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 ≤ x^2+2 y ≤ 4. Let l be the number of elements in R and m be the…

$$\begin{aligned} & \mathrm{A}=\{-3,-2,-1,0,1,2,3\} \\ & -2 \mathrm{y} \leq \mathrm{x}^2 \leq 4-2 \mathrm{y}\end{aligned}$$​A={−3,−2,−1,0,1,2,3}−2y≤x2≤4−2y​

$$\begin{array}{lll}y=-3 & 6 \leq x^2 \leq 10 & \Rightarrow x \in\{-3,3\} \\ y=-2 & 4 \leq x^2 \leq 8 & \Rightarrow x \in\{-2,2\} \\ y=-1 & 2 \leq x^2 \leq 6 & \Rightarrow x \in\{-2,2\} \\ y=0 & 0 \leq x^2 \leq 4 & \Rightarrow x \in\{-2,-1,0,1,2\} \\ y=1 & -2 \leq x^2 \leq 2 & \Rightarrow x \in\{-1,0,1\} \\ y=2 & -4 \leq x^2 \leq 0 & \Rightarrow x \in\{0\} \\ y=3 & -6 \leq x^2 \leq-2 & \Rightarrow \text { No } x \text {-Exist }\end{array}$$y=−3y=−2y=−1y=0y=1y=2y=3​6≤x2≤104≤x2≤82≤x2≤60≤x2≤4−2≤x2≤2−4≤x2≤0−6≤x2≤−2​⇒x∈{−3,3}⇒x∈{−2,2}⇒x∈{−2,2}⇒x∈{−2,−1,0,1,2}⇒x∈{−1,0,1}⇒x∈{0}⇒ No x-Exist ​

$$\begin{aligned} & \mathrm{R}=\{(-3,-3)(-3,3),(-2,-2)(-2,2)(-1,-2)(-1,2) \\ & (0,-2)(0,-1)(0,0)(0,1)(0,2)(1,-1)(1,0)(1,1) \\ & (2,0)\} \\ & \therefore \ell=15\end{aligned}$$​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)}∴ℓ=15​

To make it reflexive we will add

$$\begin{aligned} & \{(-1,-1),(2,2),(3,3)\} \quad \therefore \mathrm{m}=3 \\ & \therefore \ell+\mathrm{m}=15+3=18\end{aligned}$$​{(−1,−1),(2,2),(3,3)}∴m=3∴ℓ+m=15+3=18​