Answer : Given: A = {0, 1, 2} and B = {3, 5, 7, 9}
f = {(x, y): x ∈ A, y ∈ B and y = 2x + 3}
For x = 0 y = 2x + 3
y = 2(0) + 3
y = 3 ∈ B For x = 1 y = 2x + 3
y = 2(1) + 3
y = 5 ∈ B For x = 2 y = 2x + 3
y = 2(2) + 3
y = 7 ∈ B
∴ f = {(0, 3), (1, 5), (2, 7)}
(0, 5), (0, 7), (0, 9), (1, 3), (1, 7), (1, 9), (2, 3), (2, 5), (2, 9) are not the members of ‘f’
because they are not satisfying the given condition y = 2x + 3 Now, we have to show that f is a function from A to B Function:
- all elements of the first set are associated with the elements of the second
- An element of the first set has a unique image in the second f = {(0, 3), (1, 5), (2, 7)}
Here, (i) all elements of set A are associated with an element in set B.
(ii) an element of set A is associated with a unique element in set B.
∴ f is a function. Dom (f) = 0, 1, 2
Range (f) = 3, 5, 7