高中复习- 合理构造函数解导数问题
例1: 已知函数749a988fd0fcaaa43d2ded5e13de122c.png.
(1) 若6ca8c824c79dbb80005f071431350618.png为f4c7a893604bc6ecb2fed03958976357.png的极值点,求实数0cc175b9c0f1b6a831c399e269772661.png的值;
(2) 若f4c7a893604bc6ecb2fed03958976357.png在c17f3d6c0e9a86fb5f9e7f21111e1ae4.png上增函数,求实数0cc175b9c0f1b6a831c399e269772661.png的取值范围;
(3) 若ba77036f543f07a168320c280152a8ad.png时,方程ea4cb5a74a2492acba28a9b50efaa9bf.png有实根,求实数92eb5ffee6ae2fec3ad71c777531578f.png的取值范围。
解:(1)因为4e77a783daf447361658f525cebd4ead.png是函数的一个极值点,所以2d0b4157be8283cffdfbcebf67b0cb55.png,进而解得:ded681eaa02d11064c9a469dd1b3e04c.png,经检验是符合的,所以79739438408054c228825520f737ae5f.png
(2)显然30b062ead6d0e57b38863df9bf9e40a1.png结合定义域知道46c7a65e77903e00adc7422feb2346a9.png在b1a240f257cf4fbe395f28b56a6d3d49.png上恒成立,所以77b4599594def71ab1fd041bbd4e3bc7.png且398063331484dcae9ec9c464670ac800.png。同时9a40cb49c856d6579493e3cbd9cb74ac.png此函数是63d6de3633750c814e26b8510dd43d17.png时递减,78a4cdad446d2b24f4c652d7c3ec7b19.png时递增,
故此我们只需要保证922637e9e57ef1c1f5c9f8b67f697eaf.png,解得:0d884213426cc9d0c527895de5803013.png
(3)方法一、变量分离直接构造函数
解:由于887fb68a10cbd4369b27c90bee0334d8.png,所以:5c6faf6636d840a7235ad6d9e152279a.pngc1c64000d5507341d9f2ebda94b5a8cd.png
f41a4634defcea8ce192dcc36668266a.png 6d7afc0746de1abb4c551c259f9bc419.png
当6e17756c030f9f8373cc5456cb02fe7a.png时,24db3e35484a0a8136f5f24290f090eb.png所以9e3e33f4d583830632e12df255cf82e8.png在6e17756c030f9f8373cc5456cb02fe7a.png上递增;
当7f172a5c466990528643c8c376887fe1.png时,eb564e6da222cc8c4ccafc39837a2c56.png所以9e3e33f4d583830632e12df255cf82e8.png在7f172a5c466990528643c8c376887fe1.png上递减;
又355b0498e765bdc9b0eb3976faf45ca2.png 6c8a717b9b799b8c5062406b4d786823.png
当5af4cd940cf69781b46bf9a4b5ab1f0d.png时,4a739b150090f328c8055c4b17d405e8.png所以5ee266697bf95635052ba8f5a55eedeb.png在5af4cd940cf69781b46bf9a4b5ab1f0d.png上递减;
当5876464efb1022907b68867df69608fa.png时,b3605264bd690668cd386cdd1c7610e8.png所以5876464efb1022907b68867df69608fa.png上递增;
当3d3e00e0b84ad6b64a3461fe9092698a.png时,4a739b150090f328c8055c4b17d405e8.png所以5ee266697bf95635052ba8f5a55eedeb.png在3d3e00e0b84ad6b64a3461fe9092698a.png上递减;
又当c0fa4f18886b5f43a477a87249d879f5.png时,51573f26dbca7e9da313a2b4e2f2f69c.pngb34f8831335b737ee744277ce5b8beaa.png
当c3becc575b4c9df99e99068f375e24d7.png时,02c567a393a639ee3a8c2e6f0e8e071e.png则aaffe419e140ab553623b5559198499b.png且3e0cb3bf2b7fbdb3900014364fcdbd15.png
36f2ca1e3e278ccbd762c764ff71c32e.png92eb5ffee6ae2fec3ad71c777531578f.png的取值范围为989362552aeb754093d23f47167d0ea1.png
word/media/image60.gifword/media/image61.gif
6d7afc0746de1abb4c551c259f9bc419.png,f41a4634defcea8ce192dcc36668266a.png,c0ea9f2227918fc8a772579c757516fd.png
方法二、
构造:6facc2486574c08414f2fd9c1c76c8aa.png
3f8fe20f306d6fcd12882e3fa69202cc.png
bf3b835c3c776035146bfe35b3fbc86e.png daf3de2b254d224cca019be1afae8695.png f19e9f6dd773ec4c9dc9803ca7dc85a0.png 从而4826979cc121bc42ebd9a28b8efaf127.png在7438252393c71b507edb4d7a116aa1c2.png上为增函数;
536e0bc33b078e736ec1744038f43df0.png从而4826979cc121bc42ebd9a28b8efaf127.png在2257e95d394e2d1f1b79f282585fe914.png上为减函数
534bdcd646e11b56ecab12ca0c89b45e.png 而887fb68a10cbd4369b27c90bee0334d8.png 665a9d1ee8eef7e685b284c8dab1288a.png 51a1bb09960f2989cdace99350b0251f.png
分析点评:第(3)问的两种解法难易繁杂一目了然,关键在合理构造函数上。
例2:已知函数f (x)=eca3281ad39ca975e9fc92baa0a9649f.png+aln(x-1),其中n是正整数,a是常数,若a=1时,求证:当x≥2时,f (x)≤x-1.
证法一:当a=1时,f (x)=eca3281ad39ca975e9fc92baa0a9649f.png+ln(x-1),构造函数F(x)= (x-1)-f (x),下证:当x≥2时,F(x)=(x-1)-eca3281ad39ca975e9fc92baa0a9649f.png-ln(x-1)≥0恒成立.
F´(x)=1-7fb66a4ddf1979a123ed8d54c6caac1b.png=aa6120b9defaedf50e87117c0f10f8e4.png-e2dc3117d29a94a1fc92067864964b8d.png( x≥2).
①若n为偶数,∵x≥2,∴aa6120b9defaedf50e87117c0f10f8e4.png≥0,1-x<-1<0,f397c983e636d83c35655a424b1fa99a.png<0,-e2dc3117d29a94a1fc92067864964b8d.png>0,
所以:当x≥2时,F´(x)>0.∴F(x)min=F(2)=(2-1)-f23b568979ee86c9c4ab60abb02d54ee.png-ln(2-1)=0,所以:当x≥2,且n为偶数时,F(x)=(x-1)-eca3281ad39ca975e9fc92baa0a9649f.png-ln(x-1)≥0恒成立.
②若n为奇数,要证eca3281ad39ca975e9fc92baa0a9649f.png+ln(x-1)≤x-1,∵x≥2,∴eca3281ad39ca975e9fc92baa0a9649f.png<0,所以只需证:
ln(x-1)≤x-1(下略).
小结2:含有正整数“n”的表达式的符号、数值判断,“对n分奇、偶讨论”是一种重要的方法.在数列中运用很多.
证法二:∵当x≥2时,eca3281ad39ca975e9fc92baa0a9649f.png≤1,∴只需要证明1+ln(x-1)≤x-1.构造函数F(x)= (x-1)-[1+ln(x-1)],即F(x)= x-2-ln(x-1),则F´(x)=aa6120b9defaedf50e87117c0f10f8e4.png(下略).
小结3:证法一是直接作“差函数”(直接构造新函数),然后分奇、偶讨论;证法二是先适当放缩,然后构造新函数.解题时,要有敏锐的观察力.
