If $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{mx + 1,}&{{\rm{ if }}\ x \le \frac{\pi}{2}}\\{(\sin x + n),}&{{\rm{ if }}\ x > \frac{\pi}{2}}\end{array}} \right.$ is continuous at $x = \frac{\pi}{2}$, find the relation between $m$ and $n$.

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{mx + 1,}&{{\rm{ if }}x \le \frac{\pi }{2}}\\{(\sin x + n),}&{{\rm{ if }}x > \frac{\pi }{2}}\end{array}} \right.$ is continuous at $x = \frac{\pi }{2}$

therefore,${\rm{LHL}} = \mathop {\lim }\limits_{x \to \frac{{{\pi ^ - }}}{2}} (mx + 1) = \mathop {\lim }\limits_{h \to 0} \left[ {m\left( {\frac{\pi }{2} - h} \right) + 1} \right] = \frac{{m\pi }}{2} + 1$

and ${\rm{RHL}} = \mathop {\lim }\limits_{x \to \frac{{{\pi ^ + }}}{2}} (\sin x + n) = \mathop {\lim }\limits_{h \to 0} \left[ {\sin \left( {\frac{\pi }{2} + h} \right) + n} \right]$

$= \mathop {\lim }\limits_{h \to 0} \cos h + n = 1 + n$
therefore,${\rm{LHL}} = {\rm{RHL}}$ [to be continuous at $x = \frac{\pi }{2}$]

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

therefore,$n = m \cdot \frac{\pi }{2}$