The function f(x)= e1 x -1 e1 x +1 ,&x 0 0,&x=0 - Vidyadip

MCQ

The function $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},&\text{x}\neq0\\0,&\text{x}=0\end{cases}$

A
is continuous at $x = 0$
✓
is not continuous at $x = 0$
C
is not continuous at $x = 0,$ but can be made continuous at $x = 0$
D
None of these.

✓

Answer

Correct option: B.

is not continuous at $x = 0$

Given, $\text{f(x)=}\begin{cases}\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1},\text{x}\neq0\\0,\text{x}=0\end{cases}$
We have
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\text{e}\frac{1}{\text{x}}-1}{\text{e}\frac{1}{\text{x}}+1}\Bigg)$
if $\text{e}^\frac{1}{\text{x}}=\text{t},$ then
$\text{x}\rightarrow0, \text{t}\rightarrow\infty$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{t}\rightarrow\infty}\Big(\frac{\text{t}-1}{\text{t}+1}\Big)$
$=\lim\limits_{\text{t}\rightarrow\infty}\Bigg(\frac{1-\frac{1}{\text{t}}}{1+\frac{1}{\text{t}}}\Bigg)=\frac{1-0}{1+0}=1$
Also, $f(0) = 0$
$\because\ \lim\limits_{\text{x}\rightarrow0}\text{f(x)}\neq\text{f}(0)$
Hence, $f(x)$ is discontinuons at $x = 0.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Continuity and Differentiability→Continuity and Differentiability — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The lines $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ and $\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$ are:

The area of the region bounded by the curves $=x e^x, y=x e^{-x}$ and the line $x=1$ is:

Choose the correct answer from the given four options. If $\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix},$ then the value of $x + y$ is :

Let $[x]$ denote the greatest integer $\leq x$, where $x \in R$. If the domain of the real valued function $\mathrm{f}(\mathrm{x})=\sqrt{\frac{[\mathrm{x}] \mid-2}{\sqrt{[\mathrm{x}] \mid-3}}}$ is $(-\infty, \mathrm{a}) \cup[\mathrm{b}, \mathrm{c}) \cup[4, \infty), \mathrm{a}\,<\,\mathrm{b}\,<\,\mathrm{c}$, then the value of $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is:

For every integer $n$, let $a_n$ and $b_n$ be real numbers. Let function $f: I R \rightarrow$ $IR$ be given by $f(x)=\left\{\begin{array}{ll}a_n+\sin \pi x, & \text { for } x \in[2 n, 2 n+1] \\ b_n+\cos \pi x, & \text { for } x \in(2 n-1,2 n)\end{array}\right.$, for all integers $n$.

If $f$ is continuous, then which of the following hold$(s)$ for all $n$ ?

$(A)$ $a_{n-1}-b_{n-1}=0$ $(B)$ $a_n-b_n=1$ $(C)$ $a_n-b_{n+1}=1$ $(D)$ $a_{n-1}-b_n=-1$

A coin is tossed $4$ times. The probability that at least one head turns up is:

If $A$ and $B$ are matrices of order $3 \times 2$ and $C$ is of order $2 \times 3$, then which of the following matrices is not defined :

If $y = \log {\left( {{{1 + x} \over {1 - x}}} \right)^{1/4}} - {1 \over 2}{\tan ^{ - 1}}x,$ then ${{dy} \over {dx}} = $

Solution of the differential equation $\cos x\;dy = y\left( {\sin x - y} \right)dx,0 < x < \frac{\pi }{2}$ is

Let $x \in \left( {0,1} \right)$. The set of all $x$ such that ${\sin ^{ - 1}}\,x > {\cos ^{ - 1}}\,x$, is the interval