2018 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)

Question 1: [10 × 2]

(i) The binary operation $\ast : R \times R \ \rightarrow R$ is defined as $a \ast b = 2a+ b$ Find $(2 \ast 3) \ast 4$ .

(ii) If $\begin{bmatrix} 5 & a \\ b & 0 \end{bmatrix}$ and $A$ is symmetric matrix, show that $a = b$

(iii) Solve : $3 \ \tan^{-1}x + \cot^{-1}x = \pi$

(iv) Without expanding at any stage, find the value of:

$\left| \begin{array}{ccc} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{array} \right|$

(v) Find the value of constant $k$ so that the function $f (x)$ defined as:

$f(x) = \Bigg\{$ $\begin{array}{ccc} \frac{a}{b}, x \neq -1 \\ \ \\ k, x = -1 \end{array}$

is continuous at $x=-1$ .

(vi) Find the approximate change in the volume $V$ of a cube of side $x$ meters caused by decreasing the side by $1\%$ .

(vii) Evaluate : $\int \limits_{}^{}$ $\frac{x^3+5x^2+4x+1}{x^2}$ $dx$

(viii) Find the differential equation of the family of concentric circles $x^2+y^2 = a^2$

(ix) If $A$ and $B$ are events such that $P(A) =$ $\frac{1}{3}$ $, P(B) =$ $\frac{1}{6}$ and $P(A \cap B) =$ $\frac{1}{4}$ , then find:

(a) $P(A/ B)$

(b) $P(B / A)$

(x) In a race, the probabilities of A and B winning the race are $\frac{1}{3}$ and $\frac{1}{6}$ respectively. Find the probability of neither of them winning the race.

Answer:

(i) Given $a * b = 2 a + b$

$(2 * 3) *4 = (4+ 3) * 4$

$= 7 * 4$

$= 14+ 4 = 18$

(ii) Since $\begin{bmatrix} 5 & a \\ b & 0 \end{bmatrix}$ is symmetric, therefore

$\begin{bmatrix} 5 & a \\ b & 0 \end{bmatrix}^T = \begin{bmatrix} 5 & a \\ b & 0 \end{bmatrix}$

$\begin{bmatrix} 5 & b \\ a & 0 \end{bmatrix}^T = \begin{bmatrix} 5 & a \\ b & 0 \end{bmatrix}$

We can compare corresponding terms. We get

$a = b$

(iii) $3 \ \tan^{-1}x + \cot^{-1}x = \pi$

$2 \ \tan^{-1}x + \tan^{-1}x + \cot^{-1}x = \pi$

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

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

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

$x = \tan \Big($ $\frac{\pi}{2}$ $-$ $\frac{\pi}{4}$ $\Big)$

$x = \tan \Big($ $\frac{\pi}{4}$ $\Big)$

$x = 1$

(iv) $\left| \begin{array}{ccc} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{array} \right|$

Applying $R_1 \rightarrow R_1 + R_3$ and $R_2 \rightarrow R_2 - R_3$ we get

$\left| \begin{array}{ccc} a+x & b+y & c+z \\ a+x & b+y & c+z \\ x & y & z \end{array} \right| = 0$ since $R_1 = R_3$

(v) Given $f (x)$ is continuous at $x =-1$

Therefore $f(-1) = \lim \limits_{x \to -1} f(x)$

Therefore $k = \lim \limits_{x \to -1}$ $\frac{x^2 - 2x - 3}{x+1}$

$= \lim \limits_{x \to -1}$ $\frac{(x-3)(x+1)}{(x+1)}$

Since $x \rightarrow -1 = x+ 1 \neq 0$

$= -1 -3 = -4$

Therefore $k = -4$

(vi) Volume of a cube

$V = x^3$

$\frac{dV}{dx}$ $= 3x^2$

Therefore $\delta x = x .$ $\frac{1}{100}$ $=$ $\frac{-x}{100}$

Hence change in volume

$\delta V =$ $\frac{dV}{dx}$ $\delta x = (3x^2) \Big($ $\frac{-x}{100}$ $\Big) = - \Big($ $\frac{3}{100}$ $\Big) x^3 = -$ $\frac{3}{100}$ $V$

Hence change in volume decrease by $3\%$

(vii) $I = \int \limits_{}^{}$ $\frac{x^3+5x^2+4x+1}{x^2}$ $dx$

$= \int \limits_{}^{}$ $\Big($ $\frac{x^3}{x^2}$ $+$ $\frac{5x^2}{x^2}$ $+$ $\frac{4x}{x^2}$ $+$ $\frac{1}{x^2}$ $\Big)$ $dx$

$= \int \limits_{}^{}$ $\Big( x+ 5 +$ $\frac{4}{x}$ $+$ $\frac{1}{x^2}$ $\Big)$ $dx$

$=$ $\frac{x^2}{2}$ $+ 5x + 4 \log |x| -$ $\frac{1}{x}$ $+ c$

(viii) Family of concentric circles is $x^2 + y^2 = a^2$

Therefore Differential w.r.t. $x$

$2x + 2y$ $\frac{dy}{dx}$ $= 0$

Therefore $y$ $\frac{dy}{dx}$ $+ x = 0$

(ix) $P(A) =$ $\frac{1}{2}$ $, P(B) =$ $\frac{1}{3}$

$P (A / B) =$ $\frac{P(A \cap B)}{P(B)}$ $=$ $\frac{\frac{1}{4}}{\frac{1}{3}}$ $=$ $\frac{3}{4}$

$P (B / A) =$ $\frac{P(A \cap B)}{P(A)}$ $=$ $\frac{\frac{1}{4}}{\frac{1}{2}}$ $=$ $\frac{1}{2}$

(x) Let $A$ win the race be $E_1$ $B$ win the race be $E_2$

$P(A) =$ $\frac{1}{3}$ $, P(B) =$ $\frac{1}{6}$

$P({E'}_1 \cap {E'}_2) = P({E'}_1).P({E'}_2) = [1-P(E_1)].[1-P(E_2)]$

$= (1-$ $\frac{1}{3}$ $)(1-$ $\frac{1}{6}$ $)=$ $\frac{2}{3}$ $\times$ $\frac{5}{6}$ $= \frac{5}{9}$

$\\$

Question 2: If the function $f (x) = \sqrt{2x-3}$ is invertible then find its inverse. Hence prove that $(fof^{-1})(x) = x$ . [4]

Answer:

