91Ó°ÊÓ

To find the critical points of the function, we need to find the first derivative of the given function \(f(x)\). We have: \(f(x) = \left(4 \sin^2 x - 1\right)^n \left(e^x - x + 1\right)\) Using Chain rule and Product rule, we get: \(f'(x) = \left(4 \sin^2 x - 1\right)^{n-1}\left(\left(e^x - x + 1\right)n(4 \sin x\cos x)(4\sin x\cos x - 1) + \left(4 \sin^2 x - 1\right)(e^x - 1) \right)\)

A critical point occurs when the first derivative of the function is zero or does not exist. In this case, we have: \(f'(x) = \left(4 \sin^2 x - 1\right)^{n-1}\left(\left(e^x - x + 1\right)n(4 \sin x\cos x)(4\sin x\cos x - 1) + \left(4 \sin^2 x - 1\right)(e^x - 1) \right)\) Since \(n\) is a natural number and \(4\sin^2x-1\neq0\) and \(e^x\neq0\) We know that the derivative exists for every x. Then, we want to search for the values of x in which the derivative is zero. Set \(f'(x) = 0\): \(\left(e^x - x + 1\right)n(4 \sin x\cos x)(4\sin x\cos x - 1) + \left(4 \sin^2 x - 1\right)(e^x - 1) = 0\) As we're given in problem statement that \(x=\frac{\pi}{6}\) is the local minimum, so this should satisfy the above equation.

Substitute \(x=\frac{\pi}{6}\) into the equation: \(\left(e^{\frac{\pi}{6}} - \frac{\pi}{6} + 1\right)n(4 \sin \frac{\pi}{6}\cos \frac{\pi}{6})(4\sin \frac{\pi}{6}\cos \frac{\pi}{6} - 1) + \left(4 \sin^2 \frac{\pi}{6} - 1\right)(e^{\frac{\pi}{6}} - 1) = 0\) By simplifying the equation, we get: \((\ldots)n(\ldots)=-\ldots(\ldots)\) From this equation, we can see that all the terms are nonzero. So there is no restriction on the value of \(n\). Therefore, \(n\) can be any natural number. The correct answer is (D) n can be any natural number.