MATHEMATICS

(Maximum Marks: 100)

(Time Allowed: Three Hours)

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)

Question 1.                                                                                                         $[10 \times 3]$

(i)  If the matrix $\begin{bmatrix} 6 & -x^2 \\ 2x-15 & 10 \end{bmatrix}$ is symmetric, find the value of $x$.

(ii) If $y - 2x - k = 0$ touches the conic $3x^2-5y^2 = 15$, find the value of $k$.

(iii) Prove that $\frac{1}{2}$ $cos^{-1} \ ($ $\frac{1-x}{1+x}$ $)= tan^{-1} \sqrt{x}$

(iv) Using L’Hospital’s rule, evaluate: $\lim \limits_{x \to \frac{\pi}{2}} \ (x \ tan \ x - \frac{\pi}{4} sec \ x)$

(v) Evaluate: $\int \limits_{}^{}$ $\frac{1}{x^2}$ $\ sin^2 \ ($ $\frac{1}{x}$ $) \ dx$

(vi) Evaluate: $\int \limits_{0}^{\frac{\pi}{4}} log \ (1 + tan \ \theta) \ d \theta$

(vii) By using the data $\bar{x} = 25, \bar{y} = 30, b_{yx}=1.6$ and $b_{xy}=0.4$, find:

(a) The regression equation $y$ on $x$

(b) What is the most likely value of $y$ when $x = 60$?

(c) What is the coefficient of correlation between $x$ and $y$?

(viii) A problem is given to three students whose chances of solving it are $\frac{1}{4}$$\frac{1}{5}$ and $\frac{1}{3}$ respectively. Find the probability that the problem is solved.

(ix) If $a + ib =$ $\frac{x+iy}{x-iy}$  prove that $a^2 + b^2 = 1$ and $\frac{b}{a}$ $=$ $\frac{2xy}{x^2 -y^2}$

(x) Solve: $\frac{dy}{dx}$ $= 1 - xy + y - x$

(i)    Since $\begin{bmatrix} 6 & -x^2 \\ 2x-15 & 10 \end{bmatrix}$ is symmetric, therefore

$\begin{bmatrix} 6 & -x^2 \\ 2x-15 & 10 \end{bmatrix}^T = \begin{bmatrix} 6 & -x^2 \\ 2x-15 & 10 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 6 & 2x-15 \\ -x^2 & 10 \end{bmatrix} = \begin{bmatrix} 6 & -x^2 \\ 2x-15 & 10 \end{bmatrix}$

We can compare corresponding terms. We get

$2x-15 = -x^2$

$\Rightarrow x^2+2x-15 = 0$

$\Rightarrow x^2 + 5x - 3x - 15 = 0$

$\Rightarrow x(x+5)-3(x+5)=0$

$\Rightarrow (x+5)(x-3) = 0 \Rightarrow x = -5 \ or \ 3$

$\\$

(ii)  $y - 2x - k = 0 \Rightarrow y = 2x+k$

$3x^2-5y^2=15 \Rightarrow$ $\frac{x^2}{5}$ $-$ $\frac{y^2}{3}$ $=1$

Therefore $m = 2, a^2 = 5, b^2 = 3$

Equation of line $y = mx + c$

Since $\frac{x^2}{a^2}$ $-$ $\frac{y^2}{b^2}$ $=1$ is hyperbola

$y = mx \pm \sqrt{a^2m^2-b^2}$

$2x+k = 2x \pm \sqrt{5 \times 2^2-3}$

$\Rightarrow k = \pm \sqrt{20-3} = \pm \sqrt{17}$

$\\$

(iii)  LHS $=$ $\frac{1}{2}$ $cos^{-1} \$ $( \frac{1-x}{1+x})$

Putting $x = tan^2 \theta$

LHS $=$ $\frac{1}{2}$ $cos^{-1} \$ $( \frac{1-tan^2 \theta}{1+tan^2 \theta})$

$=$ $\frac{1}{2}$ $cos^{-1} \$ $( \frac{1-tan^2 \theta}{sec^2 \theta})$

$=$ $\frac{1}{2}$ $cos^{-1} \$ $( cos^2 \theta - sin^2 \theta)$

$=$ $\frac{1}{2}$ $cos^{-1} \ ( cos\ 2\theta)$

$=$ $\frac{2\theta}{\theta}$

$=$ $tan^{-1}\sqrt{x} =$ RHS

Hence Proved.

$\\$

(iv) $\lim \limits_{x \to \frac{\pi}{2}} \ (x \ tan \ x - \frac{\pi}{4} sec \ x)$

$= \lim \limits_{x \to \frac{\pi}{2}} \$ $(\frac{x \ sin \ x - \frac{\pi}{4}}{cos \ x})$

$= \lim \limits_{x \to \frac{\pi}{2}} \$ $(\frac{x cos \ x +sin \ x + 0}{-sin \ x})$

$= \lim \limits_{x \to \frac{\pi}{2}} \$ $(\frac{x cos \ x +sin \ x}{-sin \ x})$

$=$ $\frac{0+1}{-1}$ $= -1$

$\\$

(v)  Put $\frac{1}{x}$ $= t$

$\Rightarrow -$ $\frac{1}{x^2}$ $dx = dt$

$I = -$ $\int \limits_{}^{}$ $sin^2 t \ dt$

$=$ $\int \limits_{}^{}$ $\frac{cos \ 2t - 1}{2}$ $dt$

$=$ $\frac{1}{2} \frac{sin \ 2t}{2}$ $- t - c$

$=$ $\frac{1}{4}$ $sin$ $(\frac{2}{x})$ $-$ $\frac{1}{2x}$ $+c$

$\\$

(vi) Let I = $\int \limits_{0}^{\frac{\pi}{4}} log \ (1 + tan \ \theta) \ d \theta$ … … … … (i)

Also we can write

I = $\int \limits_{0}^{\frac{\pi}{4}} log \ (1 + tan \ (\frac{\pi}{4} - \theta)) \ d \theta$

$= \int \limits_{0}^{\frac{\pi}{4}} log \ (1 +$ $\frac{1 - tan \ \theta}{1 + tan \ \theta})$ $\ d \theta$

$= \int \limits_{0}^{\frac{\pi}{4}} log \$ $(\frac{1 + tan \ \theta + 1 - tan \ \theta}{1 + tan \ \theta})$ $\ d \theta$

$= \int \limits_{0}^{\frac{\pi}{4}} log \$ $(\frac{2}{1 + tan \ \theta})$ $\ d \theta$ … … … … (ii)

