全国100所名校最新高考模拟示范卷5.已知an=2n-1,bn2an,n为偶数,若数列{bn}的前n项和为Sm,则S2man,n为奇数A.3n2+nB.4n2-3n+4C.3n2+2nD.2n2+3n6,如图，在△ABC中，E是AB的中点，BD-2D心，F心-号A市，EF与AD交于点M,则AMA8AB+号ACBABAC面个西第的，面四的音所常非四留雄长察回来料一角指竹独网，增魅精MC号Ai+8AC其公游个四分半比的州红烟现，S,S比高的)一A州面四馆内面四客通D.号AB+4AC7.已知函数f(x)是(0，十o∞)上的单调函数，且f(f(x)一x-log2x)=5,则f(x)在[1,8]上的值域为A.[2,10]B.[3,10]C.[2,13]mD.[3,13]8已知f）=naa7y5cas,若对任意实数：都有/)厂Asm(ax+p+,其中A,ω∈R,p∈[0,3π)，则P的所有可能的取值有A.2个B.4个C.6个D.8个二、选择题：本题共4小题，每小题5分，共20分.每小题有多个选项符合题目要求，全部选对的得5分，部分选对的得2分，有选错的得0分，9.甲、乙两个盒子中各装有4个相同的小球，甲盒子中小球的编号依次为1,2,3,4，乙盒子中小球的编号依次为5,6,7,8，同时从两个盒子中各取出1个小球，记下小球上的数字.记事件A为“取出的数字之和为偶数”，事件B为“取出的数字之和等于9”,事件C为“取出的数字之和大于9”，则下列结论正确的是A.A与B是互斥事件B.B与C是对立事件C.A与C不是相互独立事件D.A与B是相互独立事件10.为了向社会输送优秀毕业生，中等职业学校越来越重视学生的实际操作（简称实操)能力的培养.中职生小王在对口工厂完成实操产品100件，质检人员测量其质量（单位：克），将所得数据分成5组：[95,97)，[97,99)，[99,101)，[101,103)，数学卷（二）第2页（共8页）【23·(新高考)ZX·MNW·数学·Y)

全围100所名税爱所韦考横拟示花卷答素○、数学卷参考答案（七）第8页（共8页）【23·新言老1X,W,数学，月

y=kx+m,(2)由得(k2+2)x2+2mkx+m2-2故f()在x=3处取得极大值，且极大值为f(罗）=6mln3驷-150，无极小值(5分)价=0，则41=4m2k2-4(k2+2)(m2-2)=0,(2)证明：当m=2时，f(x)=24lnx-x2-2x,整理得m2=k十2.卧期数五5出(7分)设g(x)=(18-x)lnx-f(x)=t+2x-6lnx一y=kx十m,xIn x,+y=1由得(2k+1)x2+4mkx+2m2-2则g(x)=2x-6-1nx+1(x>0),=0,则4=16m'k2-4(2k2+1)(2m2-2)>0,令-go.测o-2+号-1280整理得k>1.Amk2m2-2则十2k+1,022k2+1(9分)又-号，所以名十号8m2k2所以h(x)在区间(0，+∞)内单调递增，(2k2+1)(m2-1)-2又h(1)=-3<0,h(2)=2-1n2>0,0日00海所以h(x)存在唯一的零点x0,且x6∈(1,2).16k+8=k2十2，解得2=2或4，由h(x)=2x。-6-1n+1=0,8,则m2ok2+8(11分)得nx=2红，-6+1，(8分)解得k=一√2或=√2或k=2或k=一2.(12分)当x∈(0，xo)时，h(x)<0,即g(x)<0,g(x)单调22.(1)解：由题得f(x)=6mlnx一x2Tmx的定义域递诚；为(0，+∞)，且f(x)=6m22x-m=当x∈(x,+o∞时，h(x>0,即g(x)>0,g(x)单调递增.(9分)2x2+m.x-6m2=(x+2m)(2x-3m)(2分)所以g(x)≥g(xo)=x6十2xo-6lnx0-xoln xo=当m=0时，f(x)=一2x<0,f(x)单调递减，所以6+2a-(6+)(2.8+1）=-云-11f(x)无极值；(3分)当m0时，令f<0,得>3令f(x)>0,+36To得0-2-11×2+9所以g(x)=(18-x)nx-f(x)>-8,,+3m内单调递减，即(18-x)lnx十8>f(x).400,(12分)

数学爱好者七年级同步训练下册（华师大版）。SHUXUE AIHAO ZHE解这个方程，得b=-1.解这个方程，得y=-6.所以k的值为3，b的值为一1.x=2,把代入原题中两个方程组，得8.解：因为|2x-4+3y+(3x-2y+7)2=0,(y=-6且2x-4+3y≥0，(3x-2y+7)2≥0，2m+6n=-4,③2x-4+3y=0,①-6m+2n=-8.④所以3x-2y+7=0.②由③-④得，8m+4n=4,由①×3-②×2，得13y-26=0,整理，得2m+n=1,解这个方程，得y=2.所以(2m+n)2023=12023=1.把y=2代入①，得2x-4+3×2=0,C组解这个方程，得x=一1.13.解：.x￥y=ax+by+2xy,2*1=9,所以x=(-1)2=1.3￥(-2)=-8，B组(2a+b=5,9.123a-2b=4.思路点拨：两个方程相加，得m十n=2,两个方程a=2,解方程组，得相减，得m-n=6,所以(m十n)·(m-n)=12.b=1.10.8∴.x￥y=2x+y+2xy,思路点拨：由①+②，得3a=9k,解得a=3k,把号*6-2×号+6+2×2×6=18.a=3k代入①，得b=-k,再把a=3k,b=-k代入a+2b=8,得k=8.课时3用代入法和加减消元法解x=1,(a-b)x+y=4,11.解：把代入方程组得二元一次方程组y=2bx+ay=1,「a-b+2=4,①必备知识b+2a=1.②消元一元一次方程代入消元法加减消元法由①+②，得3a+2=5,课堂互动解这个方程，得a=1.3.x+4y=25,①例1解：(1)把a=1代入①，得1-b+2=4,2x-y=2;②解这个方程，得b=-1.由②，得y=2x-2,③所以a2022+3b2023=1202+3×(-1)2023=-2.把③代入①，得3x+4(2x-2)=25,3.x+2y=-6,①解这个方程，得x=3,12.解：由题意可知，方程组的解5.x-2y=22②把x=3代入③，得y=4.是原题中两个方程组的解，x=3,所以由①+②，得8x=16,y=4.解这个方程，得x=2,(2x+4y=5,①(2)把x=2代人①，得6+2y=-6,(3.x-4y=5.②·16·

