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

Question

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

Accepted Answer

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

For $y=-3$:

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

For $y=-2$:

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

For $y=-1$:

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

For $y=0$:

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

For $y=1$:

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

For $y=2$:

Solving ${x}^{2}+2(2)\le 4$, 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 $x\in A$. 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 $l+m=15+3=18$.

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