2019 ISC (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) If $f : R \rightarrow R, f(x)= x^3$ and $g : R \rightarrow R, g(x) = 2x^2+1$ , and $R$ is the set of real numbers, then find $fog (x)$ and $gof (x)$ .

ii) Solve: $\sin (2 \tan^{-1} x) = 1$

iii) Using determinants, find the values of $k$ , if the area of triangle with vertices $(-2, 0), (0, 4)$ and $(0, k)$ is $4$ square units.

iv) Show that $(A + A')$ is symmetric matrix, if $A = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}$

v) $f(x) =$ $\frac{x^2-9}{x-3}$ is not defined at $x = 3$ . What value should be assigned to $f(3)$ for continuity of $f(x)$ at $x = 3$ ?

vi) Prove that the function $f(x) = x^3 - 6x^2 + 12x + 5$ is increasing on $R$ .

vii) Evaluate: $\int \limits_{}^{}$ $\frac{\sec^2 x}{\mathrm{cosec}^2 x}$ $dx$

viii) Using L’Hospital’s Rule, evaluate: $\lim \limits_{x \to 0}$ $\frac{8^x-4^x}{4x}$

ix) Two balls are drawn from an urn containing $3$ white, $5$ red and $2$ black balls, one by one without replacement. What is the probability that at least one ball is red?

x) If events $A$ and $B$ are independent, such that $P(A) =$ $\frac{3}{5}$ and $P(B) =$ $\frac{2}{3}$ , find $P(A \cup B)$ .

Answer:

i) Given $f : R \rightarrow R, f(x)= x^3$ and $g : R \rightarrow R, g(x) = 2x^2+1$

$\therefore f(g(x)) = ((g(x))^3 = (2x^2+1)^3 = 8x^6+ 12x^4+6x^2+1$

Also $g(f(x)) = 2(f(x))^2 + 1 = 2(x^3)^2 + 1 = 2x^6 + 1$

ii) $\sin (2 \tan^{-1} x) = 1$

$\because \sin 2\theta =$ $\frac{2 \tan \theta}{1 + \tan^2 \theta}$

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

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

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

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

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

$\Rightarrow x-1 = 0$

$x = 1$

iii) $Area \ ( \triangle ABC) = 4$

$\frac{1}{2}$ $\left| \left| \begin{array}{ccc} 1 & 1 & 1 \\ -2 & 0 & 0 \\ 0 & 4 & k \end{array} \right|\right|= 4$

$1 (0- 0) - 1 ( -2k) + 1 (-8 -0) = \pm 8$

$2k-8 = \pm 8$

$2k = \pm 8 + 8$

$\therefore k =$ $\frac{16}{2}$ or $k =$ $\frac{0}{2}$

$\Rightarrow k = 8$ or $k = 0$

iv) $A = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}$

$A^T = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$

$A + A^T = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ 7 & 10 \end{bmatrix}$

$\therefore (A + A^T)^T = \begin{bmatrix} 4 & 7 \\ 7 & 10 \end{bmatrix} = (A + A^T)$

$\therefore A + A^T$ is symmetric matrix

v) To be continuous at $x = 3$

$f(3) =$ $\lim \limits_{x \to 3}$ $\frac{x^2-9}{x-3}$

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

$=$ $\lim \limits_{x \to 3} (x+3) = 6$

$f(3) = 6$

vi) $f(x) = x^3 - 6x^2 + 12x + 5 ; \hspace*{1.0cm} x \in R$

$f'(x) = 3x^2 - 12x + 12 ; \hspace*{1.0cm} x \in R$

$f'(x) = 3 ( x^2 - 4x + 4) = 3 (x-2)^2 ; \hspace*{1.0cm} x \in R$

$\therefore f'(x) \geq 0 ; \hspace*{1.0cm} x \in R$

$\therefore f(x) =$ is increasing for $x \in R$

vii) $I =$ $\int \limits_{}^{}$ $\frac{\sec^2 x}{\mathrm{cosec}^2 x}$ $dx$

$I =$ $\int \limits_{}^{}$ $\frac{\sin^2 x}{\cos^2 x}$ $dx$

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

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

$I = \tan x - x + C$

viii) $\lim \limits_{x \to 0}$ $\frac{8^x-4^x}{4x}$

$= \lim \limits_{x \to 0}$ $\frac{8^x \log 8 -4^x \log 4}{4 \times 1}$

$= \lim \limits_{x \to 0}$ $\frac{8^0 \log 8 -4^0 \log 4}{4 \times 1}$

$=$ $\frac{1}{4}$ $\log 2$

ix) $P ($ at least one ball is Red $) = 1 - P($ no ball is Red $)$

$= 1- P ($ no Red ball in first draw $) \times P ($ no Red ball in second draw $)$

$= 1 -$ $\frac{5}{10}$ $\times$ $\frac{4}{9}$

$= 1 -$ $\frac{20}{90}$ $=$ $\frac{7}{9}$

x) Since A and B are independent,

$P(A \cup B) = P(A) + P(B)-P(A) \times P(B)=$ $\frac{3}{5}$ $+$ $\frac{2}{3}$ $-$ $\frac{3}{5}$ $\times$ $\frac{2}{3}$ $=$ $\frac{9+10}{15}$ $-$ $\frac{6}{15}$ $=$ $\frac{13}{15}$

$\\$

Question 2: If $f : A \rightarrow A$ and $A = R -\{$ $\frac{8}{5} \}$ show that the function $f(x) =$ $\frac{8x+3}{5x-8}$ is one $-$ one onto. Hence, find $f^{-1}$ . [ 4 ]

Answer:

Given $f(x) =$ $\frac{8x+3}{5x-8}$

Therefore $f(x_1) =$ $\frac{8x_1+3}{5x_1-8}$ and $f(x_2) =$ $\frac{8x_2+3}{5x_2-8}$

$f(x_1) = f(x_2)$

$\frac{8x_1+3}{5x_1-8}$ $=$ $\frac{8x_2+3}{5x_2-8}$

$(8x_1+3)(5x_2-8) = (5x_1-8)(8x_2+3)$

$40 x_1 x_2 - 64 x_1 + 15 x_2 - 24 = 40 x_1 x_2 - 64 x_2 + 15x_1 - 24$

$40 x_1 x_2 - 40 x_1 x_2 - 64 x_1 + 64 x_2 + 15x_2 - 15 x_1 = 0$

$64(x_2-x_1)+ 15(x_2 -x_1) = 0$

$79(x_2-x_1)=0$

$\therefore x_2 = x_1$

$\therefore f(x)$ is one $-$ one function.

Let $f (x)= y$

$y =$ $\frac{8x+3}{5x-8}$

$5yx - 8y = 8x + 3$
$5yx - 8x = 3+ 8y$
$x(5y - 8)= 8y + 3$

$x =$ $\frac{8y-3}{5y-8}$ $; y \in R - \{$ $\frac{8}{5}$ $\}$

Therefore Co-domain $=$ Range

$f (x)$ is onto function

$\because$ $\frac{8y-3}{5y-8}$

Replacing $x$ by $y$ and $y$ by $x$ .

