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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 2 - MCQExams.com
CBSE
Class 12 Commerce Maths
Continuity And Differentiability
Quiz 2
Find $$\displaystyle\frac{dy}{dx}$$ if $$x=a(\theta-\sin{\theta})$$ and $$y=a(1-\cos{\theta})$$.
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$$\cot{\left(\displaystyle\frac{\theta}{2}\right)}$$
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$$\tan{\left(\displaystyle\frac{\theta}{2}\right)}$$
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$$-\cos{\left(\displaystyle\frac{\theta}{2}\right)}$$
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None of these
Explanation
We have $$x=a(\theta-\sin{\theta})$$ and $$y=a(1-\cos{\theta})$$
Differentiate w.r.t $$\theta $$
$$\therefore\displaystyle\frac{dx}{d\theta}=a(1-\cos{\theta})$$ and $$\displaystyle\frac{dy}{d\theta}=a\sin{\theta}$$
or $$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{d\theta}}{\displaystyle\frac{dx}{d\theta}}$$
$$=\displaystyle\frac{a\sin{\theta}}{a(1-\cos{\theta})}=\frac{2\sin{\left(\displaystyle\frac{\theta}{2}\right)}\cos{\left(\displaystyle\frac{\theta}{2}\right)}}{2\sin^{2}{\left(\displaystyle\frac{\theta}{2}\right)}}=\cot{\left(\displaystyle\frac{\theta}{2}\right)}$$
The derivative of $$f(\tan x)$$ w.r.t. $$g (\sec x)$$ at $$x=\displaystyle \frac{\pi }{4},$$ where $${f}'(1)=2$$ and $${g}'(\sqrt{2})=4,$$ is
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$$\displaystyle \frac{1}{\sqrt{2}}$$
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$$\sqrt{2}$$
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$$1$$
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none of these
Explanation
$$y = f(\tan x)$$ and $$z = g (\sec x)$$
$$\dfrac{dy}{dx}= f'(\tan x) . \sec^{2}x$$ and $$\dfrac{dz}{dx}= g'(\sec x) . \sec x \tan x$$
$$\therefore \dfrac{dy}{dz}= \dfrac{f'(\tan x).\sec^{2} x}{g'(\sec x).\sec x \tan x}$$
$$\left.\begin{matrix}\dfrac{dy}{dz}\end{matrix}\right|_{x=\tfrac{\pi }{4}}=\dfrac{f'(1).(\sqrt{2})^{2}}{g'(\sqrt{2}).(\sqrt{2})}=\dfrac{1}{\sqrt{2}}$$
If $$r=\sqrt{x^{2}+y^{2}+z^{2}}$$ and $$x=2\sin 3t,y=2\cos 3t,z=8t$$ then $$\displaystyle \frac{dr}{dt}=$$
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$$\displaystyle \frac{32t}{\sqrt{1+16t^{2}}}$$
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$$\displaystyle \frac{16t}{\sqrt{1+16t^{2}}}$$
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$$\displaystyle \frac{t}{\sqrt{1+16t^{2}}}$$
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$$\displaystyle \frac{4t}{\sqrt{1+16t^{2}}}$$
Explanation
Given :
$$x=2\sin 3t,y=2\cos 3t,z=8t$$
$$r=\sqrt{x^{2}+y^{2}+z^{2}}$$
$$=\sqrt{(2\sin 3t)^{2}+(2 \cos 3t)^{2}+(8t)^{2}}$$
$$=\sqrt{4\sin^{2}3t + 4\cos^{2}3t+64t^{2}}$$
$$=\sqrt{4+64t^{2}}$$
$$\implies r^{2}=4+64t^{2}$$
Differentiating above equation we get
$$2r\dfrac{dr}{dt}=2\times 64\times t$$
$$\implies \dfrac{dr}{dt}=\dfrac{64\times t}{r}$$
$$\implies \dfrac{dr}{dt}=\dfrac{64\times t}{\sqrt{4+64t^{2}}}=\dfrac{64t}{2\sqrt{1+16t^{2}}}=\dfrac{32t}{\sqrt{1+16t^{2}}}$$
If $$y^x=x^y$$, then find $$\displaystyle\frac{dy}{dx}$$.
