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) Find the matrix $X$ for which:

$\begin{bmatrix} 5 & 4 \\ 1 & 1 \end{bmatrix} \ X = \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}$

(ii) Solve for $x$, if : $tan \ (cos^{-1} \ x) =$ $\frac{2}{\sqrt{5}}$

(iii) Prove that the line $2x-3x=9$ touches the conic $y^2 = -8x$. Also, find the point contact.

(iv) Using L’Hospitals rule, evaluate: $\lim \limits_{x \to 0}$ $(\frac{1}{x^2} - \frac{cot \ x}{x})$

(v) Evaluate: $\int \limits_{}{} tan^3 \ x \ dx$

(vi) Using properties of definite integrals, evaluate: $\int \limits_{0}^{\frac{\pi}{2}}$ $\frac{sin \ x - cos \ x}{1 + sin \ x \ cos \ x}$ $dx$

(vii) The two lines of regression are $x+2y-5=0$ and $2x+3y-8=0$ and the variance of $x \ is \ 12$. Find the variance of y and the coefficient of correlation.

(viii) Express $\frac{2+i}{(1+i)(1-2i)}$ in the form of $a+ib$. Find its modulus and argument.

(ix) A pair of dice is thrown. What is the probability of getting an even number on the first die or a total of $8$?

(x) Solve the differential equation: $x$ $\frac{dy}{dx}$ $+ y = 3x^2-2$

(i)  $\begin{bmatrix} 5 & 4 \\ 1 & 1 \end{bmatrix} \ X = \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}$

$\Rightarrow X = \begin{bmatrix} 5 & 4 \\ 1 & 1 \end{bmatrix}^{-1} \times \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}$

$\Rightarrow X = \begin{bmatrix} 1 & -4 \\ -1 & 5 \end{bmatrix} \times \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix}$

$\Rightarrow X = \begin{bmatrix} -3 & -14 \\ 4 & 7 \end{bmatrix}$

(ii) We know, $cos^{-1}x = tan^{-1}$ $(\frac{\sqrt{1-x^2}}{x})$

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

$\Rightarrow$ $tan \ (tan^{-1}$ $(\frac{\sqrt{1-x^2}}{x})$ $) =$ $\frac{2}{\sqrt{5}}$

Squaring on both sides

$\frac{1-x^2}{x^2}$ $=$ $\frac{4}{5}$

$\Rightarrow 5-5x^2 = 4x^2$

$\Rightarrow 9x^2=5$

$\Rightarrow x =$ $\frac{\sqrt{5}}{3}$

(iii) Line $3y = 2x-9$

$y =$ $\frac{2}{3}$ $x - 3$

$m =$ $\frac{2}{3}$ $, c = -3$

$y^2 = -8x$

$\Rightarrow a = -2$

The condition $a = mc$

$\Rightarrow -2 =$ $\frac{2}{3}$ $\times (-3)$

$\Rightarrow -2 = -2$. Therefore the lines touches the parabola

$(\frac{2x}{3}$ $-3$ $)^2$ $= -8x$

$(2x-9)^2 = -72x$

$\Rightarrow 4x^2 - 36x +81 = -72x$

$\Rightarrow 4x^2 +36x +81 = 0$

$\Rightarrow (2x+9)^2 = 0$

$\Rightarrow x = -$ $\frac{9}{2}$

and $y =$ $\frac{2}{3}$ $x - 3 \Rightarrow y =$ $\frac{2}{3} (-\frac{3}{2})$ $- 3 = -6$

Therefore the point of contact is $( -$ $\frac{9}{2}$ $, -6)$

(iv)  $\lim \limits_{x \to 0}$ $(\frac{1}{x^2} - \frac{cot \ x}{x})$

$= \lim \limits_{x \to 0}$ $\frac{sin \ x - x \ cos \ x}{x^2 sin \ x}$

$= \lim \limits_{x \to 0}$ $\frac{sin \ x - x \ cos \ x}{x^3 }$ $\lim \limits_{x \to 0}$ $\frac{x}{sin \ x}$

$= \lim \limits_{x \to 0}$ $\frac{sin \ x - x \ cos \ x}{x^3 }$ $. 1$

Since $\Big[$ $\lim \limits_{x \to 0}$ $\frac{x}{sin \ x}$ $= 1 \Big]$

$= \lim \limits_{x \to 0}$ $\frac{cos \ x - (cos \ x - x sin \ x)}{3x^2 }$

$\Big[$ $\frac{0}{0}$ form, applying L’Hospital’s Rule $\Big]$

$= \lim \limits_{x \to 0}$ $\frac{x \ sin \ x}{3x^2}$

$= \lim \limits_{x \to 0}$ $\frac{sin \ x}{3x}$

$\Big[$ $\frac{0}{0}$ form, applying L’Hospital’s Rule $\Big]$

$= \lim \limits_{x \to 0}$ $\frac{cos \ x}{3}$

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

(v)  $\int \limits_{}{} tan^3 \ x \ dx$

$=$ $\int \limits_{}{} tan \ x . tan^2 \ x \ dx$

$=$ $\int \limits_{}{} tan \ x . (sec^2 \ x -1) \ dx$

$=$ $\int \limits_{}{} tan \ x . sec^2 \ dx$ $-$ $\int \limits_{}{} tan \ x \ dx$

In the first integral, put $tan \ x = u$

$\frac{du}{dx}$ $= sec^2 x$

$=$ $\int \limits_{}{} u .du$ $-$ $\int \limits_{}{} tan \ x \ dx$

$=$ $\frac{u^2}{2}$ $- log |sec \ x| + c$

$=$ $\frac{tan^2 \ x}{2}$ $- log |sec \ x| + c$

(vi)  Let $I =$ $\int \limits_{0}^{\frac{\pi}{2}}$ $\frac{sin \ x - cos \ x}{1 + sin \ x \ cos \ x}$ $dx$

