Differentiation Test 19

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Differentiation Test 19

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Question 1
1 / -0

Differentiate $$\displaystyle x^{\sin^{-1}x}$$ w.r.t. $$\displaystyle \sin ^{-1}x.$$

A
$$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$

B
$$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$

C
$$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$$

D
$$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$$

Solution

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 $$z$$
$$\displaystyle \frac{1}{y}\frac{dy}{dz}=\log\sin z+z\frac{\cos z}{\sin z}$$
$$\therefore \displaystyle \frac{dy}{dz}=y\left(\log\sin z+z\frac{\cos z}{\sin z}\right)=\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$
Question 2
1 / -0

If $$y = \displaystyle (\tan x)^{\log x}$$, then $$\cfrac{dy}{dx} = $$

A
$$(\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$$

B
$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx$$

C
$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } $$

D
none of these

Solution

Let $$y = \displaystyle (\tan x)^{\log x}$$
$$\log y = \log x .\log \tan x$$
Differentiating both side w.r.t $$x$$
$$\cfrac{1}{y}.\cfrac{dy}{dx} = \cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x)$$
$$\Rightarrow \cfrac{dy}{dx} = (\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$$
Question 3
1 / -0

If $$2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x$$, then $$\displaystyle \frac { d }{ dx } f\left( x \right)$$ is

A
$$\sin{x}+\cos{x}$$

B
$$2$$

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

D
none of these

Solution

$$\displaystyle 2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x$$ ...(1)
Replace $$x$$ by $$\displaystyle \frac { \pi }{ 2 } -x$$
$$\displaystyle 2f\left( \cos { x } \right) +f\left( \sin { x } \right) =\frac { \pi }{ 2 } -x$$ ...(2)
Solving we get , $$\displaystyle 3f\left( \sin { x } \right) =\frac { \pi }{ 2 } +3x$$
$$\displaystyle \therefore f\left( x \right) =\frac { \pi }{ 6 } +\sin ^{ -1 }{ x } \quad \\\displaystyle\therefore \frac { d }{ dx } f\left( x \right) =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } $$
Question 4
1 / -0

Differentiate $$\displaystyle \tan x^{n}+\tan ^{n}x-\tan ^{-1}\frac{a+x^{n}}{1-ax^{n}}.$$

A
$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1-x^{2n} \right ) }\right ]nx^{n-1}$$

B
$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n}$$

C
$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$$

D
$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$$

Solution

Let $$y = \displaystyle \tan x^{n}+\tan ^{n}x-\tan ^{-1}\frac{a+x^{n}}{1-ax^{n}}.$$

$$\Rightarrow \displaystyle y= \tan x^{n}+\tan^{n}x-\left ( \tan^{-1}a+\tan^{-1}x^{n} \right )$$

$$\displaystyle\therefore \frac{dy}{dx}= \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-0-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right ) }
\right ]nx^{n-1}$$

$$= \left ( \sec^{2}x^{n} \right

).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right ) }

\right ]nx^{n-1}$$
Question 5
1 / -0

A curve passing through the point $$(1,1)$$ is such that the intercept made by a tangent to it on x-axis is three times the x co-ordinate of the point of tangency, then the equation of the curve is:

A
$$\displaystyle y=\frac{1}{x^{2}}$$

B
$$\displaystyle y=\sqrt{x}$$

C
$$\displaystyle y=\frac{1}{\sqrt{x}}$$

D
none

Solution

Equation of tangent on any point on the curve is,
$$\displaystyle Y-y=\frac{dy}{dx}\left ( X-x \right )$$
Thus intercept on the x-axis is,$$\displaystyle

I_{x}=x-\frac{y}{dy/dx}$$
Now using given condition,
$$\displaystyle

I_{x}=x-\frac{y}{dy/dx}=3x$$
$$\displaystyle \therefore

y\frac{dx}{dy}+2x=0$$
or $$\displaystyle

\frac{dx}{2x}+\frac{dy}{y}=0$$ or $$\displaystyle \frac{1}{2}\log x+\log

y=k$$ or $$\displaystyle \log y\sqrt{x}=e^{k}=constant$$
Now given this curve passing through (1,1) $$\Rightarrow k=0$$
$$\therefore $$ Hence required curve is, $$y=\cfrac{1}{\sqrt{x}}$$
Question 6
1 / -0

Differentiate the following: $$\displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}$$

