JEE Main 2021 (Online) 27th July Evening ShiftQuestion 1
Let $\mathbf{N}$ be the set of natural numbers and a relation $R$ on $\mathbf{N}$ be defined by $R=\{(x,y)\in\mathbf{N}\times\mathbf{N}:x^3-3x^2y-xy^2+3y^3=0\}$. Then the relation $R$ is :
Sets, Relations and Function
Solution
$x^3-3x^2y-xy^2+3y^3=(x-3y)(x^2-y^2)=0\Rightarrow x=3y$ or $x=\pm y$ (natural numbers: $x=y$ or $x=3y$). Reflexive: $x=y$ works. Not symmetric: $(3,1)\in R$ but $(1,3)\notin R$. Not transitive: $(9,3),(3,1)\in R$ but $(9,1)\notin R$. Option (B).