Question

If $f(x)=\{\begin{array}{ll}mx+1,& x\le \frac{\pi }{2}\\ \sin x+n,& x>\frac{\pi }{2}\end{array}$ 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 $f(x)=\{\begin{array}{ll}mx+1,& x\le \frac{\pi }{2}\\ \sin x+n,& x>\frac{\pi }{2}\end{array}$
As given this function is continuous at $x=\frac{\pi }{2},$
So, limit of function when $x\to \frac{\pi }{2}=f\left(\frac{\pi }{2}\right)$
$\Rightarrow \underset{x\to \frac{{\pi }^{+}}{2}}{\lim }(\sin x+n)=f\left(\frac{\pi }{2}\right)$
$\Rightarrow \left(\sin \left(\frac{\pi }{2}+h\right)+n\right)=\frac{m\pi }{2}+1$
$\Rightarrow \sin \frac{\pi }{2}+n=\frac{m\pi }{2}+1$
$\Rightarrow 1+n=\frac{m\pi }{2}+1$
$\Rightarrow n=\frac{m\pi }{2}$