黑龙江省哈尔滨市第三中学2019-2020学年度下学期高一学年第一模块考试数学试卷含答案

哈尔滨三中2019-2020学年高一(下)第一次段考化学试卷(3月份)一、单选题（本大题共12小题，共42.0分）1.化学与生产、生活、社会密切相关，下列说法错误的是()A. 中国古代利用明矾溶液淸除铜镜表面的铜锈，是利用了明矾的酸性B. 碘是人体内必须的微量元素，所以富含高碘酸钾的食物适合所有人群C. “光化学烟雾”“臭氧空洞”、“硝酸型酸雨”的形成都与氮氧化合物有关D. 用加入足量氢氧化钠溶液共热的方法可以区分地沟油与矿物油2.下列气体中，不能用如图所示的装置干燥的是()A. CO2B. O2C. H2D. NH33.下列关于物质分类的说法正确的是()A. 氧化铝、氧化铁都属于碱性氧化物B. 漂白粉、石英都属于纯净物C. 硫酸铜溶液、氢氧化铁胶体都属于分散系D. 葡萄糖、蛋白质都属于高分子化合物4.下列物质中，不具有漂白作用的是()A. 活性炭B. 氯气C. 次氯酸溶液D. 二氧化碳5.下列有关物质用途的说法正确的是A. 用烧碱治疗胃酸过多B. 二氧化硫用来漂白纸浆C. 明矾用作水处理的消毒剂D. 铝制容器用来储运稀硫酸6.实验室制取下列气体的实验正确的是()A. 实验室制氯气B. 实验室制氧气C. 实验室制氨气D. 实验室制二氧化碳7.同温同压下，将盛满气体的大烧瓶和水如图所示，则溶液的物质的量浓度关系是a.氯化氢和水b.氨和水c.二氧化硫和水d.二氧化氮和水A. a=b>c=dB. a=b=c>dC. a=b=c=dD. a>b>c>d8.下列反应的离子方程式正确的是()A. 铜片加入三氯化铁溶液中：Cu+2Fe3+=2Fe2++Cu2+B. 盐酸中加入碳酸钙：CO32−+2H+=H2O+CO2↑C. AlCl3溶液中加入足量的氨水：Al3++3OH−=Al(OH)3↓D. NaHCO3溶液中加入盐酸：CO32−+2H+=H2O+CO2↑9.25℃时，溶液M中含有等物质的量浓度的I−、NH4+、H+、Al3+，其中还可能含有NO3−、S2−、Cl−中的一种或几种。

哈三中2019-2020学年度下学期高一学年第一模块考试英语试卷第一部分：阅读理解（共两节，满分40分）第一节：（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。

1. Which event requires booking in advance?A.C asa Romantica Poetry Reading Series.B.Orange County Museum of Art—Visionaries Lecture Series.C.Discovery Science Center Presents: Mexico: Festival of Toys.D.D onna O’Neill Land Conservancy—Hike Off Your Thanksgiving Feast.2. A student of art who wants to learn about art history will go to ______.A.850 San Clemente Drive, Newport BeachB.2500 N. Main St. Santa Ana, CA 92705C.the Donna O’Neill Land ConservancyD.415 Avenida Granada, San Clemente, CA 926723. A mother who wants her kid to learn about culture in a fun way will call ______ for more information.A.949-759-1122B. 714-542-2823C. 949-489-9778D. 949-498-21394. A member together with his wife and 12-year-old son wants to take a hike will pay ______.A.$20B. $25C. $15D. $10BIt was New Year time, but I wasn’t looking forward to it. That winter, my mother and my stepfather moved our family to Southern California. My brother and I were leaving our rural Alabama behind. This would be our first New Year away from Alabama. My mother took to California like a swan to a royal lake. My athletic little brother, Paul, was excited at a climate that allowed him to go to the beach whenever he wanted.I, however, was a fat child with a heavy southern pronunciation. My first day in the new class, I introduced myself in a low voice. All I said was my name and where I was from. The class burst into laughter. “He talks funnily.” I was so helpless that I went to a place to call Granny Smith after school, who was my biggest support. But I didn’t get through.On Sunday evening, the phone rang. It was Granny. She often took advantage of the discounted long-distance rates on Sundays. She said she’d shipped a New Year package. Sure enough, it arrived. Surprised at the box, large enough to hold a small refrigerator, we eagerly tore it open. The smell of Granny’s house filled the room: a combination of fried meat, sausages, furniture polish and decorations. Her house was tiny and always filled with tacky holiday decorations and homemade food before New Year. But in my childhood eyes, it was precious and fantastic.There were countless tins and containers. We opened them to discover piles of holiday treats. She even included our traditional candy bars. The box was as bottomless as a magical box. There, beneath all these, was my familiar holiday.Every New Year that we spent in California, the postal service would call and say our package arrived. Over the years, many treasures arrived in the box. For me, it’s always been the best part of the holiday.5. How did he author’s brother feel when they were moving to California?A. Unconcerned.B. Joyful.C. Grateful.D. Upset.6. Why did the author’s classmates laugh at him?A. He had a strong accent.B. He made a humorous talk.C. He looked overweight.D. He spoke in a very low voice.7. What does the underlined word “tacky” in Paragraph 3 probably mean?A. Suitable.B. Large.C. Cheap.D. Attractive.8. Which of the following can be the best title of the text?A. Granny’s Care PackageB. An Unforgettable HolidayC. Our Move to CaliforniaD. A Telephone Call from GrannyCEvery animal species carries unique viruses that have specifically adapted to infecting it. Over time, some of these have jumped to humans—these are known as “zoonotic” (动物传染的) viruses.As the population grows, we move into wild areas, which brings us into more contact with animals we don’t normally have contact with. Thus, viruses can jump from animal to humans and they can spread between humans, through close contact with body fluids (体液) like blood or urine (尿液).Because every virus has evolved to target a particular species, it’s rare for a virus to be able to jump to another species. When this does happen, it’s by chance, and it usually requires a large amount of contact with the virus.At first, the virus is usually not well-suited to its new host and doesn’t spread easily. Over time, however, it can evolve in the new host to produce variants (变体) that are better adapted.When viruses jump to a new host, they often cause more severe disease. This is because viruses and their first hosts have evolved together, and the species has time to build up resistance (抵抗力). The new host species, on the other hand, might not have evolved the ability to deal with the virus. For example, when we come into contact with bats and their viruses, we may develop rabies (狂犬病) or Ebola virus (埃博拉病毒) disease, while the bats themselves are less affected.It’s likely that bats are the original source of three recently known coronaviruses: SARS-CoV (2003), MERS-CoV (2012) and SARS-CoV-2 (whose current official name is COVID-19). All of these jump from bats to humans via an intermediate animal; in the case of SARS-CoV-2, this may have been pangolins (穿山甲)，but more research is needed.9. Why is it rare for a virus to infect another species?A. The new host can deal with the virus.B. The new species builds up resistance.C. None of the virus can adapt to the new species.D. Each virus develops to target a specific species.10. What is the fifth paragraph mainly about?A. The resistance of the old species.B. The ability of the new host species.C. The infection process of virus.D. The cause of stronger disease.11. What can we infer from the text?A. Humans should not move into wild areas.B. Wild animals are actually our best friends.C. A virus can spread from animals to humans.D. A virus can evolve to infect another species.DChildren love playing games. That is a well-known fact. Kids learn how to play games at a very early age, and once they find a game they love, they can play for hours and hours on end. In fact, kids’ games are so popular that