Limits and Derivatives Test

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Limits and Derivatives Test - 25

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Question 1
1 / -0

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

A
$$\dfrac{1}{2\sqrt{x}}\sin(3\sqrt{x})$$.

B
$$\dfrac{1}{\sqrt{x}}\sin(2\sqrt{x})$$.

C
$$\dfrac{1}{2\sqrt{x}}\sin(2\sqrt{x})$$.

D
$$\dfrac{1}{2\sqrt{x}}\sin(4\sqrt{x})$$.

Solution

$$\cfrac { d }{ dx } \sin ^{ 2 }{ \sqrt { x } } $$
$$=2\sin { \left( \sqrt { x } \right) } .\cfrac { d }{ dx } \left( \sin { \sqrt { x } } \right) .\cfrac { d }{ dx } \sqrt { x } $$
$$=2\sin { \sqrt { x } } .\cos { \sqrt { x } } .\cfrac { 1 }{ 2\sqrt { x } } $$
$$=\sin { \left( 2\sqrt { x } \right) } .\cfrac { 1 }{ 2\sqrt { x } } \quad \left[ \because \sin { 2\theta } =2\sin { \theta } \cos { \theta } \right] $$
Question 2
1 / -0

$$\lim _{ x\rightarrow { 0 }^{ + } }{ \left( { \left( x\cos { x } \right) }^{ x }+{ \left( \cos { x } \right) }^{ \frac { 1 }{ \ln { x } } }+{ \left( x\sin { x } \right) }^{ x } \right) } $$ is equal to

A
$$2$$

B
$$2+e$$

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

D
$$3$$

Solution

$$\lim_{x\rightarrow 0^+}(x\cos x)^x$$

$$\lim_{x\rightarrow 0^+} x^x.\cos ^x x$$

$$\lim_{x\rightarrow 0^+} e^{x\ln x}.\cos ^x x$$

$$ 1\times 1$$

$$1$$

$$\therefore \lim_{x\rightarrow 0^+}(x\cos x)^x=1$$
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$$\lim_{x\rightarrow 0^+}(x\sin x)^x$$

$$\lim_{x\rightarrow 0^+} x^x.\sin ^x x$$

$$\lim_{x\rightarrow 0^+} e^{x\ln x}.\sin ^x x$$

$$ 1\times 1$$

$$1$$

$$\therefore \lim_{x\rightarrow 0^+}(x\sin x)^x=1$$

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$$\lim_{x\rightarrow 0^+} (\csc x)^{\frac{1}{\ln x}}$$

$$\lim_{x\rightarrow 0^+} (\dfrac{1}{\sin x})^{\frac{1}{\ln x}}$$

$$y= \lim_{x\rightarrow 0^+} (\dfrac{1}{\sin x})^{\frac{1}{\ln x}}$$

$$\ln y =\lim_{x\rightarrow 0^+} \dfrac{1}{\ln x}(-\ln (\sin x))$$

$$\ln y =\lim_{x\rightarrow 0^+} \dfrac{1}{\frac{1}{x}}\left ( -\frac{\cos x}{\sin x} \right )$$

$$\ln y =\lim_{x\rightarrow 0^+}(-\cos x)$$

$$\ln y = -1$$

$$y=e^{-1}$$

$$y=\dfrac{1}{e}$$

$$\therefore \lim_{x\rightarrow 0^+} (\csc x)^{\frac{1}{\ln x}}=\dfrac{1}{e}$$

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$$\lim_{x\rightarrow 0^+} ((x\cos x)^x+(\csc x)^{\frac{1}{\ln x}}+(x\sin x)^x)$$

$$=1+\dfrac{1}{e}+1$$

$$=2+\dfrac{1}{e}$$
Question 3
1 / -0

$$\dfrac{\displaystyle \lim_{h \rightarrow 0}(h+1)^2}{\displaystyle \lim_{h\rightarrow 0}(1+h)^{2/h}}$$ is equal to

A
$$e^1$$

B
$$e^{-2}$$

C
$$e^2$$

D
$$e^{-1}$$
Question 4
1 / -0

$$\mathop {\lim}\limits_{x \to \frac{\pi}{2}} \tan x = $$

A
$$\infty $$

B
$$ - \infty $$

C
$$0$$

D
does not exist

Solution

$$L=\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\tan x$$
$$\Rightarrow$$ $$L=\tan\dfrac{\pi}{2}$$ [ Applying limit ]
We know that value of $$\tan\dfrac{\pi}{2}$$ is does not exist.
$$\therefore$$ $$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}\tan x=$$ does not exist
Question 5
1 / -0

$$\lim _{ x\rightarrow 0 }{ \log _{ \left( \tan ^{ 2 }{ x } \right) }{ \left( \tan ^{ 2 }{ 2x } \right) = } }$$

