If $$y = sec x^0$$ then $$\displaystyle \frac{dy}{dx} = $$

$$\displaystyle \frac{\pi}{180} sec x^0 tan x^0$$

$$\displaystyle \frac{180}{\pi} sec x^0 tan x^0$$

MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 5

$\displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\left ( 1-\tan \dfrac{x}{2} \right )\left ( 1-\sin x \right )}{\left ( 1+\tan \dfrac{x}{2} \right )\left ( \pi -2x \right )^{3}}$ is

0%
$0$
0%
$\displaystyle \frac{1}{32}$
0%
$\infty$
0%
$\displaystyle \frac{1}{8}$

Explanation

$\because \tan\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right) = \dfrac{1 - \tan\dfrac{x}{2}}{1+\tan\dfrac{x}{2}}$

$\therefore \displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )\left ( 1-\sin x \right )}{4\left ( \dfrac{\pi -2x}{4} \right )\left ( \pi -2x \right )^{2}}$

$\therefore \displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{2}-\dfrac{x}{2} \right )\left(1-\cos \left ( \dfrac{\pi }{2}-x \right )\right)}{4.\left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )\left ( \pi -2x \right )^{2}}$

$\therefore \displaystyle\lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{2}-\dfrac{x}{2}\right )\cdot2\cdot\sin ^{2}\left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )}{4\cdot\left (\dfrac{\pi }{4}-\dfrac{x}{2} \right )\cdot4^{2}\cdot\left ( \dfrac{\pi -2x}{4} \right )^{2}}$

$\displaystyle =\frac{1}{4}\times \frac{2}{16}=\frac{1}{32}$

If P(x) is a polynomial such that

$P\left ( x^{2}+1 \right )=\left \{ P\left ( x^{2} \right ) \right \}^{2}+1$ and

$P(0)=0$ then

$P^{'}(0)$ is equal to

0%
$1$
0%
$0$
0%
$-$ 1
0%
none of these

Explanation

Given,

$P\left ( x^{2}+1 \right )=\left \{ P\left ( x^{2} \right ) \right \}^{2}+1$ ...(1)
Given

$P(0)=0$
Put

$x=0$ in (1)

$\Rightarrow P(1)={P(0)}^{2}+1$

$\Rightarrow P(1)=1$
Also, put

$x=1$ in (1)

$\Rightarrow P(2)=2$
Also, since, P(x) is a polynomial so we have P(x)=x

$\Rightarrow P'(x)=1$

$P'(0)=1$

Let

$f\left ( x \right )=\begin{cases}\sin x, x\neq n\pi \\ 2, x=n\pi \end{cases}$ , where

$n\epsilon \mathbb{Z}$ and

$g\left ( x \right )=\begin{cases}x^{2}+1, x\neq 2 \\ 3, x=2 \end{cases}$ .
Then

$\displaystyle \lim_{x\to 0}g\left ( f\left ( x \right ) \right )$ is

0%
$0$
0%
$1$
0%
$3$
0%
none of these

Explanation

$\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ g\left( f\left( x \right) \right) } =\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ g\left( \sin { x } \right) } \\ =\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \left( \sin ^{ 2 }{ x } +1 \right) } =0+1=1\\ \displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ g\left( f\left( x \right) \right) } =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ g\left( \sin { x } \right) } \\ =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \left( \sin ^{ 2 }{ x } +1 \right) }=1$

$\displaystyle \frac{d}{dx}(\log_{e}\left ( \frac{1+x}{1-x} \right )^{1/4}-\frac{1}{2}\tan^{-1}x.)$

0%
$\displaystyle \frac{x^{2}}{1-x^{4}}.$
0%
$\displaystyle \frac{x^{3}}{1-x^{4}}.$
0%
$\displaystyle \frac{x^{4}}{1-x^{4}}.$
0%
$-\displaystyle \frac{x^{2}}{1-x^{4}}.$

Explanation

Let

$\displaystyle y=\frac{1}{4}\left [ \log\left ( 1+x \right )-\log\left ( 1-x \right ) \right ]-\frac{1}{2}\tan^{-1}x.$

