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[转载]2012年高考四川卷理科数学第22题评析

(2013-11-12 09:24:12)
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2012年高考四川卷理科数学第22题评析
 
《数学教学》2013年第二期
 

   (2012年高考四川卷理科数学第22题)

   评析:此题涉及导数的应用、求不等式恒成立的参数的范围以及比较大小问题(实
 
际上是证明不等式问题)。
 
 
    处理不等式恒成立,有两种常用方法:特殊值法和最值法,下面分别用这两种方法
 
求解。

   (1)特殊值法,即先取特殊值得到不等式成立的必要条件,然后再证明此条件也是
 
充分的。

   先看一些特殊值:
 

   用数学归纳法,
 

只须证 

http://data.artofproblemsolving.com/images/latex/c/5/6/c56422ed0d243550e3be28e0776030937e25d999.gif


而当http://data.artofproblemsolving.com/images/latex/6/2/9/62903ba4e68c47f4fa620a57c7c4458d8d12ed7d.gif显然成立。


http://data.artofproblemsolving.com/images/latex/3/b/c/3bc293749aaabdf2d64020facbf1bc5a3c159d63.gif


所以,当
http://data.artofproblemsolving.com/images/latex/0/6/c/06caa24880496bc6c903c01a7b9caa83fe8a195e.gif成立。


   故当http://data.artofproblemsolving.com/images/latex/7/4/9/74900ba12430d4ceb5e1a240f98d3edb8a32e930.gif

 

http://data.artofproblemsolving.com/images/latex/d/1/8/d1854cae891ec7b29161ccaf79a24b00c274bdaa.gif都成立。


   因此满足条件的最小值为http://data.artofproblemsolving.com/images/latex/c/9/1/c91824801233f9ece4dbc85e25336a1b77c221cb.gif


   也可以通过构造数列证明:
http://data.artofproblemsolving.com/images/latex/0/6/c/06caa24880496bc6c903c01a7b9caa83fe8a195e.gif


   证明:设
http://data.artofproblemsolving.com/images/latex/9/0/3/903e28b93d34db82638c5f7ea6dfa1f00c8ece37.gif


http://data.artofproblemsolving.com/images/latex/a/c/3/ac3eaf43a3aa04549d6a97327b8e49f623ffa0ba.gif


因为当http://data.artofproblemsolving.com/images/latex/d/2/4/d24742f926d9941b8bbb1c164f09acecaf8c8933.gif时,有


http://data.artofproblemsolving.com/images/latex/c/9/9/c99f9d082324c307eab06cd71e1c1b8a4b804c64.gif


http://data.artofproblemsolving.com/images/latex/f/3/c/f3cc4a7469aea6324bd5e9c959eb5a4841720541.gif


所以,frac{{x_{n + 1} }}{{x_n }} > 1 Leftrightarrow x_{n + 1} > x_n,所以数列{ x_n }递增,因此,当http://data.artofproblemsolving.com/images/latex/d/2/4/d24742f926d9941b8bbb1c164f09acecaf8c8933.gif时有,http://data.artofproblemsolving.com/images/latex/c/a/6/ca66d66fb8a69db25e47c7d079751c7268b502ee.gif

 

http://data.artofproblemsolving.com/images/latex/c/a/6/ca66d66fb8a69db25e47c7d079751c7268b502ee.gif


   (2)最值法,就是应用结论
http://data.artofproblemsolving.com/images/latex/7/f/9/7f9b147a4da1db302640684cb5ed7ffea77ca1a3.gif

http://data.artofproblemsolving.com/images/latex/5/6/b/56bf52b1cc66a4843e31fb88feeed0ebbab0c572.gif


处理不等式恒成立的方法。


   下面用最值法求出使
http://data.artofproblemsolving.com/images/latex/8/6/f/86f7e437faa5a7fce15d1ddcb9eaeaea377667b8.gif的范围。


  
http://data.artofproblemsolving.com/images/latex/4/f/7/4f7d81389762366c131edcfc1778acc037b192e4.gif时, http://data.artofproblemsolving.com/images/latex/8/6/f/86f7e437faa5a7fce15d1ddcb9eaeaea377667b8.gif 可以为任意正实数;当http://data.artofproblemsolving.com/images/latex/4/d/0/4d040f015758516f8ca352dd8f6df8922f3759f5.gif


  
http://data.artofproblemsolving.com/images/latex/5/1/1/5116c2e3ba6b187f826b55c207b649edf56a91d3.gif,则


http://data.artofproblemsolving.com/images/latex/1/7/7/177d87f3557eb1b619b7fac2705e520c50d21677.gif.