A
$$ \displaystyle \frac{1}{2}.$$

B
$$\displaystyle \frac{-1}{2}.$$

C
$$\displaystyle \frac{1}{4}.$$

D
$$\displaystyle \frac{-1}{4}.$$

Solution

Let $$y = \displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin

x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}$$

$$\displaystyle y= \cot^{-1}\frac{\left (

\cos\frac{x}{2}+\sin\frac{x}{2} \right )+\left (

\cos\frac{x}{2}-\sin\frac{x}{2} \right )}{\left ( \cos\frac{x}{2}+\sin

\frac{x}{2} \right )-\left ( \cos \frac{x}{2}-\sin \frac{x}{2} \right

)}=\cot ^{-1}\frac{2\cos \left ( x/2 \right )}{2\sin \left ( x/2 \right

)}$$

or $$\displaystyle y=\cot ^{-1}\cot \frac{x}{2}= \frac{x}{2}, $$
$$\therefore \dfrac{dy}{dx}= \dfrac{1}{2}.$$
Note: Here assumption taken that $$\tan\frac{x}{2}\geq 1$$
Question 7
1 / -0

Let $$ \displaystyle f\left ( x \right ) $$ be defined by $$ \displaystyle f\left ( x \right )=\left\{\begin{matrix}\sin 2x & \text{if } 0< x\leq \dfrac{\pi}6\\ ax+b& \text{if } \dfrac{\pi}6< x\leq 1\end{matrix}\right. $$. The values of $$a$$ and $$b$$ such that $$ \displaystyle f $$ and $$ \displaystyle {f}' $$ are continuous, are

A
$$ \displaystyle a=1,b=\dfrac1{\sqrt{2}}+\dfrac{\pi}6 $$

B
$$ \displaystyle a=\dfrac1{\sqrt{2}},b=\dfrac1{\sqrt{2}} $$

C
$$ \displaystyle a=1,b=\dfrac{\sqrt{3}}2-\dfrac{\pi}6 $$

D
None of these

Solution

For $$f$$ to be continuous
$$\displaystyle \frac { \sqrt { 3 } }{ 2 } =f\left( \frac { \pi }{ 6 } \right) =\displaystyle \lim_{ x\rightarrow \tfrac { \pi }{ 6 }^+ } f\left( x \right) =\displaystyle\lim_{ x\rightarrow \tfrac { \pi }{ 6 } + } \left( ax+b \right) =\frac { a\pi }{ 6 } +b$$
$$\displaystyle f'\left( x \right) =\begin{cases} 2\cos { 2x } \quad \quad \text{if }0<x<\dfrac { \pi }{ 6 } \\ a\quad \quad \quad \quad \quad \text{if }\dfrac { \pi }{ 6 } <x<1 \end{cases}$$
$$\displaystyle f'\left( \frac { \pi }{ 6 } + \right) =a$$ and $$\displaystyle f'\left( \frac { \pi }{ 6 } - \right) =1$$
Thus $$\displaystyle a=1,b=\frac { \sqrt { 3 } }{ 2 } -\frac { \pi }{ 6 } $$
Question 8
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A polynomial $$f(x)$$ leaves remainder $$15$$ when divided by $$(x-3)$$ and $$(2x+1)$$ when divided by $$(x-1)^2$$. When $$f$$ is divided by $$(x-3)(x-1)^2,$$ the remainder is

A
$$2x^2+2x+3$$

B
$$2x^2-2x-3$$

C
$$2x^2-2x+3$$

D
none of these

Solution

Since function $$f(x)$$ leaves remainder $$15$$ when divided by $$x-3$$, therefore $$f(x)$$ can be written as
$$f(x)=(x-3)l(x)+15$$ ...(1)
Also, $$f(x)$$ leaves remainder $$2x+1$$ when divided by $$(x-1)^2.$$
Thus, $$f(x)$$ can also be written as
$$f\left( x \right)={ \left( x-1 \right) }^{ 2 }m\left( x \right) +2x+1$$ ...(2)
If $$R(x)$$ be the remainder when $$f(x)$$ is divided by $$(x-3)(x-1)^2,$$ then we may write
$$f\left( x \right)=\left( x-3 \right) { \left( x-1 \right) }^{ 2 }n\left( x \right) +R\left( x \right) $$ ...(3)
Since $$(x-3)(x-1)^2$$ is a polynomial of degree three, the remainder has to be a polynomial of degree less than or equal to two.
Thus let $$R(x)=ax^2+bx+c$$
From (1) and (3), we have
$$f(3)=15=R(3)\Rightarrow 9a+3b+c=15$$ ...(4)
From (2) and (3), we have
$$f(1)=3=R(1)\Rightarrow a+b+c=3$$ ...(5)
From (2) and (3), we have
$$f'(1)=2=R'(1)\Rightarrow 2a+b=2$$ ...(6)
Solving equation (4),(5) and (6), we get
$$a=2,b=-2,c=3$$
Question 9
1 / -0