$\Rightarrow \displaystyle \frac{dy}{dx}=\frac{1}{4}\left [ \frac{1}{1+x}-\frac{1}{1-x}\left ( -1 \right ) \right ]-\frac{1}{2}.\frac{1}{1+x^{2}}$

$\displaystyle =\frac{1}{2\left ( 1-x^{2} \right )}-\frac{1}{2\left ( 1+x^{2} \right )}=\frac{x^{2}}{1-x^{4}}.$

Differentiate

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

0%
$\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1-x^{2n} \right ) }\right ]nx^{n-1}$
0%
$\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n}$
0%
$\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n}x.\sec^{2}x-\left [\dfrac{ 1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$
0%
$\left ( \sec^{2}x^{n} \right ).nx^{n-1}+n\tan^{n-1}x.\sec^{2}x-\left [ \dfrac{1}{\left ( 1+x^{2n} \right )} \right ]nx^{n-1}$

Explanation

Let

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

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

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

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

$\displaystyle \frac{d}{dx}(\tan ^{-1}\frac{\sin x+\cos x}{\cos x-\sin x})$

0%
$-1$
0%
$-2$
0%
$1$
0%
$2$

Explanation

Let

$\displaystyle y = \tan ^{-1}\frac{\cos x+\sin x}{\cos x-\sin x}=\tan ^{-1}\frac{1+\tan x}{1-\tan x}$

$\displaystyle \Rightarrow y =\tan ^{-1}\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}=\tan^{-1}\tan(\frac{\pi}{4}+x)=\frac{\pi}{4}+x$

$\therefore \cfrac{dy}{dx}=1$

$\displaystyle \frac{d}{dx}\tan ^{-1}\left(\frac{a \cos x-b\sin x}{b\cos x+a\sin x}\right)$

0%
$-1$
0%
$-2$
0%
$1$
0%
$2$

Explanation

Put

$\displaystyle a= r\cos \alpha , b= r\sin \alpha$

$\displaystyle \therefore r= \sqrt{a^{2}+b^{2}},\tan \alpha = \dfrac{b}{a}$

$\displaystyle \therefore y= \tan ^{-1}\frac{r\cos\left ( x+\alpha \right )}{r\sin\left ( x+\alpha \right )}=\tan^{-1}\cot \left ( x+\alpha \right )$

$\displaystyle =\tan^{-1}\tan\left [ \dfrac{\pi}{2} -\left ( x+\alpha \right )\right ]$

$\displaystyle \therefore y=\dfrac{\pi} {2}-x-\alpha$

$\therefore \dfrac{dy}{dx}=-1$

Differentiate the following:

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

0%
$\displaystyle \frac{1}{2}.$
0%
$\displaystyle \frac{-1}{2}.$
0%
$\displaystyle \frac{1}{4}.$
0%
$\displaystyle \frac{-1}{4}.$

Explanation

Let

$y = \displaystyle \cot ^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{\left ( 1+\sin x \right )}-\sqrt{\left ( 1-\sin x \right )}}$

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

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

$\therefore \dfrac{dy}{dx}= \dfrac{1}{2}.$
Note: Here assumption taken that

$\tan\frac{x}{2}\geq 1$

$\displaystyle \dfrac{d}{dx}\tan ^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)$

0%
$-\displaystyle \dfrac{1}{2}$
0%
$-\displaystyle \dfrac{1}{4}$
0%
$-\displaystyle \dfrac{1}{8}$
0%
$\displaystyle \dfrac{1}{2}$

Explanation

Let

$\displaystyle y = \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$

$\Rightarrow \displaystyle y = \tan ^{-1}\left(\dfrac{(\cos^2 \dfrac{x}{2}-\sin^2\dfrac{x}{2})}{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin \dfrac{x}{2}.\cos\dfrac{x}{2}}\right)$

$\displaystyle y=\tan ^{-1}\left(\dfrac{\cos\left ( x/2 \right )-\sin \left ( x/2 \right )}{\cos \left ( x/2 \right )+\sin \left ( x/2 \right )}\right)$

