所以f(x)在(ln(-2m),-2)上单调递减，在(-o,ln(-2m))和(-2，+o)上单调递增；…(3分)③当m=-)之时f'(x)≥0在R上恒成立，所以fx)在R上单调递增：……(4分④当m<2时，令f(<0,得-20,得x<-2或x>h(-2m),所以f(x)在(-2，ln(-2m))上单调递减，在(-o,-2)和(ln(-2m),+o)上单调递增.…(5分)(Ⅱ)当x∈[-2，+o)时，f(x-1)≥m(x2+3x)-e恒成立，则xe-1-m(x-1)+e≥0恒成立.(i)当x=1时，不等式即1+e≥0，满足条件.…(6分)》)当x>1时，原不等式可化为m≤十e,该式对任意x山，+)恒成粒设0=则ge-世-gg(x-1)2设k(x)=(x2-x-1)e*-1-e,则k'(x)=(x2+x-2)e-1=(x+2)(x-1)e-1.因为x>1,所以k'(x)>0,所以k(x)在(1，+0)上单调递增，即g'(x)在(1，+∞)上单调递增又因为g(2)=k(2)=0,所以x=2是g(x)在(1，+∞)上的唯一零点，…(8分)所以当12时，g(x)>0,g(x)在(2，+∞)上单调递增，所以当x∈(1，+∞)时，g(x)min=g(2)=3e,所以m≤3e.…(9分)(i)当-2≤x<1时，原不等式可化为m≥e+ex-1’此时对于（ⅱ）中的函数k(x),可知当-2≤x<1时，k'(x)≤0，所以k(x)在[-2,1)上单调递减，且k(-2)=5e3-e<0,所以当-2≤x<1时，k(x)≤k(-2)<0,即g(x)<0,所以g(x)在[-2,1)上单调递减，所以当e[-2.1)时g(=g(-2)2；所以m≥209(11分)综上所述，m的取值范围是[2)，3c…（12分)天一文化TIANYI CULTURE9

4妇+h的最大值是号B.2+2的最小值是4V万C.a+sinb<2D.b+Ina>111.(多选题)声音是由物体振动产生的声波，其中包含着正弦函数.纯音的数学模型是函数y=As@t,我们听到的声音是由纯音合成的，称之为复合音.若一个复合音的数学模型是函数f(x)=sinx+二sin2x,则下列结论正确的是()A.f(x)的图象关于直线x=π对称B.f(在44上是增函数C.f()的最大值为3n若f)()=名-l。=行2712.设函数∫(x)=1中X中x>0,若了)之饵成立，则满足条件的正整数k可能是()一A.2B.3C.4D.5三、填空题：本题共4小题，每小题5分，共20分.13.已知函数f(x)=x(a2+2)是偶函数，则实数a=一14.写出一个定义域为R值域为[0，]的函数15.已知a>0,b>0,直线y=x+a与曲线y=c1-2b+1相切，则2+的最小值为16.己知函数f(x)=21山x,直线1的方程为y=x+2,过函数f(x)上任意一点P作与1夹角为30的直线，交I于点A,则P4A的最小值为一·四、解答题：本题共6小题，第17题10分，第1B=22题各12分，共70分，解答应写出文字说明、证明过程或演算步骤。17.设函数f(x)=x2-(a+3)x+3a,aeR.(1)解关于x的不等式f(x)<0:(2)当xe[4,+时，不等式f(x)≥-9恒成立，求a的取值范围.

再从这6人中先后抽取2人的成绩作分析，心/频率分布直方图中的x=0.030下列结论正确的是(B.估计100名学生成绩的中位数是85C.估计100名学生成绩的80%分位数是95D.从6人中先后抽取2人作分析时，若先抽取的学生成绩位于[70,80)，则后抽取的学生成绩在[80,90)的概率是4510设数了)=lg时在a切上的最小值为m,函敬8闭=n受在0a上的最大值为M。,若M。-m。=2,则满足条件的实数a可以是(BD》A.2B.C.10√10D√1o31.已知抛物线C:x2=2y(p>0)的焦点F与圆M:x2+0+2)2=1上点的距离的最小值为2，过点F的动直线1与抛物线C交于A,B两点，以A,B为切点的抛物线的两条切线的交点为P,则下列结论正确的是（小BA.p=2年B.当与M相切时，1的斜率是士4C点P在定直线上+○，D.以AB为直径的圆与直线y=-1相切12.在中国共产党第二十次全国代表大会召开期间，某学校组织了“喜庆二十大，永远跟党走，奋进新征程”书画作品比赛如图①，本次比赛的冠军奖杯由一个铜球和一个托盘组成，若球的体积为4；如图②，托盘由边长为4的正三角形铜片沿各边中点的连线垂直向上折叠而成，则下列结论正确的是(OD)图①图②本)81A.直线AD与面BEF所成的角为6B.经过三个顶点A,B,C的球的截面圆的面积为工5C异面直线AD与CF所成的角的余弦值为8D球离球托底面DEF的最小距离为5+y6三、填空题：本题共4小题，每小题5分，共20分。国道理始高三二模数学试卷第3页（共6页）

