Concept:

Domain:

Let f : A → B be a function, then the set A is called as the domain of the function f.

Co-domain:
Let f : A → B be a function, then the set B is called as the co-domain of the function f.

Range:
Let f : A → B, then the range of the function f consists of those elements in B which have at least one preimage in A. It is denoted as f (A) i.e f (A) = {b ∈ B | f (a) = b for some a A}

Note: Range is the subset of the codomain of f.

Calculation:

Given: A = {9, 10, 11, 12, 13} and f: A → N is a function where N is the set of natural numbers such that f(n) = the highest prime factor of n

We know that, the prime factorization of: 9 = 3², 10 = 2 × 5, 11 = 11 × 1, 12 = 2² × 3 and 13 = 13 × 1.

According to the definition of the function f we have,

⇒ f(9) = 3, f(10) = 5, f(11) = 11, f(12) = 3, f(13) = 13

As we know that, if f : A → B, then the range of the function f consists of those elements in B which have at least one preimage in A. It is denoted as f (A) i.e f (A) = {b ∈ B | f (a) = b for some a A}

⇒ Range of f = {3, 5, 11, 13}

Hence, option 3 is the correct answer.