3 Marks Question

Question 13 Marks

Find the general solution of the differential equation $x\left(y^3+x^3\right) d y=\left(2 y^4+5 x^3 y\right) d x$

Answer

$ x \left( y ^3+ x ^3\right) dy =\left(2 y ^4+5 x ^3 y \right) dx$
$\frac{d y}{d x}=\frac{2 y^4+5 x^3 y}{x y^3+x^4}$
It is a homogeneous differential equation
So put $y = vx , \frac{d y}{d x}=v+x \frac{d v}{d x}$
$v+x \frac{d v}{d x}=\frac{2 v^4+5 v}{v^3+1}$
$x \frac{d v}{d x}=\frac{2 v^4+5 v}{v^3+1}-v$
$x \frac{d v}{d x}=\frac{2 v^4+5 v-v^4-v}{v^3+1}$
$\frac{v^3+1}{v^4+4 v} d v=\frac{d x}{x}$
$\int \frac{4 v^3+4}{v^4+4 v} d v=4 \int \frac{d x}{x}$
$\log \left|v^4+4 v\right|=4 \log x+\log C$
$\log \left|v^4+4 v\right|=\log (x)^4+\log C$
$\log \left|v^4+4 v\right|=\log Cx^4$
$v^4+4 v=Cx$
Put $v =\frac{y}{x}$
$\frac{y^4}{x^4}+4 \frac{y}{x}=Cx^4$
$y^4+4 yx^3=Cx$

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Question 23 Marks

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

Answer

Given$, x = a(\cos t + t \sin t)$
On differentiating both sides $\text{w.r.t t,}$ we get
$\frac{d x}{d t}=a\left[-\sin t+\frac{d}{d t}(t) \cdot \sin t+t \frac{d}{d t}(\sin t)\right]$
$($by using product rule of derivative$)$
$\Rightarrow \frac{d x}{d t}=a(-\sin t+1 \cdot \sin t+t \cos t)=\text { a t } \cos t \ldots \ldots .....(i)$
Also, given$, y = a(\sin t - t \cos t)$
On differentiating both sides $\text{w.r.t t,}$ we get
$\frac{d y}{d t}=a\left[\cos t-\frac{d}{d t}(t) \cos t-t \frac{d}{d t}(\cos t)\right]$
$($by using product rule of derivative$)$
$\frac{d y}{d t}= a (\cos t -\cos t \cdot 1+ t \sin t )$
$= at \sin t ...............(ii)$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a t \sin t}{a t \cos t}=\tan t\ ($From Eqs.$(i)$ and $(ii))$
Again, differentiating both sides $\text{w.r.t x,}$ we get
$\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d t}(\tan t) \frac{d t}{d x}=\sec ^2 t \frac{1}{d x / d t}$
$=\frac{\sec ^2 t}{a t \cos t}=\frac{\sec ^3 t}{a t}\ ($From Eq $(i))$
$\text { Also, } \frac{d^2 x}{d t^2}=\frac{d}{d t}(a t \cos t)$
$=a \frac{d}{d t}(t \cos t)$
$=a\left[\frac{d}{d t}(t) \cdot \cos t+t \frac{d}{d t}(\cos t)\right]$
$($by using product rule of derivative$)$
$=a(\cos t-\sin t)$ and $\frac{d^2 y}{d t^2}$
$=\frac{d}{d t}\left(\frac{d y}{d t}\right)=\frac{d}{d t}(a t \sin t)$
$=a(\sin t+t \cos t)$

