21.(12分)已知线段MN的长度为3,其端点M,N分别在x轴与y轴上滑动,动点P满足N2=2PM(1)求动点P的轨迹C的方程:(2)当点M坐标为35】2,0,且点P在第一象限时,设动直线1与C相交于A,B两点,且两直线PM、PB的斜率互为相反数,求直线1的斜率22.(12分)已知函数f)=eΓx+2(1)求f(x)的单调性;(2)证明fx)≥+虹2【A-023】数学试卷第4页(共4页)
IV.1.letting out an excited seream2.All around the world3.made my dream come true4.To sum up5.As a result
23.Why are people advised to include their mailing address if they donate by mail?A.This ensures they can'receive the tax receipt.…》B.This can prevent the money from being stolenC.This helps spread Ryan's message to more peopleD.This makes it easier for people to contact the organization.BAnimals are gentle and often fall victim to cruelty because they trust and don't fight back.We are so grateful to be able to save our animals and prove to them that the world can be good.Five years ago;six cows pushed through three fences and escaped from a slaughterhouse()However,sadly,they were finally rounded up and returned to the'slaughterhouse.Because it was national news,the cominunity insisted they be allowed to live and even raisedmoney for their freedom.The slaughterhouse owner agreed to release them to a sanctuary护所),but no.one came to get themn.Unwilling to see them die,Jay jumped on a red eye flight and got there in time to stoptheir slaughter.He took "The St Louis Six"to the hospital to be treated for their variouswounds and infections.Later,I flew to St Louis to meet them.The minute I saw them,I knewthey had a story to share with the world and that we had to help them.In the coming months,we found a gorgeous property in St Louis and opened a GentleBarn as home for The St Louis Six.Once scared and desperate to live,now the boys aretrusting,loving,and giving hope to humans in our Cow Hug Therapy sessions where they wraptheir necks around our guests who come to The Gentle Barn looking for hope.Animals have always been my greatest teachers.They taught me whether we are trying tosurvive,or looking for a safe home,sometimes we need to leave something behind to find abetter way of life.We may be afraid of change,but only when we are brave enough to face theunknown and head out on our own can we realize our dreams!24.Why did the slaughterhouse owner give up killing the cows?对5A.The community urged him to do so.”B.Someone had already bought the cows.C.He might get punished by the government.D.The cows ran too far away from the slaughterhouse.·….,泪25.What happened to the cows after they were treated in the hospital?A.They left St Louis for Jay and the author's hometown.9、…B.They were taken to another place and lived happily there.C.They still often felt too scared and desperate to meet humans.0D.They couldn't trust anyone even though they were treated with love.26.Which of the following best describes:Jay and the author?部:消型Ac学2授产B,gec等密天C.Wealthy.27.What can be infered about the author from the last paragraph?流·A.She wants to look for a safer home.,进B.She hates changes in life very much.8。:活C.She is inspired by the animals'bravery.。D.She suggests animals be treated as teachers.nwahoe eprodhn高中2020级英语试愿第4页(共10页)f蹈这
翻译句子。1.你回到家乡工作的原因是什么?What is your reason for returningto即学即练your home town to work?2.我上课迟到的原因是我的自行车坏了。The reason why I was late for class wasthat my bike was broken.3.Dan因和他哥哥打架而向他道歉。Dan apologised to his brother forfighting with him.
17.(本小题10.0分)计算:已知角a的络边经过点P(m2),且e0sa=,求:(1)实数m的值;(2)求cos2a-sin2a+2 sinacosa的值.18(本小题120)分已知桌合A={似y=1 m.xe[.c]},B=片,2,8,C=2m+1xm+(1)求AOB;(2)若BOC=C,求实数m的取值范围.19.(本小题12.0分)已知函数f(x)=log.(5-2x)+log.(x+1),其中0
BDM=GH,∴.AP∥GH.18.解:(1)因为SA⊥平面ABC,BCC平面ABC,所以SA⊥BC,因为AB⊥BC,AB∩SA=A,AB,SAC平面SAB,所以BC⊥平面SAB,又BCC平面SBC,所以平面SBC⊥平面SAB.(2)由(1)知BC⊥平面SAB,所以BC⊥AG,又AG⊥SB,且CB∩SB=B,所以AG⊥平面SBC,又SCC平面SBC,所以AGLSC.19.解:(1)取PD的中点E,连接EN,AE,.M,N,E分别为AB,PC,PD的中点,∴.EN∥AM且EN=2AB=AM,四边形AMNE为平行四边形,故MN∥AE,:PA⊥平面ABCD,CDC平面ABCD,∴.PA⊥CD,又.'CD⊥AD,AD∩PA=A,AD,PAC平面PAD,所以CD⊥平面PAD,CDC平面PCD,.平面PCD⊥平面PAD,.PA=AD,E为PD的中点,.AE⊥PD,又平面PCD∩平面PAD=PD,AEC平面PAD,∴.AE⊥平面PCD,'.MN⊥平面PCD.(2)由(1)可知MN∥AE,∴.∠PAE即为异面直线PA与MN所成的角,在直角三角形PAD中,PA=AD,∴△PAD为等腰直角三角形,又E为PD中点∴∠PAE=∠PAD=45°,故直线PA与MN所成的角为45°.20.解:(1)连接CP,并延长与DA的延长线交于M点,因为四边形ABCD为正方形,所以BC∥AD.散△PC△PDM,所以需部=号,又因为品-部-号所以品-品=号,所以PQ∥MD.又MD,C平面A,D,DA,PQt平面A,D,DA,故PQ∥平面A,DDA(2)当的值为时,平面PQR/平面AD,D1证明如下,因为号.即然号,所以然品所以PR/DA.又AC平面AD,DA,PR学平面A,DDA,所以PR分平面A1D1DA,又PQ∩PR=P,PQ∥平面A1D1DA,所以平面PQR∥平面A1DDA.DD、B1A、B1Q、0DUD。RB21.解:(1).ED∥AB,ED在平面ABF,ABC平面ABF,.ED∥平面ABF,.ED平面PED,平面PED∩平面ABF=FG,.ED∥FG,又ED∥AB,∴.AB∥FG:(2)设AB与CD交于点M,则MD的中点为C,由(I)知PG=3PD.易知点ME平面ABF,点M∈平面PCD,'.MG=平面ABF∩平面PCD,又H为PC与平面ABF的交点,∴H=MGnPC,在△MPD中,如图,PG=3PD,C为MD的中点,由平面几何知识易知H·43·【23新教材·DY·数学·参考答案一BSD一必修第二册一Y】 2023届高三3月模拟(二)数学注意事项:1、答卷前,考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上,2、回答选择题时,选出每小题答案后,用铅笔将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,再选涂其它答案标号,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.已知全集U=R,集合A=x2>1),B=xhN≤4},则AnB=A.(1,16]B.(1,2]C.(0,16]D.(0,2]寒底出政一用甲生不已数子的指2.若复数z满足(1一)z=|1十,则z的虚部是一A.28马的2游的领范滨出过二量加甲发(1)C.1星膜D.1,爱献宽H恒束静赛出示卖(3.下表是足球世界杯连续八届的进球总数:中出一年份19941998200220062010201420182022进球总数1411711611471451711691721则进球总数的第40百分位数是A.147B.154C.1619点于)点两8D.1654.将英文单词“rabbit'”中的6个字母重新排列,其中字母b不相邻的排列方法共有A.120种B.240种C.480种D.960种5.1+tan22.5°=A.1+21+5(食)公B.0雪1一<眼后2200十0≤xs1-(0D.2高三数学试题第1页(共4页) 全国100名较蛋新高考程州年能号、18.(12分)全围10的8世漫所多考偶拟示猫卷、20.(12分)在D(6+ca)(6c十a)bc②asin C3usCb)③2h+c)coSA如图,A,B0,-2)分别为椭圆C号+=1(a>b>0)的左顶点和下顶点,P为精ac0sC=0中选一个,补充在下面的横线中,并解答在△ABC中,内角A.B,C的对边分别为abc,且满足一圆C在第一象限上的点,精圆C的离心率为(1)求A;(1)求椭圆C的方程;《2)若内角A的角平分线交BC于D点,且AD=,√3,求△ABC的面积的最小值(2)求点P到直线AB的距离的取值范围.(注:如果选择多个条件分别解答,那么按第一个解答计分)19.(12分)如图,在圆柱O0中,O0=2,A为圆0上一定点,B为圆0上异于点A的一动点,OA=23,过点O作平面ABO1的垂线,垂足为C点(1)若OA⊥OB,求证:BCLOA.(2)若△AOB为等边三角形,求二面角A一O,B-O的余弦值.0线【23·(新高考)ZX·M,数学·Y)⊙数学卷(九)第6页(共8L【23·(新高考)Z公·MN·数学·7数学卷(九)第5页(共8页1 t'(x)在[0,+o)单调递增,则t'(x)≥t(0)=1即t(x)>0在(0,+∞)恒成立,故t(x)在[0,+o)单调递增所以t(x)≥(0)=0故h'(x)≥0在[0,+oo)恒成立.8分由h(x)在[0,+oo)单调递增,而h(0)=2,h(x)≥2,故a≤2.9分(3)取a=2时,x2+2x+2≤2e,则x2+2x+1≤2e*-11s、1所以2e-(x+W…10分1111因此2e(m+12n+)nn711分则g0+g2+g<1+-t++-、313a+0+3+34nn+14n+14.12分数学参考答案及评分标准·第14页(共14页) 第四部分写作(共两节,满分35分)第一节短文改错(共10小题;每小题1分,满分10分)假定英语课上老师要求同桌之间交换修改作文,请你修改你同桌写的以下作文。文中共有10处语言错误,每句中最多有两处。每处错误仅涉及一个单词的增加、删除或修改。增加:在缺词处加一个漏字符号(∧),并在其下面写出该加的词。删除:把多余的词用斜线(八)划掉。修改:在错的词下划一横线,并在该词下面写出修改后的词。注意:1.每处错误及其修改均仅限一词;2.只允许修改10处,多者(从第11处起)不计分。Last month,our school launched a campaign which intention was to promoteenvironmental protection.The campaign last for one week.Firstly,there was a photo displayto show the seriously pollution caused by human activities.Secondly,there was a lecture on学many small step that we could take in our daily life protect the environment.For example,熔taking the bus and using the bike-sharing system would be a good way.Last but not the least,子we students were encouraging to decorate our classrooms with recycled materials.Throughthe campaign,they have benefited a lot.We realize that it is our responsible to leave a better,cleaner and healthier planet for future generations.烟第二节书面表达(满分25分)你校将举行英语演讲比赛。请你以“Be a Person with Good Manners'”为题写一篇演讲稿够参赛,内容包括:1.阐明礼貌的重要性;閭2.谈谈如何养成有礼貌的好习惯。注意:1.词数100左右;2.可以适当增加细节,以使行文连贯。【高三英语第10页(共10页)】·22-02-361C· 中考现代文阅读与新考法答案详解详析2024版 中专必刷卷·2023年安徽中考第一轮复习卷温馨提示:数学专题(二)1.本卷考查内容为方程(组)与不等式(组):2.满分150分,建议用时120分钟.斯一、选择题(本大题共10小题,每小题4分,满分40分)每小题都给出AB,CD四个选项,其中只有-个是正确的,请把正确选须写在题后的括号内.不选、错选或多选的(不论是否写在括号内)一律得0分1.满足一x十2023>0的最大整数是A.2020B.20212.若=4是分式方程2=C.2022C.D.2023的根,则a的值为A.3(D)B.43.C.5在物理学中,导体中的电流1跟导体两端的电压U、导体的电阻R之间有以下关系:1一D.6U,去分母得R=U,那么其变形的依据是A.等式的性质1BAB.等式的性质2C.分式的基本性质丑不等式的性质24.x=3±√32+4×1×22×2是下列哪个一元二次方程的根?(D.)A.2x2-3x+1=082x2+3x十1=02x2+3x-1=0D.2x2-3x-1=05.若m>n,则下列各式不二定城立的是4,,入(舟.)A.m2>n2B.1-m<1-nC.2m+1>2n-3Q2m>m+n6.有关学生体质健康评价指标规定:握力体重指数m=(握力÷体重)×100,九年级男生的合格标准是>35.若九年级男生小虎的体重是50千克,则小虎的握力合格至少要达到(BA.17千克B.17.5千克C.18千克D.18.5千克中7.某工厂第二季度的产值比第一季度的产值增长了x%,第三季度的产值又比第二季度的产值增长了x%,则第三季度的产值比第一季度的产值增长了(A)A.2x%B.1+2x%C,(1十x%)x%D.(2十x%)x%郏8.如图所示的是一个运算程序:输入输出结果第8题图若数x需要经过三次运算才能输出结果,则x的取值范围是(C)A.x<7d-fErslD.x>-13或x<7中考必刷卷·2023年安徽中考第一轮复习卷·数学专题(二)第1页共6页 信【观察思考】第一个图案中,用了6个球珠,有2个正方形;解第二个图案中,用了9个球珠,有5个正方形;6第三个图案中,用了12个球珠,有8个正方形;10CE 4……【问题解决】PD"CBoA(1)第四个图案中,用了个球珠,有」个正方形;(2)如果用了252个球珠,那么图案中有多少个正方形18.如图,建筑物BC直立于水平地面,在其顶端C处测得气球P的仰角P为37°,CP=30m:地面上的点A与建筑物BC相距60m,在点A处测得气球P的仰角为53°.求气球P到地面的距离.(参考数据!re \ssin53°=0.8,cos3°=0.6,tan53°=43七0)PA617722.第18题图五、(本大题共2小题,每小题10分,满分20分)19.如图,直线y=一x+4与坐标轴交于A,B两点,点M(2,m)在直线AB上,点N与点M关于y轴对称.I)当点N在反比例函数y一的图像上时,求的值:8(0,4)(2)当线段MN被反比例函数y=的图像分成两部分,且这两部分长度√MM2,m的比为1:3时,求k的值第19题图20.如图,在四边形ABCD中,∠A=∠D=90°,AD=AB,以BC为直径的半⊙O与AD相切于点E.(1)求证:∠BCE=∠DCE;(2)若CD=√2,求DE的长第20题图六、(本题满分12分)21.甲、乙两个人准备自驾汽车沿一条旅游线路游览.他们收集到以下信息:(不完整)信息一,这条旅游线路上依次分布着A,B,C,D,E,F六个景点,其中A,B.C都被游客评为最美景点,信息二,沿途相邻两个景点间的里程数、汽车的油耗如下表所示:A→BB→CC→DD>EE→F里程(km)40105b油耗(L/km)0.060.080.070.070.062023年“万友”中考模拟卷·数学(一)试题卷第3页,共4页 16.在一个正三角形的三边上,分别取一个距顶点最近的十等分点,连接形成的三角形也为正三角形(如图1所示,图中共有2个正三角形).然后在较小的正三角形中,以同样的方式形成一个更小的正三角形,如此重复n次,可得到如图2所示的优美图形(图有多个正三角形),这个过程称之为迭代,也叫递推.在边长为3的正三角形三边上,分别取一个三等分点,连接成一个较小的正三角形,然后递推得到如图3所示的图形(图中共有n个正三角形),则图中至少个正三角形的面积之和超过91V527图1图2图3四、解答题(本大题共6小题,共计70分.请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤)17.如图,在正方体ABCD-AB,CD中,E为DD,的中点.DE的(1)证明:直线BD∥平面ACE:(2)求直线CD,与平面ACE所成角的正弦值.18.已知数列{an}的前n项和为Sn,且Snl=Sn+an+1,请在①a3+41s=20:②42,4,4,成等比数列:③S0=230,这三个条件中任选一个补充在上面题干中,并解答下面问题.(1)求数列{an}的通项公式:(2)若b,=an-1,求数列{2”b,}的前n项和T注:如果选择多个条件分别解答,按第一个解答计分.·19.已知函数f(x)=x3+ax2-6x+1(a∈R),且f'(①)=-6. 长品对和子的长度m与和子的的取马之明调足次的教关系芳2班子的长度有1m4码鞋子的长度为27cm,则38码程于的长度为A.23cmB.24cmD.26cm8.在同平面直角坐标系中,一次函数y一C.25ema的图象可能是本,×包点P在正比例函数工因像点A22,点B2大,2若P是直0角形,则点P的个数有A.1个B.2个D.4个10.在物理实验课上·小鹏利用滑轮组及相关器材进行实验,他把得到的拉力下N)和所悬挂C.3个物体的重力G(N)的儿组数据用电脑绘制成如下图象(不计绳重和摩擦),请你根据图象判断以下结论正确的序号有◆F/N1.3124567第10题图①物体的拉力随着重力的增加而增大;②当物体的重力G=7N时,拉力F=2.2N;圆拉力F与重力G成正比例函数关系:④当滑轮组不悬挂物体时,所用拉力为0,5NA.①②B.②④C.①④D.③④二、填空题(本大题共4小题,每小题5分,满分20分)11山.函数y一x2中x的取值范围是12.三个能够重合的正六边形的位置如图.已知点B的坐标是(一3,3),则点A的坐标是第12题图第13题图第14题图中考必刷卷·2023年安徽中考第一轮复习卷·数学专题(三)第2页共6页9787107244988定价:8.20元 全国100所名校最新高老横拟示苑卷所以出=-4,0-1,所以+16,=+>≥2√.6=8.