(1)抓住问题的实质,化简函数
例1:、已知ad7ac2fcd2e22b7a507152d28ef55c97.png是二次函数,不等式61d7aa7aa6815b85fc49950fb346453f.png的解集是0e1ff5d39cd8a597452907bc397d3c09.png,且ad7ac2fcd2e22b7a507152d28ef55c97.png在区间69c476d2445642ca0f0181d280f32f91.png上的最大值c20ad4d76fe97759aa27a0c99bff6710.png. (1)求ad7ac2fcd2e22b7a507152d28ef55c97.png的解析式;
(2)是否存在自然数6f8f57715090da2632453988d9a1501b.png,使得方程b2bee1f8d4d5a66b8dc7256dd8a33be9.png在区间2acf629249890cb0dbac6714e0812d25.png内有且只有两个不等的实数根?若存在,求出所有6f8f57715090da2632453988d9a1501b.png的值;若不存在,请说明理由。
解:(1) ce422157ec440815a576ef5eba1b7362.png
(2)假设满足要求的实数6f8f57715090da2632453988d9a1501b.png存在,则b2bee1f8d4d5a66b8dc7256dd8a33be9.png,即有:0a9a48f3b082674ea3a0b9bea7b7e901.png
8f58b858e2192f6bc49da99f44c892a5.png,即有:f3d7a0596d601983dc8407ec0622c765.png 构造函数229fe81e40a9ef70fb2b8ae0cb3ec0b8.png
word/media/image106.gif 画图分析:aebf68cc5eee6d01ba13c21e1456d0dc.png
进而检验,知d31b03cd39c7d014128bed6b36ee1027.png,所以存在实数9f33e29e7d3691483f5e9bc9180a5ea9.png使得b2bee1f8d4d5a66b8dc7256dd8a33be9.png在区间444619b1235019694b99286985e6bf28.png内有且只有两个不等的实数根。
点评:本题关键是构造了函数229fe81e40a9ef70fb2b8ae0cb3ec0b8.png,舍弃了原函数中分母0e503e6ba8df74a61723be29415bcdc5.png问题得到了简化。
例2:.已知函数76c706b09b7e5dffd97d9b5199f5b132.png
(Ⅰ)求函数50bbd36e1fd2333108437a2ca378be62.png的单调区间;(Ⅱ)求证:2bf51e01e80a1960659252518be301b5.png.
解:(Ⅰ)依题意,函数的定义域为x>0. 70826b45a04dccd56daf2948e9e050b6.png∴当a≤0时,50bbd36e1fd2333108437a2ca378be62.png的单调递增区间为b921db311612fd3665c51872c7a83455.png. 当a>0时,818ac390d1e5fc5cb749b098407909d8.png令b33c11211bcb993da0168613096c0b4b.png>0,有8c565579bb9e28bcd473d845f1b2fc60.png所以函数50bbd36e1fd2333108437a2ca378be62.png的单调递增区间为41d00b7ab4bb896e3ef10bbdd0e23f26.png令b33c11211bcb993da0168613096c0b4b.png<0,有dd0f0195a7507974415b0fb5da467c1e.png所以函数50bbd36e1fd2333108437a2ca378be62.png的单调递减区间为20722b4b25e12b05778635fbd16846a9.png
(Ⅱ)设f313c97660f6181c1b8b5003c1db1e70.png
353621ced896658a414976b033a2f58a.png
∴adb3cbffd0f63ba6cdfbd86218646743.png
∴当3d3e00e0b84ad6b64a3461fe9092698a.png时,c5301802bec9c328aaacc93b32feca69.png。
例3:已知函数2e885abe7e77cf9e0c50530e6de2e20e.png.
(Ⅰ)当3872c9ae3f427af0be0ead09d07ae2cf.png时,求函数50bbd36e1fd2333108437a2ca378be62.png在c7faa44689191029e81ed95c9a29e023.png上的最大、最小值;
(Ⅱ)求50bbd36e1fd2333108437a2ca378be62.png的单调增区间;
(Ⅲ)求证:3872c9ae3f427af0be0ead09d07ae2cf.png时,在区间[1,+∞9371d7a2e3ae86a00aab4771e39d255d.png上,函数50bbd36e1fd2333108437a2ca378be62.png的图象总在函数95fa0c941b25ffe6dcae798b52d2d1f7.png的图象的下方.
解:(I)当3872c9ae3f427af0be0ead09d07ae2cf.png时,bc35ad4cac9f24f7562a3afb5758f1a8.png,ad9615cc4634dfdd6aed2216daaca5ea.png时,8baaf0ef5eac596ebf01a45909545fc1.png,故f(x)在[1,e]上是增函数.
∴ f (x )max = f (e ) =93b05c90d14a117ba52da1d743a43ab1.pnge2 + 1;f (x )min = f (1 ) =93b05c90d14a117ba52da1d743a43ab1.png.
(II)215fe4602aa406a64dd0711c2620ced0.png ,由d0382519afa56b9ed182de680169b65f.png27ff4fc5992981f65f25099b5b0f5aa6.png,∴60464bdce5a712d5076026bce86a6ceb.png,增区间为b921db311612fd3665c51872c7a83455.png;a<0时,增区间为a751be1fb014122ef95b886e7f686f0c.png。
(III)设F (x ) =93b05c90d14a117ba52da1d743a43ab1.pngx2 + lnx-6ca8c824c79dbb80005f071431350618.pngx3,则3a272b3f2388c5a003ca09b625b7e393.png(x ) = x +afc48b56873694f3d43097841ecc3f4f.png-2x2 =235271be2d819e9eddfc08a4266f6df5.png. ∵ x>1,∴ 3a272b3f2388c5a003ca09b625b7e393.png(x )<0,故F (x )在[1,+∞]上是减函数, 又F (1) =-6c2e3e2e98abd1fd9a66519db9da8d90.png<0,∴ 在[1,+∞]上,有F (x )<0,即93b05c90d14a117ba52da1d743a43ab1.pngx2 + lnx<6ca8c824c79dbb80005f071431350618.pngx3, 故函数f (x )的图象在函数e84fec1e074026d6fa8e3155482c35c3.png =6ca8c824c79dbb80005f071431350618.pngx3的图象的下方.
(2)抓住常规基本函数,利用函数草图分析问题:
例4: 已知函数7b58566395450f057d48ba17bffe7319.png的图像在点0950fd40e59e0120135b4e4d1a861ed8.png处的切线方程为74a02e39b7685bf323916a8b8e6f7595.png
设fd65b3dff1c5e84359bd0299b3549d59.png
(1) 求证:当e3f7f32051ade969e770ab04b5a4853b.png时,71c7ccfce385c6f827f147b0c53b3c8f.png恒成立;
(2) 试讨论关于9dd4e461268c8034f5c8564e155c67a6.png的方程1e9dc43595227d201b0d7e6782db87e9.png根的个数。
解证:(1)0eebf2496050a31545305085417111a7.png
(2)方程471489e4620d51c5d40a477355b9bfff.png从而 f14d748d21c3866dff6a1275ed9e5787.png
因为35a3e49f0d55eb333b6533eedd3a1fa4.png所以方程可变为6b5c2026b0037f780e67f67ee65d87bd.png
令21d25e351c8b0c97fecad064fa4f66bd.png,得:462aa91a19be301d8e77313f19052468.png
当0ead36210507b1777f7be241776a29c5.png时,185dff5807cf30ea4e4f27543de5e996.png在13228d9205fb6a8141ee0b78025764fa.png上为增函数;
当b48bb5f91bb223c2b1da7722b2316629.png时,0e6646a5d68f4c301a072e6cd7380225.png在b48bb5f91bb223c2b1da7722b2316629.png上为减函数;
当3c5a29c6dae7df5f94c227a51bbe479f.png时,b194645e9d5a78468fe60317a2f9b6b6.png
又d2d0ced382283fe5d7f39986940af6e7.png
所以函数19747493fdbb22ee2bb44634274dcf5b.png在同一坐标系的大致图像如图所示
1 当c3c197c4f8505f31f5013819163da392.png即0fb8649d53806101ad14b4396d987d92.png时,方程无解;
2 当4cfff5b70e57c20ec221c93863f51f32.png即126b9de83489a8e5504f2dd0769d713c.png时,方程一解;
3 当723d5b05f48ecae55646f423a179813b.png即16f677be710eb375635b998d4a0dd0f9.png时,方程有2个根。
分析点评:一次函数,二次函数,指对数函数,幂函数,简单的分式根式函数,绝对值函数的图象力求清晰准确,一些综合性的问题基本上是这些函数的组合体,如果适当分解和调配就一定能找到问题解决的突破口,使问题简单化明确化。
(3)复合函数问题一定要坚持定义域优先的原则,抓住函数的复合过程能够逐层分解。
例5:已知函数75670625a4c61b40fc482335b02c5f14.png在区间05132f14fb0a837981cc8b0715c9165f.png上单调递减,在区间c5168b295c366faf86f1fdc3e98436ea.png上单调递增。
(1) 求实数0cc175b9c0f1b6a831c399e269772661.png的值.
