,图为-1所以0<2可x-因00,x+1>0同时成立即可:(2)先把f(x)化为f(x)=log(5-2x)+log。(x+1)=log(-2x2+3x+5),再由二次函数性质及对数函数的单调性可求出∫(x)的最小值,根据最小值为-2,列方程解出即可.20【答案】解:(1):函数()=是定义在(-33列上的奇函数,-y-f.即g2-gr-0-a+6办0.(或利用f(0)=0,得b=0)f0-6g2日保wa=g(2)f(x)在区间(-3,3)上是增函数.证明如下:在区间(-3,3)上任取x,x2,令-3 晚上11:15⊙:@1令06工编辑㗊【2023全国名校高考模拟信息卷·数学试题(五)参考答案第6页(共6页)】2023全国名校高考模拟信息卷数学试题(六)参考答案题号123456789101112答案AC A CA B DD AD ABD ABABD13.2sin受(答案不唯-)14.11415.64x316.321.A解析:(1十i)=[(1+i)2]3=(2i)3=-8i.2.C解析:A={xx2一x-6<0}={x|一2 2a.+1,a.为偶数8.已知数列a.调足a+一2aa,为奇数且a1∈N·,则下列说法正确的有①存在a1∈N·,使得a,=37,②若a=1,则a1=4“号③数列(an)的前n项和Sn≥(2"一1)·a1;④数列{a.)的前n项和S,≤(2-1)·(a,+1)-nA.1个B.2个C.3个D.4个二、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.已知某校共有教师300人,他们的政治面貌构成比例饼状图如图1所示,其中各类型中老年教师(45周岁以上)和中青年教师(45周岁以下,含45周岁)的百分比堆积柱状图如图2所示,则下列说法正确的有100%109圆中老年口共产党口中青年60%口无党派☐民主党派3090-共产党民主党派无党派图1图2A.该校中青年教师中,共产党员占比超过了号B.逐个抽取该校若干名教师进行调查,则被抽取的第二名教师恰好是中青年无党派教师的概率为9%C.在该校中老年教师中,根据政治面貌的不同按照等比例分层抽样抽取16名教师组成学校督学专家团,则民主党派共有2名教师入选D.依据a=0.001的独立性检验,该校教师是否为共产党员与年龄有关n(ad-bc)2附:X-a+b0c+ad6+Dn=a+6+c+da0.0500.0100.0013.8416.63510.82810.如图,P,O分别是正四棱柱ABCD-AB,CD上,下底面的中心,E是AB的中点,AC=2AA1,则下列结论正确的有点音,DA.OCLPBB.A1E∥PCC异面直线AE与PA所成角的余弦值为D.面PAD与面PBC夹角的余弦值为号【高三数学第2页(共6页)】 2023届高考冲刺押题卷数学(一)本卷满分150分,考试时间120分钟。一、单项选择题:本题共8小题,每小题5分,共40分。在每小题给出的四个选项中,只有一项是符合题目要求的。1.设集合A={xy=ln(x-2)},B={x-1 数A0w一09<5)的部分图象如图所示,若玉四、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一0,a),a>O,x≠x,C)Cx22>0恒成立,x1x217.(10分)则实数。的值可以为已知数列{an》满足a1=2,a2=4,an+2=a+1十2a·1B.(1)证明:数列(a为等比数列。现A爱(②数列6满足十层十…十公-一2.求数列6的前项和5。cD.三從维A一BCD的所有棱长均为3√2,与AC和BD都行的面分别与线皮AB,BC,CD,DA相交于E,F,G,H,则合m1月义家量与线A四能锥A一EGH体积的最大值为答@八+一0-八一0S且B.四边形EFGH为矩形01AC,三棱锥A一BCD外接球的表面积为27π江,长一显需西,资安时日手校维A标CD内切球的半径为之共.代国色是共超志三18.(12分)cos B在△ABC中,内角A,B,C的对边分别为a,6,且n品C+品C题序23母04怕56,卡78x版9事1011122cos C答案sin Asin B'空题:本题共4小题,每小题5分,共20分.个,才不面以总公(1)求C:知函数f(x)=e的图象在x=0处的切线与曲线y=nx+a也相切,则实数a(2)若a=4,b=3,点M,N分别在边CA,CB上,且MN将△ABC分成面积相等的两部分,求MN的最小值.出一个过点P(4,0)且与直线1:y=x相切的圆的方程:时单附值为图,在三棱锥A一BCD中,P是AC的中点,E,F分别为线段D,CD上的动点,BC⊥CD,AB⊥面BCD,若AB-BC=CD=一的0圆已1的1女,,3拉的h户一二则PE+号EF的最小值为面的间3高,5数列1a满足出=2+2,且a=2,41,则数列)的前,3.8o,中回油消有有,定王川为推过用十周惊anan+l,元严佰同得馆流世动小”项和S,=全国100所名校最新高考冲刺卷第4页(共8页)【23·(新高考)CC·数学(三)·门第3万(其8页)23·(新高考)CC·数学(三)·】 高考数学核按钮名师优创预测卷(一)】总体评价:本套试卷重视数学学科高考的综合性,创新性的考查,重视数学学科素养与关键能力的考查,突出理性思维的:考查,在试题的难度设计上不仅有层次性,而且体现在思维的灵活性、深刻性,方法的综合性、探究性和创造性等方面,试题有较强的区分度,全面体现数学学科高考的选拔性功能,本套试卷中没有出怪题,偏题,更不回避“必考点”,但却在命题角度、方法,题型上下足了功夫,考生感觉题型也熟悉、考的是基础,给人一种简单之感,但数起来又需要用大量时间去计算,试题既考查了考生对知识的掌握、解题技巧的运用,又考查了考生对时问的合理把握,同时也考查了考生的心理抗压能力,与高考要求是非常吻合的,此套试题从高考数学评价体系出发,贴近中学数学教学实际的一贯命题思路,整体符合高考改革的理念,对协同推进新高考综合改革、引导中学数学教学都将起到积极的作用123456789101112T-x-BD D CB DBDACD BCACD BD经w=2则g(-)-m[2(-登+)十18.1014号15fx=30号+1答案不唯-)]-1m(-ξ+)-1,所以-受+9受+2张k16.(15+1]Z即g号+2张xk∈Z.因为e<,所以g=5.所1.解:因为A={x4-x2>0}={x|一2 8LEICAIHUAW8已知椭圆C号+普-1。>b>0的右焦点为F,过坐标原点0的直线1与稀圆C交于P,Q两点,点P位于第一象限,直线PF与椭圆C另交于点A,且P吓-号F月,若c0s∠AFQ=},口FQ=2|FA,则椭圆C的离心率为品。上A呀C空3D5答可二、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.若过M作PQ的垂线,垂足为N,则称向量PM在Pd上的投影向量为P市.如图已知四边形ABCD,BCFE均为正方形,则下列结论正确的是一项是符合AC在A市上的投影向量为AB少在A上的投影向量为AAB+AC在A方上的投影向量为A它D.AB+AC在A古上的投影向量为A走10.某商场开业期间举办抽奖活动,已知抽奖箱中有30张奖券,其中有5张写有“中奖”字样.假设抽完的奖券不放回,甲抽完之后乙再抽,记A表示甲中奖,B表示乙中奖,则,则p=AB)-=品BB)-告C.P(A B)-2D.P(BIA)-22911,正三棱锥P-ABC的底面边长为3,高为6,则下列结论正确的是、XAB⊥PCB.三棱锥P-ABC的表面积为9√3,曲线的曲率C.三棱锥P一ABC的外接球的表面积为27π(x)在点(xD.三棱锥P-ABC的内切球的表面积为的曲率为xln I,x>0,12.