If $$\int_{0}^{x}(t^{2}+2t+2)$$ dt, $$2\leq x\leq 4$$

The maximum value of f(x) is $$\dfrac{136}{3}$$

The minimum value of f(x) is 10

The maximum value of f(x) is 26

MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 12

$y=\left | \cos x \right |+\left | \sin x \right |$ then

$\frac{dy}{dx}$ at

$x=\frac{2\pi }{3}$ is

0%
$\frac{1-\sqrt{3}}{2}$
0%
$0$
0%
$\frac{1}{2}\left ( \sqrt{3}-1 \right )$
0%
none of these

Explanation

$\displaystyle \dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{-cosxsinx}{|cosx|}+\dfrac{cosxsinx}{|sinx|}$

$\displaystyle \dfrac{\mathrm{d} y}{\mathrm{d} x}|_{x=\dfrac{2\pi }{3}}$

$\displaystyle =\dfrac{-cos\dfrac{2\pi }{3}sin\dfrac{2\pi }{3}}{|cos\dfrac{2\pi }{3}|}+\dfrac{cos\dfrac{2\pi }{3}sin\dfrac{2\pi }{3}}{|sin\dfrac{2\pi }{3}|}$

$\displaystyle =\dfrac{sin\dfrac{2\pi }{3}}{1}+\dfrac{cos\dfrac{2\pi }{3}}{1}$

$=\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}$

$\displaystyle =\dfrac{\sqrt{3}-1}{2}$

Let

$\tan \alpha .x+\sin \alpha .y=\alpha$ and

$\alpha \ \text{cosec} \alpha .x+\cos \alpha .y=1$ be two variable straight line,

$\alpha$ being the parameter. Let

$P$ be the point of intersection of the lines. In the limiting position when

$\alpha \rightarrow 0$ , the point

$P$ lies on the line

0%
$x=2$
0%
$x=-1$
0%
$y+1=0$
0%
$y=2$

Explanation

Solving

$\tan \alpha .x+\sin \alpha .y=\alpha$ and

$\alpha \ \text{cosec}\alpha .x+\cos \alpha .y=1$ , we get

$x=\cfrac { \alpha \cos { \alpha } -\sin { \alpha } }{ \sin { \alpha } -\alpha }$ and

$y=\cfrac { \alpha -x\tan { \alpha } }{ \sin { \alpha } }$

$\displaystyle \lim _{ \alpha \rightarrow 0 }{ x } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { \cos { \alpha } -\alpha \sin { \alpha } -\cos { \alpha } }{ \cos { \alpha } -1 } } \\ =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { \alpha \sin { \alpha } }{ 2\sin ^{ 2 }{ \cfrac { \alpha }{ 2 } } } } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \cfrac { 4{ \left( \cfrac { \alpha }{ 2 } \right) }^{ 2 }\cfrac { \sin { \alpha } }{ \alpha } }{ { \left( \sin { \cfrac { \alpha }{ 2 } } \right) }^{ 2 }2 } } =2\\ \displaystyle \lim _{ \alpha \rightarrow 0 }{ y } =\displaystyle \lim _{ \alpha \rightarrow \alpha }{ \cfrac { \alpha -x\tan { \alpha } }{ \sin { \alpha } } } =\displaystyle \lim _{ \alpha \rightarrow 0 }{ \left( \cfrac { \alpha }{ \sin { \alpha } } -\cfrac { x }{ \cos { \alpha } } \right) } =1-2=-1$
Hence

$P=\left( 2,-1 \right)$

$\displaystyle y=\sum _{ r=1 }^{ x }{ \tan ^{ -1 }{ \frac { 1 }{ 1+r+{ r }^{ 2 } } } }$ then

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

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

Explanation

We have,

$\displaystyle y=\sum _{ r=1 }^{ x }{ \tan ^{ -1 }{ \frac { 1 }{ 1+r+{ r }^{ 2 } } } } =\sum _{ r=1 }^{ x }{ \tan ^{ -1 }{ \left( \frac { \left( r+1 \right) -r }{ 1+\left( r+1 \right) r } \right) } }$

$\displaystyle =\sum _{ r=1 }^{ x }{ \left[ \tan ^{ -1 }{ \left( r+1 \right) } -\tan ^{ -1 }{ r } \right] }$

