北京師大附中2018-2019學年上學期高二年級期末考試數(shù)學試卷

北京師大附中2018-2019學年上學期高二年級期末考試數(shù)學試卷(AP)本試卷有兩大部分，考試時長120分鐘，滿分100分。第一部分：學考數(shù)學(76分)、選擇題：本大題共15小題，每小題3分，共45分。已知集合M={1,2}，集合N={0,1,3}，則MAN={1,2,3}B.{1,2}C.{0,1}D.{1}1已知函數(shù)f(x)=x+1+，則/(x)的定義域是x-2[-1,2)B.[―1,+s)C.(2,+8)D.[-1,2)u(2,+8)1函數(shù)y=x2,y=x3,y=(2)x,y=lgx中，在區(qū)間(0,+8)上為減函數(shù)的是A.y=x2B.y=x3C.y=(*”D.y=lgx如下圖，給出了奇函數(shù)f(x)的局部圖象，那么f(i)等于A.-4B.-2C.2D.4已知二次函數(shù)y=x2+mx+(m+3)有兩個不同的零點，則m的取值范圍是A.[-2,6]B.(-2,6)C.(-8,-2)U(6,+8)D.{-2,6}log2-log8等于44A.-2B.-1C.1D.2函數(shù)f(x)=x3-5的零點所在的區(qū)間是A.(1,2)B.(2,3)C.(3,4)D.(4,5)3&角0的終邊經(jīng)過點P(4,y),且sm9=-5，則tan9=A.-4/3B.4/3C.-3/4D.3/4

在AABC中，角A,B,C所對的邊分別為a,b,c,如果a=3,b=,c二2^2，那么AABC的最大內(nèi)角的余弦值為1/8B.1/4C.3/8D.1/2兀把函數(shù)y=Sinx的圖象向右平移丁個單位得到y(tǒng)=g(x)的圖象，再把y=g(x)圖4象上所有點的縱坐標伸長到原來的2倍(橫坐標不變),所得到圖象的解析式為兀兀y=2sin(x-)B.y=2sin(x+)44C.y=1sin(x-—)D.y=1sin(x+—)2424已知直線m,n,l，平面a,卩，Y,給出下面四個命題：①//①//Ya丄YJ②a〃?a//Yl丄l丄m\③卜nm//nl丄nJm//a\④m//nJn"〃a其中正確的命題是A.①B.②C.③D.④某幾何體的三視圖如下圖所示，那么該幾何體的體積是B.2—2MtotB.2—2Mtot左】視圖—2—H正￡主、就圖<5C.D.2—C.已知直線l經(jīng)過點0(0,0),且與直線x-y-3=0垂直，那么直線l的方程是A.x+y-3=0B.x-y+3=0C.C.x+y=0D.x-y=0某市共有初中學生270000人，其中初一年級，初二年級，初三年級學生人數(shù)分別為99000，90000，81000，為了解該市學生參加“開放性科學實驗活動”的意向，現(xiàn)采用分層抽樣的方法從中抽取一個容量為3000的樣本，那么應該抽取初三年級的人數(shù)為800B.900C.1000D.1100口袋中裝有大小、材質(zhì)都相同的6個小球，其中有3個紅球、2個黃球和1個白球，從中隨機摸出1個球，那么摸到紅球或白球的概率是A.B.C.D.A.B.C.D.、填空題：本大題共5小題，每小題3分，共15分已知圓M的方程是x2—6x+y2—16二0，則該圓的半徑是已知cosa=2，那么cos(—2a)等于。218.如果正方形ABCD的邊長為1,那么AC-AB等于函數(shù)y=—2x2+x+1在區(qū)間［一1,1］上的最小值為。已知200輛汽車通過某一段公路時的時速的頻率分布直方圖如下圖所示，則a=21.已知函數(shù)f(x)=^3sin2x+cos2x，xeR求f(x)的最小正周期和振幅。求f(x)的單調(diào)遞減區(qū)間。22.如下圖，在三棱錐P-ABC中，PA丄底面ABC，AB丄BC，D，E分別為PB，PC的中點。求證：BC//平面ADE;求證：BC平面PAB第二部分：SAT數(shù)學(共24分)(選擇題，共8小題，每小題3分)1.(x6)2(y5)216Inthexy-plane,thegraphoftheequationaboveisacircle.PointPisonthecircleandhcoordinates(10,-5).IfPQisadiameterofthecircle,whatarethecoordinatesofpointQ?(2,-5)(6,-1)(6,-5)(6,-9)2.Whichofthefollowingcouldbetheequationofthegraphabove?A.yx(x2)(x3)B.yx2(x2)(x3)C.yx(x2)(x3)D.yx2(x2)(x3)25.25.3.Agesof30StudentsEnrol拒d