全国教辅类一级报纸NGSH￥养LEARNING®新教材高二(B)(选择性必修3)第36期总19017期2022年2月8日星期二单位山西省教育厅主办单位山西教育教精传媒集团编辑出版学英语报社社长吕夺明总编李若编审NancyE.Riley国内统一刊号CN14-0702们)网址htp:wWbs.G0P唱（温馨提示：登录学英语报社官网http:wwW.xyybs..com下载本期配套教学PPP步课堂Unit8,(选择性必修)Book3■郑书阅读任务阅读课本第36-37页的文章，完成以下图表。Ⅱ，以下为课本第36-37页文章的缩写，请认真读，并在空白处填入1个适当的单词或括号内WhoWhen词的正确形式。I:a professorWhereAt 2.In 3.Conseil:my 1.The crash was so violent that I was thrown intoNed:a Canadiansea.I cried for help in 1.(desperate).SuddeI felt myself quickly 2.(draw)up to the surof the sea.It was Conseil,my servant,3saved me.Conseil was cool and calm.Nevertheless.An extract from 20,000 Leagues Under the Seasituation was no less terrible.I felt 4.(terrtired.ProcessConseil had to keep me up and our preservadepended on him alone.I asked him to leave me,burefused strongly.Conseil kept 5.(cry)for

安师联盟2023年中考质量检测试卷·数学9,直线41：y=kx十b和l2:y=bx一k在同一面直角坐标系中的图象可能是注意事项：1.本试卷满分为150分，考试时间为120分钟。2.本试卷包括“试题卷”和“答题卷”两部分.“试题卷”共4页，“答题卷”共6页10.在△ABC中，AB=AC,BC=8,面积为24，腰AC的垂直分线DE交边BC于点D,交边3.请务必在“答题卷”上答题，在“试题卷”上答题是无效的.4.考试结束后，请将“试题卷”和“答题卷”一并交回.于点E,若点F为BC边的中点，G为线段DE上一动点，则△CGF的周长最小值为A.8B.7C.10D.9一、选择题（本大题共10小题，每小题4分，满分40分）二、填空题（本大题共4小题，每小题5分，满分20分）b24a20每小题都给出A、B、C、D四个选项，其中只有一个是正确的，11.计算：√8+(-2023)°=4二为队乙V1.下面四个数中，比1小的数是12.若关于x的方程x2-2x十m=0有实数根，则m的值可能.(写出r个即可)A.TB.√3C.1)-22.“神舟十五号”飞船于2022年11月29日发射成功，将在远地点高度393000m的轨道上驻留613,如图，已知反比例函数y-上和y-分别过点A和点B,且AB/轴，S。学点C上任意一点，则=个月进行太空实验研究.将数据393000用科学记数法表示为®3.93×10B.0.393×105C.3.93×106D.39.3×1043.如图所示的立体图形的主视图是4.下列计算正确的是D.(2ab)3=8a3b14.二次函数y=x2+bx十c.A.2a2十a3=3a5B.2a2·4a3=6a5C.2a5÷a2=2a35.物理课上，老师用弹簧秤将铁块悬于盛有水的水槽中，然后匀速向上提起，直到铁块完全露出水(1)当一1≤x≤1时，y的取值范围是-1≤y≤1，该二次函数的对称轴为x=m,则m面一定高度，则能反映弹簧秤的读数y与铁块被提起的高度x之间的函数关系的大致图象是为(2)若只存在两个自变量值x1,x2,其中-1≤x1

3.Bybus.六、略第5、6版参考答案1-5DABCD6-10 BAACB11-15ACBAB16-20 CDDBA21-25 CEDBC26.cruel27.afford28.bottom29.advantages30.harmful31.When32.necessary33.reduce34.example35.restaurant36.problem37.collect38.safe39.saves40.teaches41.During a WWF turtle monitor training in a nearby village42.Because eating turtle meat was a tradition in his culture.43.他的一生都与大海紧密相连，包括猎杀海龟。44.让他感到高兴的是，他的一些朋友也想成为海龟监测员。45.How the WWF training changed Muakula's life46.good use47.took part in48.brought back49.The government has taken actionto protect the environment.50.carefully51.to describe52.of53.people's54.⑧55.was recycled 56.takes57.However58.difference59.less

