3 Marks Question

Question 13 Marks

If $x = a \left(\cos t+\log \tan \frac{t}{2}\right), y = a \sin t$, then evaluate $\frac{d^2 y}{d x^2}$ at $t=\frac{\pi}{3}$.

Answer

Given, $x = a \left(\cos t+\log \tan \frac{t}{2}\right).........(i)$
and $y= a \sin t ...........(ii)$
Therefore, on differentiating both sides w.r.t $t,$ we get,
$\frac{d x}{d t}=a\left[\frac{d}{d t}(\cos t)+\frac{d}{d t} \log \tan \frac{t}{2}\right]$
$=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \frac{d}{d t}\left(\tan \frac{t}{2}\right)\right] [$by using chain rule of derivative$]$
$=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \sec ^2 \frac{t}{2} \frac{d}{d t}\left(\frac{t}{2}\right)\right]$
$=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \times \sec ^2 \frac{t}{2} \times \frac{1}{2}\right]$
$=a\left[-\sin t+\frac{1}{\frac{\sin t / 2}{\cos t / 2}} \times \frac{1}{\cos ^2 \frac{t}{2}} \times \frac{1}{2}\right]$
$=a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}\right]$
$=a\left[-\sin t+\frac{1}{\sin t}\right][\because \sin 2 \theta=2 \sin \theta \cos \theta]$
$=a\left(\frac{1-\sin ^2 t}{\sin t}\right)$
$ \Rightarrow \frac{d x}{d t}=a\left(\frac{\cos ^2 t}{\sin t}\right)\left[\because 1-\sin ^2 \theta=\cos ^2 \theta\right]$
Again, on differentiating both sides of $(ii)$ w.r.t $t,$ we get
$\frac{d y}{d t}= a \cos t.....................(iv)$
Therefore, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \cos t}{a\left(\frac{\cos ^2 t}{\sin t}\right)}$ [ from Eqs $(iii)$ and $(iv)]$
$=\frac{a \cos t}{a \cos ^2 t} \times \sin t=\tan t$
Therefore, on differentiating both sides of above equation w.r.t $x,$ we get,
$ \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(\tan t)$
$=\frac{d}{d t}(\tan t) \frac{d t}{d x}\left[\because \frac{d}{d x} f(t)=\frac{d}{d t} f(t) \cdot \frac{d t}{d x}\right]$
$\Rightarrow \frac{d^2 y}{d x^2}=\sec ^2 t \times \frac{\sin t}{a \cos ^2 t} \text { [From Eq.(iii)] }$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{\sin t \sec ^4 t}{a} $
Therefore, on putting $t=\frac{\pi}{3}$, we get,
$ {\left[\frac{d^2 y}{d x^2}\right]_{t=\frac{\pi}{3}}=\frac{\sin \frac{\pi}{3} \times \sec ^4 \frac{\pi}{3}}{a}=\frac{\frac{\sqrt{3}}{2} \times(2)^4}{a}}$
$=\frac{8 \sqrt{3}}{a}$

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Question 23 Marks

Solve the Linear Programming Problem graphically: Maximize $Z = 3x + 4y$ Subject to $\begin{array}{l}2 x+2 y \leq 80 \\ 2 x+4 y \leq 120\end{array}$

Answer

We have to maximize $z = 3x + 4y$
First, we will convert the given inequations into equations, we obtain the following equations: $2x + 2y = 80, 2x + 4y = 120$
Region represented by $2 x+2 y \leq 80$ the line $2x + 2y = 80 $
Clearly $(0,0)$ satisfies the inequation $2 x+2 y \leq 80 \ C(60,0) $ and $D(0,30)$ respectively.
By joining these points we obtain the line $2 x+4 y \leq 120$

The corner points of the feasible region are $O(0,0), A(40,0) E(20,20)$ and $D(0,30)$
The values of objective function at the corner points are as follows:

Corner Points	$Z = 3x + 4y$
$O(0, 0)$	$3 x 0 + 4 x 0 = 0$
$A(40, 0)$	$3 x 40 + 4 x 0 = 120$
$F(20, 20)$	$3 x 20 + 4 x 20 = 140$
$D(0, 30)$	$10 x 0 + 4 x 30 = 120$

