CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 3 - MCQExams.com

Solve:
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\tan x-\tan 3x}{2x^{3}}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{3}{4}$$
  • $$4$$
  • $$-4$$
$$\displaystyle \lim_{x\rightarrow 0}(\frac{\sin x-x}{x})(\sin\frac{1}{x})$$ is:
  • does not exist
  • is equal to 0
  • is equal to 1
  • exists and different from 0 and 1

$$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$$=
  • $$2$$
  • $$1$$
  • $$0$$
  • $$3$$

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}$$=
  • 1
  • 0
  • -1
  • $$\displaystyle \dfrac{1}{\pi}$$

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$$=
  • $$\displaystyle \frac{1}{3}$$
  • $$\displaystyle \frac{1}{4}$$
  • $$\displaystyle \frac{1}{6}$$
  • $$\displaystyle \frac{1}{8}$$
$$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}=$$
  • $$1$$
  • $$-1$$
  • $$\dfrac{1}{3}$$
  • $$0$$

 lf $$\displaystyle { f }({ x })=\sqrt { \frac { { x }-\sin ^{ 2 }{ x }  }{ { x }+\cos { x }  }  } $$,then $$\displaystyle \lim _{ x\rightarrow \infty  } f(x)$$=
  • $$\dfrac{1}{2}$$
  • $$-1$$
  • $$0$$
  • $$1$$
lf $$ \mathrm{f}(\mathrm{x})=0$$ has a repeated root $$ \alpha$$, then another equation having $$\alpha$$ as root, is 
  • $$\mathrm{f}(2\mathrm{x})=0$$
  • $$\mathrm{f}(3\mathrm{x})=0$$
  • $$\mathrm{f}^{'}(\mathrm{x})=0$$
  • $$\mathrm{f}^{''}(\mathrm{x})=0$$

$$\displaystyle \lim_{x\rightarrow \infty }(\sin\sqrt{x+1}-\sin\sqrt{x})=$$
  • 2
  • -2
  • 0
  • None of these
$$\displaystyle \lim_{x\rightarrow \infty }x\displaystyle \cos\left(\frac{\pi}{8x}\right)\sin\left(\frac{\pi}{8x}\right)=$$
  • $$\displaystyle \pi$$
  • $$\displaystyle \frac{\pi}{2}$$
  • $$\dfrac{\pi}{8}$$
  • $$\displaystyle \frac{\pi}{4}$$

 $$\displaystyle \lim_{x\rightarrow\infty}\frac{\sin^{4}x-\sin^{2}x+1}{\cos^{4}x-\cos^{2}x+1}$$ is equal to
  • $$0$$
  • $$1$$
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{2}$$

$$\displaystyle \lim_{x\rightarrow \displaystyle \frac{\pi }{2}}\displaystyle \frac{1-\sin\theta}{\cos\theta\left(\dfrac{\pi}{2}-{\theta}\right)}=$$
  • $$1$$
  • $$-1$$
  • $$-\displaystyle \frac{1}{2}$$
  • $$\displaystyle \frac{1}{2}$$

$$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})$$
  • $$1$$
  • $$0$$
  • $$2$$
  • does not exist
Evaluate: $$\displaystyle \underset{x\rightarrow 0}{\lim}\ \ \frac{\sin3x^{2}}{\cos(2x^{2}-x)}$$
  • $$0$$
  • $$-1$$
  • $$4$$
  • $$-6$$
$$\displaystyle \lim_{x\rightarrow \infty }\frac{2x+7\sin x}{4x+3\cos x}=$$
  • $$1$$
  • $$-1$$
  • $$\dfrac{1}{2}$$
  • $$-\dfrac{1}{2}$$

