\displaystyle \textbf{MATHEMATICS}
\displaystyle \text{(Maximum Marks: 100)}
\displaystyle \text{(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{The 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 EITHER from Section B OR Section C.}
\displaystyle \text{Section A: Internal choice has been provided in three questions of four marks each and two questions of six marks each.}
\displaystyle \text{Section B: Internal choice has been provided in two questions of four marks each.}
\displaystyle \text{Section C: Internal choice has been provided in two questions of four marks each.}
\displaystyle \text{All working, including rough work, should be done on the same sheet as, and adjacent to, the rest 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.}

Section – A (80 Marks)

\displaystyle \textbf{Question 1: } \hspace{10.0cm} [10 \text{ times } 3]
\displaystyle \text{(i) If } A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}, \text{ find the values of } x \text{ and } y \text{ such that } A^2+xI_2=yA.
\displaystyle \text{Answer:}
\displaystyle \text{ Given, } A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}
\displaystyle A^2 = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}
\displaystyle = \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix}
\displaystyle \text{Now, } A^2+xI_2=yA
\displaystyle \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix}+x\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=y\begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix} 16+x & 8 \\ 56 & 32+x \end{bmatrix}=\begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix}
\displaystyle \Rightarrow y=8
\displaystyle 16+x=3y
\displaystyle \Rightarrow x=3(8)-16=8
\displaystyle \therefore x=8,\ y=8
\displaystyle \\
\displaystyle \text{(ii) Find the eccentricity and the coordinates of foci of the hyperbola } \\ 25x^2-9y^2=225.
\displaystyle \text{Answer:}
\displaystyle   25x^2-9y^2=225
\displaystyle \Rightarrow \frac{x^2}{9}-\frac{y^2}{25}=1
\displaystyle \Rightarrow a^2=9,\ b^2=25
\displaystyle \Rightarrow a=3,\ b=5
\displaystyle c=\sqrt{a^2+b^2}=\sqrt{9+25}=\sqrt{34}
\displaystyle e=\frac{c}{a}=\frac{\sqrt{34}}{3}
\displaystyle \text{Foci } = (\pm c,0) = (\pm \sqrt{34},0)
\displaystyle \\
\displaystyle \text{(iii) Evaluate: } \tan \left[2\tan^{-1}\frac{1}{2}-\cot^{-1}3\right].
\displaystyle \text{Answer:}
\displaystyle   \tan \left[2\tan^{-1}\frac{1}{2}-\cot^{-1}3\right]
\displaystyle =\tan \left[\tan^{-1}\left(\frac{2\cdot \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\right)-\tan^{-1}\frac{1}{3}\right]
\displaystyle =\tan \left[\tan^{-1}\left(\frac{1}{1-\frac{1}{4}}\right)-\tan^{-1}\frac{1}{3}\right]
\displaystyle =\tan \left[\tan^{-1}\frac{4}{3}-\tan^{-1}\frac{1}{3}\right]
\displaystyle =\tan \left[\tan^{-1}\left(\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{9}}\right)\right]
\displaystyle =\tan \left[\tan^{-1}\left(\frac{1}{\frac{13}{9}}\right)\right]
\displaystyle =\frac{9}{13}
\displaystyle \\
\displaystyle \text{(iv) Using L'Hospital's Rule, evaluate: } \lim\limits_{x \to 0}(1+\sin x)^{\cot x}.
\displaystyle \text{Answer:}
\displaystyle   \text{Let } y=(1+\sin x)^{\cot x}
\displaystyle \log y=\cot x \log(1+\sin x)
\displaystyle \log y=\frac{\log(1+\sin x)}{\tan x}
\displaystyle \lim\limits_{x \to 0}\log y=\lim\limits_{x \to 0}\frac{\log(1+\sin x)}{\tan x}
\displaystyle =\lim\limits_{x \to 0}\frac{\frac{\cos x}{1+\sin x}}{\sec^2 x}
\displaystyle =\lim\limits_{x \to 0}\frac{\cos^3 x}{1+\sin x}
\displaystyle =\frac{1}{1}=1
\displaystyle \therefore \log y=1 \Rightarrow y=e
\displaystyle \\
\displaystyle \text{(v) Evaluate: } \int e^x\frac{2+\sin 2x}{\cos^2 x}\ dx.
\displaystyle \text{Answer:}
\displaystyle   I=\int e^x\frac{2+\sin 2x}{\cos^2 x}\ dx
\displaystyle =\int e^x\left(\frac{2}{\cos^2 x}+\frac{2\sin x\cos x}{\cos^2 x}\right)dx
\displaystyle =\int e^x(2\sec^2 x+2\tan x)\ dx
\displaystyle =2\int e^x(\sec^2 x+\tan x)\ dx
\displaystyle =2\left[\int e^x\sec^2 x\ dx+\int e^x\tan x\ dx\right]
\displaystyle =2\left[e^x\tan x-\int e^x\tan x\ dx+\int e^x\tan x\ dx\right]
\displaystyle =2e^x\tan x+C
\displaystyle \\
\displaystyle \text{(vi) Using the properties of definite integrals, evaluate: } \int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\ dx.
\displaystyle \text{Answer:}
\displaystyle   I=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\ dx
\displaystyle =\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}\ dx
\displaystyle I=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\ dx
\displaystyle \text{Adding, we get}
\displaystyle 2I=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\ dx
\displaystyle 2I=\int\limits_{0}^{\frac{\pi}{2}}1\ dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}
\displaystyle I=\frac{\pi}{4}
\displaystyle \\
\displaystyle \text{(vii) For the given lines of regression, } 3x-2y=5 \text{ and } x-4y=7, \text{ find:}
\displaystyle \text{(a) regression coefficients } b_{yx} \text{ and } b_{xy} \quad \text{(b) coefficient of correlation } r(x,y).
\displaystyle \text{Answer:}
\displaystyle  \text{The two regression lines are}
\displaystyle 3x-2y=5 \hspace{1.0cm} \text{ ...(i)}
\displaystyle x-4y=7 \hspace{1.0cm} \text{ ...(ii)}
\displaystyle \text{From (i), } 3x=2y+5
\displaystyle \Rightarrow x=\frac{2}{3}y+\frac{5}{3}
\displaystyle \therefore b_{xy}=\frac{2}{3}
\displaystyle \text{From (ii), } 4y=x-7
\displaystyle \Rightarrow y=\frac{1}{4}x-\frac{7}{4}
\displaystyle \therefore b_{yx}=\frac{1}{4}
\displaystyle b_{xy}b_{yx}=\frac{2}{3}\times \frac{1}{4}=\frac{1}{6}
\displaystyle r=\sqrt{b_{xy}b_{yx}}=\sqrt{\frac{1}{6}}=\frac{1}{\sqrt{6}}
\displaystyle \therefore r=\frac{1}{\sqrt{6}}
\displaystyle \\
\displaystyle \text{(viii) Express the complex number } \frac{(1+\sqrt{3}i)^2}{\sqrt{3}-i} \text{ in the form } a+ib.
\displaystyle \text{Hence, find the modulus and argument of the complex number.}
\displaystyle \text{Answer:}
\displaystyle   z=\frac{(1+\sqrt{3}i)^2}{\sqrt{3}-i}
\displaystyle =\frac{1+2\sqrt{3}i+3i^2}{\sqrt{3}-i}
\displaystyle =\frac{-2+2\sqrt{3}i}{\sqrt{3}-i}
\displaystyle =\frac{(-2+2\sqrt{3}i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)}
\displaystyle =\frac{-2\sqrt{3}-2i+6i+2\sqrt{3}i^2}{3+1}
\displaystyle =\frac{-2\sqrt{3}-2\sqrt{3}+4i}{4}
\displaystyle =\frac{-4\sqrt{3}+4i}{4}
\displaystyle =-\sqrt{3}+i
\displaystyle \therefore z=-\sqrt{3}+i
\displaystyle |z|=\sqrt{(-\sqrt{3})^2+1^2}=\sqrt{3+1}=2
\displaystyle \text{Since } z \text{ lies in the second quadrant,}
\displaystyle \arg(z)=\pi-\tan^{-1}\frac{1}{\sqrt{3}}=\pi-\frac{\pi}{6}=\frac{5\pi}{6}
\displaystyle \\
\displaystyle \text{(ix) A bag contains } 20 \text{ balls numbered from } 1 \text{ to } 20. \text{ One ball is drawn at random} \\ \text{from the bag.}   \text{ What is the probability that the ball drawn is marked with} \\ \text{a number which is a multiple of } 3 \text{ or } 4?
\displaystyle \text{Answer:}
\displaystyle \text{Number of balls }=20 \Rightarrow n(S)=20
\displaystyle \text{Multiples of } 3 \text{ are } 3,6,9,12,15,18
\displaystyle \Rightarrow n(A)=6
\displaystyle \text{Multiples of } 4 \text{ are } 4,8,12,16,20
\displaystyle \Rightarrow n(B)=5
\displaystyle \text{Multiple of both } 3 \text{ and } 4 \text{ is } 12
\displaystyle \Rightarrow n(A\cap B)=1
\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B)
\displaystyle =6+5-1=10
\displaystyle P(A\cup B)=\frac{10}{20}=\frac{1}{2}
\displaystyle \\
\displaystyle \text{(x) Solve the differential equation: } (x+1)\ dy-2xy\ dx=0.
\displaystyle \text{Answer:}
\displaystyle   (x+1)\ dy-2xy\ dx=0
\displaystyle (x+1)\ dy=2xy\ dx
\displaystyle \frac{dy}{dx}=\frac{2xy}{x+1}
\displaystyle \frac{1}{y}\ dy=\frac{2x}{x+1}\ dx
\displaystyle \frac{1}{y}\ dy=\left(\frac{2x+2-2}{x+1}\right)dx
\displaystyle \frac{1}{y}\ dy=\left(2-\frac{2}{x+1}\right)dx
\displaystyle \int \frac{1}{y}\ dy=\int \left(2-\frac{2}{x+1}\right)dx
\displaystyle \log |y|=2x-2\log |x+1|+\log C
\displaystyle \log |y|+2\log |x+1|=2x+\log C
\displaystyle \log \left|y(x+1)^2\right|=\log Ce^{2x}
\displaystyle y(x+1)^2=Ce^{2x}
\displaystyle y=\frac{Ce^{2x}}{(x+1)^2}
\displaystyle \\

