21.(12分)包.(12分)已知函数g(x)=2e一a.x2-x(a∈R)(1)若曲线y=g(x)在点(1，g1)处的切线方程为x+y+b=0,求a,b的值；已知椭圆C后+芳=1(a>b>0)的离心率为)，椭圆C的右焦点为F1,0).(2)已知f(x)=g(x)一e+x在[0,1]上的最大值不小于2，求a的取值范围.(1)求椭圆C的标准方程；(2)过点P(4,0)且斜率为的直线I与C相交于A,B两点（点B在线段AP上），a)':g(x)=2ex-ax-x (aeR)若BF⊥L,求..9X)'=2ex20x-1)题得0=2又‘：9')=2e-20-e-102b1b-J30=e,布椭圆万程勤￥+兰引9=e-T;.1te-1十b=02:BF⊥U1 b=-eBF⊥BP7.0=e.b-e1儿P内拍圆的程为X-去+y=2)2)f(x)=9x)-ex+×与有圆℃方程耳联粒ex-ax*9>X,0)9月法y悍3X+4[是-1×月1之即X-20x十26=0首X[o.,fw最力F小于2等1厅x):e×-0X'1在XL0,门有期2860得X=06万卉X10时6万（法）在△BFP中.H点B1作厅边FP上局EB显沈X0不是解时B丫：(1EF=10-6万-1=9-6D-09即门上有辆000-001月=4(06月=@67-6kx1=eX2×f(0,]十EF月-6万十9几K(x)=xex-2ex寸4是010EPH6U5-6X4hx)=xe*-2e×+4.x6(0.月见k2=/EB/2PB广-E上EF-=IEPIEPP见1h'(x)=e*(x-1)<0;-h(×)在(0，门Vhx)3h()=4-e7o2k'(X)70,k(X)在01门个i.k(X)m0×=k(1J=e-2数学卷（六）第1页（共8页）≤e-2【23·MNJ·数学·YN·A】i,0们乖-w,e0数学卷（六）第8页（共8页）【23·MNW·数学·YN·A

设M(x,乃)，N(x2,y2),则△=-48m2+144+192k2>0,42+3’￥5=4m3-12+5=-8m4k2+38分3所以kpw+kpN=x+1x2+1即2+k+m-多Xx+x)+2m-3=0,…9分即4k2+8k+3-4m-2m=0,即(2k+1)(2k+3-2m)=0.…10分若2k+3-2m=0即m=k+2直线Mw的方程为y一号x+,此时直线MW过点P,不合题意，11分所以2k+1=0,即k=方故直线W的斜率为-212分22.解：(1)由题意可知f(x)的定义域为Rf'0=2x+ae'-(r+a+ae_-2-x+2xe-xx+a-2)(e)2(e*)2Q+.1分令f'"(x)=0,则x=0,x2=2-a2分①当a=2时，x=x2=0,'(x)≤0在R上恒成立，f(x)在R上单调递减.3分②当a>2时，x>2,x∈(-o,2-a)时，f"(x)<0,x∈(2-a,0)时，f"(x)>0,(0,+o)时，f"'(x)<0,故f(x)在(-0,2-a)单调递减，在(2-a,0)单调递增，在(0，+o∞)单调递减.4分③当a<2时，x0,x∈(2-a,+o)时，f"(x)<0,故f()在(-0,0)单调递减，在(0,2-d单调递增，在(2-a,+o)单调递减.5分(2)当x≥0时，f(x)≤2恒成立，故+a+as2,所以x+m+a≤2e,即ax+)≤2e*-x2,由x+1>0得as2e-rx+16分令h()=2e-x2x+1(x≥0)，则h=2e-2xx+)-2e-x=x(2e*-x-2)(x+1)2(x+1027分令(x)=2e*-x-2,则t(x)=2e*-1数学参考答案及评分标准·第13页（共14页）

19.(12分)20.(12分)如图，已知正方形ABCD和正方形BDEF所在的面互相垂直，AC交BD于如图所示，ABCD为圆柱OO'的轴截面，EF是圆柱上异于AD,BC的母线O点，M为EF的中点、AB=2.A且S0.8(1)证明：BE⊥面DEF.(1)求证：BM∥面ACE的面0,0(2)若AB=BC=√6，当三棱锥B-DEF的体积最大时，求二面角B-DF-E(2)求点A到面BCM的距离.到9否的正弦值。H日且，1山洱汽亡D少且临义M尚开中门E上g三2Dr斗图挡限入2m业60人M50本杨国池)OElI BM(FAP>可子0AOECOAEC52☑pBMI面AE○3NoA六炬I/FBE1FAAC1石T是@抢母代99AA人F面ABE3099,A-AAALE二面BE六FBE0AApFABF-F○9897BE上@旰0y39)51AA03o.0-90-001323新高考·ZXQH·数学·N23新高考·ZXQH·数学·N卷七第6页（共8页）

21.解析：(1)方法-：三点共线18p8即：2p卫。2卫，解得p2=4-1-40-4又p>1,∴.p=2,故抛物线的标准方程为x2=4y.…4分方法二：设抛物线的标准方程x2=2py(p>1)设直线DF的方程为y=x+号D(,.G,)以.易知名=4…2分y=+号x=2py整理得2-2p-p2=0.Xx=-p2=4,因为p>1,所以p=2,故抛物线的标准方程为x2=4y.……4分(本问可以用其它方法求解，阅卷时可酌情给分)(2)法-：设点E(x,),则x。2=4y:EM=x,2+0y。-3}=y。2-2y。+9=0y。-1}+8;…6分4在RIAAME中，有cos2∠AMEAMEM6y,-1}+8易知∠AEB=π-2∠AME.则cosA∠EB=-cogb∠AME)=1-2 oLAME=l-G,-i旷+88周a函-m西--+小l》s分设t=(6y。-1+8,t∈8+o∞)a函=6--142-20分因为y=1+32-12在18∞)上单调递增，所以1=8，即E的坐标为任2，)时.EA.EB取得最小值0，所以EA.EBe[0,+∞)：综上，EA.EB的取值范围是[0，十o).…12分

