\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 (x) choose the correct options and in subparts (xi) to (xv), answer the} \ \
\displaystyle \text{questions as instructed.} \ \

\displaystyle \text{(i) Let }L\text{ be a set of all straight lines in a plane. The relation }R\text{ on }L\text{ defined as} \ \
\displaystyle \text{'perpendicular to' is:} \ \
\displaystyle \text{(a) Symmetric and Transitive} \ \
\displaystyle \text{(b) Transitive} \ \
\displaystyle \text{(c) Symmetric} \ \
\displaystyle \text{(d) Equivalence} \ \
\displaystyle \text{Answer:}
\displaystyle \text{If a line }l_{1}\text{ is perpendicular to }l_{2},\text{ then }l_{2}\text{ is}
\displaystyle \text{also perpendicular to }l_{1},\text{ hence the relation is symmetric.}
\displaystyle \text{If }l_{1}\perp l_{2}\text{ and }l_{2}\perp l_{3},\text{ then }l_{1}\parallel l_{3},
\displaystyle \text{so }l_{1}\text{ is not perpendicular to }l_{3},\text{ hence not transitive.}
\displaystyle \text{Also, no line is perpendicular to itself, so the relation is not reflexive.}
\displaystyle \text{Therefore, the relation is only symmetric.}
\displaystyle \therefore\ \text{Correct option is (c) Symmetric.}
\\
\displaystyle \text{(ii) The order and the degree of differential equation }1+\left(\frac{dy}{dx}\right)^{2}=\frac{d^{2}y}{dx^{2}}\text{ are:} \ \
\displaystyle \text{(a) }2\text{ and }\frac{3}{2} \ \
\displaystyle \text{(b) }2\text{ and }3 \ \
\displaystyle \text{(c) }3\text{ and }4 \ \
\displaystyle \text{(d) }2\text{ and }1 \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given equation: }1+\left(\frac{dy}{dx}\right)^{2}=\frac{d^{2}y}{dx^{2}}
\displaystyle \text{The highest order derivative present is }\frac{d^{2}y}{dx^{2}},\text{ so order }=2.
\displaystyle \text{The equation is already polynomial in derivatives.}
\displaystyle \text{The highest power of the highest order derivative is }1.
\displaystyle \therefore\ \text{degree }=1.
\displaystyle \therefore\ \text{Order }=2,\ \text{Degree }=1.
\displaystyle \therefore\ \text{Correct option is (d).}
\\
\displaystyle \text{(iii) Let }A\text{ be a non-empty set.} \ \
\displaystyle \textbf{Statement 1: }\text{Identity relation on }A\text{ is Reflexive.} \ \
\displaystyle \textbf{Statement 2: }\text{Every Reflexive relation on }A\text{ is an Identity relation.} \ \
\displaystyle \text{(a) Both the statements are true.} \ \
\displaystyle \text{(b) Both the statements are false.} \ \
\displaystyle \text{(c) Statement 1 is true and Statement 2 is false.} \ \
\displaystyle \text{(d) Statement 1 is false and Statement 2 is true.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Statement 1: Identity relation on }A\text{ is Reflexive.}
\displaystyle \text{In identity relation, }R=\{(a,a)\mid a\in A\},\text{ so every element is related to itself.}
\displaystyle \therefore\ \text{Identity relation is reflexive, so Statement 1 is true.}
\displaystyle \text{Statement 2: Every reflexive relation on }A\text{ is an Identity relation.}
\displaystyle \text{A reflexive relation only requires }(a,a)\in R\ \forall a\in A,
\displaystyle \text{but it may also contain other ordered pairs.}
\displaystyle \text{Hence, every reflexive relation need not be identity relation.}
\displaystyle \therefore\ \text{Statement 2 is false.}
\displaystyle \therefore\ \text{Statement 1 is true and Statement 2 is false.}
\displaystyle \therefore\ \text{Correct option is (c).}
\\
\displaystyle \text{(iv) The graph of the function }f\text{ is shown below.} \ \
\displaystyle \text{Of the following options, at what values of }x\text{ is the function }f\text{ NOT differentiable?} \ \
\displaystyle \text{(a) At }x=0\text{ and }x=2 \ \
\displaystyle \text{(b) At }x=1\text{ and }x=3 \ \
\displaystyle \text{(c) At }x=-1\text{ and }x=1 \ \
\displaystyle \text{(d) At }x=-1.5\text{ and }x=1.5 \ \
\displaystyle \text{Answer:}
\displaystyle \text{From the graph, }f\text{ is not differentiable at points where there is a sharp corner}
\displaystyle \text{or where two curves meet non-smoothly.}
\displaystyle \text{At }x=0,\text{ the left curve and right curve do not have the same slope.}
\displaystyle \text{At }x=2,\text{ the curve changes abruptly forming a corner point.}
\displaystyle \therefore\ f\text{ is not differentiable at }x=0\text{ and }x=2.
\displaystyle \therefore\ \text{Correct option is (a).}
\\
\displaystyle \text{(v) The value of }\mathrm{cosec}\left(\sin^{-1}\left(\frac{-1}{2}\right)\right)-\sec\left(\cos^{-1}\left(\frac{-1}{2}\right)\right)\text{ is equal to:} \ \
\displaystyle \text{(a) }-4 \ \
\displaystyle \text{(b) }0 \ \
\displaystyle \text{(c) }-1 \ \
\displaystyle \text{(d) }4 \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta=\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}
\displaystyle \Rightarrow\ \mathrm{cosec}\left(\sin^{-1}\left(-\frac{1}{2}\right)\right)=\mathrm{cosec}\left(-\frac{\pi}{6}\right)=-2
\displaystyle \text{Let }\phi=\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}
\displaystyle \Rightarrow\ \sec\left(\cos^{-1}\left(-\frac{1}{2}\right)\right)=\sec\left(\frac{2\pi}{3}\right)=-2
\displaystyle \therefore\ \mathrm{cosec}\left(\sin^{-1}\left(-\frac{1}{2}\right)\right)-\sec\left(\cos^{-1}\left(-\frac{1}{2}\right)\right)=-2-(-2)=0
\displaystyle \therefore\ \text{Correct option is (b).}
\\
\displaystyle \text{(vi) The value of }\int_{1}^{\sqrt{3}}\frac{dx}{1+x^{2}}\text{ is:} \ \
\displaystyle \text{(a) }\frac{\pi}{2} \ \
\displaystyle \text{(b) }\frac{2\pi}{3} \ \
\displaystyle \text{(c) }\frac{\pi}{6} \ \
\displaystyle \text{(d) }\frac{\pi}{12} \ \
\displaystyle \text{Answer:}
\displaystyle \int_{1}^{\sqrt{3}}\frac{dx}{1+x^{2}}=\left[\tan^{-1}x\right]_{1}^{\sqrt{3}}
\displaystyle =\tan^{-1}(\sqrt{3})-\tan^{-1}(1)
\displaystyle =\frac{\pi}{3}-\frac{\pi}{4}
\displaystyle =\frac{\pi}{12}
\displaystyle \therefore\ \text{Correct option is (d).}
\\
\displaystyle \text{(vii) Assertion: Let the matrices }A=\begin{pmatrix}-3&2\\-5&4\end{pmatrix}\text{ and }B=\begin{pmatrix}4&-2\\5&-3\end{pmatrix}\text{ be such that} \ \
\displaystyle A^{100}B=BA^{100} \ \
\displaystyle \text{Reason: }AB=BA\text{ implies }A^{n}B=BA^{n}\text{ for all positive integers }n. \ \
\displaystyle \text{(a) Both Assertion and Reason are true and Reason is the correct explanation} \ \
\displaystyle \text{for Assertion.} \ \
\displaystyle \text{(b) Both Assertion and Reason are true but Reason is not the correct} \ \
\displaystyle \text{explanation 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 A=\begin{pmatrix}-3&2\\-5&4\end{pmatrix},\quad B=\begin{pmatrix}4&-2\\5&-3\end{pmatrix}
\displaystyle AB=\begin{pmatrix}-3&2\\-5&4\end{pmatrix} \begin{pmatrix}4&-2\\5&-3\end{pmatrix} =\begin{pmatrix}-12+10&6-6\\-20+20&10-12\end{pmatrix} =\begin{pmatrix}-2&0\\0&-2\end{pmatrix}
\displaystyle BA=\begin{pmatrix}4&-2\\5&-3\end{pmatrix} \begin{pmatrix}-3&2\\-5&4\end{pmatrix} =\begin{pmatrix}-12+10&8-8\\-15+15&10-12\end{pmatrix} =\begin{pmatrix}-2&0\\0&-2\end{pmatrix}
\displaystyle \therefore\ AB=BA
\displaystyle \text{Hence, }A^{n}B=BA^{n}\ \text{for all positive integers }n
\displaystyle \text{Thus, }A^{100}B=BA^{100}\text{, so Assertion is true.}
\displaystyle \text{Reason states that if }AB=BA,\text{ then }A^{n}B=BA^{n},\text{ which is true.}
\displaystyle \therefore\ \text{Both Assertion and Reason are true and Reason is the correct explanation.}
\displaystyle \therefore\ \text{Correct option is (a).}
\\
\displaystyle \text{(viii) If }\int(\cot x-\mathrm{cosec}^{2}x)e^{x}\,dx=e^{x}f(x)+c\text{ then }f(x)\text{ will be:} \ \
\displaystyle \text{(a) }\cot x+\mathrm{cosec}\ x \ \
\displaystyle \text{(b) }\cot^{2}x \ \
\displaystyle \text{(c) }\cot x \ \
\displaystyle \text{(d) }\mathrm{cosec}\ x \ \
\displaystyle \text{Answer:}
\displaystyle \int\left(\cot x-\mathrm{cosec}^{2}x\right)e^{x}dx=e^{x}f(x)+c
\displaystyle \text{Using }\int e^{x}g(x)\,dx=e^{x}\left(g(x)-g'(x)\right)+c
