Continuity and Differentiability Test

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Continuity and Differentiability Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

If $$x=a\cos ^{ 4 }{ t } ,y=b\ cosec^{ 4 }{ t } $$, then $$\cfrac { dx }{ dy } $$ at $$t=\cfrac { 3\pi }{ 4 } $$

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

B
$$\cfrac{b}{a}$$

C
$$\cfrac{a}{b}$$

D
$$\cfrac{-a}{16b}$$

Solution
Question 2
1 / -0

If $$y = \tan ^ { - 1 } \left( \cot \left( \dfrac { \pi } { 2 } - x \right) \right) ,$$ then $$\dfrac { d y } { d x } =$$

A
$$1$$

B
$$-1$$

C
$$0$$

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

Solution

$$y={ \tan }^{ -1 }\left( \cot\left( \pi /2-x \right) \right) $$
$$\Rightarrow y={ \tan }^{ -1 }\left( tanx \right) =x$$
$$\therefore$$ $$\dfrac { dy }{ dx } =1$$ [A]
Question 3
1 / -0

If $$y=\tan^{-1}\left(\dfrac{2^{x+1}}{1+2^{2x}}\right) $$ then $$\dfrac{dy}{dx}$$ at $$x=0$$ is

A
$$\dfrac{1}{10}log2$$

B
$$\dfrac{1}{5}log2$$

C
$$-\dfrac{1}{10}log2$$

D
$$log2$$

Solution

Given,
$$y=\tan^{-1}\left(\dfrac{2^{x+1}}{1+2^{2x}}\right) $$

or, $$y=\tan^{-1}\left(\dfrac{2.2^{x}}{1+(2^{x})^2}\right) $$

or, $$y=2\tan^{-1}(2^x)$$

Now, $$\dfrac{dy}{dx}=\dfrac{2}{1+(2^x)^2}\times 2^x\times \log 2$$.

And $$\left.\dfrac{dy}{dx}\right|_{x=0}=\dfrac{2}{1+1}\times \log 2=\log 2$$.
Question 4
1 / -0

If $$y=\log \sin x$$ find $$x$$ if $$y=0$$

A
$$\dfrac {\pi}2$$

B
$$\pi $$

C
$${\dfrac {\pi}3}$$

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

Solution

$$y=\log \sin x\\y=0\\\log \sin x=\log 1\\\sin x=1\\\sin x=\sin \dfrac{\pi}2\\x=\dfrac{\pi}2$$
Question 5
1 / -0

If $$f ( x ) = \tan x$$ and $$f$$ is inverse of $$g ,$$ then $$g ^ { \prime } ( x )$$is equal to

A
$$\cot x$$

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

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

D
$$\tan x$$

Solution

Given $$f(x)=\tan x$$ and $$f$$ is inverse of $$g$$
So , $$ g(x)$$ is $$\tan ^{-1} x $$
$$g^{\prime}(x)=\dfrac{d}{dx}\tan^{-1}x\\g^{\prime}(x)=\dfrac{1}{1+x^2}$$
Question 6
1 / -0

Find the values of a and b so that the function $$f(x)=\left\{\begin{matrix} x^2+3x+a, & if & x\leq 1\\ bx+2, & if & x > 1\end{matrix}\right.$$ is differentiable at each $$x\in R$$.

A
$$a=1,b=3$$

B
$$a=5,b=3$$

C
$$a=3,b=5$$

D
$$a=3,b=1$$

Solution

If the function f(x) is required to be differentiable, it must be continuous too for $$x\in R$$

But it is clearly evident that f(x) is continuous and differentiable in interval $$(-\infty,1)\cup(1,\infty)$$

So, only point need to be examined is x=1 .

For $$f(x)$$ to be continuous at $$x=1$$,

LHL of $$f(x)$$ at $$x=1$$ must be equal to RHL of $$f(x)$$ at $$x=1$$

i.e., $$(1)^2+3(1)+a =b(1)+2 \rightarrow b-a=2$$ ...............................................1

For $$f(x)$$ to be differentiable at $$x=1$$,

LHD of f(x) at x=1 must be equal to RHD of $$f(x)$$ at $$x=1$$

i.e., $$2(1) + 3 = b \rightarrow b=5$$

by putting $$b$$ value in we have $$a=3$$

Thus we get $$(a,b) = (3,5)$$
Question 7
1 / -0

The set of points where the functions f given by $$f(x)=|x-3|\cos x$$ differentiable is