W:Apples.She said apples are a good choice for food art,because they can be made into differentshapes easilyQuestions:6.What fruit does Tina like?7.Where is Shannon from?8.What does Shannon use fruit to create?9.What does Shannon encourage her kids to do?10.Why does Shannon like apples best?第三部分听短文回答问题I am a middle school student in Shenzhen.My father bought a mobile phone for me in 2019,whenI was thirteen.It was a South Korean brand but it was made in China.I used it to search forinformation at school and study online at home.However,from February 1st this year,we aren'tallowed to take mobile phones to school.Sometimes,if I carry the phone to school,I have to putit into a box in the classroom.At first,I thought the rule was boring and I didn't get used to thelife without a mobile phone at school.Now,I find it is helpful.I listen to teachers more carefullythan before and my study has improved a lot.Questions:11.Who bought the boy a mobile phone?12.How old was the boy when he got the mobile phone?13.Where was his mobile phone made?14.What did the boy use his mobile phone to do at school?15.What did the boy think of the rule at first?第四部分朗读短文略第五部分情景问答M:Hey,Gina.There's going to be a concert in our city next Saturday evening.

*8.4三元二次方程组的解法1.三，1，三2.B3.2y3z=12,2y-z=64.原方程组的解为=5,y=2=1.x=-1,5.y=2,z=36.（1)a=2,b=-3,c=-1.(2)当x=-2时，y=2×(-2)2-3×(-2）-1=13.第八章8.38.4同步测试题一、.1.B2.D3.A4.D5.B6.D二、7.三z-18.10,59.42010.57或7511.100000,8000012.1x+y=-1,①三、13.(1)x-y+z=7,②2x-y-z=0.③+③；得3x-2y7.④①×2+④，得x=1.将：=1代人①，得y-2.将x=1,y=-2代入②，得z=4所以原方程组的解为x=1,y=-2,z=4.(下转第2、3版中缝)人教版七年级

上期参考答案第十八章综合测试题-、1.D2B3.D4.D.5.D6.B7.C8.D9.D10.A二、11.答案不唯一，如A=90°12.25013.（7,3)14.415.416.5+37三、17.因为E是口ABCD边AD的中点，所以AE=DE.因为四边形ABCD是行四边形，折以AB=CD=6,AB∥CD.3-所以F=DCE.在△AEE和△DEC中，因为∠F=DC配，∠AEF=∠DEC,AEDE,3所以△AEF≌△DEC所以AF=CD=6.:所以BF=AB+AF=1218.连接A心因为四边形ABCD是菱形，所以LEAC=LFAC.在△ACE和△ACF中，因为AE=AF,2EAC=∠FAC,AC=AC,所以△ACE≌△ACF所以CE=CR19.因为四边形ABCD是正方形，所以AD=AB,∠BAD=90°.因为DEAG,所以ZAED=90所以ADE+∠DAE=90°.因为LBAF+LDAE=∠BAD=90°，所以乙ADE=∠BAF.因为BF∥DE,所以∠AFB=∠DEG=LDEA.

第19期B2版Keys及重点解析：newbands of dentine to replace its做饭技巧的小窍门。阅读方略tooth-like beak if it becomes worn or1.E。该句与上文是转折关系。空前句意为1-2DCdamaged可知答案“对于初学做饭的人来说，菜谱能很好本文是一篇说明文。科学家根据短吻2.C。推理判断题。通读全文，可知文章主鳄与河豚鱼牙齿的特点，研发出一种能让地指导他们去做饭。”要介绍了一项科学研究，该研究主要是人牙齿生长的技术，这种技术若通过临床为患有牙齿疾病的人们减轻痛苦。3.C。该空与上文存在指代关系，they指代试验，有望造福人类。特色题型people you know who cook well。与下文I.D。词义猜测题。根据前文regenerate its1-5 EFCGD存在并列关系，空后的句子进一步解释lost teeth as many as 50 times creating本文是一篇说明文。介绍了几个掌握说明和会做饭的人一起做饭的好处。

基础点5函数的基本性质高霸[37考，主要考查周期性、奇偶性、对称性及单调性]答案见P4考向1单调性与最值3*怎么解题(2021年全国甲卷)下列函数中是增函数的为A.f(x)=-xB.f(x)=()C.f(x)=x2D.f(x)=链接教材内容（新人教A版必修第一册P91练T3)根据奇偶性和单调性的定义，讨论函数f(x)=x3的单调性，。并判断其奇偶性【考查点】本题考查函数的单调性，【解题思路】A项：f(x)=-x是减函数；B项：因为0<，由指数函数性质可知fx)在R上为[1](增/减)函数；C项：函数f(x)=x2在区间(0，+0)上为[2](增/减)函数，在区间(-∞，0)上为[3](增/减)函数；D项：由幂函数性质可知(x)=在定义域R上为[4](增/减)函数.教材改编题腾远原创好题建议用时：5分钟1.(新人教A版必修第一册P160T6改编)函数f(x)=e-1的单调递减区间为A.（-∞，0)B.RC.(0,+∞)D.(-0,1)2.(新人教A版必修第一册P160T4改编)已知a>0,函数f(x)=,x<0X的最大值-ln(x+1)+3a,x≥0为3，则a=1A.B.1C.1D.242真题变式题膳远原创好题建议用时：10分钟3.(2021年全国乙卷)下列函数中最小值为4的是A.y=x2+2x+4B.y=sin x+4sin xC.y=2*+22-D.y=Inx+inx11

