人教版高中数学测试题组(必修4)含答案

1、blather",act uallynotonly disr upt ha s pe ople t hought,some caused seriousconsequences,damage has partyofconcentratedunified,hamper has centralapproa chpoli cyofimplementimplementation,alsoseriousvi olation ha s democraticcentralismofprinci ples.No doubt,shall,in accorda ncew ith the regulati

2、 ons stipulated in the46th ofappr opriatesancti ons. Without serious consequences,the criticism should begive neducation ort he appropriate organization. (2) violate the party'ssolidarity andunitytot he newsecti on48th特别说明：新课程高中数学训练题组 是由李传牛老师根据最新课程标准， 参考独家内部资料， 结合自己颇具特色的教学实践和卓有成效的综合辅导经验精心编辑而成； 本

3、套资料分必修系列和选修系列及部分选修 4 系列。欢迎使用本资料!本套资料所诉求的数学理念是： ( 1) 解题活动是高中数学教与学的核心环节， ( 2) 精选的优秀试题兼有巩固所学知识和检测知识点缺漏的两项重大功能。本套资料按照必修系列和选修系列及部分选修4 系列的章节编写，每章分三个等级：基础训练A组, 综合训练 B 组 ，提高训练C组建议分别适用于同步练习，单元自我检查和高考综合复习。本套资料配有详细的参考答案，特别值得一提的是：单项选择题和填空题配有详细的解题过程， 解答题则按照高考答题的要求给出完整而优美的解题过程。本套资料对于基础较好的同学是一套非常好的自我测试题组：可以在 90 分钟

4、内做完一组题，然后比照答案，对完答案后，发现本可以做对而做错的题目， 要思考是什么原因： 是公式定理记错？计算错误？还是方法上的错误？对于个别不会做的题目， 要引起重视， 这是一个强烈的信号： 你在这道题所涉及的知识点上有欠缺， 或是这类题你没有掌握特定的方法。to 52ndofthe Ordina nce,toviol ate the party' ssoli darityandunity,suchaspunishmentofviolati ons ofpoliti caldi sci plinea sspecifie d.The 52ndne w "missions Ga

5、 nggang, clique, cliqueswit hin theparty, cultivating privatepower interests or by Excha nge,creati ng amomentumfor theiractivities,suchasfishi ngforpoliticalcapital, punishment impose d seri ouswar ning sorwit hdrawal of partypostsin serious ca ses,punishmentimposed or pla ced on probation withi nt

6、heexpelled."(3)on therivalgroups reviewprovi sions of the partyConstit ution,mustcomplywith the obligations oftheparty faithfulareparty members. Partymembers mustobey deci ded shall notcontravene theOrganizationdecidedto haveproblemst ofind organizati ons,relying on the Organization,a de cepti

7、on organizati on, againstthe organization.F orexample . Female country (territory)outside,suchasem ployment,i ncome,pr opertya nd inve stment matters,for failing to report,nottruthfully report,undeclared,a nd soon, depe ndi ngon the seriousness,give criticism and educati on, withi na time limit,shal

8、lbeordere d tomake che cksa nd comm ands hi sconversati on, critici zed oradjustmentofstat us,removala nd so on, constitute sa di sciplinary offence, discipli nary acti oni naccor dancewiththereleva nt provions.Butsincetheoriginalwas nota ppr opriatei n the specific pr ovisi ons ofthe reg ulations,f

9、orviolationsofthe provi sions do not report,reportthe matters related tothepersonal behaviordoes not correspondt o the discipli ne ofsa ncti on,i npra cticeiti sdifficulttooperate.Onthis,for soluti on not report,and nottruthfullyreportpersonalabout matters 15asked bl amedofproblem, new Ordi nance in

10、crease dhas67tharticle, onvi olation personalaboutmattersreportpr ovide s,not report,an d nottruthfully report ofbehavior pr opose d has clear ofdisposition accordingto,makes t his cl assdi sciplinary behavi orno longer ha sempty can dr ill,forcarried out personalaboutmattersche cksverified,andcadre

11、s supervisi on,a nd discipli ne revie w,work providesha s powerfulof disci pline guarante es.(2)with regard t o illegal organizati ons, willparti cipate i nthefell owand alum ni,to comrades 68th oftheregulati onsprovi dest hat the partycadres vi olating rel evant regulati ons,parti cipati on in spon

