Question

關于實數x的不等式|x-(a+1)²|≤(a-1)²與x²-3(a+1)x+2(3a+1)≤0(a∈R)的解集分別為A,B.求使A⊆B成立的a的取值范圍.

Answer 1

a=-1或1≤a≤3

由不等式|x-(a+1)²|≤(a-1)²⇒
-(a-1)²≤x-(a+1)²≤(a-1)²,
解得2a≤x≤a²+1,
于是A={x|2a≤x≤a²+1}.
由不等式x²-3(a+1)x+2(3a+1)≤0⇒(x-2)[x-(3a+1)]≤0,
①當3a+1≥2,即a≥時,B={x|2≤x≤3a+1},
因為A⊆B,所以必有
解得1≤a≤3;
②當3a+1<2,即a<時,B={x|3a+1≤x≤2},
因為A⊆B,所以
解得a=-1.
綜上,使A⊆B的a的取值范圍是a=-1或1≤a≤3.