$y =$ $\frac{8x-3}{5x-8}$

$f^{-1} (x) =$ $\frac{8x-3}{5x-8}$

$\\$

Question 3:

a) Solve for $x: \tan^{-1} \Big($ $\frac{x-1}{x-2}$ $\Big) + \tan^{-1} \Big($ $\frac{x+1}{x+2}$ $\Big) =$ $\frac{\pi}{4}$

b) If $\sec^{-1} = \mathrm{cosec}^{-1}y$ , show that $\frac{1}{x^2}$ $+$ $\frac{1}{y^2}$ $= 1$ [ 4 ]

Answer:

a) Simplify the expression $\tan^{-1} \Big($ $\frac{x-1}{x-2}$ $\Big) + \tan^{-1} \Big($ $\frac{x+1}{x+2}$ $\Big) =$ $\frac{\pi}{4}$

We know, that $\tan^{-1}(A+B) = \tan^{-1} \Big($ $\frac{A+B}{1 - AB}$ $\Big)$

Therefore

$\tan^{-1} \Big($ $\frac{x-1}{x-2}$ $\Big) + \tan^{-1} \Big($ $\frac{x+1}{x+2}$ $\Big) =$ $\frac{\pi}{4}$

$\tan^{-1} \Bigg($ $\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1 - \frac{x-1}{x-2} . \frac{x+1}{x+2}}$ $\Bigg) =$ $\frac{\pi}{4}$

$\tan^{-1} \Bigg($ $\frac{ \frac{x^2 - x + 2x-2+x^2-2x+x-2}{x^2 - 4} }{ \frac{x^2 - 4 -x^2 -1}{x^2 - 4} }$ $\Bigg) =$ $\frac{\pi}{4}$

$\tan^{-1} \Big($ $\frac{ x^2 - x + 2x-2+x^2-2x+x-2} { x^2 - 4 -x^2 +1}$ $\Big) =$ $\frac{\pi}{4}$

$\tan^{-1} \Big($ $\frac{ 2x^2 - 4} { -3}$ $\Big) =$ $\frac{\pi}{4}$

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

$2x^2 - 4 = -3$

$2x^2 = 1$

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

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

b) Given $\sec^{-1} x = \mathrm{cosec}^{-1}y$

$\cos^{-1} \Big($ $\frac{1}{x}$ $\Big) = \sin^{-1} \Big($ $\frac{1}{y}$ $\Big)$

$\cos^{-1} \Big($ $\frac{1}{x}$ $\Big) =$ $\frac{\pi}{2}$ $- \cos^{-1} \Big($ $\frac{1}{y}$ $\Big)$

$\cos^{-1} \Big($ $\frac{1}{x}$ $\Big) + \cos^{-1} \Big($ $\frac{1}{y}$ $\Big) =$ $\frac{\pi}{2}$

$\cos^{-1} \Big($ $\frac{1}{x}$ $\times$ $\frac{1}{y}$ $- \sqrt{1 - \frac{1}{x^2}} \times \sqrt{1 - \frac{1}{y^2}} \Big) =$ $\frac{\pi}{2}$

$\Big($ $\frac{1}{x}$ $\times$ $\frac{1}{y}$ $- \sqrt{1 - \frac{1}{x^2}} \times \sqrt{1 - \frac{1}{y^2}} \Big) = \cos \Big($ $\frac{\pi}{2}$ $\Big)$

$\frac{1}{xy}$ $-$ $\sqrt{ 1 - \frac{1}{y^2} - \frac{1}{x^2} + \frac{1}{x^2y^2} }$ $= 0$

$\frac{1}{xy}$ $=$ $\sqrt{ 1 - \frac{1}{y^2} - \frac{1}{x^2} + \frac{1}{x^2y^2} }$

Squaring both sides

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

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

Hence Proved

$\\$

Question 4: Using properties of determinants prove that:

$\left| \begin{array}{ccc} x & x(x^2+1) & x+1 \\ y & y(y^2+1) & y+1 \\ z & z(z^2+1) & z+1 \end{array} \right| = (x-y)(y-z)(z-x)(x+y+z)$ [ 4 ]

Answer:

LHS $= \left| \begin{array}{ccc} x & x(x^2+1) & x+1 \\ y & y(y^2+1) & y+1 \\ z & z(z^2+1) & z+1 \end{array} \right|$

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

$= \left| \begin{array}{ccc} x-y & x^3+x-y-y^3 & x+1-y-1 \\ y-z & y^3+y-z-z^3 & y+1-z-1 \\ z & z(z^2+1) & z+1 \end{array} \right|$

We know $x^3 - y^3 + x - y = (x-y)(x^2 + xy + y^2)+ x - y = (x-y)(x^2 + xy + y^2 + 1)$

$= \left| \begin{array}{ccc} x-y & (x-y)(x^2 + xy + y^2 + 1) & x-y \\ y-z & (y-z)(y^2+z^2+yz+1) & y-z \\ z & z(z^2+1) & z+1 \end{array} \right|$

$= (x-y)(y-z) \left| \begin{array}{ccc} 1 & x^2 + xy + y^2 + 1 & 1 \\ 1 & y^2+z^2+yz+1 & 1 \\ z & z(z^2+1) & z+1 \end{array} \right|$

$R_1 \rightarrow R_1 - R_2$

$\because (x^2 + xy + y^2 + 1) - (y^2+z^2+yz+1) = (x-z)(x+z) + y(x-z) = (x-z)(x+y+z)$

$= (x-y)(y-z) \left| \begin{array}{ccc} 0 & (x-z)(x+y+z) & 0 \\ 1 & y^2+z^2+yz+1 & 1 \\ z & z(z^2+1) & z+1 \end{array} \right|$

$= (x-y)(y-z)(x-z) \left| \begin{array}{ccc} 0 & x+y+z & 0 \\ 1 & y^2+z^2+yz+1 & 1 \\ z & z(z^2+1) & z+1 \end{array} \right|$

$= (x-y)(y-z)(x-z) [ -(x+y+x)(z+1 - z) ]$

$= - (x-y)(y-z)(x-z)(x+ y + z)$

$= (x-y)(y-z)(z-x)(x+ y + z) =$ RHS

Hence Proved

$\\$

Question 5:

a) Show that the function $f(x) = | x - 4 | , x \in R$ is continuous, but not differentiable at $x = 4$ .

b) Verify the Lagrange’s mean value theorem for the function: $f(x) = x+$ $\frac{1}{x}$ in the interval $[1, 3]$ [ 4 ]

Answer:

a) $f(x) = x - 4$ for $x \geq 0$

and $f(x) = 4-x$ for $x < 0$

L.H.L. $: \ \lim \limits_{x \to 4^-} f(x) = \lim \limits_{x \to 4^-} 4-x = 4 - 4 = 0$

R.H.L. $: \ \lim \limits_{x \to 4^+} f(x) = \lim \limits_{x \to 4^+} x-4 = 4 - 4 = 0$

$f(4) = 4 - 4 = 0$

$\therefore R.H.L = L.H.L = f(4)$

$\therefore f(x)$ is continuous at $x = 4$

R.H.D $= f'(4^+) = \lim \limits_{h \to 0}$ $\frac{f(4+h) - f(4)}{(4+h) - 4}$ $= \lim \limits_{h \to 0}$ $\frac{4+h-4 - 0}{h}$ $= \lim \limits_{h \to 0} 1 = 1$

L.H.D $= f'(4^-) = \lim \limits_{h \to 0}$ $\frac{f(4) - f(4-h)}{4 - (4-h)}$ $= \lim \limits_{h \to 0}$ $\frac{0 - (4 -(4-h))}{h}$ $= \lim \limits_{h \to 0} -1 = -1$

Since L.H.D $\neq$ R.H.D, $f'(4)$ does not exist.