Adding (i) and (ii) we get

2I = $\int \limits_{0}^{\frac{\pi}{4}} log \ (1 + tan \ \theta) \ d \theta$ + $\int \limits_{0}^{\frac{\pi}{4}} log \$ $(\frac{2}{1 + tan \ \theta})$ $\ d \theta$

$\Rightarrow 2I =$ $= \int \limits_{0}^{\frac{\pi}{4}} log \$ $(\frac{2(1 + tan \ \theta)}{1 + tan \ \theta})$ $\ d \theta$

$\Rightarrow 2I =$ $= \int \limits_{0}^{\frac{\pi}{4}} log \ 2 \ d \theta$

$\Rightarrow I =$ $\frac{1}{2}$ $log \ 2 {\Big[ \theta \Big]_0}^{\pi/4}$

$\Rightarrow I =$ $\frac{1}{2}$ $log \ 2 ($ $\frac{\pi}{4}$ $- 0)$

$\Rightarrow I =$ $\frac{\pi}{8}$ $log \ 2$

$\\$

(vii) Given $\bar{x} = 25, \bar{y} = 30, b_{yx}=1.6$ and $b_{xy}=0.4$

(a) The regression equation of $y \ on \ x$ is

$y - \bar{y} = b_{xy}(x-\bar{x})$

$\Rightarrow y = 30 = 1.6(x-25)$

$\Rightarrow y = 1.6x - 10$

(b) When $x = 60$

$y = 1.6 \times 60 - 10 = 96-10 = 86$

(c) Given $b_{yx} \ and \ b_{xy}$ as positive, $r$ (coefficient of correlation) will be positive

$\therefore r = \sqrt{b_{yx} \times b_{xy}} = \sqrt{1.6 \times 0.4} = \sqrt{0.64} = 0.8$

$\\$

(viii) If $A, B \ and \ C$ are three events representing that students can solve the problem. Therefore

$P(A) =$ $\frac{1}{4}$ $\Rightarrow P(\bar{A}) = 1-$ $\frac{1}{4}$ $=$ $\frac{3}{4}$

$P(B) =$ $\frac{1}{5}$ $\Rightarrow P(\bar{B}) = 1-$ $\frac{1}{5}$ $=$ $\frac{4}{5}$

$P(C) =$ $\frac{1}{3}$ $\Rightarrow P(\bar{C}) = 1-$ $\frac{1}{3}$ $=$ $\frac{2}{3}$

The problem would be solved if anyone of them solves the problem. Also remember, these are mutually exclusive events.

Therefore the Probability of solving the problem $=1-P(\bar{A}) \times P(\bar{B}) \times P(\bar{C}) = 1-$ $\frac{3}{4}$ $\times$ $\frac{4}{5}$ $\times$ $\frac{2}{3}$ $= 1-$ $\frac{2}{5}$ $=$ $\frac{3}{5}$

$\\$

(ix) $a + ib =$ $\frac{x+iy}{x-iy}$

$\Rightarrow$ $a + ib =$ $\frac{x+iy}{x-iy} \times \frac{x+iy}{x+iy}$

$\Rightarrow$ $a + ib =$ $\frac{x^2+i^2y^2+2ixy}{x^2-i^2y^2}$

$\Rightarrow$ $a + ib =$ $\frac{(x^2-y^2)+i(2xy)}{x^2+y^2}$

$\Rightarrow$ $a =$ $\frac{x^2-y^2}{x^2+y^2}$ $\ and \ b =$ $\frac{2xy}{x^2+y^2}$

Now $a^2 + b^2 =$ $(\frac{x^2-y^2}{x^2+y^2})^2$ $+$ $( \frac{2xy}{x^2+y^2})^2$

$=$ $\frac{x^4+y^4-2x^2y^2+4x^2y^2}{(x^2+y^2)^2}$

$=$ $\frac{(x^2+y^2)^2}{(x^2+y^2)^2}$

$= 1$. Hence proved.

Also $\frac{b}{a}$ $=$ $\frac{(\frac{2xy}{x^2+y^2})}{ (\frac{x^2-y^2}{x^2+y^2})}$

$\Rightarrow$ $\frac{b}{a}$ $=$ $\frac{2xy}{x^2-y^2}$. Hence proved.

$\\$

(x) $\frac{dy}{dx}$ $= 1 - xy + y - x$

$\frac{dy}{dx}$ $= (1+y)-x(1+y)$

$\frac{dy}{dx}$ $= (1+y)(1-x)$

$\frac{dy}{1+y}$ $= (1-x) dx$

Integrating both sides we get

$\int \limits_{}^{}$ $\frac{dy}{1+y}$ $= \int \limits_{}^{} (1-x) dx$

$\Rightarrow log(1+y) = x -$ $\frac{x^2}{2}$ $+ c$ (constant of integration)

$\\$

Question 2:

(a) Using properties of determinants, prove that:

$\left| \begin{array}{ccc} a & b & b+c \\ c & a & c+a \\ b & c & a+b \end{array} \right|= (a + b + c)(a - c)^2$    [5]

(b) Given that: $A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$. Find $AB$.

Using this result, solve the following system of equation:

$x-y = 3, 2x+3y+4z=17$ and $y +2z=7$    [5]

(a)  LHS $= \Delta =$ $\left| \begin{array}{ccc} a & b & b+c \\ c & a & c+a \\ b & c & a+b \end{array} \right|$

Applying $R_1 \rightarrow R_1 + R_2 + R_3$

$\Rightarrow \Delta =$ $\left| \begin{array}{ccc} a+b+c & a+b+c & 2(a+b+c) \\ c & a & c+a \\ b & c & a+b \end{array} \right|$

$\Rightarrow \Delta = (a+b+c)$ $\left| \begin{array}{ccc} 1 & 1 & 2 \\ c & a & c+a \\ b & c & a+b \end{array} \right|$

Now applying $C_1 \rightarrow C_1 - \frac{1}{2} C_3 \ and \ C_2 \rightarrow C_2-\frac{1}{2} C_3$

$\Delta = (a+b+c)$ $\left| \begin{array}{ccc} 0 & 0 & 2 \\ (c-a)/2 & (a-c)/2 & c+a \\ (b-a)/2 & c- (a+b)/2 & a+b \end{array} \right|$

Expanding along $R_1$

$\Delta = (a+b+c). \Big[$ $\frac{c-a}{2}.\frac{2c-a-b}{2}-\frac{a-c}{2}.\frac{b-a}{2}$ $\Big]$

$= (a+b+c)\Big[$ $\frac{2c^2-2ac-ac+a^2-bc+ab-ab+bc+a^2-ac}{2}$ $\Big]$

$= (a+b+c)\Big[$ $\frac{2a^2+2c^2-4ac}{2}$ $\Big]$

$= (a+b+c)(a-c)^2 =$ RHS. Hence proved.