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$$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$$
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$$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$
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$${y(x\log{y}-y)}$$
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$$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$$
Explanation
$$y^x=x^y$$
Taking log on both sides
$$\Rightarrow$$ $$\log{y^x}=\log{x^y}$$
$$\Rightarrow$$ $$x\log{y}=y\log{x}$$
$$\Rightarrow$$ $$\displaystyle x\frac{1}{y}\frac{dy}{dx}+\log{y}\times 1=\frac{y}{x}+\log{x}\frac{dy}{dx}$$
$$\Rightarrow$$ $$\displaystyle\left(\frac{x}{y}-\log{x}\right)\frac{dy}{dx}=\frac{y}{x}-\log{y}$$
$$\Rightarrow$$ $$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{y}{x}-\log{y}}{\displaystyle\frac{x}{y}-\log{x}}=\frac{y(y-x\log{y})}{x(x-y\log{x})}=\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$
If $$\displaystyle x=\frac{2t}{1+t^2}$$, $$\displaystyle y=\frac{1-t^2}{1+t^2}$$, then $$\displaystyle\frac{dy}{dx}$$ at $$t=2$$ is
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$$\displaystyle\frac{4}{3}$$
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$$\displaystyle -\frac{4}{3}$$
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$$\displaystyle\frac{3}{4}$$
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$$\displaystyle -\frac{3}{4}$$
Explanation
Given that
$$\displaystyle x=\frac{2t}{1+t^2}$$, $$\displaystyle y=\frac{1-t^2}{1+t^2}$$
By differentiating w.r. to $$t$$, we get
$$\displaystyle \frac { dx }{ dt } =\frac { \left( \left( 1+{ t }^{ 2 } \right)\displaystyle \frac { d }{ dt } \left( 2t \right) -2t\displaystyle \frac { d }{ dt } \left( 1+{ t }^{ 2 } \right) \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } $$
And $$\displaystyle \frac { dy }{ dt } =\frac { \left( \left( 1+{ t }^{ 2 } \right) \displaystyle \frac { d }{ dt } \left( 1-{ t }^{ 2 } \right) -\left( 1-{ t }^{ 2 } \right) \displaystyle \frac { d }{ dt } \left( 1+{ t }^{ 2 } \right) \right) }{ { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } $$
$$\displaystyle\frac{dx}{dt}=\frac{(1+t^2)2-2t\times 2t}{{(1+t^2)}^2}=\frac{2-2t^2}{{(1+t^2)}^2}$$
$$\displaystyle\frac{dy}{dt}=\frac{(1+t^2)(-2t)-(1-t^2)2t}{{(1+t^2)}^2}=\frac{-4t}{{(1+t^2)}^2}$$
$$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{-4t}{2-2t^2}=\frac{2t}{t^2-1}$$
$$\therefore { \left( \displaystyle \frac { dy }{ dx } \right) }_{ t=2 }=\displaystyle \frac { 4 }{ 3 } $$
$$\displaystyle\frac{dy}{dx}$$ for $$y=x^x$$ is
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$$x^x(1-\log{x})$$
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$$x^x(1-\log{y})$$
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$$x^x(1+\log{y})$$
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$$x^x(1+\log{x})$$
Explanation
$$y=x^x$$
Taking $$\log $$ on both the sides
$$\log { y } =\log { \left( { x }^{ x } \right) } $$
$$\log { y } =x\log { x } $$
Differentiate w.r to $$x$$
$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =x\frac { d }{ dx } \left( \log { x } \right) +\left( \log { x } \right) \frac { d }{ dx } \left( x \right) $$
$$\displaystyle \frac { dy }{ dx } =y\left[ \frac { x }{ x } +\log { x } \right] $$
$$\displaystyle \frac { dy }{ dx } ={ x }^{ x }\left[ 1+\log { x } \right] $$
Let the function $$y=f(x)$$ be given by $$x=t^{5}-5t^{3}-20t+7$$ and $$y=4t^{3}-3t^{2}-18t+3$$, where $$t\epsilon \left ( -2, 2 \right )$$. Then $$f^{'}(x)$$ at $$t=1$$ is ?
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$$\displaystyle \frac{5}{2}$$
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$$\displaystyle \frac{2}{5}$$
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$$\displaystyle \frac{7}{5}$$
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none of these
Explanation
Given, $$x=t^{5}-5t^{3}-20t+7$$
$$\displaystyle \frac{dx}{dt}=5t^{4}-15t^{2}-20$$
$$\displaystyle \left (\frac{dx}{dt}\right)_{t=1}=-30$$
Also, given $$y=4t^{3}-3t^{2}-18t+3$$
$$\displaystyle \frac{dy}{dt}=12t^{2}-6t-18$$
$$\displaystyle \left (\frac{dy}{dt}\right)_{t=1}=-12$$
