Let $A=\{-3,-2,-1,0,1,2,3$,$\} . Let R$ be a relation on A defined by $x R y$ 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

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{A}=\{-3,-2,-1,0,1,2,3\} \\ & -2 \mathrm{y} \leq \mathrm{x}^2 \leq 4-2 \mathrm{y}\end{aligned}$
$\begin{array}{lll}y=-3 & 6 \leq x^2 \leq 10 & \Rightarrow x \in\{-3,3\} \\ y=-2 & 4 \leq x^2 \leq 8 & \Rightarrow x \in\{-2,2\} \\ y=-1 & 2 \leq x^2 \leq 6 & \Rightarrow x \in\{-2,2\} \\ y=0 & 0 \leq x^2 \leq 4 & \Rightarrow x \in\{-2,-1,0,1,2\} \\ y=1 & -2 \leq x^2 \leq 2 & \Rightarrow x \in\{-1,0,1\} \\ y=2 & -4 \leq x^2 \leq 0 & \Rightarrow x \in\{0\} \\ y=3 & -6 \leq x^2 \leq-2 & \Rightarrow \text { No } x \text {-Exist }\end{array}$
$\begin{aligned} & \mathrm{R}=\{(-3,-3)(-3,3),(-2,-2)(-2,2)(-1,-2)(-1,2) \\ & (0,-2)(0,-1)(0,0)(0,1)(0,2)(1,-1)(1,0)(1,1) \\ & (2,0)\} \\ & \therefore \ell=15\end{aligned}$
To make it reflexive we will add
$\begin{aligned} & \{(-1,-1),(2,2),(3,3)\} \quad \therefore \mathrm{m}=3 \\ & \therefore \ell+\mathrm{m}=15+3=18\end{aligned}$ ^

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

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