Let $y = \sqrt{2x-3}$

Squaring both sides $y^2 = 2x-3$

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

Therefore $f^{-1} (x) =$ $\frac{x^2+3}{2}$

Now, $(fof^{-1} (x)) = \sqrt{f^{-1}(x) - 3}$

$= \sqrt{ 2 \Big( \frac{x^2+3}{2} \Big) } = x$

Therefore $fo(f^{-1} (x)) = x$

$\\$

Question 3: If $\tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi$ , prove that $a + b + c = abc$ . [4]

Answer:

$\tan^{-1} a + \tan^{-1} b + \tan^{-1} c = \pi$

$\Rightarrow \tan^{-1} b + \tan^{-1} c = \pi - \tan^{-1} a$

$\Rightarrow \tan^{-1} \Big($ $\frac{b+c}{1-bc}$ $\Big) = \pi - tan^{-1}a$

$\Rightarrow$ $\frac{b+c}{1-bc}$ $= \tan (\pi - \tan^{-1} a)$

$\Rightarrow$ $\frac{b+c}{1-bc}$ $= \tan (\tan^{-1} a)$

$\Rightarrow$ $\frac{b+c}{1-bc}$ $= -a$

$\Rightarrow b+ c = -a + abc$

$\Rightarrow a + b + c = abc$

$\\$

Question 4: Use properties of determinants to solve for $x$ :

$\left| \begin{array}{ccc} x+a & b & c \\ c & x+b & a \\ a & b & x+c \end{array} \right|= 0$ and $x \neq 0$ [4]

Answer:

Given $\left| \begin{array}{ccc} x+a & b & c \\ c & x+b & a \\ a & b & x+c \end{array} \right|= 0$ and $x \neq 0$

$C_1 \rightarrow C_1 + C_2 + C_3$ we get

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

$(x+a+b+c)$ $\left| \begin{array}{ccc} 1 & b & c \\ 1 & x+b & a \\ 1 & b & x+c \end{array} \right|= 0$

$R_1 \rightarrow R_1 - R_3$ we get

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

$(x+a+b+c) \Big( 0-0-x(b-x-b) \Big) = 0$

$(x+a+b+c) x^2 = 0$

$\Rightarrow x^2 = 0$ or $(x+a+b+c) = 0$

But $x \neq 0$ , therefore $x = -(a+b+c)$

$\\$

Question 5: [4]

(a) Show that the function $f(x) = \Bigg\{$ $\begin{array}{ccc} x^2, x \leq 1 \\ \ \\ \frac{1}{2}, x > 1 \end{array}$ is continuous at $x =1$ but not differentiable.

(b) Verify Rolle’s theorem for the following function: $f(x) = e^{-x} \sin x \ on \ [0, \pi ]$

Answer:

(a) Continuity at $x = 1$

$f(x=1) = x^2 = 1$

$\lim \limits_{x \to 1^+} f(x) = \lim \limits_{x \to 1^+}$ $\frac{1}{x}$ $= 1$

$\lim \limits_{x \to 1^-} f(x) = \lim \limits_{x \to 1^-} x^2 = 1$

Therefore $f(x=1) = \lim \limits_{x \to 1^-} f(x) = \lim \limits_{x \to 1^+} f(x) = 1$

Therefore $f(x)$ is continuous at $x =1$

Now differentiate at $x = 1$

$(R.H.D \ at \ x = 1) = \lim \limits_{x \to 1^+}$ $\frac{f(x) - f(1)}{x-1}$

$= \lim \limits_{x \to 1}$ $\frac{\frac{1}{x} -1}{x-1}$

$= \lim \limits_{x \to 1}$ $\frac{-(x-1)}{x(x-1)}$

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

$(L.H.D \ at \ x = 1 = ) = \lim \limits_{x \to 1^-}$ $\frac{f(x) - f(1)}{x-1}$

$= \lim \limits_{x \to 1^-}$ $\frac{x^2 - 1}{x-1}$ $= 2$

Therefore $L.H.D \neq R.H.D$

Hence $f(x)$ is not differentiable at $x = 1$

(b) $f(x) = e^{-x} \sin x \ on \ [0, \pi ]$

(i) $f (x)$ is continuous on $[0, \pi ]$ because $e^x \ \ and \ \sin x$ are continuous function on its domain.

(ii) $e^{-x}$ and $\sin x$ is differentiable on $(0,\pi )$

(iii) $f (0) = e^{-0} . \sin 0 = 0$

$f(\pi) = e^{-x}. \sin \pi = 0$

(iv) Let $c$ be number such that $f'(c)=0$

Therefore $f'(x) = e^{-x} . \cos x + \sin x . e^{-x} (-1)$

Therefore $f'(c) = e^{-c} (\cos c - \sin c)$

Therefore $f'(c) = 0$

$e^{-c} (\cos c - \sin c) = 0 \Rightarrow \tan c = 1$

Therefore $c =$ $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}$ $, \ ...$

Therefore $\frac{\pi}{4}$ $\in [0, \pi]$

Therefore Rolle’s theorem verified

$\\$

Question 6: If $x = \tan \Big($ $\frac{1}{a}$ $log \ y \Big)$ , prove that $(1+x^2)$ $\frac{d^2y}{dx^2}$ $+(2x-a)$ $\frac{dy}{dx}$ $= 0$ [4]

Answer:

$x = \tan \Big($ $\frac{1}{a}$ $log \ y \Big)$

Therefore $\frac{1}{a}$ $\log y = \tan^{-1}x$

differentiating both sides w.r.t. $x$

$y =$ $e^{a \ \tan^{-1}x}$

$\frac{dy}{dx}$ $=$ $e^{a \ \tan^{-1}x}$ $\Big($ $\frac{a}{1+x^2}$ $\Big)$

$(1+x^2)$ $\frac{dy}{dx}$ $= ay$

Again differentiating both sides w.r.t. $x$

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

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

$\\$

Question 7: Evaluate: $\int \limits_{}^{}$ $\tan^{-1}\sqrt{x} \ dx$ [4]

Answer:

$I =$ $\int \limits_{}^{}$ $\tan^{-1}\sqrt{x} \ dx$

Put $\sqrt{x} = t$

$\frac{1}{2\sqrt{x}}$ $\ dx = dt$

$dx = 2\sqrt{x} dt \Rightarrow dx = 2t \ dt$

$I = \int \limits_{}^{}$ $2t \ \tan^{-1} t \ dt$

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

$I = 2 \Big[$ $\frac{t^2}{2}$ $\tan^{-1}t -$ $\frac{1}{2}$ $\int \limits_{}^{}$ $\Big($ $\frac{1+t^2}{1+t^2}$ $-$ $\frac{1}{1+t^2}$ $\Big) \ dt \Big]$

$I = (t^2 \ \tan^{-1} t -t + \tan^{-1} t ) + c$

$I = t^2 \ \tan^{-1} t -t + \tan^{-1} t + c$

$I = (x+1) \ \tan^{-1} \sqrt{x} - \sqrt{x}+c$

$\\$

Question 8: [4]

(a) Find the points on the curve $y = 4x^3 - 3x + 5$ at which the equation of the tangent is parallel to the x-axis.

