Question

$\mathrm{If}\mathrm{f}:\mathrm{N}\to \mathrm{N}$is defined by. Then $f$ is $\mathrm{f}\left(\mathrm{n}\right)=\left\{\begin{array}{l}\frac{\mathrm{n}+1}{2},\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{odd}\\ \frac{\mathrm{n}}{2},\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}\end{array}\right.$

• A. Onto but not one-one
• B. One-one and onto
• C. Neither one-one nor onto
• D. One-one but not onto

Answer 1

The correct option is A

Onto but not one-one

Step 1: Given information

$\mathrm{f}:\mathrm{N}\to \mathrm{N}$ is defined by

$\mathrm{f}\left(\mathrm{n}\right)=\left\{\begin{array}{l}\frac{\mathrm{n}+1}{2},\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{odd}\\ \frac{\mathrm{n}}{2},\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}\end{array}\right.$

Step 2: Determination of the equation:

When $\mathit{n}\mathbf{=}\mathbf{1}$,

$\begin{array}{rcl}& \Rightarrow & \mathrm{f}\left(1\right)=\frac{\left[1+1\right]}{2}\\ & \Rightarrow & \mathrm{f}\left(1\right)=\frac{2}{2}\\ & \Rightarrow & \mathrm{f}\left(1\right)=1\end{array}$

When $n=2,$

$\begin{array}{rcl}& \Rightarrow & \mathrm{f}\left(2\right)=\frac{2}{2}\\ & \Rightarrow & \mathrm{f}\left(2\right)=1\\ \therefore \mathrm{f}\left(1\right)& =& \mathrm{f}\left(2\right)\end{array}$

Since, the value of $f\left(1\right)$and $f\left(2\right)$ is repetitive,

Therefore, $\mathrm{f}\left(\mathrm{x}\right)$ is not one-one.

If $n$ is odd,

$$\begin{gathered} \Rightarrow \mathrm{f}\left(\mathrm{n}\right)=\frac{\left[\mathrm{n}+1\right]}{2} \\ \Rightarrow \frac{\left[\mathrm{n}+1\right]}{2}=\mathrm{y} \\ \Rightarrow 2\mathrm{y}-1=\mathrm{n} \end{gathered}$$

If $n$ is even,

$\begin{array}{rcl}\mathrm{f}\left(\mathrm{n}\right)& =& \frac{\mathrm{n}}{2}=\mathrm{y}\\ 2\mathrm{y}& =& \mathrm{n}\end{array}$

Therefore, $\mathrm{f}\left(\mathrm{x}\right)$ is onto mapping.

Hence, the correct option is option (A).