MCQ Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 13

For the function,

$f(x) = (x - \frac{1}{x})^2$ , the first derivative with respect to x is

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$2 (x - \frac{1}{x^3})$
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$2 (x - \frac{1}{x})$
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$2 (x + \frac{1}{x^2})$
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$2 (x - \frac{1}{x^2})$

$\displaystyle \lim _{ \theta \rightarrow \pi /2 }{ \dfrac { 1-\sin \theta }{ (\pi /2-\theta )\cos { \theta } } }$ is equal to

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

If x + y = sin (x + y) then

$\dfrac{dy}{dx}$ =

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

$\lim_{n\rightarrow \infty}\dfrac{1}{n^{2}}\left[\sin^{3}\dfrac{\pi}{4n}+2\sin^{3}\dfrac{2\pi}{4n}+3\sin^{3}\dfrac{3\pi}{4n}+....+n\sin^{3}\dfrac{n\pi}{4n}\right]=$

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$\dfrac{\sqrt{2}}{9\pi^{2}}\left(52-15\pi\right)$
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$\dfrac{\sqrt{2}}{9\pi^{2}}\left(52+15\pi\right)$
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$\dfrac{\sqrt{2}}{9\pi}\left(52-17\pi\right)$
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$\dfrac{\sqrt{2}}{9\pi^{2}}\left(52+17\pi\right)$

$\lim- {x\to 0}$

$\dfrac{1- cos(1 - cos4x)}{x^4}$ is equal to :

0%
4
0%
16
0%
32
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None of these

$\underset { x\rightarrow 0 }{ lim } \left( \dfrac { \left( 1+x \right) ^{ \dfrac { 1 }{ x } } }{ e } \right) ^{ \dfrac { 1 }{ sinx } }$ is equal to

0%
$\sqrt { e }$
0%
e
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$\dfrac { 1 }{ \sqrt { e } }$
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1/e

$\underset { x\rightarrow 0 }{ lim } \cfrac { { \left( 25 \right) }^{ x }-2\left( 15 \right)^ x+{ 9 }^{ x } }{ cos6x-cos2x }$ is equal to :

0%
$log \left ( \cfrac {5} {3} \right )$
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$\cfrac {1} {4} log 15$
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$- \cfrac {1} {16} \left( \cfrac {5} {3} \right) ^2$
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$log \left ( \cfrac {3} {5} \right )$

Arrange the following limits in the ascending order :
(1)

$\lim _ { x \rightarrow \infty } \left( \dfrac { 1 + x } { 2 + x } \right) ^ { x + 2 }$

(2)

$\lim _ { x \rightarrow 0 } ( 1 + 2 x ) ^ { 3 / x }$

(3)

$\lim _ { \theta \rightarrow 0 } \dfrac { \sin \theta } { 2 \theta }$

(4)

$\lim _ { x \rightarrow 0 } \dfrac { \log _ { e } ( 1 + x ) } { x }$

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

Explanation

(i)

$\underset { x\rightarrow \infty }{ lim } { \left( \dfrac { 1+x }{ 2+x } \right) }^{ x+2 }$

$=$

${ \left( \dfrac { \dfrac { 1 }{ x } +1 }{ \dfrac { 2 }{ x } +1 } \right) }^{ x+2 }$

$=\underset { x\rightarrow \infty }{ lim } { e }^{ ln{ \left( \dfrac { 1+x }{ 2+x } \right) }^{ 2+x } }$

$={ e }^{ \underset { x\rightarrow \infty }{ lt } \left( 2+x \right) ln\left( \dfrac { 1+x }{ 2+x } \right) }$

$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { ln\left( 1+x \right) -ln\left( 2+x \right) }{ \left( 1/2+x \right) } }$

$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { \dfrac { 1 }{ 1+x } -\dfrac { 1 }{ 2+x } }{ \dfrac { -1 }{ { \left( 2+x \right) }^{ 2 } } } }$