$$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}=$$
  • $${e}$$
  • $$e^{2}$$
  • $$-1$$
  • 1
The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( \pi \cos ^{ 2 }{ x }  \right)  }  }{ { x }^{ 2 } }  } $$ is
  • $$-\pi$$
  • $$\displaystyle \frac { \pi  }{ 2 } $$
  • $$\pi$$
  • $$\displaystyle \frac { 3\pi  }{ 2 } $$
$$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$$ exists and finite then $$\mathrm{a}=$$
  • $$2$$
  • $$-2$$
  • $$\displaystyle \frac{2}{3}$$
  • $$\displaystyle \frac{-2}{3}$$

$$\displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}=$$
  • $$\mathrm{e}$$
  • $$\mathrm{e}^{4}$$
  • $$\mathrm{e}^{-1}$$
  • $$\mathrm{e}^{-4}$$
lf $$f(x)=\displaystyle \frac{x}{\sqrt{1-x^{2}}},g(x)=\frac{x}{\sqrt{1+x^{2}}}$$, then $$\displaystyle \frac{d}{dx}(fog (x))=$$
  • $$1$$
  • $$0$$
  • $$-1$$
  • $$2$$
The integer $$n$$ for which $$\displaystyle \lim_{x\rightarrow 0}\frac{(\cos x-1)(\cos x-e^{x})}{x^{n}}$$ is finite non zero number is
  • $$1$$
  • $$2$$
  • $$3$$
  • $$4$$
The right-hand limit of the function $$\sec{x}$$ at $$\displaystyle x=-\frac { \pi  }{ 2 } $$ is
  • $$-\infty$$
  • $$-1$$
  • $$0$$
  • $$\infty$$
$$\displaystyle \lim_{x\rightarrow 1}(2-x)^{\displaystyle \tan( \frac{\pi x}{2})}=$$
  • $$e^{\displaystyle \frac{1}{\pi}}$$
  • $$e^{\displaystyle \frac{2}{\pi}}$$
  • $$-e^{\displaystyle \frac{2}{\pi}}$$
  • $$\mathrm{e}$$
$$\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}\cos ^{ -1 }{ \left( \frac { 1-x^{ 2 } }{ 1+x^{ 2 } }  \right)  } =$$
  • 0
  • 1
  • 2
  • does not exist
If $$f(x)=(ax+b)\cos x + (cx+d)\sin x$$ and $$f^{'}(x)=x \cos x$$, for all values of $$x\in R$$, then $$a,b,c,d$$ are given by
  • $$a = b = c = d$$
  • $$0, 1, -1, 0$$
  • $$1, 0, -1, 0$$
  • $$0, 1, 1, 0$$
If $$\displaystyle \lim_{x\to0}{\displaystyle \frac{x^n - \sin^nx}{x - \sin^nx}}$$ is non-zero finite, then $$n$$ must be equal
  • 4
  • 1
  • 2
  • 3
Assertion (A): $$\mathrm{f}(\mathrm{x})=\sin(\pi[x])$$ is differentiable every where $$[\ ]$$ is greatest integer function

Reason (R): lf $$\mathrm{x}=\mathrm{n}\pi\Rightarrow $$ $$\sin x$$ $$=0\ \forall \ \mathrm{n}\in \mathrm{Z}$$ then
  • Both (A) and (R) are true and R is correct explanation for A
  • Both (A) and (R) are true and R is not correct explanation for A
  • (A) is true (R) is false
  • (A) is false (R) is true
$$\displaystyle \lim_{x\to1}{\displaystyle \frac{1-x^2}{\sin 2\pi x}}$$ is equal to
  • $$\displaystyle \frac{1}{2\pi}$$
  • $$\displaystyle \frac{-1}{\pi}$$
  • $$\displaystyle \frac{-2}{\pi}$$
  • None of these
$$\displaystyle\lim_{x \rightarrow \infty}(1^x + 2^x + 3^x+.........+n^x)^{1/x}$$ is
  • $$\ln (n!)$$
  • $$n$$
  • $$n!$$
  • $$0$$
  • Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
  • Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
  • Assertion is correct but Reason is incorrect
  • Assertion is incorrect but Reason is correct
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