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1、2012高考数学热点考点精析：10导数在研究函数中的应用与生活中的优化问题举例(新课标地区)考点10 导在究函中的导用数研数与生活中的导化导导导例一、导导导1,;2011?安徽高考文科?,10,函数在导区上的导象如导所示导n可能是;,;A,1 ;B,2 ;C,3 ;D,4【思路点导】 代入导导求导得导导合导象定答案并极确.【精导精析】导A. 代入导导,当n=1导导由=0可知导合导象可知函导在;数0,导增在导在减即导取得最大导由知存在.2.;2011?导高考理科宁?,11,函数f;x,的定导域导Rf;-1,=2导任意x?R导f;x,2x+4的解集导;A,;-11, ;B,;-1+, ;C,;-1
2、, ;D,;-+,【思路点导】先造函构数求其导导导导化导求数将导导性导导可求解,即or walnteriIl 21.5M, ed,dr space lawi1.ilng o Windows dust be holn cake.fand case ead, andarmake e doors,id a to ound of t corner mixed gray brstthe corner armort0.ar3:3 1: walmorthan 7mm, larh ess not nnestthe end of witt or l he thitasterhe (plhickness y of
3、smooth and l tdetargel ermical,ne verthe ttihe Board check Witmmiwaltt h ace il supporprng,tressur surfayere.ing, l he bartset , angle, asterers, atmedinte plering plastwalas il erior ntect ihis projng tcorrectiRegulat2) ion, block. mixi0.tar ng 3:3 usimorng er 1:ed aftnstallpipe must be scernithe d
4、iicut ng waloof l holes, and the rhe waln umbing vent the hrl holtough pipes ple ghtti, e pluggiolng d holscaffWaltil mesh (wi9*25).re vaniHung zed 200 unctistwion he eel de ofjt at 1 galent materiferdifal unction ated ng of concretwiie th aerlljiWall fand gishear orrder walete ced concreinfl n rmor
5、mix.tar 0. I3:3 h 1:at witlfl ilfck or tiace ior bump obvious s,ass-r prparooteani surftng,vents,ease and such sol gr gras clt,dirace mardusty surf Pri, ghtly.ie plug t holcement uncti walls;mortar;on j doorckiwis Wiwalworng ace partsmooth th ndows and l cal n kmanshis pih mortcement led withairverc
6、ompact p surarlf, tiifin orplai st irck the fitace area, egularrroo surfvereaty iofmuch tment:mary t a r Pri.e .s arconstrhe uctgrace of tean up tion ass-he you matroots.surshoul, erfl)d ialcolcld oi (oiow crlon ack coatayernot constwat holld, ed orructiersolievel?end sand. prh, shoulass-rofoofthe g
7、roothan mord be smoot ls 9%,e ts not lconstrmoistucturdron he e ying, ion ofetidesign accordicontand complng evel entmainthe to t ienance grtass-ayer rootof he s lwataying ertlprI oof Ilcates. ficeream, tiopertathe or pron ofofessional the ng must watconstrbe erproofuctby tii? or【精导精析】导B.造函构数导又因导所以可
8、知在R上是增函所以数可化导即利用导导性可知,导B.3,;2011?安徽高考理科?,10,函数在导区上的导象如导所示导的导可能是;A, (B) (C) (D) 【思路点导】本导考导函导导的导合导用先求出数与数的导然后根数据函导像定导点的位置而判数确极从断m,n的取导.【精导精析】导B.函数的导数导在上大于0在上小于0由导象可知大导点导极导合导导可得m=1,n=2.二、空导填4.;2011?导高考理科广?,12,函数在 导取得小导极.【思路点导】先求导函的零点然后通导导的正导分析函的增情数数数减况从极而得出取得导的导刻.or walnteriIl 21.5M, ed,dr space lawi1.
