山东省青岛地区2021-2022高一下学期数学期中试卷及答案

2021—2022学年度第二学期期中学业水平检测高一数高一数学试题学试题2022.05本试卷共4页，22题．全卷满分150分．考试用时120分钟．注意事项：1．答卷前，考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置上，并将条形码粘贴在答题卡指定位置上。2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。如需改动，用橡皮擦干净后，再选涂其他答案标号。回答非选择题时，将答案写在答题卡上。写在本试卷上无效。3．考试结束后，请将答题卡上交。一、单项选择题：本大题共8小题．每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目 要求 对教师党员的评价套管和固井爆破片与爆破装置仓库管理基本要求三甲医院都需要复审吗 的．1．若复数z满足(2i)z3i（i为虚数单位），则复数z在复平面内对应的点位于A．第一象限B．第二象限C．第三象限D．第四象限2．若D为ABC的边BC的中点，则ACA．2ABADB．2ADABC．2ADABD．2ABAD3．已知正四棱锥的底面边长和侧棱长都为2，则该正四棱锥的 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 面积为A．43B．45C．434D．4544．已知a,b是不共线的向量，且ABa2b,AC2ab,CD3a3b，则A．A,B,C三点共线B．A,C,D三点共线C．B,C,D三点共线D．A,B,D三点共线5．已知tan2，则cos24343A．B．C．D．5555高一数学试题第1页（共4页）ππ6．将函数ysin(2x)(||)的图象向右平移个单位长度，得到的函数图象关于26y轴对称，则的值为ππππA．B．C．D．36637．一个长方体的顶点都在球面上，它的长、宽、高分别为3,4,5，则此球的体积为1252π2502π5002π10002πA．B．C．D．3333π8．已知ABC的内角A,B,C的对边分别为a,b,c，若a2,c3b,A，则ABC的6面积为A．4B．23C．2D．3二、多项选择题：本大题共4小题．每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，选对但不全的得2分，有选错的得0分．9．已知复数z满足zi(1i)（i为虚数单位），则下列结论正确的是A．z的虚部为iB．|z|2C．z1iD．z2为纯虚数10．已知向量a(1,3)，b(1,2)，c(2,4)，则下列结论正确的是πA．b//cB．|ac|50C．(ab)bD．a,b411．设函数f(x)sin2x3cos2x，则下列结论正确的是πA．f(x)的最小正周期为πB．f(x)的图象关于直线x对称12πC．f(x)的一个零点为xD．f(x)的最大值为31312．若ABC的内角A,B,C的对边分别为a,b,c，则下列结论中正确的是A．若AB，则sinAsinBB．若acosBbcosAc，则ABC为直角三角形C．若acosAbcosB，则ABC为等腰三角形AcbD．若cos2，则ABC为直角三角形22c高一数学试题第2页（共4页）三、填空题：本大题共4小题，每小题5分，共20分．y13．如图所示，正方形OABC是水平放置的一个平面CB图形的直观图，其中OA2，则AB在原图形中的对应线段AB的长度为．OAx14．在平面直角坐标系xOy中，角与角的顶点与原点O重合，均以Ox为始边，它们3的终边关于x轴对称，若cos，则cos()．315．测量塔高AB时，选取与塔底B在同一水平面内的两个测量点C与D，现测得5πππBCD，BDC，CD100，在点C测得塔顶A的仰角为，则1234塔高AB．16．已知ABC是等边三角形，点D在BC的延长线上，且BC2CD，AD27，则AB；ABACABBCACBC=．（本小题第一问2分，第二问3分）四、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．（本题满分10分）已知复数z(m1)(m2)(m2)i(mR)，其中i为虚数单位．（1）若z是纯虚数，求实数m的值；（2）若m3，z是关于x的实系数方程x2axb0的一个复数根，求实数a,b的值．18．（本题满分12分）43已知角的顶点与原点O重合，始边与x轴的非负半轴重合，它的终边过点P(,)．55（1）求sin2；π5（2）若0，sin()，求sin．213高一数学试题第3页（共4页）19．（本题满分12分）如图所示，在四边形ABCD中，AB2DC，ABBC0，|AB|23，|AD|2，E为AB中点．（1）将四边形ABCD绕着线段AB所在的直线旋转一周，A求所形成封闭几何体的表面积和体积；（2）将AED绕着线段AE所在直线旋转一周形成几何体W，ED若球O是几何体W的内切球，求球O的表面积．BC20．（本题满分12分）已知函数f(x)23sinxcosx22sin2x．（1）求函数f(x)的单调递减区间；π（2）当x(0,)时，求函数f(x)的值域；2（3）在ABC中，内角A,B,C所对的边分别为a,b,c，且f(A)1，a3，求bc的取值范围．21．（本题满分12分）13在锐角ABC中，向量AC在向量AB上的投影向量为p(,)，|AB|2，22AC(t,0)．（1）求t；（2）已知D是BC的中点，BE2EA，设EDmABnAC，求mn的值及|ED|．22．（本题满分12分）已知ABC的内角A,B,C所对的边分别为a,b,c．（1）在这三个条件中任选一个，补充在下面的横线上，并加以解答．①3sinBcosB1；②(2ca)cosBbcosA；③(ac)sinAcsinCbsinB．若，且b7，c2．（ⅰ）求角B及a的值；（ⅱ）若内角B的平分线交AC于点D，求BCD的面积．注：如果选择两个条件分别解答，按第一个解答计分．（2）若tanA,tanB,tanC均为正整数，求tanAtanBtanC．高一数学试题第4页（共4页）2021—2022学年度第二学期期中学业水平检测高一数学答案及 评分 售楼处物业服务评分营养不良炎症评分法中国大学排行榜100强国家临床重点专科供应商现场质量稽核 标准一、单项选择题：本大题共8小题．每小题5分，共40分．ABCCDBAD二、多项选择题：本大题共4小题．每小题5分，共20分．9．BD；10．AC；11．ABC；12．ABD.三、填空题：本大题共4小题，每小题5分，共20分．113．6；14．−；15．506；16．4；8．3四、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．