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Differentiation by substitution - class-XI

Description: differentiation by substitution
Number of Questions: 31
Created by:
Tags: differentiation differential calculus i: fundamentals maths differencial calculus - differenciability and methods of differnciation
Attempted 0/31 Correct 0 Score 0

Differentiate $\tan^{-1} \sqrt{\dfrac{1+\cos x}{1- \cos x}}$

  1. $\dfrac{-1}{2}$

  2. $\dfrac{1}{4}$

  3. $\dfrac{1}{8}$

  4. None of these


Correct Option: A
Explanation:

Let $y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 + \cos x}}{{1 - \cos x}}} $

$ = {\tan ^{ - 1}}\sqrt {\dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2{{\sin }^2}\dfrac{x}{2}}}} $
$ = {\tan ^{ - 1}}\cot \left( {\dfrac{x}{2}} \right)$
$ = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right)} \right]$
$ = \dfrac{\pi }{2} - \dfrac{x}{2}$
$ \Rightarrow \dfrac{{dy}}{{dx}} =  - \dfrac{1}{2}$

If $y=\dfrac{1+x^2+x^4}{1+x+x^2}$ and $\dfrac{dy}{dx}=ax+b$, then the values of $a$ and $b$ are,

  1. $a=2,b=1$

  2. $a=-2,b=1$

  3. $a=2,b=-1$

  4. $a=-2,b=-1$


Correct Option: C
Explanation:
$\displaystyle y=\frac{1+x^{2}+x^{4}}{1+x+x^{2}}$

$\displaystyle \frac{dy}{dx}= \frac{(2x+4x^{3})(1+x+x^{2})-(1+2x)(1+x^{2}+x^{4})}{(1+x^{2}+x^{4}+2x+2x^{2}+2x^{3})}$

$\displaystyle \frac{dy}{dx}=\frac{2x+2x^{2}+2x^{3}+4x^{3}+4x^{4}+4x^{5}-1-x^{2}-x^{4}-2x-2x^{3}-2x^{5}}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle \frac{dy}{dx}=\frac{2x^{5}+3x^{4}+4x^{3}+x^{2}-1}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle (\frac{dy}{dx})=\frac{(2x^{5}+4x^{4}+6x^{3}+4x^{2}+2x)-(x^{4}+2x^{3}+3x^{2}+2x+1)}{(x^{4}+2x^{3}+3x^{2}+2x+1)}$

$\displaystyle (\frac{dy}{dx})=\frac{2x[x^{4}+2x^{3}+3x^{2}+2x+1]}{(x^{4}+2x^{3}+3x^{2}+2x+1)}-1$

$\displaystyle (\frac{dy}{dx})=2x-1= ax+b$

$\Rightarrow a=2$ & $b=-1$

If $x=1(\theta+sin\,\theta), y=a(1-cos \theta)$, then at $\theta=\dfrac{\pi}{2},y'=\dfrac{2}{a}$.

  1. True

  2. False


Correct Option: B

If $f\left( x \right) =|x-2|,g\left( x \right) =f\left( f\left( x \right) \right) $, then for $x>4$, $g'(x)=$

  1. 0

  2. 1

  3. -1

  4. 2


Correct Option: A

Differential coefficient of $\log\ \sin x$ is :

  1. $\cos x$

  2. $\tan x$

  3. $\text{cosec} \,x$

  4. $\cot x$


Correct Option: D
Explanation:

We have,

$y=\log \sin x$

On differentiating w.r.t $x$, we get
$\dfrac{dy}{dx}=\dfrac{d(\log \sin x)}{dx}$
$\dfrac{dy}{dx}=\dfrac{1}{\sin x}\times \cos x$
$\dfrac{dy}{dx}=\dfrac{\cos x}{\sin x}$
$\dfrac{dy}{dx}=\cot x$

Hence, this is the answer.

