Relations and Functions – Types of Relations – JEE Main 22 Jan 2026 Shift 1

Solution:

Given the relation equation is $4y = 5x – 3$.
$$ y = \frac{5x – 3}{4} $$

We need to find pairs $(x, y)$ such that both $x, y \in M = \{1, 2, \dots, 16\}$.

Substitute possible values of $x$ to get integer values of $y$:
If $x = 3 \Rightarrow y = \frac{15 – 3}{4} = \frac{12}{4} = 3 \in M \Rightarrow (3, 3) \in R$

If $x = 7 \Rightarrow y = \frac{35 – 3}{4} = \frac{32}{4} = 8 \in M \Rightarrow (7, 8) \in R$

If $x = 11 \Rightarrow y = \frac{55 – 3}{4} = \frac{52}{4} = 13 \in M \Rightarrow (11, 13) \in R$

If $x = 15 \Rightarrow y = \frac{75 – 3}{4} = \frac{72}{4} = 18 \notin M$ (Rejected)

So, the relation $R$ in roster form is:
$$ R = \{(3, 3), (7, 8), (11, 13)\} $$

For a relation to be symmetric, if $(a, b) \in R$, then $(b, a)$ must also be in $R$.

$(3, 3)$ is already symmetric.

For $(7, 8) \in R$, we must add $(8, 7)$.

For $(11, 13) \in R$, we must add $(13, 11)$.

Therefore, the elements required to be added are $(8, 7)$ and $(13, 11)$.

Total number of elements to add is 2.

Ans. (2)