If $$f\left( x \right) $$ is a polynomial of degree $$n(>2)$$ and $$f\left( x \right) =f\left( k-x \right) ,($$ where $$k$$ is a fixed real number$$),$$ then degree of $$f'(x)$$ is

A
$$n$$

B
$$n-1$$

C
$$n-2$$

D
None of these

Solution

Clearly, $$f(x)$$ must be of the from
$$f\left( x \right) ={ a }_{ 0 }\left[ { x }^{ n }+{ \left( k-x \right) }^{ n } \right] +{ a }_{ 1 }\left[ { x }^{ n-1 }+{ \left( k-x \right) }^{ n-1 } \right] +...+{ a }_{ n-1 }\left[ x+\left( k-x \right) \right] +{ a }_{ n. }$$
It may be noted that $$n$$ must be even for otherwise $$f(x)$$ will become a polynomial of degree $$n-1.$$
Clearly, $$f'(x)$$ is a polynomial of degree $$n-1.$$
Question 10
1 / -0

If for all $$x,y$$ the function $$f$$ is defined by $$f\left( x \right)+f\left( y \right)+f\left( x \right).f\left( y \right)=1$$ and $$f\left( x \right)>0$$, then

A
$$f'\left( x \right) $$ does not exist

B
$$f'\left( x \right) =0$$ for all $$x$$

C
$$f'\left( 0 \right) <f'\left( 1 \right) $$

D
None of these

Solution

Putting $$x=0,y=0,$$ we get
$$2f(0)+{ \left\{ f\left( 0 \right) \right\} }^{ 2 }=1\Rightarrow f\left( 0 \right) =\sqrt { 2 } -1\quad \left( \because f\left( x \right) >0 \right) $$
Putting $$y=x,2f(x)+{ \left\{ f\left( x \right) \right\} }^{ 2 }=1$$
Differentiating w.r.t. $$x,$$ we get
$$2f'\left( x \right) +2f\left( x \right) .f'\left( x \right) =0\quad f'\left( x \right) \left\{ 1+f\left( x \right) \right\} =0$$
$$\Rightarrow f'\left( x \right) =0,$$ because $$f\left( x \right) >0$$

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Differentiation Test 19

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Differentiation Test 19

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Question 1

$$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$

$$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$

$$\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$$

$$-\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x\frac{\sqrt{\left ( 1+x^{2} \right )}}{x} \right ]$$

Solution

Question 2

$$(\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$$

$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } logx$$

$$\frac { 1 }{ x } { tanx }^{ logx }log(tanx)+\frac { 1 }{ tanx } \sec ^{ 2 }{ x } $$

none of these

Solution

Question 3

$$\sin{x}+\cos{x}$$

$$2$$

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

none of these

Solution

Question 4

$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1-x^{2n} \right ) }\right ]nx^{n-1}$$

$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n}$$

$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$$

$$ \left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$$

Solution

Question 5

$$\displaystyle y=\frac{1}{x^{2}}$$

$$\displaystyle y=\sqrt{x}$$

$$\displaystyle y=\frac{1}{\sqrt{x}}$$

none

Solution

Question 6

$$ \displaystyle \frac{1}{2}.$$

$$\displaystyle \frac{-1}{2}.$$

$$\displaystyle \frac{1}{4}.$$

$$\displaystyle \frac{-1}{4}.$$

Solution

Question 7

$$ \displaystyle a=1,b=\dfrac1{\sqrt{2}}+\dfrac{\pi}6 $$

$$ \displaystyle a=\dfrac1{\sqrt{2}},b=\dfrac1{\sqrt{2}} $$

$$ \displaystyle a=1,b=\dfrac{\sqrt{3}}2-\dfrac{\pi}6 $$

None of these

Solution

Question 8

$$2x^2+2x+3$$

$$2x^2-2x-3$$

$$2x^2-2x+3$$

none of these

Solution

Question 9

$$n$$

$$n-1$$

$$n-2$$

None of these

Solution

Question 10

$$f'\left( x \right) $$ does not exist

$$f'\left( x \right) =0$$ for all $$x$$

$$f'\left( 0 \right) <f'\left( 1 \right) $$

None of these

Solution

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