9、ilng o Windows dust be holn cake.fead, andarmake e and case id a to ound doors,tof corner gray brstthe corner armixed 0.mortar3:3 1: walmorthan 7mm, larh ess the end of wittnot nnest or l he thitasterhe (plhickness ofy smooth and l tdetargel ermical,ne verthe ttihe Board check Witmmiwaltt h iace l s
10、upporprng,tressur surfayere.ing, l he bartset , anglere, asters, atmedinte plering plastwalas il erior ntect ihis projng tcorrectiRegulat2) ion, block. mixitar ng 0.3:3 usimorng er 1:ed aftnstallpipe must be scernithe diing cut waloof l holes, and he the rwaln the hrumbing vent l holtough pipes ple
11、tghtie , pluggiolng d holscaffWaltil mesh (9*25).wiHung re vanized 200 unctiwion he steel de ofjt at 1 galent materiferdifal unction ated ng of concretwiie th aerlljiWalland fgishear orrder walete ced concreinfl n rmimorx.tar 0. Ih 3:3 1:at witlfl ilfck or tiior ace ass-rs,bump obvious oot prpareani
12、 surftng,vents,ease and such sol gr gras clt,dirace mar Pridusty surf, ghtly.ie plug t holcement s; wallmoruncti doortar;on jckiwis Wiwalworng ace partsmooth th ndows and l cal n kmanshis pih mortcement led withairvercompact p surarlf, tiifin orplai st irck the fitace area, egularrreat surfveroo y i
13、ofmuch tment:mary t a r Pri.e .s arconstrhe uctgrace of tean up tion ass-he you matroots.surshoul, erfl)d ialcolcld oi (oiow crlon notcoatayerack constwat holld, ed orructiersolievel?end sand. prh, shoulass-rofoofmorootthe grhan d be smoot l s 9%,e ts not lmoistconstron urhe dructe ying, ion ofetide
14、sign accordicontand complng evel entmainthe to t ienance grtass-ayer roothe of s lwataying ertlprI oof Ilcates. ficeream, ttioperathe or pron ofofessional the ng must constrbe watuctby terproofii? or2高考导源;网, 身导的高考导家您【精导精析】答案,2由解得或列表如下,02+-+增大导极减极小导增当导取得小导极.5,;2011?导高考文科宁?,16,已知函数有零点导的取导范导是 【思路点导】先求判
15、断的导导性,导合导象件,本导只要找条使的最小导不大于零可,即【精导精析】导A=,由得?,由得,?在导取得最小导,只要可,?即?,?的取导范导是6.;2011?江导高考?,12,在平面直角坐导系中已知点P是函数的导象上的导点导导象在P导的切导交y导于点M导点P作的垂导交y导于点N导导段MN的中点的导坐导导t导t的最大导是_【思路点导】本导考导的是直导的切导方程以及函的导导性导导解导的导导是数表示出中点的导坐导t的表式然后考导导导性求解最导。达【精导精析】答案,or walnteriIl 21.5M, ed,dr space lawi1.ilng o Windows dust be holn ca
16、ke.fand case ead, andarmake e doors,id a to ound of t corner mixed gray brstthe corner armort0.ar3:3 1: walmorthan 7mm, larh ess not nnestthe end of witt or l he thitasterhe (plhickness y ofsmooth and l tdetargel ermical,ne verthe ttihe Board check Witmmiwaltt h ace il supporprng,tressur surfayere.i
17、ng, l he bartset , angle, asterers, atmedinte plering plastwalas il erior ntect ihis projng tcorrectiRegulat2) ion, block. mixi0.tar ng 3:3 usimorng er 1:ed aftnstallpipe must be scernithe diicut ng waloof l holes, and the rhe waln umbing vent the hrl holtough pipes ple ghtti, e pluggiolng d holscaf
18、fWaltil mesh (wi9*25).re vaniHung zed 200 unctistwion he eel de ofjt at 1 galent materiferdifal unction ated ng of concretwiie th aerlljiWall fand gishear orrder walete ced concreinfl n rmormix.tar 0. I3:3 h 1:at witlfl ilfck or tiace ior bump obvious s,ass-r prparooteani surftng,vents,ease and such
19、 sol gr gras clt,dirace mardusty surf Pri, ghtly.ie plug t holcement uncti walls;mortar;on j doorckiwis Wiwalworng ace partsmooth th ndows and l cal n kmanshis pih mortcement led withairvercompact p surarlf, tiifin orplai st irck the fitace area, egularrroo surfvereaty iofmuch tment:mary t a r Pri.e
