If $f(x)=\left\{\begin{array}{cc}\frac{1-\sin x}{(\pi-2 x)^{2}} \cdot \frac{\log \sin x}{\log \left(1+\pi^{2}-4 \pi x+4 x^{2}\right)}, & x \neq \frac{\pi}{2} \\ k & , x=\frac{\pi}{2}\end{array}\right.$
is continuous at $x=\pi / 2$, then $\mathrm{k}=$
(a) $-\frac{1}{16}$
(b) $-\frac{1}{32}$
(c) $-\frac{1}{64}$
(d) $-\frac{1}{28}$
(c) $\frac{-1}{64}$
If $f(x)$ is continuous at $x=\frac{\pi}{2}$, then
$\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)$
If $\frac{\pi}{2}-x=t$, then
$\Rightarrow \lim _{t \rightarrow 0} f\left(\frac{\pi}{2}-t\right)=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \lim _{t \rightarrow 0}\left(\frac{1-\sin \left(\frac{\pi}{2}-t\right)}{4 t^{2}} \times \frac{\log \sin \left(\frac{\pi}{2}-t\right)}{\log \left(1+\pi^{2}-4 \pi\left(\frac{\pi}{2}-t\right)+4\left(\frac{\pi}{2}-t\right)^{2}\right)}\right)=k$
$\Rightarrow \lim _{t \rightarrow 0}\left(\frac{(1-\cos t)}{4 t^{2}} \times \frac{\log \cos t}{\log \left(1+\pi^{2}-2 \pi^{2}+4 \pi t+4\left(\frac{\pi^{2}}{4}+t^{2}-\pi t\right)\right)}\right)=k$
$\Rightarrow \lim _{t \rightarrow 0}\left(\frac{(1-\cos t)}{4 t^{2}} \times \frac{\log \cos t}{\log \left(1-\pi^{2}+4 \pi t+\left(\pi^{2}+4 t^{2}-4 \pi t\right)\right)}\right)=k$
$\Rightarrow \lim _{t \rightarrow 0}\left(\frac{(1-\cos t)}{4 t^{2}} \times \frac{\log \cos t}{\log \left(1+4 t^{2}\right)}\right)=k$
$\Rightarrow \lim _{t \rightarrow 0}\left(\frac{2 \sin ^{2} \frac{t}{2}}{16 \times \frac{t^{2}}{4}} \times \frac{\log \cos t}{\log \left(1+4 t^{2}\right)}\right)=k$
$\Rightarrow \frac{2}{16} \lim _{t \rightarrow 0}\left(\frac{\sin ^{2} \frac{t}{2}}{\left(\frac{t^{2}}{4}\right)} \times \frac{\log \cos t}{\left(\frac{4 t^{2} \log \left(1+4 t^{2}\right)}{4 t^{2}}\right)}\right)=k$
$\Rightarrow \frac{1}{8} \lim _{t \rightarrow 0}\left(\frac{\sin ^{2} \frac{t}{2}}{\left(\frac{t}{2}\right)^{2}} \times \frac{\left(\frac{\log \cos t}{4 t^{2}}\right)}{\left(\frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}\right)}\right)=k$
$\Rightarrow \frac{1}{8} \lim _{t \rightarrow 0}\left(\frac{\sin ^{2} \frac{t}{2}}{\left(\frac{t}{2}\right)^{2}} \times \frac{\left(\frac{\log \sqrt{1-\sin ^{2} t}}{4 t^{2}}\right)}{\left(\frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}\right)}\right)=k$
$\Rightarrow \frac{1}{8} \lim _{t \rightarrow 0}\left(\frac{\sin ^{2} \frac{t}{2}}{\left(\frac{t}{2}\right)^{2}} \times \frac{\left(\frac{\log \left(1-\sin ^{2} t\right)}{\left(8 t^{2}\right)}\right)}{\left(\frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}\right)}\right)=k$
$\Rightarrow \frac{1}{64} \lim _{t \rightarrow 0}\left(\frac{\sin ^{2} \frac{t}{2}}{\left(\frac{t}{2}\right)^{2}} \times \frac{\left(\frac{\log \left(1-\sin ^{2} t\right)}{t^{2}}\right)}{\left(\frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}\right)}\right)=k$
$\Rightarrow \frac{1}{64}\left(\lim _{t \rightarrow 0}\left(\frac{\sin \frac{t}{2}}{\left(\frac{t}{2}\right)}\right)^{2} \times \frac{\lim _{t \rightarrow 0}\left(\frac{\log \left(1-\sin ^{2} t\right)}{t^{2}}\right)}{\lim _{t \rightarrow 0}\left(\frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}\right)}\right)=k$
$\Rightarrow \frac{1}{64}\left(1 \times \lim _{t \rightarrow 0} \frac{\left(-\sin ^{2} t\right) \log \left(1-\sin ^{2} t\right)}{t^{2}\left(-\sin ^{2} t\right)}\right)=k$
$\Rightarrow \frac{-1}{64}\left(\lim _{t \rightarrow 0} \frac{\left(\sin ^{2} t\right) \log \left(1-\sin ^{2} t\right)}{t^{2}\left(-\sin ^{2} t\right)}\right)=k$
$\Rightarrow \frac{-1}{64}\left(\lim _{t \rightarrow 0}\left(\frac{\sin t}{t}\right)^{2} \lim _{t \rightarrow 0} \frac{\log \left(1-\sin ^{2} t\right)}{\left(-\sin ^{2} t\right)}\right)=k$
$\Rightarrow \frac{-1}{64}\left(\lim _{t \rightarrow 0}\left(\frac{\sin t}{t}\right)^{2} \lim _{t \rightarrow 0} \frac{\log \left(1-\sin ^{2} t\right)}{\left(-\sin ^{2} t\right)}\right)=k$
$\Rightarrow k=\frac{-1}{64} \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}=1\right]$