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If $f(x)=\frac{(1-\sin x)}{\log \left(1+\pi^2-4 \pi x+4 x^2\right)}$ is continuous at $x=\frac{\pi}{2}$, then $f\left(\frac{\pi}{2}\right)=$
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The correct answer is:
$\frac{1}{8}$
$\because \mathrm{f}(\mathrm{x})$ is continous at $\mathrm{x}=\frac{\pi}{2}$
$$
\begin{aligned}
& \therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\
& \therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\log \left(1+\pi^2-4 \pi x+4 x^2\right)} \\
& \therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{-\cos x}{\left(1+\pi^2-4 \pi x+4 x^2\right)}\right) \times(-4 \pi+8 x) \\
& \therefore f\left(\frac{\pi}{2}\right)=-\lim _{x \rightarrow \frac{\pi}{2}} \frac{(\cos x)\left(1+\pi^2-4 \pi x+4 x^2\right)}{(-4 \pi+8 x)} \\
& \therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(\cos x)(-4 \pi+8 x)+\left(1+\pi^2-4 \pi x+4 x^2\right)(-\sin x)}{8} \\
& \therefore f\left(\frac{\pi 8}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left(1+2 \pi^2-2 \pi^2\right)}{8}=\frac{1}{8} \\
& \therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left(1+\pi^2-4 \pi \times \frac{\pi}{2}+4 \times \frac{\pi^2}{4}\right)(+1)}{8}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\
& \therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\log \left(1+\pi^2-4 \pi x+4 x^2\right)} \\
& \therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{-\cos x}{\left(1+\pi^2-4 \pi x+4 x^2\right)}\right) \times(-4 \pi+8 x) \\
& \therefore f\left(\frac{\pi}{2}\right)=-\lim _{x \rightarrow \frac{\pi}{2}} \frac{(\cos x)\left(1+\pi^2-4 \pi x+4 x^2\right)}{(-4 \pi+8 x)} \\
& \therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(\cos x)(-4 \pi+8 x)+\left(1+\pi^2-4 \pi x+4 x^2\right)(-\sin x)}{8} \\
& \therefore f\left(\frac{\pi 8}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left(1+2 \pi^2-2 \pi^2\right)}{8}=\frac{1}{8} \\
& \therefore f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left(1+\pi^2-4 \pi \times \frac{\pi}{2}+4 \times \frac{\pi^2}{4}\right)(+1)}{8}
\end{aligned}
$$
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