$\therefore f(x)$ is continuous at $x = 4$ but is non differentiable at $x=4$ .

b) Given $f(x) = x+$ $\frac{1}{x}$ and $x \in [1, 3]$

$f(x)$ is continuous for $x \in [1, 3]$

$f(x)$ is differentiable for $x \in [1, 3]$

Therefore Lagrange’s mean value theorem is applicable.

$f(1) = 1+$ $\frac{1}{1}$ $= 2$

$f(3) = 3+$ $\frac{1}{3}$ $=$ $\frac{10}{3}$

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

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

$\therefore f'(c) =$ $\frac{f(b) - f(a)}{b-a}$

$\Rightarrow 1 -$ $\frac{1}{c^2}$ $=$ $\frac{\frac{10}{3} - 2}{3-1}$ $=$ $\frac{4}{6}$ $=$ $\frac{2}{3}$

$\Rightarrow 1-$ $\frac{2}{3}$ $=$ $\frac{1}{c^2}$

$\Rightarrow c^2 = 3$

$\Rightarrow c = \pm \sqrt{3}$

We know, $c = \sqrt{3} \in [1, 3]$

Hence Lagrange’s mean value theorem is verified.

$\\$

Question 6: If $y = e^{\sin^{-1} x}$ and $z = e^{- \cos^{-1} x}$ , prove that $\frac{dy}{dz}$ $= e^{\frac{\pi}{2}}$ [ 4 ]

Answer:

Given $y = e^{\sin^{-1} x}$ and $z = e^{- \cos^{-1} x}$

Therefore $\frac{y}{z}$ $=$ $\frac{e^{\sin^{-1} x}}{e^{- \cos^{-1} x}}$ $=$ $e^{\cos^{-1} x +\sin^{-1} x }$

$\frac{y}{z}$ $=$ $e^{\frac{\pi}{2}}$

$y = e^{\frac{\pi}{2}} z$

$\frac{dy}{dz}$ $=$ $e^{\frac{\pi}{2}}$ $\frac{dz}{dz}$

$\frac{dy}{dz}$ $=$ $e^{\frac{\pi}{2}}$

$\\$

Question 7: A $13$ m long ladder is leaning against a wall, touching the wall at a certain height from the ground level. The bottom of the ladder is pulled away from the wall, along the ground, at the rate of $2$ m/s. How fast is the height on the wall decreasing when the foot of the ladder is $5$ m away from the wall? [ 4 ]

Answer:

$x^2 + y^2 = (13)^2$ … … … … … i)

The bottom of the ladder is begin pulled, so these distances are changing with time,

$2x$ $\frac{dx}{dt}$ $+ 2y$ $\frac{dy}{dt}$ $= 0$ … … … … … ii)

$\frac{dy}{dt}$ $= -$ $\frac{x}{y}$ $\frac{dx}{dt}$ … … … … … iii)

$x = 5m ,$ $\frac{dx}{dt}$ $= 2$ $\frac{m}{s}$

If $x = 5$ m, then $y = 12$ m from i) (Pythagoras theorem)

From equation iii) we get $\frac{dy}{dt}$ $= -$ $\frac{5}{12}$ $\times 2 = -$ $\frac{5}{6}$ $\frac{m}{s}$

$-ve$ sign indicates the height is decreasing

$\\$

Question 8:

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

b) Evaluate: $\int \limits_{-6}^{3}$ $|x+3| \ dx$ [ 4 ]

Answer:

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

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

Let $1 + x^2 = t$

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

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

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

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

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

$I =$ $\frac{1}{4}$ $\Bigg($ $\int \limits_{}^{}$ $\frac{(2t-2) }{t^2 - 2t+2}$ $dt + 2$ $\int \limits_{}^{}$ $\frac{2 }{t^2 - 2t+2}$ $\ dt \Bigg)$

$I =$ $\frac{1}{4}$ $\log | t^2 - 2t + 2 | +$ $\frac{2}{4}$ $\int \limits_{}^{}$ $\frac{1}{(t-1)^2+ 1}$ $\ dt + C_1$

$I =$ $\frac{1}{4}$ $\log | t^2 - 2t + 2 | +$ $\frac{1}{2}$ $\tan^{-1} (t-1) + C_1$

$\because x^2 + 1 = t$

$I =$ $\frac{1}{4}$ $\log | (x^2 + 1)^2 - 2(x^2 + 1) + 2 | +$ $\frac{1}{2}$ $\tan^{-1} (x^2 + 1-1) + C$

$I =$ $\frac{1}{4}$ $\log | x^4 + 1 + 2x^2 - 2x^2 - 2 + 2 | +$ $\frac{1}{2}$ $\tan^{-1} (x^2) + C$

$I =$ $\frac{1}{4}$ $\log | x^4 + 1 | +$ $\frac{1}{2}$ $\tan^{-1} (x^2) + C$

b) Given $\int \limits_{-6}^{3}$ $|x+3|dx$

$f(x) = | x+3|$

Therefore

$f(x) = x+ 3 ; x+3 \geq 0$ for all $x \geq -3$

$f(x) = -x- 3 ; x+3 < 0$ for all $x < -3$

$I =$ $\int \limits_{-6}^{-3}$ $(-x-3) \ dx +$ $\int \limits_{-3}^{3}$ $(x+3) \ dx$

$I = - \Big($ $\frac{x^2}{2}$ $+ 3x \Big)_{-6}^{-3} + \Big($ $\frac{x^2}{2}$ $+ 3x \Big)_{-3}^{3}$

$I = - \Bigg[ \Big($ $\frac{9}{2}$ $- 9 \Big) - \Big($ $\frac{36}{2}$ $- 18 \Big) \Bigg] + \Bigg[ \Big($ $\frac{9}{2}$ $+ 9 \Big) - \Big($ $\frac{9}{2}$ $- 9 \Big) \Bigg]$

$I = - \Big[ -$ $\frac{9}{2}$ $\Big] + 18 =$ $\frac{9}{2}$ $+ 18 =$ $\frac{45}{2}$

$\\$

Question 9: Solve the differential equation: $\frac{dy}{dx}$ $=$ $\frac{x+y+2}{2(x+y)-1}$ [ 4 ]

Answer:

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

Let $x + y = t$

$\Rightarrow$ $1 +$ $\frac{dy}{dx}$ $=$ $\frac{dt}{dx}$ $\Rightarrow$ $\frac{dy}{dx}$ $=$ $\frac{dt}{dx}$ $- 1$

