Limits and Derivatives Test

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Limits and Derivatives Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

$$\lim _{ x\rightarrow \infty }{ \frac { 1 }{ x } \int _{ 0 }^{ x }{ \left( \sqrt { { t }^{ 2 }+5t } -t \right) dt } }$$

A
$$0$$

B
$$1$$

C
$$\dfrac{5}{2}$$

D
$$5$$

Solution
Question 2
1 / -0

The value of $$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b } }{ { x }^{ 2 }-{ a }^{ 2 } } } \left( a>b \right) $$ is

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

B
$$\dfrac {1}{a\sqrt {a-b}}$$

C
$$\dfrac {1}{2a\sqrt {a-b}}$$

D
$$\dfrac {1}{4a\sqrt {a-b}}$$

Solution

We have,
$$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b } }{ { x }^{ 2 }-{ a }^{ 2 } } }$$

This is the $$\dfrac{0}{0}$$ form.
Apply L-hospital rule,
$$\lim_{x\to a}\dfrac{\dfrac{1}{2\sqrt {x-b}}-0}{2x-0}$$
$$\lim_{x\to a}\dfrac{1}{4x\sqrt {x-b}}$$
$$=\dfrac{1}{4a\sqrt {a-b}}$$

Hence, this is the answer.
Question 3
1 / -0

$$\lim _ { x \rightarrow 0 } \dfrac { ( 27 + x ) ^ { 1 / 3 } - 3 } { 9 - ( 27 + x ) ^ { 2 / 3 } }$$ equals :

A
$$- 1 / 6$$

B
$$1 / 6$$

C
$$1 / 3$$

D
$$- 1 / 3$$

Solution

$$\underset { x\rightarrow 0 }{ lt } \dfrac { { \left( 27+x \right) }^{ 1/3 }-3 }{ 9-{ \left( 27+x \right) }^{ 2/3 } } $$
$$=\underset { x\rightarrow 0 }{ lt } \dfrac { \dfrac { 1 }{ 3 } { \left( 27+x \right) }^{ -2/3 } }{ \dfrac { -2 }{ 3 } { \left( 27+x \right) }^{ -1/3 } } $$
$$=\underset { x\rightarrow 0 }{ lt } \dfrac { -1 }{ 2 } \dfrac { { \left( 27+x \right) }^{ 1/3 } }{ { \left( 27+x \right) }^{ 2/3 } } $$
$$=\dfrac { -1 }{ 2 } \left( \dfrac { 3 }{ 9 } \right) ,\quad \dfrac { -1 }{ 6 } $$ [A]
Question 4
1 / -0

A curve in the $$1^{st}$$ quadrant passes through $$(1,1)$$. Its drifferential equation is $$(y-xy^{2})dx+(x+x^{2}y^{2})dy=0$$. Hence the equation of the curve is

A
$$y-\dfrac{1}{xy}=\ln y$$

B
$$y-\dfrac{1}{xy}=\ln x$$

C
$$y-xy=\ln y$$

D
$$y-xy=\ln x$$

Solution
Question 5
1 / -0

if cos y = x cos (a + y ), then $$\dfrac{dy}{dx}$$ =

A
$$\dfrac{cos^2 (a + y)}{sina}$$

B
$$\dfrac{cos^2(a+y)}{cosa}$$

C
$$\dfrac{cosa}{sin^2(a + y)}$$

D
$$\dfrac{cos(a - y)}{sina}$$

Solution
Question 6
1 / -0

If $$(cos x)^y = (sin y)^x$$, then $$\dfrac{dy}{dx}$$ =

A
$$\dfrac{log(siny) + y tan x}{ log (cos x) - x cot y}$$

B
$$\dfrac{log(siny) - y tan x}{ log (cos x) + cot y}$$

C
$$\dfrac{log(siny) }{ log (cos x) }$$

D
$$\dfrac{log(cos x) }{ log (sin y) }$$

Solution
Question 7
1 / -0

The derivative of $$\tan^{-1}\left[\dfrac{\sin x}{1+\cos x}\right]$$ with respect to $$\tan^{-1}\left[\dfrac{\cos x}{1+\sin x}\right]$$ is

