Limits and Derivatives Test

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Limits and Derivatives Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f$$ be a differentiable function such that $$f'(x) = 7- \dfrac{3}{4}\dfrac{f(x)}{x}, (x > 0)$$ and $$f(1) \neq 4$$.
Then $$\underset{x\to 0^+}{\lim} xf \left(\dfrac{1}{x}\right) $$:

A
Exists and equals 4

B
Does not exist

C
Exist and equals

D
Exists and equals $$\dfrac{4}{7}$$

Solution

$$f'(x) = 7 -\dfrac{3}{4}\dfrac{f(x)}{x}$$ $$(x > 0)$$

Given $$f(1) \neq 4$$ $$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = ?$$

$$\dfrac{dy}{dx} + \dfrac{3}{4}\dfrac{y}{x} = 7$$ (This is LDE)

IF $$=e^{\int \tfrac{3}{4x}dx} = e^{\tfrac{3}{4}ln|x|} = X^{\tfrac{3}{4}}$$

$$y.x^{\frac{3}{4}} =\int 7.x^{\frac{3}{4}} dx$$

$$y.x^{\frac{3}{4}} = 7.\dfrac{x^{\frac{7}{4}}}{\frac{7}{4}} + C$$

$$f(x) = 4x+C.x^{-\frac{3}{4}}$$

$$f\left(\dfrac{1}{x}\right) = \dfrac{4}{x} + C.x^{\frac{3}{4}}$$

$$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = \underset{x\to 0^+}{\lim}\left(4+C.x^{\frac{7}{4}}\right) = 4$$
Question 2
1 / -0

$$\underset{x\to 0}{\lim} \left(\dfrac{3x^2+2}{7x^2+2}\right)^{1/x^2}$$ is equal to:

A
$$\dfrac{1}{e^2}$$

B
$$\dfrac{1}{e}$$

C
$$e^2$$

D
$$e$$

Solution

$$L = \underset{x \to 0}{\lim} \left(\dfrac{3x^2 + 2}{7x^2 + 2}\right)^{\tfrac{1}{x^2}}$$

As it is a $$1^{\infty}$$ form

We know the formula,

$$\Rightarrow$$$$\displaystyle\lim_{x \to 0} f(x)^{g(x)}= \displaystyle e^{\underset{x\to 0}{\lim} g(x) \left (f(x)-1\right)}$$

$$= e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{3x^2+2}{7x^2 + 2}\displaystyle-1\right)}$$

$$ = e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{3x^2+2-7x^2-2}{7x^2 + 2}\right)}$$

$$ = e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{-4x^2}{7x^2 + 2}\right)}$$

$$=e^{\tfrac{-4}{2}}$$

$$=e^{-2}$$

$$=\dfrac1{e^{2}}$$

$$\boxed{L=\dfrac1{e^{2}}}....Answer$$

Hence option $$'A'$$ is the answer.
Question 3
1 / -0

$$\lim _{ { x\rightarrow \pi /4 } } \dfrac { \cot ^{ { 3 } } x-\tan x }{ \cos (x+\pi /4) } $$ is

A
$$4$$

B
$$8 \sqrt { 2 }$$

C
$$8$$

D
$$4 \sqrt { 2 }$$

Solution
Question 4
1 / -0

If $$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$$, then $$\displaystyle \lim_{x\rightarrow 0} \dfrac {f'(x)}{x}$$.

A
Exists and is equal to $$-2$$

B
Does not exist

C
Exist and is equal to $$0$$

D
Exists and is equal to $$2$$

Solution

$$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$$
$$=\cos x(x^{2}-2x^{2})-x(2\sin x-2x\tan x)+1(2x\sin x-x^{2}\tan x)$$
$$=-x^{2}\cos x-2x\sin x+2x^{2}\tan x+2x\sin x-x^{2}\tan x$$
$$=x^{2}\tan x-x^{2}\cos x$$
$$=x^{2}(\tan x-\cos x)$$
$$f'(x)=2x(\tan x-\cos x)+x^{2}(\sec^{2}x+\sin x)$$
$$\therefore \displaystyle \lim_{x\rightarrow 0} \dfrac{f'(x)}{x}=\lim_{x\rightarrow 0}\dfrac{2x(\tan x-\cos x)+x^{2}(\sec^2{x}+\sin x)}{x}$$
$$=\displaystyle \lim_{x\rightarrow 0} 2(\tan x-\cos x)+x(\sec^{2}x+\sin x)$$
$$=2(0-1)+0=-2$$
Hence, $$\therefore \displaystyle \lim_{x\rightarrow 0} \dfrac{f'(x)}{x}=-2$$
Question 5
1 / -0

$$\displaystyle \frac{d}{dx}\left(\frac{\sin x}{x}\right)$$

A
$$ \dfrac{x\cos x-\sin x}{x^{2}}.$$

B
$$ \dfrac{x\cos x+\sin x}{x^{2}}.$$

C
$$ \dfrac{x\cos x+\sin x}{x^{3}}.$$

D
$$ \dfrac{x\cos x-\sin x}{x^{3}}.$$

Solution

$$\displaystyle y=\dfrac{\sin x}{x}$$
$$\displaystyle \dfrac{dy}{dx}=

\lim_{\delta x\rightarrow 0}\dfrac{1}{\delta x}\left [ \dfrac{\sin\left (

x+\delta x \right )}{x+\delta x}-\dfrac{\sin x}{x} \right ]$$

$$\displaystyle

= \lim_{\delta x\rightarrow 0}\dfrac{x\sin\left ( x+\delta x \right

)-\left ( x+\delta x \right )\sin x }{x\left ( x+\delta x \right

).\delta x}$$

$$\displaystyle =\lim_{\delta x\rightarrow 0}\left [

\dfrac{1}{x+\delta x}.\dfrac{\sin \left ( x+\delta x \right )-\sin

x}{\delta x}-\dfrac{\sin x}{x\left ( x+\delta x \right )} \right ]$$

$$\displaystyle = \dfrac1x\dfrac{d(\sin x)}{dx} - \dfrac{\sin x}{x^2}$$

$$\displaystyle = \dfrac{1}{x}.\cos x-\dfrac{\sin x}{x^{2}}= \dfrac{x\cos x-\sin x}{x^{2}}.$$
Question 6
1 / -0

