Question

If $f\left(x\right)=\left\{\begin{array}{ll}mx+1,& x\le \frac{\mathrm{\pi }}{2}\\ \sin x+n,& x>\frac{\mathrm{\pi }}{2}\end{array}\right.$is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
(a) m = 1, n = 0

(b) $m=\frac{n\mathrm{\pi }}{2}+1$

(c) $n=\frac{m\mathrm{\pi }}{2}$

(d) $m=n=\frac{\mathrm{\pi }}{2}$

Answer 1

(c) $n=\frac{\mathrm{m}\mathrm{\pi }}{2}$n=mπ2


Here,

$f\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{m\pi }}{2}+1$

We have
(LHL at $x=\frac{\mathrm{\pi }}{2}$) = $\underset{\mathit{}\mathit{}\mathit{}\mathit{}x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }f\left(x\right)=\underset{\mathit{}\mathit{}\mathit{}h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}-h\right)=\underset{\mathit{}\mathit{}\mathit{}h\to 0}{\lim }m\left(\frac{\mathrm{\pi }}{2}-h\right)+1=\frac{\mathrm{m\pi }}{2}+1$

(RHL at $x=\frac{\mathrm{\pi }}{2}$) = $\underset{\mathit{}\mathit{}\mathit{}\mathit{}x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }f\left(x\right)=\underset{\mathit{}\mathit{}\mathit{}h\to 0}{\lim }f\left(\frac{\mathrm{\pi }}{2}+h\right)=\underset{\mathit{}\mathit{}\mathit{}h\to 0}{\lim }\left[\sin \left(\frac{\mathrm{\pi }}{2}+h\right)+n\right]=n+1$

Thus,
If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
$\underset{\mathit{}\mathit{}\mathit{}\mathit{}x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }f\left(x\right)=\underset{\mathit{}\mathit{}\mathit{}\mathit{}x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }f\left(x\right)$


$$\begin{gathered} \Rightarrow \frac{\mathrm{m\pi }}{2}+1=n+1 \\ \Rightarrow \frac{\mathrm{m\pi }}{2}=n \end{gathered}$$