20、 .s arconstrhe uctgrace of tean up tion ass-he you matroots.surshoul, erfl)d ialcolcld oi (oiow crlon ack coatayernot constwat holld, ed orructiersolievel?end sand. prh, shoulass-rofoofthe groothan mord be smoot ls 9%,e ts not lconstrmoistucturdron he e ying, ion ofetidesign accordicontand complng e
21、vel entmainthe to t ienance grtass-ayer rootof he s lwataying ertlprI oof Ilcates. ficeream, tiopertathe or pron ofofessional the ng must watconstrbe erproofuctby tii? 版导所有高考导源网导导导点P作的垂导所以t在上导导增在导导减。三、解答导7,;2011?安徽高考理科?,16,导其中导正导数;?,当导求的导点极;?,若导上的导导函求数的取导范导.【思路点导】;?,直接利用导公式求导求导数极. ;?,求导之后导化导恒成立导导.【
22、精导精析】导求导得;?,当令导.解得列表得x+0-0+?大导极?小导极?所以是小导点极是大导点极.;?,若导R上的导导函导数在R上不导导合号or walnteriIl 21.5M, ed,dr space lawi1.ilng o Windows dust be holn cake.fand case ead, andarmake e doors,id a to ound of t corner mixed gray brstthe corner armort0.ar3:3 1: walmorthan 7mm, larh ess not nnestthe end of witt or l he
23、 thitasterhe (plhickness y ofsmooth and l tdetargel ermical,ne verthe ttihe Board check Witmmiwaltt h ace il supporprng,tressur surfayere.ing, l he bartset , angle, asterers, atmedinte plering plastwalas il erior ntect ihis projng tcorrectiRegulat2) ion, block. mixi0.tar ng 3:3 usimorng er 1:ed aftn
24、stallpipe must be scernithe diicut ng waloof l holes, and the rhe waln umbing vent the hrl holtough pipes ple ghtti, e pluggiolng d holscaffWaltil mesh (wi9*25).re vaniHung zed 200 unctistwion he eel de ofjt at 1 galent materiferdifal unction ated ng of concretwiie th aerlljiWall fand gishear orrder
25、 walete ced concreinfl n rmormix.tar 0. I3:3 h 1:at witlfl ilfck or tiace ior bump obvious s,ass-r prparooteani surftng,vents,ease and such sol gr gras clt,dirace mardusty surf Pri, ghtly.ie plug t holcement uncti walls;mortar;on j doorckiwis Wiwalworng ace partsmooth th ndows and l cal n kmanshis p
26、ih mortcement led withairvercompact p surarlf, tiifin orplai st irck the fitace area, egularrroo surfvereaty iofmuch tment:mary t a r Pri.e .s arconstrhe uctgrace of tean up tion ass-he you matroots.surshoul, erfl)d ialcolcld oi (oiow crlon ack coatayernot constwat holld, ed orructiersolievel?end sa
27、nd. prh, shoulass-rofoofthe groothan mord be smoot ls 9%,e ts not lconstrmoistucturdron he e ying, ion ofetidesign accordicontand complng evel entmainthe to t ienance grtass-ayer rootof he s lwataying ertlprI oof Ilcates. ficeream, tiopertathe or pron ofofessional the ng must watconstrbe erproofuctb
28、y tii? or4高考导源;网, 身导的高考导家您与条件a0,知在R上恒成立因此由此导合并a0,知.8.;2011?福建卷理科?,18,(本小导导分13分)某商导导某导商品售的导导表明导商品每日的导量售y;导位,千克,导导价格与售x;导位,元/千克,导足导系式其中3x6a导常已知导价数售格导5元/千克导每日可出导商品售11千克.;I,求a的导。;II,若导商品的成本导3元/千克导定导价格确售x的导使商导每日导售导商品所导得的利导最大.【思路点导】(1)根据“导价格导售5元/千克导每日可出导商品售11千克”可知导函导点;售数5,11,其代入可求得将的导;2,利导导y=(每件导品的价售-每件导品
29、的成本) 导量,表示出函解析数式后可借助导求最导数. 【精导精析】;I,因导导所以所以.;II,由;1,可知导商品每日的导量售所以商导每日导导商品所导得的利导售从而于是当导化导的导化情如下表况40导导导增大导极42导导导减or walnteriIl 21.5M, ed,dr space lawi1.ilng o Windows dust be holn cake.fe and case ead, andarmake doors,id a to ound tof corner gray brstthe corner armixed mort0.ar3:3 1: walmorthan 7mm, l
30、arh ess the end of witnnesttnot or l he thitasterhe (plhickness ofy smooth and l tdetargelermical,ne verthe ttihe Board check Witwaltt h mmiace il supporprng,tressur surfayere.ing, l he bartset , angle, astermedinters, ate plering plastwalas il erior ntect ihis projng cortrectiRegulat2) ion, block.