If $f\left( x \right) =\sqrt { { x }^{ 2 }-2x+1 } $, then

  1. $f^{ ' }\left( x \right) =1,\forall x$

  2. $f^{ ' }\left( x \right) =1, \forall x\ge 1$

  3. $f^{ ' }\left( x \right) =1, \forall x\le 1$

  4. $f^{ ' }\left( x \right) =1,if\quad x>1\quad and\quad f^{ ' }\left( x \right) =-1\quad if\quad x<1$


Correct Option: D

Derivative of $(\sin x)^x + \sin^{-1} \sqrt{x}$ with respect to $x$ is

  1. $(x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$

  2. $(x \cot x + \log \sin x) + \dfrac{1}{\sqrt{x - x^2}}$

  3. $(\sin x)^x (x \cot x + \log \,x) + \dfrac{1}{\sqrt{x - x^2}}$

  4. $(\sin x)^x (x \cot x + \log \sin x) + \dfrac{1}{2\sqrt{x - x^2}}$


Correct Option: D
Explanation:

Let $y=(\sin x)^x$

$\Rightarrow \log y=x \log (\sin x)$
Now differentiating both sides with respect to $x$ 
$\dfrac{1}{y}\dfrac{dy}{dx}=x\cot x+\log \sin x$
or, $\dfrac{dy}{dx}=(\sin x )^x{x\cot x+\log \sin x}$........(1).
Again let $z=(\sin x)^x+\sin^{-1}\sqrt{x}$
Now differentiating both sides with respect to $x$.
$\dfrac{dz}{dx}=\dfrac{dy}{dx}+\dfrac{1}{\sqrt{1-x}}.\dfrac{1}{2\sqrt{x}}$
$\dfrac{dz}{dx}=(\sin x)^x{x\cot x+\log \sin x}+\dfrac{1}{2\sqrt{x-x^2}}$ [Using (1)]

Let f(x) be a differentiable function satisfying $f(x+y)=f(x)+f(y)\forall x, y \in R$ and $f(0)=1$ then $\displaystyle\lim _{x\rightarrow 0}\dfrac{2^{f(\tan^2x)}-2^{f(\sin^2x)}}{x^3f(\sin x)}$ equals to?

  1. $\dfrac{1}{2} ln2$

  2. $ln 2$

  3. $\dfrac{1}{4}ln 2$

  4. $\dfrac{1}{8} ln2$


Correct Option: A

If $t={ \sin {  }  }^{ -1 }{ 2 }^{ s }$ Then $\dfrac { ds }{ dt }$ is equal to

  1. $\dfrac { \log { 2 } }{ \sqrt { 1-t^{ 2 } } }$

  2. $\dfrac { \sin { t } }{ \log { 2 } }$

  3. $\dfrac { \cot { t } }{ \log { 2 } }$

  4. None of these


Correct Option: D

If $u=e^{x}(xcosy-ysiny)$ then $\frac{d^{2}y}{dx^{2}}+\frac{d^{2}u}{dy^{2}}=0$.

  1. True

  2. False


Correct Option: A

If $U=tan^{-1}(\dfrac{x^3+y^3}{x+y})$ , then $x\dfrac{du}{dx}+y\dfrac{du}{dy}=sinu$.

  1. True

  2. False


Correct Option: B

If $y=\sqrt{x}-\dfrac{1}{\sqrt{x}}$, then $2x\dfrac{dy}{dx}+y$=

  1. $\sqrt{x}$

  2. $2\sqrt{x}$

  3. $3\sqrt{x}$

  4. $\dfrac{\sqrt{x}}{2}$


Correct Option: B
Explanation:

$y=\sqrt x-\dfrac{1}{\sqrt x}\ \dfrac{dy}{dx}=\dfrac{d}{dx}(x)^{\dfrac{1}{2}}-\dfrac{d}{dx}(\dfrac{1}{\sqrt x})=\dfrac{1}{2\sqrt x}-(-\dfrac{1}{2})x^{-\dfrac{3}{2}}=\dfrac{1}{2\sqrt x}+\dfrac{x^{-\dfrac{3}{2}}}{2}\ \dfrac{2x dy}{dx}+y=2x[\dfrac{1}{2\sqrt x}+\dfrac{x^{-\dfrac{3}{2}}}{2}]+\sqrt x-\dfrac{1}{\sqrt x}\ \quad =\sqrt x+\dfrac{1}{\sqrt x}+\sqrt x-\dfrac{1}{\sqrt x}\ \quad=2\sqrt x$


