Let S={1,2,3, …, 10}. Suppose M is the set of all the subsets of S, then the relation
R={(A, B): A ∩ B ≠ φ ; A, B ∈ M} is :

Solution

Maths sir · 4 steps

Break each property with one counterexample: the empty set kills reflexive, a chain of sets kills transitive.

1. Reflexive fails. Reflexive needs (A, A) ∈ R for every A ∈ M. The empty set φ is a subset of S, so it is in M, but φ ∩ φ = φ. So (φ, φ) ∉ R.
2. Symmetric holds. If A ∩ B ≠ φ, then B ∩ A ≠ φ too, because intersection does not depend on order. So (A, B) ∈ R always gives (B, A) ∈ R.
3. Transitive fails. Take A = {1, 2}, B = {2, 3}, C = {3, 4}. Then A ∩ B = {2} and B ∩ C = {3}, so both pairs are in R. But A ∩ C = φ, so (A, C) ∉ R.
4. So. R is symmetric but neither reflexive nor transitive. The answer is (a), symmetric only.

The idea to keep. To prove a property, argue for every element. To break it, one counterexample is enough. With subsets, always check the empty set first.

Answer from the NTA final answer key. Solution by PrepWiser.