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Maths Question 12 – JEE-MAIN 2025

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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:

Recall the definitions of reflexive, symmetric, and transitive relations.

🥷
Ninja Strategy•Systematic Property Check

Systematically check reflexivity, symmetry, and transitivity. Reflexivity and symmetry are quick to verify, narrowing down options significantly.

● Video Walkthrough
Step 1: Check for Reflexivity✦ Active

For R to be reflexive, (f,f)∈R must hold for all functions f∈A. According to the definition of R, this means f(0)=f(1) and f(1)=f(0). This simplifies to f(0)=f(1). However, this is not true for all functions f:Z→Z. For example, consider f(x)=x. Then f(0)=0 and f(1)=1, so f(0)≠f(1). Thus, R is not reflexive.

Step 2: Check for Symmetry○ Expand

For R to be symmetric, if (f,g)∈R, then (g,f)∈R. Assume (f,g)∈R. By definition, we have f(0)=g(1) and f(1)=g(0). For (g,f)∈R, we need g(0)=f(1) and g(1)=f(0). These are precisely the conditions given by (f,g)∈R. Therefore, R is symmetric.

Step 3: Check for Transitivity○ Expand

For R to be transitive, if (f,g)∈R and (g,h)∈R, then (f,h)∈R. Let's assume (f,g)∈R and (g,h)∈R:

(f,g)∈R⟹f(0)=g(1) and f(1)=g(0)(1)
(g,h)∈R⟹g(0)=h(1) and g(1)=h(0)(2)

For (f,h)∈R, we would need f(0)=h(1) and f(1)=h(0). From (1) and (2), we can deduce:

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

So, we have f(0)=h(0) and f(1)=h(1). For transitivity to hold, we would need h(0)=h(1) for any h that can be part of such a chain. This is not generally true. Consider a counterexample: Let f(x)=x, g(x)=1−x, and h(x)=x. Then:

1. (f,g)∈R: f(0)=0,f(1)=1. g(0)=1,g(1)=0. f(0)=g(1)⟹0=0 (True). f(1)=g(0)⟹1=1 (True). So (f,g)∈R.

2. (g,h)∈R: g(0)=1,g(1)=0. h(0)=0,h(1)=1. g(0)=h(1)⟹1=1 (True). g(1)=h(0)⟹0=0 (True). So (g,h)∈R.

3. Is (f,h)∈R? We need f(0)=h(1) and f(1)=h(0). f(0)=0,h(1)=1. So f(0)=h(1)⟹0=1 (False). Thus, (f,h)∉R. Therefore, R is not transitive.

Combining the results, R is symmetric but neither reflexive nor transitive.

💡 Teacher's Secret Hint

Carefully construct counterexamples for properties that do not hold. A single counterexample is sufficient to disprove a property.

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