Question

Let $A = \\{ 9,10,11,12,13\\}$ and $f:A \to N$ be defined by $f\left( n \right)$=highest prime factor of $n$, then its range is
A. \left\\{ {13} \right\\}
B. \left\\{ {3,5,11,13} \right\\}
C. \left\\{ {11,13} \right\\}
D. \left\\{ {2,3,5,11} \right\\}

Answer 1

Consider the set is given $A = \\{ 9,10,11,12,13\\}$ and the function is defined as $f:A \to N$ and its range $f\left( n \right)$ contain the highest prime factor of $n$.
First, find the factors of elements of the set $A = \\{ 9,10,11,12,13\\}$.
Find the highest prime factor of each element.
The set of the highest prime factors of the individual elements is the range $f\left( n \right)$

Complete step-by-step solution:
The function is defined $f:A \to N$ where, $A = \\{ 9,10,11,12,13\\}$.
Take the values of $n = \\{ 9,10,11,12,13\\}$, find the factor of each element.
The factors of $9$ =$3 \times 3$
The factors of $10$ =$2 \times 5$
The factors of $11$ =$1 \times 11$
The factors of $12$ =$2 \times 2 \times 3$
The factors of $13$ =$1 \times 13$
$f\left( n \right)$=highest prime factor of $n$
$f\left( 9 \right)$=the highest prime factor of $9$
$\Rightarrow$ $f\left( 9 \right) = 3$
$f\left( {10} \right)$=the highest prime factor of $10$
$\Rightarrow$ $f\left( {10} \right) = 5$
$f\left( {11} \right)$=the highest prime factor of $11$
$\Rightarrow$ $f\left( {11} \right) = 11$
$f\left( {12} \right)$=the highest prime factor of $12$
$\Rightarrow$ $f\left( {12} \right) = 3$
$f\left( {13} \right)$=the highest prime factor of $13$
$\Rightarrow$ $f\left( {13} \right) = 13$
The range of $f$ is the set of all $f(n)$ =\left\\{ {3,5,11,13} \right\\}
The range $f\left( n \right)$ is \left\\{ {3,5,11,13} \right\\}.

Option B is the correct answer.

Note: Here are some basic ideas about the domain and range of a function.
Domain: The set of possible input values, the values go into a function is called domain.
Range: The set of possible output values and all the values that come out is called range.
For example: If the set A = \left\\{ {1,2,3} \right\\} and $f(x) = {x^2}$ then,
$f(1) = {1^2}$
$\Rightarrow f(1) = 1$
$f(2) = {2^2}$
$\Rightarrow f(2) = 4$
$f(3) = {3^2}$
$\Rightarrow f(3) = 9$
The range is set of all elements \left\\{ {1,4,9} \right\\} and domain is \left\\{ {1,2,3} \right\\}.