◆/cm0A.1=0时刻，质点M运动方向沿y轴正方向B.该简谐波的最小频率为2HzC.该简情波的最小传播速度为4msD.1=0.8s时，质点M位移可能为0【答案】CD(3)根据实验获得的T2-L图像如图所示，T为周期，L为摆长，实验中应该在0.40.60.81.01.2m【解析】根据同侧法可知，1=0时刻，质点M运动方向浴y轴负方向，A错误；由图可知nT（选填“OAAB旷4=05s,则或“BC)区间多采集数据，由图像求得当地重力加速度为【答案](1)BC(2分)(2)195(2分)(3)BC(2分)986(9.70-9.90均可)(3分)n1ms2(保留三位有效数字)。了天子网时河大家资0.B【解析】ama左细ca-a(1)为了减小空气阻力的影响，摆球应选用质量大、体积小的球，不能选用泡沫小球，故A错误；摆球摆动时要保证摆球在同一竖直面内摆动，不能形成圆维摆，故B正确；单摆摆长应保持不变，单摆悬线的上端时，质点M位移为0，D正确。不可随意卷在横杆上，应夹紧在铁夹中，故C正确：第11卷（非选择题共54分）二非选择题，共5小题，共54分，考生根据要求作答。(2)单摆做30次金振动所用的时问为1=585，则这单摆的网期为T=-5855=1.95s:日A303011.(6分)(3)为减小实验误差，单摆摆米应适当长些，由图像可知，实脸中应该在BC区间多米集数据。由单摆周期公在“用单摆测重力加速度的实验中，某同学的操作步骤如下：取一根细线，下端系住直径为d中1.0cm的金属小球，上端固定在铁架台上：2合T盈*+加故m:料地t雪h.用米尺量得固定点到小球上端之向细线的长度=98.2cm;g=元2m/s2≈9.86m/s2。c.在摆线偏离竖直方向小于5的位置静止释放小球：13.(12分)d用秒表记录小球完成50次全振动的总时间=99.8s。(1)用秒表测时间时为尽量减少误差，应从摆球通过弹簧振子的运动示意图如图所示，质点经口点向右运动2s后到达b点，又经15到达c点时速度为0。、b(2)根据上述测量数据，可以得到当地的力加速度g=(选填“最高点”或“最低点”)时开始计时；中点0是衡位置。已知ab-20cm,bc4,1cm。试求：(1)振动的振幅和频率；【答案】(1)最低点(3分)（2)9.770(3分)二ms2(小数点后保留三位小数)。(2)质点经a点向右运动17s末的位移：【解析】(3)质点经a点向右运动17s末的路程。()摆球在最低点时的速度最快，在最低点作为计时起点误差最小，故选择摆球通过最低点时开始计时；日a o b c（2）由随老可物，单摆摆长L=1+号=987cm=0987m,羊括周期7-0n50=1.996s,由单摆用期【答案】(1)A=14.1cm,∫=0.125Hz(2)x=10cm,方向沿x轴正方向(3)s=122.8cm【解析】会7=2T得重力g=4970n。2(9分)(1)质点经4点向右运动2s后到达b点，a、b中点O是衡位置，根据简谐振动对称性可知，质点经衡位点出T置0达到，点b的时问为15，而从b点经1s到达c点时速度为0，故从衡位置到最右端的时间为25T=2s×4-8s(2分)在用单摆测量重力加速度”的实验中：(1)下列关于该实验的操作，正确的是-0125批(2分)A摆球应选用泡沫小球中，上是腰武相食上失长在来车-0,20,0振幅A=0b+bc=b+bc=14.1em(2分)(2)用台表记录弹县做0次全纸动所用的时间为585：，则该单摆运动的周期下，（保留三位有效(2)由(1)中分析可知质点从a点向右运动13后到达衡位置0点17s=2T+1s(2分)数字)故质点经口点向右运动17s末到达衡位置，位移大小x=10cm,方向沿x轴正方向(2分)(3)由(2)中分析可知质点经a点向右运动17s末的路程3=4A×2+0a=122.8cm(2分)教师用卷·物理·第110页共139页教师用卷·物理·第111页共139页

天利卷6答案.pdf1/6即nt-}mB=0<月<月=青或装6分()、a=}wnB=}w=1,(8分)当B=时.由余蓝定理得三0+-2B=因为P1⊥Y面ABCD.所以∠1即为线P与面D所成的角.m/1=是-号2m-3m=4(2-y3)因为AC。75,所以P1=60.当且仅当n=c=2时，b有最小值2V(2-3):则B80,0.0),C(45,60,0),D0,60.0),P(0.0.60(10分)(8分)当=若时，由余弦定理得8=。2+2-2B≥所以Di=(80,-60,0),P元=(45,60，-60).Pi=(0,60,-60).2ar+、3ac=4(2+3).由(I)知，BD⊥而PMC,当且仅当a=c=2时，b有最小值2√(2+3).则面P1C的个法向量m=(4,-3,0),(12分)设面PCD的法向量为n=(x,y,z),18【名师指导】木题考查直线与而垂直的判定定理、Jn元=45x+60-60:=0,得=0.直线与面所成角、二面角，考查运算求解能力、空山nP币=60y-60:=0.Ly=z.间想象能力，考查逻辑推理、直观想象核心素养取y=-1,则n=(0.-1,-1).(10分)(1)利用勾股定理逆定理证明D1CD,B⊥D,山设面A-2℃-D的面角为0，行线的件质求出AE,DE的长，在△DE中年今皮332定理逆定理证明BDLAC,再由线面垂直的判定定理则0=5×210即可证明：(1)以点A为坐标原点，AB,D,P所在由图易得二面角A-PC-D为锐角直线分别为x,刀，：轴建立空问直角坐标系，分别求所以=面角A-心-D的余弦值为沿(12分)出面PAC和面PCD的一个法向量，利用空向向量的夹角公式即可求解。19【名师指导】本题考在频率分布直方图、离散型随机解：(I)证明：内为P1⊥面ABC),BDC面变量的分布列和数学期煤，考查运算求解能力、推理(2分)论证能力，考查数据分析，逻排推理核心索养，ABCD,所以PABD.(I)根据各组频率和为1求出m的值，再求出数据在△ACD中，AC=D+CD,的均数即可：（Ⅱ）用期亭估计慨率，得出X的所有所以AD⊥CD可能取值，再运用二明分布的概室公式求X的分布因为AB∥CD,所以AB⊥AD,列和数学期望。所以BD=√AB+AD=1O0.解：(1)因为(0.006+0.01+m+0.072+0.061+设AC与BD交点为E,0.016)×5=1.解得m=0.032,(2分)则“是品器台各组数据的颜率分别为0.03,0.05,0.16,0.36，0.32,0.08,所以A5=48,D水=36.所以均数为2.5×0.03+7.5×0.05+12.5×0.16+在△ADE中，AD=AE+DE,17.5×0.36+22.5×0.32+27.5×0.08=18.15.所以A5LDE,即BDAC(4分)(5分)又因为PA∩AC=A,(6分)(1)因为在校教师每周坐公交的次数在20次以上的所以BD⊥面PAC(Ⅱ)以A为坐标原点，AB,AD,P所在直线分别为频序为0.4，X-B(4,0.4),X的所有可能取值为0、1,2,3,4,(6分】x,y,轴建立空向直角坐标系如图所示·数学（里科）答30·100%