there is an entire industry devoted to designing and selling them. There seems to be a fun game for kids for every occasion. Christmas and birthdays are often celebrated with the family tradition of gifting a new game.Old games handed down from generation to generation include such favorites such as hopscotch, red rover, London Bridge and duck-duck-goose; they are always big hits and still played at many schools during recreation time.In recent years, though, Internet games and gaming systems seem to have taken the kid s’world by storm. Kids’ online games have quickly become a very popular way to pass the time with small children and teens alike. Online games are a great way to have fun, and they can be quite educational as well. Many online games created for specific age groups encourage grade appropriate development and educational skills.Hundreds of Internet sites offer fun and educational games for kids. Some games need them to use brains, while others involve hands and feet. Many of these sites are created by big names, such as National Geographic, which we can trust. They offer free games that parents know are appropriate for kids. Parents are encouraged to go online and research online games appropriate for their kids’ age before allowing their kids to play these games. Parents want to have the ability to set or monitor their kids’ play time, which is a problem National Geographic is working to solve.With so many options to choose from, parents and children will find it easy to find fun games and activities on the Internet. And nowadays, being able to use a computer is a very important skill for your little one to learn, so playing on the computer is definitely both educational and fun. Your child is sure to learn new skills about any subject with the right, trusted website.12. The industry of kids’ games is prosperous probably because _______.A. there are all kinds of games for childrenB. children are mostly crazy about gamesC. people would like to play games togetherD. parents love to buy gifts for their child13. We can infer from the third paragraph that online games are _______.A. popular and practicalB. complicated and expensiveC. similar and educationalD. funny and instructive14. Which of the following is true of National Geographic?A. The news is updated every day.B. There is a large amount of information.C. The games are good for kids’ mental and physical health.D. It can help parents to control kids’ play time.15. According to this passage, the author’s attitude to playing computer games is _______.A. supportiveB. objectiveC. negativeD. pessimistic第二节：（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

哈三中2019-2020学年度下学期高一学年第一模块考试莫语试卷第一部分＜阅读理解（共两节•满分40分）第- V/：（共15小题：毎小题2分.满分30分）阅读卜列短文，从每题所给的以个选项（A. B、C^II））中•选出⅛M圭选项•井在答题卡I •将该项涂然•岛一英IS第I页共10天I ・ WhiCh event requires booking In advance9A. CaSa ROnWntiCa PoCtr) RCading SCriCs.B Orange COUnty MIlSCIlm Of Art ViSiOnarieS IxCtUre SeneSC DiSCOVei> SCiCnCC CCnICr PrCSClns: Mexico: FeStiVal OfTOySD. DOnna O'Neil) Land Conscnancy—Hike OtT YOllr ThankSgiVIng FeaSt-2. A StUdent Of an UhO WantS (o Ieanl alx)ut art hιstoπ, WilI go to 一.A. 850 San CIemente Drive, NeWPOrt BCaChB. 2500 N. MaIn St. Santa Ana. CΛ 92705C the Donn<ι O'Neill l>an<l CoiiscrvancyD. 415 AVenida Granada, San Clemente, CA 926723. A Inother WhO VVantS her kid to Ieam about CUhUre ir a fun VVay Will CaIl _____________ f or moreinfbπnationA. 949.759∙1122 B 714-542-2823 C. 949∙489∙9778 D 949∙498∙21394. A IneInber IOgeIIIer WitI) his VVite aικl 12-year-old SOn WalnS to take a IIIke Wlll Pay ___________ ・Λ $20 B. S25 C.SI5 D.SIOBIl WaS NeW Year time・ but I WaSlVt IOOking forward to it. That winter, Iny mother and Iny StePfather moved OUr family to SOUthem CaIifbmia. My brother and I Were IeaVing OUr rural Alabama behind ThiS WoUId be Ollr flret NCW YCar away from Alabama. My mother took to CalifbnIia Iike a SWau to a Ioyal Iake- My atlιle(ic IilIle brotlιer. Paul, WaS excited at a CIiIlIate that allowed him to go to the beach WheneVer he Hanted 1, IIOWeVer, was a Iat Chlld ∖vιlħ a heavy SOUtll ell) prominciation My first day In Ihe new CIaSS. 1 ImrodUCCd myself In a IOVV VOiCe- All I SaId WaS ιnv name and WherC 1 WaS from. Γhc CIaSS buιst into Iauglitcr. U He talks fuιmily.v 1 WaS So helpless Ihat 1 WenI to a PIaCe 10 CalI Granny Smlth after SChOOLwhG VVaS Iny biggest SuPIX)It BUt I didn't get throιιglι.On SUIKIay evening, the PhonC rang. It was Granny. ShC OftCn took advantage Of the discounted IOiIg-distance rates On SUndayS- She SaId She d SlnPPed a NeW Yeaf PaCkage・ SlIle enough, it arrived SUrPriSed at the box. Iarge enough to hold a SmaII refrigerator, We eagerly tore It open. The Smell Of Granny W S house filled Ihe room: a CombInatlOn Of fried meat, sausages, furniture PoIiSh and decorations. HCr house WaS tiny and always tilled WIth hκk∖holiday decoιations and IIOnICInade food before Nev∖ Year. BUt in Iny ClIiIdhOod eyes, It WaS PreCioUS and fantasticTherC Were COUntIeSS tins and containers. We OPCnCd them to discover PlIeS Of IlolI(Iay treats. She CVen included OUr traditional cand> bars The box UaS as bottomless as a magical box. ThCrC・ beneath all these. WaS my familiar holiday・l⅛ ^tfi第2页共IC页EVery NeW Year that We SPent in California, the POStal SeniCe WOllId CalI and Say Ollr ∣x>ckagc aπ,ivcd Over Ihe years, main treasures am∖ed in the boκ. FOr me, it's ak¼ays been the best part Of the holiday5. IlOU did he author's Broilier IeeI Uhen tlιe∖ Uere moving to California?Λ. UnCOnCemCd. B. JOyfUI C. GratefUL D. Upset.6. Why did the ClUthOl S ClaSSmateS IaUgh at him?A He had a StrOng accentB He Inade A humorous talkC. 1 Ic IOOkCd overweight.D. He SPOke In a Very IOW voice・7. What does the UfKlCrlined Hord U taCky M In ParagTaPh ? probably mean oA. SUitable.B. Large.C. ChCaP.D. AttraCtiVC-& WhlCh Of Ihe （OIlOVVlng Can be the best title Of Ihe text9Λ Granny5s Care PaCkage B An UnfbReuable HOlidayC. OUr MOVe to CalItbmIaD. A Telephone Call from GrannyCE∖ery animal SPeCiCS CarrieS UnIqUe VinlSeS that have SPeCIfiCally adapted to infecting it. OVer time, SOme Of IheSe have jumped to humans——these are known as "zoonotic （动物传染的）VirUSCS ・AS the POPllIatIOn grows. VVC move IntO Wlld areas. NhlCh bπngs US into InOlC contact Wlth animals v⅜c don't ικ）∏nally IIaVe COntaCt VVith l ThUS. VIrIlSeS Can jump from animal to huπιans and they Can SPlCad between humans・ through ClOSC COnIaCt Wlth body fluids （体液）like blood Or uruιe （尿液）.BeCaUSe every virus has evolved to target a PartiCUIar species, it's rare for a VinlS to be able IOJUmP to another SPeCieS、WhCn this does happen, it's by ChanCC・ and It USUaIly requires a Iarge amount Of COntaCt