(2) 若关于9dd4e461268c8034f5c8564e155c67a6.png的方程b969d936f20dc789e8af348ac0b1fddd.png有3个不同的实数解,求实数6f8f57715090da2632453988d9a1501b.png的取值范围.
(3) 若函数6c883dd749eadc076f4ce6258ed7944b.png的图像与坐标轴无交点,求实数83878c91171338902e0fe0fb97a8c47a.png的取值范围。
解:(1)利用5452bae9f1e9781d9829274c6a15da26.png 得:f5d35420b0365189cf20d9fa818c53b0.png
(2)因为0b97dec8f8c6bebdbb021ed102ca1e19.png
得 4ae2b5b2d234b9dc41c67cfd33a39595.png 列表得
word/media/image210.gif60cf11173160ddd51b6b8f3104129115.png
word/media/image212.gif393256c48d5ccd80605ea369eb815afd.png
word/media/image214.gifword/media/image212.gifd0ad80af4a45a80ac9f74acac4798357.png
因此ad7ac2fcd2e22b7a507152d28ef55c97.png有极大值6424816a52cbb5134bd06f6e5e8105ba.png极小值e0a6a6b30a37b7ff60fedb068c2bb084.png作出ad7ac2fcd2e22b7a507152d28ef55c97.png的示意图,
word/media/image220.gif如图:
因为关于9dd4e461268c8034f5c8564e155c67a6.png的方程b969d936f20dc789e8af348ac0b1fddd.png有3个不同的实数解,令4a241271279d0b11e2ce6ce42e07185e.png
即关于e358efa489f58062f10dd7316b65649e.png的方程507a3cd8ade19ec6de4a17375c6e7e24.png在570bbe675fa418394c49ec5b3c62e845.png上有3个不同的实数解,
所以600ffa23a9ae895869be00da1b8e8083.png的图像与直线c20e256d116adc2fa6a59beb6f6139cf.png在570bbe675fa418394c49ec5b3c62e845.png上有3个不同的交点。
而600ffa23a9ae895869be00da1b8e8083.png的图像与f4c7a893604bc6ecb2fed03958976357.png的图像一致。即053dd924b7ff9c4916a09b483635081d.png
(3)函数6c883dd749eadc076f4ce6258ed7944b.png的图像与坐标轴无交点,可以分以下2种情况:
①当函数6c883dd749eadc076f4ce6258ed7944b.png的图像与9dd4e461268c8034f5c8564e155c67a6.png轴无交点时,则必须有7d9b8521dc0a70d12cafb01374603af8.png无解,而
c0ba310a8bcdb23e5ac38e12d04ab68d.png函数2c14e02bcd7765326cc93a820d5c9365.png的值域为b99cf00a9412dbab8a3f549b5d434114.png所以4a9422e2194a8ddc2b1289a29648392c.png
解得a483b66fe79bc482f4a9d89820b09590.png
②当函数6c883dd749eadc076f4ce6258ed7944b.png的图像与415290769594460e2e485922904f345d.png轴无交点时,则必须有0f4e33581c50af1e7839a56d17457b13.png不存在,即0ff81584b4548a91f9994fdf28f13c9d.png或6f6d1cf1cd1fdf4e70914668e5e44306.png,有意义,所以f22fe439b55dd82c5971dac62516f159.png,解得40e7e2e9525c9dca49598c217ecb5041.png.
③由函数存在,可知494b72b72254c58a72ca49261402bded.png有解,解得92f42f77fb6ee1d9f290da51db85fc71.png,故实数83878c91171338902e0fe0fb97a8c47a.png的取值范围为55318a33bc979f2296c7a1b14f77047e.png
分析点评:复合函数尤其是两次复合,一定要好好掌握,构造两种函数逐层分解研究,化繁为简,导数仍然是主要工具。
专题:构造函数法练习
1.设函数f4b53eaf0f76d76758f5a11edce9bdd7.png若对所有的14bdb1ea4683521ea7de464bc9fe1046.png都有b5cab03570961b2cfc3d8bd9df62d169.png成立,求实数0cc175b9c0f1b6a831c399e269772661.png的取值范围。
2.设函数e759dd7de92162cacd3f369647d6da3b.png,其中b593070d4214e5cfb1365a41f0d8d9ba.png.
(Ⅰ)当0657e252e02b67c8748b88cc0ff208b1.png时,判断函数50bbd36e1fd2333108437a2ca378be62.png在定义域上的单调性;求函数50bbd36e1fd2333108437a2ca378be62.png的极值点;
(Ⅱ)证明对任意的正整数7b8b965ad4bca0e41ab51de7b31363a1.png,不等式8a6fb501366fde8b1c00b65c62e3ef1a.png都成立.
(Ⅲ)求证对任意不小于3的正整数7b8b965ad4bca0e41ab51de7b31363a1.png,不等式63d59a0d24845b1d6c73f400a754571e.png都成立
3.已知ee7c6743698614077391e527aef181ba.png,703fa333ae3a162f16ba9c694a6db9ca.png(e6384b0ee3faddc93a135be3e78067ab.png),直线2db95e8e1a9267b7a1188556b2013b33.png与函数50bbd36e1fd2333108437a2ca378be62.png、e84fec1e074026d6fa8e3155482c35c3.png的图像都相切,且与函数50bbd36e1fd2333108437a2ca378be62.png的图像的切点的横坐标为1.
(Ⅰ)求直线2db95e8e1a9267b7a1188556b2013b33.png的方程及6f8f57715090da2632453988d9a1501b.png的值;
(Ⅱ)若34b848c5c67a6e61fd0d404743cb5fbe.png(其中620d7872524e44a56669d72441609ab4.png是e84fec1e074026d6fa8e3155482c35c3.png的导函数),求函数ca8e608169b20a94570ac837e8ba0833.png的最大值;
(Ⅲ)当9ef3e0b2955c863f7e34fa38c96bdd07.png时,求证:aaf3dc6006ca87628f7f491dcbcc1eb3.png.
4.已知cbef1b11de08f059b9be95b079f8438e.png
(1)求函数d0872f2dbf41bb2b22b279f0a022dd76.png上的最小值;
(2)对一切cce9ca0106c47c73f9a14c40ea682476.png恒成立,求实数0cc175b9c0f1b6a831c399e269772661.png的取值范围;
(3)证明:对一切03a701ad75b944b688fc49694f7dc8a1.png,都有2bee0529c9ae94c23cd51f85b11257c6.png成立.
5.已知二次函数2e253c792f57321c41825ca3950ff7db.png为常数);7f677022bea29eff384f3945eb548e89.png.若直线2db95e8e1a9267b7a1188556b2013b33.png1、2db95e8e1a9267b7a1188556b2013b33.png2与函数f(x)的图象以及2db95e8e1a9267b7a1188556b2013b33.png1,y轴与函数f(x)的图象所围成的封闭图形如阴影所示.
