Let $\mathrm{A}=\{1,2,3, \ldots, 10\}$ and $f: \mathrm{A} \rightarrow \mathrm{A}$ be defined as $f(\mathrm{k})=\left\{\begin{array}{cl}\mathrm{k}+1 & \text { if } \mathrm{k} \text { is odd } \\ \mathrm{k} & \text { if } \mathrm{k} \text { is even }\end{array}\right.$
Then the number of possible functions $\mathrm{g}: \mathrm{A} \rightarrow \mathrm{A}$ such that go $f=f$ is
Correct Option: 1
$f(\mathrm{x})=\left\{\begin{array}{cl}\mathrm{x}+1, & \text { if } \mathrm{x} \text { is odd } \\ \mathrm{x}, & \text { if } \mathrm{x} \text { is even }\end{array}\right.$
$\because \mathrm{g}: \mathrm{A} \rightarrow \mathrm{A}$ such that $\mathrm{g}(f(\mathrm{x}))=f(\mathrm{x})$
$\Rightarrow$ 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 $\mathrm{g}(\mathrm{x})=\mathrm{x}$ if $\mathrm{x}$ is even
$\Rightarrow$ If $x$ is odd then $g(x)$ can take any value in set $\mathrm{A}$
so number of $\mathrm{g}(\mathrm{x})=10^{5} \times 1$
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