已知函数f(x)=函数g(x)=[f(x)]2一(a一1)f(x)一a,则下列结论-x2-2x+1,x≤0,10不正确的是0小舍a<-。则g)恰有2个零点B.若1≤a<2,则g(x)恰有4个零点C若g(x)恰有3个零点,则a的取值范围是[0,l)Dy若g(x)恰有2个零点,则a的取值范围是(一∞,-U(2,+eo)【高三数学第2页(共4页)】·23-401C··23-401C 三步一体高效训练灯②asin C十√3 ccos A=0这两个条件中选择一个作为已知,求:(1)A的大小;(2)△ABC的面积注:如果条件①和条件②分别解答,按第一个解答计分.解析:若选条件①:(1)因为(a+b+c)(c+b-a)=bc,所以+2-a2=-bc,所以2kc0sA=-k,即c0sA=-2,因为A∈(0,,所以A=牙(2)因为a=26,b什c=6,所以cosA=cos2红=+-a2=h+c)2-2ce-a2」32bc2bc所以-之-S一杂8解得么=8,所以△AC的面积为2hmA=号X8×号-2点.2bc若选条件②:(1)因为asin C+√3 ccos A=0,所以sin Asin C-+√3 sin Ccos A=0.因为C∈(0,π),所以sinA十√3cosA=0,即tanA=-√3,因为A∈(0,π),所以A=受(2)因为a=26,b叶c=6,所以cosA=os2红=+c-d2=+c)2-2c-a】32bc2bc所以-=36一张28,解得=8,所以△ABC的面积为之simA=之X8X号=23,2bc222.(12分)已知△ABC的内角A,B,C的对边分别为a,b,c,且b=4.(1)若acos C=4,求A;(2)若sinA=2 sin Bcos B,且△ABC的边长均为正整数,求△ABC的周长.解析:(1)因为acos C=4,所以acos C=b,所以a.+2-C=b,2ab所以a2=+c2,所以A=受(2)sinA=2 sin Bcos B,所以a=2 bcos B,因为osB=心+C-B,所以a=26.心+-止,因为6=4,2ac2ac所以a2(c-4)=4(c2-16).若c=4,则B=C,因为A=2B,A十B十C=π,所以B=C=,A=受,此时a=4V2,不合题意,所以a=2√c十4,又c-a
亲薮安微省2023年中考仿真极品试卷·数学(三)注意事项:1.本试卷满分为150分,考试时间为120分钟2.本试卷包括“试题卷”和“答题卷”两部分.“试题卷”共4页,“答题卷”共6页.3.请务必在“答题卷”上答题,在“试题卷”上答题是无效的4.考试结束后,请将“试题卷”和“答题卷”一并交回、选择题(本大题共10小题,每小题4分,满分40分)年小题都给出A,B,C,D四个选项,其中只有个是正确的.个帕心下列四个数中,最小的数是A,-2B.-1C.0D.12据报道:芯片被誉为现代工业的掌上明珠,芯片制造的核心是光刻技术,我国的光刻技术水已突破28nm,已知1nm=109m,则28nm用科学记数法表示是A.28×10-9mB.2.8×10-9mC,2.8×10-8mD.2.810-10m3,如图所示的几何体的俯视图为卫不列计算结果正偏的是正面A.a8÷a4=a2B.(-2ab2)3=-8a3b5C,人a3)2=aD.(1+2a)2=4a2+2a+15,如图,在方格纸中,点P,Q,M的坐标分别记为(0,2),(3,0),(1,4),若MN∥PQ,则点N的坐标可能是A.2,3)B.(3,3)C.42)D.(5,1)6若关于x的一元二次方程x2一2x一k=0没有实数根,则k的值可以是○A.-2B.-1C,D.1学V型明四Y8牛名一术希gC辽居文A(xx2)(y-y2)>0B.(x1-x2)(y1-y2)<0肉是权年级二班清经行天演许比赛,斤,乙丙人速过辅签的方式只定中场证作,则出杨顺子行C(x1-x2)(y1-y2)≥0D.(x1-x2)(y1-y2)≤0好是甲、乙、丙的概率为A.6B.3C.8).如图,在矩形ABCD中,AB=6,AD二4,点E,F分别为BC,CD的中点,BF,DE相交于点G过点E作EH∥CD,交BF于点H,则线段GH的长度是DB.1055@,6为文数且清足a天16上一1,设d。”产N=。十己则下列结论。b10.①产ab=1时,M=V;②当ab>1时,M>V;①当ab<1时,M 马鞍山市2023年高三第三次教学质量监测数学试题注意事项:【,苓卷前,考生务必将自己的准考证号、姓名和座位号填在答题卡上。将条形码横贴在答题卡“条形码粘贴处”2作答选择题时,选出每小题答策后,用2B铅笔在答题卡上对应题目选项的答案信息点涂黑;如需要改动,用橡皮擦干净后,再选涂其他答案。答策不能答在试卷上3,非逃择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡各题目指定区战内相应位置上;如需改动,先划掉原来的答案,然后再写上斯答聚;不准使用铅笔和涂改液。不按以上要求作答无效。4,考生必须保证答题卡的整洁,考试结束后,监考员将试题卷和答题卡一并收回.一、选择题:本大题共8小题,每小题5分,共40分。在每小题给出的四个选项中,只有一项是符合题目要求的。1.设集合M=x0 b(以11ob01)0+00g。马修M哈师大附中2023年高三第三次模拟考试数学注意事项1、答卷前,考生务必将自己的姓名、准考证号填写在答题卡上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第I卷(选择题共60分)、选择题(本大题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项符合题目要求的)1.设集合A=1<2<8,B=y=-2x+8,则AnB=2A.[-4,3]B.(0,2]C.[-4,0)D.「2,3)2.已知复数则12-7A.1+iB.1C.1-iD.i3.行四边形ABCD中,点M在边AB上,AM=3MB,记CA=a,CM=b,则AD=A.3a-3B含0D.ga-3b4.记a,b,c,d为1,2,3,4的任意一个排列,则使得(a+b)(c+d)为奇数的排列个数为CA.8B.12C.16AD.185.已知函数f(x)=x2,面区域内的点P(x,y)满足(x)+f()<1,(V)+f(√yT)>1、则0的面积为DA-1C.TD.m-26.已知四棱锥P-ABCD的底面为正方形,PD上底面ABCD,PD=AD,点E是线段PB上的动点、则直线DE与面PBC所成角的最大值为2●A.oB.D.57,如图,阴影正方形的边长为1,以其对角线长为边长,各边均经过阴影正方形的顶点,作第2个正方形:然后再以第2个正方形的对角线长为边长,各边均经过第2个正方形的顶点,作第3个正方形;依此方法一直继续下去,若视阴影正方形为第1个正方形,第”个正方形的面积为。,则[cs(nm)·1bg,a1=02B数学试卷第1页(共4页) 多日删地地★/任华的20.(12分)已知数列(a)的前n项和为S,且a,=S十1。(aw·bn)".2,首项为1的正项数列6.满是b,·b,·b,··b,(1)求数列{am)、{bn)的通项公式;(2)求数列((21一1)bn)的前1项和Tw21.(12分)已知圆C过点(-3,0),(-1,2),(1,0),抛物线C,2=2px(p>0)过点A(片1(1)求圆C,的方程以及抛物线C2的方程;(2)过点A作抛物线C?的切线1与圆C,交于P,Q两点,点B在圆C,上,且直线BP,BQ均为抛物线C2的切线,求满足条件的所有点B的坐标.22.(12分)已知函数f(x)=e一a.x2.(1)若函数f(x)在[1,3]上有两个零点,求实数a的取值范围(2)探究:是否存在正数a,使得F(.x)=f(x)十asin一(1+a)x在R上单调递增,若存在,求出a的值,若不存在,请说明理由。 