Let $$f\left( n \right)$$ denote the number of different ways in which the positive integer $$n$$ can be expressed as the sum of $$1s$$ and $$2s$$ . For example, $$f\left( 4 \right) = 5$$ , since $$4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1$$ . Note that the order of $$1s$$ and $$2s$$ is important.$$f:N \to N\;$$ isA. One-one and ontoB. One-one and intoC. Many-one and ontoD. Many-one and into

Question

Let $$f\left( n \right)$$ denote the number of different ways in which the positive integer $$n$$ can be expressed as the sum of $$1s$$ and $$2s$$ . For example, $$f\left( 4 \right) = 5$$ , since $$4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1$$ . Note that the order of $$1s$$ and $$2s$$ is important.$$f:N \to N\;$$  isA. One-one and ontoB. One-one and intoC. Many-one and ontoD. Many-one and into

Vedantu Content Team · Accepted Answer

Hint: In the above given question, we are given a function $$f$$ of natural numbers that is $$f:N \to N\;$$ . The function $$f$$ here is defined as the number of ways a positive integer $$n$$ can be expressed as the sum of $$1s$$ and $$2s$$ where the order of  $$1s$$ and $$2s$$ is important. We have to determine the nature of the function $$f$$ if it is one-one and onto or not. Complete step by step answer: Given function is $$f:N \to N\;$$, where $$f\left( n \right)$$ is the number of ways $$n$$ can be expressed as the sum of $$1s$$ and $$2s$$. We have to check the injectivity and surjectivity of the given function $$f:N \to N\;$$  whether it is one-one and onto or not. For a function to be one-one or injective, there must be the condition such that each element of one set, say $$N$$ here, is mapped with a unique element of another set, $$N$$ .That is. if $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$$ then $${x_1} = {x_2}$$Here we have, $$f\left( 4 \right) = 5$$ since$$ \Rightarrow 4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1$$Similarly, we can write the values for $$f\left( 1 \right)$$ , $$f\left( 2 \right)$$ , $$f\left( 3 \right)$$ , etc as,Hence we get $$f\left( 1 \right) = 1$$ as,$$ \Rightarrow 1 = 1$$And $$f\left( 2 \right) = 2$$ as,$$ \Rightarrow 2 = 2 = 1 + 1$$And $$f\left( 3 \right) = 3$$ as$$ \Rightarrow 3 = 1 + 1 + 1 = 1 + 2 = 2 + 1$$As we can see that the value of $$f\left( n \right)$$ is increasing with the increasing value of $$n$$ ,Hence for each value of $$n$$ we have a distinct i.e. unique value of $$f\left( n \right)$$.Therefore, $$f:N \to N\;$$ is one-one.Now, a function $$f:A \to B$$ is surjective if and only if for all the elements in $$B$$ , there exists an element in $$A$$.That is, $$\forall y \in B$$ we have $$x \in A$$ such that $$f\left( x \right) = y$$ .But in $$f:N \to N\;$$ , we have $$4 \in N$$ but there does not exist any $$n \in N$$ such that $$f\left( n \right) = 4$$ .Therefore, $$f:N \to N\;$$ is not onto but into.Hence, $$f:N \to N\;$$ is one-one and into.So the correct option is B.Note:When a function $$f:A \to B$$ is both injective and surjective, i.e. both one-one and onto, then the function is known as a bijective function or one-one onto function. Such functions are also known as an invertible function because the inverse of a function exists if and only if that function is both one-one and onto, i.e. is a bijective function.