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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 3
lf $$f(x)=\displaystyle \begin{cases}\dfrac{a^{2[x]+\{x\}}-1}{2[x]+\{x\}};x\neq 0 \\ \log a;x=0 \end{cases}$$ where $$[.\ ]$$ and $$\{.\ \}$$ denote integral and fractional part respectively, then
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$$f(x)$$ is continuous at $$x=0$$
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$$f(x)$$ is discontinuous at $$x=0$$
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$$f(x)$$ is continuous $$\forall x\in R$$
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$$f(x)$$ is differentiable at $$x=0$$
Explanation
Given definition of $$f(x)$$ can be written as
$$f(x)=\displaystyle \begin{cases}\dfrac{a^{[x]+x}}{[x]+x};x\neq 0 \\ \log a;x=0 \end{cases}$$ ($$\because \{x\}=x-[x]$$)
To check continuity of $$f(x)$$ at $$x=0$$
$$f(0^{+})=f(0^{-})=f(0)$$
$$f(0^{+})=\lim_{h\rightarrow 0}\dfrac{a^{h+[h]}}{\left [ h \right ]+h}$$
$$=\lim_{h\rightarrow 0}\dfrac{a^{h}-1}{h}=\log(a)$$
$$f(0^-)=\lim_{h\rightarrow 0}\dfrac{a^{[-h] -h}}{[-h]-h}$$
$$=\lim_{h\rightarrow 0}\dfrac{a^{-1-h}-1}{-1-h}$$
$$=1-\dfrac{1}{a}$$
$$f(0^{+})\neq f(0^{-})$$
So, $$f(x)$$ is discontinuous at $$x=0$$
lf $$\displaystyle { f }({ x })=\sqrt { \frac { { x }-\sin ^{ 2 }{ x } }{ { x }+\cos { x } } } $$,then $$\displaystyle \lim _{ x\rightarrow \infty } f(x)$$=
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$$\dfrac{1}{2}$$
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$$-1$$
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$$0$$
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$$1$$
Explanation
$$\displaystyle \lim _{ x\rightarrow \infty } f(x)$$
$$\displaystyle = \lim _{ x\rightarrow \infty }\sqrt { \dfrac { { x }-\sin ^{ 2 }{ x } }{ { x }+\cos { x } } }=\lim _{ x\rightarrow \infty }\sqrt { \dfrac { 1-\dfrac{\sin^2x}{x}}{ 1+\dfrac{\cos x}{x} }}=\sqrt{\dfrac{1-0}{1-0}}=1$$
The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( \pi \cos ^{ 2 }{ x } \right) } }{ { x }^{ 2 } } } $$ is
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$$-\pi$$
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$$\displaystyle \frac { \pi }{ 2 } $$
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$$\pi$$
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$$\displaystyle \frac { 3\pi }{ 2 } $$
Explanation
$$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( \pi \cos ^{ 2 }{ x } \right) } }{ { x }^{ 2 } } }=\lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi (1-\sin ^{ 2 }{ x }) \right] } }{ { x }^{ 2 } } } $$
$$\quad =\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi-\pi\sin ^{ 2 }{ x } \right] } }{ { x }^{ 2 } } }=\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi\sin ^{ 2 }{ x } \right] } }{ { x }^{ 2 } } }, [\because \sin(\pi-\theta)=\sin\theta]$$
$$\quad \displaystyle =\pi\lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi\sin ^{ 2 }{ x } \right] } }{ \pi\sin^2x } }\left( \frac{\sin x}{x}\right)^2 =\pi\cdot 1\cdot 1=\pi$$
Let $$f(x)=\cos2x.\cot\left (\displaystyle \frac{\pi }{4}-x \right )$$ If $$f$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$ then the value of $$f(\displaystyle \frac{\pi}{4})$$ is equal to
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$$2$$
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$$-2$$
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$$\displaystyle \frac{-1}{2}$$
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$$\displaystyle \frac{1}{2}$$
Explanation
$$f(x)=cos2x.cot\left ( \frac{\pi }{4}-x \right )$$
Given $$f(x)$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$
$$\displaystyle \lim _{ x\rightarrow \frac { \pi }{ 4 } } f(x)=f(\frac { \pi }{ 4 } )$$
$$\displaystyle=\lim _{ x\rightarrow \frac { \pi }{ 4 } } \frac { \cos { 2x } }{ \tan { (\frac { \pi }{ 4 } -x) } } $$
It is of the form $$\displaystyle \frac{0}{0}$$ , so applying L-Hospital's rule
$$\displaystyle=\lim _{ x\rightarrow \frac { \pi }{ 4 } } \frac { -2\sin { 2x } }{ -\sec ^{ 2 }{ (\frac { \pi }{ 4 } -x) } } $$
$$=2$$
$$\displaystyle \lim_{x\rightarrow \infty }x\displaystyle \cos\left(\frac{\pi}{8x}\right)\sin\left(\frac{\pi}{8x}\right)=$$
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$$\displaystyle \pi$$
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$$\displaystyle \frac{\pi}{2}$$
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$$\dfrac{\pi}{8}$$
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$$\displaystyle \frac{\pi}{4}$$
Explanation
As $$ x \rightarrow \infty , \cos \left (\dfrac{\pi}{8x}\right) \rightarrow 1 $$
Looking at the rest of the part,
$$\underset{ x \rightarrow \infty}{\lim} \dfrac{\pi}{8x} \rightarrow 0 $$
$$\underset{ x \rightarrow \infty}{\lim} x.\sin (\dfrac{\pi}{8x}) $$
$$=\underset{ x \rightarrow \infty}{\lim} \dfrac{\sin (\dfrac{\pi}{8x})}{\dfrac{1}{x}} $$
We multiply and divide by $$ \dfrac{\pi}{8} $$
$$=\underset{ x \rightarrow \infty}{\lim} , \dfrac{\sin \left (\dfrac{\pi}{8x}\right)}{\dfrac{\pi}{8x}} \times \dfrac{\pi}{8x} $$
Now using the property of a limit,
$$\underset{ y \rightarrow 0}{\lim} , \dfrac{\sin y}{y} $$ $$= 1$$
We get, applying the limit,