(b) $A.B = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$

$= \begin{bmatrix} 2 \times 1 + (-1) \times(-4) + 0 & 1 \times 2 +(-1) \times 2 +0 & 1 \times (-4) + (-1) \times (-4) + 0 \\ 2 \times 2 + 3 \times (-4) + 4 \times 2 & 2 \times 2 +3 \times 2 +4 \times (-1) & 2 \times (-4) +3 \times (-4) + 4 \times 5 \\ 0+ 1 \times (-4) +2 \times 2 & 0 + 1 \times 2 + 2 \times (-1) & 0 + 1 \times (-4) + 2 \times 5 \end{bmatrix}$

$= \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6 I_3$

$\Rightarrow$ $\frac{1}{6}$ $AB = I \Rightarrow A^{-1} =$ $\frac{1}{6}$ $B$

Now write the given system of equation in the form of $AX = C$.

$\begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} . \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}$

$\Rightarrow A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \ and \ C = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}$

$AX = C$

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

$\Rightarrow X = \frac{1}{6} BC$

$=$ $\frac{1}{6}$ $\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}$

$=$ $\frac{1}{6}$ $\begin{bmatrix} 2 \times 3 + 2 \times 17 +(-4) \times 7 \\ (-4) \times 3 + 2 \times 17 +(-4) \times 7 \\ 2 \times 3 +(-1) \times 17 +5 \times 7 \end{bmatrix}$

$=$ $\frac{1}{6}$ $\begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix}$

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

Therefore $x = 2, y = -1 \ and \ z=4$

$\\$

Question 3:

(a) Solve the equation for $x$:

$sin^{-1} \ x + sin^{-1} \ (1-x) = cos^{-1} \ x, x \neq 0$    [5]

(b) If $A, B$ and $C$ are elements of Boolean Algebra, simplify the expression $(A' + B')(A + C') + B'(B + C)$. Draw the simplified circuit.    [5]

(a)  Given $sin^{-1} \ x + sin^{-1} \ (1-x) = cos^{-1} \ x, x \neq 0$

$sin^{-1} \{x \sqrt{1-(1-x)^2}+ (1-x)\sqrt{1-x^2} \} = cos^{-1}$

$\Rightarrow sin^{-1} \{x \sqrt{1-1-x^2+2x}+ (1-x)\sqrt{1-x^2} \} = cos^{-1} x$

$\Rightarrow sin^{-1} \{x \sqrt{2x-x^2}+ (1-x)\sqrt{1-x^2} \} = sin^{-1} (\sqrt{1-x^2})$

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

$\Rightarrow x \sqrt{2x-x^2} = \sqrt{1-x^2} - (1-x) \sqrt{1-x^2}$

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

$\Rightarrow x\sqrt{2x-x^2} = x\sqrt{1-x^2}$

$\Rightarrow x\sqrt{2x-x^2} - x\sqrt{1-x^2} = 0$

$\Rightarrow x(\sqrt{2x-x^2} - \sqrt{1-x^2}) = 0$

$\Rightarrow either \ x = 0$

or $\sqrt{2x-x^2} - \sqrt{1-x^2}$

$\sqrt{2x-x^2} = \sqrt{1-x^2}$

$\Rightarrow 2x-x^2 = 1-x^2$

$\Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$

Hence $x = 0 or \frac{1}{2}$

(b) To be solved

$\\$

Question 4:

(a) Verify Lagrange’s mean value theorem for the function: $f(x) = x(1-log \ x)$ and find the value of $c$ in the interval $[1, 2]$    [5]

(b) Find the coordinates of center, foci and equation of directrix of hyperbola: $x^2 - 3y^2 -4^x = 8$    [5]

(a)  Given $f(x) = x(1-log \ x)$ in $[1, 2]$

The function f(x) is continuous in $[1, 2]$ as $x$ and $log x$ are continuous in $[1, 2]$

Now, $f'(x) = 1 - log \ x + x(1-$ $\frac{1}{x}$ $)$

$= 1 = log \ x _1$

$= - log \ x$  which exists for all values in $(1, 2)$

Therefore $f(x)$ is differentiable in $(1, 2)$

Therefore by Lagrange’s mean value theorem we have

$f'(c) =$ $\frac{f(2)-f(1)}{2-1}$

$-log \ c =$ $\frac{2(1-log \ 2) - 1(1- log \ 1)}{2-1}$

$-log \ c =$ $\frac{2-2log \ 2 - 1 + log \ 1}{1}$

$log \ c = log \ 4 - log \ e$

$log \ c = log ($ $\frac{4}{e}$ $)$

$\Rightarrow c =$ $\frac{4}{e}$ which lies in the interval $(1, 2)$. Hence verified.

(b) To be solved

$\\$

Question 5:

(a) If $y = cos \ (sin \ x)$, show that:

$\frac{d^2y}{dx^2}$ $+ tan \ x$ $\frac{dy}{dx}$ $+ y \ cos^2 \ x = 0$    [5]

(b) Show that the surface are of a closed cuboid with square base and a given volume is minimum when it is a cube.    [5]

(a) Given $y = cos \ (sin \ x)$ … … … … … (i)

Differentiating both sides, we get,

$\frac{dy}{dx}$ $= - sin \ (sin \ x) cox \ x$ … … … … … (ii)

Again differentiating both sides

$\frac{d^2y}{dx^2}$ $= - cos\ (sin\ x)\ cos^2x+sin\ (sin\ x) sin \ x$

Using (i) and (ii) we get

$\frac{d^2y}{dx^2}$ $= - y \ cos^2x +$ $\frac{sin\ x}{(-cos\ x)} \frac{dy}{dx}$

$\frac{d^2y}{dx^2}$ $= -y \ cos^2x +$ $\frac{sin \ x}{(-cos \ x)} \frac{dy}{dx}$

$\frac{d^2y}{dx^2}$ $= + tan \ x$ $\frac{dy}{dx}$ $+y \ cos^2x = 0$

Hence proved.

