JEE Mathematics 2023 Question with Solution

Given: $A = \{0, 3, 4, 6, 7, 8, 9, 10\}$ and $R = \{(x, y) : x - y \text{ is odd positive integer or } x - y = 2\}$.

Find: The minimum number of ordered pairs that must be added so that $R$ becomes symmetric.

A relation is symmetric if whenever $(x, y) \in R$, then $(y, x) \in R$ must also be in $R$.

From the extracted solution:

1. Define the relation on the given set $A$.
2. Check each pair in $R$ for the presence of its reverse pair.
3. Count the missing reverse pairs.

The solution states that:

• $15$ pairs are missing corresponding to pairs with odd positive difference.
• $4$ pairs are missing corresponding to pairs with $x - y = 2$.

Hence, the minimum number of pairs to be added is

$$15 + 4 = 19$$

Therefore, the minimum number of elements that must be added is $19$.

Given: $R$ contains those ordered pairs for which $x - y$ is an odd positive integer or $x - y = 2$.

Find: How many reverse ordered pairs are absent.

To make $R$ symmetric, each ordered pair $(x, y)$ already in $R$ must be accompanied by $(y, x)$.

So the task is to count how many such reverse pairs are not already present in $R$. According to the provided solution, these missing pairs are counted category-wise:

$$\text{Missing pairs} = 15 + 4 = 19$$

Thus, the relation can be made symmetric by adding $19$ ordered pairs. The correct numerical answer is $19$.