If $$f(x)=\left | x \right |+\left | \cos x \right |$$, then

$$ f^{'}( \dfrac{\pi }{2} )$$ does not exist

$$ f^{'}( \dfrac{\pi }{2} )=2$$

$$ f^{'}( \dfrac{\pi }{2} )=0$$

$$ f^{'}( \dfrac{\pi }{2} )=1$$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 11

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

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

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$\displaystyle\frac{y}{x}((\ln{x^{\displaystyle x-1}})+2\ln{x}\ln{(\ln{x})})$
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$\displaystyle\frac{y}{x}{(\log{x})}^{\displaystyle\log{(\log{x})}}(2\log{(\log{x})}+1)$
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$\displaystyle\frac{y}{x\ln{x}}[{(ln{x})}^2+2\ln{(\ln{x})}]$
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$\displaystyle\frac{y}{x}\frac{\log{y}}{\log{x}}[2\log{(\log{x})}+1]$

Explanation

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

Taking log of both sides, we get

$\therefore\log{y}=(\log{x}){(\log{x})}^{\displaystyle\log{(\log{x})}}$ ... (1)

Taking log of both sides, we get

$\log{(\log{y})}=\log{(\log{x})}+\log{(\log{x})}\log{(\log{x})}$

Differentiating w.r.t.

$x$ , we get

$\displaystyle\frac{1}{\log{y}}.\frac{1}{y}\frac{dy}{dx}=\frac{1}{x\log{x}}+\frac{2\log{(\log{x})}}{\log{x}}\frac{1}{x}$

$=\displaystyle\frac{2\log{(\log{x})}+1}{x\log{x}}$

$\displaystyle\frac{dy}{dx}=\frac{y}{x}.\frac{\log{y}}{\log{x}}[2\log{(\log{x})}+1]$

Substituting the value of

$y$ from (1), we get

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

$f(x)={|x|}^{|\sin{x}|}$ , then

$\displaystyle f^\prime\left(-\frac{\pi}{4}\right)=$

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$\displaystyle {\left(\frac{\pi}{5}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{5}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$
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$\displaystyle {\left(\frac{\pi}{4}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{4}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$
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$\displaystyle {\left(\frac{\pi}{3}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{3}{\pi}}-\frac{3\sqrt{3}}{\pi}\right)$
0%
None of these

Explanation

In the neighbourhood of

$\displaystyle -\frac{\pi}{4}$ , we have

$f(x)={(-x)}^{\displaystyle-\sin{x}}=e^{\displaystyle-\sin{x}\log{(-x)}}$

$\therefore \displaystyle f^\prime(x)=e^{\displaystyle-\sin{x}\log{(-x)}}\left(-\cos{x}.\log{(-x)}-\frac{\sin{x}}{x}\right)$

$\displaystyle =(-x)^{\displaystyle-\sin{x}}\left(-\cos{x}.\log{(-x)}-\frac{\sin{x}}{x}\right)$

$\therefore \displaystyle f^\prime\left(-\frac{\pi}{4}\right)={\left(\frac{\pi}{4}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{-1}{\sqrt{2}}\log{\frac{\pi}{4}}+\frac{4}{\pi}\times\left(\frac{-1}{\sqrt{2}}\right)\right)$

$\displaystyle={\left(\frac{\pi}{4}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{4}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$

$x=\sin^{-1}t$ and

$y=\log(1-{t}^{2})$ , then

$\displaystyle \left .\frac{{d}^{2}y}{{d}{x}^{2}}\right|_{{t}=\frac{1}{2}} =$

0%
$-\dfrac{8}{3}$
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$\dfrac{8}{3}$
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$\dfrac{3}{4}$
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$-\dfrac{3}{4}$

Explanation

Given,

$x=\sin^{-1}t$ or

$\sin x=t$

and

$y=\log(1-t^{2})$

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

$=\log(\cos^{2}x)$

Hence,

$y=2\log(\ cosx)$

Therefore,

$\dfrac{dy}{dx}=\dfrac{-2\sin x}{\cos x}=-2\tan x$

$\dfrac{d^{2}y}{dx^{2}}=-2\sec^{2}x=\dfrac{-2}{\cos^{2}x}$

$=\dfrac{-2}{1-\sin^{2}x}=\dfrac{-2}{1-t^{2}}$

Hence,

$\dfrac{d^{2}y}{dx^{2}}|_{t=\frac{1}{2}}=\dfrac{-2}{1-\left(\dfrac{1}{2}\right)^{2}}=\dfrac{-2}{1-\dfrac{1}{4}}=\dfrac{-8}{4-1}=\dfrac{-8}{3}$ .

If a function is represented parametrically by the equations

$\displaystyle x=\frac{1+\log_e{t}}{t^2}$ ;

$\displaystyle y=\frac{3+2\log_e{t}}{t}$ , then which of the following statements are true?

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$y^{\prime\prime}(x-2xy^\prime)=y$
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$yy^\prime=2x{(y^\prime)}^2+1$
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$xy^\prime=2y{(y^\prime)}^2+2$
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$y^{\prime\prime}(y-4xy^\prime)={(y^\prime)}^2$

Explanation

$\displaystyle x=\frac{1+\log_e{t}}{t^2}$ ;

$\displaystyle y=\frac{3+2\log_e{t}}{t}$

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

$\displaystyle=\frac{\displaystyle\frac{\displaystyle t\left(\frac{2}{t}\right)-(3+2\log_e{t})}{t^2}}{\displaystyle\frac{\displaystyle t^2\left(\frac{1}{t}\right)-(1+\log_e{t})2t}{t^4}}$

$\displaystyle =\left(\frac{-1-2\log_e{t}}{-1-2\log_e{t}}\right).t=t$

Eliminating

$\log_e{t}$ term from

$y$ , we get

$\displaystyle y=\frac{1+2t^2x}{t}=\frac{1+2{(y^\prime)}^2x}{y^\prime}$

or

$yy^\prime=1+2x{(y^\prime)}^2$
or

$yy^{\prime\prime}+{(y^\prime)}^2=4xy^\prime.y^{\prime\prime}+2{(y^\prime)}^2$
or

$yy^{\prime\prime}=4xy^\prime.y^{\prime\prime}+{(y^\prime)}^2$
Hence, options B and D are correct.

