Mathematics · 2021
JEE Main · 27 July 2021, Shift 2 · Q74
Let N be the set of natural numbers and a relation R on N be defined by R ={(x, y) ∈ N × N: x^3-3 x^2 y-x y^2+3 y^3=0}. Then the relation R is:
Let $\displaystyle \mathbf{N}$ be the set of natural numbers and a relation R on $\displaystyle \mathbf{N}$ be defined by $\displaystyle \mathrm{R}=\left\{(x, y) \in \mathbf{N} \times \mathbf{N}: x^3-3 x^2 y-x y^2+3 y^3=0\right\}$. Then the relation R is :
Official answer
(2)
reflexive but neither symmetric nor transitive
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.