佛山市第一中学2019届高三上学期期中考试

2019-2020学年广东省佛山一中高三（上）期中语文试卷试题数：27，满分：1501.（单选题，2分）下列对加点词解释有误的一项是（）A.未尝不叹息痛恨．于桓灵也（遗憾）既．克，公问其故（已经）B.胡．人不敢南下而牧马（什么）齐与楚从亲，惠王患．之（疾病）C.而或．长烟一空（有时候）君子疾．夫舍曰“欲之”而必为之辞（厌恶）D.项王即．日因留沛公与饮（当）君美甚，徐公何能及．公也（比得上）2.（单选题，2分）下列对加点词解释有误的一项是（）A.何竟日默默在此，大类．女郎也（似）距关，毋内．诸侯（接纳）B.可怜．体无比（同情）舟首尾长约八分有奇．（余）C.保民而王，莫．之能御也（没有人）予左迁．九江郡司马（调动官职）D.家祭无忘告乃．翁（你的）奉之弥．繁，侵之愈急（更加）3.（单选题，2分）下列对加点词解释有误的一项是（）A.且举世而誉之而不加劝．（勉励）贫贱有此女，始适．还家门（出嫁）B.纵一苇之所如．（到）沛公不胜．杯勺（受得住）C.若．属皆且为所虏（你）却．匈奴七百余里（打退）D.旧时栏楯亦遂增胜．（优美）素善．留侯张良（好的）4.（单选题，2分）下列对加点词解释有误的一项是（）A.使．六国各爱其人（让）数．罟不入池（细密）B.彼且奚适．也（到……去）无乃尔是．过与（宾语前置的标志）C.如其克．谐，天下可定也（能够）则胜负之数．，存亡之理（命运）D.其故封识．具存（记号）范增数．目项王（屡次）5.（单选题，2分）下列对加点词解释有误的一项是（）A.举酒属．客（嘱咐）此吾所以敢率．性就死不顾汝也（遵循）B.范增说．项羽曰（劝说）十三能织素．（白色的绸绢）C.有朋自远方来，不亦说．乎（喜欢）媵人持汤．沃灌（热水）D.吾妻之美我者，私．我也（偏爱）吾属．今为之虏矣（类）6.（单选题，2分）下列对加点词解释有误的一项是（）A.涕．落百余行（鼻涕）沛公今事有急，亡．去不义（死）B.暮而果大亡．其财（丢失）先破秦入咸阳者王．之（使……为王）C.郯子之徒．（类）寡人虽亡．国之余，不鼓不成列（灭亡）D.河曲智叟亡．以应（没有）沛公欲王．关中（统治）7.（单选题，2分）下列对加点词解释有误的一项是（）A.夜雪初霁，荞麦弥望．（视野）岁恶．不入，请卖爵子（收成不好）B.壬戌之秋，七月既望．（看见）人有恶．苏秦于燕王者（诋毁）C.彼且恶．乎待哉（什么）敢用是为怨望．，又投间蔑污使君（怨恨）D.望．秋先陨（接近）恶．!是何言也!（相当于“啊”）8.（单选题，2分）下列对加点词“相”“谢”的解释都有误的一项是（）A.好自相．扶将（互相）阿母谢．媒人（感谢）B.伯乐学相．马（察看）谢．家来贵门（辞别）C.至于幽暗昏惑而无物以相．之（辅助）旦日不可不蚤自来谢．项王（道歉）D.颠而不扶，则将焉用彼相．矣（引盲人行走的人）多谢．后世人（告诫）9.（单选题，2分）下列对加点字的解释有误的一项是（）A.缦立远视，而望幸．焉（皇帝到某处）作《师说》以贻．之（送给）B.又重之以修．能（美好的）淡妆浓抹总相宜．（合适）C.潭中鱼可百许．头（大约的数量）备他盗之出入．．与非常也（进入）D.以其乃华山之阳．名之也（南边）张良出，要．项伯（要求）10.（单选题，2分）下列对加点字的解释有误的一项是（）A.径造．庐访成（到……去）如今人方为刀俎，我为鱼肉．．（鱼的肉）B.吾与之共知．越国之政（了解）以手抚膺坐．长叹（坐着）C.投诸．渤海之尾（兼词，之于）流连信宿．．，不觉忘反（住两晚））D.非利足也，而致．千里（达到）烈士．．暮年，壮心不已（有节操抱负的人）11.（问答题，10分）阅读下列文字，完成下列问题。

广东省佛山市第一中学2019届高三上学期期中考试二、选择题：本题共8小题，每小题6分。

在每小题给出的四个选项中，第14-18题只有一项符合题目要求，第19-21题有多项符合题目要求。

全部选对的得6分，选对但不全的得3分，有选错的得0分。

1.甲、乙两物体从同一点出发且在同一条直线上运动，它们的位移-时间（x-t）图象如图所示，由图象可以得出在0-4s内（）A. 甲、乙两物体始终同向运动B. 4s时甲、乙两物体间的距离最大C. 甲的平均速度等于乙的平均速度D. 甲、乙两物体间的最大距离为6m2.如图所示，A、B、C三颗人造地球卫星绕地球做匀速圆周运动，已知三颗卫星的质量关系为m A=m B＜m C，轨道半径的关系为r A＜r B=r C，则三颗卫星相关物理量关系正确的是（）A. 线速度的大小B. 加速度的大小C. 向心力的大小D. 周期的大小3.如图所示为用绞车拖物块的示意图。

