问答题(2023年新高考Ⅰ

已知函数f(x)=a(ex+a)-x.

(1)讨论f(x)的单调性;

(2)证明:当a>0时,f(x)>2lna+3/2.

答案解析

(1)对f(x)求导得:f' (x)=aex-1,① 当a≤0时,f' (x)<0,f(x)单调递减;② 当a>0时,令f' (x)=0,得:aex-1=0,解得:x=-lna.当x<-lna时,f' (x)<0,f(x)单调递减;当x>-lna时,f' (x)>0,f(x)单调递增.(2)由(1)知,当a>0时,f(x)先减后增,且在x=-lna处取最小值,故只需证f(-lna)>2lna+3/2即可.令g(a)=f(-lna)-2ln...

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