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

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

Explanation

Solution

It is given that the domain of the function f(n)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(n)f\left( n \right).

Complete step-by-step answer:
Given data: A=9,10,11,12,13A = \\{ 9,10,11,12,13\\}
f:A→Nf:A \to N,f(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=3×3×19 = 3 \times 3 \times 1
10=2×5×110 = 2 \times 5 \times 1
11=11×111 = 11 \times 1
12=2×2×3×112 = 2 \times 2 \times 3 \times 1
13=13×113 = 13 \times 1
From the above prime factorization of the elements of A, we can conclude that
⇒f(9)=3\Rightarrow f(9) = 3
⇒f(10)=5\Rightarrow f(10) = 5
⇒f(11)=11\Rightarrow f(11) = 11
⇒f(12)=3\Rightarrow f(12) = 3
⇒f(13)=13\Rightarrow f(13) = 13
From the above, we can say that the range set of the function f(n)f(n)
i.e. Rf(n)=3,5,11,13{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)f(n)on substituting two different values will result in the same value i.e. 3.
Therefore they write the range set asRf(n)=3,3,5,11,13{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 Rf(n)=3,5,11,13{R_{f(n)}} = \\{ 3,5,11,13\\}