Let \[N\] denote the set of all the 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. both ${R_1},{R_2}$ are transitive relations.
B. both ${R_1},{R_2}$ are symmetric relations.
C. Range of ${R_2}$ is $\{ 1,2,3,4\} $
D. Range of ${R_1}$ is $\{ 2,4,8\} $

Hint: In this question, we are given two binary relations on $N$ which are ${R_1},{R_2}$. Now firstly we find all the $(x,y) \in N \times N$ satisfying ${R_1}$ and then for ${R_2}$. And then, we will solve according to the given options.

Complete step-by-step answer:
In the above question, we are given that $N$ is the set of all the natural numbers and${R_1},{R_2}$ are the two binary operations on$N$.
${R_1} = \{ (x,y) \in N \times N:2x + y = 10\} $$ - - - - - (1)$
${R_2} = \{ (x,y) \in N \times N:x + 2y = 10\} $$ - - - - (2)$
Now firstly we find all the $(x,y) \in N \times N$ satisfying ${R_1}$ and then for ${R_2}$.
We will consider the relation ${R_1}$.
We know that
${R_1} = \{ (x,y) \in N \times N:2x + y = 10\} $
$2x + y = 10$, $(x,y) \in N$
$x = \dfrac{{10 - y}}{2}$
So as $x \in N \Rightarrow \dfrac{{10 - y}}{2} \in N$
Now if $\dfrac{{10 - y}}{2} \in N$, $y \in N$
Then $10 - y,y \in N$ belongs to even number.
So we can say that the possible values of$y \in N$ for $x \in N$ will be
$y = \{ 2,4,6,8\} $
So we can say the range of the${R_1} = \{ 2,4,6,8\} $
Now considering the relation ${R_2}$ on $N$
${R_2} = \{ (x,y) \in N \times N:x + 2y = 10\} $
$x + 2y = 10$, $(x,y) \in N$
$x = 10 - 2y$
Now for $x \in N$
$10 - 2y \in N$
Now if $10 - 2y \in N$ is the natural number, then
$y \in \{ 1,2,3,4\} $
So we can say that the range of ${R_2} = \{ 1,2,3,4\} $

Now from options given, we can say that option C is correct.

Note: In the above question, when we considered the relation ${R_1}$, we have taken $x$ in the terms of $y$. We can also find $y$ in the term of $x$.
$2x + y = 10$
$y = 10 - 2x$
From this also, we can find the possible values of $x,y$. Then we will use the value of $x$ to find the range of ${R_1}$ which is $y.$