3 Marks Question

Question 13 Marks

Evaluate the integral: $\int_0^\pi \frac{x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$

Answer

We have,
$I=\int_0^\pi \frac{x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
Using property of definite integral we have,
$=\int_0^\pi \frac{(\pi-x)}{a^2 \cos ^2(\pi-x)+b^2 \sin ^2(\pi-x)} d x$
$=\int_0^\pi \frac{\pi-x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x \ldots \text { (ii) }$
Adding $(i)$ and $(ii)$
$2 I=\int_0^\pi \frac{x+\pi-x}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
$=\pi \int_0^\pi \frac{1}{a^2 \cos ^2 x+b^2 \sin ^2 x} d x$
$=\pi \int_0^\pi \frac{\sec ^2 x}{a^2+b^2 \tan ^2 x} d x \ldots\ ($Dividing numerator and denominator by $\cos ^2 x) $
$=2 \pi \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{a^2+b^2 \tan ^2 x} d x \ldots[$Using $\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x]$
Put $\tan x = t$
$\Rightarrow \sec ^2 x d x=d t$
When $x \rightarrow 0 ; t \rightarrow 0$
and $x \rightarrow \frac{\pi}{2} ; t \rightarrow \infty$
$\therefore 2 I=2 \pi \int_0^{\frac{x}{2}} \frac{d t}{a^2+b^2 t^2}$
$\Rightarrow I=\frac{\pi}{b^2} \int_0^{\frac{\pi}{2}} \frac{d t}{\frac{a^2}{b^2}+t^2}$
$=\frac{\pi}{b^2} \times \frac{b}{a}\left[\tan ^{-1}\left(\frac{b t}{a}\right)\right]_0^{\infty}$
$=\frac{\pi}{a b}\left[\frac{\pi}{2}-0\right]$
$=\frac{\pi}{a b} \times \frac{\pi}{2}$
$=\frac{\pi^2}{2 a b}$
Hence, $I=\frac{\pi^2}{2 a b}$

View full question & answer→

Question 23 Marks

If $\vec{a}=(\hat{i}-\hat{j}), \vec{b}=(3 \hat{j}-\hat{k})$ and $\vec{c}=(7 \hat{i}-\hat{k})$, find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and for which $\vec{c} \cdot \vec{d}=1$

Answer

$\text { Let } \vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$
$\vec{d} \perp \vec{a}, \vec{d} \cdot \vec{a}=0$
$ \Rightarrow\left(a_1 \hat{\imath}+a_2 \hat{\jmath}+a_3 \hat{x}\right)(\hat{i}-\hat{j})=0$
$\Rightarrow a_1-a_2=0 \ldots \text { (i) }$
$\vec{d} \perp \vec{b}, \vec{a} \cdot \vec{b}=0$
$ \Rightarrow\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)(3 j-\hat{k})=0$
$\Rightarrow 3 a_2-a_2=0 \ldots \text { (ii) }$
$\vec{d} \vec{c}=1 $
$\Rightarrow\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)(7 \hat{i}-\hat{k})=1$
$\Rightarrow 7 a_1-a_3=1 \ldots \text { (iii) }$
Solving equation $(i)$ and $(ii)$ we get $3 a _1- a _3=0$.
Again solving equation $(iii) \ (iv)$ we get $a_1=\frac{1}{4}$
From equation $(i), \ a_1-a_2=0$ or $a_1=a_2=\frac{1}{4}$
From equation $(ii), 3 a _2- a _2=0$
$ \Rightarrow 3 \cdot \frac{1}{4}=a_3 $
$\Rightarrow a_3=\frac{3}{4}$
Hence, $\vec{d}=\frac{1}{4} \hat{i}+\frac{1}{4} \hat{j}+\frac{3}{4} \hat{k}$

View full question & answer→

Question 33 Marks

Find $\frac{d y}{d x}$ of the function $(\cos x )^{ y }=(\cos y )^{ x }$.

