Limits and Derivatives Test

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Limits and Derivatives Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

$$\underset { x\rightarrow 0 }{ \lim } \dfrac { { 3 }^{ 2x }-{ 2 }^{ 3x } }{ x } $$ is equal to

A
$$\log\dfrac { 3 }{ 2 } $$

B
$$1$$

C
$$\log\cfrac { 9 }{ 8 } $$

D
$$0$$

Solution

We have,

$$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{3}^{2x}}-{{2}^{3x}}}{x}$$

Applying L’ Hospital rule and we get,

$$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{3}^{2x}}\log 3\times2-{{2}^{3x}}\log 2\times(3)}{1}$$

Taking limit and we get,

$$ ={{3}^{2\times 0}}\log 3^2-{{2}^{3\times 0}}\log 2^3 $$

$$ =\log 9-\log 8 $$

$$ =\log \dfrac{9}{8} $$

Hence, this is the answer.
Question 2
1 / -0

Let $$f(x)=\dfrac{ax+b}{x+1},lim_{x\rightarrow 0} f(x)=2$$ and $$lim_{x\rightarrow \infty} f(x)=1$$ then $$f(-2)=$$

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$0$$

Solution

$$\underset{x\rightarrow 0}{lim} f(x)=2$$
$$\Rightarrow \underset{x\rightarrow 0}{lim} \dfrac{ax+b}{x+1}=2$$
$$\Rightarrow \dfrac{b}{1}=2$$
$$\therefore b=2$$

Its is also given that
$$\underset{x\rightarrow \infty}{lim} f(x)=1$$
$$\Rightarrow \underset{x\rightarrow \infty}{lim} \dfrac{ax+b}{x+1}=1$$

$$\Rightarrow \underset{x\rightarrow \infty}{lim}\dfrac{a+\dfrac{b}{x}}{1+\dfrac{1}{x}}=1$$

$$\Rightarrow \dfrac{a+0}{1+0}=1$$

$$\therefore a=1$$
Substituting the values of a and b in $$f(x)=\dfrac{ax+b}{x+1}$$, we get,

$$f(x)=\dfrac{x+2}{x+1}$$

$$\therefore f(-2)=\dfrac{-2+2}{-2+1}=0$$
Question 3
1 / -0

$$\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } - x + 1 } - a x - b \right) = 0,$$ then the values of $$a$$ and $$b$$ are given by

A
$$a = - 1 , b = 1 / 2$$

B
$$a = 1 , b = 1 / 2$$

C
$$a = 1 , b = - 1 / 2$$

D
None of these

Solution

We have,
$$\begin{array}{l} { \lim }_{ x\to \infty }\left( { \sqrt { { x^{ 2 } }-x+1 } -ax-b } \right) =0 \\ Put\, \, x=\cfrac { 1 }{ t } \\ { \lim }_{ t\to { 0^{ + } } }\left( { \sqrt { \cfrac { 1 }{ { { t^{ 2 } } } } -\cfrac { 1 }{ t } +1 } -\cfrac { a }{ t } -6 } \right) =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { { { \left( { 1-t+{ t^{ 2 } } } \right) }^{ \cfrac { 1 }{ 2 } } }-a-bt } }{ t } =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { \left( { 1-\cfrac { t }{ 2 } +\cfrac { { { t^{ 2 } } } }{ 2 } } \right) -a-bt } }{ t } =0 \\ { \lim }_{ x\to { 0^{ + } } }\cfrac { { \left( { 1-a } \right) +t\left( { -b-\cfrac { 1 }{ 2 } } \right) } }{ t } =0 \\ 1=a \\ and\, \, b=-\cfrac { 1 }{ 2 } \\ Option\, \, \, C\, \, \, is\, \, correct\, \, answer. \end{array}$$
Question 4
1 / -0

Let $$U_{ { n } }=\dfrac { n! }{ (n+2)! } $$ where $$n \in N .$$ If $$S_{ { n } }=\sum _{ { n-1 } }^{ { n } } U_{ { n } }$$ then $$\lim _ { n \rightarrow \infty } \mathrm { S } _ { n }$$ equals :

A
$$2$$

B
$$1$$

C
$$1/2$$

D
$$1/3$$

Solution

$$\begin{array}{l} { U_{ n } }=\frac { { n! } }{ { \left( { n+2 } \right) ! } } =\frac { 1 }{ { \left( { n+1 } \right) \left( { n+2 } \right) } } =\frac { { \left( { n+1 } \right) -\left( { n+1 } \right) } }{ { \left( { n+1 } \right) \left( { n+2 } \right) } } \\ =\frac { 1 }{ { n+1 } } -\frac { 1 }{ { n+2 } } \\ { S_{ n } }=\sum _{ n=1 }^{ n }{ { U_{ n } } } =\left[ { \left( { \frac { 1 }{ 2 } -\frac { 1 }{ 3 } } \right) +\left( { \frac { 1 }{ 3 } -\frac { 1 }{ 4 } } \right) +............+\left( { \frac { 1 }{ { n+1 } } -\frac { 1 }{ { n+2 } } } \right) } \right] \\ { S_{ n } }=\frac { 1 }{ 2 } -\frac { 1 }{ { n+2 } } \\ { \lim }_{ n\to \infty }{ S_{ n } }=\frac { 1 }{ 2 } \\ Option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$$
Question 5
1 / -0

