# Mark (√) against the correct answer in the following:

Question:

Mark (√) against the correct answer in the following:

Let $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}: \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\frac{1}{2}(\mathrm{n}+1), \text { when } \mathrm{n} \text { is odd } \\ \frac{\mathrm{n}}{2}, \text { when } \mathrm{n} \text { is even. }\end{array}\right.$

Then, $f$ is

A. one - one and into

B. one - one and onto

C. many - one and into

D. many - one and onto

Solution:

$f(1)=1$

$f(2)=1$

$f(3)=2$

$f(4)=2$

$f(5)=3$

$f(6)=3$

Since at different values of $x$ we get same value of $y \therefore f(n)$ is many -one

And range of $f(n)=N=N$ (codomain)

$\therefore$ the function $f: N \rightarrow Z$, defined by

$f: N \rightarrow N: f(x)=\left\{\begin{array}{l}\frac{1}{2}(n+1), \text { when } n \text { is odd } \\ \frac{n}{2}, \text { when } n \text { is even. }\end{array}\right.$ is both many - one and onto.