{n_{2}}} {|\overrightarrow{n_{1}}||\overrightarrow{n_{2}}|}=\frac{3}{\sqrt{6}\cdot\sqrt{6}}=\frac{3}{6}=\frac{1}{2}
\displaystyle \therefore\ \theta=\frac{\pi}{3}
\displaystyle \therefore\ \text{the correct option is (b)}
\displaystyle \textbf{(iii) } \text{To show that }P(-2,3,5),\ Q(1,2,3)\text{ and }R(7,0,-1)\text{ are collinear,}
\displaystyle \text{find }\overrightarrow{PQ}\text{ and }\overrightarrow{QR}
\displaystyle \overrightarrow{PQ}=(1-(-2),\,2-3,\,3-5)=(3,-1,-2)
\displaystyle \overrightarrow{QR}=(7-1,\,0-2,\,-1-3)=(6,-2,-4)
\displaystyle \overrightarrow{QR}=2\overrightarrow{PQ}
\displaystyle \text{Since }\overrightarrow{PQ}\text{ and }\overrightarrow{QR}\text{ are parallel, the points are collinear}
\displaystyle \therefore\ P,\ Q\text{ and }R\text{ are collinear}
\displaystyle \text{(iv) Given }\overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda(2\widehat{i}+3\widehat{j}+4\widehat{k})
\displaystyle \text{(a) Comparing components,}
\displaystyle x=1+2\lambda,\quad y=2+3\lambda,\quad z=3+4\lambda
\displaystyle \therefore\ \lambda=\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}
\displaystyle \therefore\ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}
\displaystyle \text{(b) Since the paths are parallel, the direction ratios are }2,3,4
\displaystyle \text{Equation of the required line passing through }(2,4,5)
\displaystyle \frac{x-2}{2}=\frac{y-4}{3}=\frac{z-5}{4}
\\
\displaystyle \textbf{Question 16} \ \
\displaystyle \textbf{(i) } \text{Find the equation of the plane passing through the points }(2,2,-1),\ (3,4,2)\text{ and} \ \
\displaystyle (7,0,6). \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{Find the equation of the plane passing through the points }(2,3,1),\ (4,-5,3)\text{ and} \ \
\displaystyle \text{parallel to }x\text{-axis.} \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{Let }A(2,2,-1),\ B(3,4,2),\ C(7,0,6)
\displaystyle \overrightarrow{AB}=(3-2,\ 4-2,\ 2-(-1))=(1,2,3)
\displaystyle \overrightarrow{AC}=(7-2,\ 0-2,\ 6-(-1))=(5,-2,7)
\displaystyle \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&2&3\\5&-2&7\end{vmatrix}
\displaystyle \overrightarrow{n}=\widehat{i}(14+6)-\widehat{j}(7-15)+\widehat{k}(-2-10)
\displaystyle \overrightarrow{n}=20\widehat{i}+8\widehat{j}-12\widehat{k}
\displaystyle \therefore\ \text{a normal vector is }(5,2,-3)
\displaystyle \text{Equation of the plane through }(2,2,-1)\text{ is}
\displaystyle 5(x-2)+2(y-2)-3(z+1)=0
\displaystyle 5x-10+2y-4-3z-3=0
\displaystyle 5x+2y-3z-17=0
\displaystyle \therefore\ \text{the required plane is }5x+2y-3z-17=0
\displaystyle \textbf{(ii)}
\displaystyle \text{Let }P(2,3,1)\text{ and }Q(4,-5,3)
\displaystyle \text{Since the plane is parallel to the }x\text{-axis, it contains the vector }(1,0,0)
\displaystyle \overrightarrow{PQ}=(4-2,\ -5-3,\ 3-1)=(2,-8,2)
\displaystyle \text{Hence two direction vectors in the plane are }(1,0,0)\text{ and }(2,-8,2)
\displaystyle \overrightarrow{n}=(1,0,0)\times(2,-8,2)
