Question

$f\left(x\right)=\{\begin{array}{c}x\sin x.if0<x\le \frac{\pi }{2}\\ \frac{\pi }{2}\sin (\pi +x),if\frac{\pi }{2}<x<\pi \end{array}$ then $f\left(x\right)$ is discontinuous at

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{2\pi }{3}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is B $\frac{\pi }{2}$

As $x\sin x$ is continuous in $x\in (0,\frac{\pi }{2})$ and $\frac{\pi }{2}\sin (\pi +x)$ in $x\in (\frac{\pi }{2},\pi )$, we have to check continuity at $x=\frac{\pi }{2}$

L.H.L $\underset{x\to \frac{\pi }{2}-}{lim}f\left(x\right)=\underset{x\to 0}{lim}(\frac{\pi }{2}-h)\sin (\frac{\pi }{2}-h)=\frac{\pi }{2}$

R.H.L$=\underset{x\to 0}{lim}f(\frac{\pi }{2}+h)=\frac{\pi }{2}\sin (\pi +\frac{\pi }{2}+h)=\frac{\pi }{2}(-\cos h)=-\frac{\pi }{2}$

$\Rightarrow f$ is discontinuous at $x=\frac{\pi }{2}$.