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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 13 - MCQExams.com
CBSE
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
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$$1$$
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$$-1$$
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$$1/2$$
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$$-1/2$$
If x + y = sin (x + y) then $$\dfrac{dy}{dx}$$ =
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$$\dfrac{1}{2}$$
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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 :
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4
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16
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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
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$$\sqrt { e } $$
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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 :
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$$ log \left ( \cfrac {5} {3} \right ) $$
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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) $$\lim _ { x \rightarrow 0 } \dfrac { \log _ { e } ( 1 + x ) } { x }$$
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$$1,2,3,4$$
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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
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$$\frac { 3 }{ 2 } $$
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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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$$1$$
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$$\dfrac{1}{2}$$
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$$-\dfrac{1}{2}$$
If $$\mathop {\lim }\limits_{x \to 0} \frac{{x\left( {1 + a\cos x} \right) - b\sin x}}{{{x^3}}} = 1,$$ 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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$$1$$
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$$2$$
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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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$$0$$
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$$-1$$
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$$1$$
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does not exist
If $$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { \left( \sin { nx } \right) \left[ (a-n)nx-tanx \right] }{ { x }^{ 2 } } } =0$$, then the value of $$a$$
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$$\dfrac { 1 }{ n }$$
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$$n-\dfrac { 1 }{ n } $$
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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
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1/2
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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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$$3$$
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$$-1$$
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$$0$$
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$$1$$
$$\displaystyle\lim_{x \to \pi/2} (sec x +tan x)$$ is equal to
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$$1$$
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$$-1$$
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$$\dfrac{1}{2}$$
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$$0$$
The value f $$\lim_{x\rightarrow \pi/4}\dfrac{\sqrt{1-\sqrt{\sin 2x}}}{\pi-4x}=$$
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$$-\dfrac{1}{4}$$
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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?
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$$0$$
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e
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$$1$$
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$$\infty$$
$$\underset { x\rightarrow \pi/2 }{ lim } \left(\dfrac{cosec x-1}{cot^2x}\right)= $$
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$$0$$
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$$-\dfrac{1}{2}$$
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$$\dfrac{1}{2}$$
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$$1$$
If $$\displaystyle \lim_{x\rightarrow 0}\dfrac {ae^{-x}-b\cos x-\dfrac {1}{2}cx}{x\cos x}=2$$ then the value of $$a+b+c$$ is-
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$$4$$
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$$-4$$
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$$2$$
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$$-2$$
$$\underset{x \rightarrow 2}{lim} \dfrac{\sqrt[3]{60 + x^2} - 4}{\sin (x - 2)}$$ equals
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$$\dfrac{1}{4}$$
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$$0$$
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$$\dfrac{1}{12}$$
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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
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$$\dfrac{3}{4}$$
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$$\dfrac{1}{6}$$
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$$\dfrac{1}{12}$$
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$$\dfrac{5}{12}$$
$$\lim _ { x \rightarrow 0 } \frac { 1 - \cos x } { x \log ( 1 + x ) } =$$
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$$1$$
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$$0$$
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$$-1$$
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$$\frac { 1 } { 2 }$$
The value of $$\displaystyle\int^{\pi/2}_0ln|\tan x+\cot x|dx$$ is equal to?
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$$\pi ln 2$$
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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
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$$e^a$$
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$$e^{ab}$$
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$$e^b$$
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$$e^{a/b}$$
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?
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$$1$$
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$$0$$
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$$-1$$
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$$\infty$$
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 $$\underset{x\to 0^+}{\lim} xf \left(\dfrac{1}{x}\right) $$:
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Exists and equals 4
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Does not exist
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Exist and equals
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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)
IF $$=e^{\int \tfrac{3}{4x}dx} = e^{\tfrac{3}{4}ln|x|} = x^{\tfrac{3}{4}}$$
$$y.x^{\frac{3}{4}} =\int 7.x^{\frac{3}{4}} dx$$
$$y.x^{\frac{3}{4}} = 7.\dfrac{x^{\frac{7}{4}}}{\frac{7}{4}} + C$$
$$f(x) = 4x+C.x^{-\frac{3}{4}}$$
$$f\left(\dfrac{1}{x}\right) = \dfrac{4}{x} + C.x^{\frac{3}{4}}$$
$$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = \underset{x\to 0^+}{\lim}\left(4+C.x^{\frac{7}{4}}\right) = 4$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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