If A={(x,y):x2+y2≤1;x,yϵR} and B={(x,y):x2+y2≥4;x,yϵR}, then
A
A−B=ϕ
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B
B−A=ϕ
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C
A∩B≠ϕ
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D
A∩B=ϕ
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Solution
The correct option is DA∩B=ϕ A is the set of all points on the inner circle x2+y2=1.B is the set of all points on the outer circle x2+y2=4. ∴A−B=A,B−A=B,A∩B=ϕ.