Let N denote the set of all natural numbers. | vedclass

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(C) Given relations on $N$ are:$R_1 = \{(x,y) \in N 	imes N : 2x + y = 10\}$$R_2 = \{(x,y) \in N 	imes N : x + 2y = 10\}$For $R_1$,we solve $y = 10

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Range of $R_2$ is $\{1, 2, 3, 4\}$

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(C) Given relations on $N$ are:$R_1 = \{(x,y) \in N 	imes N : 2x + y = 10\}$$R_2 = \{(x,y) \in N 	imes N : x + 2y = 10\}$For $R_1$,we solve $y = 10 - 2x$ for $x, y \in N$:If $x=1, y=8$; if $x=2, y=6$; if $x=3, y=4$; if $x=4, y=2$.So,$R_1 = \{(1,8), (2,6), (3,4), (4,2)\}$.The range of $R_1$ is $\{2, 4, 6, 8\}$. Thus,option $D$ is incorrect.$R_1$ is not symmetric because $(1,8) \in R_1$ but $(8,1) 
otin R_1$.$R_1$ is not transitive because $(3,4) \in R_1$ and $(4,2) \in R_1$,but $(3,2) 
otin R_1$.For $R_2$,we solve $x = 10 - 2y$ for $x, y \in N$:If $y=1, x=8$; if $y=2, x=6$; if $y=3, x=4$; if $y=4, x=2$.So,$R_2 = \{(8,1), (6,2), (4,3), (2,4)\}$.The range of $R_2$ is $\{1, 2, 3, 4\}$.Thus,option $C$ is correct.$R_2$ is not symmetric because $(8,1) \in R_2$ but $(1,8) 
otin R_2$.$R_2$ is not transitive because $(4,3) \in R_2$ and $(3,y) 
otin R_2$ (as no $y$ exists for $x=3$ in $R_2$).

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

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