MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 3

Solve:

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\tan x-\tan 3x}{2x^{3}}$

0%
$\dfrac{1}{4}$
0%
$\dfrac{3}{4}$
0%
$4$
0%
$-4$

Explanation

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3\tan x-\tan 3x}{2x^{3}}$

$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3(x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+......)-(3x+\dfrac{(3x)^3}{3}+\dfrac{2(3x)^5}{15}+....)}{2x^{3}}=\dfrac{1-9}{2}=-4$

$\displaystyle \lim_{x\rightarrow 0}(\frac{\sin x-x}{x})(\sin\frac{1}{x})$ is:

0%
does not exist
0%
is equal to 0
0%
is equal to 1
0%
exists and different from 0 and 1

Explanation

$sin x = x -\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}....(1)$

$sin\ x-x =-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}$

$\dfrac{sin\ x-x}{x} =-\dfrac{x^{2}}{3!}+\dfrac{x^{4}}{5!}-\dfrac{x^{6}}{7!}....=0\ (as\ x\ found\ to\ 0)$
sin

$(\dfrac{1}{x})$ is an oscillatory function means same finite no.

$0\times finite\ no.=0$

$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$ =

0%
$2$
0%
$1$
0%
$0$
0%
$3$

Explanation

$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$

$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{1-\tan x}{(\frac{\pi}{4}-x)(1+\tan x)}$ .................(dividing the num and deno by cosx)

$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\tan\left(\frac{\pi}{4}-x\right) }{(\frac{\pi}{4}-x)} = \lim_{h\to 0}\frac{\tan h}{h} = 1$

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}$ =

0%
1
0%
0
0%
-1
0%
$\displaystyle \dfrac{1}{\pi}$

Explanation

$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { (\dfrac { \pi }{ 2 } -x)\sec x }{ \csc { x } }$

$\displaystyle=\lim _{ h\rightarrow 0 } \dfrac { (\dfrac { \pi }{ 2 } -\dfrac { \pi }{ 2 } +h)\sec { (\dfrac { \pi }{ 2 } -h) } }{ \csc { (\dfrac { \pi }{ 2 } -h } ) }$

$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\csc { h } }{ \sec { h } }$

$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\cos { h } }{ \sin { h } } =1$

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$ =

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{6}$
0%
$\displaystyle \frac{1}{8}$

Explanation

$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$

$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\sin (\dfrac { \pi }{ 2 } -h) }{ (\pi -2(\dfrac { \pi }{ 2 } -h))^{ 2 } }$ ......(x-->h-pi/2)

$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\cos { h } }{ 4h^{ 2 } }$

$\displaystyle =\dfrac { 1 }{ 8 }$

$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}=$

0%
$1$
0%
$-1$
0%
$\dfrac{1}{3}$
0%
$0$

Explanation

$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}$

$=\displaystyle \lim_{x\rightarrow \infty }\frac{\displaystyle 1+\frac{\sin x}{x}}{\displaystyle 1+ \frac{\cos x}{x}}$

$=\displaystyle \frac{\displaystyle 1+\frac{\text{Any finite number between -1 and 1}}{\infty}}{\displaystyle 1+ \frac{\text{Any finite number between -1 and 1}}{\infty}}=\frac{1+0}{1+0}=1$

$\displaystyle { f }({ x })=\sqrt { \frac { { x }-\sin ^{ 2 }{ x } }{ { x }+\cos { x } } }$ ,then

$\displaystyle \lim _{ x\rightarrow \infty } f(x)$ =

0%
$\dfrac{1}{2}$
0%
$-1$
0%
$0$
0%
$1$

Explanation

$\displaystyle \lim _{ x\rightarrow \infty } f(x)$

$\displaystyle = \lim _{ x\rightarrow \infty }\sqrt { \dfrac { { x }-\sin ^{ 2 }{ x } }{ { x }+\cos { x } } }=\lim _{ x\rightarrow \infty }\sqrt { \dfrac { 1-\dfrac{\sin^2x}{x}}{ 1+\dfrac{\cos x}{x} }}=\sqrt{\dfrac{1-0}{1-0}}=1$

