Question

集合A={x|x²-[2a+(a²+1)]x+2a(a²+1)≤0},B=｛x|(x-2)[x-(3a+1)]≤0｝当实

Answer 1

①

(x-2a)[x-(a²+1)]≤0;

∵a²+1≥2a;

∴ 2a≤x≤a²+1;

②当a>1/3;

(x-2)[x-(3a+1)]≤0;

2≤x≤3a+1;

∴2a≥2;

3a+1≥a²+1

∴1≤a≤3

③ 当a≤1/3;

(x-2)[x-(3a+1)]≤0;

3a+1≤x≤2;

∴2a≥3a+1;

2≥a²+1;

∴a=-1;

∴综上得：

a=-1或1≤a≤3