Continuity and Differentiability Test

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Continuity and Differentiability Test - 37

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Question 1
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If $$y=x^{\left(x^{ x}\right)}$$, then $$\displaystyle\frac{dy}{dx}$$ is

A
$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x^{\displaystyle x}\right]$$

B
$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x\right]$$

C
$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x^{\displaystyle x-1}\right]$$

D
$$y\left[x^{\displaystyle x}\left(\log_e{x}\right)\log{x}+x^{\displaystyle x-1}\right]$$

Solution

$$y={ x }^{ \left( { x }^{ x } \right) }$$
Take log on both sides
$$\log y=\log { { x }^{ \left( { x }^{ x } \right) } } $$
$$\log { y } ={ x }^{ x }\log { x } $$ .... $$(i)$$

Let $${ x }^{ x }=z$$
$$\therefore x\log { x } =\log { z } $$
Differentiate both sides with respect to $$x$$

$$\cfrac { 1 }{ z } \cfrac { dz }{ dx } =\log { x } +\cfrac { x }{ x } $$

$$\cfrac { dz }{ dx } =z\left[ 1+\log { x } \right] $$

$$\cfrac { dz }{ dx } ={ x }^{ x }\left[ \log { e } +\log { x } \right] $$

$$\cfrac { dz }{ dx } ={ x }^{ x }\log { ex } $$

Equation $$(i)\Longrightarrow \log { y } =z\log { x } $$
Differentiate both sides with respect to $$x$$

$$\cfrac { 1 }{ y } \cfrac { dy }{ dx } =\log { x\cdot \cfrac { dz }{ dx } } +\cfrac { z }{ x } $$

$$\therefore \cfrac { dy }{ dx } =y\left[ { x }^{ x }\left( \log { ex } \right) \log { x+{ x }^{ x-1 } } \right] $$
Question 2
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If $$f(x)={|x|}^{|\sin{x}|}$$, then $$\displaystyle f^\prime\left(-\frac{\pi}{4}\right)=$$

A
$$\displaystyle {\left(\frac{\pi}{5}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{5}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$$

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

C
$$\displaystyle {\left(\frac{\pi}{3}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{3}{\pi}}-\frac{3\sqrt{3}}{\pi}\right)$$

D
None of these

Solution

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)$$

$$\displaystyle={\left(\frac{\pi}{4}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{4}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$$
Question 3
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If $$x=\sin^{-1}t$$ and $$y=\log(1-{t}^{2})$$ , then $$\displaystyle \left .\frac{{d}^{2}y}{{d}{x}^{2}}\right|_{{t}=\frac{1}{2}} = $$

A
$$-\dfrac{8}{3}$$

B
$$\dfrac{8}{3}$$

C
$$\dfrac{3}{4}$$

D
$$-\dfrac{3}{4}$$

Solution

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

A
$$1$$

B
$$5$$

C
$$8$$

D
$$31$$

Solution

$$(2t^{2}+3)(t^{2}+t+2)$$
$$=2t^{4}+2t^{3}+4t^{2}+3t^{2}+3t+6$$
$$W=2t^{4}+2t^{3}+7t^{2}+3t+6$$
Hence
$$\dfrac{dW}{dt}$$
$$=8t^{3}+6t^{2}+14t+3$$
At $$t=1$$ we get
$$\dfrac{dW}{dt}$$ at $$t=1$$
$$=8+6+14+3$$
$$=31$$.
Question 5
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$$\displaystyle y=\left ( 1+\frac{1}{x} \right )^{x}+x^{1+\frac{1}{x}}.$$
Differentiate

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

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

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

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

Solution

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]$$
Question 6
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Directions For Questions

If $$y=f(x)$$ be a differentiable function of x whose second, third, ..., n$$^{th}$$ order derivatives exist & the n$$^{th}$$ derivative of $$y$$ is denoted by $$y_{n}$$, $$\displaystyle \frac{d^{n}y}{dx^{n}}$$, $$D^{n}y$$, $$y^{n}$$, $$f^{n}\left ( x \right )$$
$$\Rightarrow $$   $$\displaystyle \frac{d^{n}y}{dx^{n}}=\lim_{h\rightarrow 0}\frac{f^{n-1}\left ( x+h \right )-f^{n-1}\left ( x \right )}{h}$$
On the basis of above information answer the following questions.

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If $$x=\sin t$$, $$y=\sin kt$$ then value of $$\left ( 1-x^{2} \right )y_{2}-xy_{1}$$ is

A
$$-k^{2}y$$

B
$$k^{2}y$$

C
$$ky^{2}$$

D
$$-ky^{2}$$

Solution

$$\because $$   $$x=\sin t$$, $$y=\sin kt$$
$$\Rightarrow \dfrac{dx}{dt}=\cos t, \dfrac{dy}{dt}=k\cos kt$$
$$\therefore $$   $$\displaystyle y_{1}=\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{k\cos kt}{\cos t}$$
$$\Rightarrow $$   $$y_{1}\cos t=k\cos kt$$
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
$$\left ( 1-x^{2} \right )\left ( 2y_{1}y_{2} \right )+y{_{1}}^{2}\left ( -2x \right )=k^{2}\left ( -2yy_{1} \right )$$
$$\Rightarrow $$   $$\left ( 1-x^{2} \right )y_{2}-xy_{1}=-k^{2}y$$
Question 7
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If $$f(x)=\left | x \right |+\left | \cos x \right |$$, then