So, $$\displaystyle \left (\frac{dy}{dx}\right)_{t=1}=\frac{2}{5}$$
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect
Explanation
As $$\displaystyle f\left ( x \right )= g\left ( x \right ).\sin x.$$
$$ f'\left( x \right)=g\left( x \right)\cos { x } +g'\left( x \right)\sin { x } $$
Putting $$x=0$$
$$ f'\left( 0 \right)=g\left( 0 \right)=0 $$
Now, $$ \displaystyle f'\left( 0 \right)=\lim _{ x\rightarrow 0 }{ \dfrac { f'\left( x \right)-f'\left( 0 \right) }{ x } } $$
$$ \displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } +g'\left( x \right)\sin { x } -g\left( 0 \right) }{ x } } $$
$$ \displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ x } } +\lim _{ x\rightarrow 0 }{ \dfrac { g'\left( x \right)\sin { x } }{ x } } $$
$$ \displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ x } } $$
$$ \displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { g\left( x \right)\cos { x } -g\left( 0 \right) }{ \sin { x } } } $$
$$ =\lim _{ x\rightarrow 0 }{ \left( g\left( x \right)\cot { x-g\left( 0 \right)\csc { x } } \right) } $$
$$\displaystyle\frac{dy}{dx}$$ at $$\displaystyle t=\frac{\pi}{4}$$ for $$\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]$$ and $$y=a\sin{t}$$ is
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$$1$$
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$$0$$
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$$-1$$
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$$ \displaystyle \frac { 1 }{ 2 }$$
Explanation
$$\displaystyle x=\frac{\pi}{4}$$ for $$\displaystyle
x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]$$ and
$$y=a\sin{t}$$
Differentiating w.r.t $$t$$, we get
$$\displaystyle \frac { dx }{ dt } =a\left[ \frac { d }{ dt } \cos { t } +\frac { 1 }{ 2 } \frac { d }{ dt } \left( \log { \tan ^{ 2 }{ \frac { t }{ 2 } } } \right) \right] $$
$$\displaystyle\frac{dx}{dt}=a\left[-\sin{t}+\frac{1}{\displaystyle\tan{\frac{t}{2}}}\sec^2{\frac{t}{2}}\times\frac{1}{2}\right]$$
$$\displaystyle \frac { dx }{ dt } =a\left[ -\sin { t } +\frac { 1 }{ 2\displaystyle \frac { \sin { { t }/{ 2 } } }{ \cos { { t }/{ 2 } } } .\cos ^{ 2 }{ { t }/{ 2 } } } \right]$$
$$\displaystyle =a\left[-\sin{t}+\frac{1}{\displaystyle 2\sin{\frac{t}{2}}\cos{\frac{t}{2}}}\right]$$
$$\displaystyle =a\left[-\sin{t}+\frac{1}{\sin{t}}\right]$$
$$=a\left[ \displaystyle \frac { 1-\sin ^{ 2 }{ t } }{ \sin { t } } \right] =a\left[ \displaystyle \frac { \cos ^{ 2 }{ t } }{ \sin { t } } \right] $$
$$\displaystyle\frac{dy}{dt}=a\cos{t}$$
$$\displaystyle\therefore\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{a\cos{t}}{\displaystyle\frac{a\cos^2{t}}{\sin{t}}}=\tan{t}$$
At $$\displaystyle t=\frac{\pi}{4}$$, $$\displaystyle\frac{dy}{dx}=1$$
The relation between the parameter '$$t$$' and the angle $$\alpha$$ between the tangent to the given curve and the $$x-$$axis is given by, '$$t$$' equals to
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$$\displaystyle \dfrac {\pi}{2}-\alpha$$
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$$\displaystyle \dfrac {\pi}{4}+\alpha$$
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$$\alpha-\displaystyle \dfrac {\pi}{4}$$
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$$\displaystyle \dfrac {\pi}{4}-\alpha$$
Explanation
$$x=e^t \cos t$$ and $$y=e^t \sin t$$
$$\displaystyle \dfrac { dy }{ dt } =\displaystyle \dfrac { d }{ dt } \left( { e }^{ t }\sin { t } \right) $$
$$\displaystyle \dfrac { dy }{ dt } ={ e }^{ t }\left( \cos { t } +\sin { t } \right) $$
$$\displaystyle \dfrac { dx }{ dt } =\displaystyle \dfrac { d }{ dt } \left( { e }^{ t }\cos { t } \right) $$
$$\displaystyle \dfrac { dx }{ dt } ={ e }^{ t }\left( \cos { t } -\sin { t } \right) $$
$$\therefore \displaystyle \dfrac { dy }{ dx } =\displaystyle \dfrac { \cos { t } +\sin { t } }{ \cos { t } -\sin { t } } $$
$$\therefore \tan { \left( \displaystyle \dfrac { \pi }{ 4 } +t \right) =\tan { \alpha } } $$
$$\displaystyle \dfrac { \pi }{ 4 } +t=\alpha $$
$$t=\alpha -\displaystyle \dfrac { \pi }{ 4 } $$
If $$y={(\tan{x})}^{\displaystyle{(\tan{x})}^{\displaystyle\tan{x}}}$$, then find $$\displaystyle\frac{dy}{dx}$$ at $$\displaystyle x=\frac{\pi}{4}$$.
