设集合A={x||x-a|<1,x∈R},B={x||x-b|>2,x∈R}.若A⊆B,则实数a,b必满足( ) A.|a+b|≤3 B.|a+b|≥3 C.|a-b|≤3 D.|a-b|≥3
问题描述:
设集合A={x||x-a|<1,x∈R},B={x||x-b|>2,x∈R}.若A⊆B,则实数a,b必满足( )
A. |a+b|≤3
B. |a+b|≥3
C. |a-b|≤3
D. |a-b|≥3
答
∵A={x|a-1<x<a+1},B={x|x<b-2或x>b+2},
因为A⊆B,所以b-2≥a+1或b+2≤a-1,
即a-b≤-3或a-b≥3,
即|a-b|≥3.
故选D.