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) \[\left( 0,\frac{5\pi }{4} \right)\]
Correct Answer: C
Solution :
\[\because \] \[f(x)={{\tan }^{-1}}(\sin x+\cos x)\] 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\[=\left( 0,\frac{\pi }{4} \right)\]
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