$I =$ $\int \limits_{0}^{\frac{\pi}{2}}$ $\frac{sin \ x}{1 + sin \ x \ cos \ x}$ $dx$ $-$ $\int \limits_{0}^{\frac{\pi}{2}}$ $\frac{sin \ x}{1 + sin \ x \ cos \ x}$ $dx$

Now, $a + b -x =$ $\frac{\pi}{2}$. Using the property

$\int \limits_{0}^{\frac{\pi}{2}}$ $f(x) \ dx =$ $\int \limits_{0}^{\frac{\pi}{2}}$ $f(a+b-x) \ dx$ in the first integral

$I =$ $\int \limits_{0}^{\frac{\pi}{2}}$ $\frac{sin \ (\frac{\pi}{2} - x)}{1 + sin \ (\frac{\pi}{2} - x) \ cos \ (\frac{\pi}{2} - x)}$ $dx$ $-$ $\int \limits_{0}^{\frac{\pi}{2}}$ $\frac{cos \ x}{1 + sin \ x \ cos \ x}$ $dx$

$I =$ $\int \limits_{0}^{\frac{\pi}{2}}$ $\frac{cos \ x}{1 + sin \ x \ cos \ x}$ $dx$ $-$ $\int \limits_{0}^{\frac{\pi}{2}}$ $\frac{cos \ x}{1 + sin \ x \ cos \ x}$ $dx$

$I = 0$

(vii) Two lines of regression are:

$x+2y-5=0$

$\Rightarrow y =$ $\frac{-x}{2}$ $+$ $\frac{5}{2}$

$\Rightarrow y = -0.5x + 2.5$ (regression line of $y \ on \ x$)

Therefore $b_{yx} = r$ $\frac{\sigma_y}{\sigma_x}$ $= -0.5$

$2x+3y-8=0$

$\Rightarrow x =$ $\frac{-3y}{2}$ $+$ $\frac{8}{2}$

$\Rightarrow x = -1.5y+4$ (regression line of $x \ on \ y$)

Therefore $b_{xy} = r$ $\frac{\sigma_x}{\sigma_y}$ $= -1.5$

Therefore $r^2 = b_{yx} . b _{xy} = (-0.5).(-1.5) = 0.75$

$\Rightarrow r = \sqrt{0.75} = \pm$ $\frac{\sqrt{3}}{2}$

Since both  $b_{yx}$  and  $b _{xy}$ are negative, $r$ is also negative. Hence Correlation Coefficient $(r) = -$ $\frac{\sqrt{3}}{2}$

Variance of $x: {\sigma^2}_x = 12$

$\Rightarrow \sigma_x = \sqrt{12} = 2\sqrt{3}$

Therefore $r$ $\frac{\sigma_x}{\sigma_y}$ $= -0.5$

$\Rightarrow -$ $\frac{\sqrt{3}}{2} . \frac{\sigma_y}{2\sqrt{3}}$ $= -0.5$

$\Rightarrow \sigma_y =$ $\frac{0.5 \times 2 \sqrt{3}}{\frac{\sqrt{3}}{2}}$ $= 2$

Therefore Variance of $y; {\sigma}^2_y = 4$

(viii) $\frac{2+i}{(1+i)(1-2i)}$

$=$ $\frac{2+i}{1+i-2i-i^2}$

$=$ $\frac{2+i}{3-i}$ $\times$ $\frac{3+i}{3+i}$

$=$ $\frac{6+3i+2i+i^2}{9-i^2}$

$=$ $\frac{5+5i}{10}$

$=$ $\frac{1+i}{2}$

$|z| =$ $\sqrt{\frac{1}{4}+\frac{1}{4}} = \sqrt{\frac{1}{2}}$

Argument of $z = tan^{-1} \Big[ \frac{\frac{1}{2}}{\frac{1}{2}} \Big] = tan^{-1} 1$

$\Rightarrow z =$ $\frac{\pi}{4}$

(ix) Total number of possible outcomes $n(S) = 36$

Even $A$: Getting a total of $8$

Favorable outcomes $= (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$

Therefore $n(A) = 5$

Hence $P(A) =$ $\frac{n(A)}{n(S)}$ $=$ $\frac{5}{36}$

Event $B$: Getting even number on first die

Favorable outcomes $= (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(4, 1), \\ (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$

Therefore $n(B) = 18$

Hence $P(B)$ $=$ $\frac{n(B)}{n(S)}$ $=$ $\frac{18}{36}$ $=$ $\frac{1}{2}$

Therefore $P(A \cap B) =$ $\frac{n(A \cap B)}{n(S)}$ $=$ $\frac{3}{36}$ $=$ $\frac{1}{12}$

Hence $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$= \frac{5}{36}$ $+$ $\frac{1}{2}$ $-$ $\frac{1}{12}$ $=$ $\frac{5+18-3}{36} = \frac{5}{9}$

(x)  $x$ $\frac{dy}{dx}$ $+ y = 3x^2-2$

$\frac{dy}{dx}$ $+$ $\frac{1}{x}$ $.$ $y = 3x -$ $\frac{2}{x}$

$e^{\int \limits_{}{} \frac{1}{x} dx }$ $= e^{log \ x} = x$

Therefore  $x.$ $\frac{dy}{dx}$ $+$ $\frac{y}{x}$ $. x = (3x -$ $\frac{2}{x}$ $).x$

$x.$ $\frac{dy}{dx}$ $+ y = 3x^2 -2$

Integrating

$xy = \int \limits_{}{} (3x^2 -2) \ dx +c$

$\Rightarrow xy = x^3-2x+c$

$\\$

Question 2:

(a) Using properties of determinants, prove that:

$A = \left| \begin{array}{ccc} b+c & a & a \\ b & a+c & b \\ c & c & a+b \end{array} \right|= 4abc$    [5]

(b) Solve the following system of linear equations using matrix method:

$3x+y+z = 1, \ \ 2x+2z=0, \ \ 5x+y+2z=2$    [5]

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

Operating: $R_1 \rightarrow R_1+R_2+R_3$

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

Now applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$

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

Expanding along $R_1$, we will get

$\Delta = 2 \Big[ (b+c)(-a)^2 - (a+c)(0-ab) + (a+b)(ac) \Big]$

$= 2 (-a^2b-a^2c+a^2b+abc+a^2c + abc )$

$= 4abc$. Hence proved.