\displaystyle \textbf{Question 2:}
\displaystyle \text{(a) Using properties of determinants, prove that:}
\displaystyle \left| \begin{array}{ccc} a^2+1 & ab & ac \\ ba & b^2+1 & bc \\ ca & cb & c^2+1 \end{array} \right|=a^2+b^2+c^2+1 \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) } \Delta=\left| \begin{array}{ccc} a^2+1 & ab & ac \\ ba & b^2+1 & bc \\ ca & cb & c^2+1 \end{array} \right|
\displaystyle \text{Applying } R_1\rightarrow \frac{1}{a}R_1,\ R_2\rightarrow \frac{1}{b}R_2,\ R_3\rightarrow \frac{1}{c}R_3
\displaystyle \Delta=abc\left| \begin{array}{ccc} a+\frac{1}{a} & b & c \\ a & b+\frac{1}{b} & c \\ a & b & c+\frac{1}{c} \end{array} \right|
\displaystyle \Delta=\left| \begin{array}{ccc} a^2+1 & b^2 & c^2 \\ a^2 & b^2+1 & c^2 \\ a^2 & b^2 & c^2+1 \end{array} \right|
\displaystyle \text{Applying } C_1\rightarrow C_1+C_2+C_3
\displaystyle \Delta=\left| \begin{array}{ccc} 1+a^2+b^2+c^2 & b^2 & c^2 \\ 1+a^2+b^2+c^2 & b^2+1 & c^2 \\ 1+a^2+b^2+c^2 & b^2 & c^2+1 \end{array} \right|
\displaystyle \Delta=(1+a^2+b^2+c^2)\left| \begin{array}{ccc} 1 & b^2 & c^2 \\ 1 & b^2+1 & c^2 \\ 1 & b^2 & c^2+1 \end{array} \right|
\displaystyle \text{Applying } R_2\rightarrow R_2-R_1 \text{ and } R_3\rightarrow R_3-R_1
\displaystyle \Delta=(1+a^2+b^2+c^2)\left| \begin{array}{ccc} 1 & b^2 & c^2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right|
\displaystyle \text{Expanding along } C_1,\text{ we get}
\displaystyle \Delta=(1+a^2+b^2+c^2)\left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right|
\displaystyle \Delta=1+a^2+b^2+c^2
\displaystyle \therefore \left| \begin{array}{ccc} a^2+1 & ab & ac \\ ba & b^2+1 & bc \\ ca & cb & c^2+1 \end{array} \right|=a^2+b^2+c^2+1
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \text{(b) Solve the following system of linear equations using matrix method:}
\displaystyle x-2y=10,\quad 2x+y+3z=8,\quad -2y+z=7 \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given system of equations is}
\displaystyle x-2y=10
\displaystyle 2x+y+3z=8
\displaystyle -2y+z=7
\displaystyle \Rightarrow A=\begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix},\ X=\begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B=\begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}
\displaystyle |A|=\left| \begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array} \right|
\displaystyle =1(1+6)+2(2-0)+0
\displaystyle =7+4=11\neq 0
\displaystyle \therefore A^{-1}\text{ exists.}
\displaystyle \text{Therefore, the system has the unique solution } X=A^{-1}B
\displaystyle A_{11}=\left| \begin{array}{cc} 1 & 3 \\ -2 & 1 \end{array} \right|=7,\quad A_{12}=-\left| \begin{array}{cc} 2 & 3 \\ 0 & 1 \end{array} \right|=-2,\quad A_{13}=\left| \begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array} \right|=-4
\displaystyle A_{21}=-\left| \begin{array}{cc} -2 & 0 \\ -2 & 1 \end{array} \right|=2,\quad A_{22}=\left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right|=1,\quad A_{23}=-\left| \begin{array}{cc} 1 & -2 \\ 0 & -2 \end{array} \right|=2
\displaystyle A_{31}=\left| \begin{array}{cc} -2 & 0 \\ 1 & 3 \end{array} \right|=-6,\quad A_{32}=-\left| \begin{array}{cc} 1 & 0 \\ 2 & 3 \end{array} \right|=-3,\quad A_{33}=\left| \begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array} \right|=5
\displaystyle \therefore {adj}\  A=\left[\begin{array}{ccc} 7 & -2 & -4 \\ 2 & 1 & 2 \\ -6 & -3 & 5 \end{array}\right]^T
\displaystyle =\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right]
\displaystyle A^{-1}=\frac{{adj}\  A}{|A|}
\displaystyle \Rightarrow A^{-1}=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right]
\displaystyle \text{Now, } AX=B \Rightarrow X=A^{-1}B
\displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{11}\left[\begin{array}{ccc} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{array}\right]\begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{11}\begin{bmatrix} 70+16-42 \\ -20+8-21 \\ -40+16+35 \end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{11}\begin{bmatrix} 44 \\ -33 \\ 11 \end{bmatrix}
\displaystyle \Rightarrow x=4,\ y=-3,\ z=1
\displaystyle \\

