3 Marks Question

Question 13 Marks

If with reference to the right handed system of mutually perpendicular unit vectors $\hat{i}, \hat{j}$ and $\hat{k}, \vec{\alpha}=3 \hat{i}-\hat{j.}$
$\vec{\beta}=2 \hat{i}+\hat{j}-3 \hat{k.}$ then express $\vec{\beta}$ in the form $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$, where $\vec{\beta}_1$ is $\|$ to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$.

Answer

Let $ \vec{\beta}_1=\lambda \vec{\alpha}\left[\because \vec{\beta}_1 \| t \ o \ \vec{\alpha}\right]$
$\vec{\beta}_1=\lambda(3 \hat{i}-\hat{j})$
$=3 \lambda \hat{i}-\lambda \hat{j}$
$\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$
$=(2-3 \lambda) \hat{i}+(1+\lambda) \hat{j}-3 \hat{k}$
$\vec{\alpha} \cdot \vec{\beta}_2=0\left[\because \vec{\beta}_2 \perp \vec{\alpha}\right]$
$3(2-3 \lambda)-(1+\lambda)=0$
$\lambda=\frac{1}{2}$
$\vec{\beta}_1=\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j}$
$\vec{\beta}_2=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}$

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Question 23 Marks

Evaluate $\int_0^{\pi / 2} \frac{x+\sin x}{1+\cos x} d x$

Answer

$\text { Given } I =\int_0^{\pi / 2} \frac{x+\sin x}{1+\cos x} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{x+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x $
$ {\left[\begin{array}{l}\because \sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \\ \text { and } 1+\cos x=2 \cos ^2 \frac{x}{2}\end{array}\right]}$
$\Rightarrow I=\frac{1}{2} \int_0^{\pi / 2} \underset{I}{x} \sec _{I I}^2 \frac{x}{2} d x+\int_0^{\pi / 2} \tan \frac{x}{2} d x$
$\Rightarrow I=\frac{1}{2}\left\{\left[x \int \sec ^2 \frac{x}{2} d x\right]_0^{\pi / 2}-\int_0^{\pi / 2}\left[\frac{d}{d x}(x) \int\left(\sec ^2 \frac{x}{2} d x\right)\right] d x\right\}+\int_0^{\pi / 2} \tan \frac{x}{2} d x$
$\Rightarrow I=\frac{1}{2}\left\{\left[x \cdot \frac{\tan \frac{x}{2}}{\frac{1}{2}}\right]_0^{\pi / 2}-\int_0^{\pi / 2} \frac{\tan \frac{x}{2}}{\frac{1}{2}} d x\right\}$
$+\int_0^{\pi / 2} \tan \frac{x}{2} d x$
$[$Integration by parts$]$
$=\left[x \cdot \tan \frac{x}{2}\right]_0^{\pi / 2}-\int_0^{\pi / 2} \tan \frac{x}{2} d x+\int_0^{\pi / 2} \tan \frac{x}{2} d x$
$=\frac{\pi}{2} \cdot \tan \frac{\pi}{4}-0$
$\therefore I=\frac{\pi}{2}\left[\because \tan \frac{\pi}{4}=1\right]$

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Question 33 Marks

A factory has two machines $A$ and $B$. Past records show that the machine $A$ produced $60\%$ of the items of output and machine $B$ produced $40\%$ of the items. Further $2\%$ of the items produced by machine $A$ were defective and $1\%$ produced by machine $B$ were defective. If an item is drawn at random, what is the probability that it is defective?

