JEE Main 2025 Apr 2 Shift 1, Mathematics Q12: Relations

Reflexivity. $(f,f)\in R$ would require $f(0)=f(1)$. This is not generally true for $f\in A$ (e.g. take $f(n)=n$). So $R$ is not reflexive.

Symmetry. Suppose $(f,g)\in R$, i.e. $f(0)=g(1)$ and $f(1)=g(0)$. Then immediately $g(0)=f(1)$ and $g(1)=f(0)$, which is exactly the condition $(g,f)\in R$. So $R$ is symmetric.

Transitivity. Suppose $(f,g),(g,h)\in R$:

From $(f,g)$: $f(0)=g(1)$, $f(1)=g(0)$.

From $(g,h)$: $g(0)=h(1)$, $g(1)=h(0)$.

Combining: $f(0)=g(1)=h(0)$ and $f(1)=g(0)=h(1)$.

But for $(f,h)\in R$ we need $f(0)=h(1)$ and $f(1)=h(0)$ — these are different equalities. A counterexample with $f(0)\ne f(1)$ shows $(f,h)\notin R$ in general. So $R$ is not transitive.

$\therefore$ $R$ is symmetric only. Option (2) is correct.