f(x)=ln[(1/2)+(ax/2)]+x^2-ax
则f'(x)=(a/2)/[(1/2)+(ax/2)]+2x-a
=[a/(ax+1)]+2x-a
=[a+(ax+1)(2x-a)]/(ax+1)
=(2ax^2-a^2*x+2x)/(ax+1)
=x*(2ax-a^2+2)/(ax+1)
可见,当0则,f'(x)＞0
所以,f(x)为增函数.