全国100所名枚最所高考機拟示范卷众14.已知直线1:kx一y一2k十1=0被圆C:x2十(y十1)2=12所截得的弦长为整数,则满足条件的一个k的值为15.已知数列{an}满足a1=1,a2=4,且a品一am-1a+1=3m-1(n≥2,n∈N"),则an=,记数列{am}的前n和为Sn,若Sm≥2023,则n的最小取值为16.已知正四棱锥P一ABCD的各顶点都在体积为号x的球0的球面上,过A,B,0三点的面a与侧棱PC有交点,则正四棱锥P一ABCD的侧面积的最大值为四、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)记数列{an}的前n项和为Sm,且a2=2,2Sn=nam十4(n≥2).(1)求数列{an}的通项公式;(2)对所有正整数,若am<3 2023届高三年级5月模拟(五)注意事项:数学·答卷前,考试务必将自已的姓名,准考证号等信息填写在答题卡指定位置上又回答选择题时,法出每个小华案后,用绍笔把答题卡上对应题日的答家标号套黑如署政动:用檬皮擦干净后,再选涂尖他登案标号,回答非选择题时,将答案写在答斑卡上。写在本试卷上无效,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.1.已知集合4={l,l24,B={x-1≥},则4nCB=A.{-1,2,4B/1,2}C.{-1,2}D.{1}2.已知等腰梯形ABCD满足AB∥CD,AC与BD交于点P,且AB=2CD=2BC,则下列结论错误的是A4C-写而+号西B.而+C.IAP=21PDID.AP=2PC3.已知抛物线M:2=16x的焦点为F,倾斜角为60的直线1过点F交M于A,B两点(A在第一象限),O为坐标原点,过点B作x轴的行线,交直线A0于点D,则点D的横坐标A.-1B.-2c.-4D.-84.某医院安排3名男医生和2名女医生去甲、乙、丙三所医院支援,每所医院安排一到两名医生,其中甲医院要求至少安排一名女医生,则不同的安排方法有A.66种B.54种C.30种D18种5.三棱锥S-ABC中,SA⊥面ABC,AB⊥BC,SA=AB=BC.过点A分别作AE⊥SB,AF⊥SC交SB、SC于点E、F,记三棱锥S-FAE的外接球表面积为S,三棱锥S-ABC的外接球表面积为5,SB.22c5D.36.在面直角坐标系内,已知A(-3,4),B-3,),动点P(x,)满足IPA=2引PB1,则(x-1)2+y-)2(teR)的最小值是A.16B.4.2D.2数学试卷第1页(共4页) .△ODF≌△OBE(ASA),.DF=BE,.DE=EB=BF=DF,四边形DEBF为菱形18.解:设每个A型扫地机器人的进价为x元,则每个B型扫地机器人的进价为(2x-400)元,96000168000依题意得:2x-400解得:x=1600,经检验,x=1600是原方程的解,且符合题意,.2x-400=2×1600-400=2800.答:每个A型扫地机器人的进价为1600元,每个B型扫地机器人的进价为2800元.19.解:(1)根据乒乓球所七的比例和人数可得,40抽取的人数为-=100(人),400%.参加篮球的人数有:100-40-10-25-5=20(人),补全条形统计图如图所示:+人数4040100乒乓球摄影书法篮球足球项目参加摄影的人数为10人,10×100%=10%,100.m=10:根据扇形图可得:1-40%-5%-25%-10%=20%.n=20:(2)根据统计图可知“书法”所占25%,∴.2000×25%=500(人),∴.若该校有2000名学生,估计该校参加“书法”活动的学生有500人:(3)根据条形统计图和扇形统计图可知,参加乒乓球的学生人数是最多的,其次是书法、篮球,参加摄 (2)由题知g()=2sin(4x-零),由方程g)-号,得8in(4x晋)-号,即sin(4x-子)-3.....。。。......。6分因为x[语竖1,可得4一骨∈[受6m]。设0=4-吾,其中0e[受6m],即sim0=4…7分结合正弦函数y=sin0的图象,如图yy=00,5元可得方程sin0=子在区间[受,6m]有5个解,即n=5,….…9分其中0十02=3π,02十0s=5π,0十04=7π,04十0=9π,即4知-子+4-号=3m,4-子+4,-号=5m,4-号+4-号=7,4-晋+4x-号=9m解得十十=竖十云=瓷么十=登12…11分所以a+2+2十…十2x4十=(十)+(2+)+(十x)+(x4十)=23·…12分2解,0旧为。n28西-27所以乙N08-吾,A81=-2.∠0B产-ABI4OA=4sin0,……2分10B到=4sim(3F-0)=4sim(开+0).……4分(2)因为AB引=22,∠MAB=∠MBA=T,所以AM=BM=2,…5分在△OMB中,由余弦定理易得OM2=|OB+|BM2-2 OBBMI cos(于+0),|oM2=16sim(年+0)+4-16sim(年+0)cos(F+0)=16simr(年+9)+4-16sin(至+0)cos(至+0)=8[1-cos(+20)]+4-8sin(5+20)=8sim20-8c0s29+12=82sin(20-T)+12,…7分因为0(不,受),所以29-∈(,F)sin(20-T)(91……8分当sim(20-)=1,即9=时,1OM2取最大值12+8√2,OM取最大值2十22,…10分此时10A=4sin=2V2+厄,OBl=4sin(于+F)=2V2+厄,故当|OA|=|OB|=2√2+√2时,OM取最大值2+2√2.12分【高一下学期第二次阶段性考试·数学参考答案第4页(共4页)】23096A 对于60在少上给墨有的藏为-汤号放c结灵:对于D,(b-4a)·b=b2-4a·b=0,又b-4a与b均不是零向量,所以(b-4a)⊥b,故D正确.12.答案ACD命题意图本题考查立体几何中的相关计算.解析对于A,由已知得AA,⊥面ABP,PBC面APB,所以AM,⊥PB,又因为AB是底面圆的直径,P在圆周上,所以BP⊥AP,又AA∩AP=A,所以PB⊥面A,AP,故A正确;对于B,因为AA,⊥面ABP,所以直线A,P与面ABP所成的角为∠A,PA,计算易得∠PAO=30°,所以PB=1,PA=万,AM,=3,放m∠APA,=格=,故直线A,P与面ABP所成的角的正切值为万,故B错误;对于C,连接B,P,由已知得AB∥A,B1,所以直线A,P与直线AB所成的角为∠B,A,P,在△A,B,P中,A,P=√AP2+A=√(W3)2+32=25,B,P=√BP2+B,B=√+3=√0,所以c0s∠B,A,P=2+22-套故c正确:2×2×25对于D,设点A到面APB的距离为A,则-m=4-n,即写·Sam·A=号·Sm·M,又Sag分P~B即-写=AP:P8=万,所以太=号放D正确三、填空题:本题共4小题,每小题5分,共20分13.答案√2命题意图本题考查斜二测画法的基本概念,解析根据题意可得0'A=2,在△MB0中,0B=0'B'=1,0A=20'A=2,D,所以△AB0的面积为S=之×1×22=2.014.答案i03命题意图本题考查古典概型的概率计算解析不考虑顺序,列举可得总的样本点的个数为10,事件“他们加入的都是球类运动社团”包含的样本点有3个,故所求概率为品15.答案(-1,号)命题意图本题考查面向量的性质,—3— 基础题与中考新考法·八年级·上·数学微专题1双角分线模型79考建议用时:20分钟教材原题改编练改编2将两条内角分线均改为外角分线如图,在△ABC中,外角∠CBD的分线与原题(教材P29第11题)如图,在△ABC中,外角∠BCE的分线相交于点P,探索∠P∠ABC和∠ACB的分线BE,CF相交于与∠A之间的数量关系。点G求证,)∠BC=1802(2AC+LAG8:(2)∠BGC=902
