Limits and Derivatives Test

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Limits and Derivatives Test - 34

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Question 1
1 / -0

If $$\displaystyle {f}'(x) = sin\,x + sin\,4x .\, cos \,x $$ then $$\displaystyle {f}'(x) \left (2x^{2} + \dfrac{\pi}{2} \right ) $$ at $$ x = \sqrt{\dfrac{\pi}{2}} $$ is equal to

A
$$ -1 $$

B
$$ 0 $$

C
$$\displaystyle -2\sqrt{2\pi} $$

D
None of these

Solution
Question 2
1 / -0

If $$\displaystyle sin\, y = x\, sin ( a + y) $$ and
$$\displaystyle \dfrac{dy}{dx} = \dfrac{A}{ 1 + x^{2} - 2x \, cos a } $$ then the value of $$ A $$ is

A
$$ 2 $$

B
$$ cos\, a $$

C
$$ sin\,a $$

D
None of these

Solution
Question 3
1 / -0

$$ \displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}} $$ is equal to

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

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

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

D
Does not exist

Solution

a. $$\displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{\sqrt{8 x^{2}+7 x+1}}= \displaystyle \lim _{x \rightarrow-\infty} \dfrac{x^{2} \tan \dfrac{1}{x}}{-x \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}} $$
$$ =-\displaystyle \lim _{x \rightarrow-\infty} \dfrac{\tan \dfrac{1}{x}}{\dfrac{1}{x} \sqrt{8+\dfrac{7}{x}+\dfrac{1}{x^{2}}}}=-\dfrac{1}{2 \sqrt{2}} $$
Question 4
1 / -0

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x} $$ is equal to

A
0

B
$$\infty$$

C
-2

D
2

Solution

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{x\left(e^{x}-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \dfrac{2 x\left(e^{x}-1\right)}{4 \sin ^{2} \dfrac{x}{2}} $$
$$ =2 \displaystyle \lim _{x \rightarrow 0}\left[\dfrac{(x / 2)^{2}}{\sin ^{2} \dfrac{x}{2}}\right]\left(\dfrac{e^{x}-1}{x}\right)=2 $$
Question 5
1 / -0

$$ \displaystyle \lim _{x \to \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right] $$ is equal to

A
1

B
-1

C
0

D
$$None \ of \ these$$

Solution

$$ \displaystyle \lim _{x \rightarrow \pi / 2}\left[x \tan x-\left(\dfrac{\pi}{2}\right) \sec x\right] $$

$$ =\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{2 x \sin x-\pi}{2 \cos x} \quad\quad\quad \left(\dfrac{0}{0}form\right)$$

$$ =\displaystyle \lim _{x \rightarrow \pi / 2} \dfrac{[2 \sin x+2 x \cos x]}{-2 \sin x} $$
(Applying L'Hopital's rule) $$ =-1 $$
Question 6
1 / -0

If $$ f(x)=\dfrac{\cos x}{(1-\sin x)^{1 / 3}}, $$ then

A
$$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty } $$

B
$$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty } $$

C
$$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty } $$

D
none of these

Solution

d. $$\lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{\cos x}{(1-\sin x)^{1 / 3}}=\lim _{t \rightarrow 0} \dfrac{-\sin t}{(1-\cos t)^{1 / 3}}$$
$$=-\lim _{t \rightarrow 0} \dfrac{2 \sin \dfrac{t}{2} \cos \dfrac{t}{2}}{\left(2 \sin ^{2} \dfrac{t}{2}\right)^{1 / 3}}$$
$$ =-\lim _{t \rightarrow 0} 2^{2 / 3} \cos \dfrac{t}{2}\left(\sin \dfrac{t}{2}\right)^{1 / 3}=0 $$
Question 7
1 / -0

$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}},(m<n) $$ is equal to

A
1

B
0

C
n/m

D
None of these

Solution

$$ \displaystyle \lim _{x \rightarrow 0} \dfrac{\sin x^{n}}{(\sin x)^{m}}=\lim _{x \rightarrow 0}\left(\dfrac{\sin x^{n}}{x^{n}}\right)\left(\dfrac{x^{n}}{x^{m}}\right)\left(\dfrac{x}{\sin x}\right)^{m} $$
$$ =\displaystyle \lim _{x \rightarrow 0} x^{n-m}=0 \quad[\because m<n] $$
Question 8
1 / -0