  下面求出当http://data.artofproblemsolving.com/images/latex/f/1/c/f1c2d165d7250f2fd4ebc5eeb10b8a20bcf8d131.gif的最大值。

 

   先证明:当http://data.artofproblemsolving.com/images/latex/4/7/7/4777e4bda3ab8a441ab13289cab8cb6e7e8c9479.gif时, http://data.artofproblemsolving.com/images/latex/4/4/f/44f9b5e21c8353d8282dcf93c2d3c6dd4c5442ff.gif 递减, 只须证,对http://data.artofproblemsolving.com/images/latex/4/7/7/4777e4bda3ab8a441ab13289cab8cb6e7e8c9479.gif,有g(n) > g(n + 1)

 


http://data.artofproblemsolving.com/images/latex/0/b/1/0b1f5cd7ae0821fbb42b9685bf4c80f0c6a69d0b.gif


http://data.artofproblemsolving.com/images/latex/6/3/d/63d886650c7e7c575c46b7ad3d2b6c4d3c72b939.gif (http://data.artofproblemsolving.com/images/latex/4/7/7/4777e4bda3ab8a441ab13289cab8cb6e7e8c9479.gif),x_2 = frac{{17^3 }}{{55^2 }} > frac{{16^3 }}{{55^2 }} = left( {frac{{64}}{{55}}} right)^2 > 1


http://data.artofproblemsolving.com/images/latex/8/4/2/8420564c58024d24f42039a76f604befa9970349.gif,有

 

http://data.artofproblemsolving.com/images/latex/1/d/2/1d2875683c0508d9c929611de3c2121f1dcc1cef.gif
http://data.artofproblemsolving.com/images/latex/d/1/9/d1909dfa2664b1a4767455be0e7fd9623d01a5d8.gif


而当http://data.artofproblemsolving.com/images/latex/8/4/2/8420564c58024d24f42039a76f604befa9970349.gif时,有


http://data.artofproblemsolving.com/images/latex/1/4/7/14756e452ed8545cb825ceee2d923284fda0c3c3.gif


http://data.artofproblemsolving.com/images/latex/8/a/b/8ab3aded2cfccf33dbfeeb302827b27a6286cafe.gif

 

所以


http://data.artofproblemsolving.com/images/latex/7/8/f/78fd99eb6a3b94c4b2e459566fd9270e86572ddf.gif


http://data.artofproblemsolving.com/images/latex/c/3/a/c3af463c197837ca3a36632647c4b0e9bb0f786c.gif


因此


http://data.artofproblemsolving.com/images/latex/3/8/c/38c3b44d9d5ea5c310b633fe81c623088f000690.gif


{ x_n }是递增数列,所以当http://data.artofproblemsolving.com/images/latex/4/7/7/4777e4bda3ab8a441ab13289cab8cb6e7e8c9479.gif时,有http://data.artofproblemsolving.com/images/latex/c/a/6/ca66d66fb8a69db25e47c7d079751c7268b502ee.gif,因此,对http://data.artofproblemsolving.com/images/latex/4/7/7/4777e4bda3ab8a441ab13289cab8cb6e7e8c9479.gif,有

 

g(n) < g(n + 1),即http://data.artofproblemsolving.com/images/latex/4/4/f/44f9b5e21c8353d8282dcf93c2d3c6dd4c5442ff.gif递减,由此得到,当http://data.artofproblemsolving.com/images/latex/4/7/7/4777e4bda3ab8a441ab13289cab8cb6e7e8c9479.gif时,有g(n) le g(2) = sqrt {17}

 
所以,当
a ge sqrt {17}时,对http://data.artofproblemsolving.com/images/latex/4/7/7/4777e4bda3ab8a441ab13289cab8cb6e7e8c9479.gif均有,a ge left( {2n^3 + 1} right)^{frac{1}{n}} Leftrightarrow a^n ge 2n^3 + 1;此时,对

 

http://data.artofproblemsolving.com/images/latex/0/6/c/06caa24880496bc6c903c01a7b9caa83fe8a195e.gif


  故当
http://data.artofproblemsolving.com/images/latex/d/1/8/d1854cae891ec7b29161ccaf79a24b00c274bdaa.gif都成立成立。

 

  即当http://data.artofproblemsolving.com/images/latex/d/1/8/d1854cae891ec7b29161ccaf79a24b00c274bdaa.gif都成立。


(III). 因
http://data.artofproblemsolving.com/images/latex/7/2/d/72dea082d929a17ebc17b683ba29ec017097eff8.gif,则


http://data.artofproblemsolving.com/images/latex/b/6/f/b6fa5d3836819d0fa2fa63761b0351814c5dbcdc.gif


http://data.artofproblemsolving.com/images/latex/c/9/b/c9b3f89f33709b94de91319c682f556f61e14338.gif


  先取特殊值探路:


 
http://data.artofproblemsolving.com/images/latex/9/2/e/92ee913214330816c21ed1490df01843b5cb2075.gif时,


http://data.artofproblemsolving.com/images/latex/6/b/f/6bfa7abce7a3083f1abe83cfcd885de97febc514.gif


 
http://data.artofproblemsolving.com/images/latex/9/a/3/9a3240ad86b2b1d076825686aaf98565beeaac04.gif时,

http://data.artofproblemsolving.com/images/latex/7/a/1/7a1fadc7c4f9c5a66d0b9de53eca29407353e593.gif


http://data.artofproblemsolving.com/images/latex/6/6/d/66d8d8186fa7ed4c6f16049f4e1fd8cd1c5f49fb.gif http://data.artofproblemsolving.com/images/latex/b/f/7/bf7fb809c7b4cd053644a5390c3405ad7c2a1f4e.gif


http://data.artofproblemsolving.com/images/latex/c/3/8/c3843eaf7ca5138b9b3474ccb7190feef3940669.gif,则


http://data.artofproblemsolving.com/images/latex/b/d/0/bd0f1e381b7dce1529171ac0b4556d6e3f801b52.gif


