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

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

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

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

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

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

$\left (\dfrac {dy}{dx}\right )_{x = 2} + \left (\dfrac {dy}{dx}\right )_{x = 3}$ .

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

Explanation

$y=\tan ^{ -1 }{ \left( \cfrac { \log { \cfrac { e }{ { x }^{ 2 } } } }{ \log { e{ x }^{ 2 } } } \right) } +\tan ^{ -1 }{ \left( \cfrac { 3+2\log { x } }{ 1-6\log { x } } \right) } \\ \quad =\tan ^{ -1 }{ \left( \cfrac { \log { e } -\log { { x }^{ 2 } } }{ \log { e } +\log { { x }^{ 2 } } } \right) } +\tan ^{ -1 }{ \left[ \cfrac { 3+2\log { x } }{ 1-3-2\log { x } } \right] } \\ \quad =\tan ^{ -1 }{ \left[ \cfrac { 1-\log { { x }^{ 2 } } }{ 1+\log { { x }^{ 2 } } } \right] + } \tan ^{ -1 }{ 3 } +\tan ^{ -1 }{ 2\log { x } } \\ \quad =\tan ^{ -1 }{ 1 } -\tan ^{ -1 }{ \log { { \left( x \right) }^{ 2 } } +\tan ^{ -1 }{ 3 } } +\tan ^{ -1 }{ \log { { \left( x \right) }^{ 2 } } } \\ \quad =\tan ^{ -1 }{ 1+\tan ^{ -1 }{ 3 } } \\ y=\tan ^{ -1 }{ 1 } +\tan ^{ -1 }{ 3 }$ is constant

So, $\cfrac { dy }{ dx } =0$

Answer C

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

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

Explanation

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

$\dfrac{d{y}}{d{x}}=\dfrac{2\text{tan}^{-1}x}{1+x^{2}}$

$(1+x^{2})\dfrac{d{y}}{d{x}}=2\text{tan}^{-1}x$

$(1+x^{2})\dfrac{d^{2}{y}}{d{x^{2}}}+2{x}\dfrac{d{y}}{d{x}}=2\dfrac{1}{1+x^{2}}$

$(1+x^{2})^{2}\dfrac{d^{2}{y}}{d{x^{2}}}+2{x}(1+x^{2})\dfrac{d{y}}{d{x}}=2\implies k=2$

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

0%
differentiable at $x=1$
0%
differentiable at $x=2$
0%
differentiable at $x=1$ and $x=2$
0%
not differentiable at $x=10$

Explanation

Differentiability check at 1

$\lim _{ h\rightarrow 0 }{ \cfrac { f(1+h)-f(1) }{ h } } \\ \Rightarrow \cfrac { 2-(1+h)-(2-1) }{ h } \\ \Rightarrow \cfrac { -h }{ h } =-1\\ \lim _{ h\rightarrow 0 }{ \cfrac { f(1)-f(1-h) }{ h } } \\ \Rightarrow \cfrac { 1-(1-h) }{ h } \\ \Rightarrow \cfrac { h }{ h } =1\\$

Not differentiable at 1.

Differentiability check at 2

$\lim _{ h\rightarrow 0 }{ \cfrac { f(2+h)-f(2) }{ h } } \\ \Rightarrow \cfrac { (-2+3(2+h)-{ (2+h) }^{ 2 }) }{ h } -0\\ \Rightarrow \cfrac { -2+6+3h-4-{ h }^{ 2 }-4h }{ h } \\ \Rightarrow \cfrac { -{ h }^{ 2 }-h }{ h } \\ \Rightarrow \cfrac { -h(h+1) }{ h } =-1\\ \lim _{ h\rightarrow 0 }{ \cfrac { f(2)-f(2-h) }{ h } } \\ \Rightarrow \cfrac { 0-(2-(2-h)) }{ h } \\ \Rightarrow \cfrac { -2+2-h }{ h } \Rightarrow -1$

Differentiable at 2.

Which of the following is not differentiable in the interval

$(-1,2)$ ?

0%
$\displaystyle\int\limits_{x}^{2x}(\log x)^2 dx$
0%
$\displaystyle\int\limits_{x}^{2x}\large{\frac{\sin x}{x}} dx$
0%
$\displaystyle\int\limits_{x}^{2x}\large{\frac{1-t+t^2}{1+t+t^2}} dt$
0%
none of these

Explanation

For

$x\in (-1,2)$ ,

$\displaystyle\int\limits_{x}^{2x}f(x) dx$ is differentiable in

$(-1,2)$ if

$f(x)$ is continuous in

$(-1,2)$ .

