Limits and Derivatives Test

Result

Limits and Derivatives Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

if $$x=\cos^{3}\theta,y=\sin^{3}\theta$$, then $$\sqrt{1+\left(\dfrac{dy}{dx}\right)^{2}}=$$

A
$$\tan^{2}\theta$$

B
$$\sec^{2}\theta$$

C
$$\sec\theta$$

D
$$\left|\sec\theta\right|$$

Solution
Question 2
1 / -0

$$\displaystyle \lim _{ x\rightarrow 2 }{ \frac { \sqrt [ 3 ]{ 60+{ x }^{ 2 } } -4 }{ \sin { \left( x-2 \right) } } } $$

A
$$\dfrac {1}{4}$$

B
$$0$$

C
$$\dfrac {1}{12}$$

D
$$Does\ not\ exist$$

Solution
Question 3
1 / -0

The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \csc^{ 4 }{ x } \int _{ 0 }^{ { x }^{ 2 } }{ \frac { ln\left( 1+4t \right) }{ { t }^{ 2 }+1 } } dt } $$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 4
1 / -0

$$\displaystyle \lim_{x\rightarrow 0^{+}}{(\csc x)^{1/\log x}}$$=

A
$$e$$

B
$$e^{-1}$$

C
$$e^{2}$$

D
$$1$$

Solution
Question 5
1 / -0

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

A
$$0$$

B
$$-\dfrac{\pi^{2}}{2}$$

C
$$-\dfrac{\pi^{2}}{4}$$

D
$$-\dfrac{\pi^{2}}{8}$$

Solution
Question 6
1 / -0

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

A
$$ \cfrac {1} {24} $$

B
$$ \cfrac {1} {16} $$

C
$$ \cfrac {1} {8} $$

D
$$ \cfrac {1} {4} $$

Solution

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,

$$=\dfrac{1}{16}$$
Question 7
1 / -0

$$\underset { x\rightarrow \frac { \pi }{ 2 } }{ lim } \frac { (1-sinx)({ 8x }^{ 2 }-{ \pi }^{ 3 })cosx }{ { (\pi -2x) }^{ 4 } } $$

A
$$-\cfrac { { \pi }^{ 2 } }{ 16 } $$

B
$$\cfrac { { 3\pi }^{ 2 } }{ 16 } $$

C
$$\cfrac { { \pi }^{ 2 } }{ 16 } \quad $$

D
$$-\cfrac { 3{ \pi }^{ 2 } }{ 16 } \quad $$

Solution
Question 8
1 / -0

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

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

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

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

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

Solution
Question 9
1 / -0

If $$ \alpha \quad and \beta $$ are the roots of the equation $$ {ax}^{2}+bx+c=0 $$, then
$$ \underset { x\rightarrow \cfrac { \pi }{ 2 } }{ lim } \cfrac { tan\left[ \left( \alpha +\beta \right) x \right] }{ sin\left[ \left( \alpha \beta \right) x \right] } $$ is equal to :

A
$$ \cfrac {c} {b} $$

B
$$ - \cfrac {b} {c} $$

C
$$ \cfrac {a} {b} $$

D
$$- \cfrac {a} {b} $$

Solution
Question 10
1 / -0

If $$L = \underset{x \rightarrow 0}{lim} \dfrac{a \, sin \, x - sin \, 2x}{tan^3 x}$$ is finite, then the value of L is :

A
1

B
2

C
3

D
-1

Solution