Question

设集合A={(x,y)|m/2≤(x-2)²+y²≤m²,x,y∈R}

B={(x,y)|2m≤x+y≤2m+1,x,y∈R},若A交B≠∅，则m的取值范围是

Answer 1

A表示圆环，B表示两平行线间的区域.
A∩B≠∅，
<==>大圆与平行线之一有公共点，
<==>|2-(2m+1)|/√2<=|m|,或|2-2m|/√2<=|m|,
<==>|2m-1|<=√2|m|,或√2|m-1|<=|m|,
平方得4m^2-4m+1<=2m^2,或2(m^2-2m+1)<=m^2,
整理得2m^2-4m+1<=0,或m^2-4m+2<=0,
解得（2-√2)/2<=m<=(2+√2)/2,或2-√2<=m<=2+√2，
求并集得（2-√2)/2<=m<=2+√2，为所求.