A范与向量BC的夹角为餐，所以∠A=,所以A位.AC-(A店+号A)·(A店+AD)AB+号AB.Ai+2AD=4+号×2×2X(-2)+号×4=3.11.BC如图示，∠DAB=45°，∠DBE=60°，∠DBC=∠CBE=30°，↑F∠ADB=15,又sin15°=sin(45°-30°)=62，所以在△ABD4ABBD中，sin∠ADB-sin DAB~解得BD=2a,在△BDC中，D=BDA东+BC2-2BD·BCcos,∠DBC=a,所以DC2十BC=BD,则∠C=90°，所以D在C的北偏西30°方向，且D,C相距akm.12.ABD由棱柱的定义知，选项A正确；对于选项B,由于AD1∥BC,BC∥FG,所以AD1∥FG,且AD1不在水面所在面内，所以棱A1D1与水面所在面行，选项B正确；对于选项C,在图(1)中，SrGH=FG·EF=BC·AB=4,在图(2)中，SGH=FG·EF>AB·BC=4,选项C错误；对于选项D.Vk=2X2X1=合·BE·BF·B,所以BE·BF=4E=BE十B≥2BE·BF=8,当且仅当BE=BF=2时，等号成立，所以EF的最小值为2√2，选项D正确.故选ABD.13.8√2直观图的面积为号×4×2=4，则原图形的面积为4÷2-8√2.414受1设a,6的火角为9.则m0=日治=得0=吾，所以a-0=信-a00=1.D取E为AB的中点，易知DC∥ME,所以∠NME为直线MN和ADC1所成的角.设正方体ABCD一AB1CD1的棱长为2，则ME=√2，NE=2,MN=6,所以ME+NE=MN,所以cos∠NME=2-E√6316.2闲为ACD为矩形，所以MP1AP.MP1DP,所以MP1百ADP,又S=X4XV3-2=25,SAw=2X4X3=6,点P到面MAD的距离为，Vr-m=【高一数学·参考答案第2页（共5页）】·23-375A·

202/-2022学年第一学期喝吉州期末质量检测试卷高二年级英话考试时间：100分钟满分：/20注意事项：1,答题前填涂好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上第一部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分0分）阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项。YWhether you are hoping to relax or you are looking forward to staying active in vacation,seeking out great hikes is a goodchoice.If you are planning a trip to southern California,or you live in San Diego,now it is time to search for new hiking trails()Here are some of the most popular hikes in San Diego.Cedar Greek FallsFor this hike,you'll hike from the Cedar Creek Falls to the Devil's Punch Bowl,which is a pool at the base of CedarCreek Falls.It can really heat up in the summer,and it is hard to hike,so be sure to pack lots of water and protein-filledsnacksCuyamaca Peak(顶峰)The peak is in Cuyamaca Rancho State Park,and you'll see plenty of snow-capped trees if you do this hike in thewinter.There are 37 trails in the state park,but you'll want to stick to Azalea Glen Loop.It's a difficult seven-milehike that will bring you right to Cuyamaca Peak.Cowles MountainIt is well known that Cowles Mountain is the highest point in San Diego-so it's worth visiting for the fact thatyou can see Mexico from the peak.The three-mile hike is quite difficult.If you are willing to wake up before the sun,CowlesMountain is a beautiful spot to watch the sunrise.It's one of San Diego's most popular hikes,so starting early mightalso help avold the crowds.Swami's BeachTake the trail from Swami's Beach to Cardiff Beach and you will experience impressive ocean views and the smoothsound of waves crashing onto the sand.As an easy out-and-back hike,the path is accessible (from Swami'sBeach and is about three miles in total.1.What do you need to take if you choose the Cedar Creek Falls?试卷第2页，共3页

ENGLiSHLEARNING国数辅类一级报纸外研高二基砧第13期总18994期2021年9月28日星期■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.Riey国内统-刊号CN140702I(F)网址htp:/MTEST YOURSELF女经外语教学与研究出版社授权在全国范围使用★服务号《学英语》订阅号资源码：tce634Module 6,Book 5方二维码关注报社微信公公众号中回复上方资源码ClassNameMarks(满分150分)单“资源码”输入资源码■试题命制：，《学英语》报社试题测评组第一部分听力（共两节，满分30分）(共5小题；每小题1.5分，满分7.5分)7.What does the man's father probably do?15.How does the woman feel?A.Worried.B.Tired.A.An actor.B.A lawyer.ow much is the shirt?C.Confident.￡19.15B.￡9.18.C.￡9.15.C.A businessman.听第7段材料，回答第8至10题16.What will the woman do next?C。8.How many people will go shopping?A.Serve a customer.B.Continue her paper.A.Two.B.Three.C.Four.at does the man like best?9.How will the woman get to Moon Town?C.Talk to her boss.B.By bus.听第10段材料，回答第17至20题B.Poems.Novels.A.On foot.17.Why did Frank come to Ramsey's school?omics.C.By subway.re are the speakers?10.When are they going to meet tomorrow?A.To give a talk.At school.030B.At10:00.B.To donate some books.

高二英话答案听力原文1.W:Hey,Jim.We have run out of salt.Please buy me some at the shopdownstalrs.M:Oh,Mum.I'm preparing for tomorrow's English test.Maybe Jane isfree.She's watching TV in her bedroom.2.W:My husband and I come to see our parents this evening.Their house isnear.You see the red one?It's theirs.M:Sorry,but you'll have to drive your car away.You cannot park the car here.3.M:Is $150 enough for the watch?W:I'm afraid not.50 more dollars should be given4.W:Hi,Buzz.Oh,you're injured.What happened?M:After withdrawing some money in the bank,I was going to the supermarketto buy something for Marley when I fell off my bike.You know,she is inhospital these days.5.M:How's Peter doing?W:Oh,he's fine.He just had a birthday,and...M:How old is he,nine?W:Yeah.He just turned nine.6.W:Excuse me,Sir.Can I ask you a question?M:Yes,please.What is it?W:How many times do you take the plane every week?

