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
$f\left(n\right)=\{\begin{array}{cc}\frac{n+1}{2},& \text{if}n\text{is odd}\\ \frac{n}{2},& \text{if}n\text{is even}\end{array}$
and $g\left(n\right)=n-(-1{)}^n$. Then $f\circ g$ is :

• A. both one-one and onto function
• B. one-one but not onto function
• C. onto but not one-one function
• D. neither one-one nor onto function

Answer 1

The correct option is C onto but not one-one function

$g\left(n\right)=n-(-1{)}^n$
$\Rightarrow g\left(n\right)=\{\begin{array}{cc}n+1,& \text{if}n\text{is odd}\\ n-1,& \text{if}n\text{is even}\end{array}$
Given,
$f\left(n\right)=\{\begin{array}{cc}\frac{n+1}{2}& \text{if}n\text{is odd}\\ \frac{n}{2}& \text{if}n\text{is even}\end{array}$
Now from observation,
$f\left(g\right(1\left)\right)=f\left(2\right)=1(\because g(1)=2)$
$f\left(g\right(2\left)\right)=f\left(1\right)=1(\because g(2)=1)$
$f\left(g\right(3\left)\right)=f\left(4\right)=2(\because g(3)=4)$
$f\left(g\right(4\left)\right)=f\left(3\right)=2(\because g(4)=3)$

$\Rightarrow f\left(g\right(x\left)\right)$ is many one function

Now,
$(f\circ g)\left(x\right)$, when $x$ is even
$\Rightarrow f\left(g\right(2m\left)\right)=f(2m-1)=m(\text{where}m\in N)$
$(f\circ g)\left(x\right)$, when $x$ is odd
$\Rightarrow f\left(g\right(2m+1\left)\right)=f(2m+2)=m+1\in N$
$\Rightarrow f\left(g\right(x\left)\right)$ is onto function
$\therefore f\left(g\right(x\left)\right)$ is onto but not one-one function.