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$$0$$
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$$1$$
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$$-1$$
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$$2$$
Explanation
Taking $$\log$$ on both sides, we get
$$\log{y}={(\tan{x})}^{\displaystyle\tan{x}}\log{\tan{x}}$$
Again taking $$\log$$ on both the sides
$$\log{\log{y}}=[\tan{x}\log{\tan{x}}]+\log{\log{\tan{x}}}$$
Differentiating w.r.t. $$x$$, we get
$$\displaystyle\frac{1}{y\log{y}}\frac{dy}{dx}=\log { \tan { x } \frac { d }{ dx } \left( \tan { x } \right) +\tan { x } \frac { d }{ dx } \left( \log { \tan { x } } \right) +\frac { d }{ dx } \left( \log { \log { \tan { x } } } \right) } $$
$$\displaystyle \frac{dy}{dx}=\log { \tan { x } .\sec ^{ 2 }{ x } +\tan { x } .\frac { \sec ^{ 2 }{ x } }{ \tan { x } } +\frac { \sec ^{ 2 }{ x } }{ \tan { x } .\log { \tan { x } } } } $$
$$\displaystyle \frac { dy }{ dx } =y\log { y.\sec ^{ 2 }{ x } \left[ \log { \tan { x } +1+\frac { 1 }{ \tan { x } .\log { \tan { x } } } } \right] } $$
At $$\displaystyle x=\frac{\pi}{4}$$, $$y=1$$ and $$\log{y}=0$$
So, putting this values in the equation of $$\displaystyle \frac { dy }{ dx } $$, we get
$$\displaystyle\therefore{\left(\frac{dy}{dx}\right)}_{\displaystyle x=\frac{\pi}{4}}=0$$
If $$x=a\sec^{3}{\theta}$$ and $$y=a\tan^{3}{\theta}$$, then $$\displaystyle\frac{dy}{dx}$$ at $$\theta=\displaystyle\frac{\pi}{3}$$ is
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$$\displaystyle\frac{\sqrt{3}}{2}$$
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$$\displaystyle\frac{\sqrt{5}}{2}$$
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$$\displaystyle\frac{\sqrt{2}}{3}$$
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$$\displaystyle\frac{\sqrt{4}}{3}$$
Explanation
We have $$x=a\sec^{3}{\theta}$$ and $$y=a\tan^{3}{\theta}$$
Differentiate w. r to $$\theta $$
$$\displaystyle\frac{dx}{d\theta}=3a\sec^{2}{\theta}\displaystyle\frac{d}{d\theta}(\sec{\theta})=3a\sec^{3}{\theta}\tan{\theta}$$
$$\displaystyle\frac{dy}{d\theta}=3a\tan^{2}{\theta}\displaystyle\frac{d}{d\theta}(\tan{\theta})=3a\tan^{2}{\theta}\sec^{2}{\theta}$$
$$\therefore \displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{d\theta}}{\displaystyle\frac{dx}{d\theta}}=\frac{3a\tan^{2}{\theta}\sec^{2}{\theta}}{3a\sec^{3}{\theta}\tan{\theta}}=\frac{\tan{\theta}}{\sec{\theta}}=\sin{\theta}$$
$$\displaystyle{\left(\frac{dy}{dx}\right)}_{\displaystyle\theta=\frac{\pi}{3}}=\sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2}$$
If $$\displaystyle y^{x}=x^{\sin y} $$, find $$\cfrac{dy}{dx}$$.
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$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$
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$$\displaystyle \frac{y}{x}\left [ \frac{x \log y+\sin y}{y \:\log x\: \cos y+x} \right ]$$
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$$\displaystyle \frac{-y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y-x} \right ]$$
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$$\displaystyle \frac{y}{x}\left [ \frac{x \log y-\sin y}{y \:\log x\: \cos y+x} \right ]$$
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
$$\displaystyle \log y+x.\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \sin y+ (\log x)\cos y .\frac{dy}{dx}$$
$$\displaystyle \left ( \log y - \frac{\sin y}{x} \right )=\frac{dy}{dx}\left [ \cos y \log x-\frac{x}{y} \right ]$$
$$\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}\left [ \frac{x \log y-\sin y}{y \log x \cos y-x}\right ]$$
Discuss the applicability of Rolle's theorem to $$\displaystyle f(x)=\log \left[\frac{x^{2}+ab}{(a+b)x}\right],$$ in the interval$$ [a,b].$$
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Yes Rolle's theorem is applicable and the stationary point is $$x=\sqrt { ab } $$
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No Rolle's theorem is not applicable due to the discontinuity in the given interval
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Yes Rolle's theorem is applicable and the stationary point is $$x= ab $$
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none of these
Explanation
We have $$\displaystyle f(a)=\log \left [
\frac{a^{2}+ab}{(a+b)a} \right ]=\log
1=0$$ and $$\displaystyle f(b)=\log \left [
\frac{b^{2}+ab}{(a+b)b} \right ]=\log1=0$$
$$\Rightarrow f(a)=f(b)=0.$$ Also, it
can be easily seen that $$f(x)$$ is continuous on $$[a,b]$$ and differentiable on
$$[a,b]$$.
Thus all the three conditions of Rolle's
theorem are satisfied. Hence $$\displaystyle
f^{'}(x)=0 $$for at last one value of x in $$[a,b]$$
Now $$\displaystyle f^{'}(x)=0 =
\frac{2x}{x^{2}+ab}-\frac{1}{x}=0\Rightarrow \displaystyle
2x^{2}-(x^{2}+ab)=0=x^{2}=ab$$ or $$\displaystyle
x=\sqrt{ab}$$ which is also known as stationary point.
Verify the Rolle's theorem for the function $$\displaystyle f(x)=x^{2}-3x+2$$ on the interval[1,2]
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No Rolle's theorem is not applicable in the given interval
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Yes Rolle's theorem is applicable in the given interval and the stationary point $$x=\frac { 5 }{ 4 } $$
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Yes Rolle's theorem is applicable in the given interval and the stationary point $$x=\frac { 3 }{ 2 } $$
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nnone of these
Explanation
It can be easily seen that $$f(x)=x^{2}-3x+2$$ is continuous as differentiable on R (being a polynomial) $$\Rightarrow f(x)$$ is continous in (1,2) and differentiable in [1,2]. Also, we have
$$f(1)=f(2)=0$$.