带上所有需要的装备，即提前做好准备。【补充解释】be in bad shape(para.3:in a bad state of health or in a bad condition健康状况不佳take chances(para.4):behave riskily冒险C篇主题语境：人与社会一一文学本文是新闻报道。约翰·巴勒斯自然保护区将扩建一块地方，让更多的观光者来此感受巴勒斯在自然保护和自然文学历史上的重要贡献。28.C。理解具体信息。由第一段中的John Burroughs should be a household name可知，作者认为，Burroughs应当被更多的人所熟知。29.A。理解目的。由第二段中的through insightful nature essays...during his lifetime可知，Burroughs的自然随笔见解深刻，使他在生前广受欢迎，段末列举出前米拜访他的各界名人是为了强调他的声望。30.C。理解具体信息。由第三段中的Slabsides now is the centerpiece of the John BurroughsNature Sanctuary可知，Burroughs的乡间小屋位于自然保护区的中心。31.B。理解具体信息。由第四段中的The goal of the project.Burroughs'importance可知，该项目的目的之一是增加人们对Burroughs及其成就的了解，承认他在自然保护与自然文学方面的重要地位。【补充解释】John Burroughs出生于1837年，本人曾当过农民、教师、专栏作家、演讲经纪人及政府职员，然而真正令他倾心的事业是：体验自然，书写自然。他被公认为是那个年代最受欢迎的作家。他的著作多以描述自然，尤其是鸟类为主。《醒来的森林》是他的第一部自然散文集，被誉为自然文学中的经典之作。D篇主题语境：人与社会一一文学本文是新闻报道。J.B.Priestley的作品重新受到青睐，作为他的遗稿管理人，Priestley的儿子力图向大众展示这位作家最真实的风格。32.D。理解目的。由第一段中的But now the...remarkable renewal可知，本段提到人们对Priestley写作风格的负面评价和他从大众视野中淡出是为后文作铺垫，引出文章主旨，即：Priestley的作品又重新得到了大家的关注。33.B。理解观点、态度。由第二段中的It was a great surprise.his works on可知，越来越多的制片人开始咨询Priestley的作品，这让他的儿子很吃惊。34.D。推断。由第三段中的He didn't reach.stay there可推知，Priestley追求自我的突破，喜欢尝试新事物，是一个勇于挑战自我的人。35.A。理解具体信息。由最后一段中的I really believe.he once was可知，每确定重新演绎一部作品之后，Priestley的儿子总会请演员们观看Priestley曾经的一段采访，目的是帮助他们了解真正的Priestley是什么样子。【补充解释】flavour of the month(para.2):somebody or something that is very popular at a particular time风靡一时的人或物第二节主题语境：人与社会一一文学本文是说明文。文章介绍了创作自然诗歌的几点建议。36.「。设空处承上启下，起过渡作用。设空处上文指出在诗歌中生动地抓住自然的本质并

(C)26.What does the underlined word "stunned"第二部分mean in paragraph 3?(O23.What might be the purpose of the learningA.Encouraged.B.Satisfied.record?第二部C.Surprised.D.Moved.A.To explain how to learn English well.(A)27.How will Friedman deal with the piano?B.To find out the students'learning methods.A.He will let more people see it.C.To check the students'learning of the unit.B.He will put it in his own home.读读C.He will use it to play piano music.D.To show what the students have done atD.He will sell it to a musician.school.BC第Thomas Edison,the famous inventor,hadEight-year-old Tyler Stallings has beenhearing problems.As someone played the piano,he第helping homeless veterans(退伍军人)since he waswould move close to the instrument and bite ()it.four years old,and so far,this little“superhero'部has raised over $50,000 for his causePressing his teeth into the wood of pianos helped“It's been a journey,but it's worth it,”Edison experience the music in his skull.Or in hisTyler's mother Andrea Blackstone said."I justown words,it allowed him to "hear through mywant to encourage parents to listen to their smartteeth”阅读kids because you never know where their ideas canFriedman,who makes a living by buying andg0.读selling pianos,bought a piano for about $45,000 inWhen Tyler was four,Blackstone wanted toearly January 2021.His friend Charles Frommer isteach him about the importance of veterans.Sheturned to the Internet to show her son videos abouta musician and recording-history lover.veterans and the things they do for the country.He came to Friedman's house in New York'sTyler watched videos of veterans holding signs but no第Hudson Valley to tune(给..调音)the piano.第one replying to their cry for help.He thought thisFrommer noticed the marks and told him "thosewasn't right and asked his mother,"If they're heroes,are Edison's bite marks'why should they be on the street?"Then Tyler put"I've seen thousands of pianos over 50 years,forward a wild idea to buy wood to build houses for分but I've never seen that anywhere,"he said aboutthe possible tooth marks.Friedman said he was分the veterans.However,his mother explained to himthey couldn't afford to do it,and then she wonderedwhat Tyler could do to help homeless veterans.stunned by the tooth marks left by Edison.Blackstone got Tyler in touch with Maryland's阅读Edison bought the piano in 1890 for $725.Paperwork with the sale includes the wordssuggesting it was sent读Governor Larry Hogan,and with his help,Tyler wasgiven a kid's grant(t补助金)of $100.With the help ofoffice furniture(家具)，”the grant and a website,Tyler decided to collectto his laboratory in New Jersey.There is no knownclothes,bedding and food to give homeless veteransphoto of Edison biting this piano.care bags.PBut he was known to bite into pianos as hisTyler began to volunteer at the Maryland第hearing worsened.His daughter once第Center for Veteran Education and Trainingremembered that a guest cried at the sight of(MCVET)as the designated(指定的)“superhero'”of their shelter.Tyler's goal is to make people部分Edison biting on to a piano as someone played it.Friedman is now looking for the right home努better understand the veterans'contribution andbring“hero packages'”for veterans who arefor the instrument."I think it should belong tohomeless or need basic things to help them getsomewhere many people can see it,"Friedmanback on their feet.Over the past four years,Tyler图said.Friedman does not want the piano to gohas raised thousands of dollars for them.back into private hands because of its connection读The veterans appreciated it very much.Thereto Edison.are a lot of smiles and letters of thanks,while Tylerlearns a lot about humanity(人性)and how tog厚connect with people and be kind.(B)24.Why did Edison bite the piano?第A.To tune the piano.第(A)28.What can be inferred from AndreaB.To enjoy piano musicBlackstone's words in the second paragraph?C.To make the piano unique.D.To learn to play the piano.部A.She thinks parents should be open-minded.B.Tyler's actions have won him a lot of分(A)25.What is the main idea of the secondparagraph?money.A.How Friedman found the tooth marks onC.Tyler went around the whole country tothe piano.help veterans.读B.When and where Friedman bought the读D.She thinks parents should help achievepiano.their kids'dreams.C.Friedman wasn't good at tuning the piano.D.Friedman had a very special piano.