Answer

We have, $(\cos x)^y=(\cos y)^x$
On taking $\log$ both sides, we get
$\log (\cos x)^y=\log (\cos y)^x$
$\Rightarrow y \log (\cos x)=x \log (\cos y)$
On differentiating both sides $\text{w.r.t x},$ we get
$y \cdot \frac{d}{d x} \log (\cos x)+\log \cos x \cdot \frac{d}{d x}(y)$
$=x \frac{d}{d x} \log (\cos y)+\log (\cos y) \frac{d}{d x}(x)\ [$by using product rule of derivative$]$
$\Rightarrow y \cdot \frac{1}{\cos x} \frac{d}{d x}(\cos x)+\log (\cos x) \frac{d y}{d x}$
$=x \cdot \frac{1}{\cos y} \frac{d}{d x}(\cos y )+\log \cos y \cdot 1$
$\Rightarrow y \cdot \frac{1}{\cos x}(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x}$
$=x \cdot \frac{1}{\cos y}(-\sin y ) \frac{d y}{d x}+\log \cos y \cdot 1$
$\Rightarrow- y \tan x +\log (\cos x ) \frac{d y}{d x}$
$=- x \tan y \frac{d y}{d x}+\log (\cos y )$
$\Rightarrow[ x \tan y +\log (\cos x )] \frac{d y}{d x}$
$=\log (\cos y )+ y \tan x $
$\therefore \frac{d y}{d x}=\frac{\log (\cos y)+y \tan x}{x \tan y+\log (\cos x)}$

View full question & answer→

Question 43 Marks

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5$ and each one of them being $\perp$ to the sum of the other two, find $|\vec{a}+\vec{b}+\vec{c}|$

Answer

$\vec{a} \cdot(\vec{b}+\vec{c})=0, \vec{b} \cdot(\vec{c}+\vec{a})=0, \vec{c} \cdot(\vec{a}+\vec{b})=0$
$($Given$) |\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})$
$=\vec{a} \cdot \vec{a}+\vec{a} \cdot(\vec{b}+\vec{c})+\vec{b} \cdot \vec{b}+\vec{b} \cdot(\vec{a}+\vec{c})+\vec{c} \cdot \vec{c}+\vec{c} \cdot(\vec{a}+\vec{b})$
$=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2$
$=9+16+25$
$=50$
$|\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}$
$=5 \sqrt{2}$

View full question & answer→

Question 53 Marks

Find the particular solution of the differential equation $\left(x e^{x / y}+y\right) d x=x\ d y$, given that $y(1)=0$

Answer

The given differential equation can be rewritten as,
$x e^{\frac{y}{x}}-y+x \frac{d y}{d x}=0$
$\Rightarrow x \frac{d y}{d x}=y-x e^{\frac{y}{x}}$
$\Rightarrow \frac{d y}{d x}=\left(\frac{y}{x}\right)-e^{\frac{y}{z}}$
$\Rightarrow \frac{ dy }{ dx }= f \left(\frac{ y }{ x }\right)$
$\Rightarrow$ the given differential equation is a homogenous equation.
The solution of the given differential equation is:
Put $y = vx$
$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$v+x \frac{d v}{d x}=\left(\frac{v x}{x}\right)-e^{\frac{w x}{x}}$
$\Rightarrow x \frac{ dv }{ dx }=- e ^{ v }$
$\Rightarrow \frac{ dv }{ e ^{ x }}=\frac{- d x}{ x }$
Integrating both the sides we get:
$\Rightarrow \int \frac{ d v}{ e ^{ v }}=-\int \frac{ d x}{ x }+c$
$\Rightarrow- e ^{- v }=-\operatorname{In}| x |+ c $
Resubstituting the value of $y = vx$ we get
$\Rightarrow- e ^{-\left(\frac{y}{x}\right)}=-\ln |x|+c$
Now, $y\ (1)=0$
$\Rightarrow- e ^{-(0)}=-\ln |1|+ c$
$\Rightarrow c =-1$
$\Rightarrow \log | x |+ e ^{- y / x }=1$

View full question & answer→

Question 63 Marks

Solve the following differential equation $\frac{d y}{d x}=1+x^2+y^2+x^2 y^2,$ given that $y =1, $ when $x =0$.

Answer

According to the question,
Given differential equation is,
$\frac{d y}{d x}=1+x^2+y^2+x^2 y^2$
$\Rightarrow \frac{d y}{d x}=1\left(1+x^2\right)+y^2\left(1+x^2\right)$
$\Rightarrow \frac{d y}{d x}=\left(1+x^2\right)\left(1+y^2\right)$
$\Rightarrow \frac{d y}{1+y^2}=\left(1+x^2\right) d x$
On integrating both sides, we get
$\int \frac{d y}{1+y^2}=\int\left(1+x^2\right) d x$
$\Rightarrow \tan ^{-1} y=x+\frac{x^3}{3}+C .$
Given that $y = 1$ when $x = 0$
On putting $x = 0$ and $y = 1$ in Eq. $(i),$ we get
$\tan ^{-1} 1= C$
$\Rightarrow \tan ^{-1}(\tan \pi / 4)=C \quad\left[\because \tan \frac{\pi}{4}=1\right]$
$\Rightarrow C=\frac{\pi}{4}$
On putting the value of $C$ in Eq. $(i),$ we get
$\tan ^{-1} y=x+\frac{x^3}{3}+\frac{\pi}{4}$
$\therefore y=\tan \left(x+\frac{x^3}{3}+\frac{\pi}{4}\right)$
which is the required solution of differential equation.

