Limits and Continuity Test 24

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Limits and Continuity Test 24

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Question 1
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 2
1 / -0

Evaluate: $$\underset { x\rightarrow 0 }{ \lim } \dfrac { x\tan2x-2x\tan x }{ \left( 1-\cos2x \right) ^{ 2 } } $$

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

B
$$1$$

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

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

Solution

$$\underset{x\rightarrow 0}{lt}\dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}$$

$$=\underset{x\rightarrow 0}{lt}\dfrac{x\dfrac{2\tan x}{1-\tan^2x}-2x\tan x}{(1-1+2\sin^2x)^2}$$ (since $$\tan 2\theta =\dfrac{2\tan \theta}{1-\tan^2\theta}$$)

$$=\underset{x\rightarrow 0}{lt}\dfrac{2x\tan x(1-1+\tan^2x)}{(1-\tan^2x)y\sin^4x}=\underset{x\rightarrow 0}{lt}\dfrac{x\tan^3x}{2\sin^4x}\underset{x\rightarrow 0}{lt}\dfrac{1}{1-\tan^2x}$$

$$=\underset{x\rightarrow 0}{lt}\dfrac{\sin^3x/\cos^3x}{2\sin^3x(\sin x/x)}\times 1=\underset{x\rightarrow 0}{lt}\dfrac{1}{2\cos^3x}\underset{x\rightarrow 0}{lt}\dfrac{1}{\sin x/x}=\dfrac{1}{2}\times 1$$

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

Evaluate: $$\underset { x\rightarrow { 0 } }{ lim } \dfrac { x\tan 2x-2x \tan\ x\quad }{ (1-\cos2x)^{ 2 } } $$

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

B
$$1$$

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

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

Solution

$$\rightarrow \underset{x \rightarrow 0}{lt} \dfrac{x (\tan 2x - 2 \tan x)}{(1 - \cos 2x)^2}$$

$$= \underset{x \rightarrow 0}{lt} \dfrac{x\left[\dfrac{2 \tan x}{1 - \tan^2 x} - 2 \tan x\right]}{(2 \sin^2 x)^2}$$

$$= \underset{x \rightarrow 0}{lt} \dfrac{2x \tan x \left[\dfrac{1}{1 - \tan^2 x} -1 \right]}{4 \sin^4 x}$$

$$= \underset{x \rightarrow 0}{lt} \dfrac{\dfrac{x \tan x}{1 - \tan^2 x} [x - 1 + \tan^2 x]}{2 \sin^4x}$$

$$= \underset{x \rightarrow 0}{lt} \dfrac{x \tan^3 x \left(\dfrac{\sin^3 x}{\cos^3 x} \right) / x^4}{(1 - \tan^2 x). 2 \sin^4 x/ xy}$$

$$= \underset{x \rightarrow 0}{lt} \dfrac{\left(\dfrac{\sin x}{x} \right)^3 . \dfrac{1}{\cos^3x}}{2 \left(\dfrac{\sin x}{x} \right)^4 (1 - \tan^2 x)} = \dfrac{1.1}{2.1 (1 - 0)} = \dfrac{1}{2}$$

Given $$\underset{x \rightarrow 0}{lt} \dfrac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2} = \dfrac{1}{2}$$
Question 4
1 / -0

$$\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \cot x - \cos x } { ( \pi - 2 x ) ^ { 3 } }$$ equals:

A
$$\frac { 1 } { 24 }$$

B
$$\frac { 1 } { 16 }$$

C
0

D
$$\frac { 1 } { 4 }$$

Solution
Question 5
1 / -0

The function $$f\left( x \right) = \dfrac{{4 - {x^2}}}{{4x - {x^3}}}$$ is

A
discontinuous at only one point

B
discontinuous exactly at two points

C
discontinuous exactly at three points

D
None of these

Solution

$$f\left(x\right)= \dfrac{4-x^2}{4x-x^3}$$

$$=\dfrac{4-x^2}{x\left(4-x^2\right)}$$

$$for 4-x^2\neq 0$$

$$x\neq\pm 2$$

$$ f\left(x\right)=\dfrac{1}{x}$$

so this function will be discontinue at $$x=0$$

$$x=\pm2$$

three Points
Question 6
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 7
1 / -0

$$\underset {x\rightarrow0}{ lim } \dfrac{x \tan 2 x - 2 x \tan x}{(1 - \cos 2 x)^2}$$ equals

