Question

Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f,g:N\to N$ such that :
$f\left(n\right)=\{\begin{array}{cc}\frac{n+1}{2}& \text{if n is odd}\\ \frac{n}{2}& \text{in n is even}\end{array}$
and $g\left(n\right)=n-(-1{)}^n$. The fog is:

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

Answer 1

The correct option is D onto but not one-one

$fx=\{\begin{array}{cc}\frac{n+1}{2}& \text{n is odd}\\ \frac{n}{2}& \text{n is even}\end{array}$
$g\left(x\right)=n-(-1{)}^n\{\begin{array}{c}n+1;\text{n is odd}\\ n-1;\text{n is even}\end{array}$

$f\left(g\right(n\left)\right)=\{\begin{array}{cc}\frac{n}{2};& \text{n is even}\\ \frac{n+1}{2};& \text{n is odd}\end{array}$

$\therefore$ onto but not one-one