(Ⅰ)求0cc175b9c0f1b6a831c399e269772661.png、b、c的值
(Ⅱ)求阴影面积S关于t的函数S(t)的解析式;
(Ⅲ)若36bc3d2d21751bcecafd546c80f6fca6.png问是否存在实数m,使得y=f(x)的图象与y=g(x)的图象有且只有两个不同的交点?若存在,求出m的值;若不存在,说明理由.
6.已知函数f(x)=baff015d9b4bfb074c409733e75986b5.png
(1) 若h(x)=f(x)-g(x)存在单调增区间,求a的取值范围;
(2) 是否存在实数a>0,使得方程57b1617dcb9014301e4e19b782ab5a24.png在区间01db524505ba6238c94b98a8419845c7.png内有且只有两个不相等的实数根?若存在,求出a的取值范围?若不存在,请说明理由。
7.已知数列3d0299a906f22a56ae7e72f5cb3590bf.png满足:4683592a3b18fb3bf42dddf8245189ba.png,07b37fc601ad1f2a5482f390af57e9e2.png(其中e1671797c52e15f763380b45e841ec32.png为自然对数的底数).
(1)求数列3d0299a906f22a56ae7e72f5cb3590bf.png的通项9ded7825070b255e7bc092cdc2c8e98a.png;
(2)设8f706790a2886e8afeb80d288aaa43c3.png,a248fe28c5fecfc299020f3910f44f90.png,求证:3487456075a3f91bea192f0640d5fee7.png, b303b6646f96f2efdd1c8b3240502608.png.
8.已知函数word/media/image299_1.pngRword/media/image300_1.png, word/media/image301_1.png.
(1)求函数的单调区间;
(2)若关于word/media/image303_1.png的方程word/media/image305_1.png为自然对数的底数)只有一个实数根, 求的值
9.已知函数aa533be7e61bb46a105d5834d1814193.png在97b7c0cb884b483e3107f779bdcadd64.png是增函数,c584098caf15e0d25176cc1cd56692ef.png在(0,1)为减函数.
(I)求50bbd36e1fd2333108437a2ca378be62.png、e84fec1e074026d6fa8e3155482c35c3.png的表达式;
(II)求证:当887fb68a10cbd4369b27c90bee0334d8.png时,方程b9eb8a00cf5929e6c8b701e5a0beb253.png有唯一解;
(III)当fc2e8ec8dcf30bb9fbb8aca436bd661b.png时,若384c1970c7654f14a88384120b9d3b7c.png在9dd4e461268c8034f5c8564e155c67a6.png∈8d66c401c5121f5675ad8f55eaa5bc08.png内恒成立,求92eb5ffee6ae2fec3ad71c777531578f.png的取值范围.
10.已知函数32faf6e347ae549fee91da66516daadd.png(e1671797c52e15f763380b45e841ec32.png为自然对数的底数).
(1)求函数50bbd36e1fd2333108437a2ca378be62.png的最小值;
(2)若f6e696c1952cff44affab43c8beaec2f.png,证明:86d1f598a1d0a5249496d9b41d451c2c.png.
11.已知函数ff0e230602871b480e30c304442bd676.png,其中323c5f97105643bc61e288fe596194ca.png。
(1)若a255512f9d61a6777bd5a304235bd26d.png是函数952b14c79876fd6d92cc3eb52eb536f5.png的极值点,求实数0cc175b9c0f1b6a831c399e269772661.png的值;
(2)若对任意的1dbfb2a9d0eedea2d8dec2d3a1126148.png(e1671797c52e15f763380b45e841ec32.png为自然对数的底数)都有223b88c33b439ad7a1d441e20e520432.png成立,求实数0cc175b9c0f1b6a831c399e269772661.png的取值范围。
12.已知函数word/media/image333_1.png在[1,+∞)上为增函数,且θ∈(0,π),,m∈R.
(1)求θ的值;
(2)若word/media/image335_1.png在[1,+∞)上为单调函数,求m的取值范围;
(3)设,若在[1,e]上至少存在一个word/media/image337_1.png,使得成立,求word/media/image339_1.png的取值范围.
构造函数法(理科)答案
1、解析:令b83d45f8c9a11175502dc1af7c2f3b4b.png
对g(x)求导得b7c2c3ca09343d079369110b6ef5fdd2.png
令0eaeb573c4a0b15b627a38eab554e229.png
当3c2e5aaa95f2f9fcdef893f929370d51.png时,对所有的x>0都有bdb1b826acd593af45089831b06e7f4b.png,所以811a5d366ca73c259f8d0c2285bc5c74.png上为单调增函数
又g(0)=0,所以对c5e92a1f755ee967d7c92b6dc344ef11.png 即当3fa279a839f3e4ecf22455377ceaf264.png所以3c2e5aaa95f2f9fcdef893f929370d51.png成立
当a>1时,对于cdf703eac9659238e03b1c2274233033.png 所以g(x)在9eee8940f22dae4ed77c3230b8768a29.png 所以对于bf7442eccb721244c48dda6e18471ae3.png
即f(x)所以当a>1时b5cab03570961b2cfc3d8bd9df62d169.png不一定成立
综上所述可知a的取值范围是450db401bb9024f15c73fb0f6c89d167.png
本题主要考察了函数的导数和利用导数判断函数的单调性,涉及分类讨论的数学思想难度较大
2解:(Ⅰ)由题意知,50bbd36e1fd2333108437a2ca378be62.png的定义域为d1f4372ca07aff2ab405b5c6b38be364.png,c7188e06e6894fd673a0d4c2c4817fe0.png
设d9263c61584842c7ad8e1311c152b48f.png,其图象的对称轴为ca77ecc9c71129409b44ebba5781627f.png,
44d908b51874f3b65e53632a32bc7cbd.png.
当0657e252e02b67c8748b88cc0ff208b1.png时,a100dbcb9f787e89c1b6d793ad7bd24a.png,即45a3a7014b3bc27bc42003e73e3c346f.png在d1f4372ca07aff2ab405b5c6b38be364.png上恒成立,
95e029696a8e77db6f75665e6464c095.png当44d8533cedbaddf0df9a2490bcc7059e.png时,8baaf0ef5eac596ebf01a45909545fc1.png95e029696a8e77db6f75665e6464c095.png当0657e252e02b67c8748b88cc0ff208b1.png时,函数50bbd36e1fd2333108437a2ca378be62.png在定义域d1f4372ca07aff2ab405b5c6b38be364.png上单调递增.
(Ⅱ)①由(Ⅰ)得,当0657e252e02b67c8748b88cc0ff208b1.png时,函数50bbd36e1fd2333108437a2ca378be62.png无极值点.
②0ac73d60ed62472f5a3f4ebb94ce54cf.png时,0ac9f61f12e23d322884ca0f6b40a109.png有两个相同的解5fd9120495733ddd85fddc936e4c4dc4.png,
0df44af2678253b57b7365eb6d86d835.png时,8baaf0ef5eac596ebf01a45909545fc1.png,c76e36c650294494af26baaee8f89f94.png时,8baaf0ef5eac596ebf01a45909545fc1.png,
8e102afd05059997286ec7c17386ba42.png时,函数50bbd36e1fd2333108437a2ca378be62.png在d1f4372ca07aff2ab405b5c6b38be364.png上无极值点.
