3 Marks Question

Question 13 Marks

If $x=a \sin 2 t(1+\cos 2 t)$ and $y=b \cos 2 t(1-\cos 2 t)$, then find $\frac{d y}{d x}$ at $t=\frac{\pi}{4}$.

Answer

We know that, $x=a \sin 2 t(1+\cos 2 t)$ and $y=b \cos 2 t(1-\cos 2 t)$
$\Rightarrow \frac{d y}{d t}= a [-2 \sin 2 t \sin 2 t +2 \cos 2 t (1+\cos 2 t )] \ldots$ (I)
and $\frac{d x}{d t}= b [2 \sin 2 t \cos 2 t -2 \sin 2 t (1-\cos 2 t )] \ldots( ii )$
$\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{b[2 \sin 2 t \cos 2 t-2 \sin 2 t(1-\cos 2 t)]}{a\left[-2 \sin ^2 2 t+2 \cos 2 t(1+\cos 2 t)\right]}$. [Using (i) and (ii)]
$\left.\Rightarrow \frac{d y}{d x}\right]_{\text {at } t=\pi / 4}=\frac{b\left[2 \sin \frac{\pi}{2} \cos \frac{\pi}{2}-2 \sin \frac{\pi}{2}\left(1-\cos \frac{\pi}{2}\right)\right]}{a\left[-2 \sin ^2\left(\frac{\pi}{2}\right)+2 \cos \frac{\pi}{2}\left(1+\cos \frac{\pi}{2}\right)\right]}=\frac{b}{a} \cdot \frac{(0-1)}{(-1-0)}=\frac{b}{a}$

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Question 23 Marks

Solve the following $\text{LPP}$ by graphical method:
Minimize $Z = 20x + 10y$
Subject to $x+2 y \leq 40 , 3 x+y \geq 30 , 4 x+3 y \geq 60$ and $ x, y \geq 0$

Answer

Converting the given inequations into equations, we obtain the following equations:
$x+2 y=40,3 x+y=30,4 x+3 y=60, x=0$ and $y=0$
Region represented by $x+2 y \leq 40$ :
The line $x+2 y=40$ meets the coordinate axes at $A_1(40,0)$ and $B_1(0,20)$ respectively. Join these points to obtain the line $x+2 y$=40
Clearly, $(0,0)$ satisfies the inequation $x+2 y \leq 40$. So, the region in $x y$-plane that contains the origin represents the solution set of the given inequation.
Region represented by $3 x+y \geq 30$ :
The line $3 x+y=30$ meets $x$ and $y$ axes at $A_2(10,0)$ and $B_2(0,30)$ respectively. Join these points to obtain this line.
We find that the point $O(0,0)$ does not satisfy the inequation $3 x+y \geq 30$.
So, that region in $xy-$plane which does not contain the origin is the solution set of this inequation.
Region represented by $x \geq 0, y \geq 0$ :
Clearly, the region represented by the non-negativity restrictions $x \geq 0$ and $y \geq 0$ is the first quadrant in $xy-$plane.
The shaded region $A _3 A_1 QP$ in a figure represents the common region of the regions represented by the above inequations.
This region represents the feasible region of the given $\text{LPP}.$

The coordinates of the corner points of the shaded feasible region are $A_3(15,0), A_1(40,0), Q(4,18)$ and $P(6,12)$ These points
have been obtained by solving the equations of the corresponding intersecting lines, simultaneously.
The values of the objective function at these points are given in the following table

Point $(x, y)$	Value of the objective function $z = 20x + 10y$
$A_3(15, 0)$	$Z = 20 x 15 + 10 x 0 = 300$
$A_1(40, 0)$	$Z = 20 x 40 + 10 x 0 = 800$
$P(6,12)$	$Z = 20 x 4 + 0 x 18 = 260$
$P(6,12)$	$Z = 20 x 6 + 10 x 12 = 240$

Out of these values of $Z,$ the minimum value is $240$ which is attained at point $P (6,12).$ Hence$, x = 6 y = 12$ the optimal solution of the given $\text{LPP}$ and the optimal value of $Z$ is $240$.

