QJEEM24Ma8a01: JEE Main 2024 Mathematics - Class 11 JEEM24Ma8a01 Solution: Let and . Let be a relation defined on by if and only if . Then …

To find the number of elements in the relation $R$ defined on $A \times B$ , we need to determine all pairs $((a_1, b_1), (a_2, b_2))$ such that $a_1 + a_2 = b_1 + b_2$ , where $a_1, a_2 \in A$ and $b_1, b_2 \in B$ .

First, consider all possible sums of pairs from set $A$ and set $B$ .

Possible sums from set $A = \{2, 3, 6, 7\}$ :

• $2 + 2 = 4$
• $2 + 3 = 5$
• $2 + 6 = 8$
• $2 + 7 = 9$
• $3 + 2 = 5$
• $3 + 3 = 6$
• $3 + 6 = 9$
• $3 + 7 = 10$
• $6 + 2 = 8$
• $6 + 3 = 9$
• $6 + 6 = 12$
• $6 + 7 = 13$
• $7 + 2 = 9$
• $7 + 3 = 10$
• $7 + 6 = 13$
• $7 + 7 = 14$

Possible sums from set $B = \{4, 5, 6, 8\}$ :

• $4 + 4 = 8$
• $4 + 5 = 9$
• $4 + 6 = 10$
• $4 + 8 = 12$
• $5 + 4 = 9$
• $5 + 5 = 10$
• $5 + 6 = 11$
• $5 + 8 = 13$
• $6 + 4 = 10$
• $6 + 5 = 11$
• $6 + 6 = 12$
• $6 + 8 = 14$
• $8 + 4 = 12$
• $8 + 5 = 13$
• $8 + 6 = 14$
• $8 + 8 = 16$

Now, identify the common sums from both sets:

Common sums: $8, 9, 10, 12, 13, 14$

For each common sum, count the pairs from set $A$ and set $B$ that produce these sums:

• Sum = 8: From $A$ : {(2,6), (6,2)} - 2 pairs; From $B$ : {(4,4)} - 1 pair; Hence, 2 * 1 = 2 pairs
• Sum = 9: From $A$ : {(2,7), (3,6), (6,3), (7,2)} - 4 pairs; From $B$ : {(4,5), (5,4)} - 2 pairs; Hence, 4 * 2 = 8 pairs
• Sum = 10: From $A$ : {(3,7), (7,3)} - 2 pairs; From $B$ : {(4,6), (5,5), (6,4)} - 3 pairs; Hence, 2 * 3 = 6 pairs
• Sum = 12: From $A$ : {(6,6)} - 1 pair; From $B$ : {(4,8), (6,6), (8,4)} - 3 pairs; Hence, 1 * 3 = 3 pairs
• Sum = 13: From $A$ : {(6,7), (7,6)} - 2 pairs; From $B$ : {(5,8), (8,5)} - 2 pairs; Hence, 2 * 2 = 4 pairs
• Sum = 14: From $A$ : {(7,7)} - 1 pair; From $B$ : {(6,8), (8,6)} - 2 pairs; Hence, 1 * 2 = 2 pairs

Adding all these, we get the number of elements in the relation $R$ :

2 + 8 + 6 + 3 + 4 + 2 = 25

Thus, the number of elements in $R$ is 25.