Thank you for reporting, we will resolve it shortly
Q.
The global maximum value of $f(x)=\cot x-\sqrt{2} \csc x$ in interval $(0, \pi)$ is equal to
Application of Derivatives
Solution:
We have $f(x)=\cot x-\sqrt{2} \operatorname{cosec} x=\frac{\cos x-\sqrt{2}}{\sin x}$
So, $ f^{\prime}(x)=\frac{\sin x(-\sin x)-(\cos x-\sqrt{2}) \cos x}{\sin ^2 x}=\frac{-1+\sqrt{2} \cos x}{\sin ^2 x}$
$\therefore f ^{\prime}( x )=0 \Rightarrow \cos x =\frac{1}{\sqrt{2}} \Rightarrow x =\frac{\pi}{4} \in(0, \pi)$
$\therefore f ( x ) \uparrow$ on $\left(0, \frac{\pi}{4}\right)$ and $f ( x ) \downarrow$ on $\left(\frac{\pi}{4}, \pi\right)$
Graph of $f ( x )-\cot x -\sqrt{2} \operatorname{cosec} x$ in $(0, \pi)$
Also, $\operatorname{Lim}_{x \rightarrow 0^{+}} f(x)=\operatorname{Lim}_{x \rightarrow 0^{+}}\left(\frac{\cos x-\sqrt{2}}{\sin x}\right) \rightarrow-\infty$ and $\operatorname{Lim}_{x \rightarrow \pi^{-}} f(x)=\operatorname{Lim}_{x \rightarrow \pi^{-}}\left(\frac{\cos x-\sqrt{2}}{\sin x}\right) \rightarrow-\infty$ and $f(x)$ is also continuous on $(0, \pi)$
Clearly $f ( x )<0 \forall x \in(0, \pi)$
$\therefore $ At $x=\frac{\pi}{4}$ (a local maximum point), so $f(x)$ takes its absolute maximum value also at $x=\frac{\pi}{4}$.
Hence, absolute maximum value of $f \left( x =\frac{\pi}{4}\right)=1-\sqrt{2}(\sqrt{2})=1-2=-1$.