当且仅当一4时等号成x)立.所以|AF|+161BF1≥25.7.D【命题意图】本题考查几何体的体积,要求考生熟悉棱柱体积的计算公式,能用公式解决简单的实际问题.【解题分折过点E作F/AC交BC于点F.连接FC(图路).设器-△APC的面积为S.三棱柱ABC-A,B,C的高为h,所以△BEF的面积为x2S,所以三棱台A,B:C,一EBF的体积为时(S+S+√S·S)M=言(1十2十x)SM.因为平面ACE把三核柱ABC-A,BG分成体积比为31的两部分所以后1++)S以-230解得一-1片成二16(合22去).因为AB=2,所以BE=6-1.8,B【命题意图】本题考查实数的大小比较,要求考生能利用导数研究函数的单调性。【保这分行玲=n用:广是一学出了0积1所以数(x在(1,卜o上单调递增,所以f(2)≤f(3),即1n2√ 19.(8分)某同学在解方程2,1-寸4-2去分母时,方程右边的-2没有乘3,因而求得的方程20.(8分)某校分配学生住宿,如果每间住5人,就有30人没有宿舍住,如果每间住6人,就可空出3320个床位.该校有多少间宿舍?有多少住校学生?的解为x=2,试求a的值,并求出原方程的正确的解。第5页(共8页)第6页(共8页) 第6章一元一次方程吉祥物,冰墩墩是一只熊猫,它的外表给人元.问小明购买了钢笔和签字笔各多一种朴实的感觉,雪容融是一个灯笼,它的少支?外表总能给人温暖.钥匙扣、手办两用冰墩墩和雪容融立体挂件在奥林匹克官方旗舰店销售异常火爆.开售第一天,旗舰店共花费84000元从授权生产厂家购进两种挂件各1000件,其中1件雪容融挂件成本比1件冰墩墩挂件成本少6元,则1件雪容融挂件成本和1件冰墩墩挂件成本分别是多少元?B组能力提升9.某个体商贩同时售出两件不同的大衣,每件都以150元售出,按成本核算,其中一件盈利25%,另一件亏损25%,那么这次经营活动中该商贩A.不赔不赚B.赔20元C.赚20元D.赚18元10.某商场在“庆元旦”的活动中将某种服装打折销售,如果每件服装按标价的6折出售将亏10元,而按标价的9折出售将赚50元,则每件服装的标价是元.8.某校开展校园艺术节系列活动,派小明到11.为迎春节,某商家将文具按进价60%提文体超市购买若干个文具袋作为奖品.这高后标价,销售时按标价打折销售,最后种文具袋标价每个10元,结账时老板对小相对于进价仍获利12%,则这件文具销明说:“如果你再多买一个,就可以全部打售时打折。八五折,省17元!”,小明说:“那就多买一12.为喜迎新春,某水果店现购进水果篮40个吧,谢谢!”个和坚果礼盒20个,已知每个水果篮的(1)求小明原计划购买文具袋多少个?进价比每个坚果礼盒的进价便宜10%,(2)学校决定,再次购买钢笔和签字笔共水果篮每个售价110元,坚果礼盒每个售50支作为补充奖品,其中钢笔标价每价150元,支8元,签字笔标价每支6元.经过沟(1)春节期间水果店促销,坚果礼盒按售通,这次老板给予八折优惠,合计272价八折出售,水果篮按原价销售.某公·19 参考答案点拨:列方程组解应用题7.解:设有x辆车,人数为y人,上月买4斤萝卜2斤排骨共花了34元y=3(x-2),由题意,得今天买4斤萝卜,2斤排骨共花了42元y=2x+9.变式练习x=15,解这个方程组,得解:符合安全规定,理由如下:y=39.设平均每分钟一道正门可以通过x名学生,一道侧答:有15辆车,39人.门可以通过y名学生,由题意,得8.解:设购买1个温馨提示牌需要x元,1个垃圾箱[x+y=240,需要y元,3x+2×3y=1020.4x+3y=360,依题意,得x=140,y-x=50.解这个方程组,得y=100.x=30,解这个方程组,得∴.5分钟可以通过的学生人数为5×(1一20%)×y=80.(140×2+100×2)=1920(名),答:购买1个温馨提示牌需要30元,1个垃圾箱这栋大楼最多有学生50×8×4=1600(名).需要80元..1920>1600,B组.建造的这4道门符合安全规定.9.D思路点拨:解这个题的实质是列方程组解应用题,思路点拨:设现在儿子的年龄是x岁,现在父亲的关键是抓两个等量关系.先求出每道门每分钟通过y=3x,年龄是y岁,则可列方程组解方的学生人数,再求出5分钟内4道门通过的学生y-7=5(x-7).人数x=14,程组,得设m年后父亲的年龄是儿子的方法总结y=42.2.等量年龄的2倍.则2(14+m)=42+m,解该方程随堂演练即可.A组10.解:设购进甲种节能灯x只,乙种节能灯y只,1.C2.B3.Ax+y=100,由题意,得30x+40y=3300.4.695.246.解:设安排x人加工桌子,y人加工椅子,由题x=70,解这个方程组得意,得y=30.∴.70(45-30)+30(50-40)=70×15+30×(x+y=28,4×3x=2010=1350(元).y.答:该商场售完100只节能灯后,能获利x=10,1350元.解这个方程组,得y=18.思路点拨:先列方程组求出甲、乙两种节能灯的答:安排10人加工桌子,18人加工椅子只数,再求其获利.·19· 周周自测先锋卷(33)测试范围:第十八章18.2-、1.C2.D3.C4.B5.A6.D7.C8.19.C10.C提示:7.连接AC交BD于点O因为四边形AECF是菱形,所以AC⊥BD,AO=OC,OE=OF.又因为E,F为线段BD的三等分 当0 49920:06g00…米⊙月令15©明光一模(数学)参考答案.pdf文件预览明光巾W5牛儿牛级弟一伙惧以亏风·双子1/6参考答案及评分标准一、选择题(本大题共10小题,每小题4分,满分40分)题号12345678910答案ADCDC9.解析:根据题意画图,记AB和OM的交点为D,连接OA,OB.AB垂直平分半径OM,OD=2OM=2OA,在R1△AOD中,cos∠AOD=OD=号OA=2∠AOD=60°,∴.∠AOB=2∠AOD=120°,当点C和O在AB同侧时,∠ACB=1∠AOB=60°;当点C和O在AB异侧时,ACB气120故选C3x(抛物线开口向上):当1<≤3时,点Q在DC上,PQ=5y-(线段):安微大字版祥当3 教学全国©0所名校单元测试示范卷札记全国@⊙所名校单元测试示范卷·数学第七单元第三次综合测试(120分钟150分)概率统计是高考必考知识点,客观题和主观题都会考;统计案例、排列扫码看微课视频高考对接点获取复习资料包命题组合和二项式定理是轮考知识点,常考客观题,统计案例还会考主观题▣视点单元疑难点概率、统计案例的应用命题情境点生活实践情境题:4、7、17、21题序123456789101112答案BAACD BCD ADAD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.某中学开展主题为“学习宪法知识,弘扬宪法精神”的知识竞赛活动,甲同学答对第一道题的概率为号,连续答对两道题的概率为号,用事件A表示“甲同学答对第一道题”,事件B表示“甲同学答对第二道题”,则P(B|A)AB.c解折:由己起可知PCAB》=号所以PBA=-喜-号答案:C2.设随机变量X~B(,p),若E(X)-号,D(X)-号,则D=A日B司cD号解析:由XBm.B00=专,X0=号,可得m广专m1-)=8,解得=子4答案:B3.已知x,y的取值如下表:x2346y3.24.87.3若y与x线性相关,且经验回归方程为y=2.14x-1.29,则实数m的值为A.9.1B.9.7C.9.3D.9.5解折:因方7-2士3时+5=名-3.2士48时13士四-1返,又经登园归直线方程方=21-128小44所以153十0=2.14×号-1.29,解得m=9.5。4答案:D4.某校为创建文明校园,从学生会中的8名女生干部与4名男生干部中随机选取5名学生干部组成“文明校园督察队”,则组成3女2男的“文明校园督察队”的概率为A是普c装D.CC0【23新教材·DY·数学-RA-选择性必修第三册-N】 (ii)当-0,符合题意:②当a<0时,因x>0,则e->0,则e'->alnr-a,即c>+lnr-I)a,设m计n-小,则m意生过所以m(x)在(0,1)上单调递减,在(1,+∞)上单调递增.所以m(x)≥m(1)=0,所以,当a<0时,e>0≥+lnr-1)a,即lf(x)l>alnx-a成立,即a<0合题意:③当a>0时,由(1)可知,hx)一a=xe-a,在(0,+∞)上单调递增.又h(0)-a=-a<0,h(a)-a=a(ea-l)>0,所以3xo∈(0,a),使h(xo)-a=xoexo-a=0,iD当x∈(0,o时,xe-a<0,即e-<0,设g)=g-e--alnx+-a>0,则g)=一是一e一-<0,所以g在0,x0上单调递减,所以x∈(0,xo)时,gx)>gxo)=-alnxo-+a:i)当x∈(xo,+∞)时,xe-a>0,即e-8>0,设w)=e-:-alnx+a>0,因为t6=e*+是-是=on-感,Y2令p(x)=x2ex+a-ax,x∈(x,+o),则p'(x)=(x2+2x)ex-a,又令n(x)=(x2+2x)ex-a,x∈(xo,+∞),则n'(x)=(x2+4x+2)ex>0,得n(x)在(xo,+∞)上单调递增.有p'(x)=n(x)≥n(xo)=(x+2x)e*o-a=axo+a>0,得p(x)在(xo,+∞)上单调递增,有p(x)≥p(xo)=x行exo+a-axo=a>0.则t'()=四>0,得t(x)在(x,+∞)上单调递增.则x∈(xo,+∞)时,t(x)≥t(xo)=-alnx0+a.又x∈(0,xo)时,g(x)>g(x)=-alnx0+a,得当a>0时,lf(xl>almx-a时,-alnxo+a>0→0 15.What does the last paragraph focus on about Chunhui Xu's experiment?A.Its limit.B.Its equipment.C.Its condition.D.Its significance.第二节(共5小题,每小题2.5分,满分12.5分)阅读下面短文,从短文后的选项中选出可以填人空白处的最佳选项。选项中有两项为多余选项Commuter towns,also called exurbs,are a phenomenon of the development of high-speedhighways,extensive public transportation system high population rates in cities,and vari-ous other factors.16 People prefer to live in arens with less crime,better school dis-tricts,and larger homes at cheaper prices.Commuter towns may start as an area on the fringe of a suburb,usually in rural areas.A量housing projeet,with maybe as few as 100 homes,could define what is calledan exub.If the exurb begins to grow,more housing developments will be built.17In order for a commuter town to survive,grow and prosper,quick access to high-speed轮freeways or public transportation like high-speed trains is required.Commuter towns mayspring up along relatively rural areas close to a highway to provide less expensive housing.1When this occurs,towns farther away from a central work location may be built,butstill with quick access to a highway or freeway.19With many of the homes occupied by families,children can easily become latch-key kids,or remain in after school day care.Commuting can become expensive,contributing长to pollution when people make lengthy commutes in cars alone.The commuter town may still be preferable to city living,especially when neighborhoodshave good schools,large homes and low crime rates.20 So they are willing to sacrifice☒the extra hours in their day it takes to commute.A.People may feel safer living in commuter towns.B.Some people find a job in fairly large commuter towns.郡C.When commuter towns grow,there are several problems.D.This makes it hard for people to escape the commute lifestyle.E.As commuter towns grow,they may become too big or expensive to live in.F.Then it can evolve into a town with public resources like schools and stores.杯G.They show people's desire to live in a place different from where they work.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)期阅读下面短文,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。I enjoyed music and have always ad-21.A.preparationB.predictionmired guitar playing.However,it wasn'tuntil my cousin Lisa got out an old guitarC.interestD.promisefrom her basement that my intangible(难以22.A.excitementB.explanation确定的)21became a real goal.SeeingLisa teach herself to play guitar was enoughC.encouragementD.improvement22 for me to try.23.A.successB.effortsI made great 23 in the process.Acommon misunderstanding is that learningC.progressD.choicesthe guitar is easy.But becoming accus-24.A.surprisingB.amusingtomed to this foreign practice is probablythe most 24 things I have ever done.IC.satisfyingD.discouragingwas so determined to be as good as my icons全国100所名校最新高考模拟示范卷第5页(共8页)【22·(新高考)ZX·MN·英语(三)·SD】 答案A
解题分析A是典型的同物之境,诗句经过了移情作用,执情
强物,具有“以物观物”的特征。 唐山市2023年普通高等学校招生统一考试第二次模拟演练数学参考答案选择题:14.BDBA5~8.CCBD二.选择题:9.BC10.ABD11.AD12.BCD三.填空题:13.1214.215.五边形,23+V216.[日,+∞第15题第一空2分,第二空3分四.解答题:(若有其他解法,请参照给分)17.解:(1)因为2 sin Asin Bcos C=sin2C,由正弦定理得,2 abcos C=c2,…2分由余弦定理得,a2+b2-c2=c2,…2分整理得,a2+b2c2=2.…1分(2)S=absin C…1分因为c=2,由(1)可得cosC=房…1分则c-V…1分又2c2=8=a2+b2≥2ab,即ab≤4,…1分于是S分a6-4s216-4-V5所以S的最大值为W3,…1分18.解:(1)①采桑不采桑合计患皮炎426未患皮炎11819合计52025…2分②零假设为H:患皮炎与采桑之间无关联.根据列联表中的数据,经计算得到x2-25X4×18-2x126×19×5×20…2分1225114≈10.746>7.879=x005.…1分根据小概率值α=0.005的独立性检验,我们推断Ho不成立,即认为患皮炎与采桑之间有关联,此推断犯错误的概率不大于0.005.…1分(2)用X表示抽取的4人中采桑的工作人员人数,X的取值为:2,3,4.Px2》餐景x)图CC3 2C4C91,PX=4)=Cg=i5…3分高三数学参考答案第1页(共4页) 8.在平面直角坐标系中,P为圆2+y2:16上的动点,定点A(-3,2).现将y轴左侧半圆所在坐标平面沿y轴翻折,与y轴右侧半圆所在平面成2的二面角,使点A翻折至A,P仍在右侧半圆和折起的左侧半圆上运动,则A',P两点间距离的取值范围是A.[13,35]B.[4-√13,7]C.[4-√/13,35]1D.[√/13,7]二、多项选择题:(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.在△ABC中,若anA扌B=sinC,则下列论断正确的是2A.tanA =1B.sinA+sinB≤2tanBC.sin2A cos2B =1D.cos2A co82B sin2C10.阅读数学材料:“设P为多面体M的一个顶点,定义多面体M在点P处的离散曲率为1-2Q,P0,+∠0,P0,++0P0.+∠0.P0)其中0(i=1,2k,≥3)为多面体M的所有与点P相邻的顶点,且平面Q,PQ2,平面Q,PQ,,平面Q-PQ.和平面QPQ,为多面体M的所有以P为公共点的面.”解答问题:已知在直四棱柱ABCD-A,B,C,D,中,底面ABCD为菱形,AA1=AB,则下列结论正确的是A.直四棱柱ABCD-A,B,C,D,在其各顶点处的离散曲率都相等B.若AC=BD,则直四棱柱ABCD-A,BCD,在顶点A处的离散曲率为C.若四面体A,ABD在点A,处的离散曲率为7,则AC,1平面A,BDD.若直四棱柱ABCD-A,B,C,D,在顶点A处的离散曲率为},则BC,与平面ACC,的夹角为好11.定义在R上的函数f(x)=x+2x3+4x2+ax+1,则A存在唯一实数,使函数)图象关于直线x=-?对称B.存在实数a,使函数f(x)为单调函数C.任意实数a,函数f(x)都存在最小值D.任意实数a,函数f(x)都存在两条过原点的切线12已知直线1:y-:+m与辆圆C号+号-1交于AB两点,点P为椭圆C的下瓶点。则下列结论正确的是A.当m=1时,3k∈R,使得1F1+1F1=3B.当m=1时,keR,1F+FE1>2C.当k=1时,3m∈R,使得1F1+1F1=4D.当k=1时,meR,1F+FB1>9数学试卷第2页(共4页) 21.解析:(1)因为双曲线的右焦点为F(2,0),所以a2十b2=4,因为双曲线的渐近线方程为y-士x,所以。-5,即6-尽a,所以a2十b2=4a2=4,所以a=1,b=√3.(3分)所以双曲线C的方程为x2一号=1.3…(5分)(2)证明:设直线BP的斜率为k(k≠0),则直线BP的方程为y=k(x一1),又CQ∥PB,所以直线CQ的方程为y=k,Q点的坐标为(号,号).y=k(x-1),联立直线BP与双曲线C的方程消去y得(k2-3)x2-2kx十b2+3=0,k2所以十23则M点的坐标为2二323.……(8分)又F2,0,易得,直线QF的斜率为一专直线OM的斜*为是,由于一台·月1,则直线OM与直线Qr垂直设直线OM与直线QF的交点为G(x,y),则GO·G=0,G0=(-x,-y),Gf=(2-x,-y),则G0.Gi=x2-2x+y2=0,即直线OM与直线QF的交点G在曲线(x一1)2十y2=1上.…(12分)22.解析:(1)f(x)=ex(sinx十c0sx),x∈[-π,π].…(1分)令f'(x)=0,得a=一至w=7当x∈[-,-开)或x∈(3开,π]时f'(x)<0;当x(-T,3)时,f'(x)>0.所以fx)的单调递减区间是[一,一平],[不],单调递增区间是[一至,1。…(3分)又f-)=号e)-e放心的极小值为-号。i,极大值为号c。…(5分)2023届高三“一起考”大联考(模拟三)·数学参考答案9 高考模拟信息卷15.{x一√2 14.What does Lucy think of communicating with classmates or friends?15.How many good habits are mentioned by Lucy?第四部分朗读短文略第五部分情景问答M:Would you like to have dinner with me later,Betty?W:Sorry,I promised my mom that I would eat at home tonight.M:All right.What food will your mother cook?W:Dumplings,my favorite food.M:I see.Do you often eat at home?W:Yes,three times a week.I like talking about my recent life with my parents at table.Questions:1.What does the man want to do with Betty?2.What is Betty's favorite food?3.How often does Betty eat at home?第六部分话题简述It's important for everyone to have enough sleep.I go to bed at ten o'clock every night.I oftendrink a glass of warm milk before that.It can make me sleep better.参考答案-、1-5 BBACA二、6-10 ACBBA三、11-15 ACACB四、略 )选考题:共10分.请考生在第22、23题中任选一题作答.如果多做,则按所做的第一题计分,A9面平音了点,8A9面平10A,中O814剂同本,示被图破[选修44:坐标系与参数方程](10分)3A=8,0=91.98)19号在平面直角坐标系x0中,曲线C的参数方程为4十20os“。为参数),以坐y-2sin a标原点为极点,x轴的正半轴为极轴建立极坐标系,直线1的极坐标方程为cos(0+外3(1)求曲线C的普通方程和直线1的直角坐标方程;密(②)若直线1与曲线C相交于A,B两点,点P的极坐标为8,经),求PA十PB的值.3.[选修4一5:不等式选讲](10分)已知函数f(x)=x一1+x十2.1)求不等式f(x)≤号x+号的解集:2)若fx的最小值为m,正数a6请是a+6=m,证明:a+6叶名≥架封 22.解:(1)f'(x)==1-x-a-0-0'-ae,x<).1分e*1-x (1-x)e*①当a≤0时,f'(x)>0,此时,f(x)在(-o,1)单调递增:.2分②当a>0时,令g(x)=(1-x)2-ae,可以判断g(x)在(-o,1)是减少的注意到:g(-a-1)=(a+2)2-aea-l>(a+2)2-a>0g(I)=-ae<0则必存在∈(-o,l)使得g(x)=0,即(1-x)2-ae。=04分且当x∈(-0,x)时,g(x)>0,于是f'(x)>0,此时f(x)在(-0,x)单调递增:当x∈(x,)时,g(x)<0,于是f'(x)<0,此时f(x)在(x,l)单调递减:…5分(2)当a>0时,令h(x)=f'(x),则:h)=红-16-2》-ae<0于是:了在(-o,)是减少的(x-1)2ex7分对于给定的x2∈(-0,0),令p(x)=f(x+x2)-f(x)-f(x2),2∈(-0,0)则p'(x)=f'(x+x2)-f'(x)因为x+x2 0时,对于给定的x2∈(-o,0),令p(x)=f(x+x2)-f(x)-f(x2),x2∈(-0,0)则p=fx+x)-f(田=1--五-a-1-xae+51-x-x2e*1-x5-e+,11)>0第5页因此p(x)在(-0,0)是增加的于是,p(x)