no parking at all.At last,I'd like to recommendyou some street food.The first is...and参考答案1-5 ACBCB6-10 CBACC11-15ABABAset16-20 BAABC21-25 CDBBC26-30ABDAA36-40 CFBAD41-45 BDCACwe31-35 CACDBthe46-50 DABDC51-55ABDCA56-60 CBABD61.themselves62.at63.but64.who65.constantly66.battling67.is68.third69.disagreement70.healthyke短文改错Spring had come.Our teacher informs us thatinformedes,adwe were going to have a class outing.The newswas so much cheering that we jumped excitedly.We set off at 8 o'clock on a brighter Saturday.Webrightacewalked in lines with a student hold a flag in frontholdingof us.They were many colourful flowers and greenTherefftrees along the road.We admired the beauty ofspring,as well A breathing the fresh air in aasthecountryside.After an hour,we reachedourdestination,that had fantastic scenery,and decidedwhichto eaten our picnic lunch by the river.The foodeatwas surprisingly delicious after such a long walkTiring as we were,the trip was worth it.Tired书面表达One possible version:Dear Mr O'Leary,I have read your job advertisement in amagazine and I believe I am qualified for the job.First,1 will have a two-month summer holidayafter sitting the college entrance examinations,which means your working time is suitable for me.Secondly,I am experienced in dealing withchildren.I have an 11-year-old brother and we getalong very well.Thirdly,I like outdoor activitiesand often go hiking at the weekend.Finally,I hostan on-air English learning program in our school,so I canEnglish.communicate fluently with others inteam.I look forward to the opportunity to join your

trousers"over skinny trousers any day.Parents have noticed this"retro and reimagined"phenomenon,too.Four infiveparents say they often see their kids wearing a set of clothes similar to one theywould have worn when they were in school.This back-to-school season,retro styles.DIY self-expression,andpersonalization are becoming key trends,and 99%of parents say it's at leastsomewhat important to consider their kids'opinions on purchasing decisions when itcomes to back to school shopping.More than half (51)also think their child'sgeneration actually has a better fashion sense than their own.Yet,the survey,conducted by OnePoll on behalf of Zulily,found that it's notjust retro styles that are trending this back-to-school season.A second phenomenonshowed by the survey shows that kids see style as a way to express their confidence.Ninety percent think being able to express their personal style helps them feel moreproductive at school,and nearly half of kids (48%)think a cool or fashionable suitis vital to their success at school.28.What can we learn from Paragraph 2?A.Kids can make their own closets.B.Wearing“mom trousers'”is a retro fashion.C.Kids show no interest in styles in the past.D.Kids can draw inspiration from school uniforms.29.What does the underlined word "retro"refer to?A.Old.B.Simple.C.Unique.D.Popular.30.What's most parents'opinion on their kids'daily clothes?A.It is important for kids to DIY their daily clothes.B.Kids tend to wear their parents'old clothes to school.英语试题（七）第8页（共16页）

又1P2=V2+1Vx-x)2=Vk2+1V(x+x)2-4xx2,4k242-31=8+,8分所以PQP=e+)-G+x广-4=+2+-42+」(2k2+1)2(RNF--2K+1-ww=0+.9分2k2IPRP+IORP-IPQP-21RNP-IPQP-242水+2k2+20%+k》-42+132k2+1'(2k2+1)22器岛4k2+102=2x2+22-4221=0+4+22-42k2+12k2+12k2+1=2x,2+2,2-4,+1+4k+4,-5,…10分2k2+15所以4=0,解符-子此时1R+1QF-PO=2.R0=名…1分4x0-5=0,y0=0,②当e的斜*不在时，P阳的方程为：=，时方Q1分：PRP+IORP-IPOP=2RP.RO-_8综上，可知存在定点R,0),即O=OF.…12分解法五：（1)同解法三.…4分(2)因为点F是以AB为直径的圆上AB的中点，所以点F在x轴上，不妨设点FL,O).…5分假设存在满足条件的点R(x,y)·①当PQ的斜率存在时，设PQ的方程为y=k(x-1),P(x,y),Q(x2,y2),y=k(x-1),由x22+y2=1,消去y,得x2+2k2(x-1)2=2,高三数学试题参考答案第15页（共15页)

22.(12分)已知函数f(x)=e一ax+a有两个零点，(1)求实数a的取值范围；(2)设x1,x2是函数f(x)的两个零点，证明：x·<工十2的动甲爱麻设市代角/光，/比人学用中人封关己可0,9,0n青线全雷100所名校最新高考中刺卷第8页（共8页）【23·（新高考）高考样卷·数学（二）一门

2023届高三信息押题卷（四）数学试题注意事项：1答卷前，考生务必将自己的姓名、考场号、座位号、准考证号填写在答题卡上。之回答选择题时，选出每小题答案后，用B铅笔把答题卡上对应题目的答案标号涂黑。如需改动，用橡皮擦于净后，再越涂其他答案标号。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，将本试卷和答题卡一并交回。考试时间为120分钟，满分150分一、选择题：本题共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的。1.设复数满足-=2，在复面内对应的点为(x,y),则A.(gp1)2+y=4B.(x-1)2+y2=4C,x+(y+1)=4D.x2+(y-1)2=42.设集合A=xx'≤9}，B=(x2x-a≤0，且A∩B=(x一3≤x≤1)，则a=A.-2B.1C.2D.33.如图，在△ABC中，点D在BC的延长线上，BD=3DC,如果AD=xAB+yAC,那么A.2=2y=23B.x=-2y13C.x=-2=2D.x=2y=-34.已知直线1：x+y-3=0上的两点A,B,且AB=1,点P为圆D:x十y+2x-3=0上任意一点，则△PAB的面积的最大值为A.√2+1B.2N2+2C.2-1D.2/2-25.“太极生两仪”最早出自《易经》，太极生两仪，即阴阳，阴用“一一”表示，阳用“一”表示两仪分别生阴阳，形成四象，四象继续生阴阳，形成人卦，从两仪到四象人卦的过程中，始终用“一”、…一”两种符号进行描述（太极图如下图所示）.进制，是计算技术中广泛采用的一种数制，由德国数理暂学大师菜布尼茨于1679年发明，二进制数是用0和1两个数字来表示的数，可以表达从0到无穷大的数字当前的计算机系统基本上是二进制系统，计算机通过识别由0和1组成的代码进行运算，从而实现各种操作，可以说0和1组成的数字，形成了计算机的世界，阴和阳描述了我们自然界的万物0和1形成了丰富的计算机世界，两仪四象八卦的生成过餐为从下到上，二进制的推列从左到右，用0代表阴，1代表阳，我们看一下四象人单对位的数字信惠押难卷（隆）戴学式题第」真（并年系