- 所以g(2π)≥0.10分所以(5)-a-e<0,解得-e-2m≤a 22.(0f0=a+1-2e(-1,0),所以a>1:…3分(Ⅱ)f(x)在(0,+∞)单调递增,且f()<0,f(2)>0,所以由零点存在定理,得∫(x)在(1,2)内有唯一零点,即函数∫(x)存在唯一零点;6分(I)证明:若f(=0,则1e(1,2)→1+1e(2,3),所以了2)=a+2-1 short arrye.oit really is a nice area wth a special placein my heart.第2期B版测试题参考答案1-5 BBACA6-10BCABC11-15 ABCCA16-20 CBACA21-25 BCAAA26-30CCDBC31-35 ACDBC36-40 FCDAG41-45 BADBA46-50 DCCDC51-55 ABADC56-65 (One possible version)56.working57.carefully58.to reduce59.strategies60.where61.effective62.0m63.responsibility64.will be pushed65.a写作(One possible version)第一节Contribution WantedTo promote knowledge of China'sachievements and progress in science andest animal hospital.technology,our School English Neuspaperis launching a China's Science and Technology Development column.We sincerely invite all students and sto contribute related articles.The articlescan cover any techological developmntChina,past and presentIt'sxptd that第your contributions should be clear in languageand within 500 words.Pasyokto studentsnewspapershend of next mont and incde basic pinformation in your emaii.Lookingfoardtohrticipaionof students and teachers.Sw第二节School English NewspaperskiParagraph 1:are 若=则oa-土,所以s血,-士V一日-士号放D正确,故应选AD.)11.ABC(A选项,圆心为(k,),一定在直线y=x上,A正确;B选项,将(3,0)代人得:2k一6k十5=0,其中△=-4<0,方程无解,即所有圆C,均不经过点(3,0),B正确;C选项,将(2,2)代入得:2一4k十2=0,其中△=16一8=8>0,故经过点(2,2)的圆C有两个,故C正确;D选项,所有圆的半径为2,面积为4π,故D错误故应选ABC.)12.AD(函数fD)=ar+(ab≠0),定义域(-m,0)U(0,+w).且f(-)=a(-0+0=一ax一b=一f(x),函数f(x)是奇函数,A选项正确;设直线y=x,联立方程:a以+名-bz,得(函-a)2-b=0,k-0≠0,△=46(k-a)≠0,直线y=b虹不可能是f(x)的一条切线,B选项错误,者u)=f子则a+名=u,+乌得-名2即in(sin十2)=名由snr的有界性,显然smr(snr+2)-。不一定有解,C选项错误当a>0,b<0时,f(x)在(-∞,0)上为增函数.由f(x)=0,得x=-,x=-Q6,叉f(x)为奇函数,则f(x)一0在R上有2个零点,D选项正确故应选AD.)三、填空题:本大题共4小题,每小题5分,共20分.把答案填在答题卡的相应位置13.一90(展开式的通项公式为T+1=C(3x)5·(-y)-3-(-1)Cxy,令-3,则T4=32(-1)3Cx2y3=-90x2y,故答案为:一90.)14.2(设f(r)-2+2x+十sinx=1+2x十sinxx2十tx2+t设g(x)=2x++sinax2+t因为t>0,所以函数g(x)=2x十s1n2的定义域为全体实数,x2+t因为g(-x)--2x-sinzx2+t=-g(x),所以函数g(x)=2x十sinz是奇函数,它的图象关于原点对称,x2十t因此g(c)max十g(x)min=0,因为f(x)ox=t+g(x)nmx=M,f(x)mm=t+g(x)mm=N,所以由M+N=4→2t=4→t=2.)15.5(因为sing-2c0a=1,2sia+c09=V2,数学参考答案第3页 听第9段材料,回答第13至16题。听力材料Text1013.When will the visitors come?One of the joys in life that the blind people cannotexperience conveniently might be reading,but a new听In March.B.In April.C.In May.bookstore in Shanghai has made that possible.(17)14.How many visitors are coming?Niguang 226,a new bookstore,is a 37-square-A.8.货10.C.20.meter space designed for blind people.Besides being15.What will the visitors do on the second day?the smallest Xinhua Bookstore in Shanghai,it is alsoA.Attend a party.B.Visit Scotland.the first of its kind in the city.Visit the school.To help the blind find their way to the store,16.How will the visitors probably go sightseeing?special paths to Niguang 226 have been laid.The铃0 n foot.B.By bus.lamp posts outside the store also have signs for theC.By shipblind.Beside them,there are seats for guide dogs tohave a rest.(18)听力材料Text9Inside the shop are more than 300 books andM:Hi,Cindy.Do you know when the visitors fromcultural products made by disabled people.Toensure the safety of its visitors,Niguang 226 has no听Australia are coming?W:We offer them three choices:the end of March,sharp corners.