JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Sets Relations and Functions

Since $(1,2)\in R$ and $(2,3)\in R$, transitivity forces $(1,3)\in R$. Reflexivity forces $(1,1),(2,2),(3,3)\in R$. So every valid relation must contain


$R_0$ itself is reflexive, transitive, and not symmetric (since $(2,1)\notin R_0$ though $(1,2)\in R_0$) — one valid relation.

The only pairs that could still be added are $(2,1),(3,2),(3,1)$.

Adding $(2,1)$ alone: transitivity needs $(2,1),(1,2)\Rightarrow (2,2)$ (already there) and $(2,1),(1,3)\Rightarrow (2,3)$ (already there); consistent, and the relation stays non-symmetric since $(3,1)\notin R$ while $(1,3)\in R$. Valid.

Adding $(3,2)$ alone: transitivity needs $(3,2),(2,3)\Rightarrow (3,3)$ (already there) and $(1,3),(3,2)\Rightarrow (1,2)$ (already there); consistent and non-symmetric. Valid.

Adding $(3,1)$ alone: transitivity needs $(3,1),(1,2)\Rightarrow (3,2)$, which is not present, so this fails unless $(3,2)$ is also added.

Adding both $(2,1)$ and $(3,2)$: transitivity then forces $(3,2),(2,1)\Rightarrow (3,1)$, and with $(3,1)$ alongside $(1,3)$, $(2,1)$ alongside $(1,2)$, $(3,2)$ alongside $(2,3)$, the relation becomes fully symmetric — not allowed.

Exactly three relations work: $R_0$, $R_0\cup \{(2,1)\}$, and $R_0\cup \{(3,2)\}$.

The count is 3.