2013年高考理科数学分类汇编 - 函数与导数大题目 2 - 图文

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12.（2013年辽宁卷22题）（本小题满分12分）

已知f0(x)?xn，fk(x)?fk??1(x)，其中k≤n(n，k?N+)．设 fk?1(1)01knF(x)?Cnf0(x2)?Cnf1(x2)?…?Cnfk(x2)?…?Cnfn(x2)，x???11，?．

（I）写出fk(1)；

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（II）证明：对任意的x1，x2???11 ，?，恒有F(x1)?F(x2)≤2n?1(n?2)?n?1．

（22）本小题主要考查导数的基本计算，函数的性质，绝对值不等式及组合数性质等基础知识，考查归纳推理能力以及综合运用数学知识分析问题和解决问题的能力．满分12分．

（I）解：由已知推得fk?x???n?k?1?xn?k，从而有fk?1??n?k?1． · 3分 （II）证法一：当?1≤x≤1时，

1F?x??x2n?nCnx2?n?1?2??n?1?Cnx2?n?2?k????n?k?1?Cnx2?n?k?n?12???2Cnx?1，

当x?0时，F??x??0，所以F?x?在?01，?上是增函数． 又F?x?是偶函数，所以F?x?在??10，?上是减函数．

所以对任意的x1，x2???11····· 7分 ，?，恒有F?x1??F?x2?≤F?1??F?0?．·

012kn?1 F?1??F?0??Cn?nCn??n?1?Cn????n?k?1?Cn???2Cnn?1n?2n?k10． ?nCn??n?1?Cn????n?k?1?Cn???2Cn?Cnn?kn?kn?k??n?k????n?k?1?Cn??n?k?Cn?Cn?n?1?!?Ck n!k?Cn?n??n?k?!k!?n?1?k?!k!nkk·································································· 10分 ?nCn，2，?，n?1?， ·?1?Cn?k?112n?112n?10?F?1??F?0??n?Cn?1?Cn?1???Cn?1???Cn?Cn???Cn??Cn

?n?2n?1?1??2n?1?2n?1?n?2??n?1．

因此结论成立．······························································································ 12分 证法二：当?1≤x≤1时，

1F?x??x2n?nCnx2?n?1?2??n?1?Cnx2?n?2?k????n?k?1?Cnx2?n?k?n?12???2Cnx?1，

当x?0时，F??x??0，所以F?x?在?01，?上是增函数． 又F?x?是偶函数，所以F?x?在??10，?上是减函数．

所以对任意的x1，x2???11····· 7分 ，?，恒有F?x1??F?x2?≤F?1??F?0?．·

012kn?1， F?1??F?0??Cn?nCn??n?1?Cn????n?k?1?Cn???2Cn12n?10又?F?1??F?0??2Cn， ?3Cn???nCn?Cn 15

12n?1?2?F?1??F?0????n?2??Cn?Cn???Cn·································· 10分 ??2，·

?F?1??F?0??112n?1?Cn???Cn?1 ?n?2??Cn?22n?2??n?2???1?2n?1?n?2??n?1．因此结论成立． ··························· 12分

2证法三：当?1≤x≤1时，

1F?x??x2n?nCnx2?n?1?2??n?1?Cnx2?n?2?k????n?k?1?Cnx2?n?k?n?12???2Cnx?1，

当x?0时，F??x??0，所以F?x?在?01，?上是增函数． 又F?x?是偶函数，所以F?x?在??10，?上是减函数．

所以对任意的x1，x2???11····· 7分 ，?，恒有F?x1??F?x2?≤F?1??F?0?．·

n?kkk2?n?k???n?k?CnxCnfk?x2???n?k?1?Cnx2?n?k?n?k?Cnx2?n?k?2，?，n-1?， ?k?1，由?n?k?Cn?knn?1?!?n!n?1?k，得 ??n?k???n??nCn?1n?k!k!n?1?k!k!????2?n?2?n?3?Cn?1x2?n?3?0n?12?n?1?0?2n?Cn???Cnx???Cn ?1??xn?2F?x??nx2?Cn??1x2?nx2?1?x????n?1?x2?n?1????1?x2?n． ··························································· 10分

???F?1??F?0??n?2n?1?1??2n?1?2n?1?n?2??n?1．

因此结论成立．······························································································ 12分

证法四：当?1≤x≤1时，

1F?x??x2n?nCnx2?n?1?2??n?1?Cnx2?n?2?k????n?k?1?Cnx2?n?k?n?12???2Cnx?1，

当x?0时，F??x??0，所以F?x?在?01，?上是增函数． 又F?x?是偶函数，所以F?x?在??1,0?上是减函数． 所以对任意的x1，x2???11，?，恒有

F?x1??F?x2?≤F?1??F?0?． ··································································· 7分

1n?12n?2kn?kn?22n?1?x??1?x??xn??x?Cnx?Cnx???Cnx???Cnx?Cnx?1??? ??n1n2n?1kn?k?1n?23n?12?Cnx?Cnx???Cnx???Cnx?Cnx?x

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