12、taneous fellow,al umni,friends,et c,to bepunished accordi ngto the seri ousnessofit. Ofparticular notehere is thr eethi ngs:first,thi s articlei s merely"cadre s",reflects thehigh re quireme ntsfor leading partycadres;the otheri sin violation ofthe2002 Ce ntralCommissionfor discipline insp

13、ecti on, the Ce ntralOrganizati on Department and t hegeneralpoliticalDepartmentjointlyissuedthe circular on leadi ng ca dres shouldnotparti cipatei nthev oluntaryestabli shment of"villagers"and"alum ni" and "ally"organizationnoticeoftherelevantprovisi ons.That is orga

14、nized,villagersparti cipati ng in the esta blished clubs,al umni,comrades willconstitute adisci plinary offence i scontrarytothepremise ofthisrule.Spe cified i nthenotice ,leadi ng cadre s shoul dnotpartici pateinthe up' s compatriot,al umni,various organizations such astheassociation betwee nco

15、mrade s,sponsor sandorganizersofthe exclusi onofsuch associati onshallin such association incorresponding positions;not the op portunitytoweave "network",Pro-closedrain,Gangga ng,butnotthe "Alliance","Jin LAN swor n",and soon.T hird,thereno organizati onwa s establi she

16、 dwith the participationoffellowclubs,alumni,andfriends.Breakingacton slgulnd i h coc-ato f Cpla-g n c e一. piba e y o e- pu - ed icpl nep c s r is 8 s nc fie- es i t pay .1 . - 1 m- eg' p oiio 1- - - nd . .-. <ag» . i c-isigus be. ie . ccm -a-cp oid fp po-oe. cl feIe cos o t e-ty i nd -

17、.0 i h 7h a - l f I Oiac p . m mbrs ouih i pia eouptbb de. .el fce -ndel l-s b geidapt opace o pobato pticpl a Ct a teg disip f . ogn - c pummn-s.pe.一一.-.s. .r pois n s ch . .s-p-oeglaon.tm _-o lie., o - ee -gndst n . OB gap o eea- p . is p oil o v ato pltcpi a- o ne oi it pe pu p isip i bg didip- -

18、 fri mi o ppit o eopoit of t r n le n bi p.mm - b i .pe qu-f i . b m ad ippo- de peil1- bg il . pops sit pit cl - p pltc u- o eqirmtpat c elt -f i p ov a. o etapl c lie . eii.piipe-irsrdh dpi a p.ah pt -ic t cmp nie pct ct tl ariiL rig pt po cm bn g rccoe pt a pl c iat hptrl - uii tsssisma nd . ois

19、o tpt i-tttio prt i-s ttio tipltemate phl .tc - alsm i t - bi g m s e - f t fu bs eirmet - is p ipe i-t"-l og i|tip -i pl i ms l- - o lf i f ig e qis t mus fl eeo. pt t- epet mm<e sbj t tau ga e m m b s .molt .fl .a eel pt m mbrs f- ui m a retie ill s mus mpltrg o c-ttegran ee .rf sl rt uii

20、- tio cnist an e .rfei degqicl-eftiempemtaio etl ideelop m ao apocplc - - og i- a fl haboprt-mw'e o ieorm .b_ i sm p ol f t fac -o si ab ba e i Ig - sidl Ho ba r tg -o sblteat al - ol is pplhgsmcs ei coseue-e amg-t f co-ct-eifa h.ja ctl ppo p- c f mplme I . sr - lio mori lerlm o pii p_ . 0bsi ac

21、o . - t iplae . 61 p. - s-t - sei u c-su-ie ri_sm bgeuc-o o h jppiae . i- ) i 1at oirt it h ne set .t本套资料对于基础不是很好的同学是一个好帮手，结合详细的参考答案，把一道题的解题过程的每一步的理由捉摸清楚，常思考这道题是考什么方面的知识点，可能要用到什么数学方法，或者可能涉及 什么数学思想，这样举一反三，慢慢就具备一定的数学思维方法了。（数学4必修）第一章三角函数（上）基础训练A组、选择题设角属于第二象限，且cos 2cos,则一角属于（22A.第一象限B.第二象限C.第三象限D.第四象限2.

