Limits and Derivatives Test

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Limits and Derivatives Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

If $$y=\sec(\tan^{-1}x)$$, then $$\displaystyle\frac{dy}{dx}$$ at $$x=1$$ is equal to.

A
$$\displaystyle\frac{1}{\sqrt{2}}$$

B
$$\displaystyle\frac{1}{2}$$

C
$$1$$

D
$$\sqrt{2}$$

Solution

Given, $$y=\sec(\tan^{-1}x)$$
$$=\sec(\sec^{-1}\sqrt{1+x^2})$$
$$\Rightarrow y=\sqrt{1+x^2}$$
$$\therefore \displaystyle\frac{dy}{dx}=\frac{2x}{2\sqrt{1+x^2}}$$
$$\Rightarrow \displaystyle\frac{dy}{dx}=\frac{x}{\sqrt{1+x^2}}$$
$$\Rightarrow \left(\displaystyle\frac{dy}{dx}\right)_{x=1}=\displaystyle\frac{1}{\sqrt{2}}$$.
Question 2
1 / -0

If $$\displaystyle \lim _{ n\rightarrow \infty }{ \cfrac { n.{ 3 }^{ n } }{ n{ \left( x-2 \right) }^{ n }+n.{ 3 }^{ n+1 }-{ 3 }^{ n } } } =\cfrac { 1 }{ 3 } $$, then the range of $$x$$ is (When $$n\in N$$)

A
$$[2,\ 5)$$

B
$$\left( 1,\ 5 \right) $$

C
$$\left( -1,\ 5 \right) $$

D
$$\left( -\infty ,\ \infty \right) $$

Solution

$$\displaystyle\lim_{n\rightarrow \infty}\dfrac{n.3^n}{n(x-2)^n+n3^{n+1}-3^n}=\displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{\left(\dfrac{x-2}{3}\right)^n+3-\dfrac{1}{n}}=\dfrac{1}{3}$$
We know $$\displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{n}=0$$
then we want $$\dfrac{|x-2|}{3} < 1$$ $$\rightarrow \because$$ if $$a < 1, a^n=0$$ for $$n\rightarrow \infty$$
$$x < 5$$
$$x > -1$$
$$\therefore (-1, 5)$$.
Question 3
1 / -0

If $$y = |\cos x| + |\sin x|$$, then $$\dfrac {dy}{dx}$$ at $$x = \dfrac {2\pi}{3}$$ is

A
$$\dfrac {1 - \sqrt {3}}{2}$$

B
$$0$$

C
$$\dfrac {1}{2}(\sqrt {3} - 1)$$

D
None of these

Solution

According to question,
$$|\cos x| = -\cos x$$
$$|\sin x| = \sin x$$
$$\therefore \dfrac {dy}{dx} = \sin x + \cos x$$
$$\therefore \left (\dfrac {dy}{dx}\right )_{\dfrac {2\pi}{3}} = \dfrac {\sqrt {3}}{2} - \dfrac {1}{2} = \dfrac {\sqrt {3} - 1}{2}$$.
Question 4
1 / -0

The derivative of $$f\left( \tan { x } \right) $$ with respect to $$g(\sec x)$$ at $$\quad x=\cfrac { \pi }{ 4 } $$, where $$f'(1)=2;\quad g'(\sqrt { 2 } )=4$$ is

A
$$\cfrac { 1 }{ \sqrt { 2 } } $$

B
$$\sqrt { 2 } $$

C
$$1$$

D
$$0$$

Solution

$$\cfrac { df\left( \tan { x } \right) }{ dg\left( \sec { x } \right) } =\cfrac { f'\left( \tan { x } \right) \sec { x } }{ g'\left( \sec { x } \right) \tan { x } } =\cfrac { f'\left( \tan { x } \right) }{ g'\left( \sec { x } \right) } .\cfrac { 1 }{ \cos { x } } $$
$$=\cfrac { f'\left( \tan { x } \right) }{ g'\left( \sec { x } \right) } .\cfrac { 1 }{ \sin { x } } $$
$$\therefore { \left[ \cfrac { df\left( \tan { x } \right) }{ dg\left( \sec { x } \right) } \right] }_{ x=\pi /4 }^{ }=\cfrac { f'(1) }{ g'(\sqrt { 2 } ) } .\sqrt { 2 } =\cfrac { 2 }{ 4 } \times \sqrt { 2 } =\cfrac { 1 }{ \sqrt { 2 } } $$
Question 5
1 / -0

