解:(Ⅰ)求導(dǎo)函數(shù),可得
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,
∵曲線y=f(x)在點(diǎn)(2,f(2))處的切線方程為y=3
∴f(2)=3,f′(2)=0
∴
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,
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∴
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或
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,
由于m,n∈Z,所以
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,則
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. (4分)
(Ⅱ)由(Ⅰ)得F(x)=aln(x-1)+
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,定義域?yàn)椋?,+∞),F(xiàn)′(x)=
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,由于a>0,
令F′(x)=0,得
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,
當(dāng)x∈
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時(shí),F(xiàn)′(x)<0,知F(x)在x∈
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時(shí)單調(diào)遞減,
同理,F(xiàn)(x)在x∈
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時(shí)單調(diào)遞增
所以F(x)min=F
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=a-alna
令a-alna<0,即a>e時(shí),函數(shù)F(x)=0有兩個(gè)實(shí)數(shù)根
所以a的取值范圍是(a,+∞)
分析:(Ⅰ)求導(dǎo)函數(shù),利用曲線y=f(x)在點(diǎn)(2,f(2))處的切線方程為y=3,建立方程,可求函數(shù)f(x)的解析式;
(Ⅱ)由(Ⅰ)得F(x)=aln(x-1)+
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,定義域?yàn)椋?,+∞),F(xiàn)′(x)=
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,確定函數(shù)的單調(diào)性,求得函數(shù)的最小值,由此即可求出實(shí)數(shù)a的取值范圍.
點(diǎn)評:本題考查導(dǎo)數(shù)知識(shí)的運(yùn)用,考查導(dǎo)數(shù)的幾何意義,考查函數(shù)的單調(diào)性與最值,解題的關(guān)鍵是正確求導(dǎo).