First let us define what range and domain are.

Range: The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually) after we have substituted the domain. In plain English, the definition means: The range is the resulting y-values we get after substituting all the possible x-values.

Domain: The domain of definition of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or value for each member of the domain.

We have been given that, R is a relation defined on N.

N = set of natural numbers

R = {(x, y): x, y ∈ N, 2x + y = 41}

We have the function as

2x + y = 41

⇒ y = 41 – 2x

As y ∈ N (Natural number)

⇒ 41 – 2x ≥ 1

⇒ –2x ≥ 1 – 41

⇒ –2x ≥ –40

⇒ 2x ≤ 40

⇒ x ≤ 20

So, the domain is first 20 natural numbers.

As, 2x + y = 41

⇒ 2x = 41 – y

As x ∈ N (Natural number)

⇒ 41 – y ≥ 2

⇒ –y ≥ 2 – 41

⇒ –y ≥ –39

⇒ y ≤ 39

So, the range is first 39 natural numbers.

We have relation R defined on set N.

R = {(x, y): x, y ∈ N, 2x + y = 41}

Check for Reflexivity:

∀ x ∈ N

If (x, x) ∈ R

⇒ 2x + x = 41

⇒ 3x = 41

⇒ x = 13.67

But, x ≠ 13.67 as x ∈ N.

⇒ (x, x) ∉ R

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ 2x + y = 41 …(i)

Now, replace x by y and y by x, we get

⇒ 2y + x = 41 …(ii)

Take x = 20 and y = 1.

Equation (i) ⇒ 2(20) + 1 = 41

⇒ 40 + 1 = 41

⇒ 41 = 41, holds true.

Equation (ii) ⇒ 2(1) + 20 = 41

⇒ 2 + 20 = 41

⇒ 22 = 41, which is not true as 22 ≠ 41.

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, then (y, x) ∉ R.

∀ x, y ∈ N

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ 2x – y = 41 and 2y – z = 41

⇒ 2x – z = 41, may be true or not.

Let us sole these to find out.

We have

2x – y = 41 …(iii)

2y – z = 41 …(iv)

Multiply 2 by equation (i), we get

4x – 2y = 82 …(v)

Adding equation (v) and (iv), we get

(4x – 2y) + (2y – z) = 82 + 41

⇒ 4x – z = 123

⇒ 2x + 2x – z = 123

⇒ 2x – z = 123 – 2x

Take x = 40 (as x ∈ N)

⇒ 2x – z = 123 – 2(40)

⇒ 2x – z = 123 – 80

⇒ 2x – z = 43 ≠ 41

⇒ (x, z) ∉ R

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.

∀ x, y, z ∈ N

⇒ R is not transitive.

∴ R is not transitive.