If $$\displaystyle y^{x}=x^{\sin y} $$, find $$\cfrac{dy}{dx}$$.

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]$$

$$\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$

$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]$$

If $$y = \displaystyle (\tan x)^{\log x}$$, then $$\cfrac{dy}{dx} = $$

$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx$$

$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } $$

If $$\displaystyle y=x^{n} log x+x(log x)^{n}$$, find $$dy/dx.$$

$$x^{n}(1+n log x)+(log x)^{n}[n+log x]$$

$$x^{n-1}(1+(n-1) log x)+(log x)^{n-1}[n-1+log x]$$

MCQ Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 5

If P(x) is a polynomial such that

$P\left ( x^{2}+1 \right )=\left \{ P\left ( x^{2} \right ) \right \}^{2}+1$ and

$P(0)=0$ then

$P^{'}(0)$ is equal to

0%
$1$
0%
$0$
0%
$-$ 1
0%
none of these

Explanation

Given,

$P\left ( x^{2}+1 \right )=\left \{ P\left ( x^{2} \right ) \right \}^{2}+1$ ...(1)
Given

$P(0)=0$
Put

$x=0$ in (1)

$\Rightarrow P(1)={P(0)}^{2}+1$

$\Rightarrow P(1)=1$
Also, put

$x=1$ in (1)

$\Rightarrow P(2)=2$
Also, since, P(x) is a polynomial so we have P(x)=x

$\Rightarrow P'(x)=1$

$P'(0)=1$

$\displaystyle y^{x}=x^{\sin y}$ , find

$\cfrac{dy}{dx}$ .

0%
$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$
0%
$\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]$
0%
$\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$
0%
$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]$

Explanation

Given

$\displaystyle y^{x}\:=x^{\sin \:y}$ .

Take log both sides.

$\displaystyle x \log y = \sin y \: \log \:x.$

Differentiate w.r.t. x

$\displaystyle \log y+x.\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \sin y+ (\log x)\cos y .\frac{dy}{dx}$

$\displaystyle \left ( \log y - \frac{\sin y}{x} \right )=\frac{dy}{dx}\left [ \cos y \log x-\frac{x}{y} \right ]$

$\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}\left [ \frac{x \log y-\sin y}{y \log x \cos y-x}\right ]$

$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$ .Find dy/dx

0%
$\sin x(\cot x)^{ \sin x-1 }(-cosec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(-sinx) }$
0%
$\sin x(\cot x)^{ \sin x-1 }(-co\sec ^{ 2 } x)+\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x }$
0%
$\sin x(\cot x)^{ \sin x-1 }(-sec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x+1 } }co\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(sinx) }$
0%
None of these

Explanation

Given

$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$

Let

$u=(\cot x)^{\sin x}$

Taking log on both sides

$\log u=\sin x \log(\cot x)$

Differentiating w.r.t. x, we get,

$\dfrac{1}{u}\dfrac{du}{dx}=\dfrac{\sin x}{\cot x}(-cosec^2 x)+\cos x \log \cot x$

$\Rightarrow \dfrac{du}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x$

Now,

$v=(\tan x)^{\cos x}$

Taking log on both sides

$\log v=\cos x \log(\tan x)$

Differentiating w.r.t. x, we get,

$\dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{\cos x}{\tan x}(\sec^2 x)-\sin x \log \tan x$

$\Rightarrow \dfrac{dv}{dx}=\cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$

Since,

$y=u+v$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$

$\Rightarrow \dfrac{dy}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x+$

$\cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$

$\displaystyle x^{2}+y^{2}= 1$ then

$\displaystyle \left ( where\ y'= \frac{dy}{dx}, y''= \frac{d^{2}y}{dx^{2}}\right )$

0%
$\displaystyle yy''-2y'^{2}+1= 0$
0%
$\displaystyle yy''+y'^{2}+1= 0$
0%
$\displaystyle yy''-y'^{2}-1= 0$
0%
$\displaystyle yy''+2y'^{2}+1= 0$