$=\left[ \tan ^{ -1 }{ 2 } -\tan ^{ -1 }{ 1 } +\tan ^{ -1 }{ 3 } -\tan ^{ -1 }{ 2 } +...+\tan ^{ -1 }{ x } -\tan ^{ -1 }{ \left( x-1 \right) } +\tan ^{ -1 }{ \left( x+1 \right) } -\tan ^{ -1 }{ x } \right]$

$=\left[ \tan ^{ -1 }{ \left( x+1 \right) } -\tan ^{ -1 }{ 1 } \right]$

$\displaystyle \therefore \frac { dy }{ dx } =\frac { 1 }{ 1+{ \left( x+1 \right) }^{ 2 } } .$

Let f and g be differentiable function such that

${f}'\left ( x \right )=2g\left ( x \right )$ and

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

$T\left ( x \right )=\left ( f\left ( x \right ) \right )^{2}-\left ( g\left ( x \right ) \right )^{2}$ . Then

${T}'\left ( x \right )$ is equal to

0%
$T(x)$
0%
$0$
0%
$2f(x)g(x)$
0%
$6f(x)g(x)$

Explanation

The given equation is:

$T(x) = f(x)^2 - g(x)^2$

$\Rightarrow T(x) = (f(x) - g(x))(f(x) + g(x))$

Differentiating once w.r.t to x we get,

$\Rightarrow T'(x) = (f'(x) + g'(x))(f(x)-g(x)) + (f'(x)-g'(x))(f(x)+g(x))$

$\Rightarrow T'(x) = (2g(x)-f(x))(f(x)-g(x)) + (2g(x)+f(x))(f(x)+ g(x))$

$\Rightarrow T'(x) = 2g(x)f(x) - 2g(x)^2 -f(x)^2 + f(x)g(x) +2g(x)f(x) + 2g(x)^2 +f(x)^2 + f(x)g(x)$

$\Rightarrow T'(x)= 6f(x)g(x)$ .....Answer

The value of

$\displaystyle \underset { n\rightarrow \infty }{ lim } \left( \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 6n } \right)$ is

0%
$\displaystyle \log { 2 }$
0%
$\displaystyle \log { 6 }$
0%
$\displaystyle 1$
0%
$\displaystyle \log { 3 }$

If the prime sign (') represents differentiation w.r.t.

$x$ and

$f^{'}=\sin x+\sin 4x.\cos x$ , then

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

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

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

Explanation

Given

$f^{'}(x)=\sin x+\sin 4x.\cos x$

Now,

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

$=f^{ ' }\left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) \dfrac { d }{ dx } \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right)$

$=\left[ \sin \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) +\sin 4\left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) .\cos \left( 2x^{ 2 }+\dfrac { \pi }{ 2 } \right) \right] 4x$

$=\left[ \cos 2x^{ 2 }-\sin 8x^{ 2 }.\sin 2x^{ 2 } \right] 4x$

$=\left[ \cos \pi -\sin 4\pi .\sin \pi \right] 4\sqrt { \dfrac { \pi }{ 2 } }$

$=-2\sqrt{2\pi}$

For

$n\epsilon N$ , let

$f\left ( x \right )=min\:\left \{ 1-\tan ^{n}x, 1-\sin ^{n}x, 1-x^{n} \right \}$ ,

$x\epsilon \left ( -\cfrac {\pi}{2}, \cfrac {\pi}{2} \right )$ . The left hand derivative of

$f$ at

$x=\cfrac {\pi}{4}$ is

0%
$-2n$
0%
$-2(n+1)$
0%
$\displaystyle -n\frac{\pi }{4}$
0%
$\displaystyle -n\left ( \frac{\pi }{4} \right )^{n-1}$

Explanation

Given,

$f\left ( x \right )=min\:\left \{ 1-\tan ^{n}x, 1-\sin ^{n}x, 1-x^{n} \right \}$
For

$x\in \left ( 0, \cfrac {\pi}{4} \right )$ , we have

$\tan x> x> \sin x$

$\Rightarrow 1-\tan ^{n}x<1 -x^{n}< 1-\sin ^{n}x$

$\Rightarrow f\left ( x \right )=1-\tan ^{n}x$

So,

$\displaystyle f\left (\dfrac{ \pi }{4} \right )=min\:\left \{ 0, 1-\left ( \frac{1}{\sqrt{2}} \right )^{n}, 1-\left ( \frac{\pi }{4} \right )^{n} \right \}$

$\Rightarrow f(\dfrac{\pi}{4})=0$

Now,

$\displaystyle {f}'\left ( \dfrac{\pi}{4}^- \right )=\lim_{h\rightarrow 0^-}\frac{f\left ( \dfrac{\pi}{4}+h \right )-f\left ( \dfrac{\pi}{4} \right )}{h}$

