QJEEM23Ma03b5: JEE Main 2023 Mathematics - Class 11 JEEM23Ma03b5 Solution: Let A = { 1 , 2 , 3 , 4 , … . , 10 } and B = { 0 , 1 , 2 , 3 ,…

Given sets :

A={1,2,3,4, ............,10}

B={0,1,2,3,4}

We are looking for pairs $(a,b) \in A \times A$ such that :

$2(a-b)^2 + 3(a-b) \in B$

Let's break down the relation :

Case 1 : $a-b = 0$

$2(a-b)^2 + 3(a-b) = 0$

Pairs : $(1,1), (2,2), (3,3), \ldots, (10,10)$ which gives 10 pairs.

Case 2 : $a-b = 1$

$2(a-b)^2 + 3(a-b) = 2(1) + 3(1) = 5$

But 5 is not in B, so no pairs for this case.

Case 3 : $a-b = -1$

$2(a-b)^2 + 3(a-b) = 2(1) - 3(1) = -1$

This value is not in B, so no pairs for this case.

Case 4 : $a-b = 2$

$2(a-b)^2 + 3(a-b) = 2(4) + 3(2) = 8+6 = 14$

Again, 14 is not in B, so no pairs for this case.

Case 5 : $a-b = -2$

$2(a-b)^2 + 3(a-b) = 2(4) - 3(2) = 8 - 6 = 2$

Pairs : $(1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8,10)$ which gives 8 pairs.

For any other $a-b$ value, the quadratic will grow larger than the maximum value in B, so we don't need to consider them.

In total, we have $10 + 8 = 18$ pairs in the relation $R$ .

Therefore, the number of elements in the relation $R$ is 18.