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

B
1

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

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

Solution
Question 8
1 / -0

Solve:
$$\underset{x \rightarrow 2}{Lt} \dfrac{x^{\sqrt{2}} - 2^{\sqrt{2}}}{x - 2} =$$

A
$$\sqrt{2}.2^{\sqrt{2}}$$

B
$$2^{\sqrt{2}- 1}$$

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

D
$$2^{\sqrt{2}}$$

Solution

$$\underset{x \rightarrow 2}{Lt} \dfrac{x^{\sqrt{2}} - 2^{\sqrt{2}}}{x - 2} \Rightarrow $$
L' Hospital -
$$\underset{x \rightarrow 2}{Lt}\frac{\sqrt{2}.x^{\sqrt{2}-1}}{1}$$
$$ = \sqrt{2}\times (2)^{\sqrt{2}-1}$$
$$ = \frac{\sqrt{2}\times 2^{\sqrt{2}}}{2}$$
$$ = (2)^{\sqrt{2}-\frac{1}{2}}$$
Question 9
1 / -0

Let $$f(x)=\begin{cases} { x }^{ 2 }+k,\quad \quad when\quad x\ge 0 \\ -{ x }^{ 2 }-k,\quad \quad when \quad x<0 \end{cases}$$. If the function $$f(x)$$ be continous at $$x=0$$, then $$k=$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$-2$$

Solution

We have,

$$ {{x}^{2}}+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,when,\,x\ge 0\, $$

$$ -{{x}^{2}}-k\,\,\,\,\,\,\,\,\,\,\,\,\,when,x\le 0 $$

Given that,

$$L.H.L=R.H.L=f\left( x \right)$$

At point $$x=0$$

$$ {{x}^{2}}+k=-{{x}^{2}}-k=0 $$

$$ \Rightarrow {{x}^{2}}+k=0 $$

$$ \Rightarrow k=-{{x}^{2}} $$

$$ \Rightarrow k=0 $$

Hence, this is the answer.
Question 10
1 / -0

Evaluate: $$\displaystyle\lim_{x\to 10}\dfrac{x^2-100}{x-10}$$

A
$$10$$

B
$$-5 $$

C
$$20$$

D
$$5$$

Solution

$$\displaystyle\lim_{x\to 10}\dfrac{x^2-100}{x-10}$$

$$\displaystyle\lim_{x\to 10}\dfrac{(x-10)(x+10)}{x-10}$$

$$\displaystyle\lim_{x\to 10}{x+10}$$

$$=10+10=20$$

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Limits and Continuity Test 24

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Limits and Continuity Test 24

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SHARING IS CARING

Question 1

$$ \pi /2$$

$$ 2 /\pi$$

$$-\pi/2$$

$$-2/ \pi$$

Solution

Question 2

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

$$1$$

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

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

Solution

Question 3

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

$$1$$

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

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

Solution

Question 4

$$\frac { 1 } { 24 }$$

$$\frac { 1 } { 16 }$$

0

$$\frac { 1 } { 4 }$$

Solution

Question 5

discontinuous at only one point

discontinuous exactly at two points

discontinuous exactly at three points

None of these

Solution

Question 6

$$\sqrt {2}$$

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

$$-\sqrt {2}$$

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

Question 7

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

1

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

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

Solution

Question 8

$$\sqrt{2}.2^{\sqrt{2}}$$

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

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

$$2^{\sqrt{2}}$$

Solution

Question 9

$$0$$

$$1$$

$$2$$

$$-2$$

Solution

Question 10

$$10$$

$$-5 $$

$$20$$

$$5$$

Solution

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