Question

The function $f\left(x\right)={\tan }^{-1}(\sin x+\cos x)$ is strictly increasing function in

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

Answer 1

The correct option is C $(0,\frac{\pi }{4})$

$\begin{array}{cc}f\left(x\right)={\tan }^{-1}\left(\sin x+\cos x\right)& \\ forfntobeincrea\sin g& f^{\prime }\left(x\right)>0\\ or\frac{df\left(x\right)}{dx}=\frac{d}{dx}{\tan }^{-1}\left(X\right)& X=\sin x+\cos x\\ =\frac{1}{1+X^2}\frac{d\left(X\right)}{dx}& \\ or\frac{1}{1+\left({\sin }^{2}x+{\cos }^{2}x\right)+2\sin x\cos x}\frac{d}{dx}\left(\sin x+\cos x\right)& \\ or\frac{\cos x-\sin x}{2+\sin 2x}& \\ or2+\sin 2xisalways+ve\&\cos x-\sin x\ge 0& \\ or\tan x<1& \end{array}$

from graph of $tanx$, the above condition is possible only when $x$ lies between $0to\frac{\pi }{4}$

$0<x<\frac{\pi }{4}$