$\displaystyle y=\left ( 1+\frac{1}{x} \right )^{x}+x^{1+\frac{1}{x}}.$
Differentiate

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

Explanation

Let $y = y_1+y_2$
$\Rightarrow \cfrac{dy}{dx}=\cfrac{dy_1}{dx}+\cfrac{dy_2}{dx}$
Now $y_1 = \left(1+\frac{1}{x}\right)^x \Rightarrow \log y_1 =x \log\left(1+\frac{1}{x}\right)$

Differentiating both side w.r.t $x$

$\Rightarrow \cfrac{1}{y_1}\cfrac{dy_1}{dx} = \log(1+\frac{1}{x})+x.\cfrac{1}{1+\frac{1}{x}}.\left(\frac{-1}{x}^2\right)$

$\Rightarrow \cfrac{dy_1}{dx} = y_1\left[\log(1+\frac{1}{x})-\cfrac{1}{1+x}\right]$
And $y_2 = x^{1+\frac{1}{x}} \Rightarrow \log y_2 = (1+\frac{1}{x}) \log x$

Differentiating both sides,

$\Rightarrow \cfrac{1}{y_2}\cfrac{dy_2}{dx} = (1+\frac{1}{x})\cfrac{1}{x}+(\log x)(\frac{-1}{x}^2)$

$\Rightarrow \cfrac{dy_2}{dx} = y_2\left[\cfrac{(1+x)- \log x}{x^2}\right]$
$\therefore \cfrac{dy}{dx} = \left(1+\cfrac{1}{x}\right)\left[\log(1+\frac{1}{x})-\cfrac{1}{1+x}\right]+x^{1+\frac{1}{x}}\left[\cfrac{(1+x)- \log x}{x^2}\right]$

$x=\sin t$ ,

$y=\sin kt$ then value of

$\left ( 1-x^{2} \right )y_{2}-xy_{1}$ is

0%
$-k^{2}y$
0%
$k^{2}y$
0%
$ky^{2}$
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$-ky^{2}$

Explanation

$\because$

$x=\sin t$ ,

$y=\sin kt$

$\Rightarrow \dfrac{dx}{dt}=\cos t, \dfrac{dy}{dt}=k\cos kt$

$\therefore$

$\displaystyle y_{1}=\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{k\cos kt}{\cos t}$

$\Rightarrow$

$y_{1}\cos t=k\cos kt$

On squaring both sides, we have

$y_{1}^{2}\cos^{2} t=k^{2}\cos^{2} kt$

$y_{1}^{2}\left ( 1-\sin ^{2}t \right )=k^{2}\left ( 1-\sin ^{2}kt \right )$

$\Rightarrow$

$y_{1}^{2}\left ( 1-x^{2} \right )=k^{2}\left ( 1-y^{2} \right )$
Differentiating both sides w.r to x, we have

$\left ( 1-x^{2} \right )\left ( 2y_{1}y_{2} \right )+y{_{1}}^{2}\left ( -2x \right )=k^{2}\left ( -2yy_{1} \right )$

$\Rightarrow$

$\left ( 1-x^{2} \right )y_{2}-xy_{1}=-k^{2}y$

$\displaystyle y=x^{\ln x^{\ln \left ( \ln x \right )}}$ , then

$\displaystyle \frac{dy}{dx}$ is equal to:

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$\displaystyle \frac{y}{x}\left ( \ln x^{\ln x-1}+2\ln x\ln (\ln x) \right )$
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$\displaystyle \frac{y}{x}\ln x^{\ln \left ( \ln x \right )}(2\ln (\ln x)+1)$
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$\displaystyle \frac{y}{x\ln x}\ln x^{2}+2\ln (\ln x)$
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$\displaystyle \frac{y\ln x}{x\ln x}(2\ln (\ln x)+1)$

Explanation

Given

$y=x^{\ln x^{\ln \left ( \ln x \right )}}$

Taking log on both sides,

$\displaystyle lny=ln(lnx)(lnx)^{2}$

$\dfrac{y'}{y}=ln(lnx)(2lnx.\dfrac{1}{x})+(lnx)^{2}(\dfrac{1}{xlnx})$

$\Rightarrow \displaystyle \dfrac{y'}{y}=ln(lnx)(2lnx.\dfrac{1}{x})+(lnx)(\dfrac{1}{x})$

$\Rightarrow \displaystyle y'=y(lnx)(\dfrac{1}{x})(2ln(lnx)+1)$

$\Rightarrow \displaystyle y'=\dfrac{y}{x}(lnx^{lnlnx})(2ln(lnx)+1)$

We know,

$\Rightarrow \displaystyle lny=lnx^{lnlnx}.lnx$

So,

$y'=\dfrac{ylny}{xlnx}(2ln(lnx)+1)$

$f(x)=\left | x \right |+\left | \cos x \right |$ , then

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$f^{'}( \dfrac{\pi }{2} )=2$
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$f^{'}( \dfrac{\pi }{2} )=0$
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$f^{'}( \dfrac{\pi }{2} )=1$
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$f^{'}( \dfrac{\pi }{2} )$ does not exist