\displaystyle \text{Let }g(x)=\cot x\Rightarrow g'(x)=-\mathrm{cosec}^{2}x
\displaystyle \Rightarrow g(x)-g'(x)=\cot x+\mathrm{cosec}^{2}x
\displaystyle \text{But required integrand is }\cot x-\mathrm{cosec}^{2}x
\displaystyle \text{Let }f(x)=\cot x\Rightarrow \frac{d}{dx}\left(e^{x}f(x)\right)=e^{x}\left(f(x)+f'(x)\right)
\displaystyle =e^{x}\left(\cot x-\mathrm{cosec}^{2}x\right)
\displaystyle \therefore\ \int\left(\cot x-\mathrm{cosec}^{2}x\right)e^{x}dx=e^{x}\cot x+c
\displaystyle \therefore\ f(x)=\cot x
\displaystyle \therefore\ \text{Correct option is (c).}
\\
\displaystyle \text{(ix) In which one of the following intervals is the function }f(x)=x^{3}-12x \ \
\displaystyle \text{increasing?} \ \
\displaystyle \text{(a) }(-2,2) \ \
\displaystyle \text{(b) }(-\infty,-2)\cup(2,\infty) \ \
\displaystyle \text{(c) }(-2,\infty) \ \
\displaystyle \text{(d) }(-\infty,2) \ \
\displaystyle \text{Answer:}
\displaystyle f(x)=x^{3}-12x
\displaystyle f'(x)=3x^{2}-12=3(x^{2}-4)=3(x-2)(x+2)
\displaystyle f'(x)>0\ \text{when }x<-2\ \text{or }x>2
\displaystyle \therefore\ f(x)\text{ is increasing on }(-\infty,-2)\cup(2,\infty)
\displaystyle \therefore\ \text{Correct option is (b).}
\\
\displaystyle \text{(x) If }A\text{ and }B\text{ are symmetric matrices of the same order, then }AB-BA\text{ is:} \ \
\displaystyle \text{(a) Skew-symmetric matrix} \ \
\displaystyle \text{(b) Symmetric matrix} \ \
\displaystyle \text{(c) Diagonal matrix} \ \
\displaystyle \text{(d) Identity matrix} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ and }B\text{ be symmetric matrices, so }A^{T}=A,\ B^{T}=B
\displaystyle (AB-BA)^{T}=B^{T}A^{T}-A^{T}B^{T}
\displaystyle =BA-AB
\displaystyle =-(AB-BA)
\displaystyle \therefore\ AB-BA\text{ is skew-symmetric.}
\displaystyle \therefore\ \text{Correct option is (a).}
\\
\displaystyle \text{(xi) Find the derivative of }y=\log x+\frac{1}{x}\text{ with respect to }x. \ \
\displaystyle \text{Answer:}
\displaystyle y=\log x+\frac{1}{x}
\displaystyle \frac{dy}{dx}=\frac{1}{x}-\frac{1}{x^{2}}
\\
\displaystyle \text{(xii) Teena is practising for an upcoming Rifle Shooting tournament. The probability of} \ \
\displaystyle \text{her shooting the target in the 1st, 2nd, 3rd and 4th shots are 0.4, 0.3, 0.2 and 0.1} \ \
\displaystyle \text{respectively. Find the probability of at least one shot of Teena hitting the target.} \ \
\displaystyle \text{Answer:}
\displaystyle P(\text{at least one hit})=1-P(\text{no hit in all 4 shots})
\displaystyle =1-(1-0.4)(1-0.3)(1-0.2)(1-0.1)
\displaystyle =1-(0.6)(0.7)(0.8)(0.9)
\displaystyle =1-0.3024
\displaystyle =0.6976
\\
\displaystyle \text{(xiii) Which one of the following graphs is a function of }x\text{?} \ \
\displaystyle \text{Graph A \qquad Graph B} \ \ \displaystyle \text{Answer:}
\displaystyle \text{A graph represents a function of }x\text{ if it satisfies the vertical line test.}
\displaystyle \text{Graph A: Every vertical line cuts the graph at most at one point.}
\displaystyle \text{Hence, Graph A represents a function.}
\displaystyle \text{Graph B: Some vertical lines intersect the graph at more than one point.}
\displaystyle \text{Hence, Graph B does not represent a function.}
\displaystyle \therefore\ \text{Correct answer is Graph A.}
\\
\displaystyle \text{(xiv) Evaluate: }\int_{0}^{6}|x+3|\,dx \ \
\displaystyle \text{Answer:}
\displaystyle \int_{0}^{6}|x+3|\,dx
\displaystyle \text{Since }x+3>0\text{ for }x\in[0,6],\ |x+3|=x+3
\displaystyle \therefore\ \int_{0}^{6}(x+3)\,dx
\displaystyle =\left[\frac{x^{2}}{2}+3x\right]_{0}^{6}
\displaystyle =\left(\frac{36}{2}+18\right)-0
\displaystyle =18+18
\displaystyle =36
\\
\displaystyle \text{(xv) Given that }\frac{1}{y}+\frac{1}{x}=\frac{1}{12}\text{ and }y\text{ decreases at a rate of }1\ \mathrm{cm\,s}^{-1}\text{, find the rate of change} \ \
\displaystyle \text{of }x\text{ when }x=5\ \mathrm{cm}\text{ and }y=1\ \mathrm{cm} \ \
\displaystyle \text{Answer:}
\displaystyle \frac{1}{y}+\frac{1}{x}=\frac{1}{12}
\displaystyle \text{Differentiate w.r.t. }t
\displaystyle -\frac{1}{y^{2}}\frac{dy}{dt}-\frac{1}{x^{2}}\frac{dx}{dt}=0
\displaystyle \Rightarrow\ \frac{1}{x^{2}}\frac{dx}{dt}=-\frac{1}{y^{2}}\frac{dy}{dt}
\displaystyle \Rightarrow\ \frac{dx}{dt}=-\frac{x^{2}}{y^{2}}\frac{dy}{dt}
\displaystyle \text{Given }\frac{dy}{dt}=-1,\ x=5,\ y=1
\displaystyle \therefore\ \frac{dx}{dt}=-\frac{25}{1}(-1)=25
\displaystyle \therefore\ \frac{dx}{dt}=25\ \text{cm s}^{-1}
\\

\displaystyle \textbf{Question 2} \ \
\displaystyle \text{(i) Let }f:R-\left\{\frac{-1}{3}\right\}\rightarrow R-\{0\}\text{ be defined as }f(x)=\frac{5}{3x+1}\text{ is invertible. Find }f^{-1}(x) \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) If }f:R\rightarrow R\text{ is defined by }f(x)=\frac{2x-7}{4}\text{, show that }f(x)\text{ is one-one and onto.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Finding }f^{-1}(x)\text{:}
\displaystyle \text{Given }y=\frac{5}{3x+1}
\displaystyle \Rightarrow\ y(3x+1)=5
\displaystyle \Rightarrow\ 3xy+y=5
\displaystyle \Rightarrow\ 3xy=5-y
\displaystyle \Rightarrow\ x=\frac{5-y}{3y}
\displaystyle \text{Replacing }y\text{ by }x,\ \therefore\ f^{-1}(x)=\frac{5-x}{3x}
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) Showing that }f(x)=\frac{2x-7}{4}\text{ is one-one and onto:}
\displaystyle \text{One-one:}
\displaystyle \text{Let }f(a)=f(b)
\displaystyle \Rightarrow\ \frac{2a-7}{4}=\frac{2b-7}{4}
\displaystyle \Rightarrow\ 2a-7=2b-7
\displaystyle \Rightarrow\ 2a=2b
\displaystyle \Rightarrow\ a=b
\displaystyle \therefore\ f\text{ is one-one.}
\displaystyle \text{Onto:}
\displaystyle \text{Let }y\in R\text{ be any real number.}
\displaystyle \text{We need to find }x\in R\text{ such that }f(x)=y
\displaystyle \Rightarrow\ \frac{2x-7}{4}=y
\displaystyle \Rightarrow\ 2x-7=4y
\displaystyle \Rightarrow\ 2x=4y+7
\displaystyle \Rightarrow\ x=\frac{4y+7}{2}
\displaystyle \text{Since }\frac{4y+7}{2}\in R,\text{ there exists }x\in R\text{ for every }y\in R
\displaystyle \therefore\ f\text{ is onto.}
\displaystyle \therefore\ f(x)=\frac{2x-7}{4}\text{ is one-one and onto.}
\\
\displaystyle \textbf{Question 3} \ \
\displaystyle \text{Find the value of the determinant given below, without expanding it at any stage.} \ \
\displaystyle \left|\begin{matrix}\beta\gamma&1&\alpha(\beta+\gamma)\\\gamma\alpha&1&\beta(\gamma+\alpha)\\\alpha\beta&1&\gamma(\alpha+\beta)\end{matrix}\right| \ \
\displaystyle \text{Answer:}
\displaystyle D=\begin{vmatrix}\beta\gamma&1&\alpha(\beta+\gamma)\\ \gamma\alpha&1&\beta(\gamma+\alpha)\\ \alpha\beta&1&\gamma(\alpha+\beta)\end{vmatrix}
\displaystyle \text{Apply }C_{3}\rightarrow C_{3}-\alpha C_{1}-\beta C_{1}-\gamma C_{1}\ \text{(rearranged row-wise)}
\displaystyle \text{Observe each row: }\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma,\ \beta(\gamma+\alpha)=\beta\gamma+\beta\alpha,\ \gamma(\alpha+\beta)=\gamma\alpha+\gamma\beta
\displaystyle \text{Thus }C_{3}=C_{1}+C_{1}\ \text{(cyclic combinations give linear dependence)}
\displaystyle \text{Hence columns are linearly dependent}
\displaystyle \therefore\ D=0
\\
\displaystyle \textbf{Question 4} \ \
\displaystyle \text{(i) Determine the value of }k\text{ for which the following function is continuous at }x=3. \ \
\displaystyle f(x)=\left\{\begin{matrix}\frac{(x+3)^{2}-36}{x-3}\ ;&x\neq 3\\k\ ;&x=3\end{matrix}\right. \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) Find a point on the curve }y=(x-2)^{2}\text{ at which the tangent is parallel to the line} \ \
\displaystyle \text{joining the chord through the points }(2,0)\text{ and }(4,4). \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) For continuity at }x=3\text{:}
\displaystyle f(x)=\frac{(x+3)^{2}-36}{x-3},\ x\neq 3
\displaystyle =\frac{x^{2}+6x+9-36}{x-3}
\displaystyle =\frac{x^{2}+6x-27}{x-3}
\displaystyle =\frac{(x-3)(x+9)}{x-3}
\displaystyle =x+9,\ x\neq 3
\displaystyle \text{For continuity at }x=3,\ \lim_{x\to 3}f(x)=f(3)=k
\displaystyle \therefore\ k=\lim_{x\to 3}(x+9)=12
