JEE Mathematics 2025 Question with Solution

Given: $A$ is the set of all functions $f: \mathbb{Z} \to \mathbb{Z}$ and $R$ on $A$ is defined by

$$R = \{(f,g): f(0)=g(1) \text{ and } f(1)=g(0)\}$$

Find: Whether $R$ is reflexive, symmetric, and transitive.

To test the relation, we use the standard definitions of these three properties.

Reflexive: For reflexivity, we need $(f,f) \in R$ for every $f \in A$. This requires

$$f(0)=f(1) \quad \text{and} \quad f(1)=f(0)$$

So we must have $f(0)=f(1)$ for every function in $A$. This is false because $A$ contains all functions from $\mathbb{Z}$ to $\mathbb{Z}$. For example, if $f(x)=x$, then $f(0)=0$ and $f(1)=1$, so $f(0) \ne f(1)$. Hence, $R$ is not reflexive.

Symmetric: Assume $(f,g) \in R$. Then

$$f(0)=g(1) \quad \text{and} \quad f(1)=g(0)$$

For $(g,f) \in R$, we need

$$g(0)=f(1) \quad \text{and} \quad g(1)=f(0)$$

These are exactly the same equalities written in reverse order. Therefore, whenever $(f,g) \in R$, we also have $(g,f) \in R$. Hence, $R$ is symmetric.

Transitive: Assume $(f,g) \in R$ and $(g,h) \in R$. Then

$$f(0)=g(1), \quad f(1)=g(0)$$

and

$$g(0)=h(1), \quad g(1)=h(0)$$

From these, we get

$$f(0)=h(0) \quad \text{and} \quad f(1)=h(1)$$

But for $(f,h) \in R$, we would need

$$f(0)=h(1) \quad \text{and} \quad f(1)=h(0)$$

This does not follow.

A counterexample given is:

• $f(0)=1, \; f(1)=2$
• then choose $g$ so that $g(1)=1, \; g(0)=2$
• then choose $h$ so that $h(1)=2, \; h(0)=1$

Now $fRg$ and $gRh$ hold, but for $fRh$ we need $f(0)=h(1)$. Here $f(0)=1$ and $h(1)=2$, so this fails. Hence, $R$ is not transitive.

Therefore, the relation is symmetric but neither reflective nor transitive. The correct option is B.

Given: $R$ is defined by swapping the values at $0$ and $1$:

$$fRg \iff f(0)=g(1) \text{ and } f(1)=g(0)$$

Find: The nature of the relation.

Think of $R$ as a relation that matches the pair $\big(f(0), f(1)\big)$ with the reversed pair $\big(g(1), g(0)\big)$.

• Not reflexive: We would need every function to satisfy $f(0)=f(1)$, which is not true for all functions.
• Symmetric: If the pair for $f$ matches the reversed pair for $g$, then the pair for $g$ also matches the reversed pair for $f$.
• Not transitive: Reversing twice brings us to equality of corresponding values, not to the swapped condition required again.

Therefore, $R$ is symmetric but neither reflective nor transitive. The correct option is B.