If $x\sqrt {1+y}+y\sqrt {1+x}=0$ then  $\dfrac {dy}{dx}=\dfrac {1}{(1+x)^{2}}$

  1. True

  2. False


Correct Option: A

Let $f(x)$ be a function continuous on $[1, 2]$ and differentiable on $(1, 2)$ satisfying $f(1)=2, f(2)=3$ and $f'(x)\ge 1\forall x\in (1, 2)$. Define $g(x)=\displaystyle \int _{1}^{x}{f(t)dt}\forall x\in [1, 2]$ then the greatest value of  $g(x)$ on $[1, 2]$ is-

  1. $3$

  2. $5$

  3. $\dfrac{5}{2}$

  4. $\dfrac{3}{2}$


Correct Option: C

Given $y = x \sqrt{x^2+1}, \dfrac{dy}{dx}$=

  1. $ \sqrt{x^2+1}$

  2. $\dfrac{2x^2+1}{ \sqrt{x^2+1}}$

  3. $\dfrac{3x^2+1}{ \sqrt{x^2+1}}$

  4. $\dfrac{3x^2+2}{ \sqrt{x^2+1}}$


Correct Option: B
Explanation:
Given,

$y=x\sqrt{x^2+1}$

$\Rightarrow \dfrac{d}{dx}y=\dfrac{d}{dx}x\sqrt{x^2+1}$

$=\dfrac{d}{dx}\left(x\right)\sqrt{x^2+1}+\dfrac{d}{dx}\left(\sqrt{x^2+1}\right)x$

$=1\cdot \sqrt{x^2+1}+\dfrac{x}{\sqrt{x^2+1}}x$

$=\dfrac{2x^2+1}{\sqrt{x^2+1}}$

If $\sin { { y+e }^{ -x\cos { y }  } } =e\quad then\quad \frac { dy }{ dx } \quad at\quad (1,\pi )$ is equal to 

  1. $\sin { y } $

  2. $-x\cos { y } $

  3. $e$

  4. $\sin { y } -x\cos { y } $


Correct Option: C
Explanation:
Given,

$\sin y+e^{-x\cos y}=e$

$\cos y \dfrac{dy}{dx}+e^{-x\cos y}\left [ x\sin y \dfrac{dy}{dx}-\cos y \right ]=0$

$\Rightarrow \cos y \dfrac{dy}{dx}+e^{-x\cos y} x\sin y \dfrac{dy}{dx} -\cos ye^{-x\cos y}=0$

$\dfrac{dy}{dx}[\cos y+x\sin y e^{-x\cos y}]=\cos y e^{-x\cos y}$

$\dfrac{dy}{dx}=\dfrac{\cos y e^{-x\cos y}}{\cos y+x\sin y e^{-x\cos y}}$

$\left [ \dfrac{dy}{dx} \right ] _{(1, \pi )}=\dfrac{\cos \pi  e^{-\cos \pi}}{\cos \pi+1 \sin \pi e^{-\cos \pi }}$

$\left [ \dfrac{dy}{dx} \right ] _{(1, \pi )}=\dfrac{-1 \times e}{-1+0 \times e}=e$

If $\displaystyle \cos^{-1}\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )=\log a$ then $\displaystyle \frac{dy}{dx}$ is equal to

  1. $\dfrac{y}{x}$

  2. $\dfrac{x}{y}$

  3. $\displaystyle\dfrac{ x^{2}}{y^{2}}$

  4. $\displaystyle\dfrac{ y^{2}}{x^{2}}$


Correct Option: A
Explanation:
$\displaystyle \cos^{-1}\left ( \frac{x^{2}-y^{2}}{x^{2}+y^{2}} \right )=\log a$ 
$\Rightarrow \displaystyle \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\cos \log a=A$ (say)
Putting $u=\dfrac{y}{x}$ and applying componendo and dividendo, we have
$\left ( \dfrac{y}{x} \right )^{2}=u^{2}=\left ( 1-A \right )\left ( 1+A \right )$
$\Rightarrow \dfrac{y}{x}=\sqrt{\left ( 1-A \right )\left ( 1+A \right )}\Rightarrow x\dfrac{dy}{dx}-y=0$
$\Rightarrow $   $\dfrac{dy}{dx}=\dfrac{y}{x}$