17.(本题8分)如图，在菱形ABCD中，对角线AC和BD相交于点O.(1)实践与操作：过点D作DE∥AC交BC的延长线于点E.(要求：尺规作图，保留作图痕迹，不写作法，标明字母)(2)猜想与证明：试猜想线段DC与BE之间的数量关系，并证明你的猜想，A18.(本题7分)在今年植树节期间，运城市某学校计划在校园每天种植相同数量的树木，该活动开始后，实际每天比原计划多植树25棵，实际植树200棵所需要的时间与原计划150棵所需的时间相同，求实际每天植树多少棵.19.(本题8分)为落实“双减提质”，进一步深化“数学提升工程”，提升学生数学核心素养，某学校拟开展“双减”背景下的初中数学活动型作业成果展示现场会，为了解学生最喜爱的项目，现随机抽取若干名学生进行调查，并将调查结果绘制成如下两幅不完整的统计图：（注：记A为测量，B为七巧板，C为调查活动，D为无字证明，E为数学园地设计)↑人数5045B20%30…20201030%0A B C D E项目根据以上信息，解答下列问题：(1)参与此次抽样调查的学生人数是,在扇形统计图中，扇形B的圆心角度数为(2)把条形统计图补充完整（要求写出计算过程，并在条形统计图上方注明人数）(3)若参加成果展示活动的学生共有1600人，请估计其中最喜爱“℃.调查活动”项目的学生人数【数学第4页（共6页）】

way possible,with thousands of volunteers from all walks of life fighting the fires withprofessionals.Under a hot sun,some residents cut down trees to make fire barriers ()Others rodemotorcycles on newly-formed paths to take firefighters and supplies up a mountain,while somevolunteers carried food and drinks on their backs on foot.Women and children helped organizesupplies and clear the ground,and performed other important tasks.On Thursday night,as wildfiresmoved towards a man-made barrier on Jinyun Mountain in Beibei District,a total of more than 1,000 firefighters and volunteers stood alongside the barrier with fire extinguishers ()Seenfrom overhead,their lights formed a silver wall against the increasing fire line.The residents'stories have touched many people."Chongqing people are really amazing,"said Du Hailang,head of the Beijing Emergency Rescue Association.Du and 10 other members ofthe association joined the fire rescue team in Chongqing."The man-made wall was unbelievable,but shocking.It really showed Chinese people's determination to defend their homeland,Dusaid.Zou Yu,from the Chongqing Emergency Management Bureau,said at a news conference thatby Friday morning,the wildfires in the city had been put out and efforts were being made toprevent new ones from breaking out.He said Chongqing had experienced several forest wildfires insummer due to heat wave and drought since 1961.The city has experienced record-high localtemperatures,with those in Beibei District reaching 45C.21.What does paragraph 2 mainly focus on?A.The ways to fight the wildfire.B.The quick spread of the wildfireC.The determination of the firefighters.D.The fighting efforts of Chongqing people.22.What caused the fire?A.Someone's smokingB.The hot and dry weather.C.No fire prevention workD.The shortage of fire barriers.23.What type of writing is the text?A.A research paper.B.A book reviewC.A news report.D.A travel diary.小When Thomas Panek lost his eyesight more than 25 years ago,he doubted whether he wouldever follow his lifelong dream again."I was too scared to run,he told CBS this morning.Indeed,although Panek had been running since high school,the idea of running blind seemed tooscared.But he did manage to keep his dream alivewith help from human guides who assisted himon each run.Even so,the real joy of the run-the excitement of independence that comes frompassing a course on your own terms-doesn't exist."When you're tied to another person,it's nolonger your own race,the 48-year-old told CBS."The independence isn't quite there."But Panek found a friend-indeed,man's best friend-who would help him regain that senseof purpose.He started running with a guide dog named Gus.Not only did Panek rediscover his lovefor running,but also found his best dog friend.Gus remained a great helper at Panek's side formany races.And,last Sunday,the old dog ran across the finish line with Panek at the New YorkCity Half Marathon.At that moment.they were both recorded in the history books.Panek,who第3页