Unth the ViruaΛl first, Ihe VinIS IS USUalIy not WeII-SInICd Io its new host and doesn't SPread easily, （）∖cr time. ho\、c、ci\ it Can CVOlVC in the new host to PrOdUCC VariantS （变M i） that arc better adaptedWhCn VirUSeS jump to a new host, they Often CaUSe InOre SeVere disease・ TlIiS is because VirUSeS aικl theu IlrSt IK）StS IIaVe e∖ohed IOgether, aικl IhC SPeCIeS has time to build UP resistance （抵抗Z/）・Tħc new host SlXCieS. On Ihe Other hand. miglιt not have evolved IhC ability to deal VVlth Ihe virus. FOr example, When VVe come IntO COntaCt Wlth bats and their viruses, we may develop rabies （壮犬炳）Eboh MnlS （埃博荷柄毒）disease. Whll C the bats themselves arc ICSS a∏ected.It s IIkeIy that bats are Ihe OrigInal SOUrCe Of three recently known coronaviruses: SARS- COV （2003）. MERS-COV （2012） and SARS-COV-2 （WhOSC CUrrCnI Onkial name is COvTD-19）. All Of these JUnIP ITonI bats to humans Vla an Intennedia:e animal; In the CaSe Of SARS-CoV-2, IhIS may have been PangOIinS （穿山r P）, but more research IS needed∣⅛英语第J页共10页9. Why IS It rare for a VInIS to InfeCt another species?人TIIe new host Can deal With the virus.B The new SPeCieS builds UP resistance.C. NOIIC Of the VinlS Can adapt to the new SPCCiCS.D EaCh VHUS d已∖ek^∣)s to Iaf gel a SPeCifiC species.10. What is the fifth ParagraPh mainly about?A. ThC rcs∣sta∣κx Ofthe Old SPyClCs.ThC ability OflhC new host SPCClCS・C l he InfeCtlOn PrOCeSS Of VlnISD IIIe CaUSe Of Stronger disease・11. WIIat Can We Infel from Ihe text?Λ HUmanS ShOIIId not move into Wild areasB Wlld animals are actuallv OlIr best friends.C. A VlnIS Can SPread from animals to hιπnansO. A VinlS Call CVOIVe io infect another species.DClnIdlen IOVe PlaJing games. Ihat IS a Uell-known fact. KldS Ieanl IIOW to Play games at a very* early age, and OnCe they find a game they IOVe. they Can Play for hours and IIollrS on end. In tact, kids games arc SO POPUIal e that there IS an CntirC mdust∏f dcx,otcd to dcsigιmιg and SClIing them. There ScCmS to be a iun game for kιdδ for every OCCaSlon. ChnStnlaS and birthdays arc Often Celebrated Wi(I) IIle family (radilion Of gi∏ing a new game.OId games handed down IrOnl generation to gcnσation iικlude SllCh fa∖oriιes SUCh as hops cotch. FCd rover. LondOn BrldgC and (IUCk-(IUCk-g<M>sc. they arc always big hits and StiIl Played at Hlany SChoOlS during recreation tnne.In recent years, though. Intemet games and gaming S)StClnS SeCm to have taken Ihe kids, WOrld by StOrm. Kids' Online games ha∖ e quickly become a VerY POPUIar Way to PaSS the tune With Small ChildrCn and teens alike OnImC games arc a great VVay to have hιn. and they Can be quite CdUCatiOnal as WeIl Many Online games CIealed Rir SpeCifiC age groups CnCoUragC gιade appropπa te de∖elopιnent and educational skills.HUndrCdS Of IntemCt SltCS OffCr fiιn and CdllCatiOnal games fbr kids SomC games need them to USe brains, WhlIe OtherS InVOh e hands and ieet. Many Of IheSe SlteS are Created by big names. SIICh as NatiOlIaI GeOgraPhi e, WhiCh We CaJI tnιst ThCy Ofler free games that ParClItS know arc appropriate Ibr kids. ParentS are encouraged to go Online and research Online games appropriate fbr their kids' age before allowing their kids to PIay these games ParentS Want to have the ability to Set Or InOnItOr their kids* PIay time, WhiCh is a PrObIeln NationaI GeOgraPhiC is UOrking to SolVeWlth So Inany OPtiO<ιs to ChOOse from. ParCntS and CInIdrCn will find it easy to find IiIn games and activities On the InternCt And nowadays, bein g able to USC a COmPlltCr is a Wy iιn∣x>rtant SkiIl for your Iiltle One to Iearn a SO PIaying on Ihe COn)PUter is definitely both educational and fun. YOUr Chlld is SUre to Ieanl neu- SkiIIS about any SUbjeCt Wilh Ihe right, IrUSIed WebSIte岛一英ifi第J页共IC页12 ITie IndlIStn r OfkidL games is PrOSPerOlIS ProbabIy because ___________Λ. there are all kinds Of games ∏)r ChiIdrenB. ChIICIren are mostly CraZy about gamesC IxX)PlC WOUId IikC to PIay games IOgCthCrL). paients IOVe to buy gifts for IheiI CiUld13. WC Can inlcι fr om IhC third ParagraPh that OnIinC games arc __________ ・Λ. POPUIar and PraCtlCaI B. COfnPliCatCd and CXPCnSiVCC. SimiIar and edιιcarional D fiιnnv and instπιctive■14 WhlCh Of ll>e IbIIovMtig IS true Of NatIonal GeOgraPhICeA. The news is UPdated e∖ery dayB ThCrC is(i IargC (ImOIInt OfinfbmuUionC. I he games are good for kids' InenIal aικl PhySICal l⅞ealthD. It c<ιn help ParelItS to COntIOl kids, PIay time.15. According to IhIS passage, IhC author's altitude to Playlng ConlPUtCr games isA. SUPPOrtiVe B ObjeCtiVe C. negative D・ PeSSimiStiC第二节：(共5小每小魅2分•満分1()分)HiiKm文内容•从短文后的选项中选出能填入空白处的最佳选项•选项中有两顶为多余选项.TIK OXfbrd EngliSll DiCtiOnary iικlιιdes InOrC than 170,000 EngIiSh entries (词条).LUCkil∖. you don't need to ICam SO Tnany WordS BUt as a Rircign bngιιagc Ieamcr. you have to ICam a CCnalIl number Of WOrdS to communicate CflfeCtiVCIy. 16 ThC following are the most InIPortaDt ones.NarrOW it do、Vn. The first SteP is to identify what you Want to focus On and Set goals17 PerhaPS you VVant to read about ⑷CCer Or Ieam how to COok IIke a native Chef The best learners are motivated to StUdJ f because they ChooSC ∖ocabulafy based On their IntereStS and needs・18 Iiley refer to tlιe InOSI WideIy USed WOrdS in any kind Of SPeeCll or writing. Generally, there arc four CategOriCS Of words in any language: high-frequency words, academic words, technical words aικi IOW・fbequency words. 19 These are the WOrdS PeOPIe USe most IU CVeAday Wntmg and speech. SO IeafuUIg them CaU IUakC you UUdCrStalId What you Ilear aud respond appropriately. An OIIllne SearCh for the PhniSe **tlιe InOSt COnHnon WOrdS In X IangUage ' Can help you identity a IlSt Of high-frequency WordS to focus On Seek OUt OI)POrtllnitlCS to USe ι>eu WOrdS Learning the ∖xxzabulary, involves two processes: lo be able to recognize a U Ord and to Use It We IyPiCally Iearn to recogmze ∖¼ords In Untlng Or SPCCCh before WC re able to USe them It takes additional. COnCCntratCd PraCtiCe to ICam a new UOrd 20 _ SuCh OPPortUnIlieS include Wnting CmaiIS to and Chatling With native SPCakCrSA. DlVIde UOldS IntO dιilerent categories.B. FOCUS On the most COtnnlOn WOrdS髙一英语3B页共】0页C. If you IIaVC a sirong interest in a PartiCUlar subject. Start With thatD. YOU will Ieanl new v¼ords better if you SPeak rather tlιan StInPIy read them.E. With SomC StrategieS・you Can ICanl IaStCr Iind IllOrC CaSiIy th<ιn you have IllIaginCd I∖ High-frequency WOrdS help to quickly ImPrOVC your ability to COmmlmicate.G. That's v∖hy iΓs important to find OUt πιore ChanCeS OrPnXhlCtiVe Iangllage PraCliCe第二部分英语知识运用第一节，完型填空(共2U小题，毎小题IS分，满分30分)In 1982 SteVen Callahan WaS CrOSSing the Ailanlic alone in his SailbOal When it StrUCk Something and