JEE Main 2023 · 8 April · evening · Single correct · +4 / −1
Let A={1,2,3,4,5,6,7}. Then the relation R={(x, y) ∈ A × A: x+y=7} is :
- (a) reflexive but neither symmetric nor transitive
- (b) transitive but neither symmetric nor reflexive
- (c) symmetric but neither reflexive nor transitive
- (d) an equivalence relation
Answer and solution
Answer (c) symmetric but neither reflexive nor transitive
Solution
Maths sir · 3 steps
Write out every pair first. Once you see the six pairs, each property takes one line to check.
- List the pairs. We need x + y = 7 with both numbers in A. That gives R = {(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)}. The element 7 has no partner, because its partner would have to be 0.
Property Test Result Reflexive Need (x,x) for every x, so 2x = 7, which gives x = 3.5 ∉ A fails Symmetric If x + y = 7, then y + x = 7 as well holds Transitive (1,6) and (6,1) are in R, but (1,1) is not, since 1 + 1 ≠ 7 fails - One counterexample. To show a property fails, a single counterexample is enough. The pairs (1,6) and (6,1) break transitivity, and (1,1) ∉ R breaks reflexivity.
- So. R is symmetric but neither reflexive nor transitive, which is option (c).
The idea to keep. For a symmetric relation, take any pair (a,b) with its reverse (b,a). Transitivity then forces (a,a), so check whether (a,a) is in the relation.
Answer from the NTA final answer key. Solution by PrepWiser.