P(x) be a polynomial of degree 3 satisfying P(-1) =10 , P(1) =-6 and p(x) has maxima at x = -1 and p(x) has minima at x=1 then The value of P(2) is

Options

A. -15
B. -16
C. -17
D. -22

Answer

3

Explanation

P(x) is polynomial of degree 3
Given p(x) has minima at x=1
`therefore p(x) =a(x-1)`
`therefore p(x)=a(x^(2)/(2)-x)+b`
Given p(x) has maximum at x =-1
`therefore p(-1)=a(x^(2)/(2)-x)+b`
Given p(x) has maxima at x =-1
`therefore b=-3/2a`
`therefore p(x)=a(x^(3)/(6)-(x^(2)/(3)x)+C`
Given p(-1)=10
`therefore a(-1/6-1/2+3/21)+C=10`
`therefore 5a+6c=60`
`therefore p(1)=a(1/6-1/2-3/2)+c=-6`
solving (1) alnd (2) we get
a=6 an c=5
`therefore p(x)=x^(3)-3x^(2)-9x+5`
`therefore p(2)=8-12-18+5=-17`
p(x)=`3x^(2)-6x-9`
`therefore p(1)=3-6=-12`