Limits and Derivatives Test

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Limits and Derivatives Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)$$ is the integral of $$\dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0$$. Find $$\lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right) } $$, where $$f^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}$$

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

B
$$1$$

C
$$\dfrac{1}{3}$$

D
$$2$$

Solution

$$f(x)=\dfrac{2\sin x-\sin 2x}{x^3}$$ $$x\neq 0$$

$$f'(x)=\dfrac{df(x)}{dx}=\dfrac{2\sin x-\sin 2x}{x^3}$$

$$=\dfrac{2\sin x}{x}\left(\dfrac{1-\cos x}{x^2}\right)$$

$$\Rightarrow \displaystyle \lim_{x\rightarrow 0}{f'(x)}=\displaystyle \lim_{x\rightarrow 0}{2\left(\dfrac{\sin x}{x}\right)}\left( \dfrac{2\sin^2(x/2)}{x^2}\right)$$

$$\Rightarrow \displaystyle 4\cdot 1 \lim_{x\rightarrow 0}{\left(\dfrac{\sin^2(x/2)}{4\times (x/2)^2}\right)}$$

$$=1 $$
Question 2
1 / -0

$$\underset{x\rightarrow 0}{lim} \dfrac{\sqrt{a^2 -ax+x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} -\sqrt{a-x}}$$ is equal to (a > 0)

A
$$\sqrt{a}$$

B
$$-\sqrt{a}$$

C
$$a \sqrt{a}$$

D
$$-a\sqrt{a}$$

Solution

$$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{(\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2})(\sqrt{a+x}+\sqrt{a-x})}{(a+x)-(a-x)}$$
$$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\sqrt{a}}{2x}(\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2})$$
$$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a}}{x}\left(\dfrac{(a^2-ax+x^2)-(a^2+ax+x^2)}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}\right)$$
$$\Rightarrow L=\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a}}{x}\times \dfrac{(-2ax)}{2\sqrt{a}}=-\sqrt{a}$$
$$\Rightarrow (B)$$.
Question 3
1 / -0

If $$\dfrac{dx}{tan(x+y)} = \dfrac{dy}{cot(x+y)} = \dfrac{dz}{I}$$ then z in term of x & y can be expressed as

A
$$Z =\dfrac{sin(2x+2y)}{4} + c$$

B
$$Z =-\dfrac{sin(2x+2y)}{4} + c$$

C
$$Z =-\dfrac{cos(2x+2y)}{4} + c$$

D
None of these
Question 4
1 / -0

The value of $$\displaystyle lim_{x\to 0} \dfrac{cos (sin x) - cos x}{x^4} $$ is equal to

A
$$1/5$$

B
$$1/6$$

C
$$1/4$$

D
$$1/2$$

Solution

Given,

$$\lim _{x\to \:0}\left(\dfrac{\cos \left(\sin \left(x\right)\right)-\cos \left(x\right)}{x^4}\right)$$

apply L-Hospital's rule

$$=\lim _{x\to \:0}\left(\dfrac{-\sin \left(\sin \left(x\right)\right)\cos \left(x\right)+\sin \left(x\right)}{4x^3}\right)$$

again apply L-Hospital's rule

$$=\lim _{x\to \:0}\left(\dfrac{-\cos ^2\left(x\right)\cos \left(\sin \left(x\right)\right)+\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos \left(x\right)}{12x^2}\right)$$

once again apply L-Hospital's rule

$$=\lim _{x\to \:0}\left(\dfrac{\frac{3\sin \left(2x\right)\cos \left(\sin \left(x\right)\right)+2\cos ^3\left(x\right)\sin \left(\sin \left(x\right)\right)+2\cos \left(x\right)\sin \left(\sin \left(x\right)\right)-2\sin \left(x\right)}{2}}{24x}\right)$$

upon simplification, we get,

$$=\lim _{x\to \:0}\left(\dfrac{3\sin \left(2x\right)\cos \left(\sin \left(x\right)\right)+2\cos ^3\left(x\right)\sin \left(\sin \left(x\right)\right)+2\cos \left(x\right)\sin \left(\sin \left(x\right)\right)-2\sin \left(x\right)}{48x}\right)$$