如纵回州脚么图3帐塔e扫提地钻解园·FH城中m丽号单望以坦恒1D3以示听第10段材料，回答第17至20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17.What will Todd Messegee be responsible for?C.Starring in a playZ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A.Writing a play.B Directing a play.18.What will participants mainly do by the fourth day?周至县2023、2024学年度高考第三次模拟考试出5秒钟的作答时间。每段对话或独白读两遍。C.Perform on stage formally.(A Practice the play.B.Attend a lecture on acting.听第6段材料，回答第67题。19.What is suggated by the speaker?英语试题6.Where did the man go for his summer vacation?(B.To omo famous citiesC.Toa fishing villageReceiving top-level training.A.To the mountains.县市区7.What did the man enjoy most in his holiday?B.Developing complex characters注意事项：B.Catching fiahon the beachC.Signing up early.A Swimming周身1.本试卷共10页，全卷调分150分，答题时间为120分钟。2.答卷前，考生须准确填写自己的姓名、准考证号，并认真核准条形码上的姓名、准考证号听第7段材料，回答第89题。20.What is the speaker doing?A.Making a plan for a meeting.3、回答进择恩时，选出每小题答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑。如需8.What is the man'atitude towards the womnB.Announcing arrangements fora play学校改动，用橡皮擦干净后，再进涂其它答案标号。回答非进择圆时，将答案写在答题卡上。写在本试A Concerned9.What will the man do next?Cecruiting actor for a playet his ticket checked.B.Make a call.C.Book a ticket第二部分：阅读理解（共两节，满分40分）参上无效。用至，中4.考试结束后，监考员将答题卡按顺序收回，装袋整理：试题不回收。听第8段材料，回答第10至12。第一节（共15小题，每小题2分，满分30分）10.What does the woman have to do for the party?阅读下列四篇短文，从每题所给的四个选项(A,B,C和D)中，选出最佳选项，并在答题卡上将第一部分：听力（共两节，满分30分）A.Buy paintings._（)Prepare candies.C.Design costumes该项涂黑。A姓名做题时，先将答案标在试卷上。听力部分结束前，你将有两分钟的时间将试卷上的答案转涂到11.Why is the man having the party?答题卡上。A.To make Lucy happy.Museums in Baltimore第一节（共5小题，每小题1.5分，满分7.5分】B.To make some new friends.The Walters Art Museum听下面5R对话。每R对话后有一个小题，从题中所给的A,B,C三个进项中进出最佳选项。听完The Walter Art Museum contins 36,000 objects from around the world.Walking through the每R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。每及对话仅读一迪。12.What does the woman think of the man?ounter a stunning panorama (of thousands of years of art蓝级A Kind.B.StrictC.Patient.南工60跑例：How much is the shirt?听第9段材料，回答第13至16圆。from romantic 17th-century images of French gardens to fascinating Ethiopian icons()ancientC￡9.15.Roman sarcophegi(石t棺)，and peacefulimages of the Buddha.A.￡19.15.B.E9.1813.How does the woman deal with her old clothTickets:$9.5.Free for children aged 7 and under.答案是C。A.By throwing them as waste.Baltimore Museum of Industry试场B.By donating them toa charity.Where does the probably take place?C.In the classroomBy putting them into the reeyoling box.The Baltimore Museum of Industry celebrates the innovators,entrepreneurs and workerswhoAAt the dentist's.At a washroom.romoed this por city nt the industriae.Fromgment makingtoairplnemnufturngi214.What do we know about the man?2.What is the woman?the discover how their pioneering the region'(B A guide.C.A waitress.AHe likes to buy big furnitureA.A secretary.B.He often buys second-hand furniture.l860 s tuna(金t枪鱼）canning factory on a five-acre water front campus,the BMI offers dynamieexhibitions and hands-on activities for guests of all ages.