三、填空题：本题共4小题，每小题5分，共20分20.(12分)13.已知角a满足3osa=8sina,则sn(受-2a)三已知正方形ABCD的边长为4，E、F分别为ADBC的中点，以EF为棱将正方形AB.14.(1+之)(2x-)°的展开式中含2项的系数为CD折成如图所示的60的二面角，点M在线段AB上.D15.已知抛物线C:y2=2px(p>0)经过x轴上方一点M(1,b),F为C的焦点且IMF1=2,直线1：y=k(x-2)与C交于A,B两点（其中点A在x轴上方），若1AF1=)1FB引，则△ABM的面积为延意16若西数)-以-。”+2在：与和与两处取到极值，日2≥2，则实数。的取M值范围是(1)若M为AB的中点，且直线MF与由A,D,E三点所确定面的交点为O,试确定四、解答题：本题共6小题，共70分。解答应写出文字说明、证明过程或演算步骤。点O的位置，并证明直线OD∥面EMC;17.（10分)(2)是否存在点M,使直线DE与面EMC所成的角为60°？若存在，求此时二面角已知等差数列{an}的前n项和为Sn,且,n∈N°.数列1bn}中，b1=2,bm+n=M-EC-F的正弦值；若不存在，说明理由.bmb.0共，伦2浸小海，补？共醋本球式，从①a.+1>a.,a4·a5=63,S8=64,②S6=4S3,a2n=2a.+1,③S10=100,a,=13,这21.(12分)三个条件中任选一个补充在上面的问题中，并解答下列问题(1)求数列{a.},{bn}的通项公式；已知频圆c话+卡1(a>b>0),四点6(a,9,6(0,2,5(1,5.5,4.rbn,n为奇数，9)中怡有三点在精圆C上(2)设cn=,n为偶数，求数列c的前2n+1项的和21LanQn+2(1)求椭圆C的方程；(2)若A,B分别为C的左、右顶点，P为C上任一点（不与A、B重合），直线1过点B18.(12分)且垂直于x轴，延长AP交I于点N,以BN为直径的圆交BP于点M,设O为坐标原点，求在△ABC中，设角A,B,C的对边分别为a,b,c,且b=2 acosAcosC+2ccos2A.证：0、M、N三点共线.(1)求角A;22.(12分)o)者△aC的百职号求：9时的最小酒已知函数f(x)=ax-sinx,x∈[0，+o),(aeR):每体量的面9△(1)若f(x)≥0，求a的取值范围；19.(12分)甲、乙两人进行对抗赛，每场比赛均能分出胜负.已知本次比赛的主办方提供8000元(2)证明：当x≥0时，2xe+sin2x+cos2x字4sinx+1.天而克克学量乙奖金，并规定：①若其中一人赢的场数先达到4场，则比赛终止，同时这个人获得全部奖的质科里负的出用年大的上更竿是厚起，反零具了爱有的节园金；②若比赛意外终止时无人先赢4场，则按照比赛继续进行各自赢得全部奖金的概率之性最上京的是小的)卧发个县，费地讯雪工北可漫收部比给甲、乙分配奖金.已知每场比赛甲赢的概率为p(0

根据题意，得AE=EF,∠A=∠BFE-90°，∠D-90°.ED∴.∠DF∠EFM-90°.在Rt△EFM和Rt△EDM中，,'FM=DM,EM=EM,∴.Rt△EFM≌Rt△EDM.(第22题答图3)∴EF=ED,…11分AE=ED.点E是AD的中点AD=6,AE-3.综上所述，4E的长是2或3，…12分>23.(本题13分)解：(1)抛物线y=r2+bx+4过点4A(-2,0)和B(4,0),「4a-2b+4=0,…】分16a+4b+4=0数学（二）参考答案第5页（共6页）1解，得a=--2…3分b=1∴抛物线的函数表达式是y=-+x+4.…4分22)当0时，y=+x+44.C(0,4…5分A-2,0),B(4,0),∴.OB=OC=4,AB=5.…6分∠B0C-90，·∠ABC-}580°-∠B0C=45°.……7.△BCD≌△BCA,∴.∠DBC=∠ABC=45°，BD=BA=5.∴.∠ABD=∠DBC+∠ABC=90°.…8分∴BDLx轴于点B.点D的坐标是(4,5)·……9分(3)点E的坐标是(1+5，-3)，(1-5，-3)，(1+√3,3)和(1-√5,3).…13分评分说明：解答题的其它解法参照上述标准评分.

2在Rt△D0B中，OD2+OG2=DG2,rr=2,⊙0的半径为2.44444444…(10分)六、（本大题满分12分）21.解：(1)a=2×6+5×743×8+6×9+3x10+1x1183444444444444444(4分)）20(2）b=8.5,444444444444444444444444444444444444444440444444444444444444…(7分)(3)八年级学生课外阅读时长的中位数，众数均比七年级的高…(11分)∴，八年级学生课外阅读积极性更高…(12分)七、（本大题满分12分）22.解：(1)证明：：∠ABC=120°，BD分∠ABC,∠ABE=∠EBG=60°，AB=BE在△ABF和△EBG中，∠ABE=∠EBG,∴.△MBF≌△EBG(SMS)FB=GB∠FAB=∠BEG,444444444444444*4444444”(3分),∠EFH=∠AFB,,∠EHF=∠ABE=60°…(4s分)(2),∠EHF=∠EBG=60°，∠FEH=∠BEG,∴.△EFH∽△EGB·(5分):坠-ESEH,BG=EF,EBEB EG由(1)得：△MBF2△EBG∴，EG=AF…(8分)又：若H为BG中点H=号BG=号AF,2AB,AF=F,B即：A.2EF,EB4444400中4转4004040404004444中卡404中408转4号44中年号44行(10分)2(3)∠ACE=∠ACB.详解如下：AB=BE在△DMB和△CEB中，∠ABE=∠EBG,,.△DMB≌△CEB(SAS).∠DAB=∠BEC,BD BC又：若AC分∠DAB,.∠DAC=∠CAB