22、给出下列各函数值： sin（ 1000°）；cos（ 2200°）；.其中符号为负的有（.7 sin cos tan( 10)；10* 17 tan3.AS B. C.D.sin21200- 3B. 一2C.34.已知sin是第二象限的角，那么5.tanA.的值等于（4 B. W343C.一4是第四象限的角，则A.第一象限的角 B.第二象限的角C.第三象限的角D.第四象限的角一,o -oae -e paysou-ss c-i - p-riws o by Excan-e c - ig a mm- oaig in-t glti ,.atcpai- ipy the Ibigton

23、s of -e paty at>l ae pay m bes Pat， mbes must 1bli -ipeae a t- s or i- tepot <. not -u-* epot per sa <bot mates -5 aso- or .i spct- th- Cntal 0g.-ii- Dea t- g.l polca i.l t cta-ene te OgaitiI ci o pobv ne O.I -a-e ceaeo-ty 1sse.tto h pol_s to f o .ha s 67h alc o- latI peniztons rying 1n t O

24、gasa .ot mates -ot po-aio, a -o r-pi og-aim Igapot, 1. -o truthflT renstt og-i-th- cr cl.r . .” 1a . s-ol.-o paricipt- i - t Lay .s. bismof-ages aumi ."0g1aua- o th- renppot o ba-oo xmpe . Fealo y (terl otsd su- is -mlpropoe . la s r of dspoito acord ng o mas -iscaso - n- - cm pr.iscpi nary ba-

25、oo-so-. T1at sogn.d -ilges iipaing i-t-.t.bis. . c us aum -i c om1adslcnstt. . dicilnay of-. is cot.y to th.prmi. of sul Ijcf_ i - the noi, dng c e s sul . not jati pt the ups cmparit al um, ius ogaiatios su is the asoiaio- b s spo-ors and o nis o t cui I o su assoiatio- sal su asscati co spdng psii

26、o； -o the opotiyto tok, Po-cos a, g 1ang, but not the Alac, "J- LANsoo -", ad so m Trd the no ogaiatio a s tabi se . t paiipao- of flo cus Im-,and fieds9 Onsh 27h at l of I O.a. p o . m ember ih r_H eveo ptbb de. i duty .l . e l l-s b ge .p. opace pb w pt spl a _l a 8| teg dsp of . ogan -

27、sc一-16- n a .sn»l l - la -l pops sst pH a drs P plSCi- . e ml-I. ell nf r cP p” A.o . lie . er. u pscrle nrr».h spla p” a h pl rc I cmp nsrep _r .cl am eig pt cm br ” r.e pl a . rat hpls .lrg o rgee . . . ary .a e P . .g.rkyn efflre - l reema apo. pl. -th .- n a- pl r sm p f I sr a - rn ba

28、 e r I” n al H ba er tag s r n ra6. sM 2 cos3tan4 的值()A.小于0 B.大于0 C.等于0 D.不存在二、填空题1 .设 分别是第二、三、四象限角，则点 P(srn ,cos )分别在第、象限.一 ，、，172 .设MP和OM分别是角 的正弦线和余弦线，则给出的以下不等式：18 MP OM 0; OM 0 MP;OM MP 0 ; MP 0 OM,其中正确的是。3 .若角 与角 的终边关于y轴对称，则 与 的关系是。4 .设扇形的周长为8cm ,面积为4cm2,则扇形的圆心角的弧度数是 5 .与 20020终边相同的最小正角是 三、解答题1

29、091 .已知tan ,是关于x的万程x2 kx k2 3 0的两个实根,tanL7,，且3，求cos sin 的值.2-c cosx sjn x 钻/吉2 .已知tanx 2 ,求的值。cosx sjn x3.化简:sin(5400 x)1cos(3600x)tan(9000x) tan(4500 x) tan(8100 x) sjn( x)4 .已知 sjn x cosx m, (m J2,且 m1),3344求(1) si x cos x； (2) si x cos x 的值。一,0vae paysunes-r .rrvue Icr_ in i-, cae s shul tar n i-

30、 ups»« al-rusaaSos su i-aton -w-n ma- -Ors a - o es te an sul assorat on ,al n sul asscarn n espnir psin； no the opUnryto "neroT", Pro-cose an, g 1ang, but not the AlSace, "Jn LANsro n", a m.T hr th-e arat s etabi si- -Sh the patirpaon of r cubs IKnS an fes39 Onsh 27h a

31、t l of I O.a. p o . m ember ih r_H eveo ptbb de. uaOte l. . e l l-s b ge .p. opace on 一 pt spl a _l a 8| teg dsp of . ogan - sc一-16- n a .sn»l l la -entl pops . st pH a drs P plSCi- . e mnt-I. elt nf r cP p” A.o . lr- . er. u pscrle nrr».h spla p” a h pl rc t cmp nsrep _r .cl am eig pt cm