$\Rightarrow$ $\frac{dt}{dx}$ $- 1 =$ $\frac{t+2}{2t-1}$

$\Rightarrow$ $\frac{dt}{dx}$ $=$ $\frac{t+2}{2t-1}$ $+ 1$

$\Rightarrow$ $\frac{dt}{dx}$ $=$ $\frac{t+2+ 2t-1}{2t-1}$

$\Rightarrow$ $\frac{dt}{dx}$ $=$ $\frac{3t+1}{2t-1}$

$\Rightarrow$ $\Big($ $\frac{2t-1}{3t+1}$ $\Big) dt = dx$

$\Rightarrow$ $\frac{2}{3}$ $\Bigg($ $\frac{3(2t-1)}{2(3t+1)}$ $\Bigg) dt = dx$

$\Rightarrow$ $\frac{2}{3}$ $\int \limits_{}^{}$ $\frac{6t-3}{6t+2}$ $\ dt=$ $\int \limits_{}^{}$ $\ dx$

$\Rightarrow$ $\frac{2}{3}$ $\int \limits_{}^{}$ $\frac{(6t+2) - 5}{6t+2}$ $\ dt= x + C_1$

$\Rightarrow$ $\frac{2}{3}$ $\int \limits_{}^{}$ $\ dt -$ $\frac{10}{3}$ $\int \limits_{}^{}$ $\frac{1}{6t+2}$ $\ dt = x + C_1$

$\Rightarrow$ $\frac{2}{3}$ $\int \limits_{}^{}$ $\ dt -$ $\frac{5}{3}$ $\int \limits_{}^{}$ $\frac{1}{3t+1}$ $\ dt = x + C_1$

$\Rightarrow$ $\frac{2}{3}$ $t -$ $\frac{5}{3}$ $\times$ $\frac{\log |3t+1|}{3}$ $= x + C$

$\Rightarrow$ $\frac{2}{3}$ $t -$ $\frac{5}{9}$ $\times \log |3t+1| = x + C$

$\Rightarrow$ $\frac{2}{3}$ $(x + y) -$ $\frac{5}{9}$ $\times \log |3(x + y)+1| = x + C$

$\Rightarrow$ $\frac{2}{3}$ $(x + y) -$ $\frac{5}{9}$ $\times \log |3x + 3y+1| = x + C$

$\\$

Question 10: Bag $A$ contains $4$ white balls and $3$ black balls, while Bag $B$ contains $3$ white balls and $5$ black balls. Two balls are drawn from Bag $A$ and placed in Bag $B$ . Then, what is the probability of drawing a white ball from Bag $B$ ? [ 4 ]

Answer:

Bag $A$ contains $4$ white balls and $3$ black balls

Bag $B$ contains $3$ white balls and $5$ black balls

Case I : Let Both white balls are transferred from Bag $A$ to Bag $B$ and then a white ball is drawn from Bag $B$ .

Required probability $=$ $\frac{^4C_2}{^7C_2}$ $\times$ $\frac{^5C_1}{^{10}C_1}$ $=$ $\frac{ \frac{4!}{2! 2!} \times \frac{5!}{4! 1!} }{ \frac{7!}{5! 2!} \times \frac{10!}{9! 1!} }$ $=$ $\frac{4 \times 3}{7 \times 6}$ $\times$ $\frac{5}{10}$ $=$ $\frac{2}{7}$ $\times$ $\frac{1}{2}$ $=$ $\frac{1}{7}$

Case II: Let both black balls are transferred from Bag $A$ to Bag $B$ and then a white ball is drawn from Bag $B$ .

Required probability $=$ $\frac{^3C_2}{^7C_2}$ $\times$ $\frac{^3C_1}{^{10}C_1}$ $=$ $\frac{ \frac{3!}{1! 2!} \times \frac{3!}{2! 1!} }{ \frac{7!}{5! 2!} \times \frac{10!}{9! 1!} }$ $=$ $\frac{3 \times 2}{7 \times 6}$ $\times$ $\frac{3}{10}$ $=$ $\frac{1}{7}$ $\times \frac{3}{10}$ $=$ $\frac{3}{70}$

Case III: 1 white and 1 black ball is transferred and then a white ball is drawn from Bag $B$ .

Required probability $=$ $\frac{^4C_1 \times ^3C_1}{^7C_2}$ $\times$ $\frac{^4C_1}{^{10}C_1}$ $=$ $\frac{ \frac{4!}{3! 1!} \times \frac{3!}{2! 1!} }{ \frac{7!}{5! 2!} }$ $\times$ $\frac{\frac{4!}{3! 1!}}{\frac{10!}{9! 1!}}$ $=$ $\frac{4 \times 3}{7 \times 3}$ $\times$ $\frac{4}{10}$ $=$ $\frac{4}{7}$ $\times$ $\frac{3}{10}$ $=$ $\frac{16}{70}$

Total probability $= Case \ I + Case \ II + Case \ III =$ $\frac{1}{7}$ $+$ $\frac{3}{70}$ $+$ $\frac{16}{70}$ $=$ $\frac{29}{70}$

$\\$

Question 11: Solve the following system of linear equations using matrix method: [ 6 ]

$\frac{1}{x}$ $+$ $\frac{1}{y}$ $+$ $\frac{1}{z}$ $= 9$

$\frac{2}{x}$ $+$ $\frac{5}{y}$ $+$ $\frac{7}{z}$ $= 52$

$\frac{1}{x}$ $+$ $\frac{1}{y}$ $+$ $\frac{1}{z}$ $= 0$

Answer:

Let $\frac{1}{x}$ $= X,$ $\frac{1}{y}$ $= Y,$ $\frac{1}{z}$ $= Z$

$X + Y + Z = 9$ … … … … … i)

$2X + 5Y + 7Z = 52$ … … … … … ii)

$2X + Y - Z = 0$ … … … … … iii)

$AX = B$

$\begin{bmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} =$ $\begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix}$

$R_2 \rightarrow R_2 - 2 R_1$ and $R_3 \rightarrow R_3 - 2 R_1$

$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 3 & 5 \\ 0 & -1 & -3 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} =$ $\begin{bmatrix} 9 \\ 34 \\ -18 \end{bmatrix}$

$R_2 \rightarrow R_2 + 3 R_3$

$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & -4 \\ 0 & -1 & -3 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} =$ $\begin{bmatrix} 9 \\ -20 \\ -18 \end{bmatrix}$

$\begin{bmatrix} X+Y+Z \\ -4Z \\ -Y-3Z \end{bmatrix} = \begin{bmatrix} 9 \\ -20 \\ -18 \end{bmatrix}$

$\therefore -4Z = - 20 \Rightarrow Z = 5 \Rightarrow z =$ $\frac{1}{5}$

$\therefore -Y-3Z = -18 \Rightarrow Y = 18 - 3 (5) = 3 \Rightarrow y=$ $\frac{1}{3}$

$\therefore X + Y + Z = 9 \Rightarrow X = 9 - 5 - 3 = 1 \Rightarrow x =$ $\frac{1}{1} = 1$

$\\$

Question 12:

(a) The volume of a closed rectangular metal box with a square base is $4096 \ cm^3$ . The cost of polishing the outer surface of the box is Rs. $4 \ per \ cm^2$ . Find the dimensions of the box for the minimum cost of polishing it.