(b)  We have to write the given equations in the form $AX = B$

$A = \begin{bmatrix} 3 & 1 & 1 \\ 2 & 0 & 2 \\ 5 & 1 & 2 \end{bmatrix}$

$|A| = \left| \begin{array}{ccc} 3 & 1 & 1 \\ 2 & 0 & 2 \\ 5 & 1 & 2 \end{array} \right|= 3(0-2) -1(4-10) +1(2-0) = 2$ This shows that A is a non-singular matrix. Therefore the system has a unique solution defined by $X = A^{-1} B$

$A^{-1} = \frac{1}{2} \begin{bmatrix} -2 & -1 & 2 \\ 6 & 1 & -4 \\ 2 & 2 & -2 \end{bmatrix}$

$X = A^{-1} B = \frac{1}{2} \begin{bmatrix} -2 & -1 & 2 \\ 6 & 1 & -4 \\ 2 & 2 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}$

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

Hence $x = 1, y = -1, \ and \ z = -1$

$\\$

Question 3:

(a) If $sin^{-1} \ x + tan^{-1} \ x =$ $\frac{\pi}{2}$, prove that: $2x^2 + 1 = \sqrt{5}$     [5]

(b) Write the Boolean function corresponding to the switching circuit given below:

$A, \ B \ and \ C$ represent switches in ‘on’position and $A', \ B' \ and \ C'$ represent them in ‘off’ position. Using Boolean algebra, simplify the function and construct an equivalent circuit.     [5]

(a)  Given $sin^{-1} \ x + tan^{-1} \ x =$ $\frac{\pi}{2}$

$\Rightarrow tan^{-1} \ x =$ $\frac{\pi}{2}$ $- sin^{-1} \ x$

$\Rightarrow x = tan \Big($ $\frac{\pi}{2}$ $- sin^{-1}x \Big)$

$\Rightarrow x = cot \ (sin^{-1}x)$

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

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

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

$\Rightarrow x^4 = 1-x^2$

$\Rightarrow x^4+x^2-1 = 0$

Therefore $x =$ $\frac{-1 \pm \sqrt{5}}{2}$. But $x^2$ cannot be negative, hence $x^2 =$ $\frac{-1 + \sqrt{5}}{2}$ or $2x^2 + 1 = \sqrt{5}$. Hence proved.

(b)  $F(A, B, C) = A (A'+B) + A'B + (A+B')C$

$= AA' + AB + A'B + AC + B'C$

$= O B(A+A') +AC + B'C$

$= B + B'C + AC$

$= (B+B')(B+C)+AC$

$= B +C + AC$

$= B + C$

$\\$

Question 4:

(a) Verify the conditions of Rolle’s Theorem for the following function:

$f(x) = \log (x^2 + 2) - \log 3 \ on \ [ -1, 1 ]$

Find a point in the interval, where the tangent to the curve is parallel to $x-axis$.     [5]

(b) Find the equation of a standard ellipse, taking its axes as the coordinate axes, whose minor axis is equal to the distance between the foci and whose lengths of latus rectum is $10$. Also find its eccentricity.     [5]

(a)   Since $x^2 + 2 > 0 \ for \ x \in [-1, 1]$, therefore the function $f(x) = \log (x^2 + 2) - \log 3$ is continuous in $[-1, 1]$

$f'(x) =$ $\frac{1}{x^2+2}$ $. 2x =$ $\frac{2x}{x^2+2}$ which exists for all  $x$. Therefore the function is differentiable in $[-1, 1]$

$f(-1) = \log (1+2) - \log 3 = 0$

$f(1) = \log (1+2) - \log 3 = 0$

Therefore $f(-1) = f(1)$

Thus $f(x)$ satisfies all the conditions of Rolle’s Theorem.

Therefore there must exists at least one value of $x$ say $c$ in the open interval $(-1, 1)$ such that $f'(c) = 0$

$\Rightarrow f'(c) =$ $\frac{2c}{c^2+2}$ $= 0 \Rightarrow c = 0 \in (-1, 1)$

Hence Rolle’s Theorem is verified.

When $c = 0, f(0) = \log 2 - \log 3 = \log ($ $\frac{2}{3}$ $)$

So, there exist a point $(0, \log ($ $\frac{2}{3}$ $) )$ on the given curve $f(x) = \log (x^2 + 2) - log \ 3$ where the tangent to the curve is parallel to $x-axis$

(b)  Let the equation of ellipse be $\frac{x^2}{a^2}$ $+$ $\frac{y^2}{b^2}$ $= 1$

$a > b$

Given $2b = 2ae \Rightarrow b = ae$ … … … … … (i)

and $\frac{2b^2}{a} = 10 \Rightarrow b^2 = 5a$ … … … … … (ii)

We also know that $b^2 = a^2(1-e^2)$ … … … … … (iii)

Substituting (i) and (ii) into (iii)

$5a = a^2 - b^2$

$\Rightarrow 5a = a^2 - 5a$

$\Rightarrow 10 a = 10a^2$

$\Rightarrow a = 10 \ and \ b^2 = 50$

$a^2 = 100$

Therefore the equation of the ellipse: $\frac{x^2}{100}$ $+$ $\frac{y^2}{50}$ $= 1 \ or \ x^2 + 2y^2 = 100$

Also $b^2 = a^2(1-e^2)$

$50 = 100(1-e^2)$

$\Rightarrow e^2 =$ $\frac{1}{2}$

$\Rightarrow e =$ $\frac{1}{\sqrt{2}}$

$\\$

Question 5:

(a) If $log \ y = tan^{-1} \ x$, prove that:

$(1+x^2)$ $\frac{d^2y}{dx^2}$ $+ (2x-1)$ $\frac{dy}{dx}$ $= 0$     [5]

(b) A rectangle is inscribed in a semicircle of radius $r$ with one of its sides on the diameter of the semicircle. Find the dimensions of the rectangle to get maximum area. Also, find the maximum are.      [5]

(a)  Given: $log \ y = tan^{-1} \ x$

Differentiating both sides with respect to $x$

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

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

Differentiating both sides with respect to $x$

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

$(1+x^2)$ $\frac{d^2y}{dx^2}$ $+(2x-1)$ $\frac{dy}{dx}$ $= 0$  Hence proved.

