Let $A$ be the set of all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ and $R$ be a relation on $A$ such that $R=\{(f, g): f(0)=g(1)$ and $f(1)=g(0)\}$. Then $R$ is:

(A) Symmetric and transitive but not reflexive

(C) Reflexive but neither symmetric nor transitive

(D) Transitive but neither reflexive nor symmetric

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Sets Relations and Functions

Let

A

be the set of all functions

f : Z \to Z

and

R

be a relation on

A

such that

R = {(f, g) : f (0) = g (1)

and

f (1) = g (0)}

. Then

R

is:

Answer: (B)

For reflexivity,

(f, f) \in R

would require

f (0) = f (1)

, which fails for any function with

f (0) \neq = f (1)

, so

R

is not reflexive. For symmetry, if

f (0) = g (1)

and

f (1) = g (0)

, then

g (0) = f (1)

and

g (1) = f (0)

, which are exactly the conditions for

(g, f) \in R

, so

R

is symmetric. For transitivity, take

(f, g) \in R

and

(g, h) \in R

; then

f (0) = g (1) = h (0)

and

f (1) = g (0) = h (1)

, whereas

(f, h) \in R

would demand

f (0) = h (1)

and

f (1) = h (0)

, which need not hold. Thus

R

is symmetric but neither reflexive nor transitive.

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option