$$= 1 \times $$ $$ \dfrac{\pi}{8} $$
$$=$$ $$ \dfrac{\pi}{8} $$
$$\displaystyle \lim_{x\rightarrow \infty }(\sin\sqrt{x+1}-\sin\sqrt{x})=$$
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2
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-2
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0
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None of these
Explanation
$$\displaystyle \lim _{ x\rightarrow \infty }{ \left( \sin { \sqrt { x+1 } } -\sin { \sqrt { x } } \right) } $$
$$=\displaystyle \lim _{ x\rightarrow \infty }{ 2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x } }{ 2 } \right) \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } \right) } } } $$
Now for
$$\displaystyle \lim _{ x\rightarrow \infty }{ { \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } \right) } } } $$
$$=\displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } \right) } }{ \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } } \times \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } } $$ (sandwich theorem)
$$=\displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } } $$
$$=\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow \infty }{ \cfrac { \sqrt { x+1 } -\sqrt { x } }{ \sqrt { x+1 } +\sqrt { x } } \cdot \cfrac { \sqrt { x+1 } +\sqrt { x } }{ 1 } } $$
$$=\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow \infty }{ \cdot \cfrac { 1 }{ \sqrt { x+1 } +\sqrt { x } } } =\cfrac { 1 }{ 2 } \times 0=0$$
Now, $$2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x } }{ 2 } \right) } $$ is always finite and lies between $$\left[ -2,2 \right] $$
$$\therefore \displaystyle \lim _{ x\rightarrow \infty }{ 2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x } }{ 2 } \right) \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x } }{ 2 } \right) } } } =finite\times 0$$
$$=0$$
$$\displaystyle \lim_{x\rightarrow\infty}\frac{\sin^{4}x-\sin^{2}x+1}{\cos^{4}x-\cos^{2}x+1}$$ is equal to
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$$0$$
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$$1$$
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{2}$$
Explanation
For all $$x$$,
$$ {sin}^{4} x = {({sin}^{2} x)}^{2} $$
$$= {(1 - {cos}^{2} x)}^{2} $$
$$= 1 - 2 {cos}^{2} x + {cos}^{4} x $$
Thus numerator $$= 1 - 2 {cos}^{2} x + {cos}^{4} x + {cos}^{2} x = 1 - {cos}^{2} x + {cos}^{4} x $$
Hence, for all x, numerator $$=$$ denominator.
Thus the limit $$= 1$$ for all $$x$$
$$f(x)=\left\{\begin{matrix}[x]+[-x], & \\ \lambda ,& \end{matrix}\right.\begin{matrix}x\neq 2 & \\ x=2& \end{matrix},$$ then f(x) is continuous at $$x=2$$ provided $$\lambda $$ is:
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$$-1$$
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$$0$$
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$$1$$
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$$2$$
Explanation
$$RHL=\lim_{x\rightarrow 2^{+}} [x]+[-x]$$
$$=\lim_{h \rightarrow 0}[2+h]+[-(2+h)]$$
$$= 2 - 3$$
$$\Rightarrow RHL= - 1$$
Now, $$LHL=\lim_{x\rightarrow 2^{-}} [x]+[-x]$$
$$=\lim_{h\rightarrow 0}[2-h]+[-(2-h)] $$
$$= 1 -2$$
$$\Rightarrow LHL= -1$$
Since, $$f(x)$$ is continuous at $$x=2$$
$$LHL=RHL=f(2)$$
$$\Rightarrow \lambda =-1$$
$$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})$$
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$$1$$
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$$0$$
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$$2$$
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does not exist
Explanation
$$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})=\lim_{x\rightarrow \infty }\frac{\sin(2^{x})}{2^x}=\frac{\mbox{Any finite value between -1 and 1}}{\infty} = 0$$
Let $$f : R\rightarrow R$$ be any function, Define
$$g:R\rightarrow R$$ by $$g(x)=|f(x)|\forall x$$, then
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$$g$$ is continuous if $$f$$ is not continuous
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$$g$$ is not continuous if $$f$$ is not continuous
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$$g$$ is continuous if $$f$$ is continuous
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$$g$$ is differentiable if $$f$$ is differentiable
Explanation
From the property of the modulus function, it is clear that it is continuous along the entire number line.
Thus, the function defined by $$g(x) = |f(x)|$$ is continuous if $$f(x)$$ is a continuous function.
$$\displaystyle \lim_{x\rightarrow \infty }\frac{2x+7\sin x}{4x+3\cos x}=$$
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$$1$$
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$$-1$$
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$$\dfrac{1}{2}$$
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$$-\dfrac{1}{2}$$
Explanation
We divide both the numerator and denominator by x.
Since, from the property of trigonometric functions we know that $$sin x$$ and $$cos x$$ can only have their values between $$-1$$ and $$1$$
Thus, the expression transforms to,
$$lim _{x \rightarrow \infty } \dfrac{2 + 7\dfrac{sin x}{x}}{4 + 3\dfrac{cos x}{x}} $$
Now applying the limit,
$$= \dfrac{2 +0}{4+0} $$
$$= \dfrac{1}{2} $$
Hence, option 'C' is correct.
