f(x) = 4 ${\log }_{\mathrm{e}}(\mathrm{x}-1)-2{\mathrm{x}}^2+4\mathrm{x}+5,\mathrm{x}>1$which one of the following is NOT correct ?

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=4{\log }_{\mathrm{e}}(\mathrm{x}-1)-2{\mathrm{x}}^2+4\mathrm{x}+5,\mathrm{x}>1\\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\frac{4}{\mathrm{x}-1}-4(\mathrm{x}-1)\end{array}$
For $1<\mathrm{x}<2\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)>0$
For $\mathrm{x}>2\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)<0$    (option 1 is correct)
f(x) = -1 has two solution    (option 2 is correct)
$\begin{array}{l}\mathrm{f}\left(\mathrm{e}\right)>0\\ \mathrm{f}(\mathrm{e}+1)<0\\ \mathrm{f}\left(\mathrm{e}\right)\cdot \mathrm{f}(\mathrm{e}+1)<0 \text{ (option }4\text{ is correct) }\\ {\mathrm{f}}^{ \text{'}}\left(\mathrm{e}\right)-{\mathrm{f}}^{ \text{'}\text{'}}\left(2\right)=\frac{4}{\mathrm{e}-1}-4(\mathrm{e}-1)+8>0\end{array}$
(option C is incorrect)