\displaystyle \textbf{Question 3:}
\displaystyle \text{(a) If } \cos^{-1}x+\cos^{-1}y+\cos^{-1}z=\pi,\text{ prove that } x^2+y^2+z^2+2xyz=1. \hspace{0.5cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } \cos^{-1}x+\cos^{-1}y+\cos^{-1}z=\pi
\displaystyle \cos^{-1}x+\cos^{-1}y=\pi-\cos^{-1}z
\displaystyle \Rightarrow \cos^{-1}\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)=\pi-\cos^{-1}z
\displaystyle \Rightarrow xy-\sqrt{1-x^2}\sqrt{1-y^2}=\cos\left(\pi-\cos^{-1}z\right)
\displaystyle \Rightarrow xy-\sqrt{1-x^2}\sqrt{1-y^2}=-\cos(\cos^{-1}z)
\displaystyle \Rightarrow xy-\sqrt{1-x^2}\sqrt{1-y^2}=-z
\displaystyle \Rightarrow xy+z=\sqrt{1-x^2}\sqrt{1-y^2}
\displaystyle \text{Squaring both sides, we get}
\displaystyle (xy+z)^2=(1-x^2)(1-y^2)
\displaystyle \Rightarrow x^2y^2+z^2+2xyz=1-x^2-y^2+x^2y^2
\displaystyle \Rightarrow x^2+y^2+z^2+2xyz=1
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \text{(b) } P,Q \text{ and } R \text{ represent switches in 'on' position and } P',Q' \text{ and } R' \text{ represent} \\ \text{switches in off position. }   \text{Construct a switching circuit representing the polynomial } \\ PR+Q(Q'+R)(P+QR).
\displaystyle \text{Using Boolean algebra, simplify the polynomial expression and construct the} \\ \text{simplified circuit.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} \displaystyle \text{(b) The given Boolean expression is } PR+Q(Q'+R)(P+QR)
\displaystyle PR + Q(Q' + R)(P + QR)
\displaystyle = PR + (QQ' + QR)(P + QR)
\displaystyle = PR + (0 + QR)(P + QR)
\displaystyle = PR + QR(P + QR)
\displaystyle = PR + PQR + QR
\displaystyle = PR + QR
\displaystyle = R(P + Q)
\displaystyle \text{Hence, the simplified circuit is represented by } R(P+Q).
\displaystyle \\

\displaystyle \textbf{Question 4:}
\displaystyle \text{(a) Verify Rolle's Theorem for the function } f(x)=e^x(\sin x-\cos x) \text{ on } \left[\frac{\pi}{4},\frac{5\pi}{4}\right].
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given function is } f(x)=e^x(\sin x-\cos x) \text{ on } \left[\frac{\pi}{4},\frac{5\pi}{4}\right]
\displaystyle \text{The function } f(x) \text{ is continuous on } \left[\frac{\pi}{4},\frac{5\pi}{4}\right] \text{ and differentiable on } \left(\frac{\pi}{4},\frac{5\pi}{4}\right)
\displaystyle f'(x)=e^x(\cos x+\sin x)+e^x(\sin x-\cos x)
\displaystyle =e^x(\cos x+\sin x+\sin x-\cos x)
\displaystyle =2e^x\sin x
\displaystyle f\left(\frac{\pi}{4}\right)=e^{\frac{\pi}{4}}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)=0
\displaystyle f\left(\frac{5\pi}{4}\right)=e^{\frac{5\pi}{4}}\left(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=0
\displaystyle \therefore f\left(\frac{\pi}{4}\right)=f\left(\frac{5\pi}{4}\right)
\displaystyle \text{Therefore, all conditions of Rolle's Theorem are satisfied.}
\displaystyle \text{Now, } f'(c)=0
\displaystyle \Rightarrow 2e^c\sin c=0
\displaystyle \Rightarrow \sin c=0
\displaystyle \Rightarrow c=n\pi
\displaystyle \text{Since } c=\pi \text{ lies in } \left(\frac{\pi}{4},\frac{5\pi}{4}\right), \text{ Rolle's Theorem is verified.}
\displaystyle \\
\displaystyle \text{(b) Find the equation of the parabola with latus rectum joining points } (4,6) \text{ and } (4,-2).
\displaystyle \text{Answer:}
\displaystyle \text{(b) The endpoints of the latus rectum are } (4,6) \text{ and } (4,-2)
\displaystyle \text{Midpoint of the latus rectum is the focus.}
\displaystyle S=\left(\frac{4+4}{2},\frac{6+(-2)}{2}\right)=(4,2)
\displaystyle \text{Length of latus rectum}=\sqrt{(4-4)^2+(6+2)^2}=8
\displaystyle \therefore 4a=8
\displaystyle \Rightarrow a=2
\displaystyle \text{Since the latus rectum is vertical, the parabola is of the form } (y-k)^2=4a(x-h)
\displaystyle \text{Its focus is } (h+a,k)=(4,2)
\displaystyle \Rightarrow h+2=4,\ k=2
\displaystyle \Rightarrow h=2,\ k=2
\displaystyle \therefore (y-2)^2=8(x-2)
\displaystyle \text{Hence, the equation of the parabola is } (y-2)^2=8(x-2)
\displaystyle \\