(b) Water is dripping out from a conical funnel of semi-vertical angle $\frac{\pi}{4}$ at the uniform rate of $2 \ cm^2/sec$ in the surface, through a tiny hole at the vertex of the bottom. When the slant height of the water level is 4 cm, find the rate of decrease of the slant height of the water.

Answer:

(a) $y = 4x^3 - 3x + 5$ … … … … … (i)

$\frac{dy}{dx}$ $= 12x^2 - 3$

Given that lines is parallel to $x-axis$

Therefore $\frac{dy}{dx}$ $= 0$

$12x^2 - 3 = 0 \Rightarrow 12x^2 = 3 \Rightarrow x^2 =$ $\frac{1}{4}$

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

Put $x = \pm$ $\frac{1}{2}$ in equation (i)

When $x =$ $\frac{1}{2}$ then $y = 4\Big($ $\frac{1}{2}$ $\Big)^3 - 3 \Big($ $\frac{1}{2}$ $\Big) + 5 = 4$

Therefore Point $($ $\frac{1}{2}$ $, 4)$

When $x =$ $\frac{-1}{2}$ , then $y = 4\Big($ $\frac{-1}{2}$ $\Big)^3 - 3 \Big($ $\frac{-1}{2}$ $\Big) + 5 = 6$

Therefore Point $($ $\frac{-1}{2}$ $, 6)$

Therefore Points $(x, y) = ($ $\frac{1}{2}$ $, 4)$ and $($ $\frac{-1}{2}$ $, 6)$

(b) Let $r$ be the radius, $h$ be the height and $V$ be the volume of the funnel at any time $t$ .

$V =$ $\frac{1}{3}$ $\pi r^2 h$ … … … … … (i)

Let $I$ be the slant height of the funnel
Given : Semi-vertical angle $= 45^o$ in the $\triangle ADE$ :

$sin \ 45^o =$ $\frac{DE}{AE}$ $\Rightarrow$ $\frac{1}{\sqrt{2}}$ $=$ $\frac{r}{l}$

$cos \ 45^o =$ $\frac{AD}{AE}$ $\Rightarrow$ $\frac{1}{\sqrt{2}}$ $=$ $\frac{h}{l}$

$r =$ $\frac{1}{\sqrt{2}}$ $\ and \ h =$ $\frac{1}{\sqrt{2}}$ … … … … … (ii)

therefore the equation (i) can be rewritten as:

$V =$ $\frac{1}{3}$ $\pi \times \Big($ $\frac{I}{\sqrt{2}}$ $\Big)^2 \times \Big($ $\frac{I}{\sqrt{2}}$ $\Big) = \Big($ $\frac{\pi}{3 \times 2 \times \sqrt{2}}$ $\Big) \times I^3$

$V =$ $\frac{\pi}{6\sqrt{2}}$ $I^3$

Differentiate w.r.t. $t$ :

$\frac{dV}{dt}$ $=$ $\frac{\pi}{6\sqrt{2}}$ $\times 3I^2 \times$ $\frac{dl}{dt}$

$\frac{dV}{dt}$ $=$ $\frac{\pi}{2\sqrt{2}}$ $\times I^2 \times$ $\frac{dl}{dt}$

$\frac{dl}{dt}$ $=$ $\frac{2\sqrt{2}}{\pi l^2}$ $\frac{dV}{dt}$

Since it is given that rate of change (decrease) of volume of water w.r.t. $t$ is

$\frac{dV}{dt}$ $= -2$ $\frac{cm^3}{sec}$

Thereofore

$\frac{dl}{dt}$ $=$ $\frac{2\sqrt{2}}{\pi l^2}$ $\times (-2) = -$ $\frac{4\sqrt{2}}{\pi l^2}$

${\frac{dl}{dt}}_{|_{at \ l = 4}}$ $= -$ $\frac{4\sqrt{2}}{\pi \times 4^2}$ $= -$ $\frac{\sqrt{2}}{4\pi}$ $\frac{cm}{sec}$

$\\$

Question 9: [4]

(a) Solve: $\sin x$ $\frac{dy}{dx}$ $- y = \sin x. \tan$ $\frac{x}{2}$

(b) The population of a town grows at the rate of $10\%$ per year. Using differential equation, find how long will it take for the population to grow $4$ times.

Answer:

(a) $\sin x$ $\frac{dy}{dx}$ $- y = \sin x. \tan$ $\frac{x}{2}$

$\Rightarrow$ $\frac{dy}{dx}$ $- y . cosec \ x = \tan$ $\frac{x}{2}$

Compare with $\frac{dy}{dx}$ $+Py = Q$

$P = - cosec \ x, Q = \tan$ $\frac{x}{2}$

$I.F =$ $e^{\int \limits_{}^{} P dx}$

$I.F =$ $e^{\int \limits_{}^{} -cosec \ x \ dx}$

$I.F =$ $e^{-\log_e(cosec \ x - \cot x)}$

$I.F =$ $e^{\log_e(cosec \ x - \cot x)^{-1}}$

$I.F = (cosec \ x - \cot x )^{-1} = (cosec \ x + \cot x)$

Therefore solution of the linear differential equation

$y.IF = \int \limits_{}^{} Q.IF. dx$

$y (cosec \ x + \cot x) = \int \limits_{}^{} \tan$ $\frac{x}{2}$ $\Big($ $\frac{1+\cos x}{\sin x}$ $\Big) dx$

$y (cosec \ x - \cot x) = \int \limits_{}^{}$ $\frac{\sin \ \frac{x}{2}}{\cos \frac{x}{2}}$ $.$ $\frac{2 \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2}.\cos \frac{x}{2}}$ $dx = \int \limits_{}^{} 1 dx$

$y (cosec \ x + \cot x) = x + c$

(b) $\frac{dx}{dt}$ $\propto x$ (Since increase in population speeds up with increase in population) and let $x$ be the population at anytime $t$ .

