Mathematics · Sets and Relations

JEE Main 2026 — 23 January, Morning Shift — Question 6

Let $\mathrm{A}=\{-2,-1,0,1,2,3,4\}$ . Let R be a relation on A defined by xRy if and only if $2 \mathrm{x}+\mathrm{y} \leq 2$ . Let $l$ 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 $l+\mathrm{m}+\mathrm{n}$ is equal to :

Option A:
32
Option B:
34
Option C:
33
Correct
Option D:
35

Answer: C

Step-by-step solution

$\mathrm{R}\{(-2, \mathrm{a}),(-1, \mathrm{~b}),(0, \mathrm{c}),(1, \mathrm{~d}),(2, \mathrm{e})\}$

$\mathrm{a}=\{-2,-1,0,1,2,3,4\} ; \mathrm{b}=\{-2,-1,0,1,2,3,4\}$

$\mathrm{c}=\{-2,-1,0,1,2\} ; \mathrm{d}=\{-2,-1,0\}$

$\mathrm{e}=\{-2\}$ ∴ No. of elements in R $=7+7+5+3+1=23=\ell$

Minimum number of element to be added to make it reflexive $=m=4 \Rightarrow\{(1,1),(2,2),(3,3),(4,4)\}$

minimum number of element to be added to make it symmetric $=\mathrm{n}=6$ for ' n ' $\Rightarrow \mathrm{R}=\{(3,-2),(4,-2),(2,-1),(2,0),(3,-1),(4,-1)\}$

$\therefore \ell+\mathrm{m}+\mathrm{n}=23+4+6=33$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Sets and Relations
Topic: Number of Relations