If the function $f(x)=\left(4, \sin ^2 x-1\right)^n\left(x^2-x+1\right)$, $n \in N$, has a local maximum at $x=\pi / 6$, then 'n'

can be any even natural number

We have,

$f(x) =\left(4 \sin ^2 x-1\right)^n\left(x^2-x+1\right)$

$\Rightarrow f'(x) =n\left(4 \sin ^2 x-1\right)^{n-1} 4 \sin 2 x\left(x^2-x+1\right) +\left(4 \sin ^2 x-1\right)^n(2 x-1)$

$\Rightarrow f'(x)= \left(4 \sin ^2 x-1\right)^{(n-1)}\left[4 n\left(x^2-x+1\right) \sin 2 x+(2 x-1)\left(4 \sin ^2 x-1\right)\right]$

If $f(x)$ has a local maximum at $x=\frac{\pi}{6}$, then

$f'\left(\frac{\pi}{6}\right)=0 \Rightarrow n-1>0 \Rightarrow n>1$

Also, $f'(x)>0$ in the left neighbourhood of $x=\pi / 6$ and $f'(x)<0$ in the right neighbourhood of $x=\pi / 6$. This is possible only when $(n-1)$ is an odd natural number.

Hence, n is an even natural number.