An extreme value of $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ in $(0, \pi / 2)$ is

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Aicharya · Accepted Answer

$\because f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ $\Rightarrow f^{\prime}(x)=-\frac{4 \cos x}{\sin ^2 x}+\frac{\cos x}{(1-\sin x)^2}$ For critical points: $f^{\prime}(x)=0$ $\begin{aligned} & \Rightarrow-\frac{4 \cos x}{\sin ^2 x}+\frac{\cos x}{(1-\sin x)^2}=0 \ & \Rightarrow-\cos x\left[\frac{4(1-\sin x)^2+\sin ^2 x}{\sin ^2 x(1-\sin x)^2}ight]=0 \ & \Rightarrow 4(1-\sin x)^2+\sin ^2 x=0 \ & \Rightarrow 3 \sin ^3 x-8 \sin x+4=0 \ & \Rightarrow(3 \sin x-2)(\sin x-2)=0 \ & \Rightarrow 3 \sin x-2=0 	ext { or } \sin x=2(	ext { not possible) } \ & \Rightarrow \sin x=\frac{2}{3}\end{aligned}$ So, the only critical value is $x=\sin ^{-1}\left(\frac{2}{3}ight)$ $	herefore f(x)$ has extremum at $x=\sin ^{-1}\left(\frac{2}{3}ight)$ So, the extreme value is $\begin{aligned} & f\left(\sin ^{-1} \frac{2}{3}ight)=\frac{4}{\sin \left(\sin ^{-1} \frac{2}{3}ight)}+\frac{1}{1-\sin \left(\sin ^{-1} \frac{2}{3}ight)} \ & =4 	imes \frac{3}{2}+3=9\end{aligned}$

An extreme value of $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ in $(0, \pi / 2)$ is

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