入FEBRUARY(二)L.A细节理解题。根据第一段首句“When Maddi,aI5-year-old girl,went to an evening party with some boys withoutpermission”中的without permission可知，Maddi违犯了家规，所以受到惩处，面临交手机或社交媒体两种选择，她58选择了交出社交媒体由家长代管半个月。5192.C细节理解题。根据第二段最后一句“But that was only the beginning,and by her dad's third post,the teen was【综合训练·英语参考答案第1页（共8页）E】already begging her parents to take away her phone for a month.”可知，Maddi的父亲接管她的社交媒体TikTok后，在上面发了一些滑稽的自拍照，这让她感到很难堪，因此她后悔了，想再选择上交手机。故选C。3.B推理判断题。根据第三段Mddi的妈妈接受记着采访所讲内容可推断，通过这种惩罚方式教育子女的结果让人忍俊不禁，她妈妈也觉得好笑，因此说话的语气显得轻松开心。故选B。4.D推理判断题。根据语篇内容，本文主要讲述了家长对犯错孩子的一种新颖特别的惩罚一父亲管理女儿社交媒体闹出的笑话，而Dad Took over Daughter's TikTok符合题意。故选D。5.which 6.is considered 7.to 8.depressed 9.worrying 10.unfortunately 11.media/mediums12.to conserve13.has applied 14.the /this15.years 16.who 17.her 18.had survived 19.illness 20.to give 21.smiling 22.a23.misunderstood 24.thought

第4章三角函数学生用人书②求扇形的弧长及该弧所在弓形的面积，考点了，：静都三角函数的定义例4(1)已知点P叫-停9)是角。的终边与单位圆的交点，则sin2a=()A号B一C,-22号(2)若角α的顶点在坐标原点，始边与x轴的非负半轴重合，终边经过点P(一1,2)，则sina-cosa十tana=[小结]用定义法求三角函数值的两种情况(1)已知角α终边上一点P的坐标，则可先求出点P到原点的距离r,然后用三角函数的定义求解；(3)已知一扇形的周长为20cm,当这个扇形的(2)已知角α的终边所在的直线方程，则可先设面积最大时，半径r的值为出终边上一点的坐标，求出此，点到原点的距离，然后用三角函数的定义来求解>●[小结]1.孤度制下1=a,S=号x,此时a为2例5(1)已知角a的终边过点P(-8m,弧定扇形面极公式号-，扇形中孩长公式-6sin30),且cosa=-手，则m的值为2in号，扇形孤长公式l=am.在角度制下，孤长微扇形面积S-,此时n为角度，它们之A-司B、③2间有着必然的联系.2.在解决孤长、面积及弓形面积时要注意合理应cD厚用圆心角所在的三角形.(2)已知角a的终边经过点(3a一9，a十2)，且3.当扇形周长一定时，其面积有最大值，最大值的求法是把面积S转化为r的函数，函数思想、cosa≤0，sina>0,则实数a的取值范围是转化为方程的思想是解决数学问题的常用思想.()A.(-2,3]B.(-2,3)砂训练巩固C.[-2,3)D.[-2,3]3.若扇形的圆心角是a=120°，弦长AB=12cm,则弧长1等于([小结]判定三角函数值的符号，关键是要搞清三角函数中的角是第几象限角，再根据正、余弦函cmB83πcm数值在各象限的符号确定值的符号.如果角不能确定所在象限，那就要进行分类讨论求解.C.4/3 cmD.8√3cm95

