If $f(x) = \begin{cases} x \sin x, & 0 < x \le \frac{\pi}{2} \\ \frac{\pi}{2} \sin(\pi + x), & \frac{\pi}{2} < x < \pi \end{cases}$,then

  • A
    $f(x)$ is discontinuous at $x = \pi/2$
  • B
    $f(x)$ is continuous at $x = \pi/2$
  • C
    $f(x)$ is continuous at $x = 0$
  • D
    None of these

Explore More

Similar Questions

Let $S_n = 1 + 3x + 9x^2 + 27x^3 + \ldots$ ($n$ terms) and $-\frac{1}{3} < x < \frac{1}{3}$. If $\lim_{n \rightarrow \infty} S_n = f(x)$,then $f(x)$ is discontinuous at the point $x =$

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 $x=3$?

Let $f(x) = \frac{1-\tan x}{4x-\pi}$ for $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{1}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is

Let $[x]$ denote the greatest integer less than or equal to $x$. Then $f(x) = \frac{1 + \sin([\cos x])}{\cos([\sin x])}$ is

If $f(x) = \begin{cases} \frac{x - |x|}{x}, & x \ne 0 \\ 2, & x = 0 \end{cases}$,then

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