$\displaystyle =\tan^{-1}\left(\frac{1-\tan\left ( x/2 \right )}{1+\tan \left ( x/2 \right )}\right)$

$\displaystyle =\tan^{-1}\left(\tan \left ( \pi /4-x/2 \right )\right)=\pi/4-x/2$

$\displaystyle \therefore \dfrac{dy}{dx}=-\dfrac{1}{2}$

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

0%
$\displaystyle a=1,b=\dfrac1{\sqrt{2}}+\dfrac{\pi}6$
0%
$\displaystyle a=\dfrac1{\sqrt{2}},b=\dfrac1{\sqrt{2}}$
0%
$\displaystyle a=1,b=\dfrac{\sqrt{3}}2-\dfrac{\pi}6$
0%
None of these

Explanation

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

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

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

0%
$2x^2+2x+3$
0%
$2x^2-2x-3$
0%
$2x^2-2x+3$
0%
none of these

Explanation

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)

$R(x)$ 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

$a=2,b=-2,c=3$

$f\left( x \right)$ 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

0%
$n$
0%
$n-1$
0%
$n-2$
0%
None of these

Explanation

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

$n-1.$

$2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x$ , then

$\displaystyle \frac { d }{ dx } f\left( x \right)$ is

0%
$\sin{x}+\cos{x}$
0%
$2$
0%
$\displaystyle \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$
0%
none of these

Explanation

$\displaystyle 2f\left( \sin { x } \right) +f\left( \cos { x } \right) =x$ ...(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 ,

$\displaystyle 3f\left( \sin { x } \right) =\frac { \pi }{ 2 } +3x$

$\displaystyle \therefore f\left( x \right) =\frac { \pi }{ 6 } +\sin ^{ -1 }{ x } \quad \\\displaystyle\therefore \frac { d }{ dx } f\left( x \right) =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$

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

0%
$f'\left( x \right)$ does not exist
0%
$f'\left( x \right) =0$ for all $x$
0%
$f'\left( 0 \right) <f'\left( 1 \right)$
0%
None of these

Explanation

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

$f\left( x \right) >0$

If the functions

$\displaystyle f\left ( x \right )=\sin \left ( x+a \right )$ and

$\displaystyle g\left ( x \right )=b\sin x+c\cos x$ satisfy

$\displaystyle f\left ( 0 \right )=g\left ( 0 \right )$ and

$\displaystyle {f}'\left ( 0 \right )={g}'\left ( 0 \right )$ then

0%
$\displaystyle b=\dfrac{\pi}2$
0%
$\displaystyle b=\cos a$
0%
$\displaystyle c=\sin a$
0%
$\displaystyle c=\cos a$

Explanation

Given,

$f(x)=\sin(x+a)\quad$

and

$g(x)=b\sin x+c\cos x\\ f(0)=g(0)\Rightarrow \sin a=c\\ f'(x)=\cos(x+a)\\ g'(x)=b\cos x-c\sin x\\ f'(0)=g'(0)\Rightarrow \cos a=b$

$\displaystyle {f}'\left ( x \right )=g\left ( x \right )$ and

$\displaystyle {g}'\left ( x \right )=-f\left ( x \right )$ and

$\displaystyle f\left ( 2 \right )=4={f}'\left ( 2 \right )$ then

$\displaystyle f^{2}\left ( 16 \right )+g^{2}\left ( 16 \right )$ is

0%
16
0%
32
0%
64
0%
None of these

Explanation

$\displaystyle \frac { d }{ dx } \left( { f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right) \right) \\ =2\left[ f\left( x \right) f'\left( x \right) +g\left( x \right) g'\left( x \right) \right] \\ =2\left[ f\left( x \right) -g\left( x \right) f\left( x \right) \right] =0$
Thus

${ f }^{ 2 }\left( x \right) +{ g }^{ 2 }\left( x \right)$ is constant.
Therefore

