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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 8 - MCQExams.com

Find the value of limit lim.
  • 0
  • 3
  • -3
  • 1
If f(x+y)=2f(x).f(y) for all x,y, where f'(0)=3 and f'(4)=2 then f'(4)=3 is equal to
  • 6
  • 12
  • 4
  • 3
If y=(x+\sqrt{x^{2}+a^{2}})^{n} then \dfrac{dy}{dx}=
  • y
  • ny
  • \dfrac{ny}{\sqrt{x^{2}+a^{2}}}
  • \dfrac{y}{\sqrt{x^{2}+a^{2}}}
If f(x) is the integral of \dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0. Find \lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right)  } , where f^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}
  • \dfrac{1}{2}
  • 1
  • \dfrac{1}{3}
  • 2
\lim\limits_{x\rightarrow0}\dfrac{x\tan 2x-2x\tan x}{\left(1-\cos 2x\right)^{2}}=
  • 2
  • \dfrac{1}{2}
  • -2
  • -\dfrac{1}{2}
\displaystyle\lim_{h\rightarrow 0}\dfrac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}=__________.
  • \cos\sqrt{x}
  • \dfrac{1}{2\sin \sqrt{x}}
  • \dfrac{\cos \sqrt{x}}{2\sqrt{x}}
  • \sin\sqrt{x}
If 2f(\sin x)+\sqrt {2}f(\cos x)=\tan x,\ (x> 0), then \displaystyle \lim _{ x\rightarrow 1 }{ \sqrt { 1-x } f\left( x \right) = } 
  • \sqrt {2}
  • \dfrac {1}{\sqrt {2}}
  • -\sqrt {2}
  • -\dfrac {1}{\sqrt {2}}
\dfrac{d}{dx}\left[\dfrac{tan x - cot x}{tan x + cot x}\right]=
  • 2 sin2x
  • -2 sin2x
  • 2 cos 2x
  • -2cos 2x
If \underset { x\rightarrow { 0 } }{ lim } \dfrac { \left( ax+b \right) -\sqrt { 4+\sin x }  }{ \tan\quad x } =\dfrac { 27 }{ 4 } ~where ~a,b\in R then the value of 
  • a = 2~ and~ b = 7
  • a = -2~and~ b = 7
  • a = 7~ and~ b = 2
  • a=7~ and ~b = -2
Evaluate: \underset { x\rightarrow 0 }{ \lim } \dfrac { x\tan2x-2x\tan x }{ \left( 1-\cos2x \right) ^{ 2 } }  
  • \dfrac { 1 }{ 4 }
  • 1
  • \dfrac { 1 }{ 2 }
  • -\dfrac { 1 }{ 2 }
\lim _ { x \rightarrow \frac { \pi } { 2 } } \frac { \cot x - \cos x } { ( \pi - 2 x ) ^ { 3 } } equals:

  • \frac { 1 } { 24 }
  • \frac { 1 } { 16 }
  • 0
  • \frac { 1 } { 4 }
The solution of the differential equation  \left( \dfrac { d y } { d x } \right) ^ { 2 } - 3 x \left( \dfrac { d y } { d x } \right) - 2 y = 8  is
  • y = 2 x ^ { 2 } + 4
  • y = 2 x ^ { 2 } - 4
  • y = 2 x + 4
  • y = 2 x - 4
\underset {x\rightarrow0}{ lim } \dfrac{x \tan 2 x - 2 x \tan x}{(1 - \cos 2 x)^2} equals 
  • \dfrac{1}{4}
  • 1
  • \dfrac{1}{2}
  • -\dfrac{1}{2}
Let f(x)=\begin{cases} { x }^{ 2 }+k,\quad \quad  when\quad x\ge 0 \\ -{ x }^{ 2 }-k,\quad \quad when \quad x<0 \end{cases}. If the function f(x) be continous at x=0, then k=
  • 0
  • 1
  • 2
  • -2
The value of \displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) }  where f(x)=\dfrac {\cos (\sin x)-\cos x}{x^{4}}, is
  • 2
  • \dfrac {1}{6}
  • \dfrac {2}{3}
  • -\dfrac {1}{3}
The value of \displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b }  }{ { x }^{ 2 }-{ a }^{ 2 } }  } \left( a>b \right)  is