\displaystyle \therefore\ k=12
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) Point on }y=(x-2)^{2}\text{ where tangent is parallel to the chord:}
\displaystyle \text{The chord joins }(2,0)\text{ and }(4,4)
\displaystyle \text{Slope of chord }=\frac{4-0}{4-2}=\frac{4}{2}=2
\displaystyle y=(x-2)^{2}
\displaystyle \Rightarrow\ \frac{dy}{dx}=2(x-2)
\displaystyle \text{For tangent parallel to chord, }\frac{dy}{dx}=2
\displaystyle \Rightarrow\ 2(x-2)=2
\displaystyle \Rightarrow\ x-2=1
\displaystyle \Rightarrow\ x=3
\displaystyle y=(3-2)^{2}=1
\displaystyle \therefore\ \text{The required point is }(3,1)
\\
\displaystyle \textbf{Question 5} \ \
\displaystyle \text{Evaluate: }\int_{0}^{2\pi}\frac{1}{1+e^{\sin x}}\,dx \ \
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{2\pi}\frac{dx}{1+e^{\sin x}}
\displaystyle \text{Let }I=\int_{0}^{2\pi}\frac{dx}{1+e^{\sin x}}
\displaystyle \text{Also, }I=\int_{0}^{2\pi}\frac{dx}{1+e^{-\sin x}}
\displaystyle \text{Adding, }2I=\int_{0}^{2\pi}\left(\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}\right)dx
\displaystyle =\int_{0}^{2\pi}1\ dx
\displaystyle =2\pi
\displaystyle \therefore\ I=\pi
\\
\displaystyle \textbf{Question 6} \ \
\displaystyle \text{Evaluate: }P(A\cup B)\text{ if }2P(A)=P(B)=\frac{5}{13}\text{ and }P(A|B)=\frac{2}{5} \ \
\displaystyle \text{Answer:}
\displaystyle P(A)=\frac{5}{13},\quad P(B)=\frac{5}{13},\quad P(A\mid B)=\frac{2}{5}
\displaystyle P(A\mid B)=\frac{P(A\cap B)}{P(B)}
\displaystyle \Rightarrow\ \frac{2}{5}=\frac{P(A\cap B)}{\frac{5}{13}}
\displaystyle \Rightarrow\ P(A\cap B)=\frac{2}{5}\cdot\frac{5}{13}=\frac{2}{13}
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)
\displaystyle =\frac{5}{13}+\frac{5}{13}-\frac{2}{13}
\displaystyle =\frac{8}{13}
\\
\displaystyle \textbf{Question 7} \ \
\displaystyle \text{If }y=3\cos(\log x)+4\sin(\log x)\text{, show that }x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+y=0 \ \
\displaystyle \text{Answer:}
\displaystyle y=3\cos(\log x)+4\sin(\log x)
\displaystyle \frac{dy}{dx}=3\left(-\sin(\log x)\right)\frac{1}{x}+4\cos(\log x)\frac{1}{x}
\displaystyle \frac{dy}{dx}=\frac{-3\sin(\log x)+4\cos(\log x)}{x}
\displaystyle x\frac{dy}{dx}=-3\sin(\log x)+4\cos(\log x)
\displaystyle \frac{d}{dx}\left(x\frac{dy}{dx}\right)=\frac{d}{dx}\left(-3\sin(\log x)+4\cos(\log x)\right)
\displaystyle x\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}=\frac{-3\cos(\log x)-4\sin(\log x)}{x}
\displaystyle x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=-3\cos(\log x)-4\sin(\log x)
\displaystyle x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=-(3\cos(\log x)+4\sin(\log x))
\displaystyle x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}=-y
\displaystyle \therefore\ x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}+y=0
\\
\displaystyle \textbf{Question 8} \ \
\displaystyle \text{(i) Solve for }x:\ \sin^{-1}\left(\frac{x}{2}\right)+\cos^{-1}x=\frac{\pi}{6} \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) If }\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi\text{, show that }x^{2}-y^{2}-z^{2}+2yz\sqrt{1-x^{2}}=0 \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Solve for }x\text{:}
\displaystyle \sin^{-1}\left(\frac{x}{2}\right)+\cos^{-1}x=\frac{\pi}{6}
\displaystyle \text{Let }\sin^{-1}\left(\frac{x}{2}\right)=\theta
\displaystyle \Rightarrow\ \frac{x}{2}=\sin\theta\ \Rightarrow\ x=2\sin\theta
\displaystyle \text{Also, }\cos^{-1}x=\frac{\pi}{6}-\theta
\displaystyle \Rightarrow\ x=\cos\left(\frac{\pi}{6}-\theta\right)
\displaystyle \therefore\ 2\sin\theta=\cos\left(\frac{\pi}{6}-\theta\right)
\displaystyle =\cos\frac{\pi}{6}\cos\theta+\sin\frac{\pi}{6}\sin\theta
\displaystyle =\frac{\sqrt{3}}{2}\cos\theta+\frac{1}{2}\sin\theta
\displaystyle \Rightarrow\ 4\sin\theta=\sqrt{3}\cos\theta+\sin\theta
\displaystyle \Rightarrow\ 3\sin\theta=\sqrt{3}\cos\theta
\displaystyle \Rightarrow\ \tan\theta=\frac{1}{\sqrt{3}}
\displaystyle \Rightarrow\ \theta=\frac{\pi}{6}
\displaystyle \therefore\ x=2\sin\frac{\pi}{6}=1
\displaystyle \therefore\ x=1
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) To show: }x^{2}-y^{2}-z^{2}+2yz\sqrt{1-x^{2}}=0\text{:}
\displaystyle \sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi
\displaystyle \Rightarrow\ \sin^{-1}x=\pi-\left(\sin^{-1}y+\sin^{-1}z\right)
\displaystyle \text{Taking sine on both sides,}
\displaystyle x=\sin\left(\sin^{-1}y+\sin^{-1}z\right)
\displaystyle =y\sqrt{1-z^{2}}+z\sqrt{1-y^{2}}
\displaystyle \text{Now, }\sqrt{1-x^{2}}=\cos(\sin^{-1}x)
\displaystyle =-\cos\left(\sin^{-1}y+\sin^{-1}z\right)
\displaystyle =yz-\sqrt{1-y^{2}}\sqrt{1-z^{2}}
\displaystyle \text{Hence, }\sqrt{1-y^{2}}\sqrt{1-z^{2}}=yz-\sqrt{1-x^{2}}
\displaystyle \text{Squaring, }(1-y^{2})(1-z^{2})=y^{2}z^{2}+1-x^{2}-2yz\sqrt{1-x^{2}}
\displaystyle \Rightarrow\ 1-y^{2}-z^{2}+y^{2}z^{2}=y^{2}z^{2}+1-x^{2}-2yz\sqrt{1-x^{2}}
\displaystyle \Rightarrow\ -y^{2}-z^{2}=-x^{2}-2yz\sqrt{1-x^{2}}
\displaystyle \Rightarrow\ x^{2}-y^{2}-z^{2}+2yz\sqrt{1-x^{2}}=0
\displaystyle \therefore\ \text{Proved.}
\\
\displaystyle \textbf{Question 9} \ \
\displaystyle \text{(i) Evaluate: }\int x^{2}\cos x\,dx \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) Evaluate: }\int\frac{x+7}{x^{2}+4x+7}\,dx \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i)} \int x^{2}\cos x\,dx
\displaystyle =x^{2}\sin x-\int 2x\sin x\,dx
\displaystyle =x^{2}\sin x-2\left(-x\cos x+\int \cos x\,dx\right)
\displaystyle =x^{2}\sin x+2x\cos x-2\sin x+c
\displaystyle \text{OR} \ \
\displaystyle \text{(ii)} \int \frac{x+7}{x^{2}+4x+7}\,dx
\displaystyle \text{Let }I=\int \frac{x+7}{x^{2}+4x+7}\,dx
\displaystyle =\int \frac{x+2}{x^{2}+4x+7}\,dx+\int \frac{5}{x^{2}+4x+7}\,dx
\displaystyle =\frac{1}{2}\int \frac{2x+4}{x^{2}+4x+7}\,dx+5\int \frac{dx}{(x+2)^{2}+3}
\displaystyle =\frac{1}{2}\log(x^{2}+4x+7)+\frac{5}{\sqrt{3}}\tan^{-1}\left(\frac{x+2}{\sqrt{3}}\right)+c
\\
\displaystyle \textbf{Question 10} \ \
\displaystyle \text{A jewellery seller has precious gems in white and red colour which he has put in three} \ \
\displaystyle \text{boxes. The distribution of these gems is shown in the table given below:} \ \
\displaystyle \begin{array}{|c|c|c|}\hline \text{Box}&\multicolumn{2}{c|}{\text{Number of Gems}}\\\hline &\text{White}&\text{Red}\\\hline \text{I}&1&2\\\hline \text{II}&2&3\\\hline \text{III}&3&1\\\hline \end{array} \ \
\displaystyle \text{He wants to gift two gems to his mother. So, he asks her to select one box at random and} \ \
\displaystyle \text{pick out any two gems one after the other without replacement from the selected box. The} \ \
\displaystyle \text{mother selects one white and one red gem.} \ \
\displaystyle \text{Calculate the probability that the gems drawn are from Box II.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let }E\text{ be the event of getting one white and one red gem.}
\displaystyle \text{Since one box is selected at random, }P(\mathrm{I})=P(\mathrm{II})=P(\mathrm{III})=\frac{1}{3}
\displaystyle P(E\mid \mathrm{I})=\frac{1}{3}\cdot\frac{2}{2}+\frac{2}{3}\cdot\frac{1}{2}=\frac{2}{3}
\displaystyle P(E\mid \mathrm{II})=\frac{2}{5}\cdot\frac{3}{4}+\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{5}
\displaystyle P(E\mid \mathrm{III})=\frac{3}{4}\cdot\frac{1}{3}+\frac{1}{4}\cdot\frac{3}{3}=\frac{1}{2}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(\mathrm{II}\mid E)=\frac{P(\mathrm{II})P(E\mid \mathrm{II})}{P(\mathrm{I})P(E\mid \mathrm{I})+P(\mathrm{II})P(E\mid \mathrm{II})+P(\mathrm{III})P(E\mid \mathrm{III})}
\displaystyle =\frac{\frac{1}{3}\cdot\frac{3}{5}}{\frac{1}{3}\cdot\frac{2}{3}+\frac{1}{3}\cdot\frac{3}{5}+\frac{1}{3}\cdot\frac{1}{2}}
\displaystyle =\frac{\frac{1}{5}}{\frac{2}{9}+\frac{1}{5}+\frac{1}{6}}
\displaystyle =\frac{\frac{1}{5}}{\frac{20+18+15}{90}}
\displaystyle =\frac{\frac{1}{5}}{\frac{53}{90}}
\displaystyle =\frac{18}{53}
\displaystyle \therefore\ \text{The required probability is }\frac{18}{53}.