\displaystyle \overrightarrow{n}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\1&0&0\\2&-8&2\end{vmatrix}
\displaystyle \overrightarrow{n}=0\widehat{i}-2\widehat{j}-8\widehat{k}
\displaystyle \therefore\ \text{a normal vector is }(0,1,4)
\displaystyle \text{Equation of the plane through }(2,3,1)\text{ is}
\displaystyle 0(x-2)+1(y-3)+4(z-1)=0
\displaystyle y-3+4z-4=0
\displaystyle y+4z-7=0
\displaystyle \therefore\ \text{the required plane is }y+4z-7=0
\\
\displaystyle \textbf{Question 17} \ \
\displaystyle \text{Consider the position vectors of }A,\ B\text{ and }C\text{ as }\overrightarrow{OA}=2\hat{i}-2\hat{j}+\hat{k},\ \overrightarrow{OB}=\hat{i}+2\hat{j}-2\hat{k}\text{ and} \ \
\displaystyle \overrightarrow{OC}=2\hat{i}-\hat{j}+4\hat{k} \ \
\displaystyle \textbf{(i) } \text{Calculate }\overrightarrow{AB}\text{ and }\overrightarrow{BC}. \ \
\displaystyle \textbf{(ii) } \text{Find the projection of }\overrightarrow{AB}\text{ on }\overrightarrow{BC}. \ \
\displaystyle \textbf{(iii) } \text{Find the area of the triangle }ABC\text{ whose sides are }\overrightarrow{AB}\text{ and }\overrightarrow{BC}. \ \
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{OA}=2\widehat{i}-2\widehat{j}+\widehat{k},\quad \overrightarrow{OB}=\widehat{i}+2\widehat{j}-2\widehat{k},\quad \overrightarrow{OC}=2\widehat{i}-\widehat{j}+4\widehat{k}
\displaystyle \textbf{(i) } \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}
\displaystyle \overrightarrow{AB}=(\widehat{i}+2\widehat{j}-2\widehat{k})- (2\widehat{i}-2\widehat{j}+\widehat{k})
\displaystyle \overrightarrow{AB}=-\widehat{i}+4\widehat{j}-3\widehat{k}
\displaystyle \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}
\displaystyle \overrightarrow{BC}=(2\widehat{i}-\widehat{j}+4\widehat{k})- (\widehat{i}+2\widehat{j}-2\widehat{k})
\displaystyle \overrightarrow{BC}=\widehat{i}-3\widehat{j}+6\widehat{k}
\displaystyle \textbf{(ii) } \text{Projection of }\overrightarrow{AB}\text{ on }\overrightarrow{BC} =\frac{\overrightarrow{AB}\cdot\overrightarrow{BC}}{|\overrightarrow{BC}|}
\displaystyle \overrightarrow{AB}\cdot\overrightarrow{BC}=(-1)(1)+(4)(-3)+(-3)(6)
\displaystyle \overrightarrow{AB}\cdot\overrightarrow{BC}=-1-12-18=-31
\displaystyle |\overrightarrow{BC}|=\sqrt{1^{2}+(-3)^{2}+6^{2}}=\sqrt{46}
\displaystyle \therefore\ \text{projection of }\overrightarrow{AB}\text{ on } \overrightarrow{BC}=-\frac{31}{\sqrt{46}}
\displaystyle \textbf{(iii) } \text{Area of triangle }ABC=\frac{1}{2} \left|\overrightarrow{AB}\times\overrightarrow{BC}\right|
\displaystyle \overrightarrow{AB}\times\overrightarrow{BC}= \begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\-1&4&-3\\1&-3&6\end{vmatrix}
\displaystyle \overrightarrow{AB}\times\overrightarrow{BC}= \widehat{i}(24-9)-\widehat{j}(-6+3)+\widehat{k}(3-4)
\displaystyle \overrightarrow{AB}\times\overrightarrow{BC}= 15\widehat{i}+3\widehat{j}-\widehat{k}
\displaystyle \left|\overrightarrow{AB}\times\overrightarrow{BC}\right|= \sqrt{15^{2}+3^{2}+(-1)^{2}}=\sqrt{235}