Thus, $$f(x)$$ satisfies all the conditions of Rolle's theorem in $$[1,2]$$ $$\Rightarrow \displaystyle \exists$$ at least one number, say $$x$$ in $$[1,2]$$ such that $$\displaystyle
f^{'}(c)=0.$$ Now, $$\displaystyle f^{'}(x)=2x-3=0\Rightarrow x=\frac{3}{2}$$ Since, the root (stationary point) $$\displaystyle x=\frac{3}{2}$$ lies in the interval(1,2).
Hence Rolle's theorem is verified.
Differentiate $$\displaystyle x^{\sin^{-1}x}$$ w.r.t. $$\displaystyle \sin ^{-1}x.$$
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$$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$
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$$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$
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$$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$$
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$$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$$
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 $$z$$
$$\displaystyle \frac{1}{y}\frac{dy}{dz}=\log\sin z+z\frac{\cos z}{\sin z}$$
$$\therefore \displaystyle \frac{dy}{dz}=y\left(\log\sin z+z\frac{\cos z}{\sin z}\right)=\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$
$$\displaystyle y=(\cot x)^{\sin x}+(\tan x)^{\cos x}$$.Find dy/dx
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$$\sin x(\cot x)^{ \sin x-1 }(-cosec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(-sinx) } $$
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$$\sin x(\cot x)^{ \sin x-1 }(-co\sec ^{ 2 } x)+\cos { x } { (\tan { x } ) }^{ \cos { x-1 } }\sec ^{ 2 }{ x } $$
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$$\sin x(\cot x)^{ \sin x-1 }(-sec ^{ 2 } x)+(\cot x)^{ \sin x }(log\cot x)\cos x+\\\cos { x } { (\tan { x } ) }^{ \cos { x+1 } }co\sec ^{ 2 }{ x } +{ (\tan { x } ) }^{ cosx }(\log { tanx)(sinx) } $$
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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, $$y=u+v$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}$$
$$\Rightarrow \dfrac{dy}{dx}=\sin x(\cot x)^{\sin x -1}(-cosec^2 x)+(\cot x)^{\sin x} \cos x \log \cot x+$$
$$ \cos x(\tan x)^{\cos x -1}(\sec^2 x)+(\tan x)^{\cos x}(-\sin x) \log \tan x$$
Verify Rolle's theorem for $$\displaystyle f(x)=x(x+3)e^{-x/2}$$ in $$(-3,0)$$
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Yes Rolle's theorem is applicable and the stationary point is $$x=-2$$
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Yes Rolle's theorem is applicable and the stationary point is $$x=-1$$
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No Rolle's theorem is not applicable in the given interval
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Both A and B
Explanation
We have $$\displaystyle f(x)=x(x+3)e^{-x/2}$$
$$\displaystyle \therefore f^{'}(x)=(2x+3)e^{-x/2}+(x^{2}+3x)e^{-x/2}\left ( -\frac{1}{2} \right )$$
$$\displaystyle =e^{-x/2}\left [ 2x+3-\frac{1}{2}[x^{2}+3x] \right ]=-\frac{1}{2}[x^{2}-x-6]e^{-x/2}$$
$$f^{'}(x)$$ exist for every value of x in the interval[-3,0]
Hence, $$f(x)$$ is differentiable and hence, continous in the interval $$[-3,0]$$
Also, we have $$\displaystyle f(-3)=f(0)=0\Rightarrow $$All the three conditions of Rolle's theorem are satisfied.
So $$\displaystyle f^{'}(x)=0\Rightarrow \frac{1}{2}(x^{2}-x-6)e^{-x/6}=0\Rightarrow x^{2}-x-6=0 \Rightarrow x=3,-2$$
Since, the value $$x=-2$$ lies in the open interval $$[-3,0]$$ the Rolle's theorem is verified.
Verify Rolle's theorem the function $$\displaystyle f(x)=x^{3}-4x $$ on $$ [-2,2].$$ If you think it is applicable in the given interval then find the stationary point ?
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Yes Rolle's theorem is applicable and stationary point is $$x=\pm \dfrac{2}{\sqrt{3}}$$
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No Rolle's theorem is not applicable
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Yes Rolle's theorem is applicable and $$x=2\ or\ -2 $$
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none of these
Explanation
The function $$\displaystyle f(x)=x^{3}-4x$$ is a polynomial and so it is continuous and differentiable at all x$$\displaystyle \epsilon $$ R.
In particular it is continuous in the closed interval $$[-2,2]$$ .
Also $$f(-2)]=0=f(2)$$. Thus , $$f(x)$$ satisfies all three conditions of Rolle's theorem in $$(-2,2)$$.
Therefore , there must exist at least one real number $$'x'$$ in the
open interval $$(-2,2)$$ for which $$f'(x)=0$$
Also $$\displaystyle f'(x)= 3x^{2}-4$$
Now $$\displaystyle f^{'}(x)=0$$ gives $$3x^{2}-4=0$$ or $$x=\pm \dfrac{2}{\sqrt{3}}$$ which is also known as stationary point.
