Question

If R = {(x, y) : x, y ∈ Z, x² + y² ≤ 4} is a relation on Z, then the domain of R is
(a) [0, 1, 2]
(b) [0, −1, −2]
(c) {−2, −1, 0, 1, 2]
(d) None of these

Answer 1

(c) {−2, −1, 0, 1, 2}
R = {(x, y) : x, y ∈ Z, x² + y² ≤ 4}
We know that,
$$\begin{gathered} {\left(-2\right)}^2+0^2\le 4 \\ \Rightarrow {\left(2\right)}^2+0^2\le 4 \\ \Rightarrow {\left(-1\right)}^2+0^2\le 4 \\ \Rightarrow {\left(1\right)}^2+0^2\le 4 \\ \Rightarrow {\left(-1\right)}^2+{\left(1\right)}^2\le 4 \\ \Rightarrow 0^2+0^2\le 4 \\ \Rightarrow {\left(1\right)}^2+{\left(1\right)}^2\le 4 \\ \Rightarrow {\left(-1\right)}^2+{\left(-1\right)}^2\le 4 \end{gathered}$$)2+02 ≤4(2)2+02 ≤4(−1)2+02 2)2+02 ≤4(2)2+02 ≤4(−1)2+02 (−2)2+02 ≤4(2)2+02 ≤4(−1)2+02 ≤4(1)2+02 ≤4(−1)2+(1)2 ≤4
Hence, domain (R) = {$-$2,$-$1,0,1,2,}