(b)   Please refer to the figure on the right.

Consider the rectangle $PQRS:$

$OQ = y = r \ cos \ \theta$

$QR = x = r \ sin \ \theta$

Therefore the area of the rectangle $= PQ \times QR = 2r \ cos \ \theta . r \ sin \ \theta = r^2 \ sin\ 2\theta$

Differentiating both sides with respect to $\theta$

$\frac{dA}{d \theta}$ $= 2r^2 \ cos \ 2\theta$

For maximizing the area of $PQRS,$ $\frac{dA}{d\theta}$ $= 0$

$\Rightarrow 2r^2 \ cos \ 2\theta = 0 \Rightarrow \theta =$ $\frac{\pi}{4}$

Hence the area is maximum when $\theta =$ $\frac{\pi}{4}$

$\Rightarrow x = r \ sin \$ $\frac{\pi}{4}$ $=$ $\frac{r}{\sqrt{2}}$

and $y = r \ cos \$ $\frac{\pi}{4}$ $= r\sqrt{2}$

Hence sides are $r\sqrt{2},$ $\frac{r}{\sqrt{2}}$

Therefore the area of $PQRS =$ $\frac{r}{\sqrt{2}}$ $\times 2\sqrt{2} = r^2$ units.

$\\$

Question 6:

(a) Evaluate: $\int \limits_{}{}$ $\frac{sin \ x + cos \ x}{\sqrt{9 + 16 \ sin \ 2x}}$ $\ dx$     [5]

(b) Find the area of the region bounded by the curve $y = 6x-x^2$ and $y = x^2-2x$     [5]

(a) Let $I =$ $\int \limits_{}{}$ $\frac{sin \ x + cos \ x}{\sqrt{9 + 16 \ sin \ 2x}}$ $\ dx$

$=$ $\int \limits_{}{}$ $\frac{sin \ x + cos \ x}{\sqrt{9 + 16 \ sin \ 2x + 16 - 16}}$ $\ dx$

$=$ $\int \limits_{}{}$ $\frac{sin \ x + cos \ x}{ \sqrt{25 - 16(1-\ sin \ 2x) }}$ $\ dx$

$=$ $\int \limits_{}{}$ $\frac{sin \ x + cos \ x}{\sqrt{25 - 16 ( sin^2 x + cos^2 x - \ sin \ 2x)}}$ $\ dx$

$=$ $\int \limits_{}{}$ $\frac{sin \ x + cos \ x}{\sqrt{25 - 16 ( sin x - cos x)^2}}$ $\ dx$

Now let $(sin \ x - cox \ x) = u$. Differentiating both sides with respect to $u$

$\Rightarrow (cos \ x + sin \ x) dx = du$

$\Rightarrow I =$ $\int \limits_{}{}$ $\frac{du}{(5)^2 - (4u)^2}$

$=$ $\frac{1}{4}$ $sin^{-1}$ $( \frac{4u}{5} )$ $+ c$

$=$ $\frac{1}{4}$ $sin^{-1}$ $\frac{4(sin \ x - cox \ x)}{5}$ $+ c$

(b)  Given: $y = 6x-x^2 \Rightarrow y = -(3-x)^2 +9$. This represents a parabola with vertex at (3, 9) and it opens downwards

Similarly for $y = x^2-2x \Rightarrow y = (x-1)^2-1$. This represents a parabola with vertex $(1, -1)$  and it opens upwards.

Both the curves passes through origin $(0,0)$. Solving the two equation:

$x^2-2x=6x-x^2 \Rightarrow x-2 = 6-x \Rightarrow x = 4. \therefore y = 2x = 8$

Hence the integration point is $(4, 8)$

Therefore the enclosed area is:

$=$ $\int \limits_{0}^{4}$ $(6x^2-x^2) \ dx -$ $\int \limits_{0}^{4}$ $(x^2 - 2x) \ dx$

$=$ $\int \limits_{0}^{4}$ $(6x^2-x^2 - x^2 + 2x) dx$

$=$ $\int \limits_{0}^{4}$ $(8x-2x^2) dx$

$=$ ${{\Big[ \frac{8x^2}{2} - \frac{2x^3}{3} \Big]}_0}^4$

$= (64-$ $\frac{128}{3}$ $) =$ $\frac{64}{3}$ $\ sq. \ units$

$\\$

Question 7:

(a) Calculate Carl Pearson’s coefficient of correlation between $x \ and \ y$ for the following data and interpret the result:

$(1,6), (2, 5), (3, 7), (4,9), (5,8), (6,10), (7, 11), (8,13), (9,12)$     [5]

(b) The marks obtained by 10 students in English and Mathematics are given below:

 Marks in English 20 13 18 21 11 12 17 14 19 15 Marks in Mathematics 17 12 23 25 14 8 19 21 22 19

Estimate the probable score for Mathematics of the marks obtained in English are 24.     [5]

(a)

 $x$ $y$ $xy$ $x^2$ $y^2$ 1 6 6 1 36 2 5 10 4 25 3 7 21 9 49 4 9 36 16 81 5 8 40 25 64 6 10 60 36 100 7 11 77 49 121 8 13 104 64 169 9 12 108 81 144 $\Sigma x = 45$ $\Sigma y = 81$ $\Sigma xy = 462$ $\Sigma x^2 = 285$ $\Sigma y^2 = 789$

$r =$ $\frac{n \Sigma xy - \Sigma x . \Sigma y}{\sqrt{ \Big( n \Sigma x^2 - (\Sigma x)^2 \Big) . \Big( n \Sigma y^2 - (\Sigma y)^2 \Big) }}$

$=$ $\frac{9 \times 462 - 45 \times 81}{\sqrt{\Big( 9 \times 285 - (45)^2 \Big) \Big( 9 \times 789 - (81)^2 \Big) }}$

$=$ $\frac{513}{\sqrt{540 \times 540}}$ $= 0.95$

Hence there is high correlation between $x$ and $y$.

