f(x)=(1/2a)x^2-lnx a=1
f(x)=(1/2)x^2-lnx f(x)′=x-(1/x)=(x^2-1)/x
f(x)′＞0,x＜-1或x＞1 ∵x＞0∴x＞1此时f(x)单调递增
f(x)′≤0,-1≤x＜0或0＜x≤1 ∴0＜x≤1,此时f(x)单调递减
极值为f(1)=1/2