JEE Main 2021 · 27 July · evening · Single correct · +4 / −1
Let N be the set of natural numbers and a relation R on N be defined by R = { (x,y) ∈ N × N:x³ - 3x²y - xy² + 3y³ = 0}. Then the relation R is :
- (a) symmetric but neither reflexive nor transitive
- (b) reflexive but neither symmetric nor transitive
- (c) reflexive and symmetric, but not transitive
- (d) an equivalence relation
Answer and solution
Answer (b) reflexive but neither symmetric nor transitive
Solution
Maths sir · 3 steps
Factorise the cubic first: it turns into x = y or x = 3y, and each property is a quick check.
- Factorise. Group the terms: x²(x - 3y) - y²(x - 3y) = (x - 3y)(x² - y²). This splits further into three linear factors.
x³ - 3x²y - xy² + 3y³ = (x - 3y)(x - y)(x + y)
- Read the condition. For natural numbers x + y > 0, so that factor is never zero. Hence (x, y) ∈ R exactly when x = y or x = 3y.
Property Test Result Reflexive x = y always works, so (x, x) ∈ R for every x Yes Symmetric (3, 1) ∈ R since 3 = 3 × 1, but for (1, 3) we would need 1 = 3 or 1 = 9 No Transitive (9, 3) ∈ R and (3, 1) ∈ R, but for (9, 1) we would need 9 = 1 or 9 = 3 No - So. R is reflexive but neither symmetric nor transitive, which is option (b).
The idea to keep. When a relation is defined by a polynomial equation, factorise it first. The factors turn it into plain conditions like x = y or x = 3y, and one small counterexample settles each property.
Answer from the NTA final answer key. Solution by PrepWiser.