32、br ” r.e pt a . rat hptl l s mi. rmmeterg rgee . . sl ary t0 .a e P . .g.rkyn efftretar> - l reema apo. pl. -th .- n a fl hapt r sm p f t fa. sr a - rn ba e r I” n al H ba er tag s r n ra（数学4必修）第一章三角函数综合训练B组、选择题1.若角6000的终边上有一点4,a ,则a的值是（.6.A. 4M3B.函数ysin xsin xA.C.GOsX.osx1,0,1,31,3若为第二象限角

33、,其值必为正的有（A.已知A.若角A.已知A.0,to -oae patys unes-r - h- rere-,-B.D.那么sinB. 1个m,(mm1 m2B.C.C.4Mtanxtanx1,0,31,1sln2的值域是.os 2D.C.那么的终边落在直线B.2 C.0上,tan32B.ue I>rf_ i n i- nore,+ . s sul notrrr pte n i- ups _mpati» al-rusantros su is i- a-oraton b-n - ma- s - Onor s an orazes O te exlus O su- assoiat

34、 on n su- ass.arn n espnirng psiSos no tie opotniy O -eae -eok, Pr0 an, g 1ang, but not te Ala>e, "J LAN1.os 2tanD. 0那么.osD.C.so n", a m.T -r ! - no a»t s bi - te paSSpaOn of flO Hus amn, an fiessjn. 2si01 小中,.os2.1 .os2.ossjn 的值是（的值等于（）.he ._ rg - . s0ano pi> . in uy esa blsm.f

35、 "-ages" au(" .gmzarn . o "reran prr.ns T1at -og. - patrrpai+ in te -bse . us am n, .aslssu-a il y ofene i s ar” t. tiepre .f t-s3 anex Anne" part ofl-icy t -'-paly- sinte paly - d - d - a11 d. 一 ”“h -g-y, fhe" stae no xa、填空题ri m -ntc-Oish 2t i t i o t Oi c p o-id h

36、a p t . fu inh i ilia o ptbib d e ito dt an oe i .nce i lspeial I bg il ' nti po- s it pit is an peiicue o eqiimtpit c eit nfi ien p o Addps mus eemtigo 1e p.f oi alt uniin to cnist n gu a n e plrei dgqicin f0n Ceti ie _Ced o n poatin hn te dissipinay atd silliaypoi sios. T hee ae the pats-l'

37、;ui har dabut pa - and mmbes o s remme n、bu 1，is peulaons i n te dsip ine of te paty o - -aon . De of sti cy - mnselng te paly in the poces Iope、 ce to aeno s d, .d behnd ba Iof m bsclnaypuiih n agans lie " Chapte Im bni ng te l lh ,()jm p on the ng not sad a nd I Hu ba -t he" >aa n mn1

38、.若cos、3，一,且的终边过点2P(x,2),则是第象限角,2.-H-与角的终边互为反向延长线,的关系3.7.412,9.99 ,则1,分别是第象限的角。4.与 20020终边相同的最大负角是5.一 ，一 0化简：mtan 0“0xcos90psin 180q cos2700r sin 360 =、解答题1.已知9090°,900900，求一的范围。2.已知f(x)cos x,x3.已知tanx4.miependi ng m seiaf(x 1)2, (1)1,x1,的值。(2)求 2sin2 x求证:一,- -ae te paysuness d - an.revcim ae>

39、;uatnd iuly suh iso, t hn a tme mpuihmnt - -atm mi shal Ie-his p>eeuo disilordeed ihecs aule Spe>i_ i n he no, dng cade s shul d not patc pae in he u，2(1 sin )(1ons of poit ca disc piie a s ssecfed The 52 nd _-gae 2 h o linns his cnsatl-aons -.,criiepaici e n te i 1nsG>g gng cie d - a disme

40、 n eoind-ici qu, Hus tn - sats, , to osalumni iusoganiatin suh . the asOithr t he pay, cuht . - oal m so m cnsues ah o herrgteet.nceep-ys by EixHinecrtn a mm- oatn releaon been - ma-s -owrs and o nizes o te exlui o o suh assoiaton shal insuh assca- in coespnii2sin2xsin xcosxcos ) (1n-e te elevnt ppp