For the given functions

$f(x) =\log x$ is not defined for

$x\leq 0$ so it is not continuous in

$(-1,2)$ .

So, option

$(A)$ is not differentiable in

$(-1,2)$ .

But remaining two functions are continuous in

$(-1,2)$ hence option

$(C)$ and

$(D)$ are differentiable in

$(-1,2)$ .

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

$f(1) \ne 1$

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

Explanation

$f'(x)=2-\dfrac{f(x)}{x}\implies x f'(x)+f(x)=2{x}$

$d(x f(x))=2{x} d{x}$

Integrating on both sides

$x f(x)=x^{2}+c\implies f(x)=x+\dfrac{c}{x}$

$f'(x)=1-\dfrac{c}{x^2}$

$\displaystyle \lim _{x\to 0} f'\bigg(\dfrac{1}{x}\bigg)=\lim _{x\to 0} (1-c x^2)=1$

$\displaystyle\lim _{x\to 0} x f\bigg(\dfrac{1}{x}\bigg)=\lim _{x\to 0} (1+x^2)=1$

$\displaystyle\lim _{x\to 0} x^2 f'(x)=\lim_{x\to 0} (x^{2}-c)=-c$

$AM\ge GM$

$\implies |f(x)|\ge 2\sqrt{c}$

$x=a{ t }^{ 2 },y=2at$ , then

$\cfrac { dy }{ dx } =$ _____;

$t\neq 0$

0%
$at$
0%
$\cfrac { t }{ 2 }$
0%
$\cfrac { 2 }{ t }$
0%
$\cfrac { 1 }{ t }$

Explanation

$x=at^{2}$

Differentiating both sides with respect to t

$\dfrac{dx}{dt}=\dfrac{d}{dt}(at^{2})$

$\dfrac{dx}{dt}=a\dfrac{d}{dt}(t^{2})$

$\dfrac{dx}{dt}=2at$ (1)

$y=2at$

Differentiating both sides with respect to t

$\dfrac{dy}{dt}=\dfrac{d}{dt}(2at)$

$\dfrac{dy}{dt}=2a\dfrac{d}{dt}(t)$

$\dfrac{dy}{dt}=2a$ (2)

Now

$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

From (1) and (2)

$\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}$

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?

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

Explanation

${f}'(x)=\sin\left ( \cfrac{1}{x-1} \right )+(x-1)\cos\left ( \cfrac{1}{x-1} \right )\left ( \cfrac{-1}{(x-1)^{2}} \right )$

${f}'(0)=\sin(-1)+(-1)\cos(-1)(-1)$

$=\cos1-\sin1$

$x\rightarrow 1 ,{f}'(1)\rightarrow \infty$

$\left ( \because \cfrac{1}{x-1}\rightarrow \infty \right )$

$\therefore f(x)$ is differentiable at

$x=0$ , but not at

$x=1$

If the function

$f(x)=\dfrac{e^{x^{2}}-\cos x}{x^{2}}$ for

$x \neq 0$ continuous at

$x=0$ then

$f(0)=$

0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{2}$
0%
$2$
0%
$\dfrac{1}{3}$

Explanation

Here , function is continuous ,

So, left limit and right limit boh are equal and equals to the value of

$f(x)$ at

$x=0$

Therefore

$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{e^{x^2}-cosx}{x^2}$

Solving limit by derivative approach,

=>

$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{e^{x^2}.2x+sinx}{2x}$

=>

$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{2e^{x^2}+e^{x^2}.x^2+cosx}{2x}$

Putting

$x=0$ , we get

=>

$f(0)=\dfrac{2+0+1}{2}=\dfrac{3}{2}$

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

0%
0
0%
1
0%
2
0%
-1

Explanation

Given

$f(x)=\dfrac{1}{x}-\dfrac{2}{e^{2x}-1}$

For

$f(x)$ to be continuous at

$x=0$ , its limit should exists at

$x=0$ and must be finite.

