高二,不等式问题~~~~

来源：百度知道 编辑：UC知道 时间：2024/05/22 07:49:33
关于实数x的不等式|x-(a+1)^2/2|<=(a+1)^2/2,x^2-3(a+1)x+2(3a+1)<=0的解集分别为A与B,若使A包含于B,求a的取值.
答案:1<=a<=3或a=-1

要过程,谢谢!!!!

由|x-(a+1)^2/2|<=(a-1)^2/2可得
解集A=＜ x!2a<=x<=a2+1 ＞
由x^2-3(a+1)x+2(3a+1)<=0可得
解集B=＜ x!3a+1<=x<=2 ＞当且仅当a<=1/3时成立
或B=＜ x!2<=x<=3a+1 ＞当且仅当a>=1/3时成立
因为A是B的子集，所以
当a<=1/3时，B=＜ x!3a+1<=x<=2 ＞
满足 （ 3a+1<=2a
＜ 解得a=-1
（ a2+1<=2
当a>=1/3时,B=＜ x!2<=x<=3a+1 ＞
满足 （ 2<=2a
＜ 解得1<=a<=3
（ a2+1<=3a+1
综上所得a的取值范围是a=-1或1<=a<=3