Continuity and Differentiability Test

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Continuity and Differentiability Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \frac{d}{dx}\left ( x^{\log x} \right )$$ is equal to

A
$$\displaystyle 2x^{\log x-1}\log x$$

B
$$\displaystyle x^{\log x-1}$$

C
$$\dfrac 23 (\log x)$$

D
$$\displaystyle x^{\log x-1}.\log x$$

Solution

$$\displaystyle \dfrac { d }{ dx } \left( x^{ \log x } \right) =\dfrac { d }{ dx } \left( { e }^{ \log { x } \log { x } } \right) $$
$$\displaystyle =\dfrac { d }{ dx } \left( { \left( \log { x } \right) }^{ 2 } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$$
$$\displaystyle =2\log { x } \dfrac { d }{ dx } \left( \log { x } \right) { e }^{ { \left( \log { x } \right) }^{ 2 } }$$
$$\displaystyle ={ e }^{ { \left( \log { x } \right) }^{ 2 } }2\log { x } \left( \dfrac { d }{ dx } \left( \log { x } \right) \right) $$
$$\displaystyle =\dfrac { 1 }{ x } 2\log { x } { e }^{ { \left( \log { x } \right) }^{ 2 } }=2x^{ \log { x } -1 }\log { x } $$
Question 2
1 / -0

If $$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$$, then $$y'(0)$$ is equal to

A
$$-1$$

B
$$1$$

C
$$2$$

D
Non existant

Solution

$$\displaystyle y= \frac {\sqrt[3]{1+3x}\sqrt[4]{1+4x}\sqrt[5]{1+5x}}{\sqrt[7]{1+7x}\sqrt[8]{1+8x}}$$
Take log both sides,
$$\displaystyle \log y = \frac{1}{3}\log(1+3x)+\frac{1}{4}\log(1+4x)+\frac{1}{5}\log(1+5x)-\frac{1}{7}\log(1+7x)-\frac{1}{8}\log(1+8x)$$
Differentiating both side w.r.t $$x$$
$$\displaystyle

\frac {1}{y}\times \frac {dy}{dx}=\frac {1}{(1+3x)}+\frac

{1}{1+4x}+\frac {1}{1+5x}-\frac {1}{1+7x}-\frac {1}{1+8x}$$
at $$x=0, y=1$$,
$$\therefore \displaystyle \frac {dy}{dx}=1+1+1-1-1=1$$
Question 3
1 / -0

Find $$ \displaystyle \frac{dy}{dx}$$ if $$ y=x^{x} $$

A
$$ \displaystyle x^{x}\left ( lnx+1 \right ) $$

B
$$ \displaystyle x^{x}\left ( lnx-1 \right ) $$

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

D
$$ \displaystyle x^{x-1}\left ( lnx+1 \right ) $$

Solution

$$y = x^x$$
Take log both sides,
$$\log y = x\log x$$
Now differentiate both side w.r.t $$x$$
$$\Rightarrow \cfrac{1}{y}\cfrac{dy}{dx} = 1+\log x$$
$$\Rightarrow \cfrac{dy}{dx} =x^x(1+\log x)$$
Question 4
1 / -0

Find $$ \displaystyle \frac { dy }{ dx }$$ at $$t =\displaystyle \frac { \pi }{ 4 }$$ if $$y =\displaystyle \cos ^{ 4 }{ t } $$ & $$x =\displaystyle \sin ^{ 4 }{ t } $$.

A
1

B
0

C
-1

D
4

Solution

$$\cfrac{dx}{dt} = 4\sin^3t .\cos t$$ and $$\cfrac{dy}{dt} =-4\cos^3t. \sin t$$
$$\Rightarrow \cfrac{dy}{dx} = -\cfrac{\cos^2t}{\sin^2t}$$
Thus at $$t = \cfrac{\pi}{4}$$, $$\cfrac{dy}{dx} = -1$$
Question 5
1 / -0

Derivative of sin x w.r.t. cos x is

A
cos x

B
cot x

C
- cot x

D
tan x

Solution

Let $$ cos\,x = t $$
than, $$ dt = -sinx\,dx...(1) $$
let $$ y = sinx $$
$$ dy = cosx\,dx...(2) $$
then (2) $$\div$$ (1)
$$ \dfrac{dy}{dx} = \dfrac{cosx\,dx}{-sinx\,dx} = -cotx $$
Question 6
1 / -0

If $$y=\tan ^{ -1 }{ \sqrt { \dfrac { 1-\sin { x } }{ 1+\sin { x } } } } $$, then the value of $$\dfrac { dy }{ dx } $$ at $$x=\dfrac { \pi }{ 6 } $$ is

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

B
$$\dfrac { 1 }{ 2 } $$

C
$$1$$

D
$$-1$$

Solution

Given, $$y=\tan ^{ -1 }{ \sqrt { \dfrac { 1-\sin { x } }{ 1+\sin { x } } } } $$
$$=\tan ^{ -1 }{ \sqrt { \dfrac { 1-\cos { \left( \dfrac { \pi }{ 2 } -x \right) } }{ 1+\cos { \left( \dfrac { \pi }{ 2 } -x \right) } } } } $$
$$=\tan ^{ -1 }{ \left| \tan { \left( \dfrac { \pi }{ 4 } -\dfrac { x }{ 2 } \right) } \right| } $$
$$=\dfrac { \pi }{ 4 } -\dfrac { x }{ 2 } $$ $$\left[ \because x=\dfrac { \pi }{ 6 } \right] $$
$$\Rightarrow \dfrac { dy }{ dx } =-\dfrac { 1 }{ 2 } $$
Question 7
1 / -0