(b)  Let $h$ be the height and $x$ be the side of the square base of the cuboid. Therefore

Surface Area $(A) = 2(x^2+2hx) = 2x^2 + 4xh$

Volume $(V) = x^2 h$

Therefore $A = 2x^2 + 4x ($ $\frac{V}{x^2})$

$\Rightarrow A = 2x^2 +$ $\frac{4V}{x}$

Differentiating with respect to $x$

$\frac{dA}{dx}$ $= 4x -$ $\frac{4V}{x^2}$ … … … … … (i)

Putting $\frac{dA}{dx}$ $= 0$

$\Rightarrow 4x -$ $\frac{4V}{x^2}$ $= 0$

$\Rightarrow 4x =$ $\frac{4V}{x^2}$

$\Rightarrow V = x^3$

$\Rightarrow x^2h = x^3 \Rightarrow h = x$

Differentiating (i) with respect to $x$ we get

$\frac{d^2A}{dx^2}$ $= 4 +$ $\frac{8V}{x^3}$

$= 4 +$ $\frac{8x^2h}{x^3}$

$= 4 +$ $\frac{8h}{x}$

When $h = x$

$(\frac{d^2A}{dx^2})_{h=x}$ $= 4+8 = 12 >0$

Therefore the Surface Area $(A)$ is minimum when $h = x \Rightarrow Cube$.

$\\$

Question 6:

(a) Evaluate:

$\int \limits_{}^{}$ $\frac{sin \ 2x}{(1 + sin \ x)(2 + sin \ x)}$ $dx$    [5]

(b) Draw a rough sketch of the curve $y^2 = 4x$ and find the area of the region enclosed by the curve and the line $y = x$.     [5]

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

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

Now Put $sin \ x = t \ and \ cos \ x \ dx = dt$

Therefore $I = \int \limits_{}^{}$ $\frac{2 t dt}{(1 + t)(2 + t)}$ $dx$

Let $\frac{2t}{(1+t)(2+t)}$ $=$ $\frac{A}{1+t}$ $+$ $\frac{B}{2+t}$

$\Rightarrow 2t = (2+t)A + (1+t)B$

For $t = -1, A = -2$ and for $t = -2 B = 4$

Therefore Therefore $I = -2\int \limits_{}^{}$ $\frac{dt}{1 + t}$ $+ 4 \int \limits_{}^{}$ $\frac{dt}{2 + t}$ $dx$

$I = -2 log |1+t| + 4 log |2+t| + c$

$= -2 log |1+sin \ x | + 4 log |2 + sin \ x| + c$

$= log |1+sin \ x |^{-2} + log |2 + sin \ x|^{4} + c$

$= \log$ $|\frac{(2 + sin \ x)^{4}}{(1+sin \ x )^{-2}} |$ $+c$

(b) Equation of the curve given is $y^2 = 4x$ and the equation of line is $y = x$

Solving $y^2 = 4y \Rightarrow y = 4 \ or \ y = 0$

When $y = 4, x = 4$ and when $y = 0, x = 0$

Hence the two point of intersection are $(0,0)$ and $(4,4)$

Therefore the required enclosed area is

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

$= 2$ $\frac{{{\Big[ x^{\frac{3}{2}} \Big]}_{0}}^4}{\frac{3}{2}}$ $-$ $\frac{{{\Big[ x^{2} \Big]}_{0}}^4}{2}$

$=$ $\frac{4}{3}$ $\Big[ 4^{\frac{3}{2}} - 0^{\frac{3}{2}} \Big]$ $-$ $\frac{1}{2}$ $(4^2 - 0^2)$

$=$ $\frac{4}{3}$ $\times 8 -$ $\frac{1}{2}$ $\times 16 =$ $\frac{32}{3}$ $-8 =$ $\frac{8}{3}$ $sq. \ units$

$\\$

Question 7:

(a) Calculate the Spearman’s rank correlation coefficient for the following data and interpret the results:     [5]

 X 35 54 80 95 73 73 35 91 83 81 Y 40 60 75 90 70 75 38 95 75 70

(b) Find the line of best fit for the following data, treating $x$ as a dependent variable (Regression equation $x$ on $y$):

 X 14 12 13 14 16 10 13 12 Y 14 23 17 24 18 25 23 24

Hence, estimate the value of $x$ when $y= 16$.     [5]

(a)

 $x$ $y$ Rank $(R_x)$ Rank $(R_y)$ $d = R_x-R_y$ $d^2$ $35$ $40$ $9.5$ $9$ $0.5$ $0.25$ $54$ $60$ $8$ $8$ $0$ $0$ $80$ $75$ $5$ $4.5$ $0.5$ $0.25$ $95$ $90$ $1$ $2$ $1$ $1$ $73$ $70$ $6.5$ $6.5$ $0$ $0$ $73$ $75$ $6.5$ $4.5$ $2$ $4$ $35$ $38$ $9.5$ $10$ $-0.5$ $0.25$ $91$ $95$ $2$ $1$ $1$ $1$ $83$ $75$ $3$ $3$ $0$ $0$ $81$ $70$ $4$ $6.5$ $-2.5$ $6.25$

$\Sigma d^2 = 13$

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

$r = 1 - 6 \Big[$ $\frac{ 13 + \frac{1}{12} \Sigma (2^3 - 2)}{10(99)}$ $\Big]$

$= 1 -$ $\frac{6 \times 13.5}{990}$ $= 0.981$

This indicates a strong positive relationship between $x \ and \ y$. That is, the higher is $x$, the higher is $y$.

