If $$y=(\tan x)^{\cot x}$$ then $$\dfrac{dy}{dx}=?$$

$$(\tan x)^{\cot x}.\csc^2 x(1-\log \tan x)$$

$$\cot x.(\tan x)^{\cot x-1}.\sec^2x$$

If $$y=\sqrt{x\sin x}$$ then $$\dfrac{dy}{dx}=?$$

$$\dfrac{(x\cos x+\sin x)}{2\sqrt{x\sin x}}$$

$$\dfrac{1}{2}(x\cos x+\sin x).\sqrt{x\sin x}$$

If $$y=\sqrt{\dfrac{1+\sin x}{1-\sin x}}$$ then $$\dfrac{dy}{dx}=?$$

$$\dfrac{1}{2}\sec^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$$

$$\dfrac{1}{2}\csc \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\cot \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$$

If $$ f(x)=\dfrac{\cos x}{(1-\sin x)^{1 / 3}}, $$ then

$$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty } $$

$$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty } $$

$$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty } $$

MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 10

Differentiate the following function with respect to x.

$(2x^2-3)\sin x$ .

0%
$4x\sin x+(2x^2+3)\cos x$ .
0%
$4x\sin x+(2x^2-3)\sin x$ .
0%
$4x\sin x+(2x^2-3)\cos x$ .
0%
$4x\cos x+(2x^2-3)\cos x$ .

Explanation

Given expression

$(2x^2-3)\sin x$

differentiating w.r.t.

$x$ , we get

$=\dfrac d{dx} ((2x^2-3)\sin x)$

$=\dfrac d{dx}(2x^2-3)\sin x+2x^2-3\dfrac d{dx}(\sin x)$

$= 4x\sin x+(2x^2-3)\cos x$

$f$ is differentiable at

$x = 1$ and

$\underset{h \rightarrow 0}{\lim} \dfrac{1}{h} f (1 + h) = 5, f'(1) =$

0%
$0$
0%
$1$
0%
$3$
0%
$4$
0%
$5$

Explanation

$f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)-f(1)}{h}$ ; function is differentiable.

and

$\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5$ [Given function is continuous]

$\Rightarrow f(1)=0$

Hence,

$f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5$

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

$\dfrac{dy}{dx}=?$

0%
$\dfrac{\sin x}{(2y-1)}$
0%
$\dfrac{\cos x}{(y-1)}$
0%
$\dfrac{\cos x}{(2y-1)}$
0%
$none\ of\ these$

$y=(\tan x)^{\cot x}$ then

$\dfrac{dy}{dx}=?$

0%
$\cot x.(\tan x)^{\cot x-1}.\sec^2x$
0%
$-(\tan x)^{\cot x}.\csc^2x$
0%
$(\tan x)^{\cot x}.\csc^2 x(1-\log \tan x)$
0%
$none\ of\ these$

$y=\sqrt{x\sin x}$ then

$\dfrac{dy}{dx}=?$

0%
$\dfrac{(x\cos x+\sin x)}{2\sqrt{x\sin x}}$
0%
$\dfrac{1}{2}(x\cos x+\sin x).\sqrt{x\sin x}$
0%
$\dfrac{1}{2\sqrt{x\sin x}}$
0%
$none\ of\ these$

$x=a(\cos\theta+\theta\sin\theta)$ and

$y=a(\sin\theta-\theta\cos\theta)$ then

$\dfrac{dy}{dx}=?$

0%
$\cot\theta$
0%
$\tan\theta$
0%
$a\cot\theta$
0%
$a\tan\theta$

$x=a\sec\theta, y=b\tan\theta$ then

$\dfrac{dy}{dx}=$

0%
$\dfrac{b}{a}\sec\theta$
0%
$\dfrac{b}{a} \ cosec\theta$
0%
$\dfrac{b}{a}\cot\theta$
0%
$none\ of\ these$

$y=\sqrt{\dfrac{1+\sin x}{1-\sin x}}$ then

$\dfrac{dy}{dx}=?$

0%
$\dfrac{1}{2}\sec^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
0%
$\dfrac{1}{2}\csc^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
0%
$\dfrac{1}{2}\csc \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\cot \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
0%
$none\ of\ these$

Explanation

$y=\left\{\dfrac{1+\cos \left(\dfrac{\pi}{2}-x\right)}{1-\cos \left(\dfrac{\pi}{2}-x\right)}\right\}^{\dfrac{1}{2}}=\left\{\dfrac{2\cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}{2\sin^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}\right\}^{\dfrac{1}{2}}=\cot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$ .