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Question 33 Marks

Find $\int \frac{x^2-3 x+1}{\sqrt{1-x^2}} d x$

Answer

Let $I=\int \frac{x^2-3 x+1}{\sqrt{1-x^2}}$
$=(-1) \int \frac{-x^2+3 x-1}{\sqrt{1-x^2}} d x$
$=(-1) \int \frac{-x^2+3 x-1+1-1}{\sqrt{1-x^2}} d x$
$=(-1) \int \frac{1-x^2+3 x-2}{\sqrt{1-x^2}} d x$
$=(-1) \int\left[\frac{1-x^2}{\sqrt{1-x^2}}+\frac{3 x-2}{\sqrt{1-x^2}}\right] d x$
$=(-1) \int\left[\sqrt{1-x^2}+\int \frac{3 x-2}{\sqrt{1-x^2}}\right] d x$
$=(-1)\left[\int \sqrt{1-x^2} d x+\int \frac{3 x-2}{\sqrt{1-x^2}} d x\right]$
$=(-1)\left(I_1+I_2\right) \ldots \ldots \text { (i) }$
$\text { consider, } I_1=\int \sqrt{1-x^2} d x$
$=\frac{1}{2}\left[x \sqrt{1-x^2}+\sin ^{-1}(x)\right]+C_1 \ldots \text { (ii) }\left[\because \int \sqrt{a^2-x^2} d x=\frac{1}{2}\left[x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right]+C\right)$
$\text { consider } I_2=\int \frac{3 x-2}{\sqrt{1-x^2}} d x$
$=\int \frac{3 x}{\sqrt{1-x^2}} d x-2 \int \frac{d x}{\sqrt{1-x^2}}$
$=-\frac{3}{2} \int \frac{-2 x}{\sqrt{1-x^2}} d x-2 \int \frac{d x}{\sqrt{1-x^2}}$
$=-\frac{3}{2} \times 2 \sqrt{1-x^2}-2 \sin ^{-1}(x)+C_2\left[\because \int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x=2 \sqrt{f(x)}+C\right]$
$=-3 \sqrt{1-x^2}-2 \sin ^{-1}(x)+C_2 \ldots \text { (iii) }$
From Equations $(i), (ii)$ and $(iii),$ we get
$I=(-1)\left(\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}(x)+C_1-2 \sin ^{-1}(x)-3 \sqrt{1-x^2}+C_2\right)$
$I=\frac{3}{2} \sin ^{-1}(x)-\frac{x}{2} \sqrt{1-x^2}+3 \sqrt{1-x^2}+C\ ($where $C=-C_1-C_2)$

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Question 43 Marks

Three groups of children contain $3$ girls and $1$ boy; $2$ girls and $2$ boys; $1$ girl and $3$ boys respectively. One child is selected at random from each group. Find the chance that the three selected comprise one girl and $2 $ boys.

Answer

One girl and $2$ boys can be selected in the following mutually exclusive ways:

	Group $1$	Group $2$	Group $3$
$(I)$	Girl	Boy	Boy
$(II)$	Boy	Girl	Boy
$(III)$	Boy	Boy	Girl

Therefore, if we define $G_1G_2G_3 $ as the events of selecting a girl from first, second and third group respectively and $B_1 B_2 B_3:$ as the events of selecting a boy from first, second and third group respectively.
Then $B_1, B_2 B_3, G_1 G_2G_3$ are independent events such that
$P\left(G_1\right)=\frac{3}{4}, P\left(G_2\right)=\frac{2}{4}, P\left(G_3\right)=\frac{1}{4}$
$P\left(B_1\right)=\frac{1}{4}, P\left(B_2\right)=\frac{2}{4}, P\left(B_3\right)=\frac{3}{4}$
Therefore, required probability is given by,
$P( S$ Selecting $1$ girl and $2$ boys$)$
$=(\text { I or II or III) }$
$=P( I \cup II \cup III )$
$=P\left(\left(G_1 \cap B_2 \cap B_3\right) \cup\left(B_1 \cap G_2 \cap B_3\right) \cup\left(B_1 \cap B_2 \cap G_3\right)\right)$
$=P\left(G_1 \cap B_2 \cap B_3\right)+P\left(B_1 \cap G_2 \cap B_3\right)+P\left(B_1 \cap B_2 \cap G_3\right)$
$= P \left( G _1\right) P \left( B _2\right) P \left( B _3\right)+ P \left( B _1\right) P \left( B _2\right) P \left( G _3\right)$
$=\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{2}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}=\frac{9}{32}+\frac{3}{32}+\frac{1}{32}=\frac{13}{32}$
$P \left( B _3\right)+ P \left( B _1\right) P \left( G _2\right)$

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Question 53 Marks

Find $\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4 \cos ^2 \theta\right)} d \theta$.

