MATHEMATICS

\displaystyle \textbf{(Maximum Marks: 100)}

\displaystyle \textbf{(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}
\displaystyle \text{EITHER from Section B OR Section C.}

\displaystyle \text{Section A: Internal choice has been provided in three questions of four marks each}
\displaystyle \text{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,}
\displaystyle \text{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 \times  3]

\displaystyle \text{(i) If } A=\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}, \text{ find } x \text{ such that } A^2=xA-2I. \text{ Hence find } A^{-1}.
\displaystyle \text{Answer:} 
\displaystyle A^2=\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}=\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} 
\displaystyle xA-2I=\begin{bmatrix} 3x-2 & -2x \\ 4x & -2x-2 \end{bmatrix} 
\displaystyle \therefore 3x-2=1\Rightarrow x=1 
\displaystyle \therefore A^2=A-2I 
\displaystyle A^2=A-2I\Rightarrow A^2-A=-2I 
\displaystyle A(A-I)=-2I 
\displaystyle \therefore A^{-1}=-\frac{1}{2}(A-I) 
\displaystyle =-\frac{1}{2}\left(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}-\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) 
\displaystyle =-\frac{1}{2}\begin{bmatrix} 2 & -2 \\ 4 & -3 \end{bmatrix}=\begin{bmatrix} -1 & 1 \\ -2 & \frac{3}{2} \end{bmatrix} 
\\

\displaystyle \text{(ii) Find the value of } k, \text{ if the equation } 8x^2-16xy+ky^2-22x+34y=12 \text{ represents an ellipse.}
\displaystyle \text{Answer:} 
\displaystyle \text{For }ax^2+2hxy+by^2+2gx+2fy+c=0\text{ to represent an ellipse, }h^2<ab 
\displaystyle \text{Here }a=8,\quad 2h=-16\Rightarrow h=-8,\quad b=k 
\displaystyle h^2<ab\Rightarrow (-8)^2<8k 
\displaystyle 64<8k\Rightarrow k>8 
\displaystyle \therefore \text{the equation represents an ellipse if }k>8 
\\

\displaystyle \text{(iii) Solve for } x:\ \sin\left(2\tan^{-1}x\right)=1.
\displaystyle \text{Answer:} 
\displaystyle \text{Let }\theta=\tan^{-1}x\Rightarrow \tan\theta=x 
\displaystyle \sin 2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2x}{1+x^2} 
\displaystyle \frac{2x}{1+x^2}=1\Rightarrow x^2-2x+1=0 
\displaystyle (x-1)^2=0\Rightarrow x=1 
\\

\displaystyle \text{(iv) Two regression lines are represented by } 2x+3y-10=0 \text{ and } 4x+y-5=0.
\displaystyle \text{Answer:} 
\displaystyle \text{The two regression lines intersect at }(\overline{x},\overline{y}). 
\displaystyle 2x+3y-10=0,\quad 4x+y-5=0 
\displaystyle \text{Solving, }x=\frac{1}{2},\quad y=3 
\displaystyle \therefore \overline{x}=\frac{1}{2},\quad \overline{y}=3 
\displaystyle 2x+3y-10=0\Rightarrow y=-\frac{2}{3}x+\frac{10}{3}\Rightarrow b_{yx}=-\frac{2}{3} 
\displaystyle 4x+y-5=0\Rightarrow x=-\frac{1}{4}y+\frac{5}{4}\Rightarrow b_{xy}=-\frac{1}{4} 
\displaystyle r=\pm\sqrt{b_{xy}b_{yx}}=\pm\sqrt{\frac{1}{6}} 
\displaystyle \therefore r=-\frac{1}{\sqrt{6}}\text{, since both regression coefficients are negative.} 
\\

\displaystyle \text{(v) Evaluate: } \int \frac{\mathrm{cosec}\ x}{\log \tan \left(\frac{x}{2}\right)}\ dx.
\displaystyle \text{Answer:} 
\displaystyle \text{Let }t=\log \tan\left(\frac{x}{2}\right) 
\displaystyle \therefore \frac{dt}{dx}=\mathrm{cosec}\ x 
\displaystyle \int \frac{\mathrm{cosec}\ x}{\log \tan \left(\frac{x}{2}\right)}\ dx=\int \frac{dt}{t} 
\displaystyle =\log t+C 
\displaystyle =\log\left\{\log \tan\left(\frac{x}{2}\right)\right\}+C 
\\

\displaystyle \text{(vi) Evaluate: } \lim\limits_{y\to 0}\frac{y-\tan^{-1}y}{y-\sin y}.
\displaystyle \text{Answer:} 
\displaystyle L=\lim\limits_{y\to 0}\frac{y-\tan^{-1}y}{y-\sin y} 
\displaystyle \text{Using L'Hospital's rule, }L=\lim\limits_{y\to 0}\frac{1-\frac{1}{1+y^2}}{1-\cos y} 
\displaystyle =\lim\limits_{y\to 0}\frac{\frac{y^2}{1+y^2}}{1-\cos y} 
\displaystyle =\lim\limits_{y\to 0}\frac{y^2}{1-\cos y}\cdot\frac{1}{1+y^2}=2 
\\