$\displaystyle =\lim_{h\rightarrow 0^-}\frac{1-\tan ^{n}\left ( \dfrac{\pi}{4}+h \right )}{h}$

$\displaystyle=\lim _{ h\rightarrow 0^{ - } } \dfrac { 1-\left( \frac { 1+\tan { h } }{ 1-\tan { h } } \right) ^{ n } }{ h }$

$\displaystyle =\lim_{h\rightarrow 0^-}\frac{\left ( 1-\tan h \right )^{n}-\left ( 1+\tan h \right )^{n}}{h\left ( 1-\tan h \right )^{n}}$

$=\displaystyle \lim _{ h\rightarrow 0^{ - } } \frac { -2\left[ n\tan { h } +\frac { n(n-1) }{ 2 } \tan ^{ 3 }{ h } +..... \right] }{ h\left( 1-n\tan h+\frac { n(n-1) }{ 2 } \tan ^{ 2 }{ h } +.... \right) }$

As,

$\underset { h\rightarrow 0 } \lim { \tan h } =0$

So, using this result , we get

$=\underset { h\rightarrow 0^{ - } } \lim \cfrac { -2n\tan { h } }{ h }$

$=-2n$

$\left (\underset { h\rightarrow 0 } \lim \dfrac{ \tan h } { h }=1\right)$

$f '$ (0) = 0 and f(x) is a differentiable and increasing function,then lim

$x \rightarrow 0$

$\frac {x.f ' (x^2)}{f ' (x)}$

0%
is always equal to zero
0%
may not exist as left hand limit may not exist
0%
may not exist as left hand limit may not exist
0%
right hand limit is always zero

The left-handed derivative of

$f\left( x \right) =\left| x \right| \sin { \left( \pi x \right) }$ at

$x=K$ where

$K$ is an integer, is :

0%
${ \left( -1 \right) }^{ K }\left( K-1 \right) \pi$
0%
${ \left( -1 \right) }^{ K-1 }\left( K-1 \right) \pi$
0%
${ \left( -1 \right) }^{ K }\ K\pi$
0%
${ \left( -1 \right) }^{ K-1 }\ K\pi$

$\sum _{ r=1 }^{ k }{ \cos ^{ -1 }{ \beta } } =\cfrac { k\pi }{ 2 }$ for any

$k\ge 1$ and

$A=\sum _{ r=1 }^{ k }{ { \left( { \beta }_{ r } \right) }^{ r } }$ , then

$\lim _{ x\leftarrow A }{ \cfrac { { \left( 1+x \right) }^{ 1/3 }-{ \left( 1-2x \right) }^{ 1/4 } }{ x+{ x }^{ 2 } } }$ is equal to

0%
$0$
0%
$\cfrac{1}{2}$
0%
$\cfrac { \pi }{ 2 }$
0%
$\cfrac { 5 }{ 6 }$

The value of

$\lim _{ x\rightarrow 0 }{ \left( { \left( \sin { x } \right) }^{ 1/x }+{ \left( 1+x \right) }^{ \sin { x } } \right) }$ whre

$x> 0$ is

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

Explanation

Given

$\lim _{ x\rightarrow 0 }{ \left( { \left( sinx \right) }^{ { \frac { 1 }{ x } } }+{ \left( 1+x \right) }^{ sinx } \right) }$

$=\lim _{ x\rightarrow 0 }{ { \left( sinx \right) }^{ { \frac { 1 }{ x } } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } } }$

$={ \left( sin0 \right) }^{ { \frac { 1 }{ 0 } } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } }$

$={ \left( 0 \right) }^{ { \infty } }+\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ sinx } }$

$=0+{ e }^{ \lim _{ x\rightarrow 0 }{ \left[ log{ \left( 1+x \right) }^{ sinx } \right] } }$

$\left( \because { e }^{ loga }=a \right)$

$={ e }^{ \lim _{ x\rightarrow 0 }{ \left[ sinx.log{ \left( 1+x \right) } \right] } }$

$={ e }^{ \lim _{ x\rightarrow 0 }{ sinx } \times \lim _{ x\rightarrow 0 }{ log\left( 1+x \right) } }$

