Continuity and Differentiability Test

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Continuity and Differentiability Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

If $$y=1-\cos\theta,x=1-\sin\theta$$, then $$\dfrac{dy}{dx}$$ at $$\theta=\dfrac{\pi}{4}$$ is

A
$$-1$$

B
$$1$$

C
$$12$$

D
$$-12$$

Solution
Question 2
1 / -0

If $$x=\sqrt{2^{cosec^{-1}t}}$$ and $$y=\sqrt{2^{sec^{-1}t}}(|t|\geq 1)$$ then $$\dfrac{dy}{dx}$$ is equal to:

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

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

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

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

Solution
Question 3
1 / -0

For $$x > 1 ,$$ if $$( 2 x ) ^ { 2 y } = 4 e ^ { 2 x - 2 y } ,$$ then $$\left( 1 + \log _ { \mathrm { e } } 2 \mathrm { x } \right) ^ { 2 } \dfrac { \mathrm { dy } } { \mathrm { dx } }$$ is equal to :

A
$$\log _ { e } 2 x$$

B
$$\dfrac { x \log _ { e } 2 x + \log _ { e } 2 } { x }$$

C
$$x \log _ { e } 2 x$$

D
$$\dfrac { x \log _ { e } 2 x - \log _ { e } 2 } { x }$$

Solution

$$( 2 x ) ^ { 2 y } = 4 e ^ { 2 x - 2 y }$$

$$2 y \ln 2 x = \ln 4 + 2 x - 2 y$$

$$\mathrm { y } = \dfrac { \mathrm { x } + \mathrm { \ln } 2 } { 1 + \mathrm { \ln } 2 \mathrm { x } }$$

$$y ^ { \prime } = \dfrac { ( 1 + \ln 2 x ) - ( x + \ln 2 ) \dfrac { 1 } { x } } { ( 1 + \ln 2 x ) ^ { 2 } }$$

$$\mathrm { y } ^ { \prime } ( 1 + \mathrm { \ln } 2 \mathrm { x } ) ^ { 2 } = \left[ \dfrac { \mathrm { x } \mathrm { \ln } 2 \mathrm { x } - \mathrm { \ln } 2 } { \mathrm { x } } \right]$$
Question 4
1 / -0

$$\dfrac{d}{dx}\left[\tan h^{-1}\left(\dfrac{2x}{1+x^2}\right)\right]=?$$

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

B
$$\dfrac{2}{x^2-1}$$

C
$$\dfrac{2}{1+x^2}$$

D
$$\dfrac{-2}{x^2+1}$$

Solution
Question 5
1 / -0

A differential function satisfies equation $$f(x)=\int_{0}^{x}(f(t)\cos\ t-\cos(t-x))dt$$ then

A
$${ f }^{ \prime \prime }\left( \dfrac { \pi }{ 2 } \right) =e$$

B
$$\lim _{ x\rightarrow -\infty }{ f\left( x \right) =1 }$$

C
$$f(x) has minimum value 1-e^{-1}$$

D
$${ f }^{ \prime}(0)=-1$$

Solution
Question 6
1 / -0

If $$f ( x )$$ and $$g ( x )$$ are differentiable functions in $$[ 0,1 ]$$ such that $$f ( 0 ) = 2 , f ( 1 ) = 6 , g ( 0 ) = 0 , g ( 1 ) =2$$ then there exists $$0 < c < 1$$ such that

A
$$\mathrm { f } ^ { \prime } ( \mathrm { c } ) = \mathrm { g } ^ { \prime } ( \mathrm { c } )$$

B
$$f ^ { \prime } ( c ) = - g ^ { \prime } ( c )$$

C
$$f ^ { \prime } ( c ) = 2 g ^ { \prime } ( c )$$

D
$$2 f ^ { \prime } ( c ) = g ^ { \prime } ( c )$$

Solution
Question 7
1 / -0

If $$ y = \sin ^ { - 1 } x$$ then $$ \frac { d y } { d x } $$ is equal to

A
$$\sec y$$

B
$$\cos x$$

C
$$\tan x$$

D
$$1$$

Solution

$$y=\sin ^{-1}x\\x=\sin y \\\dfrac{dx}{dx}=\dfrac{d}{dx}\sin y\\1=\cos y \dfrac{dy}{dx}\\\dfrac{dy}{dx}=\sec y$$
Question 8
1 / -0

If $$f(x)=0$$ for $$x < 0$$ and $$f(x)$$ is differential at $$x=0$$ then for $$x\ge 0, f(x)$$ may be

A
$$x^{2}$$

B
$$x$$

C
$$-x$$

D
$$-x^{3/2}$$

Solution
Question 9
1 / -0

If $$y={ \tan }^{ -1 }{ ax}$$ then, which of the following is true

A
$$\displaystyle \frac { dy }{ dx } =\frac { a }{ 1+{ x }^{ 2 } }$$

B
$$\displaystyle \frac { dy }{ dx } =\frac { a }{ 1-{(ax) }^{ 2 } }$$

C
$$\displaystyle \frac { dy }{ dx } =\frac { a }{ 2+{ (ax) }^{ 2 } }$$

D
$$\displaystyle \frac { dy }{ dx } =\frac { a}{ 1+{ (ax) }^{ 2 } }$$

Solution

Hint:- For a composite function $$f\left( g\left( x \right) \right)$$ the chain rule for the derivative is expressed as $$f'\left( x \right)g'\left( x \right)$$.
The derivative of the function $$y = {\tan ^{ - 1}}x$$ is given as $${{dy} \over {dx}} = {1 \over {1 + {x^2}}}$$.

Given:-
The function is given as $$y = {\tan ^{ - 1}}\left( {ax} \right)$$.

Step 1: Find the derivative of the function by applying chain rule.
$$ \Rightarrow {{dy} \over {dx}} = \left[ {{{\tan }^{ - 1}}\left( {ax} \right)} \right]'$$
$$ \Rightarrow {{dy} \over {dx}} = {1 \over {1 + {{\left( {ax} \right)}^2}}} \times \left( {ax} \right)'$$

Step 2: Find the derivative of the term $$\left( {ax} \right)'$$ and simplify the expression.
$$ \Rightarrow {{dy} \over {dx}} = {1 \over {1 + {{\left( {ax} \right)}^2}}} \times a$$
$$ \therefore {{dy} \over {dx}} = {a \over {1 + {{\left( {ax} \right)}^2}}}$$

Final step: The derivative of the function $$y = {\tan ^{ - 1}}\left( {ax} \right)$$ is given by the relation
$$ \ {{dy} \over {dx}} = {a \over {1 + {{\left( {ax} \right)}^2}}}$$
Question 10
1 / -0

If $$f(x)$$ is a differentiable function $$\forall \ x\ \epsilon \ R$$ so that, $$f(2)=4,\ f'(x)\ge 5\ \forall \ x\ \epsilon \ [2,6]$$, then, $$f(6)$$ is :

A
$$\ge 24$$

B
$$\le 24$$

C
$$\ge 9$$

D
$$\le 9$$

Solution