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1.Relation and Function
hard
Let $A=\{0,3,4,6,7,8,9,10\} \quad$ and $R$ be the relation defined on A such that $R =\{( x , y ) \in A \times A : x - y \quad$ 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 $...........$.
A
$18$
B
$19$
C
$17$
D
$16$
(JEE MAIN-2023)
Solution
$A =\{0,3,4,6,7,8,9,10\} \quad 3,7,9 \rightarrow \text { odd }$
$R =\{ x – y =\text { odd }+ \text { ve or } x – y =2\} 0,4,6,8,10 \rightarrow \text { even }$
${ }^3 C _1 \cdot{ }^5 C _1=15+(6,4),(8,6),(10,8),(9,7)$
$Min ^{ m }$ ordered pairs to be added must be :
$15+4=19$
Standard 12
Mathematics