\\
\displaystyle \textbf{Question 11} \ \
\displaystyle \text{A furniture factory uses three types of wood namely, teakwood, rosewood and satinwood} \ \
\displaystyle \text{for manufacturing three types of furniture, that are, table, chair and cot.} \ \
\displaystyle \text{The wood requirements (in tonnes) for each type of furniture are given below:} \ \
\displaystyle \begin{array}{|c|c|c|c|}\hline &\text{Table}&\text{Chair}&\text{Cot}\\\hline \text{Teakwood}&2&3&4\\\hline \text{Rosewood}&1&1&2\\\hline \text{Satinwood}&3&2&1\\\hline \end{array} \ \
\displaystyle \text{It is found that }29\text{ tonnes of teakwood, }13\text{ tonnes of rosewood and }16\text{ tonnes of satinwood} \ \
\displaystyle \text{are available to make all three types of furniture.} \ \
\displaystyle \text{Using the above information, answer the following questions:} \ \
\displaystyle \text{(i) Express the data given in the table above in the form of a set of simultaneous} \ \
\displaystyle \text{equations.} \ \
\displaystyle \text{(ii) Solve the set of simultaneous equations formed in subpart (i) by matrix method.} \ \
\displaystyle \text{(iii) Hence, find the number of table(s), chair(s) and cot(s) produced.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }x,y,z\text{ be the number of tables, chairs and cots respectively.}
\displaystyle 2x+3y+4z=29
\displaystyle x+y+2z=13
\displaystyle 3x+2y+z=16
\displaystyle \text{(ii) Writing in matrix form:}
\displaystyle \begin{pmatrix}2&3&4\\1&1&2\\3&2&1\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix}= \begin{pmatrix}29\\13\\16\end{pmatrix}
\displaystyle \text{Solving,}
\displaystyle x+y+2z=13\Rightarrow x=13-y-2z
\displaystyle \text{Substitute in }2x+3y+4z=29
\displaystyle 2(13-y-2z)+3y+4z=29
\displaystyle 26-2y-4z+3y+4z=29
\displaystyle y=3
\displaystyle \text{Substitute }y=3\text{ in }x=13-y-2z\Rightarrow x=10-2z
\displaystyle \text{Substitute in }3x+2y+z=16
\displaystyle 3(10-2z)+6+z=16
\displaystyle 30-6z+6+z=16
\displaystyle 36-5z=16
\displaystyle z=4
\displaystyle x=10-2(4)=2
\displaystyle \text{(iii) Hence, number of tables }=2,\ \text{chairs }=3,\ \text{cots }=4.
\\
\displaystyle \textbf{Question 12} \ \
\displaystyle \text{(i) Mrs. Roy designs a window in her son's study room so that the room gets maximum} \ \
\displaystyle \text{sunlight. She designs the window in the shape of a rectangle surmounted by an} \ \
\displaystyle \text{equilateral triangle. If the perimeter of the window is }12\text{ m, find the dimensions of the} \ \
\displaystyle \text{window that will admit maximum sunlight into the room.} \ \ \displaystyle \text{OR} \ \
\displaystyle \text{(ii) Sumit has bought a closed cylindrical dustbin. The radius of the dustbin is }r\text{ cm and} \ \
\displaystyle \text{height is }h\text{ cm. It has a volume of }20\pi\ \mathrm{cm}^{3}\text{.} \ \ \displaystyle \text{(a) Express }h\text{ in terms of }r\text{, using the given volume.} \ \
\displaystyle \text{(b) Prove that the total surface area of the dustbin is }2\pi r^{2}+\frac{40\pi}{r} \ \
\displaystyle \text{(c) Sumit wants to paint the dustbin. The cost of painting the base and top of the} \ \
\displaystyle \text{dustbin is Rs }2\text{ per }\mathrm{cm}^{2}\text{ and the cost of painting the curved side is} \ \
\displaystyle \text{Rs }25\text{ per }\mathrm{cm}^{2}\text{. Find the total cost in terms of }r\text{, for painting the outer surface} \ \
\displaystyle \text{of the dustbin including the base and top.} \ \
\displaystyle \text{(d) Calculate the minimum cost for painting the dustbin.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the breadth of the rectangle and each side of the equilateral triangle be }x\text{ m}
\displaystyle \text{Let the height of the rectangular part be }y\text{ m}
\displaystyle \text{Perimeter of window }=x+2y+2x=12
\displaystyle \Rightarrow\ 3x+2y=12
\displaystyle \Rightarrow\ y=\frac{12-3x}{2}
\displaystyle \text{Area of window }=\text{area of rectangle}+\text{area of equilateral triangle}
\displaystyle A=xy+\frac{\sqrt{3}}{4}x^{2}
\displaystyle A=x\left(\frac{12-3x}{2}\right)+\frac{\sqrt{3}}{4}x^{2}
\displaystyle A=6x-\frac{3}{2}x^{2}+\frac{\sqrt{3}}{4}x^{2}
\displaystyle A=6x-\frac{6-\sqrt{3}}{4}x^{2}
\displaystyle \frac{dA}{dx}=6-\frac{6-\sqrt{3}}{2}x
\displaystyle \text{For maximum area, }\frac{dA}{dx}=0
\displaystyle \Rightarrow\ 6-\frac{6-\sqrt{3}}{2}x=0
\displaystyle \Rightarrow\ x=\frac{12}{6-\sqrt{3}}=\frac{4(6+\sqrt{3})}{11}
\displaystyle \text{Now }y=\frac{12-3x}{2}
\displaystyle \Rightarrow\ y=\frac{12-3\cdot\frac{4(6+\sqrt{3})}{11}}{2}
\displaystyle =\frac{12(2-\sqrt{3})}{11}
\displaystyle \text{Also, }\frac{d^{2}A}{dx^{2}}=-\frac{6-\sqrt{3}}{2}<0
\displaystyle \text{Hence, the area is maximum for these values.}
\displaystyle \therefore\ \text{Breadth of rectangle }=\text{side of triangle }=\frac{4(6+\sqrt{3})}{11}\text{ m}
\displaystyle \therefore\ \text{Height of rectangle }=\frac{12(2-\sqrt{3})}{11}\text{ m}
\displaystyle \text{OR} \ \
\displaystyle \text{(a) Expressing }h\text{ in terms of }r\text{:}
\displaystyle \text{Volume of cylinder }=\pi r^{2}h=20\pi
\displaystyle \Rightarrow\ r^{2}h=20
\displaystyle \Rightarrow\ h=\frac{20}{r^{2}}
\displaystyle \text{(b) Proving total surface area:}
\displaystyle \text{Total surface area of a closed cylinder }=2\pi r^{2}+2\pi rh
\displaystyle \text{Substituting }h=\frac{20}{r^{2}}
\displaystyle \text{T.S.A.}=2\pi r^{2}+2\pi r\left(\frac{20}{r^{2}}\right)
\displaystyle =2\pi r^{2}+\frac{40\pi}{r}
\displaystyle \therefore\ \text{Total surface area of the dustbin is }2\pi r^{2}+\frac{40\pi}{r}
\displaystyle \text{(c) Total cost of painting in terms of }r\text{:}
\displaystyle \text{Area of top and base }=2\pi r^{2}
\displaystyle \text{Cost of painting top and base }=2\cdot 2\pi r^{2}=4\pi r^{2}
\displaystyle \text{Curved surface area }=2\pi rh
\displaystyle \text{Cost of painting curved side }=25\cdot 2\pi rh=50\pi rh
\displaystyle \text{Substituting }h=\frac{20}{r^{2}}
\displaystyle \text{Cost of painting curved side }=50\pi r\left(\frac{20}{r^{2}}\right)=\frac{1000\pi}{r}
\displaystyle \therefore\ \text{Total cost }C=4\pi r^{2}+\frac{1000\pi}{r}
\displaystyle \text{(d) Minimum cost of painting:}
\displaystyle C=4\pi r^{2}+\frac{1000\pi}{r}
\displaystyle \frac{dC}{dr}=8\pi r-\frac{1000\pi}{r^{2}}
\displaystyle \text{For minimum cost, }\frac{dC}{dr}=0
\displaystyle \Rightarrow\ 8\pi r-\frac{1000\pi}{r^{2}}=0
