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

Given function is,  $f\left(x\right)=4{\log }_e(x-1)-2x^2+4x+5,x>1$
Differentiate w.r.t ‘x’ 
$f\text{'}\left(x\right)=\frac{4}{x-1}-4(x-1)\mathrm{................}\left(i\right)$

Now, check option wise 
Take  $1<x<2\Rightarrow f\text{'}\left(x\right)>0$
Take  $x>2\Rightarrow f\text{'}\left(x\right)<0$
So, option (a) is correct. 
Take $f\left(x\right)=-1$
We have   
${\log }_e{(x-1)}^2=(x-3)(x+1)$

Therefore, it has two solutions 
So option (b) is correct.
Take  $f\left(e\right)>0,f(e+1)<0$
 $f\left(e\right).f(e+1)<0$
So, option (d) is correct. 
Now, put  $x=e$ in eq. (i) and again diff. eq. (i).
 $f\text{'}\left(e\right)-f"\left(2\right)=\frac{4}{e-1}-4(e-1)+8>0$
Therefore, option (c)  is incorrect.