Both these value lie in the open interval $$(-2,2)$$ and thus the conclusion of Rolle's theorem is verified.
Verify the Rolle's theorem for the function $$\displaystyle f(x)=x^{2}$$ in $$(-1,1)$$
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Yes Rolle's theorem is applicable and the stationary point is $$x=\frac { 1 }{ 2 } $$
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Yes Rolle's theorem is applicable and the stationary point is $$x=0 $$
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No Rolle's theorem is not applicable in the given interval
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Both A and B
Explanation
$$f\left( x \right) ={ x }^{ 2 }$$
$${ x }^{ 2 }$$ is continuous in $$[-1,1]$$ since it is quadratic equation
$$f'\left( x \right) =2x$$
$$2x$$ is defined in $$(-1,1)$$
$$\Rightarrow f\left( x \right) $$ is differentiable on $$(-1,1)$$
$$f\left( -1 \right) =f\left( 1 \right) =1$$
There exists a $$c,a\le c\le b$$ such that
$$f'\left( c \right) =0$$
$$2c=0$$
$$\Rightarrow c=0$$ which lies in $$(-1,1)$$
Verify Rolle's theorem for the function $$\displaystyle f(x)=10x-x^{2}$$ in the interval [0,10]
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Yes Rolle's theorem is applicable and the stationary point is $$x=5$$
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Yes Rolle's theorem is applicable and the stationary point is $$x=4$$
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No Rolle's theorem is not applicable in the given interval
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none of these
Explanation
$$f\left( x \right) =10x-{ x }^{ 2 }\quad \quad \left[ 0,10 \right] $$
$$10x-{ x }^{ 2 }$$ is continuous in $$\left[ 0,10 \right] $$ since it is polynomial function.
$$f'\left( x \right) =10-2x$$ is defined for all values of x in $$(0,10)$$
$$\Rightarrow f\left( x \right) $$ is differentiable on $$(0,10)$$
$$f\left( 0 \right) =f\left( 10 \right) =0$$
$$\therefore $$ There exists a $$c,a\le c\le b$$
$$0\le c\le 10$$
Such that
$$f'\left( c \right) =0$$
$$10-2x=0$$
$$x=5$$ $$5$$ lies in $$\left[ 0,10 \right] $$
If $$y = \displaystyle (\tan x)^{\log x}$$, then $$\cfrac{dy}{dx} = $$
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$$(\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$$
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$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx$$
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$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } $$
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none of these
Explanation
Let $$y = \displaystyle (\tan x)^{\log x}$$
$$\log y = \log x .\log \tan x$$
Differentiating both side w.r.t $$x$$
$$\cfrac{1}{y}.\cfrac{dy}{dx} = \cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x)$$
$$\Rightarrow \cfrac{dy}{dx} = (\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$$
Verify Lagrange's mean value for the function $$f(x)=\displaystyle \sin x$$ in $$\displaystyle \left [ 0,\frac{\pi }{2} \right ]$$
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No Lagrange's theorem is not applicable
0%
Yes Lagrange's theorem is applicable and $$\displaystyle c =\cos ^{-1}\left ( \frac{2}{\pi } \right )$$
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Yes Lagrange's theorem is applicable and $$\displaystyle c =\sin ^{-1}\left ( \frac{4}{\pi } \right )$$
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none of these
Explanation
The function $$f\left( x \right)=\sin { x } $$ is continuous and differentiable for all $$x\in R$$.
In particular it is continuous in the closed interval $$\displaystyle \left[ 0,\frac { \pi }{ 2 } \right] $$ and differentiable in the open interval $$\displaystyle \left( 0,\frac { \pi }{ 2 } \right) $$ as is required for the application of lagrange's mean value theorem for it.
By lagrange's mean value theorem, there must exist at least one value of $$'c'$$ of $$x$$ lying in the open interval $$\displaystyle \left( 0,\frac { \pi }{ 2 } \right) $$ such that
$$\displaystyle \frac { f\left( \frac { \pi }{ 2 } \right) -f\left( 0 \right) }{ \left( \frac { \pi }{ 2 } \right) -0 } =f'\left( c \right) $$ ...(1)
Let us verify it, we have
$$\displaystyle f\left( 0 \right) =\sin { \left( 0 \right) } =0;f\left( \frac { \pi }{ 2 } \right) =\sin { \left( \frac { \pi }{ 2 } \right) } =1$$
Also $$f'\left( x \right) =\cos { x } $$ gives $$f'\left( c \right) =\cos { c } $$.