Answer

According to the question, $I=\int \frac{\cos \theta}{\left.\left(4+\sin ^2 \theta\right)\left(5-4 \cos ^2 \theta\right)\right)} d \theta$
$=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4\left(1-\sin ^2\theta\right)\right)} d \theta\left(\therefore \cos ^2 \theta=1-\sin ^2 \theta\right)$
$=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(5-4+4 \sin ^2 \theta\right)} d \theta$
$=\int \frac{\cos \theta}{\left(4+\sin ^2 \theta\right)\left(1+4 \sin ^2 \theta\right)} d \theta$
Let $\sin \theta=t $
$\Rightarrow \cos \theta d \theta=d t$
Then, $I=\int \frac{d t}{\left(4+t^2\right)\left(1+4 t^2\right)}$
let, $\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{A}{4+t^2}+\frac{B}{1+4 t^2}$
using partial fractions
At $t =0, \frac{A}{4}+\frac{B}{1}=\frac{1}{4 \times 1} $
$\Rightarrow A+4 B=1$
At $t=1, \frac{A}{5}+\frac{B}{5}=\frac{1}{5 \times 5} $
$\Rightarrow 5 A+5 B=1$
On solving Equations $(i)$ and $(ii),$ we get
$A=\frac{-1}{15}$ and $B=\frac{4}{15}$
$\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{-\frac{1}{15}}{4+t^2}+\frac{\frac{4}{15}}{1+4 t^2}$
$\Rightarrow \frac{1}{\left(4+t^2\right)\left(+4 t^2\right)}=\frac{-1}{15\left(4+t^2\right)}+\frac{4}{15\left(1+4 t^2\right)}$
Integrating both sides $\text{w.r.t. t,}$
$\Rightarrow \int \frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)} d t=\frac{-1}{15} \int \frac{1}{4+t^2} d t+\frac{4}{15} \int \frac{1}{1+4 t^2} dt$
$=\frac{-1}{15} \int \frac{1}{2^2+t^2}+\frac{4}{15 \times 4} \int \frac{1}{\left(\frac{1}{2}\right)^2+t^2} d t$
$=\frac{-1}{15} \cdot \frac{1}{2} \tan ^{-1} \frac{t}{2}+\frac{1}{15} \cdot \frac{1}{1 / 2} \tan ^{-1} \frac{t}{1 / 2}+C\left[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c\right]$
put $t=\sin \theta$
$=\frac{-1}{30} \tan ^{-1} \frac{\sin \theta}{2}+\frac{2}{15} \tan ^{-1} 2 \sin \theta+C$

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Question 63 Marks

If $\vec{a}=3 \hat{i}-\hat{j}$ and $\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$, then express $\vec{b}$ in the form $\vec{b}=\vec{b}_1+\vec{b}_2$, where $\vec{b}_1 \| \vec{a}$ and $\vec{b}_2 \perp \vec{a}$.

Answer

According to the question,
$\vec{a}=3 \hat{i}-\hat{j}$ and $\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$
Let $\overrightarrow{b_1}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}$ and $\overrightarrow{b_2}=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}$
$\overrightarrow{b_1}+\overrightarrow{b_2}=\vec{b}, \overrightarrow{b_1} \| \vec{a}$ and $\overrightarrow{b_2} \perp \vec{a}$
Consider, $\overrightarrow{b_1}+\overrightarrow{b_2}=\vec{b}$
$\Rightarrow \left(x_1+x_2\right) \hat{i}+\left(y_1+y_2\right) \hat{j}+\left(z_1+z_2\right) \hat{k}=2 \hat{i}+\hat{j}-3 \hat{k}$
On comparing the coefficient of $\hat{i}, \hat{\jmath}$ and $\hat{k}$ both sides; we get
$\Rightarrow x_1+x_2=2$
$y_1+y_2=1 \ldots \text { (ii) }$ and $z_1+z_2=-3$
Now, consider $\overrightarrow{b_1} \| \vec{a}$
$\Rightarrow \frac{x_1}{3}=\frac{y_1}{-1}=\frac{z_1}{0}=\lambda($say$)$
$\Rightarrow x_1=3 \lambda, y_1=-\lambda$ and $z_1=0$
On substituting the values of $x,y$ and $z,$ form Eq. $(iv)$ to Eq. $(i), (ii)$ and $(iii),$ respectively, we get
$x_2=2-3 \lambda, y_2=1+\lambda$ and $z_2=-3$
Since, $\overrightarrow{b_2} \perp \vec{a}$, therefore $\vec{b}_2 \cdot \vec{a}=0$
$\Rightarrow 3 x_2-y_2=0$
$\Rightarrow 3(2-3 \lambda)-(1+\lambda)=0$
$\Rightarrow 6-9 \lambda-1-\lambda=0$
$\Rightarrow 5-10 \lambda=0 $
$\Rightarrow \lambda=\frac{1}{2}$
On substituting $\lambda=\frac{1}{2}$ in Eqs. $(iv)$ and Eqs. $(iv)$ and $(v),$ we get
$x_1=\frac{3}{2}, y_1=\frac{-1}{2}, z_1=0$ and $x_2=\frac{1}{2}, y_2=\frac{3}{2}$ and $z_2=-3$
Hence$, \vec{b}=\vec{b}_1+\vec{b}_2=\left(\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j}\right)+\left(\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}\right)$
$=2 \hat{i}+\hat{j}-3 \hat{k}$

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Question 73 Marks

If $\vec{a}=\hat{i}+j+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}$, then find a vector $\vec{c}$, such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$.

