2015 ICSE (Class 12) Board Paper Solution: Mathematics

MATHEMATICS

(Maximum Marks: 100)

(Time Allowed: Three Hours)

(Candidates are allowed additional 15 minutes for only reading the paper.

They must NOT start writing during this time)

The Question Paper consists of three sections A, B and C.

Candidates are required to attempt all questions from Section A and all question EITHER from Section B OR Section C

Section A: Internal choice has been provided in three questions of four marks each and two questions of six marks each.

Section B: Internal choice has been provided in two question of four marks each.

Section C: Internal choice has been provided in two question of four marks each.

All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.

The intended marks for questions or parts of questions are given in brackets [ ].

Mathematical tables and graphs papers are provided.

Section – A (80 Marks)

$\displaystyle \text{Question 1: } \hspace{10.0cm} [10 \times 3]$

$\displaystyle \text{(i) Find the value of } k \text{ if } M = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \text{ and } M^2-kM-I_2=0$

$\displaystyle \text{(ii) Find the equation of an ellipse whose latus rectum is 8 and eccentricity is } \frac{1}{3}$

$\displaystyle \text{(iii) Solve: } \cos^{-1} (\sin \cos^{-1} x) = \frac{\pi}{6}$

$\displaystyle \text{(iv) Using L'Hospital's rule, evaluate: } \lim \limits_{x \to 0} \frac{x- \sin x}{x^2 \sin x}$

$\displaystyle \text{(v) Evaluate: } \int \limits_{}^{}\frac{2y^2}{y^2+4} \ dy$

$\displaystyle \text{(vi) Evaluate: } \int \limits_{0}^{3} f(x) \ dx , \text{ where } f(x) = \Bigg \{ \begin{matrix} \cos \ 2x, 0 \leq x \leq \frac{\pi}{2} \\ \\ 3, \ \ \ \ \ \ \frac{\pi}{2} \leq x \leq 3 \end{matrix}$

$\displaystyle \text{(vii) The two lines of regressions are } 4x+2y-3=0 \text{ and } \displaystyle 3x+6y+5 = 0 .$

$\displaystyle \text{Find the correlation co-efficient between } x \text{ and } y .$

(viii) A card is drawn from a well shuffled pack of playing cards. What is the probability that it is either a spade or an ace or both?

$\displaystyle \text{(ix) If } 1, \omega \ and \ \omega^2 \text{ are the cube roots of unity, prove that } \frac{a + b\omega + c\omega^2}{c+a\omega + b\omega^2} = \omega^2$

$\displaystyle \text{(x) Solve the differential equation: } \sin^{-1} (\frac{dy}{dx}) = x + y$

Answer:

$\displaystyle \text{(i) } M^2-kM-I_2=0$

$\displaystyle \Rightarrow \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} - k \times \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} - \begin{bmatrix} k & 2k \\ 2k & 3k \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 5-k & 8-2k \\ 8-2k & 13-3k \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle \Rightarrow 5 - k = 1 \Rightarrow k = 4$

Similarly, $\displaystyle 13-3k = 1 \Rightarrow k = 4$

$\displaystyle \text{(ii) } \frac{2b^2}{a} =8$

$\displaystyle e = \frac{1}{3}$

$\displaystyle b^2 = 4a$

$\displaystyle b^2 = a^{2(1-e^2)}$

$\displaystyle \Rightarrow 4a = a^{2(1-\frac{1}{9})}$

$\displaystyle \Rightarrow a = \frac{9}{2}$

$\displaystyle b^2 = 18$

$\displaystyle \text{Equation of ellipse: } \frac{4x^2}{81} + \frac{y^2}{18} = 1 \Rightarrow 8x^2 + 9y^2 = 162$

$\displaystyle \text{(iii) } \cos^{-1} (\sin \cos^{-1} x) = \frac{\pi}{6}$

$\displaystyle \sin (\cos^{-1} x) = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$

$\displaystyle \sqrt{1-x^2} = \frac{\sqrt{3}}{2}$

$\displaystyle 1-x^2 = \frac{3}{4}$

$\displaystyle \Rightarrow x^2 = \frac{1}{4}$

$\displaystyle x = \pm \frac{1}{2}$

$\displaystyle \text{(iv) } \lim \limits_{x \to 0} \frac{x- \sin x}{x^2 \sin x}$

Apply L’Hospital’s rule

$\displaystyle =\lim _{x\to \:0} \left(\frac{1-\cos x}{2x\sin x+\cos x \ x^2}\right)$

$\displaystyle =\lim _{x\to \:0} \left(\frac{1-\cos x}{x \left(x\cos x+2\sin x \right)}\right)$

Apply L’Hospital’s rule

$\displaystyle =\lim _{x\to \:0} \left(\frac{\sin x}{-x^2\sin x+4x\cos x+2\sin x}\right)$

$\displaystyle =\lim _{x\to \:0} \left(\frac{-\sin x}{\sin x \ x^2-4\cos x \ x-2\sin x}\right)$

Apply L’Hospital’s rule

$\displaystyle =\lim _{x\to \:0} \left(\frac{-\cos x}{x^2\cos x+6x\sin \left(x\right)-6\cos x}\right)$

$\displaystyle \mathrm{Plug\:in\:the\:value}\:x=0$

$\displaystyle = \frac{-\cos 0}{0^2\cos 0+6\cdot \:0\cdot \sin 0-6\cos 0}$

$\displaystyle = \frac{1}{6}$

$\displaystyle \text{(v) } \int \limits_{}^{}\frac{2y^2}{y^2+4} \ dy$

$\displaystyle = 2 \int \limits_{}^{} \Big(1 - \frac{4}{y^2+4} \Big)$

$\displaystyle = 2y - 8 . \frac{1}{2} \tan^{-1} \frac{y}{2} + c$

$\displaystyle = 2y - 4 \tan^{-1} \frac{y}{2} + c$

$\displaystyle \text{(vi) } \int \limits_{0}^{3} f(x) \ dx , \text{ where } f(x) = \Bigg \{ \begin{matrix} cos \ 2x, 0 \leq x \leq \frac{\pi}{2} \\ \\ 3, \ \ \ \ \ \ \frac{\pi}{2} \leq x \leq 3 \end{matrix}$

