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JEE Mains · Maths · STD 12 - 1. relation and function

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

  1. A \(24\)
  2. B \(23\)
  3. C \(25\)
  4. D \(26\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(25\)

Step-by-step Solution

Detailed explanation

Given : \(4 x \leq 5 y\) then \(\mathrm{R}=\{ \)\( (1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4) \) \( (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\}\) i.e. \(16\) elements. i.e. \(\mathrm{m}=16\) Now to make \(\mathrm{R}\) a symmetric relation add…
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