力Han Ying,honorary manager of the shop,(19)the middle of April and the beginning of May.used to be a teacher.Years ago,Han became blindThey've chosen the earliest one that is goodafter an illness.But she has been living positivelyactually with exams coming up in May.(13)and tried to help many people in need.By openingM:And how many are coming?Did you say aboutthe bookstore,she wanted the blind to feel the love20?and warmth of society,and other people can get toW:Yes,they said 20 at first,but reduced by halflearn more about the disabled.Han hoped thatthis morning.(14)there would be more places like this bookstore inthe futureM:Good,we have eight weeks to prepare.Here are第一节Amy suggestions:On the first day,we will have aAre you travelling in Honolulu,Hawaii?If yes,welcome party for them,and then show themwhy not visit some of the following restaurants and听around our school the next day.(15)阅have a good meal?W:That will be very nice.Waiahole Poi FactoryHawaiians Charlene and Calvin Hoe bought anM:And on the last day,they would do someactual poi factory in 1971,using it primarily as ansightseeing.We could take them on a tour ofart gallery,and then began serving food in 2009.London by bus but many of them may have解Today,it's also one of the few places that serve freshalready been there,and Scotland is too far away.pa'i'ai,cooked taro(芋头)pounded(捣碎)on a longW:Why not take them for a walk along the coast?wooden board.You'll have to call in advance toreserve some.(16)It should be interesting.●48-140 Kamehameha Hwy,Kaneohe,,HⅡ96744M:Good idea!Cindy,I'm sure they'll like it.(16)●(808)239-2222听第10段材料,回答第17至20题。Over Easy17.Who is the bookstore specially designedHonolulu loves breakfast,and few places do itfor?阅better than Over Easy,a warm,happy familyA.Those who can't walk.operation anyone would be proud to support.力Fancy,golden,tasty pancakes are the star of sweetB.Those who can't speak.Those who can't see.理dishes,but don't leave without trying the bacon-cabbage soup poured over a bowl of rice and18.Why did the store put seats near the lampPortuguese sausage.posts?418 Kuulei Rd #103,Kailua,HI 96734To let guide dogs relax.●(808)260-1732B.To put up helpful signs on them.C.To encourage more people to read.Helena's19.What does Han Ying do now?If you're only going to one spot for traditionalA.A teacher.A manager.阅Hawaiian food,make it Helena's.Locals have beenC.A writer.lining up since the day Helena's opened in 1946.力20.What is the speaker mainly talkingFirst-timers should order set menu D,which comesabout?理with kalua pig,lomi salmon,and squid lu'au (aA.Some tips on helping the disableddelicious dish of octopus and taro leaves in coconutpeople.解milk),along with rice.An introduction to a warm bookstore.1240 N School St,Honolulu,HI 96817●(808)845-8044C.The importance of reading books 1,A【解析】本题考查集合,考查数学运算的核心素养。因为MUN=1,2,3},所以C(MUN)={4.5.2.D【解析】本题考查复数,考查数学运算的核心素养。20-号+1-2i(1-2i)(1+21)53.D【解析】本题考查面向量的数量积,考查数学运算的核心素养。因为(a+2b)·(a-b)=a2-2b+a·b=-号,所以a…b-号4.C【解析】本题考查用样本估计总体,考查数据分析的核心素养。由频率分布直方图可得,质量在区间[1.55,1.65)内的柚子数量是100×2.5×0.1=25.5.D【解析】本题考查椭圆,考查逻辑推理及数学运算的核心素养.易知Pc台.A(-a,0,B0,60.ke-台kr-ac因为AB/OP,所以w=km,则之-左,即b=C,a=F+元-2c,a ac所以e=S=区a 26.B【解析】本题考查三角恒等变换,考查数学运算的核心素养。因为sina十cosa=Esin(a+)-2cosB,所以sin(a+产)=cosB=sin(交-).因为a∈[,受]8[,受].