Continuity and Differentiability Test

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Continuity and Differentiability Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

If $$y \tan^{-1} \left ( \sqrt{\frac{a - x}{a + x}} \right )$$, where -a < x < a, then $$\frac{dy}{dx} =$$.....

A
$$\frac{x}{\sqrt{a^2 - x^2}}$$

B
$$\frac{a}{\sqrt{a^2 - x^2}}$$

C
$$- \frac{1}{2 \sqrt{a^2 - x^2}}$$

D
$$\frac{1}{2 \sqrt{a^2 - x^2}}$$

Solution

Put $$x = a \cos \theta$$
Question 2
1 / -0

If $$y = \tan^{-1} \left ( \frac{x}{1 + \sqrt{1 - x^2}} \right ) + \sin \left [ 2 \tan^{-1} \left ( \sqrt{\frac{1 - x}{1 + x}} \right ) \right ]$$ then $$\frac{dy}{dx} = $$...........

A
$$\frac{x}{\sqrt{1 - x^2}}$$

B
$$\frac{1 - 2x}{\sqrt{1 - x^2}}$$

C
$$\frac{1 - 2x}{2 \sqrt{1 - x^2}}$$

D
$$\frac{1 - 2x^2}{\sqrt{1 - x^2}}$$

Solution

$$y = \tan^{-1} \left ( \frac{x}{1 + \sqrt{1 - x^2}} \right ) + \sin \left [ 2 \tan^{-1} \left ( \sqrt{\frac{1 - x}{1 + x}} \right ) \right ]$$
Put $$x = \cos \theta.$$ Then $$\theta = \cos^{-1} x$$
$$\therefore y = \tan^{-1} \left ( \frac{\cos \theta}{1 + \sqrt{1 - \cos \theta^2}} \right ) + \sin \left [ 2 \tan^{-1} \left ( \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \right ) \right ]$$
$$= \tan^{-1} \left ( \frac{\cos \theta}{1 + 1 + \tan \theta} \right ) + \sin \left [ 2 \tan^{-1} \left ( \sqrt{\frac{2 \sin^2 (\frac{\theta}{2})}{2 \cos^2(\frac{\theta}{2})}} \right ) \right ]$$
$$= \tan^{-1} \left [ \frac{\sin (\frac{\pi}{2} - \theta)}{1 + \cos (\frac{\pi}{2} - \theta)} \right ] + \sin [2 \tan^{-1} (\tan \frac{\theta}{2})]$$
$$= \tan^{-1} \left [ \frac{2 \sin (\frac{\pi}{4} - \frac{\theta}{2}) \cdot \cos (\frac{\pi}{4} - \frac{\theta}{2})}{2 \cos^2 (\frac{\pi}{4} - \frac{\theta}{2})} \right ] + \sin (2 \times \frac{\theta}{2})$$
$$= \tan^{-1} [\tan (\frac{\pi}{4} - \frac{\theta}{2})] + \sin \theta$$
$$= \frac{\pi}{4} - \frac{\theta}{2} + \sqrt{1 - \cos^2 \theta}$$
$$= \frac{\pi}{4} - \frac{2}{2} \cos^{-1} x + \sqrt{1 - x^2}$$
$$\frac{dy}{dx} = 0 - \frac{1}{2} \times \frac{1}{\sqrt{1 - x^2}} + \frac{1}{2 \sqrt{1 - x^2}} \times (-2x)$$
$$= \frac{1}{2 \sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}}$$
$$= \frac{1 - 2x}{2 \sqrt{1 - x^2}}$$
Question 3
1 / -0

If $$y = \sin(2 \sin^{-1} x)$$, then dx = .......

A
$$\frac{2 - 4x^2}{\sqrt{1 - x^2}}$$

B
$$\frac{2 + 4x^2}{\sqrt{1 - x^2}}$$

C
$$\frac{4x^2 - 1}{\sqrt{1 - x^2}}$$

D
$$\frac{1 - 2x^2}{\sqrt{1 - x^2}}$$

Solution

$$y = \sin(2 \sin^{-1} x)$$
Put $$x = \sin \theta$$. Then $$\theta = \sin^{-1} x$$
$$\therefore y = \sin 2 \theta$$
$$\therefore \frac{dy}{dx} = \frac{dy}{d \theta} \times \frac{d \theta}{dx} = 2 \cos 2 \theta \times \frac{1}{\sqrt{1 - x^2}}$$
$$= \frac{2(1 - 2 \sin^2 \theta)}{\sqrt{1 - x^2}} = \frac{2 - 4x^2}{\sqrt{1 - x^2}} $$
Question 4
1 / -0

If function $$f(x)=\dfrac{x^2-9}{x-3}$$ is continuous at $$x=3$$, then value of $$(3)$$ will be:

A
$$6$$

B
$$3$$

C
$$1$$

D
$$0$$

Solution

$$f(x)=\dfrac{x^2-9}{x-3}$$
Left hand limit
$$f(3+0)=\displaystyle \lim_{h \rightarrow 0}f(3+h)$$
$$=\displaystyle \lim_{h \rightarrow 0} \dfrac{(3+h)^2-9}{3+h-3}$$
$$=\displaystyle \lim_{h \rightarrow 0} \dfrac{h(6+h)}{h}$$
$$=\displaystyle \lim_{h \rightarrow 0} (6+h)$$
$$=6$$
Function is continous at $$x=3$$, so
$$f(3)=f(3+0)$$
$$f(3)=6$$
Hence, option $$(a)$$ is correct.
Question 5
1 / -0

If $$f(x)=\begin{cases} \begin{matrix} \dfrac{\log (1+mx)- \log (1-nx)}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} k; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$$
is continuous at $$x=0$$ then the value of $$k$$ will be:

A
$$0$$

B
$$m+n$$

C
$$m-n$$

D
$$m.n$$

Solution

$$\therefore$$ Function is continous at $$x=0$$
$$\therefore \displaystyle \lim_{x \rightarrow 0} f(x)=f(0)$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{\log (1+mx)- \log (1-nx)}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{ [mx-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3} - ...]- [nx-\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} - ...]}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \dfrac{x \left[ m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... \right]}{x}=k$$
$$\Rightarrow \displaystyle \lim_{x \rightarrow 0} m-\dfrac{m^2 x^2}{2}+ \dfrac{m^3 x^3}{3}-...+n+\dfrac{n^2 x^2}{2}+ \dfrac{n^3 x^3}{3} +... =k$$
$$\Rightarrow m+n=k$$
$$\Rightarrow k=m+n$$
Hence, option $$(b)$$ is correct.
Question 6
1 / -0

If function $$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$$
is continuous at $$x=2$$ then value of $$m$$ will be:

A
$$3$$

B
$$1/3$$

C
$$1$$

D
$$0$$

Solution

$$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$$
$$f(0)=m$$
$$f(0+0)=\displaystyle \lim_{h \rightarrow 0}f(0+h)$$
$$\displaystyle \lim_{h \rightarrow 0} \dfrac{\sin 3 (0+h)}{0+h}$$
$$\displaystyle \lim_{h \rightarrow 0}3 \dfrac{\sin 3 h}{3h}$$
$$=3 \times 1$$
$$=3$$
At $$x=a$$ function is continuous.
So, $$f(0)=f(0+0)$$
$$m=3$$
Hence, option $$(a)$$ is correct.
Question 7
1 / -0

Let $$x={f}''(t) cost +{f}'(t) sint$$ and $$y={-f}''(t) sint+{f}'(t) cost.$$ Then $$\displaystyle \int \left [ \left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2 \right ]^{\frac{1}{2}} dt$$ equals
(Note : $$f(x), f'(x), f''(x) , f'''(x) >0$$ )

A
$${f}'(t)+{f}''(t)+c$$

B
$${f}''(t)+{f}'''(t)+c$$

C
$$f(t)+{f}''(t)+c$$

D
$${f}'(t)-{f}''(t)+c$$

Solution

$$x=f''(t)\cos { t } +f'(t)\sin { t } \\ \dfrac { dx }{ dt } =f'''(t)\cos { t } -f''(t)\sin { t } +f''(t)\sin { t } +f'(t)\cos { t } \\ \dfrac { dx }{ dt } =f'''(t)\cos { t } +f'(t)\cos { t } \\ \dfrac { dx }{ dt } =(f'''(t)+f'(t))\cos { t } ..........(1)\\ y=-f''(t)\sin { t } +f'(t)\cos { t } \\ \dfrac { dy }{ dt } =-f'''(t)\sin { t } -f''(t)\cos { t } +f''(t)\cos { t } -f'(t)\sin { t } \\ \dfrac { dy }{ dt } =-f'''(t)\sin { t } -f'(t)\sin { t } \\ \dfrac { dy }{ dt } =-\sin { t } (f'''(t)+f'(t))........(2)\\ add\quad squares\quad of\quad eq.1\quad and\quad 2\quad \\ { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 }={ (f'''(t)+f'(t)) }^{ 2 }\\ { \left( { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 } \right) }^{ \dfrac { 1 }{ 2 } }=f'''(t)+f'(t)\\ integrate\quad both\quad sides\quad \\ \int { { \left( { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 } \right) }^{ \dfrac { 1 }{ 2 } }dt } =f''(t)+f(t)+c$$
Question 8
1 / -0

The set of all points where the function $$f(\displaystyle \mathrm{x})=\frac{x}{1+|x|}$$ is differentiable is

A
$$(-\infty, \infty)$$

B
$$(0,\infty)$$

C
$$(-\infty ,0)\cup (0,\infty )$$

D
$$(-\infty, 0)$$

Solution

$$f(x) = \dfrac{x}{1+|x|} $$

For $$x\geq 0 $$, $$f(x) = \dfrac{x}{1+x} $$

For $$x<0 $$, $$f(x) = \dfrac{x}{1-x}$$

$$f(x)$$ is continuous and differentiable at all points except perhaps at $$x=0 $$.

