Title: The known function f (x) = 2lnx-x 2. If the image of function g(x)=f(x)-ax intersects the X axis at two points A(x 1, 0), B(x2, 0) and 0.

Analysis is easy to use reduction to absurdity.

Let's set the midpoint C(xo, 0) of AB,

Then xo=(x 1+x2)/2, from P+Q = 1, Q >；; =p, and both p and q are positive real numbers, so it is easy to get 0.

Then (px1+qx2)-XO = px1+(1-p) x2-(x1+x2)/2 = (x1-x2) (2p-/kloc. =0，(0 & ltx 1 & lt； x2)

Get (px 1+qx2) >: =xo .

The first derivative g'(x)=2/x-2x-a is obtained by g (x) = 2lnx-x 2-ax.

Then take the derivative of g'(x) and get its second derivative g "(x) =-2/x2-2.

Knowing that g'(x) is in x>0, and g' (px 1+qx2) < =g'(xo),

So it is necessary to prove G' (px 1+qx2).

Let's prove it by reducing to absurdity

Suppose g' (XO) >; =0 holds.

Combine what is known to be available

2lnx 1-x 1^2-ax 1=0.....( 1),

2lnx2-x2^2-ax2=0......(2),

2/XO-2xo-a & gt； =0......(3),

xo=(x 1+x2)/2......(4),

There are 0

ln(x2/x 1)-2(x2-x 1)/(x2+x 1)& lt； =0,

That is ln (x2/x1)-2 [(x2/x1)-1]/[(x2/x1)+1]

Let x2/x 1 = t, (t >；; 1) and h (t) = LNT-2 (t-1)/(t+1), (t >；; 1)

Derivation is easy to get H' (t) = (t-1) 2/[t (t+1) 2] > 0, (t> 1).

Then h (t) at t > 1 increases monotonically, and h(t) can be continuous when t= 1.

So h (t) gt; H( 1)=0, (t> 1) is LNT-2 (t-1)/(t+1) > 0.

That is ln (x2/x1)-2 [(x2/x1)-1]/[(x2/x1)+1] > 0.

But it contradicts the formula (5), so g' (XO) >: Hypothesis =0 is not valid.

Then there is g' (XO) < 0, so g' (px 1+qx2).

Therefore g' (px 1+qx2)

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