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.
$f:N \to N\;$ is
A. One-one and onto
B. One-one and into
C. Many-one and onto
D. Many-one and into

Answer 1

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.