已知两集合A={x|x=t^2+(a+1)t+b,x∈R},B={x|x=-t^2-(a-1)t-b,t∈R},且A∩B={x|-1≤x≤2}.

题目

已知两集合A={x|x=t^2+(a+1)t+b,x∈R},B={x|x=-t^2-(a-1)t-b,t∈R},且A∩B={x|-1≤x≤2}.
求常数a、b的值

答案

A={x|x=t^2+(a+1)t+b,x∈R}
x=t^2+(a+1)t+b
x' = 2t+(a+1) =0
t = -(a+1)/2
x'' = 2 >0 (min)
min x = x(-(a+1)/2)
=(a+1)^2/4 - (a+1)^2/2 +b
= -(a+1)^2/4 +b
A = { x | x ≥-(a+1)^2/4 +b }
B={x|x=-t^2-(a-1)t-b,t∈R}
x= -t^2-(a-b)t-b
x' =-2t -(a-b)=0
t = -(a-b)/2
x''= -2 >0 (max)
max x = x( -(a-1)/2)
= -(a-1)^2/4 +b
B={x|x≤-(a-1)^2/4 +b }
A∩B={x|-1≤x≤2}
=>
-(a+1)^2/4 +b =-1 (1) and
-(a-1)^2/4 +b = 2 (2)
(2)-(1)
-(a-1)^2/4 +(a+1)^2/4 =3
a= 3
b=3