Putting these values in (1), we have
$$\displaystyle \frac { 1-0 }{ \frac { \pi }{ 2 } } =\cos { \left( c \right) \Rightarrow \cos { \left( c \right) } } =\frac { 2 }{ \pi } \Rightarrow c=\cos ^{ -1 }{ \left( \frac { 2 }{ \pi } \right) } $$
Since $$\displaystyle 0<\frac { 2 }{ \pi } <1,$$ therefore the principal value of $$\displaystyle c=\cos ^{ -1 }{ \left( \frac { 2 }{ \pi } \right) } $$ lies in the open interval $$\displaystyle \left( 0,\frac { \pi }{ 2 } \right) $$ and so is the required value of $$c$$
This verifies lagrange's mean value theorem
Find 'c' of the mean value theorem, if $$\displaystyle f(x)=x(x-1)(x-2);a=0, b=1/2$$
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$$\displaystyle c=\frac{1+\sqrt{21}}{6}$$
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$$\displaystyle c=\frac{1-\sqrt{21}}{6}$$
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$$\displaystyle c=\frac{1+\sqrt{6}}{3}$$
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$$\displaystyle c=\frac{1-\sqrt{6}}{3}$$
Explanation
We have $$f\left( a \right) =f\left( 0 \right) =0$$ and $$\displaystyle f\left( b \right) =f\left( \frac { 1 }{ 2 } \right) =\frac { 3 }{ 8 } $$
$$\displaystyle \therefore \frac { f\left( b \right) -f\left( a \right) }{ b-a } =\frac { \frac { 3 }{ 8 } -0 }{ \frac { 1 }{ 2 } -0 } =\frac { 3 }{ 4 } $$
Now
$$f\left( x \right) ={ x }^{ 3 }-3{ x }^{ 2 }+2x\\ \Rightarrow f'\left( x \right) ={ 3x }^{ 2 }-6x+2\\ \Rightarrow f'\left( c \right) ={ 3c }^{ 2 }-6c+2$$
Putting all these value in lagrange's mean value theorem
$$\displaystyle \frac { f\left( b \right) -f\left( a \right) }{ b-a } =f'\left( c \right) ,\left( a<c,b \right) $$
We get $$\displaystyle \frac { 3 }{ 4 } ={ 3c }^{ 2 }-6c+2\Rightarrow c=1\pm \frac { \sqrt { 21 } }{ 6 } $$
Hence $$\displaystyle c=\frac { 1-\sqrt { 21 } }{ 6 } $$ lies in the open interval $$\displaystyle \left( 0,\frac { 1 }{ 2 } \right) $$
Therefore it is the required value
Discuss the applicapibility of Rolle's Theorem to the function $$\displaystyle f(x)=x^{2/3}$$ in (-1,1).
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Yes Rolle's theorem is applicable and the stationary point $$x=0$$
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Yes Rolle's theorem is applicable and the stationary point $$x=\frac { 1 }{ 2 } $$
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No Rolle's theorem is not applicable in the given interval
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none of these
Explanation
$$\displaystyle f\left( x \right)={ x }^{ \frac { 2 }{ 3 } }\Rightarrow f'\left( x \right) =\frac { 2 }{ x } { x }^{ -\frac { 1 }{ 3 } }$$
$$\displaystyle\therefore \lim _{ x\rightarrow 0 }{ f'\left( x \right) } =\lim _{ x\rightarrow 0 }{ \frac { 2 }{ 3 } { \left( 0+h \right) }^{ -\frac { 1 }{ 3 } } } =\infty$$
$$\displaystyle Rf'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } \right\} } =\lim _{ h\rightarrow 0 }{ \left\{ \frac { { h }^{ \frac { 2 }{ 3 } }-1 }{ h } \right\} } =+\infty$$
$$\displaystyle Lf'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ -h } \right\} } =f'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { { \left( -h \right) }^{ \frac { 2 }{ 3 } } }{ h } \right\} } =-\infty$$
$$\displaystyle \therefore Lf'\left( 0 \right) \neq Rf'\left( 0 \right)$$
$$\displaystyle\therefore f'\left( 0 \right)$$ does not exist showing that $$f'(x)$$ does not exists in the open interval $$(-1,1)$$
Hence, Rolle's Theorem is not applicable
although $$f(-1)=f(1)=1$$ and $$f(x)$$ is continuous in the closed interval $$(-1,1)$$
If $$\displaystyle x=2 \cos t-cos 2t, y=2\sin t-\sin 2t,$$ find the value of $$dy/dx$$.