Explanation

$f\left( x \right) =\left| x \right| +\left| \cos { x } \right|$

$f'\left( x \right) =\cfrac { \left| x \right| }{ x } +\cfrac { \left| \cos { x } \right| }{ \cos { x } }$

$x\rightarrow { \cfrac { \pi }{ 2 } }^{ - }\quad \cfrac { \left| x \right| }{ x } =1$

Since

$\cos { x } >0$ from

$x=0$ to

$\cfrac { \pi }{ 2 }$

$\cfrac { \left| \cos { x } \right| }{ \cos { x } } =1$

$\Rightarrow \displaystyle \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ f'\left( x \right) } =1+1=2$

$x\rightarrow \cfrac { \pi }{ 2 } ^{ + }\quad \cfrac { \left| x \right| }{ x } =1$

Since

$\cos { x } <0$ from

$x=\cfrac { \pi }{ 2 }$ to

$\pi$

$\cfrac { \left| \cos { x } \right| }{ \cos { x } } =-1$

$\displaystyle \lim _{ x\rightarrow \frac { \pi }{ 2 } ^{ + } }{ f'\left( x \right) } =1-1=0$

LHL

$\neq$ RHL

$\therefore \displaystyle \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ f'\left( x \right) } =f'\left( \cfrac { \pi }{ 2 } \right)$ does not exist.

Given the parametric equations

$x=f(t),y=g(t).$ Then

$\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ equals

0%
$\displaystyle \dfrac { \dfrac { { d }^{ 2 }y }{ d{ t }^{ 2 } } .\dfrac { dx }{ dt } -\dfrac { dy }{ dt } .\dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } }{ { \left( \dfrac { dx }{ dt } \right) }^{ 2 } }$
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$\displaystyle \dfrac { \dfrac { { d }^{ 2 }y }{ d{ t }^{ 2 } } .\dfrac { dx }{ dt } -\dfrac { dy }{ dt } .\dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } }{ { \left( \dfrac { dx }{ dt } \right) }^{ 3 } }$
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$\displaystyle \dfrac { \dfrac { { d }^{ 2 }y }{ d{ t }^{ 2 } } }{ \dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } }$
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None of these

Explanation

We have

$\displaystyle \dfrac { dy }{ dx } =\dfrac { \dfrac { dy }{ dt } }{ \dfrac { dx }{ dt } }$

Again differentiating both sides w.r.t x,

we get

$\displaystyle \dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\dfrac { d }{ dt } \left( \dfrac { \dfrac { dy }{ dt } }{ \dfrac { dx }{ dt } } \right) \dfrac { dt }{ dx }$

$\displaystyle =\dfrac { \dfrac { { d }^{ 2 }y }{ { dt }^{ 2 } } \left( \dfrac { dx }{ dt } \right) -\left( \dfrac { { d }^{ 2 }x }{ { dt }^{ 2 } } \right) \dfrac { dy }{ dt } }{ { \left( \dfrac { dt }{ dt } \right) }^{ 2 } } \times \dfrac { 1 }{ \dfrac { dx }{ dt } }$

$\displaystyle =\dfrac { \dfrac { { d }^{ 2 }y }{ { dt }^{ 2 } } .\left( \dfrac { dx }{ dt } \right) -\left( \dfrac { { d }^{ 2 }x }{ { dt }^{ 2 } } \right) .\left( \dfrac { dy }{ dt } \right) }{ { \left( \dfrac { dx }{ dt } \right) }^{ 3 } }$

Derivative of

${(x\cos{x})}^x$ with respect to

$x$ is

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$\displaystyle {(x\cos{x})}^x\left[(\log{x}+1)-\left\{\log{\cos{x}}+\frac{x}{\cos{x}}.(\sin{x})\right\}\right]$
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$\displaystyle {(x\cos{x})}^x\left[(\log{x}+1)+\left\{\log{\cos{x}}+\frac{x}{\cos{x}}.(-\sin{x})\right\}\right]$
0%
$\displaystyle {(x\cos{x})}^x\left[(\log{x}+1)+\left\{\log{\sin{x}}+\frac{x}{\cos{x}}.(\cos{x})\right\}\right]$
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None of these

Explanation

Let

$y={(x\cos{x})}^x$ .
Taking logarithm on both sides, we get

$\log{y}=\log{{(x\cos{x})}^x}$

$=x\log{(x\cos{x})}$

$=x\log{x}+x\log{\cos{x}}$
Differentiating both sides with respect to

$x$ , we get

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

$\displaystyle \frac { dy }{ dx } =y\left[ \left( 1+\log { x } \right) +\left( \log { \cos { x\quad +\displaystyle \frac { x }{ \cos { x } } \displaystyle \frac { d }{ dx } \left( \cos { x } \right) } } \right) \right]$

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

The value of

$y''(1)$ if

${ x }^{ 3 }-2{ x }^{ 2 }{ y }^{ 2 }+5x+y-5=0$ when

$y(1)>1,$ is equal to

0%
$\displaystyle \frac { 22 }{ 7 }$
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$\displaystyle -7\frac { 21 }{ 28 }$
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$8$
0%
$\displaystyle -8\frac { 22 }{ 27 }$

Explanation

Given expression is

${ x }^{ 3 }-2{ x }^{ 2 }{ y }^{ 2 }+5x+y-5=0$ ...(1)

Differentiating the given expression, we get

$3{x}^{2}-4x{y}^{2}-4{x}^{2}yy'+5+y'=0$ ....(2)

