JEE Main 2023 — Sets And Relations Question with Solution

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

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.