The function $$y=3\sqrt{x}-|x-1|$$ is continuous at
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$$x<0$$
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$$x\geq 1$$
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$$0\leq x\leq 1$$
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$$x\geq 0$$
Explanation
Since, $$|x|$$ is a function that is continuous along the entire number line, therefore, $$|x-1|$$ is continuous for all $$x$$
Now, we know that a negative value cannot go inside the square root sign. So $$x >= 0$$
Thus the function $$f(x)$$ defined by $$ 3{x}^{0.5} - |x -1| $$ is continuous for all $$x >= 0$$
lf $$\mathrm{f}(\mathrm{x})=\left\{\begin{matrix}1+x &x\leq 1 \\ 3-ax^{2}& x>1\end{matrix}\right.$$ is continuous at $${x}=1$$ then $${a}=({a}>0)$$
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$$1$$
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$$2$$
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$$0$$
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$$3$$
Explanation
For the function to be continuous at $$x = 1$$
The LHL should be equal to RHL
LHL
$$\lim_{ x \rightarrow {1}^{-}} f(x) $$
$$= \lim_{ x \rightarrow {1}^{-}} 1 + x $$
$$= 2$$
RHL
$$\lim_{x \rightarrow {1}^{+}} f(x) $$
$$=\lim_{ x \rightarrow {1}^{+}} 3 - a{x}^{2} $$
$$= 3-a$$
Since, LHL = RHL
So, $$2 = 3-a$$
Hence, $$a = 1 $$
$$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}=$$
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$${e}$$
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$$e^{2}$$
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$$-1$$
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1
Explanation
$$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}$$
$$\log { k } =Lt_{ x\rightarrow 0^+ }\tan { x } \log { \sin { x } } $$
$$k={ e }^{ Lt_{ x\rightarrow 0^+ }\tan { x } \log { \sin { x } } }$$
$$ k={ e }^{ Lt_{ x\rightarrow 0^+ }\displaystyle \frac { \log { \sin { x } } }{ \cot { x } } }$$
It is of the form $$\displaystyle \frac{\infty}{\infty}$$, so applying L-Hospital's rule
$$k={ e }^{ Lt_{ x\rightarrow 0^+ }\displaystyle \frac { \cot { x } }{ -\csc ^{ 2 }{ x } } }$$
$$k={ e }^{ 0 }=1$$
The function $$\displaystyle \mathrm{f}({x})=\frac{1+\sin x-\cos x}{1-\sin x-\cos x}$$ is not defined at $${x}=0$$. The value of $$\mathrm{f}(\mathrm{0})$$ so that $$\mathrm{f}({x})$$ is continuous at $${x}=0$$ is
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$$1$$
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$$-1$$
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$$0$$
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$$2$$
Explanation
Since, on applying the limit, the function is of $$ \dfrac{0}{0} $$ form,
we apply the L-Hospital's Rule and differentiate both the numerator and the denominator individually.
Thus, the question transforms to,
$$\displaystyle\lim_{x \rightarrow 0} \dfrac{cos x +sin x}{sin x - cos x} $$
Now applying the limit, we get
$$\Rightarrow \dfrac{0+1}{0-1} = -1$$
Thus, for the function to be continuous at $$x = 0$$, $$f(0) = -1$$
lf the function $$\displaystyle \mathrm{f}({x})=\frac{e^{x^{2}}-\cos {x}}{x^{2}}$$ for $$x \neq 0$$ is continuous at $${x}=0$$ then $$\mathrm{f}(\mathrm{0})=$$
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$$\dfrac{1}{2}$$
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$$\dfrac{3}{2}$$
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$$2$$
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$$\dfrac{1}{3}$$
Explanation
$${ f }({ x })=\dfrac { e^{ x^{ 2 } }-{ c }{ o }{ s }{ x } }{ x^{ 2 } } $$
Given f(x) is continuous at $$x=0$$
$$\displaystyle \lim _{ x\rightarrow 0 } f(x)=f(0)$$
$$\displaystyle=\lim _{ x\rightarrow 0 } \dfrac { e^{ x^{ 2 } }-{ c }{ o }{ s }{ x } }{ x^{ 2 } } $$
It is of the form $$\displaystyle \dfrac{0}{0}$$ , so applying L-Hospital's rule
$$\displaystyle =\lim _{ x\rightarrow 0 } \dfrac { 2xe^{ x^{ 2 } }+\sin { x } }{ 2x } $$
$$\displaystyle \lim_{x\to 0} e^{x^2}+\lim_{x\to 0} \dfrac{\sin x}{2x}$$
$$\displaystyle =1+\dfrac{1}{2}=\dfrac{3}{2}$$
$$\displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}=$$
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$$\mathrm{e}$$
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$$\mathrm{e}^{4}$$
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$$\mathrm{e}^{-1}$$
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$$\mathrm{e}^{-4}$$
Explanation
Let, $$\text{L}= \displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}$$
Put $$y=\pi-x\Rightarrow x\to \pi\Leftrightarrow y\to 0$$
$$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1- 4 \tan \mathrm{(\pi-y)} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{(\pi-y)}}$$
$$\quad \quad = \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{-\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{y}}[\because \tan (\pi-y)=-\tan y]$$
Clearly form of the limit is $$1^{\infty}$$
$$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{\dfrac{1}{4\tan y}\times(-4)}=e^{-4}$$
$$\displaystyle \lim_{n\rightarrow \infty }(\pi n)^{2/n}=$$
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0
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1
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2
0%
3
Explanation
Let $$l = \displaystyle \lim_{n\rightarrow \infty }(\pi n)^{2/n}$$
Take log both sides,
$$\displaystyle \log l = 2\lim_{n\to \infty}\frac{\log(\pi n)}{n}=2\lim_{n\to \infty}\frac{\log(n)+\log(\pi)}{n}$$
Clearly form of the limit is $$\dfrac{\infty}{\infty}$$
Applying L'Hospital's rule,
$$\Rightarrow \displaystyle \log l =2\lim_{n\to \infty}\frac{1/n}{1} =0$$
$$\Rightarrow l =e^0=1$$
