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CBSE Questions for Class 12 Commerce Maths Continuity And Differentiability Quiz 14 - MCQExams.com

If f(x) is a differentiable function in the interval (0,) such that f(1)=1 and  limtxt2f(x)x2f(t)tx=1, for each x>0, then f(32) is equal to:
  • 259
  • 136
  • 2318
  • 3118
Let f(x)=x|x|,g(x)=sinx and h(x)=(gof)(x). Then
  • h(x)is differentiable at x=0
  • h(x) is continuous at x=0 but is not differentiable at x=0
  • h(x) is differentiable at x=0 but h(x) differentiable is not continuous at x=0
  • h(x) is not differentiable at x=0
If f(x)=0 for x<0 and f(x) is differentiable at x=0, then for x0,f(x) may be
  • x2
  • x
  • x
  • x3/2
The value of f(4) is
  • 160
  • 240
  • 200
  • None of these
If y=sin1x+sin11x2, then dxdx
  • 0
  • 1
  • 2x
  • yx
y=x20,findd2ydx2
  • 381x18
  • 380x18
  • 389x18
  • 370x18
If x=θ1θ,y=θ+1θ then dydx=
  • xyy2+2
  • y/x
  • -x/y
  • -y/x
If y=\cos ^{ -1 }{ (\frac { x-{ x }^{ -1 } }{ x+{ x }^{ -1 } } ) } \quad then\quad \frac { dy }{ dx } =
  • \frac { -2 }{ 1+{ x }^{ 2 } }
  • \frac { 2 }{ 1+{ x }^{ 2 } }
  • \frac { 1 }{ 1+{ x }^{ 2 } }
  • \frac { -1 }{ 1+{ x }^{ 2 } }
dx\left\{ sin{  }^{ -1 }(\frac { 5x+12\sqrt { 1-x{  }^{ 2 } }  }{ 13 } ) \right\} =
  • \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }
  • \frac { -1 }{ \sqrt { 1+{ x }^{ 2 } } }
  • \frac { 2 }{ \sqrt { 1-{ x }^{ 2 } } }
  • 0
Let fbe twice differentiable function such that { g }^{ 1 }\left( x \right) =-f\left( x \right) and\quad \quad \quad \quad \quad { f }^{ 1 }\left( x \right) =g\left( x \right) ,\\ h(x)={ \left( f\left( x \right)  \right)  }^{ 2 }+{ \left( g\left( x \right)  \right)  }^{ 2 }.\quad Ifh(5)=11,\quad thenh(10)\quad is
  • 22
  • 11
  • 0
  • 8
If y=\tan ^{ -1 }{ [x+\sqrt { 1+{ x }^{ 2 } } ] }  then \frac { dy }{ dx } =
  • \frac { 1 }{ 1+{ x }^{ 2 } }
  • \frac { 1 }{ 2(1+{ x }^{ 2 }) }
  • \frac { 2 }{ 1+{ x }^{ 2 } }
  • None of these
\frac { d[\sec ^{ -1 }{ (\sin { x } +{ x }^{ 2 }) } ] }{ dx } =
  • \frac { \sin { 2x } +{ x }^{ 2 } }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } }
  • \frac { \cos { x } +2x }{ \sin { x } +{ x }^{ 2 }()\sqrt { { (\sin { x } +{ x }^{ 2 }) }^{ 2 }-1 } }
  • \frac { 2x }{ \sin { x } \sqrt { \sin { x } -1 } }
  • None of these
f(x) = \log_{1 - 2x}(1 + 2x)    for x \ne 0
          = k                              for x = 0
is continuous at x = 0, find k.
