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

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

Let the relations R_1 and R_2 on the set X=\ 1,2,3, , 20\ be given...

Let the relations R1R_1 and R2R_2 on the set X={1,2,3,…,20}X=\{1,2,3, \ldots, 20\} be given by R1={(x,y):2x−3y=2}R_1=\{(x, y): 2 x-3 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+NM+N equals

Options

  1. A12
  2. B16
  3. C8
  4. D10

Correct answer

D. 10

Step-by-step solution

x={1,2,3,……..20}R1={(x,y):2x−3y=2}R2={(x,y):−5x+4y=0}\begin{aligned} & \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\}\end{aligned}R1={(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)}R2={(4,5),(8,10),(12,15),(16,20)}\begin{aligned} & \mathrm{R}_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\} \\ & \mathrm{R}_2=\{(4,5),(8,10),(12,15),(16,20)\} \end{aligned}inR16elementneededinin R_1 6 element needed inR_2 4elementneededSo,totalelement needed So, total6+4=10$ element
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