$\mathrm{f}(\mathrm{x})=0$ has a repeated root

$\alpha$ , then another equation having

$\alpha$ as root, is

0%
$\mathrm{f}(2\mathrm{x})=0$
0%
$\mathrm{f}(3\mathrm{x})=0$
0%
$\mathrm{f}^{'}(\mathrm{x})=0$
0%
$\mathrm{f}^{''}(\mathrm{x})=0$

Explanation

Repeated roots mean that the curve will touch the

$x$ -axis.
The curve will have a convex or concave part at the repeated root

$\alpha$ or a point of inflection.
In any case

$f'(x)$ will always be zero at the point of repeated root

$\alpha$ .

$\displaystyle \lim_{x\rightarrow \infty }(\sin\sqrt{x+1}-\sin\sqrt{x})=$

0%
2
0%
-2
0%
0
0%
None of these

Explanation

$\displaystyle \lim _{ x\rightarrow \infty }{ \left( \sin { \sqrt { x+1 } } -\sin { \sqrt { x } } \right) }$

$=\displaystyle \lim _{ x\rightarrow \infty }{ 2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x } }{ 2 } \right) \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } \right) } } }$

Now for

$\displaystyle \lim _{ x\rightarrow \infty }{ { \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } \right) } } }$

$=\displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } \right) } }{ \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } } \times \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } }$ (sandwich theorem)

$=\displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } }$

$=\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { \sqrt { x+1 } -\sqrt { x } }{ \sqrt { x+1 } +\sqrt { x } } \cdot \cfrac { \sqrt { x+1 } +\sqrt { x } }{ 1 } }$

$=\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow \infty }{ \cdot \cfrac { 1 }{ \sqrt { x+1 } +\sqrt { x } } } =\cfrac { 1 }{ 2 } \times 0=0$

Now,

$2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x } }{ 2 } \right) }$ is always finite and lies between

$\left[ -2,2 \right]$

$\therefore \displaystyle \lim _{ x\rightarrow \infty }{ 2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x } }{ 2 } \right) \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } \right) } } } =finite\times 0$

$=0$

$\displaystyle \lim_{x\rightarrow \infty }x\displaystyle \cos\left(\frac{\pi}{8x}\right)\sin\left(\frac{\pi}{8x}\right)=$

0%
$\displaystyle \pi$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\dfrac{\pi}{8}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

$x \rightarrow \infty , \cos \left (\dfrac{\pi}{8x}\right) \rightarrow 1$
Looking at the rest of the part,

$\underset{ x \rightarrow \infty}{\lim} \dfrac{\pi}{8x} \rightarrow 0$

$\underset{ x \rightarrow \infty}{\lim} x.\sin (\dfrac{\pi}{8x})$

$=\underset{ x \rightarrow \infty}{\lim} \dfrac{\sin (\dfrac{\pi}{8x})}{\dfrac{1}{x}}$
We multiply and divide by

$\dfrac{\pi}{8}$

$=\underset{ x \rightarrow \infty}{\lim} , \dfrac{\sin \left (\dfrac{\pi}{8x}\right)}{\dfrac{\pi}{8x}} \times \dfrac{\pi}{8x}$
Now using the property of a limit,

$\underset{ y \rightarrow 0}{\lim} , \dfrac{\sin y}{y}$

$= 1$
We get, applying the limit,

$= 1 \times$

$\dfrac{\pi}{8}$

$=$

$\dfrac{\pi}{8}$

$\displaystyle \lim_{x\rightarrow\infty}\frac{\sin^{4}x-\sin^{2}x+1}{\cos^{4}x-\cos^{2}x+1}$ is equal to

0%
$0$
0%
$1$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$

Explanation

For all

$x$ ,

${sin}^{4} x = {({sin}^{2} x)}^{2}$

$= {(1 - {cos}^{2} x)}^{2}$

$= 1 - 2 {cos}^{2} x + {cos}^{4} x$
Thus numerator

$= 1 - 2 {cos}^{2} x + {cos}^{4} x + {cos}^{2} x = 1 - {cos}^{2} x + {cos}^{4} x$
Hence, for all x, numerator