8:16ll令61数学511A答案(1).pdfES=E0-S0'=号EG-1=?EG-号L=反x-2,所以IS=EP-ES-√FK2-ES=√一4x+8,又-4x十8>0，所以12则后+2k<2x十吾<+2x∈刀，得x

Ⅷ。阅读理解（共20小题；每小题2分，满分40分）阅读下列短文，从每小题所给的A、B、C、D四个选项中选出C.Where Peter is studyingD.Why Peteris late for schoo.A.Dick's trousers are一个最佳选项。B.One of Dick's T-shAAnnie has got some messages on her mobile phone.Mike is at his grandparents'home.It's in a small village and far fromthe cityC.Dick often plays soD.Dick buys the sweaJennyAe I'm going to have a birthday party at homecould we eat in the restaurant today?"Mike asks.Tto eat fish."Grandpa says no.He takes Mike to the north of the villageDear Zhao Min.next Saturday.Can you come?Thank you for yourBuy some sandwiches and milk on your way home.Andand gets a fish in the river.Then he brings it to Grandma."We can haveJackisha homeGrandpa sayNow let me tell you somesome fruit.Nothing for dinner today!"Grandpa,I want to play computer games,"Mike says.Grandpa saysI go to school frommoming and two in theSteveH,Annie!I have fixed your bicye You an comeadno.He takes Mike to the village park.Some boys are plaving baskethall"oandayith themGrandpain the moming.We learit now.AliceWhereare you,Annie?ne of y students is looking forGrandpa,canonhar condio(开空调2I'maiand P.E.My favorite sucold,"Mike asks.Grandpa says no.He takes Mike to the vegetableclassmates.There are alplay baseball,basketbayou in your school office.I teld him to come back later.We can also learn to plA56.Who will have a birthday party soon?deehind th houseWork with me hereand yo will feel hotGrandpa says.Now I am in the art clulA.Jenny..Jack"Howis your weekend,Mike?"Mom calls and asks.I like my new scheC.Steve.A57.D.Alice.A63."I have a great time here,"says Mike.What do we know about Steve?Grandpa gets a fishA.He can fix bikes.A.in the riverB.in the kitchen70.What is Cindy's fC.He loves sandwichesB.He is good at cooking.D.He is looking for Alice64.C.in the restaurantD.in the supermarketA.EnglishWhat does Annie do?Grandpa takes Mike to the park so that he canC.Music.A,A student.A.play computer games71What kind of speC.A doctor.B.Ateacher.B.play with some boysA.BaseballD.A reporter.C.VolleyballBC.give a call to his momD.have a talk with the boys7Little Peter is a boy ofnine.He is in Grade Three now.He lives nearWhy is Cindy in65.Grandpa asks Mike to work in the vegetable garden becaA.Because shethe school but he's often late for class.He likes watching TV in theA.Mike feels coldB.Because sheevening and goes to bed late,so he can't get up on time in the morning.C.Because sheThis term Mrs.Black,Peter's aunt,works in Peter's school.SheB.he needs Mike's helpD.Because sheteaches math in Peter's grade.She's strict,and she often tells the boy toC.Mike has to exercise to keep healthyfollow the school rules and come to school on time.Yesterday morning,D.he doesn't want Mike to stay at home阅读下面短文Peter got up late.When he hurried to school,it was a quarter past eight.D66.Which is the best title of the text?A.Mike's grandpaB.Mike's weekendam a 14-yHis aunt was waiting for him at the school gate.C.Let's have funD.Life in the villageProvince.Like ma"You're fifteen minutes late for the first class,Peter."Mrs.Blacklive with my grandsaid angrily,"why are you often late for class?"Hello!My name is Dick.This is my wardrobe.Thereare manyThey tell me to lis"Every time,when I get to the street corner,I always see a guidepostclothes in it.I like sports very much.So I have many clothes for sports.month to ask abou(路标).It says,.SCHOOL一GO SLOW!”C59.How old is Peter?Look!These are black shorts.I wear them when I play basketball.TheseJune 17th isA.Eleventrousers are nice.They are only fourteen yuan.I have many T-shirts,too.But my parents caB.Ten.They are in different colors.One is yellow.Two of them are blue and theold and they don'C.Nine.D.Eight.others are white.Whose green sweater is this?It's too big for me!I can'tfor me.There are4.Mrs.Black is a(n)wear it.It's a surprise for my father!Oh!Today is Father's Day I buy itI reafly missA.English teacherB.math teacherC.history teacherfor my father.73How old isA61 PeterD.music teacher7.What does the underlined word"wardrobe"mean in Chinese?minutes late for school.A.抽屉B.衣柜74What doesA.15B.20C.桌子D.书房CakeC.30D.40What do th62.What does the story mainly tell us?68.How many colors of T-shirts does Dick have?A.What Peter's favorite subject is.A.Only oneB.Two.me eagC.Three.D.Four.B.What subject Mrs.Black teachesWhich of the following is TRUE?2023年安徽中考第一轮复试卷·英语试题（一）第3页共4页