If $$|x| < 1$$, then $$\dfrac{d}{dx}\left[1+\dfrac{p}{q}x+\dfrac{p(p+q)}{2!}\left(\dfrac{x}{q}\right)^2+\dfrac{p(p+q)(p+2q)}{3!}\left(\dfrac{x}{q}\right)^3....\infty\right]=$$

A
$$\dfrac{p}{q(1-x)^{\frac{p}{q}+1}}$$

B
$$\dfrac{p}{q(1-x)^{\frac{p}{q}}}$$

C
$$(1-x)^{-pq-1}$$

D
$$(1-x)^{pq+1}$$

Solution

$$\dfrac{d}{dx}\left\{1+\dfrac{px}{q}+\dfrac{p(p+q)}{2!}\dfrac{x^2}{q^2}+\dots\right\}=\dfrac{p}{q}+\dfrac{p(p+q)}{2!}\dfrac{2x}{q^2}+\dfrac{p(p+q)(p+2q)}{3!}\dfrac{3x^2}{q^3}+\dots$$
$$=\dfrac{p}{q}\left(1+\dfrac{(p+q)}{1!}\dfrac{x}{q}+\dfrac{(p+q)(p+2q)}{2!}\dfrac{x^2}{q^2}+\dots\right)$$
Now, consider $$(1-a)^n=1-na+\dfrac{n(n-1)}{2!}x^2+\dots$$
Put $$a=x$$ and $$n=-\left(1+\dfrac{p}{q}\right)$$
$$\therefore (1-x)^{-\left(1+\frac{p}{q}\right)}=1+\dfrac{(p+q)}{q}x+\left[-\dfrac{(p+q)}{q}\left(-\dfrac{(p+q)}{q}-1\right)\right]\dfrac{x^2}{2!}\dots$$
$$=1+\dfrac{(p+q)}{q}x+\left[\dfrac{(p+q)}{q}\left(\dfrac{(p+q+q)}{q}\right)\right]\dfrac{x^2}{2!}\dots$$
$$=1+\dfrac{(p+q)}{q}x+\dfrac{(p+q)(p+2q)}{q^2}\dfrac{x^2}{2!}\dots$$
Hence,
$$\dfrac{d}{dx}\left\{1+\dfrac{px}{q}+\dfrac{p(p+q)}{2!}\dfrac{x^2}{q^2}+\dots\right\}=\dfrac{p}{q}(1-x)^{-\left(1+\frac{p}{q}\right)}$$
This is the required answer.
Question 6
1 / -0

$$\lim _{ x\rightarrow 3 }{ \left( { x }^{ 3 }-4 \right) /\left( x+1 \right) } =$$

A
$$(4/23)$$

B
$$(2/23)$$

C
$$(1/8)$$

D
$$(23/4)$$
Question 7
1 / -0

Differentiate the following w.r.t. $$x$$.
$$\sin x\ log x$$.

A
$$\dfrac{\sin x}{x}-\cos x\,\, log \,x$$

B
$$\dfrac{\sin x}{x}+\cos x\,\, log \,x$$

C
$$\dfrac{\cos x}{x}+\cos x\,\, log \,x$$

D
$$\dfrac{\tan x}{x}+\cos x\,\, log \,x$$

Solution

$$\cfrac { d }{ dx } \sin { x } \log { x } $$
$$=\left\{ \cfrac { d }{ dx } \sin { (x) } \right\} \log { x } +\sin { x } .\cfrac { d }{ dx } \left( \log { \left( x \right) } \right) $$
$$=\log { x } .\cos { x } +\sin { x } .\cfrac { 1 }{ x } $$
$$=\cfrac { \sin { x } }{ x } +\log { x } .\cos { x }$$
Question 8
1 / -0

Differentiate the following w.r.t. $$x$$.
$$\tan^{3}x$$.