Putting

$x=1$ in the given expression, we get

$3-2{y}^{2}+5+y-5=0\Rightarrow 2{y}^{2}-y-1=0$

$\Rightarrow y=1$ or

$\displaystyle y=-\frac{1}{2}$

Now

$\because y(1)>0\Rightarrow y(1)=1$

Differentiating again we get

$6x-4{ y }^{ 2 }-8xyy'-8xyy'-4x2{ \left( y' \right) }^{ 2 }-4{ x }^{ 2 }yy''+y''=0$

Putting

$x=1,y=1$ and

$\displaystyle y'(1)=\frac{4}{3},$ we get

$\displaystyle 6-4-8\left( \frac { 4 }{ 3 } \right) -8\left( \frac { 4 }{ 3 } \right) -4\left( \frac { 16 }{ 9 } \right) -3y''\left( 1 \right) =0$

$\displaystyle \Rightarrow y''\left( 1 \right) =-8\frac { 22 }{ 27 }$

The weight

$W$ of a certain stock of fish is given by

$W = nw$ , where n is the size of stock and

$w$ is the average weight of a fish. If

$n$ and

$w$ change with time

$t$ as

$n = 2t^2 + 3$ and

$w = t^2 - t + 2$ , then the rate of change of

$W$ with respect to

$t$ at

$t = 1$ is.

0%
$1$
0%
$5$
0%
$8$
0%
$31$

Explanation

$(2t^{2}+3)(t^{2}+t+2)$

$=2t^{4}+2t^{3}+4t^{2}+3t^{2}+3t+6$

$W=2t^{4}+2t^{3}+7t^{2}+3t+6$
Hence

$\dfrac{dW}{dt}$

$=8t^{3}+6t^{2}+14t+3$
At

$t=1$ we get

$\dfrac{dW}{dt}$ at

$t=1$

$=8+6+14+3$

$=31$ .

$x=log(1+t^2)$ and

$y=t-tan^{-1}t$ . Then,

$\dfrac{dy}{dx}$ is equal to

0%
$e^x-1$
0%
$t^2-1$
0%
$\dfrac{\sqrt{e^x-1}}{2}$
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$e^x-y$

Explanation

Differentiating wrt 't'

$\dfrac{dx}{dt}=\dfrac{d}{dt}(log (1+t^2))=\dfrac{1}{1+t^2}2t=\dfrac{2t}{1+t^2}$

$\dfrac{dy}{dt}=\dfrac{d}{dt}(t-\tan^{-1}t)=1-\dfrac{1}{1+t^2}=\dfrac{t^2}{1+t^2}$

$\dfrac{dy}{dx}=\dfrac{t^2/(1+t^2)}{2t/(1+t^2)}=\dfrac{t}{2}$

$x= log (1+t^2)$

$e^x=1+t^2$

$t=\sqrt{e^x-1}$

$\dfrac{t}{2}=\dfrac{\sqrt{e^x-1}}{2}$

$\therefore \dfrac{dy}{dx}=\dfrac{\sqrt{e^x-1}}{2}$ .

The function

$f(x)=\begin{cases} ax(x-1)+b\quad \quad when\quad x<1 \\ x-1\quad \quad \quad when\quad \quad 1\le x\le 3 \\ p{ x }^{ 2 }+qx+2\quad \quad when\quad x>3 \end{cases}$ Find the values of the constants

$a,b,p,q$ so that

$(i)f(x)$ is continuous for all

$x$

$(ii)f'(1)$ does not exist

$(iii)f'(x)$ is continuous at

$x=3$

0%
$a=1, b=0, p=1/3, q=-1$
0%
$a\ne 1, b=0, p=1/3, q=-1$
0%
$a\ne 1, b=0, p=1/3, q=1$
0%
$a=1, b=0, p=1/3, q=1$

Explanation

Making f(x) to be continuous at

$x = 1$ and

$3$ , we have

$a(0) + b = 0$ and

$3 - 1 = 9p + 3q + 2$

$\Rightarrow b = 0$ and

$q + 3p = 0$
Now,

$f'(x) = \begin{cases} 2ax - a, x < 1 \\ 1, 1 \le x \le 3 \\ 2px + q, x > 3 \end{cases}$
The given condition implies

$2a(1) - a \neq 1$ or

$a \neq 1$
The last condition says

$1 = 6p + q = 6p - 3p = 3p \quad \quad \quad$ (...substituting the value of q in terms of p)

$\Rightarrow p = \cfrac{1}{3}$ and

$q = -1$

Length of the subtangent at

$(x_l, y_l)$ on

$x^n y^m = a^{m+n}, m, n > 0,$ is

0%
$\dfrac{n}{m}x_l$
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$\dfrac{m}{n}|x_l|$
0%
$\dfrac{n}{m}|y_l|$
0%
$\dfrac{n}{m}|x_l|$

Explanation

The subtangent is given by

$\cfrac{y}{\tfrac{dy}{dx}}$

Here,

$x^ny^m = a^{m + n}$

$\therefore y^m = a^{m + n}x^{-n}$

$\therefore y =\left( a^{\tfrac{m + n}{m}}\right)\left(x^{\tfrac{-n}{m}}\right)$

$\Rightarrow \cfrac{dy}{dx} = \left(a^{\tfrac{m + n}{m}}\right) \times \left(\tfrac{-n}{m} \right)\times \left(x^{\tfrac{-n - m}{m}}\right)$

$\Rightarrow$ subtangent at

$(x_l,y_l) = a^{\tfrac{m + n}{m}}x_l^{\tfrac{-n}{m}} \times \cfrac{1}{\left(a^{\tfrac{m + n}{m}} \right)\times \left( \tfrac{-n}{m}\right) \times ( x_l^{\tfrac{-n - m}{m}})}$

