Question

設(shè)函數(shù)f(x)=-lnx+ln(x+1).

(1)求f(x)的單調(diào)區(qū)間和極值.

(2)是否存在實(shí)數(shù)a,使得關(guān)于x的不等式f(x)≥a的解集為(0,+∞)?若存在,求a的取值范圍;若不存在,試說(shuō)明理由.

Answer 1

本題主要考查函數(shù)的導(dǎo)數(shù)、單調(diào)性、極值,不等式等基礎(chǔ)知識(shí),考查綜合利用數(shù)學(xué)知識(shí)分析問(wèn)題、解決問(wèn)題的能力.

解:(1)f′(x)=.                          

故當(dāng)x∈(0,1)時(shí),f′(x)＞0,

x∈(1,+∞)時(shí),f′(x)＜0.

所以f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減.                                 

由此知f(x)在(0,+∞)上的極大值為f(1)=ln2,沒(méi)有極小值.                             

(2)①當(dāng)a≤0時(shí),

由于f(x)=

=＞0,

故關(guān)于x的不等式f(x)≥a的解集為(0,+∞).                                      

②當(dāng)a＞0時(shí),由f(x)=+ln(1+)知f(2ⁿ)=+ln(1+),其中n為正整數(shù),且有l(wèi)n(1+)＜n＞-log₂( -1).                                           

又n≥2時(shí),=＜=,

且＜n＞+1.

取整數(shù)n₀滿足n₀＞-log₂(-1),n₀＞+1,且n₀≥2,

則f()=+ln(1+)＜+=a,

即當(dāng)a＞0時(shí),關(guān)于x的不等式f(x)≥a的解集不是(0,+∞).

綜合①②知,存在a,使得關(guān)于x的不等式f(x)≥a的解集為(0,+∞),且a的取值范圍為(-∞,0］.