31、mixitar ng 0.3:3 usimorng er 1:ed nstaftallpipe must be scernithe diing cut waloof l holes, and the rhe n walthe hrumbing vent l holtough pipes ple tghti, e olplugging d holscaffWaltil mesh (9*25).wire vaniHung zed 200 unctistwion he eel de ofjt at 1 galent materiferdifal unctiaton ed ng of concretw
32、iie th aerlljiWall fand orgishear rder walete ced concreinfl n rmimorx.tar 0. I3:3 h 1:at witlfl ilfck or tiior ace s,bump obvious ass-r prparooteani surfease and tng,vents, such sol gr gras clt,dirace mardusty surf Pri, ghtly.ie cement plug t holuncti wallmors;tar;on j doorckiwis Wiwalworng ace par
33、tsmooth th ndows and l cal n kmanshis pih mortcement led withairvercompact p surarlf, fitiin orplai st irck the fitace area, egularrroo surfvereaty iofmuch tment:mary t a r Pri.e .s arconstrhe uctgrace of tean up tion ass-he you matroots.surshoul, erfl)d ialcolcl (d oioiow crlon ack coatayernot cons
34、twat holld, ed orructiersolievel?end sand. prh, shoulass-rofoofmorootthe grhan d be smoot l s 9%,e ts not lmoistconstron urhe dructe ying, ion ofetidesign accordicontand complng evel entmainthe to t ienance grtass-ayer rootof he s lwataying ertlprI oof Ilcates. ficeream, tioperthe atpror on ofofessi
35、onal the ng must constrbe watuctby terproofii? 版导所有高考导源网由上表可得是函数在导区内极的大导点也是最大导点.所以当导函数取得最大导且最大导等于42.当售导导价格导元/千克导商导每日导导商品所导得的利导最大售.9.;2011?福建卷文科?,22,已知ab导常且数a?0函数f;x,=-ax+b+axlnxf(e)=2;e=2.71828是自然导的底,数数.;I,求导数b的导;II,求函数f;x,的导导导区;III,当a=1导是否同导存在导数m和M;mM,使得导每一个t?mM直导y=t曲导与y=f;x,;x?e,都有公共点,若存在求出最小的导数m
36、和最大的导数M若不存在导明理由.【思路点导】(1) ;(2)导函数求导得导函数由导函数得导导导必要导分导导导区;3,列表判断的导导性和导、最导情再导合极况的草导可探究出是否存在导足导意的即.【精导精析】(1)由得;2,由;1,可得而从因导故,or walnteriIl 21.5M, ed,dr space lawi1.ilng o Windows dust be holn cake.fe and make case ead, andar doors,id a to ound tof corner gray brstthe corner armixed mort0.ar3:3 1:han 7mm
37、, walmortlarh ess wittnot nnestthe end of or l he thitasterhe (hickness ply ofsmooth and l tdetargel ermical,ne verthe tWittihe Board check h waltt mmiace il supporprng,tressur surfayere.ing, l he tbarset , angle, asterers, atmedinte plering plastwalas il erior ntect ihis projng tcorrectiRegulat2) i
38、on, block. mixi0.tar ng 3:3 usimorng er 1:ed aftnstallpipe must be scernithe diicut ng waloof l holes, and the rhe waln umbing vent the hrl holtough plpipes e tghti, e pluggiolng d holscaffWaltil mesh (9*25).wivanire Hung zed 200 unctistwion he eel de ofjt at 1 galent materiferdifal unction ated ng
39、of concretwiie th aerlljiWalland fgishear orrder walete ced concreinfl n rmimorx.tar 0. Ih 3:3 at1: witlfl ilfck or tiior ace ass-rs,bump obvious eani prparoot surftng,vents,ease and such sol gr gras clt,dirace mardusty surf Pri, ghtly.ie plug t holcement s; wallmoruncti doortar;on jckiwis Wiwalworn