$\Rightarrow \dfrac{dy}{dx}=-\csc^2\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)\times\dfrac{dy}{dx}\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)$

$\therefore \dfrac{dy}{dx}=\dfrac12\csc^2\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)$

$\lim _{ x\rightarrow 0 }{ \cfrac { x\left( 1+a\cos { x } \right) -b\sin { x } }{ { x }^{ 3 } } } =1$ then

0%
$a=-5/2$ , $b=-1/2$
0%
$a=-3/2$ , $b=-1/2$
0%
$a=-3/2$ , $b=-5/2$
0%
$a=-5/2$ , $b=-3/2$

The value of

$\displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x}$ is

0%
1/3
0%
2/3
0%
-1/4
0%
3/2

Explanation

We have

$\displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x}$

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

$=\displaystyle \lim _{x \rightarrow \pi} \dfrac{1-\cos x+\cos ^{2} x}{1-\cos x}=\dfrac{1+1+1}{1+1}=\dfrac{3}{2}$

$\displaystyle {f}'(x) = sin\,x + sin\,4x .\, cos \,x$ then

$\displaystyle {f}'(x) \left (2x^{2} + \dfrac{\pi}{2} \right )$ at

$x = \sqrt{\dfrac{\pi}{2}}$ is equal to

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

$\displaystyle sin\, y = x\, sin ( a + y)$ and

$\displaystyle \dfrac{dy}{dx} = \dfrac{A}{ 1 + x^{2} - 2x \, cos a }$ then the value of

$A$ is

0%
$2$
0%
$cos\, a$
0%
$sin\,a$
0%
None of these

$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}},(m<n)$ is equal to

0%
1
0%
0
0%
n/m
0%
None of these

Explanation

$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}}=\lim _{x \rightarrow 0}\left(\dfrac{\sin x^{n}}{x^{n}}\right)\left(\dfrac{x^{n}}{x^{m}}\right)\left(\dfrac{x}{\sin x}\right)^{m}$

$=\displaystyle \lim _{x \rightarrow 0} x^{n-m}=0 \quad[\because m<n]$

$The \ value \ of \displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}}$ is

0%
$e^{-2 \pi}$
0%
$e^{1 / \pi}$
0%
$e^{2 /\pi}$
0%
$e^{-1 / \pi}$

Explanation

$\displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}}$

$=\displaystyle \lim _{x \rightarrow 1}\{1+(1-x)\}^{\tan \dfrac{\pi x}{2}}$

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

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

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

$=e^{\dfrac{2}{\pi}\displaystyle \lim_{x \to 1}{\dfrac{\dfrac{\pi}{2}(1-x)}{tan \left(\dfrac{\pi}{2}(1-x)\right)}}}$

$=e^{\dfrac{2}{\pi}}$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x}$ is equal to

0%
0
0%
$\infty$
0%
-2
0%
2

Explanation

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \dfrac{2 x\left(e^{x}-1\right)}{4 \sin ^{2} \dfrac{x}{2}}$

$=2 \displaystyle \lim _{x \rightarrow 0}\left[\dfrac{(x / 2)^{2}}{\sin ^{2} \dfrac{x}{2}}\right]\left(\dfrac{e^{x}-1}{x}\right)=2$

$\displaystyle \lim _{x \to \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right]$ is equal to

0%
1
0%
-1
0%
0
0%
$None \ of \ these$

Explanation

$\displaystyle \lim _{x \rightarrow \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right]$

$=\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{2 x \sin x-\pi}{2 \cos x} \quad\quad\quad \left(\dfrac{0}{0}form\right)$

$=\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{[2 \sin x+2 x \cos x]}{-2 \sin x}$
(Applying L'Hopital's rule)

$=-1$

$\displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}}$ is equal to

0%
$-\dfrac{1}{2 \sqrt{2}}$
0%
$\dfrac{1}{2 \sqrt{2}}$
0%
$\dfrac{1}{\sqrt{2}}$
0%
Does not exist

Explanation

$\displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}}= \displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{-x \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}}$

$=-\displaystyle \lim _{x \rightarrow-\infty} \dfrac{\tan \dfrac{1}{x}}{\dfrac{1}{x} \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}}=-\dfrac{1}{2 \sqrt{2}}$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}$ is equal to

0%
1
0%
0
0%
2
0%
None of these

Explanation

$\dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}$

$=\dfrac{x^{4}\left(1+\tan ^{2} x+\tan ^{4} x\right)}{\tan ^{4} x\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\dfrac{x^{4}}{\tan ^{4} x}, x \neq 0$

$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}\left(\cot ^{4} x-\cot ^{2} x+1\right)}{\left(\tan ^{4} x-\tan ^{2} x+1\right)}=\displaystyle \lim _{x \rightarrow 0} \dfrac{x^{4}}{\tan ^{4} x}=1$