(b) Find the point on the straight line $2x + 3y = 6$ , which is closest to the origin. [ 6 ]

Answer:

a) Let the base of the box be $x$ and height be $y$ .

Therefore Volume $= x^2y = 4096 \ cm^3$

Hence $y =$ $\frac{4096}{x^2}$ … … … … … i)

Total surface area $(S) = 2x^2 + 4xy = 2x^2 + 4x \Big($ $\frac{4096}{x^2}$ $\Big)$

Therefore the cost function $C(x) = 4 \Bigg( 2x^2 + 4x \Big($ $\frac{4096}{x^2}$ $\Big) \Bigg)$ Rs. … … ii)

Differentiating w.r.t. $x$ we get,

$\frac{dc}{dx}$ $= 4\Big( 4x -$ $\frac{16384}{x^2}$ $\Big)$ … … … … … iii)

Let $\frac{dc}{dx}$ $= 0 \Rightarrow 4x =$ $\frac{16384}{x^2}$

$\Rightarrow x^3 = 4096 \Rightarrow x = 16$

Differentiating equation (iii) w.r.t. $x$ we get,

$\frac{d^2c}{dx^2}$ $= 4\Big( 4 +$ $\frac{2 \times 16384}{x^3}$ $\Big)$

$\therefore$ $\frac{d^2c}{dx^2}$ at $(x = 16) = 4\Big( 4 +$ $\frac{2 \times 16384}{4096}$ $\Big) = 48 > 0$

Also $y =$ $\frac{4096}{16^2}$ $= 16$ cm

Therefore The cost for polishing the surface area is minimum when length of base is $16$ cm and height of box is $16$ cm.

b) The equation of line is given as $2x + 3y = 6$

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

Therefore The point of the line can be taken as $P = \Big( x,$ $\frac{6-2x}{3}$ $\Big)$

Distance from origin $OP = \sqrt{ (x-0)^2 + \Big( \frac{6-2x}{3} - 0 \Big)^2 }$

$OP = \sqrt{ x^2 + \Big( \frac{6-2x}{3} \Big)^2 }$

$OP^2 = x^2 +$ $\frac{4x^2}{9}$ $+ 4 -$ $\frac{8x}{3}$

$OP^2 =$ $\frac{13x^2}{9}$ $-$ $\frac{8x}{3}$ $+ 4$

Let $OP^2 = f(x) , OP$ is minimum when $OP^2$ is minimum

$\therefore f(x) =$ $\frac{13x^2}{9}$ $-$ $\frac{8x}{3}$ $+ 4$ … … … … … i)

Differentiating equation (i) w.r.t. $x$ we get,

$f'(x) =$ $\frac{26x}{9}$ $-$ $\frac{8}{3}$ … … … … … ii)

Let $f'(x) = 0 \Rightarrow$ $\frac{26x}{9}$ $=$ $\frac{8}{3}$

$\Rightarrow x =$ $\frac{24}{26}$ $=$ $\frac{12}{13}$

Differentiating equation (ii) w.r.t. $x$ we get,

$f''(x) =$ $\frac{26}{9}$ $> 0$

$\therefore OP$ is minimum at $x =$ $\frac{12}{13}$

$\therefore y = 2 -$ $\frac{2x}{3}$ $= 2 -$ $\frac{2}{3}$ $\times$ $\frac{12}{13}$ $=$ $\frac{18}{13}$

$\therefore$ The closest point on the line $2x +3y = 6$ with origin is $\Big($ $\frac{12}{13}$ $,$ $\frac{18}{13}$ $\Big)$

$\\$

Question 13: Evaluate: $\int \limits_{0}^{\pi}$ $\frac{x \tan x}{\sec x + \tan x}$ $dx$ [ 6 ]

Answer:

$I =$ $\int \limits_{0}^{\pi}$ $\frac{x \tan x}{\sec x + \tan x}$ $dx$ … … … … … i)

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

$I =$ $\int \limits_{0}^{\pi}$ $\frac{ (\pi - x) \tan (\pi - x)}{\sec (\pi - x) + \tan (\pi - x)}$ $dx$

$\because \sec (\pi - x) = - \sec x \ and \ \tan (\pi - x) = - \tan x$

$I =$ $\int \limits_{0}^{\pi}$ $\frac{ (\pi - x) \tan x}{\sec x + \tan x }$ $dx$ … … … … … ii)

Adding i) and ii) we get

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

$2I = \pi$ $\int \limits_{0}^{\pi}$ $\tan x(\sec x - \tan x) \ dx$

$2I = \pi$ $\int \limits_{0}^{\pi}$ $(\tan x \sec x - \tan^2 x) \ dx$

$2I = \pi$ $\int \limits_{0}^{\pi}$ $(\tan x \sec x - \sec^2 x + 1) \ dx$

$2I = \pi \Big (\sec x - \tan x + x \Big)_{0}^{\pi}$

$2I = \pi [ (\sec \pi - \tan \pi + \pi ) - (\sec 0 - \tan 0 + 0 ) ]$

$I =$ $\frac{\pi}{2}$ $[ -1-0+\pi-(1-0+0)]$

$I =$ $\frac{\pi}{2}$ $(-1 +\pi - 1)$

$I =$ $\frac{\pi}{2}$ $(\pi - 2) = \pi \Big($ $\frac{\pi}{2}$ $-1 \Big)$

$\\$

Question 14:

a) Given three identical Boxes $A, B$ and $C$ , Box $A$ contains $2$ gold and $1$ silver coin, Box $B$ contains $1$ gold and $2$ silver coins and Box $C$ contains $3$ silver coins. A person chooses a Box at random and takes out a coin. If the coin drawn is of silver, find the probability that it has been drawn from the Box which has the remaining two coins also of silver.

b) Determine the binomial distribution where mean is 9 and standard deviation is $\frac{3}{2}$ . Also, find the probability of obtaining at most 1 success. [ 6 ]

Answer:

a) Given Box $A$ contains $2$ gold and $1$ silver coin.

Box $B$ contains $1$ gold and $2$ silver coins.

Box $C$ contains $3$ silver coins.