\displaystyle \text{(vii) Evaluate: } \int\limits_{0}^{1}\frac{xe^x}{(1+x)^2}\ dx.
\displaystyle \text{Answer:} 
\displaystyle \frac{d}{dx}\left(\frac{e^x}{1+x}\right)=\frac{e^x(1+x)-e^x}{(1+x)^2}=\frac{xe^x}{(1+x)^2} 
\displaystyle \therefore \int\limits_{0}^{1}\frac{xe^x}{(1+x)^2}\ dx=\left[\frac{e^x}{1+x}\right]_{0}^{1} 
\displaystyle =\frac{e}{2}-1 
\\

\displaystyle \text{(viii) Find the modulus and argument of the complex number } \frac{2+i}{4i+(1+i)^2}.
\displaystyle \text{Answer:} 
\displaystyle z=\frac{2+i}{4i+(1+i)^2}=\frac{2+i}{4i+2i}=\frac{2+i}{6i} 
\displaystyle z=\frac{(2+i)(-i)}{6}=\frac{1-2i}{6} 
\displaystyle \therefore |z|=\sqrt{\left(\frac{1}{6}\right)^2+\left(-\frac{2}{6}\right)^2}=\frac{\sqrt{5}}{6} 
\displaystyle \arg z=-\tan^{-1}2 
\\

\displaystyle \text{(ix) A word consists of } 9 \text{ different alphabets, in which } 4 \text{ are consonants and } 5 \text{ are vowels.}
\displaystyle \text{Three alphabets are chosen at random. What is the probability that more than one vowel will be selected?}
\displaystyle \text{Answer:} 
\displaystyle \text{Required probability}=P(2\text{ vowels})+P(3\text{ vowels}) 
\displaystyle =\frac{{}^{5}C_2\cdot{}^{4}C_1}{{}^{9}C_3}+\frac{{}^{5}C_3}{{}^{9}C_3} 
\displaystyle =\frac{10\cdot4}{84}+\frac{10}{84}=\frac{50}{84}=\frac{25}{42} 
\\

\displaystyle \text{(x) Solve the differential equation: } \frac{dy}{dx}=e^{x+y}+x^2e^y.
\displaystyle \text{Answer:} 
\displaystyle \frac{dy}{dx}=e^y(e^x+x^2) 
\displaystyle e^{-y}\frac{dy}{dx}=e^x+x^2 
\displaystyle -\frac{d}{dx}(e^{-y})=e^x+x^2 
\displaystyle e^{-y}=-e^x-\frac{x^3}{3}+C 
\displaystyle \therefore e^{-y}+e^x+\frac{x^3}{3}=C 
\\

\displaystyle \textbf{Question 2:}

\displaystyle \text{(a) Using the properties of determinants, show that } p\alpha^2+2q\alpha+r=0, \text{ given that } p,q \text{ and } r
\displaystyle \text{are not in G.P. and } \left| \begin{array}{ccc} 1 & \frac{p}{q} & \alpha+\frac{q}{p} \\ 1 & \frac{q}{r} & \alpha+\frac{r}{q} \\ p\alpha+q & q\alpha+r & 0 \end{array} \right|=0. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \Delta=\left| \begin{array}{ccc} 1 & \frac{p}{q} & \alpha+\frac{q}{p} \\ 1 & \frac{q}{r} & \alpha+\frac{r}{q} \\ p\alpha+q & q\alpha+r & 0 \end{array} \right|=0 
\displaystyle \text{Expanding along }C_3,\text{ we get} 
\displaystyle \left(\alpha+\frac{q}{p}\right)\left|\begin{array}{cc}1&\frac{q}{r}\\p\alpha+q&q\alpha+r\end{array}\right|-\left(\alpha+\frac{r}{q}\right)\left|\begin{array}{cc}1&\frac{p}{q}\\p\alpha+q&q\alpha+r\end{array}\right|=0 
\displaystyle \left(\alpha+\frac{q}{p}\right)\left(q\alpha+r-\frac{q}{r}(p\alpha+q)\right)-\left(\alpha+\frac{r}{q}\right)\left(q\alpha+r-\frac{p}{q}(p\alpha+q)\right)=0 
\displaystyle \frac{(q^2-pr)(p\alpha^2+2q\alpha+r)}{pqr}=0 
\displaystyle \text{Since }p,q,r\text{ are not in G.P., }q^2-pr\neq0 
\displaystyle \therefore p\alpha^2+2q\alpha+r=0 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(b) Solve the following system of equations using the matrix method:} \hspace{0.2cm} [5]
\displaystyle \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4
\displaystyle \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1
\displaystyle \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2
\displaystyle \text{Answer:} 
\displaystyle \text{Let }u=\frac{1}{x},\quad v=\frac{1}{y},\quad w=\frac{1}{z} 
\displaystyle \therefore 2u+3v+10w=4,\quad 4u-6v+5w=1,\quad 6u+9v-20w=2 
\displaystyle \begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}\begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}4\\1\\2\end{bmatrix} 
\displaystyle \text{Solving by matrix method, }\begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\\\frac{1}{3}\\\frac{1}{5}\end{bmatrix} 
\displaystyle \therefore \frac{1}{x}=\frac{1}{2},\quad \frac{1}{y}=\frac{1}{3},\quad \frac{1}{z}=\frac{1}{5} 
\displaystyle \therefore x=2,\quad y=3,\quad z=5 
\\

