Daily questions _191102

Known function \ (F (X) = X {\ LN} X + \ dfrac {. 1} {2} AX ^. 3-AX ^ 2 \) , \ (A \ in \ mathbb {R & lt} \) .
\ (( 1) \) when \ (a = 0 \) , the request \ (f (x) \) monotonous range;
\ ((2) \) if the function \ (g (x) = \ dfrac {f (x) } {x} \) there are two extreme points \ (x_1, x_2 \) , seeking \ (G (x_1) + G (x_2) \) . the range
analysis:
\ ((1) \) when \ (a = 0 \) , the on \ (f (x) \) derivative available \ [f '(x) =
1 + {\ ln} x, x> 0. \] At this time \ (f (x ) \) in \ (\ left (0, \ dfrac {1} {\ mathrm {e}} \ right) \) decrease monotonically in \ (\ left [\ dfrac { 1} {\ mathrm {e}}, + \ infty \ right) \) monotonically.
\ ((2) \) from the title\ [G (x) = { \ ln} x + \ dfrac {1} {2} ax ^ 2-ax, x> 0. \] to \ (g (x) \) derivative available \ [g '( x) = \ dfrac {1}
{x} + ax-a = \ dfrac {ax ^ 2-ax + 1} {x}, x> 0. \] To that \ (g '(x) \ ) in \ ((0, + \ infty ) \) there are two numbers becomes zero, and only need to \ (G '\ left (\ dfrac. 1} {2} {\ right) <0 \) , i.e. \ (a> 4 \) . The Vieta theorem available \ [x_1 + x_2 = \ dfrac
{1} {2}, x_1x_2 = \ dfrac {1} {a}. \] so \ [\ begin {split} & g (x_1) + g (x_2) \\ = & {\ ln} (x_1x_2) + \ dfrac {1} {2} a \ cdot \ left (x_1 ^ 2 + x_2 ^ 2 \ right) -a \ left (x_1 + x_2 \ right) \\ = & {\ ln } (x_1x_2) + \ dfrac {1} {2} a \ cdot \ left [\ left (x_1 + x_2 \ right) ^ 2-2x_1x_2 \ right] -a \ left (x_1 + x_2 \ right) \\ = & - {\ ln} a + \ dfrac {1} {2} a \ cdot \ left (\ dfrac {1} {4} - \ dfrac {2} {a} \ right) - \ dfrac {a} {2}
\\ = & -. {\ ln} a- \ dfrac {3} {8} a-1 \ end {split} \] obviously,The above expression is about \ (a \)A monotonically decreasing function, wherein \ (a> 4 \) and therefore of the desired expression range of. \ (\ Left (- \ infty, -2 {\ 2- LN} \ {2}. 5 dfrac {} \ right) \) .