所以a十晋∈[受,开],受-c[o,],所以a++受-月=,则aB-及7.C【解析】本题考查函数的应用,考查数学建模的核心素养,100℃的物块经过tmin后的温度a=20+80e÷,60℃的物块经过tmin后的温度a=20十40e.要使得这两块物体的温度之差不超过10℃,则20+80e÷一(20+40e)≤10,解得1>8ln2=5.52.8.A【解析】本题考查导数在研究函数中的应用,考查逻辑推理及数学运算的核心素养.设函数fx)=nx+上-1,f(x)=宁,所以f(x)在(0,1)上单调递减,在(1,十o)上单调递增则f(x)>≥f1)=0,所以n≥1-,当且仅当x=1时,等号成立.令x-号则1n号>号设函数g(x)=lnx-二g'(x)=,所以g(x)在(0,e)上单调递增,在(e,十∞)上单er【高三数学·参考答案第1页(共7页)】·HEN· 第一节阅读下面短文,从每题所给的A、B、C、D四个选第第三部分项中选出可以填入空白处的最佳选项。My son had just turned five.My husband and I were 41 as to which30.What does the underlined word“strides”inschool to send him to.We started 42 and visiting schools.Still,we felt部分paragraph 4 probably mean?43 about which school to choose.So I called my friend Shannon.She'sA.Adjustments.B.Connections.the one I 44 most when I struggle with parenting decisions.Having a sonC.Arrangements.Improvements.31.What does Brik say about the new technology?ten years older than mine,she's gone through all the 45 I'm still goingIt has a promising future.through.读语言运B.It still has a long way to goA 41.A.concerned B.curious C.excited D.confidentC.It can solve the water crisis worldwide.D 42.A.attending B.funding C.leaving D.investigatingD 43.A.unreasonable B.unwise C.uncomfortable D.uncerD.It will not be supported by the governmentB 44.A.come across B.rely on C.wait for D.vote forC45.A.interviews B.postsC.stagesD.pages第D"Jeanette,I know you have narrowed it down to two choices and32.What does the study find out?A.There are many ways to reduce screen time.第三部both are good schools.Your love is more important than the 46 aboutthe perfect school,"Shannon said."I remember when I was doing some47.Worried about my 48,I was stuck between two cans of beans.One部分B.Time outdoors definitely helps reducecan was twenty cents less than the other,but it wasn't quite as good as thesmartphone use.other.I stood there 49 for at least twenty minutes.It finally 50 me thatVisiting cities'green areas can't shorten screentime.言A46.A.choiceB.debateC.environment D.evidenceD.People are becoming increasingly dependentB47.A.cookingB.shoppingC.cleaningD.planningC48.A.preference B.scheduleC.budgetD.strengthon smartphones.用D49.A.fearlessB.breathless C.unexpectedD.undecidedA50.A.struckB.bothered C.surprisedD.annoyedg33.What suggestion might Kelton Minor give?it's just a can of beans.Either can 51 my care for my family.In the long第A.Taking long trips without phones.run,whether I picked brand A or brand B can of beans,it didn't 52."Going out into the wild more often.Shannon made me think deeply."Now,you are good parents,so whicheverC.Setting mobile phones to silent.第三部分school you pick is the right school.Besides,you have a(n)53 if one分D.Visiting urban locations with friendsdoesn'tft,”she added34.What do the researchers hope to do?A.Call on more companies to join the research.语言运B 51.A.refreshedB.displayedC.enhancedB.Use technology to improve people's healthD.examined读C 52.A.happenB.existC.Do more research on people's mental health.C.matterD.workHelp people use digital products in a healthB 53.A.lossB.alternativeway.原C.responsibilityD.rewardI didn't feel stressed out anymore after I 54.We were able to pick第a good school for our son.55 it's just a can of beans,and we have tomove forward and make our decision.And if it turns out to be the wrong35.Which of the following would be the best title for部分the text?