Therefore

$\displaystyle = \int \limits_{0}^{\frac{\pi}{2}} \cos 2x dx + \int \limits_{\frac{\pi}{2}}^{3} 3 dx$

$\displaystyle = \Big[ \frac{\sin 2x}{2} \Big]_{0}^{\frac{\pi}{2}} + \Big[ 3x \Big]_{\frac{\pi}{2}}^{3}$

$\displaystyle = (0-0) + 3(3 - \frac{\pi}{2} )$

$\displaystyle = 9 - \frac{3\pi}{2}$

(vii) Let the line of regression of $\displaystyle x$ on $\displaystyle y$ be

$\displaystyle 4x+2y -3 = 0 \Rightarrow x = \frac{-1}{2} y + \frac{3}{4}$

$\displaystyle \Rightarrow b_{xy} = \frac{-1}{2}$

Let the line of regression of $\displaystyle y$ on $\displaystyle x$ be

$\displaystyle 3x+6y+5 = 0 \Rightarrow y = \frac{-1}{2} x - \frac{5}{6}$

$\displaystyle \Rightarrow b_{yx} = \frac{-1}{2}$

$\displaystyle \text{Therefore } r^2 = b_{yx} \times b_{xy} = \frac{-1}{2} \times \frac{-1}{2} = \frac{1}{4}$

$\displaystyle \text{Hence } r = - \frac{1}{2} \text{ since both } b_{xy} \text{ and } b_{yx} \text{ are negative. }$

(viii) Probability P(E) = Probability of drawing a spade + Probability of drawing an ace – Probability of drawing ace of spade

$\displaystyle = \frac{13}{52} + \frac{4}{52} - \frac{1}{52}$

$\displaystyle = \frac{16}{52} = \frac{4}{13}$

$\displaystyle \text{(ix) } \frac{a + b\omega + c\omega^2}{c+a\omega + b\omega^2}$

Multiplying numerator and denominator by $\displaystyle \omega^2$

$\displaystyle = \frac{\omega^2(a + b\omega + c\omega^2)}{\omega^2(c+a\omega + b\omega^2)}$

$\displaystyle = \frac{\omega^2(a + b\omega + c\omega^2)}{(c\omega^2+a\omega^3 + b\omega^4)}$

$\displaystyle \text{We know } \omega^3 = 1, \omega^4 = \omega^3.\omega = \omega$

$\displaystyle = \frac{\omega^2(a + b\omega + c\omega^2)}{(c\omega^2+a + b\omega)}$

$\displaystyle = \omega^2$

$\displaystyle \text{(x) } \sin^{-1} (\frac{dy}{dx}) = x + y$

$\displaystyle \frac{dy}{dx} = \sin (x+y)$

$\displaystyle \text{Let } x + y = v$

$\displaystyle \frac{d(v-x)}{dx} = \sin v$

$\displaystyle \frac{dv}{dx} -1 = \sin v$

$\displaystyle \frac{dv}{dx} = 1+ \sin v$

$\displaystyle \int \limits_{}^{} \frac{dv}{1+ \sin x} = \int \limits_{}^{} dx$

$\displaystyle \int \limits_{}^{} \frac{(1- \sin x) dv}{1- \sin^2 x} = \int \limits_{}^{} dx$

$\displaystyle \int \limits_{}^{} \frac{(1- \sin x) dv}{\cos^2 x} = \int \limits_{}^{} dx$

$\displaystyle \int \limits_{}^{} \sec^2 v - \tan v \sec v \ dx = \int \limits_{}^{} dx$

$\displaystyle \tan v - \sec v = x + c$

$\displaystyle \Rightarrow \tan (x+y) - \sec (x+y) = x + c$

$\displaystyle \\$

Question 2:

(a) Using properties of determinants, prove that:

$\displaystyle \left| \begin{array}{ccc} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array} \right|= (1+a^2+b^2)^3 \hspace{5.0cm} [5]$

(b) Given two matrices $\displaystyle A \text{ and } B$

$\displaystyle A = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix} \text{ and } B = \begin{bmatrix} 11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix} . \text{ Find } AB.$

Using this result, solve the following system of equation:

$\displaystyle x-2y+3z=6, \ x+4y+z=12, \ x-3y+2z=1 \hspace{7.0cm} [5]$

Answer:

$\displaystyle \text{(a) LHS = } \left| \begin{array}{ccc} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array} \right|$

$\displaystyle C_1 \rightarrow C_1 - b C_3$

$\displaystyle = \left| \begin{array}{ccc} 1+a^2+b^2 & 2ab & -2b \\ 0 & 1-a^2+b^2 & 2a \\ b(1+a^2+b^2) & -2a & 1-a^2-b^2 \end{array} \right|$

$\displaystyle = (1+a^2+b^2) \left| \begin{array}{ccc} 1 & 2ab & -2b \\ 0 & 1-a^2+b^2 & 2a \\ b & -2a & 1-a^2-b^2 \end{array} \right|$

$\displaystyle C_2 \rightarrow C_2+ a C_3$

$\displaystyle = (1+a^2+b^2) \left| \begin{array}{ccc} 1 & 0 & -2b \\ 0 & 1+a^2+b^2 & 2a \\ b & -a(1+a^2+b^2) & 1-a^2-b^2 \end{array} \right|$

$\displaystyle = (1+a^2+b^2)^2 \left| \begin{array}{ccc} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1-a^2-b^2 \end{array} \right|$

$\displaystyle R_3 \rightarrow R_3 - bR_1$

$\displaystyle = (1+a^2+b^2)^2 \left| \begin{array}{ccc} 1 & 0 & -2b \\ 0 & 1 & 2a \\ 0 & -1 & 1-a^2+b^2 \end{array} \right|$

$\displaystyle = (1+a^2+b^2)^2 (1 - a^2 + b^2 + 2a^2)$

$\displaystyle = (1+a^2+b^2)^3$

$\displaystyle =$ RHS. Hence proved.

