Let $\mathbf{N}$ be the set of natural numbers and a relation $R$ on $N$ be defined by
$\mathrm{R}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbf{N} \times \mathbf{N}: \mathrm{x}^{3}-3 \mathrm{x}^{2} \mathrm{y}-\mathrm{xy}^{2}+3 \mathrm{y}^{3}=0\right\}$
Then the relation $R$ is :
Correct Option: , 2
$x^{3}-3 x^{2} y-x y^{2}+3 y^{3}=0$
$\Rightarrow x\left(x^{2}-y^{2}\right)-3 y\left(x^{2}-y^{2}\right)=0$
$\Rightarrow(x-3 y)(x-y)(x+y)=0$
Now, $x=y \quad \forall(x, y) \in N \times N$ so reflexive
But not symmetric \& transitive
See, $(3,1)$ satisfies but $(1,3)$ does not. Also $(3,1) \&$ $(1,-1)$ satisfies but $(3,-1)$ does not
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