\displaystyle \textbf{Question 5:}
\displaystyle \text{(a) If } y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}},\text{ prove that } (1-x^2)\frac{dy}{dx}=x+\frac{y}{x}. \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}
\displaystyle y\sqrt{1-x^2}=x\sin^{-1}x
\displaystyle \text{Differentiating both sides with respect to } x,\text{ we get}
\displaystyle y\left(\frac{-x}{\sqrt{1-x^2}}\right)+\sqrt{1-x^2}\frac{dy}{dx}=x\frac{1}{\sqrt{1-x^2}}+\sin^{-1}x
\displaystyle \text{Multiplying throughout by } \sqrt{1-x^2},\text{ we get}
\displaystyle -xy+(1-x^2)\frac{dy}{dx}=x+\sqrt{1-x^2}\sin^{-1}x
\displaystyle \text{From } y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}},\text{ we get } \sin^{-1}x=\frac{y\sqrt{1-x^2}}{x}
\displaystyle \Rightarrow -xy+(1-x^2)\frac{dy}{dx}=x+\sqrt{1-x^2}\cdot \frac{y\sqrt{1-x^2}}{x}
\displaystyle \Rightarrow -xy+(1-x^2)\frac{dy}{dx}=x+\frac{y(1-x^2)}{x}
\displaystyle \Rightarrow -xy+(1-x^2)\frac{dy}{dx}=x+\frac{y}{x}-xy
\displaystyle \Rightarrow (1-x^2)\frac{dy}{dx}=x+\frac{y}{x}
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \text{(b) A wire of length } 50 \text{ m is cut into two pieces. One piece is bent in the shape } \\  \text{of a square and the other in the shape of a circle. What should be the length of each } \\ \text{piece so that the combined area is minimum? } [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given length of the wire }=50\text{ m}
\displaystyle \text{Let the length of wire used for the square be } x\text{ m}
\displaystyle \therefore \text{Length of wire used for the circle }=(50-x)\text{ m}
\displaystyle \text{Let the side of the square be } a
\displaystyle \therefore 4a=x \Rightarrow a=\frac{x}{4}
\displaystyle \text{Let the radius of the circle be } r
\displaystyle \therefore 2\pi r=50-x \Rightarrow r=\frac{50-x}{2\pi}
\displaystyle \text{Combined area } A=a^2+\pi r^2
\displaystyle A=\left(\frac{x}{4}\right)^2+\pi\left(\frac{50-x}{2\pi}\right)^2
\displaystyle A=\frac{x^2}{16}+\frac{(50-x)^2}{4\pi}
\displaystyle \text{Differentiating with respect to } x,\text{ we get}
\displaystyle \frac{dA}{dx}=\frac{2x}{16}+\frac{2(50-x)(-1)}{4\pi}
\displaystyle \frac{dA}{dx}=\frac{x}{8}+\frac{x-50}{2\pi}
\displaystyle \frac{dA}{dx}=\frac{\pi x+4x-200}{8\pi}
\displaystyle \frac{dA}{dx}=\frac{x(\pi+4)-200}{8\pi}
\displaystyle \text{For maximum or minimum, } \frac{dA}{dx}=0
\displaystyle \Rightarrow x(\pi+4)-200=0
\displaystyle \Rightarrow x=\frac{200}{\pi+4}
\displaystyle \frac{d^2A}{dx^2}=\frac{\pi+4}{8\pi}>0
\displaystyle \therefore A \text{ is minimum when } x=\frac{200}{\pi+4}
\displaystyle \text{Length of wire used for the square }=\frac{200}{\pi+4}\text{ m}
\displaystyle \text{Length of wire used for the circle }=50-\frac{200}{\pi+4}
\displaystyle =\frac{50(\pi+4)-200}{\pi+4}
\displaystyle =\frac{50\pi}{\pi+4}\text{ m}
\displaystyle \\

\displaystyle \textbf{Question 6:}
\displaystyle \text{(a) Evaluate: } \int \frac{x+\sin x}{1+\cos x}\ dx. \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) } I=\int \frac{x+\sin x}{1+\cos x}\ dx
\displaystyle =\int \frac{x+\sin x}{2\cos^2\frac{x}{2}}\ dx
\displaystyle =\int \frac{x}{2\cos^2\frac{x}{2}}\ dx+\int \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}\ dx
\displaystyle =\frac{1}{2}\int x\sec^2\frac{x}{2}\ dx+\int \tan\frac{x}{2}\ dx
\displaystyle =\frac{1}{2}\left[x\int \sec^2\frac{x}{2}\ dx-\int \left(\frac{d}{dx}x\right)\left(\int \sec^2\frac{x}{2}\ dx\right)dx\right]+\int \tan\frac{x}{2}\ dx
\displaystyle =\frac{1}{2}\left[x\cdot 2\tan\frac{x}{2}-\int 2\tan\frac{x}{2}\ dx\right]+\int \tan\frac{x}{2}\ dx
\displaystyle =x\tan\frac{x}{2}-\int \tan\frac{x}{2}\ dx+\int \tan\frac{x}{2}\ dx
\displaystyle =x\tan\frac{x}{2}+C
\displaystyle \\
\displaystyle \text{(b) Sketch the graph of the curves } y^2=x \text{ and } y^2=4-3x \text{ and find the area} \\ \text{enclosed between them. } [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given curves are } y^2=x \text{ and } y^2=4-3x\displaystyle y^2=4-3x
\displaystyle \Rightarrow y^2=-3\left(x-\frac{4}{3}\right)
\displaystyle \text{The vertex of this parabola is } \left(\frac{4}{3},0\right)
\displaystyle \text{It intersects the } y\text{-axis at } A(0,2) \text{ and } B(0,-2)
\displaystyle \text{For points of intersection, } x=4-3x
\displaystyle \Rightarrow 4x=4
\displaystyle \Rightarrow x=1
\displaystyle \therefore y^2=1
\displaystyle \Rightarrow y=\pm 1
\displaystyle \text{Therefore, the points of intersection are } P(1,1) \text{ and } Q(1,-1)
\displaystyle \text{Required area }=2\left[\int\limits_{0}^{1}\sqrt{x}\ dx+\int\limits_{1}^{\frac{4}{3}}\sqrt{4-3x}\ dx\right]
\displaystyle =2\left[\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)_{0}^{1}+\left(\frac{2(4-3x)^{\frac{3}{2}}}{-9}\right)_{1}^{\frac{4}{3}}\right]
\displaystyle =2\left[\frac{2}{3}-0-\frac{2}{9}(0-1)\right]
\displaystyle =2\left[\frac{2}{3}+\frac{2}{9}\right]
\displaystyle =2\left[\frac{6+2}{9}\right]
\displaystyle =2\left(\frac{8}{9}\right)
\displaystyle =\frac{16}{9}\text{ sq. units}
\displaystyle \\

