Question

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

• A. $(0,\pi )$
• B. $\left(0,\frac{\pi }{2}\right)$
• C. $\left(0,\frac{\pi }{4}\right)$
• D. $\left(0,\frac{3\pi }{4}\right)$
• E. $\left(0,\frac{5\pi }{4}\right)$

Answer 1

Answer: (C)

$\because$ $f(x)={\tan }^{-1}(\sin x+\cos x)$
On differentiating w.r.t. $x,$ we get
$f(x)=\frac{1}{1+{(\sin x+\cos x)}^2}(\cos x-\sin x)$
$=\frac{1}{1+1+2\sin x\cos x}(\cos x-\sin x)$
$=\frac{\cos x-\sin x}{2(1+\sin x\cos x)}$
For function to be increasing $f(x)>0$
$\Rightarrow$ $\cos x-\sin x>0$
$\Rightarrow$ $\tan x<1$
$\therefore$ Required interval $=\left(0,\frac{\pi }{4}\right)$