\displaystyle \therefore\ \text{Area of triangle }ABC=\frac{1}{2}\sqrt{235}
\\
\displaystyle \textbf{Question 18} \ \
\displaystyle \textbf{(i) } \text{The equation }y=4-x^{2}\text{ represents a parabola.} \ \
\displaystyle \text{(a) Make a rough sketch of the graph of the given function.} \ \
\displaystyle \text{(b) Determine the area enclosed between the curve, the }x\text{-axis, the lines } \\ x=0\text{ and}  x=2. \ \
\displaystyle \text{(c) Hence, find the area bounded by the parabola and the }x\text{-axis.} \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{A farmer has a field bounded by three lines } \\ x+2y=2,\ y-x=1,\ 2x+y=7. \ \
\displaystyle \text{Using integration, find the area of the region bounded by these lines.} \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle y=4-x^{2}
\displaystyle \text{(a) Since the coefficient of }x^{2}\text{ is negative, the parabola opens downward}
\displaystyle \text{Vertex }=(0,4)
\displaystyle \text{It cuts the }x\text{-axis where }4-x^{2}=0
\displaystyle x^{2}=4
\displaystyle x=\pm 2
\displaystyle \text{So the }x\text{-intercepts are }(-2,0)\text{ and }(2,0)
\displaystyle \text{Thus the rough sketch is a downward opening parabola with vertex }(0,4)\text{ and}
\displaystyle \text{intercepts }(-2,0)\text{ and }(2,0)
\displaystyle \text{(b) Required area }=\int_{0}^{2}(4-x^{2})\,dx
\displaystyle =\left[4x-\frac{x^{3}}{3}\right]_{0}^{2}
\displaystyle =\left(8-\frac{8}{3}\right)-0
\displaystyle =\frac{24-8}{3}
\displaystyle =\frac{16}{3}\ \text{square units}
\displaystyle \text{(c) The parabola is symmetric about the }y\text{-axis}
\displaystyle \text{Hence, total area bounded by the parabola and the }x\text{-axis}
\displaystyle =2\cdot\frac{16}{3}
\displaystyle =\frac{32}{3}\ \text{square units}
\displaystyle \textbf{OR} 
\displaystyle \textbf{(ii)}
\displaystyle \text{Given lines }x+2y=2,\ y-x=1,\ 2x+y=7
\displaystyle \text{Write them as }y=1-\frac{x}{2},\ y=x+1,\ y=7-2x
\displaystyle \text{Find the points of intersection:}
\displaystyle \text{Between }x+2y=2\text{ and }y-x=1
\displaystyle y=x+1
\displaystyle x+2(x+1)=2
\displaystyle 3x=0
\displaystyle x=0,\ y=1
\displaystyle \text{So one point is }A(0,1)
\displaystyle \text{Between }y-x=1\text{ and }2x+y=7
\displaystyle y=x+1
\displaystyle 2x+x+1=7
\displaystyle 3x=6
\displaystyle x=2,\ y=3
\displaystyle \text{So another point is }B(2,3)
\displaystyle \text{Between }x+2y=2\text{ and }2x+y=7
\displaystyle x+2(7-2x)=2
\displaystyle x+14-4x=2
\displaystyle -3x=-12
\displaystyle x=4,\ y=-1
\displaystyle \text{So the third point is }C(4,-1)
\displaystyle \text{For }0\leq x\leq 2,\ \text{upper curve is }y=x+1\text{ and lower curve is }y=1-\frac{x}{2}
\displaystyle \text{For }2\leq x\leq 4,\ \text{upper curve is }y=7-2x\text{ and lower curve is }y=1-\frac{x}{2}
\displaystyle \therefore\ \text{Area }=\int_{0}^{2}\left[\left(x+1\right)-\left(1-\frac{x}{2}\right)\right]dx
\displaystyle \qquad\qquad\qquad +\int_{2}^{4}\left[\left(7-2x\right)-\left(1-\frac{x}{2}\right)\right]dx
