1. Given relations

On $N$ (natural numbers), define: $R_1 = \{(x,y) \in N \times N : 2x+y=10\}$ $R_2 = \{(x,y) \in N \times N : x+2y=10\}$

We find the ordered pairs first.

2. Find $R_1$

From $2x+y=10 \implies y=10-2x$

Since $x,y \in N$, both must be natural numbers.

Try natural values of $x$:

• If $x=1$, then $y=8$
• If $x=2$, then $y=6$
• If $x=3$, then $y=4$
• If $x=4$, then $y=2$
• If $x=5$, then $y=0$ which is not in $N$ (taking natural numbers as positive integers)

So, $R_1=\{(1,8),(2,6),(3,4),(4,2)\}$

Hence the range of $R_1$ is $\{8,6,4,2\}$

So option A is false, because it states an incorrect set.

3. Find $R_2$

From $x+2y=10 \implies x=10-2y$

Try natural values of $y$:

• If $y=1$, then $x=8$
• If $y=2$, then $x=6$
• If $y=3$, then $x=4$
• If $y=4$, then $x=2$
• If $y=5$, then $x=0$ which is not in $N$

So, $R_2=\{(8,1),(6,2),(4,3),(2,4)\}$

Hence the range of $R_2$ is $\{1,2,3,4\}$

So option B is true.

4. Check symmetry

A relation is symmetric if whenever $(a,b) \in R$, then $(b,a) \in R$ also.

For $R_1$:

$(1,8) \in R_1$, but $(8,1)$ is not in $R_1$ because $2(8)+1=17 \neq 10$ So $R_1$ is not symmetric.

For $R_2$:

$(8,1) \in R_2$, but $(1,8)$ is not in $R_2$ because $1+2(8)=17 \neq 10$ So $R_2$ is not symmetric.

Therefore option C is false.

5. Check transitivity

A relation is transitive if whenever $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$.

For $R_1$:

Elements are: $\{(1,8),(2,6),(3,4),(4,2)\}$ For transitivity, we need chains like $(a,b)$ and $(b,c)$ both in $R_1$.

We have $(4,2)$ and $(2,6)$ both in $R_1$. Then transitivity would require $(4,6) \in R_1$. But $2(4)+6=14 \neq 10$ So $(4,6) \notin R_1$. Hence $R_1$ is not transitive.

For $R_2$:

Elements are: $\{(8,1),(6,2),(4,3),(2,4)\}$ We have $(6,2)$ and $(2,4)$ in $R_2$. Then transitivity would require $(6,4) \in R_2$. But $6+2(4)=14 \neq 10$ So $(6,4) \notin R_2$. Hence $R_2$ is not transitive.

Therefore option D is false.

6. Conclusion

Only option B is correct.

$\boxed{\text{B}}$