(I)函数f(x)的定义域为(0,+∞),由题意k=f′(x0)=≤在(0,+∞)上恒成立.,所以a≥(-+x0)max,当x0=1时, (-+x0)max=,∴a≥ (Ⅱ)当a=0时,F(x)=f(1+ex)-g(x)=ln(1+ex)-x,(x∈R),设x1,x2∈R,且x1<x2, 则F(x1)+F(x2)-2F()=ln(1+ex1)+ln(1+ex2)-x1-x2-2[ln(1+e)-] =ln(1+ex1)(1+ex2)-ln(1+e)2 =ln(1+ex1+ex2+ex1+x2)-ln(1+2e+ex1+x2), ∵ex1>0,ex2>0,且x1≠x2,∴+ex1+ex2>2=2e, 1+ex1+ex2+ex1+x2)>1+2e+ex1+x2), ln(1+ex1+ex2+ex1+x2)>ln(1+2e+ex1+x2), ∴F(x1)+F(x2)-2F()>0 即F()<; (Ⅲ)当a=0时,方程m[f(x)+g(x)]=x2(m>0)有唯一解,即为x2-2mlnx-2mx=0有唯一解, 设(x)=x2-2mlnx-2mx,则H′(x)=,令H′(x)=0,则x2-mx-m=0,m>0,x>0,∴x1=<0(舍去),x2=, 当x∈(0,x2)时,H′(x)<0,H(x)在(0,x2)上单调递减 当x∈(x2,+∞)时,H′(x)>0,H(x)在(x2,+∞)上单调递增. 当x=x2时,H(x)取最小值H(x2),则即两式相减得2mlnx2+mx2-m=0,∵m>0,∴2lnx2+x2-1=0①, 设M(x)=2lnx+x-1,∵x>0,M(x)是增函数,∴M(x)=0至多有一解.∵M(1)=0,∴方程①的解为x2=1, 即x2==1,解得m=. |