$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { { \left( 2+x \right) }^{ 2 }\left[ \left( 2+x \right) -\left( 1+x \right) \right] }{ \left( 2+x \right) -\left( 1+x \right) \left( 2+x \right) \left( 1+x \right) } }$

$={ e }^{ \underset { x\rightarrow \infty }{ lt } \dfrac { \left( 2+x \right) }{ 1+x } }$

$={ e }^{ 1 }$

(ii)

$\underset { x\rightarrow 0 }{ lim } { \left( 1+2x \right) }^{ 3/x }$

$={ e }^{ \underset { x\rightarrow 0 }{ lt } \left( 2x \right) 3/x }$

$={ e }^{ 6 }$

(iii)

$\underset { \theta \rightarrow 0 }{ lt } \dfrac { \sin\theta }{ 2\theta } =\underset { \theta \rightarrow 0 }{ lt } \dfrac { \cos\theta }{ 2 } =\dfrac { 1 }{ 2 }$

(iv)

$\underset { x\rightarrow 0 }{ lt } \dfrac { { log }_{ e }\left( 1+x \right) }{ x } .\underset { x\rightarrow 0 }{ lt } \dfrac { 1 }{ 1+x } =1$

$\therefore$

$3<4<1<2$ .

If z = z(x) and

$(2cosx)\frac { dz }{ dx } +(sinx)z=sinx$ , z(0) = 3, then

$z(\frac { \pi }{ 2 } )$ equals :

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1
0%
$\frac { 3 }{ 2 }$
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$\frac { 5 }{ 2 }$
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$\frac { 1 }{ 2 }$

$\underset { x\rightarrow 0 }{ lim } \dfrac { x\tan { 2x } -2\tan { 2x } }{ { \left( 1-cos2x \right) } }$ equals:

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

$\mathop {\lim }\limits_{x \to 0} \frac{{x\left( {1 + a\cos x} \right) - b\sin x}}{{{x^3}}} = 1,$ then

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$a = \frac{5}{2}$
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$b = \frac{{ - 5}}{2}$
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$a + b = 4$
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$a + b = -4$

$lim_{n\to \infty} \Sigma^n_{r=1} \dfrac{\pi}{n} sin(\dfrac{\pi r}{n})$ is equal to

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

Evaluate :

$\displaystyle\lim _{ x\rightarrow 0 }{ \left( \dfrac { { e }^{ x\ell n\left( { 3 }^{ x }-1 \right) }-\left( { 3 }^{ x }-1 \right) ^{ x }\sin { x } }{ { e }^{ x\ell nx } } \right) }$ is equal to

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$\dfrac{1}{e}\ell n3$
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$e\ \ell n\ 3$
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$3$
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$\dfrac{1}{3}$

The value of

$\displaystyle\lim_{x\to 0} |x|^{sinx}$ equals

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$0$
0%
$-1$
0%
$1$
0%
does not exist

$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { \left( \sin { nx } \right) \left[ (a-n)nx-tanx \right] }{ { x }^{ 2 } } } =0$ , then the value of

$a$

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$\dfrac { 1 }{ n }$
0%
$n-\dfrac { 1 }{ n }$
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$n+\dfrac{1}{n}$
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$None\ of\ these$

$\displaystyle \lim_{x\rightarrow0 }{\dfrac{(\cos\alpha)^{x}-(\sin\alpha)^{x}-\cos 2\alpha}{(x-4)}}, \alpha\in \left(0, \dfrac{\pi}{2}\right)$ is equal to

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$\cos^{4}\alpha.\log(\cos\alpha)-\sin^{4}\alpha.\log(\sin\alpha)$
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$\sin^{4}\alpha.\log(\cos\alpha)-\cos^{4}\alpha.\log(\sin\alpha)$
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$\sin^{4}\alpha.\log(\cos\alpha)+\cos^{4}\alpha.\log(\sin\alpha)$
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$None\ of\ the \ above$

$\underset { x\rightarrow 0 }{ Lt } \cfrac {tanx-x}{x^2tanx}$ equals:

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1
0%
1/2
0%
1/3
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None of these

evaluate

$\underset { x\rightarrow 0 }{ lim } \frac { x-\int _{ 0 }^{ x }{ { cost }^{ 2 }dt } }{ { x }^{ 3 }-6x }$

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$3$
0%
$-1$
0%
$0$
0%
$1$

$\displaystyle\lim_{x \to \pi/2} (sec x +tan x)$ is equal to

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

The value f

$\lim_{x\rightarrow \pi/4}\dfrac{\sqrt{1-\sqrt{\sin 2x}}}{\pi-4x}=$

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

$\displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{x+1}{2x+1}\right)^{x^2}$ equals?

0%
$0$
0%
e
0%
$1$
0%
$\infty$

$\underset { x\rightarrow \pi/2 }{ lim } \left(\dfrac{cosec x-1}{cot^2x}\right)=$

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

$\displaystyle \lim_{x\rightarrow 0}\dfrac {ae^{-x}-b\cos x-\dfrac {1}{2}cx}{x\cos x}=2$ then the value of

$a+b+c$ is-

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

$\underset{x \rightarrow 2}{lim} \dfrac{\sqrt[3]{60 + x^2} - 4}{\sin (x - 2)}$ equals

0%
$\dfrac{1}{4}$
0%
$0$
0%
$\dfrac{1}{12}$
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Does not exist

$lim_{x\to \dfrac{\pi}{2}} tan^2x(\sqrt{2sin^2x + 3 sin x +4} - \sqrt{sin^2x + 6 sin x+2})$ is equal to

0%
$\dfrac{3}{4}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{12}$
0%
$\dfrac{5}{12}$

$\lim _ { x \rightarrow 0 } \frac { 1 - \cos x } { x \log ( 1 + x ) } =$

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

The value of

$\displaystyle\int^{\pi/2}_0ln|\tan x+\cot x|dx$ is equal to?

0%
$\pi ln 2$
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$-\pi ln 2$
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$\dfrac{\pi}{2} ln 2$
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$-\dfrac{\pi}{2} ln 2$

$\underset { x\rightarrow 0 }{ lim } (\cos x+a\sin b{ x) }^{ \frac { 1 }{ x } }$ is equal to

0%
$e^a$
0%
$e^{ab}$
0%
$e^b$
0%
$e^{a/b}$

The values of

$\displaystyle\lim_{n\rightarrow \infty}\dfrac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt[2]{n^3+1}}$ is?

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

Let

$f$ be a differentiable function such that

$f'(x) = 7- \dfrac{3}{4}\dfrac{f(x)}{x}, (x > 0)$ and

$f(1) \neq 4$ .
Then

$\underset{x\to 0^+}{\lim} xf \left(\dfrac{1}{x}\right)$ :

0%
Exists and equals 4
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Does not exist
0%
Exist and equals
0%
Exists and equals $\dfrac{4}{7}$

Explanation

$f'(x) = 7 -\dfrac{3}{4}\dfrac{f(x)}{x}$

$(x > 0)$

Given

$f(1) \neq 4$

$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = ?$

$\dfrac{dy}{dx} + \dfrac{3}{4}\dfrac{y}{x} = 7$ (This is LDE)

$=e^{\int \tfrac{3}{4x}dx} = e^{\tfrac{3}{4}ln|x|} = x^{\tfrac{3}{4}}$

$y.x^{\frac{3}{4}} =\int 7.x^{\frac{3}{4}} dx$

$y.x^{\frac{3}{4}} = 7.\dfrac{x^{\frac{7}{4}}}{\frac{7}{4}} + C$

$f(x) = 4x+C.x^{-\frac{3}{4}}$

$f\left(\dfrac{1}{x}\right) = \dfrac{4}{x} + C.x^{\frac{3}{4}}$

$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = \underset{x\to 0^+}{\lim}\left(4+C.x^{\frac{7}{4}}\right) = 4$

CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 13 - 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 13 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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