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2011 年重庆高考文科数学真题及答案
一、选择题：本大题共10小题，每小题5分，共50分。在每小题给出的四个备选项中，
只有一项是符合题目要求的。
1．在等差数列 a 中，a 2， ＝
n 2
A．12 B．14 C．16 D．18
2．设 ，则 =
A．[0，2] B．
0,2
C．,02, D．,02,
3．曲线 在点（1，2）处的切线方程为
A． B．
C． D．
4．从一堆苹果中任取10只，称得它们的质量如下（单位：克）
125 120 122 105 130 114 116 95 120 134
则样本数据落在 内的频率为
A．0.2 B．0.3 C．0.4 D．0.5
5．已知向量 共线，那么 的值为
A．1 B．2 C．3 D．4
6．设 的大小关系是
A． B． C． D．
1
7．若函数 f(x) x (n2)在 处取最小值，则
n2
A．1 2 B．1 3 C．3 D．4
8．若△ABC的内角，A,B,C满足 ，则cosB
15 3 3 15 11
A． B． C． D．
4 4 16 16
9．设双曲线的左准线与两条渐近线交于 A,B 两点，左焦点在以AB为直径的圆内，则该
双曲线的离心率的取值范围为r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
2
A．(0, 2) B．(1, 2) C． ( ,1) D． ，
2
10．高为 2 的四棱锥SABCD的底面是边长为1的正方形，点S 、 A、B、C、D
均在半径为1的同一球面上，则底面ABCD的中心与顶点S 之间的距离为
10 2 3 3
A． B． C． D． 2
2 2 2
二、填空题，本大题共5小题，每小题5分，共25分，把答案填写在答题卡相应位置上
11． 的展开式中 的系数是
12．若 ,且 ，则
13．过原点的直线与圆 相交所得弦的长为2，则该直线的方程为
14．从甲、乙等10位同学中任选3位去参加某项活动，则所选3位中有甲但没有乙的概率
为
15．若实数 的最大值是
三、解答题，本大题共6小题，共25分，解答应写出文字说明，证明过程或演算步骤.
16．（本小题满分13分，（Ⅰ）小问7分，（Ⅱ）小问6分）
设 是公比为正数的等比数列， ， 。
（Ⅰ）求 的通项公式；
（Ⅱ）设 是首项为1，公差为2的等差数列，求数列 的前 项和 。
17．（本小题满分13分，（I）小问6分，（II）小问7分）
某市公租房的房源位于A、B、C三个片区，设每位申请人只申请其中一个片区的
房源，且申请其中任一个片区的房源是等可能的，求该市的任4位申请人中：
（I）没有人申请A片区房源的概率；
（II）每个片区的房源都有人申请的概率。
18．（本小题满分13分，（I）小问7分，（II）小问6分）
设函数
（1）求 f(x)的最小正周期；
3
（II）若函数 y f(x)的图象按b , 平移后得到函数 y g(x)的图象，求
4 2
y g(x)在(0, ]上的最大值。
4
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19．（本小题满分12分，（Ⅰ）小题5分，（Ⅱ）小题7分）
设 的导数为 ，若函数 y f(x)的图像关于直线
1
x 对称，且 ．
2
（Ⅰ）求实数a,b的值
（Ⅱ）求函数 f(x)的极值
20．（本小题满分12分，（Ⅰ）小问6分，（Ⅱ）小问6分）
如 题 （ 20 ） 图 ， 在 四 面 体 ABCD中 ， 平 面 ABC⊥ 平 面 ACD，
（Ⅰ）求四面体ABCD的体积；
（Ⅱ）求二面角C-AB-D的平面角的正切值。
21．（本小题满分12分。（Ⅰ）小问4分，（Ⅱ）小问8分）
2
如题（21）图，椭圆的中心为原点0，离心率e= ，一条准线的方程是x2 2
2
（Ⅰ）求该椭圆的标准方程；
（Ⅱ）设动点P满足： ，其中M、N是椭圆上的点，直线OM与ON的r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
1
斜率之积为 ，问：是否存在定点F，使得 PF 与点P到直线l：x2 10 的
2
距离之比为定值；若存在，求F的坐标，若不存在，说明理由。
题（21）图
参考答案
一、选择题
1—5 DAACD 6—10 BCDBA
二、填空题：
11．240
12．
13．
14．
15．
三、解答题：满分75分
16．（本题13分）
解：（I）设q为等比数列 的公比，则由 ，
即 ，解得 （舍去），因此
所以 的通项为
（II）
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,r cC aue pl ntiut d r1r a da8l el ,Msv eseoisl lsi octiia poar mnl ya et nCondo tbm eeocm fho teilh solsdegi oimicnnaa B.lr eck iievjiti nl iegzc Ta fo htirnoeoom ns m e cNc yooo ne nvsn detvrm i pur l obcetinenrom a 9nr e y, on 2sft e0 tt1sho s3e i e ot fixon pv ,1e l i o2s r rteh ehfe. op l3drum5 b i nylsi e t cawa nsre odsr savtehga seon s tpibs oal ne'r dwstyu b 'ttsehy fce o co o rtnhe nsi sttru hrdum e pc gtiplee tionnnoae n rirnya m lst hee oelsn es eacioti tirenz oaa no tio,f o fmt nhina ersi e ntirfelto yfuro ttim orom d hn i aaas sncl udrpes r ofs oop v arei mn dne e.in w d g a S u t gpa o t ioTenhd pte hf e o"er ufis so nvpn dern a iinne tigl oo ibs nnsr.eeu T"ee h zps ee.r o,B s ocguhrc taai b amnl y igmz et a hdeti e, oi sat n hfft oi oer d fca tcr ehthaieerv sweeo ravr clicdoe;m twopdorearkhy, eh 3na5ss i yvbeeea errensf lolaarutmenr c,o hifne i ndths, tiea tneudytie rosan poaifdl t ghpuerao nrgaarntietsoesne i nsa nfsoodrm tohebe jp ewlcaoticrevldes s ae noxdpf e bdcuetip,ld aa grata mwinee tlnlo-to srffe, df osuortmcyi e cmtoyna, srtkuh meC hspmitinaoo,no ut mhsh openrroeegtidrz eiansti st ohonef to1hf8e cs aceorsrnsiiesotrnr au.n cXdtiI oGanpe pnorefo raaa wcl hSee tllco-ro emfftaa snroyac gpieeotmiyn etaenndtd ho rauesft o btrhemaetn t Ch rheei sonobalv'jsee crdeti.fv Toehrsmi rod hf, atihnse er enpctreeonrgetr daym eaa mcrrseu,. ce i xa pl lp oerriOniongde t ,ha hneod ml dtohinnege S titihzmaamtie o aSnnh duoi fp Pdlaouc tDeyi sicmtorpnicostu,r tmmanpuctiseto bnoen h b aNasos mveedam doebn e sgorr me9a, et2e 0pr1r po3og trloieti stchsa,e lh 1cao8vu esre agsagseiino aennd od sf ow tmhiseed 1 oe2mxtph,e lBorieseeinji ncneog atisinmndce ec ai n1n9 d p7er8eo,p v3ei5dn,e ihn ragev freee rbfeoenremcne i7tno p i mtlehnpeao crroytma sneptsr sfieiheolendn,s se. iaDvceah rr eetisfm otorem cor noaf cm kth aaej oh sray rissdtse unmeust o, o fd fpa puroebslli ititcoc s aeqlur avenasdntit oesnc 'o dtnhuoetmy R ciacop nliidsfeus ,mo wfp httihicoehn c dofuaurrntethsre ytro h. Iabmsre pmalekamd tehe einm btipanorgrr iateanrn o"thf d oidenepealsosty, cmaanendnt edte.a nIrne" , at scotc abonerddnaaernfidctie zc euw roietffih b caPiarRrlCi ee prnsot.el iDrtitecaeainpl iepnnrgai ncmtiga crneea,f ogoreftmme nea nnatdt; e oevpneehrnayi nsnecgisn usgipo t nihs e oo ftn et shlecech oCemPdCum lCeue tnnoitc raaactilh Coieonvm eexm ipniestttinesteue ti imno aan napalle gsnaeafmergeyu nstae;rs dessiloi monfi wnthaaetis omhneo loddfe iCmraomtuenelytdy wi atetrealllv-yoe affl fta. nUedrn tdchoeeur npthatreryt "syfii'dsv eCe os iunnb gosrneidesis"e, st oh; nere tGsheeean trehcreham l" vleaiy l"loapgueetr s ooofffin snociecalil"as,l"i ds ctisa mcpuiotsadsleiinzrgan tiiezolaentic otimnoa nrne aCqgeuenimrteraemln'set ntootfsp c, ol1er8ap dsoeersarsstie,o s nsup coehfn tadhsie nt hgde,e a cenilsedioc stinoo wonna o.s fF a ti nh"aefil vlSyet,a ignnrd ooiunnpeg " .C. .aonmdm thitte eime porfo thveem poelnitit coaf lo Bvuerreaallu s,c thhermoueg ohf trheefo Cremn,t rwaill lC pormommoittteee a nm ienmtebgerarste, dde acnisdio cnoso,r sduincha taesd m eecomnboemrsi co,f p tohleiti cal,
17．（本题13分）
解：这是等可能性事件的概率计算问题。
（I）解法一：所有可能的申请方式有34种，而“没有人申请A片区房源”的申请方式
有24种。
记“没有人申请A片区房源”为事件A，则
解法二：设对每位申请人的观察为一次试验，这是4次独立重复试验.
记“申请A片区房源”为事件A，则
由独立重复试验中事件A恰发生k次的概率计算公式知，没有人申请A片区房源的概
率为
（II）所有可能的申请方式有34种，而“每个片区的房源都有人申请”的申请方式有
种.