(b)

 $x$(English) $y$(Math) $dx = x-16$ $dy = y-18$ $dx.dy$ $(dx)^2$ $(dy)^2$ 20 17 4 -1 -4 16 1 13 12 -3 -6 18 9 36 18 23 2 5 10 4 25 21 25 5 7 35 25 49 11 14 -5 -4 20 25 16 12 8 -4 -10 40 16 100 17 19 1 1 1 1 1 14 21 -2 3 -6 4 9 19 22 3 4 12 9 16 15 19 -1 1 -1 1 1 $\Sigma x \\ = 160$ $\Sigma y \\ = 180$ $\Sigma dx.dy \\ = 125$ $\Sigma(dx)^2 \\ = 100$ $\Sigma (dy)^2 \\ = 254$

$\overline{x} =$ $\frac{160}{10}$ $=16$

$\overline{y} =$ $\frac{180}{10}$ $= 18$

Therefore equation of line: $y - \overline{y} =$ $\frac{\Sigma dx.dy}{\Sigma(dx)^2}$ $(x- \overline{x})$

$\Rightarrow y - 18 =$ $\frac{125}{110}$ $(x-16)$

When $x = 24$ we get $y - 18 =$ $\frac{125}{110}$ $(24-16)$

$\Rightarrow 22y -396 = 200$

$\Rightarrow y =$ $\frac{596}{22}$ $= 27.09 \ or \ 27$ marks

$\\$

Question 8:

(a) A committee of $4$ person has to be chosen from $8$ boys and $6$ girls, consisting at least of one girl. Find the probability that the committee consists of more girls than boys.      [5]

(b) An urn contains $10$ white and $3$ black balls while another urn contains $3$ white and $5$ black balls. Two balls are drawn from the first urn and put into the second urn and then a ball is drawn from the second urn. Find the probability that the ball drawn from the second urn is a white ball.    [5]

(a) The various possibilities of forming a committee:

 Boys Girls Committee Size 4 0 4 3 1 4 2 2 4 1 3 4 0 4 4

All possible combinations where a committee can be formed with at least one girl: $n(s) = ^14C_4 - ^8C_4 \times ^6C_0 = 1001 - 70 = 931$

Number of committees where the number of girls are more than boys $n(A) = ^6C_4 \times ^8C_0 + ^6C_3 \times ^8C_1 = 15+160 = 175$

Hence the $P(E) =$ $\frac{n(A)}{n(s)}$ $=$ $\frac{175}{931}$ $= 0.188$

(b)  When one transfers 2 balls from first urn to another, three possible scenarios can happen:

1. Both the balls transferred can be white.
2. Both the balls transferred can be black.
3. One of the ball is white, while the other is black.

P (transferring 2 white balls & drawing 1 white ball) $=$ $\frac{^{10}C_2 \times ^5C_1}{^{13}C_2 \times ^{10}C_1}$ $=$ $\frac{45}{156}$

P (transferring 2 black balls & drawing 1 white ball) $=$ $\frac{^3C_2 \times ^3C_1}{^{13}C_2 \times ^{10}C_1}$ $=$ $\frac{9}{780}$

P (transferring 2 white balls & drawing 1 white ball) $=$ $\frac{^{10}C_1 \times ^3C_1 \times ^4C_1}{^{13}C_2 \times ^{10}C_1}$ $=$ $\frac{24}{156}$

Hence the probability is $=$ $\frac{45}{156} + \frac{9}{780} + \frac{24}{156}$ $=$ $\frac{1}{156}$ $(45+$ $\frac{9}{5}$ $+ 24)$ $=$ $\frac{59}{130}$

$\\$

Question 9:

(a) Find the locus of a complex number, $z=x+iy$, satisfying $|\frac{z-3i}{z+3i} |$ $\leq \sqrt{2}$.

Illustrate the locus of $z$ in the Argand plane.     [5]

(b) Solve the following differential equation:

$x^2 \ dy + (xy+y^2) \ dx = 0$, when $x = 1 \ and \ y = 1$   [5]

(a)  $z=x+iy$,

$|\frac{z-3i}{z+3i} |$ $\leq \sqrt{2}$

$\Rightarrow \sqrt{x^2+(y-3)^2} \leq \sqrt{2} \sqrt{x^2+(y+3)^2}$

Squaring $(x^2+y^2+9-6y) \leq 2(x^2+y^2+9+6y)$

$\Rightarrow x^2+y^2 + 18 y + 9 \geq 0$

$\Rightarrow x^2 + (y+9)^2-81+9 \geq 0$

$\Rightarrow x^2 +(y+9)^2 \geq 72$

The locus is the region outside the circle with center $(0, -9)$  and radius $\sqrt{72}$

(b)  $x^2 \ dy + (xy+y^2) \ dx = 0$

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

Put $y = vx \Rightarrow$ $\frac{dy}{dx}$ $= v + x$ $\frac{dv}{dx}$

Therefore $v + x$ $\frac{dv}{dx}$ $=$ $\frac{-vx(x+vx)}{x^2}$

$x$ $\frac{dv}{dx}$ $= -v(v+2)$

$\Rightarrow$ $\frac{dv}{v(v+2)}$ $= -$ $\frac{1}{x}$ $dx$

Integrating both sides we get

$\int \limits_{}{}$ $\frac{dv}{v(v+2)}$ $= -$ $\int \limits_{}{}$ $\frac{1}{x}$ $dx$