We see that the maximum value of the objective function $Z$ is $140$
Which is at $E(20, 20)$ that means at $x = 20$ and $y = 20$
Thus, the optimal value of objective function $z$ is $140.$

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Question 33 Marks

In Fig, the feasible region $($shaded$)$ for a $\text{LPP}$ is shown. Determine the maximum and minimum value of $Z =x+2y$

Solve the Linear Programming Problem graphically:
Maximize $Z = 3x + 4y$ Subject to
$2 x+2 y \leq 80$
$2 x+4 y \leq 120$

Answer

From the shaded bounded region, it is clear that the coordinates of corner points are$\left(\frac{3}{13}, \frac{24}{13}\right),\left(\frac{18}{7}, \frac{2}{7}\right),\left(\frac{7}{2}, \frac{3}{4}\right)$ and $\left(\frac{3}{2}, \frac{15}{4}\right)$ Also, we have to determine maximum and minimum value of $Z = x +2 y$

Corner Points	Corresponding value of $Z$
$\left(\frac{3}{13}, \frac{24}{13}\right)$	$\frac{3}{13}+\frac{48}{13}=\frac{51}{13}=3 \frac{12}{13}$
$\left(\frac{18}{7}, \frac{2}{7}\right)$	$\frac{18}{7}+\frac{4}{7}=\frac{22}{7}=3 \frac{1}{7}($ Minimum $)$
$\left(\frac{7}{2}, \frac{3}{4}\right)$	$\frac{7}{2}+\frac{6}{4}=\frac{20}{4}=5$
$\left(\frac{3}{2}, \frac{15}{4}\right)$	$\frac{3}{2}+\frac{30}{4}=\frac{36}{4}=9 ($Maximum$)$

Hence, the maximum and minimum value of are $9$ and $3 \frac{1}{7}$ respectively

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Question 43 Marks

In the differential equation show that it is homogeneous and solve it : $x^2 \frac{d y}{d x}= x ^2+ xy + y ^2$

Answer

The given differential equation is,
$ x^2 \frac{d y}{d x}=x^2+x y+y^2$
$\Rightarrow \frac{d y}{d x}=\frac{x^2+x y+y^2}{x^2}$
$\Rightarrow \frac{d y}{d x}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^2$
$\Rightarrow \frac{d y}{d x}=f\left(\frac{y}{x}\right)$
$\Rightarrow$ the given differential equation is a homogenous equation.The solution of the given differential equation is:
Put $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x \frac{dv}{dx}$
$v+x \frac{d v}{d x}=1+\frac{v x}{x}+\left(\frac{v x}{x}\right)^2=1+v+(v)^2$
$\Rightarrow x \frac{d v}{d x}=1+v+(v)^2-v=1+(v)^2$
$\Rightarrow \frac{d v}{1+(v)^2}=\frac{d x}{x} $
Integrating both the sides we get:
$ \Rightarrow \int \frac{d v}{1+(v)^2}=\int \frac{d x}{x}+c$
$\Rightarrow \tan ^{-1} v=\ln |x|+c $
Resubstituting the value of $y=v x$, we get,
$\Rightarrow \tan ^{-}\left(\frac{y}{x}\right)=\ln |x|+c$

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Question 53 Marks

Find the general solution of the differential equation $\left(1+x^2\right) \frac{d y}{d x}-x=2 \tan ^{-1} x$

Answer

The given differential equation can be rewritten as
$\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{\left(1+x^2\right)}+\frac{x}{\left(1+x^2\right)}$
$\Rightarrow d y=\left\{\frac{2 \tan ^{-1} x}{\left(1+x^2\right)}+\frac{x}{\left(1+x^2\right)}\right\} d x [$separating the variables$]$
$\Rightarrow \int d y=\int \frac{2 \tan ^{-1} x}{\left(1+x^2\right)} d x+\int \frac{x}{\left(1+x^2\right)} d x+C$
where $C$ is an arbitrary constant
$ \Rightarrow y=2 \int t d t+\frac{1}{2} \int \frac{2 x}{\left(1+x^2\right)} d x+C$
$[$put $\tan ^{-1} x = t$ and $\frac{1}{\left(1+ x ^2\right)} d x=d t$ in $1^{st}$ integral$]$
$\Rightarrow y=t^2+\frac{1}{2} \log \left|1+x^2\right|+C$
$\Rightarrow y=\left(\tan ^{-1} x\right)^2+\frac{1}{2} \log \left|1+x^2\right|+C $
Therefore, $y=\left(\tan ^{-1} x\right)^2+\frac{1}{2} \log \left|1+x^2\right|+C$ is the required solution.