故函数f(x)=2x一x2零,点个数多于2个,故A错误;对于B,由题意得∫(x)=1十≥0在(1,十∞)上恒成立,即≥-x2在(1,十∞)上恒成立,因为在(1,十∞)上,y=一x2<一1,故a≥一1,故实数a的取值范围是[一1,十∞),B正确;对于C,令x=-3,则由f(x十6)=f(x)十f(3),得f(3)=f(-3)+f(3)=2f(3),故f(3)=0,而由f(3)=0得:f(6十x)=f(x),故f(x)以6为周期.又f(x)为偶函数即关于直线x=0对称,故直线x=一6是函数y=f(x)的图象的一条对称轴,C正确;对于D,由题意得2=m有解,其中y=2∈[1,十o∞),故实数m的取值范围是[1,十o∞),D错误.故选BC.12.ACD【解析】对于A,若x∈Q,则一x∈Q,满足f(x)=f(-x);若x∈CRQ,则一x∈CRQ,满足f(x)=f(一x).故函数f(x)为偶函数,选项A正确;对于B,取x1=π∈CRQ,x2=一π∈CRQ,则f(x1十x2)=f(0)=1,f(x1)十f(x2)=0,故选项B错误;对于C,若x∈Q,则x十T∈Q,满足f(x)=f(x十T);若x∈CRQ,则x十T∈RQ,满足f(x)=f(x十T),故选项C正确,文化对于D,△ABC要为等腰直角三角形,只可能存在如下四种情况:①直角顶,点A在y=1上,斜边在x轴上,此时点B,点C的横坐标为无理数,则BC中点的横坐标仍然为无理数,那么点A的横坐标也为无理数,这与点A的纵坐标为1矛盾,故不成立;y本AB②直角顶,点A在y=1上,斜边不在x轴上,此时点B的横坐标为无理数,则,点A的横坐标也应为无理数,这与点A的纵坐标为1矛盾,故不成立;B0③直角顶点A在x轴上,斜边在y=1上,此时点B,点C的横坐标为有理数,则BC中点的横坐标仍然为有理数,那么,点A的横坐标也应为有理数,这与点A的纵坐标为0予盾,故不成立;ytBC④直角顶,点A在x轴上,斜边不在y=1上,此时点A的横坐标为无理数,则点B的横坐标也应为无理数,这与点B的纵坐标为1矛盾,故不成立.高二数学参考答案一3 .AB=√AC2+BC2=3V5∠ABD=∠ACD,∠DEB=∠AEC,∠DEB=∠DBE,∴.∠ACE=-∠AEC.AC=AE=√5,.BE=AB-AE=2V5,.EF=VBE2+BF2=6:AJ⊥CE,∴.∠AJE=90°,∴.∠AJE=∠ABF,又,∠AEJ=∠BEF,.△4JE∽△BFE小品指9“2V56AC=AE,AJ LCE,.CE =2EJ=2,∴.BF=EF+BE=6+2=8.…10分23.(1)y=x2-2x+m2-2m-2,当m=2时,y=x2-4x-2=(x-2)2-6,顶点坐标为(2,-6):…3分(2)①y=x2-2m+m2-2m-2经过点(0,-3)则m2-2m-2=-3,解得m=1,.y=x2-2x-3令y=0时,x2-2x-3=0,解得x1=-1,x2=3∴.A(-1,0),B(3,0),C(0,-3),直线BC解析式为y=x-3设P(a,d2-2a-3,过点P向x轴作垂线,交直线BC于点N,则点N(a,a-3),过点A向x轴作垂线,交直线BC于点M,则点M(-1,-4).AM∥PN,PD PN,又pN=(a-3)-(a2-2a-3)=-a2+3a,AM=4AD AMAD AM4.当a=时·PD此时点P的坐标为(侵-》…8分AD②n≤-2.…12分抛物线与x轴交于两点,∴.△=(-2m)2-4(m2-2m-2)>0,得m>-1当x=n,y=n2-2mn+m2-2m-2,当x=n+2,y=(n+2)2-2m(n+2)+m2-2m-2片>2,∴片-2=4m-4n-4>0,n 第七单元第二次综合测试1.B直线m∥平面a,直线nC平面a,可知m与n平行或异面,2.D根据斜二测画法的规则可知AB=4,OA=1,故周长为2(OA十AB)=10.3.C由圆锥的轴截面是边长为1的等边三角形可知,圆锥的底面圆半径r=,母线长1=1,所以圆维的侧面积为S=l=牙4.Dm∥a,n∥a,则m,n可能平行、相交、异面,故A项错误;m∥a,a∥B,则可能mC3,故B项错误;m∥a,aLB,则可能mC3,也可能m∥B,故C项错误;根据两条平行线中的一条直线垂直一个平面,则另一条也垂直该平面,故D项正确.5.C四棱锥的高为√5一1=2,若圆柱的一个底面的周圆经过四棱锥四条侧棱的中点,圆柱的底面半径为?,一个底面的圆心为四棱锥底面的中心,则圆柱的高为1,故圆柱的体积为x×(2)×1=T6.A如图所示,连接BD,因为BD1在平面ABCD上的投影为BD,故作PQ⊥BD交于点Q,且PQ⊥平面ABCD,连接AQ,则AP与平面ABCD所成角为∠PAQ,因为PB-2PD,故PQ-号DD号,且B02D0.D所以AQ-√(AB2+(号AD)2-5所以AP与平面ABCD所成角的正切值为an∠PAQ-PQ 3932V557.D设AC-BC=CG=3,CP=CC-1,CQ-号0=1,PQ=1,由AM-2MC,BN-2NG,可得CM-3AC-3X32=√2.又PN∥CB,PN⊥CC1,可得CN=√CP2+PN2=√/4+I=√5.由BC⊥AC,BC⊥CC1,AC∩CC1=C,AC,CC1C平面ACC1A1,可得BC⊥平面ACCA1,则NP⊥平面ACC1A1,PMC平面ACC1A1,A所以NP⊥PM,因为MQ⊥CQ,所以MN=√/MQ+PQ+NP=√1+1+1=√3,所以CM+MN2-CN2,·29【23新教材·DY·数学·参考答案一RB一必修第四册一QG】 21、(12分)已知抛物线E:y2=2px(p>0)的焦点为F,点P(2,o)在E上,且PF=4,22.(12分)(1)求抛物线E的标准方程.已知椭圆E,文=1(a>b>0),且直线l:x十my-1=0过E的右焦点F.当(2)直线:y=k(x一)与曲线E交于A,B两点,直线4:y=k(x-2)与曲线m=1时,椭圆E的长轴长是其下顶点到直线l的距离的2倍.E相交于C,D两点,M,N分别为AB,CD的中点,若+k=2,且≠k1,证(1)求椭圆E的离心率。明:直线MN恒过定点.(2)设过点Q(0,一3)且斜率为k的直线1与椭圆E交于不同的两点M,N,且MN=8号求大的值置分等立写出文字地期、过相或演年步度线卷八第7页(共8页)23·ZXQH·数学文科卷八第8页(共8页)23·ZXQH·数学文科 Greek.She was believed to have learned Egyptian,Hebrew,and Troglodyte.She was also believed tohave spoken the languages of the Ethiopians,Syrians,Armenians,Medes,and Parthians as well.35 The famous Serbian-American engineer and inventor of the alternating-current electrical systemwas fluent in eight different languages,including Serbo-Croatian,English,French,Czech,German,Hungarian,Italian,and Latin.A.Another historical polyglot is Nikola Tesla.B.Polyglots are certainly rare and interesting peopleC.Many don't even know that she is from a different country.D.However,there's a record-holder for most languages spoken.E.She even has a Bachelor of the Arts degree in Psychology from HarvardF.If we dig a little deeper through history,there are many famous polyglots that few know about.G.There are many famous polyglots known for other things besides speaking multiple languages.第三部分:语言运用(共两节,满分45分)第一节完形填空(共20小题;每小题1.5分,满分30分〉阅读下面短文,从短文后各题所给的A、B、C和D四个选项中,选出可以填入空白处的最佳选项,并在答题卷上将该项涂黑。"I can't decide!"Carter told his mom.He had a big 36 to make,and he was 37."You have to."Mom took a sip of her coffee."It's not 38 for both boys to think youare coming to their parties.I think you should go to Tyler's party.He invited you 39 andyou told him that you were going.40,it will be at the lake!You love to 41.""I know.But,Josh is having a go-kart (party!That would be 42!"Four days later,Carter was riding a bike for Josh's party.He43 told Josh or Tyler thathe would not be 44 their party.They were both 45 him to attend.He 46 to tellTyler that he wasn't coming,but he felt so 47 about it.Tyler was his best friend.Butgo-karts!How could he possibly 48 this golden chance?As he arrived,Josh's sister,Kayla told him they had already left for the party.Carter stood at the door,stunned.He 49 the party!It started one hour earlier than hethought.No go-karts!How could he have made that kind of 50 Maybe he could still 51it to Tyler's party.Sweat trickled down his sides as he pedaled home.His T-shirt clung to hisback.He would have to 52 clothes before he left for Tyler's house.When he pedaled home,the door was locked."Mom!"he yelled outside.No 53Weary,he dragged his feet over to the porch (and plopped down.Right now,hisfriends were54 racing around a go-kart track or swimming at the lake.He had to wait forhis family to 55 home.高一英语试题(A卷)第7页(共10页) 上期参考答案方程(组与不等式综合题1.(1)30元,50元.(2)20根.2.(1)120元,80元(2)100,个日方程与函数综合题1.B0g2.(1)20件,10件,(2)40件A款40件B款,1080元.一次函数与反比例函数综合题1.A2-62.(107是(2)Sex三l.3(3m=6(2)a的值为3或-11.一次函数气二次函数综合题1.(①2-分-1(2)(-1,0.(3)-1 2023年名校之约·中考导向总复习模拟样卷二轮·数学(六)参考答案一、选择题(本大题共10小题,每小题4分,满分40分)题号2358910答案BDCDBD10.[提示]当点P在CM上时,PC+PM=CM=1不符合题意;当点P在BM上时,如答图1,点P越靠近点B的PC+PM值越大.当点P与B重合时,PC+PM的最大值是7>6,即在BM边上存在一点P,使PC+PM=6;同理当点P在AC上时,如答图2,点P与A重合时,PC+PM的最大值是4+√17>6,即在AC边上存在一点P,使PC+PM=6:当点P在AB边上时,如答图3,作点C关于AB的对称点D,连接CD,DP,则PC=PD,PC+PM=PD+PM,连接DM,DM与AB的交点为点P时,PC+PM的值最小,最小值为DM的长,连接BD,易求DM=5<6,故AB边上存在两个点P,使PC+PM=6.综上,共有4个点,故选C.图2图3第10题答案图二、填空题(本大题共4小题,每小题5分,满分20分)11.23、12.2y(x+2y)(x-2g)13.114.(1)(3,-13)(2分)(2)4或-4(3分)14.(2)[提示]抛物线y=2-2x-4=(x-k)2-k2-4,即点A(k,-2-4):OA=OB,点A,B在直线y=mx上,∴.点AB关于原点O对称,则点B(-k,k2+4)…点B(-k,2+4)也在抛物线y=x2-2hx-4上,∴.(-k)2-2k·(-k)-4=k2+4,解得k=±2.当k=2,即点A(2,-8)时,将点A(2,-8)代入y=mx,得2m=-8,解得m=-4.当k=-2,即点A(-2,-8)时,将点A(-2,-8)代入y=mx,得-2m=-8,解得m=4.综上m的值为4或-4.三、(本大题共2小题,每小题8分,满分16分).1115.解:原式=1+」…4分24548分16.解:设5月、6月这两个月利润的平均增长率是x,根据题意,得(1+44%)(1+21%)=(1+x)2.…4分解得x1=0.32=32%,x1=-2.32(舍去).答:5、6月份这两个月利润的平均增长率为32%.…8分数学总复习模拟样卷(六)·参考答案第1页 3.应用了较多的语法结构和词汇;4.具备较强的语言运用能力,语法或词汇方面有些许错误,但为尽力使用较复杂结构或词汇所致;5.有效地使用了语句间的连接成分,使全义结构紧凑;6.完全达到了预期的写作目的。第四档(好)(1620分)1.完全完成了试题规定的任务;2.虽漏掉一两个次重点,但覆盖所有主要内容;3.应用的语法结构和词汇能满足任务的要求;4.语法结构或词汇方面应用基本准确,些许错误主要是因尝试较复杂语法结构或词汇所致;5.应用简单的语句间的连接成分,使全文结构紧凑;6.达到了预期的写作目的。第三档(适当)(11一15分)1.基本完成了试题规定的任务:2.虽漏掉一些内容,但覆盖所有主要内容;3.应用的语法结构和词汇能满足任务的要求;4.有一些语法结构或词汇方面的错误,但不影响理解;5.应用简单的语句间的连接成分,使全文内容连贯;6.整体而言,基本达到了预期的写作目的。第二档(较差)(6一10分)1.未恰当完成试题规定的任务;2.漏掉或未描述清楚一些主要内容,写了一些无关内容;3.词法结构单调,词汇项目有限;4.有一些语法结构或词汇方面的错误,影响了对写作内容的理解;5.较少使用语句间的连接成分,内容缺少连贯性;6.信息未能清楚地传达给读者。第一档(差)(1一5分)1.未完成试题规定的任务;2.明显遗漏主要内容,写了一些无关内容,原因可能是未理解试题要求;3.语法结构单调,词汇项目有限;4.较多语法结构或词汇方面的错误,影响对写作内容的理解;5.缺乏语句间的连接成分,内容不连贯;6.信息未能传达给读者。不得分(0分)未能传达给读者任何信息:内容太少,无法评判,写的内容均与所要求内容无关或所写内容无法看清。注意事项:1.评分时,先根据文章的内容和语言初步确定其所属档次,然后以该档次的要求来衡量、确定或调整档次,最后给分;2.词数少于80和多于120的,从总分巾减去2分;3.评分时,应注意的主要内容为:内容要点,应用词汇和语法结构的数量和准确性,上下文的连贯性及语言的得体性;4.拼写与标点符号是语言准确性的一个方面,评分时,应视其对交际的影响程度予以考虑。英、美拼写均可接受;5.如书写较差,以致影响交际,将分数降低一个档次;6.内容要点可用不同方式表达,对紧扣主题的适当发挥不予扣分。—5 平遥县20222023学年度第二学期八年级期中教学质量监测试题(卷)数学(满分:100分时间:90分钟)题号三总分得分一、选择题:(本大题10个小题,每小题3分,共30分)每个小题都给出了代号为A、B、C、D的四个答案,其中只有一个是正确的,请将正确答案的代号填在表格中.题号23678910褽答案的1.下列图形中,既是轴对称图形又是中心对称图形的是(毁兴米4D长2.公元前500年,古希腊毕达哥拉斯(Pythagoras)学派的弟子希帕索斯(Hippasus)发现了一个惊人的事实:边长为1的正方形的对角线的长度是不可度量的,即不能表示成两个☒整数之比.这个发现是基于一个表述直角三角形三条边长之间关系的定理,请问这个定理瑞被称为()A.勾股定理B.韦达定理C.费马大定理D.阿基米德折弦定理布3.在求解一个关于x的一元一次不等式组的解集时,在数轴上表示为如图所示,则它的解集期是()A.x≥2B.x>2-2-1034C.x≥-1D.x<-184.如图,在平面内将RABC绕着直角顶点C逆时针旋转90°得到REFC.若AB=5,BC=3,则线段BE的长为()摇A.5B.6C.7D.85.等腰三角形的周长为20cm,已知其一边长为5cm,则其腰长为(A.5cmB.5cm或7.5cmC.7.5cmD.以上都不对八年级数学第1页(共8页) 因为P(0,0,4),C(4,4,0),D(0,4,0),F(1,0,3),所以PC=(4,4,-4),CD=(-4,0,0),CF=(-3,-4,3),…(8分)设平面CDF的法向量为n=(x,y,2),n.Ci=(x,y,x)·(-4,0,0)=-4.x=0,x=0,由n·C7=(x,y,z)·(-3,-4,3)=-3x-4y+3z=0解得3」y=42,令之=4,得平面CDF的一个法向量为n=(0,3,4).…(10分)设直线PC与平面CDF所成的角为O,nlo2=(4,4,-4)·(0,3,4)√34√3×515放直线PC与平西CDF所成角的E孩值为…(12分)21.【解析J1)因为猫圆C若+苦=1a>6>0)的左、右顶,点分别为A(-22,0,B(2.0),所以=2W2.…(们分)因为|DF2|=2-√2,D在F2的左方,所以c一2a=2-2,解得c=2,…(2分)则b=√(2√2)222=2,。..........。g。。(3分)所以猫圆C的标准方程为写十置=1.(4分)(2)由题意设直线MN:x=my+√2,M(x1,y1),N(x2,y2),A(-2W2,0),。。。。。。。。。(5分)yO D/F.B xx=my+√2,联立王+=1,消去x,得(m2+2)y2+2√2my-6=0,(6分)、8T42√2m6则y1十y2=7m2干2M次=一m+2,(7分)所以1·2=:·yy1y2x1十2W2x2+2W2x1x2+22(x1十x2)十8yiy2(my+√2)(my2+√2)+2W2(my1+√2+my2+√2)+8y1y2m2y1y2+3V2m(y+y2)+18数学参考答案-5 第二档(6-10分)未恰当完成试题规定的任务。一漏掉或未描述清楚一些主要内容,写了一些无关内容。一语法结构单调,词汇知识有限。一有一些语法结构或词汇方面的错误,影响了对写作内容的理解。一较少使用语句间的连接成分,内容缺少连贯性。信息未能清楚地传达给读者。第一档(15分)未完成试题规定的任务。一明显遗漏主要内容,写了一些无关内容,原因可能是未理解试题要求。