Limits and Derivatives Test

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Limits and Derivatives Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

$$\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{2{x^2} + x - 3}} = $$

A
$$\dfrac{1}{{10}}$$

B
$$\dfrac{{ - 1}}{{10}}$$

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

D
$$ - \dfrac{2}{5}$$

Solution

Now,
$$\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{2{x^2} + x - 3}} $$

$$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(2{x^2} +3 x-2x - 3)}} $$

$$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(x-1)(2x + 3)}} $$

$$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x - 1} \right)}}{{(\sqrt x-\sqrt 1)(\sqrt x+\sqrt 1)(2x + 3)}} $$

$$=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)}}{{(\sqrt x+\sqrt 1)(2x + 3)}} $$

$$=\dfrac{-1}{2\times5}$$

$$=-\dfrac{1}{10}$$.
Question 2
1 / -0

If $$a{x^2} + 2hxy + b{y^2} = 0$$ then $$\frac{{dy}}{{dx}}$$ is equal to

A
$$\frac{y}{x}$$

B
$$\frac{x}{y}$$

C
$$ - \frac{x}{y}$$

D
none of these

Solution
Question 3
1 / -0

$$\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{\cot x - \cos x}{\left(\dfrac{\pi}{2} -x \right)^3} = $$

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

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

C
$$2$$

D
$$-2$$

Solution
Question 4
1 / -0

Solve:
$$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}dx=$$

A
$$\dfrac{4}{3}(\sqrt{2}+1)$$

B
$$\dfrac{4}{3}(\sqrt{2}-1)$$

C
$$\dfrac{3}{4}(\sqrt{2}-1)$$

D
$$\dfrac{3}{4}(\sqrt{2}-2)$$

Solution

$$\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}}$$

$$=\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}\times\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}}$$

$$=\displaystyle\int_{0}^{1}{\dfrac{\left(\sqrt{x+1}+\sqrt{x}\right)dx}{x+1-x}}$$

$$=\displaystyle\int_{0}^{1}{\left(\sqrt{x+1}+\sqrt{x}\right)dx}$$

$$=\left[\dfrac{{\left(x+1\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+\dfrac{{\left(x\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{0}^{1}$$

$$=\left[2\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+2\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$$

$$=2\left[\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}$$

$$=\dfrac{2}{3}\left[\left({2}^{\frac{3}{2}}-1\right)-\left(1-0\right)\right]$$

$$=\dfrac{2}{3}\left[2\sqrt{2}-2\right]$$

$$=\dfrac{4}{3}\left(\sqrt{2}-1\right)$$
Question 5
1 / -0

Solve

$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 5x}}{{\tan 3x}}$$

A
$$-\dfrac{5}{3}$$

B
$$\dfrac{5}{3}$$

C
$$\dfrac{7}{3}$$

D
None of these

Solution

Let$$L=\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}$$

$$= \displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}\times \dfrac {3x}{\tan x}\times \dfrac {5}{3}$$

we know that $$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}=1$$ and $$\displaystyle \lim_{x\rightarrow 0}\dfrac {3x}{\tan 3x}=1$$

$$L=1\times 1\times \dfrac {5}{3}$$

$$\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}= \dfrac {5}{3}$$
Question 6
1 / -0

If $${z_r} = \cos \dfrac{{r\alpha }}{{{n^2}}} + i\sin \dfrac{{r\alpha }}{{{n^2}}}$$, where $$ r= 1, 2, 3, ....n$$, then $$\mathop {\lim }\limits_{n \to \infty } \left( {{z_1}.{z_2}.....{z_n}} \right)$$ is equal to

A
$$\cos \alpha + i\sin \alpha $$

B
$$\sin \frac{\alpha }{2} + i\cos \frac{\alpha }{2}$$

C
$${e^{i\alpha /2}}$$

D
$$\sqrt {{e^{i\alpha }}} $$

Solution

Solution -
$$ z_{r} = \dfrac{cosr\alpha }{n^{2}}+i\dfrac{sinr\alpha }{n^{2}} $$

Comparing with $$ cos \theta +isin \theta$$

$$\theta =\dfrac {r\alpha}{n^2}$$

$$\therefore z_r= e^{\dfrac{ir\alpha }{n^{2}}}$$

$$ \lim_{n\rightarrow \infty } (z_{1},z_{2},z_{3}....z_n) = e^{\dfrac{i\alpha }{n^{2}}}e^{\dfrac{2i\alpha }{n^{2}}} e^{\dfrac{3i\alpha }{n^{2}}}...e^{\dfrac{ni\alpha }{n^{2}}}$$

$$ = \lim_{n\rightarrow \infty } e^{i\alpha (\dfrac{1}{n^{2}}+\dfrac{2}{n^{2}}+...)}$$

$$ = \lim_{n\rightarrow \infty } e^{i\alpha \dfrac{1}{n^2}(1+2+3)}$$

$$ = \lim_{n\rightarrow \infty } e^{i\alpha (\dfrac{n^{2}+n}{2n^{^{2}}})} \quad \dots S_{n}=\dfrac{n}{2}\left [ 2a+(n-1)d \right ]$$

at $$ n\rightarrow \infty $$

$$ = e^{i\alpha }\times \dfrac{1}{2}$$

$$\lim_{n\rightarrow \infty } (z_{1},z_{2},z_{3}....z_n) = \sqrt{e^{i\alpha }}$$
Question 7
1 / -0

