Mathematics · Sets and Relations

JEE Main 2025 — 3 April, Morning Shift — Question 24

Let $\mathrm{A}=\{-3,-2,-1,0,1,2,3$ , $\}$ . Let $R$ be a relation on A defined by xRy if and only if $0 \leq x^{2}+2 y \leq 4$ . Let $l$ be the number of elements in R and $m$ be the minimum number of elements required

to be added in R to make it a reflexive relation. then $l+m$ is equal to

Option A:
19
Option B:
20
Option C:
17
Option D:
18
Correct

Answer: D

Step-by-step solution

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

-2y \leq x^2 \leq 4 - 2y

\begin{array}{rcll} y = -3 & \implies & 6 \leq x^2 \leq 10 & \implies x \in \{-3, 3\} \\ y = -2 & \implies & 4 \leq x^2 \leq 8 & \implies x \in \{-2, 2\} \\ y = -1 & \implies & 2 \leq x^2 \leq 6 & \implies x \in \{-2, 2\} \\ y = 0 & \implies & 0 \leq x^2 \leq 4 & \implies x \in \{-2, -1, 0, 1, 2\} \\ y = 1 & \implies & -2 \leq x^2 \leq 2 & \implies x \in \{-1, 0, 1\} \\ y = 2 & \implies & -4 \leq x^2 \leq 0 & \implies x \in \{0\} \\ y = 3 & \implies & -6 \leq x^2 \leq -2 & \implies \text{No x-Exist} \end{array}

R = \{(-3, -3), (-3, 3), (-2, -2), (-2, 2), (-1, -2), (-1, 2), (0, -2),

(0, -1), (0, 0), (0, 1), (0, 2), (1, -1), (1, 0), (1, 1), (2, 0)\}

\therefore \ell = 15

\text{To make it reflexive we will add } \{(-1, -1), (2, 2), (3, 3)\} \quad \therefore m = 3

\therefore \ell + m = 15 + 3 = 18

Answer key and solution verified before publishing.

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