$={ e }^{ sin\left( 0 \right) \times log\left( 1+0 \right) }$

$={ e }^{ 0\times log\left( 1 \right) }$

$={ e }^{ 0\times 0 }={ e }^{ 0 }$

$=1$

The value of

$\displaystyle \lim_{x \rightarrow 1^{-}}\dfrac {1 - \sqrt {x}}{(\cos^{-1}x)^{2}}$

0%
$\dfrac{-1}{4}$
0%
$1/2$
0%
$2$
0%
None of these

Explanation

${ lim }_{ x\rightarrow 1 }\left( \frac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) \\ { \Rightarrow lim }_{ x\rightarrow 1 }1-\sqrt { x } =0\quad ,\quad { \Rightarrow lim }_{ x\rightarrow 1 }(cos^{ -1 }x)^{ 2 }=0\\ { lim }_{ x\rightarrow 1 }\left( \frac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) =\frac { 0 }{ 0 } \\ Now\quad using\quad l'hospital's\quad rule\\ { lim }_{ x\rightarrow 1 }\left( \dfrac { 1-\sqrt { x } }{ (cos^{ -1 }x)^{ 2 } } \right) ={ lim }_{ x\rightarrow 1 }\left( \dfrac { \frac { d(1-\sqrt { x } ) }{ dx } }{ \frac { d(cos^{ -1 }x)^{ 2 } }{ dx } } \right) ={ lim }_{ x\rightarrow 1 }\dfrac { 1-\frac { 1 }{ 2\sqrt { x } } }{ 2cos^{ -1 }x*\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } } \\ Now\quad { lim }_{ x\rightarrow 1 }1-\dfrac { 1 }{ 2\sqrt { x } } =1-1/2=1/2\\ And\quad { lim }_{ x\rightarrow 1 }\dfrac { -2cos^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } =\frac { 0 }{ 0 } \\ Again\quad using\quad l'hospital's\quad rule\\ \\ { lim }_{ x\rightarrow 1 }\dfrac { -2cos^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } ={ lim }_{ x\rightarrow 1 }\dfrac { -2d(cos^{ -1 }x)/dx }{ d(\sqrt { 1-{ x }^{ 2 } } )/dx } ={ lim }_{ x\rightarrow 1 }\dfrac { -2(\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } )\quad }{ \frac { -2x }{ 2\sqrt { 1-{ x }^{ 2 } } } } ={ lim }_{ x\rightarrow 1 }\dfrac { -2 }{ x } =-2\\ Therefore,\quad { lim }_{ x\rightarrow 1 }\dfrac { 1-\frac { 1 }{ 2\sqrt { x } } }{ 2cos^{ -1 }x*\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } } =\dfrac { 1/2 }{ -2 } =-1/4\\$

$\displaystyle \lim_{x \rightarrow 0} \frac{ae^x + b cos x + c. e^{-x}}{sin^2 x} = 4$ then b =

0%
2
0%
4
0%
-2
0%
-4

$\int_{0}^{x}(t^{2}+2t+2)$ dt,

$2\leq x\leq 4$

0%
The maximum value of f(x) is $\dfrac{136}{3}$
0%
The minimum value of f(x) is 10
0%
The maximum value of f(x) is 26
0%
None of these

$x\sqrt {1 + y} + y\sqrt {1 + x} = 0$ , then

$\dfrac {dy}{dx}$ is equal to

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

Explanation

Given

$x\sqrt{1+y}+y\sqrt{1+x}=0$

$x\sqrt{1+y}=-y\sqrt{1+x}$

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

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

$\cfrac{1+y}{y^{2}}=\cfrac{1+x}{x^{2}}$

$x^{2}+x^{2}y=y^{2}+y^{2}x$

$x^{2}-y^{2}=y^{2}x-x^{2}y$

$(x+y)(x-y)=xy(y-x)$

$\therefore x+y+xy=0\Rightarrow y=\cfrac{-x}{1+x}$

$(\because x\neq y)$

Differentiate on both sides w.r.t x

$1+\cfrac{\mathrm{d} y}{\mathrm{d} x}+x\cfrac{\mathrm{d} y}{\mathrm{d} x}+y=0$

$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{-(1+y)}{1+x}$

$=\cfrac{-\left ( 1-\cfrac{x}{1+x} \right )}{1+x}$

$=\cfrac{-1}{(1+x)^{2}}$

$\therefore \cfrac{\mathrm{d} y}{\mathrm{d} x}=\cfrac{-1}{(1+x)^{2}}$

$\mathop {\lim }\limits_{x \to {a^ + }} {{\left\{ x \right\}\sin \left( {x - a} \right)} \over {{{\left( {x - a} \right)}^2}}}$

is equal to (where {.} denotes the fraction
part of x and $a \in N$

0%
0
0%
1
0%
does not exist
0%
none of thes

Consider the following statements:

$S_1: \lim_\limits{x \to 0} \dfrac{[x]}{x}$ is an indeterminate form (where [.] denotes greatest integer function).

