【答案】
分析:(1)首先設(shè)出直線y=2x-2與曲線y=g(x)的切點(diǎn),把切點(diǎn)代入兩曲線方程后聯(lián)立可求得b的值,解出g(x)后把f(x)和g(x)的解析式代入f(x)≥g(x),分離變量a后對(duì)函數(shù)進(jìn)行兩次求導(dǎo)得到函數(shù)在區(qū)間[1,+∞)內(nèi)的最小值,則實(shí)數(shù)a的范圍可求;
(2)當(dāng)a=1時(shí)可證得函數(shù)f(x)在[e,3]上為增函數(shù),而g(x)也是增函數(shù),把不等式左邊放大取最大值,右邊取最小值,代入后即可求解最大的正整數(shù)k;
(3)該命題是與自然數(shù)有關(guān)的不等式,采用數(shù)學(xué)歸納法證明,由歸納假設(shè)證明n=k+1成立時(shí),穿插運(yùn)用分析法.
解答:解:(1)設(shè)點(diǎn)(x
,y
)為直線y=2x-2與曲線y=g(x)的切點(diǎn),則有2lnx
+bx
=2x
-2①
∵

,∴

②
由②得,2x
-2=bx
,代入①得x
=1,所以b=0,則g(x)=2lnx.
由f(x)≥g(x),即

,整理得

,
∵x≥1,∴要使不等式f(x)≥g(x)恒成立,必須a≤x
2-2xlnx恒成立.
設(shè)h(x)=x
2-2xlnx,

,
∵

,∴當(dāng)x≥1時(shí),h''(x)≥0,則h'(x)是增函數(shù),
∴h'(x)≥h'(1)=0,∴h(x)是增函數(shù),則h(x)≥h(1)=1,∴a≤1.
又a>0,因此,實(shí)數(shù)a的取值范圍是0<a≤1.
(2)當(dāng)a=1時(shí),

,∵

,∴f(x)在[e,3]上是增函數(shù),
f(x)在[e,3]上的最大值為

.
要對(duì)[e,3]內(nèi)的任意k個(gè)實(shí)數(shù)x
1,x
2,…,x
k,都有f(x
1)+f(x
2)+…+f(x
k-1)≤16g(x
k)成立,
必須使得不等式左邊的最大值小于或等于右邊的最小值,∵當(dāng)x
1=x
2=…=x
k-1=3時(shí)不等式左邊取得最大值,
x
k=e時(shí)不等式右邊取得最小值.∴(k-1)f(3)≤16g(3),即

,解得k≤13.
因此,k的最大值為13.
(3)證明:1°當(dāng)n=1時(shí),左邊=

,右邊=ln3,
根據(jù)(1)的推導(dǎo)有,x∈(1,+∞)時(shí),f(x)>g(x),即

.
令x=3,得

,即

.
因此,n=1時(shí)不等式成立.
2°假設(shè)當(dāng)n=k時(shí)不等式成立,即

,
則當(dāng)n=k+1時(shí),

,
要證n=k+1時(shí)命題成立,即證

,
即證

.
在不等式

中,令

,得

.
∴n=k+1時(shí)命題也成立.
綜上所述,不等式

對(duì)一切n∈N
*成立.
點(diǎn)評(píng):本題主要考查函數(shù)的性質(zhì)、導(dǎo)數(shù)運(yùn)算法則、導(dǎo)數(shù)的幾何意義及其應(yīng)用、不等式的求解與證明、數(shù)學(xué)歸納法等綜合知識(shí),考查學(xué)生的計(jì)算推理能力及分析問題、解決問題的能力及創(chuàng)新意識(shí),屬難題.