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

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 
  • 6
  • 12
  • 4
  • 3
If $$f(x) = \cos^{-1}  (\cos x)$$, then at the point, where $$f$$ is differentiable , $$f ' (x)$$ equals 
  • $$1$$
  • $$-1$$
  • $$\text{sng} (\sin x)$$
  • $$-\text{sng} (\sin x)$$
If $$f\left( x \right) =\left\{ \dfrac { x }{ { e }^{ { 1 }/{ x } }+1 }  \right\} $$ when $$x\neq 0,$$ then 0,when x=0
  • lim
    $$x\rightarrow 0+$$

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

    $$f\left( x \right) =1$$
  • $$f\left( x \right) $$ is continuous at x=0
  • None of the above
If $$f(x)=a|\sin x|+ be^{|x|}+c|x|^{3}$$, where $$a,b,c\in R$$, is differentiable at $$x=0$$, then 
  • $$a=0,b$$ and $$c$$ are any real numbers
  • $$c=0,a=0 ,b$$ is any real number
  • $$b=0,c=0 ,a$$ is any real number
  • $$a=0,b=0 ,c$$ is any real number
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 $$y=\tan{x}$$, then $$\dfrac{d^{2}y}{dx^{2}}=$$
  • $$x\dfrac{dy}{dx}$$
  • $$2x\dfrac{dy}{dx}$$
  • $$2y\dfrac{dy}{dx}$$
  • $$y\dfrac{dy}{dx}$$
The function $$f\left( x \right) = \dfrac{{4 - {x^2}}}{{4x - {x^3}}}$$ is
  • discontinuous at only one point
  • discontinuous exactly at two points
  • discontinuous exactly at three points
  • None of these
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
  • $$1$$
  • $$2$$
  • $$0$$
  • $$None$$ $$of$$ $$these$$
The function $$f(x)=\dfrac{x^3+3x+5}{x^3-3x+2}$$ is :
  • Continuous on R
  • Discontinuous at one point on R
  • Discontinuous at two points on R
  • 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 :
  • $$\dfrac{3}{1+9x^{3}}$$
  • $$\dfrac{9}{1+9x^{3}}$$
  • $$\dfrac{3x\sqrt{x}}{1-9x^{3}}$$
  • $$\dfrac{3x}{1-9x^{3}}$$
If $$y = \exp \left\{ {{{\sin }^2}x + {{\sin }^4}x + {{\sin }^6}x + ....} \right\}$$ then $$\frac {dy}{dx}=$$
  • $${e^{{{\tan }^2}x}}$$
  • $${e^{{{\tan }^2}x}}{\sec ^2}x$$
  • $$2{e^{{{\tan }^2}x}}\tan x{\sec ^2}x$$
  • 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? 
  • a + b =$$-\frac { 3 }{ 4 } $$
  • a + b = $$-\frac { 5 }{ 4 } $$
  • ab = $$\frac { 1 }{ 4 } $$
  • ab = $$-\frac { 1 }{ 4 } $$
If $$y = \cos \left( {m{{\cos }^{ - 1}}x} \right)$$, then $$\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} = $$
  • $$my$$
  • $$-my$$
  • $${m^2}y$$
  • $${-m^2}y$$
If $$\dfrac {dy}{dx}=(e^ {y}-x)^ {-1}$$ where $$y(0)=0$$ then $$y$$ is expressed explicity as 
  • $$0.5\log_{e}(1+x^{2})$$
  • $$\log_{e}(1+x^{2})$$
  • $$\log _{ e } \left( x+\sqrt { 1+{ x }^{ 2 } } \right) $$
  • $$\\ \log _{ e } \left( x+\sqrt { 1-{ x }^{ 2 } } \right) $$
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
  • $$[0,1]$$
  • $$[-1,2]$$
  • $$[-2,3]$$
  • $$\left[ 0,\dfrac { 2 }{ 3 } \right] $$
$$\dfrac{d}{dx}$$ { $$\cot^{-1} \dfrac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}}$$ } =
  • $$\dfrac{1}{\sqrt{1-x^2}}$$
  • $$\dfrac{-1}{2\sqrt{1-x^2}}$$
  • $$\dfrac{1}{1+x^2}$$
  • None of these
in $$\left[ a,b \right] $$ 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
  • $$\sqrt { a } $$
  • $$\sqrt { b } $$
  • $$\sqrt { ab } $$
  • $$-\sqrt { ab } $$
