# Let $A={1,2,3,......10}$ and $f: A\rightarrow A$ be defined as $f(k)=\left \{ \begin{matrix} k+1 & if\;k\; is\; odd\\ k &if \;k \;is\; even \end{matrix} \right.$ Then the number of possible functions $g: A\rightarrow A$ such that gof =f is: Option: 1 $^{10}C_5$ Option: 3 $10^5$ Option: 5 $5!$ Option: 7 $5^5$

$f(x)=\left\{\begin{array}{cc} x+1, & \text { if } x \text { is odd } \\ x, & \text { if } x \text { is even } \end{array}\right.$

\begin{aligned} &\because \mathrm{g}: \mathrm{A} \rightarrow \text { A such that } \mathrm{g}(f(\mathrm{x}))=f(\mathrm{x})\\ &\Rightarrow \text { If } x \text { is even then } g(x)=x\;\;\;\;\;\;\;\;\;\;\;\ldots(1) \end{aligned}

$\\\text{If } x\text { is odd then }g(x+1)=x+1 \quad \ldots(2)\\ \text{from (1) and (2) we can say that}\\\text{g(x) = x if x is even}\\\Rightarrow \text{If x is odd then g(x) can take any value in set A}$

$\text { so number of } \mathrm{g}(\mathrm{x})=10^{5} \times 1$

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