$=$ denominator.
Thus the limit

$= 1$ for all

$x$

$\displaystyle \lim_{x\rightarrow \displaystyle \frac{\pi }{2}}\displaystyle \frac{1-\sin\theta}{\cos\theta\left(\dfrac{\pi}{2}-{\theta}\right)}=$

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

Explanation

$Put\ \theta = \dfrac{\pi }{2}-\phi$

$\begin{matrix}Lim\\\phi \rightarrow 0 \end{matrix}\ \dfrac{1-cos\phi}{\dfrac{sin\phi}{\phi}\phi ^{2}}=\dfrac{1}{2}$

$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})$

0%
$1$
0%
$0$
0%
$2$
0%
does not exist

Explanation

$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})=\lim_{x\rightarrow \infty }\frac{\sin(2^{x})}{2^x}=\frac{\mbox{Any finite value between -1 and 1}}{\infty} = 0$

Evaluate:

$\displaystyle \underset{x\rightarrow 0}{\lim}\ \ \frac{\sin3x^{2}}{\cos(2x^{2}-x)}$

0%
$0$
0%
$-1$
0%
$4$
0%
$-6$

Explanation

Given,

$\lim _{x\to \:0}\left(\dfrac{\sin \left(3\right)x^2}{\cos \left(2x^2-x\right)}\right)$

$=\dfrac{\sin \left(3\right)\cdot \:0^2}{\cos \left(2\cdot \:0^2-0\right)}$

$=0$

$\displaystyle \lim_{x\rightarrow \infty }\frac{2x+7\sin x}{4x+3\cos x}=$

0%
$1$
0%
$-1$
0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$

Explanation

We divide both the numerator and denominator by x.
Since, from the property of trigonometric functions we know that

$sin x$ and

$cos x$ can only have their values between

$-1$ and

$1$
Thus, the expression transforms to,

$lim _{x \rightarrow \infty } \dfrac{2 + 7\dfrac{sin x}{x}}{4 + 3\dfrac{cos x}{x}}$
Now applying the limit,

$= \dfrac{2 +0}{4+0}$

$= \dfrac{1}{2}$
Hence, option 'C' is correct.

$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}=$

0%
${e}$
0%
$e^{2}$
0%
$-1$
0%
1

Explanation

$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}$

$\log { k } =Lt_{ x\rightarrow 0^+ }\tan { x } \log { \sin { x } }$

$k={ e }^{ Lt_{ x\rightarrow 0^+ }\tan { x } \log { \sin { x } } }$

$k={ e }^{ Lt_{ x\rightarrow 0^+ }\displaystyle \frac { \log { \sin { x } } }{ \cot { x } } }$

It is of the form

$\displaystyle \frac{\infty}{\infty}$ , so applying L-Hospital's rule

$k={ e }^{ Lt_{ x\rightarrow 0^+ }\displaystyle \frac { \cot { x } }{ -\csc ^{ 2 }{ x } } }$

$k={ e }^{ 0 }=1$

The value of

$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( \pi \cos ^{ 2 }{ x } \right) } }{ { x }^{ 2 } } }$ is

0%
$-\pi$
0%
$\displaystyle \frac { \pi }{ 2 }$
0%
$\pi$
0%
$\displaystyle \frac { 3\pi }{ 2 }$

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( \pi \cos ^{ 2 }{ x } \right) } }{ { x }^{ 2 } } }=\lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi (1-\sin ^{ 2 }{ x }) \right] } }{ { x }^{ 2 } } }$

$\quad =\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi-\pi\sin ^{ 2 }{ x } \right] } }{ { x }^{ 2 } } }=\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi\sin ^{ 2 }{ x } \right] } }{ { x }^{ 2 } } }, [\because \sin(\pi-\theta)=\sin\theta]$

$\quad \displaystyle =\pi\lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi\sin ^{ 2 }{ x } \right] } }{ \pi\sin^2x } }\left( \frac{\sin x}{x}\right)^2 =\pi\cdot 1\cdot 1=\pi$

$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$ exists and finite then

$\mathrm{a}=$

0%
$2$
0%
$-2$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{-2}{3}$

Explanation

$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$

$Lt_{ x\rightarrow 0 }\dfrac { \sin x(2\cos { x } +a) }{ x^{ 3 } }$

$=Lt_{ x\rightarrow 0 }\dfrac { (2\cos { x } +a) }{ x^{ 2 } }$

$x\rightarrow 0$ , denominator tends to 0, so the numerator also tends to 0.

$\Rightarrow Lt_{ x\rightarrow 0 } 2\cos { x } +a=0$

$\Rightarrow a=-2$

Hence, option 'B' is correct.

$\displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}=$

0%
$\mathrm{e}$
0%
$\mathrm{e}^{4}$
0%
$\mathrm{e}^{-1}$
0%
$\mathrm{e}^{-4}$

Explanation

Let,

$\text{L}= \displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}$
Put

$y=\pi-x\Rightarrow x\to \pi\Leftrightarrow y\to 0$

$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1- 4 \tan \mathrm{(\pi-y)} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{(\pi-y)}}$

$\quad \quad = \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{-\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{y}}[\because \tan (\pi-y)=-\tan y]$
Clearly form of the limit is

$1^{\infty}$

$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{\dfrac{1}{4\tan y}\times(-4)}=e^{-4}$

$f(x)=\displaystyle \frac{x}{\sqrt{1-x^{2}}},g(x)=\frac{x}{\sqrt{1+x^{2}}}$ , then

$\displaystyle \frac{d}{dx}(fog (x))=$

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

Explanation

$\displaystyle f(g(x)) = \frac{g(x)}{\sqrt{1-(g(x))^2}}$ provided

$g(x) \neq 1$

$\displaystyle \quad = \frac{\dfrac{x}{\sqrt{1+x^2}}}{\sqrt{1-\dfrac{x^2}{1+x^2}}} =x$

$\therefore \dfrac{d}{dx}(fog(x)) = \dfrac{d}{dx}(x) =1$

The integer

$n$ for which

$\displaystyle \lim_{x\rightarrow 0}\frac{(\cos x-1)(\cos x-e^{x})}{x^{n}}$ is finite non zero number is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { (\cos { x } -1)\left( \cos { x } -{ e }^{ x } \right) }{ { x }^{ n } } } =\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { \left( \cos { x } -1 \right) }{ { x }^{ 2 } } \right] } \times \cfrac { { \cos { x } -{ e }^{ x } } }{ { x }^{ n-2 } }$

$=-\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { \cos { x } -{ e }^{ x } } }{ { x }^{ n-2 } } } =\cfrac { 0 }{ 0 }$ form

$=-\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { -\sin { x } -{ e }^{ x } } }{ (n-2){ x }^{ n-3 } } } =\cfrac { 1 }{ 2(n-2) } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { 1 } }{ { x }^{ n-3 } } }$

$\therefore$ for the above limit to be finite

$n-3=0$

$\Rightarrow n=3$

The right-hand limit of the function

$\sec{x}$ at

$\displaystyle x=-\frac { \pi }{ 2 }$ is

0%
$-\infty$
0%
$-1$
0%
$0$
0%
$\infty$

Explanation

Function

$\displaystyle f\left( x \right)=\sec { x }$ and point

$\displaystyle x=-\left( \frac { \pi }{ 2 } \right)$ .