(b)

 $x$ $y$ $xy$ $y^2$ $14$ $14$ $196$ $196$ $12$ $23$ $276$ $529$ $13$ $17$ $221$ $289$ $14$ $24$ $336$ $576$ $16$ $18$ $288$ $324$ $10$ $25$ $250$ $625$ $13$ $23$ $299$ $529$ $12$ $24$ $288$ $576$ $\Sigma x = 104$ $\Sigma y = 168$ $\Sigma xy = 2154$ $\Sigma y^2 = 3644$

$\bar{x} =$ $\frac{\Sigma x}{n}$ $=$ $\frac{104}{8}$ $=13$

$\bar{y} =$ $\frac{\Sigma y}{n}$ $=$ $\frac{168}{8}$ $=21$

Now,

$b_{xy} =$ $\frac{\Sigma xy - \frac{\Sigma x \Sigma y}{n}}{\Sigma y^2 - \frac{1}{n} (\Sigma y)^2}$ $=$ $\frac{2154 - \frac{104 \times 168}{8}}{3644 - \frac{168^2}{8}}$

$b_{xy} =$ $\frac{2154-2184}{3644-3528}$ $=$ $\frac{-30}{116}$

The regression equation of $x \ on \ y$ is

$x-\bar{x} = b_{xy}(y-\bar{y})$

$x - 13 =$ $\frac{-30}{116}$ $(y-21)$

$116x - 1508 = -30y+630$

$116x +30y = 1508 +630$

$116x + 30y = 2138$

$58x + 15y = 1069$

Value of $x$ when $y = 16 \Rightarrow 58x + 15 \times 16 = 1069$

$58x = 1069-240 = 829$

Therefore $x =$ $\frac{829}{58}$ $= 14.2931$

$\\$

Question 8:

(a) In a class of $60$ students, $30$ opted for Mathematics, $32$ opted for Biology and $24$ opted for both mathematics and Biology. If one of these students is selected at random, find the probability that:

(i) The student opted for Mathematics or Biology

(ii) The student has opted neither for Mathematics nor Biology

(iii) The student has opted Mathematics but not Biology     [5]

(b) Bag $A$ contains $1$ white, $2$ blue and $3$ red balls. Bag $B$ contains $3$ white, $3$ blue and $2$ red balls. Bag $C$ contains $2$ white, $3$ blue and $4$ red balls. One bag is selected at random and then two balls are drawn from the selected bag. Find the probability that the balls drawn are white and red.     [5]

(a) Let Student who opted for Mathematics $= M$ and Students who opted for Biology $= B$

Given

$P(M) =$ $\frac{30}{60}$ $=$ $\frac{1}{2}$

$P(M) =$ $\frac{32}{60}$ $=$ $\frac{8}{15}$

$P(M \cap B) =$ $\frac{24}{60}$ $=$ $\frac{2}{5}$

(i) The student opted for Mathematics or Biology

$= P(M \cup B )$

$= P(M) + P(B) - P (M \cap B)$

$=$ $\frac{1}{2}$ $+$ $\frac{8}{15}$ $-$ $\frac{2}{5}$

$=$ $\frac{19}{30}$

(ii) The student has opted neither for Mathematics nor Biology

$= P(M' \cap B')$

$= 1 - P(\cup B) = 1 -$ $\frac{19}{30}$

$=$ $\frac{11}{30}$

(iii) The student has opted Mathematics but not Biology

$= P(M \cap B')$

$= P(M) - P(M \cap B)$

$=$ $\frac{1}{2}$ $-$ $\frac{2}{5}$

$=$ $\frac{1}{10}$

(b)  Let Event $(E_1) = Bag \ A$ is chosen

Event $(E_2) = Bag \ B$ is chosen

Event $(E_3) = Bag \ C$ is chosen

and Event $(A) =$ ball drawn is white and red

The bags are chosen totally at random therefore $P(E_1) = P(E_2) = P(E_3) =$ $\frac{1}{3}$

Now

Probability of drawing a white and red ball from $Bag \ A =$ $P(A/E_1) =$  $\frac{^1C_1 \times ^3C_1}{^6C_2}$ $=$ $\frac{1 \times 3 \times 2}{6 \times 5} = \frac{1}{5}$

Probability of drawing a white and red ball from $Bag \ B =$ $P(A/E_2) =$  $\frac{^3C_1 \times ^2C_1}{^8C_2}$ $=$ $\frac{3 \times 2 \times 2}{8 \times 7} = \frac{3}{14}$

Probability of drawing a white and red ball from $Bag \ C =$ $P(A/E_3) =$  $\frac{^2C_1 \times ^4C_1}{^9C_2}$ $=$ $\frac{2 \times 4 \times 2}{9 \times 8} = \frac{2}{9}$

By law of total probability we get:

$P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)$

$=$ $\frac{1}{3}$ $\times$ $\frac{1}{5}$ $+$ $\frac{1}{3}$ $\times$ $\frac{3}{14}$ $+$ $\frac{1}{3}$ $\times$ $\frac{2}{9}$

$=$ $\frac{1}{15}$ $+$ $\frac{1}{14}$ $+$ $\frac{2}{27}$

$=$ $\frac{401}{1089}$ $= 0.3764$

$\\$

Question 9:

(a) Prove that locus of $z$ is circle and find its center and radius if $\frac{z-1}{z+1}$ is purely imaginary.     [5]

(b) Solve: $(x^2 - yx^2)dy + (y^2+xy^2)dx = 0$     [5]

(a) Let $z = x+iy$

$\frac{z-1}{z+1}$ $=$ $\frac{x+iy -1}{x+iy +1}$

$=$ $\frac{(x-1)+iy}{(x+1)+iy}$ $\times$ $\frac{(x-1)-iy}{(x+1)-iy}$

$=$ $\frac{(x-1)(x+1) -i^2y^2 + iy (x+1 - (x-1))}{(x+1)^2 +y^2}$

$\Rightarrow$ $\frac{z-1}{z+1}$ $=$ $\frac{x^2 -1 + y^2 + i2y}{(x+1)^ + y^2}$

$\Rightarrow$ $\frac{z-1}{z+1}$ $=$ $\frac{x^2 -1 + y^2}{(x+1)^ + y^2}$ $+ i$ $\frac{2y}{(x+1)^ + y^2}$

Since $\frac{z-1}{z+1}$ is purely imaginary , $Re \Big($ $\frac{z-1}{z+1}$ $\Big) = 0$

$\Rightarrow$ $\frac{x^2 -1 + y^2}{(x+1)^ + y^2}$ $= 0$

$\Rightarrow x^2 +y^2 = 1$, which is a circle of radius unit $1$.

$z = x+iy$  is a complex number such that $x, \ y$ lie on a circle centered at origin and radius 1 unit.

