Solve this

Given: $A=\{0,1,2\}$ and $B=\{3,5,7,9\}$

$f=\{(x, y): x \in A, y \in B$ and $y=2 x+3\}$

For $x=0$

$y=2 x+3$

$y=2(0)+3$

$y=3 \in B$

For $x=1$

$y=2 x+3$

$y=2(1)+3$

$y=5 \in B$

For $x=2$

$y=2 x+3$

$y=2(2)+3$

$y=7 \in B$

$\therefore f=\{(0,3),(1,5),(2,7)\}$

(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 $B$.

$\therefore \mathrm{f}$ is a function.

Dom (f) = 0, 1, 2

Range (f) = 3, 5, 7