Evaluate
$$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}$$

A
$$4$$

B
$$2$$

C
$$1$$

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

Solution

We have,
$$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}$$

We know that
$$\cos 2x=1-2\sin^2x$$

Therefore,
$$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - (1-2\sin^2 x))}}{{{x^4}}}$$

$$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (2\sin^2 x)}}{{{x^4}}}$$

$$\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - (1-2\sin^2 (\sin^2 x))}}{{{x^4}}}$$

$$\mathop {\lim }\limits_{x \to 0} \cfrac{{2\sin^2 (\sin^2 x)}}{{{x^4}}}$$

$$\mathop {\lim }\limits_{x \to 0} \cfrac{2\sin^2 (\sin^2 x)}{\sin^4 x}\times \dfrac{\sin^4 x}{x^4}$$

$$2\mathop {\lim }\limits_{x \to 0} \left(\cfrac{\sin^2 (\sin^2 x)}{(\sin^2 x)}\right)^2\times \dfrac{\sin^4 x}{x^4}$$

We know that
$$\lim_{x\to\ 0}\dfrac{\sin x}{x}=1$$

Therefore,
$$\Rightarrow 2\times 1\times 1$$

$$\Rightarrow 2$$

Hence, this is the answer.
Question 6
1 / -0

If $$y ( x ):$$ Solution of a $$D.E.$$

$$( x \log x ) \dfrac { d y } { d x } + y = 2 x \log x,$$ $$( x , 1 )$$
$$y ( e ) = ? \quad x = e$$

A
$$e$$

B
$$ 0$$

C
$$ 2$$

D
$$ 2e$$

Solution

$$\begin{array}{l} From\, \, the\, given\, ,\, data \\ { e^{ \int { \frac { { dx } }{ { x\log x } } } } }\alpha \left( { \frac { { dy } }{ { dx } } +\frac { y }{ { x\log x } } =2 } \right) \, \, ---\left( 1 \right) \\ Now,\, solving \\ I=\int { \frac { { dx } }{ { x\log x } } } \, \, \, let\, \, \, \log x=t,\, dt=\frac { { dx } }{ x } \\ I=\int { \frac { { dt } }{ t } } =\log t \\ hence,\, \left( 1 \right) \, becomes\, \\ \frac { d }{ { dx } } \left( { { e^{ \log t } }y } \right) =2{ e^{ \log t } } \\ \frac { d }{ { dx } } \left( { y\log x } \right) =2\log x \\ \int { d\left( { y\log x } \right) } =2\int { \log x } dx \\ y\log x=2\left( { x\log x-x } \right) \\ Hence,\, \\ y\left( x \right) =f\left( x \right) =\frac { { 2\left( { x\log x-x } \right) } }{ { \log x } } \\ Now, \\ y\left( e \right) =\frac { { 2e\left( { \log e } \right) -e } }{ { \log e } } =e \\ Hence,\, the\, option\, A\, is\, the\, \, correct\, answer. \end{array}$$
Question 7
1 / -0

The value of $$\lim_{x\rightarrow \infty }$$ y In $$(\frac{sin (x+1/y)}{sin x})$$ when $$0 < x < \pi /2$$ is

A
cos x

B
$$\infty$$

C
cot x

D
does not exist
Question 8
1 / -0

$$\displaystyle\lim_{n\rightarrow\infty}\left\{\dfrac{n!}{(kn)^n}\right\}^{\dfrac{1}{n}}, k\neq 0$$, is equal to?

A
$$\dfrac{k}{e}$$

B
$$\dfrac{e}{k}$$

C
$$\dfrac{1}{ke}$$

D
None of these

Solution
Question 9
1 / -0

Let $$f(x)=\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{r=0}\dfrac{x}{(rx+1)\{(r+1)x+1\}}$$, then?

A
$$f(x)$$ is continuous but not differentiable at $$x=0$$

B
$$f(x)$$ is both continuous and differentiable at $$x=0$$

C
$$f(x)$$ is neither continuous nor differentiable at $$x=0$$

D
$$f(x)$$ is a periodic function

Solution
Question 10
1 / -0

$$\displaystyle \lim _{ x\rightarrow \infty }{ \left[\dfrac{n}{n^{2}+1^{2}}+\dfrac{n}{n^{2}+2^{2}}+\dfrac{n}{n^{2}+3^{2}}+....+\dfrac{1}{n^{5}}\right] }$$

A
$$\pi/4$$

B
$$\tan^{-1}{(2)}$$

C
$$\pi/2$$

D
$$\tan^{-1}{(3)}$$

Solution

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 31

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Limits and Derivatives Test - 31

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SHARING IS CARING

Question 1

$$\log\dfrac { 3 }{ 2 } $$

$$1$$

$$\log\cfrac { 9 }{ 8 } $$

$$0$$

Solution

Question 2

$$1$$

$$2$$

$$-1$$

$$0$$

Solution

Question 3

$$a = - 1 , b = 1 / 2$$

$$a = 1 , b = 1 / 2$$

$$a = 1 , b = - 1 / 2$$

None of these

Solution

Question 4

$$2$$

$$1$$

$$1/2$$

$$1/3$$

Solution

Question 5

$$4$$

$$2$$

$$1$$

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

Solution

Question 6

$$e$$

$$ 0$$

$$ 2$$

$$ 2e$$

Solution

Question 7

cos x

$$\infty$$

cot x

does not exist

Question 8

$$\dfrac{k}{e}$$

$$\dfrac{e}{k}$$

$$\dfrac{1}{ke}$$

None of these

Solution

Question 9

$$f(x)$$ is continuous but not differentiable at $$x=0$$

$$f(x)$$ is both continuous and differentiable at $$x=0$$

$$f(x)$$ is neither continuous nor differentiable at $$x=0$$

$$f(x)$$ is a periodic function

Solution

Question 10

$$\pi/4$$

$$\tan^{-1}{(2)}$$

$$\pi/2$$

$$\tan^{-1}{(3)}$$

Solution

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