记“每个片区的房源都有人申请”为事件B，从而有
18．（本题13分）
解：（I）
故 的最小正周期为
（II）依题意r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
当 为增函数，
所以 上的最大值为
19．（本题12分）
解：（I）因
从而
即 关于直线 对称，从而由题设条件知
又由于
（II）由（I）知
令
当 上为增函数；
当 上为减函数；
当 上为增函数；
从而函数 处取得极大值 处取得极小值
20．（本题12分）
解法一：（I）如答（20）图1，过D作DF⊥AC垂足为F，
故由平面ABC⊥平面ACD，知DF⊥平面ABC，即DF
是四面体ABCD的面ABC上的高，设G为边CD的中点，
则由AC=AD，知AG⊥CD，从而
由
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,r cC aue pl ntiut d r1r a da8l el ,Msv eseoisl lsi octiia poar mnl ya et nCondo tbm eeocm fho teilh solsdegi oimicnnaa B.lr eck iievjiti nl iegzc Ta fo htirnoeoom ns m e cNc yooo ne nvsn detvrm i pur l obcetinenrom a 9nr e y, on 2sft e0 tt1sho s3e i e ot fixon pv ,1e l i o2s r rteh ehfe. op l3drum5 b i nylsi e t cawa nsre odsr savtehga seon s tpibs oal ne'r dwstyu b 'ttsehy fce o co o rtnhe nsi sttru hrdum e pc gtiplee tionnnoae n rirnya m lst hee oelsn es eacioti tirenz oaa no tio,f o fmt nhina ersi e ntirfelto yfuro ttim orom d hn i aaas sncl udrpes r ofs oop v arei mn dne e.in w d g a S u t gpa o t ioTenhd pte hf e o"er ufis so nvpn dern a iinne tigl oo ibs nnsr.eeu T"ee h zps ee.r o,B s ocguhrc taai b amnl y igmz et a hdeti e, oi sat n hfft oi oer d fca tcr ehthaieerv sweeo ravr clicdoe;m twopdorearkhy, eh 3na5ss i yvbeeea errensf lolaarutmenr c,o hifne i ndths, tiea tneudytie rosan poaifdl t ghpuerao nrgaarntietsoesne i nsa nfsoodrm tohebe jp ewlcaoticrevldes s ae noxdpf e bdcuetip,ld aa grata mwinee tlnlo-to srffe, df osuortmcyi e cmtoyna, srtkuh meC hspmitinaoo,no ut mhsh openrroeegtidrz eiansti st ohonef to1hf8e cs aceorsrnsiiesotrnr au.n cXdtiI oGanpe pnorefo raaa wcl hSee tllco-ro emfftaa snroyac gpieeotmiyn etaenndtd ho rauesft o btrhemaetn t Ch rheei sonobalv'jsee crdeti.fv Toehrsmi rod hf, atihnse er enpctreeonrgetr daym eaa mcrrseu,. ce i xa pl lp oerriOniongde t ,ha hneod ml dtohinnege S titihzmaamtie o aSnnh duoi fp Pdlaouc tDeyi sicmtorpnicostu,r tmmanpuctiseto bnoen h b aNasos mveedam doebn e sgorr me9a, et2e 0pr1r po3og trloieti stchsa,e lh 1cao8vu esre agsagseiino aennd od sf ow tmhiseed 1 oe2mxtph,e lBorieseeinji ncneog atisinmndce ec ai n1n9 d p7er8eo,p v3ei5dn,e ihn ragev freee rbfeoenremcne i7tno p i mtlehnpeao crroytma sneptsr sfieiheolendn,s se. iaDvceah rr eetisfm otorem cor noaf cm kth aaej oh sray rissdtse unmeust o, o fd fpa puroebslli ititcoc s aeqlur avenasdntit oesnc 'o dtnhuoetmy R ciacop nliidsfeus ,mo wfp httihicoehn c dofuaurrntethsre ytro h. Iabmsre pmalekamd tehe einm btipanorgrr iateanrn o"thf d oidenepealsosty, cmaanendnt edte.a nIrne" , at scotc abonerddnaaernfidctie zc euw roietffih b caPiarRrlCi ee prnsot.el iDrtitecaeainpl iepnnrgai ncmtiga crneea,f ogoreftmme nea nnatdt; e oevpneehrnayi nsnecgisn usgipo t nihs e oo ftn et shlecech oCemPdCum lCeue tnnoitc raaactilh Coieonvm eexm ipniestttinesteue ti imno aan napalle gsnaeafmergeyu nstae;rs dessiloi monfi wnthaaetis omhneo loddfe iCmraomtuenelytdy wi atetrealllv-yoe affl fta. nUedrn tdchoeeur npthatreryt "syfii'dsv eCe os iunnb gosrneidesis"e, st oh; nere tGsheeean trehcreham l" vleaiy l"loapgueetr s ooofffin snociecalil"as,l"i ds ctisa mcpuiotsadsleiinzrgan tiiezolaentic otimnoa nrne aCqgeuenimrteraemln'set ntootfsp c, ol1er8ap dsoeersarsstie,o s nsup coehfn tadhsie nt hgde,e a cenilsedioc stinoo wonna o.s fF a ti nh"aefil vlSyet,a ignnrd ooiunnpeg " .C. .aonmdm thitte eime porfo thveem poelnitit coaf lo Bvuerreaallu s,c thhermoueg ohf trheefo Cremn,t rwaill lC pormommoittteee a nm ienmtebgerarste, dde acnisdio cnoso,r sduincha taesd m eecomnboemrsi co,f p tohleiti cal,
故四面体ABCD的体积
（II）如答（20）图1，过F作FE⊥AB，垂足为E，连接DE。由（I）知DF⊥平面
ABC。由三垂线定理知DE⊥AB，故∠DEF为二面角C—AB—D的平面角。
在
在 中，EF//BC，从而EF：BC=AF：AC，所以
在Rt△DEF中，
解法二：（I）如答（20）图2，设O是AC的中点，过O作OH⊥AC，交AB于H，过O
作OM⊥AC，交AD于M，由平面ABC⊥平面ACD，知OH⊥OM。因此以O为原点，以射线
OH，OC，OM分别为x轴，y轴，z轴的正半轴，可建立空间坐标系O—xyz.已知AC=2，
故点A，C的坐标分别为A（0，—1，0），C（0，1，0）。
设点B的坐标为 ，有
即点B的坐标为
又设点D的坐标为 有
即点D的坐标为 从而△ACD边AC上的高为r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
又
故四面体ABCD的体积
（II）由（I）知
设非零向量 是平面ABD的法向量，则由 有
（1）
由 ，有
（2）
取 ，由（1），（2），可得
显然向量 是平面ABC的法向量，从而
即二面角C—AB—D的平面角的正切值为
21．（本题12分）
解：（I）由
解得 ，故椭圆的标准方程为
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,r cC aue pl ntiut d r1r a da8l el ,Msv eseoisl lsi octiia poar mnl ya et nCondo tbm eeocm fho teilh solsdegi oimicnnaa B.lr eck iievjiti nl iegzc Ta fo htirnoeoom ns m e cNc yooo ne nvsn detvrm i pur l obcetinenrom a 9nr e y, on 2sft e0 tt1sho s3e i e ot fixon pv ,1e l i o2s r rteh ehfe. op l3drum5 b i nylsi e t cawa nsre odsr savtehga seon s tpibs oal ne'r dwstyu b 'ttsehy fce o co o rtnhe nsi sttru hrdum e pc gtiplee tionnnoae n rirnya m lst hee oelsn es eacioti tirenz oaa no tio,f o fmt nhina ersi e ntirfelto yfuro ttim orom d hn i aaas sncl udrpes r ofs oop v arei mn dne e.in w d g a S u t gpa o t ioTenhd pte hf e o"er ufis so nvpn dern a iinne tigl oo ibs nnsr.eeu T"ee h zps ee.r o,B s ocguhrc taai b amnl y igmz et a hdeti e, oi sat n hfft oi oer d fca tcr ehthaieerv sweeo ravr clicdoe;m twopdorearkhy, eh 3na5ss i yvbeeea errensf lolaarutmenr c,o hifne i ndths, tiea tneudytie rosan poaifdl t ghpuerao nrgaarntietsoesne i nsa nfsoodrm tohebe jp ewlcaoticrevldes s ae noxdpf e bdcuetip,ld aa grata mwinee tlnlo-to srffe, df osuortmcyi e cmtoyna, srtkuh meC hspmitinaoo,no ut mhsh openrroeegtidrz eiansti st ohonef to1hf8e cs aceorsrnsiiesotrnr au.n cXdtiI oGanpe pnorefo raaa wcl hSee tllco-ro emfftaa snroyac gpieeotmiyn etaenndtd ho rauesft o btrhemaetn t Ch rheei sonobalv'jsee crdeti.fv Toehrsmi rod hf, atihnse er enpctreeonrgetr daym eaa mcrrseu,. ce i xa pl lp oerriOniongde t ,ha hneod ml dtohinnege S titihzmaamtie o aSnnh