\displaystyle =\int_{0}^{2}\frac{3x}{2}\,dx+\int_{2}^{4}\left(6-\frac{3x}{2}\right)dx
\displaystyle =\frac{3}{2}\left[\frac{x^{2}}{2}\right]_{0}^{2}+\left[6x-\frac{3x^{2}}{4}\right]_{2}^{4}
\displaystyle =\frac{3}{2}\cdot 2+\left[(24-12)-(12-3)\right]
\displaystyle =3+(12-9)
\displaystyle =3+3
\displaystyle =6\ \text{square units}

\displaystyle \textbf{SECTION C - 15 MARKS} \ \

\displaystyle \textbf{Question 19} \ \
\displaystyle \text{In subparts (i) and (ii) choose the correct options and in subpart (iii), answer the questions} \ \
\displaystyle \text{as instructed.} \ \
\displaystyle \textbf{(i) } \text{The total revenue received from the sale of }'x'\text{ units of a product is} \ \
\displaystyle R(x)=36x+3x^{2}+5.\ \text{Then, the actual revenue for selling the 10th item will be:} \ \
\displaystyle \text{(a) }27 \ \
\displaystyle \text{(b) }90 \ \
\displaystyle \text{(c) }93 \ \
\displaystyle \text{(d) }33 \ \
\displaystyle \textbf{(ii) } \text{Read the following statements and choose the correct option.} \ \
\displaystyle \text{(I) The correlation coefficient and the regression coefficients are of the same sign.} \ \
\displaystyle \text{(II) The correlation coefficient is the arithmetic mean between the regression} \ \
\displaystyle \text{coefficients.} \ \
\displaystyle \text{(III) The product of two regression coefficients is always equal to }1. \ \
\displaystyle \text{(IV) Both the regression coefficients cannot be numerically greater than unity.} \ \
\displaystyle \text{(a) Only (IV) is correct.} \ \
\displaystyle \text{(b) Only (I) and (II) are correct.} \ \
\displaystyle \text{(c) Only (I) and (IV) are correct.} \ \
\displaystyle \text{(d) Only (III) and (IV) are correct.} \ \
\displaystyle \textbf{(iii) } \text{Consider the following data:} \ \
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x&1&2&3&6\\\hline y&6&5&4&1\\\hline x-\bar{x}&&&&\\\hline y-\bar{y}&&&&\\\hline \end{array} \ \
\displaystyle \text{(a) Calculate }\bar{x}\text{ and }\bar{y} \ \
\displaystyle \text{(b) Complete the table.} \ \
\displaystyle \text{(c) Calculate }b_{xy} \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i) } \text{Total revenue }R(x)=36x+3x^{2}+5
\displaystyle \text{Actual revenue from selling the }10^{\text{th}}\text{ item}=R(10)-R(9)
\displaystyle R(10)=36(10)+3(10)^{2}+5=360+300+5=665
\displaystyle R(9)=36(9)+3(9)^{2}+5=324+243+5=572
\displaystyle R(10)-R(9)=665-572=93
\displaystyle \therefore\ \text{the correct option is (c)}
\displaystyle \textbf{(ii) } \text{(I) The correlation coefficient and the regression coefficients are of the same sign}
\displaystyle \text{This is true}
\displaystyle \text{(II) The correlation coefficient is the arithmetic mean between the regression coefficients}
\displaystyle \text{This is false, because }r=\pm\sqrt{b_{xy}b_{yx}}\text{, not the arithmetic mean}
\displaystyle \text{(III) The product of two regression coefficients is always equal to 1}