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Question 33 Marks

Show that the solution set of the linear constraints is empty $x-2 y \geq 0,2 x-y \leq-2, x \geq 0$ and $y \geq 0$

Answer

Here, it is given that the equations
$x-2 y \geq 0 \ldots \ldots \text { (i) }$
$2 x-y \leq-2 \ldots \ldots . \text { (ii) }$
$x \geq 0 \ldots(iii)$
$y \geq 0 \ldots . .(iv)$
and $x=2$
$\Rightarrow 2-2 y=0=y=1$
$(0,0)$ and $(2,1)$ are on the line $(v)$
$(0,1)$ is not on this line and it lies in the half
plane of $(i)$ which is not true The line corresponding to $(ii)$ is
$2 x-y=-2 \ldots \ldots( vi )$
on the line $(vi)$ put $x=0$
$\Rightarrow 0-y=-2$
$\Rightarrow y=2$
and $y =0$
$\Rightarrow2 x-0=-2$
$\Rightarrow x=-1$
The inequation $x > 0$ represents the closed half$-$plane on the right of the $y-$axis.
The inequation $y > 0$ represent the closed half$-$plane above the $x-$axis
The graph of the given system is the intersection of half$-$planes of the inequation
The intersection of half$-$planes is empty
The solution set of the given inequation is empty. This is the required solution

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Question 43 Marks

Solve : $\frac{d y}{d x}+\frac{y}{x}=\cos x +\frac{\sin x}{x}$

Answer

We have, $\frac{d y}{d x}+\left(\frac{1}{x}\right) y=\cos x+\frac{\sin x}{x} \ldots$
This is a linear differential equation of the form
$\frac{d y}{d x}+ Py = Q$, where $P =\frac{1}{x}$ and $Q =\cos x +\frac{\sin x}{x}$
$\therefore I. F=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x$
Multiplying both sides of $(i)$ by $I.F. = x,$
we get $x \frac{d y}{d x}+y=x \cos x+\sin x$
Integrating both sides with respect to $x,$ we get
$y x=\int(x \cos x+\sin x) d x+C\left[\right.$ Using: $y=( I.F. )=\int Q( I.F. \left.) d x+C\right]$
$\Rightarrow xy =\int \underset{I}{x} \cos xdx +\int \sin x dx + C$
$\Rightarrow x y=x \sin x-\int \sin x d x+\int \sin x d x+C [$Integrating $1^{st}$ integral by parts$]$
$\Rightarrow x y=x \sin x+C$
$\Rightarrow y=\sin x+\frac{C}{x}$

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Question 53 Marks

Find the particular solution of the differential equation $\left(1+ y ^2\right)+\left( x -e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$ given that y = 0 when x =1

Answer

Given differential equation can be written as
$\frac{d x}{d y}+\frac{x}{1+y^2}=\frac{e^{\tan ^{-1} y}}{1+y^2}$
L.F $=e^{\int \frac{d y}{1+y^2}}=e^{\tan ^{-1} y}$
Solution is given by
$x e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y}}{1+y^2} \times e^{\tan ^{-1} y} d y=\int \frac{e^{2 \tan ^{-1} y}}{1+y^2} d y$
or $x e^{\tan ^{-1} y}=\frac{e^{2 \tan ^{-1} y}}{2}+ c$
when $x =1, y =0$ or $c =\frac{1}{2}$
$\therefore$ Solution is given by $x e^{\tan ^{-1} y}=\frac{1}{2} e^{2 \tan ^{-1} y}+\frac{1}{2}$
or $x=\frac{1}{2}\left(e^{\tan -1} y+e^{-\tan ^{-1} y}\right)$

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Question 63 Marks

Find $\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x$.

Answer

Given $I=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x=\int \frac{\sqrt{x}}{\sqrt{\left(a^{3 / 2}\right)^2-\left(x^{3 / 2}\right)^2}} d x$
$\text { Let, } x^{3 / 2}=a^{3 / 2} t$
$\Rightarrow \frac{3}{2} x^{1 / 2} d x=a^{3 / 2} d t$
$\Rightarrow \sqrt{x} d x=\frac{2}{3} a^{3 / 2} d t$
$\therefore I=\int \frac{\frac{2}{3} a^{3 / 2}}{\sqrt{\left(a^{3 / 2}\right)^2-\left(a^{3 / 2} t\right)^2}} d t$
$=\frac{2}{3} a^{3 / 2} \int \frac{d t}{a^{3 / 2} \sqrt{1-t^2}}$
$=\frac{2}{3} \int \frac{d t}{\sqrt{1-t^2}}=\frac{2}{3} \sin ^{-1}\left(\frac{t}{1}\right)+C$
$\left[\because \int \frac{d x}{a^2-x^2}=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]$
$=\frac{2}{3} \sin ^{-1}\left(\frac{x^{3 / 2}}{a^{3 / 2}}\right)+C \left[\text { put } t=\frac{x^{3 / 2}}{a^{3 / 2}}\right]$
$=\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)+C$

