Let function $f\left(x\right)$  be thrice differentiable and satisfies $f\left(f\left(x\right)\right)=1-xforallx\in [0,1]$. If  $J=\underset{0}{\overset{1}{\int }}f\left(x\right)dx$ and $f"\left(\frac{4}{5}\right)=0$ , then which of the following is/are TRUE?

 $$\begin{gathered} f\left(f\left(x\right)\right)=1-x \\ f\left(f\left(f\left(x\right)\right)\right)=1-f\left(x\right) \\ f(1-x)=1-f\left(x\right) \\ f\left(x\right)+f(1-x)=1 \\ f\text{'}\left(x\right)-f\text{'}(1-x)=0 \\ f\text{'}\text{'}\left(x\right)+f\text{'}\text{'}(1-x)=0 \\ f\text{'}\text{'}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)+f\text{'}\text{'}\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)=0 \\ J=\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}f(1-x)dx \\ 2J=\underset{0}{\overset{1}{\int }}\left(f\left(x\right)+f(1-x)\right)dx=\underset{0}{\overset{1}{\int }}1dx \\ f\text{'}\text{'}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)=0 \end{gathered}$$

$f\text{'}\text{'}\text{'}\left(x\right)=0$ has atleast one root in  $x\in (\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.)$