大一轮复学案数学能力强化练17.已知函数f(x)=xm-x|(x∈R),且f(4)=0.(1)求实数m的值；13.(2023辽宁葫芦岛模拟)函数f(x)=(2)作出函数f(x)的图象；的大致图象为(3)若方程f(x)=a只有一个实数根，求a的取值范围。14.多选题(2022山东青岛一模)已知函数f(x)是定义在R上的奇函数，当x<0时，f(x)=e(x+1),则下列命题正确的是A.当x>0时，f(x)=-e(x-1)B.方程f代x)=0有3个解Cf(x)<0的解集为(-0，-1)U(0,1)》D.Hx1,x2∈R,都有f(x1)-f(x2)1<2(In x,x>0,15.已知函数f(x)=若3x0∈（-0,0)，使g(x),x<0,得f代x)+f(-x)=0成立，请写出一个符合条件的函数g(x)的表达式：16.（2023福建三明模拟)已知函数f（x)=(sin,0≤x≤1若实数a,b,c互不相等，且f(a)(log2 022x,x>1,=f(b)=f(c),求a+b+c的取值范围，素养综合练18.若面直角坐标系内的A,B两点满足：①点A,B都在f(x)的图象上；②点A,B关于原点对称，则称点(A,B)与点(B,A)是函数f(x)的一个“和谐点对”.x2+2x(x<0),已知函数f(x)=(x≥0)，2则f(x)的“和谐点对”有()A.1个B.2个C.3个D.4个·286·

为x轴、y轴、之轴建立空间直角坐标系（图略），则则A(-1,1,0),C(1,0,0),G(2,0w3),DA1=(2,0,2),DE=(0,2,1),则面A1DE的一个法所以CG=(1,0W3),AC=(2,-1,0).向量为n=(2,1,-2).设M(x,2,x),则AM=(x一2,2，设面ACGD的法向量为n=(x,y,z),z).由AM·n=0,得2(x-2)十2-2z=0→x-之=1，故点M的轨迹为以BC,BB,的中点为端点的线段，长为交n=0'即z+5x=0,则AC.n=0,2x-y=0.√1+1=√2.故选B.所以可取n=(3,6,-3).9②xD2解析：如图，连接B1D1,又面BCGE的法向量可取m=(0,1,0),A易知△B1CD1为正三角形，所n·m√3所以cosn,m〉=nm=2以BD1=CD1=2.分别取因此面ACGD与面BCGE夹角的大小为30°B1C1,BB1,CC1的中点M,G,【方法导航】折叠问题中的行与垂直关系的处理关键H,连接D1M,D1G,D1H,则A是结合图形弄清折叠前后变与不变的关系，尤其是隐含易得D1G=D1H=√22+1=√5，DM⊥B,C1,且的垂直关系.一般地，翻折后还在同一个面上的性质D1M=√3.由题意知G,H分别是BB1,CC1与球面的交不发生变化，不在同一面上的性质发生变化点.在侧面BCCB1内任取一点P,使MP=√2，连接11.解：(1)证明：因为PD⊥底面ABCD,所以PD⊥AD.又底面ABCD为正方形，所以AD⊥DC,DP,则DP=√DM+MP=√(W3)2+(W2)2=√5，因此AD⊥面PDC.连接MG,MH,易得MG=MH=√2，故可知以M为圆因为AD∥BC,AD中面PBC,所以AD∥面PBC心，√2为半径的圆弧GH为球面与侧面BCC,B,的交由已知得L∥AD,线.由∠B1MG=∠C,MH=45°知∠GMH=90°，所以因此l⊥面PDCG升的长为}×2xX2-=(2)以D为坐标原点，DA的方ZA2向为x轴正方向，建立如图所示【方法导航】立体几何中空间动，点轨迹的判断或求轨迹的的空间直角坐标系Dxyz,则D长度，一般是根据线、面行，线、面垂直的判定定理和性(0,0,0),C(0,1,0),B(1,1,0),质定理，结合圆或圆锥曲线的定义推断出动点的轨迹（还P(0,0,1),DC=(0,1,0),可以利用空间向量的坐标运算求出动，点的轨迹方程).PB=(1,1,-1).10.解：(1)证明：由已知得AD/BE,CG∥BE,所以AD/CG,由(1)可设Q(a,0,1),所以AD,CG确定一个面，则DQ=(a,0,1).从而A,C,G,D四点共面.n·DQ=0,由已知得AB⊥BE,AB⊥BC,且BE∩BC=B,设n=(x,y,z)是面QCD的法向量，则n·Dd=0,所以AB⊥面BCGE.又ABC面ABC,所以面ABC⊥面BCGE即/ax+z=0可取n=(-1,0,a),y=0,(2)作EH⊥BC,垂足为H.-1-a因为面BCGE⊥面ABC,面BCGE∩面所以cosn,Pi)=n·P响n|PB1√5.√1+aABC=BC,所以EH⊥面ABC.设PB与面QCD所成角为，由已知，菱形BCGE的边长为2，∠EBC=60°，可求得则sin9=5.la+1L_53h+2aBH=1,EH=3.3√1+aa2+1以H为坐标原点，元的方向为x轴的正方向，建立如图所示的空间直角坐标系Hxyz,因为识月十年1≤3，当且仅当a=1时，等号成立，1+DZA所以PB与面QCD所成角的正弦值的最大值为53【方法导航】利用向量求线面角的两种方法(I)分别求出斜线和它所在面内的射影直线的方向向量，转化为求两个方向向量的夹角（或其补角）；教学笔记数学·参考答案/67