A
$$1$$

B
$$2$$

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

D
$$Does\ not\ exist$$

Solution

$$\mathop {\lim }\limits_{x \to 0} {\log _{{{\tan }^2}x}}\left( {{{\tan }^2}2x} \right)$$

Let,

$$\begin{array}{l} y={ \log _{ { { \tan }^{ 2 } }x } }{ \tan ^{ 2 } }2x \\ =\log { \tan ^{ 2 } } 2x={ \left( { { { \tan }^{ 2 } }x } \right) ^{ y } } \\ =2\log \tan 2x=2y\log \tan x \\ \Rightarrow y=\dfrac { { \log \tan 2x } }{ { \log \tan x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \log \tan 2x } }{ { \log \tan x } } \end{array}$$

Using L-Hospital Rule

$$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \dfrac { 1 }{ { \tan 2x } } \times { { \sec }^{ 2 } }2x\times 2 } }{ { \dfrac { 1 }{ { \tan x } } { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2{ { \sec }^{ 2 } }2x } }{ { \tan 2x } } \times \dfrac { { \tan x } }{ { { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\tan x } }{ { \tan 2x } } \, \, \, \, \, \, \left( { { \lim }_{ x\to 0 }\sec x=4 } \right) \end{array}$$

$$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\dfrac { { \tan x } }{ x } \times x } }{ { \dfrac { { \tan 2x } }{ { 2x } } \times 2x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\, \, \, 1\, \, \left( { { \lim }_{ x\to 0 }\dfrac { { \tan x } }{ x } =1 } \right) \\ \Rightarrow { \lim }_{ x\to 0 }{ \log _{ { { \tan }^{ 2 } }x } }\left( { { { \tan }^{ 2 } }2x } \right) =1 \end{array}$$
Question 6
1 / -0

$$\displaystyle \lim_{n \rightarrow \infty} {^{n}C_{c}}\left(\dfrac {m}{n}\right)^{x}\left(1-\dfrac {m}{n}\right)^{n-x}$$ equal to

A
$$\dfrac {M^{x}}{x\ !}.e^{-m}$$

B
$$\dfrac {M^{x}}{x\ !}.e^{m}$$

C
$$0$$

D
$$\dfrac {m^{x+1}}{me^{m}x\ !}$$

Solution

Given $$\displaystyle\lim_{n\rightarrow \infty}{^{n}}C_x \bigg(\dfrac{m}{n}\bigg)^{x}\bigg(1-\dfrac{m}{n}\bigg)^{n-x}=$$coefficient of $$x$$ in $$\displaystyle\lim_{n\rightarrow\infty}\bigg(\dfrac{m}{n}+1-\dfrac{m}{n}\bigg)^{n}$$
$$=$$coeeficient of $$x$$ in $$\displaystyle\lim_{n\rightarrow\infty}\bigg(1\bigg)^{n}=$$coefficient of $$x$$ in $$1=0$$
Question 7
1 / -0

$$\underset{h \rightarrow 0}{lim} \dfrac{\sqrt{x + h} -\sqrt{x}}{h}$$ is equal to

A
$$\sqrt{x}$$

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

C
$$2 \sqrt{x}$$

D
$$\dfrac{1}{\sqrt{x}}$$

Solution

The given limit can be written as,

$$\underset { h\rightarrow 0 }{ lim } \dfrac { \sqrt { x+h } -\sqrt { x } }{ h } =\underset { x+h\rightarrow x }{ lim } \dfrac { \sqrt { x+h } -\sqrt { x } }{ (x+h)-x } $$

We know that,
$$\underset { x\rightarrow a }{ lim } \dfrac { { x }^{ n }-{ a }^{ n } }{ x-a } =n{ a }^{ n-1 }$$

So, the given limit is,
$$=\dfrac { 1 }{ 2 } { x }^{ \left( \dfrac { 1 }{ 2 } -1 \right) }=\dfrac { 1 }{ 2\sqrt { x } } $$
Question 8
1 / -0