again applying L-Hospital's rule,

$$=\lim _{x\to \:0}\left(\dfrac{3\left(2\cos \left(2x\right)\cos \left(\sin \left(x\right)\right)-\sin \left(\sin \left(x\right)\right)\cos \left(x\right)\sin \left(2x\right)\right)+2\left(-3\cos ^2\left(x\right)\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos ^4\left(x\right)\cos \left(\sin \left(x\right)\right)\right)+2\left(-\sin \left(x\right)\sin \left(\sin \left(x\right)\right)+\cos ^2\left(x\right)\cos \left(\sin \left(x\right)\right)\right)-2\cos \left(x\right)}{48}\right)$$

substituting $$x=0$$

$$\dfrac{3\left(2\cos \left(2\cdot \:0\right)\cos \left(\sin \left(0\right)\right)-\sin \left(\sin \left(0\right)\right)\cos \left(0\right)\sin \left(2\cdot \:0\right)\right)+2\left(-3\cos ^2\left(0\right)\sin \left(0\right)\sin \left(\sin \left(0\right)\right)+\cos ^4\left(0\right)\cos \left(\sin \left(0\right)\right)\right)+2\left(-\sin \left(0\right)\sin \left(\sin \left(0\right)\right)+\cos ^2\left(0\right)\cos \left(\sin \left(0\right)\right)\right)-2\cos \left(0\right)}{48}$$

we get,

$$=\dfrac{1}{6}$$
Question 5
1 / -0

The derivative of $$\cos^{-1}\dfrac{1-x^{2}}{1+x^{2}} w.r.t\ \tan^{-1}\dfrac{2x}{1-x^{2}}$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$1/2$$
Question 6
1 / -0

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

A
$$ \dfrac{1}{24} $$

B
$$ \dfrac{1}{16} $$

C
$$ \dfrac{1}{8} $$

D
$$ \dfrac{1}{4} $$
Question 7
1 / -0

The value of $$\lim_{x \rightarrow 1} \sec \dfrac{\pi}{2x} \log x$$ is-

A
$$ \pi /2$$

B
$$ 2 /\pi$$

C
$$-\pi/2$$

D
$$-2/ \pi$$

Solution
Question 8
1 / -0

$$\lim _ { x \rightarrow 1 } \{ 1 - x + [ x + 1 ] + [ 1 - x ] \} , \text { where } [ x ]$$ denotes greatest integer function is

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution
Question 9
1 / -0

If $$2f(\sin x)+\sqrt {2}f(\cos x)=\tan x,\ (x> 0)$$, then $$\displaystyle \lim _{ x\rightarrow 1 }{ \sqrt { 1-x } f\left( x \right) = }$$

A
$$\sqrt {2}$$

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

C
$$-\sqrt {2}$$

D
$$-\dfrac {1}{\sqrt {2}}$$
Question 10
1 / -0

$$ \underset { x\rightarrow 0 }{ lim } \cfrac { 1+cos\left( \pi x \right) }{ \left( 1-x \right)^ 2 } $$ is equal to :

A
$$ \cfrac { \pi} {2} $$

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

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

D
$$ \cfrac { \pi} {3} $$

Solution

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 28

Result

Limits and Derivatives Test - 28

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SHARING IS CARING

Question 1

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

$$1$$

$$\dfrac{1}{3}$$

$$2$$

Solution

Question 2

$$\sqrt{a}$$

$$-\sqrt{a}$$

$$a \sqrt{a}$$

$$-a\sqrt{a}$$

Solution

Question 3

$$Z =\dfrac{sin(2x+2y)}{4} + c$$

$$Z =-\dfrac{sin(2x+2y)}{4} + c$$

$$Z =-\dfrac{cos(2x+2y)}{4} + c$$

None of these

Question 4

$$1/5$$

$$1/6$$

$$1/4$$

$$1/2$$

Solution

Question 5

$$0$$

$$1$$

$$2$$

$$1/2$$

Question 6

$$ \dfrac{1}{24} $$

$$ \dfrac{1}{16} $$

$$ \dfrac{1}{8} $$

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

Question 7

$$ \pi /2$$

$$ 2 /\pi$$

$$-\pi/2$$

$$-2/ \pi$$

Solution

Question 8

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 9

$$\sqrt {2}$$

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

$$-\sqrt {2}$$

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

Question 10

$$ \cfrac { \pi} {2} $$

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

$$\cfrac { \pi ^2} {2} $$

$$ \cfrac { \pi} {3} $$

Solution

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