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

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

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

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

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

If $$y = \exp \left\{ {{{\sin }^2}x + {{\sin }^4}x + {{\sin }^6}x + ....} \right\}$$ then $$\frac {dy}{dx}=$$

$$2{e^{{{\tan }^2}x}}\tan x{\sec ^2}x$$

$${e^{{{\tan }^2}x}}{\sec ^2}x$$

If $$y={ \tan }^{ -1 }{ ax}$$ then, which of the following is true

$$\displaystyle \frac { dy }{ dx } =\frac { a}{ 1+{ (ax) }^{ 2 } }$$

$$\displaystyle \frac { dy }{ dx } =\frac { a }{ 1+{ x }^{ 2 } }$$

$$\displaystyle \frac { dy }{ dx } =\frac { a }{ 1-{(ax) }^{ 2 } }$$

$$\displaystyle \frac { dy }{ dx } =\frac { a }{ 2+{ (ax) }^{ 2 } }$$

MCQ Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 8

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

0%
6
0%
12
0%
4
0%
3

Explanation

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

$f(x) = y \Rightarrow y^{1} = 6y$

$\dfrac{dy}{y} = 6x\Rightarrow lny = \frac{6x^{2}}{2}+c$

$x = 4,y = 2 \Rightarrow ln2 = 48+c$

$\Rightarrow C = ln2-48$

$\Rightarrow lny = 3x^{2}+ln2-48$

$\Rightarrow at x = 4,lny = ln2 \Rightarrow y = 2$

$\Rightarrow y^{1} = 6y = 12 at x = 4 \Rightarrow (B)$

$f(x) = \cos^{-1} (\cos x)$ , then at the point, where

$f$ is differentiable ,

$f ' (x)$ equals

0%
$1$
0%
$-1$
0%
$\text{sng} (\sin x)$
0%
$-\text{sng} (\sin x)$

$f\left( x \right) =\left\{ \dfrac { x }{ { e }^{ { 1 }/{ x } }+1 } \right\}$ when

$x\neq 0,$ then 0,when x=0

0%
lim
$x\rightarrow 0+$

$f\left( x \right) =1$
0%
lim
$x\rightarrow 0-$

$f\left( x \right) =1$
0%
$f\left( x \right)$ is continuous at x=0
0%
None of the above

$f(x)=a|\sin x|+ be^{|x|}+c|x|^{3}$ , where

$a,b,c\in R$ , is differentiable at

$x=0$ , then

0%
$a=0,b$ and $c$ are any real numbers
0%
$c=0,a=0 ,b$ is any real number
0%
$b=0,c=0 ,a$ is any real number
0%
$a=0,b=0 ,c$ is any real number

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

$\dfrac{dy}{dx}=$

0%
$\dfrac{\log (\sin y)+y \tan x}{\log (\cos x)- x \cot y}$
0%
$\dfrac{\log (\sin y)-y \tan x}{\log (\cos x)+ x \cot y}$
0%
$\dfrac{\log (\sin y)}{\log (\cos x)}$
0%
$\dfrac{\log (\cos x)}{\log (\sin y)}$

Explanation

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,

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

$y=\tan{x}$ , then

$\dfrac{d^{2}y}{dx^{2}}=$

0%
$x\dfrac{dy}{dx}$
0%
$2x\dfrac{dy}{dx}$
0%
$2y\dfrac{dy}{dx}$
0%
$y\dfrac{dy}{dx}$

Explanation

Given

$y=\tan x$

$\therefore \dfrac{dy}{dx}=\sec^2\;x\;\;\;\;\; .......\left( 1\right)$

$\dfrac{d^2y}{dx^2}=2\sec x.\left( \sec x .\tan x \right)=2\sec^2x \tan x$

$.....\left( 2\right)$

substituting the value of

$\sec^2x$ from

$\left(1\right)$ in

$\left( 2\right)$ we get,

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

Also we have

$y=\tan x,$ substituting we get

$\Rightarrow \dfrac{d^2y}{dx^2}=2y\dfrac{dy}{dx}$

Hence, the answer is

$2y\dfrac{dy}{dx}.$

The function

$f\left( x \right) = \dfrac{{4 - {x^2}}}{{4x - {x^3}}}$ is

0%
discontinuous at only one point
0%
discontinuous exactly at two points
0%
discontinuous exactly at three points
0%
None of these

