2003年高考题(江苏卷)21题的思路与解法

导读:2003年全国高考(江苏卷)21题的思路与解法,2003年全国高考(江苏卷)第21题内容新、题型新,解答的思路和方法较多,这里给出不同层次的若干思路和方法供参考.,思路一:用导数定义.,思路二:直接使用二项式定理和多项式求导公式xm????mxm?1.,n思路1:用导数证明f?x??x??x?a?单调性进而证明(*)式.n,符合命题思路.,nnn?1n?2n?2n?1思路2:利用x?y??x?

2003年高考题(江苏卷)21题的思路与解法

本文发表于《中学数学月刊》2003年第8期

2003年全国高考(江苏卷)21题的思路与解法

215006 苏州市第一中学 刘祖希(第二作者)

2003年全国高考(江苏卷)第21题内容新、题型新,集中考察了导数和不等式证明等知识,解答的思路和方法较多,这里给出不同层次的若干思路和方法供参考.

已知a?0,n为正整数.

(I)设y??x?a?,证明y??n?x?a?

nnn?1; n(II)设fn?x??x??x?a?,对任意n?a,证明fn?1??n?1???n?1?fn??n?.

先解答第(I)题:

思路一:用导数定义.

证法1:∵y??x?a?, n

f?x??x??f?x?x??x?a???x?a??∴y??lim ?lim?x?0?x?0?x?xnn

?lim?x?a??x???x?a?

?xnn?x?0?lim?Cnk?x?a?k?0nn?k??xk??x?a?n?x?0?x

?lim?Cnk?x?a?

k?1nn?k??xk

?x?0?x?0?x

n?1nk?lim?Cn?x?a?k?1nn?k??xk?1

?n?x?a?

?n?x?a?k?lim?Cn?x?a??x?0k?2n?k??xk?1 n?1,

n?1即y??n?x?a?.

n证法2:∵y??x?a?,

f?x??x??f?x?x??x?a???x?a??∴y??lim ?lim?x?0?x?0?x?xnn

?lim??x?a??x??x?0?n?1??x?a??x?n?2?x?a?????x?a?

n?1n?1? ??x?a?n?1??x?a?

n?1n?2?x?a?????x?a? ?n?x?a?,

n?1即y??n?x?a?.

思路二:直接使用二项式定理和多项式求导公式xm????mxm?1.

证法3:∵?n?k?Cn??n?k?kn! n?k!k!n?1?!?k?n??nCn?1,k?1,2,?,n?1. n?1?k!k!对y??x?a?有,y??

nn??Ck?0nknxn?k??a?k? ?kn?k?1???n?k?Cnx??a?

k?0

n?1?k?kn?1?k??nCnx??a? ?1

k?0

n?1??k??kn?1?k??nCn??a? ?1x

k?0k

?n?x?a?n?1.

再解答第(II)题:

易得fn?1??n?1???n?1?n?1??n?1??n?1?a?,

n?1nnn?n?n?a??n?1?fn??n???n?1???

命题转化为证明?n?1?n?1?, ?nn?1??n?1??n?1?a???n?1??nn?n?n?a??

n?1?, ??n?1???n?1?a?nn?nn?n?n?a?

nn, nn加强命题,证明?n?1???n?1?a??n??n?a?.(*)

n思路1:用导数证明f?x??x??x?a?单调性进而证明(*)式. n

证法1:∵f??x??n?xn?1

???x?a?n?1??0,(x?a) ?

n∴f?x??x??x?a?在?a,???上递增.

n∴f?n?1??f?n?,即?n?1???n?1?a??n??n?a?. nnnn

这种方法简单明了,前后呼应,符合命题思路.

nnn?1n?2n?2n?1思路2:利用x?y??x?y?x?xy???xy?y分解因式证明(*)式. ??

证法2:∵?n?1???n?1?a??a

n?1nn??n?1?k?0kn?1n?1?k??n?1?a?, k及n??n?a??ann?n

k?0n?1?k??n?a?,

又因为对任意k,?n?1?

nnn?1?k?n?1?a?nk?nn?1?k?n?a?, kn∴?n?1???n?1?a??n??n?a?.

n另外,(*)式等价于?n?1???n?a??n??n?1?a?,其本质是以下的一般命题: nnn

对自然数n?2,若0?M?P?Q?N且M?N?P?Q,则Mn?Nn?Pn?Qn.(**) 思路3:用数学归纳法证明(**)式.

证法3:n?2时,由2M?N

2?22???M?N???N?M?222, 2?P2?Q2???P?Q???Q?P?,显然M2?N2?P2?Q2.

假设n?k及n?k?1时,(**)式成立,

则n?k?1时,Mk?1?Nk?1??M?N??Mk?Nk??MN?Mk?1?Nk?1?

Pk?1?Qk?1??P?Q??Pk?Qk??PQ?Pk?1?Qk?1?,

?M?N???N?M?∵MN?4

∴Mk?122?P?Q???Q?P??PQ?422, ?Nk?1?Pk?1?Qk?1.

由上知(**)成立.

取M?n?a,P?n,Q?n?1?a,N?n?1,

n得?n?1???n?a??n??n?1?a?(下同从略). nnn

思路4:用增量法及二项式定理证明(**)式.

