Continuity and Differentiability Test

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Continuity and Differentiability Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f$$ differentiable at $$x=0$$ and $$f'\left( 0 \right) = 1$$. Then $$\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( h \right) - f\left( { - 2h} \right)}}{h} = $$=

A
3

B
2

C
1

D
6

Solution

$$\begin{matrix} { \lim }_{ h\to 0 }\dfrac { { F\left( h \right) -f\left( { -2h } \right) } }{ h } \\ \Rightarrow { \lim }_{ h\to 0 }\dfrac { { f'\left( h \right) +2f'\left( { -2h } \right) } }{ 1 } \, \, \left[ { \because f\left( 0 \right) =1 } \right] \\ \to f'\left( 0 \right) +2f'\left( 0 \right) \\ =1+2+3=6 \\ \end{matrix}$$
Question 2
1 / -0

Let $$f(x)$$ be differentiable function such that $$f\left(\dfrac{x+y}{1-xy}\right)=f(x)+f(y) \forall x$$ and $$y$$. If $$\underset { x\rightarrow 0 }{ lt } \dfrac { f\left( x \right) }{ x } =\dfrac { 1 }{ 3 }$$ then $$f(1)$$ equals

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{6}$$

C
$$\dfrac{1}{12}$$

D
$$\dfrac{1}{8}$$

Solution

We have,
$$f\left( {\dfrac{{x + y}}{{1 - xy}}} \right) = f\left( x \right) + f\left( y \right)\,\,\,\forall {x^2}y$$
$$ \Rightarrow f\left( x \right) = k{\tan ^{ - 1}}\left( x \right)$$
$$\mathop {\lim }\limits_{x \to 0} \dfrac{{k{{\tan }^{ - 1}}x}}{x} = \dfrac{1}{3}$$
$$k = \dfrac{1}{3}$$
$$f\left( x \right) = \dfrac{1}{3}{\tan ^{ - 1}}x$$
$$f\left( 1 \right) = \dfrac{1 }{{12}}$$

Hence, this is the answer.
Question 3
1 / -0

If $$f (x)$$ is differentiable everywhere, then:

A
$$| f |$$ is differentiable everywhere

B
$$| { f } | ^ { 2 }$$ is differentiable everywhere

C
$$ f | f |$$ is not differentiable at some point

D
$$ f + | f |$$ is differentiable everywhere

Solution

Given: $$f(x)$$ is differentiable everywhere

$$|f|$$ is differentiable everywhere

Let $$f(x)=x$$

$$\Rightarrow |f(x)|=|x|$$

$$\therefore |f|$$ is not differentiable everywhere

$$|f|^2$$ is differentiable everywhere

Let $$f(x)=x$$

$$\Rightarrow |f(x)|^2=x^2$$

$$f(x)=e^x$$

$$|f(x)|^2=e^{2x}$$

Differentiable everywhere

$$f|f|$$ is not differentiable at same points

$$f(x)=e^x\Rightarrow |f|=e^x$$

$$f|f|=e^{2x}\rightarrow$$ differentiable everywhere

$$\Rightarrow$$ This option is wrong

$$f+|f|$$ is differentiable everywhere

$$f(x)=x$$, $$|f|=|x|$$

$$f+|f|=\left\{\begin{matrix} 0, x < 0\\ 2x, x\geq 0\end{matrix}\right\}$$ not
differentiable at $$x=0$$

Options (ii), (iii), (iv) are wrong.

$$\Rightarrow$$ (ii) is correct.
Question 4
1 / -0

If y = sin $$^{-1} (x. \sqrt{1-x}+ \sqrt{x}\sqrt{1-x^2})$$, then $$\dfrac{dy}{dx} =$$

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

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

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

D
None of these

Solution

$$y=\sin^{-1}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2})\\ \quad=\sin^{-1}x+\sin^{-1}\sqrt x\\ \cfrac{dy}{dx}=\cfrac{1}{\sqrt{1-x^2}}+\cfrac{1}{\sqrt{1-x}}\times\cfrac{1}{2\sqrt{x}}\\ \quad=\cfrac{1}{\sqrt{1-x^2}}+\cfrac{1}{2\sqrt x\sqrt{1-x}}$$
Question 5
1 / -0

