JEE Main 2024 — 9 April, Shift 1 — Question 30

Let $x = (a_1, b_1)$ and $y = (a_2, b_2)$ be elements in $A \times B$. The given condition is:

$$a_1 + a_2 = b_1 + b_2 \implies a_1 - b_1 = -(a_2 - b_2)$$

Let $k = a - b$ for each ordered pair $(a, b) \in A \times B$. The condition simplifies to finding pairs of elements $(x, y)$ such that:

$$k_1 = -k_2$$

Group all 16 elements of $A \times B$ by their $k$ value By calculating $k = a - b$ for every combination, we get the following frequency distribution: $k = 3$: 1 pair $\rightarrow \{(7, 4)\}$ $k = 2$: 2 pairs $\rightarrow \{(6, 4), (7, 5)\}$ $k = 1$: 2 pairs $\rightarrow \{(6, 5), (7, 6)\}$ $k = 0$: 1 pair $\rightarrow \{(6, 6)\}$ $k = -1$: 2 pairs $\rightarrow \{(3, 4), (7, 8)\}$ $k = -2$: 3 pairs $\rightarrow \{(2, 4), (3, 5), (6, 8)\}$ $k = -3$: 2 pairs $\rightarrow \{(2, 5), (3, 6)\}$ $k = -4, -5, -6$: 1 pair each (No positive counterparts exist)

Calculate ordered pairs $(x, y)$ satisfying $k_1 = -k_2$ Since $(x, y)$ is an ordered relation, a set of $m$ elements with value $k$ and $n$ elements with value $-k$ yields $m \times n$ ordered pairs. Similarly, its converse matching yields another $n \times m$ ordered pairs. For $k = 0$: Matches with itself $\implies 1 \times 1 = 1$ pair. For $|k| = 1$: $k_1 = 1, k_2 = -1 \implies 2 \times 2 = 4$ pairs. $k_1 = -1, k_2 = 1 \implies 2 \times 2 = 4$ pairs. For $|k| = 2$: $k_1 = 2, k_2 = -2 \implies 2 \times 3 = 6$ pairs. $k_1 = -2, k_2 = 2 \implies 3 \times 2 = 6$ pairs. For $|k| = 3$: $k_1 = 3, k_2 = -3 \implies 1 \times 2 = 2$ pairs. $k_1 = -3, k_2 = 3 \implies 2 \times 1 = 2$ pairs. Summing all valid directed relations: Total elements in $R = 1 + (4 + 4) + (6 + 6) + (2 + 2) = 1 + 8 + 12 + 4 = 25$

Final Answer: The number of elements in $R$ is $25$