Verified Solution • 2025–26 Session moderate

QJEEM23Ma95f4: JEE Main 2023 Mathematics - Class 11 JEEM23Ma95f4 Solution: Let and . Then the number of elements in the relation divides…

Let A={2,3,4}\mathrm{A}=\{2,3,4\} and B={8,9,12}\mathrm{B}=\{8,9,12\} . Then the number of elements in the relation R={((a1, b1),(a2, b2))∈(A×B,A×B):a1\mathrm{R}=\left\{\left(\left(a_{1}, \mathrm{~b}_{1}\right),\left(a_{2}, \mathrm{~b}_{2}\right)\right) \in(A \times B, A \times B): a_{1}\right. divides b2\mathrm{b}_{2} and a2\mathrm{a}_{2} divides b1}\left.\mathrm{b}_{1}\right\} is :

Multiple Choice Options

  • A 18
  • B 24
  • C 36
  • D 12

Step-by-Step Verified Solution

Given sets :
A=2,3,4A = {2,3,4}
B=8,9,12B = {8,9,12}

We want to find the number of elements of the form ((a1,b1),(a2,b2))( (a_1, b_1), (a_2, b_2) ) such that :

  1. a1a_1 divides b2b_2
  2. a2a_2 divides b1b_1

For the first condition :
a1a_1 divides b2b_2
Given a1∈Aa_1 \in A and b2∈Bb_2 \in B , we can list the pairs:
(a1,b2)∈(2,8),(2,12),(3,9),(3,12),(4,8),(4,12)(a_1, b_2) \in {(2,8),(2,12),(3,9),(3,12),(4,8),(4,12)}
This gives 6 pairs.

For the second condition, the pairs are the same, because it's just the reversed relation. So :
a2a_2 divides b1b_1
Again has 6 valid pairs.

Now, for every pair from the first condition, we can have any pair from the second condition. This leads to :
6×6=366 \times 6 = 36 relations.

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