A) even
B) odd
C) any natural number
D) None of these
Correct Answer: A
Solution :
\[f(x)={{(2\sin x+1)}^{n}}{{(2\sin x-1)}^{n}}({{x}^{2}}-x+1)\] If n is odd \[f(x)\] changes sign about \[x=\frac{\pi }{6}\]. i.e., \[f\left( \frac{\pi }{6}+h \right)>f\left( \frac{\pi }{6} \right)\] and \[f\left( \frac{\pi }{6}-h \right)<f\left( \frac{\pi }{6} \right)\] \[\Rightarrow \] \[f(x)\] has neither maxima nor minima at \[\pi /6\].
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