\displaystyle \textbf{Question 7:}
\displaystyle \text{(a) A psychologist selected a random sample of } 22 \text{ students. He grouped them in } \\  11 \text{ pairs so that}   \text{the students in each pair have nearly equal scores in an intelligence} \\ \text{test. In each pair, one student was taught}   \text{by method } A \text{ and the other by method } \\ B \text{ and }\text{examined after the course. The marks obtained are as follows:}
\displaystyle \begin{array}{|c|ccccccccccc|} \hline \text{Pairs} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline \text{Method } A & 24 & 29 & 19 & 14 & 30 & 19 & 27 & 30 & 20 & 28 & 11 \\ \text{Method } B & 37 & 35 & 16 & 26 & 23 & 27 & 19 & 20 & 16 & 11 & 21 \\ \hline \end{array}
\displaystyle \text{Calculate Spearman's Rank correlation.} \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) The calculation table is as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Pairs} & A & B & \text{Rank } A & \text{Rank } B & D & D^2 \\ \hline 1 & 24 & 37 & 6 & 1 & 5 & 25 \\ 2 & 29 & 35 & 3 & 2 & 1 & 1 \\ 3 & 19 & 16 & 8.5 & 9.5 & -1 & 1 \\ 4 & 14 & 26 & 10 & 4 & 6 & 36 \\ 5 & 30 & 23 & 1.5 & 5 & -3.5 & 12.25 \\ 6 & 19 & 27 & 8.5 & 3 & 5.5 & 30.25 \\ 7 & 27 & 19 & 5 & 8 & -3 & 9 \\ 8 & 30 & 20 & 1.5 & 7 & -5.5 & 30.25 \\ 9 & 20 & 16 & 7 & 9.5 & -2.5 & 6.25 \\ 10 & 28 & 11 & 4 & 11 & -7 & 49 \\ 11 & 11 & 21 & 11 & 6 & 5 & 25 \\ \hline \multicolumn{6}{|r|}{\Sigma D^2} & 225 \\ \hline \end{array}
\displaystyle \text{Here, } n=11,\ \Sigma D^2=225
\displaystyle \text{There are three ties, each of length } 2
\displaystyle \therefore \text{Correction factor }=\frac{2^3-2}{12}+\frac{2^3-2}{12}+\frac{2^3-2}{12}
\displaystyle =\frac{6}{12}+\frac{6}{12}+\frac{6}{12}=\frac{3}{2}
\displaystyle R=1-\frac{6\left[\Sigma D^2+\Sigma \frac{m^3-m}{12}\right]}{n(n^2-1)}
\displaystyle =1-\frac{6\left[225+\frac{3}{2}\right]}{11(11^2-1)}
\displaystyle =1-\frac{6(226.5)}{11(120)}
\displaystyle =1-\frac{1359}{1320}
\displaystyle =1-1.0295
\displaystyle =-0.0295
\displaystyle \therefore R\approx -0.03
\displaystyle \text{Hence, there is a very low negative correlation.}
\displaystyle \\
\displaystyle \text{(b) The coefficient of correlation between } X \text{ and } Y \text{ is } 0.5. \text{ The mean of } X \text{ is } 3
\displaystyle \text{and that of } Y \text{ is } 5. \text{ Their standard deviations are } 5 \text{ and } 4 \text{ respectively. Find:}
\displaystyle \text{(i) the two lines of regression}
\displaystyle \text{(ii) the expected value of } Y,\text{ when } X \text{ is given as } 14
\displaystyle \text{(iii) the expected value of } X,\text{ when } Y \text{ is given as } 9. \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, } \overline{x}=3,\ \overline{y}=5,\ \sigma_x=5,\ \sigma_y=4,\ r=0.5
\displaystyle b_{yx}=r\frac{\sigma_y}{\sigma_x}=0.5\times \frac{4}{5}=0.4
\displaystyle b_{xy}=r\frac{\sigma_x}{\sigma_y}=0.5\times \frac{5}{4}=0.625
\displaystyle \text{(i) Regression equation of } y \text{ on } x \text{ is}
\displaystyle y-\overline{y}=b_{yx}(x-\overline{x})
\displaystyle y-5=0.4(x-3)
\displaystyle y=0.4x+3.8
\displaystyle \text{Regression equation of } x \text{ on } y \text{ is}
\displaystyle x-\overline{x}=b_{xy}(y-\overline{y})
\displaystyle x-3=0.625(y-5)
\displaystyle x=0.625y-0.125
\displaystyle \text{(ii) When } x=14,
\displaystyle y=0.4(14)+3.8
\displaystyle y=5.6+3.8=9.4
\displaystyle \text{(iii) When } y=9,
\displaystyle x=0.625(9)-0.125
\displaystyle x=5.625-0.125=5.5
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(a) In a college, } 70\% \text{ students pass in Physics, } 75\% \text{ pass in Mathematics and } 10\% \\ \text{ students fail in both.}   \text{One student is chosen at random. What is the probability that:}
\displaystyle \text{(i) He passes in Physics and Mathematics}
\displaystyle \text{(ii) He passes in Mathematics given that he passes in Physics}
\displaystyle \text{(iii) He passes in Physics given that he passes in Mathematics.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let } P \text{ be the event that a student passes in Physics and }  \\ M \text{ be the event that a student passes in Mathematics.}
\displaystyle P(P)=\frac{70}{100},\quad P(M)=\frac{75}{100}
\displaystyle P(P'\cap M')=\frac{10}{100}
\displaystyle \therefore P(P\cup M)=1-P(P'\cap M')
\displaystyle =1-\frac{10}{100}=\frac{90}{100}
\displaystyle \text{Now, } P(P\cup M)=P(P)+P(M)-P(P\cap M)
\displaystyle \frac{90}{100}=\frac{70}{100}+\frac{75}{100}-P(P\cap M)
\displaystyle P(P\cap M)=\frac{145}{100}-\frac{90}{100}=\frac{55}{100}
\displaystyle \text{(i) Probability that he passes in Physics and Mathematics}
\displaystyle =P(P\cap M)=\frac{55}{100}=\frac{11}{20}
\displaystyle \text{(ii) Probability that he passes in Mathematics given that he passes in Physics}
\displaystyle P(M/P)=\frac{P(M\cap P)}{P(P)}
\displaystyle =\frac{\frac{55}{100}}{\frac{70}{100}}=\frac{55}{70}=\frac{11}{14}
\displaystyle \text{(iii) Probability that he passes in Physics given that he passes in Mathematics}
\displaystyle P(P/M)=\frac{P(P\cap M)}{P(M)}
\displaystyle =\frac{\frac{55}{100}}{\frac{75}{100}}=\frac{55}{75}=\frac{11}{15}
\displaystyle \\
\displaystyle \text{(b) A bag contains } 5 \text{ white and } 4 \text{ black balls and another bag contains } 7 \text{ white and } 9 \\ \text{ black balls.}   \text{A ball is drawn from the first bag and two balls are drawn from the } \\ \text{second bag. What is the probability of drawing } 1 \text{ white and } 2 \text{ black balls?} \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) To get } 1 \text{ white and } 2 \text{ black balls, the following two cases are possible:}
\displaystyle \text{Case I: From Bag I, } 1 \text{ black ball and from Bag II, } 1 \text{ white and } 1 \text{ black ball.}
\displaystyle \text{Case II: From Bag I, } 1 \text{ white ball and from Bag II, } 2 \text{ black balls.}
\displaystyle P(\text{black from Bag I})=\frac{{}^{4}C_1}{{}^{9}C_1}=\frac{4}{9}
\displaystyle P(\text{one white and one black from Bag II})=\frac{{}^{7}C_1\cdot {}^{9}C_1}{{}^{16}C_2}=\frac{63}{120}=\frac{21}{40}
\displaystyle P(\text{white from Bag I})=\frac{{}^{5}C_1}{{}^{9}C_1}=\frac{5}{9}
\displaystyle P(\text{two black from Bag II})=\frac{{}^{9}C_2}{{}^{16}C_2}=\frac{36}{120}=\frac{3}{10}
\displaystyle \text{Required probability}
\displaystyle =\frac{4}{9}\cdot \frac{21}{40}+\frac{5}{9}\cdot \frac{3}{10}
\displaystyle =\frac{84}{360}+\frac{15}{90}
\displaystyle =\frac{7}{30}+\frac{1}{6}
\displaystyle =\frac{7}{30}+\frac{5}{30}
\displaystyle =\frac{12}{30}=\frac{2}{5}
\displaystyle \\

\displaystyle \textbf{Question 9:}
\displaystyle \text{(a) Using De Moivre's theorem, find the least positive integer } n \text{ such that } \\ \left(\frac{2i}{1+i}\right)^n \text{ is a positive integer.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) We have,}
\displaystyle \frac{2i}{1+i}=\frac{2i}{1+i}\times \frac{1-i}{1-i}
\displaystyle =\frac{2i(1-i)}{1-i^2}
\displaystyle =\frac{2i-2i^2}{2}
\displaystyle =1+i
\displaystyle =\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)
\displaystyle =\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)
\displaystyle \therefore \left(\frac{2i}{1+i}\right)^n=\left[\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\right]^n
\displaystyle =2^{\frac{n}{2}}\left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right)
\displaystyle \text{This is a positive integer when } \sin\frac{n\pi}{4}=0 \text{ and } \cos\frac{n\pi}{4}=1
\displaystyle \therefore \frac{n\pi}{4}=2k\pi,\text{ where } k\in I
\displaystyle \Rightarrow n=8k
\displaystyle \text{For the least positive integer, take } k=1
\displaystyle \therefore n=8
\displaystyle \\
\displaystyle \text{(b) Solve the following differential equation: }(3xy+y^2)dx+(x^2+xy)dy=0. \ \ \  [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) } (3xy+y^2)dx+(x^2+xy)dy=0
\displaystyle \frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy}
\displaystyle \text{Put } y=vx
\displaystyle \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=-\frac{3x(vx)+v^2x^2}{x^2+x(vx)}
\displaystyle v+x\frac{dv}{dx}=-\frac{3v+v^2}{1+v}
\displaystyle x\frac{dv}{dx}=-\frac{3v+v^2}{1+v}-v
\displaystyle x\frac{dv}{dx}=\frac{-3v-v^2-v-v^2}{1+v}
\displaystyle x\frac{dv}{dx}=-\frac{2v^2+4v}{1+v}
\displaystyle \frac{1+v}{2v^2+4v}\ dv=-\frac{1}{x}\ dx
\displaystyle \int \frac{1+v}{2v^2+4v}\ dv=-\int \frac{1}{x}\ dx
\displaystyle \frac{1}{4}\int \frac{2v+2}{v^2+2v}\ dv=-\int \frac{1}{x}\ dx
\displaystyle \frac{1}{4}\log|v^2+2v|=-\log|x|+\log C
\displaystyle \log|v^2+2v|=\log\frac{C}{x^4}
\displaystyle v^2+2v=\frac{C}{x^4}
\displaystyle \left(\frac{y}{x}\right)^2+2\left(\frac{y}{x}\right)=\frac{C}{x^4}
\displaystyle \frac{y^2}{x^2}+\frac{2y}{x}=\frac{C}{x^4}
\displaystyle x^2y^2+2x^3y=C
\displaystyle \\