证法4:记N?Q?P?M???0,

∴M?N??P?????Q????nnnnkk?n?kkn?k? CQ??P?????n??k?0n

kk?n?kk?Q?P??CnQ??Pn?k?????, ??k?1nnn

若k为偶数,?Qn?k

?k?k?Pn?k??????0; ?

若k为奇数,?Qn?k

?k?k?Pn?k???????Qn?k?Pn?k??k?0, ?

∴M?N?P?Q.

证法5:记M?N?P?Q?2s?0,Q?s?s?P?t1,N?s?s?M?t2,0?t1?t2, ∵?s?t???s?t??nnnnnnkk?n?kkn?k? Cst?s?t???n??k?0n

2n?224n?44?2sn?2Cnst?2Cnst???0,

∴f?t???s?t???s?t?

nnnn?s?t?0?是关于t的增函数. nn∴?s?t2???s?t2???s?t1???s?t1?,

即M?N?P?Q. nnnn

注:f?t???s?t???s?t?nn?s?t?0?是关于t的增函数也可用导数证明. 思路5:用凸函数性质证明(**)式.

证法6:取函数f?x??xn,∵f??x??nxn?1?0,f???x??n?n?1?xn?2?0 ∴f?x??xn在?0,???上是下凸函数,

f?tx1??1?t?x2??tf?x1???1?t?f?x2?对任意t??0,1?成立,

取P?tM??1?t?N,则Q?M?N?P??1?t?M?tN,

∴Pn?ftM??1?t?N?tf?M???1?t?f?N??tMn??1?t?Nn, ??

Qn?f??1?t?M?tN???1?t?f?M??tf?N???1?t?Mn?tNn, 相加得,M?N?P?Q.

思路6:用均值不等式证明(**)式.

证法7:取P?tM??1?t?N,则Q?M?N?P??1?t?M?tN, 利用均值不等式nnnn?a

k?1nnk?na1a2?an,ak?0,k?1,2,?,n,

得Mn??n?1?Pn?nMPn?1,①

Nn??n?1?Pn?nNPn?1,②,

①?t?②??1?t?,得tM??1?t?N??n?1?P?nP, nnnn

∴tM??1?t?N?P,③ nnn

同理, ?1?t?M?tN?Q,④ nnn

③?④得,M?N?P?Q.

思路7:用柯西不等式推广式证明(**)式.

引理:设ak,bk?R?,则?a1?b1??a2?b2???an?bn??a1a2?an?b1b2?bn.(柯西不等式推广) 事实上, nnnna1a2?anan?a21?a1?, ????????a1?b1a2?b2?an?bnn?a1?b1a2?b2an?bn?b1b2?bnbn?b21?b1?, ????????a1?b1a2?b2?an?bnn?a1?b1a2?b2an?bn?相加得,?a1?b1??a2?b2???an?bn??a1a2?an?1b2?bn. 证法8:取P?tM??1?t?N,则Q?M?N?P??1?t?M?tN, 由上述引理,

?tMn??1?t?Nn?t??1?t???n?1??tM??1?t?N?, n

n?1同理,tNn??1?t?Mn?t??1?t??????tN??1?t?M?, n

相加得,Mn?Nn?Pn?Qn.

思路8:用拉格朗日中值定理证明(*)式.

拉格朗日中值定理:

函数f?x?在闭区间?a,b?上可导,则至少存在一点???a,b?, 使得f?b??f?a??f?????b?a?.

证法9:函数f?x??xn在R上可导,

若a?1,则n?a?n?1?a?n?n?1,

则存在一点?1??n?a,n?1?a?,使得?n?1?a???n?a??f???1?;

n且存在一点?2??n,n?1?,使得?n?1??n?f???2?; nnn

∵f??x??nxn?1在R上递增,?1?n?1?a?n??2,

n∴f???1??f???2?,?n?1?a???n?a???n?1??n,

n即?n?1???n?a??n??n?1?a?; nnnnnn

若0?a?1,则n?a?n?n?1?a?n?1,

n同理可证n??n?a???n?1???n?1?a?,

n即?n?1???n?a??n??n?1?a?. nnnnnn

思路9:用贝努利不等式证明(*)式.

贝努利不等式:设x??1且x?0,对任意n?1,有?1?x??1?nx. 证法10:对x?1,∵

nn11?0??1,???1,由贝努利不等式, xxnnn?1??1?n∴?1???1?,?1???1?,两边同乘x, xx?x??x?nn?1nn?1∴?x?1??x?nx,?x?1??x?nx,

nn?1n?1n∴?x?1??x?nx且nx?x??x?1?. nnnn

若a?1,分别取x?n及x?n?1?a,

nn?1则?n?1??n?nn?n?n?1?a?

nnnn?1??n?1?a???n?a?, nnnn即得(*)式:?n?1???n?1?a??n??n?a?;

若0?a?1,分别取x?nn?1?an及x?, aann?1?n?1?a??n?1?a??n?1?a?则??1????n????a??a??a?

nn?n??n???a?nn?1?n??n???????1?, ?a??a?nnnn同乘a即得(*)式:?n?1???n?1?a??n??n?a?.

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