设函数f(x)=InX-1/2ax^2-bx

F'(x)=1/x-a/x²=0
x²-2x+2a>=0
(x-1)²>=1-2a
(x-1)²>=0
所以1-2a==1/2
2mf(x)=x²有唯一实数解
x²-2mlnx-2mx=0有唯一实数解
设g(x)=x²-2mlnx-2mx
g'(x)=(2x²-2mx-2m)/x
g'(x)=0,x²-mx-m=0
m>0,x>0
x1=[m-√(m²+4m)]/20
设h(x)=2lnx+x-1
x>0时,h(x)是增函数,h(x)至多有一解
h(1)=0,
2lnx2+x2-1=0,x2=1
[m+√(m²+4m)]/2=1
m=1/2