MCQ Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 3

$f(x)=\displaystyle \begin{cases}\dfrac{a^{2[x]+\{x\}}-1}{2[x]+\{x\}};x\neq 0 \\ \log a;x=0 \end{cases}$ where

$[.\ ]$ and

$\{.\ \}$ denote integral and fractional part respectively, then

0%
$f(x)$ is continuous at $x=0$
0%
$f(x)$ is discontinuous at $x=0$
0%
$f(x)$ is continuous $\forall x\in R$
0%
$f(x)$ 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

$x=0$

$\displaystyle { f }({ x })=\sqrt { \frac { { x }-\sin ^{ 2 }{ x } }{ { x }+\cos { x } } }$ ,then

$\displaystyle \lim _{ x\rightarrow \infty } f(x)$ =

0%
$\dfrac{1}{2}$
0%
$-1$
0%
$0$
0%
$1$

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

0%
$-\pi$
0%
$\displaystyle \frac { \pi }{ 2 }$
0%
$\pi$
0%
$\displaystyle \frac { 3\pi }{ 2 }$

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

0%
$2$
0%
$-2$
0%
$\displaystyle \frac{-1}{2}$
0%
$\displaystyle \frac{1}{2}$

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

$\displaystyle=\lim _{ x\rightarrow \frac { \pi }{ 4 } } \frac { -2\sin { 2x } }{ -\sec ^{ 2 }{ (\frac { \pi }{ 4 } -x) } }$

$=2$

$\displaystyle \lim_{x\rightarrow \infty }x\displaystyle \cos\left(\frac{\pi}{8x}\right)\sin\left(\frac{\pi}{8x}\right)=$

0%
$\displaystyle \pi$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\dfrac{\pi}{8}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

$x \rightarrow \infty , \cos \left (\dfrac{\pi}{8x}\right) \rightarrow 1$
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
0%
-2
0%
0
0%
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

0%
$0$
0%
$1$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$

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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$-1$
0%
$0$
0%
$1$
0%
$2$

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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$1$
0%
$0$
0%
$2$
0%
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

0%
$g$ 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
0%
$g$ 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.

$\displaystyle \lim_{x\rightarrow \infty }\frac{2x+7\sin x}{4x+3\cos x}=$

0%
$1$
0%
$-1$
0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$

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

0%
$x<0$
0%
$x\geq 1$
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$0\leq x\leq 1$
0%
$x\geq 0$

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

$x >= 0$

$\mathrm{f}(\mathrm{x})=\left\{\begin{matrix}1+x &x\leq 1 \\ 3-ax^{2}& x>1\end{matrix}\right.$ is continuous at

${x}=1$ then

${a}=({a}>0)$

0%
$1$
0%
$2$
0%
$0$
0%
$3$

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,

$a = 1$

$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}=$

0%
${e}$
0%
$e^{2}$
0%
$-1$
0%
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

0%
$1$
0%
$-1$
0%
$0$
0%
$2$

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

$\mathrm{f}(\mathrm{0})=$

0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{2}$
0%
$2$
0%
$\dfrac{1}{3}$

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

$\displaystyle =\lim _{ x\rightarrow 0 } \dfrac { 2xe^{ x^{ 2 } }+\sin { x } }{ 2x }$

$\displaystyle \lim_{x\to 0} e^{x^2}+\lim_{x\to 0} \dfrac{\sin x}{2x}$

$\displaystyle =1+\dfrac{1}{2}=\dfrac{3}{2}$

$\displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}=$

0%
$\mathrm{e}$
0%
$\mathrm{e}^{4}$
0%
$\mathrm{e}^{-1}$
0%
$\mathrm{e}^{-4}$

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

$1^{\infty}$

$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{\dfrac{1}{4\tan y}\times(-4)}=e^{-4}$

$\displaystyle \lim_{n\rightarrow \infty }(\pi n)^{2/n}=$

0%
0
0%
1
0%
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

0%
$10$
0%
$15$
0%
$35$
0%
$80$

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

$\displaystyle \lim _{ x\rightarrow 2 }{ f\left( x \right) } =f\left( 2 \right) \Rightarrow k=80.$

$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\mathrm{a}^{2}\cos^{2}\mathrm{x}+\mathrm{b}^{2}\sin^{2}\mathrm{x},\mathrm{x}\leq 0\\\mathrm{e}^{\mathrm{a}\mathrm{x}+\mathrm{b}},\mathrm{x}>0\end{array}\right.$ is continuous at

$\mathrm{x}=0$ then

0%
$2\log|\mathrm{a}|=\mathrm{b}$
0%
$2\log|\mathrm{b}|=\mathrm{e}$
0%
$\log a=2\log|\mathrm{b}|$
0%
$\mathrm{a}=\mathrm{b}$

