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  • Consider <img alt="f(x)=\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}." src="https://entrancecorner.oncodecogs.com/gif.la

Consider f(x)=\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}. What must be the value of  f\left(\frac{\pi}{2}\right) such that it is continuous at x=\frac{\pi}{2} ?

 

 

Option: 1

\frac{1}{64}
 


Option: 2

\frac{1}{32}
 


Option: 3

\frac{-1}{64}
 


Option: 4

\frac{1}{8}


Answers (1)

best_answer

For continuity at \mathrm{x=\frac{\pi}{2} }, we must have
\mathrm{\begin{aligned} & \lim _{x \rightarrow \pi / 2} f(x)=f(\pi / 2) \\ & \text { Now, } f(x)=\frac{1-\sin x}{4\left(\frac{\pi}{2}-x\right)^2} \cdot \frac{\log \sin x}{\log \left\{1+4\left(\frac{\pi}{2}-x\right)^2\right\}} \end{aligned} }

\mathrm{Put\, \frac{\pi}{2}-x=t \Rightarrow x=\frac{\pi}{2}-t \, As\, \, x \rightarrow \frac{\pi}{2}, t \rightarrow 0 }
\mathrm{\begin{aligned} & \therefore \lim _{x \rightarrow \pi / 2} f(x)=\lim _{t \rightarrow 0} \frac{1-\cos t}{4 t^2} \cdot \lim _{t \rightarrow 0} \frac{\log \cos t}{\log \left(1+4 t^2\right)} \\ & =\lim _{t \rightarrow 0} \frac{2 \sin ^2 \frac{t}{2}}{4 t^2} \cdot \lim _{t \rightarrow 0} \frac{\log \left(1-\frac{t^2}{2 !}+\ldots\right)}{\log \left(1+4 t^2\right)} \\ & =\lim _{t \rightarrow 0} \frac{2}{4} \cdot \frac{\left(\sin \frac{t}{2}\right)^2}{t^2} \cdot \lim _{t \rightarrow 0} \frac{-\frac{t^2}{2}-\frac{1}{2}\left(\frac{t^2}{2}\right)^2 \ldots}{4 t^2-\frac{1}{2}\left(4 t^2\right)^2 \ldots} \end{aligned} }

\mathrm{=\frac{1}{8}\left(-\frac{1}{8}\right)=-\frac{1}{64} }

\mathrm{Hence, f\left(\frac{\pi}{2}\right)=-\frac{1}{64}. }

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Sayak

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