Verified Solution • 2025–26 Session moderate

QJEEM21Ma5eb4: JEE Main 2021 Mathematics - Class 11 JEEM21Ma5eb4 Solution: Let N be the set of natural numbers and a relation R on N be…

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}R = \{ (x,y) \in N \times N:{x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0\} . Then the relation R is :

Multiple Choice Options

  • 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

Step-by-Step Verified Solution

x3−3x2y−xy2+3y3=0{x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0

⇒x(x2−y2)−3y(x2−y2)=0\Rightarrow x({x^2} - {y^2}) - 3y({x^2} - {y^2}) = 0

⇒(x−3y)(x−y)(x+y)=0\Rightarrow (x - 3y)(x - y)(x + y) = 0

Now, x = y ∀\forall (x, y) ∈\in N ×\times N so reflexive but not symmetric & transitive.

See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) & (1, −- 1) satisfies but (3, −- 1) does not.

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