討論
(1)a≤0.x>0,則2ax-1<0令f'(x)=(2ax-1)lnx<0,可得當(dāng)0
所以f(x)在(0,1)上遞增,在(1,+∞)遞減
(2)0
所以當(dāng)x屬于(0,1)時(shí),f‘(x)>0,f(x)在(0,1)單調(diào)遞增；當(dāng)x屬于(1,1/2a)時(shí),f‘(x)<0,f(x)在(1,1/2a)單調(diào)遞減；當(dāng)x>1/2a時(shí),f‘(x)>0,f(x)在(1/2a,+∞)單調(diào)遞增.
(3)a>1/2,令f'(x)=(2ax-1)lnx=0得x=1/(2a)或x=1,當(dāng)a>1/2時(shí),1/(2a)<1,
所以當(dāng)x屬于(0,1/2a)時(shí),f‘(x)>0,f(x)在(0,1/2a)單調(diào)遞增；當(dāng)x屬于(1/2a,1)時(shí),f‘(x)<0,f(x)在(1/2a,1)單調(diào)遞減；當(dāng)x>1時(shí),f‘(x)>0,f(x)在(1,+∞)單調(diào)遞增.