$\displaystyle \text{(b) } A = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix}$

$\displaystyle B = \begin{bmatrix} 11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix}$

$\displaystyle AB = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix} \times \begin{bmatrix} 11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix} = \begin{bmatrix} -8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -8 \end{bmatrix} = -8 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle AB = -8I$

Now,

$\displaystyle A = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix} X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} C = \begin{bmatrix} 6 \\ 12 \\ 1 \end{bmatrix}$

$\displaystyle X = A^{-1} C$

$\displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix} = - \frac{1}{8} \begin{bmatrix} 11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix} \times \begin{bmatrix} 6 \\ 12 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Hence $\displaystyle x = 1, y = 2 \ and \ z = 3$

$\displaystyle \\$

Question 3:

(a) Solve the equation for $\displaystyle x$ :

$\displaystyle \sin^{-1} \frac{5}{x} + \sin^{-1} \frac{12}{x} = \frac{\pi}{2} , x \neq 0. \hspace{13.0cm} [5]$

(b) $\displaystyle A, B \text{ and } C$ represent switches in ‘on’ position and $\displaystyle A', B' \text{ and } C'$ represent them in ‘off’ position. Construct a switching circuit representing the polynomial $\displaystyle ABC + AB'C + A'B'C$ . Using Boolean Algebra, prove that the given polynomial can be simplified to $\displaystyle C(A + B')$ . Construct an equivalent switching circuit. [5]

Answer:

$\displaystyle \text{(a) } \sin^{-1} \frac{5}{x} + \sin^{-1} \frac{12}{x} = \frac{\pi}{2}$

$\displaystyle \sin^{-1} \frac{5}{x} = \frac{\pi}{2} - \sin^{-1} \frac{12}{x}$

$\displaystyle \frac{5}{x} = \sin \Big( \frac{\pi}{2} - \sin^{-1} \frac{12}{x} \Big)$

$\displaystyle \frac{5}{x} =\cos \Big( \sin^{-1} \frac{12}{x} \Big)$

$\displaystyle \frac{5}{x} = \cos \Bigg( \cos^{-1} \sqrt{1 - (\frac{12}{x})^2} \Bigg)$

$\displaystyle \frac{5}{x} = \sqrt{1 - ( \frac{12}{x})^2}$

$\displaystyle \Rightarrow \frac{25}{x^2} = 1 - ( \frac{12}{x} )^2$

$\displaystyle \Rightarrow \frac{25}{x^2} = 1 - \frac{144}{x^2}$

$\displaystyle \Rightarrow \frac{169}{x^2} = 1$

$\displaystyle \Rightarrow x = \pm 13$

$\displaystyle \text{(b) } ABC+AB'C+A'B'C$

$\displaystyle = ACB + ACB' + A'B'C$

$\displaystyle = AC (B+B') + A'B'C \text{ Since } (B+B') = 1$

$\displaystyle = AC + A'B'C$

$\displaystyle = (A+A')(A+B')C$

$\displaystyle = (A+B')C$

$\displaystyle \\$

Question 4:

(a) Verify Lagrange’s Mean Value Theorem for the following function:

$\displaystyle f(x) = 2 \sin x + \sin 2x \ on \ [0, \pi] \hspace{6.0cm} [5]$

(b) Find the equation of the hyperbola whose foci are $\displaystyle (0, \pm \sqrt{10})$ and passing through the point $\displaystyle (2, 3). \hspace{10.0cm} [5]$

Answer:

$\displaystyle \text{(a) } f(x) = (2 \sin x + \sin 2x) \text{ is continuous in } [0, \pi]$

$\displaystyle f(0) = 0, f(\pi) = 0$

$\displaystyle f'(x) \text{ exists in } (0, \pi)$

$\displaystyle f'(x) = 2 \cos x + 2 \cos 2x$

All the conditions of Lagrange’s Mean Value theorem satisfied there exists $\displaystyle 'c'$ in $\displaystyle (0, \pi)$ such that $\displaystyle f'(c) = \frac{f(b) - f(a)}{b-a}$

$\displaystyle 2 \cos c + 2 \cos 2c = 0$

$\displaystyle 2 \cos^2 c + \cos c - 1 = 0$

$\displaystyle \cos c = -1 \Rightarrow \cos c = \cos \pi \Rightarrow c = \pi$ (not possible)

$\displaystyle \text{or } \cos c = \cos \frac{\pi}{3} \Rightarrow c = \frac{\pi}{3} \in (0, \frac{\pi}{3} )$ which lies between $\displaystyle 0 \text{ and } \pi .$

Hence Lagrange’s Mean Value theorem is verified.

$\displaystyle \text{(b) Foci } = (0, \pm \sqrt{10})$

$\displaystyle be = \sqrt{10}$

$\displaystyle a^2 = b^2 (e^2 - 1) = b^2 e^2 - b^2$

$\displaystyle a^2 = 10 - b^2$

$\displaystyle \text{Let the equation be } : \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$

$\displaystyle \Rightarrow \frac{9}{10-a^2} - \frac{4}{a^2} = 1$

$\displaystyle \Rightarrow 9a^2 - 40 + 4a^2 = 10a^2 - a^4$

$\displaystyle \Rightarrow a^4 + 3a^2 - 40 = 0$

$\displaystyle \Rightarrow (a^2+8) (a^2 - 5) = 0$

$\displaystyle \Rightarrow a^2 = -8 ($ not possible as $\displaystyle a^2$ cannot be negative $\displaystyle )$ or $\displaystyle a^2 = 5$

$\displaystyle \text{Therefore } a^2 = 5$ and $\displaystyle b^2 = 10-5 = 5$

$\displaystyle \text{Hence the equation is } \frac{y^2}{5} - \frac{x^2}{5} = 1 \Rightarrow y^2 -x^2 = 5$

$\displaystyle \\$

Question 5:

$\displaystyle \text{(a) If } y = e^{m \cos^{-1} x} , \text{ prove that } (1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = m^2y \hspace{5.0cm} [5]$

(b) Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius $\displaystyle 10 \ \text{ cm }$ is a square of side $\displaystyle 10\sqrt{2} \hspace{13.0cm} [5]$

Answer:

(a) $\displaystyle y = e^{m \cos^{-1} x}$

Differentiating with respect to $\displaystyle x$

$\displaystyle \frac{dy}{dx} = e^{m \cos^{-1} x} \times \frac{-m}{\sqrt{1-x^2}}$

$\displaystyle \sqrt{1-x^2} \frac{dy}{dx} = -me^{m \cos^{-1} x}$

$\displaystyle \sqrt{1-x^2} \frac{dy}{dx} = -my$

Differentiating again with respect to $\displaystyle x$

$\displaystyle \Rightarrow \sqrt{1-x^2} \frac{d^2y}{dx^2} - \frac{2x}{2\sqrt{1-x^2}} \frac{dy}{dx} = -m \frac{dy}{dx}$

$\displaystyle (1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = - m \sqrt{1-x^2} \frac{dy}{dx} = -m (-my) = m^2y$

(b) Please refer to the adjoining figure

$\displaystyle AB = 2x$ and $\displaystyle BC = 2y$

$\displaystyle (2x)^2+(2y)^2 = (20)^2 \Rightarrow x^2 + y^2 = 100$

$\displaystyle P = 4x + 4y = 4x + 4\sqrt{100-x^2}$

Differentiating with respect to $\displaystyle x$

$\displaystyle \frac{dP}{dx} = 4 - \frac{4x}{\sqrt{100-x^2}} = 0$

$\displaystyle \Rightarrow 100 - x^2 = x^2$

$\displaystyle \Rightarrow x^2 = 50$

$\displaystyle \Rightarrow x = 5\sqrt{2}$

Differentiating with respect to $\displaystyle x$

$\displaystyle \frac{d^2P}{dx^2} = -4 \Big[ \frac{\sqrt{100-x^2} - \frac{x(-x)}{ \sqrt{100-x^2}} }{100-x^2} \Big] = \frac{-4 \times 100}{(100-x^2)^{\frac{3}{2}}} < 0$

Hence the perimeter is maximum when $\displaystyle x = 5\sqrt{2}$

Therefore $\displaystyle y = 5\sqrt{2} \Rightarrow x = y$

Therefore $\displaystyle ABCD$ is a square with side of $\displaystyle 10\sqrt{2}$

$\displaystyle \\$

Question 6:

$\displaystyle \text{(a) Evaluate: } \int \frac{ \sec x}{1+ cosec \ x} \ dx \hspace{6.0cm} [5]$

$\displaystyle \text{(b) Find the smaller area enclosed by the circle } x^2+y^2 \text{ and the line } \\ \\ x+y = 2 \hspace{5.0cm} [5]$

Answer:

$\displaystyle \text{(a) } I = \int \frac{ \sec x}{1+ \mathrm{cosec} x} \ dx$

$\displaystyle I = \int \frac{ \sin x}{\cos x (1+ \sin x)} \ dx$

$\displaystyle I = \int \frac{ \sin x \cos x}{\cos^2 x (1+ \sin x)} \ dx$

$\displaystyle \text{Now put } \sin x = t \Rightarrow dt = \cos x dx$

$\displaystyle I = \int \frac{ t }{(1-t^2)(1+t)} dt$

$\displaystyle I = \int \frac{ t }{(1-t)(1+t)^2} dt$

$\displaystyle \text{Now } \frac{ t }{(1-t^2)(1+t)} = \frac{A}{1-t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$

$\displaystyle \Rightarrow t = A(1+t)^2 + B(1-t)(1+t) + C(1-t)$

$\displaystyle \Rightarrow t = A(1+t^2 + 2t) + B(1-t^2) + (C(1-t)$

$\displaystyle \Rightarrow t = t^2(A-B) + t(2A-C) + (A+B+C)$

$\displaystyle \Rightarrow A-B = 0, 2A-C = 1 \ and \ A+B+C = 0$

$\displaystyle \Rightarrow A = \frac{1}{4} , B = \frac{1}{4} , C = - \frac{1}{2}$

$\displaystyle \text{Thus} I = \int \frac{A}{1-t} + \frac{B}{1+t} + \frac{C}{(1+t)^2} dt$

$\displaystyle I = \frac{1}{4}\int \Big[ \frac{1}{1-t} + \frac{1}{1+t} - \frac{2}{1+t^2} \Big] \ dt$

$\displaystyle = \frac{1}{4} | - \log |1-t| + \log |1+t| + \frac{2}{1+t} | + c$

$\displaystyle = \frac{1}{4} | \log | \frac{1+t}{1-t} | + \frac{2}{1+t} | + c$

$\displaystyle \text{Therefore } = \frac{1}{4} | \log | \frac{1+\sin x}{1-\sin x} | + \frac{2}{1+\sin x} | + c$

(b) We have to calculate

$\displaystyle \int \limits_{0}^{2} \sqrt{4-x^2} \ dx - \int \limits_{0}^{2} 2-x \ dx$

$\displaystyle \text{First Integral: } \int \limits_{0}^{2} \sqrt{4-x^2} \ dx$

$\displaystyle \text{Let } x = 2 \sin u \Rightarrow dx = 2 \cos u \ du$

$\displaystyle = \int \limits_{0}^{2} \sqrt{4- 4\sin^2 u} \ 2 \cos u \ du = \int \limits_{0}^{2} 4 \sqrt{1- \sin^2 u} \ \cos u \ du = 4 \int \limits_{0}^{2} \cos^2 u \ du$

$\displaystyle \text{Now } \cos^2 u = \frac{1+\cos 2u}{2}$ , Therefore

$\displaystyle = 4 \int \limits_{0}^{2} \frac{1+\cos 2u}{2} \ du = 2 \int \limits_{0}^{2} 1 + \cos 2u \ du = \Big[ 2(u + \frac{1}{2} \sin 2u) \Big]_{0}^{2}$

$\displaystyle \text{Now substituting back} x = 2 \sin u \Rightarrow u = \sin^{-1} (\frac{x}{2})$

$\displaystyle = \Big[ 2(\sin^{-1} (\frac{x}{2}) + \frac{1}{2} \sin (2 \sin^{-1} (\frac{x}{2})) \Big]_{0}^{2}$

$\displaystyle = \Big[ 2(\sin^{-1} (1) + \frac{1}{2} \sin (2 \sin^{-1} (1)) \Big] - \Big[ 2(\sin^{-1} (0) + \frac{1}{2} \sin (2 \sin^{-1} (0)) \Big] = \pi$

