Let `f(x) = (x^2 - 1)^(n+1) + (x^2 + x + 1)`. Then f(x) has local extremum at `x = 1`, when n is (A) `n = 2` (C) `n = 4` (B) `n = 3 ` (D)` n = 5 `

Options

A. `n=2`
B. `n=6`
C. `n=4`
D. `n=5`

Answer

D

Explanation

We have
`f(x)=(x^(2)-1)^(1+1)(x^(2)+x+1)`
On differentiating w.r.t. x, we get
`f'(x)=[(n+1)(x^(2)-1)^(1)(2x) (x^(2)+x+1)+(x^(2)-1)^(1+1)(2x+1)]`
If n is even, then derivative does not change its sign when `xx` passes through 1. But f'(x) changes its sign from negative to positive when n is odd.
Hence f(x) has local extremum at `x=1,`when n=5