Answer

According to the question,
Given vectors are $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and
$\begin{array}{l}\vec{b}=\hat{j}-\hat{k}
\\ \text { Let } \vec{c}=x \hat{i}+y \hat{j}+z \hat{k}\end{array}$
Then,$\vec{a} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z\end{array}\right|$
$=\hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x)$
Given that $\vec{a} \times \vec{c}=\vec{b}$.
$\Rightarrow \hat{i}(z-y)+\hat{j}(x-z)+\hat{k}(y-x)=0 \hat{i}+1 \hat{j}+(-1) \hat{k}(\because \vec{b}=\hat{j}-\hat{k})$
On comparing the coefficients of i, j, and k. from both sides, we get
$z - y = 0$
$x - z = 1$ and
$y - x = - 1$
$x - y =1...(i)$
Also given that, $\vec{a} \cdot \vec{c}=3$
$\Rightarrow(\hat{i}+\hat{j}+\hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})=3$
$\Rightarrow x + y + z =3$
$\Rightarrow x +2 y =3(\because y = z ) \ldots \text { (ii) }$
On subtracting Eq. $(i)$ from Eq. $(ii),$ we get $3y = 2$
$\Rightarrow y=\frac{2}{3}=z(\because y=z)$
From Eq. $(i), x=1+y=1+\frac{2}{3}=\frac{5}{3}$
$\therefore \vec{c}=\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}$

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Question 83 Marks

Find a particular solution of $x \frac{d y}{d x}- y =\log x$, given that $y = 0$ when $x = 1$

Answer

The given differential equation is,
$x \frac{d y}{d x}-y=\log x$
$\Rightarrow \frac{d y}{d x}+\left(-\frac{1}{x}\right) y=\frac{1}{x} \log x$
This is of the form $\frac{d y}{d x}+P y=Q$
Where, $P=-\frac{1}{x}, Q=\frac{1}{x} \cdot \log x$
Here $I . F=e^{\int\left(-\frac{1}{x}\right) d x}=e^{-\log x}=\frac{1}{x}$
$\therefore y \cdot(I F)=\int(I F) Q d x+C$
$\Rightarrow y \cdot \frac{1}{x}=\int \frac{1}{x} \cdot \frac{1}{x} \cdot \log x d x+C \Rightarrow \frac{y}{x}=\int \frac{1}{x^2} \cdot \log x d x+C$
$\Rightarrow \frac{y}{x}=\log x \int \frac{1}{x^2} d x-\int\left\{\frac{d}{d x}(\log x) \int \frac{1}{x^2} d x\right\} d x+C$
$\Rightarrow \frac{y}{x}=-\frac{1}{x} \cdot \log x+\int \frac{1}{x} \cdot \frac{1}{x} d x+C$
$\Rightarrow \frac{y}{x}=-\frac{1}{x} \cdot \log x-\frac{1}{x}+C \ldots \text { (i) }$
Putting $x = 1$ and $y = 0$ we get,
$0 = - \log(1) - 1 + C$
$C = 1$
Putting $C = 1$ in equation $(i)$ we have $\frac{y}{x}=-\frac{1}{x} \cdot \log x-\frac{1}{x}+1$
$\Rightarrow y=x-1-\log x$

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Question 93 Marks

Evaluate: $\int_0^\pi \frac{1}{5+4 \cos x} d x$

Answer

Let $I=\int_0^\pi \frac{1}{5+4 \cos x} d x$. Then
$I=\int_0^\pi \frac{1}{5+4\left(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)} d x=\int_0^\pi \frac{1+\tan ^2 \frac{x}{2}}{5\left(1+\tan ^2 \frac{x}{2}\right)+4\left(1-\tan ^2 \frac{z}{2}\right)} d x$
$\Rightarrow I=\int_0^\pi \frac{1+\tan ^2 \frac{2}{2}}{9+\tan ^2 \frac{z}{2}} d x=\int_0^\pi \frac{\sec ^2\frac{x}{2}}{9+\tan ^2 \frac{x}{2}} d x$
By using substitution
Let $\tan \frac{x}{2}=t$. Then, $d\left(\tan \frac{x}{2}\right)=d t $
$\Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t$
$ \Rightarrow d x=\frac{2 d t}{\sec ^2 \frac{x}{2}}$
Also, $x=0 \Rightarrow t=\tan 0=0$ and $x=\pi \Rightarrow t=\tan \frac{\pi}{2}=\infty$
$\therefore I=\int_0^{\infty} \frac{\sec ^2 \frac{x}{2}}{9+t^2} \times \frac{2 d t}{\sec ^2 \frac{x}{2}}$
$\Rightarrow I=2 \int_0^{\infty} \frac{d t}{3^2+t^2}=\frac{2}{3}\left(\tan ^{-1} \frac{t}{3}\right)_0^{\infty}
=\frac{2}{3}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)=\frac{2}{3}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{3}$

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