So g(x)= into * 5 (ax)-4 x = into * 2 x-4 x 0.

Let t = 2 x, 0

According to the properties of quadratic function, just enter /2.

(2)M * xlnx/2 & lt； =x^2-cx+ 12(x >； 0)M = 2 & lt； = = & gtcx & lt=x^2-xlnx+ 12(x >； 0)& lt； = = & gtc & lt=x-lnx+ 12/x

The problem of constant establishment is transformed into finding the function y = x-lnx+12/x (x >); 0) Scope problem.

y'= 1- 1/x- 12/x^2(x >； 0) 1- 1/x- 12/x^2 & gt； = 0(x & gt； 0)= = = & gt； (3/x+ 1)(4/x- 1)& lt； = 0(x & gt； 0)= = = & gt； 4/x- 1 & lt； = 0x & gt； 0

= = = & gtX & gt=4

Therefore, the function y=x-lnx+ 12/x monotonically increases in the interval [4,+infinity] and monotonically decreases in (0,4).

Function y = x-lnx+12/x (x >); The minimum value of 0) is ymin=4-ln4+ 12/4=7-2ln2.

So c < =7-2ln2.