③当5cee229875c88e240da1c6fe42695f32.png时,06605d0b94674429f3ab62bec2350d50.png有两个不同解,dfa44695647849915d8b4232ed70c867.png,93bc788c5a46719249e92029de61d138.png,
b5a26bf0028bb182d6d345af8db2827f.png时,1e764f5b5a9b380f85224ea0a30ded3c.png,a99481607fc7c172be8a71f03db26814.png,
即c3e57122611e35a1a168586bc464cb31.png,47e2d5d849b981bf6fb70835c924c738.png.94be6119691c36e369a113de17f7ddf2.png时,d267392b36cc3ca35802359790ea76e6.png,50bbd36e1fd2333108437a2ca378be62.png随9dd4e461268c8034f5c8564e155c67a6.png的变化情况如下表:
由此表可知:e117be13e3711019f9681197ed90f543.png时,50bbd36e1fd2333108437a2ca378be62.png有惟一极小值点dfa44695647849915d8b4232ed70c867.png,
当8c87e98e867d1dcf4904865c31127098.png时,3250779f2dd3be6e1159ece0a9e55e08.png,6a7e1eefd439a73fb73ac813d71e12cb.png,
此时,d267392b36cc3ca35802359790ea76e6.png,50bbd36e1fd2333108437a2ca378be62.png随9dd4e461268c8034f5c8564e155c67a6.png的变化情况如下表:
由此表可知:8c87e98e867d1dcf4904865c31127098.png时,50bbd36e1fd2333108437a2ca378be62.png有一个极大值dfa44695647849915d8b4232ed70c867.png和一个极小值点93bc788c5a46719249e92029de61d138.png;
综上所述:e117be13e3711019f9681197ed90f543.png时,50bbd36e1fd2333108437a2ca378be62.png有惟一最小值点be6b2d2af931c94300325edad3f9eeb9.png;8c87e98e867d1dcf4904865c31127098.png时,50bbd36e1fd2333108437a2ca378be62.png有一个极大值点62aaced6e784a6a5b344b43850f98398.png和一个极小值点d1be9c1d0b50253b37378ee2e627d1fc.png;1770d1fc70b4b0bb0f094d74b2ba461c.png时,50bbd36e1fd2333108437a2ca378be62.png无极值点.
(Ⅲ)当4fa2c30c74730418523e9ad5cd4ac6ac.png时,函数865c546469cfd584b45f52fc48731336.png,
令函数2c4e998e7df6b212b0caf20bc70d153c.png,则69297a7b4b994b96947d080db482c4d6.png.
95e029696a8e77db6f75665e6464c095.png当b065c246672c30c6090519247cc14e84.png时,8baaf0ef5eac596ebf01a45909545fc1.png,所以函数ca8e608169b20a94570ac837e8ba0833.png在5c13dd19f7cfd32e0db625dd6f2cad06.png上单调递增,又352383e4e798148b05ec354cff23c231.png.
9cb7b49e121f4c629b355e8622ae58e8.png时,恒有ee50f5e3e9412f2cfdb20aa2d7b87574.png,即5e283cbf60c9c00a2c030709b586db36.png恒成立.
故当03a701ad75b944b688fc49694f7dc8a1.png时,有ae767b541c2ea2faa811b9363641df1a.png.
对任意正整数7b8b965ad4bca0e41ab51de7b31363a1.png取0307c469d4dd2d7af31ded7ccbe1a421.png,则有8a6fb501366fde8b1c00b65c62e3ef1a.png.
所以结论成立.
(3)由(2)可知当4fa2c30c74730418523e9ad5cd4ac6ac.png时,函数a63283f4c4429c4d15fcfbf89bd6aecc.png, ………………(10分)
此时50bbd36e1fd2333108437a2ca378be62.png有惟一极小值点:db93d570b146fed13dd13c4003e09d3f.png且89c295542f4a1a741b7df82403d44c0d.png.………(12分)
ab50904089abe6f7cb303a62fb4193ce.png
∴dba98c5bb17c17aa3e79665fb287f726.png.
∴e28bcc7bc03a6b08c2154edc37ce4d65.png,6f212411758476e3cc6d7bf4cae0bbab.png. ……………………(14分)
3.解:(Ⅰ)依题意知:直线2db95e8e1a9267b7a1188556b2013b33.png是函数ee7c6743698614077391e527aef181ba.png在点ef678978f5f19957dd157b03ca76516e.png处的切线,故其斜率142519c15d686d0193e3d9b7b58a8900.png,
所以直线2db95e8e1a9267b7a1188556b2013b33.png的方程为de1edca1d87138fa1b61b2997e47ca84.png.又因为直线2db95e8e1a9267b7a1188556b2013b33.png与e84fec1e074026d6fa8e3155482c35c3.png的图像相切,所以由
7827505fc6b13ad560ca692933426c0b.png,
得a565550111844cf6d5f6a4832a7ac33b.png(cc29e84152f151181d4813b7ea8adb81.png不合题意,舍去);
(Ⅱ)因为f0205e46f405b743f04ae16e0b5867b5.png(7ae27cecfbd92cbd679b00187dac70d1.png),所以91738286e1e12f0ea60ff83ef1cf2ddd.png.
当349e030b5c1333d483a7136eed991809.png时,5c1c0727f238086db0ed17d69d6ae0f0.png;当887fb68a10cbd4369b27c90bee0334d8.png时,2e75f73a775b8649817f1314b2b6ea48.png.因此,ca8e608169b20a94570ac837e8ba0833.png在b310ee4083e1f905b701ed7ba7aadb29.png上单调递增,在02bacf0de82c8d2b9eea0463575750da.png上单调递减.因此,当e11729b0b65ecade3fc272548a3883fc.png时,ca8e608169b20a94570ac837e8ba0833.png取得最大值523fc7ea19a7f5935c36d66cf0041600.png;
(Ⅲ)当9ef3e0b2955c863f7e34fa38c96bdd07.png时,ba5165a0f5a46fb3e0425d523512fd5c.png.由(Ⅱ)知:当349e030b5c1333d483a7136eed991809.png时,c12b4256ebbea1fb676f54c91eae4794.png,即f26faa91ffb2f735b32a87f87b15f01c.png.因此,有90ee03e3576796b49cebde8a0157e4e2.png.
