Continuity and Differentiability Test

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Continuity and Differentiability Test - 29

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Question 1
1 / -0

If $$f(x)=a|\sin x|+ be^{|x|}+c|x|^{3}$$, where $$a,b,c\in R$$, is differentiable at $$x=0$$, then

A
$$a=0,b$$ and $$c$$ are any real numbers

B
$$c=0,a=0 ,b$$ is any real number

C
$$b=0,c=0 ,a$$ is any real number

D
$$a=0,b=0 ,c$$ is any real number

Solution
Question 2
1 / -0

If $$y=\tan{x}$$, then $$\dfrac{d^{2}y}{dx^{2}}=$$

A
$$x\dfrac{dy}{dx}$$

B
$$2x\dfrac{dy}{dx}$$

C
$$2y\dfrac{dy}{dx}$$

D
$$y\dfrac{dy}{dx}$$

Solution

Given $$y=\tan x$$
$$\therefore \dfrac{dy}{dx}=\sec^2\;x\;\;\;\;\; .......\left( 1\right)$$
$$\dfrac{d^2y}{dx^2}=2\sec x.\left( \sec x .\tan x \right)=2\sec^2x \tan x$$ $$.....\left( 2\right)$$
substituting the value of $$\sec^2x$$ from $$\left(1\right) $$ in $$\left( 2\right)$$ we get,
$$\Rightarrow \dfrac{d^2 y}{dx^2}=2\left( \dfrac{dy}{dx}\right) \tan x$$
Also we have $$y=\tan x,$$ substituting we get
$$\Rightarrow \dfrac{d^2y}{dx^2}=2y\dfrac{dy}{dx}$$
Hence, the answer is $$2y\dfrac{dy}{dx}.$$
Question 3
1 / -0

Let $$g$$ is the inverse function of $$f$$ and $$f'(x)=\dfrac {x^{10}}{(1+x^{2})}$$. If $$g(2)=a$$ then $$g'(2)$$ is equal to

A
$$\dfrac {5}{2^{10}}$$

B
$$\dfrac {1+a^{2}}{a^{10}}$$

C
$$\dfrac {a^{10}}{1+a^{2}}$$

D
$$\dfrac {1+a^{10}}{a^{2}}$$

Solution

g is inverse of function f
$$ f'(x) = \frac{x^{10}}{(1+x^{2})}$$
$$ g(2) = a$$
$$ g'(2) = (1)$$
$$ \rightarrow $$ g is inverse of function f, then
$$ f(g(a)) = x$$
differentiate with respect to x
$$ f'(g(x)). g'(x) = 1 $$
$$ g'(x) = \frac{1}{f'(g.(x))}$$
$$ \rightarrow g'(2) = \frac{1}{f'(g(2))}$$
$$ = \frac{1}{f'(a)}$$
$$ = \frac{1}{\frac{a^{10}}{1+a^{2}}}$$
$$ g'(2) = \frac{1+a^{2}}{a^{10}}$$ Ans
Question 4
1 / -0

If $$f\left( x \right) =\left\{ \dfrac { x }{ { e }^{ { 1 }/{ x } }+1 } \right\} $$ when $$x\neq 0,$$ then 0,when x=0

A
lim
$$x\rightarrow 0+$$

$$f\left( x \right) =1$$

B
lim
$$x\rightarrow 0-$$

$$f\left( x \right) =1$$

C
$$f\left( x \right) $$ is continuous at x=0

D
None of the above
Question 5
1 / -0

If $$y=\sin^{-1}\dfrac{2x}{1+x^{2}}$$ then $${ \dfrac {dy}{dx} }$$ is :

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

B
$$\dfrac{2}{\sqrt{5}}$$

C
$$-\dfrac{2}{5}$$

D
$$none$$

Solution
Question 6
1 / -0

If for $$x \in \left(0, \dfrac{1}{4}\right)$$, the derivative $$\tan^{-1} \left(\dfrac{6x\sqrt{x}}{1-9x^{3}}\right)$$ is $$\sqrt{x}.g(x)$$, then $$g(x)$$ equals :

