If $R={(x,y)|x, y \in Z, x^2+y^2≤4}$ is a relation on Z, then domain of R is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\{-2, -1, 0, 1, 2\}$
$x^2+y^2≤4$
Representing on graph
all crosses are points ∈ R
all points are unit distanced
$⇒ x∈\{-2,-1,0,1,2\}$
so $|x|≤|\sqrt{4-y^2}|$