拴接物块的细线被缠绕在轮轴上，轮轴逆时针转动从而拖动物块。

已知轮轴的半径R=0.5m，细线始终保持水平；被拖动的物块初速度为零，质量m=1kg，与地面间的动摩擦因数μ=0.5；轮轴的角速度随时间t变化的关系是ω=kt，k=2rad/s2，g取10m/s2，以下判断正确的是()A. 物块的加速度逐渐增大B. 细线对物块的拉力逐渐增大C. 前2秒，细线对物块做的功为2JD. t=2s，细线对物块的拉力的瞬时功率为12W4.如图甲所示，当A、B两物块放在光滑的水平面上时，用水平恒力F作用于A的左端，使A、B一起向右做匀加速直线运动时的加速度大小为a1，A、B间的相互作用力的大小为N1．如图乙所示，当A、B两物块放在固定光滑斜面上时，此时在恒力F作用下沿斜面向上做匀加速时的加速度大小为a2，A、B间的相互作用力的大小为N2，则有关a1，a2和N1、N2的关系正确的是（）A.B.C.D.5.如图所示，小球以v0正对倾角为θ的斜面水平抛出，若小球到达斜面的位移最小，则以下说法正确的是（重力加速度为g）（）A. 小球空中运动时间为B. 小球的水平位移大小为C. 由于不知道抛出点位置，位移大小无法求解D. 小球的竖直位移大小为6.如图甲所示，在水平路面上有一质量为1×103kg的汽车(可视为质点)以10m/s的速度向右匀速行驶，由于路面发生变化，BC段阻力变小。

佛山市第一中学2019届高三上学期期中考试理综2018.11.13 本试卷共14页，38小题，满分300分。

考试用时150分钟。

注意事项：1．本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。

2．回答第Ⅰ卷时，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。

写在本试卷上无效。

3．回答第Ⅱ卷时，将答案写在答卷上。

写在本试卷上无效。

4．考试结束后，上交答卷。

可能用到的相对原子质量：H-1 C-12 N-14 O-16 Cu-64 Ag-108 Pb-207第Ⅰ卷一、选择题：本题共13小题，每小题6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．下列关于细胞结构与成分的叙述，正确的是A．细胞间的通讯都与细胞膜表面的糖蛋白有关B．核糖体是细胞内蛋白质的“装配机器”，主要由蛋白质和tRNA组成C．细胞核是遗传信息库，遗传信息的表达在细胞核中完成D．细胞骨架是由蛋白质纤维组成的网架结构，能保持细胞形态2．下列有关细胞生命历程的叙述，错误的是A．细胞分裂能力随分化程度的提高而增强B．癌变细胞的形态结构会发生显著变化C．细胞凋亡有助于机体维持自身的相对稳定D．衰老的细胞，细胞新陈代谢的速率减慢3．图Z2-3表示可在植物细胞内发生的一些生理过程，下列分析错误的是A．不同条件下发生a过程的场所可能不同B．a、c、d均发生在生物膜上C．在光照充足的条件下，a<b，c<dD．根尖分生区细胞内能发生a、c过程4．XY型性别决定的生物，群体中的性别比例为1：1，原因是：A．雌配子：雄配子=1：1 B．含X的精子：含Y的精子=1：1C．含X的配子：含Y的配子=1：1 D．含X的卵细胞：含Y的卵细胞=1：15．据你所知道杂交选育新品种之外，杂交的另一个结果是获得：A．纯种B．染色体变异C．基因突变D．杂交表现的优势6．下列有关生物多样性和进化的叙述中，不正确的是A．自然选择能定向改变种群的基因频率，决定了生物进化的方向B．新物种的形成通常要经过突变和基因重组、自然选择及隔离三个基本环节C．细菌在接触青霉素后会产生抗药性的突变个体，青霉素的选择作用使其生存D．蜂鸟细长的喙与倒挂金钟的筒状花萼是它们长期共同进化形成的相互适应特征7．下列说法正确的是A．金属钛性能非常优越，广泛应用于医疗及日常用品中。

2018-2019年度（上）高三级期中考试理科综合试题命题：陈学良 彭余凤 周志杰 审题：陈巨华 许荣喜 刘成刚2018.11.13本试卷共14页，38小题，满分300分。

考试用时150分钟。

二、选择题：本题共8小题，每小题6分。

在每小题给出的四个选项中，第14-18题只有一项符合题目要求，第19-21题有多项符合题目要求。

全部选对的得6分，选对但不全的得3分，有选错的得0分。

14.甲、乙两物体从同一点出发且在同一条直线上运动，它们的位移时间图象如图所示，由图象可以得出在0~4s 内（ ） A. 甲、乙两物体始终同向运动 B. 4s 时甲、乙两物体间的距离最大 C. 甲的平均速度等于乙的平均速度 D. 甲、乙两物体间的最大距离为6m15.如图所示，A 、B 、C 三颗人造地球卫星绕地球做匀速圆周运动，已知三颗卫星的质量关系为m A =m B ＜m C ，轨道半径的关系为r A ＜r B =r C ，则三颗卫星相关物理量关系正确的是（ ）A.线速度的大小C B A v v v =<B.加速度的大小C B A a a a =>C.向心力的大小C B A F F F <=D.周期的大小C B A T T T =>16.如图所示为用绞车拖物块的示意图。