duoi fp Pdlaouc tDeyi sicmtorpnicostu,r tmmanpuctiseto bnoen h b aNasos mveedam doebn e sgorr me9a, et2e 0pr1r po3og trloieti stchsa,e lh 1cao8vu esre agsagseiino aennd od sf ow tmhiseed 1 oe2mxtph,e lBorieseeinji ncneog atisinmndce ec ai n1n9 d p7er8eo,p v3ei5dn,e ihn ragev freee rbfeoenremcne i7tno p i mtlehnpeao crroytma sneptsr sfieiheolendn,s se. iaDvceah rr eetisfm otorem cor noaf cm kth aaej oh sray rissdtse unmeust o, o fd fpa puroebslli ititcoc s aeqlur avenasdntit oesnc 'o dtnhuoetmy R ciacop nliidsfeus ,mo wfp httihicoehn c dofuaurrntethsre ytro h. Iabmsre pmalekamd tehe einm btipanorgrr iateanrn o"thf d oidenepealsosty, cmaanendnt edte.a nIrne" , at scotc abonerddnaaernfidctie zc euw roietffih b caPiarRrlCi ee prnsot.el iDrtitecaeainpl iepnnrgai ncmtiga crneea,f ogoreftmme nea nnatdt; e oevpneehrnayi nsnecgisn usgipo t nihs e oo ftn et shlecech oCemPdCum lCeue tnnoitc raaactilh Coieonvm eexm ipniestttinesteue ti imno aan napalle gsnaeafmergeyu nstae;rs dessiloi monfi wnthaaetis omhneo loddfe iCmraomtuenelytdy wi atetrealllv-yoe affl fta. nUedrn tdchoeeur npthatreryt "syfii'dsv eCe os iunnb gosrneidesis"e, st oh; nere tGsheeean trehcreham l" vleaiy l"loapgueetr s ooofffin snociecalil"as,l"i ds ctisa mcpuiotsadsleiinzrgan tiiezolaentic otimnoa nrne aCqgeuenimrteraemln'set ntootfsp c, ol1er8ap dsoeersarsstie,o s nsup coehfn tadhsie nt hgde,e a cenilsedioc stinoo wonna o.s fF a ti nh"aefil vlSyet,a ignnrd ooiunnpeg " .C. .aonmdm thitte eime porfo thveem poelnitit coaf lo Bvuerreaallu s,c thhermoueg ohf trheefo Cremn,t rwaill lC pormommoittteee a nm ienmtebgerarste, dde acnisdio cnoso,r sduincha taesd m eecomnboemrsi co,f p tohleiti cal,
（II）设 ，则由
得
因为点M，N在椭圆 上，所以
，
故
设 分别为直线OM，ON的斜率，由题设条件知
因此
所以
所以P点是椭圆 上的点，该椭圆的右焦点为 ，离心
率 是该椭圆的右准线，故根据椭圆的第二定义，存在定点
，使得|PF|与P点到直线l的距离之比为定值。r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
2011年普通高等学校招生全国统一考试
文科数学
（重庆卷）
一、选择题(共10小题，每小题5分，满分50分)
1、(2011•重庆)在等差数列{a}中，a=2，a=4，则a=( )
n 2 3 10
A、12 B、14
C、16 D、18
考点：等差数列的通项公式。
专题：计算题。
分析：根据所给的等差数列的两项做出等差数列的公差，写出等差数列的第十项的表示式
用第三项加上七倍的公差，代入数值，求出结果．
解答：解：∵等差数列{a}中，a=2，a=4，
n 2 3
∴d=a﹣a=4﹣2=2，
3 2
∴a=a+7d=4+14=18
10 3
故选D．
点评：本题考查等差数列的公差求法，考查等差数列的通项公式，这是一个等差数列基本
量的运算，是一个数列中最常出现的基础题．
2、(2011•重庆)设U=R，M={a|a2﹣2a＞0}，则CM=( )
U
A、[0，2] B、(0，2)
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,r cC aue pl ntiut d r1r a da8l el ,Msv eseoisl lsi octiia poar mnl ya et nCondo tbm eeocm fho teilh solsdegi oimicnnaa B.lr eck iievjiti nl iegzc Ta fo htirnoeoom ns m e cNc yooo ne nvsn detvrm i pur l obcetinenrom a 9nr e y, on 2sft e0 tt1sho s3e i e ot fixon pv ,1e l i o2s r rteh ehfe. op l3drum5 b i nylsi e t cawa nsre odsr savtehga seon s tpibs oal ne'r dwstyu b 'ttsehy fce o co o rtnhe nsi sttru hrdum e pc gtiplee tionnnoae n rirnya m lst hee oelsn es eacioti tirenz oaa no tio,f o fmt nhina ersi e ntirfelto yfuro ttim orom d hn i aaas sncl udrpes r ofs oop v arei mn dne e.in w d g a S u t gpa o t ioTenhd pte hf e o"er ufis so nvpn dern a iinne tigl oo ibs nnsr.eeu T"ee h zps ee.r o,B s ocguhrc taai b amnl y igmz et a hdeti e, oi sat n hfft oi oer d fca tcr ehthaieerv sweeo ravr clicdoe;m twopdorearkhy, eh 3na5ss i yvbeeea errensf lolaarutmenr c,o hifne i ndths, tiea tneudytie rosan poaifdl t ghpuerao nrgaarntietsoesne i nsa nfsoodrm tohebe jp ewlcaoticrevldes s ae noxdpf e bdcuetip,ld aa grata mwinee tlnlo-to srffe, df osuortmcyi e cmtoyna, srtkuh meC hspmitinaoo,no ut mhsh openrroeegtidrz eiansti st ohonef to1hf8e cs aceorsrnsiiesotrnr au.n cXdtiI oGanpe pnorefo raaa wcl hSee tllco-ro emfftaa snroyac gpieeotmiyn etaenndtd ho rauesft o btrhemaetn t Ch rheei sonobalv'jsee crdeti.fv Toehrsmi rod hf, atihnse er enpctreeonrgetr daym eaa mcrrseu,. ce i xa pl lp oerriOniongde t ,ha hneod ml dtohinnege S titihzmaamtie o aSnnh duoi fp Pdlaouc tDeyi sicmtorpnicostu,r tmmanpuctiseto bnoen h b aNasos mveedam doebn e sgorr me9a, et2e 0pr1r po3og trloieti stchsa,e lh 1cao8vu esre agsagseiino aennd od sf ow tmhiseed 1 oe2mxtph,e lBorieseeinji ncneog atisinmndce ec ai n1n9 d p7er8eo,p v3ei5dn,e ihn ragev freee rbfeoenremcne i7tno p i mtlehnpeao crroytma sneptsr sfieiheolendn,s se. iaDvceah rr eetisfm otorem cor noaf cm kth aaej oh sray rissdtse unmeust o, o fd fpa puroebslli ititcoc s aeqlur avenasdntit oesnc 'o dtnhuoetmy R ciacop nliidsfeus ,mo wfp httihicoehn c dofuaurrntethsre ytro h. Iabmsre pmalekamd tehe einm btipanorgrr iateanrn o"thf d oidenepealsosty, cmaanendnt edte.a nIrne" , at scotc abonerddnaaernfidctie zc euw roietffih b caPiarRrlCi ee prnsot.el iDrtitecaeainpl iepnnrgai ncmtiga crneea,f ogoreftmme nea nnatdt; e oevpneehrnayi nsnecgisn usgipo t nihs e oo ftn et shlecech oCemPdCum lCeue tnnoitc raaactilh Coieonvm eexm ipniestttinesteue ti imno aan napalle gsnaeafmergeyu nstae;rs dessiloi monfi wnthaaetis omhneo loddfe iCmraomtuenelytdy wi atetrealllv-yoe affl fta. nUedrn tdchoeeur npthatreryt "syfii'dsv eCe os iunnb gosrneidesis"e, st oh; nere tGsheeean trehcreham l" vleaiy l"loapgueetr s ooofffin snociecalil"as,l"i ds ctisa mcpuiotsadsleiinzrgan tiiezolaentic otimnoa nrne aCqgeuenimrteraemln'set ntootfsp c, ol1er8ap dsoeersarsstie,o s nsup coehfn tadhsie nt hgde,e a cenilsedioc stinoo wonna o.s fF a ti nh"aefil vlSyet,a ignnrd ooiunnpeg " .C. .aonmdm thitte eime porfo thveem poelnitit coaf lo Bvuerreaallu s,c thhermoueg ohf trheefo Cremn,t rwaill lC pormommoittteee a nm ienmtebgerarste, dde acnisdio cnoso,r sduincha taesd m eecomnboemrsi co,f p tohleiti cal,
C、(﹣∞，0)∪(2，+∞) D、(﹣∞，0]∪[2，+∞)
考点：补集及其运算。
专题：计算题。
分析：根据已知中M={a|a2﹣2a＞0}，我们易求出M，再根据集合补集运算即可得到答案．
解答：解：∵M={a|a2﹣2a＞0}={a|a＜0，或a＞2}，
∴CM={a|0≤a≤2}，
