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1.Relation and Function
hard
Let $\mathrm{A}=\{1,2,3,4,5\}$. Let $\mathrm{R}$ be a relation on $\mathrm{A}$ defined by $x R y$ if and only if $4 x \leq 5 y$. Let $m$ be the number of elements in $\mathrm{R}$ and $\mathrm{n}$ be the minimum number of elements from $\mathrm{A} \times \mathrm{A}$ that are required to be added to $\mathrm{R}$ to make it a symmetric relation. Then $m+n$ is equal to:
A
$24$
B
$23$
C
$25$
D
$26$
(JEE MAIN-2024)
Solution
Given : $4 x \leq 5 y$
then
$\mathrm{R}=\{ $$ (1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4) $
$ (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\}$
i.e. $16$ elements.
i.e. $\mathrm{m}=16$
Now to make $\mathrm{R}$ a symmetric relation add
$\{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)\}$
i.e. $\mathrm{n}=9$
So $\mathrm{m}+\mathrm{n}=25$
Standard 12
Mathematics