Explanation

$f\left(x\right)= \dfrac{4-x^2}{4x-x^3}$

$=\dfrac{4-x^2}{x\left(4-x^2\right)}$

$for 4-x^2\neq 0$

$x\neq\pm 2$

$f\left(x\right)=\dfrac{1}{x}$

so this function will be discontinue at

$x=0$

$x=\pm2$

three Points

Let

$f:R\rightarrow R$ be continuous quadratic function such that

$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,$ If

$f(0)=0$ then number of points of non-differentiablity of

$y=\left| f(x)-2 \right|$ is

0%
$1$
0%
$2$
0%
$0$
0%
$None$ $of$ $these$

The function

$f(x)=\dfrac{x^3+3x+5}{x^3-3x+2}$ is :

0%
Continuous on R
0%
Discontinuous at one point on R
0%
Discontinuous at two points on R
0%
Discontinuous at three points on R

If for

$x \in \left(0, \dfrac{1}{4}\right)$ , the derivative

$\tan^{-1} \left(\dfrac{6x\sqrt{x}}{1-9x^{3}}\right)$ is

$\sqrt{x}.g(x)$ , then

$g(x)$ equals :

0%
$\dfrac{3}{1+9x^{3}}$
0%
$\dfrac{9}{1+9x^{3}}$
0%
$\dfrac{3x\sqrt{x}}{1-9x^{3}}$
0%
$\dfrac{3x}{1-9x^{3}}$

Explanation

$y=\tan ^{ -1 }{ \left( \cfrac { 6x\sqrt { x } }{ 1-9{ x }^{ 3 } } \right) }$

and

$x\in \left( 0,\cfrac { 1 }{ 9 } \right)$

$y=\tan ^{ -1 }{ \left( \cfrac { 6x\sqrt { x } }{ 1-9{ x }^{ 3 } } \right) } =y=\tan ^{ -1 }{ \left( \cfrac { 2\left( 3{ x }^{ 3/2 } \right) }{ 1-{ \left( 3{ x }^{ 3 } \right) }^{ 3/2 } } \right) } =2\tan ^{ -1 }{ { \left( 3{ x }^{ 3/2 } \right) }^{ } } \left[ 2\tan ^{ -1 }{ \left( \cfrac { 2x }{ 1-{ x }^{ 2 } } \right) } \right]$

$\cfrac { dy }{ dx } 2\tan ^{ -1 }{ { \left( 3{ x }^{ 3/2 } \right) }^{ } } =2x\cfrac { 1 }{ 1+9{ x }^{ 3 } } \times 3\times \cfrac { 3 }{ 2 } \times { x }^{ 1/2 }=\cfrac { 9 }{ 1+{ x }^{ 3 }\sqrt { 3 } } =\cfrac { 9 }{ 1+9{ x }^{ 3 } }$

$y = \exp \left\{ {{{\sin }^2}x + {{\sin }^4}x + {{\sin }^6}x + ....} \right\}$ then

$\frac {dy}{dx}=$

0%
${e^{{{\tan }^2}x}}$
0%
${e^{{{\tan }^2}x}}{\sec ^2}x$
0%
$2{e^{{{\tan }^2}x}}\tan x{\sec ^2}x$
0%
none

If Rolle's therorem holds for

$f(x)=x\left( { x }^{ 2 }+ax+b \right) +2atx=\frac { 1 }{ 2 }$ in the interval (-1 , 1), then which of the following (S) is/are corect?

