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Mathematics · Sets and Relations

JEE Main 2026 — 21 January, Evening Shift — Question 17

Let A={2,3,5,7,9}\mathrm{A}=\{2,3,5,7,9\}. Let R be the relation on A defined by x Ry if and only if 2x≤3y2 \mathrm{x} \leq 3 \mathrm{y}. Let ℓ\ell 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 symmetric relation. Then ℓ+m\ell+\mathrm{m} is equal to:

  1. Option A:

    23

  2. Option B:

    25

    Correct
  3. Option C:

    21

  4. Option D:

    27

Answer: B

Step-by-step solution

A={2,3,5,7,9}A = \{2,3,5,7,9\} y≥2x3y \ge \frac{2x}{3} x=2:  y=2,3,5,7,9x=3:  y=2,3,5,7,9x=5:  y=5,7,9x=7:  y=5,7,9x=9:  y=7,9⇒ℓ=18\begin{aligned} x=2 &:\; y = 2,3,5,7,9 \\ x=3 &:\; y = 2,3,5,7,9 \\ x=5 &:\; y = 5,7,9 \\ x=7 &:\; y = 5,7,9 \\ x=9 &:\; y = 7,9 \end{aligned} \qquad \Rightarrow \ell = 18

To make it symmetric, elements to be added are

 {(5,2),(7,2),(9,2),(5,3),(7,3),(9,3),(9,5)}\text{ } \{(5,2),(7,2),(9,2),(5,3),(7,3),(9,3),(9,5)\} m=7m = 7 ∴  ℓ+m=25\therefore \; \ell + m = 25

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Evening Shift
Subject
Mathematics
Chapter
Sets and Relations
Topic
Types of Relations

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