Sank. He ∖sas OUt Of IlK ShiPPing IaneS and floating in a Ilfe rail. 21 HiS SUPPIieS Were ftnv. HiS 22 WerC SnIalI YCt Uhen three fishermen found him Sevent)-SiX days Iaier4 he WaS much SkInnICr (han he UaS When he 23 、but aliveΓhe thing that 24 my eye WaS how he managed to keep himself going When all hope SeeIned IOSL When there Seemed no ______________________ 25 _ in COntinUing Ihe struggle, WlIen he WaS SUftenng greatly. When his IIfeboal WaS broken Ulth a hole and after more Ihan a Week StrUggllng Wnh his WCak body to _26_ it, It WilS Stlll IeakIng Water Ilc WaS _27_ and thoroughly CXhaUSted・28 VVOUId have SeenIed the Only wise ChOiCeWhen PCOPlC SUn Z iVC these kinds Of circumstances, they do SOlnCthlng With their _29_ that gives them the CoUragC to keep going Many PCoPIC In 30 desperate ClrCUnlStanCCS give in Or go mad・ SOmCthing Ihe SUlXiVOrS do WitlI Iheir J)OuglltS helps IIIeIn find IhC 31 to carr> on.T tell InySclf I Can 32 it/' WrOtC CaIIahan in his account. w4Comparcd to What OthCrS IIaS r e gone 33_, Γm Ibrtwnate. 1 tell mysell* IhCSe things OVer and over. _34 my deteπninalion to keep going.,,1 Wrote IIlat down after I Iead it. It 35 InC as SOmet hmg important And Γve IOId myself the SamC thing WhCn my Own goals SCClnCd 36 off. And CYCry time I Say it. 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SIrUCk C. disappointed D. PrCSCrVCd36 A. far B IoW C heavy D bright37. A. StatUS B- CrOSSrOadS C. IlOnIeS L>. SelISeS38 A. CXPenment B IHirnCane C (IOOd D SUniVal39. Λ. HOWeVeI B. Whoe∖er C. WhateVer D. WhiCheVer40 A. SlnOOth B.rough C. CaSy D ColnfbrtabIe第二节：单项选择（共30个小趣■每小SSl分■満分30分）从下列乞題中所给的四个选项 5、Ik C和D〉中，选出最佳选项，并在答题卡上将该项涂黑C41. WC II InCet again ill IlIe IUOrumg aιιd We Can Where WC Ieft off.A. PiCk UP B PUt down C take In D CUt OfT42. OlIr PIaHe didn t take Off On time, because OnC Of the CUgineS •A. VVere On tireB. CaIlgllt fire C HCiS CaUglIt fire D Set On fire43 He Said he UaS goinι* (o build a Plane in his back yard, bill I didifl take Ililn !A. ObViOUSlVW B. SeliOUSIy C CntlrCly D heavily44 NOt UntiI T Ixgan to WOrk how much time T had wasted.A. didn’t 1 realize B did 1 realize C had I realized L) I realized45. Don,ι IeaVe IOOIS ______ ; PUt IIIem in order.A .lying about B. being Iaid alx>ut C. IayIng about D Io Iie about46. 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Will See 59. Tibet IieSthe WeSt Of ChiIla WhiIe JaPan IieS the CaSt Of China. A. to; toB. in. inC. in; toD. to; in 60 Mr WhIte is _________ OlIr EngliSh Ieachcr. He OnCllhClpS IIS to deal With SOIne PerSOnaI affairsA in Other WOrdS B. On the Olher hand C. as a Inaner Of fact D. for anotherΛ. TO face B. IIaVing faced C. FaCCdD. FaCing 66. 一It's many years SinCe I SaW you last. 1you at all. I WoUIdrft have. either, if SomCOne _________ you by the name.A dιdn t recognize; hadn't CaIledB. didn't recognize; didn't CaII C IIaVen t recognized; (Iidn t CaIID haven^t recognized; hadιft CaIled 67 If I K ： his ICgS in the Iabt training, he the COming WOrId Cup, WhiCh he hasbeen IOnging to COlnPete inΛ. didn't hurt; WoUId go in R )rB. hadn't hurt, WoUld IlaVe taken Part in C hadn't hurt; UOUld join in D didn t hurt; UOUld have taken Parl in岛-英讯第8更共ID 页68 It's time youthe fact Ihat it's a dangerous WOrICi OlIt there Be brave and OPtimiStiC A I)Ut UP WithB. WOke Ul) toC. t∞k the PlaCe OfD. IoOked down UPOnΛ no more than B OthCr than C. mon: Ihan D rather tlιan61 ・ I Want to SCIl IhC house, butCan t bear Ihe thouglπ Of moving. 62. I'll tiever agree What he SaId IhaI point.A. with; withB WitIE On C. on: On 63. ChHla has POPUIatiOn Of 1.1 bιllιoik 80 PCrCCnt Of Wholnhome during the epidemic (疫情)Of COronaVinis.A. a; WaSB. /; WaSC. a; WeTe 64 ThC PnCCS Of the sh<x^> in this ShoPPmg mall S25 to $100 A. range betweenB. range IrOlnC. Vary in 65. D OnI With ____ a sked to Stay atD. the; Were DVary at With a difficult SitUation, ArnOld decided to ask his IeaClIer for advice69. J haven't SeCn you for a COUPIC Of days. What have you been UP ________ ?A in B. to C. Wth D. for70 DUe to the O(ItbrCak Of the CPKlenliC in China. a kιrge (IUantity Of foreign IeaChCrS andStUdentS abroad.A StUCkB Were SnICk C. WaS S IlCk D. have StlICk第三节：单句语法填空(共W个小题，每小βl.5分，满分15分)阅读下面材料•在空门处塡入适匕的内容或括号刚丫I词的正确形式•71 After years Of WOrklng With children, Man (PrOVe) qualified as a teacher.72. 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三中2019—2020学年度上学期 高一学年第二模块数学考试试卷考试说明：（1）本试卷分第I 卷（选择题）和第II 卷（非选择题）两部分，满分150分．考试时间为120分钟；（2）第I 卷，第II 卷试题答案均答在答题卡上，交卷时只交答题卡．第I 卷 （选择题, 共60分）一、选择题：本大题共12小题，每小题5分，共60分. 在每小题给出的四个选项中，只有一项是符合题目要求的.1. 已知}01|{2≥-=x x A ,}|{xe y y B ==，则=⋂B A),0.(+∞A ]1,.(-∞B ),1.[+∞C ),1[]1.(+∞⋃--∞D2. =︒480cos21.A 21.-B 23.C 23.-D3. 已知x x x f 2cos 32sin )(+=，则)(x f 的周期为π.A π2.B 1.C 2.D4. 已知扇形的周长为cm 6，圆心角为41，则扇形面积为 22.cm A 298.cm B 289.cm C21.cm D5. 方程2log 2=+x x 的解所在的区间为)1,0.(A )2,1.(B )3,2.(C )4,3.(D6. 已知ααπαπsin )cos(2)23sin(=-+-，则22sin sin cos ααα-=1021.A 23.B 23.C 2.D7. 比较133log 2a =,51)31(-=b ,51)32(-=c 的大小a b c A <<. b a c B <<. c b a C <<. b c a D <<.8. 为了得到)62sin(π-=x y 的图象，可以将x y 2sin =的图象.A 向左平移1112π个单位 .B 向左平移12π个单位 .C 向右平移6π个单位 .D 向右平移3π个单位9. 已知函数)2tan(2)(ϕ+-=x x f ，)20(πϕ<<，其函数图象的一个对称中心是)0,12(π，则该函数的一个单调递减区间是)6,65.(ππ-A )3,6.(ππ-B )6,3.(ππ-C )12,125.(ππ-D10. 已知函数2()sin()2cos 2264f x x x πππ=+--，则)(x f 在]23,0[上的最大值与最小值之和为29.-A 27.-B 0.C 211.-D11. 已知sin y x ω=的图象在[0,1]上存在10个最高点，则ω的范围3741.[,)22A ππ .[20,22)B ππ 3741.[,]22C ππ .(20,22)D ππ12. 定义在R 上的奇函数)(x f 满足)()4(x f x f -=-，且当]2,0[∈x 时，2)(x x f =，则方程m x f =)()41(<<m 在]2019,2019[-上的所有根的和为1004.-A 3028.B 2019.C 2020.D第Ⅱ卷 （非选择题, 共90分）二、填空题：本大题共4小题，每小题5分，共20分．将答案填在答题卡相应的位置上． 13. tan 22tan 23tan 22tan 23︒+︒+︒⋅︒=14. 已知)1(log ax y a -=在)2,1(上单调递增，则a 的范围是15. 函数b x A y ++=)sin(ϕω，其中0>A ,0>ω,2||πϕ<的图象如图所示，求y 的解析式16. 已知⎪⎩⎪⎨⎧+∞∈-∈--=),2()2(21]2,0[|1|1)(x x f x x x f ，，，若x k x f ≤)(对任意的0>x 恒成立，则k 的取值范围为三、解答题：本大题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．17．求下列各式的值（1）22cos 101sin 35cos35- （2）cos 7cos8cos15cos 23cos8cos15--18．已知函数()2sin(2)13f x x π=++（1）写出函数()f x 单调递减区间和其图象的对称轴方程；（2）用五点法作图，填表并作出()f x 在5[,]66ππ-的图象.19．已知22()log 2axf x x +=-为奇函数，且0a > （1）求a 的值；（2）判断()f x 在()2,+∞上的单调性，并用单调性定义证明．20.2cos 2sin 2sin 62sin 62sin 2)(2x x x x x x f ++⎪⎭⎫⎝⎛-⎪⎭⎫ ⎝⎛+-=ππ（1）若21)4(2>-πx f ,求x 的范围； （2）若1023)(=αf ，1312)43cos(-=-πβ，且434παπ<<，4743πβπ<<，求)s i n (βα-.21. 设函数()()R a x x x a x f ∈--⎪⎭⎫ ⎝⎛+=,52cos 212cos 2sin )(2.（1）求函数)(x f 在R 上的最小值；（2）若方程0)(=x f 在⎪⎭⎫ ⎝⎛650π，上有四个不相等的实根，求a 的范围.22. 设函数()()222()log 2log a a f x x ak x a =---()01,0a a k >≠>且（1）若函数)(x f 存在零点，求实数k 的最小值；（2）若函数)(x f 有两个零点分别是sin θ，cos θ且对于任意的()0,1x ∈时1212421x x x m a+--+<恒成立，求实数m 的取值集合.哈三中2019—2020学年度上学期 高一学年第二模块数学考试试卷第I 卷 （选择题， 共60分）一．选择题 1-5 CBABB 6-10 ADADD 11-12 AD第Ⅱ卷 （非选择题， 共90分）二．填空题 13. 1;14. 102a <≤; 15. 14sin()223y x π=++;16. 32k ≥.三．解答17. （1）2; （2）1-. 18. （1）对称轴方程：()122k x k Z ππ=+∈;对称中心的坐标： (,1)()62k k Z ππ-+∈.（2）略.19. （1）1a =;(2) 略. 20. （1）())4f x x π=+,5{|22,}66x k x k k Z ππππ+<<+∈; （2）16sin()65αβ-=. 21.（1）min 2minmin 1.2,()23;2.22,()2;43.2,() 3.a f x a a a f x a a f x <-=+-≤<=-++≥=(2)3(,22-- 22. (1).()222222222222222222222log (2)log ();(2)log (2)log ()20;2034(1)0;2()34(1);1612(1)0;()021232a a a a x ak x a x ak x ax ak x a x ak x ak x a k akx a k akx f x k akx a k a k a k k k ak f ak ak -=-⎧-=-⎧-=-⎪⇔⇔->⎨⎨->⎩⎪-⎩⎧-++=⎪⇒⎨>⎪⎩=-++∆=-+≥∴≤≥⎧≥⎪⎨>令()min 22()0;0202; 2.222()0;202;2k ak f k k x k k akf k k k -<≤⎫⎪⇒≥⇒⎬>⎭⎪⎪⎩<≤≤≤>>⎫<⇒⎬>⎭>∴=⎤⎦或(2)2222222222212221234(1)0;2416sin cos 12sin cos ;39(1)sin cos 39;106933201;3480 1.2122;21321(0,1);227.6x x xx m x x x x x k akx a k k ak a k a k a k k a a a a m m x m θθθθθθ-⋅-+-++=⎧⎪<<⎪⎪+=⇒+⋅=⎨⎪⎪+⋅=⎪⎩⇒=-<<∴<<<<∴<<-<⇒>⋅-+⨯-∴<∈+∴≤。