拴接物块的细线被缠绕在轮轴上，轮轴逆时针转动从而拖动物块。

已知轮轴的半径R=0.5m ，细线始终保持水平；被拖动的物块初速度为零，质量m=1kg ，与地面间的动摩擦因数μ=0.5；轮轴的角速度随时间t 变化的关系是ω=kt ，k=2rad/s 2，g 取10m/s 2，以下判断正确的是( ) A. 物块的加速度逐渐增大B. 细线对物块的拉力逐渐增大C. 前2秒，细线对物块做的功为2JD. t=2s ，细线对物块的拉力的瞬时功率为12W17.如图甲所示，当A 、B 两物块放在光滑的水平面上时，用水平恒力F 作用于A 的左端，使A 、B 一起向右做匀加速直线运动时的加速度大小为a 1，A 、B 间的相互作用力的大小为N 1．如图乙所示，当A 、B 两物块放在固定光滑斜面上时，此时在恒力F 作用下沿斜面向上做匀加速时的加速度大小为a 2，A 、B 间的相互作用力的大小为N 2，则有关a 1，a 2和N 1、N 2的关系正确的是（ ）A.2121,N N a a >>B.2121,N N a a <>甲乙C.2121,N N a a ==D.2121,N N a a =>18.如图所示，小球以v 0正对倾角为θ的斜面水平抛出，若小球到达斜面的位移最小，则以下说法正确的是（重力加速度为g ）（ ） A. 小球空中运动时间为θtan 0g vB. 小球的水平位移大小为θtan 220g vC. 由于不知道抛出点位置，位移大小无法求解D. 小球的竖直位移大小为θ220tan 4g v 19.如图甲所示，在水平路面上有一质量为1×103kg 的汽车(可视为质点)以10m/s 的速度向右匀速行驶，由于路面发生变化，BC 段阻力变小。

1广东省佛山市第一中学2019届高三上学期期中考试理科综合参考答案及评分标准物理答案选择题实验题（共15分：22题每空2分，共8分；23题：第1空，1分，其它三个空，每空2分，共7分） 22.（1）2.01 （2）A 、B 、C （3）2)(21v m m m gh m C B A C ++= （4）①由于空气的阻碍作用；②纸带和打点计时器的摩擦作用；③纸带的重力影响（只要原因合理，均可给分）【解析】（1）铁环C 运动到圆环D 处时的速度大小即为铁环被挡住后，A 、B 匀速运动的速度，由纸带可以求得：s m s m v /01.2/02.0310)20.1928.31(2=⨯⨯-=- （2、3）要验证机械能守恒，铁环C 从开始下落到运动到D 处过程中，有： 2)(21)(v m m m gh m gh m m C B A A C B ++=-+， 由于A 、B 质量相等，因此要验证的表达式为：2)(21v m m m gh m C B A C ++=， 因此还需要测量的量为：A 、B 、C 。

（4）看答案23.（1）C （2）ON m OM m OP m 211+= （3）AC （4）2124h h s =【解析】P 为碰前入射小球落点的平均位置，M 为碰后入射小球的位置，N 为碰后被碰小球的位置，碰撞前入射小球的速度为： t OP v /1=碰撞后入射小球的速度为：t OM v /2=碰撞后被碰小球的速度为：t ON v /3=若322111v m v m v m +=则表明通过该实验验证了两球碰撞过程中动量守恒，由于小球做平抛运动的时间相同，可得： ON m OM m OP m 211+=故需要测量的工具有刻度尺和天平。

