Question

Let \[A = \left\{ {12,13,14,15,16,17} \right\}\]and \[f:{\text{ }}A \to Z\] be a function given by f(x) \[ = \] highest common factor of x.
Find range of f.

Answer 1

Hint: To solve this problem, i.e., to find the range of the function. We will find the highest common factor of x, where we will consider the set\[A = \left\{ {12,13,14,15,16,17} \right\}\]and the condition\[f:{\text{ }}A \to Z\]. Now since, the range of f is the set of all f(x), thus the set of highest common factors of x, will be our required answer.

Complete step-by-step answer:
We have been given that, \[A = \left\{ {12,13,14,15,16,17} \right\}\] and \[f:{\text{ }}A \to Z\] be a function given by f(x) equals to highest common factor of x. We need to find the range of f.
So, it is given that, \[A = \left\{ {12,13,14,15,16,17} \right\}\]. To find the range of f, we will find the highest common factor of the elements of the set A.
\[\Rightarrow f\left( {12} \right) = \]highest prime factor of \[12 = 2 \times 2 \times 3 = 3\]
\[\Rightarrow f\left( {13} \right) = \]highest prime factor of \[13 = 13\]
\[\Rightarrow f\left( {14} \right) = \]highest prime factor of \[14 = 2 \times 7 = 7\]
\[\Rightarrow f\left( {15} \right) = \]highest prime factor of \[15 = 3 \times 5 = 5\]
\[\Rightarrow f\left( {16} \right) = \]highest prime factor of \[16 = 2 \times 2 \times 2 \times 2 = 2\]
\[\Rightarrow f\left( {17} \right) = \]highest prime factor of \[17 = 17\]
We know that, range of f is the set of all f(x), where \[x \in A.\] Thus, range of \[f = \left\{ {3,13,7,5,2,17} \right\}.\]

Note: In the question, we are asked about the range, let us understand it in detail.
In relation and function, all of the values that can go into a relation or function, i.e., the input) are called the domain. And the domain is the set of all first elements of ordered pairs, i.e., the x-coordinates. Whereas the range is the set of all second elements of ordered pairs i.e., the y-coordinates. So, we can say that only those elements which are used by the relation or function constitute the range.