考号3.How will the wman probably go to the libra?C.By car.C.His sofa has been changed three times.4.On foot.By subway15.Why did the woman buy a new TV?Tickets:$15.Half price for children.4.Why does the woman call?A.The previous one was out of dateBaltimore Museum of ArtA.To cancel a meeting.B.The previous one was damaged by the lightening88888B To rearrange an appointment.The previous one was destroyed by her childrerand contemprryr Ithsorkofar-including the larest hling fby HenrMC.To seek advice for an emergeney16.How does the man sound in the end?world.The the mui-year project5.When will Mary bave a job interview?B.Annoyed.C.Satisfied个9：00amB.At 10:00 am.C.A3:00pm.网至县商考英酒第三次模位考试-1（共10）周至县高考英语第三次模拟考试-2-（共0页）周至县高考英语第三次模拟考试-3-（共1心页）88888

6:30总656l8④猜题卷C_物理卷_答案.pdf3Q添加标签佩佩教育·2024年普通高中学业水选择性考试糊南四大名校名师团队猜题卷(C)物理参考答案1.A【解析】原子核发生一次B衰变后，由电荷数守恒可知，新核的电荷数要增加1，选项A正确；原子核发生一次α衰变后，由质量数守恒可知，新核的质量数要减少4，选项B错误；用放射性同位素的Y射线照射食品可以延长食品的保存期，选项C错误；一种元素的各种同位素都有相同的化学性质，选项D错误。故正确选项为A。2.C【解析】由波形图得到波长X=4m,故选项A错误：由题意可知：0.5s=nT,得到周期T=0,5s(n=1,2,3,…),最大周期为Tm=0.5s,最小频率fm=05Hz=2H,故选项B错误：波速为v=宁=8mm/s,当n=1时，m有最小值为8m/s,故选项C正确：1=0到1=0.2s时刻经历的周期数为号，当n取5的整数倍时，x=1m处的质点经过衡位置，故选项D错误。3.D【解析】在图示位置，线框和磁感线垂直，磁通量最大，磁通量变化率为零，选项A错误；线框中产生的感应电动势最大值E.=NBLa=25厄V,选项B错误：电动势有效值E-号E。=25V,流过溶新器的电流为1=是-号A=0,4A,选项C错误：由U=25V.U=20V,可知变压暑厚，副我图E数之比为是光-号，选项D正确，4.B【解析】两小球在C点发生弹性碰撞后，小球2静止在C点，小球1从A点到C点做抛运动，若将小球1置于C点右侧少许，则小球1在A点速度v要稍小，即弹簧压缩量设定为比△x稍小时，也可能撞到小球2，游戏也可能成功，选项A错误；若将小球1置于C点左侧少许，则小球1在A点速度v要稍大，即弹簧压缩量设定为比△x稍大时，也可能撞到小球2，游戏也可能成功，选项B正确：若只将两小球均换成质量稍小或稍大的相同小球，则小球2到达C点时速度将变大或变小，则游戏不可能成功，故选项C和D都错误。5.C【解析】对该行星和地球有：p一4M元R,对它们的同步卫星均有：GM=m禁。所（尽）广c,代入两星的密度比和自转周期比可解得该行星的同步卫星轨道半径与该行星半径比值约为4.95。正确选项为C。6.A【解析】设粒子的质量为m,电量为q,从P点到O点的时间为t,到达O点的速度大小为0,与正方向的夫角为0。则有2L=,L-号5，m-二=可求得0=30，设粒子从0到M运动的轨道半径为，则有b=四，由几何关系可知，=2L。联立以上各式可求得唱-得，故正确装须为A7.AD【解析】因小球C与天花板间的细线沿坚直方向，对小球C进行受力分析可知，BC间的细线弹力为零，故B能静止在斜面的条件是u≥tan0:对AB整体受力分析可知，斜面体A受地面的摩擦力为零。故正确选项为AD。8.BC【解析】由几何关系可知射向C点的激光入射角为20，折射角为0，该圆柱对射向C点的激光的折射率为n=血29sin A2os0,而射向D点的激光对应的OD>0,该圆柱对射向D点的激光的折射率小于该圆柱对射向C点的激光的折射率，故射向C点的激光是绿色的，选项A错误，选项C正确：射向D点的激光在圆柱中的速度比射向C点的激光大，选项B正确；根据对称性，能从D点射入的激光一定能从B点射出，不会发生全反射，选项D错误。故正确选项为B和C。9.BD【解析】小包裹P先做匀加速直线运动，下滑的加速度a1=gsim37°+出gc0s37°=12m/s2,与传送带共速时经过的时问为一。号=0.2s,当与传送带共建后，因为A=075=m37,则P将与传送带相对静止：故在1=0.4s时刻小包素P和Q之间相距△1-十(t一)=0.72m,选项A错误。小包裹Q开始下滑的加速度a,=gsin37°+g00s37°-10m/s,与传送带共速时经过的时间为t2=卫=0.24s:当与传送带共速后，又以加速度ag=gsin37°-gcos37°=2m/s2,继续加速下滑，故在t=0.64s时刻小包裹P和Q在传送带上的距离最远，选项B正确。在t=0.64s时刻小包裹P和Q在传送带上的距离为△n=△十(t-)=1.008m,小包裹P到达B处的时刻为t=十受)=1.6s,在t=0.64s到t=1.6s(时间间隔y=.96S)内小包裹Q和P的相对位移为△r3=y+@Y-A=0.96m<△2，故小包裹Q在传送带上一定不能追上P。故选项C错误，选项D正确。物理试题参考答案第1页10.AC【解折1由题意可知：Uu-"-3V.Uc-=-9,令9=0，所以m=-3V=9V,求得E一e0-12V.将电场强度E分解为水分量E,(洛CA方向)和整直分量E(沿DB方向)，则E-光75V/m,E,-箭=100V/m,故E=√E+区-125V/m,故选项A正确：设E与CA方向的夹角为0，则E以U0入m0一是-专，而CB与CA方向的夹角的正切为？，电场程度的方向不沿CB方向，放选项B错误：因为电场强度不沿CB方向，也不沿DB方向，所以电子从C到B或从D到B都不能是直线运动，到达B点时应还有动能，根据能量守恒可知，一个初动能为I2eV的电子从C点出发不可能到达B点，一个初动能为l5©V的电子从D点出发不可能到达B点，故选项C正确，选项D错误。☑刀0o▣0标注全屏格式转换全部工具