Let $$y=a\cos t+b\sin t$$ then $$\dfrac{d^2y}{dt^2}=$$

A
$$y$$

B
$$-y$$

C
$$2y$$

D
none of these

Solution

Given,
$$y=a\cos t+b\sin t$$......(1).
Now differentiating both sides with respect to $$t$$ we get,
$$\dfrac{dy}{dt}=-a\sin t+b\cos t$$
Again differentiating both sides with respect to $$t$$ we get,
$$\dfrac{d^2y}{dt^2}=-a\cos t-b\sin t$$
or, $$\dfrac{d^2y}{dt^2}=-y$$. [ Using (1)]
Question 9
1 / -0

The difference of slopes of lines represent by $${y^2} - 2xy{\sec ^2}\alpha + \left( {3 + {{\tan }^2}\alpha } \right)\left( {{{\tan }^2}\alpha - 1} \right){x^2} = 0$$ is

A
$$3$$

B
$$4$$

C
$$0$$

D
$$2$$

Solution

$$y^2-2xy\sec ^2\alpha +(3+\tan^2\alpha)(\tan ^2\alpha -1)x^2=0$$
Divide by $$x^2$$ & substitute $$m=y/x$$
$$m^2-2m\sec ^2 \alpha+(3+\tan^2\alpha)(\tan^2\alpha -1)=0$$m_1
Let $$m_1$$ & $$m_2$$ be note of equation
$$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2\left[\begin{matrix}m_1+m_2=-b/a\\m_1m_2=c/a \end{matrix}\right.$$
$$(m_1-m_2)^2(-2\sec\alpha)^2-4(3\tan ^2\alpha -3+\tan ^4 \alpha -\tan ^2\alpha)$$
$$=4\sec ^4\alpha -4(\tan ^4\alpha +2 \tan ^2 \alpha -3)$$
$$=4(\sec ^4 \alpha -\tan ^4\alpha )-8\tan ^2\alpha +12$$
$$=\dfrac{(\sec^2\alpha -\tan ^2\alpha)}{1}(\sec ^2\alpha +\tan ^2\alpha )-8\tan ^2\alpha +12$$
$$4\sec^2\alpha +4\tan ^2\alpha -4\tan ^2\alpha -4\tan ^2\alpha +12$$
$$4(\sec ^2\alpha -\tan ^2\alpha)+12$$
$$4(1)+12$$
$$(m_1-m_2)^2=16$$
$$m_1-m_2=4$$
Question 10
1 / -0

$$\lim\limits_{x\to 0}\dfrac{1-\cos x }{x^2}=$$

A
$$4$$

B
$$2$$

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

D
$$1$$

Solution

Now,
$$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}$$
$$=\lim\limits_{x\to 0}\dfrac{2\sin^2\dfrac{x}{2}}{x^2}$$
$$=\dfrac{1}{2}\lim\limits_{x\to 0}\dfrac{\sin^2\dfrac{x}{2}}{\dfrac{x^2}{4}}$$
$$=\dfrac{1}{2}\left(\lim\limits_{\dfrac{x}{2}\to 0}\dfrac{\sin\dfrac{x}{2}}{\dfrac{x}{2}}\right)^2$$
$$=\dfrac{1}{2}.1$$
$$=\dfrac{1}{2}$$.

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 25

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Limits and Derivatives Test - 25

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SHARING IS CARING

Question 1

$$\dfrac{1}{2\sqrt{x}}\sin(3\sqrt{x})$$.

$$\dfrac{1}{\sqrt{x}}\sin(2\sqrt{x})$$.

$$\dfrac{1}{2\sqrt{x}}\sin(2\sqrt{x})$$.

$$\dfrac{1}{2\sqrt{x}}\sin(4\sqrt{x})$$.

Solution

Question 2

$$2$$

$$2+e$$

$$2+\dfrac { 1 }{ e }$$

$$3$$

Solution

Question 3

$$e^1$$

$$e^{-2}$$

$$e^2$$

$$e^{-1}$$

Question 4

$$\infty $$

$$ - \infty $$

$$0$$

does not exist

Solution

Question 5

$$1$$

$$2$$

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

$$Does\ not\ exist$$

Solution

Question 6

$$\dfrac {M^{x}}{x\ !}.e^{-m}$$

$$\dfrac {M^{x}}{x\ !}.e^{m}$$

$$0$$

$$\dfrac {m^{x+1}}{me^{m}x\ !}$$

Solution

Question 7

$$\sqrt{x}$$

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

$$2 \sqrt{x}$$

$$\dfrac{1}{\sqrt{x}}$$

Solution

Question 8

$$y$$

$$-y$$

$$2y$$

none of these

Solution

Question 9

$$3$$

$$4$$

$$0$$

$$2$$

Solution

Question 10

$$4$$

$$2$$

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

$$1$$

Solution

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