JEE Main 2024 (Online) 1st February Evening ShiftQuestion 1
Consider the relations $R_1$ and $R_2$ defined as $aR_1b\Leftrightarrow a^{2}+b^{2}=1$ for all $a,b\in\mathbf{R}$ and $(a,b)R_2(c,d)\Leftrightarrow a+d=b+c$ for all $(a,b),(c,d)\in\mathbf{N}\times\mathbf{N}$. Then
Sets, Relations and Function
Solution
$R_1$: $a=0$ gives $0+0\ne1$, not reflexive. $R_2$: $a+d=b+c$ is equivalent to $a-b=c-d$, an equivalence relation. Option (B).