JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 19)
Let $$A=\{0,3,4,6,7,8,9,10\}$$ and $$R$$ be the relation defined on $$A$$ such that $$R=\{(x, y) \in A \times A: x-y$$ is odd positive integer or $$x-y=2\}$$. The minimum number of elements that must be added to the relation $$R$$, so that it is a symmetric relation, is equal to ____________.
Answer
19
Explanation
We have, $A=\{0,3,4,6,7,8,9,10\}$
Case I : $x-y$ is odd, if one is odd and one is even and $x>y$.
$\therefore$ Possibilites are $\{(3,0),(4,3),(6,3),(7,6),(7,4)$, $(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10$, $7),(10,3)\}$
No. of cases $=15$
Case II : $x-y=2$
$\therefore$ Possibilities are $\{(6,4),(8,6),(9,7),(10,8)\}$
$\therefore$ No. of cases $=4$
So, minimum ordered pair to be added $=15+4=19$
Case I : $x-y$ is odd, if one is odd and one is even and $x>y$.
$\therefore$ Possibilites are $\{(3,0),(4,3),(6,3),(7,6),(7,4)$, $(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10$, $7),(10,3)\}$
No. of cases $=15$
Case II : $x-y=2$
$\therefore$ Possibilities are $\{(6,4),(8,6),(9,7),(10,8)\}$
$\therefore$ No. of cases $=4$
So, minimum ordered pair to be added $=15+4=19$
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