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Question 73 Marks

Evaluate: $\int_0^{1 / \sqrt{2}} \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x$

Answer

Let $\sin ^{-1} x=\theta$ or, $x=\sin \theta$.
Then, $d x=d(\sin \theta)=\cos \theta d \theta$
Now, $ x=0$
$\Rightarrow \sin \theta=0$
$\Rightarrow \theta=0$ and $x=\frac{1}{\sqrt{2}}$
$\Rightarrow \sin \theta=\frac{1}{\sqrt{2}}$
$\Rightarrow \theta=\frac{\pi}{4}$
$\therefore I=\int_0^{1 / \sqrt{2}} \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x$
$\Rightarrow I=\int_0^{\pi / 4} \frac{\theta}{\cos ^3 \theta} \cos \theta d \theta$
$=\int_0^{\pi / 4} \theta \sec ^2 \theta d \theta$
Now using integratio by parts.
$\Rightarrow I=\left[\theta \tan \theta \right]_0^{\pi / 4}+[\log \cos \theta]_0^{\pi / 4}$
$=\frac{\pi}{4}+\left\{\log \left(\frac{1}{\sqrt{2}}\right)-\log 1\right\}$
$=\frac{\pi}{4}-\frac{1}{2} \log 2$

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Question 83 Marks

Bag I contains $3$ white and $4$ black balls, while Bag $II$ contains $5$ white and $3$ black balls. One ball is transferred at random from Bag $I$ to Bag $II$ and then a ball is drawn at random from Bag $II.$ The ball so drawn is found to be white. Find the probability that the transferred ball is also white.

Answer

Let $E_{1:}$ Transferred ball is white
$E_2:$ Transferred ball is black
$A:$ white ball is found
$\text { Here, } P \left( E _1\right)=\frac{3}{7}, P \left( E _2\right)=\frac{4}{7}$
$P \left( A / E _1\right)=\frac{6}{9}, P \left( A / E _2\right)=\frac{5}{9}$
Using Baye's theorem
$P\left(E_1 / A\right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)}$
$=\frac{\frac{3}{7} \times \frac{6}{9}}{\frac{3}{7} \times \frac{6}{9}+\frac{4}{7} \times \frac{5}{9}}$
$=\frac{18}{18+20}=\frac{9}{19}$

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Question 93 Marks

Evaluate: $\int \frac{x+2}{\sqrt{x^2+2 x+3}}$

Answer

Let $I =\int \frac{x+2}{\sqrt{x^2+2 x+3}}$
$x +2=A \frac{d}{d z}\left[ x ^2+2 x +3\right]+ B$
$\Rightarrow x +2=2 Ax +2 A+ B$
Comparing the coefficients,
we have, $2A = 1$ and $2A + B = 2$
$\Rightarrow A =\frac{1}{2}$
Substituting the value of $A$ in $2 A+B=2$,
we have, $2 \times \frac{1}{2}+B=2$
$\Rightarrow 1+B=2$
$\Rightarrow B=2-1$
$\Rightarrow B=1$
Thus we have, $x +2=\frac{1}{2}[2 x+2]+1$
Hence, using values of $A,$ and $B,$ we have
$I=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x$
$\left.=\int \frac{\left[\frac{1}{2}[2 x+2]+1\right]}{\sqrt{x^2+2 x+3}}\right] d x$
$=\int \frac{\left[\frac{1}{2}[2 x+2]\right]}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
$=\frac{1}{2} \int \frac{[2 x+2]}{\sqrt{x^2+2 x+3}} d x+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
Substituting $t = x ^2+2 x +3$ and $dt =2 x +2$
in the first integrand,
we have, $I =\frac{1}{2} \int \frac{d t}{\sqrt{t}}+\int \frac{d x}{\sqrt{x^2+2 x+3}}$
$=\frac{1}{2} \times 2 \sqrt{t}+\int \frac{d x}{\sqrt{x^2+2 x+1+2}}+C$
$=\sqrt{t}+\int \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}}+c$
$ I =\sqrt{x^2+2 x+3}+\log \left[| x +1|+\sqrt{(x+1)^2+(\sqrt{2})^2}\right]+C$
$\Rightarrow I =\sqrt{x^2+2 x+3}+\log \left[| x +1| \sqrt{x^2+2 x+3}\right]+ c $

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