If $$f(x)=x^{n}\sin \dfrac {1}{x},f(0)=0$$

A
$$f$$ is continuous differentiable at $$x=0$$ for any $$n< 2, n\in N$$

B
$$f$$ is continuous differentiable for all $$x$$ for any $$n\ge 2, n\in N$$

C
$$f$$ is discountinuous at $$x=0$$ if $$n=0$$

D
$$f'(x)$$ is discountinuous at $$x=0$$ if $$n=0$$

Solution
Question 6
1 / -0

If $$x=\dfrac{e^t+e^{-t}}{2}, y=\dfrac{e^t-e^{-t}}{2}$$, then $$\dfrac{dx}{dy}=$$

A
$$-\dfrac{x}{y}$$

B
$$\dfrac{y}{x}$$

C
$$\dfrac{x}{y}$$

D
$$-\dfrac{y}{x}$$

Solution

$$x=\dfrac{e^t+e^{-t}}{2},$$ $$y=\dfrac{e^t-e^{-t}}{2}$$

$$\Rightarrow \dfrac{dx}{dt}=\dfrac{e^t-e^{-t}}{2}=y,\dfrac{dy}{dt}=\dfrac{e^t+e^{-t}}{2}=x$$

$$\Rightarrow \dfrac{dx}{dy}=\dfrac{dx}{dt}\times \dfrac{dt}{dy}$$
$$=y\times \dfrac{1}{x}$$
$$=\dfrac{y}{x}$$
Hence, the answer is $$\dfrac{y}{x}.$$
Question 7
1 / -0

If $$y=a^{{a}^{{x}}}$$, then $$\dfrac {dy}{dx}=$$

A
$$y.a^{x}{(\log a)}^{2}$$

B
$$y.a^{x}.\log a$$

C
$${(y.a^{x})}^{2}$$

D
$$(y.a^{x})$$

Solution

$$\quad y={ a }^{ { x }^{ y } }$$
taking logarithm on both sides
$$\log { y } ={ x }^{ y }\log { a } $$
again we take logarithm on both sides
$$\log { \log { y } } =y\left( \log { x } \right) +\log { \log { a } } $$
taking derivative on both sides w.r. t $$x$$
$$\cfrac { 1 }{ \log { y } } .\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =\cfrac { dy }{ dx } \log { x } +y.\cfrac { 1 }{ x } +0$$
$$\Rightarrow \cfrac { dy }{ dx } \left( \cfrac { 1 }{ y\log { y } } -\log { x } \right) =\cfrac { y }{ x } \Rightarrow \cfrac { dy }{ dx } \left( \cfrac { 1-y\log { x } .\log { y } }{ y\log { y } } \right) =\cfrac { y }{ x } $$
$$\Rightarrow \cfrac { dy }{ dx } =\cfrac { { y }^{ 2 }\log { y } }{ x\left( 1-y\log { x } .\log { y } \right) } $$
Question 8
1 / -0

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

A
6

B
12

C
4

D
3

Solution

Putting y = 0, f(x) = 2f(x)f(0)
$$ \Rightarrow f(0) = \frac{1}{2} $$
$$ f(x) = \lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h} $$
$$ =\underset{h\rightarrow 0} {\lim}\dfrac{2f(x)f(h)-f(x)}{h} $$
$$ =\underset {h\rightarrow 0}{\lim} f(x)[\dfrac{2f(h)-1}{h}] $$
$$ f^{1}(x) = \underset{h\rightarrow 0} {\lim}f(x)[\dfrac{2f(h)-2f(10)}{h}] $$
$$ 2f(x)\underset{h\rightarrow 0} {\lim}\frac{f(0+h)-f10}{h} $$
$$ = 2f(x)f^{1}(0) = 6f(x) $$
let $$ f(x) = y \Rightarrow y^{1} = 6y $$
$$ \dfrac{dy}{y} = 6x\Rightarrow lny = \frac{6x^{2}}{2}+c $$
At $$ x = 4,y = 2 \Rightarrow ln2 = 48+c $$
$$ \Rightarrow C = ln2-48 $$
$$ \Rightarrow lny = 3x^{2}+ln2-48 $$
$$ \Rightarrow at x = 4,lny = ln2 \Rightarrow y = 2 $$
$$ \Rightarrow y^{1} = 6y = 12 at x = 4 \Rightarrow (B) $$
Question 9
1 / -0

