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 $f$ be defined by $f(x) = \begin{cases} \frac{\tan x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$.
Statement-$1$: $x = 0$ is a point of local minima for $f$.
Statement-$2$: $f'(0) = 0$.

Difficult
View Solution

Let $f: R \rightarrow R$ be the function defined by $f(x) = \begin{cases} 5, & \text{if } x \leq 1 \\ a+bx, & \text{if } 1 < x < 3 \\ b+5x, & \text{if } 3 \leq x < 5 \\ 30, & \text{if } x \geq 5 \end{cases}$. Then $f$ is:

If the function $f(x)$ is defined as:
$f(x) = \begin{cases} 1 + \sin \frac{\pi x}{2}, & -\infty < x \leq 1 \\ ax + b, & 1 < x < 3 \\ 6 \tan \frac{x \pi}{12}, & 3 \leq x < 6 \end{cases}$
and is continuous in $(-\infty, 6)$,then the values of $a$ and $b$ are respectively.

Is the function defined by $f(x) = \begin{cases} x + 5, & \text{if } x \le 1 \\ x - 5, & \text{if } x > 1 \end{cases}$ a continuous function?

If the function $f$ defined by $f(x) = \begin{cases} \cos x, & \text{if } x \leq 0 \\ 3x + \alpha, & \text{if } 0 < x < 2 \\ \beta x + 3, & \text{if } 2 \leq x \leq 4 \\ 11, & \text{if } x > 4 \end{cases}$ where $\alpha$ and $\beta$ are real constants,is continuous on $R$,then $\alpha^2 + \beta^2 =$

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