We know that right

$-$ hand limit of the function

$f(x)$ at

$x=-\left( \frac { \pi }{ 2 } \right)$ is

$\displaystyle \lim _{ x\rightarrow { \left[ -\left( \pi /2 \right) \right] }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { \left[ -\left( \pi /2 \right) \right] } }{ \sec { x } } .$

Substituting

$\displaystyle x=h+\left( -\frac { \pi }{ 2 } \right)$ and

$h\rightarrow 0,$ we get

$\displaystyle \lim _{ h\rightarrow 0 }{ \sec { \left[ h+\left( -\frac { \pi }{ 2 } \right) \right] } = } \lim _{ h\rightarrow 0 }{ \sec { \left( \frac { \pi }{ 2 } -h \right) } } =\lim _{ h\rightarrow 0 }{ \csc { h } } =\csc { 0 } =\infty$

$\displaystyle \left[ \because \sec { \left( -\theta \right) = } \sec { \theta } \right]$

$\displaystyle \lim_{x\rightarrow 1}(2-x)^{\displaystyle \tan( \frac{\pi x}{2})}=$

0%
$e^{\displaystyle \frac{1}{\pi}}$
0%
$e^{\displaystyle \frac{2}{\pi}}$
0%
$-e^{\displaystyle \frac{2}{\pi}}$
0%
$\mathrm{e}$

Explanation

Let

$k=\lim _{ x\rightarrow 1 } (2-x)^{ \tan \left(\dfrac { \pi x }{ 2 } \right) }$

$\log { k } =\lim _{ x\rightarrow 1 } tan\left( \dfrac { \pi x }{ 2 } \right) \log { \left( 2-x \right) }$

$\displaystyle k={ e }^{ \lim _{ x\rightarrow 1 } tan\left( \displaystyle\dfrac { \pi x }{ 2 } \right) \log { \left( 2-x \right) } }$

$\displaystyle k={ e }^{ \lim _{ x\rightarrow 1 } \displaystyle \dfrac { \log { \left( 2-x \right) } }{ \cot { \left( \dfrac { \pi x }{ 2 } \right) } } }$

$k={ e }^{ \lim _{ x\rightarrow 1 } \displaystyle \dfrac { -1 }{ -(2-x)\csc ^{ 2 }{ \left( \dfrac { \pi x }{ 2 } \right) \left( \dfrac { \pi }{ 2 } \right) } } }$

$\Rightarrow k={e}^{\displaystyle\dfrac{2}{\pi}}$

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

0%
0
0%
1
0%
2
0%
does not exist

Explanation

$\lim_{x\rightarrow 0^{-}} \dfrac{-2}{\dfrac{2x}{(1+x^{2})}}\dfrac{sin^{-1}}{(1+x^{2})}\dfrac{(2x)}{1+x^{2}}=-2$

$\lim_{x\rightarrow 0^{+}} \dfrac{2}{(\dfrac{2x}{1+x^{2}})(1+x^{2})} sin^{-1}(\dfrac{2x}{1+x^{2}})=2$

$L.H.S\neq R.H.S$

$f(x)=(ax+b)\cos x + (cx+d)\sin x$ and

$f^{'}(x)=x \cos x$ , for all values of

$x\in R$ , then

$a,b,c,d$ are given by

0%
$a = b = c = d$
0%
$0, 1, -1, 0$
0%
$1, 0, -1, 0$
0%
$0, 1, 1, 0$

Explanation

Given,

$f(x) = (ax+b)\cos x + (cx+d)\sin x$
Thus

$f'(x) = (ax+b)(-\sin x) + a\cos x + c\sin x + (cx+d)\cos x$

$= (-ax-b+c)\sin x + (a+cx+d)\cos x$
Also given,

$f'(x) = x\cos x$

Comparing the coefficients, we get

$a =0, b=1, c =1, d=0$

$\displaystyle \lim_{x\to0}{\displaystyle \frac{x^n - \sin^nx}{x - \sin^nx}}$ is non-zero finite, then

$n$ must be equal

0%
4
0%
1
0%
2
0%
3

Explanation

$\displaystyle \lim_{x\to0}{\displaystyle \frac{x^n - \sin^nx}{x - \sin^nx}}$

$\mbox{For n = 0,we have }\displaystyle\lim_{x\to0}\displaystyle\frac{1-1}{x-1}=0$

$\mbox{For n = 1,}\displaystyle\lim_{x\to0}\displaystyle\frac{x-\sin x}{x- \sin x}=1\\$