(2)由题意可知λ=12m的波形向x轴负方向移1.6m,可知t2=2.3s时，质点b则周期为T=入=1,2s位于衡位置，加速度为零，B错误；设质点Q的振动方程为y=20sin(经+9)cm=20sin(经4+9）cm,由题甲图可,=0时的波形方程y=Asm(经：十p)知，4=1.5s时y,=10cm,代入方程解得p=牙，则质点当1m时经十g=0a的振动方程为y=20sin(受+号）em,当：=2.3s时，可知=一君·A可得-20n(管×2.3计号）m=20m传)m≠0，此x=2m在t。=0时的位置y:=2时质点Q不是处于衡位置，故速度没有达到最大值，C错x=2m处质点的振动方程y=Asin+》误；假设以题图甲时刻作为x=0处质点振动的初始时刻，则工=0处质点的最动方程为=20sin()cm,由于当t=0.2s时y=-21.4s=1.2s十0.2s=T+0.2s,则0.2s时，有y=所以振幅A=4cm所以z=2m质点的表动方程y=4n受+）cm。20sin(管×0.2）小cm=10v5cm,故从题图甲所示时刻开始再经过1.4s,x=0处质点通过的路程为s=4A十y'=4X微专题9220cm+10W3cm=(80+10W3)cm,D正确。真考3.A由题图乙可知t=1.0s时刻质点Q正沿y轴负方向运1.ABD由波的图像可知波的波长入=4m,由振动图像可知动，所以此时质点Q应位于波传播方向波形的上坡上，则该波的周期T=2s,所以波的传播速度0=子=2m/s,A正波沿x轴负方向传播，故A正确；该波的周期为T=2.0s,所以从t=1.0s到t=1.5s时间内，质点P从波峰位置沿确；由振动图像可知质点P在t=0时刻经衡位置沿一yy轴负方向运动四分之一个周期，则t=1.5s时，质点P运方向运动，结合波形图由同侧法可知波沿一x方向传播，即：动至衡位置，位移为零，故B错误；从t=1.0s到t=向左传播，B正确；振幅为质点离开衡位置的最大距离，故：1.5s时间内，质点Q从衡位置沿y轴负方向运动四分之由波的图像可知波的振幅A=5cm,C错误；质点P在0~一个周期，则t=1.5s时，质点Q位于波谷处，回复力方向7s时间内，即3.5个周期内通过的路程为s=3.5×4A=沿y轴正方向，故C错误；该波的周期为T=2.0s,波长为70cm,D正确。2.B由题图乙可知质点L的振动情况，该时刻质点L向y入=8m,所以波速为0=分=4m/s,故D错误。轴正方向振动，根据上下坡法或者移法可知，该横波沿x微专题93轴正方向传播，质点N该时刻向y轴负方向运动，故A错真考误，B正确；质点L只在衡位置附近沿y轴方向上下振1.C由振动图像可看出该波的周期是4s,A错误；由于Q、P动，波传播时，质点不会沿x轴方向移动，故C错误；该时刻质点K与M的速度为零，质点K加速度方向为一y方向，两个质点振动反相，可知两者间距离等于(n十2)以=6m质点M加速度方向为十y方向，故D错误。3.AC由振动图像可知波的振幅A=20cm,波的周期T=《n=0,1,2,所以0三2n十m/s(n=0,1,2,…)为12s,从振动图像上可以看出t=7s时衡位置位于原点O检验知，B错误；由P质点的振动图像可看出，在4s时P的质点相对衡位置的位移为y,=二)A,且向下运动，故质点在衡位置向上振动，C正确；由Q质点的振动图像可看出，在4s时Q质点在衡位置向下振动，D错误。B,D错误；A图该质点位移符合条件，且当波沿x轴负方向2.AC机械波的传播方向不确定，所以需要分情况讨论。若传播时向下运动，故A正确；C图该质点位移符合条件，且机械波沿x轴正方向传播，在t1=2s时O点振动方向竖直当波沿x轴正方向传播时向下运动，故C正确。真练向上，则传播时间△4=4-4，=3s满足4=T十1.BC由上下坡法可知波沿x轴负方向传播，大小为=行12nT(n=0,1,2,3,…),解得T=4n千3s(n=0,1,2,3,…),0.4m/s=60m/s,故A错误；由题图乙可得当t=0.35s时24当n=0时，解得周期T=4s,A正确，B错误；若机械波沿x轴负方向传播，在t2=5s时O点处于波谷，则△t=质点P恰好回到衡位置，故B正确；在0~0.2s时间内质点P运动的路程为s=2A=10cm,故C正确；由题图乙T+nT(n=0,12,3,…),解得T=4n千s(n=0,1,2,12可得在0~0.1s时间内，质点P先向y轴负方向运动，后向3,…),当n=0时，解得周期T=12s,C正确，D错误。y轴正方向运动，故D错误。真练2.Dx=0处质点t=1.5s时刻向y轴正方向振动，根据同1.A波长为AB中点到CD中点长度的2倍，入=2×100×4cm=侧法可知绳波沿x轴负方向传播，波的周期为1.2s,波长为2,4m,波速为口=产=2m/s,A错误：从4，=L55到8m放A正确：根据是意T+召=子a=0,1.2…,波边t2=2.3s,绳波向x轴负方向传播的距离为△x=v(t2一t1)=为v=7,当n=0时v=20m/s,当n=1时v=60m/s,故2×0.8m=1.6m,t2=2.3s时刻的波形为t1=1.5s时刻：B错误；质点振动方向与水传播方向垂直，故C、D错误。233

根的个数.令t=f(x),则方程f(f(x)-f(x)-1=0等/4+1+4×价于)2=7f)=t+1选项向量的模正误第二步：求曲线y=lnx在点(1,0)处的切线方程，得曲线y=lnt和y=t+1的交点情况1e,+2e,l=Ve,+2e=√1+4+4×2=7正确对于函数y=血x,易知当x=1时y=0,y=子yB21=(2e-e2)=4+1-=√5错误1,故曲线y=nx在点(1,0)处的切线方程为y=x-1,2e,-3e,1=V2,-3了=√4+9-2x6×2=7正确因此曲线y=nt和y=t+1无交点.（技巧：通过研究曲线y=lnx在点(1,0)处的切线，数形结合判断曲线y=ln&和D%+e=e+g-√1+9+2x3=√3错误2y=t+1的交点情况)10.BCD二项式定理+常数项+所有项的系数和+导数第三步：求方程f(t)=t+1的根，并判断该根的大致的几何意义范围将y=t+1代人y=-t2-2at,得t2+(1+2a)t+1=0,则界名师教审题思维导引44=42+4a-3,令4-0，得a=方或e=-先根据展开式的项数不超过9，得到1≤几≤8，并利用2,故当0<二项式定理写出二项展开式的通项，再根据展开式中a<2时，4<0，y=-f-2al与y=t+1无交点，作出函存在常数项求出n的所有取值，即可判断选项A;当n取最大值时求出，k的值，根据二项展开式的通项即数y=f(t)和y=t+1的大致图象如图所示，结合图象可可求出常数项，进而可判断选项B；当n取最小值时可知，方程f(t)=t+1有且仅有1个解，且此解就是方程lnt+t+1=0的解.易知函数h(x)=lnx+x+1是增函得f(x)的解析式，然后利用导数的几何意义求出a的值，最后进行检验，即可判断选项C;令x=1可得二项数，且()=是>0，a(分)=-2hn2+<0,(点拔展开式中的所有项的系数和，进而得到α的值，即可因为4=256>243=3>8，所以4>e,故2n2>子)判断选项D.