If $\displaystyle y=\sec(\tan^{-1}x)$, then $\displaystyle \frac {dy}{dx}$ at $x=1$ is equal to

  1. $\displaystyle \frac {1}{\sqrt {2}}$

  2. $\displaystyle \frac {1}{2}$

  3. $1$

  4. $\sqrt {2}$


Correct Option: A
Explanation:

Given, $\displaystyle y=\sec(\tan^{-1}x)=\sqrt {1+x^{2}}$
$\Rightarrow \displaystyle \frac {dy}{dx}=\frac {x}{\sqrt {1+x^{2}}}$
$\displaystyle\therefore  \left.\begin{matrix}\frac {dy}{dx}\end{matrix}\right| _{x=1}=\frac {1}{\sqrt {2}}$

If $y=sec(tan^{-1}x)$, then $\displaystyle\frac{dy}{dx}$ is.

  1. $\displaystyle\frac{x}{\sqrt{1+x^2}}$

  2. $\displaystyle\frac{-x}{\sqrt{1+x^2}}$

  3. $\displaystyle\frac{x}{\sqrt{1-x^2}}$

  4. None of these


Correct Option: A
Explanation:

Given, $y=sec(\tan^{-1}x)$
On differentiating w.r.t. $x,$ we get
$\displaystyle\frac{dy}{dx}=sec(\tan^{-1}x)\cdot \tan(\tan^{-1}x)\frac{1}{1+x^2}$
$=\displaystyle\frac{x}{1+x^2}\sqrt{1+x^2}$
$[\because \tan^{-1} x=sec^{-1}(\sqrt{1+x^2})]$
$=\displaystyle\frac{x}{\sqrt{1+x^2}}$

The differential equation $\dfrac {dy}{dx}=\dfrac {1}{ax+by+c}$ where a,b,c are all non zero real number ,is

  1. Linear in $y$

  2. Linear in $x$

  3. linear in both $x$ and $y$

  4. Homogeneous equation


Correct Option: B
Explanation:

$\dfrac {dx}{dy}= ax+by+c$
$\dfrac {dx}{dy}-a=by+c$
Linear in $x$

If ${ x }^{ 2 }.{ e }^{ y }+2x{ ye }^{ x }+13=0$, then $\dfrac { dy }{ dx }$ is

  1. $\dfrac { -2x{ e }^{ y-x }-2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$

  2. $\dfrac { 2x{ e }^{ x-y }+2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$

  3. $\dfrac { -2x{ e }^{ x-y }+2y\left( x+1 \right) }{ x\left( { xe }^{ y-x }+2 \right) }$

  4. $None\ of\ these$


Correct Option: A
Explanation:

Differentiating w.r.t. $x$

$2xe^y+x^2e^y\dfrac{dy}{dx}+2[\dfrac{d}{dx}(xy)e^x+xye^x]=0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2[\dfrac{d}{{dx}}(xy){e^x} + xy{e^x}] = 0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2{e^x}[y + x\dfrac{{dy}}{{dx}} + xy] = 0$

$2x{e^y} + {x^2}{e^y}\dfrac{{dy}}{{dx}} + 2y{e^x} + 2x{e^x}\dfrac{{dy}}{{dx}} + 2xy2{e^x} = 0$

$\left( {{x^2}{e^y} + 2x{e^x}} \right)\dfrac{{dy}}{{dx}} =  - \left( {2x{e^y} + 2y{e^x} + 2xy{e^x}} \right)$

$\dfrac{{dy}}{{dx}} =  - \dfrac{{\left( {2x{e^y} + 2y{e^x} + 2xy{e^x}} \right)}}{{{x^2}{e^y} + 2x{e^x}}}$

$\dfrac{{dy}}{{dx}} =  - \dfrac{{\left( {2x{e^{y - x}} + 2y + 2xy} \right)}}{{{x^2}{e^{y - x}} + 2x}}$