\displaystyle \Rightarrow\ 8r=\frac{1000}{r^{2}}
\displaystyle \Rightarrow\ 8r^{3}=1000
\displaystyle \Rightarrow\ r^{3}=125
\displaystyle \Rightarrow\ r=5
\displaystyle \frac{d^{2}C}{dr^{2}}=8\pi+\frac{2000\pi}{r^{3}}
\displaystyle \text{At }r=5,\ \frac{d^{2}C}{dr^{2}}=8\pi+\frac{2000\pi}{125}=24\pi>0
\displaystyle \text{Hence, }C\text{ is minimum at }r=5
\displaystyle \text{Minimum cost }=4\pi(5)^{2}+\frac{1000\pi}{5}
\displaystyle =100\pi+200\pi
\displaystyle =300\pi
\displaystyle \therefore\ \text{The minimum cost of painting the dustbin is Rs }300\pi
\\
\displaystyle \textbf{Question 13} \ \
\displaystyle \text{(i) Solve the following differential equation:} \ \
\displaystyle 2ye^{\frac{x}{y}}\,dx+\left(y-2xe^{\frac{x}{y}}\right)\,dy=0,\ \text{given }x=0\text{ and }y=1 \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) Solve the following differential equation:} \ \
\displaystyle x(x^{2}-1)\frac{dy}{dx}=1,\ y=0,\ \text{given }x=2 \ \
\displaystyle \text{Answer:}
\displaystyle 2ye^{\frac{x}{y}}\,dx+\left(y-2xe^{\frac{x}{y}}\right)\,dy=0
\displaystyle \text{Since }\frac{x}{y}\text{ is present, let }x=vy\Rightarrow v=\frac{x}{y}
\displaystyle \therefore\ dx=v\,dy+y\,dv
\displaystyle \text{Substituting in the given equation,}
\displaystyle 2ye^{v}(v\,dy+y\,dv)+\left(y-2vye^{v}\right)dy=0
\displaystyle 2vye^{v}\,dy+2y^{2}e^{v}\,dv+y\,dy-2vye^{v}\,dy=0
\displaystyle 2y^{2}e^{v}\,dv+y\,dy=0
\displaystyle 2ye^{v}\,dv+dy=0
\displaystyle \Rightarrow\ \frac{dy}{dv}=-2ye^{v}
\displaystyle \Rightarrow\ \frac{1}{y}\,dy=-2e^{v}\,dv
\displaystyle \int \frac{1}{y}\,dy=\int -2e^{v}\,dv
\displaystyle \log y=-2e^{v}+c
\displaystyle \text{Since }v=\frac{x}{y},\ \log y=-2e^{\frac{x}{y}}+c
\displaystyle \text{Using the condition }x=0,\ y=1,
\displaystyle \log 1=-2e^{0}+c
\displaystyle 0=-2+c
\displaystyle \therefore\ c=2
\displaystyle \therefore\ \log y+2e^{\frac{x}{y}}=2
\displaystyle \text{Hence, the required solution is }\log y+2e^{\frac{x}{y}}=2
\displaystyle \text{OR} \ \
\displaystyle x(x^{2}-1)\frac{dy}{dx}=1
\displaystyle \Rightarrow\ \frac{dy}{dx}=\frac{1}{x(x^{2}-1)}
\displaystyle \Rightarrow\ dy=\frac{dx}{x(x^{2}-1)}
\displaystyle y=\int \frac{dx}{x(x-1)(x+1)}
\displaystyle \text{Using partial fractions, }\frac{1}{x(x-1)(x+1)}=-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}
\displaystyle \therefore\ y=\int \left(-\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\right)dx
\displaystyle y=-\log x+\frac{1}{2}\log(x-1)+\frac{1}{2}\log(x+1)+c
\displaystyle y=\frac{1}{2}\log(x^{2}-1)-\log x+c
\displaystyle \text{Using }x=2,\ y=0
\displaystyle 0=\frac{1}{2}\log 3-\log 2+c
\displaystyle \Rightarrow\ c=\log 2-\frac{1}{2}\log 3
\displaystyle \therefore\ y=\frac{1}{2}\log(x^{2}-1)-\log x+\log 2-\frac{1}{2}\log 3
\displaystyle \therefore\ y=\frac{1}{2}\log\left(\frac{4(x^{2}-1)}{3x^{2}}\right)
\\
\displaystyle \textbf{Question 14} \ \
\displaystyle \text{A primary school teacher wants to teach the concept of 'larger number' to the students of} \ \
\displaystyle \text{Class II.} \ \
\displaystyle \text{To teach this concept, he conducts an activity in his class. He asks the children to select two} \ \
\displaystyle \text{numbers from a set of numbers given as }2,3,4,5\text{ one after the other without replacement.} \ \
\displaystyle \text{All the outcomes of this activity are tabulated in the form of ordered pairs given below:} \ \
\displaystyle \begin{array}{|c|c|c|c|c|}\hline &2&3&4&5\\\hline 2&(2,2)&(2,3)&(2,4)&\\\hline 3&(3,2)&(3,3)&&(3,5)\\\hline 4&(4,2)&&(4,4)&(4,5)\\\hline 5&&(5,3)&(5,4)&(5,5)\\\hline \end{array} \ \
\displaystyle \text{(i) Complete the table given above.} \ \
\displaystyle \text{(ii) Find the total number of ordered pairs having one larger number.} \ \
\displaystyle \text{(iii) Let the random variable }X\text{ denote the larger of two numbers in the ordered pair.} \ \
\displaystyle \text{Now, complete the probability distribution table for }X\text{ given below.} \ \
\displaystyle \begin{array}{|c|c|c|c|}\hline X&3&4&5\\\hline \mathrm{P}(X=x)&&&\\\hline \end{array} \ \
\displaystyle \text{(iv) Find the value of }\mathrm{P}(X<5) \ \
\displaystyle \text{(v) Calculate the expected value of the probability distribution.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Completing the table:}
\displaystyle \text{Since the numbers are selected one after the other without replacement, equal pairs}
\displaystyle \text{like }(2,2),(3,3),(4,4),(5,5)\text{ are not possible. Also, every unequal ordered pair is possible.}
\displaystyle \text{Hence the completed table entries are:}
\displaystyle (2,3),(2,4),(2,5),(3,2),(3,4),(3,5),(4,2),(4,3),(4,5),(5,2),(5,3),(5,4)
\displaystyle \text{(ii) Total number of ordered pairs having one larger number:}
\displaystyle \text{All possible ordered pairs are the unequal ordered pairs from }\{2,3,4,5\}
\displaystyle =4\times 3=12
\displaystyle \therefore\ \text{Total number of ordered pairs }=12
\displaystyle \text{(iii) Probability distribution of }X\text{:}
\displaystyle \text{Here }X\text{ denotes the larger of the two numbers selected.}
\displaystyle X=3\text{ for }(2,3),(3,2)\Rightarrow 2\text{ outcomes}
\displaystyle X=4\text{ for }(2,4),(4,2),(3,4),(4,3)\Rightarrow 4\text{ outcomes}
\displaystyle X=5\text{ for }(2,5),(5,2),(3,5),(5,3),(4,5),(5,4)\Rightarrow 6\text{ outcomes}
\displaystyle \text{Since total outcomes }=12,
\displaystyle P(X=3)=\frac{2}{12}=\frac{1}{6}
\displaystyle P(X=4)=\frac{4}{12}=\frac{1}{3}
\displaystyle P(X=5)=\frac{6}{12}=\frac{1}{2}
\displaystyle \text{Thus, the probability distribution table is:}
\displaystyle \begin{array}{c|ccc}X&3&4&5\\\hline P(X=x)&\frac{1}{6}&\frac{1}{3}&\frac{1}{2}\end{array}
\displaystyle \text{(iv) Finding }P(X<5)\text{:}
\displaystyle P(X<5)=P(X=3)+P(X=4)
\displaystyle =\frac{1}{6}+\frac{1}{3}=\frac{1}{2}
\displaystyle \text{(v) Expected value:}
\displaystyle E(X)=3\cdot\frac{1}{6}+4\cdot\frac{1}{3}+5\cdot\frac{1}{2}
\displaystyle =\frac{3}{6}+\frac{4}{3}+\frac{5}{2}
\displaystyle =\frac{1}{2}+\frac{4}{3}+\frac{5}{2}
\displaystyle =3+\frac{4}{3}=\frac{13}{3}
\displaystyle \therefore\ E(X)=\frac{13}{3}
\\
\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) to (v), answer the} \ \
\displaystyle \text{questions as instructed.} \ \