佛山一中2019届高三上学期期中考试文科数学答案一、选择题1. D2.A3. D4.B5.B6. C7.C8.B9. C 10.C 11. B 12. D二、填空题13. 14.12-n 15.π240 16.)1,0()1,( --∞ 9.函数为奇函数，排除 令， 令，则 故选三、解答题17、解：(1)由已知可得⎪⎩⎪⎨⎧+==++q a q a q 223123--------------------------------2分 解得，3=q 或4-=q (舍去)，62=a -----------------------------------------------------4分13,3-==∴n n n b n a ----------------------------------6分2)1(3+=n n S n ，)(23)1(232n n n n b b b b b S n +=+=∴--------------------------8分 n b b b b b n S S S S S T ++++=∴ 4321 [][]⎥⎦⎤⎢⎣⎡--+--=+++++++=++++++=--)3131()3131(23)333()333(23)()(232222201102222121n n n n n n b b b b b b ------------------------------------------10分 161543163112-+=++n n --------------------------------------------------------12分 18、解：(1)22241c a b =- 及正弦定理 54得C B 22sin 4121sin =-，C B 2sin 212cos =-∴-------------------------------------2分 又由A ＝π4，即B ＋C ＝34π，得－cos2B ＝sin2C ＝2sin C cos C ，-----------------------4分 C C C 2sin 21cos sin 2=∴ ---------------------------------------------------------------5分 解得tan C ＝4.------------------------------------------------------------------------------------6分（2）由tanC ＝4，C ∈(0，π)，1717cos ,17174sin ==∴C C ，----------------------8分又因为sin B ＝sin(A ＋C )＝sin(π4＋C )，34345sin =∴B -----------------------------------10分 由正弦定理得b c 524=---------------------------------------------------------------------11分 25sin 21,4==A bc A π ，25,4252==∴b b ------------------------------------12分 19、解：(1)2222212x nx x x S n -+++= -----------------------------------------------1分 64510675575329=++++=甲x ，5557858554143=++++=乙x --------------3分 656409647526452376022=-=-=∴甲S ，---------------------------------------------4分 6.17530256.32005551600322=-=-=乙S （其它方法酌情给分）------------------5分 (2)根据题中的统计数据，可得在这五天中甲城市空气质量等级为2级良的频率为35，则估计甲城市某一天的空气质量等级为2级良的概率为35.-------------------------------------7分 (3)设事件A “从题中甲城市和乙城市的统计数据中分别任取一个，这两个城市的空气质量等级相同”，由题意可知，从甲城市和乙城市的监测数据中分别任取一个，共有25个结果，分别记为：(29,43)，(29,41)，(29,55)，(29,58)，(29,78)，(53,43)，(53,41)，(53,55)，(53,58)，(53,78)，(57,43)，(57,41)，(57,55)，(57,58)，(57,78)，(75,43)，(75,41)，(75,55)，(75,58)，(75,78)，(106,43)，(106,41)，(106,55)，(106,58)，(106,78)．---------------------------------------------9分其数据表示两城市空气质量等级相同的包括同为1级优的为甲29，乙41，乙43，同为2级良的为甲53，甲57，甲75，乙55，乙58，乙78.-------------------------------------------10分则空气质量等级相同的为：(29,41)，(29,43)，(53,55)，(53,58)，(53,78)，(57,55)，(57,58)，(57,78)，(75,55)，(75,58)，(75,78)，共11个结果．---------------------------------------------11分由古典概型可得P (A )＝1125.所以这两个城市空气质量等级相同的概率为1125.-------------------------------------------12分20、（1）解： 底面ABCD 是菱形，，21、∴O 为AC ，BD 中点22、又 ，，23、∴，，-------------------------1分24、ABCD BD ABCD AC O BD AC 面，面⊂⊂=,∴底面----------------------------------------------2分PAC ∆中，AC=2，3=∴POPBD ∆中，6==PD PB ，32=BD ------------------3分23222133131=⋅⋅⋅⋅=⋅⋅=∴-ABCD ABCD P S PO V --------------------4分（2）过C 作BD CE //交AB 延长线于E ，过E 作BF EH //交PA 于H ，EH 与PB 交点为MBDF CE BDF BD BD CE 面面⊄⊂,,//BDF CE 面//∴ ----------------------------5分BDF EH BDF BF BF EH 面面⊄⊂,,//BDF EH 面//∴ ----------------------------------6分又CEM EH CEM CE E EH CE 面面⊂⊂=,,CEM BDF 面面//∴，CEM CM 面⊂----------------------------------7分BDF CM 面//∴-------------------------------------------------------------8分BE DC CE BD //,//为平行四边形四边形BECD ∴AB BE DC ==∴,B 为AE 中点-----------------------------------------9分BF EH AP AF //,41= 中点为PA H ∴---------------------------------------------------------------10分的交点与中线为中线EH PB M ∴31=∴BP BM M 为重心，---------------------------------------------------------12分 21、解：（1）[]x x e m x x e m x m x x f )1()1(1)2()(2/----=-+-+-=-------------------------2分 ①当m ->11，即0>m 时，），和（∞+--∞1)1,(m 上0)(/<x f ，)(x f 单调减；)1,1(m -上0)(/>x f ，)(x f 单调增 ---------------------------------------------------3分②当m -=11，即0=m 时，),(+∞-∞上0)(/<x f ，)(x f 单调减 -----------------------------------------------------4分③当m -<11，即0<m 时，），和（∞+--∞m 1)1,(上0)(/<x f ，)(x f 单调减；)1,1(m -上0)(/>x f ，)(x f 单调增 --------------------------------------------------5分（2）对任意的[]m x x -∈1,1,21，5)(421<+x x f 可转化为4541)(21+-<x x f ， 问题等价于[]m x x -∈1,1,21，min max )()(x g x f < ----------------------------------------6分 由（1）知，当)0,1(-∈m 时，)(x f 在[]m -1,1上单调递增，m em m f x f --=-=1max 2)1()(，-----------------------------------------------------------------------7分 )(x g 在[]m -1,1上单调递减，141)1()(min +=-=m m g x g ，------------------------------8分 即证14121+<--m e m m ，化简得[])1(5)2(41m e m m --<-- 令t m =-1，)2,1(∈t设)2,1(),1(4)5()(∈+--=t t t e t h t ， -----------------------------------------------------9分 0424)4()(/>->--=t t e t e t h ，故)(t h 在)2,1(上单调递增.---------------------------------10分 084)1()(>-=>∴e h t h ，即[])1(5)2(41m e m m --<---------------------------------------11分 故14121+<--m e m m ，得证。

佛山一中2019届高三上学期期中考试语文答案1．B（对北部和西部少数民族的泛称;忧虑、以……为患）2．B（爱;零数）3．D（与……友好）4．A（连词，假设）5．A（劝请）6．A（逃亡、逃跑；眼泪）7．B（夏历小月十五，大月十六；）8．A（动作偏指一方，她；谢绝）9．D（邀请）10．B（主持，管理；空，徒然）11.B（连接）12．D（功业）13.C（本分）14.A（见到）15.C（取得）参考译文：武灵王十九年春天正月，在信宫举行盛大朝会。

召见肥义同他议论天下大事，谈了五天才结束。

武灵王到北边巡视中山国的地界，到了房子县，又去代地，北到无穷，西到黄河，登上黄华山顶。

然后召见楼缓商议说：“我们先王趁着世事的变化，做了南边领地的君长，连接了漳水、滏水的险阻，修筑长城，又夺取了蔺城、郭狼，在荏地打败了林胡人，可是功业尚未完成。

如今中山国在我们腹心，北面是燕国，东面是东胡，西面是林胡、楼烦、秦国、韩国的边界，然而没有强大兵力的救援，这样下去国家要灭亡，怎么办呢？要取得高出世人的功名，必定要受到背离习俗的牵累。

