MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 6

If $$y = Tan^{-1} \left (\dfrac {\log (e/x^{2})}{\log ex^{2}}\right ) + Tan^{-1} \left (\dfrac {3 + 2\log x}{1 - 6\log x}\right )$$ then $$\left (\dfrac {dy}{dx}\right )_{x = 2} + \left (\dfrac {dy}{dx}\right )_{x = 3}$$.

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$$-6$$
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$$2$$
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$$0$$
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$$-2$$

If $$y={ \left( \tan ^{ -1 }{ x } \right) }^{ 2 }$$ and $${ \left( { x }^{ 2 }+1 \right) }^{ 2 }\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2x\left( { x }^{ 2 }+1 \right) \dfrac { dy }{ dx } =k$$, then the value of $$k$$ is

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$$3$$
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$$2$$
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$$1$$
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$$0$$

Let $$f(x) = \begin{cases} x & x<1 \\ 2-x & 1 \leq x \leq 2 \\ -2+3x-x^2 & x>2 \end{cases}$$
then $$f(x)$$ is

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differentiable at $$x=1$$
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differentiable at $$x=2$$
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differentiable at $$x=1$$ and $$x=2$$
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not differentiable at $$x=10$$

Which of the following is not differentiable in the interval $$(-1,2)$$?

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$$\displaystyle\int\limits_{x}^{2x}(\log x)^2 dx$$
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$$\displaystyle\int\limits_{x}^{2x}\large{\frac{\sin x}{x}} dx$$
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$$\displaystyle\int\limits_{x}^{2x}\large{\frac{1-t+t^2}{1+t+t^2}} dt$$
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none of these

Let $$f:(0,\infty ) \to R$$ be a differentiable function such that $$f'(x) = 2 - {{f(x)} \over x}$$ for all $$x \in (0,\infty )$$ and $$f(1) \ne 1$$

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$$\mathop {\lim }\limits_{x \to 0} f'\left( {{1 \over x}} \right) = 1$$
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$$\mathop {\lim }\limits_{x \to 0} xf\left( {{1 \over x}} \right) = 2$$
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$$\mathop {\lim }\limits_{x \to 0} {x^2}f'\left( x \right) = 0$$
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$$\left| {f(x)} \right| \le 2{\rm{ \ for\ all\ x}} \in {\rm{(0,2)}}$$

If $$x=a{ t }^{ 2 },y=2at$$, then $$\cfrac { dy }{ dx } =$$_____;$$t\neq 0$$

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$$at$$
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$$\cfrac { t }{ 2 } $$
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$$\cfrac { 2 }{ t } $$
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$$\cfrac { 1 }{ t } $$

Let $$f(x)=\begin{cases}
(x-1) \sin \left( \large{\frac{1}{x-1}}\right) \ &\mbox{if}& \ x \ne 1\\
0, \ &\mbox{if}& \ x \ne 1
\end{cases}$$.
Then which one of the following is true?

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$$f$$ is differentiable neither at $$x=0$$ nor at $$x=1$$
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$$f$$ is differentiable at $$x=0$$ and $$x=1$$
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$$f$$ is differentiable at $$x=0$$ but not at $$x=1$$
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$$f$$ is differentiable at $$x=1$$ but not at $$x=0$$

If the function $$f(x)=\dfrac{e^{x^{2}}-\cos x}{x^{2}}$$ for $$x \neq 0$$ continuous at $$x=0$$ then $$f(0)=$$

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$$\dfrac{1}{2}$$
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$$\dfrac{3}{2}$$
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$$2$$
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$$\dfrac{1}{3}$$

The function $$f : R /{0} \rightarrow R$$ given by $$f(x) =
\dfrac{1}{x} - \dfrac{2}{e^{2x} -1}$$ can be made continuous at $$x=0$$ by
defining $$f(0)$$ as

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0
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1
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2
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-1

The function $$f(x) =
\ sin^{-1} (\ cosx)$$ is

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Discontinuous at $$x=0$$
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continuous at $$x=0$$
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differentiable at $$x=0$$
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None of these

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

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True
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False

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

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Differentiable at $$x=1$$
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Differentiable at $$x=2$$
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Differentiable at $$x=1$$ and $$x=2$$
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Not differentiable at $$x=0$$

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

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

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

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$$ f^\prime (1) $$
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$$ f^\prime (0) $$
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$$ f^\prime (2) $$
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$$ f^{\prime \prime} $$

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

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$$\dfrac{-1}{x + 2}$$
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$$\dfrac{-x + 2}{4}$$
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$$\dfrac{-1}{x^2 + 2}$$
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$$\dfrac{x^2 - 2}{4}$$

Differentiate with respect to $$x$$.

$${x^{\cos x}} + \sin {x^{\tan x}}$$

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$$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$$
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$$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$$
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$$x^{\cos x}\left[ {\cos x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec x.\log \sin x]$$
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None

If $$x = a{\text{ (}}\cos \theta + \theta {\text{ }}\sin \theta ),{\text{ }}y = a{\text{ (}}\sin \theta - \theta {\text{ }}\cos \theta ),$$ then $$\dfrac{{{d^2}x}}{{d{\theta ^2}}} = a{\text{ (}}\cos \theta - \theta\sin \theta ),{\text{ }}\dfrac{{{d^2}y}}{{d{\theta ^2}}} = a{\text{ (}}\sin \theta + \theta \;\cos \theta )$$

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True
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False

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

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

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

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

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

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at most one point
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$$2$$
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exactly one point
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infinite

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

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$$4$$
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$$5$$
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$$6$$
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$$7$$

$$f(x)=\dfrac { \left[ x \right] +1 }{ \left\{ x \right\} +1 } $$ for $$f:\left[ 0,\dfrac { 5 }{ 2 } \right) \rightarrow \left[ \dfrac { 1 }{ 2 } ,3 \right) $$, where $$[.]$$ represent the integer function and $$\left\{ . \right\} $$ represent the fraction part of $$x$$. then which of the following is true?

