If $f(x)=\frac{1-\sin x}{(\pi-2 x)^{2}}$, when $x \neq \pi / 2$ and $f(\pi / 2)=\lambda$, then $f(x)$ will be continuous function at $x=\pi / 2$, where $\lambda=$
(a) $1 / 8$
(b) $1 / 4$
(c) $1 / 2$
(d) none of these
(a) $\frac{1}{8}$
If $f(x)$ is continuous at $x=\frac{\pi}{2}$, then
$\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)$
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{(\pi-2 x)^{2}}=f\left(\frac{\pi}{2}\right)$
Suppose $\left(\frac{\pi}{2}-x\right)=t$, then
$\lim _{t \rightarrow 0}\left[\frac{1-\sin \left(\frac{\pi}{2}-t\right)}{(2 t)^{2}}\right]=f\left(\frac{\pi}{2}\right)$ [From eq. (1)]
$\Rightarrow \lim _{t \rightarrow 0}\left[\frac{1-\cos t}{4 t^{2}}\right]=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \frac{1}{4} \lim _{t \rightarrow 0}\left[\frac{2 \sin ^{2}\left(\frac{t}{2}\right)}{t^{2}}\right]=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \frac{1}{4} \lim _{t \rightarrow 0}\left[\frac{\frac{2}{4} \sin ^{2}\left(\frac{t}{2}\right)}{\frac{t^{2}}{4}}\right]=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \frac{1}{8} \lim _{t \rightarrow 0}\left[\frac{\sin ^{2}\left(\frac{t}{2}\right)}{\frac{t^{2}}{4}}\right]=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \frac{1}{8} \lim _{t \rightarrow 0}\left[\frac{\sin \left(\frac{t}{2}\right)}{\frac{t}{2}}\right]^{2}=f\left(\frac{\pi}{2}\right)$
$\Rightarrow f\left(\frac{\pi}{2}\right)=\lambda=\frac{1}{8}$
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