Q. 85.0( 2 Votes )

Let A = {0, 1, 2}

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:


(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


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