Explanation

Given,

$\displaystyle x^{2}+y^{2}= 1$
Differentiating w.r.t x,

$2x+2yy'=0$

$\Rightarrow x+yy'=0$
Again differentiating w.r.t. x, we get

$1+yy''+y'y'=0$

$1+yy''+y'^{2}=0$

$\displaystyle \frac{d}{dx}(\log_{e}\left ( \frac{1+x}{1-x} \right )^{1/4}-\frac{1}{2}\tan^{-1}x.)$

0%
$\displaystyle \frac{x^{2}}{1-x^{4}}.$
0%
$\displaystyle \frac{x^{3}}{1-x^{4}}.$
0%
$\displaystyle \frac{x^{4}}{1-x^{4}}.$
0%
$-\displaystyle \frac{x^{2}}{1-x^{4}}.$

Explanation

Let

$\displaystyle y=\frac{1}{4}\left [ \log\left ( 1+x \right )-\log\left ( 1-x \right ) \right ]-\frac{1}{2}\tan^{-1}x.$

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

$\displaystyle =\frac{1}{2\left ( 1-x^{2} \right )}-\frac{1}{2\left ( 1+x^{2} \right )}=\frac{x^{2}}{1-x^{4}}.$

$\displaystyle x={e^{\displaystyle y+e^{\displaystyle y+e^{\displaystyle y+...\infty }}}}$ ,

$\forall x> 0$ , then

$\displaystyle \frac{dy}{dx}=$

0%
$\displaystyle \frac{1-x}{x}$
0%
$\displaystyle \frac{1}{x}$
0%
$\displaystyle \frac{x}{1+x}$
0%
$\displaystyle \frac{1+x}{x}$

Explanation

$\displaystyle x=e^{y\displaystyle+e^{y\displaystyle+e^{y\displaystyle+...}}}$

$\Rightarrow \displaystyle x=e^{y+x}$
Take logarithm on both sides and differentiating with respect to

$x$

$\therefore \displaystyle \frac{1}{x}=1+\frac{dy}{dx}$

$\Rightarrow$

$\displaystyle \frac{dy}{dx}=\frac{1-x}{x}$

$f$ is a real-valued differentiable function satisfying

$\displaystyle \left | f(x)-f(y) \right |\leq (x-y)^{2}$ for all

$x,y\:\epsilon\:R$ and

$f(0)=0$ then

$f(1)$ equals

0%
$0$
0%
$-1$
0%
$1$
0%
$2$

Explanation

Given

$\displaystyle \left| f\left( x \right) -f\left( y \right) \right| \le { \left| x-y \right| }^{ 2 }\quad x\neq y$

$\displaystyle \therefore \left| \frac { f\left( x \right) -f\left( y \right) }{ x-y } \right| \le \left| x-y \right|$

Taking limit as

$y\rightarrow x$ , we get

$\displaystyle \lim _{ y\rightarrow x }{ \left| \frac { f\left( x \right) -f\left( y \right) }{ x-y } \right| } \le \lim _{ y\rightarrow x }{ \left| x-y \right| }$

$\displaystyle\Rightarrow\left|\lim _{ y\rightarrow x }{ \frac { f\left( x \right) -f\left( y \right) }{ x-y } } \right| \le \left| \lim _{ y\rightarrow x }{ \left( x-y \right) } \right|$

$\therefore f'\left( x \right) =0\Rightarrow f\left( x \right) =c$

$\therefore h\left( x \right) =\int { f\left( x \right) } dx=\int { cdx } =cx+d$

where

$d$ is constant of integraion
Therefore

$h\left( x \right)$ is a linear function of

$x$
which is constant for all

$x.$

$f(0)=0$ ........given

$\therefore f(1)=0$

$\displaystyle \frac{d}{dx}(\tan ^{-1}\frac{\sin x+\cos x}{\cos x-\sin x})$

0%
$-1$
0%
$-2$
0%
$1$
0%
$2$

Explanation

Let

$\displaystyle y = \tan ^{-1}\frac{\cos x+\sin x}{\cos x-\sin x}=\tan ^{-1}\frac{1+\tan x}{1-\tan x}$

