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.