哈三中2018—2019学年度下学期高一学年第一模块数学考试试卷第I 卷 （选择题, 共60分）一、选择题（本大题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的） 1.已知向量()3,1a =，则||a =（ ）A. 1D. 2【答案】D 【解析】 【分析】由向量的模长公式求模长即可. 【详解】因为()3,1a =，所以()2||32a ==.故选D.【点睛】本题考查向量的模长.向量(,)a x y =的模长2||a x y =+2.ABC △的内角,,A B C 的对边分别为,,a b c ，若222b c a +-=，则A =（ ） A.6π B.5π6C.π3D.2π3【答案】A 【解析】 【分析】由余弦定理可求出cos A ，再求A .详解】由余弦定理可得222cos 2b c A bc a +===-， 又()0,πA ∈，所以π6A =. 故选A. 【点睛】本题考查余弦定理.222cos 2b c a A bc +-=，222cos 2a c b B ac +-=，222cos 2a b c C ab+-=，对于余弦定理，一定要记清公式的形式.3.在等差数列{}n a 中，若3712a a +=，则5a =（ ） A. 4 B. 6C. 8D. 10【答案】B 【解析】 【分析】由等差数列的性质可得3752a a a +=，则答案易求.【详解】在等差数列{}n a 中，因为37=52+⨯，所以3752a a a +=. 所以511262a =⨯=.故选B. 【点睛】本题考查等差数列性质的应用.在等差数列{}n a 中，若p q s t +=+，则p q s t a a a a +=+.特别地，若2p q s +=，则2p q s a a a +=.4.已知12,e e 是单位向量，若12|4|13e e-=则1e 与2e 的夹角为（ ） A. 30° B. 60︒C. 90︒D. 120︒【答案】B 【解析】 【分析】先由12|4|13e e -=12e e ，再求1e 与2e 夹角的余弦值，进而可得夹角. 【详解】因为12|4|13e e -=()212413e e -=，则22112281613e e e e -+=.由12,e e 是单位向量，可得12||=||=1e e ，2212==1e e ， 所以121=2e e .所以1212121cos ,=2||||e e e e e e =. 所以12,=60e e ︒.故选B.【点睛】本题考查平面向量的数量积、模、夹角的综合问题.利用22||a a =可以把模长转化为数量积运算.5.ABC ∆的内角,,A B C 的对边分别为,,a b c ，若cos cos 0a A b B -=，则ABC ∆的形状一定是（ ） A. 直角三角形 B. 等边三角形C. 钝角三角形D. 等腰三角形或直角三角形 【答案】D 【解析】 【分析】由已知等式结合正弦定理，可得sin 2sin 2A B =，再结合三角形中角的范围分析角,A B 的关系，进而判断三角形的形状.【详解】由cos cos 0a A b B -=结合正弦定理， 可得sin cos sin cos 0A A B B -=，则sin 2sin 2A B =. 所以22A B =或22πA B +=.所以A B =或π2A B +=. 所以ABC △是等腰三角形或直角三角形.故选D.【点睛】本题考查解三角形问题，应用正弦定理判断三角形的形状.若已知等式中各项都含有边(或角的正弦)，可以直接利用正弦定理实现边角的转化. 解三角形的问题中经常需要用到三角恒等变换，这就需要牢记并熟练运用诱导公式、和差角公式、二倍角公式等，还要结合三角形内角的取值范围，合理地进行取舍，做到不漏解也不增解.6.已知等比数列{}n a 的各项均为正数，且132a ，34a ，2a 成等差数列，则20191817a a a a +=+（ ） A. 9 B. 6C. 3D. 1【答案】A 【解析】 【分析】易得2220191817181718217a a a q a q a a a a q ++==++，于是根据已知条件求等比数列的公比即可. 【详解】设公比为q .由132a ，34a ，2a 成等差数列，可得312322a a a +=，所以2111322a a q a q +=，则2230q q --=，解1q =-(舍去)或3q =. 所以22201918171817181279a a a q a q a a a a q ++===++.故选A. 【点睛】本题考查等比数列、等差数列的基本问题.在等比数列和等差数列中，首项和公比(公差)是最基本的两个量，一般需要设出并求解.7.在等比数列{}n a 中，n S 为数列{}n a 的前n 项和，23S =，49S =，则6S =（ ） A. 12 B. 18C. 21D. 27【答案】C 【解析】 【分析】24264,,S S S S S --也成等比数列，则6S 易求.【详解】在等比数列中，可得24264,,S S S S S --也成等比数列，所以()()242264S S S S S -=-，则()()269339S -=-，解得621S =.故选C.【点睛】本题考查等比数列前n 项和的性质,也可以由1,a q 进行基本量计算来求解.若等比数列的前n 项和是n S ，则232,,,m m m m m S S S S S --(0m S ≠)也成等比数列.8.在数列{}n a 中，已知14a =，25a =，且满足21(3)n n n a a a n --=≥，则2019a =（ ）A.14B.54C.15 D.45【答案】B 【解析】 【分析】由已知的递推公式计算数列的前几项的值，发现周期规律，然后求2019a . 【详解】由21(3)n n n a a a n --=≥，可得12(3)n n n a a n a --=≥.又14a =，25a =，所以32341251,44a a a a a a ====， 同理可得567814,,4,555a a a a ====. 于是可得数列{}n a 是周期数列且周期是6. 因为201963363=⨯+，所以2019354a a ==.故选B. 【点睛】本题考查数列的表示法，递推公式和周期数列.由递推公式判断周期数列时，若递推公式是由前面两项推出后一项，则需要得到连续两项重复才能判定是周期数列.9.我国古代人民早在几千年以前就已经发现并应用勾股定理了，勾股定理最早的证明是东汉数学家赵爽在为《周髀算经》作注时给出的，被后人称为“赵爽弦图”．“赵爽弦图”是数形结合思想的体现，是中国古代数学的图腾，还被用作第24届国际数学家大会的会徽．如图，大正方形ABCD 是由4个全等的直角三角形和中间的小正方形组成的，若,AB a AD b ==，E 为BF 的中点，则AE =u u u r（ ）A.4255a b + B.2455a b + C.4233a b + D. 2433a b +【答案】A 【解析】 【分析】把向量AE 分解到,AB AD 方向，求出分解向量的长度即可得答案.【详解】设BE m =，则22AE BF BE m ===，在Rt ABE △中，可得AB =.过点E 作EH AB ⊥于点H ,则25EH m ==，EH AD ∥，AH ==. 所以42,55AH AB HE AD ==.所以42425555AE AH HE AB AD a b =+=+=+.故选A.【点睛】本题考查平面向量的基本定理，用基向量表示目标向量.平面内的任意一个向量都可以用一对基向量(不共线的两个向量)来线性表示.10.在等差数列{}n a 中，首项10a >，公差0d ≠，前n 项和为*()n S n ∈N ．有下列命题：①若315S S =，则180S =；②若315S S =，则9S 是n S 中的最大项；③若315S S =，则9100a a +=；④若910S S >，则1011S S >．其中正确命题的个数是（ ） A. 1 B. 2C. 3D. 4【答案】D 【解析】 【分析】方法一：由前n 项和公式()112n n n S na d -=+代入各命题判断是否正确. 方法二：由等差数列前n 项和的性质判断各命题是否正确. 【详解】方法一：若315S S =，则1132151431522a d a d ⨯⨯+=+，可得12170a d +=， ()1811181718921702S a d a d ⨯=+=+=，①正确； ()()21111128192171717n n n a S na a n a -⎛⎫=+-=--+⎪⎝⎭，则9S 是n S 中的最大项，②正确；910111892170a a a d a d a d +=+++=+=，③正确.若910S S >，则109100S S a -=<，又10a >，故0d <， 所以111011100S S a a d -==+<，即1011S S >，④正确. 故选D.方法二：若315S S =，则4591014150a a a a a a +++++++=，而415514910a a a a a a +=+=+，则9100a a +=，③正确；()()1811891018902S a a a a =+=+=，①正确； 若0d >，由10a >可得n S 单调递增，不合题意，故0d <， 等差数列的前n 项和是关于n 的二次函数， 由对称性可得当31592n +==时，n S 取得最大值，②正确. 若910S S >，则109100S S a -=<，又10a >，故0d <， 所以111011100S S a a d -==+<，即1011S S >，④正确. 故选D.【点睛】本题考查等差数列前n 项和的有关问题.有关等差数列、等比数列的问题一般都能够使用两种方法求解，一是用首项和公差(公比)进行基本量运算，二是利用有关性质进行解题.11.已知锐角ABC △的内角,,A B C 的对边分别为,,a b c ，若()2c a a b =+，则2c o s c o s ()AC A-的取值范围是（ ）A. ⎫⎪⎪⎝⎭B. 1,2⎛ ⎝⎭C. ⎝⎭D. 1,12⎛⎫⎪⎝⎭【答案】C 【解析】 【分析】由()2c a a b =+结合余弦定理得2cos b a C a -=，再由正弦定理并恒等变形得()sin sin C A A -=，故2C A =，于是2cos =cos cos()AA C A -且可由锐角三角形求得角A 的取值范围，进而可得答案.【详解】因为()2222cos c a a b a b ab C =+=+-,所以22cos b ab C ab -=，则2cos b a C a -=. 所以sin 2sin cos sin B A C A -=. 所以()sin 2sin cos sin A C A C A +-=. 所以cos sin sin cos sin A C A C A -=. 所以()sin sin C A A -=.又ABC △是锐角三角形，π,,0,2A B C ⎛⎫∈ ⎪⎝⎭， 所以C A A -=，即2C A =.所以2cos =cos cos()AA C A -. 由锐角三角形，可得πππ0,02,2222A A A A <<<<+>， 则ππ64A <<，所以cos A ∈⎝⎭.故选C. 【点睛】本题考查正弦定理、余弦定理在三角形问题中的运用，需要综合运用正弦定理、余弦定理和三角恒等变换进行解题.12.已知数列{}n a 与{}n b 前n 项和分别为n S ，n T ，且20,2,n n n n a S a a n >=+∈*N ,1121(2)(2)n n n n n n b a a +++=++，对任意的*,n n N k T ∈>恒成立，则k 的最小值是（ ） A. 1 B.12C.13D.16【答案】C 【解析】先由n S 与n a 的关系式求{}n a 的通项公式，于是可得{}n b 的通项公式，再由裂项相消法求出n T ，于是答案易得.【详解】因为20,2,n n n n a S a a n >=+∈*N , 所以当1n =时，2111122a S a a ==+，解得11a =； 当2n ≥时，21112n n n S a a ---=+.所以()()221112=22n n n n n n n a S S a a a a ----=+-+. 于是()()22110n n n n a a a a ---+=-.由10n n a a -+≠，可得11n n a a --=，所以{}n a 是首项为1，公差为1的等差数列，即n a n =.所以1111212111(2)(2)(2)(21)221n n n n n n n n n n n b a a n n n n ++++++===-++++++++.所以121223111112122211221223n n n n T b b n n b +=+++=-+-++++++-+++ 11311213n n +=<++-. 因为对任意的111,321n n n k T n +∈>=-++*N 恒成立， 所以13k ≥，即k 的最小值是13.故选C.【点睛】本题考查数列的综合问题，考查n a 与n S 的关系、等差数列的判定、裂项相消法求和、与数列有关的不等式恒成立问题，综合性较强.第Ⅱ卷 （非选择题, 共90分）二、填空题(本大题共4小题，每小题5分，共20分．将答案填在答题卡相应的位置上) 13.已知向量(2,1),(1,3),(3,2)a b c =-==，若()a b c λ+∥，则λ=_______． 【答案】1- 【解析】由向量平行的坐标条件求解即可.