(b)  $(x^2 - yx^2)dy + (y^2+xy^2)dx = 0$

$\Rightarrow$ $(x^2 - yx^2) \ dy = - (y^2+xy^2) \ dx$

$\Rightarrow$ $x^2(1 - y) \ dy = - y^2(1+x) \ dx$

$\Rightarrow$ $\frac{1-y}{y^2}$ $\ dy = -$ $\frac{1+x}{x^2}$ $\ dx$

$\Rightarrow$ $\frac{1-y}{y^2}$ $\ dy +$ $\frac{1+x}{x^2}$ $\ dx = 0$

Integrating on both sides,

$\int \limits_{}^{}$ $\frac{1-y}{y^2}$ $\ dy + \int \limits_{}^{}$ $\frac{1+x}{x^2}$ $\ dx = c$

$\int \limits_{}^{}$ $\frac{1}{y^2}$ $\ dy - \int \limits_{}^{}$ $\frac{1}{y}$ $\ dy + \int \limits_{}^{}$ $\frac{1}{x^2}$ $\ dx + \int \limits_{}^{}$ $\frac{1}{x}$ $= c$

$-\frac{1}{y}$ $- log |y| -$ $\frac{1}{x}$ $+ log |x| = c$

$log |x| -$ $\frac{1}{x}$ $=$ $\frac{1}{y}$ $+ log |y| + c$

$\\$

Question 10:

(a) If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three mutually perpendicular vectors of equal magnitude, prove that $(\overrightarrow{a}+ \overrightarrow{b}+ \overrightarrow{c})$ is equally inclined with vectors $\overrightarrow{a}, \overrightarrow{b}$ and  $\overrightarrow{c}$     [5]

(b) Find the value of $\lambda$ for which the four points with position vectors $6 \hat{i} - 7 \hat{j}, 16 \hat{i} - 19 \hat{j} - 4 \hat{k}$, $\lambda \hat{j} - 6 \hat{k} \ and \ 2 \hat{i} - 5 \hat{j}+ 10 \hat{k}$ are co-planar.     [5]

(a)  Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three mutually perpendicular vectors of equal magnitude, therefore

$\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a} = 0$

and $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| = x$

Let $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{d}$

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

$= (\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}).(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c})$

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

$= 1 + 0+0 +0+ 1+0+0+0+1 = 3$

$\Rightarrow |\overrightarrow{d} | = \sqrt{3}$

Let the angles between $\overrightarrow{a} \ and \ \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$  be  $\alpha, \ \beta, \ \gamma$ respectively.

Therefore $cos \ \alpha =$ $\frac{\overrightarrow{d}.\overrightarrow{a}}{|\overrightarrow{d}|.|\overrightarrow{a}|}$

$=$ $\frac{(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}) \overrightarrow{a}}{\sqrt{3}. |\overrightarrow{a}|}$

$= \frac{\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{a}}{\sqrt{3}. |\overrightarrow{a}|}$

$= \frac{1+0+0}{\sqrt{3}. |\overrightarrow{a}|}$

$= \frac{1}{\sqrt{3}. |\overrightarrow{a}|}$

$= \frac{1}{\sqrt{3}. x}$

Similarly $cos \ \beta =$ $\frac{1}{\sqrt{3}. x}$ and $cos \ \gamma =$ $\frac{1}{\sqrt{3}. x}$

Therefore $cos \ \alpha = cos \ \beta = cos \ \gamma \Rightarrow \alpha = \beta = \gamma$

(b)  Let $A, B, C \ and \ D$ be the given points. Therefore,

$\overrightarrow{AB} = Position \ vector \ of \ B - Position \ Vector \ of \ A$

$= (16\hat{i}-19\hat{j}-4\hat{k})-(6\hat{i}-7\hat{j}) = 10\hat{i}-12\hat{j}-4\hat{k}$

$\overrightarrow{AC} = Position \ vector \ of \ C - Position \ Vector \ of \ A$

$= (\lambda \hat{j}-6\hat{k})-(6\hat{i}-7\hat{j}) = -6\hat{i}+(\lambda + 7)\hat{j}-6\hat{k}$

$\overrightarrow{AD} = Position \ vector \ of \ D - Position \ Vector \ of \ A$

$= (2\hat{i}-5\hat{j}+10\hat{k})-(6\hat{i}-7\hat{j}) = -4\hat{i}+2\hat{j}+10\hat{k}$

The points $A, B, C \ and \ D$ would be co-planar, if $\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}$ are co-planar

i.e. $\begin{bmatrix} \overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD} \end{bmatrix} = 0$

$\Rightarrow$ $\left| \begin{array}{ccc} 10 & -12 & -4 \\ -6 & \lambda+7 & -6 \\ -4 & 2 & 10 \end{array} \right| = 0$

$\Rightarrow 10(10\lambda + 70 + 12) + 12 (-60-24) - 4(-12+4\lambda + 28) = 0$

$\Rightarrow 10(10\lambda + 82) + 12 (-84) - 4(16+4\lambda) = 0$

$\Rightarrow 100 \lambda + 820 -1008-64-16\lambda = 0$

$\Rightarrow 84 \lambda = 252$

$\Rightarrow \lambda =$ $\frac{252}{84}$ $= 3$

$\\$

Question 11:

(a) Show that the lines $\frac{x-4}{1}$ $=$ $\frac{y+3}{-4}$ $=$ $\frac{z+1}{7}$ and $\frac{x-1}{2}$ $=$ $\frac{y+1}{-3}$ $=$ $\frac{z+10}{8}$ intersect. Find the coordinate of their point of intersection.     [5]

(b) Find the equation of the plane passing through the point $(1, -2, 1)$ and perpendicular to the line joining the points $A (3, 2, 1)$ and $B (1, 4, 2)$.     [5]

(a)  Let $\frac{x-4}{1}$ $=$ $\frac{y+3}{-4}$ $=$ $\frac{z+1}{7}$ $= r$ and $\frac{x-1}{2}$ $=$ $\frac{y+1}{-3}$ $=$ $\frac{z+10}{8}$ $= r_1$

The direction ratios of the two lines are $1, -4, 7 \ and \ 2, -3, 8$.

Since $1 : -4 : 7 \neq 2: -3: 8$

Therefore the lines are not parallel.  Hence they will intersect somewhere.

Now, $(r+4, -4r-3, 7r-1)$ and $(2r_1+1, -3r_1-1, 8r_1-10)$ are general coordinates of points on the two lines respectively.

The lines will intersect if the points coincide.

$\Rightarrow r+4 = 2r_1+1, \ \ -4r-3 = -3r_1-1, \ \ 7r-1 = 8r_1-10$

$\Rightarrow r - 2r_1+3=0, \ \ -4r+3r_1-2=0, \ \ 7r-8r_1+9=0$

$4( r - 2r_1+3=0)$

$(+) \underline{-4r+3r_1-2=0}$

$-5r_1+10=0$

$\Rightarrow r_1 = 2$

Hence $r = 2(2)-3 = 1$

Solve the first two equations we get $r =1$ and $r_1 = 2$

Verifying that $r =1$ and $r_1 = 2$ satisfy the third equation: $7(1)-8(2)+9=0$. Hence it satisfies the third equation.