$\int \limits_{}{}$ $\frac{1}{2} \Big( \frac{1}{v} - \frac{1}{v+2} \Big)$ $= -$ $\int \limits_{}{}$ $\frac{1}{x}$ $dx$

$\Rightarrow$ $\frac{1}{2}$ $log \ |v| - \frac{1}{2} log \ |v+2| = -log \ |x|+log \ c$

$\Rightarrow log \$ $|\frac{x\sqrt{v}}{\sqrt{v+2}}|$ $= log \ c$

$\Rightarrow$ $\frac{x\sqrt{v}}{\sqrt{v+2}}$ $= c$

substituting $v=$ $\frac{y}{x}$

$\Rightarrow$ $\frac{x\sqrt{y}}{\sqrt{y+2x}}$ $= c$

$\Rightarrow x^2y = c^2(y+2x)$

For $x = 1$ and $y = 1, c^2 =$ $\frac{1}{3}$. Substituting back we get

$x^2y =$ $\frac{1}{3}$ $(y+2x) \Rightarrow 3x^2y = y + 2x$

$\\$

Section – B (20 Marks)

Question 10:

(a) For any three vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$, show that $\overrightarrow{a} - \overrightarrow{b}, \overrightarrow{b}- \overrightarrow{c}, \overrightarrow{c} - \overrightarrow{a}$ are co-planar.   [5]

(b) For a unit vector perpendicular to each of the vectors $(\overrightarrow{a}+\overrightarrow{b})$ and $(\overrightarrow{a} - \overrightarrow{b})$ where $\overrightarrow{a} = 3 \hat{i} + 2 \hat{j} + 2\hat{k}$ and $\overrightarrow{b} = \hat{i} + 2\hat{j} -2\hat{k}$   [5]

(a)  $[ \overrightarrow{a} - \overrightarrow{b}, \overrightarrow{b}- \overrightarrow{c}, \overrightarrow{c} - \overrightarrow{a} ]$

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

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

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

$= [ \overrightarrow{a} \ \overrightarrow{b} \ \overrightarrow{c} ] - [ \overrightarrow{a} \ \overrightarrow{b} \ \overrightarrow{a} ] + [ \overrightarrow{c} \ \overrightarrow{a} \ \overrightarrow{c} ] - [ \overrightarrow{c} \ \overrightarrow{a} \ \overrightarrow{a} ] + [ \overrightarrow{b} \ \overrightarrow{c} \ \overrightarrow{c} ] - [ \overrightarrow{b} \ \overrightarrow{c} \ \overrightarrow{a} ]$

$= [ \overrightarrow{a} \ \overrightarrow{b} \ \overrightarrow{c} ] - [ \overrightarrow{b} \ \overrightarrow{c} \ \overrightarrow{a} ]$

$= [ \overrightarrow{a} \ \overrightarrow{b} \ \overrightarrow{c} ] - [ \overrightarrow{a} \ \overrightarrow{b} \ \overrightarrow{c} ]$

$= 0$

$\Rightarrow \overrightarrow{a} - \overrightarrow{b}, \overrightarrow{b}- \overrightarrow{c}, \overrightarrow{c} - \overrightarrow{a}$ are co-planar

(b)  Given $\overrightarrow{a} = 3 \hat{i} + 2 \hat{j} + 2\hat{k}$ and $\overrightarrow{b} = \hat{i} + 2\hat{j} -2\hat{k}$

Therefore $\overrightarrow{a} + \overrightarrow{b} = 3 \hat{i} + 2 \hat{j} + 2\hat{k} + \hat{i} + 2\hat{j} - 2\hat{k} = 4\hat{i} + 4\hat{j} + 0\hat{k}$

$\overrightarrow{a} - \overrightarrow{b} = 3 \hat{i} + 2 \hat{j} + 2\hat{k} - \hat{i} - 2\hat{j} + 2\hat{k}= 2\hat{i} + 0\hat{j} + 4\hat{k}$

Unit vector perpendicular to $\overrightarrow{a} + \overrightarrow{b}$  and $\overrightarrow{a} - \overrightarrow{b}$ is given by

$\hat{n} =$ $\frac{(\overrightarrow{a} + \overrightarrow{b} ) \times (\overrightarrow{a} - \overrightarrow{b} )}{|(\overrightarrow{a} + \overrightarrow{b} ) \times (\overrightarrow{a} - \overrightarrow{b} )|}$

In our case

$(\overrightarrow{a} + \overrightarrow{b} ) \times (\overrightarrow{a} - \overrightarrow{b} ) = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{array} \right|$

$= \hat{i} (16-0) - \hat{j}(16-0) + \hat{k}(0-8)$

$= 16\hat{i}-16\hat{j}-8\hat{k}$

Now, $|(\overrightarrow{a} + \overrightarrow{b} ) \times (\overrightarrow{a} - \overrightarrow{b} )| = \sqrt{16^2+16^2+8^2} = \sqrt{576} = 24$

Therefore $\hat{n} =$ $\frac{16\hat{i}-16\hat{j}-8\hat{k} }{24}$ or $\hat{n} =$ $\frac{2}{3}$ $\hat{i}-$ $\frac{2}{3}$ $\hat{j}-$ $\frac{1}{3}$ $\hat{k}$

$\\$

Question 11:

(a) Find the image of the point $(2, -1, 5)$ in the line $\frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11}$. Also find the length of the perpendicular from point $(2, -1, 5)$ to the line.     [5]

(b) Find the Cartesian equation of the plane, passing through the line of intersection of the planes: $\overrightarrow{r}.(2\hat{i}+3\hat{j}-4\hat{k})+5=0$ and $\overrightarrow{r}.(\hat{i}-5\hat{j}+7\hat{k})+2=0$ and intersecting $y-axis$ at $(0,3)$.     [5]

(a)  Let $P (2, 1, -5)$ be the point and AB be the line given by equation

$\frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11}$ $= r$

Draw the diagram as shown. $PM \perp AB. PM = MP'. P'$ is the reflection of $P$ on $AB$.

