The function `f(x)=(4sin^2x-1)^n(x^2-x+1),n in N ,` has a local minimum at `x=pi/6` . Then `n` is any even number n is an odd number n is odd prime number `n` is any natural number

Options

A. can be any odd natural number
B. can only be an odd prime number d
C. can be any even natural number
D. can only be a multiple of 4.

Answer

C

Explanation

We have
`f(x)=(4 sin^2 x-1 )^n(x^2-x+1)`
`rArr f(x)=n(4 sin x -1 )^(n-1)4 sin 2x(x^2-x+1)+(4 sin ^2x -1)^n(2x-1)`
`rArr f(x)=(4 sin^2x-1)^(n-1)[4n(x^2-x+1)sin 2x +(2x-1)(4sin^2 x-1)]`
if f(x) has a local maximum at `x (pi)/(6)` then
`f((pi)/(6))=0 rArr n-1 gt 0 rArr n gt 1`
Also `f(x)gt 0` in the left neighbourhood of `x=pi//6` and `f(x) lt 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 .