所以


http://data.artofproblemsolving.com/images/latex/5/2/e/52e6b1a44cc34a2aee0e03aebb92ebee4ef4aab5.gif


  http://data.artofproblemsolving.com/images/latex/9/0/7/90790c2a8b5244d355544a298e98f72410ae0fe3.gif


   对于一般http://data.artofproblemsolving.com/images/latex/d/2/4/d24742f926d9941b8bbb1c164f09acecaf8c8933.gif的情形


http://data.artofproblemsolving.com/images/latex/d/c/8/dc8be9d9b7412442e0ae93dc6ffac671709310ef.gif


http://data.artofproblemsolving.com/images/latex/c/2/b/c2b6565923331ae14fc8ae871d16d0459d1cbd29.gif


http://data.artofproblemsolving.com/images/latex/8/c/1/8c1831ebcac1efb84b1ddc8dec2ee1fd7a380d2a.gif


在前面证明http://data.artofproblemsolving.com/images/latex/9/a/3/9a3240ad86b2b1d076825686aaf98565beeaac04.gif明,已证结论:


http://data.artofproblemsolving.com/images/latex/9/4/d/94d6eb424366bf157987346f9131416bb2b806a2.gif


及当http://data.artofproblemsolving.com/images/latex/5/1/c/51c6dd798ef5305391ba2e4955c2b860e27f212d.gif,有


http://data.artofproblemsolving.com/images/latex/4/2/9/4292a096b598e6e29d9e045b3be45f79aebe4dfc.gif


所以


http://data.artofproblemsolving.com/images/latex/a/1/b/a1be79743030299e59f3922839edeeb64b94b395.gif



http://data.artofproblemsolving.com/images/latex/e/5/a/e5a7972ba856f031387544acf2e52b0592d779dc.gif


实际上,我们可以证明稍强一点的结论:当http://data.artofproblemsolving.com/images/latex/0/c/0/0c0d7e220f4f43addc14f1e1211b8791a08a81c3.gif时,有


http://data.artofproblemsolving.com/images/latex/7/1/4/7143ca026b853fcddb3d1ae151d8f59f6c1f9ab2.gif


    (Ⅱ)、(III)两题的上述解法都是先从特殊入手,找出问题的结论或问题的解法。

 

其中(Ⅱ)题解法一用特殊值代入,猜出http://data.artofproblemsolving.com/images/latex/8/6/f/86f7e437faa5a7fce15d1ddcb9eaeaea377667b8.gif的最小值,再给出证明。而用数学归纳法给出

 

的证明,比参考答案用二项式定理给出的证明,思路自然且简单;(III)题先用等比数

 

列前http://data.artofproblemsolving.com/images/latex/6/1/1/611717e5a3b99149c591ce0194bb5df42da89b27.gif,然后通过特殊引路,找

到解一般情形的方法。需要注意的是http://data.artofproblemsolving.com/images/latex/0/2/6/0268a9e547c05923eb579abe60c7867e98469c5c.gif,逐

 

项比较http://data.artofproblemsolving.com/images/latex/b/9/e/b9eb2c8bb24d4cecbf516c96f7ad61fac5554863.gif ,有http://data.artofproblemsolving.com/images/latex/2/1/b/21b0ac9f4144d55cd9f4da0863bbb92994a2f986.gif,可以

 

按上述求差并分解因式证明得,也可以通过求导证得。

    另外,此题与2011年高考四川卷理科的压轴题:

已知函数http://data.artofproblemsolving.com/images/latex/2/3/f/23f7eab0854346e5d5fd26e5db22aa7b54cc4983.gif.

 

(Ⅰ)设函数http://data.artofproblemsolving.com/images/latex/6/1/7/617729eaf7470876a5410cad65a37f6af46517e1.gif的单调区间与极值;

 

(Ⅱ)设http://data.artofproblemsolving.com/images/latex/1/2/8/128646da94c305cb181250e34d58300d0fa0f87f.gif

(Ⅲ)试比较http://data.artofproblemsolving.com/images/latex/a/0/6/a063853bdfd761520e06382373fe270b6078994b.gif]的大小.
(2011年高考数学四川卷理科第22题)

十分类似,两题第(Ⅰ)都涉及导数的应用,12年题是求切线的斜率,11题是求函数的

 

单调区间与极值;11年题第(Ⅱ)题是解含参数的方程,12年题第(Ⅱ)题是求不等式

 

恒成立的参数的范围;两题第(Ⅲ)题都是比较大小问题(实际上是证明不等式问题)。

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