Let A = {-3, -2, -1, 0, 1, 2, 3} and R be a r | vedclass

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(B) The relation $R$ is defined on $A = \{-3, -2, -1, 0, 1, 2, 3\}$ by $2x - y \in \{0, 1\}$.Case $1$: $2x - y = 0 \implies y = 2x$.For $x = -1, y = -

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$17$

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(B) The relation $R$ is defined on $A = \{-3, -2, -1, 0, 1, 2, 3\}$ by $2x - y \in \{0, 1\}$.Case $1$: $2x - y = 0 \implies y = 2x$.For $x = -1, y = -2$; for $x = 0, y = 0$; for $x = 1, y = 2$.Pairs: $(-1, -2), (0, 0), (1, 2)$.Case $2$: $2x - y = 1 \implies y = 2x - 1$.For $x = -1, y = -3$; for $x = 0, y = -1$; for $x = 1, y = 1$; for $x = 2, y = 3$.Pairs: $(-1, -3), (0, -1), (1, 1), (2, 3)$.Thus,$R = \{(-1, -2), (0, 0), (1, 2), (-1, -3), (0, -1), (1, 1), (2, 3)\}$.The number of elements $l = 7$.For $R$ to be reflexive,we need $(x, x) \in R$ for all $x \in A$. Currently,only $(0, 0)$ and $(1, 1)$ are present. We need to add $(-3, -3), (-2, -2), (-1, -1), (2, 2), (3, 3)$. So,$m = 5$.For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x)$ must be in $R$. The elements are $(-1, -2), (1, 2), (-1, -3), (0, -1), (2, 3)$. Their inverses are $(-2, -1), (2, 1), (-3, -1), (-1, 0), (3, 2)$. None of these are in $R$. Thus,we need to add $5$ elements. So,$n = 5$.Therefore,$l + m + n = 7 + 5 + 5 = 17$.

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