$\mbox{For n = 2,}\displaystyle\lim_{x\to0}\displaystyle\frac{x^2-\sin^2x}{x-\sin^2x}=\displaystyle\lim_{x\to0}\displaystyle\frac{1-\displaystyle\frac{\sin^2x}{x^2}}{\displaystyle\frac{1}{x}-\displaystyle\frac{\sin^2x}{x^2}}.\\$

$\mbox{This does not exist.}\\$

$\displaystyle { For\ \ n=3, }\lim _{ x\to 0 } \dfrac { x^{ 3 }-\sin ^{ 3 } x }{ x-\sin ^{ 3 } x } =\lim _{ x\to 0 } \dfrac { 1-\dfrac { \sin ^{ 3 } x }{ x^{ 3 } } }{ \dfrac { 1 }{ x^{ 2 } } -\dfrac { \sin ^{ 3 } x }{ x^{ 3 } } }$
For

$n = 3$ , also given limit does not exist.
Hence,

$n = 1$

Assertion (A):

$\mathrm{f}(\mathrm{x})=\sin(\pi[x])$ is differentiable every where

$[\ ]$ is greatest integer function

Reason (R): lf

$\mathrm{x}=\mathrm{n}\pi\Rightarrow$

$\sin x$

$=0\ \forall \ \mathrm{n}\in \mathrm{Z}$ then

0%
Both (A) and (R) are true and R is correct explanation for A
0%
Both (A) and (R) are true and R is not correct explanation for A
0%
(A) is true (R) is false
0%
(A) is false (R) is true

Explanation

$[ x ]$ is an integer

$\ \forall \ x\epsilon R$
Hence, sin

$(\pi [ x ])=0 \ \forall\ x\epsilon \ R$
Hence,

$f(x) =0 \ \forall \ x\epsilon\ R$
Hence,

$f(x)$ is differentiable everywhere.

The Assertion is correct and the Reason provides the correct explanation for it.

$\displaystyle \lim_{x\to1}{\displaystyle \frac{1-x^2}{\sin 2\pi x}}$ is equal to

0%
$\displaystyle \frac{1}{2\pi}$
0%
$\displaystyle \frac{-1}{\pi}$
0%
$\displaystyle \frac{-2}{\pi}$
0%
None of these

Explanation

$\displaystyle\lim_{x\to1}\frac{1-x^2}{\sin2\pi x}= -\displaystyle \lim_{x\to1}\frac{2\pi(1-x)(1+x)}{-2\pi\sin(2\pi-2\pi x)}$

$=\displaystyle \lim_{x \to 1}\left( \frac{(2\pi-2\pi x}{\sin(2\pi-2\pi x)}\right).\frac{1+x}{-2\pi}=-\frac{1}{\pi}$

$\displaystyle\lim_{x \rightarrow \infty}(1^x + 2^x + 3^x+.........+n^x)^{1/x}$ is

0%
$\ln (n!)$
0%
$n$
0%
$n!$
0%
$0$

Explanation

Consider that

$y=\displaystyle\lim_{x\to\infty}(1^x+2^x+3^x+.........+n^x)^{1/x}$

take natural logarithm on both sides, we have

$\ln y=\displaystyle\lim_{x\to\infty}\ln\left[(1^x+2^x+3^x+.........+n^x)^{1/x}\right]$

$\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{\ln (1^x+2^x+3^x+.........+n^x)}{x}$

RHS has

$\dfrac{\infty}{\infty}$ form. Hence, use L'hospital's rule

$\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{1^x\ln 1+2^x\ln 2+3^x\ln 3+.........+n^x\ln n }{1^x+2^x+3^x+.........+n^x}$

On approximating this form, we have

$\ln y=\displaystyle\lim_{x\to\infty}\dfrac{n^x\ln n }{n^x}$

$\ln y=\ln n$

$\therefore y=n$

Hence, this is required answer.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

$y = 3\cot x + 5co\sec x$

$\cfrac { dy }{ dx } =-3{ co\sec }^{ 2 }x - 5co\sec x.\cot x = -co\sec x (3 co\sec x + 5\cot x)$

Therefore, option A is correct.

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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