If $f(x)=\frac{(1-\sin x)}{\log \left(1+\pi^2-4 \pi x+4 x^2\right)}$ is continuous at $x=\frac{\pi}{2}$,…

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 $f\left(\frac{\pi}{2}\right)=$
  1. $\frac{1}{4}$
  2. $\frac{1}{8}$
  3. $\frac{1}{16}$
  4. $\frac{1}{32}$

Solution

$\because \mathrm{f}(\mathrm{x})$ 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} $

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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