因此方程n+4+1=0的解么∈（任，忌。【解题思路】二项式(+士)”的展开式的通项为第四步：求函数g(x)的零点个数C(am)t.()=a-Cx2-,因为展开式的项数不又当x≤0时，-2-2ax≤a2<4,所以--2a=h,无超过9，所以n+1≤9，所以1≤n≤8，（注意：二项式(a+解，显然IIn xl=。有2个解，所以函数g(x)有2个零b)”中的几为正整数)点，故选B.因为展开式中存在常数项，所以2-3k=0有解，即k=e+有解，所以几能被3整除，因此n=3或几=6选项A:显然n的所有取值组成的集合中有且仅有2个y=f()元素，A错误。-2a选项B:当n取最大值时，n=6,此时k=4,故a2C:=15a2=30,解得a=±√2，B正确.选项C:当n取最小值时，n=3,此时f(x)=(ax2+'则f1)=(a+1)'f(x)=3(ax2+(2ax名师敲重点解题关键《q4求解本题的关键：(1)根据曲线y=nx在点(1,0)处之)，则f"(1)=3a+1(2a-1)=0,解得a=-1或的切线方程判断曲线y=lnt和y=6+1的交点情况；a=子当a=-1时，函数)=(a2+)》的图象在(2)求方程代t)=t+1的根，并根据函数的单调性及零点存在定理判断该根的大致范围；(3)判断f(x)的点(1，f(1))处的切线与x轴重合，不符合题意，当a=图象与直线y=,千<，<。的交点情况。2时，符合题意，C正确（易错：廊分学生求出a的两个9.AC向量的夹角+向量的模值后没有检验这两个值是否符合题意，直接得到C错误)【解题思路】12e,+e2|=√/(2e,+e2)产选项D:对于(a2+)”,令=l,则(a+1)=0,解得器器数1高考领航卷·敢学答聚一19（第3套）

根的个数.令t=f(x),则方程f(f(x)-f(x)-1=0等/4+1+4×价于)2=7f)=t+1选项向量的模正误第二步：求曲线y=lnx在点(1,0)处的切线方程，得曲线y=lnt和y=t+1的交点情况1e,+2e,l=Ve,+2e=√1+4+4×2=7正确对于函数y=血x,易知当x=1时y=0,y=子yB21=(2e-e2)=4+1-=√5错误1,故曲线y=nx在点(1,0)处的切线方程为y=x-1,2e,-3e,1=V2,-3了=√4+9-2x6×2=7正确因此曲线y=nt和y=t+1无交点.（技巧：通过研究曲线y=lnx在点(1,0)处的切线，数形结合判断曲线y=ln&和D%+e=e+g-√1+9+2x3=√3错误2y=t+1的交点情况)10.BCD二项式定理+常数项+所有项的系数和+导数第三步：求方程f(t)=t+1的根，并判断该根的大致的几何意义范围将y=t+1代人y=-t2-2at,得t2+(1+2a)t+1=0,则界名师教审题思维导引44=42+4a-3,令4-0，得a=方或e=-先根据展开式的项数不超过9，得到1≤几≤8，并利用2,故当0<二项式定理写出二项展开式的通项，再根据展开式中a<2时，4<0，y=-f-2al与y=t+1无交点，作出函存在常数项求出n的所有取值，即可判断选项A;当n取最大值时求出，k的值，根据二项展开式的通项即数y=f(t)和y=t+1的大致图象如图所示，结合图象可可求出常数项，进而可判断选项B；当n取最小值时可知，方程f(t)=t+1有且仅有1个解，且此解就是方程lnt+t+1=0的解.易知函数h(x)=lnx+x+1是增函得f(x)的解析式，然后利用导数的几何意义求出a的值，最后进行检验，即可判断选项C;令x=1可得二项数，且()=是>0，a(分)=-2hn2+<0,(点拔展开式中的所有项的系数和，进而得到α的值，即可因为4=256>243=3>8，所以4>e,故2n2>子)判断选项D.因此方程n+4+1=0的解么∈（任，忌。【解题思路】二项式(+士)”的展开式的通项为第四步：求函数g(x)的零点个数C(am)t.()=a-Cx2-,因为展开式的项数不又当x≤0时，-2-2ax≤a2<4,所以--2a=h,无超过9，所以n+1≤9，所以1≤n≤8，（注意：二项式(a+解，显然IIn xl=。有2个解，所以函数g(x)有2个零b)”中的几为正整数)点，故选B.因为展开式中存在常数项，所以2-3k=0有解，即k=e+有解，所以几能被3整除，因此n=3或几=6选项A:显然n的所有取值组成的集合中有且仅有2个y=f()元素，A错误。-2a选项B:当n取最大值时，n=6,此时k=4,故a2C:=15a2=30,解得a=±√2，B正确.选项C:当n取最小值时，n=3,此时f(x)=(ax2+'则f1)=(a+1)'f(x)=3(ax2+(2ax名师敲重点解题关键《q4求解本题的关键：(1)根据曲线y=nx在点(1,0)处之)，则f"(1)=3a+1(2a-1)=0,解得a=-1或的切线方程判断曲线y=lnt和y=6+1的交点情况；a=子当a=-1时，函数)=(a2+)》的图象在(2)求方程代t)=t+1的根，并根据函数的单调性及零点存在定理判断该根的大致范围；(3)判断f(x)的点(1，f(1))处的切线与x轴重合，不符合题意，当a=图象与直线y=,千<，<。的交点情况。2时，符合题意，C正确（易错：廊分学生求出a的两个9.AC向量的夹角+向量的模值后没有检验这两个值是否符合题意，直接得到C错误)【解题思路】12e,+e2|=√/(2e,+e2)产选项D:对于(a2+)”,令=l,则(a+1)=0,解得器器数1高考领航卷·敢学答聚一19（第3套）

根的个数.令t=f(x),则方程f(f(x)-f(x)-1=0等/4+1+4×价于)2=7f)=t+1选项向量的模正误第二步：求曲线y=lnx在点(1,0)处的切线方程，得曲线y=lnt和y=t+1的交点情况1e,+2e,l=Ve,+2e=√1+4+4×2=7正确对于函数y=血x,易知当x=1时y=0,y=子yB21=(2e-e2)=4+1-=√5错误1,故曲线y=nx在点(1,0)处的切线方程为y=x-1,2e,-3e,1=V2,-3了=√4+9-2x6×2=7正确因此曲线y=nt和y=t+1无交点.（技巧：通过研究曲线y=lnx在点(1,0)处的切线，数形结合判断曲线y=ln&和D%+e=e+g-√1+9+2x3=√3错误2y=t+1的交点情况)10.BCD二项式定理+常数项+所有项的系数和+导数第三步：求方程f(t)=t+1的根，并判断该根的大致的几何意义范围将y=t+1代人y=-t2-2at,得t2+(1+2a)t+1=0,则界名师教审题思维导引44=42+4a-3,令4-0，得a=方或e=-先根据展开式的项数不超过9，得到1≤几≤8，并利用2,故当0<二项式定理写出二项展开式的通项，再根据展开式中a<2时，4<0，y=-f-2al与y=t+1无交点，作出函存在常数项求出n的所有取值，即可判断选项A;当n取最大值时求出，k的值，根据二项展开式的通项即数y=f(t)和y=t+1的大致图象如图所示，结合图象可可求出常数项，进而可判断选项B；当n取最小值时可知，方程f(t)=t+1有且仅有1个解，且此解就是方程lnt+t+1=0的解.