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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Sets and Relations

Let A=\ 1,2,3,4,5\ . Let R be a relation on A defined by x R y if...

Let A={1,2,3,4,5}A=\{1,2,3,4,5\}. Let R\mathrm{R} be a relation on A\mathrm{A} defined by xRyx \mathrm{R} y if and only if 4x≤5y4 x \leq 5 \mathrm{y}. Let m\mathrm{m} be the number of elements in R\mathrm{R} and n\mathrm{n} be the minimum number of elements from A×A\mathrm{A} \times \mathrm{A} that are required to be added to R\mathrm{R} to make it a symmetric relation. Then m+n\mathrm{m}+\mathrm{n} is equal to :

Options

  1. A25
  2. B24
  3. C26
  4. D23

Correct answer

A. 25

Step-by-step solution

Given : 4x≤5y4 x \leq 5 y then R={(1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4)(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)}\begin{aligned} \mathrm{R}=\{ & (1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4) \\ & (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\} \end{aligned} i.e. 16 elements. i.e. m=16\mathrm{m}=16 Now to make R a symmetric relation add {(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)}\{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)\} i.e. n=9n=9
So m+n=25m+n=25
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