Therefore $\frac{dx}{dt}$ $= r x$ (where r is proportionality constant)

Therefore $\frac{dx}{x}$ $= r . dt$

integrating both sides

$\log x = rt + c$ , (where c is the integration constant)

Therefore $x = e^{rt+c}$

$x = k e^{rt}$ where $k = e^c$

Here $r$ is the rate of increase and $k$ is the initial population let $x_0$ then $t = 0$

$x_0 = ke^0$ $\Rightarrow k = x_0$

Given to find the time $t$ taken to attain $4$ times population, so $x = 4 x_0$

Therefore $x = ke^{rt}$ $\Rightarrow 4x_0 =$ $x_0 e^{0.10t}$

$4 = e^{0.1t}$ Taking log on both sides

$2 \log 2 = \log$ $e^{0.1t}$

$\Rightarrow 2 \times 0.3010 = 0.1t \Rightarrow t = 6.02$

$\\$

Question 10: [6]

(a) Using matrices, solve the following system of equations :

$2x - 3y + 5z = 11$
$3x + 2y - 4z = -5$
$x + y - 2z = -3$

(b) Using elementary transformation, find the inverse of the matrix :

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

Answer:

(a) Given, the three equations:

$2x - 3y + 5z = 11$
$3x + 2y - 4z = -5$
$x + y - 2z = -3$

We can write this in the form of $AX = B$ , i.e. as follows:

$\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} . \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

We know, $A^{-1} =$ $\frac{1}{|A|}$ $(adj. A)$

$| A|= 2(2\times - 2-1\times - 4) -(-3) (3 \times -2 -1 \times -4) + 5(3 \times 1 - 2 \times 1)$

$=2(-4+4) +3(-6+4) + 5(3-2) = 0-6 +5 = -1 \neq 0$

Hence it is a non – singular matrix. Therefore A^{-1} exists. Let us find the (adj A) by finding the minors and co-factors

$A^{-1} = \frac{1}{-1} \begin{bmatrix} 2(-2) -1(-4) & 3(-2) - 1(-4) & 3(1) - 1(2) \\ (-3)(-2) - 1(5) & 2(-2) - 1(5) & 2(1) - 1(-3) \\ (-3)(-4)-2(5) & 2(-4) - 3(5) & 2(2) - 3(-3) \end{bmatrix}$

$A^{-1} = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$

We know $AX = B$ , then $X = A^{-1} B$

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} . \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

Matrix multiplication can be done by multiplying the rows of matrix $A$ with the column of matrix $B$ .

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

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

(b) Let $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$

$|A| = 0(2-3) -1(1-9) +2 (1-6) = -2 \neq 0$

Therefore $A{-1}$ exists.

$A = IA$

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

Interchanging $R_1$ and $R_2$

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

$R_3 \rightarrow R_3 - 3 R_1$

$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{bmatrix}. A$

$R_3 \rightarrow R_3 + 5 R_2$

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

$R_2 \rightarrow R_2 - R_3$

$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ -4 & 3 & -1 \\ 5 & -3 & 1 \end{bmatrix}. A$

$R_1 \rightarrow R_1 - 2R_2$ and $R_3 \rightarrow R_3 \times \frac{1}{2}$

$\begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & -5 & 2 \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}. A$

$R_1 \rightarrow R_1 - 3R_3$

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}. A$

Therefore $A^{-1} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix} =$ $\frac{1}{2}$ $\begin{bmatrix} 1 & -1 & 1 \\ -8 & 6 & -2 \\ 5 & -3 & 1 \end{bmatrix}$

$\\$

Question 11: $A$ speaks truth in $60\%$ of the cases, while $B$ is $40\%$ of the cases. In what percent of cases are they likely to contradict each other in stating the same fact ? [4]

Answer:

$A$ speaks truth $P(A) =$ $\frac{60}{100}$ $\Rightarrow P(A') =$ $\frac{40}{100}$

$B$ speaks truth $P(B) =$ $\frac{40}{100}$ $\Rightarrow P(B') =$ $\frac{60}{100}$

They contradict each other $= P(A).P(B') + P(B).P(A')$

$=$ $\frac{60}{100}$ $\times$ $\frac{60}{100}$ $+$ $\frac{40}{100}$ $\times$ $\frac{40}{100}$

$=$ $\frac{3600 + 1600}{10000}$ $=$ $\frac{52}{100}$

$\%$ of cases they likely to contradict each other $= 52\%$

$\\$

Question 12: A cone is inscribed in a sphere of radius $12 \ cm$ . If the volume of the cone is maximum, find its height. [6]

Answer:

Let $DAB$ be a cone of greatest volume inscribed in a sphere of radius $12$ . It is obvious that for maximum volume the axis of the cone must be along a diameter of the sphere. Let $DC$ be the axis of the cone and $O$ be the center of the sphere such that $OC = x$ .

Then, $DC =DO+OC = R + x = (12+ x) =$ height of cone. Applying Pythagoras theorem,

$OA^2 = AC^2 + OC^2$

$\Rightarrow AC^2 = 12^2 - x^2$

$\Rightarrow AC^2 = 144 - x^2$

Let $V$ be the volume of the cone, then

$V =$ $\frac{1}{3}$ $\pi (AC)^2 (DC)$

$V =$ $\frac{1}{3}$ $\pi (144 - x^2)^2 (12+ x)$

$V =$ $\frac{1}{3}$ $\pi (1728 + 144x -12x^2 -x^3)$ … … … … … (i)

$\frac{dV}{dx}$ $=$ $\frac{1}{3}$ $\pi (144 - 24x-3x^2)$

$\frac{d^2V}{dx^2}$ $=$ $\frac{1}{3}$ $\pi (-24-6x) = -2\pi (4+x)$

Now, $\frac{dV}{dx}$ $= 0 \Rightarrow 144 - 24x-3x^2 = 0$

$\Rightarrow (x+12)(x-4) = 0 \Rightarrow x = -12 \ or \ x = 4$

$\Big[ \frac{d^2V}{dx^2} \Big]_{x=4}$ $= -2\pi (4+4) = -16\pi < 0$

Thus, $V$ is maximum when $x = 4$ . Putting $x = 4$ in (i), we obtain

Height of the cone $= x + R = 4 + 12 = 16 \ cm$

$\\$

Question 13: [6]