\displaystyle \textbf{Question 3:}

\displaystyle \text{(a) Prove that } 2\tan^{-1}\frac{1}{5}+\cos^{-1}\frac{7}{5\sqrt{2}}+2\tan^{-1}\frac{1}{8}=\frac{\pi}{4}. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let }A=2\tan^{-1}\frac{1}{5},\quad B=\cos^{-1}\frac{7}{5\sqrt{2}},\quad C=2\tan^{-1}\frac{1}{8} 
\displaystyle \tan A=\frac{2\cdot\frac{1}{5}}{1-\frac{1}{25}}=\frac{5}{12},\quad \tan C=\frac{2\cdot\frac{1}{8}}{1-\frac{1}{64}}=\frac{16}{63} 
\displaystyle \cos B=\frac{7}{5\sqrt{2}}\Rightarrow \sin B=\sqrt{1-\frac{49}{50}}=\frac{1}{5\sqrt{2}} 
\displaystyle \therefore \tan B=\frac{1}{7} 
\displaystyle \tan(A+B)=\frac{\frac{5}{12}+\frac{1}{7}}{1-\frac{5}{84}}=\frac{47}{79} 
\displaystyle \tan(A+B+C)=\frac{\frac{47}{79}+\frac{16}{63}}{1-\frac{47}{79}\cdot\frac{16}{63}}=1 
\displaystyle \therefore A+B+C=\frac{\pi}{4} 
\displaystyle \therefore 2\tan^{-1}\frac{1}{5}+\cos^{-1}\frac{7}{5\sqrt{2}}+2\tan^{-1}\frac{1}{8}=\frac{\pi}{4} 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(b) } P,Q \text{ and } R \text{ represent switches in 'ON' position and } P',Q' \text{ and } R' \text{ represent switches}
\displaystyle \text{in 'OFF' position. Construct a switching circuit representing the polynomial } P(P+Q)Q(Q+R').

\displaystyle \text{Using Boolean algebra, show that the above circuit is equivalent to a switching} 
\displaystyle \text{circuit in which when } P \text{ and } Q \text{are in 'ON' position, the light is on.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{The circuit for }P(P+Q)Q(Q+R')\text{ has }P,\ (P+Q),\ Q,\ (Q+R')\text{ in series.} 
\displaystyle \text{Here }(P+Q)\text{ means }P\text{ and }Q\text{ are connected in parallel.} 
\displaystyle \text{Also }(Q+R')\text{ means }Q\text{ and }R'\text{ are connected in parallel.} 
\displaystyle P(P+Q)Q(Q+R')=\{P(P+Q)\}\{Q(Q+R')\} 
\displaystyle =P\cdot Q \quad \text{by absorption law } A(A+B)=A 
\displaystyle \therefore P(P+Q)Q(Q+R')=PQ 
\displaystyle \text{Thus, the simplified circuit has only }P\text{ and }Q\text{ connected in series.} 
\displaystyle \text{Hence, the light is on only when }P\text{ and }Q\text{ are both in 'ON' position.} 
\\

\displaystyle \textbf{Question 4:}

\displaystyle \text{(a) Verify Lagrange's mean value theorem for the function } f(x)=\sin x-\sin 2x
\displaystyle \text{in the interval } [0,\pi]. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle f(x)=\sin x-\sin 2x\text{ is continuous in }[0,\pi]\text{ and differentiable in }(0,\pi). 
\displaystyle f(0)=0,\quad f(\pi)=0 
\displaystyle \therefore \frac{f(\pi)-f(0)}{\pi-0}=0 
\displaystyle f'(x)=\cos x-2\cos 2x 
\displaystyle \text{By LMVT, }f'(c)=0\text{ for some }c\in(0,\pi). 
\displaystyle \cos c-2\cos 2c=0 
\displaystyle \cos c-2(2\cos^2c-1)=0 
\displaystyle 4\cos^2c-\cos c-2=0 
\displaystyle \therefore \cos c=\frac{1\pm\sqrt{33}}{8} 
\displaystyle \text{Both values lie between }-1\text{ and }1,\text{ hence such }c\in(0,\pi)\text{ exists.} 
\displaystyle \text{Hence, Lagrange's mean value theorem is verified.} 
\\

\displaystyle \text{(b) Find the equation of the hyperbola whose foci are } (0,\pm 13) \text{ and the length}
\displaystyle \text{of the conjugate axis is } 20. \hspace{0.2cm} [5] 
\displaystyle \text{Answer:} 
\displaystyle \text{Foci are }(0,\pm13),\text{ so transverse axis is along }y\text{-axis and }c=13. 
\displaystyle \text{Equation is } \frac{y^2}{a^2}-\frac{x^2}{b^2}=1 
\displaystyle 2b=20\Rightarrow b=10,\quad b^2=100 
\displaystyle c^2=a^2+b^2\Rightarrow 13^2=a^2+100 
\displaystyle a^2=169-100=69 
\displaystyle \therefore \text{Required equation is } \frac{y^2}{69}-\frac{x^2}{100}=1 
\\

\displaystyle \textbf{Question 5:}

\displaystyle \text{(a) Evaluate: } \int \frac{x^2-5x-1}{x^4+x^2+1}\ dx. \hspace{0.2cm} [5] 
\displaystyle \text{Answer:} 
\displaystyle x^4+x^2+1=(x^2+x+1)(x^2-x+1) 
\displaystyle \frac{x^2-5x-1}{x^4+x^2+1}=\frac{x-3}{x^2-x+1}-\frac{x-2}{x^2+x+1} 
\displaystyle I=\int\left(\frac{x-3}{x^2-x+1}-\frac{x-2}{x^2+x+1}\right)dx 
\displaystyle I=\frac{1}{2}\log(x^2-x+1)-\frac{5}{2}\int\frac{dx}{x^2-x+1} 
\displaystyle \quad-\frac{1}{2}\log(x^2+x+1)+\frac{5}{2}\int\frac{dx}{x^2+x+1} 
\displaystyle I=\frac{1}{2}\log\left(\frac{x^2-x+1}{x^2+x+1}\right) 
\displaystyle \quad+\frac{5}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)-\frac{5}{\sqrt{3}}\tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C 
\\