17.(5分)如图，四边形ABCD是行四边形，请用尺规作图法作菱形ABEF,使得E,F分别在边BC和AD上.（保留作图痕迹，不写做法）解如@Sh同式：S一半.海别为2，18.(5分)如图，口ABCD的对角线AC,BD相交于点O,E,F分别是OA,OC的中点，求证：△ABE≌△CDF群：0b呢60OE-OF四边BD行m0'3=①，0山CEB-FDB即相表0在AE牛和CF中Ef合泥A,S4B-0BE-EoAE-(FA妮0E(年t0b吃9A(F.Af19.(5分)如图，在边长为1个单位长度的小正方形组成的网格中，给出了以格点（网格线的交点)为端点的线段AB.(1)求作等腰直角三角形ABC,其中∠A=90°，(2)计算△ABC的面积.o医5好2)-3X3X2支X3-左3=9-32五斜5对店时-30【数学第3页（共6页）】45励·

15.解：本题考查带分母的不等式的解法去分母，得-2x<4一(x-5)去括号，得一2x<4一x十5.移项，得一2x十x<4十5.合并同类项，得一x<9.系数化为1，得x>一9.(8分)16.解：本题考查图形的移与旋转。(1)如图，△DEF即为所求；点F的坐标为(4,1).(4分)(2)如图，△A1B,C1即为所求；点A的坐标为（一5，一4）.(8分)17.解：本题考查一次函数的应用（销售类）.(1)根据题意，得40m+30(m一25)=5550，解得m=90.(3分)(2)由(1)得m一25=65，根据题意，得甲、乙两种篮球进价为(35×90+45×65)×(1一20%)4860(元).…(4分）设甲种篮球打了x折，则两种篮球售价为35×100×品+45×75=(350x十3375)元。…(6分)根据题意，得350x+3375一4860=1665，解得x=9.答：甲种篮球打了九折.，(8分)18.解：本题考查解直角三角形的实际应用.如图，过点A作AE⊥CD于点E,过点B作BF⊥AE于点F,则四边形BCEF是矩形.,∠ABC=150°，∠FBC=90°，∴∠ABF=60°.在Rt△AFB中，BF=AB·cos∠ABF=30X号=15mm,AF=AB·Sin乙ABF=30X9-15,5mm(4分)∠DAB=60°，∠BAF=30°，∠DAE=30°由题意，得AE=AF+FE=AF+BC=(15√3+45)mm.…(6分)》15

【解析】(I)因为sim∠B4C=V∠BAC为锐角，14212所以cOS∠BAC57…2分1414因为AC=√7，AB=3,在△ABC中，由余弦定理得BC2=AC2+AB2-2AC·AB.cos.∠BAC,即BC-7+9-2xV7x3x5V万14=1,得BC=1,…4分CDAC(2)在△ADC中，由正弦定理得sin∠DAC sin∠ADCCD√万即√2i万，所以CD=1.…6分142在△4DC中，由余弦定理得cos∠ADC=D'+CD-AC2AD.CD即-1D1-7,解得AD=2.…8分因为S6ABc=2}vix3xv2131…10分（一个1分）144至，Saeo-)x1x2xsim2x-5232所以SARCD=S△ABc+Sa4CD=33.√35w3…12分42419.(12分)如图，棱柱ABCD-A1B1CD1的所有棱长都等于2，∠ABC和∠A1AC均为60°，D面AA1CC⊥面ABCDA(I)在直线CC上是否存在点P,使BP∥面DA1C？若存在，求出点P的位置，若不存D在，请说明理由.(2)求面DA1C和面PA1C1的夹角.【答案】()点P在CC的延长线上，且CP=CC:2)【解析】D(I)证明：设BD与AC交于点O,则BD⊥AC,连接A1O,在△AA10中，AA1=2,A0=1,∠A1AO=60°，DX所以A1O2=AA?+AO2-2AA1·AOc0s60°=3，所以AO2+A1O2=AA?,B所以A1O⊥AO,………2分由于面AA1CC⊥面ABCD,且面AACC∩面ABCD=AC,AOC面AA1CC,所以A1O⊥面ABCD…4分以OB,OC,OA1所在直线分别为x轴，y轴，z轴，建立如图所示的空间直角坐标系，试卷第8页，共15页

答案专期2022一2023学年四川专版（人教版）九年级第13~18期分数学用报MATHEMATICS WEEKLY周周自测先锋卷(14)-、1.A2.D3.A4.B5.C6.D7.C8.B9.10.B提示：10.过点C作CFL1B丁点F.在K△ABC中，山勾股理，可得AB=IO.由=角形的面积公式，得SA:=)AB·CF2AC·BC,即5CF=24,解得CF=4.8,所以小止方形可以放4层.设第一层小正方形上边的边所在的直线与△ABC的边AC,BC分别交于点D.E因为DE∥B.所以恶=4经g,解得DE7,整数部分是7，则第一层是7个止方形。设第二层正方形的上边的边所在的直线与△ABC的边AC,BC分别交于点G.H.同上可得吧=48：2，解得6H=5名整数部分是5，则第二层是5个止方形.同理可得，第二层是3个，第四层是1个综上所述，正方形纸片的个数是7+5+3+1=16.二、11.1：212.144013.答案不唯一，如∠B=∠E14.(3,2)15.316.100三、17.因为△ABC最长边为25，△A'BC的最长边为50，所以△4C与△4"BC的料似比-亮=之所以阳+2c=号所以A'B'+BC+A'C'=112即△A'BC的片长为1I218.因为AB=AC.所以∠B=∠C.因为LBDE=∠CAD,所以∠BED=∠AIDC所以180°-∠BED=180°-∠ADC,即∠AED=∠ADB.囚为∠BAD=∠DAE.所以△4DE∽△ABD19.囚为四边形ABCD是行四边形所以AD∥BC,AB∥CD.所以器器器欲所以器器即GF=GF·ER所以旅=A6又因为∠A=∠A所以△MBC∽△ANM所以祭=、=因为50=45.所以六F200ds解得MN=3000.答：直线隧道MW的K为30O0米21.(1)图，△A,BC,即为所求，点B,的坐标为(-1,2):(2)如图，△AB,C即为所求，点B,的坐标为(2,-4).YA第21题图