$$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} \displaystyle\frac { { x }^{ 5 }-32 }{ x-2 } , & x\neq 2 \end{matrix} \\ \begin{matrix} k, & x=2 \end{matrix} \end{cases}$$
is continuous at $$x=2$$, then the value of $$k$$ is
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$$10$$
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$$15$$
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$$35$$
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$$80$$
Explanation
Function $$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} \dfrac { { x }^{ 5 }-32 }{ x-2 } , & x\neq 2 \end{matrix} \\ \begin{matrix} k, & x=2 \end{matrix} \end{cases}$$ is continuous at $$x=2$$
We know that $$\displaystyle \lim _{ x\rightarrow 2 }{ f\left( x \right) } =\lim _{ x\rightarrow 2 }{ \left( \frac { { x }^{ 5 }-32 }{ x-2 } \right) } =\lim _{ x\rightarrow 2 }{ \left( \frac { { x }^{ 5 }-{ 2 }^{ 5 } }{ x-2 } \right) } $$
We also know that $$\displaystyle \lim _{ x\rightarrow a }{ \left( \frac { { x }^{ n }-{ a }^{ n } }{ x-a } \right) } =n{ a }^{ n-1 }$$
Therefore, $$\displaystyle \lim _{ x\rightarrow 2 }{ f\left( x \right) } =5\times { 2 }^{ 4 }=80$$
Since $$f(2)=k$$ and $$f(x)$$ is continuous at $$x=2,$$ therefore $$\displaystyle \lim _{ x\rightarrow 2 }{ f\left( x \right) } =f\left( 2 \right) \Rightarrow k=80.$$
If $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\mathrm{a}^{2}\cos^{2}\mathrm{x}+\mathrm{b}^{2}\sin^{2}\mathrm{x},\mathrm{x}\leq 0\\\mathrm{e}^{\mathrm{a}\mathrm{x}+\mathrm{b}},\mathrm{x}>0\end{array}\right.$$ is continuous at $$\mathrm{x}=0$$ then
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$$2\log|\mathrm{a}|=\mathrm{b}$$
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$$2\log|\mathrm{b}|=\mathrm{e}$$
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$$\log a=2\log|\mathrm{b}|$$
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$$\mathrm{a}=\mathrm{b}$$
Explanation
Since $$f(x)$$ is continuous at $$x = 0$$
LHL = RHL
$$\displaystyle\lim _{ x \rightarrow {0}^{-}} f(x) = \displaystyle\lim _{x \rightarrow {0}^{+}} f(x) $$
$$\displaystyle\lim _{ x \rightarrow {0}^{-}} {a}^{2}{cos}^{2} x + {b}^{2}{sin}^{2}x =\displaystyle \lim_{ x \rightarrow {0}^{+} } {e}^{ax+b} $$
$$\Rightarrow a^{2} = {e}^{b} $$
$$\Rightarrow 2log|a|=b$$
$$\displaystyle \lim_{n\rightarrow \infty }\left(\displaystyle \frac{e^{n}}{\pi}\right)^{1/n}=$$
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0
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1
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$$\mathrm{e}^{2}$$
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$$\mathrm{e}$$
Explanation
Let $$y$$ = $$\displaystyle\lim _{ n\rightarrow \infty }{ { \left(\displaystyle \frac { e^{ n } }{ \pi } \right) }^{\displaystyle \frac { 1 }{ n } } }$$
Applying logarithm on both sides
$$\log { y } $$ = $$\displaystyle\lim _{ n\rightarrow \infty }{ \log { { \left(\displaystyle \frac { e^{ n } }{ \pi } \right) }^{ {\displaystyle \frac { 1 }{ n } } } } } =\quad \lim _{ n\rightarrow \infty }{\displaystyle \frac { 1 }{ n } \log { { \left( \displaystyle\frac { e^{ n } }{ \pi } \right) } } }$$
$$\log { y } $$ = $$\lim _{ n\rightarrow \infty }{\displaystyle \frac { 1 }{ n } \left( n\log e-\log { \pi } \right) } $$
By L'Hospital Rule
$$\log y=\log e$$
$$ \Rightarrow \quad y = e$$
$$\therefore \quad \displaystyle\lim _{ n\rightarrow \infty }{ { \left(\displaystyle \frac { e^{ n } }{ \pi } \right) }^{\displaystyle \frac { 1 }{ n } } } = e$$
$$\displaystyle \lim_{x\rightarrow 1}(2-x)^{\displaystyle \tan( \frac{\pi x}{2})}=$$
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$$e^{\displaystyle \frac{1}{\pi}}$$
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$$e^{\displaystyle \frac{2}{\pi}}$$
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$$-e^{\displaystyle \frac{2}{\pi}}$$
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$$\mathrm{e}$$
Explanation
Let $$k=\lim _{ x\rightarrow 1 } (2-x)^{ \tan \left(\dfrac { \pi x }{ 2 } \right) }$$
$$\log { k } =\lim _{ x\rightarrow 1 } tan\left( \dfrac { \pi x }{ 2 } \right) \log { \left( 2-x \right) } $$
$$\displaystyle k={ e }^{ \lim _{ x\rightarrow 1 } tan\left( \displaystyle\dfrac { \pi x }{ 2 } \right) \log { \left( 2-x \right) } }$$
$$\displaystyle k={ e }^{ \lim _{ x\rightarrow 1 } \displaystyle \dfrac { \log { \left( 2-x \right) } }{ \cot { \left( \dfrac { \pi x }{ 2 } \right) } } }$$
$$k={ e }^{ \lim _{ x\rightarrow 1 } \displaystyle \dfrac { -1 }{ -(2-x)\csc ^{ 2 }{ \left( \dfrac { \pi x }{ 2 } \right) \left( \dfrac { \pi }{ 2 } \right) } } }$$
$$\Rightarrow k={e}^{\displaystyle\dfrac{2}{\pi}}$$
lf the function defined by $$\mathrm{f}({x})=\displaystyle \frac{\sin 3(x-p)}{\sin 2(x-p)}$$ for $${x}\neq {p}$$ is continuous at $${x}={p}$$ then $$\mathrm{f}({p})=$$
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$$\dfrac{3}{2}$$
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$$\dfrac{2}{3}$$
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$$6$$
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$$\dfrac{1}{6}$$
Explanation
Since, it is given the function is continuous at $$x = p,$$we calculate
$$\lim _{ x \rightarrow p } \dfrac{sin 3(x-p)}{sin 2(x-p)} $$
$$= lim _{ x \rightarrow p } \dfrac{sin 3(x-p)}{3(x-p)} \times \dfrac{2(x-p)}{sin 2(x-p)} \times \dfrac{3}{2} $$
Using the property of limits,
$$f(p) = 1 \times 1 \times \dfrac{3}{2} $$
If the function $$\displaystyle \mathrm{f}({x})=\begin{cases}\dfrac{2^{x+2}-16}{4^{x}-16}&& for {x}\neq 2\\ \mathrm{A} && x =2\end{cases}$$ is continuous at $$x =2$$, then $$\mathrm{A}=$$
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$$2$$
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$$\dfrac{1}{2}$$
