Question

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

Answer 1

（Ⅰ）f(x)在(－∞, －), (,1), (1,+∞)為增函數(shù), f(x)在(－,)為減函數(shù)；（Ⅱ）a∈(－∞,2]時(shí),對(duì)任意x∈(0,1)恒有f(x)>1。

(Ⅰ)f(x)的定義域?yàn)?－∞,1)∪(1,+∞).對(duì)f(x)求導(dǎo)數(shù)得 f '(x)= e^－ax.  
(ⅰ)當(dāng)a=2時(shí), 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).
(ⅱ)當(dāng)0<a<2時(shí), f '(x)>0, f(x)在(－∞,1), (1,+∞)為增函數(shù).
(ⅲ)當(dāng)a>2時(shí), 0<<1, 令f '(x)="0" ,解得x₁= － , x₂= .
當(dāng)x變化時(shí), 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).
(Ⅱ)(ⅰ)當(dāng)0<a≤2時(shí), 由(Ⅰ)知: 對(duì)任意x∈(0,1)恒有f(x)>f(0)=1.
(ⅱ)當(dāng)a>2時(shí), 取x₀= ∈(0,1),則由(Ⅰ)知 f(x₀)<f(0)=1
(ⅲ)當(dāng)a≤0時(shí), 對(duì)任意x∈(0,1),恒有 >1且e^－ax≥1,得
f(x)= e^－ax≥ >1. 綜上當(dāng)且僅當(dāng)a∈(－∞,2]時(shí),對(duì)任意x∈(0,1)恒有f(x)>1.