2023年河北省初中毕业生升学文化课考试个数学模拟试卷(十一)总分题号22232425262021市、区、乡得分学校班级姓名考场选一选择题答题框涂卡注意事项：1.使用考试专用扁头B涂卡铅笔填涂，或将普通2B铅笔削成扁鸭嘴状填涂。2涂卡时，将答题纸直接置于整的桌面上，或将答题纸置于硬质垫板上填涂。定不能将答题纸置于软垫或纸张上填涂3.修改时用橡皮擦干净后，重新填涂所选项4.填涂的正确方法：■错误方法：☑〖)©)力1[A][B][c][D]6[A][B][C][D]11[A][B][C][D]16[A][B][c][D]2[A][B][C][D]7[A][B][C][D]12[A][B][C][D]3[A][B][C][D]8[A][B][C][D]13[A][B][C][D]4[A][B][C][D]9[A][B][C][D]14[A][B][C][D]5[A][B][C][D]10[A][B][C][D]15[A][B][C][D]考号得分评卷人二、填空题.（本大题共3个小题，每小题3分，共9分.其中18小题第一空2分，第二空1分，19小题每空1分)座位号17.已知V2xV12=V2xV3=aV6,则b-a的值为418.如图13-1，六边形钢架ABCDEF是由6根边长为2米的钢管首尾顺次相接而成的，且相邻两钢管可以转动.现转动钢管得到如图13-2所示的三角形钢架FBD(1)在三角形钢架FBD中，点A,E之间的距离为2米；(2)六边形钢架ABCDEF的面积线(填“>”“<”或“=”)三角形钢架FBD的面积图13-1图13-219.某市园林局尝试种植A,B两种树苗.经试种后发现，种植A种树苗a(a>10)棵，种下00后成活了()a+5)棵，种植B种树苗b(b>2)棵，种下后成活了(b-2)棵上05=8-007(1)若两种树苗共种植40棵，且两种树苗的成活棵数相同，则种植A种树苗22棵；(2)若园林局第一阶段种植A种树苗2p(p>5)棵，B种树苗p棵，则种植A种树苗成活棵数比种植B种树苗成活棵数多棵；第二阶段，种植A种树苗比第上25-了数学模拟试卷〔十一）第5页（共12页)

惠州市2024届高三第一次调研考试数学试题参考答案与评分细则一、单项选择题：本题共8小题，每小题满分5分，共40分题号24678答案BCACDBDA1.【解析】由已知可得AUB=1,2,3,S},U={1,2,3,4,5,6},所以C（AUB)={4,6,故选：B.2.【销新1白题盒如x告+名1度年为1，黄志，C1-i(1-)1+)23.【解析】因为(x+2)4=a4x+ar3+a2x2+ax+ao,令x=-1,得a4-a3+a2-a1+a=1,故选：A4.【解析】由a>1得a>1或a<-1,由a2>1得a>1或a<-1,故选：C5【解折1由派长公式1-材r得：1-受，人-行之，马-=经3r,…=21r,其中r=h=1,3蚊香的长衡-4++++=0+2+3+1训=4故选D6【行1知，A-名-季R8器若所以0=-乎-号P(A)=5,放选：B7.【解折】设a=lc=5.b=万，则PK=h=5.or-a=l,csOg=3,cos∠P0r=-3由余弦定理可得，PF=|OF+|0Pf-2|FOPI-cos∠FOP=3+1-2wW3×1)所以P=石，所以吗=6.放选：DOP8.【解析】由题意可得f(2+x)=-f(x),f(4+x)=-f(x+2)=f(对)，即f(x)是周期为4的函数，且图象关于x=1对称.令g(x)=f()-πx,g'(x)="(x)-π，x∈[0，时，f'(x)>π，·.x∈[0,1]时，g'()>0函数g(w)在[0，上单调递增.当x∈[0，时，g(r2g0),而g(0)=f0)-π×0=0，·g(x)≥0即f(x)-≥0另设h(x)=sinπx-πx,x∈[0，l],h'(x)=πcosπx-π=π(cosπx-l)≤0，即函数h(r)在[0，刂上单调递减，则sinπr-πx≤0，即sinr≤，故fx)2sina在[0，】上恒成立。结合对称性可画出函数f(x)和y=sn元x在[-3,3]上的简图，由图象可知，不等式f(x)≤sinπx在[-3,3]上的解集为[-2,0]U[2,3].故选：A数学试题答案第1页，共11页

Tn=a1+a2+43+a4+…+a2m-1+42m=(a1+a3+…+a2m-1)+(a2+a4+…+a2m)根据奇数项构成等差数列可得：a1+4+…+a2m=8+6+…+（-2n+10)=-n2+9n1,当n为奇数0当n为偶数则有：飞。-n2+9n,n为偶数而又：a2+a4+…+a2m=-n2+9n+l,n为奇数故B错误：100-2T,9=T0-40m=-502+9×50-(-1)2=-2049,故C对：根据T,中的奇数项构成等差数列，而偶数项之和不是1就是0，因此根据T,特点可知：Tn的最大值在奇数项之和取得最大值的附近，T,=-32+9×3+1=19,T,=T。+a,=19+2=21,T3=-42+9×4=20,T,=T+,=20+0=20,T。=-52+9×5+1=21,T1=T0+41=19,Tn的最大值为T=To=21,故D错故选：AC13.【答案】（-o,1)14.【详解】由题意可得{4+4=a1+9）=12,则a,≠0，4=429=16逊两不可得。即940，因为9，怒得？34x+-,x≥115.【详解】因为函数f(x)=(2a-1)x-1,x<1当x之1时，有f)=x+4≥4，当且仅当x=2时等号成立.4当2a-1=0,即a=时，有f=x+-,x≥1,不满足题意：-1,x<1当2a-1<0,即a<时，f()-(2a-)x-1在(-o,)上单调递减，有f(x)>f(1)=2a-2,不满足题意；当2a-1>0,即a>2时，f()=(2a-)x-1在(-，)上单调递增，有f(x)