$f(x)=\dfrac{\cos x}{(1-\sin x)^{1 / 3}},$ then

0%
$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty }$
0%
$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty }$
0%
$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty }$
0%
none of these

Explanation

$\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{\cos x}{(1-\sin x)^{1 / 3}}=\lim _{t \rightarrow 0} \dfrac{-\sin t}{(1-\cos t)^{1 / 3}}$

$=-\lim _{t \rightarrow 0} \dfrac{2 \sin \dfrac{t}{2} \cos \dfrac{t}{2}}{\left(2 \sin ^{2} \dfrac{t}{2}\right)^{1 / 3}}$

$=-\lim _{t \rightarrow 0} 2^{2 / 3} \cos \dfrac{t}{2}\left(\sin \dfrac{t}{2}\right)^{1 / 3}=0$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}$ is equal to

0%
2
0%
-2
0%
1/2
0%
-1/2

Explanation

$\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{4 \sin ^{4} x}$

$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x}{4 \sin ^{4} x}\left[\dfrac{2 \tan x}{1-\tan ^{2} x}-2 \tan x\right]$

$=\displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan ^{3} x}{2 \sin ^{4} x\left(1-\tan ^{2} x\right)}$

$=\dfrac{1}{2} \lim _{x \rightarrow 0} \dfrac{x}{\sin x} \dfrac{1}{\cos ^{3} x} \dfrac{1}{1-\tan ^{2} x}$

$=\dfrac{1}{2} \times 1 \times \dfrac{1}{1^{3}} \times \dfrac{1}{1-0}=\dfrac{1}{2}$

$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x}$ is equal to

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

Explanation

$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x}$

$=\displaystyle \lim _{x \rightarrow 1} \dfrac{1-\cos \left(\dfrac{3 \pi}{2}-\dfrac{3 \pi x}{1+x^{2}}\right)}{1-\cos (\pi-\pi x)}$

$=\lim _{x \rightarrow 1} \dfrac{2 \sin ^{2}\left(\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}\right)}{2 \sin ^{2}\left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)}$

$=\displaystyle \lim _{x \rightarrow 1}\left(\dfrac{\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}}{\dfrac{\pi}{2}-\dfrac{\pi x}{2}}\right)^{2}$

$=\displaystyle \lim _{x \rightarrow 1} 9\left(\dfrac{\dfrac{1}{2}-\dfrac{x}{1+x^{2}}}{1-x}\right)^{2}=\lim _{x \rightarrow 1} 9\left(\dfrac{x-1}{2\left(1+x^{2}\right)}\right)^{2}=0$

$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x} \text { is equal to }$

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

Explanation

$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x}$

$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{2 \pi(1-x)(1+x)}{2 \pi \sin (2 \pi-2 \pi x)}$

$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{(2 \pi-2 \pi x)}{\sin (2 \pi-2 \pi x)} \dfrac{1+x}{2 \pi}=\dfrac{-1}{\pi}$

Let

$f(x)= \sin x+ax+b$ , then which of the following is/are true.

0%
$f(x)=0$ has on;ly one root which is possitive if $a>1, b<0$
0%
$f(x)=0$ has on;ly one root which is negative if $a>1, b<0$
0%
$f(x)=0$ has on;ly one root which is negaitive if $a < -1, b<0$
0%
None of these

Explanation

$f(x)=\sin x +ax+b$

$f(x)=0$

$\sin x +ax+b$

$f^{'}(x)=\cos x +G \rightarrow$ when

$a > l$

$3^{'}(x)> o$

$f(x) \uparrow$ increasing

$f^{''}(x)=-\sin x$

$f(x)=\sin x +ax+b$

$f(0)=0+0+b$

but

$b < 1$

when

$a> \bot$

$\cos x +a > 0$

$f^{'}(x) > 0$

hence

$f(x)$ increasing function

$f(x)=\sin x +ax+b$

$y=f(0)=\sin 0+a \times a +b$

$y=b$

when,

$b < 0$

$y=b$

when

$b > 0$ and

$a> 1$

But

$y=\sin x +ax+b$

$y=ax+\dfrac{(b+\sin x)}{\cos t}$

$y=mx+C$

A right line.

