Question

Let $A=\{1,2,3, \ldots, 10\}$ and $f: A \rightarrow A$ be

defined as $f(K)= \begin{cases} K +1 & \text{if $ K $is odd} \\[2ex] K & \text{if $ K $ is odd} \end{cases}$
Then the number of possible functions $g: A \rightarrow A$ such that go $f=f$ is

• A. $10^{5}$
• B. ${ }^{10} C _{5}$
• C. $5^{5}$
• D. $5 !$

Answer 1

Answer: (A)

$f(X)=
\begin{cases}
X+1 & \text{if $ X $is odd} \\[2ex]
X & \text{if $ X $ is odd}
\end{cases}$
$\because g : A \rightarrow A$ such that $g (f( x ))=f( x )$
$\Rightarrow $ If $x$ is even then $g(x)=x \,\,\,\, ...(1)$
If $x$ is odd then $g(x+1)=x+1 \,\,\,\, ....(2)$
from (1) and (2) we can say that
$g(x)=x$ if $x$ is even
$\Rightarrow $ If $x$ is odd then $g(x)$ can take any value in set $A $
so number of $g ( x )=10^{5} \times 1$