(1) When a=0, find the extreme value of f(x);
(2) When a < 0, find the monotone interval of f(x);
(3) When -3
(1) analysis: ∫f(x)=(2-a)lnx+ 1/x+2ax, and the domain is x>0.
Let a = 0 = =>F (x) = 2lnx+1/x = > Let f' (x) = 2/x-1/x 2 = 0 = = >; x= 1/2
f''(x)=-2/x^2+2/x^3==>； f ' '( 1/2)= 8 & gt； 0
When x= 1/2, the function f(x) takes the minimum value of 2-2ln2.
(2) analysis: do a; 0
Let f' (x) = (2-a)/x-1/x 2+2a = (2ax 2+(2-a) x-1)/x 2 = 0 = >; x 1= 1/2，x2=- 1/a
f''(x)=(a-2)/x^2+2/x^3==>； f ' '(x 1)= 4a+8； f''(x2)=-2a^2-a^3
When a < -2, f' (x)
When a & lt-2, f' (x) >; 0, f(x) takes the minimum value at x2=- 1/a; When a & gt-2, f' (x)
∴ when-1/a =1/2 = >; When a=-2 and f' (x) < =0, f(x) is reduced on the domain;
When-1/a
When-1/a >; 1/2== >-2 & lt； A<0, f(x) decreases at (0, 1/2) or (-1/a, +∞); Monotonically increasing in [1/2,-1/a];
(3) Analysis: ∵ When-3
According to (2), when -3
∴f(x)max=f( 1)= 1+2a,f(x)min=f(3)=(2-a)ln3+ 1/3+6a，
∴|f(λ 1)-f(λ2)|max= f(x)max-f(x)min = f( 1)-f(3)= 2/3-4a+(a-2)ln3
Let λ 1, λ2∈, | f (λ1)-f (λ 2) | > (m+ln3)a-2ln3 hold.
∴2/3-4a+(a-2)ln3>； (m+ln3)a-2ln3== > Ma & lt2/3-4a
∵a & lt； 0,∴m>； 2/(3a)-4
-3 & lt； a & lt-2== >-9 & lt； 3a & lt-6== >- 1/6 & lt； 1/(3a)& lt； - 1/9== >- 1/3 & lt； 2/(3a)& lt； -2/9
∴- 13/3<； 2/(3a)-4 & lt； -38/9
∴m>； =-38/9