基础题与中考新考法·八年级·上·数学,综合中考新考法9.小唯在练因式分解时，发现多项式☐x+y2被污染了，翻看答案因式分解的结果是13.(新定义型阅读理解题)在数的学过程-(6x-y)(6x+y),则☐中的数字是(中，我们总会对其中一些具有某种特性的A.6B.-6C.36D.-36数充满好奇，比如：如果一个正整数能表10.(BS教材八下P105改编)一题多变示为两个连续偶数的方差，那么称这个10.1改变式子形式比较大小正整数为“智慧数”，例如：12=42-22,20=已知△ABC的三边长分别为x,y,z,则(x-62-42,28=82-62,我们称12,20,28这三y）2-220.(填“>”“<”或“=”)个数为“智慧数”10.2改为结合非负性求三角形周长(1)36“智慧数”；（填“是”或△ABC的三边长a,b,c满足(a+b)2-c2=24,“不是”)且a+b-c=3,则△ABC的周长为(2)设两个连续偶数是2n和2n+2(其中n11.(教材P119第5题改编)分解因式：取正整数)，由这两个连续偶数构造的“智(1)(m2-2)2-4;慧数”是4的倍数吗？为什么？(3)如图，拼叠的正方形边长是从2开始的连续偶数，…，按此规律拼叠到正方形ABCD,其边长为100，求阴影部分的面积.视频讲解(2)x2-4y2+2x-4y.中考新考法题第13题图12.(2022课标P144例66改编)32-1可以被22和30之间的整数整除，求这个数L》此类题解法见讲册P5098

19.(本小题满分12分)如图，在直三棱柱ABC-A1B1C1中，D,M,N,P分别是AB,AA1,BB1,CC1的中点.(1)求证：BP面MDC;A(2)设AB=AC=CB=2,BB,=4,求异面直线C1N与CM所成角的余弦值BMN20.(本小题满分12分)如图，在△ABC中，∠BAC为钝角，D在BC上，且满足∠CAD=若，AB=3,BC=3,5,(I)若C=,求∠BAD:(2)若M是BC的中点，cos∠BAC=一2，求AM的长度.M D21.(本小题满分12分)已知函数fx)=4sin台cos(位+）+5.(1若fa)=},求csle+哥r小：(2)若不等式|f(x)一A≤f(x)+2对任意的x∈63恒成立，求入的取值范围.22.(本小题满分12分)如图，在行四边形ABCD中，∠A=60°，AD=2,AB=4,将△ABD沿BD折起到△A'BD,满足A'C=2√5，(1)求证：面A'BD⊥面BCD;(2)若在线段A'C上存在点M,使得二面角M-BD-C的大小为60°，求此时CM的长度.DB

广东省2023一2024学年新高二秋季开学考绝密★启用前数学9.注意事项1.答卷前，考生务必将门己的姓名准与证号等填写在答题卡和试卷指定位置上。2同答透标题时，进出知小题芥案后，用们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，将本试卷和答题卡一并交回一、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.已知复数1+i(2-a-ai)为纯虚数，则实数a的值为A.0B.1.2.-12已知全集U=-1,3,5,7,9,04=1-1,5,,B=3,7,9,则AnB12A.3.7B.{3,5C.3D.{931-2sin2195°=A.2B.-2D.-34已知2013年一2022年中国网民规模（单位：亿人）依次为62,65,6.9,7.3,7.8,8.3,9.1，99,103,10.7,则这组数据的第45百分位数是0图①G0A.7.3©可B.7.55华C.8D.8.055.若向量a=(,),b=(,9),且a1b,则实数x的取值集合为A.☑B.{-√3}C.3}D.{3,-5(0年夏天，国产动画电影（长安万里热陕点燃“商诗热为此其市电视台准备在该电视台举办的“我爱背唐诗”前三届参加总决赛的120名选手（假设每位选手只参加其中一届总决赛)中随规抽取24名参加一个唐诗交流会，若接前三届参加总决赛的人数比例分层随机抽样，则第一届抽取6人，若按性别比例分层随机抽样，州女选手抽取15，则下列结论错误的是A.24是样本容量B.第二届与第三届参加总决赛的选手共有90人C.120名选手中男选手有50人D.第一届参加总决赛的女选手最多有30人已知在三棱锥A-BCD中，面ABD⊥面BCD,△ABD是边长为2的正三角形，BC=CD=2,点E为AD的中点，若点A,B,C,D都在球O,的表面上，点E,B,C,D都在球O3的表面上，则球0，与球02的体积之比为A.83B.9c号9数学第1页（共4页）

所以f(x)的单调递减区间为（一∞，0），单调递增区间为(0，十∞).。4分(2)若a=0,则f(x)=2x2,此时x=0是f(x)的极小值点，故a≠0.…5分f'(x)=x-asin a.x,令函数h(x)=x-asin a.x,则h'(x)=1-u'cos ax=1一a2cos|ax.…6分令函数p(x)=1一cos al(a≠0)，可知g(x)在区间[0，牙)上单调递增，.…7分①当o(0)=1-a2≥0且a≠0，即-l≤a≤1且a≠0时，p(x)≥o(0)≥0，此时h(x)在区间[0,石)上单调递增，则h(x)≥h(0)=0,此时x=0不可能是f(x)的极大值点.…8分②当g(0)=1-<0,即a<-1或a>1时，由g(x)在区间[0，牙)上单调递增，可知存在m∈(0，a),使得当x∈[0，m)时，g(x)<0,则h(x)在[0，m)上单调递减，…9分从而h(x)≤h(0)=0,即f(x)≤0，f(x)在[0，m)上单调递减.………10分由f(-x)=cos(-ax)+2(-x)2-1=C0s4+22-1=f(x),可得f(x)为偶函数，f(x)的图象关于轴对称，此时x=0是f(x)的极大值点.…11分综上，a的取值范围为（一∞，一1）U(1,十∞)12分评分细则：【】第(1)问中，最后没有回答函数的单调区间，而是写为“f(x)在（一∞，0）上单调递减，在(0,十∞)上单调递增”不扣分，【2】第(2)问中，在说明a≠0后，也可以先讨论a>0,再根据函数的奇偶性，确定a<0中满足条件的a的范围，最后求两种情况的a的取值集合的并集，即得满足题意的a的取值范围。【高三数学·参考答案第7页（共7页）】

