#### Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 22

The correct option is (a)

Hint:

$\lim _{h \rightarrow 0} f\left(\left(\frac{\pi}{2}\right)-h\right)$

Given:

\begin{aligned} &\text { The function } f(x)=\frac{1-\sin x}{(\pi-2 x)^{2}}, \text { when } x \neq \frac{\pi}{2} \text { and } f\left(\frac{\pi}{2}\right)=\lambda, \text { then } f(x) \text { will be continuous function }\\ &\text { for } x=\frac{\pi}{2} \end{aligned}

Solution:

Step 1: Calculate, f(x) is continuous for

$x=\frac{\pi }{2}$

Therefore,

\begin{aligned} &\therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\ &=\lim _{x \rightarrow \frac{\pi}{2}(\pi-2 x)^{2}}=f\left(\frac{\pi}{2}\right) \end{aligned}

\begin{aligned} &\text { Assume, }\left(\frac{\pi}{2}-x\right)=t \text { and substitute in equation }(i)\\ &\therefore \lim _{t \rightarrow 0}\left[\frac{1-\sin \left(\frac{\pi}{2}-t\right)}{(2 t)^{2}}\right]=f\left(\frac{\pi}{2}\right) \end{aligned}

\begin{aligned} &\lim _{t \rightarrow 0}\left(\frac{1-\cos t}{4 t^{2}}\right)=f\left(\frac{\pi}{2}\right) \\ &\frac{1}{4} \lim _{t \rightarrow 0}\left[\frac{2 \sin ^{2}\left(\frac{t}{2}\right)}{t^{2}}\right]=f\left(\frac{\pi}{2}\right) \\ &\frac{1}{4} \lim _{i \rightarrow 0}\left[\frac{\frac{2}{4} \sin ^{2}\left(\frac{t}{2}\right)}{\frac{t^{2}}{4}}\right]=f\left(\frac{\pi}{2}\right) \end{aligned}

Simplify further,

\begin{aligned} &=\frac{1}{8} \lim _{t \rightarrow 0}\left[\frac{\sin ^{2}\left(\frac{t}{2}\right)}{\frac{t^{2}}{4}}\right]=f\left(\frac{\pi}{2}\right) \\ &=\frac{1}{8} \lim _{t \rightarrow 0}\left[\frac{\sin \left(\frac{t}{2}\right)}{\frac{t}{2}}\right]^{2}=f\left(\frac{\pi}{2}\right) \end{aligned}

Apply formula (i)  i.e

$\lim _{h \rightarrow 0} f\left(\left(\frac{\pi}{2}\right)-h\right)$

\begin{aligned} &=\frac{1}{8} \times(1)^{2}=f\left(\frac{\pi}{2}\right) \\ &=f\left(\frac{\pi}{2}\right)=\frac{1}{8} \\ &\therefore f\left(\frac{\pi}{2}\right)=\lambda \\ &\therefore \Rightarrow \lambda=\frac{1}{8} \end{aligned}

Hence, the value of

\begin{aligned} &\lambda \text { is }\frac{1}{8} \end{aligned}

Hence, the correct answer is option (a)