Let $[t]$ represent the greatest integer not exceeding $t$ and $C=1-2e^2$. If the function $f(x)=\begin{cases} [e^x], & x < 0 \\ ae^x+[x-2], & 0 \leq x < 2 \\ [e^{-x}]-C, & x \geq 2 \end{cases}$ is continuous at $x=2$,then $f(x)$ is discontinuous at

  • A
    $x=1$ only
  • B
    $x=0$ and $x=1$
  • C
    $x=0$ only
  • D
    $x=0, x=1$ and $x=\frac{1}{2}$

Explore More

Similar Questions

Let $f: R \rightarrow R$ be defined as:
$f(x) = \begin{cases} \frac{\lambda|x^{2}-5x+6|}{\mu(5x-x^{2}-6)}, & x < 2 \\ \mu, & x = 2 \\ e^{\frac{\tan(x-2)}{x-[x]}}, & x > 2 \end{cases}$
Where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x = 2$,then $\lambda + \mu$ is equal to:

Discuss the continuity of the function $f$,where $f$ is defined by $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$. Is it continuous at $x=3$?

If the function $f(x) = \begin{cases} -2 \sin x, & x \leq \frac{-\pi}{2} \\ A \sin x+B, & \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & x \geq \frac{\pi}{2} \end{cases}$ is continuous everywhere,then the values of $A$ and $B$ are respectively

Let $f, g: R \to R$ be two functions defined by $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & x \ne 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$.
Statement $I$: $f$ is a continuous function at $x = 0$.
Statement $II$: $g$ is a differentiable function at $x = 0$.

If $\lim _{x \rightarrow a^{+}} f(x)=p, \lim _{x \rightarrow a^{-}} f(x)=m$ and $f(a)=k$,then which one of the following is true?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo