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If R=x,y|x, yz, x2+y24 is a relation in Z, then domain of R is

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By Expert Faculty of Sri Chaitanya
a
0, 1, 2
b
0, -1, -2
c
-2, -1, 0, 1, 2
d
None of these
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detailed solution

Correct option is C

∵R=x,y|x,y∈Z, x2+y2≤4∴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.


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