()A.若PO=|PF,则双曲线的离心率e≥2B.若△POE是面积为√5的正三角形，则b2=2√5C.若4为双曲线的右顶点，PF⊥x轴，则FA=FPD.若射线FP与双曲线的一条渐近线交于点Q,则2F-2F>2a第II卷（非选择题）三、填空题（本大题共4小题，每题5分，共20分）13.己知展开式(2x-1)°=a。+ax+a,x2+…+a,x(n∈N)中，所有项的二项式系数之和为64，则a1+a2+…+a。=·(用数字作答)14.已知a,万为单位向量，且云在6方向上的投影为-分，则+2讽-15.已知抛物线y=4x的焦点为F,点P,0在抛物线上，且满足∠PFQ-行，设弦PQ的中点M到y轴的距离为d,则P巴的最小值为d+116.若函数g(x)=2x2-x-(x-)在区间[0,2]上是严格减函数，则实数1的取值范围是四、解答题（本题共6个小题，共70分）17.己知数列{an}的前n项和为Sn,满足nan1+Sn1=0,a,=1.(I)求数列{an}的通项公式：(2)设bn=S。·Sn1,数列{bn}的前n项和为Tn,证明：Tn<1.18.如图，已知面四边形ABCD存在外接圆，且AB=5,BC=2,cos∠ADC=54(I)求△ABC的面积：(2)求△ADC的周长的最大值.I9.在四棱锥P-ABCD中，侧面PCD⊥底面ABCD,PD⊥CD,底面ABCD是直角梯形，AB/DC,∠ADC=90°，AB=AD=PD=1,CD=2.(1)求证：BC∠面PBD:试卷第3页，共5页

2023一2024学年宁德一中高三年级第一次考试数学试题参考答案及评分标准一、选择题：本大题共8小题，每小题5分，共40分.1.C2.C3.B4.C5.C6.A7.B8.A二、多项选择题：本大题共4小题，每小题5分，共20分.9.BD10.BCD11.BD12.ABD三、填空题：本大题共4小题，每小题5分，共20分.13.1014.[-1,2]516.8四、解答题7.Da2时，A=x1二3<0=x20，d+2>a21447as2A={x|20.又当xe2时，2+时()2>0恒成立，所(（2+22+,x[2时恒成立，………82-1令1=2-1，则-m<+3北+2）-+51+6=1+5+5,13]时恒成立，t而1++522,5+5=26+5，当且仅当1=6，即1=6时，等号成立，….….10所以m>-2√6-5，即m的取值范围是(-2V6-5,+∞)）.……12

全国教辅类一级报纸GLSH岁并LEARNINGR新课程高二课标(选择性必修3)第36期总19705期CE19832023年3月7日星期管单位山西省教育厅主办单位山西教育教辅传媒集团编辑出版学英语报社社长吕夺明总编李若编审Nancy E.Riley国内统一刊号CN14-0702(F)网址http/www.xyybs..com(温馨提示：登录学英语报社官网http:lwww.xyybs.com下载本期配套教学PPT)TEST YOURSELF本期资源码：td5deUnits4-5,(选择性必修)Book3众号。复上ClassNanMarks试题命制：英培》报社试题测评组第一部分听力（共两节，满分30分）节（共5小题；每小题1.5分，满分7.5分）Who is the woman?例：How much is the shirt??A.A tour guide听第9段材料，回答第14至17题。A.￡19.15.B.A student.B.￡9.18C.￡9.15C.A teacher.14.What does the woman think of her current life?案是C。听第7段材料，回答第8至10题。A.It is relaxing.B.It is boring.8.What happened to the woman?C.It is busy.5.Which place does the man think highly of?What does the woman want to see?A.She had no driving license.B She went th.Key WestB.Lake Wales.

20.(12分)已知Sn为数列{a,}的前n项和，a=l,Sn1+Sn=(n+)2.(1)证明：an1+an=2n+1.(2)求{a}的通项公式.(3)若之票，求数列6)的前n项和21.（12分)已知抛物线C:y2=2px经过点(2，-2W6),直线l:y=a+m(km≠0)与C交于A,B两点（异于坐标原点O).(1)若OA0死=0，证明：直线过定点.(2)已知k=2,直线2在直线的右侧，4∥2,4与4之间的距离d=5,交C于M,N两点，试问是否存在m,使得MN-A=10?若存在，求m的值：若不存在，说明理由.

