CUET Preparation Today
If $R={(x,y)|x, y \in Z, x^2+y^2≤4}$ is a relation on Z, then domain of R is : |
{0, 1, 2} {0, -1, -2} {-2, -1, 0, 1, 2} {-1, 0, 2} |
{-2, -1, 0, 1, 2} |
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 $|x|≤|\sqrt{4-y^2}|$ |
