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 : $f(n) = \begin{cases} \frac{n+1}{2} & \quad \text{if } n \text{ is odd}\\\ \frac{n}{2} & \quad \text{if } n \text{ is even} \end{cases}$ and $g(n) = n-(-1)^n$. 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

Answer: (D)

onto but not one-one

Answer 2

$f(x) = \begin{cases} \frac{n+1}{2} & \quad n \text{ is odd}\\\ -(n+1)/2 & \quad n \text{ is even} \end{cases}$
$g(x) = n - (-1)^n \begin{cases} n+ 1 & ; \quad n \text{ is odd}\\\ n - 1 &; \quad \text{if } n \text{ is even} \end{cases}$
$f(g(n)) = \begin{cases} \frac{n}{2} ; & \quad n \text{ is even}\\\ \frac{n +1}{2} ; & \quad n \text{ is odd} \end{cases}$
$\therefore$ many one but one to one