Question

已知函數(shù) 
（Ⅰ）設(shè)，討論的單調(diào)性；
（Ⅱ）若對任意恒有，求的取值范圍

Answer 1

(Ⅰ)f(x)的定義域為(－∞,1)∪(1,+∞) 
對f(x)求導(dǎo)數(shù)得 f '(x)= e^－ax ------------------------------2分  
(ⅰ)當(dāng)a=2時, f '(x)= e^－2x, f '(x)在(－∞,0), (0,1)和(1,+ ∞)均大于0, 所以f(x)在(－∞,1), (1,+∞) 為增函數(shù)  -------------------------3分  
(ⅱ)當(dāng)0<a<2時, f '(x)>0, f(x)在(－∞,1), (1,+∞)為增函數(shù)   -----------4分  
(ⅲ)當(dāng)a>2時, 0<<1, 令f '(x)="0" ,解得x₁= － , x₂=   
當(dāng)x變化時, f '(x)和f(x)的變化情況如下表:

x	(－∞, －)	(－,)	(,1)	(1,+∞)
f '(x)	＋	－	＋	＋
f(x)	↗	↘	↗	↗

f(x)在(－∞, －), (,1), (1,+∞)為增函數(shù), f(x)在(－,)為減函數(shù)                                     -----------------------------8分  
(Ⅱ)(ⅰ)當(dāng)0<a≤2時, 由(Ⅰ)知: 對任意x∈(0,1)恒有f(x)>f(0)=1  -------------9分   
(ⅱ)當(dāng)a>2時, 取x₀= ∈(0,1),則由(Ⅰ)知 f(x₀)<f(0)=1----------------10分  
(ⅲ)當(dāng)a≤0時, 對任意x∈(0,1),恒有 >1且e^－ax≥1,得
f(x)= e^－ax≥ >1                                  -------------11分   綜上當(dāng)且僅當(dāng)a∈(－∞,2]時,對任意x∈(0,1)恒有f(x)>1

略