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

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

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

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

Solution

$$f\left( x \right) =\left| x \right| +\left| \cos { x } \right| $$
$$f'\left( x \right) =\cfrac { \left| x \right| }{ x } +\cfrac { \left| \cos { x } \right| }{ \cos { x } } $$
as $$x\rightarrow { \cfrac { \pi }{ 2 } }^{ - }\quad \cfrac { \left| x \right| }{ x } =1$$
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.
Question 8
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Given the parametric equations $$x=f(t),y=g(t).$$ Then $$\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } $$ equals

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

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

C
$$\displaystyle \dfrac { \dfrac { { d }^{ 2 }y }{ d{ t }^{ 2 } } }{ \dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } } $$

D
None of these

Solution

We have $$\displaystyle \dfrac { dy }{ dx } =\dfrac { \dfrac { dy }{ dt } }{ \dfrac { dx }{ dt } } $$
Again differentiating both sides w.r.t x,
we get $$\displaystyle \dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\dfrac { d }{ dt } \left( \dfrac { \dfrac { dy }{ dt } }{ \dfrac { dx }{ dt } } \right) \dfrac { dt }{ dx } $$
$$\displaystyle =\dfrac { \dfrac { { d }^{ 2 }y }{ { dt }^{ 2 } } \left( \dfrac { dx }{ dt } \right) -\left( \dfrac { { d }^{ 2 }x }{ { dt }^{ 2 } } \right) \dfrac { dy }{ dt } }{ { \left( \dfrac { dt }{ dt } \right) }^{ 2 } } \times \dfrac { 1 }{ \dfrac { dx }{ dt } } $$
$$\displaystyle =\dfrac { \dfrac { { d }^{ 2 }y }{ { dt }^{ 2 } } .\left( \dfrac { dx }{ dt } \right) -\left( \dfrac { { d }^{ 2 }x }{ { dt }^{ 2 } } \right) .\left( \dfrac { dy }{ dt } \right) }{ { \left( \dfrac { dx }{ dt } \right) }^{ 3 } } $$
Question 9
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Derivative of $${(x\cos{x})}^x$$ with respect to $$x$$ is

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

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

C
$$\displaystyle {(x\cos{x})}^x\left[(\log{x}+1)+\left\{\log{\sin{x}}+\frac{x}{\cos{x}}.(\cos{x})\right\}\right]$$

D
None of these

Solution

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

$$\displaystyle\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x\log{x})+\frac{d}{dx}(x\log{\cos{x}})$$
$$\displaystyle \frac { dy }{ dx } =y\left[ \left( 1+\log { x } \right) +\left( \log { \cos { x\quad +\displaystyle \frac { x }{ \cos { x } } \displaystyle \frac { d }{ dx } \left( \cos { x } \right) } } \right) \right] $$
$$\therefore \displaystyle\frac{dy}{dx}={(x\cos{x})}^x\left[(\log{x}+1)+\left\{\log{\cos{x}}+\frac{x}{\cos{x}}.(-\sin{x})\right\}\right]$$
Question 10
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If $$x=log(1+t^2)$$ and $$y=t-tan^{-1}t$$. Then, $$\dfrac{dy}{dx}$$ is equal to

A
$$e^x-1$$

B
$$t^2-1$$

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

D
$$e^x-y$$

Solution

Differentiating wrt 't'
$$\dfrac{dx}{dt}=\dfrac{d}{dt}(log (1+t^2))=\dfrac{1}{1+t^2}2t=\dfrac{2t}{1+t^2}$$

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

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

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

$$e^x=1+t^2$$

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

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

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

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Continuity and Differentiability Test - 37

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Continuity and Differentiability Test - 37

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SHARING IS CARING

Question 1

$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x^{\displaystyle x}\right]$$

$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x\right]$$

$$y\left[x^{\displaystyle x}\left(\log{ex}\right)\log{x}+x^{\displaystyle x-1}\right]$$

$$y\left[x^{\displaystyle x}\left(\log_e{x}\right)\log{x}+x^{\displaystyle x-1}\right]$$

Solution

Question 2

$$\displaystyle {\left(\frac{\pi}{5}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{5}{\pi}}-\frac{2\sqrt{2}}{\pi}\right)$$

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

$$\displaystyle {\left(\frac{\pi}{3}\right)}^{\tfrac{1}{\sqrt{2}}}\left(\frac{\sqrt{2}}{2}\log{\frac{3}{\pi}}-\frac{3\sqrt{3}}{\pi}\right)$$

None of these

Solution

Question 3

$$-\dfrac{8}{3}$$

$$\dfrac{8}{3}$$

$$\dfrac{3}{4}$$

$$-\dfrac{3}{4}$$

Solution

Question 4

$$1$$

$$5$$

$$8$$

$$31$$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$-k^{2}y$$

$$k^{2}y$$

$$ky^{2}$$

$$-ky^{2}$$

Solution

Question 7

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

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

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

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

Solution

Question 8

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

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

$$\displaystyle \dfrac { \dfrac { { d }^{ 2 }y }{ d{ t }^{ 2 } } }{ \dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } } $$

None of these

Solution

Question 9

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

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

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

None of these

Solution

Question 10

$$e^x-1$$

$$t^2-1$$

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

$$e^x-y$$

Solution

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