\displaystyle \text{This is false, because }b_{xy}b_{yx}=r^{2}\leq 1
\displaystyle \text{(IV) Both the regression coefficients cannot be numerically greater than unity}
\displaystyle \text{This is true}
\displaystyle \therefore\ \text{only (I) and (IV) are correct}
\displaystyle \therefore\ \text{the correct option is (c)}
\displaystyle \textbf{(iii) } \text{Given data: }x:1,2,3,6\quad \text{and}\quad y:6,5,4,1
\displaystyle \text{(a) } \overline{x}=\frac{1+2+3+6}{4}=\frac{12}{4}=3
\displaystyle \overline{y}=\frac{6+5+4+1}{4}=\frac{16}{4}=4
\displaystyle \text{(b) Completing the table:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x&1&2&3&6\\ \hline y&6&5&4&1\\ \hline x-\overline{x}&-2&-1&0&3\\ \hline y-\overline{y}&2&1&0&-3\\ \hline \end{array}
\displaystyle \text{(c) }b_{xy}=\frac{\sum (x-\overline{x})(y-\overline{y})}{\sum (y-\overline{y})^{2}}
\displaystyle \sum (x-\overline{x})(y-\overline{y})=(-2)(2)+(-1)(1)+(0)(0)+(3)(-3)
\displaystyle \sum (x-\overline{x})(y-\overline{y})=-4-1+0-9=-14
\displaystyle \sum (y-\overline{y})^{2}=2^{2}+1^{2}+0^{2}+(-3)^{2}=4+1+0+9=14
\displaystyle \therefore\ b_{xy}=\frac{-14}{14}=-1
\\
\displaystyle \textbf{Question 20} \ \
\displaystyle \textbf{(i) } \text{Find the regression line of best fit from the following data.} \ \
\displaystyle \sum x=24,\ \sum y=44,\ \sum xy=306,\ \sum x^{2}=164,\ \sum y^{2}=576,\ n=4 \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{Two lines of regression are given as }4x+3y+7=0\text{ and }3x+4y+8=0. \ \
\displaystyle \text{Identify the line of regression of }x\text{ on }y. \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \sum x=24,\quad \sum y=44,\quad \sum xy=306,\quad \sum x^{2}=164,\quad n=4
\displaystyle \overline{x}=\frac{\sum x}{n}=\frac{24}{4}=6,\quad \overline{y}=\frac{\sum y}{n}=\frac{44}{4}=11
\displaystyle S_{xy}=\sum xy-\frac{\sum x\sum y}{n}=306-\frac{24\cdot 44}{4}=306-264=42
\displaystyle S_{xx}=\sum x^{2}-\frac{(\sum x)^{2}}{n}=164-\frac{24^{2}}{4}=164-144=20
\displaystyle b_{yx}=\frac{S_{xy}}{S_{xx}}=\frac{42}{20}=\frac{21}{10}
\displaystyle \text{Hence, regression line of }y\text{ on }x\text{ is}
\displaystyle y-\overline{y}=b_{yx}(x-\overline{x})
\displaystyle y-11=\frac{21}{10}(x-6)
\displaystyle 10y-110=21x-126
\displaystyle 21x-10y-16=0
\displaystyle \therefore\ \text{the regression line is }21x-10y-16=0
\displaystyle \textbf{OR} 
\displaystyle \textbf{(ii)}
\displaystyle \text{Given regression lines }4x+3y+7=0\text{ and }3x+4y+8=0
\displaystyle \text{Write them as }x=-\frac{3}{4}y-\frac{7}{4}\text{ and }y=-\frac{3}{4}x-2
\displaystyle \text{The line of regression of }x\text{ on }y\text{ must be of the form }x=a+by
\displaystyle \therefore\ \text{the regression line of }x\text{ on }y\text{ is }4x+3y+7=0
\\
\displaystyle \textbf{Question 21} \ \
\displaystyle \textbf{(i) } \text{A utensil manufacturer produces }'x'\text{ dinner sets per week and sells each set at Rs }p\text{,} \ \
\displaystyle \text{where }x=\frac{600-p}{8}.\text{ The cost of production of }'x'\text{ sets is Rs }x^{2}+78x+2000 \ \
