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. n is any even integer
B. n is an odd integer
C. n is odd prime number
D. n is any natural number

Answer

1

Explanation

f(X) =`(4 sin ^(2)x-1)^(n)(x^(2)-x+1)`
`x^(2)-x+1gt0 forall x `
`f((pi)/(6))=0`
`f(pi^(+))/(6)=underset(xrarr(pi)/(6)lim(4sin^(2)x-1)^n)(x^(2)-x+1)=rarr0^(+)`
`f((pi)/(6))=underset(xrarr(pi)/(6))lim(4sin^(2)x-x+1)`
`=rarr 0^(-)(n)` (A positive value)
Thus `f((pi^(-))/(6))gt0 `if n is an even numbr