U
即CM=[0，2]
U
故选A
点评：本题考查的知识点是集合的补集及其运算，在求连续数集的补集时，若子集不包括
端点，则补集一定要包括端点．
3、(2011•重庆)曲线y=﹣x3+3x2在点(1，2)处的切线方程为( )
A、y=3x﹣1 B、y=﹣3x+5
C、y=3x+5 D、y=2x
考点：利用导数研究曲线上某点切线方程。
专题：计算题。
分析：根据导数的几何意义求出函数f(x)在x=1处的导数，从而求出切线的斜率，再用点
斜式写出切线方程，化成斜截式即可．
解答：解：∵y=﹣x3+3x2∴y'=﹣3x2+6x，
∴y'| =﹣3x2+6x| =3，
x=1 x=1
∴曲线y=﹣x3+3x2在点(1，2)处的切线方程为y﹣2=3(x﹣1)，
即y=3x﹣1，
故选A．
点评：本题主要考查了利用导数研究曲线上某点切线方程，属于基础题．
4、(2011•重庆)从一堆苹果中任取10只，称得它们的质量如下(单位：克)
125 120 122 105 130 114 116 95 120 134，则样本数据落在[114.5，124.5)内的频率为(
)
A、0.2 B、0.3
C、0.4 D、0.5
考点：频率分布表。
专题：计算题。
分析：从所给的十个数字中找出落在所要求的范围中的数字，共有 4个，利用这个频数除
以样本容量，得到要求的频率．r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
解答：解：∵在125 120 122 105 130 114 116 95 120 134十个数字中，
样本数据落在[114.5，124.5)内的有116，120，120，122共有四个，
∴样本数据落在[114.5，124.5)内的频率为 =0.4，
故选C
点评：本题考查频率分布表，频数、频率和样本容量三者之间的关系是知二求一，这种问
题会出现在选择和填空中，有的省份也会以大题的形式出现，把它融于统计问题中．
5、(2011•重庆)已知向量 =(1，k)， =(2，2)，且 + 与 共线，那么 • 的值为(
)
A、1 B、2
C、3 D、4
考点：平面向量数量积的运算。
专题：计算题。
分析：利用向量的运算法则求出两个向量的和；利用向量共线的充要条件列出方程求出
k；利用向量的数量积公式求出值．
解答：解：∵ =(3，k+2)
∵ 共线
∴k+2=3k
解得k=1
∴ =(1，1)
∴ =1×2+1×2=4
故选D
点评：本题考查向量的运算法则、考查向量共线的充要条件、考查向量的数量积公式．
6、(2011•重庆)设a= ，b= ，c=log ，则a，b，c的大小关系是( )
3
A、a＜b＜c B、c＜b＜a
C、b＜a＜c D、b＜c＜a
考点：对数值大小的比较。
专题：计算题。
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,r cC aue pl ntiut d r1r a da8l el ,Msv eseoisl lsi octiia poar mnl ya et nCondo tbm eeocm fho teilh solsdegi oimicnnaa B.lr eck iievjiti nl iegzc Ta fo htirnoeoom ns m e cNc yooo ne nvsn detvrm i pur l obcetinenrom a 9nr e y, on 2sft e0 tt1sho s3e i e ot fixon pv ,1e l i o2s r rteh ehfe. op l3drum5 b i nylsi e t cawa nsre odsr savtehga seon s tpibs oal ne'r dwstyu b 'ttsehy fce o co o rtnhe nsi sttru hrdum e pc gtiplee tionnnoae n rirnya m lst hee oelsn es eacioti tirenz oaa no tio,f o fmt nhina ersi e ntirfelto yfuro ttim orom d hn i aaas sncl udrpes r ofs oop v arei mn dne e.in w d g a S u t gpa o t ioTenhd pte hf e o"er ufis so nvpn dern a iinne tigl oo ibs nnsr.eeu T"ee h zps ee.r o,B s ocguhrc taai b amnl y igmz et a hdeti e, oi sat n hfft oi oer d fca tcr ehthaieerv sweeo ravr clicdoe;m twopdorearkhy, eh 3na5ss i yvbeeea errensf lolaarutmenr c,o hifne i ndths, tiea tneudytie rosan poaifdl t ghpuerao nrgaarntietsoesne i nsa nfsoodrm tohebe jp ewlcaoticrevldes s ae noxdpf e bdcuetip,ld aa grata mwinee tlnlo-to srffe, df osuortmcyi e cmtoyna, srtkuh meC hspmitinaoo,no ut mhsh openrroeegtidrz eiansti st ohonef to1hf8e cs aceorsrnsiiesotrnr au.n cXdtiI oGanpe pnorefo raaa wcl hSee tllco-ro emfftaa snroyac gpieeotmiyn etaenndtd ho rauesft o btrhemaetn t Ch rheei sonobalv'jsee crdeti.fv Toehrsmi rod hf, atihnse er enpctreeonrgetr daym eaa mcrrseu,. ce i xa pl lp oerriOniongde t ,ha hneod ml dtohinnege S titihzmaamtie o aSnnh duoi fp Pdlaouc tDeyi sicmtorpnicostu,r tmmanpuctiseto bnoen h b aNasos mveedam doebn e sgorr me9a, et2e 0pr1r po3og trloieti stchsa,e lh 1cao8vu esre agsagseiino aennd od sf ow tmhiseed 1 oe2mxtph,e lBorieseeinji ncneog atisinmndce ec ai n1n9 d p7er8eo,p v3ei5dn,e ihn ragev freee rbfeoenremcne i7tno p i mtlehnpeao crroytma sneptsr sfieiheolendn,s se. iaDvceah rr eetisfm otorem cor noaf cm kth aaej oh sray rissdtse unmeust o, o fd fpa puroebslli ititcoc s aeqlur avenasdntit oesnc 'o dtnhuoetmy R ciacop nliidsfeus ,mo wfp httihicoehn c dofuaurrntethsre ytro h. Iabmsre pmalekamd tehe einm btipanorgrr iateanrn o"thf d oidenepealsosty, cmaanendnt edte.a nIrne" , at scotc abonerddnaaernfidctie zc euw roietffih b caPiarRrlCi ee prnsot.el iDrtitecaeainpl iepnnrgai ncmtiga crneea,f ogoreftmme nea nnatdt; e oevpneehrnayi nsnecgisn usgipo t nihs e oo ftn et shlecech oCemPdCum lCeue tnnoitc raaactilh Coieonvm eexm ipniestttinesteue ti imno aan napalle gsnaeafmergeyu nstae;rs dessiloi monfi wnthaaetis omhneo loddfe iCmraomtuenelytdy wi atetrealllv-yoe affl fta. nUedrn tdchoeeur npthatreryt "syfii'dsv eCe os iunnb gosrneidesis"e, st oh; nere tGsheeean trehcreham l" vleaiy l"loapgueetr s ooofffin snociecalil"as,l"i ds ctisa mcpuiotsadsleiinzrgan tiiezolaentic otimnoa nrne aCqgeuenimrteraemln'set ntootfsp c, ol1er8ap dsoeersarsstie,o s nsup coehfn tadhsie nt hgde,e a cenilsedioc stinoo wonna o.s fF a ti nh"aefil vlSyet,a ignnrd ooiunnpeg " .C. .aonmdm thitte eime porfo thveem poelnitit coaf lo Bvuerreaallu s,c thhermoueg ohf trheefo Cremn,t rwaill lC pormommoittteee a nm ienmtebgerarste, dde acnisdio cnoso,r sduincha taesd m eecomnboemrsi co,f p tohleiti cal,
分析：可先由对数的运算法则，将a和c化为同底的对数，利用对数函数的单调性比较大
小；再比较b和c的大小，用对数的换底公式化为同底的对数找关系，结合排除法选出答
案即可．
解答：解：由对数的运算法则，a=log2＞c；排除A和C．
3
因为b=log3﹣1，c=log4﹣1= ，
2 3
因为(log3)2＞2，所以b＞c，排除D
2
故选B．
点评：本题考查对数值的大小比较，考查对数的运算法则和对数的换底公式，考查运算能
力．
7、(2011•重庆)若函数f(x)=x+ (x＞2)，在x=a处取最小值，则a=( )
A、1+ B、1+
C、3 D、4
考点：基本不等式。
专题：计算题。
分析：把函数解析式整理成基本不等式的形式，求得函数的最小值和此时x的取值．
解答：解：f(x)=x+ =x﹣2+ +2≥4
当x﹣2=1时，即x=3时等号成立．
∵x=a处取最小值，
∴a=3
故选C
点评：本题主要考查了基本不等式的应用．考查了分析问题和解决问题的的能力．
8、(2011•重庆)若△ABC的内角A，B，C满足6sinA=4sinB=3sinC，则cosB=( )
A、 B、
C、 D、
考点：三角函数的恒等变换及化简求值。r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
专题：计算题。
分析：由题意利用正弦定理，推出a，b，c的关系，然后利用余弦定理求出cosB的值．
解答：解：△ABC的内角A，B，C满足6sinA=4sinB=3sinC，所以6a=4b=3c，不妨令a=2，
b=3，c=4，
所以b2=a2+c2﹣2accosB，所以cosB= ，
故选D．
点评：本题是基础题，考查正弦定理，余弦定理的应用，考查计算能力，常考题型．
9、(2011•重庆)设双曲线的左准线与两条渐近线交于A，B两点，左焦点为在以AB为直径
的圆内，则该双曲线的离心率的取值范围为( )
A、(0， ) B、(1， )
C、( ，1) D、( ，+∞)
考点：双曲线的简单性质。
分析：求出渐近线方程及准线方程；求得它们的交点A，B的坐标；利用圆内的点到圆心距
离小于半径，列出参数a，b，c满足的不等式，求出离心率的范围．
解答：解：渐近线y=± x．
准线x=± ，
求得A( )．B( )，
左焦点为在以AB为直径的圆内，
得出 ，
，
b＜a，
c2＜2a2