$\displaystyle \text{Second Integral:} \int \limits_{0}^{2} 2-x \ dx = \Big[ 2x - \frac{1}{2} x^2 \Big]_{0}^{2} = 4 - 2 = 2$

$\displaystyle \text{Hence } \int \limits_{0}^{2} \sqrt{4-x^2} \ dx - \int \limits_{0}^{2} 2-x \ dx = \pi - 2$

$\displaystyle \\$

Question 7:

(a) Given that the observations are: $(9, -4), (10, -3), (11, -1), (12, 0), (13, 1), (14, 3), (15, 5), (16, 8)$ . Find the two lines of regression and estimate the value of $y$ when $x = 13.5$ . [5]

(b) In a contest the competitors are awarded marks out of 20 by two judges. The scores of the 10 competitors are given below. Calculate Spearman’s rank correlation. [5]

Competitors	$A$	$B$	$C$	$D$	$E$	$F$	$G$	$H$	$I$	$J$
Judge A	$2$	$11$	$11$	$18$	$6$	$5$	$8$	$16$	$13$	$15$
Judge B	$6$	$11$	$16$	$9$	$14$	$20$	$4$	$3$	$13$	$17$

Answer:

(a)

$x$	$y$	$xy$	$x^2$	$y^2$
$9$	$-4$	$-36$	$81$	$16$
$10$	$-3$	$-30$	$100$	$9$
$11$	$-1$	$-11$	$121$	$1$
$12$	$0$	$0$	$144$	$0$
$13$	$1$	$13$	$169$	$1$
$14$	$3$	$42$	$196$	$9$
$15$	$5$	$75$	$225$	$25$
$16$	$8$	$128$	$256$	$64$
$\Sigma x = 100$	$\Sigma y = 9$	$\Sigma xy = 181$	$\Sigma x^2 = 1292$	$\Sigma y^2 = 125$

$\displaystyle \overline{x} = \frac{100}{8} = 12.5$

$\displaystyle \overline{y} = \frac{9}{8} = 1.125$

$\displaystyle b_{yx} = \frac{\Sigma xy - \frac{1}{n} \Sigma x . \Sigma y }{\Sigma x^2 - \frac{1}{n} (\Sigma x)^2 } = \frac{181 - \frac{1}{8} 100 . 9 }{1292 - \frac{1}{8} (100)^2 } = \frac{68.5}{42} = 1.63$

$\displaystyle b_{xy} = \frac{\Sigma xy - \frac{1}{n} \Sigma x . \Sigma y }{\Sigma y^2 - \frac{1}{n} (\Sigma y)^2 } = \frac{181 - \frac{1}{8} 100 . 9 }{125 - \frac{1}{8} (9)^2 } = \frac{68.5}{114.875} = 0.596$

$\text{Line of regression of } y \text{ on } x$

$\displaystyle y - \frac{9}{8} = 1.63 (x - 12.5) \Rightarrow y = 1.63x - 19.25$

$\text{Line of regression of } x \text{ on } y$

$\displaystyle x - 12.5 = 0.596 \Big(y - \frac{9}{8} \Big) \Rightarrow x = 0.596y + 11.83$

$\displaystyle \text{When } x = 13.5, y = 1.63 \times 13.5 - 19.25 = 2.76$

(b)

Judge A	Judge B	Rank $(R_x)$	Rank $(R_y)$	$d = R_x-R_y$	$d^2$
$2$	$6$	$10$	$8$	$2$	$4$
$11$	$11$	$5.5$	$6$	$-0.5$	$0.25$
$11$	$16$	$5.5$	$3$	$2.5$	$6.25$
$18$	$9$	$1$	$7$	$-6$	$36$
$6$	$14$	$8$	$4$	$4$	$16$
$5$	$20$	$9$	$1$	$8$	$64$
$8$	$4$	$7$	$9$	$-2$	$4$
$16$	$3$	$2$	$10$	$-8$	$64$
$13$	$13$	$4$	$5$	$-1$	$1$
$15$	$17$	$3$	$2$	$1$	$1$
					$\Sigma d^2 = 196.5$

$\displaystyle r = 1 - 6 \Big[ \frac{ \Sigma d^2 + \frac{1}{12} \Sigma (m^3 - m)}{n(n^2-1)} \Big]$

$\displaystyle r = 1 - 6 \Big[ \frac{ 196.5 + \frac{1}{12} \Sigma (2^3 - 2)}{10(99)} \Big] = 1 - \frac{6 \times 197}{990} = -0.194$

$\\$

Question 8:

(a) An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into the urn. Otherwise, it is replaced with another ball of the same color. The process is repeated. Find the probability that the third ball drawn is black. [5]

(b) Three person $A, B \ and \ C$ shoot to hit a target. If $A$ hits the target 4 times out of 5 trials, $B$ hits it 3 times in 4 trials and $C$ hits is 2 times in 3 trials, find the probability that:

(i) Exactly two person hit the target.

(ii) At least two person hit the target.

(iii) Non hit the target. [5]

Answer:

(a) Please note: The number of balls in the urn change based on what color is drawn.

Every time you get a Black ball, not only the ball is put back in the urn, another black ball is put in the urn. So the number of balls increase every time you pull a black ball.

In case you draw a white ball, it is not put back in the urn. So the number of balls decrease when you pull white ball.