我要穿起胡人服装。

”楼缓说：“很好。

”可是群臣都不愿意。

当时肥义在旁侍奉，武灵王说：“简子、襄子二位主君的功业，就在于考虑到了胡、翟之利。

做臣子的，受宠时应有明孝悌、知长幼、顺从明理的德操，通达时应建立既可利民又能益君的功业，这两方面是臣子的本分。

如今我想继承襄主的事业，开拓胡人、翟人所住之地，可是找遍世间也见不到这样的贤臣。

为了削弱敌人，用力少而能取得更多的功效，可以不耗尽百姓的力气，就能继续两位先主的勋业。

凡是有高出世上功业的人，就要承受背弃习俗的牵累；有独特智谋的人，就要听任傲慢民众的埋怨。

如今我要穿胡人服装骑马射箭，并用这个教练百姓，可是世人一定要议论我，怎么办呢？”肥义说：“我听说做事犹疑就不会成功，行动犹豫就不会成名。

您既然考虑决定承受背弃风俗的责难，那末就无需顾虑天下的议论了。

追求最高道德的人不附和世俗，成就大功的人不找凡夫俗子商议。

佛山第一中学2019高三上学期期中英语第一节每题2分从每题所给的四个选项中，选出最佳选项AWalk For CharityDear friends,Please join us for our annual Walk For Charity. Starting in Weldown, you and your friends can choose adelightful 10, 20 or 30 kilometers’ route.The money raised will provide support to help people all over the world. Start collecting your sponsorsnow and then simply come along on the day. Please read the instructions below carefully, especially if yourequire transport to and from Weldown.See you on Sunday 15 April.V Jessop: Walk coordinatorPS. Well done to last year’s walkers for helpingto raise a grand total of ￡21,000. The moneyhas already been used to build a children’s playground.START TIME:10 km: 8-10 am 20 km: 8-10:30 am 30 km: 8-11:30 amThe organizers reserve the right to refuse late-comers.CLOTHING should be suitable for the weather. If rain is forecast, bring some protection and beprepared for all possibilities. It is better to wear shoes that have been worn in, rather than ones that are new.ROUTE MAPS will be available from the registration point. The route will be sign-posted and arranged. Where the route runs along the road, walkers should keep to one side in single file, facingoncoming traffic at all times. If you need help along the route, please inform one of the staff.Free car parking is available in car parks and on streets in Weldown.BUSES:For the 10 and 20 kilometers’ routes, a bus will be waiting at Fenton to take walkers back to Weldown. The bus will leave every half hour starting at midday. The service is free and there is no need to book.1.When may a walker be forbidden to join in the walk?A.Parking a car on the streets.B. Being late for the walk.C. Choosing the shortest route.D. Bringing no clothes for the weather.2. What are the walkers advised to do?A. To prepare new shoes for the walk.B. To book the bus ahead of time.C. To care about their own safety.D. To donate at the registration point.3. When will the first bus leave?A. 11:30 amB. 12:30 pmC. 13:20 pmD. 14:40 pmBGerman carmaker Audi equipped some of its 2017 models in the United States with a new technology.仪表板) toThe technology is known as V-to-I. Drivers will see a countdown on the car’s instrument panel (show w hen the traffic signal will turn green or red. The equipment was included on Audi’s “Allroad” models built after June 1, 2016.American cities. The information comes from The service began in 2016 in at least five “smart”　government agencies that collect data to predict traffic flow. A private company partners with governmentofficials to send the data to cars.The countdown message goes away a few seconds before the light changes for safety. Pom Malhotra is·1·。

广东省佛山市第一中学届高三上学期期中考试数学（文）试题一、选择题（本大题共小题，共分）.已知集合则（）. . . . 【答案】【解析】因为集合中，∈，所以当＝时，＝－＝；当＝时，＝×－＝；当＝时，＝×－＝；当＝时，＝×－＝.即＝{}．又因为＝{}，所以∩＝{}．故选..设为虚数单位，则复数的共轭复数是（）. . . . 【答案】【解析】【分析】利用复数的运算法则和共轭复数即可求得结果【详解】，则共轭复数为故选【点睛】本题主要考查了复数的运算法则和共轭复数，属于基础题.记为等差数列{}的前项和．若，，则（）. . . . 【答案】【解析】【分析】利用等差数列的性质和求和公式求解，属于容易题.【详解】设等差数列{}的公差为，则，解得，所以.故选.【点睛】等差数列的问题一般是利用基本量首项和公差来构建方程（组）求解，也可以利用性质进行简化计算..已知命题：∃∈，≥．命题：若＜，则＜，下列命题为真命题的是（）. . . .【答案】【解析】【分析】先判定命题的真假，再结合复合命题的判定方法进行判定.【详解】命题：∃∈，使≥成立．故命题为真命题；当，时，＜成立，但＜不成立，故命题为假命题，故命题∧，￢∧，￢∧￢均为假命题；命题∧￢为真命题，故选：．【点睛】本题以命题的真假判断与应用为载体，考查了复合命题，特称命题，不等式与不等关系，难度中档．.已知△是边长为的等边三角形，为中点，则（）•（）的值为（）. . . .【答案】【解析】【分析】由题意得到，进而由线性运算及数量积运算得到结果.【详解】∵是边长为的等边三角形，为中点，∴而故选：【点睛】平面向量数量积的类型及求法()求平面向量数量积有三种方法：一是夹角公式；二是坐标公式；三是利用数量积的几何意义.()求较复杂的平面向量数量积的运算时，可先利用平面向量数量积的运算律或相关公式进行化简..函数的最小正周期为π，若其图象向左平移个单位后得到的函数为奇函数，则函数（）的图象（）. 关于点对称. 关于点对称. 关于直线对称. 关于直线对称【答案】【解析】【分析】利用最小正周期为π，求出的值，根据平移得出，然后利用对称性求解.【详解】因为函数的最小正周期为π，所以，图象向左平移个单位后得到，由得到的函数是奇函数可得，即.令得，，故均不正确；令得，，时可得正确.故选.【点睛】本题主要考查三角函数的图像变换和性质.平移变换时注意平移方向和对解析式的影响，性质求解一般利用整体换元意识来处理..天气预报说，今后三天每天下雨的概率相同，现用随机模拟的方法预测三天中有两天下雨的概率，用骰子点数来产生随机数。

