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

Let the relations \(R_1\) and \(R_2\) on the set \(\mathrm{X}=\{1,2,3, \ldots, 20\}\) be given by \(\mathrm{R}_1=\{(\mathrm{x}, \mathrm{y}): 2 \mathrm{x}-3 \mathrm{y}=2\}\) and \(\mathrm{R}_2=\{(\mathrm{x}, \mathrm{y}):-5 \mathrm{x}+4 \mathrm{y}=0\}\). If \(\mathrm{M}\) and \(\mathrm{N}\) be the minimum number of elements required to be added in \(R_1\) and \(R_2\), respectively, in order to make the relations symmetric, then \(\mathrm{M}+\mathrm{N}\) equals

  1. A \(8\)
  2. B \(16\)
  3. C \(12\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(10\)

Step-by-step Solution

Detailed explanation

\( \mathrm{x}=\{1,2,3, \ldots \ldots .20\} \) \( \mathrm{R}_1=\{(\mathrm{x}, \mathrm{y}): 2 \mathrm{x}-3 \mathrm{y}=2\} \) \( \mathrm{R}_2=\{(\mathrm{x}, \mathrm{y}):-5 \mathrm{x}+4 \mathrm{y}=0\}\) \( \mathrm{R}_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\} \)…