$\displaystyle \Rightarrow y =\tan ^{-1}\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}=\tan^{-1}\tan(\frac{\pi}{4}+x)=\frac{\pi}{4}+x$

$\therefore \cfrac{dy}{dx}=1$

$\displaystyle \frac{d}{dx}\tan ^{-1}\left(\frac{a \cos x-b\sin x}{b\cos x+a\sin x}\right)$

0%
$-1$
0%
$-2$
0%
$1$
0%
$2$

Explanation

Put

$\displaystyle a= r\cos \alpha , b= r\sin \alpha$

$\displaystyle \therefore r= \sqrt{a^{2}+b^{2}},\tan \alpha = \dfrac{b}{a}$

$\displaystyle \therefore y= \tan ^{-1}\frac{r\cos\left ( x+\alpha \right )}{r\sin\left ( x+\alpha \right )}=\tan^{-1}\cot \left ( x+\alpha \right )$

$\displaystyle =\tan^{-1}\tan\left [ \dfrac{\pi}{2} -\left ( x+\alpha \right )\right ]$

$\displaystyle \therefore y=\dfrac{\pi} {2}-x-\alpha$

$\therefore \dfrac{dy}{dx}=-1$

$\displaystyle f\left( x \right) =\sqrt { 1+\sqrt { x } } , x > 0,$ then

$\displaystyle f\left ( x \right )\cdot f'\left ( x \right )$ is equal to

0%
$\displaystyle \frac{1}{2\sqrt{x}}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{4\sqrt{x}}$
0%
$\displaystyle \frac{2\sqrt{x}+1}{4\sqrt{x}}$

Explanation

Let

$y=f\left( x \right) =\sqrt { 1+\sqrt { x } }$

Then,

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

Differentiating w.r.t. x, we get,

$\displaystyle 2y\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$

$\Rightarrow f(x)f'(x)=\displaystyle \frac{1}{4\sqrt{x}}$

$\displaystyle \dfrac{d}{dx}\tan ^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)$

0%
$-\displaystyle \dfrac{1}{2}$
0%
$-\displaystyle \dfrac{1}{4}$
0%
$-\displaystyle \dfrac{1}{8}$
0%
$\displaystyle \dfrac{1}{2}$

Explanation

Let

$\displaystyle y = \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$

$\Rightarrow \displaystyle y = \tan ^{-1}\left(\dfrac{(\cos^2 \dfrac{x}{2}-\sin^2\dfrac{x}{2})}{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin \dfrac{x}{2}.\cos\dfrac{x}{2}}\right)$

$\displaystyle y=\tan ^{-1}\left(\dfrac{\cos\left ( x/2 \right )-\sin \left ( x/2 \right )}{\cos \left ( x/2 \right )+\sin \left ( x/2 \right )}\right)$

$\displaystyle =\tan^{-1}\left(\frac{1-\tan\left ( x/2 \right )}{1+\tan \left ( x/2 \right )}\right)$

$\displaystyle =\tan^{-1}\left(\tan \left ( \pi /4-x/2 \right )\right)=\pi/4-x/2$

$\displaystyle \therefore \dfrac{dy}{dx}=-\dfrac{1}{2}$

Differentiate

$\displaystyle x^{\sin^{-1}x}$ w.r.t.