$S_2: \lim_\limits{x\to\infty}\dfrac{sin(3^x)}{3^x}=0$

$S_3: \lim_\limits{x \to \infty}\sqrt{\dfrac{x- sinx}{x+cos^2x}}$ does not exist.

$S_4: \lim_\limits{n\to \infty}\dfrac{(n+2)!+(n+1)!}{(n+3)! }(n \in N=0$
State, in order, whether

$S_1, S_2, S_3, S_4$ are true or false

0%
FTFT
0%
FTTT
0%
FTFF
0%
TTFT

$If {A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}}, \,i = 1,2,3,.....n$ and

${a_1}< {a_2}< {a_3}....< {a_{n,}} \, then$

$\mathop {\lim }\limits_{x \to {a_m}} \left( {{A_1}{A_2}......{A_n}} \right), 1 \le m \le n$

0%
is equal to ${\left( { - 1} \right)^m}$
0%
is equal to ${\left( { - 1} \right)^{m + 1}}$
0%
is equal to ${\left( { - 1} \right)^{n-m}}$
0%
is equal to ${\left( { - 1} \right)^{n-m-1}}$

$l=\lim\limits_{n\to 3}\dfrac{x^2-9}{\sqrt{x^2+7}-4}$ and

$m=\lim\limits_{n\to -3}\dfrac{x^2-9}{\sqrt{x^2+7}-4}$ , then

0%
$l\ne m$
0%
$l=2m$
0%
$l=-m$
0%
$l=m$

Explanation

Given

$l=\underset{x\to3}\lim\dfrac{x^2-9}{\sqrt{x^2+7}-4}=\dfrac 00$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore,

$l=\underset {x\to3} \lim \cfrac{\tfrac{d}{dx}(x^2-9)}{\tfrac{d}{dx}(\sqrt{x^2+7}-4)}=\underset {x\to 3} \lim \cfrac{2x}{\dfrac{x}{\sqrt{x^2+7}}}=\underset{x\to3}\lim{2}{\sqrt{x^2+7}}=2(\sqrt{16})=8$

Given

$m=\underset{x\to-3}\lim\dfrac{x^2-9}{\sqrt{x^2+7}-4}=\dfrac 00$ which is the indeterminate form.

Therefore, we apply the L-hospital's rule. In L-hospital's rule, we take the derivative of the numerator and denominator separately and apply the limits.

Therefore,

$m=\underset {x\to-3} \lim \cfrac{\tfrac{d}{dx}(x^2-9)}{\tfrac{d}{dx}(\sqrt{x^2+7}-4)}=\underset {x\to -3} \lim \cfrac{2x}{\dfrac{x}{\sqrt{x^2+7}}}=\underset{x\to-3}\lim{2}{\sqrt{x^2+7}}=2(\sqrt{16})=8$

Hence,

$l=8,m=8$

Therefore,

$l= m$

$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + x + 1}}{{x + 1}} - ax - b} \right)\, = 4$ ,then

0%
$a=1,b=4$
0%
$a=1,b=-4$
0%
$a=2,b=-3$
0%
$a=2,b=3$

Let

$f(\theta) = \dfrac{1}{tan^{9}\theta} {(1+tan\theta)^{10}+(2+tan\theta)^{10}+....+(20+tan\theta)^{10}}-20tan\theta$ . The left hand limit of

$f(\theta)$ as

$\theta \rightarrow \dfrac{\pi}{2}$ is:

0%
1900
0%
2000
0%
2100
0%
2200

$f\left( x \right) = \left\{ {\matrix{ {\cos \left[ x \right],} & {x \ge 0} \cr {\left| x \right| + a,} & {x < 0} \cr } } \right\}$ Find
the value of a , given that

$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ exists,
where[.] denotes

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

$\displaystyle \lim _{ x\rightarrow \frac { \pi }{ 4 } }{ { \left( \sin { 2x } \right) }^{ \sec ^{ 2 }{ 2x } } }$ is equal to

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

$\underset { x\rightarrow \cfrac { \pi }{ 2 } }{ lim } \cfrac { cot \, x-cos\, x }{ \left( \pi -{ 2x } \right)^ 3 }$ equals