$\dfrac{{dy}}{{dx}} =  - \dfrac{{2x{e^{y - x}} - 2y\left( {x + 1} \right)}}{{x\left( {x{e^{y - x}} + 2} \right)}}$


Find: $\dfrac{d}{{\text dx}}\left( {\dfrac{{1 - \cos x}}{{\sin x}}} \right)  \,\,\,$

  1. ${\sec ^2}\dfrac{x}{2}$

  2. $\,\dfrac{1}{2}{\sec ^2}\dfrac{x}{2}\,$

  3. $\,\,2{\sec ^2}\dfrac{x}{2}$

  4. $\,\,3{\sec ^2}\dfrac{x}{2}$


Correct Option: B
Explanation:

Let $ y = \dfrac{1- \cos x }{\sin x}$


Formula: $\dfrac{d \left (\dfrac{u} {v}\right)}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}$

differentiate on both sides w.r.t $x$


$\dfrac{dy}{dx} = \dfrac{(\sin x)\sin x- (1- \cos x)\cos x}{(\sin x)^{2}}$

$= \dfrac{\sin ^{2}x -cos x + \cos ^{2}x}{(\sin x)^{2}}$

$= \dfrac{1-\cos x}{(\sin x)^{2}}$

$=  \dfrac{2 \sin^{2}\dfrac{x}{2}}{4\sin^{2}\dfrac{x}{2}\cos^{2}\dfrac{x}{2}}$

$= \dfrac{1}{2}\sec^{2}\left(\dfrac{x}{2}\right)$

Derivative of ${ \log { x }  }^{ \cos { x }  }$ with respect to $x$ is

  1. ${ \log { x } }^{ \cos { x } }\left[ \cfrac { \cos { x } }{ x\log { x } } -\sin { x } \log { \left( \log { x } \right) } \right] \quad $

  2. ${ \log { x } }^{ \cos { x } }\left[ \cfrac { \cos { x } }{ x\log { x } } -\cos { x } \log { \left( \log { x } \right) } \right] \quad $

  3. ${ \log { x } }^{ \sin { x } }\left[ \cfrac { \sin { x } }{ x\log { x } } -\sin { x } \log { \left( \log { x } \right) } \right] \quad $

  4. None of these


Correct Option: D
Explanation:
Let $y=\log x^{\cos x}$

$\Rightarrow \ y=(\cos x)(\log x)$

$\therefore \dfrac {dy}{dx}=(\cos x) \left (\dfrac {1}{x}\right)+(-\sin x)(\log x)$

$=\dfrac {\cos x}{x}-\sin x\log x$

If $xe^{xy}-y=\sin x$, then $\dfrac {dy}{dx}$ at $x=0$ is

  1. $0$

  2. $1$

  3. $-1$

  4. $None\ of\ these$


Correct Option: C
Explanation:
$x _e^ {xy}-y=\sin x$
Differentiating both sides with respect to $x$,
$x\dfrac {d}{dx} (e^{xy}) -\dfrac {d}{dy}+e^{xy}\dfrac {d}{dx}(x)=\dfrac {d}{dx} (\sin x)$
or, $xe^{xy}\dfrac {d}{dx}(xy)-\dfrac {dy}{dx}+e^{xy}=\cos x$
or, $xe^{xy}y+xe^{xy}\dfrac {dy}{dx}-\dfrac {dy}{dx} +e^{xy} +e^{xy}=\cos x$
or, $\dfrac {dy}{dx} (x e^{xy}-1)=\cos x e^{xy} -xye^{xy}$
$\therefore \ \left. \dfrac {dy}{dx}\right] _{x=0}=\dfrac {\cos x-e^{xy} -xye^{xy}}{xe^{xy}-1}=\dfrac {1-0}{0-1}=-1$

$\dfrac {d}{dx}(x^{\ell n x})$ is equal to

  1. $2x^{\ell n x-1}\ell n x$

  2. $x^{\ell n x-1}$

  3. $2/3(\ell n x)$

  4. $x^{\ell n x-1}.\ell n x$


Correct Option: A
Explanation:

Let $lnx=u$

$\therefore\ x={e}^{u}.$
$\frac { d }{ dx } \left( { x }^{ lmx } \right) =\frac { d }{ dx } \left( { e }^{ u.lnu } \right) $
$=\frac { d }{ dx } \left( { e }^{ { lnu }^{ 2 } } \right) =\frac { d }{ dx } \left( { u }^{ 2 } \right) $
$=2u.\frac { du }{ dx } $
$=2u.\frac { d }{ dx } \left( lnx \right) =\boxed{2lnx\quad { x }^{ lnx-1 }}$

If $y=(tan \, x)^{(tan\, x)^{tan\,x}}, $ then at $x=\dfrac{\pi}{4}, \dfrac{dy}{dx}$ is equal to 

  1. 0

  2. 3

  3. 2

  4. None of these


Correct Option: C
Explanation:
$y={ \left( tanx \right)  }^{ { \left( tanx \right)  }^{ \left( tanx \right)  } }$
We have to find $\dfrac { dy }{ dx } $ at $x=\dfrac { \Pi  }{ 4 } $.
Now consider:
$y={ f\left( x \right)  }^{ g\left( x \right)  }$
$ln\left( y \right) =g\left( x \right) lnf\left( x \right) $
$\Rightarrow \dfrac { 1 }{ y } \dfrac { dy }{ dx } =g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  } +{ g }^{ 1 }\left( x \right) lnf\left( x \right) $
$\Rightarrow \dfrac { dy }{ dx } =y\left[ g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  } +{ g }^{ 1 }\left( x \right) lnf\left( x \right)  \right] $
$\Rightarrow \dfrac { dy }{ dx } ={ f\left( x \right)  }^{ g\left( x \right)  }\left[ { g }^{ 1 }\left( x \right) lnf\left( x \right) +g\left( x \right) \dfrac { { f }^{ 1 }\left( x \right)  }{ f\left( x \right)  }  \right] $
So, We have
$y={ \left( tanx \right)  }^{ { \left( tanx \right)  }^{ tanx } }$
Where  $f\left( x \right) =tan\left( x \right) $
${ f }^{ 1 }\left( x \right) ={ \sec }^{ 2 }x$
$g\left( x \right) ={ \left( tanx \right)  }^{ tanx }$
${ g }^{ 1 }\left( x \right) ={ \left( tanx \right)  }^{ tanx\left[ { \sec }^{ 2 }xln\left( tan\left( x \right)  \right) +\dfrac { \tan\left( x \right) { \sec }^{ 2 }x }{ tanx }  \right]  }$
Now, $\tan\left( \dfrac { \Pi  }{ 4 }  \right) =1$ and $\sec\left( \Pi /4 \right) =\sqrt { 2 } $.
$f\left( \Pi /4 \right) =1$
${ f }^{ 1 }\left( \Pi /4 \right) =2$
$g\left( \Pi /4 \right) =1$
${ g }^{ 1 }\left( \Pi /4 \right) =1\left( 2ln\left( 1 \right) +2 \right) =2$
$\dfrac { dy }{ dx } $ at $\Pi /4$ is
$\dfrac { dy }{ dx } ={ f\left( \Pi /4 \right)  }^{ g\left( \Pi /4 \right) \left[ { g }^{ 1 }\left( \Pi /4 \right) ln\left( f\left( \Pi /4 \right)  \right) +g\left( \Pi /4 \right) \dfrac { { f }^{ 1 }\left( \Pi /4 \right)  }{ f\left( \Pi /4 \right)  }  \right]  }$
$\Rightarrow \dfrac { dy }{ dx } =1\left[ 2ln\left( 1 \right) +1\times 2 \right] =2$
$\Rightarrow \dfrac { dy }{ dx } =2$.