$= \cfrac{-m}{n} \times x_l$

$= \cfrac{m}{n}|x_l|$ , since length cannot be negative

$x$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$4$	$3$	$7$	$1$	$3$

The function f is continuous on the closed interval

$[1, 5]$ and values of the function are shown in the table above. If the values in the table are used to calculate a trapezoidal sum, the approximate value of

$\int_{1}^{5}f(x)dx$ is

0%
$14$
0%
$14.5$
0%
$15$
0%
$29$

The derivative of

$\displaystyle (\tan x)^{x}$ is equal to-

0%
$\displaystyle x(\tan x)^{x-1}$
0%
$\displaystyle (\tan x)^{x}\left [ \sec x+\tan x \right ]$
0%
$\displaystyle (\tan x)^{x}\left [ x\sec x \csc x +\log \tan x\right ]$
0%
$\displaystyle(\tan x)^{x}\left [ \sec ^{2}x+x\tan x \right ]$

Explanation

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

$\Rightarrow \log y = x \log \tan x$

$\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx}=\log { \tan { x } } +\frac { x\times 1 }{ \tan x } \times \sec ^{ 2 } x$

$\Rightarrow \displaystyle \frac{dy}{dx}=y\left [ \log \tan x+x\sec x \, cosec x \right ]$

$\Rightarrow \displaystyle \frac{dy}{dx}=(\tan x)^{x}\left [ \log \tan x+x\sec x\, cosec x \right ]$

Examine for continuity and differentiability at the points

$x=1, x=2$ , the function

$f$ defined by

$f(x)=\begin{cases} x\left[ x \right] ,\quad \quad \quad \quad 0\le x<2 \\ (x-1)\left[ x \right] ,\quad 2\le x\le 3 \end{cases}$ where

$\left[ x \right] =$ greatest integer less than or equal to

$x$

0%
discontinuous and not derivable at $x=1,2$
0%
discontinuous and not derivable at $x=1$ , continuous but not derivable at $x=2$
0%
continuous and not derivable at $x=1,2$ .
0%
continuous and not derivable at $x=1$ , discontinuous but not derivable at $x=2$ .

Explanation

$f(1^-) = \displaystyle \lim_{h \rightarrow 0} (1 - h)0 = 0$

$f(1^+) = \displaystyle \lim_{h \rightarrow 0} 1 + h = 1$
Since f is not continuous at

$x = 1$ , its also not differentiable at that point.

$f(2^-) = \displaystyle \lim_{h \rightarrow 0}(2 - h)(1) = 2$

$f(2^+) = \displaystyle \lim_{h \rightarrow 0} (2 + h - 1)2 = 2$

$f'(2^-) = \displaystyle \lim_{h \rightarrow 0} \cfrac{2 - h - 2}{-h} = 1$

$f(2^+) = \displaystyle \lim_{h \rightarrow 0} \cfrac{2 + 2h - 2}{h} = 2$
Hence, the function is continuous but not differentiable at

$x = 2.$

$y=\sqrt { \left( a-x \right) \left( x-b \right) } -\left( a-b \right) \tan ^{ -1 }{ \sqrt { \dfrac { a-x }{ x-b } } }$ , then

$\dfrac { dy }{ dx }$ is equal to

0%
$1$
0%
$\sqrt { \dfrac { a-x }{ x-b } }$
0%
$\sqrt { \left( a-x \right) \left( x-b \right) }$
0%
$\dfrac { 1 }{ \sqrt { \left( a-x \right) \left( b-x \right) } }$

Explanation

Let

$x=a\cos ^{ 2 }{ \theta } +b\sin ^{ 2 }{ \theta }$
Here

$a-x=a-a\cos ^{ 2 }{ \theta } -b\sin ^{ 2 }{ \theta } =\left( a-b \right) \sin ^{ 2 }{ \theta }$
and

$x-b=a\cos ^{ 2 }{ \theta } -b\sin ^{ 2 }{ \theta } -b=\left( a-b \right) \cos ^{ 2 }{ \theta }$
and

$y=\left( a-b \right) \sin { \theta } \cdot \cos { \theta } -\left( a-b \right) \tan ^{ -1 }{ \tan { \theta } }$

$=\dfrac { a-b }{ 2 } \sin { 2\theta } -\left( a-b \right) \theta$
Therefore,

$\dfrac { dy }{ dx } =\dfrac { { dy }/{ d\theta } }{ { dx }/{ d\theta } } =\dfrac { \left( a-b \right) \cos { 2\theta } -\left( a-b \right) }{ \left( b-a \right) \sin { 2\theta } }$

$=\dfrac { 1-\cos { 2\theta } }{ \sin { \theta } } =\tan { \theta } =\sqrt { \dfrac { a-x }{ x-b } }$

The derivative of

$y=x^{2^x}$ w.r.t x is :

0%
$x^{2^x}2^x\left(\displaystyle \frac{1}{x} + \ln x \ln 2 \right)$
0%
$x^{2^x}\left(\displaystyle\frac{1}{x} \ln x \ln 2\right)$
0%
$x^{2^x}2^x\left(\displaystyle\frac{1}{x} \ln x\right)$
0%
$x^{2^x}2^x\left(\displaystyle\frac{1}{x}+\frac{\ln x}{\ln 2}\right)$

Explanation

We have,

$y={ x }^{ { 2 }^{ x } }$

take logarithm on the both sides we get,

$\ln { y } ={ 2 }^{ x }\ln { x }$

diff. w.r.t.

