Easy MCQ +4 / -1 PYQ · JEE Mains 2021

Let N be the set of natural numbers and a relation R on N be defined by $R = \{ (x,y) \in N \times N:{x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0\}$. Then the relation R is :

  1. A symmetric but neither reflexive nor transitive
  2. B reflexive but neither symmetric nor transitive Correct answer
  3. C reflexive and symmetric, but not transitive
  4. D an equivalence relation

Solution

${x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0$<br><br>$\Rightarrow x({x^2} - {y^2}) - 3y({x^2} - {y^2}) = 0$<br><br>$\Rightarrow (x - 3y)(x - y)(x + y) = 0$<br><br>Now, x = y $\forall$(x, y) $\in$N $\times$ N so reflexive but not symmetric &amp; transitive.<br><br>See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) &amp; (1, $-$1) satisfies but (3, $-$1) does not.

About this question

Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations

This question is part of PrepWiser's free JEE Main question bank. 195 more solved questions on Sets, Relations and Functions are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →