设关于x的不等式丨x-(a+1)^2/2丨≤(a-1)^2/2的解集为A,x^2-3(a+1)x+2（3a+1）≤0的解集为B求使A∩B=B的a的取值范围.

Question

设关于x的不等式丨x-(a+1)^2/2丨≤(a-1)^2/2的解集为A,x^2-3(a +1)x+2（3a+1）≤0的解集为B
求使A∩B=B的a的取值范围.

Answer 1

-(a-1)²/2<=x-(a+1)²/2<=(a-1)²/2
-(a-1)²/2+(a+1)²/2<=x<=(a-1)²/2+(a+1)²/2
2a<=x<=a²+1
(x-2)[x-(3a+1)]<=0
a=1/3,则x=2
不满足2a<=x<=a²+1
a<1/3,3a+1<=x<=2
则有2a<=3a+1且2<=a²+1
所以a=-1
a>1/3,2<=x<=3a+1
则有2a<=2且3a+1<=a²+1
这个不鞥同时成立
所以a=1