Verified Solution • 2025–26 Session moderate

QJEEM22Ma0c3b: JEE Main 2022 Mathematics - Class 11 JEEM22Ma0c3b Solution: Let R 1 = {(a, b) N N :

Let R1 = {(a, b) ∈\in N ×\times N : |a −- b| ≤\le 13} and

R2 = {(a, b) ∈\in N ×\times N : |a −- b| ≠\ne 13}. Then on N :

Multiple Choice Options

  • A Both R1 and R2 are equivalence relations
  • B Neither R1 nor R2 is an equivalence relation
  • C R1 is an equivalence relation but R2 is not
  • D R2 is an equivalence relation but R1 is not

Step-by-Step Verified Solution

R1={(a,b)∈N×N:∣a−b∣≤13}R_{1}=\{(a, b) \in N \times N:|a-b| \leq 13\} and

R2={(a,b)∈N×N:∣a−b∣≠13}R_{2}=\{(a, b) \in N \times N:|a-b| \neq 13\}

In R1:∵∣2−11∣=9≤13R_{1}: \because|2-11|=9 \leq 13

∴(2,11)∈R1\therefore \quad(2,11) \in R_{1} and (11,19)∈R1(11,19) \in R_{1} but (2,19)∉R1(2,19) \notin R_{1}

∴R1\therefore \quad R_{1} is not transitive

Hence R1R_{1} is not equivalence

In R2:(13,3)∈R2R_{2}:(13,3) \in R_{2} and (3,26)∈R2(3,26) \in R_{2} but (13,26)∉R2(13,26) \notin R_{2} (∵∣13−26∣=13)(\because|13-26|=13)

∴R2\therefore R_{2} is not transitive

Hence R2R_{2} is not equivalence.

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