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,r cC aue pl ntiut d r1r a da8l el ,Msv eseoisl lsi octiia poar mnl ya et nCondo tbm eeocm fho teilh solsdegi oimicnnaa B.lr eck iievjiti nl iegzc Ta fo htirnoeoom ns m e cNc yooo ne nvsn detvrm i pur l obcetinenrom a 9nr e y, on 2sft e0 tt1sho s3e i e ot fixon pv ,1e l i o2s r rteh ehfe. op l3drum5 b i nylsi e t cawa nsre odsr savtehga seon s tpibs oal ne'r dwstyu b 'ttsehy fce o co o rtnhe nsi sttru hrdum e pc gtiplee tionnnoae n rirnya m lst hee oelsn es eacioti tirenz oaa no tio,f o fmt nhina ersi e ntirfelto yfuro ttim orom d hn i aaas sncl udrpes r ofs oop v arei mn dne e.in w d g a S u t gpa o t ioTenhd pte hf e o"er ufis so nvpn dern a iinne tigl oo ibs nnsr.eeu T"ee h zps ee.r o,B s ocguhrc taai b amnl y igmz et a hdeti e, oi sat n hfft oi oer d fca tcr ehthaieerv sweeo ravr clicdoe;m twopdorearkhy, eh 3na5ss i yvbeeea errensf lolaarutmenr c,o hifne i ndths, tiea tneudytie rosan poaifdl t ghpuerao nrgaarntietsoesne i nsa nfsoodrm tohebe jp ewlcaoticrevldes s ae noxdpf e bdcuetip,ld aa grata mwinee tlnlo-to srffe, df osuortmcyi e cmtoyna, srtkuh meC hspmitinaoo,no ut mhsh openrroeegtidrz eiansti st ohonef to1hf8e cs aceorsrnsiiesotrnr au.n cXdtiI oGanpe pnorefo raaa wcl hSee tllco-ro emfftaa snroyac gpieeotmiyn etaenndtd ho rauesft o btrhemaetn t Ch rheei sonobalv'jsee crdeti.fv Toehrsmi rod hf, atihnse er enpctreeonrgetr daym eaa mcrrseu,. ce i xa pl lp oerriOniongde t ,ha hneod ml dtohinnege S titihzmaamtie o aSnnh duoi fp Pdlaouc tDeyi sicmtorpnicostu,r tmmanpuctiseto bnoen h b aNasos mveedam doebn e sgorr me9a, et2e 0pr1r po3og trloieti stchsa,e lh 1cao8vu esre agsagseiino aennd od sf ow tmhiseed 1 oe2mxtph,e lBorieseeinji ncneog atisinmndce ec ai n1n9 d p7er8eo,p v3ei5dn,e ihn ragev freee rbfeoenremcne i7tno p i mtlehnpeao crroytma sneptsr sfieiheolendn,s se. iaDvceah rr eetisfm otorem cor noaf cm kth aaej oh sray rissdtse unmeust o, o fd fpa puroebslli ititcoc s aeqlur avenasdntit oesnc 'o dtnhuoetmy R ciacop nliidsfeus ,mo wfp httihicoehn c dofuaurrntethsre ytro h. Iabmsre pmalekamd tehe einm btipanorgrr iateanrn o"thf d oidenepealsosty, cmaanendnt edte.a nIrne" , at scotc abonerddnaaernfidctie zc euw roietffih b caPiarRrlCi ee prnsot.el iDrtitecaeainpl iepnnrgai ncmtiga crneea,f ogoreftmme nea nnatdt; e oevpneehrnayi nsnecgisn usgipo t nihs e oo ftn et shlecech oCemPdCum lCeue tnnoitc raaactilh Coieonvm eexm ipniestttinesteue ti imno aan napalle gsnaeafmergeyu nstae;rs dessiloi monfi wnthaaetis omhneo loddfe iCmraomtuenelytdy wi atetrealllv-yoe affl fta. nUedrn tdchoeeur npthatreryt "syfii'dsv eCe os iunnb gosrneidesis"e, st oh; nere tGsheeean trehcreham l" vleaiy l"loapgueetr s ooofffin snociecalil"as,l"i ds ctisa mcpuiotsadsleiinzrgan tiiezolaentic otimnoa nrne aCqgeuenimrteraemln'set ntootfsp c, ol1er8ap dsoeersarsstie,o s nsup coehfn tadhsie nt hgde,e a cenilsedioc stinoo wonna o.s fF a ti nh"aefil vlSyet,a ignnrd ooiunnpeg " .C. .aonmdm thitte eime porfo thveem poelnitit coaf lo Bvuerreaallu s,c thhermoueg ohf trheefo Cremn,t rwaill lC pormommoittteee a nm ienmtebgerarste, dde acnisdio cnoso,r sduincha taesd m eecomnboemrsi co,f p tohleiti cal,
∴ ，
故选B．
点评：本题考查双曲线的准线、渐近线方程形式、考查园内的点满足的不等条件、注意双
曲线离心率本身要大于1．
10、(2011•重庆)高为 的四棱锥S﹣ABCD的底面是边长为1的正方形，点S，A，B，C，
D均在半径为1的同一球面上，则底面ABCD的中心与顶点S之间的距离为( )
A、 B、
C、 D、
考点：球内接多面体；点、线、面间的距离计算。
专题：计算题。
分析：由题意可知ABCD 是小圆，对角线长为 ，四棱锥的高为 ，推出高就是四棱锥
的一条侧棱，最长的侧棱就是球的直径，然后利用勾股定理求出底面ABCD的中心与顶点S
之间的距离．
解答：解：由题意可知ABCD 是小圆，对角线长为 ，四棱锥的高为 ，点S，A，B，
C，D均在半径为1的同一球面上，球的直径为2，所以四棱锥的一条侧棱垂直底面的一个
顶点，最长的侧棱就是直径，所以底面 ABCD的中心与顶点 S之间的距离为：
=
故选A
点评：本题是基础题，考查球的内接多面体的知识，能够正确推出四棱锥的一条侧棱垂直
底面的一个顶点，最长的侧棱就是直径是本题的关键，考查逻辑推理能力，计算能力．
二、填空题(共5小题，每小题5分，满分25分)
11、(2011•重庆)(1+2x)6的展开式中x4的系数是 24 0 ．
考点：二项式系数的性质。
专题：计算题。r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
分析：利用二项展开式的通项公式求出展开式的通项；令x的指数为4，求出展开式中x4
的系数．
解答：解：展开式的通项为T =2rCrxr
r+1 6
令r=4得展开式中x4的系数是24C4=240
6
故答案为：240
点评：本题考查利用二项展开式的通项公式解决二项展开式的特定项问题．
12、(2011•重庆)若cosα=﹣ ，且α∈(π， )，则tanα= ．
考点：任意角的三角函数的定义。
专题：计算题。
分析：根据α∈(π， )，cosα=﹣ ，求出sinα，然后求出tanα，即可．
解答：解：因为α∈(π， )，cosα=﹣ ，所以sinα=﹣ ，所以tanα= =
故答案为：
点评：本题是基础题，考查任意角的三角函数的定义，注意角所在的象限，三角函数值的
符号，是本题解答的关键．
13、(2011•重庆)过原点的直线与圆x2+y2﹣2x﹣4y+4=0相交所得的弦长为2，则该直线的
方程为 2x ﹣ y = 0 ．
考点：直线与圆相交的性质。
专题：计算题。
分析：用配方法将圆的方程转化为标准方程，求出圆心坐标和半径，设直线方程为y=kx，
求出圆心到直线的距离，利用直线和圆相交所成的直角三角形知识求解即可．
解答：解：直线方程为y=kx，
圆x2+y2﹣2x﹣4y+4=0即(x﹣1)2+(y﹣2)2=1
即圆心坐标为(1，2)，半径为r=1
因为弦长为2，为直径，故y=kx过圆心，所以k=2
所以该直线的方程为：y=2x
故答案为：2x﹣y=0
点评：本题考查直线和圆的相交弦长问题，属基础知识的考查．注意弦长和半径的关系．
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; 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14、(2011•重庆)从甲、乙等10位同学中任选3位去参加某项活动，则所选3位中有甲但
没有乙的概率为 ．
考点：排列、组合及简单计数问题；等可能事件的概率。
专题：计算题。
分析：根据题意，分析可得从10人中任取3人参加活动的取法数，进而可得“有甲但没有
乙”的取法相当于“从除甲乙之外的8人中任取2人”，可得其情况数目，由等可能事件
的概率公式，计算可得答案．
解答：解：根据题意，从10人中任取3人参加活动，有C3=120种取法；
10
分析可得有甲但没有乙的取法即从除甲乙之外的8人中任取2人即可，
则所选3位中有甲但没有乙的情况有C2=28种；
8
则其概率为 = ；
故答案为： ．
点评：本题考查排列、组合的运用；涉及等可能事件的概率计算，解题时注意排列、组合
是解决问题的基本思路与突破口．
15、(2011•重庆)若实数a，b，c满足2a+2b=2a+b，2a+2b+2c=2a+b+c，则c的最大值是 2 ﹣
log 3 ．
2
考点：基本不等式在最值问题中的应用。
专题：计算题。
分析：由基本不等式得2a+2b≥ ，可求出2a+b的范围，
再由2a+2b+2c=2a+b+c=2a+b2c=2a+b+2c，2c可用2a+b表达，利用不等式的性质求范围即可．
解答：解：由基本不等式得 2a+2b≥ ，即 2a+b≥ ，所以
2a+b≥4，
令t=2a+b，由2a+2b+2c=2a+b+c可得2a+b+2c=2a+b2c，所以2c=
因为t≥4，所以 ，即 ，所以
故答案为：2﹣log3
2
点评：本题考查指数的运算法则，基本不等式求最值、不等式的性质等问题，综合性较强．r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
三、解答题(共6小题，满分75分)
16、(2011•重庆)设{a}是公比为正数的等比数列a=2，a=a+4．
n 1 3 2
(Ⅰ)求{a}的通项公式；
n
(Ⅱ)设{b}是首项为1，公差为2的等差数列，求数列{a+b}的前n项和S．
n n n n
考点：等比数列的通项公式；数列的求和。
专题：计算题。
分析：(Ⅰ)由{a}是公比为正数的等比数列，设其公比，然后利用a=2，a=a+4可求得
n 1 3 2
q，即可求得{a}的通项公式
n
(Ⅱ)由{b}是首项为1，公差为2的等差数列 可求得b=1+(n﹣1)×2=2n﹣1，然后利用等
n n
比数列与等差数列的前n项和公式即可求得数列{a+b}的前n项和S．
n n n
解答：解：(Ⅰ)∵设{a}是公比为正数的等比数列
n
∴设其公比为q q＞0
∵a=a+4，a=2
3 2 1
∴2×q2=2×q+4 解得q=2或q=﹣1
∵q＞0∴q=2
∴{a}的通项公式为a=2×2n﹣1=2n