-

-

-

报·高中数学·新果标版龙务性必修线的斜率为-1.若方程组mx+4,=2有唯一解，则两直线的斜率x+y=1不等，即-≠-1m≠4.1-a12325.5解析：由题意，得,'a>0,∴.a=5./24第4期《2.1直线的倾斜角与斜率，2.2直线的方程，23直线的交点坐标与距离公式》能力检测基础巩固1.D解析：设直线的斜率为k,则直线方程为y-2=(x-1),所以直线在轴上的藏距为1-2，令-3123，解得k<1或612.A解析：设直线ar+y-3=0的倾斜角为0，则tan0=-a,因为直线ar+-3-0的顿斜角大于牙，所以-心1或-<0，解得a<-1或a>0,所以a的取值范围是(-9，-1)U(0,+∞)解析：由-=1，得=”-由-=1，得ymxm3.Bm nmn mm,即两直线的斜率同号且互为倒数，排除A,C,D项，故选B项，4.A解析：根据题意画出图形，如图所示，直线，：x-2y+1=0与直线L2:mx+y+3=0的交点为A,且M为PQ的中点，若|AM=之|PO,则PA1QA,即，1，所以·(-m)=-1,解得m=2,mx+y+3=0x-2y+1=0x-5+3x-15.AC解析：设P(x,5-3x),则d=V2,化简V1+(-1)月得l4x-61=2,解得=1或x=2,故P1,2)或(2，-1).故选AC项6.BC解析：设经过直线L,与L,交点的直线的方程为2x+3y-8+A(x-2y+3)=0(A∈R),即(2+λ)x+(3-2λ)y+3A-8=0.由题意，得112-8λ+3λ-81=2,化简得5λ-8入-36=0，解得入=-2V(2+A)+(3-2A)月8,故直线的方程为-2或4x-3y+2=0,故选BC项7.2x-y+5=0或x+2y=0解析：直线经过点P(-2,1),当直线的斜率不存在时，直线的方程为x=-2,点A(-1,-2)到的距离d=1,不成立；当直线的斜率k存在时，直线的方程为y-1=k(x+2),即kx-y+2k+1=0,:点A(-1,-2)到l的距离等于5，1-k+2+2k+11k+3引=V5,解得k=2或k=-2…直线Vk+1的方程为y-122减-1=2).即2y45城+3-08-石解折：州的几何意义是过w,).-,-1)两点的直线的斜率.由于点M(x,y)在函数y=-2x+8的图象上，当xe[2,5]时，设该线段为AB,且A(2,4),B(5,-2),k4=56x+39.解：(1)设过点A且与直线1，垂直的直线方程为4x+3y+m=0.把点A(-2,2)代入，得-8+6+m=0,解得m=2.故所求的直线方程为4x+3y+2=0.(2)设过点A且和直线，行的直线，的方程为3x-4y+n=0.把点A(-2,2)代入，得-6-8+n=0,解得n=14.故直线，的方程为3x-4y+14=0.114-212所以行直线L,l,的距离d=V3+(-4)310.解：(1)由题意得，k,:乏，故1B边所在直线的方程为y43（x-3,即3x-2-1-0,2(2)联立3x-2y-1=0,解得所以A(1,1),则AC的中点x-4y+3=0y=1,为(3，弓)，则1G边的中线所在直线的方程为x=3,能力挑战1.C解析：直线(m-1)x-y+2m+1=0可化简为m(x+2)+(-xy+1)=0,令+2=0，解得x=-2,故无论m为何实数，直线-x-y+1=0y=3(m-1)x-y+2m+1=0恒过定点(-2,3).答案专页第4页

19.(12分)24已知△ABC的三个内角A,B,C所对的边分别为a,b,c,且2 bcsin A=3(a2--c2).(1)求角A的大小；(2)若b=2,D'为BC边上一点，DA⊥BA,且BD=4DC,求cosC酥)bcsm小=31b2)502多)、)bLt以—02c2ab啊cm2bl-(4记2-23bctan-5EL兄封2)内即和A毛Z CAD6朴6（府D中报店正旋4化险中由乎说存经5n合574Aa3又3D24P根据纺移(九△9%中a命2udS(3a松，72乃=4*6-x4)811142x2527227【24·G3DY(新高考)·数学（十）一XB一必考-Qd·77·A-29

-

their presentations will be evaluated by our judges based on content,languageproficiency,and overall delivery.I am confident that today's speeches will be both informative and inspiring,and Iwish all participants the best of luck.Now,let's begin this exciting event!第二节One possible version:For the next couple of weeks,I was determined to avoid any further embarrassment.I refused to open the encyclopedia and kept my mouth shut during classes.I feltlike I was walking on eggshells,constantly worried that I might say somethingfoolish.I was afraid to share my opinions or ask questions.As a result,I becameincreasingly isolated.I didn't participate in any activities that might show myignorance until my teacher talked with me,enlightening me patiently.My spiritslifted,and I felt cheerful once more.I began to share information with my classmates again and became more carefulabout the information.I started to double-check my sources and read articles intheir entirety before sharing any information with others,unwilling to make thesame mistake twice.It was a lesson learned the hard way,but it made me realise theimportance of confirming information before presenting it to others.And while mybologna tree story might have been a bit embarrassing at the time,it ultimatelytaught me an important lesson that has stayed with me throughout my life.部分解析阅读

B "Gas station"may no longer be an accurate description for a Circle K refueling station in Oslo,Norway,after it became the first in the world to replace all gas pumps with electric car charg ers. The Oslo Circle K is one of the company's 16,000 gas sta- tions around the world,but this location is the only one from a company to completely convert (to electric fueling.The conversion to electric fuel "fast chargers"came at the same time as the United Nations Climate Action Summit was taking place in New York. Christina Bu,secretary general of the Norwegian Electric Car Association,said in a press release that it was extra fun to

and belonging. The mood boost (of talking to strangers may seem brief,but the research on well-being,Epley says,suggests that a happy life is made up of a high frequency of positive events,and even small positive experiences make a difference."Happiness seems a little bit like a leaky tire on a car,"Epley explains."We just sort of have to keep pumping it up a bit to remain it. 28.What conclusion can we draw from the test in paragraph 1? A.Short conversations with strangers could lift moods. B.The customers at Starbucks have a sense of belonging. C.The cashier has a good skill to get along with the custom ers. D.The first half of the participants spent more money than