【详解】由(2,1),(1,3),a b =-=可得()2,13a b λλλ+=-++. 又(3,2)c =，()a b c λ+∥，所以()()313=22λλ+-+，解得=1λ-.【点睛】本题考查向量平行的条件.若向量()()1122,,,a x y b x y ==且a b ∥，则1221x y x y =，可记为“交叉相乘相等”.14.已知等比数列{}n a 满足14652,21a a a a ==-，则9a =________． 【答案】12【解析】 【分析】由等比数列的下标性质先求5a 再求9a .【详解】由等比数列的性质可得2465a a a =，于是25521a a =-，解得51a =.又2195a a a=，所以259112a a a ==.【点睛】本题考查等比数列的基本性质. 在等比数列{}n a 中，若p q s t +=+，则p q s taa aa =.特别地，若2p q s +=，则2p q s a a a =.15.已知数列{}n a 中，11,0n a a =>，前n 项和为n S．若*,n a n =∈N 2)n ≥，则数列11{}n n a a +的前15项和为_______． 【答案】1531【解析】 【分析】先由n a =是等差数列，进而求得数列{}n a 的通项公式，再由裂项相消法求数列11n n a a +⎧⎫⎨⎬⎩⎭的前15项和．【详解】因为*,n a n =∈N 2)n ≥，所以11n n n na -===.1==，所以是首项为1，公差为1n =.所以*121(,n a n n n n ==+-=-∈N 2)n ≥.又11a =也满足，所以*21()n a n n =-∈N .所以()()111111212122121n n a a n n n n +⎛⎫==- ⎪-+-+⎝⎭. 所以数列11n n a a +⎧⎫⎨⎬⎩⎭的前15项和为12231516111111111111111512132352293123131a a a a a a ⎛⎫⎛⎫⎛⎫⎛⎫+++=-+-++-=-= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭. 【点睛】本题考查数列的综合问题，考查n a 与n S 的关系、等差数列的判定、裂项相消法求和，综合性较强.已知n a 与n S 的关系式，有两种思路：一是由1n n nS S a --=消掉S 得到关于通项的关系式；二是把n a 代换成1n n S S --得到关于求和的关系式.16.已知,A B 是单位圆O 上的两点，120AOB ∠=︒，点C 是平面内异于,A B 的动点，MN 是圆O 的直径．若0AC BC ⋅=，则CM CN ⋅的取值范围是________．【答案】30,⎡⎫⎛⎤⎪ ⎢⎥⎪ ⎣⎭⎝⎦【解析】【分析】由MN 是单位圆O 的直径，可得2=1CM CN OC ⋅-，于是需求OC 的取值范围. 由0AC BC ⋅=可得点C 在以AB 为直径的圆上，于是可求出定点O 到圆上的动点的距离OC 的取值范围.【详解】因为MN 是单位圆O 的直径， 所以()()()()2221CM CN OM OCON OC OM OC OM OC OCOM OC =--=---=-=-.在AOB 中，=1OA OB =，120AOB ∠=︒，所以30OAB OBA ==︒∠∠，AB =因为0AC BC ⋅=，所以点C 在以AB 为直径的圆上，其圆心为AB 的中点H ，半径为2.易得12OH =，又点C 异于,A B ，1122OC -≤≤+且1OC ≠.所以2221111122OC ⎫⎫-≤-≤+-⎪⎪⎪⎪⎝⎭⎝⎭且210OC -≠，即3CM CN ≤⋅≤0CM CN ⋅≠.所以CM CN ⋅的取值范围是30022⎡⎫⎛⎤-⎪ ⎢⎥⎪ ⎣⎭⎝⎦，.【点睛】本题考查平面向量数量积的综合问题，考查数量积的取值范围、圆、动点等问题.通过几何意义求取值范围是一种常见的方法.三、解答题(本大题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤) 17.在等差数列{}n a 中，已知567,24a S ==． （1）求n a ；（2）若(1)nn n b a =-，求数列{}n b 的前10项和10T ．【答案】（1）23n a n =-；（2）1010T =. 【解析】 【分析】(1)设出公差，由567,24a S ==列方程解出1,a d 即可.(2)(1)nn n b a =-表示{}n b 的项负正相间，可把相邻两项结合起来再求和.【详解】(1)设等差数列{}n a 的公差为d ， 由题意得516147,61524,a a d S a d =+=⎧⎨=+=⎩解得11,2,a d =-⎧⎨=⎩所以()12123n a n n =-+-=-.(2)因为(1)nn n b a =-，所以123491010T b b b b b b =++++++()()()1234910a a a a a a =-++-+++-+d d d =+++55210d ==⨯=.【点睛】本题考查等差数列的基本问题，数列的求和.对于通项中含有()1n-，即正负相间的数列，可把相邻两项结合起来再求和.18.已知A ,B ,C 是ABC △的三个内角，向量(cos ,sin 2sin ),m B B C =-(2cos cos ,sin )n C B B =+，且m n ⊥．（1）求A ；（2）若BC =，求+AB AC 的取值范围．【答案】（1）π3A =；（2）. 【解析】 【分析】(1)由m n ⊥，得=0m n ，逐步化简可得()1cos =2B C +-，可得答案. (2)由正弦定理、三角形内角和把+AB AC 表示为一个角的函数，再求其取值范围. 【详解】(1)由m n ⊥，得=0m n ，则()()cos 2cos cos sin 2sin sin 0B C B B C B ++-=， 则()()222cos cos sin sin cos sin 0B C B C B B -++=，即()2cos 10B C ++=，故()1cos =2B C +-. 又()0,πB C +∈，所以2π=3B C +. 所以π3A =.（2）因为π3A =，BC =， 所以由正弦定理得2sin sin sin AB AC BCC B A===. 所以2π2sin ,2sin 2sin 3AB C AC B C ⎛⎫===-⎪⎝⎭. 所以2π2sin 2sin 3AB AC C C ⎛⎫+=+-⎪⎝⎭1=2sin 2cos sin 22C C C ⎛⎫++ ⎪ ⎪⎝⎭=3sin C C +1sin cos 22C C ⎫+⎪⎪⎭π6C⎛⎫+⎪⎝⎭.其中2π0,3C⎛⎫∈ ⎪⎝⎭，则ππ5π,666C⎛⎫+∈ ⎪⎝⎭，所以π1sin,162C⎛⎫⎛⎤+∈⎪ ⎥⎝⎭⎝⎦，π6C⎛⎫+∈⎪⎝⎭.所以+AB AC的取值范围是.【点睛】本题考查三角形中的综合问题，考查向量垂直的条件、正弦定理、三角恒等变换、三角函数的性质等.三角函数、平面向量、解三角形的知识联系紧密，解题时也经常综合在一起应用.19.已知ABC△中，45,cos5B AC C=︒==．（1）求边BC的长；（2）若边AB的中点为D，求中线CD的长．【答案】（1）（2【解析】【分析】(1)先由,B C求A，再由余弦定理求BC.(2)方法一：先在△ABC中由正弦定理(余弦定理也可)求AB，再在△BCD (或△ACD)中由余弦定理求CD.方法二：由()12CD CA CB=+求向量CD的模长.【详解】(1)因为cos C=，()0,πC∈,所以sin C==.又45B=︒，所以sin sin()45455510A B C=+=︒+︒=.由正弦定理得sin sin BC ACA B=，所以10sin 45BC ==︒(2)方法一： 在ABC △中，由正弦定理得sin sin AB ACC B=，所以52sin 45AB ==︒，则112BD AB ==. 在△BCD 中，由余弦定理，得2222cos 1814513CD BC BD BC BD B =+-=+-︒=，所以CD =方法二：因为边AB 的中点为D ，所以()12CD CA CB =+. 所以()()222211=244CD CA CB CA CB CA CB =+++1=101821345⎛⎫++= ⎪ ⎪⎝⎭.所以CD =【点睛】本题考查运用正弦定理、余弦定理解三角形.20.已知数列{}n a 满足112(1),2n n na a n a +=+=，设nn a b n=． （1）证明数列{}n b 为等比数列； （2）求数列{}n a 的前n 项和n S ．【答案】（1）证明见详解；（2）1(1)22n n S n +=-+.【解析】 【分析】(1)由1n n b qb +=(q 为非零常数)且10b ≠可证得{}n b 为等比数列.(2)可得2nn a n =，则可由错位相减法求和.【详解】(1)证明：由12(1),n n na a n +=+可得12+1n n a an n+=. 而nn a b n=，所以12n n b b +=. 又1121a b ==，所以数列{}n b 为等比数列. (2)由(1)得{}n b 为首项是2，公比是2的等比数列，所以1222n nn b -==.由n n a b n=可得2nn n a nb n ==. 所以1231222322n n S n =++++， 则234121222322n n S n +=++++.以上两式相减得()23111121222222222212n n n n n n n S n n n ++++--=++++-=-=---，所以()111222122n n n n S n n +++=-++=-+.【点睛】本题考查等比数列的证明和错位相减法求和.若数列{}n c 满足n n n c a b =，其中{},{}n n a b 分别是等差数列和等比数列，则可由错位相减法求数列{}n c 的前n 项和.21.数列{}n a 前n 项和为n S ，已知2112,32 2.n n n a S a ++==-+（1）求数列{}n a 的通项公式；（2）证明121111118n a a a +++<． 【答案】（1）42n nn a =- ；（2）证明见详解.【解析】 【分析】(1)由已知结合1n n n a S S -=-可得1142n n n a a ++-=，变形得1111442n n n n n a a +++⎛⎫-= ⎪⎝⎭，利用叠加法可求n a .(2)由42n nn a =-可得11114n na a +<，用放缩法证明不等式. 【详解】(1)由21322n n n S a ++=-+,得11322n n n S a +-=-+， 以上两式相减得211133322n n n n n n n a S S a a ++-+=-=--+，则1142n n n a a ++-=.两边同除以14n +，可得1111442n n n n n a a +++⎛⎫-= ⎪⎝⎭.22121442a a ⎛⎫-= ⎪⎝⎭， 332321442a a ⎛⎫-= ⎪⎝⎭， …，111442nn n n n a a --⎛⎫-= ⎪⎝⎭， 以上1n -个式子相加得23111144222nn na a ⎛⎫⎛⎫⎛⎫-=+++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭， 又12a =，则23111111422222nnn n a ⎛⎫⎛⎫⎛⎫⎛⎫=++++=- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭， 所以42n nn a =-.(2)证明：因为42n nn a =-，所以()11114244224n n n n n n n a a ++++=-=-+>.所以11114n na a +<. 记12111n nT a a a +++=，则1222112111111111711,4221842421218T T a a a ===<=+=+=<---， 当3n ≥时，12231111111171114124n nn n T T a a a a a a a -⎛⎫⎛⎫<+++++=+-- ⎪ ⎪⎝⎭⎝⎭， 可得()13711711114124412824442n n n n T a a <--=--<-， 所以1118n T <. 所以121111118n a a a +++<. 【点睛】本题考查求数列的通项公式，不等式的证明.求数列通项公式时一般需要构造等差数列或等比数列.放缩法是证明数列不等式的一种常用方法，有时需要保留前面的若干项，只把后面的各项放缩.22.设数列{}n a 的前n 项和为n S ，且2111,2n n a S S n n p +=+=++．（1）若=0p ，求234,,a a a ；（2）若数列{}n a 为递增数列，求实数p 的取值范围． 【答案】（1）2341,4,3a a a ===；（2）13,22⎛⎫⎪⎝⎭.【解析】 分析】(1)令1,2,3n =，求出234,,S S S ，然后可求出234,,a a a .(2)同(1)的方法求出234,,a a a ，由1234a a a a <<<解得p 的取值范围，由212n n S S n n p ++=++可推出112n n a a +--=(3n ≥)，进而可推证数列{}n a 为递增数列.【详解】(1)0p =时，212n n S S n n ++=+，所以2132433,8,15S S S S S S +=+=+=. 又111a S ==，所以2342,6,9S S S ===.所以2213324431,4,3a S S a S S a S S =-==-==-=， 即2341,4,3a a a ===.(2)212n n S S n n p ++=++,所以213S S p +=+，328S S p +=+，4315S S p +=+. 又111a S ==，所以2342,6,9S p S S p =+==+. 所以2341,4,3a p a p a p =+=-=+.若数列{}n a 为递增数列，则1143p p p <+<-<+， 解得1322p <<. 由212n n S S n n p ++=++，①可得()()21121n n S S n n p -+=-+-+(2n ≥),② ①-②，得121n n a a n ++=+(2n ≥)，③ 所以121n n a a n -+=-(3n ≥).④ ③-④，得112n n a a +--=(3n ≥). 于是由23a a <，可得4567,,,a a a a <<由34a a <，可得5678,,,a a a a <<即12345678n a a a a a a a a a <<<<<<<<<，即数列{}n a 为递增数列.综上所述，p 的取值范围为13,22⎛⎫⎪⎝⎭.【点睛】本题考查数列的综合问题，考查n a 与n S 的关系式的应用，递增数列的性质.要使数列{}n a 为递增数列，则一定要保证1n n a a +<(*n N ∈)恒成立，推理过程一定要严谨，不可用特殊性代替一般性.。