${ f }^{ 2 }\left( 16 \right) +{ g }^{ 2 }\left( 16 \right) ={ f }^{ 2 }\left( 2 \right) +{ g }^{ 2 }\left( 2 \right) \\ ={ f }^{ 2 }\left( 2 \right) +{ \left( f'\left( 2 \right) \right) }^{ 2 }=16+16=32$

Let

$f\left( x \right)=\sqrt { x-1 } +\sqrt { x+24-10\sqrt { x-1 } } ;1<x<26$ be a real valued function. Then

$f'(x)$ for

$1<x<26$ is

0%
$0$
0%
$\displaystyle \frac { 1 }{ \sqrt { x-1 } }$
0%
$2\sqrt { x-1 } -5$
0%
none of these

Explanation

Let,

$f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}$

$1< x< 26$ be a real valued function ,
We have,

$f(x) = \sqrt{x -1} + \sqrt{x + 24 -10\sqrt{x-1}}$
This can be written as

$f(x) = \sqrt{x -1} + \sqrt{\left ( 5 - \sqrt{x-1} \right )^{2} }$

$\because 1< x< 26$

$f(x) = \sqrt{x-1} + 5 - \sqrt{x-1}$

$f(x) = 5$
On differentiating wrt x , we get

$f{}'(x) = 0$
Hence ,Option A

$\displaystyle y=\sec ^{ -1 }{ \left( \frac { x+1 }{ x-1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) }$ then

$\displaystyle \frac{dy}{dx}$ is equal to

0%
$0$
0%
$\displaystyle x+1$
0%
$1$
0%
$-1$

Explanation

$\displaystyle y=\sec ^{ -1 }{ \left( \frac { x+1 }{ x-1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) }$

$\displaystyle =\cos ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) } +\sin ^{ -1 }{ \left( \frac { x-1 }{ x+1 } \right) } =\frac { \pi }{ 2 }$
However the above function is defined only for values of x, given by

$\displaystyle -1\le \frac { x-1 }{ x+1 } \le 1$
i.e.

$\displaystyle \frac { x-1 }{ x+1 } +1\ge 0$ and

$\displaystyle \frac { x-1 }{ x+1 } -1\le 0$
i.e.

$\displaystyle \frac { 2x }{ x+1 } \ge 0$ and

$\displaystyle \frac { 2 }{ x+1 } \ge 0$
i.e

$x<-1$ or

$\ge 0$ and

$x>-1$
i.e

$x\ge 0$
Hence we have

$\displaystyle y=\frac { \pi }{ 2 } ,\quad x\ge 0\Rightarrow \frac { dy }{ dx } =0,\quad x>0$

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:

0%
$\displaystyle y=\frac{1}{x^{2}}$
0%
$\displaystyle y=\sqrt{x}$
0%
$\displaystyle y=\frac{1}{\sqrt{x}}$
0%
none

Explanation

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,

$y=\cfrac{1}{\sqrt{x}}$

$\displaystyle \lim_{x\rightarrow 0}(f(x)\:g(x))$ exists for any functions

$f$ and

$g$ then

0%
$\displaystyle \lim_{x\rightarrow a}f(x)$ and $\displaystyle \lim_{x\rightarrow a}g(x)$ exist
0%
$\displaystyle \lim_{x\rightarrow a}f(x)$ exist but $\displaystyle \lim_{x\rightarrow a}g(x)$ may not exist
0%
$\displaystyle \lim_{x\rightarrow a}f(x)$ may not exist but $\displaystyle \lim_{x\rightarrow a}g(x)$ exist
0%
$\displaystyle \lim_{x\rightarrow a}f(x)$ and $\displaystyle \lim_{x\rightarrow a}g(x)$ may not exist

Explanation

Take

$\displaystyle f(x) = \begin{cases} 1 & x> 0 \\ -1 & x< 0 \end{cases}$ and

$\displaystyle g(x) = \begin{cases} 1 & x> 0 \\ -1 & x< 0 \end{cases}$ . Then

$f(x)\:g(x)=1$ for

$x\neq 0$ . Hence

$\displaystyle \displaystyle \lim_{x\rightarrow 0}f(x)$ and

$\displaystyle \lim_{x\rightarrow 0}g(x)$ does not exist but

$\displaystyle\lim_{x\rightarrow 0}f(x)\:g(x)=1$
Hence, option 'D' is correct.