4解:(1)324c2b7c4ad7a47dc1b7a521675e5802.png,……………………………………………………………1分
当753094f9ebe5a26d90ff41f354a99d23.png单调递减,当6273d6f7d3997ea1bac53a1c45b0519f.png单调递增 ……2分
①efae6c8ca9a6a18cad9cb6e1bd2688ae.png,没有最小值; ……………………………………………………3分
②0803d7eb463dbce16408365cc9b60624.png,即64170794ef3c6b44b336b9a3f60038f0.png时, a9d4aae64ad57e2e8c7ce2530ca1a616.png; ………4分
③3b3cf56548282fd03a45e557ab8bf9f3.png,即679c487dd297acd3df31370738e869ff.png时,d5a1a28b3a207d10d1edc183c7fff8e0.png上单调递增,ca86affb5b02f87d0e734c039cbb01c7.png…5分
所以d27bd489d45d7789c9871d557ee2007b.png ………………………………………………………6分
(2)f4e324e332f7a6b1a4ef20d7bbcd8714.png,则04f3ecb4cf531a09de6db5718fcd20ca.png,……………………………………7分
设944d84ad3799f000ad9b743c0b781ab6.png,则30f0b7dd9ff27b58657e5ac8276930e1.png,
① 9b54a4bc34fb1a66f7f8496773689638.png单调递减, ② 2918b89ac2677694298f3603b4f3425f.png单调递增,
所以bc5992f9bf2adbc8981e9134d10aa21b.png,对一切cce9ca0106c47c73f9a14c40ea682476.png恒成立,所以212a53a60bb3bfea946ee8758adff145.png;
………………………………………………………10分
(3)问题等价于证明73191e5550ef4453631bbb921b0ea58b.png, ……………………………11分
由(1)可知989b7a0719cc949abb1f42b026709f39.png的最小值是9c1c465315bab5d7b382882ded0d3f6a.png,当且仅当976f116183c228fd636b7b4bfb9362c6.png时取到,[来源:状#元#源]
设e5fc4500282c3b33145ae29b84660124.png,则35c2a1b70d1806ee716382ad6d0f91c7.png,易知
f41ebd4444c82248f0c07f2037e36e5d.png,当且仅当a255512f9d61a6777bd5a304235bd26d.png时取到, ………………………………………13分
从而对一切03a701ad75b944b688fc49694f7dc8a1.png,都有303d2196ac6f1bb9a470e3eb7d417510.png 成立 ……………………………14分
5.解:(I)由图形可知二次函数的图象过点(0,0),(8,0),并且f(x)的最大值为16
则e3f1ebbf21e26fea15c6171a958d9fef.png,∴函数f(x)的解析式为3f9648c002615c9f827b3547f5904eff.png 4分
(Ⅱ)由14ba75c05f821b20a377c94958660c90.png得874fdc8e3202f7d329fdc35f36b30d67.png
∵0≤t≤2,∴直线l1与f(x)的图象的交点坐标为(8a231db860226aef1d25f44e39461af0.png………………6分
由定积分的几何意义知:
b18d404246c78845d025904645a93b8f.png
bf1e0ad163192525a8713d8a68930db4.png368bca95a54c597940250f47ce029899.png
(Ⅲ)令f2efeb9a721e6678f03bedd92a89e322.png
因为x>0,要使函数f(x)与函数g(x)有且仅有2个不同的交点,则函数e7c774624e47a900c416698167b585d2.png的图象与x轴的正半轴有且只有两个不同的交点
02e59178fa5cd20879100934b2fb0e32.png∴x=1或x=3时,edaceafd7b4481d05cf373d4afbed117.png
当x∈(0,1)时,c594af2cec4dd7b631d825a4a6119157.png是增函数;当x∈(1,3)时,9b165aa58567acab194e58721762d4ba.png是减函数当x∈(3,+∞)时,c594af2cec4dd7b631d825a4a6119157.png是增函数∴2b63437c29fc02cf754b5bc6673a0f1a.pngf2429295098f36f7c9d1a68c92b7342d.png…12分
又因为当x→0时,df691c2b9ccb202f442563db16cf2d5c.png;当637843d1885a56df95e41967c0833e8d.png
所以要使89b516166d2adbd87d36aa40d0e9f73e.png有且仅有两个不同的正根,必须且只须c01ff07b1b0c345957167db0c513f34d.png
即18a3de5588920691948184467fc74789.png, ∴m=7或9c00ce2397f35fa81e89e92de8e23add.png
∴当m=7或9c00ce2397f35fa81e89e92de8e23add.png时,函数f(x)与g(x)的图象有且只有两个不同交点。14分
6解(1)由已知,得h(x)=a79eaabef514fd8492a27c9786b7299f.png且x>0,则hˊ(x)=ax+2-afc48b56873694f3d43097841ecc3f4f.png=6edfa95856976a52b0359da85a2e1bc4.png,(2
∵函数h(x)存在单调递增区间, ∴hˊ(x)≥0有解, 即不等式ax2+2x-1≥0有x>0的解.3分)
①当a<0时, y=ax2+2x-1的图象为开口向下的抛物线, 要使ax2+2x-1≥0总有x>0的解, 则方程ax2+2x-1=0至少有一个不重复正根, 而方程ax2+2x-1=0总有两个不相等的根时, 则必定是两个不相等的正根. 故只需Δ=4+4a>0, 即a>-1. 即-1(5分)
②当a>0 时, y= ax2+2x-1的图象为开口向上的抛物线, ax2+2x-1≥0一定有x>0的解6分)
综上, a的取值范围是(-1, 0)∪(0, +∞) (7分)
(2)方程57b1617dcb9014301e4e19b782ab5a24.png即为0c29894957d0d1d076361403fc6a7e2b.png
等价于方程ax2+(1-2a)x-lnx=0 .(8分)设H(x)= ax2+(1-2a)x-lnx,
于是原方程在区间(c860120cc64f68975c91743c21c75409.png)内根的问题, 转化为函数H(x)在区间(c860120cc64f68975c91743c21c75409.png)内的零点问题. (9分)
Hˊ(x)=2ax+(1-2a)-afc48b56873694f3d43097841ecc3f4f.png=66575d767e49d605b7c98bc7fdca565a.png (10分)
当x∈(0, 1)时, Hˊ(x)<0, H(x)是减函数;当x∈(1, +∞)时, Hˊ(x)>0, H(x)是增函数;若H(x)在(c860120cc64f68975c91743c21c75409.png)内有且只有两个不相等的零点, 只须
f829bce6c55c6c549fabc2b925d27bce.png (13分)
解得78e4b579633165afa7d0d28dd98c1b03.png, 所以a的取值范围是(1, c71878d4d9a2d32469617c41e919c6f5.png) (14分)
7.解:(1)d44d09fc630fd0214dbf1b61cbc18582.png,cb4430878871ca48bd15c1c26b07c1f9.png,即09a8c47f58ca464aad31c6fcc8c660e1.png………3分
令631c111905392828ec7c4afd00cae53b.png,则dd65cb486a07febe3cf6251c8d10e6e5.png,70a63b41af42a77c78406206366beffe.png,
因此,数列eac391c0815f72031ce6673528b827f2.png是首项为c81e728d9d4c2f636f067f89cc14862c.png,公差为c4ca4238a0b923820dcc509a6f75849b.png的等差数列.599dfeb3e57d9521f4c799713eca723d.png,5分
acf8b848445eead069947d6f880906ac.png. ……6分
(2)(方法一)先证明当f6e696c1952cff44affab43c8beaec2f.png时,d4522f66824735fce1e9be28f99c0cd8.png.设d74cbd7db4c596804de4513cf9609644.png,则24528f0390076cd99d9ce3b9f8cb195e.png,
f2eabb677c74c2da92c37f96ef6bd9a2.png当3d3e00e0b84ad6b64a3461fe9092698a.png时,8baaf0ef5eac596ebf01a45909545fc1.png,
78da653a5a76108dafd2de8273a2d712.png在a9b60edfbc89649206e22a2cad65c588.png上是增函数,则当e3f7f32051ade969e770ab04b5a4853b.png时,451a1bfec3f4080a2ec73a9cdf560665.png,即f2a7f17bbe08a113051ec939751f677d.png.………8分
因此,当f6e696c1952cff44affab43c8beaec2f.png时,d4522f66824735fce1e9be28f99c0cd8.png,089caef908ed4cefc5ea316d90b1bd29.png, …………9分
当f6e696c1952cff44affab43c8beaec2f.png时,1a6d618c9af0d7d763d16ba31beb9538.png,3a158badbeb7dacca321692e0b0fb3b4.png. …………………10分
a636c3d4c936542c8f820a156a1926aa.png.12分
e007df12a4a7774d58707a9754816909.png.………14分
(方法二)数学归纳法证明
(1)2b0491af3ae1f0ec8fa2af767591a970.png,bb38a786a44f0e124d6327b6bd348c7c.png,95e029696a8e77db6f75665e6464c095.png当6d24e2bc97c5e4283dd8e34674afe7ea.png时,3487456075a3f91bea192f0640d5fee7.png成立;
caa5b67515e220da6f281c9290d6d92b.png,2df9bb773fe48ff98d106ae2d19bf936.png,又3b47d6e8b5adf60d006e353875c44082.png,f33f2e06592e1ad44ce11ecfb60b53d0.png,
95e029696a8e77db6f75665e6464c095.png当6d24e2bc97c5e4283dd8e34674afe7ea.png时,b303b6646f96f2efdd1c8b3240502608.png成立. ……………………8分
(2)设84397256ec1d228f288198d1e5ae5959.png时命题成立,即8114b4db56e3ff642d98b45f9d594707.png,e8595cdca4d6281989a24ff96a9078c0.png,
当b7eb27d0b2cabbf70084c60f4ce9fc8a.png时,9f1c01ba1f4f5bf7c4fe0eab3ef8f00a.png,
要证fc21f4246980e1e3076a2149d2583002.png, 即证edd8b8234d88104d67c845cf335bc788.png,化简,即证8c8586b6fb99a8815eeec4ea97e6222d.png.……9分
设ac8b39aa7ae392103363d4d970c7126a.png,则ac01c50f1e12b73bcabdf3fd8f512103.png,f2eabb677c74c2da92c37f96ef6bd9a2.png当887fb68a10cbd4369b27c90bee0334d8.png时,8baaf0ef5eac596ebf01a45909545fc1.png,
78da653a5a76108dafd2de8273a2d712.png在b921db311612fd3665c51872c7a83455.png上是增函数,则当e54e5ec6ebe0d78e64c134099fbf0aa4.png时,469c2be4f43cb1e6d1a31d7ef5baf433.png,即74934792d56436d9c088b0ccd6e92d44.png.
因此,不等式773a0ddc088d1852d8563802a03de880.png成立,即当b7eb27d0b2cabbf70084c60f4ce9fc8a.png时3487456075a3f91bea192f0640d5fee7.png成立. …………………11分
当b7eb27d0b2cabbf70084c60f4ce9fc8a.png时,ea42e009f875e9b1cbe866c02cd322de.png,
要证b6113a1d7b518e9debaa48e0e73acf8c.png, 即证41354958ec797e3d159e646cd94131a2.png,化简,即证8c8586b6fb99a8815eeec4ea97e6222d.png.
根据前面的证明,不等式04bb3d875c39a4dbe8f50692ffb7633a.png成立,则b7eb27d0b2cabbf70084c60f4ce9fc8a.png时b303b6646f96f2efdd1c8b3240502608.png成立.
由数学归纳法可知,当f6e696c1952cff44affab43c8beaec2f.png时,不等式3487456075a3f91bea192f0640d5fee7.png, b303b6646f96f2efdd1c8b3240502608.png成立.……………14分
8 (1)解: 函数的定义域为word/media/image631_1.png.
∴word/media/image632_1.png.
① 当word/media/image634_1.png, 即时, 得word/media/image636_1.png,则.
∴函数word/media/image638_1.png在word/media/image631_1.png上单调递增. 2分
② 当, 即word/media/image640_1.png时, 令 得word/media/image642_1.png,
解得.
(ⅰ) 若, 则word/media/image645_1.png. ∵, ∴word/media/image647_1.png,
∴函数在word/media/image631_1.png上单调递增. …… 4分
(ⅱ)若word/media/image649_1.png,则时, word/media/image651_1.png; 时, word/media/image653_1.png,∴函数在区间word/media/image654_1.png上单调递减, 在区间word/media/image655_1.png上单调递增.…… 6分
综上所述, 当时, 函数的单调递增区间为word/media/image631_1.png;当word/media/image658_1.png时, 函数的单调递减区间为word/media/image654_1.png, 单调递增区间为word/media/image655_1.png. … 8分
(2) 解: 由, 得, 化为word/media/image660_1.png.
令, 则word/media/image662_1.png.令, 得word/media/image664_1.png.
当word/media/image665_1.png时, ; 当word/media/image667_1.png时, .∴函数word/media/image669_1.png在区间上单调递增, 在区间word/media/image671_1.png上单调递减.∴当word/media/image664_1.png时, 函数word/media/image669_1.png取得最大值, 其值为. 10分
而函数word/media/image673_1.png,
当word/media/image664_1.png时, 函数word/media/image674_1.png取得最小值, 其值为. … 12分
∴ 当word/media/image676_1.png, 即时, 方程只有一个根. …… 14分
9.解:(I)76a61a449fa05412811d9ea8fdb1aec2.png依题意d3eeb4a4aff643ce8057e3ead12321a5.png,即5e8591557be080d943ef446c5ee1c208.png,71cf1db6bbe582e196b087c954face82.png.
∵上式恒成立,∴2dfc03408510a4cb71c6a9953a129532.png① 1分
又d537fd7d5c2d61c19ef38ba4ab6668e3.png,依题意8e1df6fc667eb8464246caa5e7aff176.png,即b1484a940287bc49fed6a0e8b580de1d.png,abd8bd27a401243903fab6ca165d79b0.png.
∵上式恒成立,∴1868f2c94579c06a0856edbf234ab85e.png② …2分由①②得83a88ab12cf3296e031df84985733d33.png. 3分
∴f32503d4e94f998d3b272cf774bba681.png …4分
(II)由(1)可知,方程b9eb8a00cf5929e6c8b701e5a0beb253.png,69bb34953aa51afb93df0f24b02077eb.png
设c2a93f23138df45511545f3bb0aebeb9.png,d4d02cebfb3f7dcb9d8ad44508f9585c.png
令5c1c0727f238086db0ed17d69d6ae0f0.png,并由35a3e49f0d55eb333b6533eedd3a1fa4.png得5737b18cb2902485a1833f243704e61e.png解知81ca628b26eb0aa0666b1a7fcb9b2544.png ……………5分
令47a1532e8f219b4aaf0410405d96d639.png由f04813dfa45cded1436e365a40b7efb2.png …6分
列表分析:
知ca8e608169b20a94570ac837e8ba0833.png在a255512f9d61a6777bd5a304235bd26d.png处有一个最小值0,7分当29ff0065ce199148aa43c103978b384e.png时,ca8e608169b20a94570ac837e8ba0833.png>0,∴a877f3c76508a0ccf6fbaaa64ea62d49.png在(0,+)上只有一个解.即当x>0时,方程31460a30855e87471cbac21bc26346b4.png有唯一解. ……………………8分
(III)设477b7c15b1573b942ffd9204927b7887.png, …………9分
08e0137fb2c934e91ac4a044edc9b99a.png在668c7b55a37300c330dcd565d9e076da.png为减函数f464db8a7a023eefdaae856677859d38.png 又fc2e8ec8dcf30bb9fbb8aca436bd661b.png …………11分
所以:ca20567d0793bcaacb3db79b5aaa7b23.png为所求范围. …………………………………………12分
(本小题主要考查函数的导数、最值、等比数列等基础知识,考查分析问题和解决问题的能力、以及创新意识)
10解(1)∵32faf6e347ae549fee91da66516daadd.png,∴ac01c50f1e12b73bcabdf3fd8f512103.png.令06605d0b94674429f3ab62bec2350d50.png,得e11729b0b65ecade3fc272548a3883fc.png.
∴当887fb68a10cbd4369b27c90bee0334d8.png时,82ba1ab4d8827e260ef0f540d1922661.png,当97fdf90850f660f05349f4ad145b62dc.png时,ec329a2e6e725e343e248a7a56d580fc.png. ∴函数32faf6e347ae549fee91da66516daadd.png在区间6aebe4fc50113bb0aeaa150f7d4a291d.png上单调递减,在区间1fd5a0acabffccf10f5498511c239d84.png上单调递增.∴当e11729b0b65ecade3fc272548a3883fc.png时,50bbd36e1fd2333108437a2ca378be62.png有最小值1.
(2)证明:由(1)知,对任意实数9dd4e461268c8034f5c8564e155c67a6.png均有a9154f3a1df105c6ca0560bfdd62e3b2.png,即9f8d5fc3ad51c6e4ed9a3ef3b113699e.png.
令d182b035631eae622cc9b991ae1d200b.png(794b585a0aade482c16feb5d694dc00e.png),则a3d6bfc14df2c81c80b0b866c1ed1075.png,
∴f07f8131de3e0de99870d10ec0a7481a.png. 即b993b328c66e9cb4d92dc86c7b7cd7cf.png.
∵222544147f142ab9eaebb7b39d654259.png∴2691ed1ada5453be84e2e833bb2f34b6.png.
∵8fbe0530f07b3a219324173aa69aaa10.png,
∴ 86d1f598a1d0a5249496d9b41d451c2c.png.