If $$y=a\ \sin\ x+b\ \cos\ x$$, then $$y^{2}+\left ( \dfrac{dy}{dx} \right )^{2}$$ is

A
function of $$x$$

B
function of $$y$$

C
function of $$x$$ and $$y$$

D
constant

Solution

$$y=a\sin{x}+b\cos{x}$$

$$\dfrac{dy}{dx}=a\cos{x}-b\sin{x}$$

$${y}^{2}+{\left(\dfrac{dy}{dx}\right)}^{2}={\left(a\sin{x}+b\cos{x}\right)}^{2}+{\left(a\cos{x}-b\sin{x}\right)}^{2}$$

$$={a}^{2}{\sin}^{2}{x}+{b}^{2}{\cos}^{2}x+2ab\sin{x}\cos{x}+{a}^{2}{\cos}^{2}{x}+{b}^{2}{\sin}^{2}x-2ab\sin{x}\cos{x}$$

$$={a}^{2}\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)+{b}^{2}\left({\sin}^{2}{x}+{\cos}^{2}{x}\right)$$

$$={a}^{2}+{b}^{2}$$ since $${\sin}^{2}{x}+{\cos}^{2}{x}=1$$

$$=constant$$
Question 8
1 / -0

If $$f(x) = x\sin x$$ ,find $$f'(\pi )$$, using first principle.

A
$$0$$

B
$$π$$

C
$$-π$$

D
Can't be determined

Solution
Question 9
1 / -0

If $$f(x+y)=2f(x).f(y)$$ for all $$x,y$$, where $$f'(0)=3$$ and $$f'(4)=2$$ then $$f'(4)=3$$ is equal to

A
6

B
12

C
4

D
3

Solution
Question 10
1 / -0

If $$y=(x+\sqrt{x^{2}+a^{2}})^{n}$$ then $$\dfrac{dy}{dx}=$$

A
$$y$$

B
$$ny$$

C
$$\dfrac{ny}{\sqrt{x^{2}+a^{2}}}$$

D
$$\dfrac{y}{\sqrt{x^{2}+a^{2}}}$$

Solution

$$y={ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n }\quad $$
$$\cfrac { dy }{ dx } =n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }.\cfrac { d }{ dx } \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) $$
$$=n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }.\left( 1+\cfrac { x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \right) =n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n-1 }\left( \cfrac { \sqrt { { x }^{ 2 }+{ a }^{ 2 } } +x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } \right) $$
$$=\cfrac { n{ \left( x+\sqrt { { x }^{ 2 }+{ a }^{ 2 } } \right) }^{ n } }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } =\cfrac { n.y }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } $$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 27

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Limits and Derivatives Test - 27

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SHARING IS CARING

Question 1

$$\dfrac{1}{{10}}$$

$$\dfrac{{ - 1}}{{10}}$$

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

$$ - \dfrac{2}{5}$$

Solution

Question 2

$$\frac{y}{x}$$

$$\frac{x}{y}$$

$$ - \frac{x}{y}$$

none of these

Solution

Question 3

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

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

$$2$$

$$-2$$

Solution

Question 4

$$\dfrac{4}{3}(\sqrt{2}+1)$$

$$\dfrac{4}{3}(\sqrt{2}-1)$$

$$\dfrac{3}{4}(\sqrt{2}-1)$$

$$\dfrac{3}{4}(\sqrt{2}-2)$$

Solution

Question 5

$$-\dfrac{5}{3}$$

$$\dfrac{5}{3}$$

$$\dfrac{7}{3}$$

None of these

Solution

Question 6

$$\cos \alpha + i\sin \alpha $$

$$\sin \frac{\alpha }{2} + i\cos \frac{\alpha }{2}$$

$${e^{i\alpha /2}}$$

$$\sqrt {{e^{i\alpha }}} $$

Solution

Question 7

function of $$x$$

function of $$y$$

function of $$x$$ and $$y$$

constant

Solution

Question 8

$$0$$

$$π$$

$$-π$$

Can't be determined

Solution

Question 9

6

12

4

3

Solution

Question 10

$$y$$

$$ny$$

$$\dfrac{ny}{\sqrt{x^{2}+a^{2}}}$$

$$\dfrac{y}{\sqrt{x^{2}+a^{2}}}$$

Solution

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