\displaystyle \text{(b) Draw a rough sketch of the curves } y=(x-1)^2 \text{ and } y=|x-1|.
\displaystyle \text{Hence, find the area of the region bounded by these curves.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{The curve }y=(x-1)^2\text{ is a parabola and }y=|x-1|\text{ is a V-shaped curve.} 
\displaystyle \text{For points of intersection, }(x-1)^2=|x-1| 
\displaystyle \text{Let }t=|x-1|,\text{ then }t^2=t\Rightarrow t=0,1 
\displaystyle \therefore x=1,\ 0,\ 2 
\displaystyle \text{Required area}=\int\limits_0^2\{|x-1|-(x-1)^2\}\ dx 
\displaystyle =2\int\limits_0^1(t-t^2)\ dt 
\displaystyle =2\left[\frac{t^2}{2}-\frac{t^3}{3}\right]_0^1=2\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{1}{3} 
\displaystyle \therefore \text{Area bounded by the curves}=\frac{1}{3}\text{ square units} 
\\

\displaystyle \textbf{Question 6:}

\displaystyle \text{(a) If the sum of the length of the hypotenuse and a side of a right angled triangle is given,}
\displaystyle \text{show that the area of the triangle is maximum when the angle between them is } \frac{\pi}{3}. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let hypotenuse }=h,\text{ one side }=x,\text{ and }h+x=k\text{, a constant.} 
\displaystyle \text{Let }\theta\text{ be the angle between }h\text{ and }x,\text{ then }x=h\cos\theta. 
\displaystyle h+h\cos\theta=k\Rightarrow h=\frac{k}{1+\cos\theta} 
\displaystyle \text{Area }A=\frac{1}{2}(h\cos\theta)(h\sin\theta)=\frac{h^2}{2}\sin\theta\cos\theta 
\displaystyle A=\frac{k^2}{2}\cdot\frac{\sin\theta\cos\theta}{(1+\cos\theta)^2} 
\displaystyle \text{Let }u=\cos\theta,\text{ then }A^2\propto\frac{u^2(1-u^2)}{(1+u)^4}=\frac{u^2(1-u)}{(1+u)^3} 
\displaystyle \text{For maximum, differentiate }\log\left\{\frac{u^2(1-u)}{(1+u)^3}\right\}. 
\displaystyle \frac{2}{u}-\frac{1}{1-u}-\frac{3}{1+u}=0 
\displaystyle 2(1-u)(1+u)-u(1+u)-3u(1-u)=0 
\displaystyle 2-2u^2-u-u^2-3u+3u^2=0 
\displaystyle 2-4u=0\Rightarrow u=\frac{1}{2} 
\displaystyle \therefore \cos\theta=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{3} 
\displaystyle \therefore \text{area is maximum when the angle between them is }\frac{\pi}{3}. 
\\

\displaystyle \text{(b) If } y=x^x, \text{ prove that } \frac{d^2y}{dx^2}-\frac{1}{y}\left(\frac{dy}{dx}\right)^2-\frac{y}{x}=0. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle y=x^x\Rightarrow \log y=x\log x 
\displaystyle \frac{1}{y}\frac{dy}{dx}=\log x+1 
\displaystyle \therefore \frac{dy}{dx}=y(\log x+1) 
\displaystyle \frac{d^2y}{dx^2}=\frac{dy}{dx}(\log x+1)+\frac{y}{x} 
\displaystyle =y(\log x+1)^2+\frac{y}{x} 
\displaystyle \frac{1}{y}\left(\frac{dy}{dx}\right)^2=\frac{1}{y}\{y^2(\log x+1)^2\}=y(\log x+1)^2 
\displaystyle \therefore \frac{d^2y}{dx^2}-\frac{1}{y}\left(\frac{dy}{dx}\right)^2-\frac{y}{x}=0 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \textbf{Question 7:}

\displaystyle \text{(a) The following observations are given:}
\displaystyle (1,4),(2,8),(3,2),(4,12),(5,10),(6,14),(7,16),(8,6),(9,18).
\displaystyle \text{Estimate the value of } y \text{ when the value of } x \text{ is } 10 \text{ and also estimate the value}
\displaystyle \text{of } x \text{ when the value of } y=5. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle n=9,\quad \Sigma x=45,\quad \Sigma y=90,\quad \Sigma xy=530,\quad \Sigma x^2=285,\quad \Sigma y^2=1140 
\displaystyle \overline{x}=5,\quad \overline{y}=10 
\displaystyle S_{xy}=530-\frac{45\times90}{9}=80,\quad S_{xx}=285-\frac{45^2}{9}=60 
\displaystyle S_{yy}=1140-\frac{90^2}{9}=240 
\displaystyle b_{yx}=\frac{S_{xy}}{S_{xx}}=\frac{80}{60}=\frac{4}{3} 
\displaystyle \text{Regression equation of }y\text{ on }x\text{ is }y-10=\frac{4}{3}(x-5) 
\displaystyle \text{When }x=10,\quad y-10=\frac{4}{3}(5)=\frac{20}{3} 
\displaystyle \therefore y=\frac{50}{3}=16.67\text{ approximately} 
\displaystyle b_{xy}=\frac{S_{xy}}{S_{yy}}=\frac{80}{240}=\frac{1}{3} 
\displaystyle \text{Regression equation of }x\text{ on }y\text{ is }x-5=\frac{1}{3}(y-10) 
\displaystyle \text{When }y=5,\quad x-5=\frac{1}{3}(-5)=-\frac{5}{3} 
\displaystyle \therefore x=\frac{10}{3}=3.33\text{ approximately} 
\\