Hence, we will check the differentiability at $$x=0$$.

As $$x \rightarrow 0 $$, $$f(x) \rightarrow 0 $$. $$f(0) = 0 $$.

Hence, $$f(x) $$ is continuous at $$x=0$$.

$$RHD = \dfrac{ 1+x-x}{(1+x)^2} = 1 $$ {see that $$f(x)=\infty$$ at $$x=-1$$ but

$$-1 \notin \ [0,\infty)$$}

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

Hence, $$f(x)$$ is differentiable at $$x=0$$.

Hence, option A is correct.
Question 9
1 / -0

If $$y=x^{\displaystyle x^{\displaystyle x^{\displaystyle \dots^{\displaystyle\infty}}}}$$, find $$\displaystyle\frac{dy}{dx}$$.

A
$$\displaystyle\frac{y^2}{x(1-y\log{x})}$$

B
$$\displaystyle\frac{y}{x(1-\log{x})}$$

C
$$\displaystyle\frac{y^2}{x(y-\log{x})}$$

D
None of these

Solution

The given function may be written as,
$$y=x^{\displaystyle y}$$
Taking $$\log$$ on both sides,
$$\log{y}=\log { { x }^{ y } } $$
$$\log{y}=y\log{x}$$
Differentiate w.r. to $$x$$
$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =y\frac { d }{ dx } \left( \log { x } \right) +\log { x } \frac { d }{ dx } y$$

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

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

$$\displaystyle \frac { dy }{ dx } \left( \frac { 1 }{ y } -\log { x } \right) =\frac { y }{ x } $$

$$\displaystyle \frac { dy }{ dx } \left( \frac { 1-y\log { x } }{ y } \right) =\frac { y }{ x }
$$

$$\therefore \displaystyle \frac { dy }{ dx } =\frac { { y }^{ 2 } }{ x\left( 1-y\log { x } \right) } $$
Question 10
1 / -0

Derivative of $$({\log{x}})^{\displaystyle\cos{x}}$$ with respect to $$x$$ is

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

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

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

D
None of these

Solution

$$y={ \log { x } }^{ \cos { x } }$$
Taking log on both sides, we get
$$\log { y } =\log { \left( { \log { x } }^{ \cos { x } } \right) } $$
$$\log { y } =\cos { x } .\left[ \log { \left( \log { x } \right) } \right] $$
Differentiate w.r. to $$x$$
$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\cos { x } \frac { d }{ dx } \left[ \log { \left( \log { x } \right) } \right] +\left[ \log { \left( \log { x } \right) } \right] \frac { d }{ dx } \cos { x } $$

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

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

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

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 36

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Continuity and Differentiability Test - 36

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SHARING IS CARING

Question 1

$$\frac{x}{\sqrt{a^2 - x^2}}$$

$$\frac{a}{\sqrt{a^2 - x^2}}$$

$$- \frac{1}{2 \sqrt{a^2 - x^2}}$$

$$\frac{1}{2 \sqrt{a^2 - x^2}}$$

Solution

Question 2

$$\frac{x}{\sqrt{1 - x^2}}$$

$$\frac{1 - 2x}{\sqrt{1 - x^2}}$$

$$\frac{1 - 2x}{2 \sqrt{1 - x^2}}$$

$$\frac{1 - 2x^2}{\sqrt{1 - x^2}}$$

Solution

Question 3

$$\frac{2 - 4x^2}{\sqrt{1 - x^2}}$$

$$\frac{2 + 4x^2}{\sqrt{1 - x^2}}$$

$$\frac{4x^2 - 1}{\sqrt{1 - x^2}}$$

$$\frac{1 - 2x^2}{\sqrt{1 - x^2}}$$

Solution

Question 4

$$6$$

$$3$$

$$1$$

$$0$$

Solution

Question 5

$$0$$

$$m+n$$

$$m-n$$

$$m.n$$

Solution

Question 6

$$3$$

$$1/3$$

$$1$$

$$0$$

Solution

Question 7

$${f}'(t)+{f}''(t)+c$$

$${f}''(t)+{f}'''(t)+c$$

$$f(t)+{f}''(t)+c$$

$${f}'(t)-{f}''(t)+c$$

Solution

Question 8

$$(-\infty, \infty)$$

$$(0,\infty)$$

$$(-\infty ,0)\cup (0,\infty )$$

$$(-\infty, 0)$$

Solution

Question 9

$$\displaystyle\frac{y^2}{x(1-y\log{x})}$$

$$\displaystyle\frac{y}{x(1-\log{x})}$$

$$\displaystyle\frac{y^2}{x(y-\log{x})}$$

None of these

Solution

Question 10

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

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

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

None of these

Solution

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