Limits and Derivatives Test

Result

Limits and Derivatives Test - 40

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

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

A
$$0$$

B
$$1/2$$

C
$$1$$

D
Does not exits

Solution
Question 2
1 / -0

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

A
$$\dfrac {a-b}{3}$$

B
$$0$$

C
$$\dfrac {(a^{2}-b^{2})}{6a^{2}b^{2}}$$

D
$$\dfrac {a^{2}-b^{2}}{3a^{2}b^{2}}$$

Solution
Question 3
1 / -0

The value of following derivative:
$$\dfrac{d}{dx}\left(\dfrac{\sin x+\cos x}{\sqrt{1+\sin 2x}}\right), \left(0<x<\dfrac{\pi}{4}\right)$$ is:

A
$$1$$

B
$$0$$

C
$$-1$$

D
None of these

Solution
Question 4
1 / -0

The value of $$\displaystyle\lim_{n\rightarrow \infty}n(n\{ln (n)-ln (n+1)\}+1)$$ is?

A
e

B
$$\dfrac{1}{e}$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{1}{4}$$

Solution
Question 5
1 / -0

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

A
$$\dfrac{1}{\theta}$$

B
$$\dfrac{1}{\theta}-2\cot 2\theta$$

C
$$2\cot 2\theta$$

D
None of these

Solution
Question 6
1 / -0

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

A
$$-\dfrac {1}{2}$$

B
$$\dfrac {1}{2}$$

C
$$e^{-\dfrac {1}{2}}$$

D
$$e^{\dfrac {1}{2}}$$

Solution
Question 7
1 / -0

$$ \underset { x\rightarrow 0 }{ lim } \left[ { x }^{ 2 }cosec\quad \left( { x }^{ 2 } \right)^0 \right] $$is equal to :

A
$$\cfrac { \pi} {180} $$

B
$$ \cfrac { \pi} {90} $$

C
$$ {0} $$

D
$$\cfrac {90} { \pi } $$

Solution
Question 8
1 / -0

The value of $$\lim_{x \rightarrow -1} \dfrac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}$$ is given by

A
$$\dfrac{1}{\sqrt{2}\sqrt{\pi}}$$

B
$$\dfrac{1}{\sqrt{\sqrt{2\pi}}}$$

C
$$1$$

D
$$0$$

Solution
Question 9
1 / -0

If $$\lim_{x \rightarrow 0}\dfrac{a \sin x-bx+cx^{2}+x^{3}}{2x^{2} \log(1+x)-2x^{3}+x^{4}}$$ exists and is finite, then the value of $$a,b,c$$ are respectively

A
$$0,6,6$$

B
$$6,0,6$$

C
$$6,6,0$$

D
$$0,0,6$$

Solution

Given, $$\underset{x \rightarrow 0}{\lim} \dfrac{a \sin x - bx + cx^2 + x^3}{2x^2 \log (1 + x) - 2x^3 + x^4} = $$ finite

To find : $$a, b, c$$

$$\underset{x \rightarrow 0}{\lim} \dfrac{a \sin x - bx + cx^2 + x^3}{2x^2 \log (1 + x) - 2x^3 + x^4} \rightarrow \dfrac{0}{0}$$

$$\therefore$$ Using L'hospital rule

$$\underset{x \rightarrow 0}{\lim} \dfrac{a \cos x - b + 2 cx + 3x^2}{4x \ln (1 + x) + \dfrac{2x^2}{(1 + x)} - 6x^2 + 4x^3} \rightarrow \dfrac{a - b}{0}$$

Using L'Hospital again,

$$\underset{x \rightarrow 0}{\lim} \dfrac{-a \sin x + 2c + 6x}{4 \ln (1 + x) + \dfrac{4x}{1 + x} + \dfrac{4x (1 + x) - 2x^2}{(1 + x)^2} - 12x + 12x^2} \rightarrow \dfrac{2c}{0}$$

Using L' Hospital rule again,

$$\underset{x \rightarrow 0}{\lim} \dfrac{-a \cos x + 6}{\dfrac{(4 + 4x)(1 +x)^2 - (2 + 2x)(4x + 2x^2) - 12 + 24x}{(1 + x)^4}}$$ $$+\dfrac{4}{1 + x} + \dfrac{4 (1 + x) - 4x}{(1 + x)^2}$$

$$= \dfrac{-a + 6}{4 + 4 + 4 - 12} = \dfrac{-a + 6}{0} \Rightarrow a = 6$$

as the limit is finite.

Hence, $$a = 6 = b, c = 0$$. Option C.
Question 10
1 / -0

$$ \underset { x\rightarrow a }{ lim } \cfrac { sin\quad x-sin\quad a }{ \sqrt [ 3 ]{ x } -\sqrt [ 3 ]{ a } } $$

A
$$ \sqrt [ 3 ]{ a\quad cos\quad a } $$

B
$$2\sqrt [ 3 ]{ a } $$

C
$$ { 3a }^{ { 2 /}{ 3 } }cos\quad a $$

D
$$ \sqrt [ 2 ]{ a\quad cos\quad a } $$

Solution