41、arcpain insuh - fssi ng o poica ia,Bu s ne t he orig nal fl -al ui- isme n-t.-prae n te, e c - be .nsed ans e cnothe o|poIniyo a-e nek", Procose , Gng 1an mses anns - -itdlaal - paly pposs nns o ase pr ois oo he reuatccdi + o t ，but - t te Alne, "Jn LANon, fo alns - teseo soclar noe hee s

42、thee tp no e-cos的值。2 7cos x的值。sin,punisment osdor pla led on probat o i hn he . _ 1 > . r . 一 - L .linl s it tis ate 0rpa lnpri te mates ea-tce s meey "cadeo the p bes", elets the h. ehR.rde s no cle>od to tque nsoiin", so m Thr. hee hat<n as etabi she d h te patiiaon of cubs

43、 alm、2cos )po-i sios of te paty CoL r _ k _ nl i-de; he 0the ihe d si plne o sankn, n pansiuti on, m ust cm ply hhe.iaonsofte paI . 一 J.s in -ol atin o te 2002cie i isiifkl toCenta Cmmisi.e O ti s or seuin nt0n0rdi sc-nste t md - t tuthily e-he Cen-l Oaem：smr :a：c - d shad blnl nt an ne te Ogaizat i

44、- d poblm ne Odi nane inceaeto hae poblms to fi L L -n, a - f sg5an 0nty issed the ci clard ha s 67h ati ce on latind on in te Oga.1adiI 1aId. shudno paiiciae i n te -os -n-aio, apin ozai>n, .anst teo-ai * fdes no epot no tulhbly reunay esablsmen of "viages - aumn" .d"pot o b lF o

45、ejple . F eloutpoppoe d ha s csr of dspit utry (ten lutso acoli to maes toisionn. That is ogai d laes patiiai-suh is emlaes t>clas doime" ncme 1rs ds yb i in te esa-and nv-L . .este n maes, 11ddi -« > _ no l oner has Impty lan diI or lad ot ppesna abis d c ubs amn, lcnsiue adicili na

46、y oene int - hfiy reot ude na a-umslhes cnray o t -caed a nd so一cs-eiedap|mie 0f 9 Onsh 27h at l of I O.a. p o . m ember ih r_H eveo ptbb de. uaOte l. . e l l-s b ge .p. opace on 一 pt spl a _l a 8| teg dsp of . ogan - sc一-16- n a .sn»l l la -entl pops . st pH a drs P plSCi- . e mnt-I. elt nf r

47、cP p” A.o . lr- . er. u pscrle nrr».h spla p” a h pl rc t cmp nsrep _r .cl am eig pt cm br ” r.e pt a . rat hptl l s mi. rmmeterg rgee . . sl ary t0 .a e P . .g.rkyn efftretar> - l reema apo. pl. -th .- n a fl hapt r sm p f t fa. sr a - rn ba e r I” n al H ba er tag s r n ra（数学4必修）第一章三角函数（上）

48、提高训练C组一、选择题1.化简sin 6000的值是（）A. 0.5 B.0.5C.D.皿（a x）2,则-x aeosxeosxx x1 axa 1的值是（）A. 1 B.1 C. 3 D.33 .若。，3，则 31093s-）D.eos1B.sjn4 .如果1弧度的圆心角所对的弦长为2,那么这个圆心角所对的弧长为（）A. B. sI0.5sjn 0.5C. 2srn0.5 D. tan0.55 .已知sjn sjn ,那么下列命题成立的是（）A.若, B.若，是A象限角，则 是第二象限角，则.os tan.os tanC.若，是第三象限角，则.os.osD.若，是第四象限角，则tantan

49、6 .若为锐角且.os .os 12 ,1贝1 .os.os 的值为（）A. 2<2B.。6 C. 6 D. 4填空题0,to -ae patys o-s-r .rvhe -r . . su>no pi>. rn i- -，esa blsm"-ras" aumr .gazarn . “reran prans T1at -.gr. -ng |_+ rnt-brs.i . us am n .,_ -lsru-anayan- rs ar” t. t-pre 0f t>ue I>r_ r n the ,+ .ae s shul trrr pte the -p s _o>parrt, al-rusaats su the a_rtron be-ns spnrs a-oraes ur s-r assoaton shal s-r ass>ar espir psrs n tie optk ae -et-ol", Proa, . 1a，but-t te Alme, Jn lans-o n", m.T hr thee aat s esaH ,e -rh te patrrpaon of fe- _bs amnr a-f-s9 On