佛山市第一中学2019-2020学年高三上学期期中考试英语试卷命题人：高三英语备课组2019年11月本试卷共8页，52小题，满分120分，考试时间120分钟。

注意事项：1．答题前，考生务必用黑色笔迹的钢笔或签字笔将自己的姓名、考号填写在答题卷上。

2．每小题选出答案后，用2B铅笔把答题卷上对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案，答案不能答在试卷上。

非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卷各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

第一部分：阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。

AWalk For CharityDear friends,Please join us for our annual Walk For Charity. Starting in Weldown, you and your friends can choose a delightful 10, 20 or 30 kilometers’ route.The money raised will provide support to help people all over the world. Start collecting your sponsors now and then simply come alongon the day. Please read the instructions belowcarefully, especially if you require transport to andfrom Weldown.See you on Sunday 15 April.V Jessop: Walk coordinatorPS. Well done to last year’s walkers for helpingto raise a grand total of ￡21,000. The moneyhas already been used to build a children’s playground.START TIME:10 km: 8-10 am 20 km: 8-10:30 am 30 km: 8-11:30 amThe organizers reserve the right to refuse late-comers.CLOTHING should be suitable for the weather. If rain is forecast, bring some protection and be prepared for all possibilities. It is better to wear shoes that have been worn in, rather than ones that are new.ROUTE MAPS will be available from the registration point. The route will be sign-posted and arranged. Where the route runs along the road, walkers should keep to one side in single file, facing oncoming traffic at all times. If you need help along the route, please inform one of the staff.Free car parking is available in car parks and on streets in Weldown.BUSES:For the 10 and 20 kilometers’ routes, a bus will be waiting at Fenton to take walkers back to Weldown. The bus will leave every half hour starting at midday. The service is free and there is no need to book.1.When may a walker be forbidden to join in the walk?A.Parking a car on the streets.B. Being late for the walk.C. Choosing the shortest route.D. Bringing no clothes for the weather.2. What are the walkers advised to do?A. To prepare new shoes for the walk.B. To book the bus ahead of time.C. To care about their own safety.D. To donate at the registration point.3. When will the first bus leave?A. 11:30 amB. 12:30 pmC. 13:20 pmD. 14:40 pmBGerman carmaker Audi equipped some of its 2017 models in the United States with a new technology. The technology is known as V-to-I. Drivers will see a countdown on the car’s instrument panel (仪表板) to show when the traffic signal will turn green or red. The equipment was included on Audi’s Q7, A4 and A4 “Allroad”models built after June 1, 2016.The service began in 2016 in at least five “smart”American cities. The information comes from government agencies that collect data to predict traffic flow. A private company partners with government officials to send the data to cars.The countdown message goes away a few seconds before the light changes for safety. Pom Malhotra is the general manager of Audi’s connected cars division. He said the system was designed to provide drivers with a sense of ease and convenience. “A better informed driver means a less stressed driver,”an announcer said in an Audi video explaining the technology.The company said the idea was to let drivers know exactly the time left before a signal changes. It suggested they use the time to make sure other passengers are fine or monitor car operations.The technology raises the possibility that some drivers may use the vehicle’s countdown to talk on the telephone or send text messages. But Audi said it would never tell drivers to text while driving -- an activity that is illegal in almost all 50 states.The goal was to integrate traffic infrastructure (基础设施) with cars on many city roads. This would let traffic signals use the vehicle data to regulate traffic, suggest better driving paths, or suggest speeds needed to hit all green lights.4.What function does the new technology add to the car?A.Reminding drivers to monitor car operations.B.Preventing drivers phoning when driving.C.Providing information on traffic signals.D.Reporting traffic flow to government officials.5.Why does the countdown message disappear before the light changes?A.To avoid misleading drivers.B. To help drivers hit green lights.C. To provide the newest information.D. To make sure drivers focus on driving.6. How does the new system benefit drivers?A. Making driving safe.B. Making driving fun.C. Making drivers relaxed when driving.D. Making drivers have more spare time.7. Which of the following can replace the underlined word “integrate” in the last paragraph?A. Combine.B. Compare.C. Equip.D. Replace.C“Shake a leg” and “show a leg” are usually discussed together, but it isn’t at all clear how they are connected. Both phrases have more than one meaning.“Shake a leg” means “hurry up” according to the definition in the New York Magazine in 1904. Before that, “shake a leg” had another meaning, which was “to dance”. There are several quotations from all kinds of American and British sources from the mid-19th century that are related to dancing. For example, the Dubuque Democratic Herald, October 1863, in an advertisement for a local ball: “Nearly every man in town able to shake a leg has purchased a ticket.”