View full question & answer→

Question 73 Marks

Evaluate the definite integral: $\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$

Answer

We have,
$ I =\int_1^2 e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^2}\right) d x$
$I =\int_1^2 \frac{1}{x} \cdot e^{2 x}-\int_1^2 \frac{1}{2 x^2} \cdot e^{2 x} d x$
$\Rightarrow I = I _1- I _2$
Now, $I _1=\int_1^2 \frac{1}{x} e^{2 x} \ ($By parts we have$)$
$\Rightarrow I _1=\left[\frac{1}{x}\right]_1^2 \cdot \int_1^2 e^{2 x} d x-\int_1^2-\frac{1}{x^2} \frac{e^{2 x}}{2} d x$
$\Rightarrow I _1=\left[\frac{1}{x} \cdot \frac{e^{2 x}}{2}\right]_1^2+\int_1^2 \frac{1}{2 x^2} e^{2 x} d x$
$\Rightarrow I _1=\left[\frac{1}{2 x} e^{2 x}\right]_1^2+ I _2$
As, $I = I _1- I _2$
$\Rightarrow I =\left[\frac{1}{2 x} e^{2 x}\right]_1^2- I _2+ I _2$
$\Rightarrow I =\left[\frac{1}{2 x} e^{2 x}\right]_1^2=\frac{1}{2}\left[\frac{1}{2} e^4-e^2\right]$
$\Rightarrow I =\frac{1}{4} e ^2\left( e ^2-1\right)$

View full question & answer→

Question 83 Marks

Evaluate the definite integral $\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x$

Answer

$I=\int_0^{\pi / 4} \frac{\sin x \cdot \cos x}{\cos ^4 x+\sin ^4 x} d x$
Dividing $Nr$. and $Dr$. by $\cos^4 x$
$=\int_0^{\pi / 4} \frac{\frac{\sin x \cdot \cos x}{\cos ^4 z}}{\frac{\cos ^4 x}{\cos ^4 x}+\frac{\sin ^4 x}{\cos ^4 x}} d x$
$=\int_0^{\pi / 4} \frac{\tan x \cdot \sec ^2 x}{1+\tan ^4 x} d x$
$=\int_0^{\pi / 4} \frac{\tan x \cdot \sec ^2 x}{1+\left(\tan ^2 x\right)^2} d x$
Put $\tan ^2 x=t$
$2 \tan x \cdot \sec ^2 x d x=d t$
When $x =0, t =0$ and when $x=\frac{\pi}{4}, t=1$
$\therefore I=\frac{1}{2} \int_0^1 \frac{d t}{1+t^2}$
$=\frac{1}{2}\left[\tan ^{-1} t\right]_0^1$
$=\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}$

View full question & answer→

Question 93 Marks

In answering a question on a multiple choice questions test with four choices in each question, out of which only one is correct, a student either guesses or copies or knows the answer. The probability that he makes a guess $\frac{1}{4}$
and the probability the he copies is also $\frac{1}{4}$ The probability that the answer is correct, given that he copied it is $\frac{3}{4}$ Find the probability that he knows the answer to the question, given that he correctly answered it.

Answer

Let $E_1=$ Student guesses the answer
$E_2=$ Student copies the answer
$E_3 =$ Student knows the answer
$A =$ Student answers the question correctly.
$ P \left( E _1\right)=\frac{1}{4},$
$P \left( E _2\right)=\frac{1}{4},$
$P \left( E _3\right)=1-\left(\frac{1}{4}+\frac{1}{4}\right)=\frac{1}{2}$
$P \left( A \mid E _1\right)=\frac{1}{4}, P \left( A \mid E_2\right)=\frac{3}{4}, P \left( A \mid E_3\right)=1$
The required probability
$=P\left(E_3 \mid A\right)=\frac{P\left(E_3\right) \times P\left(A \mid E_3\right)}{\sum_{i=1}^3 P\left(E_i\right) \times P\left(A \mid E_i\right)}$
$=\frac{\frac{1}{2} \times 1}{\frac{1}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}+\frac{1}{2} \times 1}$
$=\frac{1}{\frac{1}{8}+\frac{3}{8}+1}=\frac{8}{12}=\frac{2}{3}$

View full question & answer→

MATHS 3 Marks Question

Take a timed test