Answer

Let $A, E_1$ and $E_2$ denote the events that the item is defective, machine $A$ is selected and machine $B$ is selected, respectievly. Therefore, we have,
$ P \left( E _1\right)=\frac{60}{100}$
$P \left( E _2\right)=\frac{40}{100}$
Now,we have,
$ P\left(\frac{A}{E_1}\right)=\frac{2}{100}$
$P\left(\frac{A}{E_2}\right)=\frac{1}{100}$
Using the law of total probability, we have, Required probability $= P ( A )= P \left( E _1\right) P \left( A / E _1\right)+ P \left( E _2\right) P \left( A / E _2\right)$
$=\frac{60}{100} \times \frac{2}{100}+\frac{40}{100} \times \frac{1}{100}$
$=\frac{120}{10000}+\frac{40}{10000}$
$=\frac{120+40}{10000}=\frac{160}{10000}=0.016$

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Question 43 Marks

Evaluate: $\int \frac{x+2}{\sqrt{x^2+2 x-1}} d x$

Answer

Let the given integral be,
$l=\int \frac{x+2}{\sqrt{x^2+2 x-1}} d x$
Let $x +2=\lambda \frac{d}{d x}\left( x ^2+2 x -1\right)+\mu$
$ x +2=\lambda(2 x + x )+\mu$
$x +2=(2 \lambda) x +2 \lambda+\mu$
Comparing the coefficients of like powers of $x,$
$2 \lambda=1 $
$\Rightarrow \lambda=\frac{1}{2}$
$2 \lambda+\mu=2$
$\Rightarrow 2\left(\frac{1}{2}\right)+\mu=2$
$\mu=1$
So $, I _1=\int \frac{\frac{1}{2}(2 x+2)+1}{\sqrt{x^2+2 x-1}} d x$
$=\frac{1}{2} \int \frac{1}{\sqrt{x^2+2 x-1}}(2 x +2) dx +1 \frac{1}{\sqrt{x^2+2 x+(1)^2-(1)^2-1}} d x$
$I =\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^2+2 x-1}} d x+1 \frac{1}{(x+1)^2-(\sqrt{2})^2} d x$
$I =\frac{1}{2} 2 \sqrt{x^2+2 x-1}+\log \left| x +1+\sqrt{(x+1)^2-(\sqrt{2})^2}\right|+c\left]\right.$
$I =\sqrt{x^2+2 x-1}+\log \left| x +1+\sqrt{x^2+2 x-1}\right|+ c$

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Question 53 Marks

Evaluate the integral : $\int \sqrt{\cot \theta} d \theta$

Answer

$I=\int \sqrt{\cot \theta} \ d \ \theta$
Let $\cot \theta=x^2$
$\Rightarrow-\operatorname{cosec} 2 \theta \ d \ \theta=2 x \ dx$
$\Rightarrow d\ \theta=\frac{-2 x}{\operatorname{cosec}^2 \theta} d x$
$=\frac{-2 x}{1+\cot ^2 \theta}\ d x$
$=\frac{-2 x}{1+x^4}\ d x$
$\therefore I=-\int \frac{2 x^2}{1+x^4} d x$
$=-\int \frac{2}{\frac{1}{x^2}+x^2} d x$
Dividing numerator and denominator by $x2$
$=-\int \frac{1+\frac{1}{x^2}+1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}} d x$
$=-\int \frac{\left(1+\frac{1}{x^2}\right)^{d x}}{\left(x-\frac{1}{x}\right)^2+2}-\int \frac{\left(1-\frac{1}{x^2}\right) d x}{\left(x+\frac{1}{x}\right)^2-2}$
Let $ x -\frac{1}{x}= t $
$\Rightarrow\left(1+\frac{1}{x^2}\right) dx = dt$
and $x +\frac{1}{x}= z $
$\Rightarrow\left(1-\frac{1}{x^2}\right) d x= dz$
$\Rightarrow I =-\int \frac{ dt }{t^2+2}-\int \frac{d z}{z^2-2}$
$=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{z-\sqrt{2}}{z+\sqrt{2}}\right|+ C$
$=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{x^2+1-\sqrt{2} x}{x^2+1+\sqrt{2} x}\right|+ C$
$I =-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\cot \theta-1}{\sqrt{2 \cot \theta}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{\cot \theta+1-\sqrt{2 \cot \theta}}{\cot \theta+1-\sqrt{2 \cot \theta}}\right|+ C $

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Question 63 Marks

Show that the function $f(x)$ defined by $f ( x )=\left\{\begin{array}{ll}\frac{\sin x}{x}+\cos x, & x>0 \\ 2, & x=0 \text { is continuous at } x =0 . \\ \frac{4(1-\sqrt{1-x})}{x}, & x<0\end{array}\right.$

Answer

To show that the given function is continuous at $x = 0$ we show that
$( \text{LHL} )_{x=0}=( \text{RHL} )_{x=0}=f(0) \ldots (i)$
Here, we have $f ( x )=\left\{\begin{array}{ll}\frac{\sin x}{x}+\cos x, x;0 \\ 2, x=0 \\ \frac{4(1-\sqrt{1-x})}{x}, x<0\end{array}\right.$
Now $, \text{LHL} =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{4(1-\sqrt{1-x})}{x}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1-(0-h)}}{0-h}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1+h}]}{-h}$
$=\lim _{h \rightarrow 0} \frac{4[1-\sqrt{1+h]}}{-h} \times \frac{1+\sqrt{1+h}}{1+\sqrt{1+h}}$
$=\lim _{h \rightarrow 0} \frac{4\left[(1)^2-(\sqrt{1+h})^2\right]}{-h[1+\sqrt{1+h]}}$
$=\lim _{h \rightarrow 0} \frac{4[1-(1+h)]}{-h[1+\sqrt{1+h}]}$
$=\lim _{h \rightarrow 0} \frac{-h \times 4}{-h[1+\sqrt{1+h}]}$
$=\lim _{h \rightarrow 0} \frac{4}{1+\sqrt{1+h}}$
$=\frac{4}{1+\sqrt{1}}=\frac{4}{2}=2$
and $\text{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}+\cos x\right)$
$\Rightarrow \text { RHL }=\lim _{h \rightarrow 0}\left(\frac{\sin h}{h}+\cos h\right)$
$=\lim _{h \rightarrow 0} \frac{\sin h}{h}+\lim _{h \rightarrow 0} \cos h$
$=1+\cos 0$
$=1+1$
$=2$
Also, given that $x=0, f(x)=2 $
$\Rightarrow f(0)=2$
Since, $(\text{LHL} )_{x=0}=( \text{RHL} )_{x=0}=f(0)=2$
Therefore $,f ( x )$ is continuous at $x =0$.

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Question 73 Marks

If $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-2 \hat{k}$,find the unit vector in the direction of $2 \vec{a}-\vec{b.}$

Answer

We need to find the unit vector in the direction of $2 \vec{a}-\vec{b}$.
First, let us calculate $2 \vec{a}-\vec{b}$.
As we have,
$\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$
$\overrightarrow{b}=2 \hat{i}+\hat{j}-2 a $
Then multiply equation $(a)$ by $2$ on both sides,
$2 \vec{a}=2(\hat{i}+\hat{j}+2 \hat{k})$
We can easily multiply vector by a scalar by multiplying similar components,
that is, vector's magnitude by the scalar's
magnitude.
$\Rightarrow 2 \overrightarrow{ a }=2 \hat{ i }+2 \hat{ j }+4 \hat{ k }$
Subtract $(b)$ from $(c)$. We get,
$2 \overrightarrow{ a }-\overrightarrow{ b }=(2 \hat{ i }+2 \hat{ j }+4 \hat{ k })-(2 \hat{ i }+\hat{ j }-2 \hat{ k })$
$\Rightarrow 2 \overrightarrow{ a }-\overrightarrow{ b }=2 \hat{i}-2 \hat{ i }+2 \hat{ j }-\hat{ j }+4 \hat{ k }+2 \vec{k}$
$\Rightarrow 2 \overrightarrow{ a }-\overrightarrow{ b }=\hat{ j }+6 \hat{ k }$
For finding unit vector, we have the formula:
$2 \hat{a}-\hat{b}=\frac{2 \vec{a}-\vec{b}}{|2 \vec{a}-\vec{b}|}$
Now we know the value of $2 \vec{a}-\vec{b}$,
So we just need to substitute in the above equation.
$\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{j}+6 \hat{k}}{|\hat{j}+6 \hat{k}|}$
Here, $|\hat{\jmath}+6 \hat{k}|=\sqrt{1^2+6^2}$
$\Rightarrow 2 \hat{a}-\hat{b}=\frac{\hat{j}+6 \hat{k}}{\sqrt{1^2+6^2}}$
$\Rightarrow 2 \hat{ a }-\hat{ b }=\frac{\hat{ j }+6 \hat{ k }}{\sqrt{1+36}}$
$\Rightarrow 2 \hat{a}-\hat{b}=\frac{j+6 \hat{k}}{\sqrt{37}}$
Thus, unit vector in the direction of $2 \vec{a}-\vec{b}$ is $\frac{\hat{j}+6 \hat{k}}{\sqrt{37}}$.

