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Question: If \(R = \left\{ ( x , y ) \mid x , y \in Z , x ^ { 2 } + y ^ { 2 } \leq 4 \right\}\) is a relation ...

If R={(x,y)∣x,y∈Z,x2+y2≤4}R = \left\{ ( x , y ) \mid x , y \in Z , x ^ { 2 } + y ^ { 2 } \leq 4 \right\} 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

Answer

{– 2, – 1, 0, 1, 2}

Explanation

Solution

∵R={(x,y)∣x,y∈Z,x2+y2≤4}\because R = \left\{ ( x , y ) \mid x , y \in Z , x ^ { 2 } + y ^ { 2 } \leq 4 \right\}

∴\therefore R={(−2,0),(−1,0),(−1,1),(0,−1)(0,1),(0,2),(0,−2)R = \{ ( - 2,0 ) , ( - 1,0 ) , ( - 1,1 ) , ( 0 , - 1 ) ( 0,1 ) , ( 0,2 ) , ( 0 , - 2 ) (1,0),(1,1),(2,0)}( 1,0 ) , ( 1,1 ) , ( 2,0 ) \}

Hence, Domain of R={−2,−1,0,1,2}R = \{ - 2 , - 1,0,1,2 \} .