Therefore the probability that the third ball is black is:

$\displaystyle P(E) = P (WWB) + P(WBB) + P(BWB) + P(BBB)$

$\displaystyle = \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} + \frac{2}{4} \times \frac{2}{3} \times \frac{3}{4} + \frac{2}{4} \times \frac{2}{5} \times \frac{3}{4} + \frac{2}{4} \times \frac{3}{5} \times \frac{4}{6}$

$\displaystyle = \frac{1}{6} + \frac{1}{4} + \frac{3}{20} + \frac{1}{5}$

$\displaystyle = \frac{23}{30}$

$\displaystyle \text{(b) Given: } P(A) = \frac{4}{5} , P(B) = \frac{3}{4} \text{ and } P(C) = \frac{2}{3}$

$\displaystyle \text{Therefore } P(\overline{A}) = \frac{1}{5} , P(\overline{B}) = \frac{1}{4} \text{ and } P(\overline{C}) = \frac{1}{3}$

(i) Probability of at least two hitting the target

$\displaystyle = P(AB\overline{C})+P(A\overline{B}C)+P(\overline{A}BC)$

$\displaystyle = \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} + \frac{4}{5} \times \frac{1}{4} \times \frac{2}{3} + \frac{1}{5} \times \frac{3}{4} \times \frac{2}{3}$ $\displaystyle = \frac{26}{60} = \frac{13}{30}$

(ii) At least two people hit the target

$\displaystyle = P(AB\overline{C})+P(A\overline{B}C)+P(\overline{A}BC) + P(ABC)$

(iii) None hit the target

$\displaystyle = P(\overline{A}.\overline{B}.\overline{C}) = \frac{1}{5} . \frac{1}{4} . \frac{1}{3} = \frac{1}{60}$

$\\$

Question 9:

$\displaystyle \text{(a) If } z = x + iy, \omega = \frac{2-iz}{2z-i} \text{(a) If } |\omega|=1 , \text{ find the locus of } z \\ \\ \text{ and illustrate it in Argand Plane. } \hspace{6.0cm} [5]$

(b) Solve the differential equation:

$\displaystyle e^{x/y}(1 - \frac{x}{y} ) + (1 + e^{x/y}) \frac{dx}{dy} = 0 \ when \ x=0, y=1 \hspace{5.0cm} [5]$

Answer:

$\displaystyle \text{(a) } |\omega| = 1$

$\displaystyle | \frac{2 - iz}{2z-i} | = 1$

$\displaystyle \Rightarrow |2 - iz| = |2z-i|$

$\displaystyle \text{Given } z = x + iy$

$\displaystyle |2 - i(x+iy)| = |2(x+iy)-i|$

$\displaystyle |2-ix+y| = |2x+2iy-i|$

$\displaystyle |(2+y)-ix| = |2x+i(2y-i)|$

$\displaystyle \sqrt{ (-x)^2 + (2+y)^2 } = \sqrt{(2x)^2+(2y-1)^2}$

$\displaystyle 4 + y^2 + 4y + x^2 = 4x^2 + 4y^2 + 1 - 4y$

$\displaystyle \Rightarrow 3x^2 + 3y^2 -8y -3 = 0$

$\displaystyle \Rightarrow x^2 + y^2 - \frac{8}{3} y -1 = 0$ … … … … … (i)

Now we know that the equation of a circle with center $\displaystyle (h,k)$ and radius $\displaystyle r$ is of the form $\displaystyle (x-h)^2 + (y - k)^2 = r^2$

Therefore (i) can be written as

$\displaystyle (x-0)^2 + (y- \frac{4}{3} )^2 = ( \frac{5}{3} )^2$

$\displaystyle \text{Therefore center is } (0, - \frac{4}{3} ) \text{ and radius is } \frac{5}{3} .$

$\displaystyle \text{(b) } e^{\frac{x}{y}} \Big( 1 - \frac{x}{y} \Big) + \Big( 1 + e^{\frac{x}{y}} \Big) \frac{dx}{dy} = 0$

Substitute $\displaystyle \text{ } x = vy \Rightarrow \frac{dx}{dy} = v + y \frac{dv}{dy}$

$\displaystyle e^{v}(1-v) + (1+e^v).\Big(v + y \frac{dv}{dy} \Big) = 0$

$\displaystyle (v+e^v)+(1-e^v) y . \frac{dv}{dy} = 0$

$\displaystyle \Big( \frac{1+e^v}{v+e^v} \Big) dv + \frac{dy}{y} = 0$

$\displaystyle \int \limits_{}^{} \Big( \frac{1+e^v}{v+e^v} \Big) dv = - \int \limits_{}^{} \frac{dy}{y}$

$\displaystyle \log (v+e^v) = - \log y + \log c$

$\displaystyle \log \{(v+e^v).y \} = \log c$

$\displaystyle \log \Big[ \Big( \frac{x}{y} + e^{\frac{x}{y}} \Big).y \Big] = \log c$

$\displaystyle \Big[ \Big( \frac{x}{y} + e^{\frac{x}{y}} \Big).y \Big] = c$

$\displaystyle x + y e^{\frac{x}{y}} = c$

$\displaystyle \text{When } x = 0$ and $\displaystyle y = 1$ we get $\displaystyle c = 1$

$\displaystyle \text{Hence } x + y e^{\frac{x}{y}} = 1$

$\displaystyle \\$

Section – B (20 Marks)

Question 10:

(a) Using vectors, prove that an angle in a semicircle is a right angle. [5]

(b) Find the volume of a parallelopiped whose edges are represented by the vectors: $\overrightarrow{a} = 2 \hat{i} - 3 \hat{j}- 4\hat{k}, \overrightarrow{b} = \hat{i} + 2 \hat{j} - \hat{k}, and \overrightarrow{c} = 3 \hat{i}+\hat{j} + 2\hat{k}$ [5]

Answer:

(a) In the adjoining diagram, we have a circle with center $O$ (origin) and the vertices of the triangle are $A, B$ and $C. AB$ is the diameter.

Let $\overrightarrow{OA} = \overrightarrow{a} \ and \ \overrightarrow{OC} = \overrightarrow{c}$

Therefore $\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = \overrightarrow{a} - \overrightarrow{c}$

$\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = -\overrightarrow{a} - \overrightarrow{c}$

$\overrightarrow{CA}. \overrightarrow{CB} = ( \overrightarrow{a} - \overrightarrow{c})(-\overrightarrow{a} - \overrightarrow{c})$

$= -\overrightarrow{a}.\overrightarrow{a}- \overrightarrow{a}.\overrightarrow{c} + \overrightarrow{c}.\overrightarrow{a} + \overrightarrow{c}.\overrightarrow{c}$

$= -|\overrightarrow{a}|^2 + |\overrightarrow{c}|^2$

$= - r^2 + r^2 = 0$ (since $r$ is the radius)

Therefore $\angle ACB$ is a right angle.

