Question

If $ f(x) = \begin{cases} mx + 1, & x \leq \frac{\pi}{2}\\ \sin \, x + n ,& x > \frac{\pi}{2} \end{cases}$ is continuous at
$x = \frac{\pi}{2} ,$ then which one of the following is correct ?

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

Answer 1

Answer: (C)

Given function is $ f(x) = \begin{cases} mx + 1, & x \leq \frac{\pi}{2}\\ \sin \, x + n ,& x > \frac{\pi}{2} \end{cases}$
As given this function is continuous at $x = \frac{\pi}{2} ,$
So, limit of function when $x \to \frac{\pi}{2} = f \left( \frac{\pi}{2}\right)$
$\Rightarrow \:\:\:\displaystyle\lim_{x \to \frac{\pi^+}{2}} (\sin \, x + n) = f \left( \frac{\pi}{2}\right)$
$\Rightarrow \left(\sin\left(\frac{\pi}{2}+h\right)+n\right)=\frac{m\pi}{2}+1$
$ \Rightarrow \sin \frac{\pi}{2}+n =\frac{m\pi}{2}+1$
$ \Rightarrow 1+n =\frac{m\pi}{2}+1 $
$\Rightarrow n =\frac{m\pi}{2}$