If for a non-zero

$x,$ the function

$f(x)$ satisfies the equation

$\displaystyle af\left( x \right)+bf\left( \frac { 1 }{ x } \right) =\frac { 1 }{ x } -5\left( a\neq b \right)$ then

$f'(x)$ is equal to

0%
$\displaystyle \frac { 1 }{ { b }^{ 2 }-{ a }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } +b \right)$
0%
$\displaystyle \frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } +b \right)$
0%
$\displaystyle \frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left( \frac { a }{ { x }^{ 2 } } -b \right)$
0%
none of these

Explanation

We have,

$\displaystyle af\left( x \right)+bf\left( \frac { 1 }{ x } \right) =\frac { 1 }{ x } -5$

Substituting

$\displaystyle\frac{1}{x}$ in place of

$x,$ we have

$\displaystyle af\left( \frac { 1 }{ x } \right) +bf\left( x \right)=x-5$ ...(2)

Eliminating

$\displaystyle f\left( \frac { 1 }{ x } \right)$ from equation (1) and (2), we have

$\displaystyle \left( { a }^{ 2 }-{ b }^{ 2 } \right) f\left( x \right) =a\left( \frac { 1 }{ x } -5 \right) -b\left( x-5 \right)$

$\displaystyle \Rightarrow f\left( x \right) =\frac { 1 }{ { a }^{ 2 }-{ b }^{ 2 } } \left[ \frac { a }{ x } -bx+5\left( b-a \right) \right]$

$\displaystyle \therefore f'\left( x \right) =\frac { 1 }{ { b }^{ 2 }-{ a }^{ 2 } } \left[ \frac { a }{ { x }^{ 2 } } +b \right]$

${ S }_{ n }$ denotes the sum of

$n$ terms of

$g.p$ . whose common ratio is

$r$ , then

$\displaystyle \left( r-1 \right) \frac { d{ S }_{ n } }{ dr }$ is equal to

0%
$\left( n-1 \right) { S }_{ n }+n{ S }_{ n-1 }$
0%
$\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }$
0%
$\left( n-1 \right) { S }_{ n }$
0%
None of these

Explanation

We have,

$\displaystyle {S}_{n}=\frac { a\left( { r }^{ n }+1 \right) }{ r-1 }$

$\Rightarrow \left( r-1 \right) { S }_{ n }={ ar }^{ n }-a$

Differentiating both sides with respect to

$r,$ we get

$\displaystyle \left( r-1 \right) \frac { { dS }_{ n } }{ dr } +{ S }_{ n }=na{ r }^{ n-1 }-0$

$\displaystyle \Rightarrow \left( r-1 \right) \frac { { dS }_{ n } }{ dr } =na{ r }^{ n-1 }-{ S }_{ n }$

$=n[n$ th term from of

$G.P.]$

$-{S}_{n}$

$=n\left( { S }_{ n }-{ S }_{ n-1 } \right) -{ S }_{ n }$

$=\left( n-1 \right) { S }_{ n }-n{ S }_{ n-1 }.$

Let

$y = \sqrt { x + \sqrt { x + \sqrt { x + ......\infty } } }$ then

$\displaystyle\frac { dy }{ dx }$

0%
$\displaystyle\frac { 1 }{ 2y-1 }$
0%
$\displaystyle\frac { x }{ x-2y }$
0%
$\displaystyle\frac { 1 }{ \sqrt { 1+4x } }$
0%
$\displaystyle\frac { y }{ 2x+y }$