\displaystyle \text{(a) Write the revenue function.} \ \
\displaystyle \text{(b) Write the profit function.} \ \
\displaystyle \text{(c) Calculate the number of dinner sets to be produced and sold per week to ensure} \ \
\displaystyle \text{maximum profit.} \ \
\displaystyle \textbf{OR} \ \
\displaystyle \textbf{(ii) } \text{The Average Cost of producing }'x'\text{ units of commodity is given by:} \ \
\displaystyle AC=\frac{x^{2}}{200}-\frac{x}{50}-30+\frac{5000}{x} \ \
\displaystyle \text{(a) Find the Cost function.} \ \
\displaystyle \text{(b) Find the Marginal Cost function.} \ \
\displaystyle \text{(c) Find the Marginal Average Cost function.} \ \
\displaystyle \text{(d) Verify that }\frac{d}{dx}(AC)=\frac{MC-AC}{x} \ \
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle x=\frac{600-p}{8}
\displaystyle \therefore\ 8x=600-p
\displaystyle \therefore\ p=600-8x
\displaystyle \text{(a) Revenue function }R(x)=xp
\displaystyle \therefore\ R(x)=x(600-8x)
\displaystyle \therefore\ R(x)=600x-8x^{2}
\displaystyle \text{(b) Cost function }C(x)=x^{2}+78x+2000
\displaystyle \text{Profit function }P(x)=R(x)-C(x)
\displaystyle \therefore\ P(x)=(600x-8x^{2})-(x^{2}+78x+2000)
\displaystyle \therefore\ P(x)=600x-8x^{2}-x^{2}-78x-2000
\displaystyle \therefore\ P(x)=-9x^{2}+522x-2000
\displaystyle \text{(c) For maximum profit, }P(x)\text{ must be maximum}
\displaystyle P(x)=-9x^{2}+522x-2000
\displaystyle \text{Since this is a downward opening parabola, maximum occurs at }x= -\frac{b}{2a}
\displaystyle \therefore\ x=-\frac{522}{2(-9)}
\displaystyle \therefore\ x=\frac{522}{18}=29
\displaystyle \therefore\ \text{the number of dinner sets for maximum profit is }29
\displaystyle \textbf{(ii)}
\displaystyle AC=\frac{x^{2}}{200}-\frac{x}{50}-30+\frac{5000}{x}
\displaystyle \text{(a) Average cost }AC=\frac{C(x)}{x}
\displaystyle \therefore\ C(x)=x\cdot AC
\displaystyle \therefore\ C(x)=x\left(\frac{x^{2}}{200}-\frac{x}{50}-30+\frac{5000}{x}\right)
\displaystyle \therefore\ C(x)=\frac{x^{3}}{200}-\frac{x^{2}}{50}-30x+5000
\displaystyle \text{(b) Marginal cost function }MC=\frac{dC}{dx}
\displaystyle \therefore\ MC=\frac{d}{dx}\left(\frac{x^{3}}{200}-\frac{x^{2}}{50}-30x+5000\right)
\displaystyle \therefore\ MC=\frac{3x^{2}}{200}-\frac{2x}{50}-30
\displaystyle \therefore\ MC=\frac{3x^{2}}{200}-\frac{x}{25}-30
\displaystyle \text{(c) Marginal average cost function }=\frac{d}{dx}(AC)
\displaystyle \therefore\ \frac{d}{dx}(AC)=\frac{d}{dx}\left(\frac{x^{2}}{200}-\frac{x}{50}-30+\frac{5000}{x}\right)
\displaystyle \therefore\ \frac{d}{dx}(AC)=\frac{x}{100}-\frac{1}{50}-\frac{5000}{x^{2}}
\displaystyle \text{(d) Now }MC-AC=\left(\frac{3x^{2}}{200}-\frac{x}{25}-30\right)- \left(\frac{x^{2}}{200}-\frac{x}{50}-30+\frac{5000}{x}\right)
\displaystyle \therefore\ MC-AC=\frac{2x^{2}}{200}-\frac{x}{50}-\frac{5000}{x}
\displaystyle \therefore\ MC-AC=\frac{x^{2}}{100}-\frac{x}{50}-\frac{5000}{x}
\displaystyle \therefore\ \frac{MC-AC}{x}=\frac{x}{100}-\frac{1}{50}-\frac{5000}{x^{2}}