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
  • $$2f(x)$$
  • $$f(x)/x$$
  • $$2xf(x)$$
  • $$2f(x)/x$$
Let f(x) be a polynomial in x. Then the second derivation of f$$\left( { e }^{ x } \right) $$, is:
  • $$f"\left( { e }^{ x } \right) .{ e }^{ x }+f'\left( { e }^{ x } \right) $$
  • $$f"\left( { e }^{ x } \right) .{ e }^{ x }+f'\left( { e }^{ x } \right) .{ e }^{ 2x }$$
  • $$f"\left( { e }^{ x } \right) { e }^{ 2x }$$
  • $$f"\left( { e }^{ x } \right) { e }^{ 2x }+f'\left( { e }^{ x } \right) .{ e }^{ x }$$
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?
  • $$f(x)=2^{-1/12}\left(\dfrac{\sin\left(6x+\dfrac{\pi}{4}\right)}{\cos^2\left(6x+\dfrac{\pi}{4}\right)}\right)^{1/6}$$
  • $$f(x)=2^{-1/6}\left(\dfrac{\sin \left(6x+\dfrac{\pi}{4}\right)}{\cos^2\left(6x+\dfrac{\pi}{4}\right)}\right)^{1/12}$$
  • $$f(x)=(\sec 12x)^{1/12}(\sec 12x+\tan 12x)^{1/4}$$
  • $$f(x)=(\sec 12x)^{1/4}(\sec 12x+\tan 12x)^{1/12}$$
If $$y=1-\cos\theta,x=1-\sin\theta$$, then $$\dfrac{dy}{dx}$$ at $$\theta=\dfrac{\pi}{4}$$ is 
  • $$-1$$
  • $$1$$
  • $$12$$
  • $$-12$$
If $$x=\sqrt{2^{cosec^{-1}t}}$$ and $$y=\sqrt{2^{sec^{-1}t}}(|t|\geq 1)$$ then $$\dfrac{dy}{dx}$$ is equal to:
  • $$\dfrac{y}{x}$$
  • $$\dfrac{x}{y}$$
  • $$\dfrac{-x}{y}$$
  • $$\dfrac{-y}{x}$$
If $$f(x)$$ is a four times differentiable even function, then $$\int_{-3}^{3}(x^{3}f(x)+xf''''(x)+2)dx$$    is equal to 
  • $$12f(x)+f''(x))$$
  • $$12f''(x)$$
  • $$12$$
  • $$6$$
$$\dfrac{d}{dx}\left[\tan h^{-1}\left(\dfrac{2x}{1+x^2}\right)\right]=?$$
  • $$\dfrac{2}{1-x^2}$$
  • $$\dfrac{2}{x^2-1}$$
  • $$\dfrac{2}{1+x^2}$$
  • $$\dfrac{-2}{x^2+1}$$
A differential function satisfies equation $$f(x)=\int_{0}^{x}(f(t)\cos\ t-\cos(t-x))dt$$ then
  • $${ f }^{ \prime \prime }\left( \dfrac { \pi }{ 2 } \right) =e$$
  • $$\lim _{ x\rightarrow -\infty }{ f\left( x \right) =1 }$$
  • $$f(x) has minimum value 1-e^{-1}$$
  • $${ f }^{ \prime}(0)=-1$$
If  $$f ( x )$$  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
  • $$\mathrm { f } ^ { \prime } ( \mathrm { c } ) = \mathrm { g } ^ { \prime } ( \mathrm { c } )$$
  • $$f ^ { \prime } ( c ) = - g ^ { \prime } ( c )$$
  • $$f ^ { \prime } ( c ) = 2 g ^ { \prime } ( c )$$
  • $$2 f ^ { \prime } ( c ) = g ^ { \prime } ( c )$$
If $$x=2\cos \theta-2\cos 2\theta$$ and $$y=2\sin \theta-\sin 2\theta$$, then  $$\dfrac{dy}{dx}=\tan \left( \dfrac{3\theta}{2} \right )$$.
  • True
  • False
If $$ y = \sin ^ { - 1 } x$$ then $$ \frac { d y } { d x } $$ is equal to 
  • $$\sec y$$
  • $$\cos x$$
  • $$\tan x$$
  • $$1$$
If $$y={ \tan }^{ -1 }{ ax}$$ then, which of the following is true
  • $$\displaystyle \frac { dy }{ dx } =\frac { a }{ 1+{ x }^{ 2 } }$$
  • $$\displaystyle \frac { dy }{ dx } =\frac { a }{ 1-{(ax) }^{ 2 } }$$
  • $$\displaystyle \frac { dy }{ dx } =\frac { a }{ 2+{ (ax) }^{ 2 } }$$
  • $$\displaystyle \frac { dy }{ dx } =\frac { a}{ 1+{ (ax) }^{ 2 } }$$
If $$y = Tan^{ -1 }\left(secx + tanx \right) then \dfrac { dy }{ dx }=$$
  • $$1$$
  • $$\dfrac { 1 }{ 2 }$$
  • $$-1$$
  • $$0$$
0:0:1


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