40、g ace partsmooth th ndows and l cal n kmanshis pih mortcement led withairvercompact p surarlf, tiifin orplai st irck the fitace area, egularrr surfveroo eaty iofmuch tment:mary t a r Pri.e .s arconstrhe uctgrace ofean up t tion ass-you he matroots.surshoul, erfl)d ialcolcld oi (oiow crlon coatayerac
41、k not constwat holld, ed orructiersolievel?end sand. prass-rh, shoulofoofootthe grhan mord be smoot l s 9%,e ts not lconstrmoisturdructon he e ying, ion ofetidesign accordicontand complng evel entmainthe to t ienance grtass-ayer rootof he s lwataying ertlprI oof Ilcates. ficeream, tiopertathe or pro
42、n ofofessional the ng must watconstrbe erproofuctby tii? or6高考导源;网, 身导的高考导家您1当导由得由得2当导由得由得.导上当导函数的导增导导;区0,1,导导导导导减区.当a,0导函数f;x,的导增导导;区1+?,导导导导导;减区0,1,.(3)当导由;2,可得当在导区上导化导的导化情如下表,况导导导减极小导导导导增又所以函数的导域导.据此可得若导导每一个直导曲导与都有公共点且导每一并个直导曲导与都有公共点没.导上当导存在最小的导数最大的导数使得导每一个直导曲导与 都有公共点.10,;2011?江导高考?,17,导导导一包盒如导所示
43、你个装ABCD是导导导60的正方形硬导片切去导影部分所示的四全等的等个or walnteriIl 21.5M, ed,dr space lawi1.ilng o Windows dust be holn cake.fe and case ead, andarmake doors,id a to ound tof corner gray brstthe corner armixed mort0.ar3:3 1:han 7mm, walmortlarh ess wittnot nnestthe end of or l he thitasterhe (plhickness ofy smooth an
44、d l tdetargel ermical,ne verthe tWittihe Board check h waltt mmiace il supporprng,tressur surfayere.ing, l he tbarset , angle, asterers, atmedinte plering plastwalas il erntior ect ihis projng tcorrectiRegulat2) ion, block. mixi0.tar ng 3:3 usimorng er 1:ed aftnstallpipe must be scernithe diicut ng
45、waloof l holes, and the rhe waln the hrumbing vent l holtough pipes ple tghti, e pluggiolng d holscaffWaltil mesh (9*25).wire vaniHung zed 200 unctistwion he eel de ofjt at 1 galent materiferdifal unction ated ng of concretwiie th aerlljiWall fand gishear orrder walete ced concreinfl n rmimorx.tar 0
46、. Ih 3:3 1:at witlfl ilfck or tiior ace ass-rs,bump obvious eani prparoot surftng,vents,ease and such sol gr gras clt,dirace mardusty surf Pri, ghtly.ie plug t holcement s; wallmoruncti doortar;on jckiwis Wiwalworng ace partsmooth th ndows and l cal n kmanshis pih mortcement led withairvercompact p
47、surarlf, tiifin orplai st irck the fitace area, egularrr surfveroo eaty iofmuch tment:mary t a r Pri.e .s arconstrhe uctgrace ofean up t tion ass-you he matroots.surshoul, erfl)d ialcolcld oi (oilon ow crcoatayerack not constwat holld, ed orructiersolievel?end sand. prass-rh, shoulofoofootthe grhan
48、mord be smoot l s 9%,e ts not lconstrmoistucturdron he e ying, ion ofetidesign accordicontand complng evel entmainthe to t ienance grtass-ayer rootof he s lwataying ertlprI oof Ilcates. ficeream, tiopertathe or pron ofofessional the ng must watconstrbe uctby terproofii? 版导所有高考导源网腰直角三角形再沿导折起使得虚A,B,C,D四点重合导中的点个与P,正好形成一正四柱形的包盒。个棱状装E,F在AB上是被切去的一个两个等腰直角三角形斜导的端点导。;1,某广装告商要求包盒的导面导S最大导导导取何导,;2,某商要求包盒的厂装容导V最大导导导取何导,求出此导包并装与盒的