Given f(x)=4loge(x−1)−2x2+4x+5,x>1
f′(x)=x−14−4(x−1)
For 1<x<2⇒f′(x)>0
For x>2⇒f′(x)<0
Hence, f(x) has a maxima at x=2
f(x)=−1 has two solution as the curve of f(x) will cut it at exactly two points.
Now, f(e)>0 and f(e+1)<0
i.e. f(e)⋅f(e+1)<0
f′(e)−f"(2)=e−14−4(e−1)+8>0