Section – B (20 Marks)

\displaystyle \textbf{Question 10:}
\displaystyle \text{(a) In a triangle } ABC,\text{ using vectors, prove that } c^2=a^2+b^2-2ab\cos C. \hspace{1.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let the sides of the triangle be represented by vectors } \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}
\displaystyle \therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}
\displaystyle \text{Squaring both sides, we get}
\displaystyle (\overrightarrow{a}+\overrightarrow{b})\cdot(\overrightarrow{a}+\overrightarrow{b})=(-\overrightarrow{c})\cdot(-\overrightarrow{c})
\displaystyle \overrightarrow{a}\cdot\overrightarrow{a}+\overrightarrow{b}\cdot\overrightarrow{b}+2\overrightarrow{a}\cdot\overrightarrow{b}=c^2
\displaystyle a^2+b^2+2ab\cos(\pi-C)=c^2
\displaystyle a^2+b^2-2ab\cos C=c^2
\displaystyle \therefore c^2=a^2+b^2-2ab\cos C
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \text{(b) Prove that: } \overrightarrow{a}\cdot(\overrightarrow{b}+\overrightarrow{c})\times(\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c})=[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]. \hspace{1.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) LHS}=\overrightarrow{a}\cdot(\overrightarrow{b}+\overrightarrow{c})\times(\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c})
\displaystyle =\overrightarrow{a}\cdot\left(\overrightarrow{b}\times\overrightarrow{a}+\overrightarrow{b}\times2\overrightarrow{b}+\overrightarrow{b}\times3\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}+\overrightarrow{c}\times2\overrightarrow{b}+\overrightarrow{c}\times3\overrightarrow{c}\right)
\displaystyle =\overrightarrow{a}\cdot\left(\overrightarrow{b}\times\overrightarrow{a}+3\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}+2\overrightarrow{c}\times\overrightarrow{b}\right)
\displaystyle =\overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{a})+3\overrightarrow{a}\cdot(\overrightarrow{b}\times\overrightarrow{c})+\overrightarrow{a}\cdot(\overrightarrow{c}\times\overrightarrow{a})+2\overrightarrow{a}\cdot(\overrightarrow{c}\times\overrightarrow{b})
\displaystyle =0+3[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]+0+2[\overrightarrow{a}\ \overrightarrow{c}\ \overrightarrow{b}]
\displaystyle =3[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]-2[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]
\displaystyle =[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]
\displaystyle =\text{RHS}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11:}
\displaystyle \text{(a) Find the equation of a line passing through the points } P(-1,3,2) \text{ and } Q(-4,2,-2).  \\ \text{Also, if the point } R(5,5,\lambda) \text{ is collinear with the points } P \text{ and } Q,\text{ find the value of } \lambda. \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Equation of the line passing through } P(-1,3,2) \text{ and } Q(-4,2,-2) \text{ is}
\displaystyle \frac{x+1}{-4+1}=\frac{y-3}{2-3}=\frac{z-2}{-2-2}
\displaystyle \Rightarrow \frac{x+1}{-3}=\frac{y-3}{-1}=\frac{z-2}{-4}
\displaystyle \Rightarrow \frac{x+1}{3}=\frac{y-3}{1}=\frac{z-2}{4}
\displaystyle \text{Since } R(5,5,\lambda) \text{ lies on it,}
\displaystyle \frac{5+1}{3}=\frac{5-3}{1}=\frac{\lambda-2}{4}
\displaystyle \Rightarrow 2=2=\frac{\lambda-2}{4}
\displaystyle \Rightarrow \lambda-2=8
\displaystyle \Rightarrow \lambda=10
\displaystyle \\
\displaystyle \text{(b) Find the equation of the plane passing through the points } (2,-3,1) \text{ and } (-1,1,-7)
\displaystyle \text{and perpendicular to the plane } x-2y+5z+1=0. \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) The required plane is perpendicular to the given plane}
\displaystyle x-2y+5z+1=0
\displaystyle \text{The normal vector of the given plane is } (1,-2,5)
\displaystyle \text{The required plane passes through } A(2,-3,1) \text{ and } B(-1,1,-7)
\displaystyle \therefore \overrightarrow{AB}=(-1-2,\ 1+3,\ -7-1)=(-3,4,-8)
\displaystyle \text{Hence, the required plane contains the vectors } (-3,4,-8) \text{ and } (1,-2,5)
\displaystyle \left| \begin{array}{ccc} x-2 & y+3 & z-1 \\ -3 & 4 & -8 \\ 1 & -2 & 5 \end{array} \right|=0
\displaystyle \Rightarrow (x-2)(20-16)-(y+3)(-15+8)+(z-1)(6-4)=0
\displaystyle \Rightarrow 4(x-2)+7(y+3)+2(z-1)=0
\displaystyle \Rightarrow 4x-8+7y+21+2z-2=0
\displaystyle \Rightarrow 4x+7y+2z+11=0
\displaystyle \text{Hence, the required plane is } 4x+7y+2z+11=0
\displaystyle \\