因为两圆相交，则圆心(1,0)到公共弦的距离圆心C到直线1：x+y+m=0的距离为d=d=51+m.因为△ABC为等腰直角三角形，所以2+(-2a)<2.解得a>2或a<-是故选D.r=2d,即1+m=2,解得m=1或m=-38.对于选项A.圆x2+2=1的圆心到直线3x+4y-三、15.(1)证明：圆C:(x-2)2+(y-3)?=5的圆心12=0的距离为号>1，所以直线3x+4为C(2,3),半径r=5若直线1经过点D(4,2).则直线1的方程为y=12=0与圆x2+y2=1相离，圆上的点到此直线的最短距离为-1>1，因此不是“理想型9x-3),即2-y-6=0圆.同理可得选项C中的圆也不是“理想型"圆，因为圆心C(2,3)到直线的距离为4-3-6。4+对于选项B,圆x2+2=16的圆心到直线3x+5=r,所以直线/与圆C相切4y-12=0的距离为12<4，所以直线3x+4y-(2)解：选择条件①：直线/分圆C12=0与圆x2+y2=16相交，圆上的点到此直线此时，直线1过圆心C(2,3),直线I的方程为的距离为1的点有4个，因此不是“理想型”圆.y=-96x-3).即3+y-9=0对于选项D.圆(x-4)°+(y-4)?=16的圆心(4,4)到直线3x+4y-12=0的距离为6<4，又|AB引=2r=25,点D(4,2)到直线1的距离为h=13×4+2-9.0所以直线3x+4y-12=0与圆(x-4P+(y-4)P=10216相交，圆上的点到此直线的距离为1的点恰有两个，因此是“理想型"圆.故选D.所以Sm=ABh=529.对于选项A,将圆Q:x2+y2-2x=0与圆Q:x2+选择条件2：直线1的斜率为-3，则直线1的方Y+2x-4y=0的方程作差可得4x-4y=0,则程为y=-3(x-3).即3x+y-9=0公共弦AB所在直线的方程为x-y=0,故A圆心C(2.3)在直线1上，则|AB=2=2、5.正确.又点D(4,2)到直线1的距离为对于选项B.圆0：x2+y2-2x=0的圆心为(1.0).h=13×4+2-91.10公共弦AB所在直线的方程为x-Y=0,则线段v102AB中垂线的斜率为-1，则线段AB中垂线的方程为y=-(x-1),即x+y-1=0,故B正确.所以SAm=ABh=52对于选项C,圆Q:x2+y2-2x=0的圆心(1,0)到16.解：(1)设A(,),线段AB的中点为Mx,,公共弦AB的距离d=一号又圆0+4P+(-1)2得24=+2y0=2y-2的半径为1.所以川AB=21-=2,故C错误.2对于选项D,P为圆Q上一动点，圆心Q,(1,0)因为点A在圆C:(x+2)+y2=16上运动到直线仍的距离为4=兰，半径为1.则点户所以(x+2)2+y=16,即(2x-4+2)2+(2y-2)2=16到直线AB距离的最大值为2+L,故D正确。化简得(x-1)尸+(y-1)2=4故选ABD.故轨迹H的方程为(x-1)P+(y-1)户=410.对于A,当m=2时，直线1的方程为x+y+2=(2)轨迹H是个圆.圆心为(1.1).半径r1=2:0.圆C的圆心为(0,0)，半径为2，圆心C到直圆C:(x+2)2+2=16的圆心为C(-2,0).半线1的距离d=2=2<2,所以直线1与圆C径2=4.两圆的圆心距d=0+2)+(1-0=10.相交，选项A正确。因为n-r0)选项B错误：因为点F,M都在圆上，对于C,画图分析可知，当圆C上有且仅有两个点到直线/的距离等于1时，圆心C到直线/所以33可+=户：解得么3，02+(3-b=r2r=6.的距离d满足|2-d川<1.解得1

9.Where are the two speakers?A.In a shop.B.On an island.C.On a boat.听第8段材料，回答第10至2题。10.What kind of books does the man like best?A.Historical novelsB.Adventure stories.C.iographies11.Where does the man get most of his books?A.From his friends.B.From online bookstores.C.From the library12.What is the woman doing?A.Selling books.、B/Doing a survey.C.Talking about her habits.听第9段材料，回答第13至16题。13.What's the problem of the woman's bike?A.The Start button is broken.B.is a wrong modelC.LCD sereen does not work14.What is the reservation ate or the repair?A.November 23rd.B.December 8th.C.September 20th15.What is the customer's name?A.Linda Swollen.B/Lisa Sloan.C.Liz Slogan.16.How does the woman feel towards the solution to the problem?A.Ungrateful.B.Annoyed.、C.Satisfied.听第10段材料，回答第17至20题17Why does the speaker surf the Intemet?To read comments about herself.B.To watch the world news.C.To order some food.18.What does the speaker usually do at about 3:00 pm?A.Drink a cup of coffee.B.Attend a meetingC.Have lunch.19.What is the speaker's job?A.A journalist.BA business woman.C.A singer.20.What is the speaker's daily life like?A.She usually goes home before six.B.She never enjoys going to bed late.C.She seldom feels tired for her work第二部分阅读（共两节，满分50分）第一节（共15小题：每小题2.5分.满分37.5分）阅读下列短文，从每题所给的A,B,C、D四个选项中选出最佳选项。AMusicChamber Orehestra:The Orchestra plays at Memorial Hall at 1406 Elm Street,which offers severalconcerts from March through June.Call 723-1182 for more information.Symphony Orchestra:A Mus Hall an Riverbend.For ticketacall 31-3300.Regularseaon runs from September through May at Musie Hall and in summer at Riverbend in the open air.高考全真模拟卷·英语（八）第2页（共8页）

