Question

Let $A = \{ 9,10,11,12,13\} $ and if $f:A \to N$ be defined by $f(n) = $ the highest prime factor of n. Find the range of f.

Answer 1

Hint: It is given that the domain of the function \[f\left( n \right)\] is set A, f is defined as the highest prime factor of the input, therefore using the prime factorization method we’ll find the highest prime factors of all elements of set A that will be on combining will result in range set of \[f\left( n \right)\].

Complete step-by-step answer:
Given data: $A = \{ 9,10,11,12,13\} $
$f:A \to N$,$f(n) = $the highest prime factor of n
It is given that the domain of f is A,
Using the prime factorization method, we will find the highest prime factor of the elements of set A
$9 = 3 \times 3 \times 1$
$10 = 2 \times 5 \times 1$
$11 = 11 \times 1$
$12 = 2 \times 2 \times 3 \times 1$
$13 = 13 \times 1$
From the above prime factorization of the elements of A, we can conclude that
$ \Rightarrow f(9) = 3$
$ \Rightarrow f(10) = 5$
$ \Rightarrow f(11) = 11$
$ \Rightarrow f(12) = 3$
$ \Rightarrow f(13) = 13$
From the above, we can say that the range set of the function $f(n)$
i.e. ${R_{f(n)}} = \{ 3,5,11,13\} $

Note: In the above solution 3 is the highest prime factor of two elements of set A, or we can say that function $f(n)$on substituting two different values will result in the same value i.e. 3.
Therefore they write the range set as${R_{f(n)}} = \{ 3,3,5,11,13\} $, which totally wrong as we know that a set does not contain two similar elements and hence the correct range set is ${R_{f(n)}} = \{ 3,5,11,13\} $