Explanation

$y=\sqrt { x+\sqrt { x+\sqrt { x+......\infty } } } \\ \therefore y=\sqrt { x+y } \\ Squaring\quad on\quad both\ sides,\\ y^{ 2 }=x+y\\ Differentiating\ on\ both\ sides,\\ 2y\dfrac { dy }{ dx }=1+\dfrac { dy }{ dx } \\ \therefore \dfrac { dy }{ dx }=\dfrac { 1 }{ 2y-1 } \\ \Rightarrow \dfrac { 1 }{ \sqrt { (2y-1)^{ 2 } } }=\dfrac { 1 }{ \sqrt { 4y^{ 2 }-4y+1 } }=\dfrac { 1 }{ \sqrt { 1+4x } } \\ \Rightarrow \dfrac { \sqrt { x+y } }{ \sqrt { (1 + 4x)(x+y) } } = \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { 4x^{ 2 } + 4xy + x + y } } \\\Rightarrow \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { 4x^{ 2 } + 4xy + y^{ 2 } } } = \dfrac { \sqrt { y^{ 2 } } }{ \sqrt { (2x+y)^{ 2 } } } = \dfrac { y }{ 2x + y }$

$\underset{x\rightarrow0}{lim}\displaystyle\frac{1-cos^{3}x+sin^{3}x+\ell n(1+x^{3})+\ell n(1+cos\,\,x)}{x^{2}-1+2\,cos^{2}x+tan^{4}x+sin^{3}x}$ is equal to -

0%
$\,\,\displaystyle\frac{3}{4}$
0%
$\,\,ln2$
0%
$\,\,\displaystyle\frac{ln2}{4}$
0%
$\,\,3/2$

Explanation

$\displaystyle \underset { x\rightarrow 0 }{ lim } \frac { 1-cos^{ 3 }x+sin^{ 3 }x+\ell n(1+x^{ 3 })+\ell n(1+cos\, \, x) }{ x^{ 2 }-1+2\, cos^{ 2 }x+tan^{ 4 }x+sin^{ 3 }x }$

$\displaystyle =\frac { 1-1+0+\ln { \left( 1+0 \right) } +\ln { \left( 1+1 \right) } }{ 0-1+2+0+0 } =\ln { 2 }$

$\displaystyle f^{ ' }\left( x \right) =g\left( x \right)$ and $\displaystyle g^{ ' }\left( x \right) =-f\left( x \right)$ for all real x and $\displaystyle f\left( 5 \right) =2=f^{ ' }\left( 5 \right)$ then $\displaystyle f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right)$ is -

0%
$2$
0%
$4$
0%
$8$
0%
None of these

Explanation

Given, $\displaystyle f^{ ' }\left( x \right) =g\left( x \right)$ and $g^{ ' }\left( x \right) =-f\left( x \right)$

Now $\displaystyle \frac { d }{ dx } \left[ f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right) \right] =2f\left( x \right) f^{ ' }\left( x \right) +2g\left( x \right) g^{ ' }\left( x \right)$

$\displaystyle =2f\left( x \right) g\left( x \right) -2g\left( x \right) f\left( x \right) =0$

$\displaystyle \therefore \quad f^{ 2 }\left( x \right) +g^{ 2 }\left( x \right)$ =constant

$\displaystyle f^{ 2 }\left( 5 \right) +g^{ 2 }\left( 5 \right) = 4 + 4 = 8$

$\displaystyle \therefore \quad f^{ 2 }\left( 10 \right) +g^{ 2 }\left( 10 \right)$ = 8

$\displaystyle \frac{d}{dx}\left ( \tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right ) \right )$ equals

$\displaystyle ($ for

$x\geq 0)$

0%
$\displaystyle \frac{1}{2\sqrt{x}(1+x)}-\frac{1}{1+x^{2}}$
0%
$\displaystyle \frac{1}{2\sqrt{x}(1+x)}+\frac{1}{1+x^{2}}$
0%
$\displaystyle \frac{1}{1+x}-\frac{1}{1+x^{2}}$
0%
None of these

Explanation

Let,

$\displaystyle y=\tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right )$

$\displaystyle y=\tan ^{-1}(\sqrt{x})-\tan ^{-1}x$

$\therefore \displaystyle \frac{dy}{dx}=\frac{1}{(1+x)}\times \frac{1}{2\sqrt{x}}-\frac{1}{1+x^{2}}$

Evaluate

$\displaystyle \lim_{n\rightarrow \infty }\left [ \frac{n!}{n^{n}} \right ]^{1/n}$ .