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Question 83 Marks

Find the particular solution of the differential equation $\left[x \sin ^2\left(\frac{y}{x}\right)-y\right] dx + x d y =0$, given that $y =\frac{\pi}{4}$ when $x =1$

Answer

We can rewrite the given differential equation as
$\frac{d y}{d x}=\frac{y}{x}-\sin ^2\left(\frac{y}{x}\right)$
This is of the form $\frac{d y}{d x}=f\left(\frac{y}{x}\right)$
So, it is homogeneous
Putting ,$y = vx$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$ in $(i),$ we get
$v+x \frac{d v}{d x}= v -\sin ^2 v$
$\Rightarrow x \frac{d v}{d x}=-\sin ^2 v$
$\Rightarrow-\operatorname{cosec}^2 v d v=\frac{1}{x} d x$
$\Rightarrow \int\left(-\operatorname{cosec}^2 v\right) d v=\int \frac{1}{x} d x \ [$on integrating both sides$]$
$\Rightarrow \cot v =\log | x |+ C,$ where $C$ is an arbitrary constant
$\Rightarrow \cot \frac{y}{x}=\log | x |+ C \ldots \text { (ii) }\left[\because v=\frac{y}{x}\right]$
Putting $x=1$ and $y=\frac{\pi}{4}$ in $(ii),$ we get $C=1$.
$\therefore \cot \frac{y}{x}=\log | x |+1$ is the desired solution.

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Question 93 Marks

Find the general solution for the differential equation $\left(x^2 y-x^2\right) d x+\left(x y^2-y^2\right) d y=0$

Answer

The given differential equation is,
$x^2(y-1) d x+y^2(x-1) d y=0$
$\frac{x^2}{x-1} d x+\frac{y^2}{y-1} d y=0$
Add and subtract $1$ in numerators we have,
$\frac{x^2-1+1}{(x-1)} d x+\frac{y^2-1+1}{(y-1)} d y=0$
By the identity $\left(a^2-b^2\right)=(a+b) \cdot(a-b)$
$\frac{(x+1)(x-1)+1}{(x-1)} d x+\frac{(y+1)(y-1)+1}{(y-1)} d y=0$
Splitting the terms,
$(x+1) d x+\frac{1}{(x-1)} d x+(y+1) dy +\frac{1}{(y-1)} d y=0$
Integrating, we get,
$\int(x+1) d x+\int \frac{1}{(x-1)} d x+\int(y+1) d y+\int \frac{1}{(y-1)} d y=C$
$\frac{x^2}{2}+x+\log |x-1|+\frac{y^2}{2}+y+\log |y-1|= C$
$\frac{1}{2} \cdot\left(x^2+y^2\right)+(x+y)+\log |(x-1)(y-1)|= C $
This is the required solution

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