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$$\dfrac{1}{4}$$
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$$0$$
Explanation
Since, the function is continuous at $$x = 2$$
$$A = \displaystyle\lim _{x \rightarrow 2 } f(x) $$
Since, on applying the limit this is of the $$ \dfrac{0}{0} $$ form, we apply L-Hospital's Rule
$$\Rightarrow\displaystyle\lim_{ x \rightarrow 2}\dfrac{4.{2}^{x}log 2}{{4}^{x} log 4} $$
Now applying the limit,
$$A = \dfrac{1}{2} $$
$$f(x)=x\left[3-\displaystyle \log\left(\frac{\sin {x}}{x}\right)\right]-2$$ to be continuous at $${x}=0$$, then $$\mathrm{f}({0})=$$
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$$0$$
0%
$$2$$
0%
$$-2$$
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$$3$$
Explanation
USing the property of limits, $$\underset{ x \rightarrow 0 }{Lim} \dfrac{sin x}{x} = 1 $$
Thus, applying the limit to the expression,
$$\underset{ x \rightarrow 0 }{Lim} \ \ x\left[3 - log\left(\dfrac{sin x}{x}\right)\right] -2 $$
$$= 0(3 - log 1) - 2$$
$$= 0 - 2$$
$$ = -2 $$
Let $$\displaystyle \mathrm{f}({x})=\begin{cases} \dfrac{(e^{kx}-1).\sin kx}{x^{2}} & for \ {x}\neq 0 \\ 4 & for \ {x} =0\end{cases}$$ is continuous at $${x}=0$$ then $${k}=$$
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$$\pm 1$$
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$$\pm 2$$
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0
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$$\pm 3$$
Explanation
For a given function to be continuous at $$x=0$$
$$\displaystyle \lim_{x\to 0} = f(0)\Rightarrow \lim_{x\to 0}\frac{(e^{kx}-1).\sin kx}{x^2} =4$$
$$\Rightarrow \displaystyle k^2. \lim_{x\to 0}\frac{e^{kx}-1}{kx}.\frac{\sin kx}{kx} =4$$
$$\Rightarrow k^2 = 4\Rightarrow k = \pm 2$$
lf $$f(x)=
\left\{\begin{matrix} (1+|\sin x|)^{\displaystyle \frac{a}{|\sin x|}}&-\displaystyle \frac{\pi}{6}<x<0\\ b&x=0 \\ e^{\displaystyle \frac{\tan 2x}{\tan 3x}} &0<x<\displaystyle \frac{\pi}{6}\end{matrix}\right.$$ is
continuous at $$\mathrm{x}=0$$ then
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$$a=e^{2/3},b=\dfrac{2}{3}$$
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$$a=\dfrac{2}{3},b=e^{2/3}$$
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$$a=\dfrac{1}{3},b=e^{1/3}$$
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$$a=e^{1/3},b=e^{1/3}$$
Explanation
For the function $$f(x)$$ to be continuous at $$x = 0$$
$$\lim_{ x \rightarrow {0}^{-}} f(x) = f(0) = \lim_{ x \rightarrow {0}^{+}} f(x) $$
LHL
$$\lim_{ x \rightarrow {0}^{-}} exp( |\sin x| \times \dfrac{a}{|\sin x|}) = e^a$$
$$f(0) = b$$
RHL
$$\lim_{ x \rightarrow {0}^{+} } exp(\dfrac{\tan 2x}{\tan 3x}) $$
$$= \lim_{ x \rightarrow {0}^{+}} exp(\dfrac{\tan 2x}{2x\ tan3x} \times {3x} \times \dfrac{2}{3}) = e ^{\frac{2}{3}} $$
Thus,
$$exp(a) = b = exp( \dfrac{2}{3} )$$
Hence, $$a = \dfrac{2}{3} $$ and $$b = exp( \dfrac{2}{3} )$$
lf the function $$f(x)=\begin{cases}\dfrac{k\cos x}{\pi-2x}, & x\neq\dfrac{\pi}{2}\\ 3 & at x=\dfrac{\pi}{2}\end{cases}$$is continuous at $$\displaystyle {x}=\dfrac{\pi}{2}$$ then $${k}=$$
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0%
$$2$$
0%
$$4$$
0%
$$6$$
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$$8$$
Explanation
On applying the limits, since the function is of the form $$ \dfrac{0}{0} $$, We apply L-Hospital's rule,
$$\displaystyle\lim_{ x \rightarrow \dfrac{\pi}{2}} \dfrac{-ksin x}{-2} $$
Since, the function has to be continuous at $$x = \dfrac{\pi}{2} $$
$$ \Rightarrow\dfrac{k}{2} = 3$$
$$\Rightarrow k = 6$$
The function $$f(x)=\begin{cases} 0,& \text{x is irrational }\\ 1,& \text{x is rational }\end{cases}$$ is
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continuous at $$x=1$$
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discontinuous only at $$0$$
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discontinuous only at 0,1
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discontinuous everywhere
Explanation
From the number theory, we already know that between any 2 rational numbers there exists an irrational number and vice versa.
Thus, for the function f(x) as defined above it will take both the values 0 and 1 in the neighbourhood of every point x = a.
Thus, function can never be continuous.
The function $$\displaystyle \mathrm{f}(\mathrm{x})=\frac{\cos x-\sin x}{\cos 2x}$$ is not defined at $$x=\displaystyle \frac{\pi}{4}$$ The value of $$f\left(\displaystyle \frac{\pi}{4}\right)$$ so that $$\mathrm{f}(\mathrm{x})$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$ is
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$$\displaystyle \frac{1}{\sqrt{2}}$$
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$$\sqrt{2}$$
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$$-\sqrt{2}$$
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$$1$$
Explanation
It is given that the function is not defined at x = $$ \dfrac{\pi}{4} $$
So, $$f\left( \dfrac{\pi}{4} \right)$$ =$$\underset{x \rightarrow \dfrac{\pi}{4}}{Lim}\ \dfrac{cos x - sin x}{cos 2x} $$
Since, this is of the $$ \dfrac{0}{0} $$ form, we apply L-Hospital's Rule,
$$\underset{x \rightarrow \dfrac{\pi}{4}}{Lim} \ \ \dfrac{-sin x - cos x}{-2sin 2x} $$
Now, applying the limit,
$$f\left( \dfrac{\pi}{4}\right) $$ = $$ \dfrac{1}{{2}^{0.5}} $$
The value of $$f(0)$$ so that the function
$$f(x)=\dfrac{\displaystyle \log\left(1+\dfrac{x}{a}\right)-\log\left(\begin{array}{l}1-\dfrac{x}{b}\end{array}\right)}{x}, (x\neq 0)$$ is continuous at $$x = 0$$ is :
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$$\displaystyle \frac{a+b}{ab}$$
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$$\displaystyle \frac{a-b}{ab}$$
