【答案】
分析:(Ⅰ)由f'(x)=1-2cosx=1得cosx=0,從而找出直線l與曲線S的兩個(gè)切點(diǎn),從而說明直線l與曲線S相切且至少有兩個(gè)切點(diǎn),然后根據(jù)對任意x∈R,g(x)-F(x)≥0,滿足“上夾線”的定義,從而得到結(jié)論;
(Ⅱ)推測:y=mx-nsinx(n>0)的“上夾線”的方程為y=mx+n,然后①先檢驗(yàn)直線y=mx+n與曲線y=mx-nsinx相切,且至少有兩個(gè)切點(diǎn),②檢驗(yàn)g(x)≥F(x)是否成立,從而得到結(jié)論.
解答:解(Ⅰ)由f'(x)=1-2cosx=1得cosx=0,(1分)
當(dāng)x=-
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時(shí),cosx=0,
此時(shí)
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,
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,(2分)
y
1=y
2,所以(
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,
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)是直線l與曲線S的一個(gè)切點(diǎn);(3分)
當(dāng)x=
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時(shí),cosx=0,
此時(shí)
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,
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,(4分)
y
1=y
2,,所以(
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,
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)是直線l與曲線S的一個(gè)切點(diǎn);(5分)
所以直線l與曲線S相切且至少有兩個(gè)切點(diǎn);
對任意x∈R,g(x)-F(x)=(x+2)-(x-2sinx)=2+2sinx≥0,
所以g(x)≥F(x)(6分)
因此直線l:y=x+2是曲線S:y=ax+bsinx的“上夾線”.(7分)
(Ⅱ)推測:y=mx-nsinx(n>0)的“上夾線”的方程為y=mx+n(9分)
①先檢驗(yàn)直線y=mx+n與曲線y=mx-nsinx相切,且至少有兩個(gè)切點(diǎn):設(shè):F(x)=mx-nsinx
∵F'(x)=m-ncosx,令F'(x)=m-ncosx=m,得:x=2kπ±
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(k∈Z)(10分)
當(dāng)x=2kπ-
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時(shí),F(xiàn)(2kπ-
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)=m(2kπ-
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)+n
故:過曲線F(x)=mx-nsinx上的點(diǎn)2kπ-
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,m(2kπ-
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)+n)的切線方程為:
y-[m(2kπ-
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)+n]=m[-(2kπ-
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)],化簡得:y=mx+n.
即直線y=mx+n與曲線y=F(x)=mx-nsinx相切且有無數(shù)個(gè)切點(diǎn).(12分)
不妨設(shè)g(x)=mx+n
②下面檢驗(yàn)g(x)≥F(x)
∵g(x)-F(x)=m(1+sinx)≥0(n>0)
∴直線y=mx+n是曲線y=F(x)=mx-nsinx的“上夾線”.(14分)
點(diǎn)評:本題主要考查了函數(shù)恒成立問題,以及利用導(dǎo)數(shù)研究切線等有關(guān)知識,同時(shí)考查了轉(zhuǎn)化與劃歸的思想,屬于中檔題.