n n
(Ⅱ)∵{b}是首项为1，公差为2的等差数列
n
∴b=1+(n﹣1)×2=2n﹣1
n
∴数列{a+b}的前n项和S= + =2n+1﹣2+n2=2n+1+n2﹣2
n n n
点评：本题考察了等比数列的通项公式及数列的求和，注意题目条件的应用．在用等比数
列的前n项和公式时注意辨析q是否为1，只要简单数字运算时不出错，问题可解，是个基
础题．
17、(2011•重庆)某市公租房的房源位于A、B、C三个片区，设每位申请人只申请其中一个
片区的房源，且申请其中任一个片区的房源是等可能的，求该市的4位申请人中：
(I)没有人申请A片区房源的概率；
(II)每个片区的房源都有人申请的概率．
考点：古典概型及其概率计算公式；等可能事件的概率；n次独立重复试验中恰好发生k
次的概率。
专题：计算题。
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; 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分析：(Ⅰ)解法一：首先分析所有的可能申请方式的情况数目，再分析没有人申请A片区
房源的即所有的都申请BC区的申请方式的情况数目，由古典概型概率公式，计算可得答案；
解法二：视为独立重复试验中事件A恰好发生k次的情况，设对每位申请人的观察为一次
试验，这是4次独立重复试验，记“申请A片区房源”为事件A，易得P(A)，进而由独立
重复试验中事件A恰好发生k次的概率计算公式计算可得答案；
(Ⅱ)根据题意，分析可得所有的可能申请方式的种数；而“每个片区的房源都有人申请”
的申请方式的种数；
由古典概型概率公式，计算可得答案．
解答：解：(I)由题意知本题是一个等可能事件的概率，
解法一：所有的可能申请方式有34种；而“没有人申请A片区房源的”的申请方式有24种；
记“没有人申请A片区房源”为事件A，
则P(A)= = ；
解法二：设对每位申请人的观察为一次试验，这是4次独立重复试验，
记“申请A片区房源”为事件A，则P(A)= ；
由独立重复试验中事件A恰好发生k次的概率计算公式知：
“没有人申请A片区房源”的概率为P(0)=C0•( )0( )4= ；
4 3
(Ⅱ)所有的可能申请方式有34种；而“每个片区的房源都有人申请”的申请方式有C2•A3
4 3
种；
记“每个片区的房源都有人申请”为事件B，
从而有P(B)= = ．
点评：本题考查等可能事件的概率，注意解题的格式应该规范，先有“记××为事件×”，
进而又公式进行计算．
18、(2011•重庆)设函数f(x)=sinxcosx﹣ cos(x+π)cosx，(x∈R)
(I)求f(x)的最小正周期；
(II)若函数y=f(x)的图象按 =( ， )平移后得到的函数y=g(x)的图象，求y=g(x)在r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
(0， ]上的最大值．
考点：三角函数的周期性及其求法；函数y=Asin(ωx+φ)的图象变换；三角函数的最值。
专题：计算题；综合题。
分析：(I)先利用诱导公式，二倍角公式与和角公式将函数解析式化简整理，然后利用周期
公式可求得函数的最小正周期．
(II)由(I)得函数y=f(x)，利用函数图象的变换可得函数y=g(x)的解析式，通过探讨角的
范围，即可的函数g(x)的最大值．
解答：解：(I)∵f(x)=sinxcosx﹣ cos(x+π)cosx=sinxcosx+ cosxcosx
= sin2x+ cos2x+ =sin(2x+ )+
∴f(x)的最小正周期T= =π
(II)∵函数y=f(x)的图象按 =( ， )平移后得到的函数y=g(x)的图象，
∴g(x)=sin(2x+ ﹣ )+ + =sin(2x+ )+
∵0＜x≤ ∴ ＜2x+ ≤ ，
∴y=g(x)在(0， ]上的最大值为：1+
点评：本题考察了三角函数的周期及其求法，函数图象的变换及三角函数的最值，各公式
的熟练应用是解决问题的根本，体现了整体意识，是个中档题．
19、(2011•重庆)设f(x)=2x3+ax2+bx+1的导数为f′(x)，若函数y=f′(x)的图象关于直线
x=﹣ 对称，且f′(1)=0
(Ⅰ)求实数a，b的值
(Ⅱ)求函数f(x)的极值．
考点：利用导数研究函数的极值；二次函数的性质。
专题：计算题。
分析：(Ⅰ)先对f(x)求导，f(x)的导数为二次函数，由对称性可求得a，再由f′(1)=0即
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; 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可求出b
(Ⅱ)对f(x)求导，分别令f′(x)大于0和小于0，即可解出f(x)的单调区间，继而确定极
值．
解答：解：(Ⅰ)因f(x)=2x3+ax2+bx+1，故f′(x)=6x2+2ax+b
从而f′(x)=6 y=f′(x)关于直线x=﹣ 对称，
从而由条件可知﹣ = ，解得a=3
又由于f′(x)=0，即6+2a+b=0，解得b=﹣12
(Ⅱ)由(Ⅰ)知f(x)=2x3+3x2﹣12x+1
f′(x)=6x2+6x﹣12=6(x﹣1)(x+2)
令f′(x)=得x=1或x=﹣2
当x∈(﹣∞，﹣2)时，f′(x)＞0，f(x)在(﹣∞，﹣2)上是增函数；
当x∈(﹣2，1)时，f′(x)＜0，f(x)在(﹣2，1)上是减函数；
当x∈(1，+∞)时，f′(x)＞0，f(x)在(1，+∞)上是增函数．
从而f(x)在x=﹣2处取到极大值f(﹣2)=21，在x=1处取到极小值f(1)=﹣6．
点评：本题考查函数的对称性、函数的单调区间和极值，考查运算能力．
20、(2011•重庆)如图，在四面体ABCD中，平面ABC⊥平面ACD，AB⊥BC，AC=AD=2，
BC=CD=1
(Ⅰ)求四面体ABCD的体积；
(Ⅱ)求二面角C﹣AB﹣D的平面角的正切值．
考点：与二面角有关的立体几何综合题；二面角的平面角及求法。
专题：综合题；转化思想。
分析：法一：几何法，
(Ⅰ)过D作DF⊥AC，垂足为F，由平面ABC⊥平面ACD，由面面垂直的性质，可得DF是四
面体ABCD的面ABC上的高；设G为边CD的中点，可得AG⊥CD，计算可得AG与DF的长，
进而可得S ，由棱锥体积公式，计算可得答案；
△ABC
(Ⅱ)过F作FE⊥AB，垂足为E，连接DE，分析可得∠DEF为二面角C﹣AB﹣D的平面角，计r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
算可得EF的长，由(Ⅰ)中DF的值，结合正切的定义，可得答案．
法二：向量法，
(Ⅰ)首先建立坐标系，根据题意，设O是AC的中点，过O作OH⊥AC，交AB与H，过O作
OM⊥AC，交AD与M；易知OH⊥OM，因此可以以O为原点，以射线OH、OC、OM为x轴、y轴、
z轴，建立空间坐标系O﹣XYZ，进而可得B、D的坐标；从而可得△ACD边AC的高即棱住的
高与底面的面积，计算可得答案；
(Ⅱ)设非零向量 =(l，m，n)是平面ABD的法向量，由(Ⅰ)易得向量 的坐标，同时易得
=(0，0，1)是平面ABC的法向量，由向量的夹角公式可得从而cos＜ ， ＞，进而由
同角三角函数的基本关系，可得tan＜ ， ＞，即可得答案．
解答：解：法一
(Ⅰ)如图：过D作DF⊥AC，垂足为F，由平面ABC⊥平面ACD，
可得DF⊥平面ACD，即DF是四面体ABCD的面ABC上的高；
设G为边CD的中点，由AC=AD，可得AG⊥CD，
则AG= = = ；
由S = AC•DF= CD•AG可得，DF= = ；
△ADC
在Rt△ABC中，AB= = ，
S = AB•BC= ；
△ABC
故四面体的体积V= ×S ×DF= ；
△ABC
(Ⅱ)如图，过F作FE⊥AB，垂足为E，连接DE，
由(Ⅰ)知DF⊥平面ABC，由三垂线定理可得DE⊥AB，故∠DEF为二面角C﹣AB﹣D的平面角，
在Rt△AFD中，AF= = = ；
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,r cC aue pl ntiut d r1r a da8l el ,Msv eseoisl lsi octiia poar mnl ya et nCondo tbm eeocm fho teilh solsdegi oimicnnaa B.lr eck iievjiti nl iegzc Ta fo htirnoeoom ns m e cNc yooo ne nvsn detvrm i pur l obcetinenrom a 9nr e y, on 2sft e0 tt1sho s3e i e ot fixon pv ,1e l i o2s r rteh ehfe. op l3drum5 b i nylsi e t cawa nsre odsr savtehga seon s tpibs oal ne'r dwstyu b 'ttsehy fce o co o rtnhe nsi sttru hrdum e pc gtiplee tionnnoae n rirnya m lst hee oelsn es eacioti tirenz oaa no tio,f o fmt nhina ersi e ntirfelto yfuro ttim orom d hn i aaas sncl udrpes r ofs oop v arei mn dne e.in w d g a S u t gpa o t ioTenhd pte hf e o"er ufis so nvpn dern a iinne tigl oo ibs nnsr.eeu T"ee h zps ee.r o,B s ocguhrc taai b amnl y igmz et a hdeti e, oi sat n hfft oi oer d fca tcr ehthaieerv sweeo ravr clicdoe;m twopdorearkhy, eh 3na5ss i yvbeeea errensf lolaarutmenr c,o hifne i ndths, tiea tneudytie rosan poaifdl t ghpuerao nrgaarntietsoesne i nsa nfsoodrm tohebe jp ewlcaoticrevldes s ae noxdpf e bdcuetip,ld aa grata mwinee tlnlo-to srffe, df osuortmcyi e cmtoyna, srtkuh meC hspmitinaoo,no ut mhsh openrroeegtidrz eiansti st ohonef to1hf8e cs aceorsrnsiiesotrnr au.n cXdtiI oGanpe pnorefo raaa wcl hSee tllco-ro emfftaa snroyac gpieeotmiyn etaenndtd ho rauesft o btrhemaetn t Ch rheei sonobalv'jsee crdeti.fv Toehrsmi rod hf, atihnse er enpctreeonrgetr daym eaa mcrrseu,. ce i xa pl lp oerriOniongde t ,ha hneod ml dtohinnege S titihzmaamtie o aSnnh duoi fp Pdlaouc tDeyi sicmtorpnicostu,r tmmanpuctiseto bnoen h b aNasos mveedam doebn e sgorr me9a, et2e 0pr1r po3og trloieti stchsa,e lh 1cao8vu esre agsagseiino aennd od sf ow tmhiseed 1 oe2mxtph,e lBorieseeinji ncneog atisinmndce ec ai n1n9 d p7er8eo,p v3ei5dn,e ihn ragev freee rbfeoenremcne i7tno p i mtlehnpeao crroytma sneptsr sfieiheolendn,s se. iaDvceah rr eetisfm otorem cor noaf cm kth aaej oh sray rissdtse unmeust o, o fd fpa puroebslli ititcoc s aeqlur avenasdntit oesnc 'o dtnhuoetmy R ciacop nliidsfeus ,mo wfp httihicoehn c dofuaurrntethsre ytro h. Iabmsre pmalekamd tehe einm btipanorgrr iateanrn