Answer : 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:
- all elements of the first set are associated with the elements of the second set.
- An element of the first set has a unique image in the second
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.