Question

The sum of the absolute maximum and absolute minimum values of the function $f\left(x\right)={\tan }^{-1}(\sin x-\cos x)$ in the interval $[0,\pi ]$ is

• A. ${\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi }{4}$
• B. 0
• C. ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi }{4}$
• D. $\frac{-\pi }{12}$

Answer 1

The correct option is C ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi }{4}$

$f\left(x\right)={\tan }^{-1}(\sin x-\cos x)$
Let $g\left(x\right)=\sin x-\cos x$
$=\sqrt{2}\sin (x-\frac{\pi }{4})$ and $x-\frac{\pi }{4}\in [\frac{-\pi }{4},\frac{3\pi }{4}]$
$\therefore g\left(x\right)\in [-1,\sqrt{2}]$
and ${\tan }^{-1}x$ is an increasing function
$\therefore f\left(x\right)\in \left[{\tan }^{-1}\right(-1),{\tan }^{-1}\sqrt{2}]$
$\in [-\frac{\pi }{4},{\tan }^{-1}\sqrt{2}]$
$\therefore$ Sum of $f_{max}$ and $f_{min}={\tan }^{-1}\sqrt{2}-\frac{\pi }{4}$
$={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi }{4}$