Question

The function $f\left(x\right)={\tan }^{-1}(\sin x+\cos x),x>0$ is always an increasing function on the interval

• A. $(0,\pi )$
• B. $(0,\frac{\pi }{2})$
• C. $(0,\frac{\pi }{4})$
• D. $(0,\frac{3\pi }{4})$

Answer 1

Answer: (C)

$(0,\frac{\pi }{4})$

Given, $f\left(x\right)={\tan }^{-1}(\sin x+\cos x)$
$f^{\prime }\left(x\right)=\frac{1}{1+(\sin x+\cos x{)}^2}(\cos x-\sin x)$
$=\frac{1}{1+1+2\sin x\cos x}(\cos x-\sin x)$
$=\frac{\cos x-\sin x}{2(1+\sin x\cos x)}$
For function to be increasing, $f^{\prime }\left(x\right)>0$
$\Rightarrow \cos x-\sin x>0$
$\Rightarrow \tan x<1$
$\therefore$ Required interval $=(0,\frac{\pi }{4})$