(b) The volume of a parallelopiped:

$\begin{bmatrix} 2 & - 3 & -4 \\ 1 & 2 & -1 \\ 3 & 1 & 2 \end{bmatrix}$

$= 2\begin{bmatrix} 2 & - 1 \\ 1 & 2 \end{bmatrix} + 3\begin{bmatrix} 1 & - 1 \\ 3 & 2 \end{bmatrix} - 4 \begin{bmatrix} 1 & 1 \\ 3 & 1 \end{bmatrix}$

$= 2 \times 5 + 3 \times 5 - 4 \times (-5) = 45 \text{ cubic units. }$

$\\$

Question 11:

(a) Find the equation of the plane passing through the intersection of the planes $x+y+z+1 = 0$ and $2x-3y+5z-2=0$ and the point $(-1, 2, 1)$ . [5]

(b) Find the shortest distance between the lines $\overrightarrow{r}=\hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda (2\hat{i} + 3 \hat{j} + 4 \hat{k})$ and $\overrightarrow{r}= 2\hat{i} + 4 \hat{j} + 5 \hat{k} + \mu (4\hat{i} + 6 \hat{j} + 8 \hat{k})$ [5]

Answer:

(a) Given planes $x+y+z+1 = 0$ and $2x-3y+5z-2=0$ and the point $(-1, 2, 1)$

The equation of a line passing through the intersection of the plans is given as

$(x+y+z+1) + k (2x-3y+5z-2) = 0$

$\displaystyle \Rightarrow 5k = 3 \Rightarrow k = \frac{3}{5}$

Hence the equation of line will be

$\displaystyle (x+y+z+1) + \frac{3}{5} (2x-3y+5z-2) = 0$

$\Rightarrow 5x + 5 y + 5 z + 5 + 6x -9y+15z-6 = 0$

$\Rightarrow 11x - 4y + 20 z -1 = 0$

(b) $\overrightarrow{a_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

$\overrightarrow{a_2} = 2\hat{i} + 4 \hat{j} + 5 \hat{k}$

$\overrightarrow{b} = 2\hat{i} + 3 \hat{j} + 4 \hat{k}$

$\overrightarrow{a_2} - \overrightarrow{a_1} = 2\hat{i} + 4 \hat{j} + 5 \hat{k} - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = \hat{i} + 2 \hat{j} + 2 \hat{k}$

$|\overrightarrow{b} |= \sqrt{4+9+16} = \sqrt{29}$

$(\overrightarrow{a_2} - \overrightarrow{a_1}) \times \overrightarrow{b} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 3 & 4 \end{bmatrix} = 2\hat{i} - 0 \hat{j} - \hat{k}$

$|(\overrightarrow{a_2} - \overrightarrow{a_1}) \times \overrightarrow{b} | = \sqrt{4+1}$

$\displaystyle \frac{|(\overrightarrow{a_2} - \overrightarrow{a_1}) \times \overrightarrow{b} |}{|\overrightarrow{b} |} = \sqrt{\frac{5}{29}}$

Hence the shortest distance $= 0.415$ units

$\\$

Question 12:

(a) $Box \ I$ contains $2$ white and $3$ black balls. $Box \ II$ contains $4$ white and $1$ black balls and $Box \ III$ contains $3$ white and $4$ black balls. A dice having $3$ red, $2$ yellow and $1$ green face, is thrown to select a box. If red face turns up we pick $Box \ I$ , if yellow face turns up we pick $Box \ II$ otherwise we pick $Box \ III$ . The we draw a ball from a selected box. If the ball drawn is white, what is the probability that the dice had turned up with a Red face. [5]

(b) 5 dices are thrown simultaneously. If the occurrence of an odd number in a single die is considered a success, find the probability of maximum three successes. [5]

Answer:

$\displaystyle \text{(a) Given: Probability of a red face when the dice is rolled : } P(A) = \frac{3}{6}$

$\displaystyle \text{Probability of a yellow face when the dice is rolled : } P(B) = \frac{2}{6}$

$\displaystyle \text{Probability of a green face when the dice is rolled : } P(C) = \frac{1}{6}$

Let D be the probability of drawing a White ball. Therefore

$\displaystyle \text{Probability of drawing a white ball from Bag I: } P(D/A) = \frac{2}{5}$

$\displaystyle \text{Probability of drawing a white ball from Bag II: } P(D/B) = \frac{4}{5}$

$\displaystyle \text{Probability of drawing a white ball from Bag III: } P(D/C) = \frac{3}{7}$

Therefore the probability of white ball drawn when red face show up:

$\displaystyle P(A/D) = \frac{P(A) \times P(D/A)}{P(A) \times P(D/A) + P(B) \times P(D/B) + P(C) \times P(D/C)}$

$\displaystyle = \frac{\frac{3}{6} \times \frac{2}{5}}{\frac{3}{6} \times \frac{2}{5} + \frac{2}{6} \times \frac{4}{5} + \frac{1}{6} \times \frac{3}{7}}$

(b) Number of dice $: 5$

$\displaystyle \text{P(odd number) } = \frac{1}{2}$

$\displaystyle \text{P(even number) } = \frac{1}{2}$

$\displaystyle P(x \leq 3) = 1 - P(x =4, 5)$

$\displaystyle = 1 - ^5C_4 \Big(\frac{1}{2}\Big)^2.\Big(\frac{1}{2}\Big) - ^5C_5 \Big(\frac{1}{2}\Big)^5$

$\displaystyle = 1 - \frac{5}{32} - \frac{1}{32}$

$\displaystyle = \frac{26}{32} = \frac{13}{16} = 0.81$

$\\$

Section – C (20 Marks)

Question 13:

(a) Mr. Nirav borrowed $Rs. 50000$ from the bank for $5$ years. The rate of interest is $9\%$ per annum compounded monthly. Find the payment he makes if he pays back at the beginning of each month. [5]

(b) A dietitian wishes to mix two kinds of food $X$ and $Y$ in such a way that the mixture contains at least $10$ units of vitamin $A$ , $12$ units of vitamin $B$ and $8$ units of vitamin $C$ . The vitamin contents of one kg of food is given below:

Food	Vitamin A	Vitamin B	Vitamin C
X	1 unit	2 units	3 units
Y	2 units	2 units	1 unit