“Show a leg” means either “make an appearance”, or it means “hurry up”. The second meaning isn’t commonly used, nor is it old. Whatever the source, it isn’t the original meaning of “show a leg”. Most commentators report that the phrase derives from the Royal Navy and that this was the order given to sailors to put a foot from their hammocks (吊床) and get up.The use of “show a leg” as a wake-up call is well documented. John Masefield (Poet Laureate from 1930 to 1967) was a trainee mariner on HMS Conway until 1891. He reported the full version of the morning call as: “Heave out, heave out, heave out, heave out! Away! Come all you sleepers, Hey! Show a leg.” That’s the earliest quotation of the naval call I can find, although it may have been used well before 1891.An alternative version comes from the fact that women were allowed on board Royal Navy ships in the 19th century and that they were allowed to stay asleep after the sailors had been woken. In the morning the bosun’s mates (水手长助手) had to check whoever was still asleep and did so by requiring them to show a leg over the side of the hammock. If a leg washairy, it was probably male and its owner was ordered to get up and begin work. Believe that if you will; personally, I don’t.All in all, although both of the two English slangs are sometimes used to mean “hurry up”, they are two independent phrases that were coined with different meanings.8. How many ways does the slang “show a leg” come from?A. 1.B. 2.C. 3.D. 4.9. In which sentence does “shake a leg” have a different meaning?A. I will shake a leg with him at the next ball with great pleasure.B. Nearly every man in town able to shake a leg has purchased a ticket.C. They wanted to shake a leg on a TV show.D. You must shake a leg to catch the last bus.10. What is the author’s attitude to the version mentioned in Paragraph 5?A. Doubtful.B. Supportive.C. Indifferent.D. Neutral.11. What is the author’s purpose of writing the text?A. To persuade readers to learn slangs seriously.B. To describe “shake a leg” and “show a leg”.C. To distinguish “shake a leg” and “show a leg”.D. To remark on “shake a leg” and “show a leg”.DAbout a month after I joined Facebook, I got a call from Lori Goler, a highly regarded senior director of marketing at eBay. She made it clear that this was a business call. “I want to apply to work with you at Facebook,” she said. “Instead of recommending myself, I want to ask you: What is your biggest problem, and how can I solve it?”My jaw hit the floor. I had hired thousands of people over the previous decade and no one had ever said anything remotely like that. People usually focus on finding the right role for themselves, with the implication that their skills will help the company. Lori put Facebook’s needs front and center. It was a killer approach. I responded, “Recruiting is my biggest problem. And, yes, you can solve it.”Lori never dreamed she would work in recruiting, but she jumped in. She even agreed to trade earnings for acquiring new skills in a new field. Lori did a great job running recruiting and within months was promoted to her current job, leading People@Facebook.The most common metaphor for careers is a ladder, but this concept no longer appliesto most workers. As of 2010, the average American had eleven jobs from the ages of eighteen to forty-six alone. Lori often quotes Pattie Sellers, who came up with a much better metaphor: “Careers are a jungle gym, not a ladder.”As Lori describes it, there’s only one way to get to the top of a ladder, but there are many ways to get to the top of a jungle gym. The jungle gym model benefits everyone, but especially women who might be starting careers, switching careers, getting blocked by external barriers, or reentering the workforce after taking time off. The ability to create a unique path with occasional dips, detours (弯路), and even dead ends presents great views of many people, not just those at the top. On a ladder, most climbers are stuck staring at the butt of the person above.12.Why did Lori make the call?A.She helped Facebook to solve the biggest problem.B.She wanted to make a business deal with Facebook.C.She tried to ask for a pay rise in Facebook.D.She wanted to become an employee in Facebook.13.What impressed “me” most?A.Lori was good at running recruiting.B.Lori attached great importance to Facebook’s needs.C.Lori jumped in Facebook with no adequate experience.D.Lori was skilled in marketing at eBay.14.What can we infer from the last two paragraphs?A.Now all people don’t tend to climb the ladder.B.None on the ladder can enjoy the great views.C.Jungle gyms offer limited exploration for employees.D.A pregnant woman, jobless, benefits little from the jungle gyms.15.What is the best title of the passage?A.How to get promoted at work.B. Facebook’s Biggest Problem.C. Applying for a Job in Facebook.D. It’s a Jungle Gym, not a Ladder.第二节（共5小题；每小题2分，满分10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。