易知函数h(x)=lnx+x+1是增函得f(x)的解析式，然后利用导数的几何意义求出a的值，最后进行检验，即可判断选项C;令x=1可得二项数，且()=是>0，a(分)=-2hn2+<0,(点拔展开式中的所有项的系数和，进而得到α的值，即可因为4=256>243=3>8，所以4>e,故2n2>子)判断选项D.因此方程n+4+1=0的解么∈（任，忌。【解题思路】二项式(+士)”的展开式的通项为第四步：求函数g(x)的零点个数C(am)t.()=a-Cx2-,因为展开式的项数不又当x≤0时，-2-2ax≤a2<4,所以--2a=h,无超过9，所以n+1≤9，所以1≤n≤8，（注意：二项式(a+解，显然IIn xl=。有2个解，所以函数g(x)有2个零b)”中的几为正整数)点，故选B.因为展开式中存在常数项，所以2-3k=0有解，即k=e+有解，所以几能被3整除，因此n=3或几=6选项A:显然n的所有取值组成的集合中有且仅有2个y=f()元素，A错误。-2a选项B:当n取最大值时，n=6,此时k=4,故a2C:=15a2=30,解得a=±√2，B正确.选项C:当n取最小值时，n=3,此时f(x)=(ax2+'则f1)=(a+1)'f(x)=3(ax2+(2ax名师敲重点解题关键《q4求解本题的关键：(1)根据曲线y=nx在点(1,0)处之)，则f"(1)=3a+1(2a-1)=0,解得a=-1或的切线方程判断曲线y=lnt和y=6+1的交点情况；a=子当a=-1时，函数)=(a2+)》的图象在(2)求方程代t)=t+1的根，并根据函数的单调性及零点存在定理判断该根的大致范围；(3)判断f(x)的点(1，f(1))处的切线与x轴重合，不符合题意，当a=图象与直线y=,千<，<。的交点情况。2时，符合题意，C正确（易错：廊分学生求出a的两个9.AC向量的夹角+向量的模值后没有检验这两个值是否符合题意，直接得到C错误)【解题思路】12e,+e2|=√/(2e,+e2)产选项D:对于(a2+)”,令=l,则(a+1)=0,解得器器数1高考领航卷·敢学答聚一19（第3套）

me s ouend Maon'm the四其中er teano cnance or是“First aid”。假如你是该中学学生the李华，篇英语短文，记录一次你实施急救的经quickly新防How did the accident happen?31-31Mason stood in silence with his head bowed."Edwin scored because ofHow did you give first aid?读my poor defense."Mason thoughtOvertime(加时赛)began.As time passed,the Engineers led by2How did you feel?points.Mason lost heart and felt that they would lose.Seeing this,Mason'scoach called a time-out.写作要求："Cheer up,Mason!The game is not over yet.We still have a chance to()要点齐全，可适当增加细节，以使行文连贯：win,”the coach said.(2)条理清楚，语言流畅，语句通顺，标点正确：The game went on.As Edwin dribbled,Mason followed him closely.(3)80词左右Mason was waiting or a chance.Edwin passed the ball.Right now!Mason可能用到的词汇：bandage n.绷带：apply涂，敷；bleed,流血the pass and ran to the basket.He saw hiseammate Joeand passed thecaught the ball,turned around,jumped up and dropped it into the4hree-pointes up.The Blazers won!Who can say they know theestore the game is over?g四【写作指导】三本篇写作要求写一篇短文描述自己实施急救)70.From the underlined sentences we know that Mason felt的经历。短文可分为三段：第一段介绍事件是怎B.nervousC.sorryD.angry傻71.Why dd the coach cou样发生的；第二段为主体段，描述实施急救的具A.To encourage Mason.B.To give Mason a break.C.To teach Mason some defensive skills.D.To have another player replace Mason.写体过程；第三段总结自己的感悟，强调掌握急救)72.What does the知识的必要性。nt to tell us?Practice mak©主题：人与自我之生活与学C.Teamwork is the key to victory◎子主题：安全与防护D.It's never a good idea to give up easily.©时态：一般过去时回人称：第一人称70-72CADg之并（敌水四【写作指导】三回行文布局：there would be a test.All the students were seated,waiting for the test tobegin.①事件发生的原因there were noWhen I put the tomatoes in the pan,my hand was图burned by the hot oilwhat you see there.写At the end of the class,the teacher collected all the test papers.Without②实施急救的过程exception ()each student described the dot,trying to explain itsFirst,I put my hand under cool running water.Then,I took out a clean bandage to dry the burn.Everyone focused on (the black dot.No one wrote about the whiteFinally,I applied the medicine to the burn.part.The same thing happens in our lives.We seem toonly to the black dots like money or work problems.he③表达自己的感悟compared to those we have in our lives such as love,friendship,and family.01Get your attention away from the black dots in your life and enjoy everyI have realised that having the knowledge of firmoment in life.”aid is very important.