Any point of $AB$ will be given by $(10r+11, -4r-2, -11r-8)$

Then DR’s of $MP$  is $10r+11-2, -4r-2-(-1), -11r-8-5 \ or \ 10r+9, -4r-1, -11r-13$

and DC’s of $AB$ are proportional to $10, -4, 11$

Therefore $MP \perp AB$

$10(10r+9) -4(-4r-1)-11(-11r-13)=0$

$\Rightarrow r = -1$

Therefore $M = (10(-1)+11, -4(-1)-2, -1(-1)-8) = (1, 2, 3)$

Let the coordinate of $P'$ be $(\alpha, \beta, \gamma)$

Since $M(1, 2, 3)$ is the mid point of $PP'$,

$1 =$ $\frac{\alpha + 2}{2}$ $, 2 =$ $\frac{\beta -1}{2}$ $, 3 =$ $\frac{\gamma + 5}{2}$

$\Rightarrow \alpha = 0, \beta = 5, \gamma = 1$

Therefore the coordinates of $P' = (0, 5, 1)$

Hence the distance of M from $P = \sqrt{(1-2)^2+(2+1)^2 + (3-5)^2} = \sqrt{14}$ units

(b)  A plane passing through the intersection of the given plane is

$\overrightarrow{r}.(2\hat{i}+3\hat{j}-4\hat{k})+5=0$ $+ \lambda$ $[ \overrightarrow{r}.(\hat{i}-5\hat{j}+7\hat{k})+2 ]=0$

$\Rightarrow \overrightarrow{r} [(2 + \lambda) \hat{i} + (3-5\lambda) \hat{j} +(-4+7\lambda) \hat{k} = 5-2\lambda$

Point of intersection on $y-axis$ is $(0,3)$. At $y-axis$, $x$ and $z$ will be $0$.

Therefore $\overrightarrow{r} = 0\hat{i} +3\hat{j} + 0\hat{k}$

$\Rightarrow ( 0\hat{i} +3\hat{j} + 0\hat{k} ) . [(2 + \lambda) \hat{i} + (3-5\lambda) \hat{j} +(-4+7\lambda) \hat{k} = 5-2\lambda$

$\Rightarrow 0(2+\lambda) + 3(3-5\lambda)+0(-4+7\lambda)+ 5 + 2\lambda = 0$

$\Rightarrow \lambda =$ $\frac{14}{13}$

Therefore the required plane is

$\Rightarrow (2+$ $\frac{14}{13}$ $)x + (3-5$ $\frac{14}{13}$ $)y+(-4+7$ $\frac{14}{13}$ $)z+ 5 + 2$ $\frac{14}{13}$ $= 0$

$\Rightarrow 40x - 31y + 46z + 93 = 0$. Equation of the plane.

$\\$

Question 12:

(a) In an automobile factory, certain parts are to be fixed into the chassis in a section before it moves into another section. One a given day, one of the three persons $A, B \ and \ C$ carries out this task. $A$ has $45\%$ chance, $B$ has $35\%$ change and $C$ has $20\%$ chance of doing the task. The probability that $A, B \ and \ C$ will take more than the allotted time is $\frac{1}{6}, \frac{1}{10}$  and  $\frac{1}{20}$ respectively.  If it is found that the time taken is more than the allotted time, what is the probability that A has done the task?    [5]

(b) The difference between mean and variance of a binomial distribution is $1$ and the difference of their square is $11$. Find the distribution.    [5]

(a)  Let $E_1, E_2 \ and \ E_3$ denote the events of carrying out the task by $A, B \ and \ C$ respectively. Let H denote the event of taking more time.

Then, $P(E_1) = 0.45, P(E_2) = 0.35 , P(E_3)=0.20$

$P(H/E_1) =$ $\frac{1}{6}$ $, P(H/E_2) =$ $\frac{1}{10}$ $, P(H/E_3) =$ $\frac{1}{20}$

Then, using Bayes Theorem

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

$= \frac{0.45 \times \frac{1}{6}}{0.45 \times \frac{1}{6} + 0.35 \times \frac{1}{10} + 0.20 \times \frac{1}{20}}$ $= \frac{0.075}{0.075 + 0.035 + 0.01}$ $= 0.625$

(b)  Let the binomial distribution be $(q+p)^n$

mean $= np$ and variance $= npq$

Given: $np - npq = 1 \Rightarrow np (1-q) = 1$ … … … … … (i)

$(np)^2 - (npq)^2 = 11 \Rightarrow (np)^2 (1-q^2) = 11$ … … … … … (ii)

Dividing (ii) by square of (i) we get

$\frac{(np)^2 (1-q^2)}{(np)^2 (1-q)^2}$ $=$ $\frac{11}{1}$

$\frac{(1+q)(1-q)}{(1-q)(1-q)}$ $= 11$

$\Rightarrow 1+q = 11-11q$

$\Rightarrow q =$ $\frac{5}{6}$

Therefore $p = 1 - q = 1 -$ $\frac{5}{6}$ $=$ $\frac{1}{6}$

Substituting in (i) we get

$n$ $(\frac{1}{6})(1-\frac{5}{6})$ $= 1 \Rightarrow n = 36$

Hence the binomial distribution is $\Big($ $\frac{5}{6}$ $+$$\frac{1}{6}$ $\Big)^{36}$

$\\$

Section – C (20 Marks)

Question 13:

(a) A man borrows $Rs. 20000$ at $12\%$ per annum, compounded semi-annually and agrees to pay it in $10$ equal semi-annual installments. Find the value of each installment, if the first payment is due at the end of two years.    [5]