佛山一中2019届高三上学期期中考试文科数学答案一、选择题1.D2.A3.D4.B5.B6.C7.C8.B9.C 10.C 11.B 12.D二、填空题13.5414.12-n 15.π24016.)1,0()1,( --∞9.函数为奇函数，排除令，令，则故选三、解答题17、解：(1)由已知可得⎪⎩⎪⎨⎧+==++a q a q 223123--------------------------------2分解得，3=q 或4-=q (舍去)，62=a -----------------------------------------------------4分13,3-==∴n n n b n a ----------------------------------6分2)1(3+=n n S n ，)(23)1(232n n n n b b b b b S n +=+=∴--------------------------8分nb b b b b n S S S S S T ++++=∴ 4321[][]⎥⎦⎤⎢⎣⎡--+--=+++++++=++++++=--)3131()3131(23)333()333(23)()(232222201102222121nn n n n n b b b b b b ------------------------------------------10分161543163112-+=++n n --------------------------------------------------------12分18、解：(1)2221c a b =- 及正弦定理得C B 22sin 4121sin =-，C B 2sin 212cos =-∴-------------------------------------2分又由A ＝π4，即B ＋C ＝34π，得－cos2B ＝sin2C ＝2sin C cos C ，-----------------------4分C C C 2sin 21cos sin 2=∴---------------------------------------------------------------5分解得tan C ＝4.------------------------------------------------------------------------------------6分（2）由tanC ＝4，C ∈(0，π)，1717cos ,17174sin ==∴C C ，----------------------8分又因为sin B ＝sin(A ＋C )＝sin(π4＋C )，34345sin =∴B -----------------------------------10分由正弦定理得b c 524=---------------------------------------------------------------------11分25sin 21,4==A bc A π ，25,4252==∴b b ------------------------------------12分19、解：(1)2222212x n x x x S n -+++= -----------------------------------------------1分64510675575329=++++=甲x ，5557858554143=++++=乙x --------------3分65640964752642376022=-=-=∴甲S ，---------------------------------------------4分6.17530256.32005551600322=-=-=乙S （其它方法酌情给分）------------------5分(2)根据题中的统计数据，可得在这五天中甲城市空气质量等级为2级良的频率为35，则估计甲城市某一天的空气质量等级为2级良的概率为35.-------------------------------------7分(3)设事件A “从题中甲城市和乙城市的统计数据中分别任取一个，这两个城市的空气质量等级相同”，由题意可知，从甲城市和乙城市的监测数据中分别任取一个，共有25个结果，分别记为：(29,43)，(29,41)，(29,55)，(29,58)，(29,78)，(53,43)，(53,41)，(53,55)，(53,58)，(53,78)，(57,43)，(57,41)，(57,55)，(57,58)，(57,78)，(75,43)，(75,41)，(75,55)，(75,58)，(75,78)，(106,43)，(106,41)，(106,55)，(106,58)，(106,78)．---------------------------------------------9分其数据表示两城市空气质量等级相同的包括同为1级优的为甲29，乙41，乙43，同为2级良的为甲53，甲57，甲75，乙55，乙58，乙78.-------------------------------------------10分则空气质量等级相同的为：(29,41)，(29,43)，(53,55)，(53,58)，(53,78)，(57,55)，(57,58)，(57,78)，(75,55)，(75,58)，(75,78)，共11个结果．---------------------------------------------11分由古典概型可得P (A )＝1125.所以这两个城市空气质量等级相同的概率为1125.-------------------------------------------12分20、（1）解： 底面ABCD 是菱形，，∴O 为AC ，BD 中点又 ，，∴，，-------------------------1分ABCDBD ABCD AC O BD AC 面，面⊂⊂=, ∴底面----------------------------------------------2分PAC ∆中，AC=2，3=∴PO PBD ∆中，6==PD PB ，32=BD ------------------3分23221311=⋅⋅⋅⋅=⋅⋅=∴-ABCD ABCD P S PO V --------------------4分（2）过C 作BD CE //交AB 延长线于E ，过E 作BF EH //交PA 于H ，EH 与PB 交点为M BDFCE BDF BD BD CE 面面⊄⊂,,// BDFCE 面//∴----------------------------5分BDFEH BDF BF BF EH 面面⊄⊂,,// BDF EH 面//∴----------------------------------6分又CEMEH CEM CE E EH CE 面面⊂⊂=,, CEM BDF 面面//∴，CEM CM 面⊂----------------------------------7分BDF CM 面//∴-------------------------------------------------------------8分BEDC CE BD //,// 为平行四边形四边形BECD ∴AB BE DC ==∴,B 为AE 中点-----------------------------------------9分BF EH AP AF //,41= 中点为PA H ∴---------------------------------------------------------------10分的交点与中线为中线EH PB M ∴31=∴BP BM M 为重心，---------------------------------------------------------12分21、解：（1）[]x x e m x x e m x m x x f )1()1(1)2()(2/----=-+-+-=-------------------------2分①当m ->11，即0>m 时，），和（∞+--∞1)1,(m 上0)(/<x f ，)(x f 单调减；)1,1(m -上0)(/>x f ，)(x f 单调增---------------------------------------------------3分②当m -=11，即0=m 时，),(+∞-∞上0)(/<x f ，)(x f 单调减-----------------------------------------------------4分③当m -<11，即0<m 时，），和（∞+--∞m 1)1,(上0)(/<x f ，)(x f 单调减；)1,1(m -上0)(/>x f ，)(x f 单调增--------------------------------------------------5分（2）对任意的[]m x x -∈1,1,21，5)(421<+x x f 可转化为4541)(21+-<x x f ，问题等价于[]m x x -∈1,1,21，min max )()(x g x f <----------------------------------------6分由（1）知，当)0,1(-∈m 时，)(x f 在[]m -1,1上单调递增，mm m f x f --=-=1max 2)1()(，-----------------------------------------------------------------------7分)(x g 在[]m -1,1上单调递减，141)1()(min +=-=m m g x g ，------------------------------8分即证14121+<--m e m m ，化简得[])1(5)2(41m e m m --<--令t m =-1，)2,1(∈t 设)2,1(),1(4)5()(∈+--=t t t e t h t ，-----------------------------------------------------9分0424)4()(/>->--=t t e t e t h ，故)(t h 在)2,1(上单调递增.---------------------------------10分084)1()(>-=>∴e h t h ，即[])1(5)2(41m e m m --<---------------------------------------11分故14121+<--m e m m ，得证。