Question

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

defined as $f(K)=\{\begin{array}{ll}K+1& \text{if Kis odd}\\ K& \text{if K is odd}\end{array}$
Then the number of possible functions $g:A\to A$ such that go $f=f$ is

• A. $10^5$
• B. ${}^{10}{C}_5$
• C. $5^5$
• D. $5!$

Answer 1

Answer: (A)

$f(X)=<br/><br/>\{\begin{array}{ll}<br/><br/>X+1& \text{if Xis odd}\\ <br/><br/>X& \text{if X is odd}<br/><br/>\end{array}$
$\because g:A\to A$ 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 $g(x)=10^5\times 1$