(a) Evaluate: $\int \limits_{}^{}$ $\frac{x-1}{\sqrt{x^2-x}}$ $\ dx$

(b) Evaluate: $\int \limits_{0}^{\pi / 2}$ $\frac{\cos^2 x}{1 + \sin x \ \cos x}$ $dx$

Answer:

(a) Let $I =$ $\int \limits_{}^{}$ $\frac{x-1}{\sqrt{x^2-x}}$ $\ dx$

Therefore $x-1 = A$ $\frac{d}{dx}$ $(x^2 - x) + B$

$x-1 = A(2x-1) + B$

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

and $-1 = -A + B \Rightarrow B = -1 +$ $\frac{1}{2}$ $= -$ $\frac{1}{2}$

$I =$ $\int \limits_{}^{}$ $\frac{\frac{1}{2} (2x-1)}{\sqrt{x^2-x}}$ $dx -$ $\frac{\frac{1}{2}}{\sqrt{x^2-x}}$ $dx$

$=$ $\int \limits_{}^{}$ $\Bigg[$ $\frac{\frac{1}{2} (2x-1) dx}{\sqrt{x^2-x}}$ $\Bigg]$ $-$ $\frac{1}{2}$ $\Bigg[$ $\frac{dx}{\sqrt{\Big( x- \frac{1}{2} \Big)^2 - \Big( \frac{1}{2} \Big)^2 }}$ $\Bigg]$

$=$ $\frac{1}{2}$ $\times 2\sqrt{x^2-x} -$ $\frac{1}{2}$ $\times \log$ $\lvert$ $\Big( x-$ $\frac{1}{2}$ $\Big) + \sqrt{\Big( x- \frac{1}{2} \Big)^2 - \Big( \frac{1}{2} \Big)^2 }$ $\lvert$ $+ c$

$= \sqrt{x^2-x} -$ $\frac{1}{2}$ $\log$ $\lvert$ $x -$ $\frac{1}{2}$ $+ \sqrt{x^2-x}$ $\lvert$ $+ c$

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

Using $\int \limits_{0}^{a} f(x) dx = \int \limits_{0}^{a} f(a-x) dx$

Therefore

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

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

Adding equation (i) and (ii)

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

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

$=$ $\int \limits_{0}^{\pi / 2}$ $\frac{\sec^2 x}{\sec^2 x + \tan x}$ $dx$

$2I =$ $\int \limits_{0}^{\pi / 2}$ $\frac{\sec^2 x}{1+ \tan^2 x + \tan x}$ $dx$

Put $\tan x = t$ and $\sec^2 x \ dx = dt$

When $x=0, t = 0$

When $x = \frac{\pi}{2}, t = \infty$

$2I =$ $\int \limits_{0}^{\infty}$ $\frac{dt}{t^2 + 2t \times \frac{1}{2} + \frac{1}{4} - \frac{1}{4} + 1}$

$2I =$ $\int \limits_{0}^{\infty}$ $\frac{dt}{ \Big( t + \frac{1}{2} \Big)^2 + \Big( \frac{\sqrt{3}}{2} \Big)^2 }$

$=$ $\frac{1}{\frac{\sqrt{3}}{2}}$ $\Big[ tan^{-1} \Big($ $\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}$ $\Big) \Big]_{0}^{\infty}$

$=$ $\frac{2}{\sqrt{3}}$ $\tan^{-1}$ $\Big[ \Big($ $\frac{2t+1}{\sqrt{3}}$ $\Big) \Big]_{0}^{\infty}$

$2I =$ $\frac{2}{\sqrt{3}}$ $\Big[$ $\frac{\pi}{2}$ $-$ $\frac{\pi}{6}$ $\Big]$

$I =$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{3\pi - \pi}{6}$ $\Big] =$ $\frac{1}{\sqrt{3}}$ $\Big[$ $\frac{\pi}{3}$ $\Big] =$ $\frac{\pi}{3\sqrt{3}}$

$\\$

Question 14: From a lot of $6$ items containing $2$ defective items, a sample of $4$ items are drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the sample is drawn without replacement, find :
(a) The probability distribution of $X$
(b) Mean of $X$
(c) Variance of $X$ [6]

Answer:

In $6$ items $2$ defective and $4$ non-defective. Let $P$ is the probability of defective items
Let $x =$ number of defective items. Therefore $x = 0,1, 2$

Therefore $P(x=0) =$ $\frac{^4C_4}{^6C_4}$ $=$ $\frac{1}{15}$

$P(x=1) =$ $\frac{^2C_1 \times ^4C_3}{^6C_4}$ $=$ $\frac{8}{15}$

$P(x=2) =$ $\frac{^2C_2 \times ^4C_2}{^6C_4}$ $=$ $\frac{6}{15}$

$x$	$P(x)$	$xP(x)$	$x^2P(x)$
$0$	$\frac{1}{15}$	$0$	$0$
$1$	$\frac{8}{15}$	$\frac{8}{15}$	$\frac{8}{15}$
$2$	$\frac{6}{15}$	$\frac{12}{15}$	$\frac{24}{15}$

(i) Mean $(\overline{x}) = \Sigma P_ix_i =$ $\frac{20}{15}$ $=$ $\frac{4}{3}$

(ii) Variance $(\sigma^2) = \Sigma P_i{x_i}^2 - (\Sigma P_ix_i)^2 =$ $\frac{32}{15}$ $- \Big($ $\frac{4}{3}$ $\Big)^2 =$ $\frac{16}{45}$ $= 0.35$

$\\$

SECTION B (20 Marks)

Question 15: [3 × 2]

(a) Find $\lambda$ if the scalar projection of $\overrightarrow{a} = \lambda \hat{i} + \hat{j} + 4 \hat{k}$ and $\overrightarrow{b} = 2 \hat{i} + 6\hat{j} + 3 \hat{k}$ is $4$ units.

(b) The Cartesian equation of line is : $2x - 3 = 3y +1 = 5 - 6z$ . Find the vector equation of a line passing through $(7, -5, 0)$ and parallel to the given line.

(c) Find the equation of the plane through the intersection of the planes $\overrightarrow{r} . (\hat{i}+3 \hat{j} - \hat{k})= 9$ and $\overrightarrow{r} . (2\hat{i}- \hat{j} +1\hat{k})= 3$ and passing through the origin.