11.(1)解法1:∵f2faa96ea9b5cfba4e7533dd69832573.png,其定义域为397ee1de718075d8eae9dbc08bc05f7c.png,∴8788f91f31b6cf1046c07347ef4040aa.png.
∵a255512f9d61a6777bd5a304235bd26d.png是函数bf3cc25c2d752823ab942d0912424f81.png的极值点,∴7fa1f5d07efb51ae3367b02ab3bea7ad.png,即f0aa22f5c9cdf85d50a9548042a41a74.png.∵323c5f97105643bc61e288fe596194ca.png,∴a826e4b34f75cd5ed2a911fc03dca168.png.
经检验当a826e4b34f75cd5ed2a911fc03dca168.png时,a255512f9d61a6777bd5a304235bd26d.png是函数bf3cc25c2d752823ab942d0912424f81.png的极值点,∴a826e4b34f75cd5ed2a911fc03dca168.png.
解法2:∵f2faa96ea9b5cfba4e7533dd69832573.png,其定义域为1fd5a0acabffccf10f5498511c239d84.png,∴8788f91f31b6cf1046c07347ef4040aa.png.
令fa85a0b3d1c84fa81d3cc1215ee61d87.png,即e3b68dc8398c9a84cd6328ec3406f554.png,整理,得018d219f65a55f463fdd412a983656c5.png.
∵e11f29dfe3c04d9b95e4d5bc03aba02d.png,∴fa85a0b3d1c84fa81d3cc1215ee61d87.png的两个实根e9e457658a6a8b31abd7aa9f88277782.png(舍去),75896ab26d9add59f830c950b6342cc2.png,
当9dd4e461268c8034f5c8564e155c67a6.png变化时,bf3cc25c2d752823ab942d0912424f81.png,7a8feaa49bcc1a33d921b0441825af31.png的变化情况如下表:
依题意,3524a26b7421905cf65f1cb15bf5af6e.png,即a27c8fbc522bf83e400be62fc1b4b9ba.png,∵323c5f97105643bc61e288fe596194ca.png,∴a826e4b34f75cd5ed2a911fc03dca168.png.
(2)解:对任意的1dbfb2a9d0eedea2d8dec2d3a1126148.png都有6d5e9a16f9712e8df90179ff2e6dba8a.png≥467aee9c1bbe1f4dbe2e91677bfbb26e.png成立等价于对任意的1dbfb2a9d0eedea2d8dec2d3a1126148.png都有bcee40521d2b8cdc8e6152e8890c7482.png≥8e92b1895dcba5f34d1ee6150b6e7c65.png.
当9dd4e461268c8034f5c8564e155c67a6.png3659f7cbbee83f93790ff7f8c7480168.png[1,e1671797c52e15f763380b45e841ec32.png]时,918afbec5b8a095a33b4effff6a0457b.png.∴函数3bf584b53369cabaf036aa10d3a769b1.png在c7faa44689191029e81ed95c9a29e023.png上是增函数.
∴78be451489643f5f4eefc5048a5f089a.png.
∵9decf8d3ce96922abcb282c8c5c88740.png,且9839ba07a2e9b51a955417802051563c.png,323c5f97105643bc61e288fe596194ca.png.
①当81ab5a0b5746d911e1d8f16c92f80df1.png且9dd4e461268c8034f5c8564e155c67a6.png3659f7cbbee83f93790ff7f8c7480168.png[1,e1671797c52e15f763380b45e841ec32.png]时,2324c8340e756cbd2a47bcdf4aaf8621.png,∴函数12531d504bd85bf8e97f97d65c575911.png在[1,e1671797c52e15f763380b45e841ec32.png]上是增函数,∴d8f1d86f72174a0d297e03c6b735ba5c.png.由41c06c686e1e2b0696652b1b5960b54a.png≥e0497676d88d495347fd4b7083fb71e7.png,得0cc175b9c0f1b6a831c399e269772661.png≥0cf6aefe551e74b4de6f4301b16cf55b.png,
又81ab5a0b5746d911e1d8f16c92f80df1.png,∴0cc175b9c0f1b6a831c399e269772661.png不合题意.
②当1≤0cc175b9c0f1b6a831c399e269772661.png≤e1671797c52e15f763380b45e841ec32.png时,若1≤9dd4e461268c8034f5c8564e155c67a6.png<0cc175b9c0f1b6a831c399e269772661.png,则db565dc8e3443a290295c82ef9ff11b9.png,若0cc175b9c0f1b6a831c399e269772661.png<9dd4e461268c8034f5c8564e155c67a6.png≤e1671797c52e15f763380b45e841ec32.png,则2324c8340e756cbd2a47bcdf4aaf8621.png.∴函数12531d504bd85bf8e97f97d65c575911.png在fd3f5b5e0354ad8be398bbc85f5bf2c8.png上是减函数,在c181c508aa7e18cb7070c0bf5307357d.png上是增函数.∴52b52b65c638152ebde03f2009ebdda1.png.由e36314e624d2b2ca257e1f1ecb381f93.png≥e0497676d88d495347fd4b7083fb71e7.png,得0cc175b9c0f1b6a831c399e269772661.png≥50f2bb4b5023f8620dbe083801882629.png,又1≤0cc175b9c0f1b6a831c399e269772661.png≤e1671797c52e15f763380b45e841ec32.png,∴50f2bb4b5023f8620dbe083801882629.png≤0cc175b9c0f1b6a831c399e269772661.png≤e1671797c52e15f763380b45e841ec32.png.
③当83e1720e9e1bae4e6d06e4e5e6c89624.png且9dd4e461268c8034f5c8564e155c67a6.png3659f7cbbee83f93790ff7f8c7480168.png[1,e1671797c52e15f763380b45e841ec32.png]时,db565dc8e3443a290295c82ef9ff11b9.png,∴函数12531d504bd85bf8e97f97d65c575911.png在c7faa44689191029e81ed95c9a29e023.png上是减函数.∴f03239214d375ef96dbb007c7a12d254.png.由9e9bd2cd85bdaf13b58f328dfc0dfeab.png≥e0497676d88d495347fd4b7083fb71e7.png,得0cc175b9c0f1b6a831c399e269772661.png≥0cf6aefe551e74b4de6f4301b16cf55b.png,又83e1720e9e1bae4e6d06e4e5e6c89624.png,∴83e1720e9e1bae4e6d06e4e5e6c89624.png.
综上所述,0cc175b9c0f1b6a831c399e269772661.png的取值范围为d362aedf92dff4fbe50f9c9716154cbf.png.
12.解:(1)由题意,word/media/image819_1.png≥0在上恒成立,即word/media/image821_1.png…1分
∵θ∈(0,π),∴.故在上恒成立,………2分
只须word/media/image824_1.png,即,只有word/media/image826_1.png.结合θ∈(0,π),得word/media/image827_1.png…4分
(2)由(1),得word/media/image829_1.png..…5分
∵word/media/image831_1.png在其定义域内为单调函数,
∴或者word/media/image833_1.png在[1,+∞)恒成立.……6分
等价于word/media/image835_1.png,即word/media/image836_1.png,
而 word/media/image837_1.png,()max=1,∴. ………………………8分
word/media/image840_1.png等价于,即word/media/image842_1.png在[1,+∞)恒成立,
而∈(0,1],word/media/image844_1.png.
综上,m的取值范围是. ……………………………10分
(3)构造word/media/image846_1.png,word/media/image847_1.png.
当word/media/image848_1.png时,word/media/image849_1.png,,word/media/image851_1.png,所以在[1,e]上不存在一个word/media/image337_1.png,使得成立. ……………………………12分
当时,word/media/image853_1.png.………14分
因为word/media/image849_1.png,所以,word/media/image855_1.png,所以在word/media/image849_1.png恒成立.
故在word/media/image858_1.png上单调递增,,只要,
解得. 故word/media/image862_1.png的取值范围是.…………………………16分
高三数学复习-合理构造函数解导数问题以及构造函数法