Continuity and Differentiability Test

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Continuity and Differentiability Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$(\sin x)^x(\ln (\sin x)+x\cot x)$$

B
$$(\ln (\sin x)+x\cot x)$$

C
$$(\sin x)^x(\ln (\sin x)+x\tan x)$$

D
$$(\sin x)^x(\ln (\sin x)-\cot x)$$

Solution

given $$y=(\sin x)^x$$
applying $$\ln$$ on both sides we get
$$\ln y=x\ln(\sin x)$$
differentiating both sides wrt $$x$$
$$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{d}{dx}\left(x\ln(\sin x)\right)$$
$$\dfrac{1}{y}\dfrac{dy}{dx}=\ln(\sin x)+x\dfrac{1}{\sin x}\cos x$$
$$\dfrac{dy}{dx}=y\left(\ln(\sin x)+x\cot x\right)$$
$$\therefore \dfrac{dy}{dx}=(\sin x)^x\left(x\cot x\right) + (\sin x)^x \ln(\sin x)$$
Question 2
1 / -0

If $$ f(x) = \bigg[ \frac {a+x}{1+x} \bigg]^{a+1+2x} $$ then $$ {a^{a+1}} \bigg [ 2 \ log \ a + {\frac {1-a ^2}{a}} \bigg]$$ is

A
$$ f^\prime (1) $$

B
$$ f^\prime (0) $$

C
$$ f^\prime (2) $$

D
$$ f^{\prime \prime} $$

Solution
Question 3
1 / -0

If a continuous function $$f$$ satisfies the relation
$$\overset{t}{\underset{0}{\int}} \left(f(x) - \sqrt{f'(x)}\right)dx = 0$$ and $$f(0) = \dfrac{-1}{2}$$
Then $$f(x)$$ is equal to

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

B
$$\dfrac{-x + 2}{4}$$

C
$$\dfrac{-1}{x^2 + 2}$$

D
$$\dfrac{x^2 - 2}{4}$$

Solution

Given,
$$\overset{t}{\underset{0}{\int}} \left(f(x) - \sqrt{f'(x)}\right)dx = 0$$
Now using Leinbnitz's rule of differentiation under integral sign we get,
$$f(t)-\sqrt{f'(t)}=0$$
or, $$\dfrac{f'(t)}{[f(t)]^2}=1$$
or, $$\dfrac{d[f(t)]}{[f(t)]^2}=dt$$
Now integrating both sides we get,
$$-\dfrac{1}{f(t)}=t+c$$ [ Where $$c$$ being integrating constant]......(1)
Given,
$$f(0)=-\dfrac{1}{2}$$.
Using this condition in (1) we get,
$$c=2$$.
So the solution takes the form,
$$-\dfrac{1}{f(t)}=t+2$$
or, $$f(t)=-\dfrac{1}{t+2}$$.
So, $$f(x)=-\dfrac{1}{x+2}$$.
Question 4
1 / -0

Differentiate with respect to $$x$$.
$${x^{\cos x}} + \sin {x^{\tan x}}$$

A
$$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$$

B
$$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$$

C
$$x^{\cos x}\left[ {\cos x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec x.\log \sin x]$$

D
None

Solution

let $$y=x^{\cos x} +\sin x ^{\tan x}$$
let $$h=x^{\cos x}$$ $$\dfrac{d}{dx}(f(x)g(x))=f'(x)g(x)+f'(x)f(x)$$
applying $$log$$ on both sides we get
$$\log h=\cos x\log x$$
differentiate on both sides wrt $$x$$
$$\dfrac{1}{h}\dfrac{dh}{dx}=-\sin x\log x+\dfrac{\cos x}{x}$$
$$\therefore \dfrac{dh}{dx}=x^{\cos x}\left(\dfrac{\cos x}{x}-\sin x \log x\right)$$
let $$t=\sin x^{\tan x}$$
applying $$\log $$ on both sides
$$\dfrac{1}{t}\dfrac{dt}{dx}=\sec ^2x\log {(\sin x)}+\dfrac{\tan x}{\sin x}(\cos x)$$
$$\therefore \dfrac{dt}{dx}=\sin x^{\tan x}\left(1+\sec ^2x \log (\sin x)\right)$$
$$y=h+t$$
$$\implies \dfrac{dy}{dx}=\dfrac{dh}{dx}+\dfrac{dt}{dx}$$
$$\therefore \dfrac{dy}{dx}=x^{\cos x}\left(\dfrac{\cos x}{x}-\sin x\log x\right)+\sin x^{\tan x}\left(1+\sec ^2x\log (\sin x)\right)$$
Question 5
1 / -0

Differentiate $${x^{\tan x}} + {{\mathop{\rm tanx}\nolimits} ^x}$$ with respect to $$x$$

A
$$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$$

B
$$x^{\tan x}(\sec^2x \log \sec x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$$

C
$$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})-\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$$

D
$$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x-\dfrac{2}{\sin 2x})$$

Solution
Question 6
1 / -0

If $$f:R \to R$$ be a differentiable function, such that $$f\left( {x + 2y} \right) = f\left( x \right) + f\left( {2y} \right) + 4xy$$ for all $$x,y \in R$$ then

A
$$f'\left( 1 \right) = f'\left( 0 \right) + 1$$

B
$$f'\left( 1 \right) = f'\left( 0 \right) - 1$$

C
$$f'\left( 0 \right) = f'\left( 1 \right) + 2$$

D
$$f'\left( 0 \right) = f'\left( 1 \right) - 2$$

Solution

$$f(x+2y)=f(x)+f(2y)+4xy$$
partially difference w.r.t $$x$$
$$f'(x+2y)=f'(x)+4y$$
At $$x=0$$ & $$y=1/2$$
$$f'(1)=f'(0)+2$$
$$C$$ is correct
Question 7
1 / -0

Let $$f\left( x \right) = \left\{ \begin{array}{l}\begin{array}{*{20}{c}}{ - 1\,\,\, - }&{2\,\,\, \le \,\,\,{\rm{x}}}&\rangle &0\end{array}\\\begin{array}{*{20}{c}}{{x^2}\,\, - }&{1,\,0\,\,\,\,\rangle }&{{\rm{x}}\,\,\rangle }&2\end{array}\end{array} \right.$$ and g $$\left( x \right) = \left| {f\left( x \right)\left| { + f\left| {x\left. {} \right|} \right.} \right.} \right.$$ then the number of points which g(x) is non differentiable,is

A
at most one point

B
$$2$$

C
exactly one point

D
infinite

Solution
Question 8
1 / -0

If $$f(x)$$ is twice differentiable function such that $$f(a)=0, f(b)=2, f(c)=-1, f(d)=2, f(e)=0$$, where $$a<b<c<d<e$$, then the minimum number of zeroes of $$g(x)={(f'(x))}^{2}+f''(x).f(x)$$ in the interval $$[a, e]$$ is

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution
Question 9
1 / -0

Let $$f\left( x \right) =\begin{cases} x\quad \quad \quad \quad \quad \quad \quad \quad x<1 \\ 2-x\quad \quad \quad \quad \quad \quad 1\le x\le 2 \\ -2+3x-{ x }^{ 2 }\quad \quad \quad x>2 \end{cases}$$ then $$f(x)$$ is

A
Differentiable at $$x=1$$

B
Differentiable at $$x=2$$

C
Differentiable at $$x=1$$ and $$x=2$$

D
Not differentiable at $$x=0$$

Solution

$$f\left(x \right)=\begin{Bmatrix} x \quad\qquad x<1\\ 2-x\qquad 1\le x\le 2 \\ -2+3x-x^{2} \quad x>2\end{Bmatrix}$$

At $$x=1$$
L.H.D(Left Hand differentiabilty)
$$\displaystyle \lim_{x\to\ 1^{-}}f^{'}(x)=1$$

R.H.D(Right Hand differentiability)
$$\displaystyle \lim_{x\to\ 1^{+}}f^{'}(x)=-1$$

$$\therefore L.H.D\ne R.H.D(\therefore$$ It is non-differentiable at $$x=1$$)

At $$x=2$$
L.H.D(Left Hand differentiabilty)
$$\displaystyle \lim_{x\to\ 2^{-}}f^{'}(x)=-1$$

R.H.D(Right Hand differentiability)
$$\displaystyle \lim_{x\to\ 2^{+}}f^{'}(x)=\lim_{x\to\ 2^{+}}3-2x=-1$$

$$\therefore R.H.D=L.H.D(at \,x=2)$$
$$(x)$$ is differentiable at $$x=2$$
Question 10
1 / -0

Let $$f(x)=\dfrac{1}{ax+b}$$ then $$f''(0)=$$

A
$$\dfrac{2a^3}{b^2}$$

B
$$\dfrac{2a^2}{b^3}$$

C
$$\dfrac{2a^3}{b^3}$$

D
none of these

Solution

Given,
$$f(x)=\dfrac{1}{ax+b}$$
Now differentiating both sides with respect to $$x$$ we get,
$$f'(x)=-\dfrac{a}{(ax+b)^2}$$
Again differentiating both sides with respect to $$x$$ we get,
$$f''(x)=\dfrac{2a^2}{(ax+b)^3}$$
So $$f''(0)=\dfrac{2a^2}{b^3}$$.

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Continuity and Differentiability Test - 24

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Continuity and Differentiability Test - 24

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Question 1

$$(\sin x)^x(\ln (\sin x)+x\cot x)$$

$$(\ln (\sin x)+x\cot x)$$

$$(\sin x)^x(\ln (\sin x)+x\tan x)$$

$$(\sin x)^x(\ln (\sin x)-\cot x)$$

Solution

Question 2

$$ f^\prime (1) $$

$$ f^\prime (0) $$

$$ f^\prime (2) $$

$$ f^{\prime \prime} $$

Solution

Question 3

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

$$\dfrac{-x + 2}{4}$$

$$\dfrac{-1}{x^2 + 2}$$

$$\dfrac{x^2 - 2}{4}$$

Solution

Question 4

$$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$$

$$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$$

$$x^{\cos x}\left[ {\cos x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec x.\log \sin x]$$

None

Solution

Question 5

$$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$$

$$x^{\tan x}(\sec^2x \log \sec x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$$

$$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})-\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$$

$$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x-\dfrac{2}{\sin 2x})$$

Solution

Question 6

$$f'\left( 1 \right) = f'\left( 0 \right) + 1$$

$$f'\left( 1 \right) = f'\left( 0 \right) - 1$$

$$f'\left( 0 \right) = f'\left( 1 \right) + 2$$

$$f'\left( 0 \right) = f'\left( 1 \right) - 2$$

Solution

Question 7

at most one point

$$2$$

exactly one point

infinite

Solution

Question 8

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 9

Differentiable at $$x=1$$

Differentiable at $$x=2$$

Differentiable at $$x=1$$ and $$x=2$$

Not differentiable at $$x=0$$

Solution

Question 10

$$\dfrac{2a^3}{b^2}$$

$$\dfrac{2a^2}{b^3}$$

$$\dfrac{2a^3}{b^3}$$

none of these

Solution

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