Let $A=\{-3,-2,-1,0,1,2,3\}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $0 \le x^{2}+2y \le 4$. Let $l$ be the number of elements in $R$ and $m$ be the minimum number of elements required to be added in $R$ to make it a reflexive relation. Then $l+m$ is equal to
- (1) 19
- (2) 20
- (3) 17
- (4) 18
Solution:
$$A = \{-3, -2, -1, 0, 1, 2, 3\}$$
$$xRy \Rightarrow 0 \le x^2 + 2y \le 4$$
$$-2y \le x^2 \le 4 – 2y$$
For $y = -3$:
$$6 \le x^2 \le 10 \Rightarrow x \in \{-3, 3\}$$
For $y = -2$:
$$4 \le x^2 \le 8 \Rightarrow x \in \{-2, 2\}$$
For $y = -1$:
$$2 \le x^2 \le 6 \Rightarrow x \in \{-2, 2\}$$
For $y = 0$:
$$0 \le x^2 \le 4 \Rightarrow x \in \{-2, -1, 0, 1, 2\}$$
For $y = 1$:
$$-2 \le x^2 \le 2 \Rightarrow x \in \{-1, 0, 1\}$$
For $y = 2$:
$$-4 \le x^2 \le 0 \Rightarrow x \in \{0\}$$
For $y = 3$:
$$-6 \le x^2 \le -2 \Rightarrow \text{No } x \text{ exists in } A$$
$$l = 2 + 2 + 2 + 5 + 3 + 1 = 15$$
For $R$ to be reflexive, $(a, a) \in R$ for all $a \in A$.
Pairs $(x, x)$ already in $R$: $(-3, -3), (-2, -2), (0, 0), (1, 1)$
Pairs required to be added for reflexivity: $(-1, -1), (2, 2), (3, 3)$
$$m = 3$$
$$l + m = 15 + 3 = 18$$
Ans. (4)