（本小题满分10分）解：(1)因为z是纯虚数，所以(m+1)(m−2)=0且m−20·····························2分所以m=−1····························································································4分（2）若m=3，则z=+4i········································································5分因为z是方程x2+ax+b=0的一个复数根，所以(4i)(4i)0++++=2ab···········6分1540++=ab所以(15+4a+b)+(a+8)i=0，即·······································8分a+=80解得a=−8,b=17··················································································10分18.（本小题满分12分）34解：由三角函数定义知：sin=−,cos=··············································2分553424（1）sin22sin==cos2−()=−···········································5分5525512（2）因为sin(+)=，所以cos(+)=········································7分1313因为sin=sin[(+)−]=sin(+)cos−cos(+)sin·····················9分125412356所以，当cos(+)=时sin=−（−)=························10分1313513565125412316当cos(+)=−时sin=+（−)=−····················11分1313513565π56又因为0，所以sin0，所以sin=········································12分265高一数学答案第1页（共5页）19．（本小题满分12分）解：（1）因为ABDC=2，ABBC=0，所以ABD//C，ABD=C2且ABB⊥C所以四边形ABCD为直角梯形，········································································1分所以，将四边形绕着线段AB所在的直线旋转一周所形成的封闭几何体为圆锥和圆柱的组合体······················································································2分因为||2AB3=，||2AD=，所以DC=3，BC=1所以圆锥的高为3，母线长为2，底面半径为1；圆柱的高为，底面半径为·····················································································································3分圆锥的侧面积S1=π=122π，圆柱的侧面积S2=2π=1323π，2圆柱的下底面的面积S3=π=1π，所以几何体的表面积为S=++=+2π23ππ(323)π····································5分13圆锥的体积V=π132=π，圆柱的体积V=π123=3π，133243所以几何体的体积V=π·········································································7分3（2）因为E为中点，，所以EB//DC，EB=DC所以EBDC为平行四边形，所以ED//BC因为AB=BC0，所以，所以AB⊥ED，所以AED为直角三角形所以将AED绕着线段AE所在直线旋转一周形成几何体W为圆锥·······················8分由题知，圆锥的高为，母线长为，底面半径为·····9分R1设圆锥的内切球O半径为R，则=3−R2R3所以球的半径R=··············································11分3R4所以内切球的表面积为SR==4π2π·······················12分320．（本小题满分12分）解：f(x)=23sinxcosx−2+2sin2x=3sin2x−cos2x−1π=2sin(2x−)−1·········································································3分6ππ3ππ5π（1）由2k+2x−2kπ+(kZ)得kπ+xkπ+26236高一数学答案第2页（共5页）π5π所以函数f(x)的递减区间为[kπ+,kπ+](kZ)······································4分36πππ5ππ1（2）当x(0,)时，2(,)x−−，sin(2)(,1]x−−······················5分266662所以函数的值域为(−2,1]····································································6分ππ（3）f()A=2sin(2A−)1−=1，所以sin(2A−=)166πππ因为A(0,π)，所以2A−=，A=····················································7分623abc3因为====2···························································8分sinAsinBsinC322π所以bcBCBB+=+=+−2sin2sin2sin2sin()32π2π=+−2sin2sincos2cossinBBB33π=3sinBBB+3cos=23sin(+)·························································10分62πππ5ππ1因为B(0,)，B+(,)，所以sin()(,1]B+366662所以bc+的取值范围为(3,23]······························································12分21．