2023学年高考数学大二轮复习能力升级练十六导数及其综合应用2理2.docxVIP免费

能力升级练(十六)导数及其综合应用(2)1.(2023湖北荆州质检)已知函数f(x)=ax-lnx.(1)讨论f(x)的单调性;(2)若a∈-∞,-1e2,求证:f(x)≥2ax-xeax-1.(1)解由题意得f'(x)=a-1x=ax-1x(x>0),①当a≤0时,则f'(x)<0在(0,+∞)上恒成立,∴f(x)在(0,+∞)上单调递减.②当a>0时,则当x∈1a,+∞时,f'(x)>0,f(x)单调递增,当x∈0,1a时,f'(x)<0,f(x)单调递减.综上,当a≤0时,f(x)在(0,+∞)上单调递减;当a>0时,f(x)在0,1a上单调递减,在1a,+∞上单调递增.(2)证明令g(x)=f(x)-2ax+xeax-1=xeax-1-ax-lnx,则g'(x)=eax-1+axeax-1-a-1x=(ax+1)eax-1-1x=\(ax+1\)\(xeax-1-1\)x(x>0),1设r(x)=xeax-1-1(x>0),则r'(x)=(1+ax)eax-1(x>0), eax-1>0,∴当x∈0,-1a时,r'(x)>0,r(x)单调递增;当x∈-1a,+∞时,r'(x)<0,r(x)单调递减.∴r(x)max=r-1a=-1ae2+1...