Answer:

(a) Projection on $\overrightarrow{a}$ on $\overrightarrow{b}$ is $\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b} |}$ $= 4$

Given $\overrightarrow{a} = \lambda \hat{i} + \hat{j} + 4 \hat{k}$ and $\overrightarrow{b} = 2 \hat{i} + 6\hat{j} + 3 \hat{k}$

Therefore

$\frac{(\lambda \hat{i} + \hat{j} + 4 \hat{k}).(2 \hat{i} + 6\hat{j} + 3 \hat{k})}{\sqrt{2^2+6^2+3^2}}$ $= 4$

Therefore $\frac{2\lambda + 6(1) +4(3)}{\sqrt{49}}$ $= 4$

$\Rightarrow$ $\frac{2\lambda + 18}{7}$ $= 4$

$\Rightarrow 2\lambda = 28 - 18$

$\Rightarrow \lambda = 5$

(b) Cartesian equation of a line is

$2x - 3 = 3y +1 = 5 - 6z$

i.e. $2 \Big( x -$ $\frac{3}{2}$ $\Big) = 3 \Big( y +$ $\frac{1}{3}$ $\Big) = -6 \Big( z -$ $\frac{6}{5}$ $\Big)$

Dividing by $(-6)$ throughout we get

$-\frac{1}{3}$ $\Big( x -$ $\frac{3}{2}$ $\Big) = -\frac{1}{2}$ $\Big( y -$ $\frac{1}{3}$ $\Big) = -1 \Big( z -$ $\frac{6}{5}$ $\Big)$

Therefore D.r.s of the above line is $-3, -2, 1$ . Now, equation of a line passing through point $(7, -5, 0)$ and parallel to the above line whose d.r.s. is $-3, -2, 1$ is $\overrightarrow{r} = ( 7 \hat{i} - 5\hat{j}) + \lambda (-3\hat{i}-2\hat{j}- \hat{k})$

i.e. $(x\hat{i} + y\hat{i} + z\hat{i}).(\hat{i}+3 \hat{j} - \hat{k})= 9$

$\Rightarrow x + 3y -z = 9$

$\Rightarrow x + 3y -z - 9 = 0$ … … … … … (i)

Equation of 2nd plane is: $\overrightarrow{r} . (2\hat{i}- \hat{j} +1\hat{k})= 3$

i.e. $(x\hat{i} + y\hat{i} + z\hat{i}) . (2\hat{i}- \hat{j} +1\hat{k})= 3$

$\Rightarrow 2x-y+z = 3$

$\Rightarrow 2x-y+z - 3 = 0$ … … … … … (ii)

Now, equation of a plane passing through intersection of given planes is:

$(x + 3y -z - 9) + \lambda (2x-y+z - 3) = 0$

$\Rightarrow (1+ 2 \lambda) x + (3 - \lambda) y )+ (-1 + \lambda) -9 -3\lambda = 0$

Since plane is passing through the origin $(0, 0, 0)$

$-9 -3\lambda = 0 \Rightarrow -3 \lambda = 9 \Rightarrow \lambda = -3$

$\\$

Question 16: [4]

(a) If $A, B, C$ are three non- collinear points with position vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ respectively, then show that the length of the perpendicular from $C \ on \ AB$ is $\frac{ | (\overrightarrow{a} \times \overrightarrow{b} +\overrightarrow{b} \times \overrightarrow{c} +\overrightarrow{b} \times \overrightarrow{a} ) | }{| \overrightarrow{a} - \overrightarrow{b} | }$

(b) Show that the four points A,B, C and D with position vectors $4\hat{i} + 5\hat{j} + \hat{k}, - \hat{j} - \hat{k}, 3\hat{i} + 9\hat{j} + 4\hat{k}$ and $4( -\hat{i} + \hat{j} + \hat{k})$

Answer:

(a) Let $ABC$ be a triangle and let $\overrightarrow{a} , \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of its vertices $A, B, C$ respectively. Let $CM$ be the perpendicular from $C$ on $AB$ . Then,

Area of $\triangle ABC = \frac{1}{2} AB.CM =$ $\frac{1}{2}$ $| \overrightarrow{AB} | (CM)$

Also, Area of $\triangle ABC =$ $\frac{1}{2}$ $| \overrightarrow{AB} \times \overrightarrow{AC} |$

Area of $\triangle ABC =$ $\frac{1}{2}$ $| \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a} |$

Therefore

$\frac{1}{2}$ $| \overrightarrow{AB} | (CM) =$ $\frac{1}{2}$ $| \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a} |$

$\Rightarrow CM =$ $\frac{| \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a} |}{ | \overrightarrow{AB} |}$

$\Rightarrow CM =$ $\frac{| \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a} |}{ |\overrightarrow{b} - \overrightarrow{a} |}$

(b) Given: $\overrightarrow{a} = 4\hat{i} + 5\hat{j} + \hat{k}$

$\overrightarrow{b} = - \hat{j} - \hat{k}$

$\overrightarrow{c} = 3\hat{i} + 9\hat{j} + 4\hat{k}$

$\overrightarrow{d} = 4( -\hat{i} + \hat{j} + \hat{k})$

$\overrightarrow{AB} = \overrightarrow{b} - \overrightarrow{a} = -4 \hat{i}-6\hat{j} - 2\hat{k}$

$\overrightarrow{AC} = \overrightarrow{c} - \overrightarrow{a} = - \hat{i}+4\hat{j} +3\hat{k}$

$\overrightarrow{AD} = \overrightarrow{d} - \overrightarrow{a} = -8 \hat{i}-\hat{j} +3\hat{k}$

$\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}$ are coplanar if $\begin{bmatrix} \overrightarrow{AB} & \overrightarrow{AC} & \overrightarrow{AD}\end{bmatrix} = 0$

i.e. $\overrightarrow{AB} (\overrightarrow{AC} \times \overrightarrow{AD}) = 0$

Therefore $\begin{bmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{bmatrix} = -4(15) + 6(21) -2(66) = -60+126-66 = 0.$

Therefore $\overrightarrow{AB}, \overrightarrow{AC} \ and \ \overrightarrow{AD}$ are coplanar.

Therefore Points $A, B, C$ and $D$ are coplanar

$\\$

Question 17: [4]

(a) Draw a rough sketch of the curve and find the area of the region bounded by curve $y^2 = 8x$ and the line $x =2$ .

(b) Sketch the graph of $y = |x + 4|$ . Using integration, find the area of the region bounded by the curve $y = |x + 4|$ and $x = -6$ and $x = 0$ .

