x={1,2,3,……..20}R1={(x,y):2x−3y=2}R2={(x,y):−5x+4y=0} $\begin{aligned}
& \mathrm{R}_1={(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)} \
& \mathrm{R}_2={(4,5),(8,10),(12,15),(16,20)}
\end{aligned}inR_1 6elementneededinR_2 4elementneededSo,total6+4=10$ element
Let the relations R_1 and R_2 on the set X=1,2,3, …, 20 be given by R_1=(x, y): 2 x-3 y=2 and R_2=(x, y):-5 x+4 y=0. If M and N be the minimum number…
JEE Main 2024 — Mathematics Algebra
Let the relations R1 and R2 on the set X={1,2,3,…,20} be given by R1={(x,y):2x−3y=2} and R2={(x,y):−5x+4y=0}. If M and N be the minimum number of elements required to be added in R1 and R2, respectively, in order to make the relations symmetric, then M+N equals
Held on 6 Apr 2024 · Verified 6 Jul 2026.
Options
- A
12
- B
16
- C
8
- D
10
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