一语法结构单调,词汇知识有限。较多语法结构或词汇方面的错误,影响对写作内容的理解。-缺乏语句间的连接成分,内容不连贯。信息未能传达给读者。0分未能传达给读者任何信息,内容太少,无法评判;所写内容均与所要求内容无关或所写内容无法看清。A Possible Version:Dear fellow students,Good morning!Here's good news for you!The English Association will invite Li Hua,a former student ofour school,to share his learning experience with us.Li Hua graduated as an honor student fromour school two years ago.He is currently studying in Peking University.He will introduceeffective and practical English learning strategies and recommend some learning apps to us.At the end of his speech,there will be a question-and-answer session,so you'd better prepareyour questions in advance.After the session,Li Hua will take pictures with us.Li Hua's speech will start at 4 p.m.next Wednesday in the lecture hall.All the students arerequired to attend it on time and take seats as required.Thank you!听力原文:第一节听力理解听第一段材料,回答第1至2题。W:Hi,Tom.What's your plan for the winter holiday?M:I'm learning to drive.And I hope I can pass the driver test soon.Would you like to join me?W:No.I hate driving.I wish I could have a self-driving car.M:Hey,I heard that some self-driving cars are being tested.We might not need human driversvery soon.Actually,experts say robots will take over 30%of our jobs within ten years.W:Do you think that robots will replace humans one day?M:Probably yes.Actually,robots can do a lot nowadays.They can even do many difficult jobs.SoI think robots have some advantages over humans.W:What can we do if robots take place of us?M:I am not that worried.New jobs will appear anyway.I think we should take the opportunity todo some more interesting work and view robots as teammates.Question One:What will Tom do during the holiday?Question Two:What is Tom's attitude to robots?2021~2022学年佛山市高中教学质量检测高一英语参考答案与评分标准第3页共5页 21.(本可小题满分12分)第20届女足亚洲杯于2022年1月20日在即度举行,在半决赛中,中国女足通过点球大战6:5惊险战胜日本女足,其中门将朱钰两度扑出日本队员的点球,表现神勇.在2月6日的决赛中,中国女足在两球落后的情况下,以3比2逆转击败韩国女足,成功夺得亚洲杯冠军,女足精神再次激励中华儿女、(1)点球决赛中,守门员扑点球的难度一般比较大,假设罚点球的球员会等可能地随机选择球门的左、中、右三个方向射门,门将也会等可能地随机选择球门的左、中、右三个方向来扑点球,中国门将朱钰经过刻苦训练,在方向判断准确的前提下扑出点球的概率为子.不考虑其它因素,在一次点球大战中,求门将朱钰在前五次扑出点球的个数X的分布列和期望;(2)中国女足在教练水庆霞的精心带领下,技战术水平非常成熟完美,中锋和前锋姚伟、刘艳秋、王霜、唐佳丽4名女足队员在决赛前的热身训练中,球从姚伟脚下开始,等可能地随机传向另外3人中的1人,接球者接到球后再等可能地随机传向另外3人中的1人,如此不停地传下去,假设传出的球都能接住.记第n次传球之前球在姚伟脚下的概率为p,易知p1=1,p2=0.炎德文化①证明数列考轻颗有悉的D必究②设第n次传球之前球往主霜脚下的概率为qm,比较p1o与qo的大小,22.(本小题满分12分)已知函数f(x)=e'-ax-1,a∈R(1)求函数y=f(x)的单调区间;(2)设函数F(x)=f(x)十sinx,当1≤a<2时,讨论函数F(x)零点的个数. 8.W2022年海南省高考全真模拟卷(一)天一大联考A英语91.本试卷满分150分,测试时间120分钟,共8页。10.装2.考查范围:高考全部内容。11.r第一部分听力(共两节,满分30分)订第一节(共5小题:每小题1.5分,满分7.5分)12.听下面5段对话。每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。每段对话仅读一遍。线例:How much is the shirt'?A.£19.15.B.£9.18.C.£9.15.13.答案是C。内l.Why is Ann so upset?14.A.She didn't pass one of her exams.B.She is worrying about other lessons.A15.不C.She didn't finish her math homework2.What does the woman usually eat?16.A.Junk food.B.Healthy food.KBC.Delivered food.要3.What will the man probably do to stay warm?A.Turn on the heater.B.Use a blanket钱C.Drink some hot chocolate.What are the speakers mainly talking about?AB答A.The man's hobby.B.The man's travel planC.The man's plan after graduating.SBC5.What do the speakers think of the painting?17A.Simple.B.Colorful.C.Complex.A题第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。每段对话或独白读两遍。听第6段材料,回答第6、7题。6.Where does the conversation take place?A.In an office downtown.B.In an apartment downtown.C.In an apartment in the East End7.What does the man want to do now?A.Get to work.B.Go to a restaurant.C.Visit the woman's company.高考全真模拟卷·英语(一)第1页(共8页) 所以-2,=1+2×1-…………9分所以Tn=(4n-10)×……10分18.解:(1)因为bsin2B+csin2C=(b+c)sin2A,由正弦定理得b3+c3=(b十c)a2,因为b+c>0,所以b2+c2-bc=a2,即b2+c2-a2=bc.所以c0sA=2+c2-a212bc2·……2分因为A∈(0,π),所以A=π3…3分由tanB+tanC3sin Acos C,可得inB+sinC=3sinAcos Bcos C=cos C'所以sin Bcos C+cos Bsin C=3 sin Acos B,所以sin(B+C)=3 sin Acos B,因为A+B+C=x,所以sinA=3 sin Acos B,所以cosB=3·…5分 A.Th-socie33.Why did Yin and his team choose fingertip sweat?imtoyfm oude tnthreorbodswe muhappenA.It can evaporate instantly.B.It holds abundant power.you havethyorWhyr yorrioyoriet's futurC.It can be obtained effortlessly.D.It contains much more lactate.the trail together?A.Benefits and risksninkers a34.What does the author mainly say about the device in paragraph 57B.Its display in wearable sensorsB.Strength and balanceA.Its potential in reducing energy use.C.Hiking can refreshus after a day's tiring workC.Its applications in health care.D.Its ability to generate power.D.Asimple walk increases the rate of our breathingport,builo35.What's the best title for the text?E.Consider just these benefits net time you need to be inspired tolace up your bootsvironmentA.Wearable sensors will make people hit the gym moreF.We can't have physical health without mental health,and hiking delivers on bothestions,givB.Sweaty fingertips could help power wearable sensorsG.The safest way to hike is with a friend or companion,in case one of you needs a handC.A device can turn the mechanical energy into electricity第三部分语言运用(共两节,满分30分)D.The highest concentration of sweat glands will be chalked up第一节(共15小题;每小题1分,满分15分)第二节(共5小题;每小题2.5分,满分12.5分)阅读下面短文,从每题所给的AB.C、D四个选项中选出可以填人空白处的最佳选项。阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。选项中有两项为多余It was our last class before summer41.A.clearingB.finishingD.lighting选项。break.I was 41 up the first year of anC.holdingOften when my family travels to a natural place like a national park I say to my husband,MFA program in poetry.I was exhausted-42.A.satisfiedB.lined"While we're there,I want to make sure we climb a thing."I mean I want to get out into na-and 42 with doubt.Was I progressingC.dottedD.riddledture and experience any "thing",usually through a hike.In addition to being fun and excit-as fast as everyone else?43.A.panickedB.complaineding,hiking is healthy.36.When a professor asked about our sum-ives you theD.faintedmer plans,I43.I didn't want to ap-C.echoedfor up to 6OEmotional well-beingpear idle and somehow undeserving of my44.A.titleB.grade37.Emotionally,a hike challenges us to persevere and follow a path wherever it may)【22·(新C.spotD.conditionlead,it engages our brains in map-reading and trail-following,and of course,being in natural44 in the program.45.A.foodsB.seasonsspaces inspires us to feel awe,the sense that there are forces larger than us at work in the“Gardening,”I blurted out,(脱口而7e51)Where did that come from?I knewC.plantsD.fieldsesworld.nothing about_45_!46.A.dirtB.neighbourhoodon of my●Heart healthMy professor nodded and said,"WhatD.temperatureHeart disease and high blood pressure are killers in this country,but hiking is a way toC.hobby52·1a good idea!”t'sB.properbuild cardiovascular health.38.It can benefit our hearts and lungs.Hiking on a trail thatA few days later,I was standing in line47.A.equalotions andat a grocery and picked a jalapeno plant up.C.enormousD.commonthe music Iarehas some ups and downs increases this effect by bringing our heart rate up and down.I watered and changed the 46 for my48.A.shiningB.sufferingous after aeam●39little plant.I even talked to it.Over theC.expandingD.coexistingis to get tosenMost of us don't hike on edges or cliffs.But even a stroll down a gentle wooded path re-quires us to pay attention to where our feet are landing and to keep our balance when clambesummer,it grew,not 47 but bigger.I49.A.takeB.make卷第9页(ring past rootssteanddipsnthe landscape.These actions require coreand le strength,started the second year of my MFA.C.protectD.stealwhich contribute toverall health in ways ranging from preventing back pain tokeepingIt was almost December and myjalapeno was 48:many of its leaves had50.A.fruitsB.rootsiew,fallen off.Would my little guy 49 it?IC.leavesD.stemsweight off.worked hard in school.Spring came.My51.A.submittecB.translatedOStronger relationships40.But beyond basic safety,hiking is a way to build relational health.The Nationaljalapeno plant came back to life with newC.appliedD.recitedPark Service puts t thiswaycaus hiking ange indifficulty fromn extremely challenging50.My jalapeno flowered.I 51my全国100所名校最新高考模拟际范卷第9页(共12页)【22·(新高考)ZX·MN·英语(二)·】eir hotel】全国10所名校最新高考模际范卷第8页(共12页)【2,(新高考)ZX·MN·英语(二)·灯XGK 所以Sn=n.10分18.(本小题满分12分)解:(I)依题意得sinB=sinC+cosCSnC-cosC.2分cosB所以sin Bsin C-sin BcosC=cos Bsin C+cos BcosC,所以sin(B+C)+cos(B+C)=0,4分所以tan(B+C)=-1即tanA=l,5分又因为0
20.(本题8分)阅读与思考下而是小颖的数学日记,请仔细阅读,并完成相应的任务×年×月×日星期六在圆中只用无刻度的直尺作出满足某条件的圆周角今天在数学课上,我学会了在圆中只用无刻度的直尺就可以作出满足某条件的圆周角。问题一:如图1,∠BAC是⊙O的圆周角,我们可以在⊙0中只用无刻度的直尺作一个圆周角等于∠BAC.作法:在⊙O上取一点D,连接BD和CD,则∠D=∠A(依据*).问题二:在图】的基础上,要在⊙0中只用无刻度的直尺以B为顶点作与∠A相等的圆周角,应该如何完成元?图图2图3作法:如图2所示,连接C0并延长,交⊙0于点D,连接BD,美接B0并延长,交⊙O于点E,则∠DBE即为所要求作的角.问题三:如图3,要在⊙0中只用无刻度的直尺作一个圆周角与∠A互余,应该如何完成呢?。4。。任务:(1)“问题一”中小颖的“依据*”是指(2)请说明“问题二”中小颖的作法是否正确并说明理由;(3)完成“问题三”:请在图3中只用无刻度的直尺作出满足条件的圆周角,并仿照“问题二”写出具体作法, 男生成绩的频数统计表等次频数频率A等30.3B等24C等m0,2D等0.1女生成绩是:42,60,39,56,52,39,55,39,42,56;抽取的男生和女生中考体育测试成绩的平均数、中位数、众数如下表:平均数中位数众数男生48044女生e47请根据以上信息解答下列问题:(1)a=U;b=;m=02(2)请选取一个统计量对该校九年级男生与女生的中考体育测试成绩进行评价,并说明理由;(3)若该校九年级共有680名学生,请估计这次中考体育测试成绩为A等次的人数,七、(本题满分12分)】22.已知抛物线y=一x2+bx十c(b,c为常数)经过点(-2,5)和(一6,-3).(1)求该抛物线的函数表达式;(2)将抛物线y=一x2+bx十c(b,c为常数)向右平移m(m>0)个单位长度得到一个新的抛物线,若新的抛物线的顶点关于原点O对称的点也在抛物线y=一x2+bx十c(b,c为常数)上,求m的值.A∽八、(本题满分14分)23.如图,已知等腰△ABC和等腰△ADE有公共的顶点A,且AB=AC,AD=AE,∠BAC=∠DAE,点E恰好落在边BC上(与B、C不重合),连接BD.(1)求证:BD=CE;()若AB与DE相交于点F,求证:CE·BE=CA·BF;专家9A0,AC=4,且噩-号请画出符合条件的图形,并求DE第23题图[中考必刷卷·2023年名校压轴三数学试题卷第4页(共4页) 17如国有正方形ACD中,对角线AC,BD相交于点O,点B,F是封角线AC上的两点:且四、(本大超飘2小期,每小角8分,满分16分】AE=CF连接DE,DF,BE,BF求证,四边形DEBF是菱形【解新】证明,:在正方形ABCD中,对扇线AC,BD相交于点O(3分)BDLAC.OB-OD.OA-0CAE-CF,.OA-AE=OC-CF,即OE=OF∴,四边形DEBF是平行四边形。又'BD⊥EF,∴,四边形DEBF是菱形.4”(8分18如图,在矩形ABCD中,点E在边BC上点F在BC的延长线上,且BE=CF,求延:∠BAE=∠CDF【解析】证明:,四边形APCD是矩形,∴.AB=CD,∠ABC=∠DCB=90'∴∠ABC=∠DCF=g0.(3分)在△ABE和△DCF中,(AB=DC,∠ABC=∠DCF,BE=CF,∴.△ABE≌△DCF(SAS),∠BAE=∠CDF.…(8分)五、(本大题共2小题,每小题10分,满分20分)19.如图,在△ABC中,∠ACB=90°,D是BC的中点,DE⊥BC,CE∥AD.若AC=2,CE=4,【(1)求证:四边形ACED是平行四边形;(2)求BC的长【解析】(I)证明:·∠ACB=90°,DELBC,.AC∥DE又:CE∥AD,.四边形ACED是平行四边形.(5分)(2)解::四边形ACED是平行四边形,..DE=AC=2.在Rt△CDE中,由勾股定理得CD=√CE-DE=√-=23.,D是BC的中点,.BC=2CD=4√3.…(10分)20.如图,在平行四边形ABCD中,CELAD于点E,点F在BC上,且BF=DE.(1)求证:四边形AFCE是矩形;(2)连接EF,若EF∥DC,DE=2,CE=4,求平行四边形ABCD的面积【解析】(1)证明:四边形ABCD是平行四边形,.AD∥BC,AD=BC.BF=DE,..AD-DE=BC-BF,AE=CF.又,AE∥CF,∴.四边形AFCE是平行四边形,CE⊥AD,,∠AEC=90°,.平行四边形AFCE是矩形:…(5分(2)解::四边形ABCD是平行四边形,【2023年安激中考一轮复习卷第动页〔共秘页)】 △A00△AB所以是-品则二=2博r=产2图维的体N√2+Rr【高三数学·参考答案第1页(共7页)】·23-467C·3A=号·栏2=晋h一-2+产2十≥等当且仅当=4时,等号成立9.BD【解析】本题考查圆的方程,考查直观想象的核心素养.由已知得圆C1的圆心C(0,0),半径r1=3,圆C2的圆心C2(3,4),半径r2=4,C1C2|=√(3-0)2+(4-0)2=5,r2-n1 根据小概率值α=0.010的独立性检验,我们推断H。不成立,即认为“编织巧手”与“年龄”有关,此推断犯错的概率不大于0.010.…5分(2)由题意可得这6人中年龄在40周岁以上(含40周岁)的人数是2;年龄在40周岁以下的人数是4.…7分从这6人中随机抽取2人的情况有C哈=15种,…8分其中符合条件的情况有CC2=8种,…。9分故所求概率P-。...........................10分评分细则:(1)在第(1)问中,直接补充完整2×2列联表,没有计算过程,只要答案正确,不予扣分;(2)在第(2)问中,算出40周岁以上(含40周岁)和40周岁以下的人数,得2分,求出总的基本事件数和事件包含的基本事件数的个数,各得1分;(3)若用其他解法,参照评分标准按步骤给分.18.解:(1)因为5cos2B-14cosB=7,所以5(2cos2B-1)-14cosB-7=0,…1分所以5c0s2B-7c0sB-6=0,即(5c0sB十3)(c0sB-2)=0,…3分解得cOsB=一35…4分因为0