8.易知函数y=1-cos的周期为4π，所以圆柱的底面圆的周长为4π，所以圆的直径为4，据题意可知该椭圆的短轴长为2b=4,所以6=2，又函数y=1-c0s5的最大值为2，所以椭圆的长轴长为2a=V42+22=2√5→a=√5→c=1,所以椭圆的离心率e=1555故选B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，有多项是符合题目要求的.全部选对的得5分，部分选对的得2分，有选错的得0分)题号9101112答案ADACDBCABD【解析】9.由表格数据得，x=2+3+5+9+1山=6，少=12+10+a+7+3_32+0,将样本中心点555/6,32+a代入回归直线方程=11-0,5x得，32+a=11-0.5x6,解得a=8,则样本中55心点为(6,8)，所以选项A正确；对选项B,当变量x增加，变量y相应值减少，两个变量之间呈负相关关系，所以选项B错误；对选项C,由经验回归方程y=11-0.5x,令x=7,得预测值y=7.5,而预测值不一定等于观测值，所以选项C错误；对选项D,由残差定义知，观测值减去预测值为残差.由经验回归方程y=11-0.5x,令x=11,得预测值y=5.5,则相应于(11,3)的残差为3-5.5=-2.5，所以选项D正确，故选AD10.12=3,12=4,∴.a=log123>0,b=log24>0,.a+b=l0g123+l0g124=log212=1,对于A:名e1g,4>1g,3=1,所以A正确：对于B:h<(a+-,枚Ba log232411错误；对于C:.'a2+b2=(a+b)2-2ab=1-2ab>1-2×二=，故C正确：对于D:42:a-b=2a-1>-1,2P>2=改D正确，故选ACD11.设切点为(x,y),f'(x)=3x2-m,切线的方程为y-(x-mx)=(3x-m)(x-x),代入点P(-1,1),可得1-(x-mx)=(3x-m)(-1-x),即2x+3x=m-1.因为切线过点数学参考答案·第2页（共9页）

26.(本圆满分10永问题提出前供自大天可以迪BD中P线D上动点，A=,C=调BP经市场的最小值为元时，问题探究的eADh.A心B,Ac=IwDE分粥AC的动点含瑞点).N分别为DE,EF的中点，当MN的值最小时求点F到CD的最短距离.问题解决③图阳菱形AD是-块空地，其中AB-0m∠A-=6,E为AB中点PF量小，2分别是对角线BD,边AD上的动点（含筑点，某学校在开学的，计划举办新生欢迎仪式，雷要在这块空地上沿者PE,PF铺设两条红毯.已知红毯单价是160元米，问举行咳新生欢迎仪式时，铺设红毯至少需要花费多少钱？（结果保留根号》学B0°loo蠕-0x0&图2图3所y烟的力的0o.n的不哦阻口门者倍小华半的民南门，字日m支所示50天0301山/，0=人，号0州月m月面rf0rc0yy力年级数学第6页（共6页）Q】

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■■■■■口■且BC为面CBB,C,与面ABC的交线，有B,C⊥面ABC,…(5分)而CA⊥CB,所以可以以C为原点，分别以CA,CB,CB的方向为x,y,z轴的正方向建立如图3所示的空间直角坐标系，则C(0,0,0),AL,0,0),B(0,10)B,0,0,B),因为B,C⊥面ABC,故面ABC的法向量可记为m=(0,0,1),因为BC=B,C=(0,-1,0),故C0,-1,5),BC=(0,-2,V5),CA=(1,0,0).图3设面AC,B的法向量为n=(x,y,z),则有i.8=0→-2y+5:=0、取n=0,反2》.8分n.CA=0x=0,设面AC,B与面ABC的夹角为a,则有1方方号，放cos0=2y5cosa Hcos(mmn227…(10分)718.(本小题满分12分)(1)证明：因为sin(C-A)=-1+sinB,且sinB=sin(A+C),所以sinCeos 4-cosCsin A=-1+-sin AcosC+cos4sinC,即sin AcosC=2①，由正弦定理及2 acosC=bsin B,得sinAcosC=2sin2B②，由①②得sinB=1,所以B=90°，因为sin(C-A)=-l+sinB,所以sin(C-A)=0,所以A=C,即△ABC为等腰直角三角形.…(5分)(2)解：如图4，不妨设a=c=1,所以b=√2，因为2AM=MC,且AN为中线，BM =2BA+BC,AN=-BA+BC所以丽而-（丽+c〔丽+图4数学参考答案·第5页（共9页）

