Let $N$ denote the set of all natural numbers. Define two binary relations on $N$ as $R_1 = \{(x, y) \in N \times N : 2x + y = 10\}$ and $R_2 = \{(x, y) \in N \times N : x + 2y = 10\}$. Then :
(A)Range of $R_1$ is $\{2, 4, 8\}$.
(B)Range of $R_2$ is $\{1, 2, 3, 4\}$.
(C)Both $R_1$ and $R_2$ are symmetric relations.
(D)Both $R_1$ and $R_2$ are transitive relations.
Correct Answer:
B
Explanation
For $R_1$, $2x + y = 10$ and $x, y \in N$, the possible values for $x$ and $y$ are:
$x = 1, y = 8 \implies (1, 8)$
$x = 2, y = 6 \implies (2, 6)$
$x = 3, y = 4 \implies (3, 4)$
$x = 4, y = 2 \implies (4, 2)$
Therefore, $R_1 = \{(1, 8), (2, 6), (3, 4), (4, 2)\}$.
Hence, the range of $R_1$ is $\{2, 4, 6, 8\}$.
$R_1$ is not symmetric.
$R_1$ is not transitive as $(3, 4), (4, 2) \in R_1$, but $(3, 2) \notin R_1$.
For $R_2$, $x + 2y = 10$ and $x, y \in N$, the possible values for $x$ and $y$ are:
$x = 8, y = 1 \implies (8, 1)$
$x = 6, y = 2 \implies (6, 2)$
$x = 4, y = 3 \implies (4, 3)$
$x = 2, y = 4 \implies (2, 4)$
Therefore, $R_2 = \{(8, 1), (6, 2), (4, 3), (2, 4)\}$.
Hence, the range of $R_2 = \{1, 2, 3, 4\}$.
$R_2$ is not symmetric and transitive.