Hence

$f(0)=\lim\limits_{x\to0}\dfrac{e^{2x}-1-2x}{x(e^{2x}-1)}$

$=\lim\limits_{x\to0}\dfrac{1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}.....-1-2x}{x+2x^2+\frac{4x^3}{2!}.....-x}=\dfrac{\frac{4x^2}{2!}}{2x^2}=1$ ......[neglecting higher power of

$x$ ]

The function

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

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

Explanation

We know that the cosine function, is nothing more than the sin

$\dfrac{\pi}{2}$ radians out of phase, as proved below,

$\cos(\theta-\dfrac{\pi}{2})=\cos(\theta)\cos(-\dfrac{\pi}{2})-\sin(\theta)\sin(-\dfrac{\pi}{2})$

$\cos(\theta-\dfrac{\pi}{2})=\cos(\theta).0-(-\sin(\theta)\sin(\dfrac{\pi}{2}))$

$\cos(\theta-\dfrac{\pi}{2})=\cos(\theta).1=\sin(\theta)$

So, it

$f(x)=sin^{-1}(cosx)$ is continuous at

$x=0$

So we can say that the since function, 90 degree ahead, is the cousin function.

$arc\sin(\cos(x))=arcsin(sin(x+\dfrac{\pi}{2}))$

Using the property of inverse that

$f^{-1}(f(x))=x$ , we have

$arc\sin(\cos(x))=x+\dfrac{\pi}{2}$

$y=x^x$ then

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

0%
True
0%
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

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

Explanation

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

$x=1$

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

$x=1$ )

$x=2$

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

$x=2$

$y = {\left( {\sin \,x} \right)^x}$ , then

$\dfrac{{dy}}{{dx}} =$

0%
$(\sin x)^x(\ln (\sin x)+x\cot x)$
0%
$(\ln (\sin x)+x\cot x)$
0%
$(\sin x)^x(\ln (\sin x)+x\tan x)$
0%
$(\sin x)^x(\ln (\sin x)-\cot x)$

Explanation

given

$y=(\sin x)^x$

applying

$\ln$ on both sides we get

$\ln y=x\ln(\sin x)$

differentiating both sides wrt

$x$

$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{d}{dx}\left(x\ln(\sin x)\right)$

$\dfrac{1}{y}\dfrac{dy}{dx}=\ln(\sin x)+x\dfrac{1}{\sin x}\cos x$

$\dfrac{dy}{dx}=y\left(\ln(\sin x)+x\cot x\right)$

$\therefore \dfrac{dy}{dx}=(\sin x)^x\left(x\cot x\right) + (\sin x)^x \ln(\sin x)$

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

${a^{a+1}} \bigg [ 2 \ log \ a + {\frac {1-a ^2}{a}} \bigg]$ is

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

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

0%
$\dfrac{-1}{x + 2}$
0%
$\dfrac{-x + 2}{4}$
0%
$\dfrac{-1}{x^2 + 2}$
0%
$\dfrac{x^2 - 2}{4}$

Explanation

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,

$f(x)=-\dfrac{1}{x+2}$ .

Differentiate with respect to

$x$ .

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

0%
$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$
0%
$x^{\cos x}\left[\dfrac {\cos x}{x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec^2x.\log \sin x]$
0%
$x^{\cos x}\left[ {\cos x}-\sin x.\log x\right]+\sin x^{\tan x}[1+\sec x.\log \sin x]$
0%
None

Explanation

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

$\dfrac{1}{t}\dfrac{dt}{dx}=\sec ^2x\log {(\sin x)}+\dfrac{\tan x}{\sin x}(\cos x)$

$\therefore \dfrac{dt}{dx}=\sin x^{\tan x}\left(1+\sec ^2x \log (\sin x)\right)$

$y=h+t$

$\implies \dfrac{dy}{dx}=\dfrac{dh}{dx}+\dfrac{dt}{dx}$

$\therefore \dfrac{dy}{dx}=x^{\cos x}\left(\dfrac{\cos x}{x}-\sin x\log x\right)+\sin x^{\tan x}\left(1+\sec ^2x\log (\sin x)\right)$

$x = a{\text{ (}}\cos \theta + \theta {\text{ }}\sin \theta ),{\text{ }}y = a{\text{ (}}\sin \theta - \theta {\text{ }}\cos \theta ),$ then

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

0%
True
0%
False

Explanation

Here,

$x=a(\cos \theta+\theta \sin \theta)$

$\dfrac{dx}{d\theta}=a(-\sin \theta+\sin \theta+\theta \cos \theta)=a(\theta \cos \theta)$

$\dfrac{d^2x}{d\theta ^2}=a(\cos \theta-\theta \sin \theta)$

$y=a(\sin \theta-\theta \cos \theta)$

$\dfrac{dy}{d\theta}=a(\cos \theta-\cos \theta+\theta \sin \theta)$

$\dfrac{d^2x}{d\theta ^2}=a(\sin \theta +\theta\cos \theta)$

Hence above statement is true.

