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Mathematics · Sets and Relations

JEE Main 2024 — 6 April, Shift 1 — Question 12

Let the relations R1R_{1} and R2R_{2} on the set X={1,2,3,…,20}\mathrm{X}=\{1,2,3, \ldots, 20\} be given by R1={(x,y):2x−3y=2}\mathrm{R}_{1}=\{(\mathrm{x}, \mathrm{y}): 2 \mathrm{x}-3 \mathrm{y}=2\} and

R2={(x,y):−5x+4y=0}R_{2}=\{(x, y):-5 x+4 y=0\}. If MM and NN be the minimum number of elements required to be added in R1R_{1}

and R2R_{2}, respectively, in order to make the relations symmetric, then M+N\mathrm{M}+\mathrm{N} equals

  1. Option A:

    8

  2. Option B:

    16

  3. Option C:

    12

  4. Option D:

    10

    Correct

Answer: D

Step-by-step solution

x={1,2,3,……..20}\mathrm{x}=\{1,2,3, \ldots \ldots . .20\}

R1={(x,y):2x−3y=2}\mathrm{R}_{1}=\{(\mathrm{x}, \mathrm{y}): 2 \mathrm{x}-3 \mathrm{y}=2\}

R2={(x,y):−5x+4y=0}R_{2}=\{(x, y):-5 x+4 y=0\}

R1={(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)}\mathrm{R}_{1}=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\}

R2={(4,5),(8,10),(12,15),(16,20)}\mathrm{R}_{2}=\{(4,5),(8,10),(12,15),(16,20)\}

in R16\mathrm{R}_{1} 6 element needed

in R24\mathrm{R}_{2} 4 element needed

So, total 6+4=106+4=10 element

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
6 April, Shift 1
Subject
Mathematics
Chapter
Sets and Relations
Topic
Types of Relations

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