(2)
Multiply f' (x) =1/x-a.
F(x) is in (0, 1/a) ↗ and (1/a, +∞) ↘.
∴f(x)max=ln( 1/a)-a*( 1/a)+4a? =-lna+4a? - 1
Let g(a)=f(x)max.
∴g'(a)=- 1/a+8a
G(a) in (0, √2/4)↗) and in (√2/4, +∞) ↘.
Therefore, when the maximum value M(a) of f(x) exists, a=√2/4.
∫f "(x)=-( 1/x)? & lt0
∴f'(x) in (0, +∞) ↘
As can be seen from the f'(x) diagram, the change of f'(x) is intense at first, and then slowly slows down.
The f(x) image thus obtained looks like this:
∴a 1+a2>; 2a=√2/2
Namely a1+a2 >; √2/2