(b) A company manufacturers two types of products $A$ and $B$. Each unit of $A$ requires $3$ grams of nickel and $1$ gram of chromium, while each unit of $B$ requires $1$ gram of nickel and $2$ grams of chromium. The firm can produce $9$ grams of nickel and $8$ grams of chromium. The profit is $Rs. \ 40$ on each product of type $A$ and $Rs. \ 50$ on each unit of type $B$. How  many units of each type should the company manufacture so as to earn maximum profit. Use linear programming to find the solution.    [5]

(a)  Principal amount $= 20000 \ Rs.$

Rate of interest $= 12\%$ semi-annually

Let the installment amount be $= A$

Therefore, if we were to calculate the net present value of all the installments, that should be equal to $20000 Rs.$

Hence we have:

$A \Big( 1-\frac{6}{100} \Big)^4+ A \Big( 1-\frac{6}{100} \Big)^5 + A \Big( 1-\frac{6}{100} \Big)^6 + ... + A \Big( 1-\frac{6}{100} \Big)^{13} = 2000$

$\Rightarrow A(0.94)^4 +A(0.94)^5 +A(0.94)^6 +... +A(0.94)^{13} = 20000$

$\Rightarrow A \Big($ $\frac{0.94^4 - 0.94^{14}}{0.06}$ $\Big) = 20000$

$\Rightarrow A = 3113.186 \ Rs.$

(b)

 Type of Product Total Items Profit per Unit (Rs.) A 3 1 40 B 1 2 50

Let number of units of type $A = x$

and Let number of units of type $B = y$

So we have to maximize $Z = 40x + 50y$

given constraints: $3x+y \leq 9$ … … … … … (i)

and $x+2y \leq 8$ … … … … … (ii)

Also $x> 0 \ \& y > 0$

For equation $3x+y = 9, \ x-intercept$ is $(3, 0)$ and$y-intercept$ is $(0, 9)$

Similarly, for equation $x+2y= 8 \ x-intercept$ is $(8, 0$) and $y-intercept$ is $(0, 4)$

Now plot the two lines on the graph paper

Therefore the solution set of this system is the shaded region as shown in the graph above.

When $(x, y) = (0, 0), Z = 40 \times 0 + 50 \times 0 = 0 \ Rs.$

$(x, y) = (3, 0), Z = 40 \times 3 + 50 \times 0 = 120 \ Rs.$

$(x, y) = (0, 4), Z = 40 \times 0 + 50 \times 4 = 200 \ Rs.$

$(x, y) = (2, 3), Z = 40 \times 2 + 50 \times 3 = 230 \ Rs.$

Therefore the maximum profit is $230 \ Rs$. Hence the number of units of $A$ produced  is $2$ and number of units of $B$ produced is $3$.

$\\$

Question 14:

(a) The demand function is $x =$ $\frac{24-2p}{3}$ where $x$ is the number of units demanded and $p$ is the price per unit. Find:

(i) The revenue function $R$ in terms of  $p$.

(ii) The price and the number of units demanded for which the revenue is maximum.    [5]

(b) A bill of $Rs. \ 1800$ drawn on 10th September, 2010 at 6 months was discounted for $Rs. \ 1782$ at a bank. If the rate of interest was $5\%$ per annum, on what date was the bill discounted.    [5]

(a) Given $x =$ $\frac{24-2p}{3}$

$\Rightarrow p = 12 -$ $\frac{3x}{2}$

(i) Revenue function $R(p) = p ($ $\frac{24-2p}{3}$ $) = 8p -$ $\frac{2}{3}$ $p^2$

(ii) For maximum revenue, $\frac{dR(p)}{dp}$ $= 8 -$ $\frac{4}{3}$ $p = 0 \Rightarrow p = 6$

and $\frac{d^2R(p)}{dp}$ $= -$ $\frac{4}{3}$ which is $<0$

$\Rightarrow R(p)$ is maximum revenue, price per unit is $6$

Therefore $x =$ $\frac{24 - 12}{3}$ $= 4$

Therefore Price is $6$ and no. of units is $4$ for which the revenue is maximum.

(b)  Amount $(A) = 1800 \ Rs.$

Interest $(i\%) = 5\%$

Bill Discount $(BD) = 1800 - 1782 = 18 \ Rs.$

Now $BD = Ani$

$\Rightarrow 18 = 1800 \times n \times$ $\frac{5}{100}$

$\Rightarrow n =$ $\frac{1}{5}$

$\Rightarrow n = 73 \ days$

Therefore Date of Expiry = 13 March 2011

and Date of Discounting is 30 December 2010

$\\$

Question 15:

(a) The index number by the method of aggregates for the year 2010, taking 2000 as the base year, was found to be $116$. If the sum of the price in year 2000 is $Rs. 300$, find the value of $x \ and \ y$ in the data given below.    [5]

 Commodity A B C D E F Price in Base year 2000 (Rs.) 50 $x$ 30 70 116 20 Price in the year 2010 (Rs.) 60 24 $y$ 80 120 28

(b) From the details given below, calculate the five yearly moving averages of the number of students who have studies in a school. Also, plot these and original data on the same graph paper.    [5]

 Year 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 No. of Students 332 317 357 392 402 405 410 427 405 438

(a)

 Commodity Price in Rs. 2000 2011 A 50 60 B $x$ 24 C 30 $y$ D 70 80 E 116 120 F 20 28 $286+x$ $312+y$

$286+x= 300$

$x = 14$

$P_{01} =$ $\frac{\Sigma P_1}{\Sigma P_0}$ $\times 100$

$\Rightarrow 116 =$ $\frac{12+y}{300}$ $\times 100$

$\Rightarrow y = 36$

(b)

 Year Number 5 year moving total 5 year moving average 1993 332 – – 1994 317 – – 1995 357 1800 360 1996 392 1873 374.6 1997 402 1966 393.2 1998 405 2036 407.2 1999 410 2049 409.8 2000 427 2085 417 2001 405 – – 2002 438 – –

$\\$