A.Personality also plays a role in running late.B.That was linked to it being an unpleasant experienceC.We make time estimates based on how urgent the tasks are.D.However,their estimates of travel time contracted with familiarity.E.But is there a good explanation for why some people are always late?F.A region of the brain,called hippocampus,processes some aspects of timeG.That's because people who run late tell themselves and others they can be punctual.第二部分语言运用（共两节，满分30分）第一节（共15小题；每小题1分，满分15分）阅读下面短文，从每题所给的A、B,C、D四个选项中选出可以填入空白处的最佳选项，并在答题纸上将该项涂黑。It isn't every day that the future and the past meet on national television.Peter Davies,a 100-year-old World War II veteran from Macclesfield,England struggled tofind 21 in his life after his wife passed away.22,when his daughter suggested he offer hishelp at the local elementary school,he decided to 23 it.Peter 24 at Dean Valley Primary School,Bollington,and he is a huge hit with all of thechildren he tutors.He makes reading into a fun game,and his new 25 as a grandfatherly figuremakes him feel more 26 to his community than ever before."The kids are 27 they arelike sponges(海绵).”said Peter..“I'm sure I get more out of it than the28do.I'm not thisold man who lives on his own.I'm part of the 29,which is great.Plus my street 30 isextremely good because I'll be going 31 in the village and a child will shout out Hello Mr.Davies,'and I feel 32."In 2023,Peter received a British Empire Medal (BEM)from KingCharlesⅢ！Now,33 being just famous locally,Peter is a national celebrity.He recently 34 ontelevision with the youngest recipient of the BEM,18-year-old author and environmentalist DaraMcAnulty.“He is the35,I am the past,"Peter said of Dara.21.A.meaningB.wisdomC.standardD.courage22.A.MeanwhileB.NeverthelessC.ThereforeD.Moreover23.A.proveB.exploreC.manageD.attempt24.A.worksB.volunteersC.registersD.engages英语试题第7页（共10页）

■■口■▣口口又AE∩PA=A,所以AB⊥面PAE,所以AB⊥PE.(6分)(2)解：因为AP,AB,AE两两垂直，所以以A为原点建立如图所示空间直角坐标系A-z,则A(0,0,0),B(N2,0,0),D(-√2，√2,0)，P(0,0,2),E(0,√2,0)因为F为PD的中点，则F设面AEF的法向量n=(x,Z),n.E=2y=0,则iF-2x+2+24+名=0，取1则-2,所以022m.DE=√2x2=0,设面DEF的法向量m=(x,2,2),则M.DF=v2v22X2-2+32=0,_V2取y2=1,则x2=0,222所以m=0,2）2所以cos(信m=m·22…（12分)Imnl31+21+221.（本小题满分12分)(1)解：设M(x,),由题意得x-3+y353:二+-1为点M的轨迹C的方程，…(4分)2516(2)证明：方法1：设A(x,y),B(2,y2),由题知k>0,m<0,N(x,y)且x>0,y。<0,:直线ky=x+m与圆+少2=16相切，m=4,即m2=16k2+1),Vk2+1数学参考答案·第7页（共9页)

学记全国O0所名校高三单元测试示范表A重合A与B重合，即AB1rn△AB时为等边三角形，所以g后所以后（台+2一厅0，解得&或日=一3（舍去），所以精圆C的离心率为【答案】C、选样随：本题共4小题，每小题5分，共20分在每小题给出的选项中有多项符合题日要求全部选对的得5分，部分选对的得2分，有选错的得0分9.已知圆O:x2+y-16上有且仅有三个点到直线1的距离为2，则直线1的方程可以是A.x+y+√2=0B.x十√3y+4=0C.x+2y+2√5=0D.x-y-√2=0【解题分析1由匿如，图O,十y=16的国心坐标为O0.0),羊径为4，因为园十/二16上有且仅有三个点到交点，椭圆·/PF直线1的距离为2，所以圆心到直线1的距离为2.【对于选项A,点O到直线x+y十V2=0的距离为W=1,不合题意；1+1对于选项B,点0到直线x+3y十4=0的距离为4=2，符合题意，/1+3对于造项C,点0到直线x+2+25=0的距离为2V5=2,符合题意；1+41PF的面选项D,点O国直战t一y亿=0的距高为L,不合题分桥如图，根【答案BC,,所以/PF篇圆和双曲线10,公元前6世纪，古希腊的毕达哥拉斯学派把5。称为黄金数离心率等于黄金数的倒数的双曲-a=2,所以线称为黄金双曲线若卖金双曲线Ey=1(a>0)的左、右顶点分别为A1,A2,左焦点为a十ai)=gF(-c,0),离心率为e,点B(0,2),则A.ac=4SB的面积B.△BFA2为直角三角形C.双曲线E的顶点到渐近线的距离等于黄金数的2倍=(a十a)2D.△BFA2外接圆的面积为π(W5十4)案ABC控题：本题士题分折由短知，士4，因为》所以4十4-5中解有222®0所以ac=4,故选项A正确：A(a,0),所以B萨=（一c,-2),BA,=(a,一2），所以B萨.BA=一ac十4=0，所以BF1E知圆C:BA2,所以△BFA2为直角三角形，故选项B正确；△BFA2外接圆的直径为FA2,所以△BFA2外接圆的面积为x()-x(+受)=xW5十2)，故选项D错误，双当线E的顶点到渐近线的距离为ab24/a2+bb=5-1,放题分析】因选项C正确.【答案】ABC11.已知抛物线C:x2=2y(p>0)的焦点F到准线L的距离为3，则精於率为A.过点(0，一2)恰有3条直线与抛物线C有且只有一个公共点B若T4,),P为C上的动点，则PT+PP的最小值为号C.直线2x一2y十3=0与抛物线C相交所得弦长为11黜一个D.抛物线C与圆x2+y2=7交于M,N两点，则MN=2√6题分好【解题分析】因为抛物线C:x2=2y(p>0)的焦点F到准线1的距离为3，所以p=3,从而抛物线C的方程是x2=6y.过点(0，一2)可以作2条与抛物线C相切的直线，而直线0与抛物线C相交，只有1个交点，从而过点(0，-2)恰有3条直线与抛物线C有且只有一个公共点，故选项A正确；抛物线C的准线方程是y=一，设T到准线的距高为d,则d=4十号号过点P作准线的至线26垂足为Q,则由抛物线的定义知PQ=PF,所以PT十|PF=|PT+PQI≥d,所以PT+PF的最小值为【24G3DY(新高考)数学-必考-Y】

