解:(1)由f′(x)=e
x-a,得x=lna,
當(dāng)a≤0時,f′(x)>0,f(x)的單調(diào)遞增區(qū)間為(-∞,+∞).
當(dāng)a>0時,f′(x)=e
x-a=0,得x=lna,
x∈(-∞,lna)時,f′(x)<0;x∈(lna,+∞)時,f′(x)>0;
∴f(x)的單減區(qū)間為(-∞,lna),單增區(qū)間為(lna,+∞),
所以a≤0時,f(x)只有單調(diào)遞增區(qū)間為(-∞,+∞).
a>0時,f(x)的增區(qū)間為(lna,+∞),減區(qū)間為(-∞,lna).…(5分)
(2)由(1)得f(x)的最小值為f(lna)=a-alna-1,
f(x)≥0對任意的x∈R恒成立,即f(x)
min≥0.
設(shè)g(a)=a-alna-1,所以g(a)≥0.
由g′(a)=1-lna-1=-lna=0,得a=1.
∴g(a)在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(diào)遞減,
∴g(a)在a=1處取得最大值,而g(1)=0.
因此g(a)≥0的解為a=1,∴a=1.(9分)
(3)證明:由(2)知,對任意實(shí)數(shù)x均有e
x-x-1≥0,即1+x≤e
x.
令x=-
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(n∈N
*,k=0,1,2,3,…,n-1),則0<1-
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≤e-
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.
∴(1-
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)n≤(e-
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)n=e-k.
∴(
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)
n+(
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)
n+…+(
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)
n+(
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)
n≤e-(n-1)+e-(n-2)+…+e-2+e-1+1
=
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<
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=
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.…(14分)
分析:(1)求導(dǎo)函數(shù),確定函數(shù)的單調(diào)性,從而可得f(x)在x=lna處取得極小值,且為最小值;
(2)f(x)≥0對任意的x∈R恒成立,即在x∈R上,f(x)
min≥0.由(1),構(gòu)造函數(shù)g(a)=a-alna-1,所以g(a)≥0,確定函數(shù)的單調(diào)性,即可求得實(shí)數(shù)a的值;
(3)由(2)知,對任意實(shí)數(shù)x均有e
x-x-1≥0,即1+x≤e
x,令x=-
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(n∈N
*,k=0,1,2,3,…,n-1),可得0<1-
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≤e-
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,從而有(1-
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)n≤(e-
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)n=e-k,由此即可證得結(jié)論.
點(diǎn)評:本題考查導(dǎo)數(shù)知識的運(yùn)用,考查函數(shù)的單調(diào)性與最值,考查恒成立問題,同時考查不等式的證明,解題的關(guān)鍵是正確求導(dǎo)數(shù),確定函數(shù)的單調(diào)性.