JEE Main 2025 · 2 April · morning · Single correct · +4 / −1
Let A be the set of all functions f: Z → 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 reflective
- (b) Symmetric but neither reflective nor transitive
- (c) Transitive but neither reflexive nor symmetric
- (d) Reflexive but neither symmetric nor transitive
Answer and solution
Answer (b) Symmetric but neither reflective nor transitive
Solution
Maths sir · 4 steps
The rule swaps the values at 0 and 1. Swap twice and you are back where you started, so transitivity breaks.
- Reflexive fails. For (f, f) ∈ R you need f(0) = f(1). Take f(x) = x: here f(0) = 0 and f(1) = 1, so (f, f) ∉ R.
- Symmetric holds. If (f, g) ∈ R, then f(0) = g(1) and f(1) = g(0). Read the same two equations from the other side: g(0) = f(1) and g(1) = f(0). That is exactly the condition for (g, f) ∈ R.
- Transitive fails. Take f(x) = x, g(x) = 1 - x and h(x) = x. Then (f, g) ∈ R because f(0) = 0 = g(1) and f(1) = 1 = g(0). Also (g, h) ∈ R because g(0) = 1 = h(1) and g(1) = 0 = h(0). But (f, h) = (f, f) ∉ R because f(0) ≠ f(1).
- So. R is symmetric but neither reflexive nor transitive. The answer is (b).
The idea to keep. When a relation swaps two things, it is symmetric automatically. Two swaps undo each other, so transitivity usually fails. Test each property with a simple function like f(x) = x.
Answer from the NTA final answer key. Solution by PrepWiser.