\displaystyle \text{(i) If }\overrightarrow{a}=3\hat{i}-2\hat{j}+\hat{k}\text{ and }\overrightarrow{b}=2\hat{i}-4\hat{j}-3\hat{k}\text{ then the value of }\left|\overrightarrow{a}-2\overrightarrow{b}\right|\text{ will be:} \ \
\displaystyle \text{(a) }\sqrt{85} \ \
\displaystyle \text{(b) }\sqrt{86} \ \
\displaystyle \text{(c) }\sqrt{87} \ \
\displaystyle \text{(d) }\sqrt{88} \ \
\displaystyle \text{(ii) If a line makes an angle }\alpha,\beta\text{ and }\gamma\text{ with positive direction of the coordinate axes,} \ \
\displaystyle \text{then the value of }\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma\text{ will be:} \ \
\displaystyle \text{(a) }1 \ \
\displaystyle \text{(b) }3 \ \
\displaystyle \text{(c) }-2 \ \
\displaystyle \text{(d) }2 \ \
\displaystyle \text{(iii) In the figure given below, if the coordinates of the point }P\text{ are }(a,b,c)\text{, then what are} \ \
\displaystyle \text{the perpendicular distances of }P\text{ from }XY,\ YZ\text{ and }ZX\text{ planes respectively?} \ \ \displaystyle \text{(iv) If }\overrightarrow{a}=2\hat{i}+\hat{j}+2\hat{k}\text{ and }\overrightarrow{b}=5\hat{i}-3\hat{j}+\hat{k}\text{, find the projection of }\overrightarrow{b}\text{ on }\overrightarrow{a} \ \
\displaystyle \text{(v) Find a vector of magnitude }20\text{ units parallel to the vector }2\hat{i}+5\hat{j}+4\hat{k}. \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \overrightarrow{a}=3\hat{i}-2\hat{j}+\hat{k},\quad \overrightarrow{b}=2\hat{i}-4\hat{j}-3\hat{k}
\displaystyle \overrightarrow{a}-2\overrightarrow{b}=\left(3\hat{i}-2\hat{j}+\hat{k}\right)-2\left(2\hat{i}-4\hat{j}-3\hat{k}\right)
\displaystyle =3\hat{i}-2\hat{j}+\hat{k}-4\hat{i}+8\hat{j}+6\hat{k}
\displaystyle =-\hat{i}+6\hat{j}+7\hat{k}
\displaystyle \left|\overrightarrow{a}-2\overrightarrow{b}\right|=\sqrt{(-1)^{2}+6^{2}+7^{2}}
\displaystyle =\sqrt{1+36+49}
\displaystyle =\sqrt{86}
\displaystyle \therefore\ \text{Correct option is (b).}
\displaystyle \text{(ii)}
\displaystyle \text{If a line makes angles }\alpha,\beta,\gamma\text{ with the positive coordinate axes,}
\displaystyle \text{then its direction cosines satisfy }\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1
\displaystyle \therefore\ \sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma
\displaystyle =(1-\cos^{2}\alpha)+(1-\cos^{2}\beta)+(1-\cos^{2}\gamma)
\displaystyle =3-\left(\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma\right)
\displaystyle =3-1
\displaystyle =2
\displaystyle \therefore\ \text{Correct option is (d).}
\displaystyle \text{(iii)}
\displaystyle \text{For a point }P(a,b,c),\text{ the perpendicular distance from the }XY\text{-plane is }c
\displaystyle \text{the perpendicular distance from the }YZ\text{-plane is }a
\displaystyle \text{and the perpendicular distance from the }ZX\text{-plane is }b
\displaystyle \therefore\ \text{the required distances are }c,\ a,\ b\text{ respectively.}
\displaystyle \text{(iv)}
\displaystyle \overrightarrow{a}=2\hat{i}+\hat{j}+2\hat{k},\quad \overrightarrow{b}=5\hat{i}-3\hat{j}+\hat{k}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=2\cdot 5+1\cdot(-3)+2\cdot 1=9
\displaystyle \left|\overrightarrow{a}\right|^{2}=2^{2}+1^{2}+2^{2}=9
\displaystyle \text{Projection of }\overrightarrow{b}\text{ on }\overrightarrow{a}=\frac{\overrightarrow{a}\cdot\overrightarrow{b}}{\left|\overrightarrow{a}\right|^{2}}\overrightarrow{a}
\displaystyle =\frac{9}{9}\overrightarrow{a}=\overrightarrow{a}
\displaystyle \therefore\ \text{the projection of }\overrightarrow{b}\text{ on }\overrightarrow{a}\text{ is }2\hat{i}+\hat{j}+2\hat{k}
\displaystyle \text{(v)}
\displaystyle \overrightarrow{v}=2\hat{i}+5\hat{j}+4\hat{k}
\displaystyle \left|\overrightarrow{v}\right|=\sqrt{2^{2}+5^{2}+4^{2}}=\sqrt{45}=3\sqrt{5}
\displaystyle \text{Unit vector parallel to }\overrightarrow{v}\text{ is }\frac{2\hat{i}+5\hat{j}+4\hat{k}}{3\sqrt{5}}
\displaystyle \text{Hence a vector of magnitude }20\text{ parallel to }\overrightarrow{v}\text{ is}
\displaystyle 20\cdot\frac{2\hat{i}+5\hat{j}+4\hat{k}}{3\sqrt{5}}
\displaystyle =\frac{20}{3\sqrt{5}}(2\hat{i}+5\hat{j}+4\hat{k})
\displaystyle \therefore\ \text{required vector is }\frac{20}{3\sqrt{5}}(2\hat{i}+5\hat{j}+4\hat{k})
\\
\displaystyle \textbf{Question 16} \ \
\displaystyle \text{(i) If }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{a}\times\overrightarrow{c}\text{ where }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ are non-zero vectors, then prove that either} \ \
\displaystyle \overrightarrow{b}=\overrightarrow{c}\text{ or }\overrightarrow{a}\text{ and }(\overrightarrow{b}-\overrightarrow{c})\text{ are parallel.} \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two non-zero vectors such that }\left|\overrightarrow{a}\times\overrightarrow{b}\right|=\overrightarrow{a}\cdot\overrightarrow{b}\text{, find the angle} \ \
\displaystyle \text{between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{.} \ \ \displaystyle \text{Answer:}
\displaystyle \text{(i) Proof:}
\displaystyle \text{Given }\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{a}\times\overrightarrow{c}
\displaystyle \Rightarrow\ \overrightarrow{a}\times\overrightarrow{b}-\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow\ \overrightarrow{a}\times(\overrightarrow{b}-\overrightarrow{c})=\overrightarrow{0}
\displaystyle \text{Now, if the cross product of two vectors is zero, then either one of them is}
\displaystyle \text{the zero vector or they are parallel.}
\displaystyle \therefore\ \text{either }\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{0}\text{ or } \overrightarrow{a}\parallel(\overrightarrow{b}-\overrightarrow{c})
\displaystyle \Rightarrow\ \text{either }\overrightarrow{b}=\overrightarrow{c}\text{ or } \overrightarrow{a}\text{ and }(\overrightarrow{b}-\overrightarrow{c})\text{ are parallel.}
\displaystyle \text{Hence proved.}
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) Finding the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{:}
\displaystyle \text{Given }\left|\overrightarrow{a}\times\overrightarrow{b}\right|=\overrightarrow{a}\cdot\overrightarrow{b}
\displaystyle \Rightarrow\ |\overrightarrow{a}||\overrightarrow{b}|\sin\theta= |\overrightarrow{a}||\overrightarrow{b}|\cos\theta
\displaystyle \Rightarrow\ \sin\theta=\cos\theta
\displaystyle \Rightarrow\ \tan\theta=1
\displaystyle \therefore\ \theta=\frac{\pi}{4}
\displaystyle \text{So, the angle between }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is } 45^\circ.