  • 1
  • -1
  • 0
  • \frac{1}{2}
The\, \, function\, \, g(x)=\left| { \begin{array} { *{ 20 }{ c } }{ x+b } & { x<0 } \\ { \cos  \, x } & { x\ge 0 } \end{array} } \right. \, \, can\, \, be\, \, made\, differentiable\, at\, x=0
  • if\,b\,\,is\,\,equal\,to\,zero\,
  • if\,b\,\,is\,\,not\,equal\,\,to\,zero\,
  • if\,b\,\,is\,\,not\,equal\,\,to\,zero\,
  • for\,no\,value\,of\,b
If y={ tan }^{ -1 }\left( \dfrac { 1-{ cos }^{ 2 }x }{ 1+{ cos }^{ 2 } }  \right) ,, then \dfrac { dy }{ xy } =
  • \dfrac { sinx }{ 1+{ cos }^{ 4 }x }
  • \dfrac { sinx }{ 1+{ cos }^{ 2 }x }
  • \dfrac { sin2x }{ 1+{ cos }^{ 4 }x }
  • \dfrac { sin2x }{ 1+{ cos }^{ 2 }x }
Solve \frac{d\tan ^{-1}}{dx} (\frac{5x + 1}{3 - x - 6x^2}) =
  • 0
  • 1
  • \frac{1}{1 + (3x + 2)^2} + \frac{1}{1 + (2x - 1)^2}
  • \frac{3}{1 + (3x + 2)^2} + \frac{2}{1 + (2x - 1)^2}
If \cos ^{-1}x + \cos ^{-1}y = \frac{\pi}{2} then \frac{dy}{dx}
  • \frac{x}{y}
  • \frac{-2x}{y}
  • \frac{-x}{y}
  • \frac{-y}{x}
Let f be a function defined by 2f\left(\sin x\right) + f\left(\cos x\right) = x ~\forall  x\in R, then set of points where f is not differentiable is 
  • set of all natural numbers
  • set of all irrational number
  • \left\{0,1,-1\right\}
  • \left\{1,-1\right\}
If f\left( x \right) =\sqrt { 1-\sqrt { 1-{ x }^{ 2 } }  } , then f(x) is
  • continuous on[-1,1]
  • differentiable on (-1,0)\cup (0,1)
  • both (a) and (b)
  • None of the above
y={ cos }^{ -1 }\cfrac { (9-{ x }^{ 2 }) }{ (9+{ x }^{ 2 }) } then Y'(-1) is equal to:
  • \cfrac {3 }{ 5 }
  • \cfrac { 3 }{ 5 }
  • \cfrac { 2 }{ 7 }
  • \cfrac { 3 }{ 8 }
The number of values of  x  at which  f(x) = \left| x - \frac { 3 } { 2 } \right| + | x - 2 | + ( x - 1 ) | x - 1 |  is not differentiable 
  • 2
  • 3
  • 4
  • 1
If f(x) is differentiable function and f(1) = \sin 1, f(2) = \sin 4, f(3) = \sin 9, then the minimum number of distinct solution of equation f'(x) = 2x \cos x^2 in (1, 3) is
  • 1
  • 2
  • 3
  • 4
Solve : \dfrac { d } { d x } \left[ \tan ^ { - 1 } \sqrt { 1 + x ^ { 2 } } - \cot ^ { - 1 } \left( - \sqrt { 1 + x ^ { 2 } } \right) \right] =?
  • \pi
  • 1
  • 0
  • \dfrac { 2 x } { \sqrt { 1 + x ^ { 2 } } }
The derivative of \cos^{-1} x w.r.t. \sqrt{1 - x^2} is
  • \dfrac{1}{\sqrt{1 - x^2}}
  • \sqrt{1 - x^2}
  • \dfrac{1}{2x}
  • \dfrac{1}{x}
let f be the differeniable at x=0 and f'(0)=1 then \underset { b\rightarrow 0 }{ lim } \frac { f(h)-f(-2h) }{ h } 
  • 3
  • 2
  • 1
  • -1
If  f  is twice differentiable such that   f ^ { \prime \prime } ( x ) = - f ( x )  and  f ^ { \prime } ( x ) = g ( x ).  If  h ( x )  is a twice differentiable function that  h ^ { \prime } ( x ) = ( f ( x ) ) ^ { 2 } + ( g ( x ) ) ^ { 2 } .  If  h ( 0 ) = 2 , h ( 1 ) = 4 ,  then the equation  y = h ( x )  represents :
  • a curve of degree 2
  • a curve passing through the origin
  • a straight line with slope 2
  • a straight line with y intercept equal to 2
f(x)=\begin{cases} x\left( \dfrac { a{ e }^{ \frac { 1 }{ \left| x \right|  }  }+3{ e }^{ \frac { -1 }{ x }  } }{ \left( a+2 \right) { e }^{ \frac { 1 }{ \left| x \right|  }  }-{ e }^{ \frac { -1 }{ x }  } }  \right),\ \ x\neq 0  \\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0 \end{cases} is differentiable at x=0 then [a]=...... ([\ ] denotes greatest integer function)
  • 0
  • -1
  • -2
  • None\ of\ these
If y = {\cot ^{ - 1}}\left( {\frac{{\sin x}}{{1 + \cos x}}} \right) then \frac{{dy}}{{dx}} is 
  • \frac{{ - 1}}{2}
  • \frac{{ 1}}{2}
  • 0
  • 1
Number of points in [0,\pi], where f(x)=\left[\dfrac{x\tan{x}}{\sin{x}+\cos{x}}\right] is non-differentiable is/are (where [.] denotes the greatest integer function)
  • 0
  • 4
  • 9
  • None\ of\ these
Let F(x) = \left( f\left( x \right)  \right) ^{ 2 }+\left( f\left( x \right)  \right) ^{ 2 },F\left( 0 \right) -6 where f(x) is a differential  function such that \left| f\left( x \right)  \right| \le 1\forall x\notin \left[ -1,1 \right] then choose the correct statement (s)
  • There is atleast one point in each of the intervals (-1, 0) and (0, 1) where \left| f'(x) \right| \le 2
  • there is atleast one point in a each of the interval (-1, 0) and (0, 1) where f(x)\le S
  • there is no point of local maximum of F(x) in (-1, 1)
  • For some c\in \left( -1,1 \right) , F'\left( c \right) \ge 6,F"\left( c \right) \le 0
0:0:2


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