$\displaystyle \sin ^{-1}x.$

0%
$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$
0%
$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$
0%
$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$
0%
$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$

Explanation

Let

$\displaystyle y = x^{\sin^{-1}x}$ and

$z = \displaystyle \sin ^{-1}x\Rightarrow \sin z = x$
we have to find

$\cfrac{dy}{dz}$

$\Rightarrow y = (\sin z)^z$ taking

$\log$ both side

$\log y = z\log \sin z$
Differentiating w.r.t

$z$

$\displaystyle \frac{1}{y}\frac{dy}{dz}=\log\sin z+z\frac{\cos z}{\sin z}$

$\therefore \displaystyle \frac{dy}{dz}=y\left(\log\sin z+z\frac{\cos z}{\sin z}\right)=\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$

$y = \displaystyle (\tan x)^{\log x}$ , then

$\cfrac{dy}{dx} =$

0%
$(\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$
0%
$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx$
0%
$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x }$
0%
none of these

Explanation

Let

$y = \displaystyle (\tan x)^{\log x}$

$\log y = \log x .\log \tan x$
Differentiating both side w.r.t

$x$

$\cfrac{1}{y}.\cfrac{dy}{dx} = \cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x)$

$\Rightarrow \cfrac{dy}{dx} = (\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$

Differentiate

$\displaystyle \tan x^{n}+\tan ^{n}x-\tan ^{-1}\frac{a+x^{n}}{1-ax^{n}}.$

0%
$\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1-x^{2n} \right ) }\right ]nx^{n-1}$
0%
$\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n}$
0%
$\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$
0%
$\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$

Explanation

Let

$y = \displaystyle \tan x^{n}+\tan ^{n}x-\tan ^{-1}\frac{a+x^{n}}{1-ax^{n}}.$

$\Rightarrow \displaystyle y= \tan x^{n}+\tan^{n}x-\left ( \tan^{-1}a+\tan^{-1}x^{n} \right )$

$\displaystyle\therefore \frac{dy}{dx}= \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-0-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right ) } \right ]nx^{n-1}$

$= \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right ) } \right ]nx^{n-1}$

$\displaystyle y=x^{n} log x+x(log x)^{n}$ , find

$dy/dx.$

0%
$x^{n-1}(1+n log x)+(log x)^{n-1}[n+log x]$
0%
$x^{n}(1+n log x)+(log x)^{n}[n+log x]$
0%
$x^{n-1}(1+(n-1) log x)+(log x)^{n-1}[n-1+log x]$
0%
none of these

Explanation

$\displaystyle y=x^{n} \log x+x (\log x)^{n}$

$\Rightarrow \cfrac{dy}{dx}=nx^{n-1} \log x+x^{n} .(1/x)+x.n(\log x)^{n-1}.(1.x)+1(\log x)^{n}$

$\Rightarrow \cfrac{dy}{dx} =x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]$

$\displaystyle x\sqrt{(1+y)}+y\sqrt{(1+x)}=0$ , then

$\displaystyle \frac{dy}{dx}$ =

0%
$\displaystyle \frac{1}{(1+x)^{2}}$
0%
$\displaystyle -\frac{1}{(1+x)^{2}}$
0%
$\displaystyle -\frac{1}{(1-x)^{2}}$
0%
$\displaystyle \frac{1}{(1-x)^{2}}$

Explanation

Given,

$\displaystyle x\sqrt{(1+y)}+y\sqrt{(1+x)}=0$

$\Rightarrow x\sqrt{(1+y)}=-y\sqrt{(1+x)}$
Squaring both side we get,

$x^{2}(1+y)=y^{2}(1+x)$

$\displaystyle x+y+xy=0$ or

$y=x$ (rejected)

$\displaystyle y=-\frac{x}{1+x}=\frac{1}{1+x}-1$

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

$x^{m}.y^{n}=\left ( x+y \right )^{m+n}$ , then

$\dfrac{dy}{dx}=$

0%
$\displaystyle \frac{y}{x}$
0%
$\displaystyle \frac{-y}{x}$
0%
$\dfrac {my}{x}$
0%
$\dfrac {ny}{x}$

Explanation

$x^{m}\times y^{n}=\left ( x+y \right )^{m+n}$

Taking log both sides we get

$m\log x+n\log y=\left ( m+n \right )\log \left ( x+y \right )$

Differentiating w.r.t. x we get

$\displaystyle \dfrac{m}{x}+\dfrac{n}{y}\dfrac{dy}{dx}=\dfrac{m+n}{x+y}\left ( 1+\dfrac{dy}{dx} \right )$

$\Rightarrow$

$\displaystyle \dfrac{dy}{dx}\left ( \dfrac{n}{y}-\dfrac{m+n}{x+y} \right )=\dfrac{m+n}{x+y}-\dfrac{m}{x}$

$\Rightarrow$

$\displaystyle \dfrac{dy}{dx}\left ( \dfrac{nx+ny-my-ny}{y\left ( x+y \right )} \right )=\dfrac{mx+nx-mx-my}{x\left ( x+y \right )}$

$\Rightarrow$

$\displaystyle \dfrac{dy}{dx}=\left ( \dfrac{nx-my}{nx-my} \right )\dfrac{y}{x}=\dfrac{y}{x}$

$\Rightarrow$

$\displaystyle \dfrac{dy}{dx}=\dfrac{y}{x}$

Differentiate the following:

$\displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}$

0%
$\displaystyle \frac{1}{2}.$
0%
$\displaystyle \frac{-1}{2}.$
0%
$\displaystyle \frac{1}{4}.$
0%
$\displaystyle \frac{-1}{4}.$

Explanation

Let

$y = \displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}$

$\displaystyle y= \cot^{-1}\frac{\left ( \cos\frac{x}{2}+\sin\frac{x}{2} \right )+\left ( \cos\frac{x}{2}-\sin\frac{x}{2} \right )}{\left ( \cos\frac{x}{2}+\sin \frac{x}{2} \right )-\left ( \cos \frac{x}{2}-\sin \frac{x}{2} \right )}=\cot ^{-1}\frac{2\cos \left ( x/2 \right )}{2\sin \left ( x/2 \right )}$

$\displaystyle y=\cot ^{-1}\cot \frac{x}{2}= \frac{x}{2},$

$\therefore \dfrac{dy}{dx}= \dfrac{1}{2}.$
Note: Here assumption taken that

$\tan\frac{x}{2}\geq 1$

Find the differential equation of the family of curves whose equations are

$\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}+\lambda }=1,$ where

$\displaystyle \lambda$ is parameter.

0%
$\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{2}-x^2}{a^{2}}$
0%
$\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{2}+x^2}{a^{2}}$
0%
$\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{4}-x^2}{a^{2}}$
0%
$\displaystyle -\frac{xy}{a^{2}{y}'}=\frac{a^{4}+x^2}{a^{2}}$

Explanation

Differentiating given equation w.r.t

$x$ we get

$\displaystyle \frac{2x}{a^{2}}+\frac{2y}{a^{2}+\lambda }\frac{dy}{dx}=0$

$\displaystyle \therefore \left ( a^{2}+\lambda \right )=-\frac{a^{2}y{y}'}{x}$
Putting for

$\displaystyle a^{2}+\lambda$ in the given equation , we get

$\displaystyle\Rightarrow \frac{x^{2}}{a^{2}}-\frac{y^{2}x}{a^{2}y{y}'}=1$

$\Rightarrow \displaystyle -\frac{xy}{a^{2}{y}'}=1-\frac{x^{2}}{a^{2}}=\frac{a^{2}-x^2}{a^{2}}$

Let

$\displaystyle f\left ( x \right )$ be defined by

$\displaystyle f\left ( x \right )=\left\{\begin{matrix}\sin 2x & \text{if } 0< x\leq \dfrac{\pi}6\\ ax+b& \text{if } \dfrac{\pi}6< x\leq 1\end{matrix}\right.$ . The values of

$a$ and

$b$ such that

$\displaystyle f$ and

$\displaystyle {f}'$ are continuous, are

0%
$\displaystyle a=1,b=\dfrac1{\sqrt{2}}+\dfrac{\pi}6$
0%
$\displaystyle a=\dfrac1{\sqrt{2}},b=\dfrac1{\sqrt{2}}$
0%
$\displaystyle a=1,b=\dfrac{\sqrt{3}}2-\dfrac{\pi}6$
0%
None of these