Explanation

Since

$f(x)$ is continuous at

$x = 0$
LHL = RHL

$\displaystyle\lim _{ x \rightarrow {0}^{-}} f(x) = \displaystyle\lim _{x \rightarrow {0}^{+}} f(x)$

$\displaystyle\lim _{ x \rightarrow {0}^{-}} {a}^{2}{cos}^{2} x + {b}^{2}{sin}^{2}x =\displaystyle \lim_{ x \rightarrow {0}^{+} } {e}^{ax+b}$

$\Rightarrow a^{2} = {e}^{b}$

$\Rightarrow 2log|a|=b$

$\displaystyle \lim_{n\rightarrow \infty }\left(\displaystyle \frac{e^{n}}{\pi}\right)^{1/n}=$

0%
0
0%
1
0%
$\mathrm{e}^{2}$
0%
$\mathrm{e}$

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

$\log y=\log e$

$\Rightarrow \quad y = e$

$\therefore \quad \displaystyle\lim _{ n\rightarrow \infty }{ { \left(\displaystyle \frac { e^{ n } }{ \pi } \right) }^{\displaystyle \frac { 1 }{ n } } } = e$

$\displaystyle \lim_{x\rightarrow 1}(2-x)^{\displaystyle \tan( \frac{\pi x}{2})}=$

0%
$e^{\displaystyle \frac{1}{\pi}}$
0%
$e^{\displaystyle \frac{2}{\pi}}$
0%
$-e^{\displaystyle \frac{2}{\pi}}$
0%
$\mathrm{e}$

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

$\mathrm{f}({p})=$

0%
$\dfrac{3}{2}$
0%
$\dfrac{2}{3}$
0%
$6$
0%
$\dfrac{1}{6}$

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

$\mathrm{A}=$

0%
$2$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$0$

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

$\mathrm{f}({0})=$

0%
$0$
0%
$2$
0%
$-2$
0%
$3$

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

${k}=$

0%
$\pm 1$
0%
$\pm 2$
0%
0
0%
$\pm 3$

Explanation

For a given function to be continuous at

$x=0$

$\displaystyle \lim_{x\to 0} = f(0)\Rightarrow \lim_{x\to 0}\frac{(e^{kx}-1).\sin kx}{x^2} =4$

$\Rightarrow \displaystyle k^2. \lim_{x\to 0}\frac{e^{kx}-1}{kx}.\frac{\sin kx}{kx} =4$

$\Rightarrow k^2 = 4\Rightarrow k = \pm 2$

$f(x)= \left\{\begin{matrix} (1+|\sin x|)^{\displaystyle \frac{a}{|\sin x|}}&-\displaystyle \frac{\pi}{6}<x<0\\ b&x=0 \\ e^{\displaystyle \frac{\tan 2x}{\tan 3x}} &0<x<\displaystyle \frac{\pi}{6}\end{matrix}\right.$ is

continuous at

$\mathrm{x}=0$ then

0%
$a=e^{2/3},b=\dfrac{2}{3}$
0%
$a=\dfrac{2}{3},b=e^{2/3}$
0%
$a=\dfrac{1}{3},b=e^{1/3}$
0%
$a=e^{1/3},b=e^{1/3}$

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

${k}=$

0%
$2$
0%
$4$
0%
$6$
0%
$8$

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

0%
continuous at $x=1$
0%
discontinuous only at $0$
0%
discontinuous only at 0,1
0%
discontinuous everywhere

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

0%
$\displaystyle \frac{1}{\sqrt{2}}$
0%
$\sqrt{2}$
0%
$-\sqrt{2}$
0%
$1$

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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$\displaystyle \frac{a+b}{ab}$
0%
$\displaystyle \frac{a-b}{ab}$
0%
$\displaystyle \frac{ab}{a+b}$
0%
$\displaystyle \frac{ab}{a-b}$

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

0%
$\displaystyle \frac{1}{8}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{-1}{8}$
0%
$-\displaystyle \frac{1}{4}$

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

$f(0) = \dfrac{\dfrac{-1}{2 \times 2}}{2} = \dfrac{-1}{8}$

$f(x)=\left\{\begin{array}{l}\dfrac{1-\sqrt{2}\sin x}{\pi-4x} x\neq\frac{\pi}{4}\\a,x=\frac{\pi}{4}\end{array}\right.$ is continuous at

$x=\displaystyle \frac{\pi}{4}$ then

$a=$

0%
$4$
0%
$2$
0%
$1$
0%
$\dfrac{1}{4}$

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

$\displaystyle a=\lim _{ x\rightarrow \frac { \pi }{ 4 } }{ \frac { { -\sqrt { 2 } \cos { x } } }{ -4 } }$