k=1心+ey,…8分令1=ee1,],e+e=h0)=1+},45∈l,2]4<4,)-M)=4+么+宁-4-00.周为1<4s6期4-601->0.tt,因此h)-h)<0,即h) 17:06114C 50【数学】中考必刷卷·2023年名校内部卷.…中考必刷卷名校内部卷5数学参考答案数学参考答案及评分标准一、选择题(本大题共10小题,每小题4分,共40分】题号234568910答案CDBCDBBAD10.D解析:将等边△ABC绕点A逆时针旋转得到△ACE,同时得到等边△APP',则PA=PP','PA=PA,∠PAC=∠P'AE,AC=AE,∴.△APC≌△AP'E(SAS),∴.PC=P'E∴.当DPP'E四点共线时,PA十PC十PD取得最小值,作EF⊥DC的延长线于点F在△CEF中,∴EF=CE·sin∠BCF=8×S=45,CF=CE·cos∠ECF=8X号=4,∴.DF=CD+CF=7+4=11,∴DE=√DF+EF=√112+(43)=13,即PA+PC+PD最小值为13.二、填空题(本大题共4小题,每小题5分,满分20分)11.x≤312.113.214.(1)∠AEB=60°:(2)/76或2√/3解析:(1)由题意得,点E在△ABC的外接圆上,∴.∠AEB=∠BEC=60°:(2)作CF⊥AF,DG1AE,:∠CEF-180-∠AEC-60,CF-sin60.CE-9×6-35.EF-cos60CE=专×6=3DG=之cF=9,GE=VDE-G=√6-(3,y=含∴AF=2GF=2(GE+EF)=7AC=AF+CF-√7+(33)=76.另一种情况:∴.AF=2GF=2(EF-GE)=5,.AC=√AF+CF=√5+(35)=2√/13.中考必刷卷·2023年名校内部卷五数学参考答案评分标准第1页(共4页) 湖南师大附中2022一2023学年度高一第二学期第二次大徐司数学参考答案一、选择题:本大题共8个小题,每小题5分,满分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题号12345678答案DCAABCDB1.D【解析】由题意得z=2十5i,则=2-5i,所以在复平面内对应的点为(2,一5),故选:D.2.C【解析】正棱锥的各条棱长并不是都相等,应该为正棱锥的侧棱长都相等,故A不正确:不是所有的空间几何体的表面都能展开成平面图形,例如球不能展开成平面图形,B不正确:棱台是由平行于棱锥底面的平面截棱锥得到的,故各条侧棱的延长线一定交于一点,C正确;只有用一个平行于底面的平面去截棱锥,得到的两个几何体才能一个是棱雏,另一个是棱台,故D不正确.故选:C3.A【解析】由2-x<1,可得x2-2.x<0,解得0 8规定:sin(-x)=-sinr,cos(-)=cos,cos(r十y)=osr6osy一sinsiny,给出以下四个结论:(1)s(-30)三-号;(2)cos2r=cos'x=sinr13)cos(r-v=c0 s.rcosy+sin.rsiny:(4)cosl5°-Y6Y2.其中正确的结论的个数为A.1个B.2个C.3个D.4个【答案C【解析】1)sin(-30')=-sin30°=-),故此结论正确:(2)cos2.x=c0s(.x+x)=cos.rcos7.-sin.rsinc=cos2x-sin2.x,故此结论正确:(3)cos(x-y)=cos[x+(-y)]=cos.xcos(-y)-sin.rsin(-y)=cos.rcosy+sin.rsiny,故此结论正确:cosl5cos450Eco45cos30+si45sn30xS+V2X=Y6十2故此结论错误21224∴正确的结论有3个,故选C9《九章算术》是中国古代的数学专著,它奠定了中国古代数学的基本框架,以计算为中心,密切联系实际,以解决人们生产,生活中的数学问题为目的.书中记载了这样一个问题:“今有勾五步,股十二步,问勾中容方几何.”其大意是:如图,Rt△ABC的两条直角边的长分别为5和12,则它的内接正方形CDEF的边长为(人)答BC.D号【答案】B第9题图【解析】四边形CDEF是正方形,∴CD=ED,DE∥CF,设ED=x,则CD=x,AD=5-x,:DE∥CF,∠ADE=∠C∠AED=∠B△ADEn△ACB÷-是克=5号,解得x-9故选BDE AD.10.如图,在△ACD中,AD=6,BC=5,AC=AB(AB+BC),且DABO△DCA,若AD=3AP,点Q是线段AB上的动点,则PQ的最小值是D.82【答案】A第10题图【解折J:△DABn△DCA…C-品与6BD=6,解得BD=4(负值舍去),:△DABn△DCA…S-g器-号=号AC=号AB.AC=ABAB+BG.(2AB)=AB(AB+BC,AB=4,AB=BD=4,过B作BH⊥AD于H,AH=2AD=3,∴BH=√/AB-A=√4-3=√7,AD=3AP,AD=6,∴.AP=2,当PQ⊥AB时,PQ的值最小,:∠AQP=∠AHB=90°,∠PAQ=∠BAH,.△APQ∽△ABH,中考必刷卷·2023年安徽中考第一轮复习卷数学第38页(共76而)扫描全能王创建 WriterAnd more!For mosThe Admission Section receives and evaluates applications to the undergraduate(can help begprogram,in addition to answering applicants'questions by phone.Jenny,their schooPhone:613-562-5315books andTol-free:1-877-868-8292Regular Office Hours:Monday to Friday企小公新带两共微要资同猫二清she still p“DecCSeptember to May:9 am to 4:30 pmway of iJune to August:9amto3:30pm边个四),A0的从,文设T宽help lea21.Who is the Work-Study Program intended for?into oneA.Technicians.B.Professors.DiC.Marketing managers.D.University students.in BritFUXI F22.How can you apply for a position of the program?thingsA.By letter.B.By phone.C.In person.D.Via the Internet.accep23.What should you do to get more information about the program?24.WA.Phone 613-562-5315 on weekdays.B.Phone 1-877-868-8292 on weekends.bloit nseod isr nt noiicoC.Go to the Admission Section in person.D.Ask other undergraduates for details.81Bhrons moo nso bas smi-lnd iow o25After deciding to move into a new house,Jessica,a full-time mother in London,broughtin a woman home organizer-Sharon."As a result,Sharon threw away 50 per cent of myclothes,decorations,shoes and books.The process of tidying up my house brought peace to2me and made me feel organized at heart,"Jessica told The Times.S0a000Like Jessica,more British people have started to seek for help from home organizers topursue minimalism(极简主义)lifestyles.This period has made many people appreciate everyday order and comfort.Having a comfortable living environment became more pressing forpeople who were forced to work at home and for families spending long hours together underone roof."People are fearful of the future,especially with this pandemic,"Sharon said."Willit get worse?Will it return...I think it is a good idea to clear your home as a way of clearingyour mind.This is an ideal time to tidy up and give some thought to what we have and tobethankful for it.”16【23新教材老高考·D·英】 16.已知菱形ABCD的边长为1,∠ADC=子,将△ADC沿AC翻折,当三棱锥D-ABC表面积最大时,其内切球表面积为四、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)在△ABC中,角A,B,C的对边分别为a,b,c,已知(b-V3c)sinB+csinC=asinA.(1)求角A的大小:(2)若3D=AC,∠DBC=受,c=6,求△ABC的面积.18.(12分)设数列{an}的前n项和为Sn,且满足Sn=2an+3.(1)求数列{an}的通项公式:(2)证明:数列{an}中的任意不同的三项均不能构成等差数列.19.(12分)在四棱锥P-ABCD中,平面PAD⊥平面ABCD,ABIICD,AB=2,AD=DC=CB=1,PA=PD=2.(1)设平面PAB与平面PCD的交线为l,求证:I∥平面ABCD;(2)点E在棱PB上,直线AE与平面ABCD所成角为若,求点E到平面PCD的距离.数学试卷第4页(共6页) 得分评卷人24.(本小题满分10分)如图11,有两个长度相同的滑梯(即BC=EF),左边滑梯的高度AC与右边滑梯水平方向的长度DF相等.(1)求证:△ABC≌△DEF;:(2)若滑梯的长度BC=10米,DE=8米,分别求出滑梯BC与EF的坡度;(3)在(2)的条件下,由于EF太陡,在保持EF长不变的情况下,现在将点E向下移动,点F随之向右移动.①若点E向下移动的距离为1米,求滑梯EF底端F向右移动的距离;②在移动的过程中,直接写出△DEF面积的最大值.t恤eD图11 姓名准考证号2023年九年级教学质量监测数学试题注意事项:1.本试卷共6页,满分120分,考试时间120分钟。2.答卷前,芳生务必先将自己的姓名、准考证号频驾在试卷相应位置上。3.答案全部在答题卡上完成,答在本试卷上无效。4.考试结束后,务必将答题卡交回。一、选择题(在每小题给出的四个选项中,只有一项符合题目要求,请选出并在答题卡上将该项涂黑.每小题3分,共30分)1.-2023的相反数是A.-2023B.2023C.-2023D.20232.2023年的央视春晚舞美设计以“满庭芳”为主题,将中华文明的传统美学理念与现代科技相结合,令人耳县一新.演播厅顶部的大花造型,来源于中国传统纹样“宝相花”(如图).下列选项对其对称性的表述正确的是A.轴对称图形B.既是轴对称图形又是中心对称图形C.中心对称图形D.既不是轴对称图形又不是中心对称图形3.下列运算正确的是A.a'-a=aB.√8+2=42C.6a2÷2a=3aD.(a-3)2=a2-94.眼下正值春耕备耕关键时期,中国人民银行运城市中心支行指导辖内银行业金融机构将“支持春耕备耕”作为重点工作,多措并举,加大信贷投放力度.截至目前,辖内银行业金融机构共向春耕备耕领域投放贷款5.68万户共计27.8贺亿元,把数据“27.87亿”用科学记数法表示为A.27.87×108B.27.87×109C.2.787×109D.2.787×1010x-3<2x5.不等式组x+1≥¥】的解集正确的是3A.无解集B.x≥5C.-3 由题意,即满足h(t)mx一h(t)min≤a2十l.…6分因为a>0,所以h(t)为开口朝上的二次函数,对称轴为t=1一4Aa当≥2-1时4=102∈(-0.①当=1a2∈(-0,-1DU(1,+∞)时,此时0
大兴区20222023学年度第一学期高三期末检测英语参考答案第一部分知识运用(共两节,30分)第一节(共10小题;每小题1.5分,共15分)1.B2.C3.A4.C5.D6.B7.A8.D9.B10.C第二节(共10小题;每小题1.5分,共15分)11.have/has passed12.their13.feels14.amazing15.with16.that/which17.that18.to organize19.left20.global第二部分阅读理解(共两节,38分)第一节(共14小题;每小题2分,共28分)21.c22.A23.D24.D25.B26.A27.C28.D29.A30.A31.C32.C33.B34.C第二节(共5小题;每小题2分,共10分)35.B36.G37.C38.F39.E第三部分书面表达(共两节,32分)第一节(共4小题;第40、41题各2分,第42题3分,第43题5分,共12分)40.They found that the problem caused by pieces of trash humans left was worse than expected.(2分)That the problem caused by pieces of trash humans left was worse than expected.(2)41.Other plastics,especially disposable plastics lead to the increasing wastes on the beaches.(2分)42.Individual efforts and international rules are both required to tackle the plastic pollution,andthe U.N.pays much attention to calling for international action.最大划线范围如上,最小划线范围:pays much attention,划线正确给1分。According to the passage,the U.N.isn't paying enough attention to calling for internationalaction on plastic pollution.(2)43.略第二节(20分)One possible version:Dear Jim.How's everything?I'm writing to share my study and life at home with you since you want to know.I get up early as usual and begin my online courses.Our teachers design a variety of activities to getus involved in the class and we all cooperate with teachers actively.We do rhythm gymnastics tomusic between classes,which makes us feel relaxed and full of energy.In my free time,I help mymother with the housework,such as cleaning my room and cooking meal for my family.My life is fulfilling these days.What about you?Looking forward to your reply.Yours,LiHua 答案专期2022一2023学年广东专版九年级第13~16期品数学闭报MATHEMATICS WEEKLY即A,B两舰队的距离为8海甲(2)过点P作PF⊥AQ于点F易得1F=PE=20海里,PF=AE=28海里】所以QF=AQ-AF=2N29海里.在t△PQ中,由勾股定理,得PQ=PF+QF=28+(229=30(海里)所以30÷40=(小时).答:救援船到点P处所需的最短时问为小时。 湖北省高中名校联盟2022~2023学年度下学期高二联合测评数学试卷参考答案与评分细则题号12346789101112答案ADBBCBBCCDBDBCDACD1.A【解析1由y=2x-和切点P(1,2)可知,切线的斜率k=1,倾斜角a=牙,故选A.a1十a19+a1q2=7,2.D【解析】在递增等比数列中,由解得a1=1,g=2,则a,=8,故选D.a1a2a3=8,3,B【解析由随机变量等可能地取值可知:P1≤X≤3)=PX=)+PX=2+PX=3)号则P(X=1)=P(X=2)=P(X=3)=6,有P(X=1)=P(X=2)=P(X=3)=…=P(X=n)=名,由n×日=1得n=6,故选B4B【解析】X的所有可能取值为0,1,2,则P(X=)=CCC,k=0,1,2.所以P(X=0)=30P(X=1)=3,0+1x3P(X=2)=0E(X)=0x+2X10=0.8,故选B.另解,服从超几何分布,由公式n×心2X号-0,8,法R5.C【解析】设点M为F,关于渐近线y=一名x的对称点,则直线y=一,一。x垂直平分线段MF,交点设为N.在R△OF,N中:ON1=2MF,=5,F,N=b,则6+65)=c,而S=2,所以a=5,从面6=3,曲线C号号-1,做选C6.B【解析】先从三个抽屉中选择一个空抽屉C,接下来可能将四个小球分为2十2两组放入不同编号的抽屉中,也可能将四个小球分为3十1两组放人不同编号的抽屉.因此恰好有1个抽屉为空的不同放法有:C(CA:+CA)=2种。7.B【解析】“雹程”为:10→5→16→8→4→2→1.c【墙.。N府学ee,32构造函数F(x)-(x>0).则M=F(写.N=F(3),P=F(受.由F(z)气可知当0c 第四部分写作(共两节,满分35分)第一节短文改错共10小题;每小题1分,满分10分)The day before Mother's Day,everybody around me is talking about what gift they wouldwasbuy their beloved mother.Most student would buy flowers,and some would buy chocolates.studentsTherefore,I wanted to do something different.Because my home was far from school and I couldn'tHowevergo back home on Mother's Day,so I planned to make a call to my mom to send her our warmestmywishes.After finish all my homework that evening,I started to search the Internet for somefinishingwonderful words.Then all of the sudden,I realized it was complete unnecessary to copy someoneacompletelyelse's words.Finally,I decided ^tell my mom I loved her and to thank her on all the things she had6fordone for me.