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

Last updated date: 20th Mar 2023
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Hint: Substitute each of the values given in the domain set $A$ in the given function. Write the prime factorization of each of the given numbers in the domain and choose the highest possible value of its prime factor. Do this for each element of the domain and the set of values which we get on applying the function to the domain is the range of the given function.

We have a set $A=\{9,10,11,12,13\}$ and a function acting on this set $f:A\to N$ defined by $f\left( n \right)=$ the highest prime factor of $n$. We have to find the range of the given function.
We will substitute each element of the domain set $A$ in the given function and operate the function as given. We will write prime factorization of each element of the domain and choose its largest prime factor. We will add that prime factor of the number to a new set. This new set will be the range of the function $f$.
We will begin by considering each element of the set $A$.
We will begin by element $9$. $9$ can be written in terms of its prime factor as $9=3\times 3\times 1$. So, the highest prime factor of $9$ is $3$. We will add $3$ to the range of function $f$.
We will now evaluate the value of element $10$ on the function $f$. $10$ can be written in terms of its prime factor as $10=2\times 5\times 1$. So, the highest prime factor of $10$ is $5$. We will add $5$ to the range of function $f$.
We will now evaluate the value of element $11$ on the function $f$. $11$ can be written in terms of its prime factor as $11=11\times 1$. So, the highest prime factor of $11$ is $11$. We will add $11$ to the range of function $f$.
We will now evaluate the value of element $12$ on the function $f$. $12$ can be written in terms of its prime factor as $12=2\times 2\times 3\times 1$. So, the highest prime factor of $12$ is $3$. But since $3$ is already in the range of the function, we don’t have to add it again.
We will now evaluate the value of element $13$ on the function $f$. $13$ can be written in terms of its prime factor as $13=13\times 1$. So, the highest prime factor of $13$ is $13$. We will add $13$ to the range of function $f$.
Hence, the range of $f$ is the set $\left\{ 3,5,11,13 \right\}$.