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Let A = {0, 1, 2} and B = {3, 5, 7, 9}. Let f = {(x, y) : x ϵ A, y ϵ B and y = 2x + 3}. Write f as a set of ordered pairs. Show that f is function from A to B. Find dom (f) and range (f).
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:
(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.
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