Probability of choosing a bag is $P(A) = P(B) =P(C) =$ $\frac{1}{3}$

$P(S/A) =$ $\frac{1}{3}$ $, P(S/B) =$ $\frac{2}{3}$ $, P(S/C) =$ $\frac{3}{3}$

Using Baye’s theorem,

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

$\Rightarrow P(C/S) =$ $\frac{\frac{1}{3} \times \frac{3}{3} }{\frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{3} + \frac{1}{3} \times \frac{3}{3} }$

$\Rightarrow P(C/S) =$ $\frac{1 }{\frac{1}{3} + \frac{2}{3} + \frac{3}{3} }$ $=$ $\frac{1}{2}$

b) Given $np = 9$

Since $\sigma =$ $\frac{3}{2}$ $= \sqrt{npq} \Rightarrow npq =$ $\frac{9}{4}$

$\Rightarrow q =$ $\frac{9}{4\times 9}$ $=$ $\frac{1}{4}$

$p = 1 -$ $\frac{1}{4}$ $=$ $\frac{3}{4}$

$\Rightarrow n =$ $\frac{9 \times 4}{3}$ $= 12$

$\therefore P(x) = ^{12}C_x \Big($ $\frac{3}{4}$ $\Big)^x \Big($ $\frac{3}{4}$ $\Big)^{12-x}$

$P(x \leq 1) = P(x=0) + P(x=1)$

$= ^{12}C_0 \Big($ $\frac{3}{4}$ $\Big)^0 \Big($ $\frac{3}{4}$ $\Big)^{12} + ^{12}C_1 \Big($ $\frac{3}{4}$ $\Big)^1 \Big($ $\frac{3}{4}$ $\Big)^{11}$

$= \Big($ $\frac{1}{4}$ $\Big)^{12} + 12 \times$ $\frac{3}{4}$ $\times \Big($ $\frac{1}{4}$ $\Big)^{11}$

$= \Big($ $\frac{1}{4}$ $\Big)^{12} (1 + 36)$

$=$ $\frac{37}{4^{12}}$

$\\$

SECTION B (20 Marks)

Question 15: [ 3 × 2 ]

a) If $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular vectors, $| \overrightarrow{a} + \overrightarrow{b} | = 13$ and $| \overrightarrow{a} | = 5$ . Find the value of $|\overrightarrow{b} |$ .

b) Find the length of the perpendicular from origin to the plane $\overrightarrow{r} (3i - 4j - 12k)+39 = 0$

c) Find the angle between the two lines $2x = 3y = -z$ and $6x = -y = -4z$ .

Answer:

a) Given $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular vectors, $| \overrightarrow{a} + \overrightarrow{b} | = 13$ and $| \overrightarrow{a} | = 5$

$\Rightarrow$ $|\overrightarrow{a} + \overrightarrow{b} |^2 = 13^2$

$\Rightarrow$ $(\overrightarrow{a} + \overrightarrow{b}).(\overrightarrow{a} + \overrightarrow{b}) = 169$

$\Rightarrow$ $|\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + 2 \overrightarrow{a} . \overrightarrow{b} = 169$

$\Rightarrow$ $|5|^2 + |\overrightarrow{b}|^2 + 0 = 169 \ \ \ \ \ \ ( \because \overrightarrow{a} \perp \overrightarrow{b} )$

$\Rightarrow$ $|\overrightarrow{b}|^2 = 144$

$\Rightarrow$ $|\overrightarrow{b}| = 12$

b) Given $\overrightarrow{r} (3i - 4j - 12k)+39 = 0$

$3x - 4y -12z + 39 = 0$

Perpendicular length from origin to plane

$= \Big|$ $\frac{39}{\sqrt{3^2 + 4^2 + 12^2}}$ $\Big|$ $= \Big|$ $\frac{39}{\sqrt{9 + 16 + 144}}$ $\Big|$ $= \Big|$ $\frac{39}{\sqrt{169}}$ $\Big|$ $= \Big|$ $\frac{39}{13}$ $\Big| = 3$ units

c) Given $2x = 3y = -z$ $\Rightarrow$ $\frac{x}{3}$ $=$ $\frac{y}{2}$ $=$ $\frac{z}{-6}$

and $6x = -y = -4z$ $\Rightarrow$ $\frac{x}{4}$ $=$ $\frac{y}{-24}$ $=$ $\frac{z}{-6}$

$\cos \theta = \Bigg|$ $\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2 } \sqrt{{a_2}^2+{b_2}^2+{c_2}^2 } }$ $\Bigg|$

$\cos \theta = \Bigg|$ $\frac{3 \times 4+2 \times (-24) +(-6) \times (-6)}{\sqrt{{3}^2+{2}^2+{6}^2 } \sqrt{{4}^2+{24}^2+{6}^2 } }$ $\Bigg|$

$\cos \theta = \Bigg|$ $\frac{12 - 48 + 36}{\sqrt{9+4 + 36 } \sqrt{16+ 576+36 }}$ $\Bigg|$

$\cos \theta = 0 \Rightarrow \theta = 90^o$

Therefore the angle between the two lines $2x = 3y = -z$ and $6x = -y = -4z$ is $90^o$

$\\$

Question 16:

a) If $\overrightarrow{a} = \hat{i}- 2 \hat{j}+3 \hat{k}, \overrightarrow{b} = 2 \hat{i}+3 \hat{j} -5 \hat{k}$ , prove that $\overrightarrow{a}$ and $\overrightarrow{a} \times \overrightarrow{b}$ are perpendicular.

b) If $\overrightarrow{a}$ and $\overrightarrow{b}$ are non-collinear vectors, find the value of $x$ such that the vectors $\overrightarrow{\alpha}= (x-2) \overrightarrow{a}+ \overrightarrow{b}$ and $\overrightarrow{\beta} = (3+2x) \overrightarrow{a} - 2 \overrightarrow{b}$ are collinear. [ 4 ]

Answer:

a) Given, $\overrightarrow{a} = \hat{i}- 2 \hat{j}+3 \hat{k}, \overrightarrow{b} = 2 \hat{i}+3 \hat{j} -5 \hat{k}$

$\overrightarrow{a} \times \overrightarrow{b} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{array} \right|$

$= \hat{i} (10-9) - \hat{j}(-5-6) + \hat{k} (3+4)$

$= \hat{i} + 11 \hat{j} + 7 \hat{k}$

$\overrightarrow{a} . (\overrightarrow{a} \times \overrightarrow{b} ) = (\hat{i}- 2 \hat{j}+3 \hat{k}) .(\hat{i} + 11 \hat{j} + 7 \hat{k}) = 1 - 22 + 21 = -21 + 21 = 0$

$\therefore \overrightarrow{a} \perp (\overrightarrow{a} \times \overrightarrow{b} )$

b) Given $\overrightarrow{\alpha}= (x-2) \overrightarrow{a}+ \overrightarrow{b}$ and $\overrightarrow{\beta} = (3+2x) \overrightarrow{a} - 2 \overrightarrow{b}$ are collinear.

