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) $