Sets and Relations – Types of Relations – JEE Main 2 Apr 2025 Shift 1

Solution:

Relation definition: $f R g \iff f(0) = g(1) \text{ and } f(1) = g(0)$.

1. **Reflexive:**
For $f R f$, we need $f(0) = f(1)$ and $f(1) = f(0)$.
This implies $f(0) = f(1)$. Since this is not true for all functions in $A$ (e.g., $f(x) = x$, where $0 \neq 1$), $R$ is not reflexive.

2. **Symmetric:**
Let $f R g$. Then $f(0) = g(1)$ and $f(1) = g(0)$.
We need to check if $g R f$, which requires $g(0) = f(1)$ and $g(1) = f(0)$.
These conditions are exactly the same as the given ones.
Thus, $R$ is symmetric.

3. **Transitive:**
Let $f R g$ and $g R h$.
Given:
(i) $f(0) = g(1)$ and $f(1) = g(0)$
(ii) $g(0) = h(1)$ and $g(1) = h(0)$

For $f R h$, we need $f(0) = h(1)$ and $f(1) = h(0)$.
From the given equations:
$f(0) = g(1) = h(0)$ (Substitute $g(1)$ from ii)
$f(1) = g(0) = h(1)$ (Substitute $g(0)$ from ii)

So we found $f(0) = h(0)$ and $f(1) = h(1)$.
We need $f(0) = h(1)$, but we have $f(0) = h(0)$.
This holds only if $h(0) = h(1)$, which is not always true.
Thus, $R$ is not transitive.

Conclusion: $R$ is symmetric but neither reflexive nor transitive.

Ans. (2)