CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 15 - MCQExams.com

$$\displaystyle\underset{x\rightarrow \dfrac{\pi}{4}}{Lt}\dfrac{\sqrt{2}-\cos x-\sin x}{(4x-\pi)^2}=?$$
  • $$\dfrac{1}{16\sqrt{2}}$$
  • $$\dfrac{1}{32\sqrt{2}}$$
  • $$\dfrac{1}{16}$$
  • $$\dfrac{1}{8}$$
$$ \lim _{x \rightarrow 0} \frac{1-\cos ^{3} x}{x \sin 2 x}= $$
  • 1$$ / 2 $$
  • 3$$ / 2 $$
  • 3$$ / 4 $$
  • 1$$ / 4 $$
The value of $$\underset { x\rightarrow 0 }{ lim } \left( \left( sinx \right) ^{ 1/x }+\left( \dfrac { 1 }{ x }  \right) ^{ sinx } \right) $$ equals
  • 0
  • 1
  • $$\infty $$
  • -1
$$\lim _ { x \rightarrow 0 } \frac { \ln ( \sin 3 x ) } { \ln ( \sin x ) }$$ is equal to
  • 0
  • 1
  • 2
  • none of these
$$\underset { x\rightarrow 0 }{ lim } \dfrac { 1 }{ { e }^{ 2 } } tan\left( \dfrac { \pi  }{ 4 } +x \right) ^{ 1/x }$$
  • 0
  • 1
  • -1
  • e
The value of $$\theta ,\quad is$$
$$\underset { o\rightarrow o }{ lim } \quad \frac { { cos }^{ 2 }\left\{ 1-{ cos }^{ 2 }\quad \left( 1-{ cos }^{ 2 }\quad .....\left( { cos }^{ 2 }\left\{ 1-{ cos }^{ 2 }\theta  \right\}  \right)  \right)  \right\}  }{ sin\left( \frac { \pi (\sqrt { \theta +4 } -2 }{ \theta  }  \right)  } $$
  • $$\frac { \sqrt { 2 } }{ 4 } $$
  • $$\sqrt { 2 } $$
  • 1
  • 2
$$\lim _{ x\rightarrow \infty  }{ \frac { \sin { x } \sin { (\frac { \pi  }{ 3 } +x) } \sin { (\frac { \pi  }{ 3 } -x) }  }{ x }  } =$$
  • $$\frac { 3 }{ 4 } $$
  • $$\frac { 1 }{ 4 } $$
  • $$\frac { 4 }{ 3 } $$
  • $$0$$
If $$ \alpha , \beta , \ in (-\frac{\pi}{2},0) $$ such that $$(sin \alpha +sin \beta ) +\frac{sin \alpha }{sin \beta} =0 $$ and $$(sin \alpha +sin \beta ) \frac{sin \alpha}{sin \beta }=-1 $$ and $$\lambda =\begin{matrix} lim \\ n\rightarrow \infty  \end{matrix}\frac { 1+(2sin\quad theta\quad ){  }^{ 2n } }{ (2sin\quad \quad beta\quad ){  }^{ 2n } } $$ then
  • $$2 \alpha +3 \beta = \frac{-5 \pi }{6}$$
  • $$ \lambda \pi +\alpha +\beta =\frac{5 \pi}{6}$$
  • $$ \alpha - \beta = \frac{ \pi }{3}$$
  • $$ \alpha + \beta = \frac{- \pi }{3}$$
$$\cfrac { d }{ dx } \left( \sin ^{ 5 }{ x } \sin { 5x }  \right) =$$
  • $$ \left( \sin ^{ 4 }{ x } \sin { 5x } \right)$$
  • $$5 \left( \sin ^{ 4 }{ x } \sin { 5x } \right)$$
  • $$5 \left( \sin ^{ 4 }{ x } \sin { 6x } \right)$$
  • $$-5 \left( \sin ^{ 4 }{ x } \sin { 6x } \right)$$
$$\lim _{ x\rightarrow 0 }{ \frac { { 27 }^{ x }-{ 9 }^{ x }-{ 3 }^{ x }+1 }{ \sqrt { 2 } -\sqrt { 1+\cos { x }  }  } = } $$
  • $$0$$
  • $$8\sqrt { 2 } { (\log { 3 } ) }^{ 2 }$$
  • $$8{ (\log { 3 } ) }^{ 2 }$$
  • $$1$$
$$\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 3h} \right) - 3\sin \left( {a + 2h} \right) + 3\sin \left( {a + h} \right) - \sin a}}{{{h^3}}}$$ is equal to 
  • $$\cos a$$
  • $$ - \cos a$$
  • $$\sin a$$
  • $$\sin a\cos a$$
$$L\underset { x\rightarrow \infty  }{ im } \left( \sin { \sqrt { x+1 } -\sin { \sqrt { x }  }  }  \right) =$$
  • 2
  • -2
  • 0
  • 1
$$\lim _{ x\rightarrow  }{ 0 } \{ (sinx-x)/{ x }^{ 3 })\} $$ equals: 
  • $$1/3$$
  • $$-1/3$$
  • $$1/6$$
  • $$-1/6$$
Solve 
$$\underset { x\rightarrow 0 }{ Lim } \frac { 8 }{ { x }^{ 8 } } \left( 1-cos\frac { { x }^{ 2 } }{ 2 } -cos\frac { { x }^{ 2 } }{ 4 } +cos\cfrac { { x }^{ 2 } }{ 2 } .cos\cfrac { { x }^{ 2 } }{ 4 }  \right) =$$
  • $$\cfrac { 1 }{ 16 } $$
  • $$\cfrac { 1 }{ 15 } $$
  • $$\cfrac { 1 }{ 32 } $$
  • $$1$$
$$\mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right]$$ is equal to
  • 1
  • -1
  • 0
  • None of these
$$ \lim _ { x \rightarrow 0 } \frac { \sin x } { x } = y $$
  • $$

