Let A = \ 1, 2, 3, , 10\ and f: A → A be defined as f(k) = \ cases...

f\left(x\right) = \left\{\begin{cases}x + 1 & ifxis odd \\ x & ifxis even\end{cases}\right)</p><p>∵ g: A \to$A such that$g\left(f\left(x\right)\right) = f\left(x\right)$</p><p>$⇒$If$x$is even then$g\left(x\right)$= x \ldots \left(1\right)$

If $x$ is odd then $g\left(x + 1\right) = x + 1 \ldots \left(2\right)</p><p>from \left(1\right)$ and $\left(2\right) we can say that$g\left(x\right)$= x$ if $x$ is even

$⇒$ If $x$ is odd then $g\left(x\right) can take any value$ in set A

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