JEE Main 2021 · 31 August · morning · Single correct · +4 / −1
Which of the following is not correct for relation R on the set of real numbers ?
- (a) (x, y) ∈ R ⇔ 0 < |x| - |y| ≤ 1 is neither transitive nor symmetric.
- (b) (x, y) ∈ R ⇔ 0 < |x - y| ≤ 1 is symmetric and transitive.
- (c) (x, y) ∈ R ⇔ |x| - |y| ≤ 1 is reflexive but not symmetric.
- (d) (x, y) ∈ R ⇔ |x - y| ≤ 1 is reflexive nd symmetric.
Answer and solution
Answer (b) (x, y) ∈ R ⇔ 0 < |x - y| ≤ 1 is symmetric and transitive.
Solution
Maths sir · 3 steps
Test each claim with one small counterexample; for (b), go from 0 to 1 and back to 0.
- What to check. Reflexive needs (x, x) ∈ R for every real x. Symmetric needs (y, x) ∈ R whenever (x, y) ∈ R. Transitive needs (x, z) ∈ R whenever (x, y) and (y, z) are both in R.
Option Test Result Statement is (a) 0 < |x| - |y| ≤ 1 (1, 0.5) ∈ R but (0.5, 1) ∉ R. Also (1.5, 1), (1, 0.2) ∈ R but |1.5| - |0.2| = 1.3 > 1 not symmetric, not transitive correct (b) 0 < |x - y| ≤ 1 (0, 1) ∈ R and (1, 0) ∈ R, but |0 - 0| = 0, so (0, 0) ∉ R symmetric, not transitive not correct (c) |x| - |y| ≤ 1 |x| - |x| = 0 ≤ 1. (0, 5) ∈ R since -5 ≤ 1, but (5, 0) ∉ R since 5 > 1 reflexive, not symmetric correct (d) |x - y| ≤ 1 |x - x| = 0 ≤ 1, and |y - x| = |x - y| reflexive and symmetric correct - Why (b) fails. The strict condition 0 < |x - y| forbids x = y. Going from 0 to 1 and back to 0 uses two allowed pairs, but transitivity would then demand (0, 0) ∈ R, which is not allowed.
- So. Relation (b) is symmetric but not transitive, so the statement in (b) is the one that is not correct.
The idea to keep. If a relation is symmetric but excludes (x, x), try going x → y → x. Transitivity would force (x, x) ∈ R, so the relation breaks at once.
Answer from the NTA final answer key. Solution by PrepWiser.