If the tangents at p are the same, then the slopes are the same, 1/x=2ax- 1, and a=(x+ 1)/2x? Lnx is brought into the above formula = (1-x)/2.
It is proved that the equation has only one real root and the p point is unique.
Let h(x)=lnx+x/2- 1/2 ∴h(x) derivative = 1/x+ 1/2 > 0 ∴ h (x) monotonically increases in the definition field.
(2) When the tangents are the same, a = 1 can be known from (1);
When the tangent points are different, let the tangent equation be y = kx+m.
The straight line is tangent to f(x), and k= 1/x is obtained, so that the abscissa of tangent point x= 1/k and the ordinate of tangent point y=ln( 1/k) are substituted into the straight line, and then into m = ln (1/k).
Similarly, if a straight line is tangent to g(x), x=(k+ 1)/2a can be obtained, thus, (-k? -2k- 1)/4a=-lnk- 1，
∴ 4a=(k？ +2k+ 1)/( 1+lnk)(k>0)
Let F(k)=(k? +2k+1)/(1+lnk) (k > 0), then f (k) = (k+1) (1+2lnk-1/k)/(66.
Let G(k)= 1+2lnk- 1/k, then the derivative of G(k) =2/k+ 1/k? > 0 ∴ g (k) monotonically increases at (0, positive infinity).
G( 1)=0,∴G(k) is less than 0 on (0, 1) and (1, positive infinity) > 0, so.
F(k) decreases monotonically at (0, 1) and increases monotonically at (1, positive infinity). The minimum value of ∴F(k) is F( 1)=4, that is, the minimum value of 4a is 4, and the minimum value of ∴a is 656;.
To sum up, the minimum value of positive real number A is 1.