\displaystyle \textbf{Question 12:}
\displaystyle \text{(a) In a bolt factory, three machines } A,B \text{ and } C \text{ manufacture } 25\%,35\% \text{ and } 40\%
\displaystyle \text{of the total production respectively. Of their respective outputs, } 5\%,4\% \text{ and } 2\% \text{ are} \\ \text{defective.}  \text{A bolt is drawn at random and is found to be defective. Find the probability} \\ \text{that it was manufactured by machine } C. \hspace{2.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let } E_1,E_2,E_3 \text{ be the events that the bolt is produced by machines } A,B,C \text{ respectively.}
\displaystyle E_1,E_2 \text{ and } E_3 \text{ are mutually exclusive and exhaustive events.}
\displaystyle P(E_1)=\frac{25}{100},\quad P(E_2)=\frac{35}{100},\quad P(E_3)=\frac{40}{100}
\displaystyle \text{Let } E \text{ be the event that the bolt chosen is defective.}
\displaystyle P(E/E_1)=\frac{5}{100},\quad P(E/E_2)=\frac{4}{100},\quad P(E/E_3)=\frac{2}{100}
\displaystyle \text{Required probability }=P(E_3/E)
\displaystyle P(E_3/E)=\frac{P(E_3)P(E/E_3)}{P(E_1)P(E/E_1)+P(E_2)P(E/E_2)+P(E_3)P(E/E_3)}
\displaystyle =\frac{\frac{40}{100}\times \frac{2}{100}}{\frac{25}{100}\times \frac{5}{100}+\frac{35}{100}\times \frac{4}{100}+\frac{40}{100}\times \frac{2}{100}}
\displaystyle =\frac{80}{125+140+80}
\displaystyle =\frac{80}{345}=\frac{16}{69}
\displaystyle \therefore P(E_3/E)=\frac{16}{69}
\displaystyle \\
\displaystyle \text{(b) On dialing certain telephone numbers, assume that on an average, one } \\ \text{telephone number out of five is busy.}   \text{Ten telephone numbers are randomly selected and dialed. } \\ \text{Find the  probability that at least three of them will be busy.} \hspace{1.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Probability that a telephone number is busy is } p=\frac{1}{5}
\displaystyle \text{Probability that a telephone number is not busy is } q=1-\frac{1}{5}=\frac{4}{5}
\displaystyle \text{Given, } n=10
\displaystyle \text{Required probability }=P(X\geq 3)
\displaystyle =1-P(X<3)
\displaystyle =1-[P(0)+P(1)+P(2)]
\displaystyle =1-\left[{}^{10}C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^{10}+{}^{10}C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^9+{}^{10}C_2\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^8\right]
\displaystyle =1-\left[\left(\frac{4}{5}\right)^{10}+10\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^9+45\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^8\right]
\displaystyle =1-\left(\frac{4}{5}\right)^8\left[\left(\frac{4}{5}\right)^2+10\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)+45\left(\frac{1}{5}\right)^2\right]
\displaystyle =1-\left(\frac{4}{5}\right)^8\left[\frac{16}{25}+\frac{8}{5}+\frac{9}{5}\right]
\displaystyle =1-\left(\frac{4}{5}\right)^8\left[\frac{16+40+45}{25}\right]
\displaystyle =1-\left(\frac{4}{5}\right)^8\left(\frac{101}{25}\right)
\displaystyle =1-0.6778
\displaystyle =0.3222
\displaystyle \therefore \text{Required probability }=0.3222 \text{ approximately.}
\displaystyle \\

Section – C (20 Marks)

\displaystyle \textbf{Question 13:}
\displaystyle \text{(a) A person borrows Rs. } 68962 \text{ on the condition that he will repay the money with} \\ \text{compound interest at } 5\%   \text{per annum in } 4 \text{ equal annual installments, the first one being} \\ \text{payable at the end of the first year.}  \text{Find the value of each installment.} \hspace{1.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let the annual installment be Rs. } a
\displaystyle \text{Money borrowed } P=\text{Rs. } 68962
\displaystyle \text{Time period } n=4
\displaystyle \text{Rate of interest } i=5\%=0.05
\displaystyle P=\frac{a}{i}\left[1-(1+i)^{-n}\right]
\displaystyle 68962=\frac{a}{0.05}\left[1-(1.05)^{-4}\right]
\displaystyle 68962=\frac{a}{0.05}\left[1-\frac{1}{(1.05)^4}\right]
\displaystyle 68962=\frac{a}{0.05}(0.1773)
\displaystyle \Rightarrow a=\frac{68962\times 0.05}{0.1773}
\displaystyle \Rightarrow a=19447.83
\displaystyle \therefore \text{Value of each installment }=\text{Rs. } 19447.83 \text{ approximately.}
\displaystyle \\
\displaystyle \text{(b) A company manufactures two types of toys } A \text{ and } B. \text{ A toy of type } A \\ \text{requires } 5 \text{ minutes for cutting}   \text{and } 10 \text{ minutes for assembling. A toy of type } B \text{ requires } \\ 8 \text{minutes for cutting and } 8 \text{ minutes for assembling.}   \text{There are } 3 \text{ hours available for} \\ \text{cutting and } 4 \text{ hours for assembling in a day. The profit is Rs. } 50   \text{on a toy of type } A \\ \text{and Rs. } 60 \text{ on a toy of type } B. \text{ Find the number of toys of each type to maximize profit. } [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) The given data can be written in tabular form as follows:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{} & \text{Toy } A & \text{Toy } B & \text{Time in a day} \\ \hline \text{Cutting Time} & 5 \text{ minutes} & 8 \text{ minutes} & 180 \text{ minutes} \\ \hline \text{Assembling Time} & 10 \text{ minutes} & 8 \text{ minutes} & 240 \text{ minutes} \\ \hline \text{Profit} & \text{Rs. } 50 & \text{Rs. } 60 & \text{} \\ \hline \end{array}
\displaystyle \text{Let the company manufacture } x \text{ toys of type } A \text{ and } y \text{ toys of type } B
\displaystyle \text{Profit function } Z=50x+60y
\displaystyle \text{Subject to constraints:}
\displaystyle 5x+8y\leq 180
\displaystyle 10x+8y\leq 240
\displaystyle x\geq 0,\ y\geq 0\displaystyle \text{For the line } 5x+8y=180,\text{ the points are:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{} & A & B \\ \hline x & 0 & 36 \\ \hline y & 22.5 & 0 \\ \hline \end{array}
\displaystyle \text{For the line } 10x+8y=240,\text{ the points are:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{} & C & D \\ \hline x & 0 & 24 \\ \hline y & 30 & 0 \\ \hline \end{array}
\displaystyle \text{The point of intersection of } 5x+8y=180 \text{ and } 10x+8y=240 \text{ is found as follows:}
\displaystyle 5x+8y=180 \hspace{1.0cm} \text{...(i)}
\displaystyle 10x+8y=240 \hspace{1.0cm} \text{...(ii)}
\displaystyle \text{Subtracting (i) from (ii), we get}
\displaystyle 5x=60
\displaystyle \Rightarrow x=12
\displaystyle \text{Substituting } x=12 \text{ in (i), we get}
\displaystyle 5(12)+8y=180
\displaystyle 60+8y=180
\displaystyle 8y=120
\displaystyle y=15
\displaystyle \therefore E=(12,15)
\displaystyle \text{The corner points and values of the objective function are:}
\displaystyle \begin{array}{|c|c|} \hline \text{Corner Point} & \text{Objective Function } Z=50x+60y \\ \hline O(0,0) & Z=50(0)+60(0)=0 \\ \hline D(24,0) & Z=50(24)+60(0)=1200 \\ \hline E(12,15) & Z=50(12)+60(15)=1500 \\ \hline A(0,22.5) & Z=50(0)+60(22.5)=1350 \\ \hline \end{array}
\displaystyle \text{The maximum profit is Rs. } 1500 \text{ at } E(12,15)
\displaystyle \therefore \text{The company should manufacture } 12 \text{ toys of type } A \text{ and } 15 \text{ toys of type } B.
\displaystyle \\

