Let $A = \{-3, -2, -1, 0, 1, 2, 3\}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $0 \le x + 2y \le 4$. Let $\ell$ be the number of elements in $R$ and $m$ be the minimum number of elements required to be added to $R$ to make it a reflexive relation. Then $\ell + m$ is equal to
Step-by-Step Solution
Key Concept: Translate the finite relation rule into explicit admissible ordered pairs and count the required set.
Given $A = \{-3, -2, -1, 0, 1, 2, 3\}$ and the condition $0 \le x + 2y \le 4$, we have $-2y \le x \le 4 - 2y$.
For each value of $y$:
$y = -3$: $6 \le x \le 10 \Rightarrow x \in \{\}$ (no elements in $A$)
$y = -2$: $4 \le x \le 8 \Rightarrow x \in \{\}$ (no elements in $A$)
$y = -1$: $2 \le x \le 6 \Rightarrow x \in \{2, 3\}$
$y = 0$: $0 \le x \le 4 \Rightarrow x \in \{0, 1, 2, 3\}$
$y = 1$: $-2 \le x \le 2 \Rightarrow x \in \{-2, -1, 0, 1, 2\}$
$y = 2$: $-4 \le x \le 0 \Rightarrow x \in \{-3, -2, -1, 0\}$
$y = 3$: $-6 \le x \le -2 \Rightarrow x \in \{-3, -2\}$
Therefore, $R = \{(2, -1), (3, -1), (0, 0), (1, 0), (2, 0), (3, 0), (-2, 1), (-1, 1), (0, 1), (1, 1), (2, 1), (-3, 2), (-2, 2), (-1, 2), (0, 2), (-3, 3), (-2, 3)\}$
$\ell = 17$
To make $R$ reflexive, we need to add $\{(-3, -3), (-1, -1), (3, 3)\}$.
$m = 3$
Therefore, $\ell + m = 17 + 3 = 20$
Correct Answer: 4