Question

The global maximum value of $f(x)=\cot x-\sqrt{2}\csc x$ in interval $(0,\pi )$ is equal to

• A. 1
• B. - 1
• C. 0
• D. non-existent

Answer 1

Answer: (B)

We have $f(x)=\cot x-\sqrt{2}\mathrm{cosec}x=\frac{\cos x-\sqrt{2}}{\sin x}$
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 $f(x)-\cot x-\sqrt{2}\mathrm{cosec}x$ in $(0,\pi )$
Also, ${\mathrm{Lim}}_{x\to 0^{+}}f(x)={\mathrm{Lim}}_{x\to 0^{+}}\left(\frac{\cos x-\sqrt{2}}{\sin x}\right)\to -\infty$ and ${\mathrm{Lim}}_{x\to {\pi }^{-}}f(x)={\mathrm{Lim}}_{x\to {\pi }^{-}}\left(\frac{\cos x-\sqrt{2}}{\sin x}\right)\to -\infty$ 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 $f\left(x=\frac{\pi }{4}\right)=1-\sqrt{2}(\sqrt{2})=1-2=-1$.