    y > 1

    $$
  • $$

    y < 1

    $$
  • $$

    y \geq 1

    $$
  • $$

    y \leq 1

    $$
The value of $$\mathop {{\text{Limit}}}\limits_{x \to 0} \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}}$$ is equal to 
  • 1/5
  • 1/6
  • 1/4
  • 1/2
$$\underset { x\rightarrow 0 }{ lim } \cfrac { 8 }{ { x }^{ 8 } } \left( 1-cos\cfrac { { x }^{ 2 } }{ 2 } -cos\cfrac { { x }^{ 2 } }{ 4 } +cos\cfrac { { x }^{ 2 } }{ 2 } .cos\cfrac { { x }^{ 2 } }{ 4 }  \right) =$$
  • $$\cfrac { 1 }{ 16 } $$
  • $$\cfrac { 1 }{ 15 } $$
  • $$\cfrac { 1 }{ 32 } $$
  • $$1$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \dfrac { 100\tan { x } .\sin { x }  }{ { x }^{ 2 } }  \right]  } $$ where $$[.]$$ represents greatest integer function is 
  • $$99$$
  • $$100$$
  • $$0$$
  • $$98$$
The value of  $$\lim _ { x \rightarrow \dfrac { 1 } { 2 } } \dfrac { 2 \sin ^ { - 1 } x - \dfrac { \pi } { 2 } } { 1 - 2 x ^ { 2 } }$$  is equal to
  • $$-1$$
  • $$0$$
  • $$1$$
  • $$2$$
The value of $$ \lim _ { x \rightarrow 0 } \dfrac { \sin^3 ( \sqrt { x } ) \ln ( 1 + 3 x ) } { \left( \tan ^ { - 1 } \sqrt { x } \right) ^ { 2 } \left( e ^ { 5 ( \sqrt { x } ) } - 1 \right)x } $$ is equal to
  • $$

    \frac { 1 } { 5 }

    $$
  • $$

    \frac { 3 } { 5 }

    $$
  • $$

    \frac { 2 } { 5 }

    $$
  • $$

    \frac { 4 } { 5 }

    $$
The value of $$\lim _{ x\rightarrow \frac { \pi  }{ 2 }  }{ \tan ^{ 2 }{ \left( \sqrt { 2\sin ^{ 2 }{ x } +3\sin { x } +4 } -\sqrt { \sin ^{ 2 }{ x } +6\sin { x } +2 }  \right)  }  } $$ is equal to
  • 0
  • $$\cfrac{1}{11}$$
  • $$\cfrac{1}{12}$$
  • $$\cfrac{1}{8}$$
If x sin $$(\alpha +y)=siny\quad and\quad y'\quad =\quad \frac { m }{ \left( { x }^{ 2 }+ \right) nx+ } ,$$ then
  • m-n=1
  • m+n=1
  • $${ m }^{ 2 }+{ n }^{ 2 }=1$$
  • m=n
$$\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2}\sin \left( {\dfrac{1}{x}} \right) - x}}{{1 - \left| x \right|}}} \right) = $$
  • 0
  • 1
  • -1`
  • 2
$$\mathop {Lt}\limits_{x \to 1} {\left( {1 - x} \right)^{\tan \pi x}} = $$
  • $$-1$$
  • $$0$$
  • $$1$$
  • $$2$$
Let $$a \in \left( 0 , \frac { \pi } { 2 } \right)$$, then the value of
$$ \lim _ { a \rightarrow 0 } \frac { 1 } { a ^ { 3 } } \int _ { 0 } ^ { a } \ell n (1+tan a tan x)dx$$ is equal to 
  • $$\frac { 1 }{ 3 } $$
  • $$\frac { 1 }{ 2 } $$
  • $$\frac { 1 }{ 6 } $$
  • 1
Find:
$$\underset { x\rightarrow 0 }{ lim } \quad \dfrac { 1-cos^{ 3 }x }{ xsin2x } =\quad $$
  • 1/2
  • 3/2
  • 3/4
  • 1/4
$$\underset { \rightarrow  }{ Lim } \frac { 1-{ cos }^{ 2 }x }{ xsin2x } $$
  • 1/2
  • 3/2
  • 3/4
  • 1/4
The value of $$\mathop {\lim }\limits_{x \to 0} \frac{{\sec 5x - \sec 3x}}{{\sec 3x - \sec x}}$$
  • -2
  • 1
  • 2
  • -1/2
$$y=\sec { \left( \tan ^{ -1 }{ x }  \right)  } $$ then at $$x=1$$ value of $$\dfrac { dy }{ dx } $$ equals :
  • $$\dfrac { 1 }{ 2 } $$
  • 1
  • $$\sqrt {2 }$$
  • $$\dfrac { 1 }{ \sqrt {2 } } $$
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