（本小题满分12分）13解：（1）因为p=(,)，所以||1p=······················································1分22因为p为向量AC在向量AB上的投影向量，且A为锐角所以||ACAcos=········································································2分所以ACp=|AC||p|cosA=1·································································3分t又ACt=(,0)，所以ACp=··································································4分2t所以=1，t=2····················································································5分221（2）EDEBBDABBC=+=+322111=AB+()AC−AB=AB+AC···········································8分3262112所以mn+=+=·············································································9分623高一数学答案第3页（共5页）由题意知，p是与AB方向相同的单位向量因为||2AB=，所以ABp==2(1,3)，ABAC=2···································10分11111所以||()||||EDABACABACABAC2222=+=++62364611113=++=4423646913所以||ED=····················································································12分322.（本小题满分12分）解：（1）（ⅰ）选择条件①：因为3sincBBos=+1，所以3sincBBos−=1ππ1所以2sin()B1−=，sin()B−=·························································2分662ππ5ππππ因为B(0,π)，B−(,)−，所以B−=，所以B=·····················3分666663选择条件②：abc因为==，所以(2c−=a)cosBbcosAsinABCsinsin可化为(2sinCABBA−=sin)cossincos····················································1分所以2sinCBABBAcos=+sincossincos即2sinCBABcos=+sin()······································································2分因为ABC++=π，所以sin(ABC+=)sin，又因为C(0,π)，所以sinC01所以cosB=，因为，所以················································3分2选择条件③：因为，所以(a−c)sinA+csinC=bsinB可化为()a−ca+c22=b，即a2+c2−b2=ac···············································1分a2+−c2b2ac1所以cosB===····························································2分2ac2ac2又因为，所以····································································3分因为b2=a2+c2−2accosB······································································4分所以7=4+aa2−2，即aa2−2−3=0，解得a=3或a=−1（舍）所以······························································································5分（ⅱ）因为BD是角B的平分线，所以ABD=CBD，设ABD=CBD=高一数学答案第4页（共5页）1ABBDsinSAB2所以ABD===2·······················································6分SBC13BCDBCBDsin2133因为SacB==sin····································································7分ABC22333393所以SS===·························································8分BCDABC55210（2）因为tantantantan()(1tantan)tanABCABABC++=+−+=+−++tan()tan()tantantanABABABC=tanABCtantan········································10分不妨设ABC，则0tantantanABC又因为tantantantantantan3tanABCABCC++=所以tantaABn3·················································································11分因为tan,tan,tanABC均为正整数，所以tan1A=，tan1B=或2或3若tan1B=，则2tantan+=CC，不成立若tan2B=，则3tan2+=tanCC，tan3C=，符合题意若tan3B=，则4tan3tan+=CC，tanC=2，与假设矛盾所以tantantan6ABC++=····································································12分高一数学答案第5页（共5页）