已知函数f(x)=lnx十1/2x平方十(2a一1)x+1的两个极值点为m,n求f(m)+f(n

f(x)=lnx+1/2x^2+(2a-1)x+1 有极值,
所以导数f'(x)=1/x+x+2a-1=0, x^2+(2a-1)x+1=0的两个值是m和n.
m+n=1-2a
mn=1
m^n+n^2=(m+n)^2-2mn=4a^2-4a-1
f(m)+f(n)=lnm+1/2m^2+(2a-1)m+1+lnn+1/2n^2+(2a-1)n+1
=ln(mn)+1/2(m^2+n^2)+(2a-1)(m+n)+2
=1/2(4a^2-4a-1)-(2a-1)^2+2
=-2a^2+2a+1/2