According to my thinking, I will help you discuss the monotonicity of functions in detail, generally using the derivative method.

Analysis: ∫ function f (x) = lnx+a (1-a) x2-2 (1-a) x, and its domain is x >；; 0

f'(x)= 1/x+2a( 1-a)x-2( 1-a)=[2a( 1-a)x^2-2( 1-a)x+ 1]/x

Because the function contains the character constant a, (1-a), when discussing the properties of the function, we should consider the influence of the character constant on the properties of the function. 0, so a is divided into 0 < a< 1, a= 1, and a> 1.

When a= 1

f '(x)= 1/x & gt； 0, ∴ Function f(x) monotonically increases in the domain of definition;

When 0

f '(x)=[2a( 1-a)x^2-2( 1-a)x+ 1]/x

Because of x>0, the sign of f'(x) depends on the molecule.

Let h (x) = 2a (1-a) x2-2 (1-a) x+1.

∫2a( 1-a)>0, ∴h(x) image is a parabola with an upward opening.

⊿=4( 1-a)^2-8a( 1-a)=4( 1-a)( 1-3a)

According to ⊿, yes > 0, =0,<0, the root of equation h(x)=0 can be determined.

The sign of ∵4( 1-a)>0,∴⊿ depends on the sign of (1-3a).

When 1-3a >: 0== >; a & lt 1/3，⊿>； 0

The equation h(x)=0 has two unequal real roots.

x 1 =[2( 1-a)-√⊿]/[4a( 1-a)],x2=[2( 1-a)+ √⊿]/[4a( 1-a]

And 0

(0, x 1) or (x2, +∞), f'(x) > on ∴; 0, the function f(x) monotonically increases;

On (x 1，x2)，f' (x)

When 1-3a=0== >; When a= 1/3, ⊿=0

The equation h(x)=0 has two equal real roots.

x 1 = x2 =( 1-a)/[2a( 1-a)]

∴ In the domain of definition, f' (x) >: =0, and the function f(x) monotonically increases;

When1-3a < 0 = >; 1/3 & lt； a & lt⊿ 1

Equation h(x)=0 has no solution.

∴ In the definition domain, f' (x) >: 0, and the function f(x) monotonically increases;

When a> is in 1

f '(x)=[2a( 1-a)x^2-2( 1-a)x+ 1]/x

Let h (x) = 2a (1-a) x2-2 (1-a) x+1.

∫2a( 1-a)& lt； 0, ∴h(x) image is a parabola with downward opening.

⊿=4( 1-a)^2-8a( 1-a)=4( 1-a)( 1-3a)>； 0

The equation h(x)=0 has two unequal real roots.

x 1 =[2( 1-a)-√⊿]/[4a( 1-a)],x2=[2( 1-a)+ √⊿]/[4a( 1-a]

And x2

(0, x 1) ∴, f'(x)>0, and the function f(x) increases monotonically;

∴ on (x 1，+∞)，f' (x)