Let rho be a relation defined on the set of n | vedclass

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(A) Given the relation $ho = \{(x, y) \in N 	imes N: 2x + y = 41\}$.Since $x, y \in N$, we have $x \geq 1$ and $y \geq 1$.From $2x + y = 41$, we ge

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$A \subset \{x \in N: 1 \leq x \leq 20\}$ and $B \subset \{y \in N: 1 \leq y \leq 39\}$

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(A) Given the relation $ho = \{(x, y) \in N 	imes N: 2x + y = 41\}$.Since $x, y \in N$, we have $x \geq 1$ and $y \geq 1$.From $2x + y = 41$, we get $y = 41 - 2x$.Since $y \geq 1$, we have $41 - 2x \geq 1$, which implies $2x \leq 40$, so $x \leq 20$.Thus, $x \in \{1, 2, 3, \dots, 20\}$.For each $x$, $y = 41 - 2x$. Substituting values of $x$:If $x = 1, y = 39$.If $x = 20, y = 1$.So, the domain $A = \{1, 2, 3, \dots, 20\}$ and the range $B = \{1, 3, 5, \dots, 39\}$.Both $A$ and $B$ are subsets of the sets given in option $A$.

Let $\rho$ be a relation defined on the set of natural numbers $N$, as $\rho = \{(x, y) \in N \times N: 2x + y = 41\}$. Then the domain $A$ and range $B$ are:

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