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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 15 - MCQExams.com

if the function f(x)={a+sin1(x+b),x1x,x<1 is differentiable at x=1, then ab is equal to 
  • 1
  • 0
  • 1
  • 2
The function f(x)=x1+|x| is differentiable at which of the following?
  • Every where
  • Everywhere except at x=1
  • Everywhere except at x=0
  • Everywhere except at x=0 or 1
If y=esin1(t21) & x=esec1(1t21), then dydx is equal to
  • xy
  • yx
  • yx
  • xy
If f(x)=sin1[2x1+x2],then f(x) is differentiable on 
  • [-1,1]
  • R-{-1,1}
  • R-(-1,1)
  • None of these
f(X)=|x|+|x-1| is continuous at 
  • '0' only
  • 0,1 only
  • Every where
  • No where
If fn(x)=efn1(x) for all nϵN and f0(x)=x then ddx{fn(x)} is equal to 
  • fn(x).ddx{fn1(x)}
  • fn(x).fn1(x)
  • fn(x).fn1(x).....f2(x).f1(x)
  • ni=1fi(x)
The order of the differential equation of all circles whose radius is 4, is?
  • 1
  • 2
  • 3
  • 4
If f\left( x \right) = {\left| x \right|^{\left| {\sin x} \right|}}, then {f'}\left( { - \dfrac{\pi }{4}} \right) is equals
  • {\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( { - \dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi }} \right)
  • {\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } + \dfrac{{2\sqrt 2 }}{\pi }} \right)
  • {\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{\pi }{4} + \dfrac{{2\sqrt 2 }}{\pi }} \right)
  • None of these.
Let f:(-1,1)\rightarrow R be a differentiable function satisfying 
             (f'(x))^4=16(f(x))^2 for all x\in (-1,1)
   f(0)=0
The number of such functions is 
  • 2
  • 3
  • 4
  • more than 4
If \displaystyle y = \dfrac{\sqrt{(1 + t^{2})} - \sqrt{(1 - t^{2})}}{\sqrt{(1 + t^{2})} +\sqrt{(1 - t^{2})}} and \displaystyle x = \sqrt {(1 - t^{4})} , then \dfrac{dy}{dx}
  • \displaystyle \dfrac{-1}{t^{2}\left \{1 +\sqrt{1 - t^{4}} \right \}}
  • \displaystyle \dfrac{\left \{\sqrt{(1 - t^{4})} - 1\right \}}{t^{6}}
  • \displaystyle \dfrac{1}{t^{2}\left \{1 + \sqrt{(1 - t^{4})}\right \}}
  • \displaystyle \dfrac{1 - \sqrt{( 1 - t^{4})}}{t^{6}}
f(x) is diffrentiable function and (f(x). g(x)) is differentiable a x=a , then 
  • g(x) must be differentiable at x=a
  • if g(x) is discontinuous , then f(a) =0
  • f(a) \neq 0 then g(x) must be differentiable
  • nothing can be said
Give that f(x) =xg(x) / \left | x \right | , g(0) = 0 and f(x) is continous at x=Then the value of f' (0)
  • Does not exist
  • is -1
  • is 1
  • is 0
Let f(x)=
1752636_53b9ea09cceb44beb34375cef34837d0.PNG
  • f' is differentiable
  • f is differentiable
  • f' is continuous
  • f is continuous
If f(x) =  \left\{\begin{matrix} \dfrac{x\log \cos x}{\log(1+x^2)}, & x \neq 0\\ 0, & x=0\end{matrix}\right. then
  • f(x) is not continuous at x=0.
  • f(x) is continuous at x=0.
  • f(x) is continuous at x=0 but not differentiable at x=0.
  • f(x) is differentiable at x=0.
Let f:  R\rightarrow R be a function such that f(x+y)= f(x)+f(y), \forall x,y \epsilon R . If f(x) is differentiable at x=0, then
  • f(x) is differentiable only in a finite interval containing zero
  • f(x) is continuous \forall x\epsilon R
  • f(x) is constant \forall x\epsilon R
  • f(x) is differentiable except at finitely many points
Let f(x) be a function satisfying f(x+y)  = f(x)+f(y) and f(x) = xg(x) \forall x, ~y \in R, where g(x) is a continuous function then, which of the following is true?
  • f'(x)=g'(0)
  • f'(x)=g (0)
  • f(x)=g (0)
  • f(x)=g'(0)
Which of the following is differentiable at x= 0
  • cos \left ( \left | x \right | \right )+\left | x \right |
  • cos \left ( \left | x \right | \right )-\left | x \right |
  • sin \left ( \left | x \right | \right )+\left | x \right |
  • sin \left ( \left | x \right | \right )-\left | x \right |
\cos |x| is differentiable everywhere.
  • True
  • False
If the function f:[0,8] \rightarrow R is differentiable, then for0<a, b<2, \int_{0}^{8} f(t) d t is equal to
  • 3\left[\alpha^{3} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]
  • 3\left[\alpha^{3} f(\alpha)+\beta^{3} f(\beta)\right]
  • {3}\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right]
  • 3\left[\alpha^{2} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]
Let f(x) and g(x) be differentiable for 0\leq x\leq 1, such that f(0) such that f'(c)=2g'(c), then the value of g(1) must be 
  • 1
  • 3
  • -2
  • -1
f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2) and g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)

The value of g(0)  is
  • 0
  • -3
  • 2
  • None of these
f(x) is not invertible for
  • x \in\left[-\dfrac{\pi}{2}-\tan ^{-1} 2, \dfrac{\pi}{2}-\tan ^{-1} 2\right]
  • x \in\left[\tan ^{-1} \dfrac{1}{2}, \pi+\tan ^{-1} \dfrac{1}{2}\right]
  • x \in\left[\pi+\cot ^{-1} 2,2 \pi+\cot ^{-1} 2\right]
  • None of these
Let f(0,\infty)\rightarrow R be a differentiable function such that f'(x)=2-\dfrac{f(x)}{x} for all x\epsilon (0,\infty) and f(1)\neq 1 Then
  • \underset{x\rightarrow 0+}{lim}f'(\dfrac{1}{x})=1
  • \underset{x\rightarrow 0+}{lim}xf'(\dfrac{1}{x})=2
  • \underset{x\rightarrow 0+}{lim}x^2f'(\dfrac{1}{x})=0
  • |f(x)|\le 2 for all x\epsilon(0,2)
0:0:2


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