A) \[0<x<\frac{\pi }{3}\]
B) \[-\frac{\pi }{3}<x<0\]
C) \[-\frac{\pi }{3}<x<\frac{\pi }{3}\]
D) \[x=\frac{\pi }{2}\]
Correct Answer: C
Solution :
\[f(x)=\sin x-\frac{x}{2}\Rightarrow f'(x)=\cos x-\frac{1}{2}\] \[f'(x)>0\] for increasing function Obviously it is increasing for \[-\frac{\pi }{3}<x<\frac{\pi }{3}\].
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