\displaystyle \textbf{Question 14:}
\displaystyle \text{(a) A firm has the cost function } C=\frac{x^3}{3}-7x^2+111x+50 \text{ and demand function } \\  x=100-p.
\displaystyle \text{(i) Write the total revenue function in terms of } x
\displaystyle \text{(ii) Formulate the total profit function } P \text{ in terms of } x
\displaystyle \text{(iii) Find the profit maximizing level of output.} \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Cost function } C(x)=\frac{x^3}{3}-7x^2+111x+50
\displaystyle \text{Demand function } x=100-p
\displaystyle \Rightarrow p=100-x
\displaystyle \text{(i) Revenue function } R(x)=px
\displaystyle R(x)=x(100-x)
\displaystyle R(x)=100x-x^2
\displaystyle \text{(ii) Profit function } P(x)=R(x)-C(x)
\displaystyle P(x)=100x-x^2-\left(\frac{x^3}{3}-7x^2+111x+50\right)
\displaystyle P(x)=100x-x^2-\frac{x^3}{3}+7x^2-111x-50
\displaystyle P(x)=-\frac{x^3}{3}+6x^2-11x-50
\displaystyle \text{(iii) For profit to be maximum,}
\displaystyle \frac{dP}{dx}=-x^2+12x-11
\displaystyle \frac{dP}{dx}=0
\displaystyle \Rightarrow -x^2+12x-11=0
\displaystyle \Rightarrow x^2-12x+11=0
\displaystyle \Rightarrow x^2-11x-x+11=0
\displaystyle \Rightarrow x(x-11)-1(x-11)=0
\displaystyle \Rightarrow (x-1)(x-11)=0
\displaystyle \Rightarrow x=1,\ 11
\displaystyle \frac{d^2P}{dx^2}=12-2x
\displaystyle \text{At } x=1,\quad \frac{d^2P}{dx^2}=12-2=10>0
\displaystyle \therefore x=1 \text{ gives minimum profit.}
\displaystyle \text{At } x=11,\quad \frac{d^2P}{dx^2}=12-22=-10<0
\displaystyle \therefore x=11 \text{ gives maximum profit.}
\displaystyle \text{Hence, the profit maximizing level of output is } x=11
\displaystyle \\
\displaystyle \text{(b) A bill of Rs. } 5050 \text{ is drawn on 13th April 2013. It was discounted on 4th July 2013 at }
\displaystyle 5\%  \text{per annum. If the banker's gain on the transaction is Rs. } 0.50,\text{ find the nominal} \\ \text{date of maturity of the bill.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let the unexpired period of the bill at the time of discounting be } t \text{ years.}
\displaystyle \text{Banker's Gain}=\frac{A(it)^2}{1+it}
\displaystyle \text{Here, } A=5050,\ i=0.05,\ \text{Banker's Gain}=0.50
\displaystyle 0.50=\frac{5050(0.05t)^2}{1+0.05t}
\displaystyle 0.50(1+0.05t)=5050(0.0025t^2)
\displaystyle 0.50+0.025t=12.625t^2
\displaystyle 12.625t^2-0.025t-0.50=0
\displaystyle \Rightarrow 505t^2-t-20=0
\displaystyle t=\frac{1\pm\sqrt{1+40400}}{1010}
\displaystyle t=\frac{1\pm 201}{1010}
\displaystyle t=\frac{202}{1010}=\frac{1}{5} \text{ year}
\displaystyle \frac{1}{5}\text{ year}=\frac{365}{5}=73 \text{ days}
\displaystyle \text{Therefore, the legal due date of maturity is } 73 \text{ days after 4th July 2013.}
\displaystyle \text{Legal due date }=15\text{th September 2013}
\displaystyle \text{Nominal due date }=15\text{th September 2013}-3\text{ days}
\displaystyle =12\text{th September 2013}
\displaystyle \therefore \text{The nominal date of maturity is } 12\text{th September 2013.}
\displaystyle \\

\displaystyle \textbf{Question 15:}
\displaystyle \text{(a) The price of six different commodities for years 2009 and 2011} \\ \text{are as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Commodities} & A & B & C & D & E & F \\ \hline \text{Price in 2009 (Rs.)} & 35 & 80 & 25 & 30 & 80 & x \\ \hline \text{Price in 2011 (Rs.)} & 50 & y & 45 & 70 & 120 & 105 \\ \hline \end{array}
\displaystyle \text{The index number for 2011 taking 2009 as base year was } 125. \text{ Find } x \text{ and } y,
\displaystyle \text{if the total price in 2009 is Rs. } 360. \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) The given data can be written as follows:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Commodities} & \text{Price in 2009 (Rs.)} & \text{Price in 2011 (Rs.)} \\ \hline A & 35 & 50 \\ \hline B & 80 & y \\ \hline C & 25 & 45 \\ \hline D & 30 & 70 \\ \hline E & 80 & 120 \\ \hline F & x & 105 \\ \hline \text{Total} & 250+x & 390+y \\ \hline \end{array}
\displaystyle \text{Given, } \Sigma P_0=360
\displaystyle \Rightarrow 250+x=360
\displaystyle \Rightarrow x=110
\displaystyle \text{Price Index}=\frac{\Sigma P_1}{\Sigma P_0}\times 100
\displaystyle 125=\frac{390+y}{360}\times 100
\displaystyle \Rightarrow 125\times 360=100(390+y)
\displaystyle \Rightarrow 45000=39000+100y
\displaystyle \Rightarrow 100y=6000
\displaystyle \Rightarrow y=60
\displaystyle \therefore x=110,\ y=60
\displaystyle \\
\displaystyle \text{(b) The number of road accidents in the city due to rash driving over a period of } \\ 3 \text{ years is given below:}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Year} & \text{Jan-Mar} & \text{April-June} & \text{July-Sept} & \text{Oct-Dec} \\ \hline 2010 & 70 & 60 & 45 & 72 \\ \hline 2011 & 79 & 56 & 46 & 84 \\ \hline 2012 & 90 & 64 & 45 & 82 \\ \hline \end{array}
\displaystyle \text{Calculate four quarterly moving averages and illustrate them on original figures} \\ \text{on one graph using the same axes.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Calculation of trend by four-quarterly moving averages:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Year} & \text{Quarter} & \text{Values} & \text{4-Quarter Moving Total} & \text{4-Quarter Moving Average} & \text{Centred Moving Average} \\ \hline 2010 & 1 & 70 & - & - & - \\ \hline 2010 & 2 & 60 & - & - & - \\ \hline 2010 & 3 & 45 & 247 & 61.75 & 62.875 \\ \hline 2010 & 4 & 72 & 256 & 64.00 & 63.500 \\ \hline 2011 & 1 & 79 & 252 & 63.00 & 63.125 \\ \hline 2011 & 2 & 56 & 253 & 63.25 & 64.750 \\ \hline 2011 & 3 & 46 & 265 & 66.25 & 67.625 \\ \hline 2011 & 4 & 84 & 276 & 69.00 & 70.000 \\ \hline 2012 & 1 & 90 & 284 & 71.00 & 70.875 \\ \hline 2012 & 2 & 64 & 283 & 70.75 & 70.500 \\ \hline 2012 & 3 & 45 & 281 & 70.25 & - \\ \hline 2012 & 4 & 82 & - & - & - \\ \hline \end{array}
\displaystyle \text{The four-quarter moving totals are calculated as follows:}
\displaystyle 70+60+45+72=247
\displaystyle 60+45+72+79=256
\displaystyle 45+72+79+56=252
\displaystyle 72+79+56+46=253
\displaystyle 79+56+46+84=265
\displaystyle 56+46+84+90=276
\displaystyle 46+84+90+64=284
\displaystyle 84+90+64+45=283
\displaystyle 90+64+45+82=281
\displaystyle \text{Centred moving averages are obtained by averaging two successive four-quarter moving averages.}
\displaystyle \text{These centred moving averages should be plotted along with the original values on the same graph.}

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