哈三中2019-2020学年度下学期高一学年第一模块考试英语试卷答案阅读：1-4DABC 5-8BACA 9-11DDC 12-15BACA 16-20ECBFG完型填空：21-25BDACC 26-30DABCA 31-35CADBB 36-40ADDCB单选：41-45ABBBA 46-50DBDAD 51-55BBCCB 56-60BDDCC61-65BBCBC 66-70ACBBC单句填空：71.proves/proved 72.when 73.of/about 74.It 75.being wasted 76.hopefully 77.had prepared 78.burying 79.to 80.poverty 单句改错：81.he和offered中间加was82.permitting改成permission83.most改成mostly84.because改成that85.measure改成measures86.in1980s中间加the87.valuable改成value88.closely改成close89.safe改成safer90.去掉a作文：One possible version：Dear Tom，I have received your letter and I'm sorry to hear about all your problems.I know exactly how you feel since I have had similar problems before.Therefore， I'm writing to give you some suggestions.To begin with， you should balance the time of study and rest.Too much time sitting in front of a computer will wear you out， thus putting you in awful condition.Taking some exercise between classes will help strengthen your body， relieve your stress and lighten your mood.As for the quarrels between you and your parents， communication is a good way to bridge the gap.Learn to think in each others positions and things will workout well.I hope my suggestions can help you in some way.May you relax yourself and achieve your academic goals!Yours，Li Hua。

黑龙江省哈尔滨市第三中学2019-2020学年高一数学上学期第一次阶段性验收考试试题（含解析）一、选择题（每小题5分）1.不等式2(1)0x x ->的解集为（）A. (1,0)-B. (1,1)-C. (1,0)(1,)-??D. (,1)(0,1)-∞-U【答案】C【解析】【分析】因式分解2(1)0x x ->得到(1)(1)0x x x -+>，利用穿针引线得到答案.【详解】2(1)0x x ->，(1)(1)0x x x -+>根据穿针引线得到110x x >-<<或故答案选C【点睛】本题考查了高次不等式的解法，也可以利用特殊值法得到答案.2.设{|{|A x y B y y ====则A B =I （）A. [0,)+∞B. [1,)+∞C. [2,)+∞D. ∅【答案】C【解析】【分析】分别计算集合A ，B ，再计算A B I 得到答案.【详解】{|{|2}A x y x x ===≥{|{|0}B y y y y ===≥{|2}A B x x =≥I故答案选C【点睛】本题考查了集合的交集，属于基础题型.3.已知全集21{|320},{||2|1},{|0},2x U x x x A x x B x x -=-+≥=->=>-则U A C B =I A. ∅ B. (,1)-∞ C. (3,)+∞D. (,1)(3,)-∞+∞U【答案】A【解析】【分析】先计算集合U ，A ，B 再计算U A C B ⋂得到答案. 【详解】2{|320}{|21}U x x x x x x =-+≥=≥≤或 {||2|1}{|31}A x x x x x =->=><或1{|0}{|21}2x B x x x x x -=>=><-或 {}12U C B x x x ===或=U A C B ∅I 故答案选A【点睛】本题考查了集合的交集和补集，意在考查学生的计算能力和对于集合运算的灵活运用.4.若函数y =[2,1]--上有意义，则实数a 的取值范围是（） A. 2a ≤ B. 1a ≤ C. 01a ≤≤ D. 02a ≤≤【答案】B【解析】【分析】将题目转化为10a x+≥在区间[2,1]--恒成立，计算得到答案.【详解】若函数y =[2,1]--上有意义等价于1a x +在区间[2,1]--上大于等于0 10a a x x+≥∴≤-在区间[2,1]--恒成立 1a ∴≤故答案选B【点睛】本题考查了函数的定义域，不等式恒成立问题，转化为函数的最值是解题的关键.5.已知函数()21,1()22,11,1,1x x f x x x x x⎧⎪+≤-⎪=+-<<⎨⎪⎪≥⎩若()1,f a >则实数a 的取值范围是（） A. 1(,2)(,)2-∞-⋃-+∞ B. 11(,)22- C. 1(,2)(,1)2-∞-⋃- D. 1(2,)(1,)2--⋃+∞ 【答案】C【解析】【分析】讨论a 的取值范围，分别计算得到答案.【详解】当1a ≤-时，()21()1,0f a a a =>>+或2a <- 故2a <-当11a -<<时，1221(),2a a f a =+>>-，故112a >>- 当1a ≥时，1()1,1f a a a=><，故无解 综上所诉：1(,2)(,1)2a ∈-∞-⋃- 故答案选C【点睛】本题考查了分段函数，解不等式，讨论范围得到不同不等式是常用的方法，也可以利用特殊值法排除选项得到答案.6.已知()f x 为一次函数，且[()]43,f f x x =-则(1)f 的值为（）A. 0B. 1C. 2D. 3【答案】B【解析】【分析】 设()f x kx b =+，代入[()]43,f f x x =-得到()21f x x =-或()23f x x =-+，计算得到答案.【详解】设()f x kx b =+则2[()]()()43f f x f kx b k kx b b k x kb b x =+=++=++=- 24,3k kb b =+=-2,1,()21,(1)1k b f x x f ==-=-=或2,3,()23,(1)1k b f x x f =-==-+=综上：(1)1f =故答案选B【点睛】本题考查了一次函数的计算，待定系数法是常规方法，需要灵活掌握和应用.7.已知函数(2)f x -的定义域为[0,2],则函数(21)f x -的定义域为（）A. [2,0]-B. [1,3]-C. 35[,]22D. 11[,]22- 【答案】D【解析】【分析】根据定义域得到220x -≤-≤，再计算112210,22x x -≤-≤-≤≤得到答案. 【详解】函数(2)f x -的定义域为[0,2]，则220x -≤-≤112210,22x x -≤-≤-≤≤ 故答案选D【点睛】本题考查了抽象函数定义域，抓住函数定义域的定义是解题的关键.8.下列是偶函数的是（） A. 31()f x x x =-B. ()f x =C. ()(f x x =-D. ()|25||25|f x x x =++-【答案】D【解析】【分析】利用偶函数定义逐一判断每个选项得到答案.【详解】A. 3311()(0),(),()()f x x x f x x f x f x x x=-+≠-=-=--奇函数 B. ()(11,0),()()()|2|2f x x x f x f x f x x x x ==-≤≤≠-==-----奇函数C. ()(11)f x x x =--≤<非奇非偶函数D. ()|25||25|,()|25||25||25||25|f x x x f x x x x x =++--=-++--=++-()()f x f x =-，偶函数 故答案选D【点睛】本题考查了偶函数的判断，忽略掉定义域是容易犯的错误.9.函数2()48f x x x =--的定义域为[0,]a ，值域为[12,8]--，则a 的取值范围是（） A. [2,4] B. [4,6] C. [2,6] D. [0,4]【答案】A【解析】【分析】画出函数2()48f x x x =--，根据函数图像得到答案.【详解】如图所示：函数值域为[12,8]--，(0)(4)8,(2)12f f f ==-=-则[2,4]a ∈故答案选A【点睛】本题考查了函数的定义域和值域，利用图像可以简化运算，直观简洁.10.已知集合2{|3100},{|121},A x x x B x m x m =--≤=+≤≤-若,B A ⊆则实数m 的取值范围是（）A. 23m -≤≤B. 32m -≤≤C. 2m ≥D. 3m ≤【答案】D【解析】【分析】先计算集合A ，再根据,B A ⊆讨论B 是否为空集得到答案.【详解】2{|3100}{|25}A x x x x x =--≤=-≤≤ {|121}B x m x m =+≤≤-B A ⊆当B =∅时：121,2m m m +>-<当B ≠∅时：121,2m m m +≤-≥且215,3312m m m -≤⎧-≤≤⎨+≥-⎩ 即23m ≤≤ 综上所述：3m ≤故答案选D【点睛】本题考查了根据集合关系求参数范围，忽略空集的情况是容易犯的错误.11.设函数:f R R →满足(0)1,f =且对任意,x y R ∈都有(1)()()()2,f xy f x f y f y x +=--+则(2019)f =（）A. 0B. 1C. 2019D. 2020【答案】D【解析】【分析】取0x =得到(1)2f =，取0y =得到()1f x x =+，代入数据得到答案.【详解】(1)()()()2f xy f x f y f y x +=--+，(0)1,f =取0x = 得到(1)(0)()()22f f f y f y =-+=取0y = 得到(1)()(0)(0)22f f x f f x =--+=得到()1f x x =+ (2019)2020f =故答案选D【点睛】本题考查了求函数表达式和函数值，取点是解题的关键，此题型是考试的常考题型，需要同学们熟练掌握.12.设函数2()(0),f x x x a a =++>若()0,f m <(1)f m -值为（）A. 正数B. 负数C. 非负数D. 正负不确定【答案】A【解析】【分析】根据()0,f m <得到2m m a ->+，22(1)220f m m m a m a -=-+>+>【详解】2()(0)f x x x a a =++>22()0,f m m m a m m a =++<->+222(1)(1)(1)220f m m m a m m a m a -=-+-+=-+>+>故答案选A【点睛】本题考查了函数值的正负判断，意在考查学生的计算能力，此题也可以通过函数图像，韦达定理的方法得到答案.二、填空题（每小题5分）13.集合{}1,2M =的子集．．的个数为_________. 【答案】4【解析】Q 集合{}1,2M =有2 个元素，∴集合{}1,2M =的子集的个数为224=，故答案为4.14.已知函数()f x 是定义在R 上的奇函数,当0x >时,4()f x x x =-,则当0x <时()f x =____【答案】4+x x【解析】【分析】设0x <则0x ->得到4()f x x x -=--，再利用奇函数的性质得到答案.【详解】设0x <则0x ->, 4()f x x x -=--函数()f x 是定义在R 上的奇函数 4()()f x f x x x =--=+故答案为4+x x【点睛】本题考查了利用函数的奇偶性计算函数表达式，属于常考题型.15.若集合42{0,1,3,},{1,4,,3},A m B a a a ==+其中**,,:31,m N a N f x y x ∈∈→=+,x A y B ∈∈是从定义域A 到值域B 的一个函数，则m a +=_______【答案】7【解析】【分析】根据条件得到410a =或者2310a a +=，根据*a N ∈得到2a =，再代入计算得到5m =得到答案.【详解】42{0,1,3,},{1,4,,3}A m B a a a ==+，**,,:31,m N a N f x y x ∈∈→=+ (0)1,(1)4f f ==，(3)10f =，()31f m m =+当410a =时，a =不满足当2310a a +=时，2a =或5a =-（舍去），故2a =4()3116,5f m m a m =+===7m a +=故答案为7【点睛】本题考查了函数映射，讨论对应关系是解题的关键，意在考查学生的计算能力和综合应用能力.16.下列说法正确的是_______（1）函数2()f x x =-(0,)+∞上单调递减；（2）函数2()y x x N =∈图象是一直线；（3）21(0)(),2(0)x x f x x x ⎧+≤=⎨->⎩若()10,f x =则x 的值为-3或-5； （4）若函数2(21)1y x a x =+-+的减区间是(,2],-∞则32a =-；（5）若函数()f x 满足R 上的任意实数12121212,(),()[()()]0x x x x x x f x f x ≠--<恒成立，则()f x 在R 上单调递减.【答案】(4)、(5)【解析】【分析】依次判断每个选项的正误得到答案.【详解】（1）函数2()f x x=-在(0,)+∞上单调递增，（1）错误 （2）函数2()y x x N =∈图象是间断的点，（2）错误（3）21(0)(),2(0)x x f x x x ⎧+≤=⎨->⎩若()10,f x =则x 的值为-3，（3）错误 （4）若函数2(21)1y x a x =+-+的减区间是(,2],-∞即2122a --=，则32a =-，（4）正确（5）若函数()f x 满足R 上的任意实数12121212,(),()[()()]0x x x x x x f x f x ≠--<恒成立，当1212,()()x x f x f x ><，当1212,()()x x f x f x <>，故()f x 在R 上单调递减. （5）正确 故答案为(4)、(5)【点睛】本题考查了函数的单调性，分段函数，函数图像，综合性强，意在考查学生对于函数性质的综合运用.三、解答题（本大题共6道题，17题10分，18-22每小题12分，共70分）17.已知集合3{||2|1},{|0},25x A x x B x x -=-<=≤+求A B U 和()R B C A I . 【答案】5(,)(1,)2A B =-∞-+∞U U ；()5(,)[3,)2R B A =-∞-+∞I U ð【解析】【分析】先计算集合A 和集合B ，再计算A B U 和()R B C A I【详解】{||2|1}{|13}A x x x x =-<=<<，{|31}R C A x x x =≥≤或35{|0}{|3}252x B x x x x x -=≤=≥<-+或 5(,)(1,)2A B =-∞-+∞U U ()5(,)[3,)2R B A =-∞-+∞I U ð 【点睛】本题考查了集合的运算，属于基础题型.18.已知函数()).f x a R ∈（1）若1,a =-求()y f x =的定义域；（2）若函数()y f x =定义域为R ,求实数a 的取值范围.【答案】(1)[5,1]-(2)5[0,]4【解析】【分析】（1）当1,a =-()f x ，计算2450x x --+≥得到答案.（2）讨论0a =和0a ≠两种情况，分别计算得到答案.