QJEEM23Maa689: JEE Main 2023 Mathematics - Class 11 JEEM23Maa689 Solution: Let and be a relation on the set defined by . Then the number…

$$2a+3b=4c+5d$$

Given A = {1, 2, 3, 4}, the maximum value of $2a + 3b$ is 20, when (a, b) = (4, 4), and the minimum value of $4c + 5d$ is 9, when (c, d) = (1, 1). Therefore, the possible values for $2a + 3b = 4c + 5d$ are 9, 13, 14, 17, 18, and 19.

Now, let's find the combinations of (a, b), (c, d) that satisfy the given equation:

1. $2a + 3b = 9 \Rightarrow (a, b) = (3, 1) \Rightarrow (c, d) = (1, 1)$

2. $2a + 3b = 13 \Rightarrow (a, b) = (2, 3) \Rightarrow (c, d) = (2, 1)$

3. $2a + 3b = 14 \Rightarrow (a, b) = (4, 2) \Rightarrow (c, d) = (1, 2)$

4. $2a + 3b = 14 \Rightarrow (a, b) = (1, 4) \Rightarrow (c, d) = (1, 2)$

5. $2a + 3b = 17 \Rightarrow (a, b) = (4, 3) \Rightarrow (c, d) = (3, 1)$

6. $2a + 3b = 18 \Rightarrow (a, b) = (3, 4) \Rightarrow (c, d) = (2, 2)$

There are a total of 6 elements in the relation R for the given equation with the specified values of a, b, c, and d.