Differentiate

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

$x$

0%
$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$
0%
$x^{\tan x}(\sec^2x \log \sec x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$
0%
$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})-\tan x^x(\log \tan x+\dfrac{2}{\sin 2x})$
0%
$x^{\tan x}(\sec^2x \log x+\dfrac{\tan x}{x})+\tan x^x(\log \tan x-\dfrac{2}{\sin 2x})$

$f:R \to R$ 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

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

Explanation

$f(x+2y)=f(x)+f(2y)+4xy$

partially difference w.r.t

$x$

$f'(x+2y)=f'(x)+4y$

$x=0$ &

$y=1/2$

$f'(1)=f'(0)+2$

$C$ is correct

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

0%
at most one point
0%
$2$
0%
exactly one point
0%
infinite

$f(x)$ 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

0%
$4$
0%
$5$
0%
$6$
0%
$7$

$f(x)=\dfrac { \left[ x \right] +1 }{ \left\{ x \right\} +1 }$ 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?

0%
$f(x)$ is injective discontinuous function
0%
$f(x)$ is surjective non-differntiable function
0%
$\min { \left( \lim _{ x\rightarrow { 1 }^{ - } }{ f(x) } ,\lim _{ x\rightarrow { 1 }^{ + } }{ f(x) } \right) }$
0%
$\max { \left( x\ values\ of\ point\ of\ discontinuity \right) =f(1) }$

Explanation

$f \left( x \right) = \cfrac{\left[ x \right] + 1}{\left\{ x \right\} + 1}$ for

$f : \left[ 0, \cfrac{5}{2} \right) \rightarrow \left( \cfrac{1}{2}, 3 \right]$

$f \left( 1 \right) = \cfrac{\left[ 1 \right] + 1}{\left\{ 1 \right\} + 1} = \cfrac{1+1}{0+1} = 2$

$f \left( x \right) \begin{cases} \cfrac{1}{x+1}, 0 \le x < 1 \\ \cfrac{2}{x}, 1 \le x < 2 \\ \cfrac{3}{x-1}, 2 \le x < \cfrac{5}{2} \end{cases}$

Clearly,

$f \left( x \right)$ is discontinuous and bijective function.

Now,

$\lim_{x \rightarrow {1}^{-}}{f \left( x \right)} = \lim_{x \rightarrow {1}^{-}}{\left( \cfrac{1}{x+1} \right)} = \cfrac{1}{2}$

$\lim_{x \rightarrow {1}^{+}}{f \left( x \right)} = \lim_{x \rightarrow {1}^{+}}{\left( \cfrac{2}{x} \right)} = 2$

Therefore,

$min \left( \lim_{x \rightarrow {1}^{-}}{f \left( x \right)}, \lim_{x \rightarrow {1}^{+}}{f \left( x \right)} \right) = \lim_{x \rightarrow {1}^{-}}{f \left( x \right)} = \cfrac{1}{2} \ne f \left( 1 \right)$

$max \left( 1, 2 \right) = 2 = f \left( 1 \right)$

Hence the required answer is (D)

$max \left( \text{x values of point of discontuuity} \right) = f \left( 1 \right)$

1043258_1063534_ans_568c547c41274433a1e3fe452bf0eda4.jpg

$y=(x^{x})^{x}$ then

$\dfrac {dy}{dx}=$

0%
$(x^{x})^{x}(1+2\log x)$
0%
$(x^{x})^{x}(1-2\log x)$
0%
$x(x^{x})^{x}(1+2\log x)$
0%
$x(x^{x})^{x}(1-2\log x)$

Explanation

We have,

$y={{\left( {{x}^{x}} \right)}^{x}}$

$y={{x}^{{{x}^{2}}}}$

On taking $\log$ both sides, we get

$\log y={{x}^{2}}\log x$ …… (1)

On differentiating w.r.t $x$ , we get

$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{x}+\log x\left( 2x \right)$

$\dfrac{1}{y}\dfrac{dy}{dx}=x+2x\log x$

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

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

$\dfrac{dy}{dx}=x{{\left( {{x}^{x}} \right)}^{x}}\left( 1+2\log x \right)$

Hence, this is the answer.

Let

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

$f''(0)=$

0%
$\dfrac{2a^3}{b^2}$
0%
$\dfrac{2a^2}{b^3}$
0%
$\dfrac{2a^3}{b^3}$
0%
none of these

Explanation

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,

$f''(x)=\dfrac{2a^2}{(ax+b)^3}$

$f''(0)=\dfrac{2a^2}{b^3}$ .