0%
a + b = $-\frac { 3 }{ 4 }$
0%
a + b = $-\frac { 5 }{ 4 }$
0%
ab = $\frac { 1 }{ 4 }$
0%
ab = $-\frac { 1 }{ 4 }$

$y = \cos \left( {m{{\cos }^{ - 1}}x} \right)$ , then

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

0%
$my$
0%
$-my$
0%
${m^2}y$
0%
${-m^2}y$

Explanation

$y = \cos (m \, \cos^{-1} x)$

$\dfrac{dy}{dx} = -\sin (m \cos^{-1} x) . mx \dfrac{-1}{\sqrt{1 - x^2}}$

$\dfrac{dy}{dx} = \dfrac{m}{\sqrt{1 - x^2}} \sin (m \cos^{-1} x)$

$\dfrac{d^2y}{dx^2} = m . \left[\dfrac{\sqrt{1 - x^2} cos (m \cos^{-1} x) \times m \dfrac{(-1)}{\sqrt{1 - x^2}} - \sin (m \cos^{-1} x) \times \dfrac{+1(-1x)}{2 \sqrt{1 - x^2}}}{(1 - x^2)} \right]$

$(1 - x^2) \dfrac{d^2y}{dx^2} = m \left[-m \, \cos (m \cos^{-1} x) + x \dfrac{\sin(m \cos^{-1} x)}{\sqrt{1 - x^2}} \right]$

$= -m^2 \cos (m \cos^{-1} x) + m x \dfrac{\sin (m \cos^{-1} x)}{\sqrt{1 - x^2}}$

$= -m^2 y + x \dfrac{dy}{dx}$

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

$\dfrac {dy}{dx}=(e^ {y}-x)^ {-1}$ where

$y(0)=0$ then

$y$ is expressed explicity as

0%
$0.5\log_{e}(1+x^{2})$
0%
$\log_{e}(1+x^{2})$
0%
$\log _{ e } \left( x+\sqrt { 1+{ x }^{ 2 } } \right)$
0%
$\\ \log _{ e } \left( x+\sqrt { 1-{ x }^{ 2 } } \right)$

Let

$f\left( x \right)$ be non-constant differentiable function for all real

$x$ and

$f\left( x \right) =f\left( 1-x \right)$ . Then Rolle's theorem is not applicable for

$f\left( x \right)$ on

0%
$[0,1]$
0%
$[-1,2]$
0%
$[-2,3]$
0%
$\left[ 0,\dfrac { 2 }{ 3 } \right]$

$\dfrac{d}{dx}$ {

$\cot^{-1} \dfrac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}}$ } =

0%
$\dfrac{1}{\sqrt{1-x^2}}$
0%
$\dfrac{-1}{2\sqrt{1-x^2}}$
0%
$\dfrac{1}{1+x^2}$
0%
None of these

$\left[ a,b \right]$ and differentiable in (a,b) then the value of 'c' for the pair of functions

$f\left( x \right) \sqrt { x, } \phi \left( x \right) =\dfrac { 1 }{ \sqrt { x } }$ is

0%
$\sqrt { a }$
0%
$\sqrt { b }$
0%
$\sqrt { ab }$
0%
$-\sqrt { ab }$

Let f be a differentiable function satisfying the condition

$f(\dfrac{x}{y}) = \dfrac{f(x)}{f (y)}$ for all

$x, y ( \neq 0) \epsilon R, f(y) \neq 0$ . If

$f' (1) =2$ , then

$f ' (x)$ is equal to

0%
$2f(x)$
0%
$f(x)/x$
0%
$2xf(x)$
0%
$2f(x)/x$

Let f(x) be a polynomial in x. Then the second derivation of f

$\left( { e }^{ x } \right)$ , is:

0%
$f"\left( { e }^{ x } \right) .{ e }^{ x }+f'\left( { e }^{ x } \right)$
0%
$f"\left( { e }^{ x } \right) .{ e }^{ x }+f'\left( { e }^{ x } \right) .{ e }^{ 2x }$
0%
$f"\left( { e }^{ x } \right) { e }^{ 2x }$
0%
$f"\left( { e }^{ x } \right) { e }^{ 2x }+f'\left( { e }^{ x } \right) .{ e }^{ x }$

Explanation

We have,

$f\left( x \right)$ is a polynomial.

