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

{−2,−1,0,1,2} ∵R={(x,y)|x,y∈Z,x2+y2≤4}R={(−2,0),(−1,0),(0,−1),(−1,1),(1,−1),(0,−1)(0,),(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}

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

{−2,−1,0,1,2} ∵R={(x,y)|x,y∈Z,x2+y2≤4}R={(−2,0),(−1,0),(0,−1),(−1,1),(1,−1),(0,−1)(0,),(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}

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

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$\{1, 2\}$

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

${0, 1, 2}$

${0, - 1, - 2}$

answer is C.

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Detailed Solution

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

$R = {(- 2, 0), (- 1, 0), (0, - 1), (- 1, 1), (1, - 1), (0, - 1) (0,), (0, 2), (0, - 2), (1, 0), (0, 1), (1, 1), (- 1, - 1), (2, 0), (0, 0)}$ Hence, Domain of $R = {- 2, - 1, 0, 1, 2}$