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$$\displaystyle \frac{ab}{a+b}$$
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$$\displaystyle \frac{ab}{a-b}$$
Explanation
$$\displaystyle\lim_{x\to0}f(x) = \lim_{x\to0}\dfrac{\log \left(1+\dfrac{x}{a}\right) - \log \left(1-\dfrac{x}{b}\right)}{x}$$
$$\displaystyle\because \lim_{x\to0}\dfrac{\log (1+x)}{x} = 1$$
Using this in the above limit,
$$\displaystyle\lim_{x\to0}\dfrac{\dfrac{x}{a}\cdot\dfrac{\log \left(1+\dfrac{x}{a}\right)}{\dfrac{x}{a}} + \dfrac{x}{b}\cdot\dfrac{\log \left(1-\dfrac{x}{b}\right)}{\dfrac{-x}{b}}}{x}$$
$$=\dfrac {x\left (\dfrac1a+\dfrac 1b\right )}{x}$$
$$= \dfrac{1}{a} + \dfrac{1}{b} $$
$$= \dfrac{a+b}{ab}$$
The value of $$\mathrm{f}(\mathrm{0})$$ for the function $$\mathrm{f}({x})=\displaystyle \frac{2-\sqrt{(x+4)}}{\sin 2x}, x\ne 0$$ is continuous at $${x}=0$$ is
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$$\displaystyle \frac{1}{8}$$
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$$\displaystyle \frac{1}{4}$$
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$$\displaystyle \frac{-1}{8}$$
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$$-\displaystyle \frac{1}{4}$$
Explanation
Since, on applying the limits, the function is of $$ \dfrac{0}{0} $$ form,
We apply L-Hospital's Rule,
$$f(0) = \lim_{ x \rightarrow 0} \dfrac{0 - \dfrac{1}{2{(x +4)}^{0.5}}}{2 cos 2x} $$
Now applying the limit, we get
$$f(0) = \dfrac{\dfrac{-1}{2 \times 2}}{2} = \dfrac{-1}{8} $$
lf $$f(x)=\left\{\begin{array}{l}\dfrac{1-\sqrt{2}\sin x}{\pi-4x} x\neq\frac{\pi}{4}\\a,x=\frac{\pi}{4}\end{array}\right.$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$ then $$a=$$
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$$4$$
0%
$$2$$
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$$1$$
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$$\dfrac{1}{4}$$
Explanation
Since it is continuous at $$x=\dfrac{\pi}{4}$$
Therefore, $$\displaystyle f\left( \frac { \pi }{ 4 } \right) =\lim _{ x\rightarrow \frac { \pi }{ 4 } }{ \frac { { 1-\sqrt { 2 } \sin x } }{ \pi -4x } } $$
It is of the $$\dfrac{0}{0}$$
By
L-Hospital's rule
$$\displaystyle a=\lim _{ x\rightarrow \frac { \pi }{ 4 } }{ \frac { { -\sqrt { 2 } \cos { x } } }{ -4 } } $$
$$=\dfrac { 1 }{ 4 } $$
If $$ \displaystyle f(x)=\left\{\begin{array}{ll}\dfrac{\sqrt{1+kx}-\sqrt{1-x}}{x} & \mathrm{f}\mathrm{o}\mathrm{r}-\mathrm{l} \leq x<0\\2x^{2}+3x-2 & \mathrm{f}\mathrm{o}\mathrm{r} 0\leq x\leq 1\end{array}\right.$$ is continuous at $$x = 0$$ then $$k$$ is:
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$$-4$$
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$$-3$$
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$$-5$$
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$$-1$$
Explanation
Given, $$f(x)$$ is continuous at $$x=0$$
$$\displaystyle \lim _{ x\rightarrow 0 }{ f(x) } =f(0)$$
Here $$f(0)=-2$$
$$LHL =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { \sqrt { 1+kx } -\sqrt { 1-x } }{ x } } $$
$$ =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { \sqrt { 1+kx } -\sqrt { 1-x } }{ x }\times \dfrac { \sqrt { 1+kx } +\sqrt { 1-x } }{ \sqrt { 1+kx } +\sqrt { 1-x } } } $$
$$=\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { (k+1)x }{ x }\times \dfrac { 1 }{ \sqrt { 1+kx } +\sqrt { 1-x } } }$$
$$= \dfrac { (k+1) }{ 2 } $$
Since, $$f(x)$$ is continuous at $$x=0$$
So, $$LHL=f(0)$$
$$\Rightarrow \dfrac{k+1}{2}=-2$$
$$\Rightarrow k=-5$$
Hence option $$'C'$$ is the answer.
$$f(x)=\begin{cases}\dfrac{x^{3}+x^{2}-16x+20}{(x-2)^{2}} & if\ x\neq 2\\ k & if\ x=2\end{cases}$$
$$\mathrm{f}({x})$$ is continuous at $${x}=2$$ then $$f(2)=$$
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$${k}=3$$
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$${k}=5$$
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$${k}=7$$
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$${k}=9$$
Explanation
Since, $$f$$ is continuous at $$x=2$$
Therefore, $$\lim _{ x\rightarrow 2 }{ \dfrac { x^{ 3 }+x^{ 2 }-16x+20 }{ (x-2)^{ 2 } } } =f\left( 2 \right) $$
it is $$\dfrac{0}{0}$$ form using
L-Hospital's rule
$$ \Rightarrow \lim _{ x\rightarrow 2 }{ \dfrac { 3x^{ 2 }+2x-16 }{ 2(x-2) } } =k$$
It is $$\dfrac{0}{0}$$ form using L-Hospital's rule
$$ \lim _{ x\rightarrow 2 }{ \dfrac { 6x+2 }{ 2 } } =k$$
Therefore, $$k=7$$
$$f(x)=\dfrac{p+q^{\frac{1}{x}}}{r+s^{\frac{1}{x}}},
s<1, q<1,r\neq 0, \mathrm{f}(\mathrm{0})=1$$, is left continuous at $$x =0$$ then
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$${p}=0$$
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$${p}={r}$$
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$${p}={q}$$
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$$p\neq q$$
Explanation
$$f\left( x \right) =\cfrac { p+{ q }^{ { 1 }/{ x } } }{ r+{ s }^{ { 1 }/{ x } } } $$
If $$f\left( x \right) $$ is right continuous
$$\Rightarrow \displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } =f\left( 0 \right) =1$$
$$\Rightarrow \displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { p+{ q }^{ { 1 }/{ x } } }{ r+{ s }^{ { 1 }/{ x } } } } =1$$
as $$x\rightarrow { 0 }^{ + };\cfrac { 1 }{ x } \rightarrow +\infty $$
Given $$s<1$$
$$\Rightarrow s=\cfrac { 1 }{ a } ,a>1$$
Similarly, $$q=\cfrac { 1 }{ b } ,b>1$$
$$=\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { p+{ \left( \cfrac { 1 }{ b } \right) }^{ \cfrac { 1 }{ x } } }{ r+{ \left( \cfrac { 1 }{ a } \right) }^{ \cfrac { 1 }{ x } } } } $$
$$=\cfrac { p+\cfrac { 1 }{ { b }^{ \infty } } }{ r+\cfrac { 1 }{ { a }^{ \infty } } } =\cfrac { p+0 }{ r+0 } =\cfrac { p }{ r } =1$$
$$\Rightarrow p=r$$
Question should be changed to right continuous instead of left continuous.