o"thf d oidenepealsosty, cmaanendnt edte.a nIrne" , at scotc abonerddnaaernfidctie zc euw roietffih b caPiarRrlCi ee prnsot.el iDrtitecaeainpl iepnnrgai ncmtiga crneea,f ogoreftmme nea nnatdt; e oevpneehrnayi nsnecgisn usgipo t nihs e oo ftn et shlecech oCemPdCum lCeue tnnoitc raaactilh Coieonvm eexm ipniestttinesteue ti imno aan napalle gsnaeafmergeyu nstae;rs dessiloi monfi wnthaaetis omhneo loddfe iCmraomtuenelytdy wi atetrealllv-yoe affl fta. nUedrn tdchoeeur npthatreryt "syfii'dsv eCe os iunnb gosrneidesis"e, st oh; nere tGsheeean trehcreham l" vleaiy l"loapgueetr s ooofffin snociecalil"as,l"i ds ctisa mcpuiotsadsleiinzrgan tiiezolaentic otimnoa nrne aCqgeuenimrteraemln'set ntootfsp c, ol1er8ap dsoeersarsstie,o s nsup coehfn tadhsie nt hgde,e a cenilsedioc stinoo wonna o.s fF a ti nh"aefil vlSyet,a ignnrd ooiunnpeg " .C. .aonmdm thitte eime porfo thveem poelnitit coaf lo Bvuerreaallu s,c thhermoueg ohf trheefo Cremn,t rwaill lC pormommoittteee a nm ienmtebgerarste, dde acnisdio cnoso,r sduincha taesd m eecomnboemrsi co,f p tohleiti cal,
在Rt△ABC中，EF∥BC，从而 ，可得EF= ；
在Rt△DEF中，tan∠DEF= = ．
则二面角C﹣AB﹣D的平面角的正切值为 ．
解法二：(Ⅰ)如图(2)，
设O是AC的中点，过O作OH⊥AC，交AB与H，过O作OM⊥AC，交AD与M；
由平面ABC⊥平面ACD，知OH⊥OM，
因此以O为原点，以射线OH、OC、OM为x轴、y轴、z轴，建立空间坐标系O﹣XYZ，
已知AC=2，故A、C的坐标分别为A(0，﹣1，0)，C(0，1，0)；
设点B的坐标为(x，y，0)，由 ⊥ ，| |=1；
1 1
有 ，
解可得 或 (舍)；
即B的坐标为( ， ，0)，
又舍D的坐标为(0，y，z)，
2 2
由| |=1，| |=2，有(y﹣1)2+z2=1且(y+1)2+z2=1；
2 2 2 2
解可得 或 (舍)，r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
则D的坐标为(0， ， )，
从而可得△ACD边AC的高为h=|z|=
2
又| |= ，| |=1；
故四面体的体积V= × ×| |×| |h= ；
(Ⅱ)由(Ⅰ)知 =( ， ，0)， =(0， ， )，
设非零向量 =(l，m，n)是平面ABD的法向量，则由 ⊥ 可得， l+ m=0，(1)；
由 ⊥ 可得， m+ n=0，(2)；
取m=﹣1，由(1)(2)可得，l= ，n= ，即 =( ，﹣1， )
显然 =(0，0，1)是平面ABC的法向量，
从而cos＜ ， ＞= ；
故tan＜ ， ＞= ；
则二面角C﹣AB﹣D的平面角的正切值为 ．
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,r cC aue pl ntiut d r1r a da8l el ,Msv eseoisl lsi octiia poar mnl ya et nCondo tbm eeocm fho teilh solsdegi oimicnnaa B.lr eck iievjiti nl iegzc Ta fo htirnoeoom ns m e cNc yooo ne nvsn detvrm i pur l obcetinenrom a 9nr e y, on 2sft e0 tt1sho s3e i e ot fixon pv ,1e l i o2s r rteh ehfe. op l3drum5 b i nylsi e t cawa nsre odsr savtehga seon s tpibs oal ne'r dwstyu b 'ttsehy fce o co o rtnhe nsi sttru hrdum e pc gtiplee tionnnoae n rirnya m lst hee oelsn es eacioti tirenz oaa no tio,f o fmt nhina ersi e ntirfelto yfuro ttim orom d hn i aaas sncl udrpes r ofs oop v arei mn dne e.in w d g a S u t gpa o t ioTenhd pte hf e o"er ufis so nvpn dern a iinne tigl oo ibs nnsr.eeu T"ee h zps ee.r o,B s ocguhrc taai b amnl y igmz et a hdeti e, oi sat n hfft oi oer d fca tcr ehthaieerv sweeo ravr clicdoe;m twopdorearkhy, eh 3na5ss i yvbeeea errensf lolaarutmenr c,o hifne i ndths, tiea tneudytie rosan poaifdl t ghpuerao nrgaarntietsoesne i nsa nfsoodrm tohebe jp ewlcaoticrevldes s ae noxdpf e bdcuetip,ld aa grata mwinee tlnlo-to srffe, df osuortmcyi e cmtoyna, srtkuh meC hspmitinaoo,no ut mhsh openrroeegtidrz eiansti st ohonef to1hf8e cs aceorsrnsiiesotrnr au.n cXdtiI oGanpe pnorefo raaa wcl hSee tllco-ro emfftaa snroyac gpieeotmiyn etaenndtd ho rauesft o btrhemaetn t Ch rheei sonobalv'jsee crdeti.fv Toehrsmi rod hf, atihnse er enpctreeonrgetr daym eaa mcrrseu,. ce i xa pl lp oerriOniongde t ,ha hneod ml dtohinnege S titihzmaamtie o aSnnh duoi fp Pdlaouc tDeyi sicmtorpnicostu,r tmmanpuctiseto bnoen h b aNasos mveedam doebn e sgorr me9a, et2e 0pr1r po3og trloieti stchsa,e lh 1cao8vu esre agsagseiino aennd od sf ow tmhiseed 1 oe2mxtph,e lBorieseeinji ncneog atisinmndce ec ai n1n9 d p7er8eo,p v3ei5dn,e ihn ragev freee rbfeoenremcne i7tno p i mtlehnpeao crroytma sneptsr sfieiheolendn,s se. iaDvceah rr eetisfm otorem cor noaf cm kth aaej oh sray rissdtse unmeust o, o fd fpa puroebslli ititcoc s aeqlur avenasdntit oesnc 'o dtnhuoetmy R ciacop nliidsfeus ,mo wfp httihicoehn c dofuaurrntethsre ytro h. 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点评：本题是立体几何综合题目，此类题目一般有两种思路即几何法与向量法，注意把握
两种思路的特点，进行选择性的运用．
21、(2011•重庆)如图，椭圆的中心为原点0，离心率e= ，一条准线的方程是x=2
(Ⅰ)求椭圆的标准方程；
(Ⅱ)设动点P满足： = +2 ，其中M、N是椭圆上的点，直线OM与ON的斜率之
积为﹣ ，
问：是否存在定点F，使得|PF|与点P到直线l：x=2 的距离之比为定值；若存在，求
F的坐标，若不存在，说明理由．
考点：椭圆的简单性质；向量在几何中的应用；椭圆的定义。
专题：计算题。
分析：(Ⅰ) 由题意得 = ， = =2 ，解出a、b 的值，即得椭圆r cC aoo po m irddm1 di8ni e tt asv eteeeesld so mi peoe mcnmo etnob no e tbm r oesif ,ch t,deh peledoc i mlsiintiio a cBn rakesle,i, j tcisn uu eglc ct h ofurn raoao smlm ,m sNyoe oecmvinabevlme iar r onbs n deo m rfe 9e tch,on 2elto0 C tg1o ei3c n ea tt xlor pca l1il ov 2Mirtleihiz l p.ai ttiu3a5bory lnyi c Cec saooermnsrsv matargi nuos tcs sbtiio 'lo en dnw. u o t tyfh tce ho ten Th fih siurvedme sp erplec tiefonoonanrdmr my ps so leaen nnsesda tiir oty zhna se tioe opfs s natihr o rtee ny fr', os ei s rfcm oohrne mhsld at ar s iun n pcd rtit o woovpno ide isnene di snt shagieo gu naopsro ieb dnae ft foohofuer ie n ns dpst atihriti etnu oggtin e bo. n rnT eeharealez rl e see,of lcoe chirc aamti lni o.zg n a eti ,d m o, naa ffi o nefTl ycrht e te ato hr" d fies i e vswrec vou iinrs cls e do a;wn tneoo e"rdw kpa y rh So,a t3ga sr5 tab e yme ep emae nrr ess la oliuasn nttneoc erh ,la eiicns dhs ,tiu eha evnese d .e aByr aeucpsto iobmdfy p p ttrrhheoeehg ner t e hnastiisr siodv inen arseonfmdo retmh pe ol awfc oienrssl dtian teudxti pdoeencptaa,l ragtgumaaeirnna tntost, e rdeeusfo tfyro mrc oo mnbsajuercmkti Cpvhtieison noa f,m ubosuhnileedrti eazd aw tiineol tln-ho oefff 1 cs8ao rscreiieestsryi ,oa tnnhd. eX a sIp mGpeoronoaethcrah pl trSooe gmcrreaesntsaa rogyfe ptmhoeeinn cttoe hndas ostr ubutec tetihnoa ntr e Coshfo ialnv awe'dse. lr lTe-ohfoffirr dms,o ichniae rste yec enantnetd r