Question

If $\mathrm{f}:\mathrm{N}\to \mathrm{N}$ defined by $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{x}+1$, $x$ belongs to $N$, then f is

• A. One-one and onto
• B. Many-one and onto
• C. One-one but not onto
• D. None of the above

Answer 1

The correct option is C

One-one but not onto

Step 1: Checking f for one-one function:

Given, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{x}+1$ , x belongs to $N.$

Let $x\mathrm{and}y$ be any two elements in domain $N,$ such that,

$$\begin{gathered} \therefore \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right) \\ \begin{array}{l}\Rightarrow {\mathrm{x}}^2+\mathrm{x}+1={\mathrm{y}}^2+\mathrm{y}+1 \\ \Rightarrow (\mathrm{x}+\mathrm{y})(\mathrm{x}-\mathrm{y})+(\mathrm{x}-\mathrm{y})=0 \\ \Rightarrow (\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y}+1)=0 \\ \Rightarrow \mathrm{x}-\mathrm{y}=0\end{array} \end{gathered}$$

$(\mathrm{x}+\mathrm{y}+1)$ cannot be zero because $x\mathrm{and}y$ are natural numbers.

$\Rightarrow \mathrm{x}=\mathrm{y}$

Therefore, $f$ is a one-one function.

Step 2: Checking f for onto function:

When $x=1,$

$\begin{array}{rcl}& \Rightarrow & {\mathrm{x}}^2+\mathrm{x}+1=1+1+1\\ & \Rightarrow & {\mathrm{x}}^2+\mathrm{x}+1=3\\ & \Rightarrow & {\mathrm{x}}^2+\mathrm{x}+1\ge 3,\end{array}$

for every x in N

As, $f\left(x\right)$ will not assume the values $1\mathrm{and}2$.

Therefore, $f$ is not onto function.

Hence, the correct answer is an option (C).