0%
$\displaystyle\frac{1}{e}$
0%
$\displaystyle\frac{1}{t}$
0%
$\displaystyle\frac{1}{n}$
0%
$none\ of\ above$

Explanation

Let

$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \frac{n!}{n^{n}} \right )^{1/n}$

$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \frac{1.2.3.4...n}{n.nnn....n} \right )^{1/n}$

$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \left ( \frac{1}{n} \right )\left ( \frac{2}{n} \right )\left ( \frac{3}{n} \right )...\left ( \frac{n}{n} \right ) \right )^{1/n}$

$\displaystyle\Rightarrow ln P=\lim_{n\rightarrow \infty }\frac{1}{n}\left ( \log \left ( \frac{1}{n} \right )+\log \left ( \frac{2}{n} \right )+...\log \left ( \frac{n}{n} \right ) \right )$

$\displaystyle P=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{n}\log \frac{r}{n}$

$=\int_{0}^{1}lnxdx=\left [ xlnx-x \right ]^{1}_{0}$

$=\displaystyle \left ( 0-1 \right )-\lim_{x\rightarrow 0 }\left ( xlnx \right )+0$

$=\displaystyle -1-\lim_{x\rightarrow 0 }\frac{lnx}{1/x}=-1-\lim_{x\rightarrow 0 }\frac{1/x}{\left ( -1/x^{2} \right )}$

$=\displaystyle -1-\lim_{x\rightarrow 0 }x=-1+0=-1\Rightarrow lnP=-1$

$\displaystyle P=e^{-1}=1/e$

$\displaystyle \frac{d}{dx}\left ( \tan ^{-1}\left ( \frac{a-x}{1+ax} \right ) \right )$ equals if ax > -1

0%
$\displaystyle \frac{a}{1+x^{2}}$
0%
$\displaystyle \frac{1}{1+x^{2}}$
0%
$\displaystyle- \frac{a}{1+x^{2}}$
0%
$\displaystyle -\frac{1}{1+x^{2}}$

Explanation

Let,

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

$\displaystyle y=\tan ^{-1}(a)-\tan ^{-1}(x)$

$\therefore \displaystyle \frac{dy}{dx}=-\frac{1}{1+x^2}$

$f(x) = \displaystyle \left | \cos x-\sin x \right |$ then

$\displaystyle f'\left ( \dfrac{\pi}4 \right )$ is equal to-

0%
$\displaystyle \sqrt{2}$
0%
$\displaystyle -\sqrt{2}$
0%
$0$
0%
$Does\ not\ exist$

Explanation

$f\left( x \right) =\left| \cos { x } -\sin { x } \right|$

$\displaystyle =\begin{cases} \cos { x-\sin { x } \quad \quad x<\dfrac { \pi }{ 4 } } \\ -\cos { x+\sin { x } } \quad x>\dfrac { \pi }{ 4 } \end{cases}$

$\displaystyle f'\left( x \right) =\begin{cases} -\sin { x } -\cos { x } \quad x<\dfrac { \pi }{ 4 } \\ +\sin { x+\cos { x } \quad x>\dfrac { \pi }{ 4 } } \end{cases}$
Hence

$\displaystyle f'\left( \dfrac { \pi }{ 4 } \right)$ does not exists.

$y = sec x^0$ then

$\displaystyle \frac{dy}{dx} =$

0%
$sec x tan x$
0%
$sec x^0 tan x^0$
0%
$\displaystyle \frac{\pi}{180} sec x^0 tan x^0$
0%
$\displaystyle \frac{180}{\pi} sec x^0 tan x^0$

Explanation

Give

$y = sec x^0$

$= \displaystyle sec \frac{x \pi}{180^o} \left ( \because x^o = \displaystyle \frac{x \pi}{180^o} radians \right )$

$\therefore \displaystyle \frac{dy}{dx} = \frac{\pi}{180^o} sec x^o tan x^o$

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.