Then,

The second order of $f\left( {{e}^{x}} \right)=?$

Find that,

$\dfrac{d}{dx}\left[ \dfrac{d}{dx}f\left( {{e}^{x}} \right) \right]=?$

So,

We know that,

$\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$

$So,$

$\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=g'\left( x \right)f'\left( g\left( x \right) \right)$

Now. Using formula,

$\dfrac{d}{dx}\left( I.II \right)=I.\dfrac{d}{dx}II+II.\dfrac{d}{dx}I$

So,

$\dfrac{d}{dx}\left[ f\left( {{e}^{x}} \right) \right]={{e}^{x}}.f'\left( {{e}^{x}} \right)\,\,\,......\,\,\left( 1 \right)$

Again differentiating and we get,

$\dfrac{d}{dx}\left[ \dfrac{d}{dx}f\left( {{e}^{x}} \right) \right]=\dfrac{d}{dx}\left[ f'\left( {{e}^{x}} \right).{{e}^{x}} \right]=f'\left( {{e}^{x}} \right){{e}^{x}}+{{e}^{x}}\left[ {{e}^{x}}f''\left( {{e}^{x}} \right) \right]$

$=\dfrac{d}{dx}\left[ \dfrac{d}{dx}f\left( {{e}^{x}} \right) \right]={{e}^{x}}f'\left( {{e}^{x}} \right)+{{e}^{2x}}f''\left( {{e}^{x}} \right)$

Hence, this is the answer.

Let f, g and h are function differentiable on some open interval around

$0$ and satisfy the equations

$f'=2f^2gh+\dfrac{1}{gh}, f(0)=1, g'=fg^2h+\dfrac{4}{fh}, g(0)=1$ and

$h'=3fgh^2+\dfrac{1}{fg}, h(0)=1$ . The function f is given by?

0%
$f(x)=2^{-1/12}\left(\dfrac{\sin\left(6x+\dfrac{\pi}{4}\right)}{\cos^2\left(6x+\dfrac{\pi}{4}\right)}\right)^{1/6}$
0%
$f(x)=2^{-1/6}\left(\dfrac{\sin \left(6x+\dfrac{\pi}{4}\right)}{\cos^2\left(6x+\dfrac{\pi}{4}\right)}\right)^{1/12}$
0%
$f(x)=(\sec 12x)^{1/12}(\sec 12x+\tan 12x)^{1/4}$
0%
$f(x)=(\sec 12x)^{1/4}(\sec 12x+\tan 12x)^{1/12}$

$y=1-\cos\theta,x=1-\sin\theta$ , then

$\dfrac{dy}{dx}$ at

$\theta=\dfrac{\pi}{4}$ is

0%
$-1$
0%
$1$
0%
$12$
0%
$-12$

$x=\sqrt{2^{cosec^{-1}t}}$ and

$y=\sqrt{2^{sec^{-1}t}}(|t|\geq 1)$ then

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

0%
$\dfrac{y}{x}$
0%
$\dfrac{x}{y}$
0%
$\dfrac{-x}{y}$
0%
$\dfrac{-y}{x}$

$f(x)$ is a four times differentiable even function, then

$\int_{-3}^{3}(x^{3}f(x)+xf''''(x)+2)dx$ is equal to

0%
$12f(x)+f''(x))$
0%
$12f''(x)$
0%
$12$
0%
$6$

$\dfrac{d}{dx}\left[\tan h^{-1}\left(\dfrac{2x}{1+x^2}\right)\right]=?$

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

A differential function satisfies equation

$f(x)=\int_{0}^{x}(f(t)\cos\ t-\cos(t-x))dt$ then

0%
${ f }^{ \prime \prime }\left( \dfrac { \pi }{ 2 } \right) =e$
0%
$\lim _{ x\rightarrow -\infty }{ f\left( x \right) =1 }$
0%
$f(x) has minimum value 1-e^{-1}$
0%
${ f }^{ \prime}(0)=-1$