\displaystyle \text{(b) Compute Karl Pearson's coefficient of correlation between sales and expenditure} \\ \text{of a firm for six months.} \hspace{0.2cm} [5]
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Sales (in lakhs of Rs)} & 18 & 20 & 27 & 20 & 21 & 29 \\ \hline \text{Expenditures (in lakhs of Rs)} & 23 & 27 & 28 & 28 & 29 & 30 \\ \hline \end{array}
\displaystyle \text{Answer:} 
\displaystyle n=6,\quad \Sigma x=135,\quad \Sigma y=165,\quad \Sigma xy=3749 
\displaystyle \Sigma x^2=3135,\quad \Sigma y^2=4567 
\displaystyle S_{xy}=\Sigma xy-\frac{\Sigma x\Sigma y}{n}=3749-\frac{135\times165}{6}=36.5 
\displaystyle S_{xx}=\Sigma x^2-\frac{(\Sigma x)^2}{n}=3135-\frac{135^2}{6}=97.5 
\displaystyle S_{yy}=\Sigma y^2-\frac{(\Sigma y)^2}{n}=4567-\frac{165^2}{6}=29.5 
\displaystyle r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{36.5}{\sqrt{97.5\times29.5}} 
\displaystyle r=0.681\text{ approximately} 
\\

\displaystyle \textbf{Question 8:}

\displaystyle \text{(a) A purse contains } 4 \text{ silver and } 5 \text{ copper coins. A second purse contains } 3 \text{ silver}
\displaystyle \text{and } 7 \text{ copper coins. If a coin is taken out at random from one of the purses, what is the}
\displaystyle \text{probability that it is a copper coin?} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle P(\text{choosing first purse})=\frac{1}{2},\quad P(\text{choosing second purse})=\frac{1}{2} 
\displaystyle P(C|P_1)=\frac{5}{9},\quad P(C|P_2)=\frac{7}{10} 
\displaystyle P(C)=P(P_1)P(C|P_1)+P(P_2)P(C|P_2) 
\displaystyle =\frac{1}{2}\cdot\frac{5}{9}+\frac{1}{2}\cdot\frac{7}{10}=\frac{5}{18}+\frac{7}{20}=\frac{113}{180} 
\\

\displaystyle \text{(b) Aman and Bhuvan throw a pair of dice alternately. In order to win, they have to get a sum}
\displaystyle \text{of } 8. \text{ Find their respective probabilities of winning if Aman starts the game.}\hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle P(\text{sum }8)=\frac{5}{36},\quad P(\text{not sum }8)=\frac{31}{36} 
\displaystyle P(\text{Aman wins})=\frac{5}{36}+\left(\frac{31}{36}\right)^2\frac{5}{36}+\left(\frac{31}{36}\right)^4\frac{5}{36}+\cdots 
\displaystyle =\frac{5}{36}\left[1+\left(\frac{31}{36}\right)^2+\left(\frac{31}{36}\right)^4+\cdots\right] 
\displaystyle =\frac{5}{36}\cdot\frac{1}{1-\left(\frac{31}{36}\right)^2}=\frac{36}{67} 
\displaystyle P(\text{Bhuvan wins})=1-\frac{36}{67}=\frac{31}{67} 
\\

\displaystyle \textbf{Question 9:}

\displaystyle \text{(a) Using De Moivre's theorem, find the value of: } (1+i\sqrt{3})^6+(1-i\sqrt{3})^6. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle 1+i\sqrt{3}=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right) 
\displaystyle 1-i\sqrt{3}=2\left(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}\right) 
\displaystyle (1+i\sqrt{3})^6=2^6\left(\cos2\pi+i\sin2\pi\right)=64 
\displaystyle (1-i\sqrt{3})^6=2^6\left(\cos2\pi-i\sin2\pi\right)=64 
\displaystyle \therefore (1+i\sqrt{3})^6+(1-i\sqrt{3})^6=64+64=128 
\\

\displaystyle \text{(b) Solve the following differential equation for a particular solution:}
\displaystyle y-x\frac{dy}{dx}=x+y\frac{dy}{dx}, \text{ when } y=0 \text{ and } x=1. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle y-x\frac{dy}{dx}=x+y\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{y-x}{x+y} 
\displaystyle \text{Put }y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} 
\displaystyle v+x\frac{dv}{dx}=\frac{v-1}{1+v} 
\displaystyle x\frac{dv}{dx}=-\frac{1+v^2}{1+v} 
\displaystyle \frac{1+v}{1+v^2}\ dv=-\frac{dx}{x} 
\displaystyle \tan^{-1}v+\frac{1}{2}\log(1+v^2)=-\log x+C 
\displaystyle \tan^{-1}\frac{y}{x}+\frac{1}{2}\log(x^2+y^2)=C 
\displaystyle \text{Given }y=0\text{ when }x=1,\quad C=0 
\displaystyle \therefore \tan^{-1}\frac{y}{x}+\frac{1}{2}\log(x^2+y^2)=0 
\\