Explanation

For

$f$ to be continuous

$\displaystyle \frac { \sqrt { 3 } }{ 2 } =f\left( \frac { \pi }{ 6 } \right) =\displaystyle \lim_{ x\rightarrow \tfrac { \pi }{ 6 }^+ } f\left( x \right) =\displaystyle\lim_{ x\rightarrow \tfrac { \pi }{ 6 } + } \left( ax+b \right) =\frac { a\pi }{ 6 } +b$

$\displaystyle f'\left( x \right) =\begin{cases} 2\cos { 2x } \quad \quad \text{if }0<x<\dfrac { \pi }{ 6 } \\ a\quad \quad \quad \quad \quad \text{if }\dfrac { \pi }{ 6 } <x<1 \end{cases}$

$\displaystyle f'\left( \frac { \pi }{ 6 } + \right) =a$ and

$\displaystyle f'\left( \frac { \pi }{ 6 } - \right) =1$
Thus

$\displaystyle a=1,b=\frac { \sqrt { 3 } }{ 2 } -\frac { \pi }{ 6 }$

Find the solution of

$\displaystyle \frac{dy}{dx}= \frac{2x+2y-2}{3x+y-5}.$

0%
$\displaystyle \left ( 2x+y-3 \right )= k\left ( x-y-3 \right )^{4}$
0%
$\displaystyle \left ( 2x-y-3 \right )= k\left ( x-y-3 \right )^{4}$
0%
$\displaystyle \left ( 2x+y+3 \right )= k\left ( x-y-3 \right )^{4}$
0%
$\displaystyle \left ( 2x+y-3 \right )= k\left ( 2x-y-3 \right )^{4}$

$f\left( x \right)$ is a polynomial of degree

$n(>2)$ and

$f\left( x \right) =f\left( k-x \right) ,($ where

$k$ is a fixed real number

$),$ then degree of

$f'(x)$ is

0%
$n$
0%
$n-1$
0%
$n-2$
0%
None of these

Explanation

Clearly,

$f(x)$ must be of the from

$f\left( x \right) ={ a }_{ 0 }\left[ { x }^{ n }+{ \left( k-x \right) }^{ n } \right] +{ a }_{ 1 }\left[ { x }^{ n-1 }+{ \left( k-x \right) }^{ n-1 } \right] +...+{ a }_{ n-1 }\left[ x+\left( k-x \right) \right] +{ a }_{ n. }$

It may be noted that

$n$ must be even for otherwise

$f(x)$ will become a polynomial of degree

$n-1.$

Clearly,

$f'(x)$ is a polynomial of degree

$n-1.$

$2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x$ , then

$\displaystyle \frac { d }{ dx } f\left( x \right)$ is

0%
$\sin{x}+\cos{x}$
0%
$2$
0%
$\displaystyle \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$
0%
none of these

Explanation

$\displaystyle 2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x$ ...(1)

Replace

$x$ by

$\displaystyle \frac { \pi }{ 2 } -x$

$\displaystyle 2f\left( \cos { x } \right) +f\left( \sin { x } \right) =\frac { \pi }{ 2 } -x$ ...(2)

Solving we get ,

$\displaystyle 3f\left( \sin { x } \right) =\frac { \pi }{ 2 } +3x$

$\displaystyle \therefore f\left( x \right) =\frac { \pi }{ 6 } +\sin ^{ -1 }{ x } \quad \\\displaystyle\therefore \frac { d }{ dx } f\left( x \right) =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$

Obtain the differential equation whose solutions are

$\displaystyle y=A\cos \left ( x+3 \right ),$ A being constant.