$f ( x )$ and

$g ( x )$ are differentiable functions in

$[ 0,1 ]$ such that

$f ( 0 ) = 2 , f ( 1 ) = 6 , g ( 0 ) = 0 , g ( 1 ) =2$ then there exists

$0 < c < 1$ such that

0%
$\mathrm { f } ^ { \prime } ( \mathrm { c } ) = \mathrm { g } ^ { \prime } ( \mathrm { c } )$
0%
$f ^ { \prime } ( c ) = - g ^ { \prime } ( c )$
0%
$f ^ { \prime } ( c ) = 2 g ^ { \prime } ( c )$
0%
$2 f ^ { \prime } ( c ) = g ^ { \prime } ( c )$

$x=2\cos \theta-2\cos 2\theta$ and

$y=2\sin \theta-\sin 2\theta$ , then

$\dfrac{dy}{dx}=\tan \left( \dfrac{3\theta}{2} \right )$ .

0%
True
0%
False

$y = \sin ^ { - 1 } x$ then

$\frac { d y } { d x }$ is equal to

0%
$\sec y$
0%
$\cos x$
0%
$\tan x$
0%
$1$

Explanation

$y=\sin ^{-1}x\\x=\sin y \\\dfrac{dx}{dx}=\dfrac{d}{dx}\sin y\\1=\cos y \dfrac{dy}{dx}\\\dfrac{dy}{dx}=\sec y$

$y={ \tan }^{ -1 }{ ax}$ then, which of the following is true

0%
$\displaystyle \frac { dy }{ dx } =\frac { a }{ 1+{ x }^{ 2 } }$
0%
$\displaystyle \frac { dy }{ dx } =\frac { a }{ 1-{(ax) }^{ 2 } }$
0%
$\displaystyle \frac { dy }{ dx } =\frac { a }{ 2+{ (ax) }^{ 2 } }$
0%
$\displaystyle \frac { dy }{ dx } =\frac { a}{ 1+{ (ax) }^{ 2 } }$

Explanation

Hint:- For a composite function

$f\left( g\left( x \right) \right)$ the chain rule for the derivative is expressed as

$f'\left( x \right)g'\left( x \right)$ .

The derivative of the function

$y = {\tan ^{ - 1}}x$ is given as

${{dy} \over {dx}} = {1 \over {1 + {x^2}}}$ .

Given:-

The function is given as

$y = {\tan ^{ - 1}}\left( {ax} \right)$ .

Step 1: Find the derivative of the function by applying chain rule.

$\Rightarrow {{dy} \over {dx}} = \left[ {{{\tan }^{ - 1}}\left( {ax} \right)} \right]'$

$\Rightarrow {{dy} \over {dx}} = {1 \over {1 + {{\left( {ax} \right)}^2}}} \times \left( {ax} \right)'$

Step 2: Find the derivative of the term

$\left( {ax} \right)'$ and simplify the expression.

$\Rightarrow {{dy} \over {dx}} = {1 \over {1 + {{\left( {ax} \right)}^2}}} \times a$

$\therefore {{dy} \over {dx}} = {a \over {1 + {{\left( {ax} \right)}^2}}}$

Final step: The derivative of the function

$y = {\tan ^{ - 1}}\left( {ax} \right)$ is given by the relation

$\ {{dy} \over {dx}} = {a \over {1 + {{\left( {ax} \right)}^2}}}$

$y = Tan^{ -1 }\left(secx + tanx \right) then \dfrac { dy }{ dx }=$

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

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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