Let A=-3,-2,-1,0,1,2,3.

Question

Let $\mathrm{A}=\{-3,-2,-1,0,1,2,3\}$. Let R be a relation on A defined by $x \mathrm{R} y$ if and only if $0 \leq x^2+2 y \leq 4$. 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

Accepted Answer

The relation $ R $ is defined for the set $ \mathrm{A} = \{-3, -2, -1, 0, 1, 2, 3\} $ with the condition $ 0 \leq x^2 + 2y \leq 4 $. Let's determine the pairs $(x, y)$ that satisfy this condition.

For $ y = -3 $:

Solving $ x^2 + 2(-3) \leq 4 $, we find $ x = \{3, -3\} $.

For $ y = -2 $:

Solving $ x^2 + 2(-2) \leq 4 $, we find $ x = \{-2, 2\} $.

For $ y = -1 $:

Solving $ x^2 + 2(-1) \leq 4 $, we find $ x = \{-2, 2\} $.

For $ y = 0 $:

Solving $ x^2 + 0 \leq 4 $, we find $ x = \{-2, -1, 0, 1, 2\} $.

For $ y = 1 $:

Solving $ x^2 + 2(1) \leq 4 $, we find $ x = \{-1, 0, 1\} $.

For $ y = 2 $:

Solving $ x^2 + 2(2) \leq 4 $, we find $ x = \{0\} $.

The relation $ R $ consists of the following pairs:

$ R = \{(3, -3), (-3, -3), (-2, -2), (2, -2), (-2, -1), (2, -1), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0), (-1, 1), (0, 1), (1, 1), (0, 2)\} $

Currently, $ R $ has $ l = 15 $ elements. To make $ R $ reflexive, it must include all pairs $(x, x)$ for every $ x \in \mathrm{A} $. We identify the missing reflexive pairs $(-1, -1)$, $(2, 2)$, and $(3, 3)$, which are required to satisfy reflexivity.

Thus, $ m = 3 $ more elements are needed. Therefore, the total $ l + m = 15 + 3 = 18 $.

Solutions · 1