【详解】（1）当1,a =-()f x = 2450x x --+≥即51x -≤≤ 故定义域为[5,1]-（2）函数()y f x =定义域为R当0a =时，()f x =当0a ≠时，()f x =R ，即2450ax ax ++≥恒成立2050(4)2004a a a a >⎧∴<≤⎨∆=-≤⎩综上所述：5[0,]4a ∈【点睛】本题考查了函数的定义域，忽略掉0a =的情况是容易犯的错误.19.已知二次函数2()3(0)f x ax bx a =++≠图象过点(3,0)A -,对称轴为 1.x =-（1）求()y f x =的解析式；（2）若函数()y g x =满足(21)()g x f x +=,求函数()y g x =的解析式. 【答案】(1)2()23f x x x =--+(2)215()424x x g x =--+ 【解析】【分析】（1）利用图象过点(3,0)A -,对称轴为 1.x =-解得函数解析式.（2）计算2(21)3(2)g x f x x x =--=++，设121,2t x t x -+==代入得到答案. 【详解】（1）二次函数2()3(0)f x ax bx a =++≠图象过点(3,0)A -,对称轴为 1.x =-则(3)9330f a b -=-+=，12b a-=- 解得：1,2a b =-=- 2()23f x x x =--+（2）2(21)3(2)g x f x x x =--=++ 设121,2t x t x -+== 221115()()2322424t t t t g t --=--+=--+ 215()424x x g x =--+ 【点睛】本题考查了求函数表达式，利用换元法可以简化运算，是解题的关键，也可以利用配凑法得到答案.20.()f x 是定义在R 上的函数，对一切,,x y R ∈都有()()2()(),f x y f x y f x f y ++-=⋅且(0)0.f ≠（1）求(0)f ；（2）判断函数()f x 的奇偶性【答案】(1)(0)1f =(2)偶函数【解析】【分析】（1）取0x y ==，得到22(0)2(0),(0)1f f f =∴=（2）取0x =得到()()2(0)()f y f y f f y +-=⋅，即()()f y f y =-得到答案.【详解】（1）()()2()(),f x y f x y f x f y ++-=⋅(0)0.f ≠取0x y ==，则22(0)2(0),(0)1f f f =∴=（2）()()2()(),f x y f x y f x f y ++-=⋅取0x =得到()()2(0)()f y f y f f y +-=⋅，即()()f y f y =-函数()f x 为偶函数【点睛】本题考查了求函数的值和函数奇偶性的判断，意在考查学生对于函数性质的灵活运用.21.解关于x 的不等式22(22)2(1)10()a a x a x a R ---+>∈ 【答案】答案不唯一，具体见解析【解析】【分析】讨论a 的取值范围解得答案.【详解】22(22)2(1)10()a a x a x a R ---+>∈1、当二次系数为0时：当0a =时，不等式的解集为1(,)2-∞；当1a =时，不等式的解集为R ；2、当二次系数为不为0时： 224(1)4(22)4(31)(1)a a a a a ∆=---=--当13a =时，不等式的解集为33(,)(,)22-∞+∞U ；当0a <时，不等式的解集为； 当103a <<时，不等式的解集为()-∞+∞U ； 当113a <<时，不等式的解集为R ；当1a >时，不等式的解集为. 综上所述：当0a <时，解集为2211(2222a a a a a a-+--- 当0a =时，解集为1(,)2-∞当103a <<时，解集为2211(,()2222a a a a a a --+-∞+∞--U 当13a =时，解集为33(,)(,)22-∞+∞U ； 当113a <≤时，解集为R当1a >时，解集为2211(2222a a a a a a----- 【点睛】本题考查了不等式的解法，讨论a 的范围是解题的关键.22.已知二次函数2()(,,)f x ax bx c a b c R =++∈为偶函数,且不等式2()1x f x x x ≤≤-+对一切实数x 恒成立.（1）求函数()f x 的解析式；（2）设函数()2()2,g x f x =-关于x 的不等式2(1)4()()4()x g x g m g m g x m-+≤-在3[,)2x ∈+∞有解，求实数m 的取值范围.【答案】(1)211()22+f x x =(2)m ≤且0m ≠ 【解析】【分析】（1）取1x = 得到1(1)1(1)1f f a c ≤≤∴=+=，再利用20ax x c -+≥得到14ac ≥，利用均值不等式得到14ac ≤，解得12a c ==. （2）将不等式化简为2221(41)230m x x m +---≤，设22141m t m+-=，讨论t 的范围得到83t <，代入式子得到答案. 【详解】（1）二次函数2()(,,)f x ax bx c a b c R =++∈为偶函数 0,02b b a-=∴= 2()1x f x x x ≤≤-+取1x = 得到1(1)1(1)1f f a c ≤≤∴=+=()x f x ≤即20ax x c -+≥恒成立，01(0,0)1404a ac a c ac >⎧∴∴≥>>⎨∆=-≤⎩ 114a c ac +=≥∴≤故12a c ==时成立 211()22+f x x = （2）2()2()21g x f x x =-=-2(1)4()()4()x g x g m g m g x m -+≤-即222222(1)14414(1)x x m m x m--+-≤--- 化简得到：2221(41)230m x x m +---≤ 设22141m t m +-=，即2230tx x --≤在3[,)2x ∈+∞有解 设2()23F x tx x =--，即min ()0F x <易知：当0t ≤时成立当0t >时，对称轴为1x t= 当132t ≤时，min 398()()60,243F x F t t ==-<∴<，故2833t ≤< 当132t>时，min 112()()30,F x F t t t==--<恒成立 综上所述：83t < 即2218413m m +-<解得m ≤且0m ≠ 【点睛】本题考查了函数的解析式，解不等式，计算量大，综合性强，其中通过换元法可以简化运算，意在考查学生的计算能力和对于函数，不等式知识的综合应用能力.。

哈尔滨三中2019-2020学年高二(下)第一次段考数学试卷(文科)(3月份)一、单项选择题（本大题共10小题，共50.0分）1. 已知函数f(x)=x 3−kx.从x =−2到x =2的平均变化率为3，则k 的值为( )A. 1B. 0C. 2D. −22. 函数f(x)=4x 2的导函数是( )A. f′(x)=2xB. f′(x)=4xC. f′(x)=8xD. f′(x)=16x3. 运动物体的位移s =3t 2−2t +1，则此物体在t =10时的瞬时速度为( )A. 281B. 58C. 85D. 104. 函数f(x)=x2−sinx,x ∈[0,π2]的最小值是( )A. π3+√32B. π4C. π6−√32 D. π3−√325. 设曲线y =a(e x −1)−x 在点(0,0)处的切线方程为y =x ，则a =( )A. 0B. 1C. 2D. 36. 若函数f(x)=ax 2−x +ln x 在区间(1,4)上单调递增，则a 的取值范围是 ( )A. [0,+∞)B. [14,+∞)C. [18,+∞)D. [332,+∞)7. 已知函数f(x)=x 2−alnx +1在(1,3)内不是单调函数，则实数a 的取值范围是( )A. (2,18)B. [2,18]C. (−∞,2]∪[18,+∞)D. [2,18)8. 若函数f (x )=e −x +tlnx 有两个极值点，则实数t 的取值范围是( )A. (0,1e )B. (−∞,1e )C. (−1e ,0)D. (1e ,+∞)9. 若关于x 的不等式x 3−3x 2−9x +2≥m 对任意x ∈[−2,2]恒成立，则m 的取值范围是( )A. (−∞,7]B. (−∞,−20]C. (−∞,0]D. [−12,7]10. 不等式2xln x ≥−x 2+ax −3对x ∈(0,+∞)恒成立，则实数a 的取值范围是( )A. (−∞,0)B. (−∞,4]C. (0,+∞)D. [4,+∞)二、填空题（本大题共4小题，共20.0分）11. 函数f(x)=13x 3−2x 2+3x −1的单调递增区间为____________．12.函数f(x)=√1−2x⋅e x+1的极大值为________．13.已知函数f(x)=(2x−3)e x+a有三个零点，则实数a的取值范围是______ ．x)=√2，且当x∈(0,π)时，14.已知f(x)是定义在(−π,π)上的奇函数，其导函数为f′(x)，f(π4f′(x)sinx+f(x)cosx>0，则不等式f(x)sinx<1的解集为________________三、解答题（本大题共4小题，共50.0分）15.求过原点作曲线C：y=x3−3x2+2x−1的切线方程．16.已知函数f(x)=ax3+(a−1)x2+48(a−2)x+b是奇函数．(1)求f(x)的单调区间及极值;(2)当x∈[1,5]时，求函数f(x)的最值．ax3+x2+1(a≤0)．17.求函数的单调区间f(x)=−1318.已知函数f(x)=xe x−a(x2+x)(a∈R)．2(1)当a=1时，求函数f(x)的极值；(2)讨论函数f(x)的单调性．【答案与解析】1.答案：A解析：本题考查函数的变化率，关键是掌握函数变化率的计算公式，属于基础题．根据题意，由函数的解析式计算f(2)与f(−2)的值，由变化率计算公式计算可得答案．解：函数f(x)从x=−2到x=2的平均变化率Δy Δx =f(2)−f(−2)2−(−2)=16−4k4=3，解得k=1．故选A．2.答案：C解析：解：∵f′(x)=8x，故选：C．直接求导即可得出答案．本题考查导函数，正确利用导数的运算法则是解决问题的关键．3.答案：B解析：利用导数的物理意义v=s′和导数的运算法则即可得出．本题考查了导数的物理意义v=s′和导数的运算法则，属于基础题．解：∵v=s′=6t−2，∴此物体在t=10时的瞬时速度=6×10−2=58．故选：B．4.答案：C解析：因为函数f(x)=x2−sinx,x∈[0,π2]，所以f′(x)=12−cosx，令f′(x)=0，则x=π3，可知在给定区间上x=π3取得最小值是π6−√32，选C5.答案：C解析：解：y =a(e x −1)−x 的导数为y′=ae x −1， 在点(0,0)处的切线斜率为a −1=1， 解得a =2， 故选：C ．求出导数，求得切线的斜率，由切线方程可得a +1=1，即可得到a 的值．本题考查导数的运用：求切线的斜率，注意运用导数的几何意义，正确求导是解题的关键．6.答案：C解析：本题考查了函数的单调性，最值问题，考查导数的应用以及函数恒成立问题，考查转化思想，属于中档题．求出函数的导数，问题转化为a ≥x−12x 2在(1,4)恒成立，令g(x)=x−12x 2，x ∈(1,4)，根据函数的单调性求出a 的范围即可． 解：f ′(x)=2ax −1+1x =2ax 2−x+1x，若f(x)在(1,4)递增，则2ax 2−x +1≥0在(1,4)恒成立， 即a ≥x−12x 2在(1,4)恒成立，令g(x)=x−12x 2，x ∈(1,4)，g ′(x)=2−x 2x 3，令g ′(x)>0，解得：1<x <2， 令g ′(x)<0，解得：2<x <4， 故g(x)在(1,2)递增，在(2,4)递减， 故a ≥g(x)max =g(2)=18， 故选C ．7.答案：A解析：本题主要考查函数单调性问题以及利用导数求函数单调性问题，属于中档题．求出f′(x)，分类讨论a≤0和a>0时函数的单调性，使f′(x)=0的一个解在(1,3)内即可得解．解：∵函数f(x)=x2−alnx+1，定义域为(0,+∞)，∴对其求导得f′(x)=2x−ax =2x2−ax；当a≤0时，f′(x)>0，f(x)在(0,+∞)上是增函数，不符合题意;当a>0时，在(√a2,+∞)上，f′(x)>0，f(x)单调递增；在(0,√a2)上，f′(x)<0，f(x)单调递减，∵函数f(x)=x2−alnx+1在(1,3)内不是单调函数，∴1<√a2<3，则2<a<18．故选A．8.答案：A解析：本题考查函数的导数的应用，函数的极值，考查转化思想以及计算能力．由函数f(x)=e−x+tlnx有两个极值点，可得f′(x)=−e−x+tx=0有两个正根，即t=xe−x有两个正根，令g(x)=xe−x，利用导数可得g(x)在(0,1)上单调递增，在(1,+∞)上单调递减，g(x)max=g(1)=1e，又x→+∞时，g(x)→0，x>0且x→0时，g(x)→0，可得t的取值范围．解：函数f(x)=e−x+tlnx有两个极值点，f′(x)=−e−x+tx=0有两个正根，即t=xe−x有两个正根，令g(x)=xe−x，g′(x)=e−x−xe−x，当g′(x)>0时，x<1，故y=g(x)在(0,1)上单调递增，在(1,+∞)上单调递减，g(x)max=g(1)=1e，当x→+∞时，g(x)→0，当x>0且x→0时，g(x)→0，所以t∈(0, 1e)，故选A．9.答案：B解析：解：设y=x3−3x2−9x+2，则y′=3x2−6x−9，令y′=3x2−6x−9=0，得x1=−1，x2=3，∵3∉[−2,2]，∴x2=3(舍)，列表讨论：x(−2,−1)−1(−1,2)f′(x)+ 0−f(x)↑极大值↓∵f(−2)=−8−12+18+2=0，f(−1)=−1−3+9+2=7，f(2)=8−12−18+2=−20，∴y=x3−3x2−9x+2在x∈[−2,2]上的最大值为7，最小值为−20，∵关于x的不等式x3−3x2−9x+2≥m对任意x∈[−2,2]恒成立，∴m≤−20，故选：B．设y=x3−3x2−9x+2，则y′=3x2−6x−9，令y′=3x2−6x−9=0，得x1=−1，x2=3(舍)，由f(−2)=0，f(−1)=7，f(2)=−20，知y=x3−3x2−9x+2在x∈[−2,2]上的最大值为7，最小值为−20，由此能求出关于x的不等式x3−3x2−9x+2≥m对任意x∈[−2,2]恒成立的m的取值范围．本题考查利用导数求函数在闭区间上最值的应用，解题时要认真审题，仔细解答，注意合理地进行等价转化．10.答案：B解析：本题考查恒成立问题，训练了利用导数求函数的最值，训练了分离变量法，是中档题．由已知可得，x>0，令，利用导数求出x=1时，y取最小值4，由此可得实数a的取值范围．解：由2xln x≥−x2+ax−3，得，设，则ℎ′(x)=(x+3)(x−1)．x2当x∈(0,1)时，ℎ′(x)<0，函数ℎ(x)单调递减；当x∈(1,+∞)时，ℎ′(x)>0，函数ℎ(x)单调递增，所以ℎ(x)min=ℎ(1)=4.所以a≤ℎ(x)min=4．故a的取值范围是(−∞,4]．11.答案：(−∞,1)，(3,+∞)x3−2x2+3x−1，所以f′(x)=x2−4x+3，解析：解：因为f(x)=13由f′(x)=x2−4x+3>0，得：x<1或x>3，所以原函数的单调增区间为(−∞,1)，(3,+∞)．故答案为(−∞,1)，(3,+∞)．x3−2x2+3x−1的单调递增区间，先求该函数的导函数，让导函数大于0求解x的求函数f(x)=13范围。