$\Rightarrow \alpha = k \beta$

$(x-2) \overrightarrow{a}+ \overrightarrow{b} = k [ (3+2x) \overrightarrow{a} - 2 \overrightarrow{b} ]$

On comparing coefficient of $\overrightarrow{a}$ and $\overrightarrow{b}$

$1 = - 2k \Rightarrow k = -$ $\frac{1}{2}$

Also $x-2 = k( 3+2x)$

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

$\Rightarrow 2x-4 = -3 -2x$

$\Rightarrow 4x = 1$

$\Rightarrow x =$ $\frac{1}{4}$

$\\$

Question 17:

(a) Find the equation of the plane passing through the intersection of the planes $2x + 2y - 3z - 7 = 0$ and $2x + 5y + 3z - 9 = 0$ such that the intercepts made by the resulting plane on the x-axis and the z-axis are equal.

b) Find the equation of the lines passing through the point $(2, 1, 3)$ and perpendicular to the lines

$\frac{x-1}{1}$ $=$ $\frac{y-2}{2}$ $=$ $\frac{z-3}{3}$ and $\frac{x}{-3}$ $=$ $\frac{y}{2}$ $=$ $\frac{z}{5}$ [ 4 ]

Answer:

a) Equation of intersecting plane is

$(2x + 2y - 3z - 7) +\lambda ( 2x + 5y + 3z - 9) = 0$

$(2 + 2 \lambda )x + (2 + 5 \lambda )y + (-3 + 3\lambda )z - (7 +9\lambda ) = 0$ … … … … … i)

For intercept made on x-axis, Put $y = 0$ , and $z = 0$

$(2 + 2 \lambda )x - (7 +9\lambda ) = 0$

$\Rightarrow x =$ $\frac{7+9\lambda }{2+2\lambda }$

For intercept made on z-axis Put $x = 0$ and $y = 0$

$(-3 + 3\lambda )z - (7 +9\lambda ) = 0$

$\Rightarrow y =$ $\frac{7+9\lambda }{-3+3\lambda }$

Given Intercept made on x-axis $=$ intercept made on z-axis

$\frac{7+9\lambda }{2+2\lambda }$ $=$ $\frac{7+9\lambda }{-3+3\lambda }$

$2+2\lambda = -3+3\lambda$

$\Rightarrow \lambda = 5$

Substituting in i) we get

$(2 + 2 \times 5 )x + (2 + 5 \times 5 )y + (-3 + 3 \times 5 )z - (7 +9 \times 5 ) = 0$

$12x + 27 y + 12 z -52 = 0$

b) Lines are passing through the point $(2, 1, 3) \Rightarrow ( x, y, z) = (2, 1, 3)$

Also given, that they are perpendicular to $\frac{x-1}{1}$ $=$ $\frac{y-2}{2}$ $=$ $\frac{z-3}{3}$ and $\frac{x}{-3}$ $=$ $\frac{y}{2}$ $=$ $\frac{z}{5}$

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

$\overrightarrow{b} = -3\hat{i} + 2 \hat{j} + 5 \hat{k}$

$\overrightarrow{a} \times \overrightarrow{b} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{array} \right|$

$= \hat{i} (10-6) - \hat{j} (5+9) + \hat{k} (2+6)$

$= 4\hat{i} - 14 \hat{j} + 8 \hat{k}$

Denominators of the required lines are $4, -14, 8$ or $2, -7, 4$

Therefore equation of line is $\frac{x-2}{2}$ $=$ $\frac{y-1}{-7}$ $=$ $\frac{z-3}{4}$

$\\$

Question 18: Draw a rough sketch and find the area bounded by the curve $x^2 = y$ and $x + y =2$ . [ 6 ]

Answer:

Equation of parabola is $x^2 = y$ … i)

Equation of straight line is $x + y = 2 \Rightarrow x = 2 - y$

Substituting it in i) we get

$(2-y)^2 = 2 - y$

$\Rightarrow 4 + y^2 - 4 y = y$

$\Rightarrow y^2 -5y + 4 = 0$

$\Rightarrow (y-1)(y-4) = 0$

$\Rightarrow y = 1$ or $y = 4$

When $y = 1, x = 1$ and when $y = 4, x = -2$

Hence we have points $(1, 1)$ and $(-2, 4)$

Area under line $x + y = 2$

$A_1 =$ $\int \limits_{-2}^{1}$ $y \ dx$

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

$= \Bigg[ 2x -$ $\frac{x^2}{2}$ $\Bigg]_{-2}^{1}$

$= \Big( 2 -$ $\frac{1}{2}$ $\Big) - \Big( -4 -$ $\frac{4}{2}$ $\Big)$

$=$ $\frac{3}{2}$ $+6 =$ $\frac{15}{2}$ square units

Area under curve $x^2 = y$

$A_2 =$ $\int \limits_{-2}^{1}$ $x^2 \ dx$

$= \Bigg[$ $\frac{x^3}{3}$ $\Bigg]_{-2}^{1}$

$=$ $\frac{1}{3}$ $\Big( 1 - (-8) \Big)$

$= 3$ square units

Area bounded by the curve $x^2 = y$ and $x + y = 2$

$= A_1 - A_2 =$ $\frac{15}{2}$ $- 3 =$ $\frac{9}{2}$ $= 4.5$ square units

$\\$

SECTION C (20 Marks)

Question 19: [ 3 × 2 ]

(a) A company produces a commodity with Rs. $24,000$ as fixed cost. The variable cost estimated to be $25\%$ of the total revenue received on selling the product, is at the rate of Rs. $8$ per unit. Find the break-even point.

b) The total cost function for a production is given by $C(x) =$ $\frac{3}{4}$ $x^2-7x + 27$ .

Find the number of units produced for which M.C. = A.C. (M.C.= Marginal Cost and A.C. = Average Cost.)

c) $\overline{x} =18 , \overline{y}= 100 , \sigma_x = 14 , \sigma_y = 20$ and correlation coefficient $r_{xy} = 0.8$ , find regression equation of $y$ on $x$ .

Answer:

a) Revenues $=$ Variable cost $+$ Fixed cost

Let the total revenue $= x$

$\therefore x = 0.25x + 24000$

$\Rightarrow 0.75x = 24000$

$\Rightarrow x = 32000$

Therefore total revenue $= Rs. \ 32000$

Therefore break even point (in units) $=$ $\frac{32000}{8}$ $= 4000$ units

b) Given $C (x) =$ $\frac{3}{4}$ $x^2 - 7x + 27$ … … … … … ii)

$MC =$ $\frac{dc}{dx}$ $=$ $\frac{3}{4}$ $(2x) - 7$

$\therefore MC =$ $\frac{3}{2}$ $x - 7$ … … … … … ii)

$AC =$ $\frac{C(x)}{x}$ $=$ $\frac{3}{4}$ $x - 7 +$ $\frac{27}{x}$ … … … … … iii)

Given that $MC = AC$

$\Rightarrow$ $\frac{3}{2}$ $x - 7 =$ $\frac{3}{4}$ $x - 7 +$ $\frac{27}{x}$

$\Rightarrow$ $\frac{3}{2}$ $x -$ $\frac{3}{4}$ $x =$ $\frac{27}{x}$

$\Rightarrow$ $\frac{3}{4}$ $x =$ $\frac{27}{x}$

$\Rightarrow x^2 = 4 \times 9 = 36$

$\Rightarrow x = \pm 6$

Therefore $x = 6$ units.

c) Regression line on $y$ on $x$ is

$(y - \overline{y}) = b_{yx} (x - \overline{x})$

We know, $b_{yx} =$ $\frac{ \Sigma xy}{ \Sigma x^2}$ $= r \Big($ $\frac{\sigma_y}{\sigma_x}$ $\Big)$

$b_{xy} = (0.8) \Big($ $\frac{20}{14}$ $\Big) =$ $\frac{16}{14}$ $=$ $\frac{8}{7}$