$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x} $$ is equal to

A
0

B
1

C
2

D
4

Solution

$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1+\sin \pi\left(\dfrac{3 x}{1+x^{2}}\right)}{1+\cos \pi x} $$
$$ =\displaystyle \lim _{x \rightarrow 1} \dfrac{1-\cos \left(\dfrac{3 \pi}{2}-\dfrac{3 \pi x}{1+x^{2}}\right)}{1-\cos (\pi-\pi x)} $$
$$ =\lim _{x \rightarrow 1} \dfrac{2 \sin ^{2}\left(\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}\right)}{2 \sin ^{2}\left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)} $$
$$ =\displaystyle \lim _{x \rightarrow 1}\left(\dfrac{\dfrac{3 \pi}{4}-\dfrac{3 \pi x}{2\left(1+x^{2}\right)}}{\dfrac{\pi}{2}-\dfrac{\pi x}{2}}\right)^{2} $$
$$ =\displaystyle \lim _{x \rightarrow 1} 9\left(\dfrac{\dfrac{1}{2}-\dfrac{x}{1+x^{2}}}{1-x}\right)^{2}=\lim _{x \rightarrow 1} 9\left(\dfrac{x-1}{2\left(1+x^{2}\right)}\right)^{2}=0 $$
Question 9
1 / -0

$$ The \ value \ of \displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}} $$ is

A
$$e^{-2 \pi} $$

B
$$ e^{1 / \pi} $$

C
$$ e^{2 /\pi} $$

D
$$ e^{-1 / \pi} $$

Solution

$$ \displaystyle \lim _{x \rightarrow 1}(2-x)^{\tan \dfrac{\pi x}{2}} $$
$$ =\displaystyle \lim _{x \rightarrow 1}\{1+(1-x)\}^{\tan \dfrac{\pi x}{2}} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1}(1-x) \tan \frac{\pi x}{2}} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1}(1-x) \cot \left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)} $$
$$ =e^{\displaystyle \lim _{x \rightarrow 1} \dfrac{(1-x)}{\tan \left(\dfrac{\pi}{2}-\dfrac{\pi x}{2}\right)}} $$
$$=e^{\dfrac{2}{\pi}\displaystyle \lim_{x \to 1}{\dfrac{\dfrac{\pi}{2}(1-x)}{tan \left(\dfrac{\pi}{2}(1-x)\right)}}}$$
$$=e^{\dfrac{2}{\pi}}$$
Question 10
1 / -0

$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x} \text { is equal to }$$

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

B
$$ \dfrac{-1}{\pi} $$

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

D
None of these

Solution

$$\displaystyle \lim _{x \rightarrow 1} \dfrac{1-x^{2}}{\sin 2 \pi x}$$
$$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{2 \pi(1-x)(1+x)}{2 \pi \sin (2 \pi-2 \pi x)}$$
$$=-\displaystyle \lim _{x \rightarrow 1} \dfrac{(2 \pi-2 \pi x)}{\sin (2 \pi-2 \pi x)} \dfrac{1+x}{2 \pi}=\dfrac{-1}{\pi}$$

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Limits and Derivatives Test - 34

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Limits and Derivatives Test - 34

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SHARING IS CARING

Question 1

$$ -1 $$

$$ 0 $$

$$\displaystyle -2\sqrt{2\pi} $$

None of these

Solution

Question 2

$$ 2 $$

$$ cos\, a $$

$$ sin\,a $$

None of these

Solution

Question 3

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

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

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

Does not exist

Solution

Question 4

0

$$\infty$$

-2

2

Solution

Question 5

1

-1

0

$$None \ of \ these$$

Solution

Question 6

$$\lim _{ x\rightarrow \dfrac { \pi^- }{2 } }{ f(x)=-\infty } $$

$$\lim _{ x\rightarrow \dfrac { \pi^+ }{2 } }{ f(x)=\infty } $$

$$\lim _{ x\rightarrow \dfrac { \pi }{2 } }{ f(x)=\infty } $$

none of these

Solution

Question 7

1

0

n/m

None of these

Solution

Question 8

0

1

2

4

Solution

Question 9

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

$$ e^{1 / \pi} $$

$$ e^{2 /\pi} $$

$$ e^{-1 / \pi} $$

Solution

Question 10

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

$$ \dfrac{-1}{\pi} $$

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

None of these

Solution

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