第二节书面表达满分25分)One possible version:Dear Simon,called.Yesterday an English speech contest whose theme was "Save the environment in China"was held in ourschool After a fierce competition,ten students from different classes entered the final.Eventually,my classmate LiMing won first place.All the contestants impressed me deeply with their wonderful performances.The competition made us realise that we can save the environment by starting small although we are students.For example,we can turn off the lights when we leave our classrooms.We believe that even the simplest actionssecan make a real difference to the environment.Yours,encourage the students to protect our environment.Li Hua 另一个交点在(©,+∞)内,若y=b与y=f(x)和y=g(x)共有三个不同的交点,则其中一个交点为两条曲线y=f(x)和y=g(x)的公共点,记其横坐标为x,令f(x2)=g(x)=f(nx2),则x2∈(1,e),lnx2∈(0,1),记y=b与y=f(x),y=g(x)的三个交点的横坐标从左到右依次为,x2,x4,且满足E<1< 0-…品后19.【详解】(1)设4C与BD相交于点O,连接FO因为四边形ADCD为菱形,所以AC L BD,且O为AC中点,因为FA=FC,所以AC⊥FO,又FOO BD=O,FO,BDC面BDEF,所以AC⊥平面BDEF,(2)连接DF,因为四边形BDEF为菱形,且∠DBF二60,所以△DBF为等边三角形,)【}因为O为BD中点,所以FO⊥BD,又AC⊥FO,BDOAC=O,AC,BDC面ABCD,所以FO⊥平面ABCD所以OA,OB,OF两两垂直,如图所示,建立空间直角坐标系O-z,B因为四边形ABCD为菱形,∠DAB=60,AB=2,所以BD=2,AC=2√3、因为△DBF为等边三角形,所以OF=√5,所以A5.,00,B0.L0),D(0-1,0),F0,0V5,所以D=(5,-,0,=(5,0),B=(51,0),设平面ABF的法向量为n=(名,y,2),则丽万=一5x+5=0,取,得=5列AB元=-5x+y=0以意(线0与r所为,则0w动小8品-雪 二、选择题: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。9.6个数据(x,y)构成的散点图,如图所示,采用一元线性回归模型建立经验回归方程,若在6个数据中去掉E(5,6)后,下列说法正确的是()A.解释变量x与预报变量y的相关性变强B.样本相关系数r变大C.残差平方和变小D.决定系数R2变小4(1,6),E(5,6)B2.5C6.4D4,4F(6,3)【答案】AC【解析】去掉E(5,6)后,变量x与预报变量y的相关性变强,故A正确;但由于散点的分布是从左上到右下,故变量x,y负相关,所以相关系数r变小,残差平方和变小,决定系数R2变大,C正确,D错误,故选:AC.10.若a,b>0,且a+b=1,则()A.Va+b≤√2B.+49a bc.a2+462>D.b2 a2十【答案】ABD【解析】由基本不等式:(Va+b=a+b+2Wab=1+2ab<1+a+b=2→Va+Vb≤V2,A正确:+42+4}a+)=1+4+b+4a≥5+2b.4a'b a b)a bVa b=9.B正确:,124、4a2+462=(1-b)2+4b2=5b2-2b+1=5b-5)+55,C不正确:+4+1=+aa ba b+a+b≥2a.2+2b.号-=4,D正确11.在△ABC中,角A,B,C所对的边分别为a,b,c,已知(a+b):(b+c):(c+a)=5:6:7,则下列结论正确的是()A.sinA:sinB sinC =2:3:4B.△ABC为钝角三角形C.若a=6,则△ABC的面积是6√15D.若△ABC外接圆半径是R,内切圆半径为r,则R=1【答案】BD【解析】设a+b=5t,b+c=6t,c+a=7t,则a=3t,b=2,c=4t,对于A,sinA:sinB:sinC=3:2:4,故A不正确;对于B,C的二子0,装B正商2ab对于C,若a=6,则1=2,b=4,c=8,b=10,所以c0sC=-4→sinc=i54 全国@0所名校高三单元测试示范卷札记全国@⊙所名校高三单元测试示范卷·数学第四单元指数函数、对数函数、幂函数(120分钟150分)考情分析微信扫码指数函数、对数函数在高考中是必考点,客观题、主观题都可能出现,常与导数知识◇高考对接点相结合;幂函数是冷考点单元疑难点三种函数模型的应用滚动知识点函数的概念及性质观看微课视频课外习题解析典型情境题5、15、18下载复习课件题序101112答案DCDBDC ACD CDBC ACD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.下列函数,在R上单调递减的是A.y=x3-2xB.y=xC.y=-lg xD.y=-3【解题分析】易知y=x3-2x在R上不单调,A项错误;y=x在R上单调递增,B项错误;y=-lgx的定义域为(0,十∞),不合题意,C项错误:y=一3在R上单调递减,D项正确.【答案】D2.已知实数m,n∈(一∞,0)U(0,十∞),且m 全国100所名校高三AB测试示范卷札记全国@迥所名校高三AB测试示范卷·数学第四套函数的概念及性质(A卷)(40分钟100分)眼微信扫码考情分析,口新高考对接点函数的概念及性质是高考必考点,主要考查客观题学习疑难点函数性质的应用典型情境题10观看微课视频课外习题解析下载复习课件题序12356答案DABDDBC一、选择题:本题共6小题,每小题6分,共36分1.函数f(x)=x十1的定义域为A.[-1,+o∞)B.(-1,0)U(0,+)C.(0,+∞)D.[-1,0)U(0,+∞)|x+1≥0【解题分析】由得x≥一1且x≠0,所以函数f(x)的定义域为[一1,0)U(0,十o∞)x≠0【答案】D2.设函数f(x)=十1,则下列函数为奇函数的是A.f(x)-1B.f(.x-1)C.f(.x-1)-1D.f(.x)+1【解题分析】因为f)=1-1十,所以)-1=1十上-1=上,故函数()1为奇函数.y【答案】A3.设a∈R,则“a>0”是“函数f(x)=-x2十2a.x在(一∞,1]上单调递增”的A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件【解题分析】函数f(x)=-x2十2a.x的图象的对称轴为x=a,又f(.x)在(-o∞,1上单调递增,所以a≥1,所以“a>0”是“函数f(x)=一x2十2a.x在(-oo,1]上单调递增”的必要不充分条件.【答案B4.若函数y=f(x)的大致图象如图所示,则f(.x)的解析式可能是A.f(x)-x2-1B.f)=1+x21-x2C.f(x)=1-元xx2D.f(x)=]-2【解题分析】当一1 则+2=-12a,y2=9a2,直线DR的方程y-y2=3m2-13m2-12y1(x-9,2x1aa3令y=0,得x=1y22y1_(my12a)y22y1_my1y2+2ay22y1_a(y1+y2)+2ay22y1Y2-y1Y2-y1Y2-Y1y2-y12-y5a,所以直线DR过定点(5a,0).Y2-Y12【解谷】解,(a+(a-11a0.x=女是(ta-1Dx x2①当a≤0时,f(x)<0,.函数f(x)在(0,+∞)上单调递减:②当a>0时,由f(x)>0,得x>1,由f(x)<0,得0 老法最新中考模拟训练·数学(五)22.在抛物线y=Qx2十bx十c中,函数y与自变量x的部分对应值如下表所示最新中考模拟训练·数学(五)x…01六解答题(本大题共12分)343,课本再现30-103日)如图,CD与BE相交于点A,AABC是等腰直角三角形,∠C=9o°,若DE∥BC,求证:(1)求该抛物线的解析式.△ADE是等腰直角三角形,类比探究(2)设P为抛物线上一动点,且关于点O(0,0)的对称点为P',请在下图所示的平面直角坐标系中描出相应的点P,再把相应的点P用平滑的曲线连接起来,猜想该曲线是何种曲2)①如图2,AB是等腰直角△ACB的斜边,G为边AB的中点,E是BA的延长线上一动点,过点E分别作AC与BC的垂线,垂足分别为D,F,顺次连接DG,GF,FD,得到线,并直接写出点P'的纵坐标的最大值及该曲线与x轴的交点坐标.(3)点M(x,y),N(x2,2)在抛物线y=a2+bx十c上,且对于I 【小问1详解】41取倒数得8+2=2台1-上+1.即1-11解:由、1an+22da doat dd+2即adne an 2所以a为公的外时”,商8之n+1【小问2详解】1n-解:当时,21≤k+1<2”台2”-1≤k<21-1,2所以,满足条件的整数k的个数为(21-)-(2”-1)=2”,即b,=2”,所以,=(n+1)-2>0,故数列{S}单调递增,所以,Sn=2×2°+3×2+4×22+…+(n+1)×2"-1,则2Sn=2×2+3×22+…+n×2"-1+(n+1)×2",上式-下式得-Sn=2+(2+22++2"1)-n+1)×2”=2+21-2-)n+1k2”1-2=-n×2”,所以,Sn=n2”,因为S2=7×2=896,S8=8×23=2048,则S7<2023 分10分)22.如图所示,A,B两点在河的两岸,为了测量A,B之间的距离,测量者在A的同侧选定一点C,测已AB⊥平出A,C之间的距离是100m,∠BAC=105°,∠ACB=45°,则A,B两点之间的距离为2m.MN∥平面B(39;=1,BC=M.m40545CC0L平/60三、解答题:本大题共3小题,共30分.解答应写出文字说明、证明过程或演算步骤。m/423.(本小题满分10分)-MN某校从参加高三模拟考试的学生中随机抽取60名学生,将其数学成绩(单位:分,均为整数)分成六段[90,100),[100,110),…,[140,150]后得到如图所示的频率分布直方图.根据图中的信息,回答下列问题:小题满分(1)求分数在[120,130)内的频率,并补全这个频率分布直方图;知奇函数(2)估计本次数学考试成绩的70%分位数;(3)用分层随机抽样的方法在分数段为[110,130)的学生中抽取一个容量为6的样本,将该样本1)求函赞看成一个总体,从中任取2人,求至多有1人在分数段[120,130)内的概率.2)试判ff1)ed-0.Dl-ms-o.心-频率组距0.035(3)当-0.030.0300.0250.020解2)130.0150.0100.005(30,12以:2小0901001101201140150分数C2只3以P。0,食共,代圆小阳,朝小共魔火本就空服,雪城,游数您家,出管出学,后将资式:一十际的过铁时求小日做料前西的日容个意人芳科育.识,安国上的所人生数学·模拟卷三第4页(共5页) X45X30%C DE六、(本题满分12分)21.解:1PC刘伯被分配到副主任医师诊室就诊)=名-日:…(4分)(2)用A表示主任医师诊室,B表示副主任医师诊室,C表示主治医师诊室,用树状图分析如下:B,B。C,C.A B.C.个A B B.C.C A B B:CC A B B:C.C一共有30种不同的结果,其中一人被配到副主任医师诊室、一人被配到主治医师诊室就诊有12种情况,所以P(一人被配到副主任医师诊室、一人被配到主治医师诊室就诊)-品-号…(12分)七、(本题满分12分)22.解:(1)①.点(-2,9)在二次函数y=mx2-4m2x-3的图象上,∴.9=4m十8m2-3,整理得2m2十m-3=0,解得m=1或m=一号,m>0m=1:…(4分)②由①得y=x2一4x一3=(x一2)2一7,∴.抛物线的对称轴为x=2,顶点(2,一7),当y=18时,(x-2)2-7=18,解得x=7或x=-3,.当0≤x≤a时,y的最大值为l8,∴.a=7;…(8分)》(2),二次函数y=mx2一4m2x一3,.对称轴为x=2m,抛物线与y轴的交点为(0,一3),.m>0,∴.对称轴x=2m>0,点P(xp,yp)是该函数图象上一点,当0≤x≤4时,yp≤一3,∴.当x=4时,y≤-3,即m·42-4m2·4-3≤-3,16m-16m2≤0,m>0,∴.m≥1.…(12分)八、(本题满分14分)23.解:(1)∠ECF=45°,EF⊥CE,.EC=EF,在△BDE中,∴.ED=EB,∠BED=90°=∠CEF,∴∠BED-∠DEC=∠CEF-∠DEC,∴∠BEC=∠DEF,.△BCE≌△DFE(SAS);…(4分)中考必刷卷·2023年名校押题卷数学参考答案及评分标准第3页(共4页) 教材改编题腾远原创好题建议用时:10分钟3.(新人教A版选择性必修第一册P124T4)<一题多变式3-1变情境:已知渐近线斜率及双曲线上一点的坐标已南吸数奇1e以月务等E率人好,温水,如411在双曲线上,则双曲线方程为3-2变情境:已知焦点到渐近线的距离已知双曲线兰。=1(a>0,b>0)的一条渐近线斜率为k(6>0),离心率e=25,且焦点到该新近线的距离为2,则k=4.(新人教A版选择性必修第一册P126例6)<|一题多变式4-1变情境:已知垂直关系已知双曲线C:x2-y2=4的左、右焦点为F1,F2,过F2的直线1与C在第一象限的交点为A,若AF1⊥AF2,则△AFF2的周长为4-2变设问:求弦长之积断带法注重创新性·新设问已知双曲线C:看61的两个焦点分别为R,,点M在C上,0为坐标原点,若OM=10,则MF1·MF2|=5.新考法注重创新性·新设问(新人教A版选择性必修第一册P127T3改编)已知双曲线C:x-。=1的左、右焦点分别为F,F,过F,的直线交C的左支于M,N两点,则点M到直8线n:22x-y-3=0距离的取值范围为真题变式题腾远原创好题建议用时:30分钟6(2021年金国甲卷)点(3,0)到双曲线。=1的一条渐近线的距离为169A.5986B.C.D.4555<引一题多变式6-1变设问:求离心率已知双曲线C:。示=1(a>0,b>0)的右焦点到-条渐近线的距离是右顶点到该渐近线距离的2倍,则C的离心率为A.1B.2C.3D.4121 - 7.2022年4月16日,神舟十三号载人飞船返回舱在东风着陆场成功着陆,返回舱呈钟形,将其近似地看作一个半球(上)和一个圆台(下)的组合体,其中半球的半径为1米,圆台的上底面与半球的底面重合,下底面半径为1.2米,若圆台的体积是半球的体积的2倍,则圆台的高约为8思A.1.0米B.1.1米C.1.2米D.1.3米8.将一枚质地均匀的骰子随机抛掷两次,甲表示事件“第一次点数为奇数”,乙表示事件“第二次点数为偶数”,丙表示“两次点数相同”,丁表示“两次点数之和为偶数”,则下列选项中的两个事件不相互独立的是A.甲与丙B.乙与丙C.乙与丁D.丙与丁二、多项选择题:本题共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,部分选对的得2分,有选错的得0分9.已知复数z满足z(1+)=-1-√5+(1-√5)i,则下列说法错误的是A.z的虚部为iB.z的共轭复数z=√5+iC.lal =2D.z2=4-25i10.某高中一年级共有甲、乙、丙3个班级,其中甲班40人,乙班50人,丙班40人,在某次数学月考中,甲班的及格率为50%,乙班的及格率为60%,丙班的及格率为70%,则A.若用简单随机抽样法从一年级所有学生中抽取13人,则甲班应抽取4人B.若按照各班人数比例用分层随机抽样法从一年级所有学生中抽取26人,则丙班应抽取8人C.这次一年级数学月考的平均及格率为60%D.若从这次一年级数学月考及格的学生中随机抽1人,则该学生来自丙班的概率最大11.小张于2017年底贷款购置了2018年各项支出2022年各项支出一套房子,根据家庭收入情水、电、娱乐2%气、通讯其交车贷况,小张选择了10年期的等9%其他109%10%额本息的还贷方式(每月还饮食房贷娱乐10%房贷款数额相等),2021年底贷款30%24%60%购置了一辆小汽车,且截至飘饮食2022年底,他没有再购买第23%10%二套房子.如图是2018年和2022年小张的家庭的各项支出占家庭收入的比例分配图.根据以上信息,判断下列结论中正确的是A.小张一家2022年的家庭收入比2018年增加了1倍B.小张一家2022年用于娱乐的支出费用为2018年的5倍C.小张一家2022年用于饮食的支出费用小于2018年D.小张一家2022年用于车贷的支出费用小于2018年用于饮食的支出费用数学试题第2页(共4页)扫描全能王创建 2022/2023学年度第二学期高二年级期终考试4+时的=现民开式中然数动数学试题A.1B.2C.4D,6(总分150分,考试时间120分钟)5.若抛物线y2=4x上的一点M到坐标原点O的距离为√5,则点M到该地物线焦点的注意事项:距离为1.本试卷考试时问为120分钟,试卷满分150分,考试形式闭卷.A号B.1C.2D.32.本试卷中所有试题必须作答在答题卡上规定的位置,否则不给分3.答题前,务必将自己的姓名、准考证号用0.5色米R色墨水签字笔填写在试卷及答6.某班经典阅读小组有5名成员,暑假期间他们每个人阅读的书本数分别如下:3,5,4,题卡上2,1,则这组数据的上四分位数为一、单选题:(本大题共8小题,每小题5分,计40分.每小题给出的四个选项中,只有A.2B.3C.3.5D.4一项是符合要求的,请在答题卡的指定位置填涂答案选项。)7.在坐标平面内,与点A(1,2)距窝为3,且与点B(3,2)距离为1的直线共有1.如果随机变量X~B(3,),那么E(X)等于A.1条B.2条C.3条D.4条A,38.如图所示,在矩形ABCD中,AB=2BC=4,动点M在以点C为圆心且与BD相切的B.1C.-2D.3圆上,则BM.BD的最大值是2.为了解双减政策的执行情况,某地教育主管部门安排甲、乙、丙三个人到两所学校进A.4B.-1D行调研,每个学校至少安排一人,则不同的安排方法有C.1D.12A.6种B.8种C.9种D.12种(第8题图)3.把红、黑、白、蓝4张纸牌随机地分给甲、乙2个人,每个人分得2张,事件“甲分得红牌和蓝牌”与“乙分得红牌和黑牌”是二、多选题:(本大题共4小题,每小题5分,计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,部分选对的得2分,有选错的得0分,请在A.对立事件B.不可能事件答题卡的指定位置填涂答案选项.)C.互斥但不对立事件D.以上均不对又。下列关于双曲线「·y=1的判断,正确的是高二数学试题第1页(共8页)高二数学试题第2页(共8页) Section BI.1.teaches 2.musician 3.home 4.people5.todayII.1.On the weekend2.the Students'Sports Center 3.is good with4.also want to play the drum(s)5.helps English-speaking students with6.make friends with herⅢ.1-5 ECFDBSelf Check1.Can 2.tell 3.club 4.musician 5.teaches6.easy 7.so 8.also 9.with 10.join第3一4版Unit1综合能力评估试题听力材料及参考答案听力材料I.听句子,选出与其意思相符的图片。1.I like to write in my free time.2.Ben wants to join the kung fu club this term.3.The school library has a book sale on the weekend.4.My classmate Sam can swim very well.5.Ms.Green always gets home at eight p.m.on school days.Ⅱ.听对话,选择最佳答案。6.W:Wu Fan,what club do you want to join?M:The chess club.But my mom wants me to join the English club7.M:Tracy,are you and your sister at home now?W:No,Dad.I'm in the sports center and Kelsey is in the library.8.W:Hi,Alan.Let's go to the talk show on the weekend!M:OK!How about this Sunday?I have a family trip on Saturday.9.M:Can you play the violin,Britney?W:Yes,but I'm not good at playing it.I can play the guitar well.10.W:Is the music show today?M:Yes,from six p.m.to ten p.m.The drums show is at eight.Ⅲ.听短文,完成表格。