$x=a(\cos\theta+log\ \tan\dfrac{\theta}{2})$ and

$y=a\sin\theta$ , then

$\dfrac{dy}{dx}$ is equal to

0%
$\cot\theta$
0%
$\tan\theta$
0%
$\sin\theta$
0%
$\cos\theta$

Explanation

$x=a(\cos\theta+\log\tan\cfrac{\theta}{2})$

$\cfrac{dx}{d\theta}$

$=a(-\sin\theta+\cfrac{\sec^2 \cfrac{\theta}{2}}{2\tan\cfrac{\theta}{2}})$

$=a(-\sin\theta+\cfrac{1}{\sin\theta})$

$y=a(sin\theta)$

$\cfrac{dy}{d\theta}$

$=a(cos\theta)$

$\cfrac{dy}{dx}=\cfrac{\cfrac{dy}{d\theta}}{\cfrac{dx}{d\theta}}$

$=\cfrac{a\sin\theta}{-a(\sin\theta-\cfrac{1}{\sin\theta})}$

$=\cfrac{\sin\theta}{\cos\theta}$

$=\tan\theta$

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

$f'(a)=$ ?

0%
$log a-1$
0%
$log a-a$
0%
$a log a-a$
0%
$a log a+a$

Explanation

According to question,

$f' (x)= \dfrac{ a^{x} loga x^{a} - x^{a+1} a^{x} }{ x^{2a} }$

$\Rightarrow f' (a)= \dfrac{ a^{a} loga. a^{a} - a^{a+1} a^{a} }{ a^{2a} }$

Hence,

$\Rightarrow f' (a)=loga-a$

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

$t=\sin ^{ -1 }{ \left( \cfrac { y }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } \right) }$ then

$\cfrac { dy }{ dx }$ is equal to

0%
$\cfrac { x-y }{ x+y }$
0%
$\cfrac { x+y }{ x-y }$
0%
$\cfrac { y-x }{ y+x }$
0%
$\cfrac { x-y }{ 2x+y }$

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

$\ x$

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

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

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

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

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

$y=\tan^{-1}x+\cot^{-1}x+\sec^{-1}x+\csc^{-1}x$ ,then

$\dfrac {dy}{dx}$ is equal to

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

Explanation

Consider the given equation.

$y={{\tan }^{-1}}x+{{\cot }^{-1}}x+{{\sec }^{-1}}x+{{\csc }^{-1}}x$

On differentiating w.r.t $x$ , we get

$\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}-\dfrac{1}{1+{{x}^{2}}}+\dfrac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}-\dfrac{1}{\left| x \right|\sqrt{{{x}^{2}}-1}}$

$\dfrac{dy}{dx}=0$

Hence, this is the answer.

Solve this:-

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

0%
$\operatorname{csch} x$
0%
$\operatorname{sech} x$
0%
$\sinh x$
0%
$\cosh x$

$f(x)=\sin^4x+\cos^4x$ . Then

$f$ is an increasing function in the interval

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

Explanation

$f{\left( x \right)} = \sin^{4}{x} + \cos^{4}{x}$

Diferentiating

$f{\left( x \right)}$ w.r.t.

$x$ , we have

$f'{\left( x \right)} = 4 \sin^{3}{x} \cos{x} + 4 \cos^{3}{x} \left( -\sin{x} \right)$

$f'{\left( x \right)} = 4 \sin^{3}{x} \cos{x} - 4 \cos^{3}{x} \sin{x}$

$f'{\left( x \right)} = 4 \sin{x} \cos{x} \left( \sin^{2}{x} - \cos^{2}{x} \right)$

$f'{\left( x \right)} = 2 \sin{2x} \left( - \cos{2x} \right)$

$f'{\left( x \right)} = - \sin{4x}$

For

$f{\left( x \right)}$ to be an increasing function,

$f'{\left( x \right)} > 0$

$\Rightarrow - \sin{4x} > 0$

$\Rightarrow \sin{4x} < 0$

$\Rightarrow \pi < 4x < 2 \pi$

$\Rightarrow \cfrac{\pi}{4} < x < \cfrac{\pi}{2}$

Hence

$f{\left( x \right)}$ will be increasing for all

$x \in \left[ \cfrac{\pi}{4}, \cfrac{\pi}{2} \right]$ .

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

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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