A
$$(3tan^2x)$$

B
$$(3tan^2x)(sec^2\,x)$$

C
$$(tan^2x)(sec^2\,x)$$

D
$$(tan^2x)(sin^2\,x)$$

Solution

$$\cfrac { d }{ dx } \tan ^{ 3 }{ x } $$
$$=3\tan ^{ 2 }{ x } .\cfrac { d }{ dx } \tan { x } .\cfrac { d }{ dx } \left( x \right) $$
$$=3\tan ^{ 2 }{ x }\sec ^{ 2 }{ x } $$
Question 9
1 / -0

Given $$y=\dfrac {3}{x}, \dfrac {dy}{dx}=$$

A
$$3$$

B
$$\dfrac {3}{x^{2}}$$

C
$$\dfrac {-3}{x^{2}}$$

D
$$3x$$

Solution

$$y=\dfrac{3}{x}$$
differentiating with respect to x
$$\dfrac{dy}{dx}=\dfrac{-3}{x^2}$$
Question 10
1 / -0

Differentiate the following w.r.t. $$x$$.
$$\tan x^{2}$$.

A
$$sec^2(x^2)$$

B
$$sec^2(x^2)2x$$

C
$$sec^2(x^3)2x$$

D
$$tan^2(x^2)2x$$

Solution

$$\cfrac { d }{ dx } \tan { \left( { x }^{ 2 } \right) } $$
$$=\sec ^{ 2 }{ \left( { x }^{ 2 } \right) } .\cfrac { d }{ dx } \left( { x }^{ 2 } \right) $$
$$=\sec ^{ 2 }{ \left( { x }^{ 2 } \right) } .2x$$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 24

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Limits and Derivatives Test - 24

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SHARING IS CARING

Question 1

$$\displaystyle\frac{1}{\sqrt{2}}$$

$$\displaystyle\frac{1}{2}$$

$$1$$

$$\sqrt{2}$$

Solution

Question 2

$$[2,\ 5)$$

$$\left( 1,\ 5 \right) $$

$$\left( -1,\ 5 \right) $$

$$\left( -\infty ,\ \infty \right) $$

Solution

Question 3

$$\dfrac {1 - \sqrt {3}}{2}$$

$$0$$

$$\dfrac {1}{2}(\sqrt {3} - 1)$$

None of these

Solution

Question 4

$$\cfrac { 1 }{ \sqrt { 2 } } $$

$$\sqrt { 2 } $$

$$1$$

$$0$$

Solution

Question 5

$$\dfrac{p}{q(1-x)^{\frac{p}{q}+1}}$$

$$\dfrac{p}{q(1-x)^{\frac{p}{q}}}$$

$$(1-x)^{-pq-1}$$

$$(1-x)^{pq+1}$$

Solution

Question 6

$$(4/23)$$

$$(2/23)$$

$$(1/8)$$

$$(23/4)$$

Question 7

$$\dfrac{\sin x}{x}-\cos x\,\, log \,x$$

$$\dfrac{\sin x}{x}+\cos x\,\, log \,x$$

$$\dfrac{\cos x}{x}+\cos x\,\, log \,x$$

$$\dfrac{\tan x}{x}+\cos x\,\, log \,x$$

Solution

Question 8

$$(3tan^2x)$$

$$(3tan^2x)(sec^2\,x)$$

$$(tan^2x)(sec^2\,x)$$

$$(tan^2x)(sin^2\,x)$$

Solution

Question 9

$$3$$

$$\dfrac {3}{x^{2}}$$

$$\dfrac {-3}{x^{2}}$$

$$3x$$

Solution

Question 10

$$sec^2(x^2)$$

$$sec^2(x^2)2x$$

$$sec^2(x^3)2x$$

$$tan^2(x^2)2x$$

Solution

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