Good morning,boys and girls.This is the student center.This term,we have a storybooksale in the library on March 5th.You can buy some interesting storybooks at great prices.Do you like basketball?From May 10th to May 15th,we have basketball games in thesports club.On the second Friday of June,some Chinese students have a kung fu showin the art center.Do you want to come and watch it?Please call Mr.King at 228-3601.参考答案I 1-5 CACBBⅡ.6-10 BBCBBII.11.March 12.15th 13.basketball14.Friday15.3601IV.16-20 BADDB 21-25 ACCAD 7.已知函数f)=sin2ax+爱+4sin2ox(oeN),若关于x的方程f)=2在[o,孕上6有且只有一个解,则0为A.1B.2C.3D.48.在△ABC中,若sinA+sinB=V3sinC,则cos2C的最小值是A.1B.c.-7D.-19二、多项选择题: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.在复平面内,点Z(2,1)对应的复数为z,则A.|z=5B.z+7=4C.Z=5D.1-21iz3310.由均匀材质制成的一个正12面体,每个面上分别印有0,1,2,3,4,5,6,7,8,9,V,×.投掷这个正12面体2次,把朝上一面的数字或符号作为投掷结果.则A.第一次结果为数字和第一次结果为符号互斥B.第一次结果为数字与第二次结果为符号不独立C.第一次结果为奇数的概率等于第一次结果为偶数的概率D.两次结果都为数字,且数字之和为6的概率为211.如图,函数f(x)=sin(ox+p(o>0,pK的图象交坐标轴于点B,C,D,直线BC与出线y=f(:)的另一交点为A.若C(分0),D(2,0),则A.函数f(x)在[3,4]上单调递增B.直线r=17是函数)图象的一条对称轴42C.sin∠BAD=Γ3D.将y=Co2π的图象向右平移个单位长度,能得到函数f)的图象4高一数学试卷第2页(共4页) endorsement of those demands,a practice that continues to be important today.In 2020,Democratic presidential candidate Joseph Biden made a promise to select a woman as his vice-presidential nominee()and ultimately shared electoral victory with running mate KamalaHarris,the first woman and first person of color to be elected vice president.Men's voices are important.When men speak up against gender discrimination,they not onlybecome obvious as allies who can be counted on to support industry or company rules to advanceequality,but they also improve awareness and acceptance of gender inequality as a shared problem,not a special interest.28.What does paragraph 3 mainly talk about?A.The reason why gender equality is hard to achieve.B.The reason why men are powerful at work.C.The result that gender inequality brings to men.D.The result that women's advancement causes to companies.29.Which of the following best explains "endorsement"underlined in paragraph 4?A.Responsibility.B.Support.C.Ignorance.D.Misunderstanding.30.Which can be the best title for the text?A.Support of Men Is on the Sidelines.B.Gender Equality Is Women's Business.C.Gender Inequality Is a Common ProblemD.The Secret to Achieving Gender Equality at Work:Men31.In which section of a newspaper may you find this text?A.Education.B.Society.C.Health.D.Entertainment.DDuring a ransomware(勒索软件)attack,attackers enter a target's computer system andencrypt()its data.They then demand a payment before they will free the system.Ransomware is a collective problem,and solving it will require joint action from companies,theU.S.government and international partners.In 2020 the Federal Bureau of Investigation received more than 2,400 reports of ransomwareattacks,which cost victims at least $29 million.The numbers underestimate the total impact ofransomware because not all organizations are willing to report it when they fall victim to this kind ofcrime.Even these limited statistics,however,show the increasing fearlessness of ransomwareattackers:the number of attacks in 2020 increased by 20 percent compared with that of the previousyear,and the amount of money paid out more than tripled.As long as victims keep paying,attackers will keep profiting from this type of attack.Butcybersecurity experts are divided on whether the government should prohibit the paying of ransoms.Such a ban would discourage attackers,but it would also place some organizations in a moraldilemma.For,say,a hospital,unlocking the computer systems as quickly as possible could be amatter of life and death for patients,and the fastest option may be to pay up.Other solutions are more straightforward and involve pushing organizations to protectthemselves better.Cybersecurity defenses make it harder for attackers to access systems.Segmenting()one's network means that breaking through to one part of the system does notmake all data immediately available.And regular backups allow a company to function even if itsoriginal data are encrypted.All these measures,however,require resources that not all organizations have access to.Meanwhile,ransomware people are adopting increasingly advanced techniques.Some work forweeks to gain entry to a company's network and then enter their system,finding the most vital datato hold hostage().Some groups deliberately compromise an organization's data backups.英语试题第5页(共8页) ◆0=品0方>0,于是h(t)在(0,1)上单调递增,21(t+1)221(t+1)2又hI0,所以h0=Int-二<0,从而-Int,t+1t+12得为-无-<为.10分Inx Inx2 2于是③武可化为1-1-a血-ln2a,得x+x>2加++5>2a.Xx2x1-x3x1+x2X+X2>2☑得证.……12分十校高二数学6(共6页) 5.What is the man doing?5AA.Making an invitation芳关在报社B.Offering a suggestion.C.Asking for permissionEARNING ENGLISHText 5新课程M:Have you been to the Science Museum lately?W:No,why?高二课标基础M:It has a wonderful new exhibit.I'm going againTEST YOURSELF第36期on Saturday.Would you like to join me?(5)Unit 3,(选择性必修)Book3W:Yes,I'd love to.Let's go in the morning so itwon't be too crowded.第一节第二节1.What will the man do first?1 B听第6段材料,回答第6、7题。46-7ABA.Go to his office.B.Return a book6.Why does the woman make the call?听C.See a doctor.A.To contact her friend.Text 1B.To confirm a phone number.W:Hello,Tom.Are you going to the office today?C.To seek information about the hotel.M:Yes,but I'll go to the library to return the7.What is Garden Hotel's phone number?book first,(1)and then I'll go to the doctor.A.23371600.B.23317600.What for?C.23361700.Text 6听力材料M:Hello,Garden Hotel.How can I help you?2.Where are the speakers?2CB.At a hotel.W:Oh,I'm sorry.I was trying to reach my friendA.At home.Anne.(6)C.At a department store.M:There's no one here by that name.What numberText 2were you trying to call?W:2-3-3-1-7-6-0-0.(7)W:Which floor,sir?M:That's our number.(7)Maybe you wrote downM:I'm not sure.Where can I find pillows?(2)the wrong number.You did say 2-3-3-1-7-6-0-0,W:You can find everything you need for yourdidn't you?bedroom on the fifth floor.(2)W:Yes,that's what I have.Sorry to bother you.Goodbye.3.What does the man think of Jessica's3A听第7段材料,回答第8至10题。《8-10ABBperformance in the film?8.What is the probable relationship between theA.Not very satisfactory.speakers?B.So disappointing.C.ExcellentA.Classmates.B.Mother and son.Text 3C.Doctor and patientM:How do you find Jessica's performance in the9.What does the woman try to persuade the man todo?film?A.Join the basketball teamW:Honestly speaking,she was not doing well asexpected.(3)B.Have a physical check-up.M:I'm afraid so.(3)But my brother thinks sheC.Develop a sense of teamwork.10.What day is it today?did a good job.A.Friday.B.Saturday.C.Sunday.听力材料4.How long will the man rent the house?4 CA.For a month.Text 7B.For three monthsC.For six months.M:I'm as healthy as a horse.Anyone can see that.Text 4W:John,you never know if everything is OK untilyou have checked it,do you?(9)M:How much should I pay,Mrs.Black?M:Do you really think I look sick?W:$300,if you just rent the house for a month.W:Of course not,but that's not the point.All weBut you can have a 5%discount for a quarter.want is to be sure you won't fall down suddenlyM:Oh,no,no,no.Half a year.(4)on the basketball court.(9)W:Then I'll give you a 10%discount. 【答案D【解析因为正六棱台的上下底面为正六边形,所以S=6×5×1?=3y5,S、=6×5×22=65.424V楼台=1/38+63+V2y×w3)×2W5=21,由祖暅原理知该几何体的体积也为21,故选D6.已知平面向量a,b满足b=1,a·b=-2,则3a-b在b上的投影向量为A.76B.-7bC.5bD.-5b【答案】B(3a-.6【解析】3a-b在b上.的投影向量为3a.-×6==6-1×6=-7元,故选B7.已知a=,b=in号,c=c0s3则a,6.c的大小关系为(A.cb,cos3V3所以c>a,综上c>a>b3故选C8.用长度分别为2,3,4,5,6(单位:cm)的5根细木棒围成一个三角形(允许连接,但不允许折断),能够得到的三角形的最大面积为A.8v5cm2B.6v10cmC.15v2cm2D.3v55 cm【答案B【解析】因为三角形的周长为20,所以三角形越接近等边三角形,面积越大,所以三边长为6,7,7时面积最大此时边长为6的边上的高为√72-3=2√10,面积为S=2×6×2W0=6W10,故选B二、选择题: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.已知一组数据4,2,a,10,7的平均数为5,则此组数据的A.众数为2B.中位数为4C.极差为3D.方差为4s5【答案】ABD【解析】由题意可得4+2+Q+10+7-5→4-2所以A正确:B正确,极差为10-2-8,C错误5对于Ds-4-5P+2-)+2-P+10-54(7-5-48,D正确55故选ABD10.下列条件中能推导出△ABC一定是锐角三角形的有A.AB.AC>0B.sinA:sinB:sinC=4:5:6C.cosAcos BcosC>0D.tanA·tanB=2【答案】BCD【解析】对于A,只能得到A为锐角,A不正确对于B:最大角为C,且cosC=a2+c2=+5-6>0,最大角C为锐角,所以一定是锐角三角形,B2ab2×4×5 2023北京西城高三二模数学2023.5本试卷共6页,150分。考试时长120分钟。考生务必将答案答在答题卡上,在试卷上作答无效。考试结束后,将本试卷和答题卡一并交回。第一部分(选择题共40分)一、选择题共10小题,每小题4分,共40分。在每小题列出的四个选项中,选出符合题目要求的一项。(1)复数z=i(1+i)的虚部为(A)1(B)-1(c)i(D)-i(2)已知集合A={x|-1≤x≤1},B={x|3<1},则AUB=(A)[-1,0)(B)(-0,0)(C)[-1,1](D)(-0,1](3)已知抛物线C与抛物线y2=4x关于y轴对称,则C的准线方程是(A)x=-2(B)x=2(C)x=-1(D)x=1(4)在△ABC中,AB=AC=1,∠A=90°,则AB.BC=(A)1(B)-1(C)5(D)-√2(5)设a=g子6-e3g2,c-g6,则(A)a
DAre you a good judge of character?Can you make an accurate judgment of people's personalities based only onyour first impression of them?Ironically,the answer lies as much in them as it does in you.US psychologist Henry Adams tried to identify good judges of character in 1927.His research led him toconclude that people fell into two groups-good judges of themselves and good judges of others.Adams's researchhas been widely criticized since then,but he wasn't entirely wrong about there being two clearly different types.We need to define what a good judge of character is.Is it someone who can read personality or someone whocan read emotion?Those are two different skills.Emotions such as anger or joy or sadness can generate easilyidentifiable physical signs.Most of us would probably be able to accurately identify these signs,even in astranger.As such,most of us are probably good judges of emotion.In order to be a good judge of personality,however,there needs to be an interaction with the other person,andthat person needs to be a "good target"."Good targets"are people who show related and useful clues to theirpersonality.So this means "the good judge"will only appear when reading "good targets".This is according toRogers and Biesanz in their 2019 journal entitled "Reassessing the Good Judge of Personality"."We found clear andstrong evidence that the good judge does exist",Rogers and Biesanz concluded.But their key finding is that thegood judge does not have magical gifts of observation-they are simply able to"detect and use information providedby the good target”.So,are first impressions really accurate?Well,if you're a good judge talking to a "good target",then it seemsthe answer is"yes".And now we know that good judges probably do exist,more researches can be done into howthey read personality,what kind of people they are-and whether their skills can be taught.32.What is the conclusion of Adams's research?A.Fewer people can read physical signs.B.Most people are good judges of themselves.C.First impressions have a huge effect on people.D.There are two obviously different types of people.33.What can be learned from Paragraph 4?A.There is no need to interact with the other person.B.Good judges are related to good targets.C.Good judges have magical talents for emotion.D.Good targets are persons who hide important information.34.What is the author's attitude towards the research on good judges?A.Disapproving.B.Indifferent.C.Objective.D.Pessimistic.35.What is the author's purpose in writing the text?A.To inform readers of a good judge of character.B.To highlight the importance of good character.C.To introduce some ways of define good characters.D.To explain what first impressions are.