#### Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise 5 multiple choice question 15

The correct option is (c)

Hint:

Use the given formula:

(i) A function f(x) is said to be continuous at a point x = a of its domain, if

$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$

Given:

f(x) is continuous at x = 0 and

$f(x)=\left\{\begin{array}{l} m x+1, x \leq \frac{\pi}{2} \\ \sin x+n, x>\frac{\pi}{2} \end{array}\right.$

Solution:

Using RHL

\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right) \\ &=\lim _{h \rightarrow 0}\left[\sin \left(\frac{\pi}{2}+h\right)+n\right] \\ &=\lim _{h \rightarrow 0}(\cos h+n)=1+n \end{aligned}

Using LHL

\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \\ &=\lim _{h \rightarrow 0} m\left(\frac{\pi}{2}-h\right)+1=\frac{m \pi}{2}+1 \end{aligned}

\begin{aligned} &\text { Function } f(x) \text { is continuous at } x=\frac{\pi}{2}\\ &\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)\\ &\frac{m \pi}{2}+1=n+1\\ &\frac{m \pi}{2}=n \end{aligned}

So, option (c) is correct