A
$$\dfrac{3}{1+9x^{3}}$$

B
$$\dfrac{9}{1+9x^{3}}$$

C
$$\dfrac{3x\sqrt{x}}{1-9x^{3}}$$

D
$$\dfrac{3x}{1-9x^{3}}$$

Solution

$$y=\tan ^{ -1 }{ \left( \cfrac { 6x\sqrt { x } }{ 1-9{ x }^{ 3 } } \right) } $$
and $$x\in \left( 0,\cfrac { 1 }{ 9 } \right) $$
$$y=\tan ^{ -1 }{ \left( \cfrac { 6x\sqrt { x } }{ 1-9{ x }^{ 3 } } \right) } =y=\tan ^{ -1 }{ \left( \cfrac { 2\left( 3{ x }^{ 3/2 } \right) }{ 1-{ \left( 3{ x }^{ 3 } \right) }^{ 3/2 } } \right) } =2\tan ^{ -1 }{ { \left( 3{ x }^{ 3/2 } \right) }^{ } } \left[ 2\tan ^{ -1 }{ \left( \cfrac { 2x }{ 1-{ x }^{ 2 } } \right) } \right] $$
$$\cfrac { dy }{ dx } 2\tan ^{ -1 }{ { \left( 3{ x }^{ 3/2 } \right) }^{ } } =2x\cfrac { 1 }{ 1+9{ x }^{ 3 } } \times 3\times \cfrac { 3 }{ 2 } \times { x }^{ 1/2 }=\cfrac { 9 }{ 1+{ x }^{ 3 }\sqrt { 3 } } =\cfrac { 9 }{ 1+9{ x }^{ 3 } } $$
Question 7
1 / -0

If $$y = \exp \left\{ {{{\sin }^2}x + {{\sin }^4}x + {{\sin }^6}x + ....} \right\}$$ then $$\frac {dy}{dx}=$$

A
$${e^{{{\tan }^2}x}}$$

B
$${e^{{{\tan }^2}x}}{\sec ^2}x$$

C
$$2{e^{{{\tan }^2}x}}\tan x{\sec ^2}x$$

D
none

Solution
Question 8
1 / -0

The function $$f\left( x \right) = \dfrac{{4 - {x^2}}}{{4x - {x^3}}}$$ is

A
discontinuous at only one point

B
discontinuous exactly at two points

C
discontinuous exactly at three points

D
None of these

Solution

$$f\left(x\right)= \dfrac{4-x^2}{4x-x^3}$$

$$=\dfrac{4-x^2}{x\left(4-x^2\right)}$$

$$for 4-x^2\neq 0$$

$$x\neq\pm 2$$

$$ f\left(x\right)=\dfrac{1}{x}$$

so this function will be discontinue at $$x=0$$

$$x=\pm2$$

three Points
Question 9
1 / -0

If $$\dfrac {dy}{dx}=(e^ {y}-x)^ {-1}$$ where $$y(0)=0$$ then $$y$$ is expressed explicity as

A
$$0.5\log_{e}(1+x^{2})$$

B
$$\log_{e}(1+x^{2})$$

C
$$\log _{ e } \left( x+\sqrt { 1+{ x }^{ 2 } } \right) $$

D
$$\\ \log _{ e } \left( x+\sqrt { 1-{ x }^{ 2 } } \right) $$

Solution
Question 10
1 / -0

If $$x\dfrac{dy}{dx}=y(\log y-\log x +1)$$, then the solution of the equation

A
$$\log\left(\dfrac{x}{y}\right)=cy$$

B
$$\log\left(\dfrac{y}{x}\right)=cx$$

C
$$x \log\left(\dfrac{x}{y}\right)=cy$$

D
$$y \log\left(\dfrac{x}{y}\right)=cy$$

Solution