辽宁省实验中学2023-2024学年度高考适应性测试（一）数学参考答案一、单选题（每题只有一个选项是正确答案，每题5分，共40分）2345678CACBCDBB二、多选题（每题至少有一个选项为正确答案，少选且正确得3分，每题5分，共20分）9101112ADACABDBC三、填空题（每题5分，共20分）13.√514.54015.-2sin216.四、解答题(17题10分，其余每题12分，共70分)17.【详解】(1)取PD的中点G,连接AG、EG根据中位线定理，BG/CD,且G=CD=4B又ABIICD,所以ABIIEG,AB=EG,则四边形ABEG为行四边形，÷BEIIAG,:BEC面PAD,AGC面PAD,:BE∥面PAD;(2)以D为原点，DA、DC、过D且垂直底面的直线分别为x、y、:轴建立空间直角坐标系，设AB=1,则D(0,00)、A,0,0)、B(1,0)、C(0,20).设P(x,2)】由lDP=R+y严+z=2.AP=x-ヅ+y+z=2Cr=V2+0-2y+2=2上面联立解方程组得，1=2,BE=31得到4’2’4cos(m,BEm.BEV66面ABCD的法向量为m=(0,0,1),由网41x612√66故直线BE与面ABCD所成角的正弦值为12，答案第1页，共5页

惠州市2024届高三第一次调研考试（趣数学全卷满分150分，0时间92站外钟登备司是氏d0量合气2023.10答题前考生务必将自己的姓名、准考证号、座位号、学校、班级等考生信息注意事项：2、作答单项及多项选择题时，选出每个小题答案后，用2B铅笔把答题卡上对应题日写在答题卡上。5士希所香的答案信息点涂黑。如需改动，用橡皮擦干净后，再选涂其它答案，：写在本试卷上无效3.非选择题必须用黑色字迹签字笔作答，答案必须写在答题卡各题指定的位置上，写在本试卷上无效。一、单项选择题：本题共8小题，每小题满分5分，共40分。在每小题给出的四个选项中，只有一项符合题目要求，选对得5分，选错得0分。1.已知集合A={x1≤x≤3}，B={xy=ln(2-x},则AnB=A.12)B.(L,2）)C.(,3D.,32.复数z满足z2+i,其中i为虚数单位，则山A.1B.5C.2D.53.己知向量a(-3,1),b=(m,2).若a/16,则m=A.6B.-63C.4,已知ah}6tan15(2C=1-tan215°则实数ab,c的天小关系是(A.azb>c史24B.b>c>aC.c>b>aD.axc>b65在一次篮球比赛中，某支球队共进行了8场比赛，得分分别为：29,30,38,25,37,40,42,32那么这组数据的第75百分位数为A.37.5B.38C.39D.406。金针菇采摘后会很快失去新鲜度，甚至腐烂，所以超市销售金针菇时需要采取保鲜膜封闭保存，已知金针菇失去的新鲜度乃与其采摘后时间1（天）清足的函数解析式为h二mh+aa>0.若采摘后1天，金针盘失去的新鲜度为40%，采精后3天数学试题第1页，共6页

2023-2024学年第一学期11月高三阶段测试卷“…数学考试说明：1.本试卷共150分。考试时间120分钟感，单方8)2.请将各题答案填在答题卡上。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。必火才。·令背的1.已知集合A={x|3≥27，x∈N),B={-1,0,1,2},则(CNA)∩B=A.{0,1,2,3}B.{0,1,2}C.{-1,0,1,2}D.{1,2}2.若复数z满足(1+)=i(i为虚数单位)，则之的共轭复数为A.-1+i.B.-1-iC.1+iD.1-i3.已知m:logx<2,则m成立的一个充分不必要条件是高：B.0f(2)C.ef(2)>f(0)D.e2f(2)

7.C【解析】.a1>0,{an}为递增数列，则a2>0,ag>0,…am>0恒成立.∴.am>0,则由2an+1an十a1-3a,=0(n∈N)可得a+1=2a.千1'3an又an+1>an恒成立，即Za7a3>1→0a<1相成主，所以0a>0)上一点A关于原点的对称点为B点F为来右低点，设左焦点为P..|AF|十|AF|=2a,根据对称关系知四边形AFBF为矩形，∴.|AB|=|FF|=2c.由于AF⊥BF,∠ABF=a,.|AF|=2 csin a,|AF|=2 ccos a,2csna叶2 c=2ae-&ino2sn(ar牙）】11由于E(冬，）故叶景∈（受，）260，解得力=4，B正确：4b2对于C,A泸-（号-，-），F店-(-号，)，A市-3F范，可得号-a-3(-名），-n-8,又n为=-护，-2-，4p可求出A39,W5p),B(名，82)，k==33=3,26数学试题参考答案（长郡版）一2