Section – B (20 Marks)

\displaystyle \textbf{Question 10:}

\displaystyle \text{(a) Prove that: } [\overrightarrow{a}+\overrightarrow{b}\quad \overrightarrow{b}+\overrightarrow{c}\quad \overrightarrow{c}+\overrightarrow{a}]=2[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle [\overrightarrow{a}+\overrightarrow{b}\quad \overrightarrow{b}+\overrightarrow{c}\quad \overrightarrow{c}+\overrightarrow{a}] 
\displaystyle =[\overrightarrow{a}\quad \overrightarrow{b}+\overrightarrow{c}\quad \overrightarrow{c}+\overrightarrow{a}]+[\overrightarrow{b}\quad \overrightarrow{b}+\overrightarrow{c}\quad \overrightarrow{c}+\overrightarrow{a}] 
\displaystyle =[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]+[\overrightarrow{a}\ \overrightarrow{c}\ \overrightarrow{a}]+[\overrightarrow{b}\ \overrightarrow{b}\ \overrightarrow{c}]+[\overrightarrow{b}\ \overrightarrow{c}\ \overrightarrow{a}] 
\displaystyle =[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]+0+0+[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] 
\displaystyle =2[\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(b) If } D,E \text{ and } F \text{ are mid-points of a triangle } ABC, \text{ prove by vector method that:}
\displaystyle \text{Area of triangle } DEF=\frac{1}{4}(\text{Area of triangle } ABC). \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let }D,E,F\text{ be the mid-points of }BC,CA,AB\text{ respectively.} 
\displaystyle \therefore \overrightarrow{DE}=\frac{1}{2}\overrightarrow{BA},\quad \overrightarrow{DF}=\frac{1}{2}\overrightarrow{CA} 
\displaystyle \text{Area of }\triangle DEF=\frac{1}{2}\left|\overrightarrow{DE}\times\overrightarrow{DF}\right| 
\displaystyle =\frac{1}{2}\left|\frac{1}{2}\overrightarrow{BA}\times\frac{1}{2}\overrightarrow{CA}\right| 
\displaystyle =\frac{1}{4}\cdot\frac{1}{2}\left|\overrightarrow{BA}\times\overrightarrow{CA}\right| 
\displaystyle =\frac{1}{4}(\text{Area of }\triangle ABC) 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \textbf{Question 11:}

\displaystyle \text{(a) Find the vector equation of the line passing through the point } (-1,2,1) \text{ and parallel to the line}
\displaystyle \overrightarrow{r}=2\widehat{i}+3\widehat{j}-\widehat{k}+\lambda(\widehat{i}-2\widehat{j}+\widehat{k}). \text{ Also, find the distance between the lines.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Required line passes through }-\widehat{i}+2\widehat{j}+\widehat{k}\text{ and has direction }\widehat{i}-2\widehat{j}+\widehat{k}. 
\displaystyle \therefore \overrightarrow{r}=-\widehat{i}+2\widehat{j}+\widehat{k}+\mu(\widehat{i}-2\widehat{j}+\widehat{k}) 
\displaystyle \text{For the given line, }\overrightarrow{a_1}=2\widehat{i}+3\widehat{j}-\widehat{k},\quad \overrightarrow{b}=\widehat{i}-2\widehat{j}+\widehat{k} 
\displaystyle \text{For the required line, }\overrightarrow{a_2}=-\widehat{i}+2\widehat{j}+\widehat{k} 
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=-3\widehat{i}-\widehat{j}+2\widehat{k} 
\displaystyle (\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}=3\widehat{i}+5\widehat{j}+7\widehat{k} 
\displaystyle \text{Distance}=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\times\overrightarrow{b}|}{|\overrightarrow{b}|}=\frac{\sqrt{3^2+5^2+7^2}}{\sqrt{1^2+(-2)^2+1^2}} 
\displaystyle \therefore \text{Distance}=\frac{\sqrt{83}}{\sqrt{6}}\text{ units} 
\\

\displaystyle \text{(b) Find the equation of the plane passing through the points } A(2,1,-3), \\ B(-3,-2,1) \text{ and } C(2,4,-1). \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \overrightarrow{AB}=-5\widehat{i}-3\widehat{j}+4\widehat{k},\quad \overrightarrow{AC}=3\widehat{j}+2\widehat{k} 
\displaystyle \overrightarrow{AB}\times\overrightarrow{AC}=-18\widehat{i}+10\widehat{j}-15\widehat{k} 
\displaystyle \text{Equation of plane is }-18(x-2)+10(y-1)-15(z+3)=0 
\displaystyle \therefore 18x-10y+15z+19=0 
\\