If $$x$$ is very large, then $$\dfrac {2x}{1+x}$$ is

A
close to $$0$$

B
arbitrarily large

C
lie between $$2$$ and $$3$$

D
close to $$2$$

Solution

$$\dfrac {2x}{1+x}=\dfrac {2}{1+\dfrac {1}{x}}=\dfrac{2}{1+0}=2$$

If $$x\rightarrow \alpha$$ then $$\dfrac {1}{x}\rightarrow 0$$.
Question 7
1 / -0

If $$y=2 \sin x -3x^4 + 8$$, then $$\dfrac{dy}{dx}$$ is

A
$$2\sin x -12 x^3$$

B
$$2 \cos x-12 x^3$$

C
$$2 \cos x+12 x^3$$

D
$$2 \sin x+12 x^3$$

Solution

If $$y=2 \sin x -3x^4 + 8$$
$$\Rightarrow \cfrac{dy}{dx}=2 \cos x-3(4x^{(4-1)})+0$$
$$\Rightarrow \cfrac{dy}{dx}=2 \cos x-12x^3$$
Therefore, the correct answer is $$B$$.
Question 8
1 / -0

$$\displaystyle \lim_{x\rightarrow \infty} \sin x$$ equals

A
$$1$$

B
$$0$$

C
$$\infty$$

D
does not exist

Solution

Let $$L = \displaystyle \lim_{x\rightarrow \infty} \sin x$$
Assume $$\displaystyle y = \frac{1}{x}$$ so as $$x\to \infty, y \to 0$$
$$\Rightarrow L = \displaystyle \lim_{y\rightarrow 0} \sin \frac{1}{y}$$
We know $$\sin x$$ lie between -1 to 1
so let $$p =\sin x$$ as $$x\to \infty$$
Thus left hand limit $$=L^+=\displaystyle \lim_{y\rightarrow 0^+} \sin \frac{1}{y}=p$$
and right hand limit $$=L^-=\displaystyle \lim_{y\rightarrow 0^-} \sin \frac{1}{y}=-p$$
Clearly L.H.L $$\neq $$ R.H.L $$\Rightarrow$$ Required limit does not exist.
Question 9
1 / -0

Derivative of $$2\tan x - 7\sec x$$ with respect to $$x$$ is:

A
$$2 \sec x + 7 \tan x $$

B
$$\sec x (2 \sec x + \tan x)$$

C
$$2 {\sec}^2 x + \sec x. \tan x$$

D
$$\sec x (2 \sec x - 7 \tan x)$$

Solution

Diffrentiating the given function with respect to $$x$$:
$$\displaystyle \frac{d}{dx} (2 \tan x - 7 \sec x) = 2 \sec^2 x - 7 \sec x \tan x$$
$$= \sec x (2 \sec x - 7 \tan x)$$
Question 10
1 / -0

$$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } } $$ is equal to

A
$$3$$

B
$$1$$

C
$$0$$

D
$$2$$

Solution

$$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } } $$
$$=\displaystyle \lim_{x\to 0}\dfrac{xe^x}{x}-\lim_{x\to 0}\dfrac{\sin x}{x}$$, split up
$$=\displaystyle \lim_{x\to 0}e^x-\lim_{x\to 0}\dfrac{\sin x}{x}$$
$$=e^0-1=1-1=0$$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 10

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Limits and Derivatives Test - 10

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SHARING IS CARING

Question 1

Exists and equals 4

Does not exist

Exist and equals

Exists and equals $$\dfrac{4}{7}$$

Solution

Question 2

$$\dfrac{1}{e^2}$$

$$\dfrac{1}{e}$$

$$e^2$$

$$e$$

Solution

Question 3

$$4$$

$$8 \sqrt { 2 }$$

$$8$$

$$4 \sqrt { 2 }$$

Solution

Question 4

Exists and is equal to $$-2$$

Does not exist

Exist and is equal to $$0$$

Exists and is equal to $$2$$

Solution

Question 5

$$ \dfrac{x\cos x-\sin x}{x^{2}}.$$

$$ \dfrac{x\cos x+\sin x}{x^{2}}.$$

$$ \dfrac{x\cos x+\sin x}{x^{3}}.$$

$$ \dfrac{x\cos x-\sin x}{x^{3}}.$$

Solution

Question 6

close to $$0$$

arbitrarily large

lie between $$2$$ and $$3$$

close to $$2$$

Solution

Question 7

$$2\sin x -12 x^3$$

$$2 \cos x-12 x^3$$

$$2 \cos x+12 x^3$$

$$2 \sin x+12 x^3$$

Solution

Question 8

$$1$$

$$0$$

$$\infty$$

does not exist

Solution

Question 9

$$2 \sec x + 7 \tan x $$

$$\sec x (2 \sec x + \tan x)$$

$$2 {\sec}^2 x + \sec x. \tan x$$

$$\sec x (2 \sec x - 7 \tan x)$$

Solution

Question 10

$$3$$

$$1$$

$$0$$

$$2$$

Solution

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