芜湖市2023∞2024学年度第一学期期中普通高中联考试卷高一数学参考答案一、单项选择题（每题5分，共40分）1.C2.A3.A4.D5.B6.D7.B8.C二、多项选择题（每小题5分，共20分.全部选对的得5分，部分选对的得2分，有选错的得0分.)9.AC 10.BC 11.BD 12.AD三、填空题（每小题5分，共20分.）13.414.[23)15.516.②③④四、解答题17.(10分)解：(1)当m=0时，B={x-1

acorns,So wherever you are,keep your eyes out and up.If you miss the migration this year,soon it'll be warming up and the birds will be on their journeys to their summer homes in theNorth.24.What does Bosque del Apache National Wildlife Refuge mean to the author?A.A scenery spot to have fun.B.A destination to take photos.C.A field to catch some birds.D.A place to watch birds migrate.25.Where does the author like to go to eat something after the visit?A.To Canada.B.To Alaska.圜C.To Buckhorn Tavern.D.To the backyard.26.Why are Sandhill Cranes mentioned in paragraph 3?如A.To show the reserves'value.斟B.To introduce a kind of birds.C.To stress the importance of birds.长D.To list some birds'migratory sites.27.What is paragraph 4 mainly about?区A.The habits of some migrating species.B.The author's knowledge of endangered birds.数C.The skills of taking photos of migrating birds.D.The author's efforts to watch birds'migration.都C潮Private identity is related to a person's identity where he cannot describe his feelingpublicly,and has difficulties in expressing his ideas and thoughts or socializing with others.Itis like being introverted (in front of the public.And public identity is related tosomewhat expressing one's feeling perfectly in front of the public.There are many factors for aperson to be publicly introverted,which includes different languages and cultures.Hunger of Memory:The Education of Richard Rodriguez written by Richard Rodriguezreflects his development from a disadvantaged son of Mexican immigrants living in northernCalifornia into a well-educated person.Rodriguez strongly believes that for accessing theprivileges of public life,people must learn how to be assimilated into public society anddepress their private identity.He argues that developing a public identity is directly related tomaturing into adulthood.Rodriguez states how he started his school life,from knowing only 50 words to becoming【高三英语第5页（共10页）】

可学科网密组卷船对于D,因为向量a+b,b+c,c+a是空间向量的一组基底，假设a,b,c共面，若a/b/,不妨设d≠0，设存在m,n∈R,使得b=ma,c=na,所以，a+b=(1+m)a,b+c=(m+n)a,c+a=(n+1)a,此时，向量a+b,b+c,c+a共线，与题设矛盾；若a,b,c共面，且b,c不共线，则存在元，4∈R,使得a=b+uc,则a+b=(2+1)b+uc,c+a=2b+(u+1)c,所以，a+b,b+c,c+a共面，与题设矛盾，故a,b,c也是空间向量的一组基底，D对.故选CD.11.ACD【解析】由x2+y2+4x=0,得(x+2)+y2=4,曲线C为一个圆，所以A正确；点(1,1)在圆(x+2)}+y2=4的外部，因为(1,1)到圆心(-2,0)的距离d=V(-2-1)}+1=V10,半径为2，所以圆上的点D到(1,1)的距离的范围为√10-2，V10+2,而6gV10-2,0+2,所以B不正确；直线1：kx-y+2k+1=0(k为常数)，则y-1=k(x+2),则直线过定点Q(-2,1),且点Q在圆(x+2)+y2=4内，则无论k为何值，直线1与曲线C恒有两个交点，所以C正确：假设存在这样的点P,使得P到点B与点(-2,0)的距离之和为8，则P在以点(2,0)与点(-2,0)为焦点，实轴长为8附圆上，即P在椭圆午上三1上，易忽猫阙女门=1与曲线C:(x+2)+y2=4有交点，16121612故曲线C上存在点P,使得P到点B(2,0)与点(-2,0)的距离之和为8，所以D正确.故选ACD.12.C【解折】对FA:0=写4C+号B,3而-C+2而，2而-2而-C-而.÷2BD=DC,故A错误：对于B:若Q为△ABC的重心，则QA+QB+QC=0,.3PQ=3PQ+QA+QB+QC=PA+PB+PC,故B正确：对于C:PABC=0,PCAB=0,∴.PA·BC+PC·AB=PABC+PC(AC-BC)PA·BC+PC·AC-PC.BC=(PA-PC)BC+PC.AC=CA.BC+PC.AC=AC.CB+PC.AC=AC.(CB+PC)=AC.PB=0故C正确：

(M)Judy is a teacher.She is happy with her job."Watching children make progress inlearning makes me very happy.Working with children keeps me young and energetic,"saysJudy.Sally is another teacher at Emmanuel School in London.She agrees with Judy.She says,"Young people are great to work with.They are humorous,clever and they want to make theworld a better place.”At the same time,Anne,a teacher of the deaf students at Whitehall Primary School,seesteaching as a chance to change our society.She says,"As a deaf person myself,I cancommunicate with deaf children easily.I love being around them.When they're lookingconfused or lost,I know what it means.I think I'm making a difference in their life.I want tohelp them realize their dreams and have a better future.参考答案听力部分1-5 CAABB 6-10 BCABC 11-15 ABABC16-20 CBBAA笔试部分(One possible version选择题除外)第一部分21-25 ADBBC26-30 BDADB 31-35 BBADD第二部分36-40 DADBD41-45 CDBCD第三部分46-50 CBACB51-55 DBCCB56-57DB第四部分58.praise59.certainly 60.promised 61.awake62.valuable 63.carelessness 64.list65.matter66.speech67.through第五部分68.If you have difficulty finishing the research,I will help you.69.You'd better not stay up late anymore70.Animals are forced to leave the forest.3/4

2.Did:use to have3.are taken good care of4.What is:used for5.is proud of6.When was;invented7.too:to8.didn't she9.not only;but also10.how;deal with11.No matter how hard12.divided;into13.Without doubt14.by mistake玩转语法1.What did;use to be like2.were asked to wear3.is grown4.is used to make5.Were:turned off by6.Asian7.were invented 8.of9.to pick10.thinner11.because12.widely13.pronunciation14.centuries15.It

A.School violence has been decreasing in recent years. B.However,most parents choose to send children to school. C.Good teachers are always willing to take the time to reach students. D.In a word,education at school seems to be preferable to learning at home. E.Some of them are especially gifted while the others have problems of some kind. F.For example,there are libraries,sports equipment and labs for science experiments. G.Meanwhile,parents will also be able to protect their chil- dren from harmful influences.