\displaystyle \textbf{Question 12:}

\displaystyle \text{(a) A box contains } 4 \text{ red and } 5 \text{ black marbles. Find the probability} \\ \text{distribution of red marbles}   \text{in a random draw of three marbles. Also, find the} \\ \text{mean and standard deviation of the distribution.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let }X\text{ denote the number of red marbles drawn. Then }X=0,1,2,3 
\displaystyle P(X=x)=\frac{{}^{4}C_x\cdot{}^{5}C_{3-x}}{{}^{9}C_3} 
\displaystyle \begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{10}{84} & \frac{40}{84} & \frac{30}{84} & \frac{4}{84} \\ \hline \end{array} 
\displaystyle \begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{5}{42} & \frac{10}{21} & \frac{5}{14} & \frac{1}{21} \\ \hline \end{array} 
\displaystyle \text{Mean }=\frac{3\times4}{9}=\frac{4}{3} 
\displaystyle \text{Variance}=3\cdot\frac{4}{9}\cdot\frac{5}{9}\cdot\frac{6}{8}=\frac{5}{9} 
\displaystyle \therefore \text{Standard deviation}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3} 
\\

\displaystyle \text{(b) Bag } A \text{ contains } 2 \text{ white, } 1 \text{ black and } 3 \text{ red balls. Bag } B \text{ contains } 3 \text{ white, } 2 \text{ black}
\displaystyle \text{and } 4 \text{ red balls and Bag } C \text{ contains } 4 \text{ white, } 3 \text{ black and } 2 \text{ red balls. One Bag is chosen}
\displaystyle \text{at random and two balls are drawn at random from that Bag. If the randomly drawn balls happen to be red}
\displaystyle \text{and black, what is the probability that both balls come from Bag } B? \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle P(A)=P(B)=P(C)=\frac{1}{3} 
\displaystyle P(RB|A)=\frac{{}^{3}C_1\cdot{}^{1}C_1}{{}^{6}C_2}=\frac{3}{15}=\frac{1}{5} 
\displaystyle P(RB|B)=\frac{{}^{4}C_1\cdot{}^{2}C_1}{{}^{9}C_2}=\frac{8}{36}=\frac{2}{9} 
\displaystyle P(RB|C)=\frac{{}^{2}C_1\cdot{}^{3}C_1}{{}^{9}C_2}=\frac{6}{36}=\frac{1}{6} 
\displaystyle P(B|RB)=\frac{P(B)P(RB|B)}{P(A)P(RB|A)+P(B)P(RB|B)+P(C)P(RB|C)} 
\displaystyle =\frac{\frac{1}{3}\cdot\frac{2}{9}}{\frac{1}{3}\cdot\frac{1}{5}+\frac{1}{3}\cdot\frac{2}{9}+\frac{1}{3}\cdot\frac{1}{6}} 
\displaystyle =\frac{\frac{2}{9}}{\frac{1}{5}+\frac{2}{9}+\frac{1}{6}}=\frac{\frac{2}{9}}{\frac{53}{90}}=\frac{20}{53} 
\displaystyle \therefore \text{Required probability}=\frac{20}{53} 
\\

Section – C (20 Marks)

\displaystyle \textbf{Question 13:}

\displaystyle \text{(a) The price of a tape recorder is Rs. } 1661. \text{ A person purchases it by making a cash payment}
\displaystyle \text{of Rs. } 400 \text{ and agrees to pay the balance with due interest in } 3 \text{ half-yearly equal installments.}
\displaystyle \text{If the dealer charged interest at the rate of } 10\% \text{ per annum compounded half-yearly, find the value}
\displaystyle \text{of the installments.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Balance amount}=1661-400=\text{Rs. }1261 
\displaystyle \text{Rate of interest per half-year}=\frac{10}{2}\%=5\% 
\displaystyle \text{Let each half-yearly installment be Rs. }x. 
\displaystyle 1261=\frac{x}{1.05}+\frac{x}{(1.05)^2}+\frac{x}{(1.05)^3} 
\displaystyle 1261=x\left(\frac{1}{1.05}+\frac{1}{1.1025}+\frac{1}{1.157625}\right) 
\displaystyle 1261=x(2.723248) 
\displaystyle x=\frac{1261}{2.723248}=463.04 
\displaystyle \therefore \text{Value of each installment}=\text{Rs. }463.04\text{ approximately} 
\\

\displaystyle \text{(b) A manufacturer manufactures two types of tea cups, A and B. Three machines are} \\ \text{needed for manufacturing }   \text{the tea cups. The time in minutes required for manufacturing } \\ \text{each cup on the machines is given below:} \hspace{0.2cm} [5]

\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Type of Cup} & \text{Machine I} & \text{Machine II} & \text{Machine III} \\ \hline A & 12 & 18 & 6 \\ \hline B & 6 & 0 & 9 \\ \hline \end{array}
\displaystyle \text{Answer:} 
\displaystyle \text{The question is incomplete. The available machine time and profit per cup are required} 
\displaystyle \text{to formulate and solve the linear programming problem.} 
\\

\displaystyle \textbf{Question 14:}

\displaystyle \text{(a) If the difference between the Banker's discount and True discount of a bill for } 73 \text{ days}
\displaystyle \text{at } 5\% \text{ per annum is Rs. } 10, \text{ find (i) the amount of the bill (ii) the Banker's discount.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle t=\frac{73}{365}=\frac{1}{5}\text{ year},\quad r=5\% 
\displaystyle \text{Difference between B.D. and T.D.}=\frac{T.D.\times r\times t}{100} 
\displaystyle 10=\frac{T.D.\times5\times\frac{1}{5}}{100}=\frac{T.D.}{100} 
\displaystyle \therefore T.D.=1000 
\displaystyle \text{Amount of bill}=\frac{T.D.\times100}{r\times t}+T.D. 
\displaystyle =\frac{1000\times100}{5\times\frac{1}{5}}+1000=100000+1000=101000 
\displaystyle \text{Banker's discount}=T.D.+10=1000+10=1010 
\displaystyle \therefore \text{Amount of bill}=\text{Rs. }101000,\quad \text{Banker's discount}=\text{Rs. }1010 
\\