One kg of food $X$ costs $Rs. \ 24$ and one kg of food $Y$ costs $Rs. \ 36$ . Using linear programming, find the least cost of the total mixture which will contain the required vitamins. [5]

Answer:

(a) Principal $= 50000 \ Rs.$

Annual rate of interest $= 9\% \Rightarrow$ Monthly rate of interest $= 0.0075\%$

$\displaystyle \text{Now } P = \frac{A}{i} (1+i) [1- (1+i)^{-n}]$

$\displaystyle 50000 = \frac{A}{0.0075} (1+0.0075) [1- (1+0.0075)^{-60}]$

$\displaystyle 50000 = \frac{A}{0.0075} (1.0075) [1- (1+0.0075)^{-60}]$

$\displaystyle A = \frac{50000 \times 0.0075}{(1.0075) [1- (1+0.0075)^{-60}]} = 1032.19$

Hence monthly installment $= 1032.19 \ Rs.$

(b) Let us say we have $x$ kgs of food $X$ and $y$ kgs of food $Y$ .

Therefore we have the following inequations:

$x + 2y \geq 10$

$2x + 2y \geq 12$

$3x + y \geq 8$

Now lets plot these lines. We got the following graph

We have to minimize $z = 24x + 36y$

Hence the four point of intersection from the graph are:

$(10, 0) \Rightarrow \text{ Cost } = 24 \times 10 + 36 \times 0 = 240 \text{ Rs. }$

$(2, 4) \Rightarrow \text{ Cost } = 24 \times 2 + 36 \times 4 = 192 \text{ Rs. }$

$(1, 5) \Rightarrow \text{ Cost } = 24 \times 1 + 36 \times 5 = 204 \text{ Rs. }$

$(0, 8) \Rightarrow \text{ Cost } = 24 \times 0 + 36 \times 8 = 288 \text{ Rs. }$

Hence, the $2 \ kg \ of \ X \text{ and } 4 kg \ of \ Y$ should be bought.

$\\$

Question 14:

(a) A bill of $Rs. \ 7650$ was drawn on 8th March 2013, at $7$ months. It was discounted on 18th May, 2013 and the holder of the bill received $Rs. \ 7497$ . What is the rate of interest charged by the bank? [5]

(b) The average cost function, $AC$ for a commodity is given by $\displaystyle AC = x + 5 + \frac{36}{x}$ in terms of output $x$ . Find:

(i) The total cost, $C$ and marginal cost, $MC$ as a function of $x$

(ii) The outputs for which $AC$ increases. [5]

Answer:

(a) Face value of the bill $(FV) = 7650 \ Rs.$

Discounted value of the bill $(DV) = 7497 \ Rs.$

Banker’s discount $= (7650 -7497) = 153 \ Rs.$

Nominal date due is 8th October

Legal due date of the bill is 11th October (3 days of grace period)

Number of unexpired days from 8th May to 11th October is 146 days $\displaystyle \Rightarrow n = \frac{146}{365} = \frac{2}{5}$

Banker’s discount $= Ani$

$\displaystyle \Rightarrow 153 = 7650 \times \frac{2}{5} \times r$

$\displaystyle \Rightarrow r = \frac{1}{20} = 5\%$

(b) $\displaystyle AC = x + 5 + \frac{36}{x}$

$\displaystyle \text{Cost function } C = (x + 5 + \frac{36}{x} ) x = x^2 + 5x + 36$

$\displaystyle \text{Marginal cost } = \frac{dC}{dx} = 2x + 5$

$\displaystyle \frac{dAC}{dx} = 1- \frac{36}{x^2}$

$\displaystyle \text{For AC to be increasing, } \frac{dAC}{dx} > 0$

$\displaystyle \Rightarrow 1- \frac{36}{x^2} > 0$

$\Rightarrow (x-6)(x+6) >0$

Case 1: $x-6 > 0 \ and \ x+6 > 0 \Rightarrow x > 6 \ and \ x > -6 \Rightarrow x > 6$

Case 2: $x-6 < 0 \ and \ x+6 < 0 \Rightarrow x < 6 \ and \ x < -6 \Rightarrow x < -6$

But $x$ cannot be negative. Therefore $x > 6$

Hence the average cost function increase if the output $x > 6$

$\\$

Question 15:

(a) Calculate the index number for the year 2014, with 2010 as the base year by the weighted aggregate method from the following data: [5]

Commodity	Price in Rs.		Weight
Commodity	2010	2014	Weight
A	$2$	$4$	$8$
B	$5$	$6$	$10$
C	$4$	$5$	$14$
D	$2$	$2$	$19$

(b) The quarterly profits of a small scale industry (in thousands of Rupees) is as follows: Calculate the quarterly moving averages. Display these and the original figures graphically on the same graph sheet. [5]

Year	Quarter 1	Quarter 2	Quarter 3	Quarter 4
2012	$39$	$47$	$20$	$56$
2013	$68$	$59$	$66$	$72$
2014	$88$	$60$	$60$	$67$

Answer:

(a)

Commodity	$2010$		$2014$		$p_ow$	$p_1w$
	$p_o$	$w$	$p_1$	$w$
$A$	$2$	$8$	$4$	$8$	$16$	$32$
$B$	$5$	$10$	$6$	$10$	$50$	$60$
$C$	$4$	$14$	$5$	$14$	$56$	$70$
$D$	$2$	$19$	$2$	$19$	$38$	$38$
					$\Sigma p_ow \\ = 160$	$\Sigma p_1w \\ = 200$

Index number for the year 2014 with respect to 2010 as the base year

$\displaystyle = \frac{200}{160} \times 100 = 125$

(b)

Year	Quarter	Quarterly Profits	4 years moving total	4 year average
$2012$	$1$	$39$
	$2$	$47$	$162$	$40.2$
	$3$	$20$	$191$	$47.75$
	$4$	$56$	$203$	$50.75$
$2013$	$1$	$68$	$249$	$62.25$
	$2$	$59$	$265$	$66.25$
	$3$	$66$	$285$	$71.25$
	$4$	$72$	$286$	$71.5$
$2014$	$1$	$88$	$280$	$70$
	$2$	$60$	$275$	$68.75$
	$3$	$60$
	$4$	$67$