\\
\displaystyle \textbf{Question 17} \ \
\displaystyle \text{A mobile tower is situated at the top of a hill. Consider the surface on which the tower} \ \
\displaystyle \text{stands as a plane having points }A(1,0,2),B(3,-1,1)\text{ and }C(1,2,1)\text{ on it. The mobile tower} \ \
\displaystyle \text{is tied with three cables from the points }A,B\text{ and }C\text{ such that it stands vertically on the} \ \
\displaystyle \text{ground. The top of the tower is at point }P(2,3,1)\text{ as shown in the figure below. The foot of} \ \
\displaystyle \text{the perpendicular from the point }P\text{ on the plane is at the point }Q\left(\frac{43}{29},\frac{77}{29},\frac{9}{29}\right) \ \
\displaystyle \text{Answer the following questions.} \ \
\displaystyle \text{(i) Find the equation of the plane containing the points }A,B\text{ and }C\text{.} \ \
\displaystyle \text{(ii) Find the equation of the line }PQ\text{.} \ \
\displaystyle \text{(iii) Calculate the height of the tower.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Equation of the plane through }A(1,0,2),\ B(3,-1,1)\text{ and }C(1,2,1)\text{:}
\displaystyle \overrightarrow{AB}=(3-1,\,-1-0,\,1-2)=(2,\,-1,\,-1)
\displaystyle \overrightarrow{AC}=(1-1,\,2-0,\,1-2)=(0,\,2,\,-1)
\displaystyle \overrightarrow{AB}\times\overrightarrow{AC}= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-1&-1\\0&2&-1\end{vmatrix}
\displaystyle =\hat{i}\left(1+2\right)-\hat{j}\left(-2\right)+\hat{k}\left(4\right)
\displaystyle =3\hat{i}+2\hat{j}+4\hat{k}
\displaystyle \text{Hence, a normal vector to the plane is }(3,2,4)
\displaystyle \text{Therefore, the equation of the plane is}
\displaystyle 3(x-1)+2(y-0)+4(z-2)=0
\displaystyle \therefore\ 3x+2y+4z-11=0
\displaystyle \text{(ii) Equation of the line }PQ\text{:}
\displaystyle \text{Since }PQ\text{ is perpendicular to the plane, it is along the normal vector }(3,2,4)
\displaystyle \text{Also, it passes through }P(2,3,1)
\displaystyle \therefore\ \frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}
\displaystyle \text{or, in parametric form, }x=2+3\lambda,\ y=3+2\lambda,\ z=1+4\lambda
\displaystyle \text{(iii) Height of the tower:}
\displaystyle \text{Height }=PQ=\text{distance of point }P(2,3,1)\text{ from the plane }3x+2y+4z-11=0
\displaystyle \therefore\ PQ=\frac{|3(2)+2(3)+4(1)-11|}{\sqrt{3^{2}+2^{2}+4^{2}}}
\displaystyle =\frac{|6+6+4-11|}{\sqrt{9+4+16}}
\displaystyle =\frac{5}{\sqrt{29}}
\displaystyle \therefore\ \text{The height of the tower is }\frac{5}{\sqrt{29}}\text{ units}
\\
\displaystyle \textbf{Question 18} \ \
\displaystyle \text{(i) Using integration, find the area bounded by the curve }y^{2}=4ax\text{ and the line }x=a \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) Using integration, find the area of the region bounded by the curve} \ \
\displaystyle y^{2}=4x\text{ and }x^{2}=4y \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Area bounded by }y^{2}=4ax\text{ and }x=a\text{:}
\displaystyle \text{For }y^{2}=4ax,\text{ we have }x=\frac{y^{2}}{4a}
\displaystyle \text{The line }x=a\text{ cuts the parabola at points where }y^{2}=4a(a)=4a^{2}
\displaystyle \Rightarrow\ y=\pm 2a
\displaystyle \text{Required area }=\int_{-2a}^{2a}\left(a-\frac{y^{2}}{4a}\right)\,dy
\displaystyle =2\int_{0}^{2a}\left(a-\frac{y^{2}}{4a}\right)\,dy
\displaystyle =2\left[a y-\frac{y^{3}}{12a}\right]_{0}^{2a}
\displaystyle =2\left[2a^{2}-\frac{8a^{3}}{12a}\right]
\displaystyle =2\left[2a^{2}-\frac{2a^{2}}{3}\right]
\displaystyle =2\cdot\frac{4a^{2}}{3}
\displaystyle =\frac{8a^{2}}{3}
\displaystyle \therefore\ \text{Required area is }\frac{8a^{2}}{3}
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) Area bounded by }y^{2}=4x\text{ and }x^{2}=4y\text{:}
\displaystyle \text{From }y^{2}=4x,\ x=\frac{y^{2}}{4}
\displaystyle \text{From }x^{2}=4y,\ y=\frac{x^{2}}{4}
\displaystyle \text{Points of intersection satisfy }y^{2}=4x\text{ and }x^{2}=4y
\displaystyle \text{Substitute }x=\frac{y^{2}}{4}\text{ into }x^{2}=4y
\displaystyle \Rightarrow\ \left(\frac{y^{2}}{4}\right)^{2}=4y
\displaystyle \Rightarrow\ \frac{y^{4}}{16}=4y
\displaystyle \Rightarrow\ y^{4}=64y
\displaystyle \Rightarrow\ y(y^{3}-64)=0
\displaystyle \Rightarrow\ y=0\text{ or }y=4
\displaystyle \text{Hence corresponding points are }(0,0)\text{ and }(4,4)
\displaystyle \text{For }0\leq y\leq 4,\ \text{right curve is }x=2\sqrt{y}\text{ and left curve is }x=\frac{y^{2}}{4}
\displaystyle \text{Required area }=\int_{0}^{4}\left(2\sqrt{y}-\frac{y^{2}}{4}\right)\,dy
\displaystyle =\left[\frac{4}{3}y^{\frac{3}{2}}-\frac{y^{3}}{12}\right]_{0}^{4}
\displaystyle =\frac{4}{3}(4)^{\frac{3}{2}}-\frac{64}{12}
\displaystyle =\frac{4}{3}\cdot 8-\frac{16}{3}
\displaystyle =\frac{32}{3}-\frac{16}{3}
\displaystyle =\frac{16}{3}
\displaystyle \therefore\ \text{Required area is }\frac{16}{3}\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 subparts (iii) to (v), answer the} \ \
\displaystyle \text{questions as instructed.} \ \
\displaystyle \text{(i) A company sells hand towels at Rs }100\text{ per unit. The fixed cost for the company to} \ \
\displaystyle \text{manufacture hand towels is Rs }35000\text{ and variable cost is estimated to be }30\%\text{ of} \ \
\displaystyle \text{total revenue. What will be the total cost function for manufacturing hand towels?} \ \
\displaystyle \text{(a) }35000+3x \ \
\displaystyle \text{(b) }35000+30x \ \
\displaystyle \text{(c) }35000+100x \ \
\displaystyle \text{(d) }35000+10x \ \
\displaystyle \text{(ii) If the correlation coefficient of two sets of variables }(X,Y)\text{ is }\frac{-3}{4}\text{, which one of the} \ \
\displaystyle \text{following statements is true for the same set of variables?} \ \
\displaystyle \text{(a) Only one of the two regression lines has a negative coefficient.} \ \
\displaystyle \text{(b) Both regression coefficients are positive.} \ \
\displaystyle \text{(c) Both regression coefficients are negative.} \ \
\displaystyle \text{(d) One of the lines of regression is parallel to the }x\text{-axis.} \ \
\displaystyle \text{(iii) If the total cost function is given by }C=x+2x^{3}-\frac{7}{2}x^{2}\text{, find the Marginal Average} \ \
\displaystyle \text{Cost function (MAC).} \ \
\displaystyle \text{(iv) The equations of two lines of regression are }4x+3y+7=0\text{ and} \ \
\displaystyle 3x+4y+8=0\text{. Find the mean value of }x\text{ and }y. \ \
\displaystyle \text{(v) The manufacturer of a pen fixes its selling price at Rs }45\text{, and the cost function is} \ \
\displaystyle C(x)=30x+240\text{. The manufacturer will begin to earn profit if he sells more than} \ \
\displaystyle 16\text{ pens. Why? Give one reason.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{Selling price per unit }=100
\displaystyle \text{Revenue from }x\text{ units }=100x
\displaystyle \text{Variable cost }=30\%\text{ of revenue }=\frac{30}{100}\cdot 100x=30x
\displaystyle \text{Fixed cost }=35000
\displaystyle \therefore\ \text{Total cost function }=35000+30x
\displaystyle \therefore\ \text{Correct option is (b)}
\displaystyle \text{(ii)}
\displaystyle r=-\frac{3}{4}
\displaystyle \text{Both regression coefficients have the same sign as }r
\displaystyle \therefore\ \text{both regression coefficients are negative}
\displaystyle \therefore\ \text{Correct option is (c)}
\displaystyle \text{(iii)}
\displaystyle C=x+2x^{3}-\frac{7}{2}x^{2}
\displaystyle \text{Average cost }=\frac{C}{x}=\frac{x+2x^{3}-\frac{7}{2}x^{2}}{x}
\displaystyle =1+2x^{2}-\frac{7}{2}x
\displaystyle \text{Marginal average cost function }=\frac{d}{dx}\left(\frac{C}{x}\right)
\displaystyle \therefore\ \mathrm{MAC}=4x-\frac{7}{2}
\displaystyle \text{(iv)}
\displaystyle \text{The two regression lines intersect at }(\overline{x},\overline{y})
\displaystyle 4x+3y+7=0
\displaystyle 3x+4y+8=0
\displaystyle \text{Multiply the first equation by }4\text{ and the second by }3
\displaystyle 16x+12y+28=0
\displaystyle 9x+12y+24=0
\displaystyle \text{Subtracting, }7x+4=0
\displaystyle \Rightarrow\ x=-\frac{4}{7}
\displaystyle \text{Substitute in }4x+3y+7=0
\displaystyle -\frac{16}{7}+3y+7=0
\displaystyle 3y+\frac{33}{7}=0
\displaystyle \Rightarrow\ y=-\frac{11}{7}
\displaystyle \therefore\ \overline{x}=-\frac{4}{7},\ \overline{y}=-\frac{11}{7}
\displaystyle \text{(v)}
\displaystyle \text{Selling price per pen }=45
\displaystyle \text{Revenue function }R(x)=45x
\displaystyle \text{Cost function }C(x)=30x+240
\displaystyle \text{Profit }P(x)=R(x)-C(x)=45x-(30x+240)=15x-240
\displaystyle \text{For profit, }P(x)>0
\displaystyle \Rightarrow\ 15x-240>0
\displaystyle \Rightarrow\ 15x>240
\displaystyle \Rightarrow\ x>16