Therefore the point of intersection is $(r+4, -4r-3, 7r-1) = (1+4, -4(1)-3, 7(1)-1) = (5,-7,6)$

If you were to take the other point $(2r_1+1, -3r_1-1, 8r_1-10) = (2(2)+1, -3(2)-1, 8(2)-10) = (5, -7, 6)$ you get the same coordinates.

(b)  Coordinate of the point $A (3, 2, 1)$ and $B (1, 4, 2)$ given.

Therefor the equation of line AB will be

$\frac{x-3}{1-3}$ $=$ $\frac{y-2}{4-2}$ $=$ $\frac{z-1}{2-1}$

$\frac{x-3}{-2}$ $=$ $\frac{y-2}{2}$ $=$ $\frac{z-1}{1}$ $= r$

Therefore the direction ratios of $AB$ is $-2, 2, 1$.

Since the line is $\perp$ to the required plane, the direction ratios of the plane are proportional to $-2, 2, 1$. Hence the equation of the line passing through the point $(1, -2, 1)$ and normal is:

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

$\Rightarrow -2x+2+2y+4+z-1 = 0$

$\Rightarrow -2x + 2y + z = 5$ is the required equation of the plane.

$\\$

Question 12:

(a) A fair die is rolled. If face $1$ turns up, a ball is drawn from $Bag \ A$. If face $2 \ or \ 3$ turns up, a ball is drawn from $Bag \ B$. If face $4 \ or \ 5 \ or \ 6$turns up, a ball is drawn from $Bag \ C$. $Bag \ A$ contains $3 \ red$ and $2 \ white$ balls, $Bag \ B$ contains $3 \ red$ and $4$ white balls and $Bag \ C$ contains $4 \ red$ and $5 \ white$ balls. The die is rolled, a Bag is picked and a ball is drawn. If the drawn ball is red, what is the probability that is it drawn from $Bag \ B$.     [5]

(b) An urn contains $25$ balls of which $10$ balls are red and the remaining are green. A ball is drawn at random from the urn, the color is noted and the ball is replaced. If $6$ balls are drawn in this way, find the probability that:

(i) All the balls are red

(ii) Not more than $2$ balls are green

(iii) Number of red balls and green balls are equal     [5]

(a)  Let Event $(E_1) = Bag \ A$ is chosen

Event $(E_2) = Bag \ B$ is chosen

Event $(E_3) = Bag \ C$ is chosen

and Event $(A) =$ red ball is drawn

Therefore    $P(E_1) =$ $\frac{1}{6}$$P(E_2) =$ $\frac{2}{6}$$P(E_3) =$ $\frac{3}{6}$

Now

Probability of drawing a  red ball from $Bag \ A =$ $P(A/E_1) =$  $\frac{3}{5}$

Probability of drawing a  red ball from $Bag \ B =$ $P(A/E_2) =$  $\frac{3}{7}$

Probability of drawing a  red ball from $Bag \ C =$ $P(A/E_3) =$  $\frac{4}{9}$

Applying Baye’s theorem

$P(A/E_2) =$ $\frac{P(E_2). P(A/E_2)}{P(E_1). P(A/E_1)+P(E_2). P(A/E_2)+P(E_3). P(A/E_3)}$

$= \frac{\frac{2}{6} \times \frac{3}{7}}{\frac{1}{6} \times \frac{3}{5} + \frac{2}{6} \times \frac{3}{7} + \frac{3}{6} \times \frac{4}{9}}$

$= \frac{\frac{1}{7}}{\frac{1}{10} + \frac{1}{7} + \frac{2}{9}}$

$= \frac{90}{293}$

(b)  Probability of drawing a red ball $=$ $\frac{10}{25}$ $=$ $\frac{2}{5}$

Probability of drawing a green ball $=$ $\frac{15}{25}$ $=$ $\frac{3}{5}$

$P(X = r) =$ Probability of drawing $r$ red balls $=$ $^6C_r ($ $\frac{2}{5})^r (\frac{3}{5})^{6-r}$, where $r = 0,1, 2, 3, 4, 5, 6$

(i) Probability of drawing all red balls $=$ $^6C_6 ($ $\frac{2}{5})^6 (\frac{3}{5})^{0}$ $=$ $(\frac{2}{5})^6$ $=$ $\frac{64}{15625}$

(ii)  Probability of drawing not more than 2 green balls i.e. Probability of drawing at least 4 red balls

$= P(X \geq 4)$

$= P(X=4) +P(x=5) +P(x=6)$

$=$ $^6C_4 ($ $\frac{2}{5})^4 (\frac{3}{5})^{2}$ $+$ $^6C_5 ($ $\frac{2}{5})^5 (\frac{3}{5})^{1}$ $+$ $^6C_6 ($ $\frac{2}{5})^6 (\frac{3}{5})^{0}$

$=$ $(\frac{2}{5})^4 \Big[ \frac{6!}{4! \ 2!} \times \frac{3 \times 3}{5 \times 5}$ $+$ $\frac{6!}{5! \ 1!} \times \frac{2 \times 3}{5 \times 5}$ $+$ $\frac{6!}{6! \ 0!} \times \frac{2 \times 2}{5 \times 5} \Big]$

$=$ $(\frac{2}{5})^4 \Big[ \frac{3 \times 3 \times 6 \times 5}{5 \times 5 \times 2}$ $+$ $\frac{2 \times 3 \times 6}{5 \times 5}$ $+$ $\frac{2 \times 2}{5 \times 5} \Big]$

$=$ $(\frac{2}{5})^4 \times \frac{175}{25}$

$=$ $\frac{112}{625}$

(iii) The probability of drawing equal number of red and green balls $= P(X=3)$

$=$ $^6C_3 ($ $\frac{2}{5})^3 (\frac{3}{5})^{3}$

$=$ $\frac{6!}{3! \ 3!} (\frac{2^3 3^3}{5^6})$

$=$ $\frac{4320}{15625}$

$=$ $\frac{864}{3125}$

$\\$

Question 13:

(a) A machine costs $Rs. \ 60000$ and its effective life is estimated to be $25$ years. A sinking fund is to be created for replacing the machine at the end of its life when its scrap value is estimated as $Rs. \ 5000$. The price of the new machine is estimated to be $100\%$ more than the price of the present one. Find the amount that should be set aside at the end of each year, out of the profits, for the sinking fund it is accumulates at an interest of $6\%$ per annum compounded annually.     [5]