四【写作指导】三73.Where did the story happen?(不超过5个词)⊙必背句型：73.In a classroom.Unluckily,when I put the tomatoes in the pan,my74.How did the students feel when they got the testhand was burned by the hot oil.papers?(不超过10个词)Suffering from great pain,I immediately put my读74.They felt surprised.写hand under cool running water.75.According to the teacher,what is the main lessonI found the medicine box and took out a cleanof the test'?(不超过10个词)bandage to dry the burn.75.To teach students to enjoy every moment in lifeI applied the medicine to the burnThanks to the immediate action,the burn on my1…hand healed after two days.From the experience,I have realised that havingknowledge of first aid is very important.Ⅸ.单词拼写（共5小题；每小题1分，满分5分）四根据首字母及汉语提示，完成下列单词的拼写，使句意明确，语言通四76.Mum,can you give me as(勺子)2 I can't use chopsticks wellCertainly.77.Fiona was b(失明的)at birth,so she has never seen thebeautiful world【参考范文】写78.-Bob,please f(去取来)me a chair.Ican't reach then the top shelf写Last Saturdav.my parents went out and I was at home alone.At noon,I decided to make myown lunch.Unluckily,when I put the tomatoes in the pan,my hand was burned by the hot oil.Okay,Mum!Suffering from great pain,I immediately put my hand under cool running water and washed79.Three q(四分之一)of the36 students in my class failedit for about 20 minutes to reduce the temperature.Then I found the medicine box and took out athe maths exam.clean bandage to dry the burn.Finally,I applied the medicine to the bum.(预计)youThanks to the immediate action,the burn on my hand healed after two days.From theexperience,I have realised that having the knowledge of first aid is very important.76.spoon 77.blind 78.fetch79.quarters 80.expected

me s ouend Maon'm the四其中er teano cnance or是“First aid”。假如你是该中学学生the李华，篇英语短文，记录一次你实施急救的经quickly新防How did the accident happen?31-31Mason stood in silence with his head bowed."Edwin scored because ofHow did you give first aid?读my poor defense."Mason thoughtOvertime(加时赛)began.As time passed,the Engineers led by2How did you feel?points.Mason lost heart and felt that they would lose.Seeing this,Mason'scoach called a time-out.写作要求："Cheer up,Mason!The game is not over yet.We still have a chance to()要点齐全，可适当增加细节，以使行文连贯：win,”the coach said.(2)条理清楚，语言流畅，语句通顺，标点正确：The game went on.As Edwin dribbled,Mason followed him closely.(3)80词左右Mason was waiting or a chance.Edwin passed the ball.Right now!Mason可能用到的词汇：bandage n.绷带：apply涂，敷；bleed,流血the pass and ran to the basket.He saw hiseammate Joeand passed thecaught the ball,turned around,jumped up and dropped it into the4hree-pointes up.The Blazers won!Who can say they know theestore the game is over?g四【写作指导】三本篇写作要求写一篇短文描述自己实施急救)70.From the underlined sentences we know that Mason felt的经历。