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$$ \tan\displaystyle \frac{3t}{2}$$
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$$ \tan\displaystyle \frac{-3t}{2}$$
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$$ \tan\displaystyle \frac{3t}{4}$$
0%
$$ \tan\displaystyle \frac{-3t}{4}$$
Explanation
$$\dfrac{dx}{dt}=\dfrac{d}{dt}\left(2\cos \left(t\right)-\cos \left(2t\right)\right)$$
$$=\dfrac{d}{dt}\left(2\cos \left(t\right)\right)-\dfrac{d}{dt}\left(\cos \left(2t\right)\right)$$
$$=-2\sin \left(t\right)-\left(-2\sin \left(2t\right)\right)$$
$$=-2\sin \left(t\right)+2\sin \left(2t\right)$$
$$\dfrac{dy}{dt}=\dfrac{d}{dt}\left(2\sin \left(t\right)-\sin \left(2t\right)\right)$$
$$=\dfrac{d}{dt}\left(2\sin \left(t\right)\right)-\frac{d}{dt}\left(\sin \left(2t\right)\right)$$
$$=2\cos \left(t\right)-\cos \left(2t\right)\cdot \:2$$
$$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx}=\dfrac{cos(t)-cos(2t)}{-sin(t)+sin(2t)}=\dfrac{2sin(\dfrac{3t}{2})sint}{2cos(\dfrac{3t}{2})sin(t)}=tan(\dfrac{3t}{2})$$
The function $$f$$ is defined, where $$x=2t-\left| t \right|, t\in R$$. Draw the graph of $$f$$ for the interval $$-1\le x\le 1$$. Also discuss its continuity and differentiability at $$x=0$$
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continuous and differentiable at $$x=0$$
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not continuous but differentiable at $$x=0$$
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neither continuous nor differentiable at $$x=0$$
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differentiable but not continuous at $$x=0$$
Explanation
For $$t\ge 0$$
$$x=2t-t=t$$
$$y={t}^{2}+{t}^{2}=2{t}^{2}$$
For $$t<0$$
$$x=3t, y=0$$
$$0\le x< 1 \Rightarrow 0\le t<1$$ [$$\therefore x=t$$]
$$-1\le x<0 \Rightarrow \cfrac{-1}{3}\le t<0$$ [$$\therefore x=3t$$]
$$f$$ is continuous and differentiable at $$x=0$$
The function $$f(x) = \displaystyle \sin ^{-1}(\cos x)is :-$$
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discontinuous at $$x = 0$$
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continuous at $$x = 0$$
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differentiable at $$x = 0$$
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none of these
Explanation
$$f(0-) = \displaystyle \lim_{h\to 0}\sin ^{-1}(\cos -h)=\lim_{h\to 0}\sin ^{-1}(\cos h) = \sin^{-1}1$$, Since $$\cos(-x) = \cos x$$
and $$f(0+) = \displaystyle \lim_{h\to 0}\sin ^{-1}(\cos h) = \sin^{-1}1$$
Clearly at $$x=0$$, L.H.L $$=$$ R.H.L $$=f(0) \Rightarrow f(x)$$ is continuous at $$x =0$$
Also $$\displaystyle {y}'=-\frac{\sin x}{\sqrt{1-\cos ^{2}x}}=-\frac{\sin
x}{\sqrt{\sin ^{2}x}}=-\frac{\sin x}{\left | \sin x \right |}$$
So the function is not differentiable at the points where $$\sin x=0$$,
that is, for $$x=k\pi \left ( k\in I \right )$$. In particular, $$x=0$$.
Examine the origin for continuity and derivability in case of the function $$f$$ defined by $$f(x)=x\tan ^{ -1 }{ (1/x) } $$, $$x\ne 0$$ and $$f(0)=0$$.
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continuous but not differentiable at $$x=0$$
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continuous and differentiable at $$x=0$$
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not continuous and not differentiable at $$x=0$$
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none of these
Explanation
$$f(0^+) =\displaystyle \lim_{h \rightarrow 0} (htan^{-1}(\cfrac{1}{h})) = 0 \times \cfrac{\pi}{2} = 0$$
$$f(0^-) = \displaystyle \lim_{h \rightarrow 0 }(-h tan^{-1}(\cfrac{1}{-h})) = 0 \times \cfrac{- \pi}{2} = 0$$
Hence the function is continuous at $$x = 0.$$
$$f'(0^+) = \displaystyle \lim_{h \rightarrow 0} \cfrac{htan^{-1}(\cfrac{1}{h})}{h} = \cfrac{\pi}{2}$$
$$f(0^+) = \displaystyle \lim_{h \rightarrow 0} (- h tan^{-1}(\cfrac{1}{-h})) = \cfrac{-\pi}{2}$$
Hence the function is not differentiable at $$x = 0.$$
State true or false:
If $$x=t^2+3t-8, y=2t^2-2t-5$$, then $$\dfrac {dy}{dx}$$ at $$(2, -1)$$ is $$\dfrac {6}{7}$$.
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True
0%
False
Explanation
$$\displaystyle x={ t }^{ 2 }+3t-8\Rightarrow \frac { dx }{ dt } =2t+3$$
$$\displaystyle y={ 2t }^{ 2 }-2t-5\Rightarrow \frac { dy }{ dt } =4t-2$$
$$\displaystyle \therefore \frac { dy }{ dx } =\frac { 4t-2 }{ 2t+3 } $$
At $$\left( 2,-1 \right) $$
$$2={ t }^{ 2 }+3t-8\Rightarrow { t }^{ 2 }+3t-10=0\\ \Rightarrow { t }^{ 2 }+5t-2t-10=0\\ \Rightarrow \left( t+5 \right) \left( t-2 \right) =0\Rightarrow t=-5,t=2$$
and
$$-1={ 2t }^{ 2 }-2t-5\Rightarrow { 2t }^{ 2 }-2t-4=0\\ \Rightarrow { t }^{ 2 }-t-2=0\Rightarrow { t }^{ 2 }-2t+t-2=0\\ \Rightarrow t=2,t=-1$$
$$\therefore t=2$$
Hence $$\displaystyle \frac { dy }{ dx } =\frac { 4.2-2 }{ 2.2+3 } =\frac { 6 }{ 7 } $$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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