Question

the co domain of f : a →b where a={2,3,5} f(x) = 2x-1 can be taken as
a. {3,5,7}
b. {3,5,9,11}
c. {5,9}
d. {3,5}
​

Answer 1

Answer:

b. {3,5,9,11}

Step-by-step explanation:

f(x) = 2x-1

f(2) = 2(2)-1 = 4-1 = 3

f(3) = 2(3)-1 = 6-1 = 5

f(5) = 2(5)-1 = 10-1 = 9

So the range is {3,5,9}

The only option with {3,5,9} is option b, {3,5,9,11}.

Answer 2

Answer :

b. {3 , 5 , 9 , 11}

Note :

• Function : Function is a rule/mapping between two non empty sets A and B such that every element of set A is associated with unique element in set B .

• For a function f from A to B , we write ;

f : A → B where A is called domain and B if called codomain .

• Domain : The set A is called domain of f if f : A → B . OR it is set of all possible values of x-coordinate for the function f .

• Range : The set of unique output of the function if called range . OR it is set of all possible values of y coordinate for the function f .

• Range ⊆ Codomain .

Solution :

• Given : f : A → B where A = {2 , 3 , 5 } and f(x) = 2x - 1
• To find : Codomain , B = ?

We have ,

f(x) = 2x - 1

Here , domain (f) = { 2 , 3 , 5 }

ie.

The possible x-coordinates are : 2 , 3 , 5 .

• If x = 2 , then ;

f(2) = 2×2 - 1 = 4 - 1 = 3

• If x = 3 , then ;

f(3) = 2×3 - 1 = 6 - 1 = 5

• If x = 5 , then ;

f(5) = 2×5 - 1 = 10 - 1 = 9

Thus ,

The possible y-coordinates are : 3 , 5 , 9 .

ie. Range(f) = {3 , 5 , 9} .

Also ,

We know that , Range ⊆ Codomain .

Clearly , here we have

Range(f) = {3 , 5 , 9} ⊆ {3 , 5 , 9 , 11}

Thus ,

Codomain(f) can be taken as {3 , 5 , 9 , 11}

Hence ,

The required answer is b. {3 , 5 , 9 , 11}