A
R

B
R-{$$3$$}

C
$$(0,\infty)$$

D
None of these

Solution

b
Question 8
1 / -0

lf $$f(x)=\sin^{6}x+\cos^{6}x$$ then $$f''(x)=$$

A
$$-6\sin 4\mathrm{x}$$

B
$$6\sin 4\mathrm{x}$$

C
$$-6\cos 4\mathrm{x}$$

D
$$6\cos 4\mathrm{x}$$

Solution

$$f'(x)=6 \sin ^{5}x \cos x+6 \cos ^{5}x(-\sin x)$$
$$=3 \sin 2x(\sin ^{4}x-\cos ^{4}x)$$
$$=-3 \sin 2x \cos 2x$$
$$=-\dfrac{3}{2}\sin 4x$$
$$f''(x)=-6 \cos 4x$$
Question 9
1 / -0

Let $$f(x)$$ be differentiable function such that $$f\left (\displaystyle \frac{x+y}{1-xy}\right)=f(x)+f(y)\forall x$$ and $$y$$. lf $$ \displaystyle \lim_{x \rightarrow 0}\frac{f(x)}{x}=\frac{1}{3}$$, then $$f(1)$$ equals

A
$$\displaystyle \frac{\pi}{4}$$

B
$$\displaystyle \frac{\pi}{12}$$

C
$$\displaystyle \frac{\pi}{6}$$

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

Solution

$$f(\dfrac{x+y}{1-xy})=f(x)+f(y)$$

$$f(x)=atan^{-1}x$$

$$\lim_{x\rightarrow 0} \dfrac{a tan^{-1}x}{x}=\dfrac{1}{3}$$

$$\Rightarrow a=\dfrac{1}{3}$$

$$f(x)=\dfrac{1}{3}tan^{-1}x$$

$$\Rightarrow f(1)=\dfrac{1}{3}tan^{-1}(1)$$

$$=\dfrac{\pi }{12}$$
Question 10
1 / -0

$$\displaystyle \mathrm{A}:\dfrac{d}{dx}(\sin x)\ at\ x=\frac{\pi}{2}$$

$$\displaystyle \mathrm{B}:\dfrac{d}{dx}(\tan^{-1}{x})$$ at $$ {x}=1$$

$$\displaystyle \mathrm{C}:\dfrac{d}{dx}(\mathrm{e}^{x})$$ at $${x}=0$$

$$\mathrm{D}:\dfrac{d}{dx}(x^{x})\ at\ x=e$$

Arrangement of the above values in the increasing order of the magnitude

A
B, C, A, D

B
D, A, B, C

C
D, B, C, A

D
A, B, C, D

Solution

$$A.\dfrac{d}{dx}(sinx) = cosx$$

Put $$x=\dfrac{\pi }{2}$$

$$=0$$

$$B.\dfrac{d}{dx}(tan^{-1}x)=\dfrac{1}{1+x^{2}}$$

Put$$\ x=1$$

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

$$C. \dfrac{d}{dx}(e^{x})=e^{x}$$

Put $$x=0$$

$$=1$$

$$D.\dfrac{d}{dx}(x^{x})=\dfrac{d}{dx}(e^{xLogx})=x^{x}(Log x+1)$$

$$=2e^{e}$$

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Continuity and Differentiability Test - 13

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Continuity and Differentiability Test - 13

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Question 1

$$\cfrac{-b}{a}$$

$$\cfrac{b}{a}$$

$$\cfrac{a}{b}$$

$$\cfrac{-a}{16b}$$

Solution

Question 2

$$1$$

$$-1$$

$$0$$

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

Solution

Question 3

$$\dfrac{1}{10}log2$$

$$\dfrac{1}{5}log2$$

$$-\dfrac{1}{10}log2$$

$$log2$$

Solution

Question 4

$$\dfrac {\pi}2$$

$$\pi $$

$${\dfrac {\pi}3}$$

$$\dfrac {-\pi}2$$

Solution

Question 5

$$\cot x$$

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

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

$$\tan x$$

Solution

Question 6

$$a=1,b=3$$

$$a=5,b=3$$

$$a=3,b=5$$

$$a=3,b=1$$

Solution

Question 7

R

R-{$$3$$}

$$(0,\infty)$$

None of these

Solution

Question 8

$$-6\sin 4\mathrm{x}$$

$$6\sin 4\mathrm{x}$$

$$-6\cos 4\mathrm{x}$$

$$6\cos 4\mathrm{x}$$

Solution

Question 9

$$\displaystyle \frac{\pi}{4}$$

$$\displaystyle \frac{\pi}{12}$$

$$\displaystyle \frac{\pi}{6}$$

$$\displaystyle \frac{\pi}{3}$$

Solution

Question 10

B, C, A, D

D, A, B, C

D, B, C, A

A, B, C, D

Solution

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