Question

已知函數(shù)f(x)=+aln(x-1),其中n∈N^*,a為常數(shù).
(1)當(dāng)n=2時(shí),求函數(shù)f(x)的極值;
(2)當(dāng)a=1時(shí),證明:對(duì)任意的正整數(shù)n,當(dāng)x≥2時(shí),有f(x)≤x-1.

Answer 1

（1）當(dāng)a＞0時(shí),f(x)在x=1+處取得極小值,極小值為f（1+）=（1+ln）.
當(dāng)a≤0時(shí),f(x)無(wú)極值.
（2）證明見解析

(1) 由已知得函數(shù)f(x)的定義域?yàn)閧x|x＞1},
當(dāng)n=2時(shí),f(x)=+aln(x-1),
所以f′(x)=.
①當(dāng)a＞0時(shí),由f′(x)=0,得
x₁=1+＞1,x₂=1-＜1,
此時(shí)f′(x)=.
當(dāng)x∈(1,x₁)時(shí),f′(x)＜0,f(x)單調(diào)遞減;
當(dāng)x∈(x₁,+∞)時(shí),f′(x)＞0,f(x)單調(diào)遞增.
②當(dāng)a≤0時(shí),f′(x)＜0恒成立,所以f(x)無(wú)極值.
綜上所述,n=2時(shí),
當(dāng)a＞0時(shí),f(x)在x=1+處取得極小值,極小值為f（1+）=（1+ln）.
當(dāng)a≤0時(shí),f(x)無(wú)極值.
(2) 方法一 因?yàn)閍=1,
所以f(x)=+ln(x-1).
當(dāng)n為偶數(shù)時(shí),
令g(x)=x-1--ln(x-1),
則g′(x)=1+-
=+＞0 (x≥2).
所以,當(dāng)x∈［2,+∞)時(shí),g(x)單調(diào)遞增,又g(2)=0,
因此,g(x)=x-1--ln(x-1)≥g(2)=0恒成立,
所以f(x)≤x-1成立.
當(dāng)n為奇數(shù)時(shí),要證f(x)≤x-1,由于＜0,
所以只需證ln(x-1)≤x-1,
令h(x)=x-1-ln(x-1),
則h′(x)=1-=≥0(x≥2),
所以,當(dāng)x∈［2,+∞)時(shí),h(x)=x-1-ln(x-1)單調(diào)遞增,
又h(2)=1＞0,所以當(dāng)x≥2時(shí),恒有h(x)＞0,
即ln(x-1)＜x-1命題成立.
綜上所述,結(jié)論成立.
方法二 當(dāng)a=1時(shí),f(x)=+ln(x-1).
當(dāng)x≥2時(shí),對(duì)任意的正整數(shù)n,恒有≤1,
故只需證明1+ln(x-1)≤x-1.
令h(x)=x-1-(1+ln(x-1))
=x-2-ln(x-1),x∈［2,+∞).
則h′(x)=1-=,
當(dāng)x≥2時(shí),h′(x)≥0,故h(x)在［2,+∞)上單調(diào)遞增,
因此,當(dāng)x≥2時(shí),h(x)≥h(2)=0,
即1+ln(x-1)≤x-1成立.
故當(dāng)x≥2時(shí),有+ln(x-1)≤x-1.
即f(x)≤x-1.