解:(I)由y=ln(x+1),得x=e
y-1,∴f
-1(x)=e
x-1,x∈R.…(1分)
∴g(x)=ln(x+1)-e
x+1,且x>-1,∴g′(x)=
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-e
x.…(3分)
當(dāng)x>0時(shí),
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<1<e
x,∴g′(x)<0;…(4分)
當(dāng)-1<x<0時(shí),
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>1>e
x,∴g′(x)>0.…(5分)
∴g(x)的單調(diào)遞增區(qū)間是(-1,0),單調(diào)遞減區(qū)間是(0,+∞).…(6分)
(II)設(shè)h(x)=
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x+f(e
x)-lnf
-1(x)=
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x+ln(e
x+1)-ln(e
x-1),x>0…(7分)
∵h(yuǎn)′(x)=
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+
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-
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=
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×
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,…(9分)
當(dāng)0<x<ln2時(shí),h′(x)<0,∴h(x)在(0,ln2)上是減函數(shù); …(10分)
當(dāng)x>ln2時(shí),h′(x)>0,∴h(x)在(ln2,+∞)上是增函數(shù). …(11分)
∴h(x)
min=h(ln2)=
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ln2+ln3=ln6
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,∴a<ln6
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.…(12分)
分析:(I)求出f(x)的反函數(shù)為f
-1(x).求出g(x)=f(x)-f
-1(x)的解析式,然后求出函數(shù)的導(dǎo)數(shù),令導(dǎo)數(shù)大于0求出函數(shù)的增區(qū)間,令導(dǎo)數(shù)小于0求出函數(shù)的減區(qū)間.
(II)構(gòu)造函數(shù)h(x)=
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x+f(e
x)-lnf
-1(x)=
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x+ln(e
x+1)-ln(e
x-1),求出其最小值,令a小于其最小值即可.
點(diǎn)評(píng):本題考查利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性以及不等式恒成立的問(wèn)題,求解此類問(wèn)題的關(guān)鍵是運(yùn)用導(dǎo)數(shù)的計(jì)算公式正確求出導(dǎo)數(shù),第二問(wèn)的恒成立問(wèn)題關(guān)鍵是正確轉(zhuǎn)化為求最值的問(wèn)題,轉(zhuǎn)化后易解.本題運(yùn)算量較大,易因馬虎導(dǎo)致某一步運(yùn)算出錯(cuò),導(dǎo)致后續(xù)的運(yùn)算結(jié)果全錯(cuò).運(yùn)算變形時(shí)要嚴(yán)謹(jǐn).