$x$

$\cfrac { 1 }{ y } \cfrac { dy }{ dx } =\left[ { 2 }^{ x }\log { 2. } \ln { x } +\cfrac { { 2 }^{ x } }{ x } \right]$

$\cfrac { dy }{ dx } ={ 2 }^{ x }.{ x }^{ { 2 }^{ x } }\left[ \ln { 2 } .\ln { x } +\cfrac { 1 }{ x } \right]$

$={ 2 }^{ x }.{ x }^{ { 2 }^{ x } }\left[ \cfrac { 1 }{ x } + \ln x \ln 2 \right]$

Let

$f(x) = \left\{\begin{matrix}2a - x, &if\ -a < x < a \\ 3x - 2a, &if\ a \leq x \end{matrix}\right.$ . Then, which of the following is true?

0%
$f(x)$ is discontinuous at $x = a$
0%
$f(x)$ is not differentiable at $x = a$
0%
$f(x)$ is differentiable at $x \geq a$
0%
$f(x)$ is continuous at all $x < a$

Explanation

Given,

$f(x) = \left\{\begin{matrix}2a - x, &if\ -a < x < a \\ 3x - 2a, &if\ a \leq x \end{matrix}\right.$
At

$x = a$ ,

$LHL = \displaystyle \lim_{x\rightarrow a^{-}} f(x) = \displaystyle \lim_{x\rightarrow a} (2a - x)$

$= 2a - a = a\ RHL$

$= \displaystyle \lim_{x\rightarrow a^{+}} f(x) = \displaystyle \lim_{x\rightarrow a}(3x - 2a)$

$= 3(a) - 2a = a$
and

$f(a) = 3(a) - 2a = a$

$\therefore LHL = RHL = f(a)$
Hence, it is continuous at

$x = a$
Again, at

$x = a$

$LHD = \displaystyle \lim_{h\rightarrow 0}\dfrac {f(a - h) - f(a)}{-h}$

$= \displaystyle \lim_{h\rightarrow 0} = \dfrac {2a - (a - h) - a}{-h} =-1$
and

$RHD = \displaystyle \lim_{h\rightarrow 0} \dfrac {f(a + h) - f(a)}{h}$

$= \displaystyle \lim_{h\rightarrow 0} \dfrac {3(a + h) - 2a - a}{h} = 3$

$\therefore LHD \neq RHD$
Hence, it is not differentiable at

$x = a$ .

$x = \sec \theta - \cos \theta$ and

$y = \sec^{n}\theta - \cos^{n}\theta$ , then

$\left (\dfrac {dy}{dx}\right )^{2}$ is equal to

0%
$\dfrac {n^{2}(y^{2} + 4)}{x^{2} + 4}$
0%
$\dfrac {n^{2}(y^{2} - 4)}{x^{2}}$
0%
$n\left (\dfrac {y^{2} - 4}{x^{2} - 4}\right )$
0%
$\left (\dfrac {ny}{x}\right )^{2} - 4$

Explanation

$\dfrac {dy}{d\theta} = n \sec^{n - 1}\theta \cdot \sec \theta \cdot \tan \theta - n\cdot \cos \theta^{n - 1} (-\sin \theta)$

$= n\tan \theta (\sec^{n} \theta + \cos^{n} \theta)$

$\dfrac {dx}{d\theta} = \dfrac {\sin \theta}{\cos \theta} (\sec \theta + \cos \theta)$

$= \tan \theta(\sec \theta + \cos \theta)$

$\therefore \dfrac {dy}{dx} = \dfrac {n\tan \theta (\sec^{n}\theta + \cos^{n}\theta)}{\tan \theta(\sec \theta + \cos \theta)}$

$\left (\dfrac {dy}{dx}\right )^{2} = \dfrac {n^{2}(\sec^{n}\theta + \cos^{n}\theta)^{2}}{(\sec \theta + \cos \theta)^{2}}$

$= \dfrac {n^{2}\left \{(\sec^{n}\theta - \cos^{n}\theta)^{2} + 4\right \}}{(\sec \theta - \cos \theta)^{2} + 4} = \dfrac {n^{2}(y^{2} + 4)}{x^{2} + 4}$

If a curve is given by

$x=a\cos t+\displaystyle\frac{b}{2}\cos 2t$ and

$y=\sin t +\displaystyle\frac{b}{2}\sin 2t$ , then the points for which

$\displaystyle\frac{d^2y}{dx^2}=0$ , are given by.

0%
$\sin t=\displaystyle\frac{2a^2+b^2}{3ab}$
0%
$\cos t=-\left [\displaystyle\frac{a^2+2b^2}{3ab}\right]$
0%
$\tan t=a/b$
0%
None of the above

Explanation

We have,

$x=a\cos t+\displaystyle\frac{b}{2}\cos 2t$
and

$y=a\sin t+\displaystyle\frac{b}{2}\sin 2t$
On differentiating both equations w.r.t. t, we get

$\displaystyle\frac{dx}{dt}=-a\sin t -b\sin 2t$
and

$\displaystyle\frac{dy}{dt}=a\cos t +b\cos 2t$

$\therefore \displaystyle\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos t +b\cos 2t}{-a\sin t-b\sin 2t}$

$\Rightarrow \displaystyle\frac{dy}{dx}=-\frac{(a\cos t+b\cos 2t)}{(a\sin t+b\sin 2t)}$
Now,

$\displaystyle\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\displaystyle\frac{dy}{dx}\right)=\frac{d}{dt}\left(\displaystyle\frac{dy}{dx}\right)\cdot \frac{dt}{dx}$

