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If   \mathrm{f(x)=\left(\begin{array}{c} \frac{1-\sin x}{(\pi-2 x)^2} \cdot \frac{\log \sin x}{\left(\log \left(1+\pi^2-4 \pi x+4 x^2\right)\right.} \\ k \quad x=\frac{\pi}{2} \end{array}\right.}  is  continuous at  \mathrm{x=\pi / 2 \text {, then } k=} 

Option: 1

-\frac{1}{16}


Option: 2

-\frac{1}{32}


Option: 3

-\frac{1}{64}


Option: 4

-\frac{1}{28}


Answers (1)

best_answer

sotuntion For f(x) to be continuous at  \mathrm{x=\pi / 2} , we must have

\mathrm{ \lim _{x \rightarrow \pi / 2} f(x)=f(\pi / 2) \\ }

\mathrm{ \Rightarrow \lim _{x \rightarrow \pi / 2} \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)}=k \\ }
\mathrm{\Rightarrow \lim _{h \rightarrow 0} \frac{1-\cos h}{4 h^2} \times \frac{\log \cos h}{\log \left(1+4 h^2\right)}=k \\ }
\mathrm{\Rightarrow \lim _{h \rightarrow 0} \frac{1-\cos h}{4 h^2} \times \frac{\log (1+\cos h-1))}{\cos h-1} }
\mathrm{\times \frac{4 h^2}{\log \left(1+4 h^2\right)} \times \frac{\cos h-1}{4 h^2}=k}

\mathrm{ \Rightarrow-\lim _{h \rightarrow 0}\left(\frac{1-\cos h}{4 h^2}\right)^2 \frac{\log (1+(\cos h-1))}{\cos h-1} \times \frac{4 h^2}{\log \left(1+4 h^2\right)}=k \\ }

\mathrm{ \Rightarrow-\lim _{h \rightarrow 0}\left(\frac{\sin ^2 h / 2}{2 h^2}\right)^2 \frac{\log (1+(\cos h-1))}{\cos h-1} \times \frac{4 h^2}{\log \left(1+4 h^2\right)}=k \\ }

\mathrm{ \Rightarrow-\frac{1}{64} \lim _{h \rightarrow 0}\left(\frac{\sin h / 2}{h / 2}\right)^4 \frac{\log (1+(\cos h-1))}{\cos h-1} \times \frac{4 h^2}{\log \left(1+4 h^2\right)}=k \\ }

\mathrm{ \Rightarrow-\frac{1}{64}=k }

Posted by

sudhir.kumar

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