Question

If $f\left(x\right)=\{\begin{array}{cc}mx+1,& x\le \frac{\pi }{2}\\ sinx+n,& x>\frac{\pi }{2}\end{array}$ 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. $m=n=\frac{n\pi }{2}$

Answer 1

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

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

$\Rightarrow m\frac{\pi }{2}+1=1+n\Rightarrow \frac{m\pi }{2}=n$

Hence, option C is correct.