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$$f(x)$$ is injective discontinuous function
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$$f(x)$$ is surjective non-differntiable function
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$$\min { \left( \lim _{ x\rightarrow { 1 }^{ - } }{ f(x) } ,\lim _{ x\rightarrow { 1 }^{ + } }{ f(x) } \right) } $$
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$$\max { \left( x\ values\ of\ point\ of\ discontinuity \right) =f(1) } $$

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

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$$(x^{x})^{x}(1+2\log x)$$
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$$(x^{x})^{x}(1-2\log x)$$
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$$x(x^{x})^{x}(1+2\log x)$$
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$$x(x^{x})^{x}(1-2\log x)$$

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

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$$\dfrac{2a^3}{b^2}$$
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$$\dfrac{2a^2}{b^3}$$
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$$\dfrac{2a^3}{b^3}$$
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none of these

If $$ x=a(\cos\theta+log\ \tan\dfrac{\theta}{2})$$ and $$y=a\sin\theta$$, then$$\dfrac{dy}{dx}$$ is equal to

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$$\cot\theta$$
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$$\tan\theta$$
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$$\sin\theta$$
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$$\cos\theta$$

If $$f(x)=\dfrac{a^x}{x^a}$$ then $$f'(a)=$$?

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$$log a-1$$
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$$log a-a$$
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$$a log a-a$$
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$$a log a+a$$

If $$\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ e }^{ t }$$ where $$t=\sin ^{ -1 }{ \left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right) } $$ then $$\cfrac { dy }{ dx } $$ is equal to

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$$\cfrac { x-y }{ x+y } $$
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$$\cfrac { x+y }{ x-y } $$
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$$\cfrac { y-x }{ y+x } $$
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$$\cfrac { x-y }{ 2x+y } $$

Explanation

$$\textbf{Hint: }$$ $$ \left( \dfrac { d\left( \sin ^{ -1 }{ x } \right) }{ dx } =\cfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) $$

Correct Option: $$\textbf{B}$$

Solution:

$$\textbf{Step 1:Use quotient rule of derivative}$$

We have given, $$t=\sin ^{ -1 }{ \left( \dfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right) } $$, differentiate on both sides with respect to $$\ x$$

$$\therefore \dfrac { dt }{ dx } =\dfrac { 1 }{ \sqrt { 1-{ \left( \dfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right) }^{ 2 } } } \dfrac { d\left( \dfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right) }{ dx }$$ $$ \left( \because \dfrac { d\left( \sin ^{ -1 }{ x } \right) }{ dx } =\cfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) $$

$$=\left( \dfrac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ x } \right) \left[ \dfrac { \left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right) \dfrac { dy }{ dx } -\left( \dfrac { 2x+2y\dfrac { dy }{ dx } }{ 2\sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right) y }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) } \right] $$

$$\left[ \because \dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}} \right]$$

$$=\left( \dfrac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ x } \right) \left[ \dfrac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \dfrac { dy }{ dx } -\left( xy+{ y }^{ 2 }\dfrac { dy }{ dx } \right) }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right] $$

$$=\dfrac { { x }^{ 2 }\dfrac { dy }{ dx } -xy }{ x\left( { x }^{ 2 }+{ y }^{ 2 } \right) } $$

$$\therefore \dfrac{dt}{dx}=\dfrac{x\dfrac{dy}{dx}-y}{{{x}^{2}}+{{y}^{2}}}$$ .....(1)

$$\textbf{Step 2: Differentiate the given equation with respect to }$$ $$\ x$$

Given that $$\sqrt { { x }^{ 2 }+{ y }^{ 2 } } ={ e }^{ t }$$

$${ x }^{ 2 }+{ y }^{ 2 }={ e }^{ 2t }$$ differentiate both sides with respect to $$\ x$$

$$\Rightarrow2x+2y\dfrac { dy }{ dx } ={ e }^{ 2t }\left( 2 \right) \dfrac { dt }{ dx } $$

$$\Rightarrow x+y\dfrac { dy }{ dx } =\left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( \dfrac { x\dfrac { dy }{ dx } -y }{ { x }^{ 2 }+{ y }^{ 2 } } \right) $$

$$\Rightarrow x+y=(x-y)\dfrac { dy }{ dx } $$

$$\therefore \dfrac { dy }{ dx } =\dfrac { x+y }{ x-y } $$

$$\textbf{ Hence, the correct option is B.}$$

If $$y=\tan^{-1}x+\cot^{-1}x+\sec^{-1}x+\csc^{-1}x$$,then $$\dfrac {dy}{dx}$$ is equal to

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$$-1$$
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$$\pi$$
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$$0$$
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$$1$$

Solve this:-$$\dfrac{d}{{dx}}\left( {\tan^{ - 1}\left( {\sinh \,X} \right)} \right) = $$

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$$\operatorname{csch} x$$
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$$\operatorname{sech} x$$
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$$\sinh x$$
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$$\cosh x$$

If $$f(x)=\sin^4x+\cos^4x$$. Then $$f$$ is an increasing function in the interval

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$$\left [ \dfrac{5\pi}{8},\dfrac{3\pi}{4} \right ]$$
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$$\left [ \dfrac{\pi}{2},\dfrac{5\pi}{8} \right ]$$
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$$\left [ \dfrac{\pi}{4},\dfrac{\pi}{2} \right ]$$
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$$\left [ 0,\dfrac{\pi}{4} \right ]$$

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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