Question

设集合A={(x,y)|(x-2)^2+y^2=1,xy∈R},B={(x,y)|2m≤x+y≤2m+1,xy∈R},若A∪B≠空集，则m的取值范围是

设集合A={(x,y)|(x-2)^2+y^2=1,xy∈R},B={(x,y)|2m≤x+y≤2m+1,xy∈R},若A∪B≠空集，则m的取值范围是
要过程为什么

Answer 1

集合A={(x,y)|(x-2)^2+y^2=1,xy∈R}表示以C(2，0)为圆心，1为半径的圆,
B={(x,y)|2m≤x+y≤2m+1,xy∈R}表示平行线l1:x+y=2m与l2:x+y=2m+1之间的区域,
A∪B≠空集，
<==>C到l1或l2的距离<=1,
<==>|2-2m|/√2<=1,或|2-(2m+1)|/√2<=1,
<==>-√2<=2m-2<=√2，或-√2<=2m-1<=√2，
<==>(2-√2)/2<=m<=(2+√2)/2,或(1-√2)/2<=(1+√2)/2,
求并集得(1-√2)/2<=m<=(2+√2)/2,为所求.