Question

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

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

Answer 1

Answer: (C)

$f(x)={\tan }^{-1}(\sin x-\cos x)$
$f^{\prime }(x)=\frac{\cos x+\sin x}{(\sin x-\cos x{)}^2+1}=0$
$\therefore x=\frac{3\pi }{4}$

$\text{sum}={\tan }^{-1}\sqrt{2}-\frac{\pi }{4}$
$={\cos }^{-1}\frac{1}{\sqrt{3}}-\frac{\pi }{4}$