A) \[\frac{\pi }{2}\]
B) \[2\pi \]
C) \[\pi \]
D) None of these
Correct Answer: C
Solution :
We observe that \[f(x+\pi )=\frac{|\sin (\pi +x)|-|\cos (\pi +x)|}{|\sin (x+\pi )+\cos (x+\pi )|}\] \[\Rightarrow \] \[f(x+\pi )=\frac{|\sin x|-|\cos x|}{|-\sin x-\cos x|}\] \[=\frac{|\sin x|-|\cos x|}{|\sin x+\cos x|}\] \[\Rightarrow \] \[f(x+\pi )=f(x),\,\,\forall x\in R\] Therefore, \[f(x)\] is periodic with period\[\pi \].
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