Question

關(guān)于實(shí)數(shù)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

【解析】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(dāng)3a+1≥2,即a≥時(shí),B={x|2≤x≤3a+1},

因?yàn)?/span>A⊆B,所以必有

解得1≤a≤3;

②當(dāng)3a+1<2,即a<時(shí),B={x|3a+1≤x≤2},

因?yàn)?/span>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.