短文可分为三段：第一段介绍事件是怎B.nervousC.sorryD.angry傻71.Why dd the coach cou样发生的；第二段为主体段，描述实施急救的具A.To encourage Mason.B.To give Mason a break.C.To teach Mason some defensive skills.D.To have another player replace Mason.写体过程；第三段总结自己的感悟，强调掌握急救)72.What does the知识的必要性。nt to tell us?Practice mak©主题：人与自我之生活与学C.Teamwork is the key to victory◎子主题：安全与防护D.It's never a good idea to give up easily.©时态：一般过去时回人称：第一人称70-72CADg之并（敌水四【写作指导】三回行文布局：there would be a test.All the students were seated,waiting for the test tobegin.①事件发生的原因there were noWhen I put the tomatoes in the pan,my hand was图burned by the hot oilwhat you see there.写At the end of the class,the teacher collected all the test papers.Without②实施急救的过程exception ()each student described the dot,trying to explain itsFirst,I put my hand under cool running water.Then,I took out a clean bandage to dry the burn.Everyone focused on (the black dot.No one wrote about the whiteFinally,I applied the medicine to the burn.part.The same thing happens in our lives.We seem toonly to the black dots like money or work problems.he③表达自己的感悟compared to those we have in our lives such as love,friendship,and family.01Get your attention away from the black dots in your life and enjoy everyI have realised that having the knowledge of firmoment in life.”aid is very important.四【写作指导】三73.Where did the story happen?(不超过5个词)⊙必背句型：73.In a classroom.Unluckily,when I put the tomatoes in the pan,my74.How did the students feel when they got the testhand was burned by the hot oil.papers?(不超过10个词)Suffering from great pain,I immediately put my读74.They felt surprised.写hand under cool running water.75.According to the teacher,what is the main lessonI found the medicine box and took out a cleanof the test'?(不超过10个词)bandage to dry the burn.75.To teach students to enjoy every moment in lifeI applied the medicine to the burnThanks to the immediate action,the burn on my1…hand healed after two days.From the experience,I have realised that havingknowledge of first aid is very important.Ⅸ.单词拼写（共5小题；每小题1分，满分5分）四根据首字母及汉语提示，完成下列单词的拼写，使句意明确，语言通四76.Mum,can you give me as(勺子)2 I can't use chopsticks wellCertainly.77.Fiona was b(失明的)at birth,so she has never seen thebeautiful world【参考范文】写78.-Bob,please f(去取来)me a chair.Ican't reach then the top shelf写Last Saturdav.my parents went out and I was at home alone.At noon,I decided to make myown lunch.Unluckily,when I put the tomatoes in the pan,my hand was burned by the hot oil.Okay,Mum!Suffering from great pain,I immediately put my hand under cool running water and washed79.Three q(四分之一)of the36 students in my class failedit for about 20 minutes to reduce the temperature.Then I found the medicine box and took out athe maths exam.clean bandage to dry the burn.Finally,I applied the medicine to the bum.(预计)youThanks to the immediate action,the burn on my hand healed after two days.From theexperience,I have realised that having the knowledge of first aid is very important.76.spoon 77.blind 78.fetch79.quarters 80.expected