If $$x=a\, cos^3\theta$$ and $$y=a\, sin^3\theta$$, then $$\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}$$ is equal to

A
$$|sec \,\theta |$$

B
$$sec^2\theta$$

C
$$sec\, \theta$$

D
$$| tan\, \theta|$$

Solution

$$x=a\cos^3\theta$$, $$y=0\sin^3\theta$$

diff wrt $$'\theta'$$
$$\dfrac{dx}{d\theta}=3a\cos^2\theta(-\sin\theta)$$

$$\dfrac{dy}{d\theta}=3a\sin^2\theta(+\cos \theta)$$

$$\dfrac{dy}{dx}=\dfrac{3a\sin^2\theta \cos\theta}{-3a\cos^2\theta \sin\theta}=-\dfrac{\sin\theta}{\cos\theta}=-\tan\theta$$

$$\therefore 1+\left(\dfrac{dy}{dx}\right)^2=1+\tan^2\theta =\sec^2\theta$$

$$\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}=\sqrt{\sec^2\theta}$$.

$$=|\sec \theta|$$
Question 8
1 / -0

The function represented by the following graph is.

A
Differentiable but not continuous $$x=1$$

B
Neither continuous nor Differentiable at $$x=1$$

C
Continuous but not Differentiable at $$x=1$$

D
Continuous but Differentiable at $$x=1$$

Solution

Since there is no discontinuity in the graph so clearly function is continuous,
but is not differentiable at $$x=1$$, because graph of the function has a sharp point or kink.

Note: The point on the graph of any function where it has a sharp corner or kink, function is not differentiable .
Question 9
1 / -0

If $$x = ct$$ and $$y = \dfrac {c}{t}$$, find $$\dfrac {dy}{dx}$$ at $$t = 2$$

A
$$\dfrac {1}{4}$$

B
$$4$$

C
$$-\dfrac {1}{4}$$

D
$$0$$

Solution

We have, $$x=ct\Rightarrow \dfrac{dx}{dt}=c$$
and $$y=\dfrac{c}{t}\Rightarrow \dfrac{dy}{dt}=-\dfrac{c}{t^2}$$
$$\therefore \dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\dfrac{1}{t^2}$$
Hence at $$t=2$$
$$\dfrac{dy}{dx}=-\dfrac{1}{2^2}=-\dfrac{1}{4}$$
Question 10
1 / -0

If y = cos t and x = sin t, then what is $$\dfrac{dy}{dx}$$ equal to ?

A
$$xy$$

B
$$\dfrac{x}{y}$$

C
$$-\dfrac{y}{x}$$

D
$$-\dfrac{x}{y}$$

Solution

Differentiating $$x,y$$ wrt $$t,$$ we get

$$\dfrac { dx}{ dt } =\dfrac { d }{ dt } (\cos t)=-\sin t$$ and

$$\dfrac { dx }{ dt } =\dfrac { d }{ dt } (\sin t)=\cos t\\$$
$$\therefore \dfrac { dy }{ dx } =\dfrac { \left( \dfrac { dy }{ dt } \right) }{ \left( \dfrac { dx }{ dt } \right) } =\dfrac { -\sin t }{ \cos t } =-\dfrac { x }{ y } $$

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Continuity and Differentiability Test - 18

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Continuity and Differentiability Test - 18

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Question 1

$$\displaystyle 2x^{\log x-1}\log x$$

$$\displaystyle x^{\log x-1}$$

$$\dfrac 23 (\log x)$$

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

Solution

Question 2

$$-1$$

$$1$$

$$2$$

Non existant

Solution

Question 3

$$ \displaystyle x^{x}\left ( lnx+1 \right ) $$

$$ \displaystyle x^{x}\left ( lnx-1 \right ) $$

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

$$ \displaystyle x^{x-1}\left ( lnx+1 \right ) $$

Solution

Question 4

1

0

-1

4

Solution

Question 5

cos x

cot x

- cot x

tan x

Solution

Question 6

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

$$\dfrac { 1 }{ 2 } $$

$$1$$

$$-1$$

Solution

Question 7

$$|sec \,\theta |$$

$$sec^2\theta$$

$$sec\, \theta$$

$$| tan\, \theta|$$

Solution

Question 8

Differentiable but not continuous $$x=1$$

Neither continuous nor Differentiable at $$x=1$$

Continuous but not Differentiable at $$x=1$$

Continuous but Differentiable at $$x=1$$

Solution

Question 9

$$\dfrac {1}{4}$$

$$4$$

$$-\dfrac {1}{4}$$

$$0$$

Solution

Question 10

$$xy$$

$$\dfrac{x}{y}$$

$$-\dfrac{y}{x}$$

$$-\dfrac{x}{y}$$

Solution

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