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Question 63 Marks

Evaluate: $\int \frac{x^2+1}{\left(x^2+2\right)\left(2 x^2+1\right)} d x$

Answer

Let $x^2=y$. Then,
$\frac{x^2+1}{\left(x^2+2\right)\left(2 x^2+1\right)}=\frac{y+1}{(y+2)(2 y+1)} $
Using partial fractions,
$\frac{y+1}{(y+2)(2 y+1)}=\frac{A}{y+2}+\frac{B}{2 y+1} \ldots \ldots \text { (i) }$
$\Rightarrow y+1=A(2 y+1)+B(y+2)$
Putting $y+2=0$
i.e. $y=-2$ in $(ii),$ we get
$-1=-3 A \Rightarrow A=\frac{1}{3}$
Putting $2 y +1=0$ i.e. $y =-\frac{1}{2}$ in $(ii),$ we get
$\frac{1}{2}=B\left(\frac{3}{2}\right)$
$\Rightarrow B=\frac{1}{3}$
Substituting the values of $A$ and $B$ in $(i)$, we obtain
$\frac{y+1}{(y+2)(2 y+1)}=\frac{1}{3} \cdot \frac{1}{y+2}+\frac{1}{3} \frac{1}{(2 y+1)}$
Replacing $y$ by $x^2$, we get
$ \frac{x^2+1}{\left(x^2+2\right)\left(2 x^2+1\right)}=\frac{1}{3} \cdot \frac{1}{x^2+2}+\frac{1}{3\left(2 x^2+1\right)}$
$\therefore I=\int \frac{x^2+1}{\left(x^2+2\right)\left(2 x^2+1\right)} d x=\frac{1}{3} \int \frac{1}{x^2+2} d x+\frac{1}{3} \int \frac{1}{(\sqrt{2} x)^2+1} d x$
$\Rightarrow I=\frac{1}{3} \times \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{3 \sqrt{2}} \tan ^{-1}(\sqrt{2} x)+C$
$\Rightarrow I=\frac{1}{3 \sqrt{2}}\left\{\tan ^{-1} \frac{x}{\sqrt{2}}+\tan ^{-1} \sqrt{2}x\right\}+C$

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Question 73 Marks

Evaluate $\int e^{2 x}\left(\frac{1-\sin 2 x}{1-\cos 2 x}\right) d x$

Answer

According to the question,
$I=\int e^{2 x}\left(\frac{1-\sin 2 x}{1-\cos 2 x}\right) d x$
$=\int e^{2 x}\left(\frac{1-2 \sin x \cos x}{2 \sin ^2 x}\right) d x[\because 1-\cos 2 x=2 \sin ^2 x$ and $\sin 2 x=2 \sin x \cos x]$
$=\frac{1}{2} \int e^{2 x}\left(\operatorname{cosec}^2 x-2 \cot x\right) d x$
$=\frac{1}{2} \int e_I^{2 x} \operatorname{cosec}_{I I}^2 x d x-\int e^{2 x} \cot x d x $
Using integration by parts for first integral :
$ =\frac{1}{2}\left[e^{2 x} \int \operatorname{cosec} e^2 x d x-\int\left\{\frac{d}{d x}\left(e^{2 x}\right) \int \operatorname{cosec} e^2 x d x\right\} d x\right]-\int e^{2 x} \cot x d x$
$=\frac{1}{2}\left[-e^{2 x} \cot x+\int 2 e^{2 x} \cot x d x\right]+C-\int e^{2 x} \cot x d x$
$=-\frac{e^{2 x}}{2} \cot x+\int e^{2 x} \cot x d x-\int e^{2 x} \cot x d x+C$
$I=-\frac{e^{2 z}}{2} \cot x+C$

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Question 83 Marks

Urn $A$ contains $1$ white, $2$ black and $3$ red balls; urn $B$ contains $2$ white, $1$ black and $1$ red ball; and urn $C$ contains $4$ white, $5$ black and $3$ red balls. One urn is chosen at random and two balls are drawn. These happen to be one white and one red. What is the probability that they come from urn $A$?

