解:(I)∵f′(x)=cosx,f′(0)=1,
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,g′(0)=p,
y=f(x)與y=g(x)在(0,0)處有相同的切線,
∴p=1…(3分)
(II)設(shè)F(x)=f(x)-g(x),
當(dāng)p=1時,F(xiàn)(x)=sinx-x+
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,
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,
F''(x)=-sinx+x,
當(dāng)x∈(0,1)時,sinx<x,故F''(x)>0,
從而F′(x)在(0,1)上單調(diào)增,
所以,F(xiàn)′(x)>F′(0)=0,
∴F(x)在(0,1)上單調(diào)增,
∴F(x)>f(0)=0,即f(x)>g(x)恒成立.
(III)當(dāng)x∈(0,1)時,
∵F''(x)=-sinx+x>0,
∴F(x)在(0,1)上單調(diào)增,從而F(x)在(0,1)內(nèi)不可能出現(xiàn)先增后減的情況,
∵F(0)=0,
∴要使F(x)>0在(0,1)上恒成立,
必有F(x)在(0,1)上單調(diào)遞增,
即F′(x)≥0在x∈(0,1)上恒成立,
∵F′(x)
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,
∴1-p≥0,
即p≤1.
分析:(I)由f′(x)=cosx,f′(0)=1,
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,g′(0)=p,知p=1.
(II)設(shè)F(x)=f(x)-g(x),當(dāng)p=1時,F(xiàn)(x)=sinx-x+
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,
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,F(xiàn)''(x)=-sinx+x,當(dāng)x∈(0,1)時,F(xiàn)′(x)>F′(0)=0,由此能夠證明(x)>g(x)恒成立.
(III)當(dāng)x∈(0,1)時,由F''(x)=-sinx+x>0,知F(x)在(0,1)上單調(diào)增,從而F(x)在(0,1)內(nèi)不可能出現(xiàn)先增后減的情況,由此能夠求出p的范圍.
點評:本題考查函數(shù)的恒成立問題,解題時要認真審題,仔細解答,注意導(dǎo)數(shù)的合理運用.