If R= (x , y) |. x , y ∈ Z , x2 + y2 ≤ 4 is a relation in Z , then domain of R is

Question

If R= (x , y) |. x , y ∈ Z , x2 + y2 ≤ 4 is a relation in Z , then domain of R is (A)  0 , 1 , 2  (B)  0 , - 1 , - 2  (C)  - 2 , - 1 , 0 , 1 , 2  (D) None of these

Tardigrade Admin · Accepted Answer

∵ R= (x , y)|x , y ∈ Z , x2 + y2 ≤ 4 R= (- 2 , 0) , (- 1 , 0) , (. 0 , - 1 .) , (- 1 , 1) , (. 1 , - 1 .) , (0 , - 1) (0 , 1) , (0 , 2) , (0 , - 2) , (1 , 0) , (. 0 , 1 .) , (1 , 1) , (. - 1 , - 1 .) , (2 , 0) , (. 0 , 0 .) Hence, Domain of R= - 2 , - 1 , 0 , 1 , 2

Q. If $R = {(x, y) ∣ x, y \in Z, x^{2} + y^{2} \leq 4}$ is a relation in $Z$ , then domain of $R$ is

${0, 1, 2}$

${0, - 1, - 2}$

${- 2, - 1, 0, 1, 2}$

None of these

$∵$ $R = {(x, y) ∣ x, y \in Z, x^{2} + y^{2} \leq 4}$
$R = {(- 2, 0), (- 1, 0), (0, - 1), (- 1, 1), (1, - 1), (0, - 1) (0, 1), (0, 2), (0, - 2), (1, 0), (0, 1), (1, 1), (- 1, - 1), (2, 0), (0, 0)}$
Hence, Domain of $R = {- 2, - 1, 0, 1, 2}$

Q. If R={(x,y)∣x,y∈Z,x2+y2≤4} is a relation in Z , then domain of R is

{0,1,2}

{0,−1,−2}

{−2,−1,0,1,2}