(Ⅰ)由f(x)=x(lnx+1)(x>0),得f′(x)=lnx+2(x>0), F(x)=ax2+lnx+2(x>0),∴F′(x)=2ax+=(x>0). ①当a≥0时,恒有F′(x)>0,故F(x)在(0,+∞)上是增函数; ②当a<0时, 令F′(x)>0,得2ax2+1>0,解得0<x<; 令F′(x)<0,得2ax2+1<0,解得x>; 综上,当a≥0时,F(x)在(0,+∞)上是增函数; 当a<0时,F(x)在(0,)上单调递增,在(,+∞)上单调递减; (Ⅱ)k==. 要证x1<<x2,即证x1<<x2, 等价于证1<<,令t=, 则只要证1<<t,由t>1,知lnt>0,故等价于lnt<t-1<tlnt(t>0)(*) ①设g(t)=t-1-lnt(t≥1),则g′(t)=1-≥0(t≥1), 故g(t)在[1,+∞)上是增函数, ∴当t>1时,g(t)=t-1-lnt>g(1)=0,即t-1>lnt(t-1) ②设h(t)=tlnt-(t-1)(t≥1),则h′(t)=lnt≥0(t≥1), 故h(t)在[1,+∞)上是增函数. ∴当t>1时,h(t)=tlnt-(t-1)>h(1)=0,即t-1(t>1). 由①②知(*)成立,故x1<<x2. |