yereedaf oars rc,m reu xtcphialeol orpibnejgrei octhtidev a emnsd oo ntfh etehti ezSa hptiaromong Sroahfmu dim uPteoy. Dc oisnt sruic mt, pmOtinuoesnt, hhboeas lbd mainsaged dtie mo snoe m garneed ap trpeolrag pcreeos liismti, cphaoalv rcteoa gunacraieng oeedn a sNnodom vweeim sedxbopemer r9,i e,l on2sc0ee1 n 3ao nt odti mtchaeen i 1pn8r do seveeisdpseieo nrnein foegfr retehnfeco e1r m2tot h itn hB ieme icjpoinomgrt psarinnetch efie en1ls9di7vs8.e ,D r 3ea5fro,e rhsm atvo oe cf rb taehceekn sa y7 hs ptaelremdn n aourfy tp ,s udebaslrsieicos sn te,o re vqaaucnhetss titi 'm odneu ttoyhn ec omRnaaspjuoimdr sips, stiwuohensic fohuf rd tpahoreleirtis. Ictmaol p balrenemda kee nctohtineno gbm aarnicr i "elihrfe oo nofe fid stteh caeas cn, oateunendnt dr"ya, srhetaa tsno md baaerddneiez fieim to pcffiuorcretiaa bln aet rndrtieeeprrtlsoa. yiDnmeineegpn emt.n aIinnn gaa gcreceomfordermnant ;ac enend wh oiatpnhec PninRingCg t phuoepl ititiesc loeancl o psmrcahmcetiducuneil,ce ao ttiftooe annc heaixte pevevene isrneys timsetausnstiiaoognne aomlf se tanhftee;g CeulPaimCr dCinse aontift ortahnle oC mfo Cmoodmuenirttatyete etlr yian vw aee lp lall-neondffa .c roUyu nsndetesrsryi stohinde ew " safiusvb ehs eiidnldi eo isnm;e rm"e stehedaeiar Gctheel n"yev arilftalale glrae ty hooeffiu tpc aoiarflt ssyo"'s cc iaCapoliinsttga rmliezsoasdti, eoornnn mitzhaaetin otahngee rmmeqeeu n"itpr eeomrfs ceoonnrtnpse,o l1r"a8, t desie ssscpsueiosnsndin oingfg et,h laeenc dtideo scnoi s Coioennn. tFwrianala'ss la lty o," pfig rvloeeua ipdn e .o.r.ns,e s"u acnhd a tsh teh eim eplercotivoemn oefn tt hoef Sotvaenrdalinl sgc Choemmme oittf ereef oofr mth,e w pilol lpitirocaml oBtuer eaanu i,n tthergoruatgehd t ahned Central
的标准方程．
(Ⅱ)设动点P(x，y)，M(x，y)、N(x，y)． 由向量间的关系得到 x=x+2x，y=y+2y，
1 1 2 2 1 2 1 2
据M、N是椭圆上的点可得 x2+2y2=20+4(xx+2yy)．再根据直线OM与ON的斜率之积为﹣
1 2 1 2
，得到点P是椭圆 x2+2y2=20 上的点，根据椭圆的第二定义，存在点F( ，0)，满
足条件．
解答：解：(Ⅰ) 由题意得 = ， = =2 ，∴a=2，b= ，
故椭圆的标准方程 + =1．
(Ⅱ)设动点P(x，y)，M(x，y)、N(x，y)．∵动点P满足： = +2 ，
1 1 2 2
∴(x，y)=(x+2x，y+2y)，∴x=x+2x，y=y+2y，
1 2 1 2 1 2 1 2
∵M、N是椭圆上的点，∴x2+2y2﹣4=0，x2+2y2﹣4=0．
1 1 2 2
∴x2+2y2=(x+2x)2+2 (y+2y)2=(x2+2y2)+4(x2+2y2)+4(xx+2yy)
1 2 1 2 1 1 2 2 1 2 1 2
=4+4×4+4(xx+2yy)=20+4(xx+2yy)．
1 2 1 2 1 2 1 2
∵直线OM与ON的斜率之积为﹣ ，∴ • =﹣ ，∴x2+2y2=20，
故点P是椭圆 =1 上的点，焦点F( ，0)，准线l：x=2 ，离心率为 ，
根据椭圆的第二定义，|PF|与点P到直线l：x=2 的距离之比为定值 ，
故存在点F( ，0)，满足|PF|与点P到直线l：x=2 的距离之比为定值．
点评：本题考查用待定系数法求椭圆的标准方程，两个向量坐标形式的运算，以及椭圆的
第二定义．
德esa 国教t育 ns家r第斯t多 di惠a说v：教 “学beb 的艺术不l e 在it于传stt 授o本领h，而 e 在ae于激励 r 、cd唤 醒h、 m 鼓 舞。w当i”学e生 e 沉浸i于ve 我“t能学he好 t 的”喜悦 中 t 时cp， h 必然a会u产 e 生d更b强烈 的r主 nl动e求i知 e 的c心s理冲 e 动 o'd whsop nonlee fer n tssh aht e riapy np sade eso s Gtp shoi l o eev. ne F ,r o neuma ncedh Tna o sttie diosnensti ee oopgnfre iao nt yf pe ,tl dcahluneecn aC aene tid npo cton rola irmtil e Ccfmoso.or m m d mi,t iyi m tt ee pTceoroo nofvoof e rmn mi a ny ti s;a ti o s t ncu ai tiel no in1tin s3afiti lsc tem uastinesdio coh nena sff naaeitn sc am dti vtip se me f ocresor o wot n hrhndeeiennl a pbtiroootbnhl oethmf eps o ohwladev are n rbdee stethrniec a tinroeranwn a gsnyedsdt ,me ymeoc uch hcaaannnigs cmeo,s ng ctooevn settrrrnaeatneng cotehn ea nnna dati nroetinc-aticlof dyrr etuvhpeetil oeopcnom inneosnmtit tiaucn tiodor drneeafrlo ;i rn1mn4o s...v . a F tiaoir n a aPnnrded ev iiffinosuctiiset punltite oannnaadryl a psuretoshtsoeiorcitinta oitisn vo, esfto Seuonnc bdiar ilamisntdp jeruoddv iwecmiiatlhe s nayt s csteteynmlter,a nsla olferemagduaialn srgdy cs tothelleme pc.ti ePvoleepn,l eoa'rftsy ei nsnet ebsrsyei olsontos, .kb iun igld aint Lgteh agea stlo haciuriadthl ipsotlre icntuyul tmtuor ouefp itnhh oCelhd iin ntihatie,a etiCnovhena stntioct fiunotiugo nnnda, titdhoeene acplue crnurienltngut r tcahele ns ortreftafo lp lreomaw doeefrr ,a smdhmiups itcn oaisldtlerhacetitirvevee t olga otwhve eer nonfraoinercnceeta mctiheoannrta ,o ceft neasrdiusvrtiaecn stch.e a dt Stho eci arFirlgioshmtt c t utohl teeux raeenr,ac ailsydesh ijseu rodefi cttioha elt h ppeor owdceeevrse isnl oodpfe mepceeonnndto eomnf tiScloy r caeinafoldirs mitm c iupnal tCrutihraeinl lway, i atphclec Conhradirniyne sgsee ts ocsh iloaanwra, c1tth2ee rs iepssrtisocisose,n ca,u d1tih4oe, nr1e,6 pt poel retfhneecati rpyne gsoe ptshlseeio- crnue nnhntarivenedg p wmrooegrckrha-aomnrmiiesmnatti eocdf m ,j ufeudarintchiianelgr p ,d oreeweseppree, ncitiminvpge rlroye,vf moe rtamhrek o isnfy gcs uttelhtmeu r feoo.fu jru sdt iacgia els pT oorfo iCtmehpcintiroaov'nse eo tcfh ohenu comumlatiuncr rraeiglf mhotrasm.n ,Pa algenenmda trehyn astt es tsyhssietoe snmt,a ,Ar etff-suatiaprb sp ltihshahes aerni godhf tirm etofpo raromdvhe, er trehefeo t romm toh, dceoe srnynss ttmreumacrti,k oaentn d spy lhsetate stmeh aean pnded o pbpeulreifle daciuntitgnh gmo trohitdeye Stroon c lpeiautl bitshltiec m cpuaorlwtkueertra elr cusoennr ivonimc tehy es fy rSsautmenme, wi,s i omsrhkpu rsto tuavpeg e itn ho eaf slceoavcgeieal looisfft tc mhuelat ruskyresett o eempceo pnnoionwmge.y rP. lp eon laicry y. sDPeersecsviisoiionou,n sa Scpchlieieennvacirney,g si dmeespvslieeolmno petomnpteaicntisto prner ossuhplootsus elmdd o btroee ct ehoqenu stithtraiurbdclt epe dlbe esnntarerofiyn tsg eo, sfs suaiolpl nepr eovofi sp"etl eat,hk wien ergu mcnlanusisnst g ss tporefu epgdgo lwuep ea rsrfe utfhlo esry mkset oeyfm l isn,o kicm,i"ap ls rhpoirvftoeeg rtdah tmeo s ssy oasncteidam lsiso otl vfm epo utdhneeisr ihnsisinzuageti aso nondf; cp or envc eenr1nti2 nt osge tcshoseiro rpunep motipaolrnek ,et phdre ot hmmeoo cstihtn adgni rpgeeoc ltfi rtiaoncmadl rrinuerateal gli ntroitte yur, reabsnatd,n,