0%
$\displaystyle \frac{dy}{dx}+y\tan \left ( x+3 \right )=0$
0%
$\displaystyle \frac{dy}{dx}+y\tan \left ( x-3 \right )=0$
0%
$\displaystyle -\frac{dy}{dx}+y\tan \left ( x-3 \right )=0$
0%
$\displaystyle -\frac{dy}{dx}+y\tan \left ( x+3 \right )=0$

Explanation

$y = A\cos(x+3)$ (i)

$\displaystyle \frac{dy}{dx}=-A\sin \left ( x+3 \right )$ ..(ii)
by (ii)

$\div$ (i)

$\displaystyle \frac{\frac{dy}{dx}}{y}=-\tan \left ( x+3 \right )$

$\displaystyle \frac{dy}{dx}+y\tan \left ( x+3 \right )=0$

$\displaystyle {f}'\left ( x \right )=g\left ( x \right )$ and

$\displaystyle {g}'\left ( x \right )=-f\left ( x \right )$ and

$\displaystyle f\left ( 2 \right )=4={f}'\left ( 2 \right )$ then

$\displaystyle f^{2}\left ( 16 \right )+g^{2}\left ( 16 \right )$ is

0%
16
0%
32
0%
64
0%
None of these

Explanation

$\displaystyle \frac { d }{ dx } \left( { f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right) \right) \\ =2\left[ f\left( x \right) f'\left( x \right) +g\left( x \right) g'\left( x \right) \right] \\ =2\left[ f\left( x \right) -g\left( x \right) f\left( x \right) \right] =0$
Thus

${ f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right)$ is constant.
Therefore

${ f }^{ 2 }\left( 16 \right) +{ g }^{ 2 }\left( 16 \right) ={ f }^{ 2 }\left( 2 \right) +{ g }^{ 2 }\left( 2 \right) \\ ={ f }^{ 2 }\left( 2 \right) +{ \left( f'\left( 2 \right) \right) }^{ 2 }=16+16=32$

Let

$f\left( x \right)=\sqrt { x-1 } +\sqrt { x+24-10\sqrt { x-1 } } ;1<x<26$ be a real valued function. Then

$f'(x)$ for

$1<x<26$ is

0%
$0$
0%
$\displaystyle \frac { 1 }{ \sqrt { x-1 } }$
0%
$2\sqrt { x-1 } -5$
0%
none of these

Explanation

Let,

$f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}$

$1< x< 26$ be a real valued function ,
We have,

$f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}$
This can be written as

$f(x) = \sqrt{x -1} + \sqrt{\left ( 5 - \sqrt{x-1} \right )^{2} }$

$\because 1< x< 26$

$f(x) = \sqrt{x-1} + 5 - \sqrt{x-1}$

$f(x) = 5$
On differentiating wrt x , we get

$f{}'(x) = 0$
Hence ,Option A

A curve passing through the point

$(1,1)$ is such that the intercept made by a tangent to it on x-axis is three times the x co-ordinate of the point of tangency, then the equation of the curve is:

0%
$\displaystyle y=\frac{1}{x^{2}}$
0%
$\displaystyle y=\sqrt{x}$
0%
$\displaystyle y=\frac{1}{\sqrt{x}}$
0%
none

Explanation

Equation of tangent on any point on the curve is,

$\displaystyle Y-y=\frac{dy}{dx}\left ( X-x \right )$
Thus intercept on the x-axis is,

$\displaystyle I_{x}=x-\frac{y}{dy/dx}$
Now using given condition,

$\displaystyle I_{x}=x-\frac{y}{dy/dx}=3x$

$\displaystyle \therefore y\frac{dx}{dy}+2x=0$
or

$\displaystyle \frac{dx}{2x}+\frac{dy}{y}=0$ or

$\displaystyle \frac{1}{2}\log x+\log y=k$ or

$\displaystyle \log y\sqrt{x}=e^{k}=constant$
Now given this curve passing through (1,1)

$\Rightarrow k=0$

$\therefore$ Hence required curve is,

$y=\cfrac{1}{\sqrt{x}}$

Let

$\displaystyle f$ be a function satisfying

$\displaystyle f\left ( x+y \right )=f\left ( x \right )f\left ( y \right )$ for all

$x$ and

$y$ and

$\displaystyle f\left ( 0 \right )={f}'\left ( 0 \right )=1$ then

0%
$\displaystyle f$ is differentiable for all x
0%
$\displaystyle {f}'\left ( x \right )=f\left ( x \right )$
0%
$\displaystyle f\left ( x \right )=e^{x}$
0%
$\displaystyle f$ is continuous for alI x

Explanation

We have,

$f(x+y)=f(x)f(y) ...........(1)$

Now

$f'(x) =\displaystyle\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\dfrac{f(x)f(h)-f(x)}{h}$ using (1)

$\Rightarrow f'(x) =\displaystyle\lim_{h\to 0}\dfrac{f(x)[f(h)-1)}{h}=f(x)\lim_{h\to 0}\dfrac{f(0+h)-f(0)}{h}=f(x)f'(0)=f(x)$

Integrating we get,

$\log(f(x)) = x+c$

Since at x

$=0, f(x) =1\Rightarrow$ we have,

$c=0$

Hence

$f(x) =e^x$ , which is continuous and differentiable for all

$x$

$\displaystyle f\left ( 1 \right )=3$ and

$\displaystyle {f}'\left ( 1 \right )=-\dfrac13$ then the derivative of

$\displaystyle \left ( x^{11} +f\left ( x \right )\right )^{-2}$ at

$\displaystyle x=1$ is

0%
$-\dfrac12$
0%
$-1$
0%
$1$
0%
$\displaystyle {f}'\left ( 1 \right )$

Explanation

The given equation is:

$y = (x^{11} + f(x))^{-2}$

Differentiating the equation w.r.t to x once we get,

$\Rightarrow y' = -2(x^{11} + f(x))^{-3}.(11x^{10} + f'(x))$

Value of the derivative at

$x = 1$ is given as,

$\Rightarrow y' = \dfrac{-2}{(1+3)^3}.(11 - \dfrac{1}{3})$

$\Rightarrow y' = \dfrac{-2}{64}.\dfrac{32}{3}$

$\Rightarrow y' = \dfrac{-1}{3}$ .....Answer

$\Rightarrow y' = f'(1)$

A polynomial

$f(x)$ leaves remainder

$15$ when divided by

$(x-3)$ and

$(2x+1)$ when divided by

$(x-1)^2$ . When

$f$ is divided by

$(x-3)(x-1)^2,$ the remainder is

0%
$2x^2+2x+3$
0%
$2x^2-2x-3$
0%
$2x^2-2x+3$
0%
none of these

Explanation

Since function

$f(x)$ leaves remainder

$15$ when divided by

$x-3$ , therefore

$f(x)$ can be written as

$f(x)=(x-3)l(x)+15$ ...(1)

Also,

$f(x)$ leaves remainder

$2x+1$ when divided by

$(x-1)^2.$

Thus,

$f(x)$ can also be written as

$f\left( x \right)={ \left( x-1 \right) }^{ 2 }m\left( x \right) +2x+1$ ...(2)

$R(x)$ be the remainder when

$f(x)$ is divided by

$(x-3)(x-1)^2,$ then we may write

$f\left( x \right)=\left( x-3 \right) { \left( x-1 \right) }^{ 2 }n\left( x \right) +R\left( x \right)$ ...(3)

Since

$(x-3)(x-1)^2$ is a polynomial of degree three, the remainder has to be a polynomial of degree less than or equal to two.

Thus let

$R(x)=ax^2+bx+c$

From (1) and (3), we have

$f(3)=15=R(3)\Rightarrow 9a+3b+c=15$ ...(4)

From (2) and (3), we have

$f(1)=3=R(1)\Rightarrow a+b+c=3$ ...(5)

From (2) and (3), we have

$f'(1)=2=R'(1)\Rightarrow 2a+b=2$ ...(6)

Solving equation (4),(5) and (6), we get

$a=2,b=-2,c=3$

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Differentiation Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.