第三部分can of beans,we can go back to get a different one.A.Give your digital life a breakA 54.A.hung upB.stepped upB.Make good use of mobile devices语言运C.settled downD.slowed downC.Get close to nature for inner peaceC 55.A.SubsequentlyB.InitiallyGo out into the wild to reduce screen timeC.ActuallyD.Especially⑧原第二节第二节阅读下面短文,在空白处填入1个适当的单词或括第阅读下面短文,从短文后的选项中选出可以填入空白第三部分号内单词的正确形式。处的最佳选项。选项中有两项为多余选项。Many people use smart phones in the toilet to sendtexts or check 56.thelatest videos on socialknowledge,expand your vocabulary and reduce stress.Here aremedia..However,.doctors say scrolling through(滚动浏览)some tins on how to starl a reading babit and stick to it.分One of the est aystostarteading habit is tobeginwith sortCyour phone while in the toilet could damage your physicalhealth.Initially,the conclusion 57.was drawn(draw)byingobd.Finda time tatrksyondmkeoou daily routine语Saurabh Sethi,a doctor who graduated from Harvardyou might find it hard to put the book dow ad 1A.Focus on the quality of your readina expericncUniversity.He has warned that bacteria in the bathroomutes won't seem like enough time.Find a guict and comfortable space without disturtanc读Setting a reading goal isan excellent way to motivate you to read more C.To develop a redin habit you needt make time for itincrease numerous health 58.risks (risk).DrSethiand habit.You can sct a goal ofD.Make sure you are passionate about the topics of the bookssgtagolofreadiaeEHowevet,iteanbectalengingigouseldoameadregularlyr goal youchoose,make sure it is用encouraged smart phone users 59.to break (break)thishabit.He said,"Being addicted to smart phones in the foiletStarting small will make it easier to include reading in your dailyTo make reading a habit,it is esserbooks that interesactually 60.has (have)some serious negativeconsequences.” 34.A.busy
B.happy
C.sorry
D,curious
35.A.competing
B.arriving
C.appearing
D.working 第二节(满分25分)
阅读下面材料,根据其内容和所给段落开头语续写两段,使
之构成一篇完整的短文。
September is most beautiful in Strathroy,where people
wholeheartedly enjoy the simple pleasures of life and nature.
Roland was no exception.He was a magnificent German
Shepherd(德国牧羊犬),a little lazy and overweight though.He
loved to doze in the backyard.He yawned and stretched his body
as he heard the slowly approaching footsteps of Mrs Proudfoot.
“Come and have your lunch dear!”said Mrs Proudfoot with her
usual fondness for him and he wagged his tail affectionately in re-
turn.
Roland was licking the last piece of meat in his plate when 【答案与解析】
本文主要介绍现实社会中学生考试难得高分的情况,并说明
高标准带来的结果以及美国人对成功的定位。
25.B根据画线词下文as high as necessary to ensure that some
students failed可知,考试标准再一次被提高,这让一些学生难
以达到这种标准。故画线词表示“增长”之意。
26.A
根据第2段第1句及最后一段可知,作者对考试高标准是
持批评态度的。
27.C根据全文内容可知,文章主要探讨学生不能达到考试标准
的原因。
28.B
【命题视角】本题主要考查学生根据文章的字面意思,通过逻辑
关系和细节的提示,推断作者的态度,理解文章的寓意的能力。 in two ways.First,scientists have noticed that some bees pro
duce a substance that feels similar to plastic.Imagine a natural,
non-poisonous plastic alternative coating your fast-food packa-
ging,toiletries,and furniture.Second,some bees have begun u-
sing thrown-away plastics in their nests.This instinctual
reusing and recycling could help clean up urban areas that are
already dealing with heavy plastic pollution.