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13.求值：1gV27+10W2+V27-10W2=14.已知a>0,b>0,且满足a+2b=3,则4+4+2b2+b+2的最小值为2a2b+15.已知R,E是双曲线二-卡=(>0，b>0)的左、右焦点，以FB为直径的圆与双曲线2一、二象限的交点分别为P、2,若P⑨=4b,则双曲线的离心率为】16.一条沿江公路上有18盏路灯，为节约用电，现打算关掉其中4盏路灯，为安全起见，要求公路的头尾两盏路灯不可关闭，关掉的相邻两个路灯之间至少有3盏亮着的路灯，则不同的方案总数共有种.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.17.(10分)记△MBC的内角A,B,C的对边分别为a,b,c,已知2 csinB_sinCbcos B(I)若amA=5,求anC的值4(2)若a+c=3+3√3，：△4BC内切圆的面积为π，求△ABC的面积.18.(12分)三棱锥S-ABD中，SAL面ABD,AB=AC,SAD=V2SAac,并且∠BAC是直角.、9.=8)1.-=1)1,州4个(1)求二面角C-SA-D所成角的余弦值；(2)若∠SCB=60°，S8、SC上客取一点B、,设SB8+9.83C=(0<<).当元为何值时，面AF⊥面B8}9-01.(第18题图)SBC19.(12分)已知数列{an}中，a1=5,且2am+1=an+2,Sn为其前n项的和.(1)求数列{an}的通项公式：②)求满足不等式水，-21-小水应的最小正整数：的值3)设，-(a-+,C=（色2）其中>0，若对任意m,n∈N,总有b。-C,>3成立，求元的取值范围。第3页共4页

语。故填location。71.gently考查副词。句意：她轻轻地抓住门把手，转动门把手，走进房间，把婴儿放在床上。分析句子结构可知，此处为副词gently作状语修饰动词短语took hold of。故填gently。.72.infectious考查形容词。句意：那些患有传染病的人与其他病人分开了。设空处在句中修饰后面的名词diseases,作定语，应用形容词；结合句意及提示词可知，此处应用infectious,表示“传染的；有传染性的”。故填infectious。73.generosity考查名词。句意：我要向格林先生表示感谢，没有他的慷慨，我的公司就不可能成立。关系代词whose作定语，意为..的，其后需接名词；generous对应的名词为generosity慷慨，大方”，不可数名词。故填generosity。74.What考查主语从句。句意：你所要做的事情就是在这台机器上插入你的信用卡。you have to do'”是一个主语从句，表示“所...的事情”，设空处为引导词，从句中缺do的宾语，所以填what引导主语从句。故填What。75.Dear School President,I am Li Hua,a senior student in your school.Recently,I found a lot of studentspouring lunch in the school canteen and going out to buy snacks instead.I made an investigation among the classmates and found some students aredisappointed with the flavour and style of canteen food.On the contrary,they think snacksare more delicious.In my opinion,first of all,it's shameful to waste food for the worldwidefood shortage still exists.There should be activities to promote the importance of savingfood and the harm of eating too much snacks.In addition,the school canteen food shouldbe improved both in flavour and style.Last but not least,the canteen's hygiene must beguaranteed.This is my personal suggestion and I hope the principal consider it.Thank you.Your student,本篇书面表达属于应用文，要求考生写一封信，告知校长学生倒掉午餐买零食吃的现象并提出建议。第一步：审题体裁：应用文时态：根据提示，时态应为一般现在时和一般过去时。结构：总分法总分法指把主题句作为总说，把支持句作为分说，并以这种方式安排所写内容。要求：1.告知现象2.介绍相关情况1)解释原因。(reasons)2)提出建议。(suggestions)第二步：列提纲（重点词组）

在到日和a上各有一个零点x,x2,11-tnIntt-1Inx,-1XInx=Int+nx=tIntt-11-Inx1、r-n,令g0=-Inx-1tint_1 tInt-1+1'tInt-t+1t-11-(dn--(-1-inghg'(t)=(tlnt-t+1)g-士：100(tlnt-t+1)2(tlnt-t+1)21118@在，+o)上V,且mg0=lmn=m片=1，∴g0)<1,1→1t.≥1.

27.reporter句型演练1.more serious than2.didn't choose;any3.the most carefully4.taller than5.No:;she doesn't6.Diana is good at playing the piano7.There are all kinds of animals in the zoo8.Did Jim act a magician in the talent show yesterday9.My sister is talented in dancing10.care about11.In fact12.plays an;role in13.as long as玩转语法1.best2.photos3.an4.that5.touches6.mine7.broke8.to help9.interesting10.with11.more quickly12.(the)most13.younger14.louder15.the most popular

M:Yes,I bought the T-shirt in the sale here a couple of weeks ago.I think it was .. W:I see,yes .. M:And when I looked at it yesterday I just thought,no,I don't like it.I don't think it suits me.The colour's not really right for me.Do you know what I mean? W:Well...um,do you have any receipt,or proof of purchase? M:No,no,I didn't keep the receipt,sorry. W:Well,we can't normally exchange goods without proof of purchase,you see. M:Oh,no. W:I can speak to the manager if you like,see what he says. (Text 9)