Question

已知函數(shù)f(x)=

(1)設(shè)a＞0，討論y=f(x)的單調(diào)性；

（2）若對任意x∈(0,1)恒有f(x)＞1，求a的取值范圍.

Answer 1

解：(1)f(x)的定義域為(-∞,1)∪(1,+∞),對f(x)求導(dǎo)數(shù)得f′(x)=.

①當(dāng)a=2時,f′(x)=,f′(x)在(-∞,0),(0,1)和(1,+∞)上均大于0,所以f(x)在(-∞,1),(1,+∞)上為增函數(shù).

②當(dāng)0＜a＜2時,f′(x)＞0,f(x)在(-∞,1),(1,+∞)上為增函數(shù).

③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,+∞)上為增函數(shù),f(x)在(-,)上為減函數(shù).

(2)①當(dāng)0＜a≤2時,由(1)知:對任意x∈(0,1)恒有f(x)＞f(0)=1.

②當(dāng)a＞2時,取x₀=∈(0,1),則由(1)知f(x₀)＜f(0)=1.

③當(dāng)a≤0時,對任意x∈(0,1),恒有＞1且e^-ax≥1,得f(x)=≥.

    綜上,當(dāng)且僅當(dāng)a∈(-∞,2］時,對任意x∈(0,1)恒有f(x)＞1.