A few years ago,scientist Veronica Harwood-Stevenson
went on the hunt for a natural alternative to plastic.She found
some bees create nesting material that is "non-poisonous,water-
proof,flame-resistant,and able to withstand heat".Harwood
Stevenson
founded a company,working with biomimicry
re
searchers,to figure out how to recreate the material in a cost-ef-
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2019-20年度考前模拟演练答案
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0,9-x2>0,9-x>0,∴f(x)-f(x2)<0即f(x) lny曰x>y>0,∴.“lnx>lny”是“x>y”的充分不必要条件,故A正确;x2>y2台|x>y,.“x2>y2”是“x>y”的既不充分也不必要条件,故B错误;.x3>y3曰x>y,∴.“x3>0>}是>y的既不充或/0y”是“>)y的充要条件,故C错误;>1台>0分也不必要条件,故D错误,4.C解析:令x=y=0→f(0)=f(0)十f(0)-1→f(0)=1,∴.f(x-x)=f(x)十f(-x)-1,即f(x)+f(-x)=2,∴f(x)的图象关于点(0,1)对称,故选C5A解析:由题意得=2e+=子,lbga+log:b=log:(ab) lnc,∴a>c,lnc=alnb>lnb,∴.c>b,此时a>c>b;②若a,b,c∈(1,+o∞),则lna,lnb,lnc均为正数,lna=blnc>lnc,∴.a>c,lnc=alnb>lnb,∴.c>b,此时a>c>b.综上,a>c>b.8.D解析:以A为坐标原点建立如图面直角坐标系,由题意可设|NQ|=a+c,lQR|=a-c,P(0,4),R(-3,0),.PR:4x-3y+12=0,kR=李,设M(n,1),Q(,0),则M到PR的距离d=M03十2=1,解得n=-1(舍去)或0=-子Q1=名-3√/42+32【2023全国名校高考模拟信息卷·数学试题(六)参考答案第1页(共6页)】2-1+4=a-c.设PN:k.x-y十4=0,则M到PN的距离d=、/2+7=1,化简得8□回心工具适应屏幕PDF转换分享 0)的部分图象如图所示,则f(0)=)毁A.2B.2W2三C.1+5D.√2+137.已知a=1e,b=In41则8e6A.c>b>aB.b>a>cC.a>b>cD.a>c>b8.如图,在三棱锥A-BCD中,AC=3V6,AB=BC=CD=AD=BD=6,点M在AC上,AM=2MC,过点M作三棱锥A-BCD外接球的截面,则截面圆面积的最小值为(A.12πB.10mC.8πD.4π数学1·1(全卷共4页) 0,g(x)单调递增:当5.解:根据条件随机选取3个节日的取法有C种,重阳节被x∈(0,十∞)时,g'(x)<0,g(x)单调递减透中的取法有心种则所求概率P=0.5.故选B.又g(-3)=忘,g(0)=1,且g(1)=-e<0,当x→6.解:将函数(x)的图象向左移于个单位长度,得到函一o∞时,g(x)0,且g(x)<0,则a=(一x6一xo十1)e有数y=Asmω(x十)十]的图象,再将其图象上所有2个解的充分条件为a=一吕或0<0<1,结合远项知。点的横坐标伸长为原来的2倍,得到函数g(x)=4=是不符台故选D2Am(x+子)十g的图象。9.解:对于A.因为(a+6)=a+b+2√/ab=2+2/ab≤由图象可知2A=1则A-之-吾-(-)-华2+2=4,所以√a十√6≤2,当且仅当a=b=1时取等号,A正确。数学(一)参考答案第1页(共4页) 0)的最小正周期为7,若2 g)>g(2m)≥0>))则存在唯一x1∈0,罗),使得f'(x)=0,当xe(0,x)时,f'(x)>0,fx)单调递增;当x∈(,罗)时,∫”(x)<0,x)单调递减,则f(x)在x=x取得极大值:存在唯一∈(受,2m小使得∫(x)=0,当x∈(受,)时,f"(x)<0,x)单调递减;当x∈(x2,2T)时,'(x)>0,f(x)单调递增,则f(x)在x=x2取得极小值综上,所求a的取值范围是[-e-2m,e司12分22(1):由题可知F(1,0).若直线AB,CD有一条斜率不存在,则另一条斜率为0,其中点分别为直线与x轴的交点、原点,过此两点的直线MN方程为y=0.……1分若直线AB,CD的斜率存在,设直线AB的斜率为k,则直线CD的斜率为-由题,可设直线AB的方程为y=k(x-1),直线CD的方程为)=-冬(x-1)2分ry=k(x-1),联立王t人消元y,整理得(3+4k2)x2-8k2x+(4K2-12)=0,82设A,方,g,,则+42西+)二2E一二6品3+4k2,…3分4k2-34h3从面u3十,3+4旅、即写+4·+4按)-3k4分同理N3k3k2+4’3k2+45分-3k3k所以kw=y-=+423k2+4-7k尤-xN4k244h2-43+4k23k2+4数学参考答案第6页(共7页) 1,所以g(a在(1,+w)上单调递增,所以8回>80=1-3+2,+1-2北++2”-t12-+2北-30,11+1(+)t(+1)t(t+1)即f0+<2a+2-2,所以a+2-1 - ‹‹ ‹ 77 › ››
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