Answer:

(a) Given equation is $y^2 = 8x$

Comparing with $y^2 = 4ax$

We get $4a = 8 \Rightarrow a = 2$

Given $y^2 = 4(2) x$

$y^2 = 8x \Rightarrow y = \sqrt{8x}$

Also, $x = 2$ meets $y^2 = 8x$

$y^2 =16 \Rightarrow y = \pm 4$

Therefore $(2,4)$ and $(2,-4)$ are their point of intersection.

Required are $A = 2 \int \limits_{0}^{2} \sqrt{8x} \ dx = 2\sqrt{8}\int \limits_{0}^{2} x^{1/2} \ dx$

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

$=$ $\frac{8\sqrt{2}}{3}$ $(2^{\frac{3}{2}} - 0 )$

$=$ $\frac{8\sqrt{2}}{3}$ $\times \sqrt{8}$

$=$ $\frac{8\sqrt{2} \times 2\sqrt{2}}{3}$ $=$ $\frac{32}{3}$ sq. units.

(b) $y = x + 4$ , if $x > 4$

$y = -(x + 4)$ , if $x < 4$

For $y = x + 4$

when $x = 0, y = 4$

when $y = 0, x = -4$

Points are $(0, 4)$ and $(-4, 0)$

For $y = -x - 4$

when $x = 0, y = -4$

when $y = 0, x = -4$

Points are $(0, -4)$ and $(-4, 0)$

Therefore the required area:

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

$= - \Big[$ $\frac{x^2}{2}$ $+4x \Big]_{-6}^{-4} + \Big[$ $\frac{x^2}{2}$ $+4x \Big]_{-6}^{-4}$

$= - \Big[$ $\frac{(-4)^2}{2}$ $+ 4(-4) - \Big[$ $\frac{(-6)^2}{2}$ $+ 4(-6) \Big] \Big] + \Big[ 0+0 - \Big[$ $\frac{(-4)^2}{2}$ $+ 4(-4) \Big] \Big]$

$= - \Big[$ $\frac{16}{2}$ $-16 - \Big[$ $\frac{36}{2}$ $-24 \Big] \Big] + \Big[ - \Big[$ $\frac{16}{2}$ $-16 \Big] \Big]$

$= -(-8+6) + 8$

$= 10$ sq. units

$\\$

Question 18: Find the image of a point having position vector : $3\hat{i} - 2\hat{j} + \hat{k}$ in the plane $\overrightarrow{r}.(3\hat{i} - \hat{j} + 4\hat{k}) = 2$ [6]

Answer:

Let $B$ be root of point $A (3\hat{i} - 2\hat{j} + \hat{k})$ in the plane $\overrightarrow{r}.(3\hat{i} - \hat{j} + 4\hat{k}) = 2$ can of $AB$ is

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

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

$\Rightarrow x = 3\lambda + 3, y = -\lambda-2, z = 4\lambda+1$

substitute $x, \ y$ and $z$ in plane $3x - y + 4z = 2$

$3(3\lambda+3) - (-\lambda-2)+4(4\lambda+1) = 2$

$9\lambda+ 9 + \lambda + 2 + 16\lambda + 4 = 2$

$26\lambda+13=0 \Rightarrow \lambda = -$ $\frac{1}{2}$

Therefore $x = -$ $\frac{3}{2}$ $+3 =$ $\frac{3}{2}$

$y =$ $\frac{1}{2}$ $-2 = -$ $\frac{3}{2}$

$z = -2 +1 = -1$

Therefore by mid point formula

$\frac{3}{2}$ $=$ $\frac{3+x_1}{2}$ $\Rightarrow x_1 = 0$

$-\frac{3}{2}$ $=$ $\frac{-2+y_1}{2}$ $\Rightarrow y_1 = -1$

$-1 =$ $\frac{1-z_1}{2}$ $\Rightarrow z_1 = 3$

$\\$

SECTION C (20 Marks)

Question 19: [3 × 2]

(a) Given the total cost function of $x$ units of a commodity as

$C(x) =$ $\frac{1}{3}$ $x^3 +3x^2 -16x + 2$

Find: (i) Marginal cost function (ii) Average cost function

(b) Find the coefficient of correlation from the regression lines: $x-2y+3 = 0$ and $4x-5y+1 = 0$ .

(c) The average cost function associated with producing and marketing $x$ units of an item is given by $AC = 2x - 11 +$ $\frac{50}{x}$ . Find the range of values of the output $x$ , from which $AC$ is increasing.

Answer:

(a) $C(x) =$ $\frac{1}{3}$ $x^3 +3x^2 -16x + 2$

Marginal cost function

$\frac{dC}{dx}$ $= 3 \times$ $\frac{1}{3}$ $x^2 + 6x-16$

$\Rightarrow$ $\frac{dC}{dx}$ $= x^2 + 6x-16$

Average cost function

$\overline{C}(x) =$ $\frac{\frac{1}{3} x^3 + 3x^2 - 16x + 2}{x}$

$\Rightarrow \overline{C}(x) = \frac{1}{3} x^2 + 3x - 16$

(b) Let $x - 2y + 3 = 0$ be of $x$ on $y$ .

Therefore $x - 2y + 3 = 0 \Rightarrow x = 2y - 3 \Rightarrow b_{xy} = 2$

Similarly, let $4x-5y+1 =0$ be of $y$ on $x$

Therefore $4x-5y+1 = 0 \Rightarrow 5y = 4x+1 \Rightarrow y = \frac{4}{5} x + 1 \Rightarrow b_{yx} =$ $\frac{4}{5}$

Hence correlation $r = \sqrt{b_{xy} . b_{yx}} = \sqrt{2 \times \frac{4}{5}} = 2\sqrt{0.4} = 1.265$

$\frac{dAC}{dx}$ $= 2 - 50x^{-2} = 2 -$ $\frac{50}{x^2}$

Now $\frac{dAC}{dx}$ $\geq 0$

$\Rightarrow 2 -$ $\frac{50}{x^2}$ $\geq 0$

$\Rightarrow$ $\frac{50}{x^2}$ $\leq 2$

$\Rightarrow x \geq 5$

Therefore for $x \geq 5, \ AC$ will keep increasing.

$\\$

Question 20: [4]

(a) Find the line of regression of $y$ on $x$ from the following table.

$x$	1	2	3	4	5
$y$	7	6	5	4	3

Hence estimate the value of $y$ when $x = 6$ .