\displaystyle \text{But }\frac{d}{dx}(AC)=\frac{x}{100}-\frac{1}{50}-\frac{5000}{x^{2}}
\displaystyle \therefore\ \frac{d}{dx}(AC)=\frac{MC-AC}{x}
\\
\displaystyle \textbf{Question 22} \ \
\displaystyle \text{Two different types of books have to be stacked in the shelf of a library. The first type of} \ \
\displaystyle \text{book weighs }1\text{ kg and has a thickness of }6\text{ cm. The second type of book weighs }1.5\text{ kg and} \ \
\displaystyle \text{has a thickness of }4\text{ cm. The shelf is }96\text{ cm long and can support a maximum weight of} \ \
\displaystyle 21\text{ kg.} \ \
\displaystyle \text{How should both the types of books be placed in the shelf to include the maximum number} \ \
\displaystyle \text{of books? Formulate a Linear Programming Problem and solve it graphically.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the number of books of first type and }y\text{ be the number of books of second type}
\displaystyle \text{Objective function: Maximize }Z=x+y
\displaystyle \text{Subject to the shelf-length constraint }6x+4y\leq 96
\displaystyle \therefore\ 3x+2y\leq 48
\displaystyle \text{Subject to the weight constraint }x+1.5y\leq 21
\displaystyle \therefore\ 2x+3y\leq 42
\displaystyle x\geq 0,\ y\geq 0
\displaystyle \text{Thus, the Linear Programming Problem is:}
\displaystyle \text{Maximize }Z=x+y
\displaystyle \text{subject to }3x+2y\leq 48,\ 2x+3y\leq 42,\ x\geq 0,\ y\geq 0
\displaystyle \text{Now find the corner points of the feasible region}
\displaystyle \text{On the }x\text{-axis, if }y=0,\ 3x\leq 48\Rightarrow x\leq 16,\ \text{and }2x\leq 42\Rightarrow x\leq 21
\displaystyle \therefore\ \text{the feasible point on the }x\text{-axis is }(16,0)
\displaystyle \text{On the }y\text{-axis, if }x=0,\ 2y\leq 48\Rightarrow y\leq 24,\ \text{and }3y\leq 42\Rightarrow y\leq 14
\displaystyle \therefore\ \text{the feasible point on the }y\text{-axis is }(0,14)
\displaystyle \text{Now find the intersection of }3x+2y=48\text{ and }2x+3y=42
\displaystyle 3x+2y=48
\displaystyle 2x+3y=42
\displaystyle \text{Multiplying the first equation by }3,\ 9x+6y=144
\displaystyle \text{Multiplying the second equation by }2,\ 4x+6y=84
\displaystyle \text{Subtracting, }5x=60
\displaystyle \therefore\ x=12
\displaystyle \text{Substituting in }3x+2y=48,\ 36+2y=48
\displaystyle 2y=12
\displaystyle \therefore\ y=6
\displaystyle \text{So the intersection point is }(12,6)
\displaystyle \text{Now evaluate }Z=x+y\text{ at the corner points}
\displaystyle \text{At }(0,0),\ Z=0
\displaystyle \text{At }(16,0),\ Z=16
\displaystyle \text{At }(0,14),\ Z=14
\displaystyle \text{At }(12,6),\ Z=18
\displaystyle \text{The maximum value of }Z\text{ is }18\text{ at }(12,6)
\displaystyle \therefore\ \text{the maximum number of books is }18
\displaystyle \therefore\ \text{the shelf should contain }12\text{ books of the first type and }6\text{ books of the second type}

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