设a≥0,f(x)=x-1-ln2x+2alnx(x>0). (Ⅰ)令F(x)=xf′(x),讨论F(x)在(0,+∞)内的单调性并求极值; (Ⅱ)求证:当x>1时,恒有x>ln2x-2alnx+1.

问题描述:

设a≥0,f(x)=x-1-ln2x+2alnx(x>0).
(Ⅰ)令F(x)=xf′(x),讨论F(x)在(0,+∞)内的单调性并求极值;
(Ⅱ)求证:当x>1时,恒有x>ln2x-2alnx+1.

(Ⅰ)根据求导法则有f′(x)=1−2lnxx+2ax,x>0,故F(x)=xf'(x)=x-2lnx+2a,x>0,于是F′(x)=1−2x=x−2x,x>0,∴知F(x)在(0,2)内是减函数,在(2,+∞)内是增函数,所以,在x=2处取得极小值F(2)...