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Domain of the Relation x²+y²≤4 over Integers | JEE

JEE Maths question with a full step-by-step solution.

Question
If R={(x,y)x,yZ, x2+y24}R = \{(x, y) \mid x, y \in Z,\ x^2 + y^2 \le 4\} is a relation in ZZ, then the domain of RR is:
A{0,1,2}\{0, 1, 2\}
B{0,1,2}\{0, -1, -2\}
C{2,1,0,1,2}\{-2, -1, 0, 1, 2\}correct
DNone of these
Solution
For xx to appear, some integer yy must satisfy x2+y24x^2 + y^2 \le 4. Taking y=0y = 0:
x24x{2,1,0,1,2},x^2 \le 4 \Rightarrow x \in \{-2, -1, 0, 1, 2\},
and each gives a valid point (x,0)(x, 0). \therefore Domain ={2,1,0,1,2}.= \{-2, -1, 0, 1, 2\}. Correct answer: (3)
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