For equal roots in f(x)m2−2f′(x)m+f′′(x)=0, discriminant = 0:
[f′(x)]2=f(x)f′′(x)
This gives f′(x)f′′(x)=f(x)f′(x)
Integrating: f′(x)=kf(x), so f(x)=Aekx
Using f(0)=1 and f′(0)=2: A=1, k=2
f(x)=e2x
For g(x)=f(lnx−x):
g′(x)=2e2(lnx−x)⋅(x1−1)
g′(x)>0 when x1−x>0, i.e., 0<x<1
(α,β)=(0,1)
α+β=1