$=\dfrac { 1 }{ 4 }$

$\displaystyle f(x)=\left\{\begin{array}{ll}\dfrac{\sqrt{1+kx}-\sqrt{1-x}}{x} & \mathrm{f}\mathrm{o}\mathrm{r}-\mathrm{l} \leq x<0\\2x^{2}+3x-2 & \mathrm{f}\mathrm{o}\mathrm{r} 0\leq x\leq 1\end{array}\right.$ is continuous at

$x = 0$ then

$k$ is:

0%
$-4$
0%
$-3$
0%
$-5$
0%
$-1$

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

$f(2)=$

0%
${k}=3$
0%
${k}=5$
0%
${k}=7$
0%
${k}=9$

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

0%
${p}=0$
0%
${p}={r}$
0%
${p}={q}$
0%
$p\neq q$

Explanation

$f\left( x \right) =\cfrac { p+{ q }^{ { 1 }/{ x } } }{ r+{ s }^{ { 1 }/{ x } } }$

$f\left( x \right)$ is right continuous

$\Rightarrow \displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right) } =f\left( 0 \right) =1$

$\Rightarrow \displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { p+{ q }^{ { 1 }/{ x } } }{ r+{ s }^{ { 1 }/{ x } } } } =1$

$x\rightarrow { 0 }^{ + };\cfrac { 1 }{ x } \rightarrow +\infty$

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.

$f$ :

$R\rightarrow R$ 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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$-2$
0%
$-4$
0%
$-6$
0%
$-8$

$f(x)=\displaystyle \frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}+e^{-1/x}} x\neq 0$ is continuous at

${x}=0$ , then

${f}(\mathrm{0})=$

0%
$1$
0%
$2$
0%
$0$
0%
$3$

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,

$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x(1-e^{ -2/x }) }{ 1+e^{ -2/x } } } =0$

$\Rightarrow f(0)=0$

$f$ :

$R\rightarrow R$ 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:

0%
$R$
0%
$R-\{-2\}$
0%
$R-\{-1\}$
0%
$R-\{-1,-2\}$

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

0%
$\sqrt { e }$
0%
e
0%
$\dfrac { 1 }{ \sqrt { e } }$
0%
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:

0%
discontinuous
0%
left continuous
0%
right continuous
0%
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:

0%
$\mathrm{a}=5,\ \mathrm{b}=2$
0%
$\mathrm{a}=6,\ \mathrm{b}=3$
0%
$\mathrm{a}=7,\ \mathrm{b}=4$
0%
$\mathrm{a}=8,\ \mathrm{b}=5$

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

0%
discontinuous at 0
0%
discontinuous at 1
0%
continuous on R
0%
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

$\mathrm{a}=$

0%
$2$
0%
$-2$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{-2}{3}$

Explanation

$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$

$Lt_{ x\rightarrow 0 }\dfrac { \sin x(2\cos { x } +a) }{ x^{ 3 } }$

$=Lt_{ x\rightarrow 0 }\dfrac { (2\cos { x } +a) }{ x^{ 2 } }$

$x\rightarrow 0$ , 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

0%
$1$
0%
$2$
0%
$3$
0%
$4$

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

0%
$0$
0%
$1$
0%
$-1$
0%
$2$

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:

0%
$3$
0%
$2$
0%
$1$
0%
$-3$

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:

0%
3
0%
6
0%
9
0%
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

0%
$-\infty$
0%
$-1$
0%
$0$
0%
$\infty$

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

$\displaystyle \lim _{ h\rightarrow 0 }{ \sec { \left[ h+\left( -\frac { \pi }{ 2 } \right) \right] } = } \lim _{ h\rightarrow 0 }{ \sec { \left( \frac { \pi }{ 2 } -h \right) } } =\lim _{ h\rightarrow 0 }{ \csc { h } } =\csc { 0 } =\infty$

$\displaystyle \left[ \because \sec { \left( -\theta \right) = } \sec { \theta } \right]$

$\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}\cos ^{ -1 }{ \left( \frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \right) } =$

0%
0
0%
1
0%
2
0%
does not exist

Explanation

$\lim_{x\rightarrow 0^{-}} \dfrac{-2}{\dfrac{2x}{(1+x^{2})}}\dfrac{sin^{-1}}{(1+x^{2})}\dfrac{(2x)}{1+x^{2}}=-2$

$\lim_{x\rightarrow 0^{+}} \dfrac{2}{(\dfrac{2x}{1+x^{2}})(1+x^{2})} sin^{-1}(\dfrac{2x}{1+x^{2}})=2$

$L.H.S\neq R.H.S$

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 3 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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