A right line cut

$X-$ axis

1938633_1759567_ans_eacf1ca86e1d4b98a8ca2154801ab5a6.PNG

The value of

$\displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}$ is

0%
$\dfrac{2 a}{\pi}$
0%
$-\dfrac{2 a}{\pi}$
0%
$\dfrac{4 a}{\pi}$
0%
$-\dfrac{4 a}{\pi}$

Explanation

$\displaystyle \lim _{x \rightarrow a} \sqrt{a^{2}-x^{2}} \cdot \cot \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}$

$=\displaystyle \lim _{x \rightarrow a} \dfrac{\sqrt{a^{2}-x^{2}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}$

$=\dfrac{2}{\pi} \displaystyle \lim _{x \rightarrow a} \dfrac{\dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}{\tan \dfrac{\pi}{2} \sqrt{\dfrac{a-x}{a+x}}}(a+x)=\dfrac{4 a}{\pi}$

Which of the following is not true about

$y=f(x)$ ?

0%
It is an increasing function
0%
It is a monotonic function
0%
It has infinite points of inflections
0%
None of these

Explanation

$f(x)=x+\cos x -a\Rightarrow f'(x)=1-\sin x \ge 0 \forall x\in R$ .
Thus

$f(x)$ is increasing in

$(-\infty , \infty)$ , as for

$f'(x)=0, x$ is not forming an interval.
Also

$f"(x)=-\cos x=0$

$\Rightarrow x=(2n+1)\dfrac {\pi}{2}, n\in Z$
Hence infinite points of inflection
Now

$f(0)=1-a$ .
For positive root

$1-a < 0 \Rightarrow a > 1$ . For negative root

$1-a > 0\Rightarrow a < 1$ .

$\displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{20} \cos ^{2 n}(x-10)$ is equal to

0%
0
0%
1
0%
19
0%
20

Explanation

$\because \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n} x=\begin{cases}1, x=r \pi, r \in I \\ 0, x \neq r \pi, r \in I\end{cases}$
Here, for

$x=10, \displaystyle \lim _{n \rightarrow \infty} \cos ^{2 n}(x-10)=1$
and in all other cases it is zero.

$\therefore \displaystyle \lim _{n \rightarrow \infty} \sum_{x=1}^{\infty} \cos ^{2 n}(x-10)=1$

$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin \left(x^{2}\right)}{\ln \left(\cos \left(2 x^{2}-x\right)\right)}$ is equal to

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

Explanation

$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{In(cos(2x^2-x))}$
$$\displaystyle \lim_{x \to 0}\dfrac{sin(x^2)}{log \left(1-2 sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}$$

$=\displaystyle \lim_{x \to 0} \dfrac{sin(x^2)x^2}{\dfrac{x^2log \left(1-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right)}{-2sin^2 \left(\dfrac{2x^2-x}{2}\right)}\left[-2sin^2 \left(\dfrac{2x^2-x}{2}\right)\right]}$

$=\displaystyle \lim_{x \to 0} \dfrac{x^2}{\dfrac{2sin^2 \left(\dfrac{2x^2-x}{2}\right)}{\left(\dfrac{2x^2-x}{2}\right)^2}\left(\dfrac{2x^2-x}{2}\right)^2}$

$\displaystyle \lim_{x \to 0}-\dfrac{2x^2}{(2x^2-x)^2}=\displaystyle \lim_{x \to 0}-\dfrac{2}{(2x-1)^2}=-2$

Differential coefficient of

$\sec (\tan^{-1} x)$ w.r.t. x is

0%
$\dfrac{x}{\sqrt{1+x^2}}$
0%
$\dfrac{x}{1+x^2}$
0%
$x\sqrt{1+x^2}$
0%
$\dfrac{1}{\sqrt{1+x^2}}$

The value of

$\displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin x}{x}$ is

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

Explanation

$\displaystyle \lim_{x \rightarrow \infty} \dfrac {\sin x}{x}$
Let

$x = 1/y$ or

$y = \dfrac {1}{x}$
Som

$x\rightarrow \infty \Rightarrow y \rightarrow 0$

$\therefore \displaystyle \lim_{x \rightarrow \infty} \left (\dfrac {\sin x}{x}\right ) = \displaystyle \lim_{y \rightarrow 0} y \sin \left (\dfrac {1}{y}\right )$

$= \displaystyle \lim_{y \rightarrow 0} y \cdot \displaystyle \lim_{y \rightarrow 0} \sin \dfrac {1}{y}$

$= 0\times \text {(any variable quantity)}$

$= 0$
Hence, option (A) is correct.

The value of

$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$ is

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

Explanation

$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$

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

$= \displaystyle \lim_{x\rightarrow 0} \dfrac {2\sin^{2} x/2}{x^{2}}$

$= \displaystyle \lim_{x\rightarrow 0} 2\left (\dfrac {\sin x/2}{x/2}\right )^{2} \times \dfrac {1}{4}$

$= \dfrac {1}{2}\times \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin x/2}{x/2}\right )^{2}$

$= \dfrac {1}{2} \times 1 = \dfrac {1}{2}$
Hence option (B) is correct.

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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