0%
$\cfrac {1} {24}$
0%
$\cfrac {1} {16}$
0%
$\cfrac {1} {8}$
0%
$\cfrac {1} {4}$

Explanation

Given,

$\lim _{x\to \frac{\pi }{2}}\left(\dfrac{\cot \left(x\right)-\cos \left(x\right)}{\left(\pi -2x\right)^3}\right)$

apply L-Hospital's rule

$=\lim _{x\to \frac{\pi }{2}}\left(\dfrac{-\csc ^2\left(x\right)+\sin \left(x\right)}{-6\left(\pi -2x\right)^2}\right)$

$=\lim _{x\to \frac{\pi }{2}}\left(-\dfrac{-\csc ^2\left(x\right)+\sin \left(x\right)}{6\left(\pi -2x\right)^2}\right)$

apply L-Hospital's rule

$=\lim _{x\to \frac{\pi }{2}}\left(\dfrac{-2\csc ^2\left(x\right)\cot \left(x\right)-\cos \left(x\right)}{-24\left(\pi -2x\right)}\right)$

$=\lim _{x\to \frac{\pi }{2}}\left(-\dfrac{-2\csc ^2\left(x\right)\cot \left(x\right)-\cos \left(x\right)}{24\left(\pi -2x\right)}\right)$

apply L-Hospital's rule

$=\lim _{x\to \frac{\pi }{2}}\left(\dfrac{2\left(-2\csc ^2\left(x\right)\cot ^2\left(x\right)-\csc ^4\left(x\right)\right)-\sin \left(x\right)}{-48}\right)$

$=\lim _{x\to \frac{\pi }{2}}\left(-\dfrac{1}{48}\left(2\left(-2\csc ^2\left(x\right)\cot ^2\left(x\right)-\csc ^4\left(x\right)\right)-\sin \left(x\right)\right)\right)$

$=\lim _{x\to \frac{\pi }{2}}\left(-\dfrac{2\left(-2\csc ^2\left(x\right)\cot ^2\left(x\right)-\csc ^4\left(x\right)\right)-\sin \left(x\right)}{48}\right)$

$=-\dfrac{2\left(-2\csc ^2\left(\frac{\pi }{2}\right)\cot ^2\left(\frac{\pi }{2}\right)-\csc ^4\left(\frac{\pi }{2}\right)\right)-\sin \left(\frac{\pi }{2}\right)}{48}$

upon solving, we get,

$=\dfrac{1}{16}$

$\displaystyle\underset{x\rightarrow 0}{Lt}\left(cosec x-\dfrac{1}{x}\right)=?$

0%
$0$
0%
$1/2$
0%
$1$
0%
Does not exits

The value of

$\displaystyle \lim_{x\rightarrow 0}\dfrac {1-\cos^{3}x}{x\sin x\cos x}$ is

0%
$\dfrac {2}{5}$
0%
$\dfrac {3}{5}$
0%
$\dfrac {3}{2}$
0%
$\dfrac {3}{4}$

$\displaystyle \lim_{x \rightarrow 0}\dfrac {1}{x\sqrt {x}}\left(a\ arc\ tan \dfrac {\sqrt {x}}{a}-b\ arc\ \tan \dfrac {\sqrt {x}}{b}\right)$ has the value equal to

0%
$\dfrac {a-b}{3}$
0%
$0$
0%
$\dfrac {(a^{2}-b^{2})}{6a^{2}b^{2}}$
0%
$\dfrac {a^{2}-b^{2}}{3a^{2}b^{2}}$

Integrate:

$lim_{x\rightarrow 0}\dfrac{(1-\cos{2x})^{2}}{2x\tan{x}-x\tan{2x}}$

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

$\displaystyle\lim_{n\rightarrow \infty}\left(\tan\theta +\dfrac{1}{2}\tan \dfrac{\theta}{2}+\dfrac{1}{2^2}\tan \dfrac{\theta}{2^2}+...+\dfrac{1}{2^n}\tan\dfrac{\theta}{2^n}\right)$ equals?

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

$\displaystyle \lim_{x\rightarrow \infty}{x^{2}\sin\left(\log_{e}\sqrt{\cos\dfrac{\pi}{x}}\right)}$

0%
$0$
0%
$-\dfrac{\pi^{2}}{2}$
0%
$-\dfrac{\pi^{2}}{4}$
0%
$-\dfrac{\pi^{2}}{8}$

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 12 - 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 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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