Given: A = {1, 2} and B = {2, 4, 6}

f = {(x, y): x ∈ A, y ∈ B and y > 2x + 1}

Putting x = 1 in y > 2x + 1, we get

y > 2(1) + 1

⇒ y > 3

and y ∈ B

this means y = 4, 6 if x = 1 because it satisfies the condition y > 3

Putting x = 2 in y > 2x + 1, we get

y > 2(2) + 1

⇒ y > 5

this means y = 6 if x = 2 because it satisfies the condition y > 5.

∴ f = {(1, 4), (1, 6), (2, 6)}

(1, 2), (2, 2), (2, 4) are not the members of ‘f’ because they do not satisfy the given condition y > 2x + 1

Firstly, we have to show that f is a relation from A to B.

First elements = 1, 2

All the first elements are in Set A

So, the first element is from set A

Second elements in F = 4, 6

All the second elements are in Set B

So, the second element is from set B

Since the first element is from set A and second element is from set B

Hence, F is a relation from A to B.

Function:

(i) all elements of the first set are associated with the elements of the second set.

(ii) An element of the first set has a unique image in the second set.

Now, we have to show that f is not a function from A to B

f = {(1, 4), (1, 6), (2, 6)}

Here, 1 is coming twice.

Hence, it does not have a unique (one) image.

So, it is not a function.