The solution of the differential equation, $y\,dx + \left( {x + {x^2}y} \right)dy = 0$ is

  1. $\log y = cx$

  2. $ \log y- \dfrac{1}{{xy}}  = c$

  3. $\dfrac{1}{{xy}} - \log y = c$

  4. $\dfrac{1}{{xy}} + \log y = c$


Correct Option: B
Explanation:

We have,

$ydx+\left( { x+{ x^{ 2 } }y } \right) dy=0 \ ydx=-\left( { x+{ x^{ 2 } }y } \right) dy \ ydx+xdy=-{ x^{ 2 } }ydy \ \dfrac { { ydx+xdy } }{ { { { \left( { xy } \right)  }^{ 2 } } } } =-\dfrac { { dy } }{ y }  \ \dfrac { { d\left( { xy } \right)  } }{ { { { \left( { xy } \right)  }^{ 2 } } } } =-\dfrac { { dy } }{ y } $

On taking integrating both sides

$\begin{array}{l} \dfrac { { -1 } }{ { xy } } =-\log  y+c \\ \log  y-\dfrac { 1 }{ { xy } } =c \end{array}$


Hence, this is the answer.

If $x=a\sin \theta$ and $y=b\cos\theta$, then $\displaystyle\frac{d^2y}{dx^2}$ is 

  1. $\displaystyle\frac{a}{b^2}\sec^2\theta$

  2. $\displaystyle\frac{-b}{a}\sec^2\theta$

  3. $\displaystyle\frac{b}{a^2}\sec^3\theta$

  4. $\displaystyle\frac{-b}{a^2}\sec^3\theta$


Correct Option: D
Explanation:

Given, $x=a\sin\theta$ and $y=b\cos\theta$
On differentiating w.r.t.$\theta$, we get
$\displaystyle\frac{dx}{d\theta}=a\cos\theta$
and $\displaystyle\frac{dy}{d\theta}=-b\sin \theta$
$\Rightarrow \displaystyle\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=-\frac{b}{a}\tan\theta$
Again, differentiating w.r.t. $x$, we get
$\displaystyle\frac{d^2y}{dx^2}=-\frac{b}{a}sec^2\theta\cdot\frac{d\theta}{dx}$
$\Rightarrow \displaystyle\frac{d^2y}{dx^2}=-\frac{b}{a}sec^2\theta\cdot\frac{1}{a\cos\theta}$
$=-\displaystyle\frac{b}{a^2}sec^3\theta$

The set of all points, where the function $f(x) = \sqrt {1 - e^{-x^{2}}}$ is differentiable, is

  1. $(0, \infty)$

  2. $(-\infty, \infty)$

  3. $(-\infty, 0) \cup (0, \infty)$

  4. $(-1, \infty)$


Correct Option: C
Explanation:

Given that

$ f(x) = \sqrt{1 - e^{-x^2}}$

By chain rule of differentiation, 

$ f'(x) = \dfrac{1}{2\sqrt{1 - e^{-x^2}} } \; \dfrac{d}{dx}(1 - e^{-x^2})$
           $  = \dfrac{1}{2\sqrt{1 - e^{-x^2}} } \; (-e^{-x^2}) (-2x)$
           $  = \dfrac{x \; e^{-x^2}}{2\sqrt{1 - e^{-x^2}} } $

f(x) is differentiable at a point x if the f'(x) exists at the point.

The denominator of f'(x) vanishes for 
$ 1 - e^{-x^2} = 0 $
i.e for
$ x = 0$

f(x) is not differentiable at x = 0.

Hence, the set of all points at which f(x) is differentiable:
$  (-\infty, 0) \; \cup \; (0, \infty) $

The answer: option C.

If $f(x) =x^{3} + e^{x/2}$ then $g"(1)$ is equal to, (where $f(x)$ and $g(x)$ are inverse functions of each other).

  1. $2$

  2. $-2$

  3. $-\dfrac {1}{2}$

  4. $\dfrac {1}{2}$


Correct Option: B
Explanation:
$g(f(x))=x$
$g'f(x)=1/f'(X)$
$g''f(x)=\dfrac{f''(x)}{{f'(x)}^3}$
$f(x)=1 at x=0$
$f(0)=1$
$f'(x)=3x^2+e^{(x/2)}/2$
$f'(0)=1/2$
$f''(x)=6x+e^{x/2}/4$
$f''(0)=1/4$
$g''(1)=\dfrac{-1/4}{1/8}=-2$
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