lf $$f$$ : $$R\rightarrow R$$ is defined by $$f(x)=\left\{\begin{array}{ll}\displaystyle \frac{\cos 3x-\cos x}{x^{2}} & \mathrm{f}\mathrm{o}\mathrm{r} x\neq 0\\\lambda & \mathrm{f}\mathrm{o}\mathrm{r} x=0\end{array}\right.$$and if $$\mathrm{f}$$ is continuous at $${x}=0$$ then $$\lambda=$$
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$$-2$$
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$$-4$$
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$$-6$$
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$$-8$$
lf $$f(x)=\displaystyle \frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}+e^{-1/x}} x\neq 0$$ is continuous at $${x}=0$$, then $${f}(\mathrm{0})=$$
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$$1$$
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$$2$$
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$$0$$
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$$3$$
Explanation
Given $$f(x)$$ is continuous at $$x=0$$
$$LHL=RHL=f(0)$$
$$f(x)=\displaystyle \frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}+e^{-1/x}} $$
$$f(x)=\displaystyle \frac { x(1-e^{ -2/x }) }{ 1+e^{ -2/x } } $$
So, $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x(1-e^{ -2/x }) }{ 1+e^{ -2/x } } } =0$$
$$\Rightarrow f(0)=0$$
If $$f$$ : $$R\rightarrow R$$ is defined by $$f(x)=\left\{\begin{array}{ll}\dfrac{x+2}{x^{2}+3x+2} & x\in R-\{-1,-2\}\\-1 & x=-2\\0 & x=-1\end{array}\right.$$then $$f$$ is continuous on the set:
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$$R$$
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$$R-\{-2\}$$
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$$R-\{-1\}$$
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$$R-\{-1,-2\}$$
Explanation
The function f(x) can be written as,
f(x) = $$ \dfrac{x + 2}{(x+2)(x+1)} $$
f(x) = $$ \dfrac{1}{x+1} $$
Now the only point of discontinuity can be x = -1 and x = -2 where the function changes
At,$$x = -1$$
the function is not defined hence, it is discontinuous.
At $$x = -2$$,
$$f(x) = -1$$
Thus it is continuous at x = -2
So the function is discontinuous at only 1 point ie at x = -1
$$\underset { x\rightarrow 0 }{ lim } \left( \dfrac { \left( 1+x \right) ^{ \dfrac { 1 }{ x } } }{ e } \right) ^{ \dfrac { 1 }{ sinx } }$$ is equal to
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$$\sqrt { e } $$
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e
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$$\dfrac { 1 }{ \sqrt { e } } $$
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1/e
$$f(x)=\displaystyle \frac{e^{1/x^{2}}}{e^{1/x^{2}}-1}$$ , $$x\neq 0$$, $$\mathrm{f}({0})=1$$, then $$\mathrm{f}$$ at $${x}=0$$ is:
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discontinuous
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left continuous
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right continuous
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both B and C
Explanation
Given: $$f(x)=\displaystyle \frac{e^{1/x^{2}}}{e^{1/x^{2}}-1}$$
Divide by $$e^{1/x^2}$$ to numerator and denominator
The function can be written in the form of,
$$\displaystyle\lim_{ x \rightarrow 0} \dfrac{1}{1 - {e}^{\frac{-1}{{x}^{2}}}} $$
Applying the limit,
$$\displaystyle f(0) = \lim_{ x \rightarrow 0} \dfrac{1}{1 - {e}^{\frac{-1}{{x}^{2}}}} = 1$$
Hence, the function is continuous at $$x = 0$$
So, the function is both left and right continuous.
$$\mathrm{A}$$ function $$\mathrm{f}(\mathrm{x})$$ is defined as
$$f(x)=\left\{ \begin{matrix} ax-b & x\leq 1 \\ 3x, & 1<x<2 \\ bx^{ 2 }-a & x\geq 2 \end{matrix} \right. $$ is continuous at
$$x=1, 2$$ then:
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$$\mathrm{a}=5,\ \mathrm{b}=2$$
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$$\mathrm{a}=6,\ \mathrm{b}=3$$
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$$\mathrm{a}=7,\ \mathrm{b}=4$$
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$$\mathrm{a}=8,\ \mathrm{b}=5$$
Explanation
Given $$f(x)$$ is continuous at $$x=1$$
$$\displaystyle \lim _{ x\rightarrow { 1 } } f(x)=f(1)$$
Now $$LHL=\displaystyle\lim _{ x\rightarrow { 1 }^{ - } } f(x)$$
$$=\displaystyle\lim _{ x\rightarrow { 1 }^{ - } } ax-b$$
$$LHL=a-b$$
And $$f(1)=3$$
$$\Rightarrow a-b=3$$ .....(i)
Given $$f(x)$$ is continuous at $$x=2. $$
$$\displaystyle \lim _{ x\rightarrow { 2 } } f(x)=f(2)$$
Now $$LHL=\displaystyle \lim _{ x\rightarrow { 2 }^{ - } } f(x)$$
$$=\displaystyle \lim _{ x\rightarrow { 2 }^{ - } } 3x$$
$$LHL=6$$
And $$f(2)=4b-a$$
$$\Rightarrow 4b-a=6$$ .....(ii)
Solving (i) and (ii)
$$\Rightarrow b=3, a=6$$
$$f(x)=\displaystyle \min\{x,\ x^{2}\}\forall x\in R$$. Then $$f(x)$$ is
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discontinuous at 0
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discontinuous at 1
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continuous on R
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continuous at 0, 1
Explanation
If we look at the function f(x),
We can divide the entire number line into 3 sections.
