Mathematics · Sets and Relations

JEE Main 2025 — 3 April, Evening Shift — Question 34

Let $A=\{-2,-1,0,1,2,3\}$ . Let $R$ be a relation on $A$ defined by $x R y$ if and only if $y=\max \{x, 1\}$ . Let $/$ be the number of elements in $R$ . Let $m$ and $n$ be the minimum number of elements required to be added in $R$ to make it reflexive and symmetric relations, respectively. Then $I+m+n$ is equal to

Option A:
12
Correct
Option B:
11
Option C:
14
Option D:
13

Answer: A

Step-by-step solution

$A=\{-2,-1,0,1,2,3\}$

$x R y \Leftrightarrow y=\max \{x, 1\}$

$\therefore \quad R=\{(-2,1),(-1,1),(0,1),(1,1),(2,2),(3,3)\}$

$I=6$ Elements to be added to make it reflexive: $(-2,-2),(-1,-1),(0,0)$

$\therefore m=3$

Elements to be added to make it symmetric:

$(1,-2),(1,-1),(1,0)$

$\therefore n=3$

$l+m+n=12$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Mathematics
Chapter: Sets and Relations
Topic: Relations