$\Rightarrow \displaystyle\frac{d^2y}{dx^2}=-\frac{d}{dt}\left(\displaystyle\frac{a\cos t+b\cos 2t}{a\sin t+b\sin 2t}\right)\frac{dt}{dx}$

$=\displaystyle \frac{\displaystyle \begin{bmatrix} (a\sin t+b]sin 2t)(-a\sin t-2b\sin 2t) \\ -(a\cos t+b\cos 2t)(a\cos t+2b\cos 2t)\end{bmatrix}}{(a\sin t+b\sin 2t)^2}\times \frac{1}{-a\sin t-b\sin 2t}$

$=\displaystyle\frac{\displaystyle\begin{bmatrix} a^2\sin^2t+3ab\sin t\sin 2t+2b^2\sin^22t\\ +a^2\cos^2t+3ab\cos t\cos 2t+2b^2\cos^22t\end{bmatrix}}{(a\sin t+b\sin 2t)^3}$

$=\displaystyle\frac{a^2(\sin^2t+\cos^2t)+b^2(\sin^22t+\cos^22t)+3ab\cos(2t-t)}{(a\sin t+b\sin 2t)^3}$

$\Rightarrow \displaystyle\frac{d^2y}{dx^2}=-\left[\displaystyle\frac{a^2+2b^2+3ab\cos t}{(a\sin t+b\sin 2t)^3}\right]$
Given,

$\displaystyle\frac{d^2y}{dx^2}=0$

$\Rightarrow a^2+2b^2+3ab\cos t=0$

$\Rightarrow \cos t=-\left(\displaystyle\frac{a^2+2b^2}{3ab}\right)$

$\sin ^{ -1 }{ \left( \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) } =\log { a }$ , then

$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ equals

0%
$\dfrac { x }{ y }$
0%
$\dfrac { y }{ { x }^{ 2 } }$
0%
$\dfrac { y }{ x }$
0%
$0$

Explanation

We have

$\sin ^{ -1 }{ \left( \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) } =\log { a }$

$\Rightarrow \dfrac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } =\sin { \left( \log { a } \right) }$

$\Rightarrow \dfrac { 1-\tan ^{ 2 }{ \theta } }{ 1+\tan ^{ 2 }{ \theta } } =\sin { \left( \log { a } \right) }$
(on putting

$y=x\tan { \theta }$ )

$\Rightarrow \cos { 2\theta } =\sin { \left( \log { a } \right) }$

$\Rightarrow 2\theta =\cos ^{ -1 }{ \left( \sin { \log { a } } \right) }$

$\Rightarrow \theta =\dfrac { 1 }{ 2 } \cos ^{ -1 }{ \left\{ \sin { \left( \log { a } \right) } \right\} }$

$\tan ^{ -1 }{ \left( \dfrac { y }{ x } \right) } =\dfrac { 1 }{ 2 } \cos ^{ -1 }{ \left\{ \sin { \log { a } } \right\} }$

$\Rightarrow \dfrac { 1 }{ 1+\dfrac { { y }^{ 2 } }{ { x }^{ 2 } } } \cdot \dfrac { x\dfrac { dy }{ dx } -y }{ { x }^{ 2 } } =0$

$\Rightarrow x\dfrac { dy }{ dx } -y=0$

$\Rightarrow \dfrac { dy }{ dx } =\dfrac { y }{ x }$ ...(i)

$\Rightarrow \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\dfrac { -y }{ { x }^{ 2 } } +\dfrac { 1 }{ x } \cdot \dfrac { dy }{ dx } =\dfrac { -y }{ { x }^{ 2 } } +\dfrac { 1 }{ x } \left( \dfrac { y }{ x } \right)$ ....{from (i)}

$\Rightarrow \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\dfrac { -y }{ { x }^{ 2 } } +\dfrac { y }{ { x }^{ 2 } } =0$

Find derivative of

$\tan^{-1}\dfrac{\cos x-\sin x}{\cos x+\sin x}$ w.r.t.

$x$ .

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

Explanation

We have to find the derivative of

$tan^{-1} \dfrac{cos x-sin x}{cos x+sin x}$ with respect to

$x$ .

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

Let

$u=\dfrac{cos x-sin x}{cos x+sin x}, f=tan^{-1} (u)$

Apply chain rule

$\dfrac{df(u)}{dx}=\dfrac{df}{du} \cdot \dfrac{du}{dx}$ , we get

$\dfrac{df(u)}{dx}=\dfrac{d}{du} (tan^{-1}(u)) \cdot \dfrac{d}{dx} \left(\dfrac{cos x-sin x}{cos x+sin x}\right)$

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

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

$=\dfrac{1}{u^2+1} \cdot \dfrac{-cos^2 x-sin^2 x-2cos x \: sin x-cos^2 x-sin^2 x+2cos x \: sin x}{cos^2 x+sin^2 x+2sin x\: cos x}$

$=\dfrac{1}{u^2+1} \cdot \dfrac{-2cos^2 x-2sin^2 x}{1+2sin x\: cos x}$

$=\dfrac{1}{u^2+1} \cdot \dfrac{-2(cos^2 x+sin^2 x)}{1+2sin x\: cos x}$

$=\dfrac{1}{u^2+1} \cdot \dfrac{-2}{1+2sin x\: cos x}$

Substitute

$u=\dfrac{cos x-sin x}{cos x+sin x}$ we get

$=\dfrac{1}{\left(\dfrac{cos x-sin x}{cos x+sin x}\right)^2+1} \cdot \dfrac{-2}{1+2sin x\: cos x}$

