Question

If $f\left(x\right)=\{\begin{array}{c}x\sin x;when0<x\le \frac{\pi }{2}\\ \frac{\pi }{2}\sin (\pi +x);when\frac{\pi }{2}<x<\pi \end{array}$, then

• A. $f\left(x\right)$ is discontinuous at $x=\frac{\pi }{2}$
• B. $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$
• C. $f\left(x\right)$ is continuous at $x=0$
• D. None of the above

Answer 1

Answer: (A)

$f\left(x\right)$ is discontinuous at $x=\frac{\pi }{2}$

Given : $f\left(x\right)=\{\begin{array}{c}x\sin x;when0<x\le \frac{\pi }{2}\\ \frac{\pi }{2}\sin (\pi +x);when\frac{\pi }{2}<x<\pi \end{array}$

$\underset{x\to \frac{{\pi }^{-}}{2}}{lim}f\left(x\right)=\underset{x\to \frac{{\pi }^{-}}{2}}{lim}x\sin x=\frac{\pi }{2}$
$\underset{x\to \frac{{\pi }^{+}}{2}}{lim}f\left(x\right)\underset{x\to \frac{{\pi }^{+}}{2}}{lim}\sin (\pi +x)=\frac{-\pi }{2}$
and $f\left(\frac{\pi }{2}\right)=\frac{\pi }{2}$
Hence, $f\left(x\right)$ is discontinuous at $x=\frac{\pi }{2}$