If f(x) = $\{\begin{array}{l}\mathrm{m}\mathrm{x}+1,\text{ if }\mathrm{x}\le \frac{\pi }{2}\\ \sin \mathrm{x}+\mathrm{n},\text{ if }\mathrm{x}>\frac{\pi }{2}\end{array}$, is continuous at x = $\frac{\pi }{2}$, then

Concept:

​A function f(x) is continuous at x = a, if $\underset{x\to a^{-}}{lim}$ f(x) = $\underset{x\to a^{+}}{lim}$f(x) = f(a).

Calculation:

Given: f(x) = $\{\begin{array}{l}\mathrm{m}\mathrm{x}+1,\text{ if }\mathrm{x}\le \frac{\pi }{2}\\ \sin \mathrm{x}+\mathrm{n},\text{ if }\mathrm{x}>\frac{\pi }{2}\end{array}$

f($\frac{\pi }{2}$) = m × $\frac{\pi }{2}$ + 1

Left-hand limit = $\underset{h\to 0}{lim}f(\frac{\pi }{2}-h)$

$=\underset{h\to 0}{lim}m\times (\frac{\pi }{2}-h)+1$

Applying the limits:

Left- hand limit = m × $\frac{\pi }{2}$ + 1

Right-hand limit = $\underset{h\to 0}{lim}f(\frac{\pi }{2}+h)$

$=\underset{h\to 0}{lim}\sin (\frac{\pi }{2}+h)+n$

Applying the limits:

 $=\sin \frac{\pi }{2}+n$

Right-hand limit = 1 + n

For the function to be continuous at x = $\frac{\pi }{2}$,

Left-hand limit = Right-hand limit = f(π/2)

⇒ m× $\frac{\pi }{2}$ + 1 = 1 + n

⇒ n = $\frac{m\pi }{2}$

The correct answer is n = $\frac{m\pi }{2}$ .