JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 6)
Let the relations $$R_1$$ and $$R_2$$ on the set $$X=\{1,2,3, \ldots, 20\}$$ 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 of elements required to be added in $$R_1$$ and $$R_2$$, respectively, in order to make the relations symmetric, then $$M+N$$ equals
16
12
8
10
Explanation
$$\begin{aligned} & R_1=\{(x, y): 2 x-3 y=2\} \\ & R_2=\{(x, y):-5 x+4 y=0\} \\ & 2 x-3 y=2 \end{aligned}$$
So $$2 x$$ and $$3 y$$ both has to be even or odd simultaneously and $$2 x$$ can't be odd so $$2 x$$ and $$3 y$$ both will be even
$$R_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\}$$
For symmetric we need to add 6 elements as
$$\begin{aligned} & (2,4),(4,7),(6,10),(8,13),(10,16),(12,19) \\ & M=6 \end{aligned}$$
For $$R_2-5 x+4 y=0$$
$$5 x$$ and $$4 y$$ has to be equal $$4 y$$ is always even so $$5 x$$ will also be even
$$R_2=\{(4,5),(8,10),(12,15),(16,20)\}$$
For symmetric we need to add 4 element as
$$\begin{aligned} & (5,4)(10,8)(15,12)(20,16) \\ & N=4 \\ & M+N=6+4=10 \end{aligned}$$
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