A
$$2$$

B
$$-1$$

C
$$0$$

D
$$-2$$

Solution

We have,
$$\begin{array}{l} { y_{ 1 } }={ \tan ^{ -1 } }\left( { \dfrac { { \sin x } }{ { 1+\cos x } } } \right) \\ ={ \tan ^{ -1 } }\left( { \dfrac { { 2\sin \dfrac { x }{ 2 }\cos \dfrac { x }{ 2 } } }{ { 1+2{ { \cos }^{ 2 } }\dfrac { x }{ 2 } -1 } } } \right) \\ ={ \tan ^{ -1 } }\left( { \tan \dfrac { x }{ 2 } } \right) \\ { y_{ 1 } }=\dfrac { x }{ 2 } \\ \dfrac { { d{ y_{ 1 } } } }{ { dx } } =\dfrac { 1 }{ 2 } \\ { y_{ 2 } }={ \tan ^{ -1 } }\left( { \dfrac { { \cos x } }{ { 1+\sin x } } } \right) \\ \dfrac { { d{ y_{ 2 } } } }{ { dx } } =\dfrac { 1 }{ { 2\left( { 1+\sin x } \right) } } .-\left( { 1+\sin x } \right) \\ =\dfrac { 1 }{ 2 } \\ \dfrac { { d{ y_{ 1 } } } }{ { dx } } .\dfrac { { dx } }{ { d{ y_{ 2 } } } } =\dfrac { 1 }{ 2 } \times -2 \\ \dfrac { { d{ y_{ 1 } } } }{ { d{ y_{ 2 } } } } =-1 \\ Hence, \\ option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$$
Question 8
1 / -0

$$\underset { x\rightarrow 1 }{ Lim } \left[ { \left[ \frac { 4 }{ { x }^{ 2 }-{ x }^{ -1 } } -\frac { { 1-3x+x }^{ 2 } }{ { 1-x }^{ 3 } } \right] }^{ -1 }+\frac { 3\left( { x }^{ 4 }-1 \right) }{ { x }^{ 3 }-{ x }^{ -1 } } \right] =$$

A
$$\frac { 1 }{ 3 } $$

B
3

C
$$\frac { 1 }{ 2 } $$

D
$$\frac { 3 }{ 2 } $$

Solution

$$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{4}{x^2-x^{-1}}-\dfrac{1-3x+x^2}{1-x^3}\right]^{-1}+\dfrac{3(x^4-1)}{x^3-x^{-1}}\right]$$
$$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{4x}{x^3-1}+\dfrac{1-3x+x^2}{x^3-1}\right]^{-1}+\dfrac{3(x^4-1)}{x^3-x^{-1}}\right]$$
$$\displaystyle\lim_{x\rightarrow 1}\left[\left[\dfrac{x^2+x+1}{x^3-1}\right]^{-1}+\dfrac{3(x^4-1)x}{x^4-1}\right]$$
$$\displaystyle\lim_{x\rightarrow 0}\left[\dfrac{(x-1)(x^2+x+1)}{x^2+x+1}+3x\right]$$ as $$(x^3-1)2(x-1)(x^2+x+1)$$
$$\displaystyle\lim_{x\rightarrow 1}[(x-1)+3x]=(1-1)+3(1)=3$$
Answer$$=3$$.
Question 9
1 / -0

$$\displaystyle \underset { x\rightarrow 0 }{ lim } \ \ \frac { ({ 1-\cos2x) }^{ 2 } }{ 2x \tan x-x \tan2x } $$ is :

A
$$-2$$

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

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

D
$$2$$

Solution

Let $$L=\displaystyle \lim_{x\rightarrow 0} \frac { (1 - cos 2x)^2}{2x \tan x - x \tan 2x}= \lim_{x\rightarrow 0} \frac {( 2 sin x)^2}{2x \tan x- \frac {x 2 \tan x}{1 - \tan^2 x}}$$

$$=\displaystyle \lim_{x\rightarrow 0} \frac { { 4 sin^4 x}}{\frac{1 - \tan^2 x}{2 x \tan x - 2x \tan^3 x - 2x \tan x}}$$

$$=\displaystyle \lim_{x\rightarrow 0} \frac {4 sin ^4 x ( 1 - \tan^2 x)}{- 2 x \tan^3 x}$$

$$=\displaystyle \lim_{x\rightarrow 0} \frac {2 sin^4 x}{-x \frac {sin^3 x}{cos x}}( 1 - \tan^2 x)$$

$$\displaystyle =- 2 \lim_{x\rightarrow 0} \left \{ \frac {sin x . cos^3 x}{x} (1 \tan^2 x) \right \}$$

$$ \displaystyle = -2 \left ( \lim_{x\rightarrow 0} \frac { sin x}{x} \right ) \lim_{x\rightarrow 0} \left \{ cos^3 x ( 1 - \tan^2 x) \right \}$$

$$L\displaystyle = -2(1)(1)$$

$$L\displaystyle = -2$$
Question 10
1 / -0

The differential of $$f(x)=\sqrt{\dfrac{2-x}{2+x}}$$ at $$x=0$$ and $$\delta x=0.15$$ is

A
$$0.07$$

B
$$0.075$$

C
$$-0.075$$

D
$$0.15$$

Solution