f'(x)=2ax+(2-a)-1/x
=(2ax^2+(2-a)x-1)/x
=(2x-1)(ax+1)/x
a>1
令f'(x)>=0
x=1/2
定義域是x>0
∴x>=1/2
增區(qū)間是[1/2,+∞),減區(qū)間是(0,1/2]
當(dāng)1/a>=1/2時(shí)
f(x)在區(qū)間[1/a,1]內(nèi)的最大值
=f(1)
=a+2-a-0
=2不是ln3
∴1/a2
f(x)在區(qū)間[1/a,1]內(nèi)的最大值
=f(1/a)
=a*1/a^2+(2-a)/a-ln(1/a)
=1/a+2/a-1+lna
=3/a-1+lna
=ln3
∴a=3符合a>2
綜上a=3