for x < 0 f(x) = x
for 0 < x < 1 f(x) = $$ {x}^{2} $$
for x > 1 f(x) = x
For checking the continuity, the doubtful points are x = 0 and x = 1
At x = 0
lim $$ x \rightarrow {0}^{-} f(x) =x= 0 $$
lim $$ x \rightarrow {0}^{+} f(x) =x^2= 0 $$
At x = 1
lim $$ x \rightarrow {1}^{-} f(x) = x^2=1 $$
lim $$ x \rightarrow {1}^{+} f(x) = x=1 $$
Hence, the function is continuous on the entire number line
$$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$$ exists and finite then $$\mathrm{a}=$$
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$$2$$
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$$-2$$
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$$\displaystyle \frac{2}{3}$$
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$$\displaystyle \frac{-2}{3}$$
Explanation
$$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$$
$$Lt_{ x\rightarrow 0 }\dfrac { \sin x(2\cos { x } +a) }{ x^{ 3 } } $$
$$=Lt_{ x\rightarrow 0 }\dfrac { (2\cos { x } +a) }{ x^{ 2 } } $$
As $$x\rightarrow 0$$, denominator tends to 0, so the numerator also tends to 0.
$$\Rightarrow Lt_{ x\rightarrow 0 } 2\cos { x } +a=0$$
$$\Rightarrow a=-2$$
Hence, option 'B' is correct.
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
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
$$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { (\cos { x } -1)\left( \cos { x } -{ e }^{ x } \right) }{ { x }^{ n } } } =\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { \left( \cos { x } -1 \right) }{ { x }^{ 2 } } \right] } \times \cfrac { { \cos { x } -{ e }^{ x } } }{ { x }^{ n-2 } } $$
$$=-\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { \cos { x } -{ e }^{ x } } }{ { x }^{ n-2 } } } =\cfrac { 0 }{ 0 } $$ form
$$=-\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { -\sin { x } -{ e }^{ x } } }{ (n-2){ x }^{ n-3 } } } =\cfrac { 1 }{ 2(n-2) } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { 1 } }{ { x }^{ n-3 } } } $$
$$\therefore$$ for the above limit to be finite $$n-3=0$$
$$\Rightarrow n=3$$
$$\displaystyle \lim_{x\rightarrow 1}\{1-x+[x+1]+[1-x]\}$$ , where $$[x]$$ denotes greatest integer function, is
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$$0$$
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$$1$$
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$$-1$$
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$$2$$
Explanation
Substitute $$x=1+t$$
L.H.S $$\lim_{t\rightarrow o^{-}} (-t+[2+t]+[-t])$$
$$=0+1+0=1$$
R.H.S $$\lim_{t\rightarrow o^{+}} (-t+[2+t]+[-t])$$
$$=0+2-1=1$$
L.H.S$$=$$R.H.S
Given that the function $$\mathrm{f}$$ is defined by $$f(x)=\left\{\begin{array}{l}2x-1,x>2\\k, x=2\\x^{2}-1,x<2\end{array}\right.$$is continuous at x =Then $${k}$$ is:
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$$3$$
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$$2$$
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$$1$$
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$$-3$$
Explanation
Given $$f(2)=k$$
$$RHL =\lim _{ x\rightarrow { 2 }^{ + } }2x-1$$
$$=\lim_{h\rightarrow 0}2(2+h)-1$$
$$=3$$
Given, f(x) is continuous at x=2.
$$LHL=RHL =f(2)$$
$$\Rightarrow k=3$$
lf the function $$\mathrm{f}({x})=\begin{cases}\dfrac{\sin 3x}{x} &(x\neq 0) \\ \dfrac{k}{2}&(x=0) \end{cases}$$
is continuous at $${x}=0$$, then $${k}$$ is:
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3
0%
6
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9
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2
Explanation
Since $$f(x)$$ is continuous at $$x = 0$$
LHL = RHL
$$\underset{x \rightarrow {0}^{-}}{Lim}\ f(x) = lim x \rightarrow {0}^{+} f(x) $$
$$\underset{x \rightarrow {0}^{-}}{Lim} \ \dfrac{sin 3x}{x} = lim x \rightarrow {0}^{+} \dfrac{k}{2} $$
$$\underset{ x \rightarrow {0}^{-}}{Lim} \ \dfrac{sin 3x}{3x} \times 3 = lim x \rightarrow {0}^{+} \dfrac{k}{2} $$
$$\Rightarrow 3 = \dfrac{k}{2} $$
Hence, $$k = 6$$
The right-hand limit of the function $$\sec{x}$$ at $$\displaystyle x=-\frac { \pi }{ 2 } $$ is
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$$-\infty$$
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$$-1$$
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$$0$$
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$$\infty$$
Explanation
Function $$\displaystyle f\left( x \right)=\sec { x } $$ and point $$\displaystyle x=-\left( \frac { \pi }{ 2 } \right) $$.
We know that right$$-$$hand limit of the function $$f(x)$$ at $$x=-\left( \frac { \pi }{ 2 } \right) $$ is $$\displaystyle \lim _{ x\rightarrow { \left[ -\left( \pi /2 \right) \right] }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { \left[ -\left( \pi /2 \right) \right] } }{ \sec { x } } .$$
Substituting $$\displaystyle x=h+\left( -\frac { \pi }{ 2 } \right) $$ and $$h\rightarrow 0,$$ we get
$$\displaystyle \lim _{ h\rightarrow 0 }{ \sec { \left[ h+\left( -\frac { \pi }{ 2 } \right) \right] } = } \lim _{ h\rightarrow 0 }{ \sec { \left( \frac { \pi }{ 2 } -h \right) } } =\lim _{ h\rightarrow 0 }{ \csc { h } } =\csc { 0 } =\infty $$
$$\displaystyle \left[ \because \sec { \left( -\theta \right) = } \sec { \theta } \right] $$
$$\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}\cos ^{ -1 }{ \left( \frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) } =$$
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0
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1
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2
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does not exist
Explanation
$$\lim_{x\rightarrow 0^{-}} \dfrac{-2}{\dfrac{2x}{(1+x^{2})}}\dfrac{sin^{-1}}{(1+x^{2})}\dfrac{(2x)}{1+x^{2}}=-2$$
$$\lim_{x\rightarrow 0^{+}} \dfrac{2}{(\dfrac{2x}{1+x^{2}})(1+x^{2})} sin^{-1}(\dfrac{2x}{1+x^{2}})=2$$
$$L.H.S\neq R.H.S$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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