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
$f(1)=1$
$f(2)=1$
$f(3)=2$
$f(4)=2$
$f(5)=3$
$f(6)=3$
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.
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