Let `p(x)` be a real polynomial of least degree which has a local maximum at `x=1` and a local minimum at `x=3.` If `p(1)=6a n dp(3)=2,` then `p^(prime)(0)` is_____

Options

A. 8
B. 9
C. 3
D. 6

Answer

B

Explanation

Since x=1 and x=3 are extreme points of p(x).
Therefore , x=1 and x=3 are roots of p(x) =0
`therefore p(x) =k(x-1)(x-3)=k(x^2-4x+3)`
`rArr p(x)=(x^3/3-2x^2+3x)+c`
Now p(1)=6 and p(3)=2
`rArr 6+(4k)/(3)+C and C=2`
`rArr C=2 and k=3`
`therefore p(x)=3(x^2-4x+3)`
Hence p(0)=9