Question

已知函數(shù)f(x)＝aln(1＋e^x)－(a＋1)x(其中a>0)，點(diǎn)A(x₁，f(x₁))，B(x₂，f(x₂))，C(x₃，f(x₃))從左到右依次是函數(shù)y＝f(x)圖象上三點(diǎn)，且2x₂＝x₁＋x₃.

(1)證明：函數(shù)f(x)在R上是減函數(shù)；

(2)求證：△ABC是鈍角三角形．

Answer 1

證明：(1)因?yàn)?i>f(x)＝aln(1＋e^x)－(a＋1)x，所以f′(x)＝－(a＋1)＝<0恒成立，所以函數(shù)f(x)在(－∞，＋∞)上是單調(diào)減函數(shù)．

(2)據(jù)題意A(x₁，f(x₁))，B(x₂，f(x₂))，C(x₃，f(x₃))且x₁<x₂<x₃，由(1)知f(x₁)>f(x₂)>f(x₃)，x₂＝.所以＝(x₁－x₂，f(x₁)－f(x₂))，＝(x₃－x₂，f(x₃)－f(x₂))，所以＝(x₁－x₂)·(x₃－x₂)＋[f(x₁)－f(x₂)]·[f(x₃)－f(x₂)]．

又x₁－x₂<0，x₃－x₂>0，f(x₁)－f(x₂)>0，f(x₃)－f(x₂)<0，所以＝(x₁－x₂)·(x₃－x₂)＋[f(x₁)－f(x₂)]·[f(x₃)－f(x₂)]<0.

從而B∈，即△ABC是鈍角三角形．