$\Rightarrow$ $(y - \overline{y}) =$ $\frac{8}{7}$ $(x - \overline{x})$

$\Rightarrow$ $(y - 100) =$ $\frac{8}{7}$ $(x - 18)$

$\Rightarrow$ $7y - 700 = 8x - 144$

$\Rightarrow$ $7y = 8x + 556$

$\Rightarrow$ $y = \Big($ $\frac{8}{7}$ $\Big)x + \Big($ $\frac{556}{7}$ $\Big)$

$\\$

Question 20:

a) The following results were obtained with respect to two variables $x$ and $y$ : $\Sigma x = 15, \Sigma y = 25, \Sigma xy = 83, \Sigma x^2 = 55, \Sigma y^2 = 135$ and $n = 5$

(i) Find the regression coefficient $b_{xy}$ .
(ii) Find the regression equation of $x$ on $y$ .

b) Find the equation of the regression line of y on $x$ , if the observations $(x, y)$ are as follows: $(1, 4), (2, 8), (3, 2), (4, 12), (5, 10), (6, 14), (7, 16), (8, 6), (9, 18)$ Also, find the estimated value of $y$ when $x = 14$ . [ 4 ]

Answer:

a) Given $\Sigma x = 15, \Sigma y = 25, \Sigma xy = 83, \Sigma x^2 = 55, \Sigma y^2 = 135$ and $n = 5$

i) Regression coefficient $b_{xy} =$ $\frac{\Sigma xy - n \overline{x} \overline{y}}{\Sigma y^2 - n(\overline{y})^2}$ $=$ $\frac{83 - 5 \times 3 \times 5}{135 - 5(25)}$ $= 0.80$

ii) Regression equation of $x$ on $y$

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

$x = 0.8y + 3 - 0.8 \times 5$

$x = 0.8y + 3 - 4$

$x = 0.8y - 1$

$x$	$y$	$xy$	$x^2$
1	4	4	1
2	8	16	4
3	2	6	9
4	12	48	16
5	10	50	25
6	14	84	36
7	16	112	49
8	6	48	64
9	18	162	81
45	90	530	285

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

$\overline{y} =$ $\frac{\Sigma y}{n}$ $=$ $\frac{90}{9}$ $= 10$

$\overline{x} =$ $\frac{\Sigma x}{n}$ $=$ $\frac{45}{9}$ $= 5$

$b_{xy} =$ $\frac{n \Sigma xy - \Sigma x \Sigma y}{n \Sigma x^2 - ( \Sigma x)^2}$ $=$ $\frac{9 \times 530 - 45 \times 90}{9 \times 285 - (45)^2}$ $=$ $\frac{4770 - 4050}{2565 - 2025}$ $=$ $\frac{720}{540}$ $= \frac{4}{3}$

$\therefore 3y - 30 = 4x - 20$

$\Rightarrow 4x-3y+ 10 = 0$

When $x = 14,$

$3y = 4 \times 14 + 10$

$\Rightarrow 3y = 66$

$\Rightarrow y = 22$

$\\$

Question 21:

a) The cost function of a product is given by $C(x) =$ $\frac{x^3}{3}$ $- 45x^2 - 900x + 36$ where $x$ is the number of units produced. How many units should be produced to minimize the marginal cost?

b) The marginal cost function of x units of a product is given by $MC = 3x^2 -10x + 3$ . The cost of producing one unit is Rs. $7$ . Find the total cost function and average cost function. [ 4 ]

Answer:

a) Given $C(x) =$ $\frac{x^3}{3}$ $- 45x^2 - 900x + 36$

$\frac{dC(x)}{dx}$ $= x^2 -90x -900 = M(x)$

$\frac{d^2C(x)}{dx^2}$ $= 2x - 90 =$ $\frac{dM(x)}{dx}$

$\therefore \frac{d^2M(x)}{dx^2}$ $= 2 > 0$

$\therefore \frac{dM(x)}{dx}$ is minimum

For minimum, $\frac{dM(x)}{dx}$ $= 0$

$2x-90 = 0$

$\Rightarrow x = 45$

b) Let total cost function $= C(x)$

Therefore Marginal Cost function $(MC) =$ $\frac{dC}{dx}$

$\therefore$ $\frac{dC}{dx}$ $= 3x^2 - 10x + 3$

$\therefore C(x) = \int \limits_{}^{} (3x^2 - 10x + 3) \ dx$

$\Rightarrow C(x)= 3 \Big($ $\frac{x^3}{3}$ $\Big) - 10\Big($ $\frac{x^2}{2}$ $\Big)+ 3x + A$

$\Rightarrow C(x)= x^3 - 5x^2 + 3x + A$

If $n =1, C(1) =7$

$\therefore 1-5+3+ A=7$

$\therefore A= 7 + 5 - 4=8$

$\therefore C(x)= x^3 - 5x^2 + 3x + 8$

Also, the average cost function is given by

$\frac{C(x)}{x}$ $= x^2 - 5x + 3 +$ $\frac{8}{x}$

$\\$

Question 22: A carpenter has $90, 80$ and $50$ running feet respectively of teak wood, plywood and rosewood which is used to produce product $A$ and product $B$ . Each unit of product $A$ requires $2, 1$ and $1$ running feet and each unit of product $B$ requires $1, 2$ and $1$ running feet of teak wood, plywood and rosewood respectively. If product $A$ is sold for Rs. $48$ per unit and product $B$ is sold for Rs. $40$ per unit, how many units of product $A$ and product $B$ should be produced and sold by the carpenter, in order to obtain the maximum gross income?

Formulate the above as a Linear Programming Problem and solve it, indicating clearly the feasible region in the graph. [ 6 ]

Answer:

Product	A	B
Teakwood	$2$	$1$	$\leq 90$
Plywood	$1$	$2$	$\leq 80$
Rosewood	$1$	$1$	$\leq 50$

Selling Price for Product $A = 48$ Rs./unit

Selling Price for Product $B = 40$ Rs./unit

Therefore the objective function is $Z_{max} = 48x_1 + 40 x_2$

$2x_1 + x_2 \leq 90$

$x_1 + 2x_2 \leq 80$

$x_1 + x_2 \leq 50$

$x_1, x_2 \geq 0$

$2x_1 + x_2 = 90$ … … … … … i)

$x_1 + 2x_2 = 80$ … … … … … ii)

$x_1 + x_2 =50$ … … … … … iii)

From equation iii) and ii) $x_2 = 30, x_1 = 20$

From equation i) and iii) $x_1 = 40, x_2 = 0$

We know $Z_{max} = 48 \times x_1 + 40 \times x_2$

$A(0, 40)$	$Z_{max} = 48 \times 0 + 40 \times 40$	$Z_A = 1600$
$B(20, 30)$	$Z_{max} = 48 \times 20 + 40 \times 30$	$Z_B = 2160$
$C(40, 10)$	$Z_{max} = 48 \times 0 + 40 \times 10$	$Z_C = 2320$
$D(45, 0)$	$Z_{max} = 48 \times 45 + 40 \times 0$	$Z_D = 2160$