Limits and Derivatives Test

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Limits and Derivatives Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

Evaluate $$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$$

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

B
$$1$$

C
$${ 3 }^{ 2 }$$

D
$$\cfrac { 1 }{ { 2 }^{ 2 } } $$

Solution

$$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$$

$$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{n^2} \times \dfrac{n(n+1)}{2}$$

$$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{2}(1+\dfrac{1}{n})=\dfrac{1}{2}$$
Question 2
1 / -0

If $$y=x^2\sin \dfrac{1}{x}$$ then $$\dfrac{dy}{dx}=?$$

A
$$x\sin \dfrac{1}{x}-\cos \dfrac{1}{x}$$

B
$$-\cos \dfrac{1}{x}+2x\sin \dfrac{1}{x}$$

C
$$-x\sin \dfrac{1}{x}+\cos \dfrac{1}{x}$$

D
$$none\ of\ these$$

Solution
Question 3
1 / -0

If $$(x+y)=\sin (x+y)$$ then $$\dfrac{dy}{dx}=?$$

A
$$-1$$

B
$$1$$

C
$$\dfrac{1-\cos (x+y)}{\cos^2(x+y)}$$

D
$$none\ of\ these$$

Solution
Question 4
1 / -0

If $$x=a\cos^{2}\theta, y=b\sin^{2}\theta$$ then $$\dfrac{dy}{dx}=?$$

A
$$\dfrac{-a}{b}$$

B
$$\dfrac{a}{b}\cot \theta$$

C
$$\dfrac{-b}{a}$$

D
$$none\ of\ these$$

Solution
Question 5
1 / -0

If $$y=e^{\sin \sqrt x}$$ then $$\dfrac{dy}{dx}=?$$

A
$$e^{\sin \sqrt x}.\cos \sqrt x$$

B
$$\dfrac{e^{\sin \sqrt x}\cos \sqrt x}{2\sqrt x}$$

C
$$\dfrac{e^{\sin \sqrt x}}{2\sqrt x}$$

D
$$none\ of\ these$$

Solution
Question 6
1 / -0

If $$y=\cos^2 x^3$$ then $$\dfrac{dy}{dx}=?$$

A
$$-3x^2\sin (2x^3)$$

B
$$-3x^2\sin^2x^3$$

C
$$-3x^2\cos^2(2x^3)$$

D
$$none\ of\ these$$

Solution
Question 7
1 / -0

lf $$ \displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) $$ is finite then the value of $$a,b$$ respectively are

A
$$5\, -4$$

B
$$-5,\ -4$$

C
$$-4,\ 3$$

D
$$4,\ 5$$

Solution

$$\lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) $$

As $${ x\rightarrow 0 }$$, denominator tends to 0, so the numerator also tends to 0.

$$\Rightarrow \lim _{ x\rightarrow 0 } \cos 4x+a\cos 2x+b=0$$
at $$x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1$$
$$\Rightarrow 1+a+b=0$$
$$\Rightarrow a+b=-1$$
Now, put $$a = -4 $$
$$ \Rightarrow b = 3$$
Option C satisfies above equation.
Question 8
1 / -0

$$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \frac{\pi x}{2})=$$

A
$$\pi$$

B
$$ 2\pi$$

C
$$\displaystyle \frac{\pi}{2}$$

D
$$\displaystyle \frac{2}{\pi}$$

Solution

$$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \dfrac{\pi x}{2})$$
$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { 1-x }{ \cot { (\dfrac { \pi x }{ 2 } } ) } $$
It is of the form $$\displaystyle \dfrac{0}{0}$$, so applying L-Hospital's rule
$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { -1 }{ -\dfrac { \pi }{ 2 } \csc ^{ 2 }{ (\dfrac { \pi x }{ 2 } ) } } $$
$$=\displaystyle \dfrac{2}{\pi}$$
Question 9
1 / -0

The value of $$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } $$ is:

A
$$0$$

B
$$1$$

C
$$\displaystyle \frac { \sin { \beta } }{ \beta } $$

D
$$\displaystyle \frac { \sin { 2\beta } }{ 2\beta } $$

Solution

$$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } \sin { \left( \alpha +\beta \right) } }{ \left( \alpha -\beta \right) \left( \alpha +\beta \right) } \right] } $$

$$\displaystyle =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } \times \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha +\beta \right) } }{ \left( \alpha +\beta \right) } \right] } $$

$$\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } .\left( \frac { \sin { 2\beta } }{ 2\beta } \right) $$ ....... $$[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]$$

$$\displaystyle =1.\frac { \sin { 2\beta } }{ 2\beta }$$ ..... $$[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]$$
$$=\dfrac { \sin { 2\beta } }{ 2\beta } $$
Question 10
1 / -0

Evaluate: $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$

A
$$\displaystyle \frac{3}{2}$$

B
$$\displaystyle \frac{2}{3}$$

C
$$\displaystyle \frac{1}{3}$$

D
$$\displaystyle \frac{3}{4}$$

Solution

Given,

$$\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$

$$\sec x=\dfrac{1}{\cos x}$$

$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}$$

$$=\lim_{x\rightarrow 0}\dfrac{\cos 2x-\cos 4x}{\cos x-\cos 3x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$

$$=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$

$$=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$

as $$\cos 0=1$$ and $$\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$$

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