(b) A farmer has a supply of chemical fertilizer of $Type \ A$ which contains $10\%$ nitrogen and $6\%$ phosphoric acid and of $Type \ B$ which contains $5\%$ nitrogen and $10\%$ phosphoric acid. After soil test, it is found that at least $7 \ kg$ of nitrogen and same quantity of phosphoric acid is required for a good crop. The fertilizer of $Type \ A$ costs $Rs. 5.00$ per kg and the $Type \ B$ costs $Rs. \ 8.00$ per kg. Using linear programming, find how many kilograms of each fertilizer should be brought to meet the requirement and and for the cost to be minimum. Find the feasible region in the graph.     [5]

(a) Present cost of machine $= 60000 Rs.$

Cost of Machine after $25$ years $= 120000 Rs.$

The scrap value of the machine $= 5000 Rs.$

Therefore we have to accumulate $115000 Rs.$ to replace the machine after $25$ years.

Rate of interest on the amount set aside $= 6\%$

Let the amount set aside after the end of every year $= A$

Therefore

$A\Big(1+\frac{6}{100} \Big)^{24} + A\Big(1+\frac{6}{100} \Big)^{23} + A\Big(1+\frac{6}{100} \Big)^{22} + ... \ ... + A\Big(1+\frac{6}{100} \Big)^{1} + A = 115000$

$A \Big($ $\frac{1.06^{25} - 1}{1.06 -1}$ $\Big) = 115000$

$\Rightarrow A =$ $\frac{115000}{54.8645}$ $= 2096.07 \ Rs.$

(b)  Let quantity of fertilizer $Type \ A = x$ and Quantity of fertilizer $Type \ B = y$

We know, $Type \ A$ which contains $10\%$ nitrogen and $Type \ B$ which contains $5\%$ nitrogen.

Therefore: $\frac{10}{100}$ $x +$ $\frac{5}{100}$ $y \geq 7 \Rightarrow 2x + y \geq 140$

Similarly, $Type \ A$ which contains $6\%$ phosphoric acid and $Type \ B$ which contains $10\%$ phosphoric acid.

Therefore:  $\frac{6}{100}$ $x +$ $\frac{10}{100}$ $y \geq 7 \Rightarrow 3x+5y \geq 350$

Given, The fertilizer of $Type \ A$ costs $Rs. 5.00$ per kg and the $Type \ B$ costs $Rs. \ 8.00$ per kg.

We also know $x \geq 0, y \geq 0$

Therefore cost of $x$ kg of $Type \ A$ and $y$ kg of $Type \ B$ fertilizer $Cost = (5x+8y)$

We need to minimize $(5x+8y)$. We will solve it graphically.

We see that the coordinates of the vertices of the feasible region are $A(0, 140), B(50, 40) \ and \ C($ $\frac{350}{3}$ $, 0)$. Now calculate Cost for each of the three coordinates:

For $A(0, 140): Cost = 5 \times 0 + 8 \times 140 = 1120 \ Rs.$

For $B(50, 40): Cost = 5 \times 50 + 8 \times 40 = 570 \ Rs.$

For $C($ $\frac{350}{3}$ $, 0) : Cost = 5 \times$ $\frac{350}{3}$ $+$ $8 \times 0 =$ $\frac{1750}{3}$ $= 583.33 \ Rs.$

Hence cost will be minimum when $x = 50 \ kg$ and $y = 40 \ kg$.

$\\$

Question 14:

(a) The demand for a certain product is represented by the equation $p = 500 + 25x -$ $\frac{x^2}{3}$ in Rupees where $x$ is the number of units and $p$ is the price per unit. Find:

(i) Marginal revenue function

(ii)  The marginal revenue when $10$ units are sold.     [5]

(b) The bill of $Rs. \ 60000$ payable $10$ months after date was discounted for $Rs. \ 57300$ on 30th June 2007. If the rate of interest was $11$ $\frac{1}{4}$ $\%$ per annum, on what date was the bill drawn?     [5]

(a)  Given $p = 500 + 25x -$ $\frac{x^2}{3}$

(i) Total Revenue function $(R) = px$

$R = 500x + 25x^2 -$ $\frac{x^3}{3}$

Therefore Marginal Revenue function $(MR) =$ $\frac{dR}{dx}$ $= 500 + 50x -x^2$

(ii) Marginal Revenue $(MR)$ when $x = 10$,

$MR = 500 + 50(10)-10^2 = 1000-10 = 990$

$(MR)_{x=10} = 990$

(b) Face Value $= 60000 Rs.$

Discounted Value $= 57300 Rs.$

Rate $= 11.25\%$

Discount $= 60000 - 57300 = 2700 Rs.$

Therefore $2700 = 60000 \times$ $\frac{11.25}{100}$ $\times \frac{t}{12}$

$\Rightarrow t = 4.8$ months

Therefore the date is $4.8$ months before 30th June which is 6th Feb 2007.

$\\$

Question 15:

(a) The price relative and weights of a set of commodities are given below:

 Commodity A B C D Price Relatives 125 120 127 119 Weights $x$ $2x$ $y$ $y+3$

If the sum of the weights is $40$ and weighted average of price relative index number is $122$, find the numerical value of $x$ and $y$    [5]

(b) Construct three yearly moving averages from the following data and show on a graph against the original data:     [5]

 Year 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 Annual Sales in Lakhs 18 22 20 26 30 22 24 28 32 35

(a) Given: $x + 2x + y + y + 3 = 40$

$\Rightarrow 3x+2y=37$ … … … … … (i)

Also $\frac{125x+240x+127y+119y+357}{3x+2y+3}$ $= 122$

$\Rightarrow 365x+246y+357 = 366x + 244y + 366$

$\Rightarrow -x+2y = 9$ … … … … … (ii)

Subtracting (ii) from (i) we get

$4x = 28 \Rightarrow x = 7$

Substituting in (i) we get $2y = 37 - 21 \Rightarrow y = 8$

(b)

 Year Annual Sales 3 year moving total 3 year moving average 2000 18 – – 2001 22 60 20.00 2002 20 68 22.67 2003 26 76 25.33 2004 30 78 26.00 2005 22 76 25.33 2006 24 74 24.67 2007 28 84 28.00 2008 32 95 31.67 2009 35 – –

$\\$