Set Theory & Relations – Relations – JEE Main 21 Jan 2026 Shift 2

Solution:

Given relation $R = \{(x, y) : 2x \le 3y, \quad x, y \in A\}$ where $A = \{2, 3, 5, 7, 9\}$.

Let’s find the elements of $R$ by checking the condition $y \ge \frac{2x}{3}$ for each $x \in A$.

For $x=2 \Rightarrow y \ge \frac{4}{3} = 1.33 \Rightarrow y \in \{2, 3, 5, 7, 9\}$ (5 elements).

For $x=3 \Rightarrow y \ge \frac{6}{3} = 2 \Rightarrow y \in \{2, 3, 5, 7, 9\}$ (5 elements).

For $x=5 \Rightarrow y \ge \frac{10}{3} = 3.33 \Rightarrow y \in \{5, 7, 9\}$ (3 elements).

For $x=7 \Rightarrow y \ge \frac{14}{3} = 4.66 \Rightarrow y \in \{5, 7, 9\}$ (3 elements).

For $x=9 \Rightarrow y \ge \frac{18}{3} = 6 \Rightarrow y \in \{7, 9\}$ (2 elements).

Total number of elements in $R$ is $l$:
$$l = 5 + 5 + 3 + 3 + 2 = 18$$

To make $R$ symmetric, if $(x, y) \in R$, then $(y, x)$ must be in $R$.
We identify pairs $(x, y) \in R$ for which $(y, x) \notin R$:

From $x=2$: $(2,5) \in R \Rightarrow (5,2) \notin R$ (since $10 \not\le 6$). Add $(5,2)$.
$(2,7) \in R \Rightarrow (7,2) \notin R$. Add $(7,2)$.
$(2,9) \in R \Rightarrow (9,2) \notin R$. Add $(9,2)$.

From $x=3$: $(3,5) \in R \Rightarrow (5,3) \notin R$ (since $10 \not\le 9$). Add $(5,3)$.
$(3,7) \in R \Rightarrow (7,3) \notin R$. Add $(7,3)$.
$(3,9) \in R \Rightarrow (9,3) \notin R$. Add $(9,3)$.

From $x=5$: $(5,9) \in R \Rightarrow (9,5) \notin R$ (since $18 \not\le 15$). Add $(9,5)$.

Note: $(5,7) \in R$ and $(7,5) \in R$ are both present ($14 \le 15$). Similarly for $(7,9)$ and $(9,7)$.

Total elements to be added is $m$:
$$m = 3 + 3 + 1 = 7$$

Value of $l+m$:
$$l + m = 18 + 7 = 25$$

Ans. (2)