inaCollegeClassThetableaboveshowsthedistributionofagesofthe20studentsenrolledinacollegeclass.Whichofthefollowinggivesthecorrectorderofthemean,median,andmodeoftheages?mode<median<meanmode<mean<medianmedian<mode<meanmean<mode<median4.WhichofthefollowingexpressionsisequivalenttoA.B.C.D.x—A.B.C.D.x—5—x—5—20x—310x—3x—3SpeciesoftreeGroMhfactorSpeciesoftreeGroMhfactorRedmaple45Riverbirch;15；Cottoiiwood2.0Biackwalnut|45Whitebirchi5.0Americanelm4.0Pinoak3.0ShagbarkhickoryTA5Onemethodofcalculatingtheapproximateage,inyears,ofatreeofaparticularspeciesistomultiplythediameterofthetree,ininches,byaconstantcalledthegrowthfactorforthatspecies.Thetableabovegivesthegrowthfactorsforeightspeciesoftrees.Ifawhitebirchtreeandapinoaktreeeachnowhaveadiameterof1foot,whichofthefollowingwillbeclosesttothedifference,ininches,oftheirdiameters10yearsfromnow?(1foot=12inches)A.1.01.21.31.4f(x)=2x3+6x2+4xg(x)=x2+3x+2Thepolynomialsf(x)andg(x)aredefinedabove.Whichofthefollowingpolynomialsisdivisibleby2x+3?h(x)=f(x)+g(x)P(x)=f(x)+3g(x)r(x)=2f(x)+3g(x)s(x)=3f(x)+2g(x)7.Todetermineifgenderandnumberofsiblingsarerelatedtowhether9-year-oldchildrenliketoread.Sarahselectedarandomsampleof40boyswithfewerthan3siblingsandarandomsampleof30girlswith3ormoresiblingsfromthe9-year-oldchildreninhercity.Foreachchild,sherecordedthechild'sage,gender,andwhetherthechildreportedlikingtoread.WhyisitinappropriateforSarahtodrawaconclusionfromherstudy?A.Thetwosamplesarenotofequalsize.Thetwosamplesshouldhavebeenchosenfromdifferentcities.Thereisnoupperhoundonthenumberofsiblingsofthe30girls.Sarahwillnotbeabletotellwhetheradifferenceinlikingtoreadisrelatedtothedifferenceingenderortothedifferenceinnumberofsiblings.8.Startingin1970,thepopulationofacitydoubledevery10yearsuntil2010.Thepopulationofthecitywas36,000in1970.Whichofthefollowingexpressionsgivesthepopulationofthecityin2010?A.36,000(2)436,000(2)1036,000(2)4036,000(2)(40)試題答案】、選擇題：本大題共15小題，每小題3分，共45分。1.D2.D3.C4.B5.C6.B7.A8.C9.A10.A11.B12.A13.C14.B15.D、填空題：本大題共5小題，每小題3分，共15分116.517?—218.119.-220.0.04三、解答題：共2道大題，每大題8分，共16分，請寫出解題步驟。21.（8分）（I）（4分）TOC\o"1-5"\h\z\o"CurrentDocument"31兀兀f(x)=2(sin2x+cos2x)=2(sin2xcos+cos2xsin\o"CurrentDocument"2266兀\o"CurrentDocument"=2sin（2x+）（2分）6最小正周期=兀（1分）振幅=2（1分）II）（4分）由f（x）遞減可得：TOC\o"1-5"\h\z\o"CurrentDocument"…兀c?！?兀，兀,2兀2分)2k兀+<2x+<2k兀+nk兀+<x<k兀+——2分)\o"CurrentDocument"26263冗2兀單調(diào)遞減區(qū)間：■■—+￡兀^――+￡兀,kwZ（2分）HYPE