(b) From the given data:

Variable	$x$	$y$
Mean	6	8
Standard Deviation	4	6

And the correlation coefficient: $\frac{2}{3}$ . Find

(i) Regression coefficients $b_{yx}$ and $b_{xy}$

(ii) Regression line $x$ on $y$

(iii) Most likely value of $x$ when $y = 14$

Answer:

(a)

$x$	$y$	$dx = x-1$	$dy = y-5$	$dx.dy$	$(dx)^2$	$(dy)^2$
1	7	0	2	0	0	4
2	6	1	1	1	1	1
3	5	2	0	0	4	0
4	4	3	-1	-3	9	1
5	3	4	-2	-8	16	4
$\Sigma x \\ = 15$	$\Sigma y \\ = 25$	$\Sigma dx \\ = 10$	$\Sigma dy \\ = 0$	$\Sigma dx.dy \\ = -10$	$\Sigma(dx)^2 \\ = 30$	$\Sigma (dy)^2 \\ = 10$

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

$=$ $\frac{10 \times (-10) - 10 \times 0}{\sqrt{\Big( 5 \times 30 - (10)^2 \Big) \Big( 5 \times 10 - (0)^2 \Big) }}$

$=$ $\frac{-50}{\sqrt{50 \times 50}}$ $= -1$

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

$\overline{x} =$ $\frac{15}{5}$ $=3$

$\overline{y} =$ $\frac{25}{5}$ $= 5$

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

$\Rightarrow y - 5 =$ $\frac{-10}{30}$ $(x-3)$

When $x = 6$ we get $y - 5=$ $\frac{-10}{30}$ $(6-3)$

$\Rightarrow 3y = 12-6$

$\Rightarrow y =$ $\frac{6}{2}$ $= 2$

(b) To Be Solved

$\\$

Question 21: [4]

(a) A product can be manufactured at a total cost $C(x) =$ $\frac{x^2}{100}$ $+ 100x + 40$ , where $x$ is the number of units produced. The price at which each unit can be sold is given by $P = \Big(200 -$ $\frac{x}{400}$ $\Big)$ . Determine the production level $x$ at which the profit is maximum.What is the price per unit and profit at the level of production.

(b) A manufacturer’s marginal cost function is $\frac{500}{\sqrt{2x + 25}}$ . Find the cost involved to increase production from $100$ units to $300$ units.

Answer:

(a) Total Cost $C(x) =$ $\frac{x^2}{100}$ $+ 100x + 40$

Price $P = \Big(200 -$ $\frac{x}{400}$ $\Big)$

If $x$ units are produced, then

$Profit(x) = x (200-$ $\frac{x}{100}$ $) - ($ $\frac{x^2}{100}$ $+ 100 x + 40 )$

$= 200x -$ $\frac{x^2}{400}$ $-$ $\frac{x^2}{100}$ $- 100x - 40$

$= 100x -$ $\frac{5}{400}$ $x^2 - 40$

For maximizing profits $\frac{dProfit}{dx}$ $= 0$

$\Rightarrow 100 -$ $\frac{10x}{400}$ $= 0$

$\Rightarrow x =$ $\frac{100 \times 400}{100}$ $= 4000$ units.

$P(4000) = 200 -$ $\frac{1000}{400}$ $= 190$ units

(b) $\frac{500}{\sqrt{2x + 25}}$

The cost involved to increase production from $100$ units to $300$ units:

$= \int \limits_{100}^{300}$ $\frac{500}{\sqrt{2x + 25}}$

$= 500 \int \limits_{100}^{300} (2x+25)^{-\frac{1}{2}}$

$= 500 \times$ $\frac{2}{3}$ $(2x+25)^{\frac{3}{2}}$ $|$ $_{100}^{300}$

$=$ $\frac{1000}{3}$ $\Big[ (2 \times 300 + 25)^{\frac{3}{2}} - (2 \times 100 + 25)^{\frac{3}{2}} \Big]$

$=$ $\frac{1000}{3}$ $\Big[ 625^{\frac{3}{2}} - 225^{\frac{3}{2}} \Big]$

$=$ $\frac{1000}{3}$ $\Big[ 25^3 - 15^3 \Big]$

$=$ $\frac{1000}{3}$ $\times 12250 = 4083333$ $\frac{1}{3}$ units

$\\$

Question 22: A manufacturing company produces two type of teaching aids $A$ and $B$ of Mathematics for Class X. Each type of $A$ requires $9$ hours for fabricating and $1$ hour for finishing. Each type of $B$ requires $12$ hours for fabricating and $3$ hours for finishing. For fabricating and finishing, the maximum labor hours available for a week are $180$ and $30$ respectively. The company makes a profit of $Rs. 80$ on each piece of type $A$ and $Rs. 120$ on each piece of type $B$ . How many pieces of type $A$ and type $B$ should be manufactured per week to get a maximum profit? Formulate this as a linear programming problem and solve it. Identify the feasible region from the rough sketch. [6]

Answer:

Let quantity of teaching aid $Type \ A = x$ and Quantity of teaching aid $Type \ B = y$

Each type of $A$ requires $9$ hours for fabricating and $1$ hour for finishing.

Therefore $9x+12y \leq 180 \Rightarrow 3x + 4y \leq 60$ … … … … … (i)

Each type of $B$ requires $12$ hours for fabricating and $3$ hours for finishing.

Therefore $x+3y \leq 30$ … … … … … (ii)

We know the company makes a profit of $Rs. 80$ on each piece of type $A$ and $Rs. 120$ on each piece of type $B$ .

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

Therefore Profit $= 80x + 120 y$ … … … … … (iii)

We need to maximize $(80x + 120 y)$ . We will solve it graphically.

We see that the coordinates of the vertices of the feasible region are $(0, 10), (20, 0) \ and \ (12,6)$ . Now calculate Profit for each of the three coordinates:

For $(0, 10): Profit = 80 \times 0 + 120 \times 10 = 1200 \ Rs.$

For $(20,0): Profit = 80 \times 20 + 120 \times 0 = 1600 \ Rs.$

For $(12,6): Profit = 80 \times 12 + 120 \times 6 = 1680 \ Rs.$

Therefore, the quantity of teaching aid $Type \ A = 12$ and Quantity of teaching aid $Type \ 6 = y$

$\\$

2 thoughts on “2018 ICSE (Class 12) Board Paper Solution: Mathematics”

Shreyansh Dwivedi says:

December 8, 2019 at 8:18 pm

Solution of Q.no. 19(b) is wrong, because r² <= 1 (always) and yours is greater.

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1. ICSE CBSE Mathematics Student-Assistant says:
  
  December 13, 2019 at 6:23 am
  
  i will check… thanks
  
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2018 ICSE (Class 12) Board Paper Solution: Mathematics

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