Question

关于实数x的不等式|x-|≤与x²-3(a+1)x+2(3a+1)≤0(a∈R)的解集依次为A与B,求使AB的a的取值范围.

Answer 1

解析:解不等式|x-|≤,

-≤x-≤,

∴A={x|2a≤x≤a²+1,a∈R }.

解不等式x²-3(a+1)x+2(3a+1)≤0,［x-(3a+1)］(x-2)≤0,

当a＞时(即3a+1＞2),得B={x|2≤x≤3a+1}.

当a≤时(即3a+1≤2),得B={x|3a+1≤x≤2}.

当a＞时,要满足AB,必须

故1≤a≤3.

当a≤时,要满足AB,必须即

∴a=-1.∴a的取值范围是{a∈R |a=-1或1≤a≤3}.