Question

集合A={-2≤x≤5},区间B={m+1,2m-1},若B∪A=A,则实数m的取值范围为多少？

Answer 1

if B∪A=A
=> B is subset of A
case 1: if B不是空集
ie 2m-1> m+1
m > 2
B is subset of A
=> m+1> -2 and 2m-1<5
=> m > -3 and m<3
ie -3<m<3
solution of case 1:
-3<m<3 and m >2
ie 2<m<3
case 1: if B是空集
ie m <= 2
B is subset of A

B∪A=A,则实数m的取值范围
case1 or case2
2<m<3 or m<=2
ie m < 3

Answer 2

BUA=A说明B是A的子集，又题中说区间B，有m+1≥-2且2m-1≤5且有2m-1＞m+1，得2＜m≤3。