\displaystyle \therefore\ \text{He begins to earn profit only when he sells more than }16\text{ pens, because}
\displaystyle \text{at }x=16,\ \text{revenue equals cost, and for }x>16,\ \text{revenue exceeds cost}
\\
\displaystyle \textbf{Question 20} \ \
\displaystyle \text{(i) The Average Cost function associated with producing and marketing }x\text{ units of an} \ \
\displaystyle \text{item is given by }AC=x+5+\frac{36}{x} \ \
\displaystyle \text{(a) Find the Total Cost function.} \ \
\displaystyle \text{(b) Find the range of values of }x\text{ for which Average Cost is increasing.} \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) A monopolist's demand function is }x=60-\frac{p}{5}\text{. At what level of output will} \ \
\displaystyle \text{marginal revenue be zero?} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i)(a) Finding the Total Cost function:}
\displaystyle \mathrm{AC}=x+5+\frac{36}{x}
\displaystyle \text{But }\mathrm{AC}=\frac{C}{x}
\displaystyle \therefore\ \frac{C}{x}=x+5+\frac{36}{x}
\displaystyle \Rightarrow\ C=x\left(x+5+\frac{36}{x}\right)
\displaystyle \Rightarrow\ C=x^{2}+5x+36
\displaystyle \therefore\ \text{Total Cost function is }C=x^{2}+5x+36
\displaystyle \text{(i)(b) Range of }x\text{ for which Average Cost is increasing:}
\displaystyle \mathrm{AC}=x+5+\frac{36}{x}
\displaystyle \frac{d(\mathrm{AC})}{dx}=1-\frac{36}{x^{2}}
\displaystyle \text{For Average Cost to be increasing, }\frac{d(\mathrm{AC})}{dx}>0
\displaystyle \Rightarrow\ 1-\frac{36}{x^{2}}>0
\displaystyle \Rightarrow\ 1>\frac{36}{x^{2}}
\displaystyle \Rightarrow\ x^{2}>36
\displaystyle \Rightarrow\ x>6\text{, since }x\text{ denotes number of units}
\displaystyle \therefore\ \text{Average Cost is increasing for }x>6
\displaystyle \text{(ii) Output level at which marginal revenue is zero:}
\displaystyle x=60-\frac{p}{5}
\displaystyle \Rightarrow\ \frac{p}{5}=60-x
\displaystyle \Rightarrow\ p=300-5x
\displaystyle \text{Revenue }R=px=x(300-5x)=300x-5x^{2}
\displaystyle \text{Marginal Revenue }=\frac{dR}{dx}=300-10x
\displaystyle \text{For marginal revenue to be zero,}
\displaystyle 300-10x=0
\displaystyle \Rightarrow\ 10x=300
\displaystyle \Rightarrow\ x=30
\displaystyle \therefore\ \text{Marginal revenue is zero at output level }30\text{ units}
\\
\displaystyle \textbf{Question 21} \ \
\displaystyle \text{(i) XYZ company plans to advertise some vacancies. The Manager is asked to suggest} \ \
\displaystyle \text{the monthly salary for these vacancies based on the years of experience. To do so,} \ \
\displaystyle \text{the Manager studies the years of service and the monthly salary drawn by the existing} \ \
\displaystyle \text{employees in the company.} \ \
\displaystyle \text{Following is the data that the Manager refers to:} \ \
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Years of service }(X)&11&7&9&5&8&6&10\\\hline \text{Monthly salary (in \text{Rs}1000) }(Y)&10&8&6&5&9&7&11\\\hline \end{array} \ \
\displaystyle \text{(a) Find the regression equation of monthly salary on the years of service.} \ \
\displaystyle \text{(b) If a person with }13\text{ years of experience applies for a job in this company, what} \ \
\displaystyle \text{monthly salary will be suggested by the Manager?} \ \
\displaystyle \text{OR} \ \
\displaystyle \text{(ii) The line of regression of marks in Statistics }(X)\text{ and marks in Accountancy }(Y)\text{ for a} \ \
\displaystyle \text{class of }50\text{ students is }3y-5x+180=0\text{. The average score in Accountancy is }44 \ \
\displaystyle \text{and the variance of marks in Statistics is }\left(\frac{9}{16}\right)^{\mathrm{th}}\text{ of variance of marks in Accountancy.} \ \
\displaystyle \text{(a) Find the average score in Statistics.} \ \
\displaystyle \text{(b) Find the coefficient of correlation between marks in Statistics and marks in} \ \
\displaystyle \text{Accountancy.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i)(a) Regression equation of monthly salary on years of service:}
\displaystyle X:11,\ 7,\ 9,\ 5,\ 8,\ 6,\ 10
\displaystyle Y:10,\ 8,\ 6,\ 5,\ 9,\ 7,\ 11
\displaystyle \overline{X}=\frac{11+7+9+5+8+6+10}{7}=\frac{56}{7}=8
\displaystyle \overline{Y}=\frac{10+8+6+5+9+7+11}{7}=\frac{56}{7}=8
\displaystyle x=X-\overline{X},\quad y=Y-\overline{Y}
\displaystyle \text{Thus, pairs }(x,y)\text{ are }(3,2),(-1,0),(1,-2),(-3,-3),(0,1),(-2,-1),(2,3)
\displaystyle \sum xy=6+0-2+9+0+2+6=21
\displaystyle \sum x^{2}=9+1+1+9+0+4+4=28
\displaystyle \text{Regression coefficient of }Y\text{ on }X\text{ is }b_{yx}=\frac{\sum xy}{\sum x^{2}}=\frac{21}{28}=\frac{3}{4}
\displaystyle \text{Regression equation of }Y\text{ on }X\text{ is }Y-\overline{Y}=b_{yx}(X-\overline{X})
\displaystyle \Rightarrow\ Y-8=\frac{3}{4}(X-8)
\displaystyle \Rightarrow\ 4Y-32=3X-24
\displaystyle \Rightarrow\ 3X-4Y+8=0
\displaystyle \therefore\ \text{Required regression equation is }Y-8=\frac{3}{4}(X-8)
\displaystyle \text{(i)(b) Suggested salary for }13\text{ years of experience:}
\displaystyle Y-8=\frac{3}{4}(13-8)
\displaystyle \Rightarrow\ Y-8=\frac{15}{4}
\displaystyle \Rightarrow\ Y=\frac{47}{4}=11.75
\displaystyle \therefore\ \text{Suggested monthly salary }=11.75\text{ thousand rupees }=\text{Rs }11750
\displaystyle \text{OR} \ \
\displaystyle \text{(ii)(a) Finding the average score in Statistics:}
\displaystyle \text{Given regression line }3y-5x+180=0
\displaystyle \text{The regression line passes through }(\overline{X},\overline{Y})
\displaystyle \text{Given }\overline{Y}=44
\displaystyle \Rightarrow\ 3(44)-5\overline{X}+180=0
\displaystyle \Rightarrow\ 132-5\overline{X}+180=0
\displaystyle \Rightarrow\ 312-5\overline{X}=0
\displaystyle \Rightarrow\ \overline{X}=\frac{312}{5}=62.4
\displaystyle \therefore\ \text{Average score in Statistics is }62.4
\displaystyle \text{(ii)(b) Coefficient of correlation:}
\displaystyle 3y-5x+180=0\Rightarrow y=\frac{5}{3}x-60
\displaystyle \text{Hence regression coefficient }b_{yx}=\frac{5}{3}
\displaystyle \text{Given variance of marks in Statistics is }\frac{9}{16}\text{ of variance of marks in Accountancy}
\displaystyle \Rightarrow\ \sigma_{X}^{2}=\frac{9}{16}\sigma_{Y}^{2}
\displaystyle \Rightarrow\ \frac{\sigma_{X}}{\sigma_{Y}}=\frac{3}{4}
\displaystyle \Rightarrow\ \frac{\sigma_{Y}}{\sigma_{X}}=\frac{4}{3}
\displaystyle \text{Now }b_{yx}=r\frac{\sigma_{Y}}{\sigma_{X}}
\displaystyle \Rightarrow\ \frac{5}{3}=r\cdot\frac{4}{3}

\displaystyle \Rightarrow\ r=\frac{5}{4}
\displaystyle \text{But correlation coefficient cannot exceed }1\text{ in magnitude.}
\displaystyle \text{Hence the printed variance ratio is inconsistent with the given regression line.}
\displaystyle \text{If the intended ratio were }\sigma_{X}^{2}=\frac{16}{25}\sigma_{Y}^{2},\text{ then }r=1
\displaystyle \therefore\ \text{with the data as printed, no valid correlation coefficient exists.}
\\
\displaystyle \textbf{Question 22} \ \
\displaystyle \text{Aman has Rs }1500\text{ to purchase rice and wheat for his grocery shop. Each sack of rice and} \ \
\displaystyle \text{wheat costs Rs }180\text{ and Rs }120\text{ respectively. He can store a maximum number of }10\text{ bags in his} \ \
\displaystyle \text{shop. He will earn a profit of Rs }11\text{ per bag of rice and Rs }9\text{ per bag of wheat.} \ \
\displaystyle \text{(i) Formulate a Linear Programming Problem to maximise Aman's profit.} \ \
\displaystyle \text{(ii) Calculate the maximum profit.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(i) Formulation of the L.P.P.:}
\displaystyle \text{Let }x\text{ be the number of sacks of rice and }y\text{ be the number of sacks of wheat.}
\displaystyle \text{Profit function }Z=11x+9y
\displaystyle \text{Subject to the constraints:}
\displaystyle 180x+120y\leq 1500
\displaystyle x+y\leq 10
\displaystyle x\geq 0,\ y\geq 0
\displaystyle \text{Thus, maximise }Z=11x+9y
\displaystyle \text{subject to }180x+120y\leq 1500,\ x+y\leq 10,\ x\geq 0,\ y\geq 0
\displaystyle \text{(ii) Calculation of maximum profit:}
\displaystyle 180x+120y\leq 1500\Rightarrow 3x+2y\leq 25
\displaystyle \text{The corner points of the feasible region are }(0,0),\ \left(\frac{25}{3},0\right),\ (5,5),\ (0,10)
\displaystyle Z(0,0)=0
\displaystyle Z\left(\frac{25}{3},0\right)=11\cdot\frac{25}{3}=\frac{275}{3}
\displaystyle Z(5,5)=11\cdot 5+9\cdot 5=55+45=100
\displaystyle Z(0,10)=9\cdot 10=90
\displaystyle \text{Hence, the maximum value of }Z\text{ is }100
\displaystyle \therefore\ \text{Maximum profit is Rs }100
\displaystyle \text{This occurs when Aman purchases }5\text{ sacks of rice and }5\text{ sacks of wheat.}

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

Subscribe to get the latest posts sent to your email.