Answer

Let $E _1, E _2, E _3$ be the events that the balls are drawn from urn $A,$ urn $B$ and urn $C$ respectively, and let $E$ be the event that the balls drawn are one white and one red.
Therefore, we have,
$P \left( E _1\right)= P \left( E _2\right)= P \left( E _3\right)=\frac{1}{3}$
Since $E_1, E_2, E_3$ are mutually exclusive and exhaustive, therefore, we have,
$P\left(\frac{E}{E_1}\right)=$probability that the balls drawn are one white and one red, given that the balls are from urn $A$
$=\frac{1 C_1 \times{ }^3 C_1}{{ }^6 C_2}=\frac{3}{15}=\frac{1}{5}$
$P\left(\frac{E}{E_2}\right)=$probability that the balls drawn are one white and one red, given that the balls are from urn $B$
$=\frac{{ }^2 C_1 \times{ }^1 C_1}{{ }^4 C_2}=\frac{2}{6}=\frac{1}{3}$
$P\left(\frac{E}{E_3}\right)=$probability that the balls drawn are one white and one red, given that the balls are from urn $C$
$=\frac{{ }^4 C_1 \times{ }^3 C_1}{{ }^{12} C_2}=\frac{12}{66}=\frac{2}{11}$
Therefore, we have,
Probability that the balls drawn are from urn $A,$ it being given that the balls drawn are one white and one red
$=P\left(\frac{E_1}{E}\right)$
$=\frac{P\left(\frac{E}{E_1}\right) \cdot P\left(E_1\right)}{P\left(\frac{E}{E_1}\right) \cdot P\left(E_1\right)+P\left(\frac{E}{E_2}\right) \cdot P\left(E_2\right)+P\left(\frac{E}{E_3}\right) \cdot P\left(E_3\right)} [$by Bayes's theorem$]$
$=\frac{\left(\frac{1}{5} \times \frac{1}{3}\right)}{\left(\frac{1}{5} \times \frac{1}{3}\right)+\left(\frac{1}{3} \times \frac{1}{3}\right)+\left(\frac{2}{11} \times \frac{1}{3}\right)}$
$=\left(\frac{1}{15} \times \frac{495}{118}\right)$
$=\frac{33}{118}$
Hence, the required probability is $\frac{33}{118}$

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Question 93 Marks

Evaluate: $\int \frac{x}{\sqrt{x+z} \sqrt{x+z}} d x$

Answer

Rationalize the given integrand we get
$ \Rightarrow \int \frac{x}{\sqrt{x+a}-\sqrt{x-b}} \times \frac{\sqrt{x+a}+\sqrt{x-b}}{\sqrt{x+a}+\sqrt{x-b}} dx$
$\Rightarrow \int \frac{x(\sqrt{x+a}-\sqrt{x-b})}{x+a-x-b} d x$
$\Rightarrow \int \frac{x(\sqrt{x+a}-\sqrt{x-b})}{a-b} d x$
$\Rightarrow \frac{1}{a-b} \int x(\sqrt{x+a}-\sqrt{x-b}) d x$
Assume $x =\sqrt{t}$
$\Rightarrow dx=\frac{dt}{2 \sqrt{t}}$
Substituting values of $t$ , and $dt$
$\Rightarrow \int \sqrt{t} \frac{(\sqrt{\sqrt{t}+a}-\sqrt{\sqrt{t}-b)}}{2 \sqrt{t}(a-b)} d t$
$\Rightarrow \frac{1}{2(a-b)} \int(\sqrt{\sqrt{t}+a}-\sqrt{\sqrt{t}-b}) d t$
$\Rightarrow \frac{1}{2(a-b)} \int(\sqrt{t}+a)^{1 / 2} d t-\int(\sqrt{t}-b)^{1 / 2} d t$
$\Rightarrow \frac{1}{2(a-b)}\left(\frac{4}{3}\left(\sqrt{t}+a^2\right)^{\frac{3}{2}}-\frac{4}{3}\left(t-a^2\right)^{\frac{3}{2}}\right) $
now replacing $x =\sqrt{t}$
$\Rightarrow \frac{1}{2(a-b)}\left(\frac{2}{3}(x+a)^{\frac{3}{2}}-\frac{2}{3}(x-b)^{\frac{3}{2}}\right)$

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Mathematics 3 Marks Question

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