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

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1.Relation and Function

easy

If $R = \{ (x,\,y)|x,\,y \in Z,\,{x^2} + {y^2} \le 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

Solution

(c) $\because \,R = \{ (x,\,y)|x,\,y \in Z,\,{x^2} + {y^2} \le 4\} $

$\therefore R = \{ ( – 2,\,0),( – 1,\,0),( – 1,\,1),(0,\, – 1)(0,\,1),(0,\,2),(0, – 2)$

$(1,\,0),(1,\,1),(2,\,0)\} $

Hence, Domain of $R = \{ – 2,\, – 1,\,0,\,1,\,2\,\} $.

Standard 12

Mathematics

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