If $$(2+\sin x)\dfrac{dy}{dx}+(y+1)\cos x=0$$ and $$y(0)=1$$, then $$y\left(\dfrac{\pi}{2}\right)$$ is equal to?

A
$$\dfrac{1}{3}$$

B
$$-\dfrac{2}{3}$$

C
$$-\dfrac{1}{3}$$

D
$$\dfrac{4}{3}$$

Solution
Question 10
1 / -0

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

A
$$\dfrac{\log (\sin y)+y \tan x}{\log (\cos x)- x \cot y}$$

B
$$\dfrac{\log (\sin y)-y \tan x}{\log (\cos x)+ x \cot y}$$

C
$$\dfrac{\log (\sin y)}{\log (\cos x)}$$

D
$$\dfrac{\log (\cos x)}{\log (\sin y)}$$

Solution

Given,
$$(\cos x)^y=(\sin y)^x$$
Now taking logarithm with respect to the base $$e$$ we get,
$$y\log (\cos x)=x\log(\sin y)$$
Now differentiating both sides with respect to $$x$$ we get,
$$\log(\cos x)\dfrac{dy}{dx}+y.\dfrac{(-\sin x)}{\cos x}$$$$=\log(\sin y)+x.\dfrac{(\cos y)}{\sin y}\dfrac{dy}{dx}$$
or, $$(\log(\cos x)-x\cot y)\dfrac{dy}{dx}=\log (\sin y)+y\tan x$$
or, $$\dfrac{dy}{dx}=\dfrac{\log (\sin y)+y\tan x}{\log(\cos x)-y\cot x}$$

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 28

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Continuity and Differentiability Test - 28

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SHARING IS CARING

Question 1

3

2

1

6

Solution

Question 2

$$\dfrac{1}{4}$$

$$\dfrac{1}{6}$$

$$\dfrac{1}{12}$$

$$\dfrac{1}{8}$$

Solution

Question 3

$$| f |$$ is differentiable everywhere

$$| { f } | ^ { 2 }$$ is differentiable everywhere

$$ f | f |$$ is not differentiable at some point

$$ f + | f |$$ is differentiable everywhere

Solution

Question 4

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

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

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

None of these

Solution

Question 5

$$f$$ is continuous differentiable at $$x=0$$ for any $$n< 2, n\in N$$

$$f$$ is continuous differentiable for all $$x$$ for any $$n\ge 2, n\in N$$

$$f$$ is discountinuous at $$x=0$$ if $$n=0$$

$$f'(x)$$ is discountinuous at $$x=0$$ if $$n=0$$

Solution

Question 6

$$-\dfrac{x}{y}$$

$$\dfrac{y}{x}$$

$$\dfrac{x}{y}$$

$$-\dfrac{y}{x}$$

Solution

Question 7

$$y.a^{x}{(\log a)}^{2}$$

$$y.a^{x}.\log a$$

$${(y.a^{x})}^{2}$$

$$(y.a^{x})$$

Solution

Question 8

6

12

4

3

Solution

Question 9

$$\dfrac{1}{3}$$

$$-\dfrac{2}{3}$$

$$-\dfrac{1}{3}$$

$$\dfrac{4}{3}$$

Solution

Question 10

$$\dfrac{\log (\sin y)+y \tan x}{\log (\cos x)- x \cot y}$$

$$\dfrac{\log (\sin y)-y \tan x}{\log (\cos x)+ x \cot y}$$

$$\dfrac{\log (\sin y)}{\log (\cos x)}$$

$$\dfrac{\log (\cos x)}{\log (\sin y)}$$

Solution

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