$=\dfrac{(cos x+sin x)^2}{(cos x-sin x)^2+(cosx+sin x)^2} \cdot \dfrac{-2}{1+2sin x\: cos x}$

$=\dfrac{cos^2 x+sin^2 x+2cos x sin x}{cos^2 x+sin^2 x-2cos x sin x+cos^2x+sin^2 x+2sin x cos x} \cdot \dfrac{-2}{1+2sin x\: cos x}$

$=\dfrac{1+2cos x sin x}{1-2cos x sin x+1+2sin x cos x} \cdot \dfrac{-2}{1+2sin x\: cos x}$

$=\dfrac{1+2 sin x cos x}{2} \cdot \dfrac{-2}{1+2sin x\: cos x}$

$=-1$

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

$y=\tan ^{ -1 }{ \left( \cfrac { 4x }{ 1+5{ x }^{ 2 } } \right) } +\tan ^{ -1 }{ \left( \cfrac { 2+3x }{ 2-3x } \right) }$ , then

$\cfrac { dy }{ dx }$ is

0%
$\cfrac { 6 }{ 1+4{ x }^{ 2 } }$
0%
$\cfrac { 3 }{ 1+4{ x }^{ 2 } }$
0%
$\cfrac { 5 }{ 1+{ 25x }^{ 2 } }$
0%
$\dfrac{5}{(1+25x^{2})} - \dfrac{1}{(1+x^{2})} - \dfrac{1.5}{(1+2.25x^{2})}$

Explanation

Use properties of inverse trigonometric functions-

$\tan ^{ -1 }{ 5x}+\tan^{-1}{(-x)}$ +

$\tan^{-1}{1} +\tan^{-1}{(-1.5x)}$

$\Rightarrow$ y=

$\tan ^{ -1 }{ 5x}-\tan^{-1}{x}$ +

$\tan^{-1}{1} -\tan^{-1}{1.5x}$

Differentiate with respect to x,

$\therefore$

$\dfrac{d(tan^{-1}x)}{dx}$ =

$\dfrac{1}{(1+x^{2})}$

So,

$\therefore$

$\dfrac{dy}{dx} = \dfrac{5}{(1+25x^{2})} - \dfrac{1}{(1+x^{2})} + 0 - \dfrac{1.5}{(1+2.25x^{2})}$

$\therefore$

$\dfrac{dy}{dx} = \dfrac{5}{(1+25x^{2})} - \dfrac{1}{(1+x^{2})} - \dfrac{1.5}{(1+2.25x^{2})}$

$x = A\cos 4t + B\sin 4t$ , then

$\dfrac {d^{2}x}{dt^{2}}$ is equal to

0%
$-16x$
0%
$16x$
0%
$x$
0%
$-x$

Explanation

Given,

$x = A\cos 4t + B\sin 4t$
On differentiating w.r.t.,

$t$ , we get

$\dfrac {dx}{dt} = -4A\sin 4t + 4B\cos 4t$
Again, differentiating w.r.t.

$t$ , we get

$\dfrac {d^{2}x}{dt^{2}} = -6A\cos 4t - 16B \sin 4t$

$\Rightarrow \dfrac {d^{2}x}{dt^{2}} = -16(x)$ .

Let

$f\left( x \right)$ and

$g\left( x \right)$ be defined and differentiable for

$x\ge { x }_{ 0 }$ and

$f\left( { x }_{ 0 } \right) =g\left( { x }_{ 0 } \right) , f^{ ' }\left( x \right) >g^{ ' }\left( x \right) for\ x>{ x }_{ 0 }$ then

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$f\left( x \right) < g\left( x \right)$ for some $x>{ x }_{ 0 }$
0%
$f\left( x \right) =g\left( x \right)$ for some $x>{ x }_{ 0 }$
0%
$f\left( x \right) >g\left( x \right)$ for all $x>{ x }_{ 0 }$
0%
None of these

Find the derivative of

$\dfrac{\tan^{-1}x}{1+\tan^{-1}x}$ w.r.t.

$\tan^{-1}x$ .

0%
$\dfrac{1}{\sec^{-1}x}$
0%
$\dfrac{1}{(\tan^{-1}x)^{2}}$
0%
$\dfrac{1}{1+\tan^{2}x}$
0%
$\dfrac{1}{(1+\tan^{-1}x)^{2}}$

Explanation

let

$\tan^{-1}x = t$ and

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

then,

$\dfrac{dy}{dt} = \dfrac{d}{dt} \dfrac{t}{1+t}$

$\dfrac{dy}{dt} = \dfrac{(1+t)1 - t(1)}{(1+t)^2}$

$\dfrac{dy}{dt} = \dfrac{1}{(1+t)^2}$

$= \dfrac{1}{(1+\tan^{-1}x)^2}$

If the function

$f:\left[ 0,8 \right] \rightarrow R$ is differentiable, then for

$0<a,b<2,\int _{ 0 }^{ 8 }{ f(t) } dt$ is equal to

0%
$$2\left[ { \alpha }^{ 3 }f({ \alpha }^{ 2 })+{ \beta }^{ 2 }f({ \beta }^{ 2 }) \right] $
0%
$3\left[ { \alpha }^{ 3 }f({ \alpha }^{ })+{ \beta }^{ 2 }f({ \beta }^{ }) \right]$
0%
$3\left[ { \alpha }^{ 2 }f({ \alpha }^{ 3 })+{ \beta }^{ 2 }f({ \beta }^{ 3 }) \right]$
0%
$3\left[ { \alpha }^{ 2 }f({ \alpha }^{ 2 })+{ \beta }^{ 2 }f({ \beta }^{ 2 }) \right]$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 11 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 11 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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