Download App

Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
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→Continuity — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The value of $f$ at $x = 0$ so that the function $f(x) = \frac{{{2^x} - {2^{ - x}}}}{x},x \ne 0$, is continuous at $x = 0$, is
→
If $f(x) = (ax^2 – b)^3,$ then the function g such that $f\{g(x)\} = g\{f(x)\}$ is given by:
→
If  $y = \sqrt {\sec x + \sqrt {\sec x + \sqrt {\sec x + ......\infty } } } \,,$ then value of  $\int\limits_0^{\pi /3} {\left( {(2y - 1)\frac{{dy}}{{dx}}} \right)} \,dx$ is equal to $(\sec x > 0)$ -
→
If $f ‘ (x) =$ $\left| {\begin{array}{*{20}{c}}{mx}&{mx - p}&{mx + p}\\n&{n + p}&{n - p}\\{mx + 2n}&{mx + 2n + p}&{mx + 2n - p}\end{array}} \right|$  then $y = f(x)$ represents
→
The taxi fare in a city is as follows. For the first $\ km$ the fare is $Rs.10$ and subsequent distance is $Rs.6 \ km.$Taking the distance covered as $x \ km$ and fare as $Rs. y,$ write a linear equation.
→
A unit vector perpendicular to the vector $4i - j + 3k$ and $ - 2i + j - 2k$ is
→
The inverse of $\left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&2\\0&0&1\end{array}} \right]$ is
→
If $f(x) = x^3-x^2+100\,x \, +1001\,;$ then
→
Choose the correct answer from given four options in each of the Exercise:
If $\text{f(x)}==\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b} \\\text{x}+\text{a} &0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
→
If $u = {({x^2} + {y^2} + {z^2})^{3/2}}$, then ${\left( {{{\partial u} \over {\partial x}}} \right)^2} + {\left( {{{\partial u} \over {\partial y}}} \right)^2} + {\left( {{{\partial u} \over {\partial z}}} \right)^2} = $
→