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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  R where f' (0) = 3 and f (4) =2, then f'(4) is equal to 
  • 6
  • 12
  • 4
  • 3
If f(x)=cos1(cosx), then at the point, where f is differentiable , f(x) equals 
  • 1
  • 1
  • sng(sinx)
  • sng(sinx)
If f(x)={xe1/x+1} when x0, then 0,when x=0
  • lim
    x0+

    f(x)=1
  • lim
    x0

    f(x)=1
  • f(x) is continuous at x=0
  • None of the above
If f(x)=a|sinx|+be|x|+c|x|3, where a,b,cR, 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 (cosx)y=(siny)x, then dydx=
  • log(siny)+ytanxlog(cosx)xcoty
  • log(siny)ytanxlog(cosx)+xcoty
  • log(siny)log(cosx)
  • log(cosx)log(siny)
If y=tanx, then d2ydx2=
  • xdydx
  • 2xdydx
  • 2ydydx
  • ydydx
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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