  • \dfrac {1}{4a}
  • \dfrac {1}{a\sqrt {a-b}}
  • \dfrac {1}{2a\sqrt {a-b}}
  • \dfrac {1}{4a\sqrt {a-b}}
\lim _ { x \rightarrow 0 } \dfrac { ( 27 + x ) ^ { 1 / 3 } - 3 } { 9 - ( 27 + x ) ^ { 2 / 3 } }  equals :
  • - 1 / 6
  • 1 / 6
  • 1 / 3
  • - 1 / 3
A curve in the 1^{st} quadrant passes through (1,1). Its drifferential equation is (y-xy^{2})dx+(x+x^{2}y^{2})dy=0. Hence the equation of the curve is 
  • y-\dfrac{1}{xy}=\ln y
  • y-\dfrac{1}{xy}=\ln x
  • y-xy=\ln y
  • y-xy=\ln x
If (cos x)^y = (sin y)^x, then \dfrac{dy}{dx} =
  • \dfrac{log(siny) + y tan x}{ log (cos x) - x cot y}
  • \dfrac{log(siny) - y tan x}{ log (cos x) + cot y}
  • \dfrac{log(siny) }{ log (cos x) }
  • \dfrac{log(cos x) }{ log (sin y) }
\underset { x\rightarrow 1 }{ Lim } \left[ { \left[ \frac { 4 }{ { x }^{ 2 }-{ x }^{ -1 } } -\frac { { 1-3x+x }^{ 2 } }{ { 1-x }^{ 3 } }  \right]  }^{ -1 }+\frac { 3\left( { x }^{ 4 }-1 \right)  }{ { x }^{ 3 }-{ x }^{ -1 } }  \right] =
  • \frac { 1 }{ 3 }
  • 3
  • \frac { 1 }{ 2 }
  • \frac { 3 }{ 2 }
Let  U_{ { n } }=\dfrac { n! }{ (n+2)! }   where  n \in N .  If  S_{ { n } }=\sum _{ { n-1 } }^{ { n } } U_{ { n } }  then  \lim _ { n \rightarrow \infty } \mathrm { S } _ { n }  equals :
  • 2
  • 1
  • 1/2
  • 1/3
\displaystyle \underset { x\rightarrow 0 }{ lim } \ \ \frac { ({ 1-\cos2x) }^{ 2 } }{ 2x \tan x-x \tan2x } is :
  • -2
  • \dfrac { -1 }{ 2 }
  • \dfrac { 1 }{ 2 }
  • 2
The differential of f(x)=\sqrt{\dfrac{2-x}{2+x}} at x=0 and \delta x=0.15 is
  • 0.07
  • 0.075
  • -0.075
  • 0.15
The value of \displaystyle \lim_{x\rightarrow 0} \dfrac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x} is
  • 10/3
  • 3/10
  • 6/5
  • 5/6
Let f(x)=\dfrac{ax+b}{x+1},lim_{x\rightarrow 0} f(x)=2 and lim_{x\rightarrow \infty} f(x)=1 then f(-2)=
  • 1
  • 2
  • -1
  • 0
Evaluate the limit, \mathop {\lim }\limits_{x \to 0} \frac{{x({{(1 + x)}^{1/x}} - e)}}{{x({{(1 + {x^2})}^{1/{x^2}}} - e)}}
  • 0
  • 1
  • 2
  • DNE
Evaluate
\mathop {\lim }\limits_{x \to 0} \cfrac{{1 - \cos (1 - \cos 2x)}}{{{x^4}}}
  • 4
  • 2
  • 1
  • \dfrac{1}{2}
\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } - x + 1 } - a x - b \right) = 0,   then the values of  a  and  b  are given by
  • a = - 1 , b = 1 / 2
  • a = 1 , b = 1 / 2
  • a = 1 , b = - 1 / 2
  • None of these
\lim _{ { x\rightarrow \pi /4 } } \dfrac { \cot ^{ { 3 } } x-\tan  x }{ \cos  (x+\pi /4) }   is
  • 4
  • 8 \sqrt { 2 }
  • 8
  • 4 \sqrt { 2 }
Let f(x)=\displaystyle\lim_{n\rightarrow \infty}\sum^{n-1}_{r=0}\dfrac{x}{(rx+1)\{(r+1)x+1\}}, then?
  • f(x) is continuous but not differentiable at x=0
  • f(x) is both continuous and differentiable at x=0
  • f(x) is neither continuous nor differentiable at x=0
  • f(x) is a periodic function
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