\displaystyle \text{(b) Given that the total cost function for } x \text{ units of a commodity is:}
\displaystyle C(x)=\frac{x^3}{3}+3x^2-7x+16
\displaystyle \text{(i) Find the Marginal Cost (MC)}
\displaystyle \text{(ii) Find the Average Cost (AC)}
\displaystyle \text{(iii) Prove that: Marginal Average Cost (MAC)}=\frac{x(MC)-C(x)}{x^2}. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle C(x)=\frac{x^3}{3}+3x^2-7x+16 
\displaystyle \text{(i) }MC=\frac{dC}{dx}=x^2+6x-7 
\displaystyle \text{(ii) }AC=\frac{C(x)}{x}=\frac{x^2}{3}+3x-7+\frac{16}{x} 
\displaystyle \text{(iii) }MAC=\frac{d}{dx}(AC)=\frac{d}{dx}\left(\frac{C(x)}{x}\right) 
\displaystyle MAC=\frac{xC'(x)-C(x)}{x^2} 
\displaystyle \text{Since }C'(x)=MC,\quad MAC=\frac{x(MC)-C(x)}{x^2} 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \textbf{Question 15:}

\displaystyle \text{(a) The price quotations of four different commodities for 2001 and 2009 are} \\ \text{given as below.}   \text{Calculate the index for 2009 with 2001 as the base year by using} \\ \text{weighted average of price relative method.} \hspace{0.2cm} [5]
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Commodity} & \text{Weight} & \text{Price in 2009 (Rs.)} & \text{Price in 2001 (Rs.)} \\ \hline A & 10 & 9.00 & 4.00 \\ \hline B & 49 & 4.40 & 5.00 \\ \hline C & 36 & 9.00 & 6.00 \\ \hline D & 4 & 3.60 & 2.00 \\ \hline \end{array}

\displaystyle \text{Answer:} 
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Commodity} & w & p_1 & p_0 & R=\frac{p_1}{p_0}\times100 \\ \hline A & 10 & 9.00 & 4.00 & 225 \\ \hline B & 49 & 4.40 & 5.00 & 88 \\ \hline C & 36 & 9.00 & 6.00 & 150 \\ \hline D & 4 & 3.60 & 2.00 & 180 \\ \hline \end{array} 
\displaystyle \Sigma w=99,\quad \Sigma wR=10(225)+49(88)+36(150)+4(180)=12682 
\displaystyle \text{Index number}=\frac{\Sigma wR}{\Sigma w}=\frac{12682}{99}=128.10 
\displaystyle \therefore \text{Price index for }2009=128.10 
\\

\displaystyle \text{(b) The profit of a soft drink firm (in thousand of Rupees) during each month} \\ \text{of the year is as given below:}   \text{Calculate the four monthly moving averages and plot these} \\ \text{and the original data on a graph sheet.}

\displaystyle \begin{array}{|c|c|} \hline \text{Months} & \text{Profits (in thousands of Rupees)} \\ \hline \text{January} & 3.6 \\ \hline \text{February} & 4.3 \\ \hline \text{March} & 4.3 \\ \hline \text{April} & 3.4 \\ \hline \text{May} & 4.4 \\ \hline \text{June} & 5.4 \\ \hline \text{July} & 3.4 \\ \hline \text{August} & 2.4 \\ \hline \text{September} & 3.4 \\ \hline \text{October} & 1.8 \\ \hline \text{November} & 0.8 \\ \hline \text{December} & 1.2 \\ \hline \end{array}
\displaystyle \text{Calculate four months moving averages and plot these and the original data} \\ \text{on a graph sheet.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \begin{array}{|c|c|c|} \hline \text{Months} & \text{4-month total} & \text{4-month moving average} \\ \hline \text{Jan-Apr} & 15.6 & 3.900 \\ \hline \text{Feb-May} & 16.4 & 4.100 \\ \hline \text{Mar-Jun} & 17.5 & 4.375 \\ \hline \text{Apr-Jul} & 16.6 & 4.150 \\ \hline \text{May-Aug} & 15.6 & 3.900 \\ \hline \text{Jun-Sep} & 14.6 & 3.650 \\ \hline \text{Jul-Oct} & 11.0 & 2.750 \\ \hline \text{Aug-Nov} & 8.4 & 2.100 \\ \hline \text{Sep-Dec} & 7.2 & 1.800 \\ \hline \end{array} 
\displaystyle \text{Since the period is }4,\text{ the moving averages are centred.} 
\displaystyle \begin{array}{|c|c|} \hline \text{Month} & \text{Centred 4-month moving average} \\ \hline \text{March} & 4.0000 \\ \hline \text{April} & 4.2375 \\ \hline \text{May} & 4.2625 \\ \hline \text{June} & 4.0250 \\ \hline \text{July} & 3.7750 \\ \hline \text{August} & 3.2000 \\ \hline \text{September} & 2.4250 \\ \hline \text{October} & 1.9500 \\ \hline \end{array} 
\displaystyle \text{Plot the original data and the centred moving averages on the graph sheet.} 
\\

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