第二节(共5小题;每小题2.5分,满分12.5分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。选项中有两项为多余选项。高二(上)期末联合检测试卷(英语)第5页共14页 2022~2023学年八年级下学期期未质量检测试题数学注意事项:1.满分120分,答题时间为120分钟。2.请将各题答案填写在答题卡上。一、选择题(本大题共10小题,每小题3分,满分30分)每小题都给出A,B.C、D四个选项,其中只有一个是符合题目要求的,1.下列函数中,y是x的一次函数的是A.y=224z231192.下列各式计算正确的是B.y=x2+2超C.y=3+2xD.y=-5A.√4÷√2=2啟C.23-3=2B.(W5-1)(W5+1)=4戡D.√/-4)×(-25)=√-4X√-25长3.直角三角形的三边长分别为3,4,x,则x的值可能为A.5B.7C.7D.5或√/7K4.在平行四边形ABCD中,∠A=45°,则∠C的度数为A.30°B.455.八年级(1)班的期末综合成绩按照课堂表现、作业成绩、考试成绩2:3:5的比例计算,小明C.60°D.135的课堂表现为80分,作业成绩为90分,考试成绩为85分,那么小明的期末综合成绩为A.85分B.85.5分骗6.如图,菱形ABCD中,E,F分别是AB,AC的中点,若菱形ABCD的周长为24,则EF的长为C.86分D.86.5分A.6B.8C.3相D.4yar+b16m21415m2第6题图第7题图7.一次函数y=a.x十b的图象如图所示,当ax十b>0时,x的取值范围是第8题图A.x>1C.x>0B.x<1D.x<0三16.器&.如图所示的是丽丽家正方形后院的示意图,丽丽家打算在正方形后院打造一个15m的正方形游泳池和一个6m的正方形花园,剩下阴影部分铺满瓷砖,则阴影部分的面积为A.6√/10mB.21m2C.2√15m2D.46 m2【数学第1页(共6页)】·23-CZ2326· 2l.DShi和她的丈夫(husband)Li全心全意为他们的学生们提供优质(quality)教育。22.C见上题解析。23.A他们鼓励学生们多参加与科技创新相关的(related to)课外活动。24.C在过去的十几年里,Shi指导(led)学生参加各种竞赛。25.B其中,有部分学生获得了奖项(awards)。26.DLi的履历同样(equally)惊人。27.A他在一所大学任教,教授(teaching)煤炭开采。28.B他经常深人煤矿中最危险的(dangerous)地方开展(conducting)科研工作。29.D见上题解析。30.B他测量数据、收集样本,回到(returns)地面后继续完成数据分析(analysis)和核算,为眼下存在的问题制定解决方案。31.C见上题解析32.C在专注工作(work)、不畏困难方面,他为学生们树立了良好的榜样(sets an example)。33.A见上题解析。34.B鼓励学生们参加(enter)各种竞赛是Shi和她丈夫激励(inspire)学生的一种方法。35.D见上题解析。36.A学生们以此获得成就感,他们的成就(success)也激励他们更加努力学习。37.DShi说教育就像一次寻宝旅程(journey)。38.B她认为(believes)对教师而言,发现并欣赏(appreciate)学生的兴趣和长处至关重要,以此激励学生为自己梦想的精彩(fantastic)生活而奋斗。39.C见上题解析。40.A见第38题解析。【答案与解析】本文是一篇说明文。研究表明,每日饮茶有助于降低心脏病等引发的死亡风险4l.is考查动词的时态和主谓一致。本段是一般性陈述,采用一般现在时,又因为主语是Drinking two ormore cups of black tea a day(为动名词短语作主语),故此处用所给be动词的第三人称单数形式。42.published考查非谓语动词。分析该句成分可知,所填词在此处作后置定语,修饰study,且publish与study之间存在逻辑上的动宾关系,故用过去分词published。43.in考查介词。participate in意为“参与.”,为固定搭配。44.their考查代词。根据空后的own和句意可知,此处用所给代词的形容词性物主代词修饰空后的名词短语tea drinking habits。45.frequency考查词形转换。根据句意及空后what they added to their cup可知,此处用所给形容词的名词形式frequency.46.regularly考查词形转换。此处用所给形容词的副词形式修饰前面的动词drank。47.original考查词形转换。根据空前的the和空后的survey可知,此处用所给名词的形容词形式original。48.to die考查非谓语动词。be likely to do sth意为“有可能做某事”,为固定搭配。49.an考查冠词。此处是泛指,又exciting为元音音素开头,故用不定冠词an。50.that/which考查定语从句。分析该句结构可知,该句是一个限制性定语从句,从句中缺少主语,又因为先行词为work,故用关系代词that或which。短文改错Yesterday I took my brother to a Father's Day event in the city library.We quickly checked in and foundtwo seat.At A beginning,a lady shared a popular origin of Father's Day with us.A few minutes late,sheseatsthelatertold a true story.It's about a selfish father who sacrificed herself in the flood for the life of his daughter.Itselflesshimselfmoved to us deeply.Finally,she gave each of us a bag of colored paper or taught us what to make a beautifulandhowcard.On received the gift,my father say nothing but gave us a big hug,with tears rolling down.receivingsaid书面表达One possible version:A Meaningful Fire DrillLast Friday,our school held a fire drill to raise the awareness of protecting ourselves in a fire.It took placein the classrooms and on the sports ground.With thick smoke rising from the windows,the sirens sounded,and every student was guided to rundownstairs towards the sports ground.We were told to cover our noses and mouths with wet towels or the liketo avoid breathing in smoke.After everyone turned up safe and sound,an officer from the fire service gave us aspeech,from which we further knew about the fire rescue work.I believe everyone will benefit a lot from such a drill.【高三开学质量检测考试·英语试卷参考答案第2页(共2页)】233402Z - treat sb to sth意为“使某人获得某种享受”。 世超金奥售假乐园自我评价年月☆☆☆☆☆六、找朋友(连一连)。456÷4905÷5612÷2285÷5156÷4814÷23064075711418139七、解决问题。工厂新进了89片扇叶,每台电扇需装3片。这些扇叶可以装多少台电扇?八、动脑筋,想一想,并把题补充完整。36☐465610 18.解,1)油Dsa十B》=8osos月5 in asin-6 o一青-3,…4分1+1=7得cos acos3十4=126分7+1=5(2)因为cos(a一B)=cosac0sB+sin asin月-i2十1=6,…9分所以0s(2a-29》=6os2a-》=2asa-0-1=2×器-1=S7…12分20.解:(1)由正弦定理得sin Acos B=sinB+sinC,22sin Acos B=sin B+2sin C.2分因为sinC=sin(A+B)=sin Acos B+cos Asin B,3分所以SinB十2c0 s Asin B=0,…4分因为snB≠0,所以c0sA=一号,5分又A∈(0,),所以A=236分(2)因为2sinB=sinC,所以2b=c,7分由Sam十SaD=Saw,得2c·AD·sin∠BAD+b·AD·sin∠DAC=?inA,8分得c十b7bc.又2b=C,解得b=3,0=6,10分则a=VB+2-2bcc0sA=√9+36-2X3X6Xc0s2=3V7,.11分3所以△ABC的周长为6+3+3√7=9+3√7.…12分21.解:(1)如图1,取HE的中点N,连接IN.如图2,连接EG,HF,设EG,HF的交点为O,连接PO.H图1图2由题意得HI=EI,'.PE=IE=√TN2十EN2=√5.1分易得四棱锥P一EFGH为正四棱锥,∴.POL平面EFGH,…3分.PE与底面EFGH所成的角为∠PEO.…4分【高一数学·参考答案第3页(共4页)】·23-526A· (Text 1)M:Good evening,I booked a room under the name of Tom Smith.I'm ready to check in now.W:OK,I've found it.Checking out on the 27th before 12 noon.Sir,here's your key.Your room is on the 7thfloor and on the left,Room 781.(Text 2)M:Hey,where are you going?W:I left my dictionary in the reading room.I'm going to find it.M:Hurry up!The dorm door is closing.(Text 3)W:Is the football match on now,John?M:No,it begins at 9:00 after the puzzle show.There is a film starting now.(Text 4)W:Excuse me.Do you happen to have change for five pounds?M:I've just used all my change for daily supplies.(Text 5)M:Mum,where shall I put the bowl?W:Here,give it to me.It goes in this cupboard!What's the butter doing in the cupboard?M:I bet Dad put it there.Now give it to me.I'll put it in the fridge.第一节到此结束。第二节听下面5段对话或独白。每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。每段对话或独白读两遍。听下面一段对话,回答第6和第7两个小题。现在,你有10秒钟的时间阅读这两个小题。(Text 6)W:Hi,Jack!What's the matter??Wake up!M:I stayed up late last night.My friend had a party.I only slept for about four hours.W:Why didn't you stay in bed this morning?M:I have to meet my study group in the library.We'll have a big test next week.W:A big test?I have read a report.It says that if you don't get enough sleep after you study,you may forget30%of what you've studied!听下面一段对话,回答第8和第9两个小题。现在,你有10秒钟的时间阅读这两个小题。(Text 7)W:So,are you still living with your parents?M:Yes,but I wish I had my own apartment.W:Why?Don't you like living at home?M:No.My parents are always asking me to be home before midnight.I wish they'd stop worrying about me.W:Yeah,parents are always like that.M:And they expect me to help around the house.I hate housework.I wish life weren't so difficult.W:So,why don't you move out?M:I wish I could,but I don't have much money and I don't want to work hard.Where else can I get free roomand food?W:Hey,you should depend on yourself.【高三英语·参考答案第2页(共4页)】·23-323C· 4已知复数:盖足:+351,其中为版单位。则:的共镜复故在复平面内对学的5不91的解集为1∠017了子点在第一象限tanz16.下图是西数/代)=4sin(@x+pA>0o0,回之d)的分图象.则f2023)=2022w):An月2m0A702029张四、解答题:本题共6小题,共0分.解答应写出文字说明、证明过程或演算步骤.17.(10分)已知ml1π-)-cos(-)=3,求值分1-C033-11)-sincosa)24n(1)tana;(2)sin 2a+cos 2a3-C05-603-如图,空间几何体ABCDE中,四边形BCD是矩形.DE⊥平面1BCD.平面4BE日D广乙18.(12分)平面CDE=I.(1)求证:CD⊥AE;(2)求证:C、AB∥19.(12分)已知函数f(x)=5sin2x+2cos2x.0y701)求函数f(x)的单调增区间:乙口132)若f(a)=59π5元1212求tan2a的值.高一数学·3·(共4页) 所以1=C2_1CP=3120.(本题满分12分)1981年,在大连召开的第一届全国数学普及工作会议上,确定将数学竞赛作为中国数学会及各省、市、自治区数学会的一项经常性工作,每年10月中旬的第一个星期日举行“全国高中数学联合竞赛”,竞赛分为一试(满分120分)和二试(满分180分),在这项竞赛中取得优异成绩的学生有资格参加由中国数学会奥林匹克委员会主办的“中国数学奥林匹克(CM0)暨全国中学生数学冬令营”(每年11月).已知某地区有50人参加全国高中数学联赛,其取得的一试成绩绘制成如图所示的频率分布直方图.频率(1)根据频率分布直方图估计学生成绩个组距0.032的平均数α和中位数b(同一组数据用该组区0.024间的中点值代替);(2)若成绩在100分及以上的试卷需要0.0120.008主委会抽样进行二次审阅,评审员甲根据上表0.004>成绩在此地区100分以上的试卷中根据分层抽样的60708090100110120原则抽取3份进行审阅.已知A同学的成绩是第20题图105分,E同学的成绩是111分,求这两位同学的试卷同时被抽到的概率.解:(1)由上表可知,0.12+0.24+0.32+10m+0.08+0.04=1,解得m=0.02,平均数a=65×0.12+75×0.24+85×0.32+95×0.2+105×0.08+115×0.04=85中位数b∈(80,90),由题意可知,0.12+0.24+(b-80)×0.032=0.5,解得b=84.375,即平均数a=85,中位数b=84.375.(2)由图可知,成绩在[100,110有50×0.08=4人,成绩在I110,120有50×0.04=2人,根据分层抽样的原则,成绩在100,110)抽2份,成绩在[110,120抽1份,设A,B,C,D四位同学的成绩在100110),E,F两位同学的成绩在110,120,高一期末检测数学参考答案第5页(共8页) 学生用书名师导学·新高考第-一轮总复习·数学心训练巩固6.已知函数f(x)=lnx+(a-2)x-2a+4(a>0),若有且只有两个整数,x2使得3.已知函数f(x)=2sinx-ax在[0,]上单f(x)>0,且f(x2)>0,则a的取值范围是调递减,则实数a的取值范围为()4.已知函数f(x)=x2(x-a).A.(1n3,2)B.[0,2-ln3)(1)若f(x)在(2,3)上单调,则实数a的取值C.(0,2-ln3)D.(0,2-ln3]范围是考点4,函数单调性的综合应用问题(2)若f(x)在(2,3)上不单调,则实数a的取值范围是例4已知函数f0=号+xasx-ar考点3,:静学函数单调性的应用问题snx2 0,g(x)=2-号x2-2sinx例3(I)(多选)已知e是自然对数的底数,一3 acos x(a为常数,a∈R)则下列不等关系中不正确的是()(1)讨论函数f(x)的单调性;A.In 22B.1n3<3(2)设函数h(x)=f(x)-g(x),证明:当0 受生用人书名师导学·新高考第一轮总复习·数学考点4,;三角函数模型的应用吵训练巩固鬓6.如图是一半径为2米的水轮,水轮的圆心O例5某实验室一天的温度(单位:℃)随时间距离水面1米,已知水轮自点M开始以1分t(单位:h)的变化近似满足函数关系:f(t)=10钟旋转4圈的速度顺时针旋转,点M距水面-3cos -sin[,24).的高度y(米)与时间x(秒)满足函数关系式y=Asin(ax+)+1(A>0,00,<(1)求实验室这一天的最大温差;(2)若要求实验室温度不高于11℃,则在哪段),则A=时间实验室需要降温?2米1米走进高考高考真题1.(2021·全国乙卷)把函数y=f(x)图象上所有点的横坐标缩短到原来的),纵坐标不变,再把所得曲线向右平移个单位长度,得到函数y=sin(x-牙)的图象,则f(x)=()Asin(登-1阅B.sim(受+8)C.sin(2)D.sin()2.(多选)(2022·新高考全国Ⅱ卷)已知函数f(x)=sin(2x十p)(0
参考答案学生月八书将函数f(x)的图象向左平移m个单位长度后得到g(x)=5sin(2x十=f(x)吾+2m)+号的图象,又西致g)的图家关于点(告,号)对张,即h()=3sin(2x+若+2m)的图象关于点(受,0)对称,故vsn(y=a+音+2m)=0,即语+2m=kmxc20,故m=经-登k∈2》.令=1,则m=音令=2,则m=瓷当0心