Question

If $f\left(x\right)=\{\begin{array}{cc}mx+1,& x\le \frac{\pi }{2}\\ \sin x+n,& x>\frac{\pi }{2}\end{array}$ is continuous at $x=\frac{\pi }{2}$, then

• A. $m=1,n=0$
• B. $m=\frac{n\pi }{2}+1$
• C. $n=\frac{m\pi }{2}$
• D. $m=n=\frac{\pi }{2}$

Answer 1

The correct option is C $n=\frac{m\pi }{2}$

Since, $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$
$\therefore \underset{x\to \frac{{\pi }^{-}}{2}}{lim}(mx+1)=\underset{x\to \frac{{\pi }^{+}}{2}}{lim}(\sin x+n)$
$\Rightarrow m\frac{\pi }{2}+1=\sin \frac{\pi }{2}+n$
$\Rightarrow \frac{m\pi }{2}=n$.