高二第一学期期中试卷(附答案)
天津市2024-2025学年高二上学期期中考试英语试题(含答案,无听力原文及音频)
天津市2024-2025学年度第一学期期中学情调研高二年级英语学科本试卷分共100分,考试时间为100分钟。
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祝各位同学考试顺利!第Ⅰ卷 (共65分)第一部分:听力理解 (共15 小题;每小题0.5分,满分7.5分)第一节听下面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,并标在试卷的相应位置。
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。
1. What does the man want to know?A. Where the woman works out.B. How the woman stays fit.C. How to stay healthy.2. What is the man interested in?A. Whether people in China bargain everywhere.B. How to get a better price when doing the shopping in China.C. Where Chinese people usually go shopping.3. What's the most probable relationship between the two speakers?A. Old friends.B. Boss and secretary.C. Colleagues.4. What do we know about the woman?A. She is severely stressed.B. She is the man's doctor.C. She falls asleep easily.5. When was the woman scheduled to go to China at first?A. This Friday.B. This Saturday.C. This Sunday第二节听下面几段材料。
高中语文高二上学期期中考试试卷附参考答案.doc
1.选出注音全都正确的一项(A.踟蹰(zhī chú)..B.侘傺 (chà chì)..C.伶俜 (ling pīng)..D.凝噎 (yē).高二(上)期中考试语文试卷3 分)()转徙 (xǐ) 迁谪 (zhé) 纶 (guāng)巾...荠 (jì)麦幢幢 (chuáng) 窒 (zhì)闷b5E2RGbCAP ....罗衾 (qīn) 沙渚 (chǔ) 三匝 (zhā) ...溘 (kè)死神妪 (yū) 落蕊 (ru ì)p1EanqFDPw ...2.选出有两个错别字的一组( 3 分)()A.祭祀轮廓虔诚宠姬佞臣爽心悦目 DXDiTa9E3dB.恶梦幽远萧瑟缕心刻骨变幻莫测C.翌日苍桑嘲哳清俊飘逸雾失楼台 RTCrpUDGiTD.憧憬懊悔霄禁一椽破屋千帆竞发3.选出解释有错误的一组( 3 分)()A.落寞(冷落,寂寞)开济(扶助)漫(不要)嗟荣辱夙(早上)兴夜寐.....B.宁谧(安静)盎(洋溢)然翠峰如簇(丛聚)可堪(能忍受)回首....C.啜(喝)茗颔(点头)首信誓旦旦(诚恳的样子)久负(享有)盛名.....D.陶(快乐)然倏(极快地)然山居秋暝(晚)游目骋(放开)怀....4.选出成语运用不正确的一项( 3 分)()A.他从旧营垒中冲出来,但一尘不染,始终保持着崇高的品质。
B.这是一架电子分析天平,它能精确地称量微量物质,其准确性可以说是毫厘不爽。
C.如果我们不修这么多水利,遇到这样严重的水旱灾害,其后果是不可思议的。
D.这些人对个人利益斤斤计较,对广大群众的疾苦却置若罔闻。
5.选出下列各组中加点字意义相同的两项( 4 分)()A.三日断五匹,大人故.嫌迟B.既欲结大义,故遣来贵门.①C.故作不良计,勿复怨鬼神.D.弟走从军阿姨死,暮去朝来颜色故.A.父母且不顾,何言子与妻.B.且放白鹿青崖间,须行即骑访名山.②C.誓不相隔卿,且.暂还家去D.磐石方且厚,可以卒千年.③A.自可断来信,徐徐更谓之.B.海客谈瀛洲,烟涛微茫信.难求C.胜似闲庭信步.D.低眉信手续续弹,说尽心中无限事.A.共事二三年,始尔.未为久B.同是被逼迫,君尔.妾亦然①C.媒人下床去,诺诺复尔.尔D.尔其无忘乃父之志.6.下面有关文学常识的表述,不正确的一项是( 3 分)()A.《诗经》是我国最早的诗歌总集,原本只称诗,儒家列为经典之一,故称《诗经》,编成于公元前 6 世纪的春秋时期,共 305 篇,又称“诗三百”。
最新高二年级第一学期语文期中考试试卷(含答案)
最新高二年级第一学期语文期中考试试卷(含答案)考生注意:1.本试卷满分150分,考试时间150分钟。
2.所有答案必须写在答题纸上,写在试卷上无效。
一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。
中国传统文化中的“礼”“礼”是中国传统文化的核心概念之一,它在中国历史的发展中扮演着重要的角色,深刻地影响着中国人的价值观和行为方式。
“礼”的内涵十分丰富。
首先,“礼”强调秩序和规范。
在中国传统文化中,社会的各个层面都有相应的礼仪规范,这些规范规定了人们在不同场合下的行为举止,从而维护了社会的秩序。
其次,“礼”注重道德修养。
礼仪不仅仅是外在的形式,更是内在道德的体现。
通过遵守礼仪,人们可以培养自己的品德,提高自己的道德境界。
最后,“礼”倡导和谐与包容。
礼仪的实施有助于协调人与人之间的关系,促进社会的和谐发展。
不同的文化和习俗都可以在“礼”的框架下得到尊重和包容。
“礼”在中国传统文化中具有重要的价值。
一方面,它有助于维护社会的稳定。
在一个有礼的社会中,人们遵守规范,尊重他人,矛盾和冲突就会减少,社会秩序得以维护。
另一方面,“礼”对于个人的成长和发展也具有积极的意义。
它可以培养人的自律、尊重他人和责任感等品质,提高个人的综合素质。
在当今社会,“礼”仍然具有重要的现实意义。
随着社会的发展和进步,人们的生活方式和价值观念发生了很大的变化,但是“礼”所倡导的秩序、道德和和谐等价值观念依然具有重要的指导意义。
我们应该继承和发扬“礼”的传统,将其融入到现代社会的建设中,促进社会的和谐发展。
1.下列关于原文内容的理解和分析,正确的一项是()(3分)A.“礼”是中国传统文化的唯一核心概念,贯穿中国历史发展始终。
B.中国传统文化认为,“礼”只强调外在形式,与内在道德无关。
C.“礼”思想有助于促进社会和谐稳定,对个人成长也有积极意义。
D.在当今社会,“礼”已经完全失去了现实意义。
2023-2024学年山东省烟台市高二(上)期中化学试卷+答案解析(附后)
一、单选题:本大题共11小题,共26分。
1.关于合成氨工业的说法中错误的是( )A. 混合气进行循环利用遵循绿色化学思想B. 工业上采用高温条件的目的是为了提高平衡转化率C. 对原料气进行压缩可增大原料气的转化率D. 2023-2024学年山东省烟台市高二(上)期中化学试卷使用催化剂能加快该反应速率的原因是降低了活化能2.是制备硫酸的重要反应。
下列叙述错误的是( )A. 催化剂不改变该反应的逆反应速率B. 使用催化剂改变活化能,但不改变反应热C. 该反应是放热反应,升高温度能缩短反应达到平衡的时间D. 恒温、恒压条件下,充入He ,化学反应速率减慢,平衡逆向移动3.下列说法中正确的是( )A. 非自发反应在任何条件下都不能发生B. 熵增加且放热的反应一定是自发反应C. 熵减小或不变的反应一定是非自发反应D. 凡是放热反应都是能自发进行的反应,而吸热反应都是非自发进行的反应4.在一定温度下可逆反应:⇌。
温度时,向密闭容器中通入X ,部分实验数据见下表:时间50100150下列说法中错误的是( )A. 50s 内,B. 温度下的平衡常数,平衡时X 的转化率为C. 温度下的平衡常数为,温度下的平衡常数为,若,则D.时,⇌的平衡常数为5.少量铁片与的稀盐酸反应盐酸过量,为了加快此反应速率而不改变产生的量,如下方法可行的是( )A. 加入少量铁粉B. 通入少量的HCl气体C. 加入50mL NaCl溶液D. 滴入几滴硫酸铜溶液6.下列有关叙述正确的是( )A. 用pH试纸测得某氨水的B. 稀醋酸加水稀释后,溶液中变大C. 溶液和溶液均显中性,两溶液中水的电离程度相同D.两种氨水的pH分别为a和,物质的量浓度分别为和,则7.在下列条件下,一定能大量共存的离子组是( )A.无色透明的水溶液中,、、、B.的的水溶液中,、、、C.的溶液中,、、、D. 由水电离产生的的溶液中,、、、8.常温下,在水中的沉淀溶解平衡曲线如图所示。
2023-2024学年上海华二附中高二上学期期中数学试卷及答案(2023.11)
1华二附中2023学年第一学期高二年级数学期中2023.11一、填空题(本大题共有12题,满分54分,第16~题每题4分,第7-12题每题5分)1.直线3310x y +-=的倾斜角为______.2.抛物线2y x =的准线方程为______.3.已知12,F F 是椭圆22:132x y C +=的两个焦点,椭圆C 上的两个动点P 、Q 与1F 满足三点共线,则2PQF △的周长是______.4.平行直线210x y +-=与2430x y ++=的距离为______.5.已知双曲线2221(0)4x y m m -=>的一条渐近线方程是520x y -=,则m =______.6.已知椭圆2222:1(0)x y C a b a b +=>>,点12,F F 是两个焦点,若椭圆上存在点P ,使得12120F PF ∠=︒,则该椭圆的离心率的取值范围是______.7.斜率为1的直线l 过抛物线22(0)y px p =>的焦点F ,若l 与圆22(5)8x y -+=相切,则P 等于______.8.在平面直角坐标系xOy 中,A 为直线:2l y x =上在第一象限内的点,(5,0)B ,以AB为直径的圆C 与直线l 交于另一点D .若0AB CD ⋅=,则点A 的横坐标为______.9.如图,发电厂的冷却塔外形是由双曲线的一部分绕其虚轴所在直线旋转所得到的曲面,该冷却塔总高度为170米,水平方向上塔身最窄处的半径为20米,最高处塔口半径25米,塔底部塔口半径为202米,则该双曲线的离心率为______.210.已知,x y 为实数,代数式22221(2)9(3)y x x y +-++-++的最小值是______.11.如图,从双曲线22135x y -=的左焦点F 引圆223x y +=的切线FP 交双曲线右支于点,P T 为切点,M 为线段FP 的中点,O 为坐标原点,则||||MO MT -=______.12.已知,P Q 分别是圆22:(4)8C x y -+=与圆22:(4)5D x y +-=上的点,O 是坐标原点,则2||||2PQ PO +的最小值为______.二、选择题(共4题,共18分,13-14题每题4分,15-16每题5分)13.椭圆22152x y +=的长轴长为()A .25B .5C .4D .214.过抛物线24y x =的焦点F 的直线交该抛物线于,A B 两点,若,A B 两点的横坐标之和为3,则||AB =()A .5B .143C .133D .415.骑自行车是一种能有效改善心肺功能的耐力性有氧运动,深受大众喜爱,下图是某一自行车的平面结构示意图,已知图中的圆A (前轮),圆D (后轮)的半径均为3,ABE △,,BEC ECD △△均是边长为4的等边三角形.设点P 为后轮上的一点,则在骑行该自行车的过程中,AB BP ⋅的最大值为()A .43B .12C .123D .3316.把方程||||194x x y y +=-表示的曲线作为函数()y f x =的图象,则下列结论正确的是()①()f x 在R 上单调递减;②()y f x =的图像关于原点对称;③函数()3()2g x f x x =+不存在零点;④()y f x =的图象上的点到坐标原点的距离的最小值为2;A .①②③B .①②④C .①③④D .②③④三、解答题(共5题,共78分)17.(14分)直线1:2110l x y +-=与直线2:2100l x y +-=相交于点P ,直线l 经过点P (1)若直线2l l ⊥,求直线l 的方程:(2)若直线l 在坐标轴上的截距相等,求直线l 的方程.18.(14分)已知抛物线2:2C y px =(p 为常数,0p >)的焦点F 与椭圆22195y x +=的右焦点重合,过点F 的直线与抛物线交于,A B 两点.(1)求抛物线C 的标准方程;(2)若直线AB 的斜率为1,求||AB .419.(14分)如图,OM ,ON 是某景区的两条道路(宽度忽略不计,OM 为东西方向),Q 为景区内一景点,A 为道路OM 上一游客休息区,已知tan 3,6MON OA ∠=-=(百米),Q 到直线ON ,ON 的距离分别为3(百米)(百米).现新修一条自A 经过Q 的有轨观光直路并延伸至道路ON 于点B ,并在B 处修建一游客休息区.(1)求有轨观光直路AB 的长;(2)已知在景点Q 的正北方6百米的P 处有一大型组合音乐喷泉,喷泉表演一次的时长为9分钟,表演时,喷泉喷洒区域以P 为圆心,r 为半径变化,且t 分钟时,2r=(百米)(09,01)t a ≤<<<,当喷泉表演开始时,一观光车s (大小忽略不计)正从休息区B 沿(1)中的轨道BA(百米/分钟)的速度开往休息区A .问:观光车在行驶途中是否会被喷泉喷洒到,并说明理由.20.(18分)设双曲线2222:1(0,0)x yC a ba b-=>>的一个焦点坐标为,离心率,e A B=是双曲线上的两点,AB的中点(1,2)M.(1)求双曲线C的方程:(2)求直线AB方程:(3)如果线段AB的垂直平分线与双曲线交于C、D两点,问A、B、C、D四点是否共圆?若共圆求之,若不共圆给予充分理由.5621.(18分)如图,D 为圆22:1O x y +=上一动点,过点D 分别作x 轴y 轴的垂线,垂足分别为,A B ,连接BA 并延长至点W ,使得||1WA =,点W 的轨迹记为曲线C .(1)求曲线C 的方程:(2)若过点(2,0)K -的两条直线12,l l 分别交曲线C 于,M N 两点,且12l l ⊥,求证:直线MN 过定点:(3)若曲线C 交y 轴正半轴于点S ,直线0x x =与曲线C 交于不同的两点,G H ,直线SH ,SG 分别交x 轴于,P Q 两点.请探究:y 轴上是否存在点R ,使得2ORP ORQ π∠+∠=?若存在,求出点R坐标;若不存在,请说明理由.7参考答案一、填空题1.120;2.14x =-;3.;4.52; 5.5;6.,12⎫⎪⎪⎣⎭;7.218或;8.3;;11.如图,从双曲线22135x y -=的左焦点F 引圆223x y +=的切线FP 交双曲线右支于点,P T 为切点,M 为线段FP 的中点,O 为坐标原点,则||||MO MT -=______.-【解析】设双曲线的右焦点为1F ,因为O 为1FF 中点,M 为PF 中点,所以MO 为三角形1PFF 的中位线,11,2MO PF =又1122MT PT PM PF FT PF PF FT=-=--=-所以()112MO MT PF PF FT FT a-=-+=-a FT ===又所以MO MT -=.-.12.已知,P Q 分别是圆22:(4)8C x y -+=与圆22:(4)5D x y +-=上的点,O 是坐标原点,则2||||2PQ PO +的最小值为______.【解析】由()2248x y -+=得22880x x y -++=,于是22222828,x x y x y -++=+从而()22221442x x y x y -++=+,=8等于点P 到点()2,0M 的距离.所以PQ PQ PM MQ =+ ,而min ||MQ =-=所以PQ +二、选择题13.A14.A 15.A 16.C15.骑自行车是一种能有效改善心肺功能的耐力性有氧运动,深受大众喜爱,下图是某一自行车的平面结构示意图,已知图中的圆A (前轮),圆DABE △,,BEC ECD △△均是边长为4的等边三角形.设点P 为后轮上的一点,则在骑行该自行车的过程中,AB BP ⋅的最大值为()A.B .12C.D .3【答案】A【解析】以D 为坐标原点,AD 为x 轴,过D 做AD 的垂线为y 轴,建立如图所示的平面直角坐标系,则()((8,0,,A B C ---.圆D 的方程为223x y +=,可设)Pαα,所以(,AB BP αα==+- .故126sin 126AB BP πααα⎛⎫⋅=++-=+≤ ⎪⎝⎭故选:A.16.把方程||||194x x y y +=-表示的曲线作为函数()y f x =的图象,则下列结论正确的是()9①()f x 在R 上单调递减;②()y f x =的图像关于原点对称;③函数()3()2g x f x x =+不存在零点;④()y f x =的图象上的点到坐标原点的距离的最小值为2;A .①②③B .①②④C .①③④D .②③④【答案】C【解析】由方程方程1x x y y +=-,当0x ,0y 时不成立;当0,0x y ><时,22149y x -=;当0,0x y <>时,22194x y -=;当0,0x y 时,22194x y +=;如下图示:由图判断函数在R 上单调递减,故(1)正确,(2)错误;当()320f x x +=,即()23f x x =-,函数()()32g x f x x =+的零点,就是函数()y f x =和23y x =-的交点,而23y x =-是曲线221,049y x x -=>,0y <和221,0,094x y x y -=<>的渐近线,所以没有交点.由图知,23y x =-和221,094x y x += ,0y 没有交点,所以函数()()32g x f x x =+不存在零点,故(3)正确;由图,()y f x =上的点到原点距离的最小值点应在0,0x y 的图象上,即满足22194x y +=,设(),P x y,PO ===,当0x =时取最小值2,故(4)正确.故选:C .三.解答题1017.(1)250x y -+=(2)43070x y x y -=+-=或18.(1)28y x =(2)16AB =19.(1)AB =(2)不会,理由略20.(18分)设双曲线2222:1(0,0)x y C a b a b-=>>的一个焦点坐标为,离心率,e A B =是双曲线上的两点,AB 的中点(1,2)M .(1)求双曲线C 的方程:(2)求直线AB 方程:(3)如果线段AB 的垂直平分线与双曲线交于C 、D 两点,问A 、B 、C 、D 四点是否共圆?若共圆求之,若不共圆给予充分理由.【答案】(1)2212y x -=(2)1y x =+(3)略【解析】(1)依题意得c ce a ⎧=⎪⎨==⎪⎩,解得1a =.所以222312b c a =-=-=,故双曲线C 的方程为2212y x -=.(2)设()()1122,,,A x y B x y ,则有221122221212y x y x ⎧-=⎪⎪⎨⎪-=⎪⎩.两式相减得:()()()()1212121212x x x x y y y y -+=-+,由题意得121212,2,4x x x x y y ≠+=+=,所以()1212121221x x y y x x y y +-==-+,即 1.AB k =故直线AB 的方程为1y x =+.11(3)假设A B C D 、、、四点共圆,且圆心为P .AB 为圆P 的弦,所以圆心P 在AB 垂直平分线CD 上,又CD 为圆P 的弦且垂直平分AB ,圆心P 为CD 中点M .下面只需证CD 的中点M 满足MA MB MC MD ===即可.由22112y x y x =+⎧⎪⎨-=⎪⎩,得:()1,0A -,()3,4B .由(1)得直线CD 方程:3y x =-+,由22312y x y x =-+⎧⎪⎨-=⎪⎩得:(3C -+(63D ---+,()3,6CD M ∴-⋅的中点2,MA MB MC MD MA MB MC MD ======∴=== 即A B C D 、、、四点在以点()3,6M -为圆心,为半径的圆上.21.(18分)如图,D 为圆22:1O x y +=上一动点,过点D 分别作x 轴y 轴的垂线,垂足分别为,A B ,连接BA 并延长至点W ,使得||1WA =,点W 的轨迹记为曲线C .(1)求曲线C 的方程:(2)若过点(2,0)K -的两条直线12,l l 分别交曲线C 于,M N 两点,且12l l ⊥,求证:直线MN 过定点:(3)若曲线C 交y 轴正半轴于点S ,直线0x x =与曲线C 交于不同的两点,G H ,直线SH ,SG 分别交x 轴于,P Q 两点.请探究:y 轴上是否存在点R ,使得2ORP ORQ π∠+∠=若存在,求出点R 坐标;若不存在,请说明理由.12【答案】(1)2214x y +=(2)见解析(3)存在,理由见解析【解析】(1)设()()00,,,W x y D x y ,则()()00,0,0,A x B y ,由题意知1AB =,所以WA AB = ,得()()000,,x x y x y --=-,所以00,2,x x y y ⎧=⎪⎨⎪=-⎩因为22001x y +=,得2214x y +=,故曲线C 的方程为2214x y +=.(2)由题意可知,直线12,l l 不平行坐标轴,则可设1l 的方程为:2x my =-,此时直线2l 的方程为12x y m=--.由22214x my x y =-⎧⎪⎨+=⎪⎩消去x 得:()22440m y my +-=,解得:244m y m =+或0y =(舍去),所以222428244m m x m m m -=⋅-=++,所以222284,44m m M m m ⎛⎫- ⎪++⎝⎭,同理可得:222284,4141m m N m m ⎛⎫-- ⎪++⎝⎭.①当1m ≠±时,直线MN 的斜率存在,13()222224222244455556441,:,2828161644445441MN MN m m m m m m m m k l y x m m m m m m m ++⎛⎫++====+ ⎪-----⎝⎭-++所以直线MN 过定点6,05⎛⎫- ⎪⎝⎭.②当1m =±时,直线MN 斜率不存在,此时直线MN 方程为:65x =-,也过定点6,05⎛⎫- ⎪⎝⎭,综上所述:直线MN 过定点6,05⎛⎫-⎪⎝⎭.(3)存在点。
上外版英语高二上学期期中试卷与参考答案
上外版英语高二上学期期中模拟试卷与参考答案一、听力第一节(本大题有5小题,每小题1.5分,共7.5分)1、Listen to the conversation and choose the best answer to the question you hear. Question: What is the main topic of the conversation?A. The man’s favorite sports team.B. The upcoming school sports day.C. The woman’s participation in a sports club.Answer: BExplanation: The conversation revolves around the woman’s plan to participate in the school sports day, indicating that the main topic is indeed the upcoming event.2、Listen to the dialogue and answer the question you hear.Question: How does the man feel about the weather?A. He is excited about the sunny weather.B. He is not happy with the rainy weather.C. He doesn’t mind the weather as long as it’s not too hot.Answer: BExplanation: The man expresses his disappointment in the rainy weather, suggesting that he is not happy with it.3、What does the woman imply when she says “You know, it’s not as easyas it looks.”A)The man should not try to do it.B)The task is more complicated than it seems.C)It will take longer than they expect.D)It’s not worth the effort.Answer: BExplanation: The phrase “not as easy as it looks” suggests that something is more difficult than it appears to be. Therefore, the woman implies that the task is more complicated than it seems.4、Why does the man say he will have to pass the exam even if he is not a student?A)He needs the credit for his college degree.B)He is taking the exam for his job.C)He is supporting his family through his studies.D)He is attending the exam as a hobby.Answer: AExplanation: The man’s statement about passing the exam implies that it is part of a requirement for his college degree. Therefore, he needs the credit for his academic progress.5、You will hear a conversation between two students about their weekend plans. Listen carefully and answer the question.Question: How many hours will the girl spend on her part-time job on Saturday?Answer: 7 hoursExplanation: In the conversation, the girl mentions that she has to work for 7 hours on Saturday, which is the duration of her part-time job.二、听力第二节(本大题有15小题,每小题1.5分,共22.5分)1、Listen to the conversation and answer the question.A. The man is a student in his first year at Shanghai International Studies University.B. The woman is the student’s tutor at the university.C. They are discussing the student’s performance in the first semester.D. They are planning a trip to Shanghai Disneyland.Question: Who is the woman in the conversation?Answer: B. The woman is the student’s tutor at the university.Explanation: The key to this question is in the context of the conversation. The woman addresses the man as “you” and refers to herself as “your tutor,” which indicates that she is in a mentoring role, likely a tutor.2、Listen to the dialogue and complete the sentence with the missing word.W: I can’t believe it’s already November. It feels like just yesterday we were saying goodbye to the summer.M: Yeah, time flies. Speaking of which, how are you doing with your midterm exams?W: (pause) I’m feeling a bit nervous about them. I’m not sure if I’ve prepared well enough.Question: How does the woman feel about the midterm exams?Answer: Nervous.Explanation: The woma n explicitly states, “I’m feeling a bit nervous about them,” which directly answers the question about her feelings regarding the exams.3.You will hear a conversation between two students discussing their weekend plans. Listen to the conversation and answer the following question.Question: What does the male student plan to do on Saturday?A. Go to the movies.B. Visit a museum.C. Stay at home and relax.Answer: C. Stay at home and relax.Explanation: In the conversation, the male student mentions that he doesn’t have any plans for Saturday and would rather stay at home and relax.4.You will hear a short interview with a famous author. Listen to the interview and answer the following question.Question: Why does the author prefer to write in the morning?A. Because she is more creative in the morning.B. Because she has less distractions in the morning.C. Because she is less tired in the morning.Answer: B. Because she has less distractions in the morning.Explanation: During the interview, the author explains that she prefers towrite in the morning because it’s when she has fewer distractions, allowing her to focus better on her writing.5.You hear a conversation between two students, Alice and Bob, discussing their midterm exam preparation. Listen and answer the following question: Question: How does Alice feel about the midterm exam?A. ExcitedB. NervousC. IndifferentD. ConfidentAnswer: B. NervousExplanation: In the conversation, Alice mentions, “I’m really nervous about the midterm,” indicating her anxiety regarding the exam.6.You hear a teacher, Mr. Green, explaining a concept to his class before the midterm. Listen and answer the following question:Question: What is Mr. Green’s main advice to the students for the upcoming exam?A. To focus on the most difficult topics.B. To review the material regularly.C. To study late into the night.D. To avoid studying altogether.Answer: B. To review the material regularly.Explanation: Mr. Green says, “The most important thing is to review thematerial regularly so you’re not overwhelmed on the day of the exam,” which suggests that consistent review is his recommended approach.7.You will hear a conversation between two students discussing a school project. Listen to the conversation and answer the question.What is the main topic of the school project?A. A science fair experimentB. A literature review on a famous novelC. A research paper on climate changeD. A historical reenactmentAnswer: BExplanation: The conversation mentions that they are working on a “comprehensive literature review of the various interpretations of the novel,” which indicates that the main topic is a literature review on a famous novel.8.You will hear a short lecture about the importance of teamwork in the workplace. Listen to the lecture and answer the question.Which of the following is NOT mentioned as a benefit of teamwork according to the lecture?A. Increased creativityB. Improved problem-solving abilitiesC. Enhanced communication skillsD. Reduced individual workloadAnswer: DExplanation: The lecture discusses how teamwork leads to increased creativity, improved problem-solving, and enhanced communication skills. However, it does not mention that teamwork reduces the individual workload as a benefit.9.You will hear a conversation between two students discussing a school project. Listen and answer the question.Question: What is the project about?A) A science fairB) A cultural exchangeC) A history presentationD) A music festivalAnswer: B) A cultural exchangeExplanation: Th e conversation mentions “we are planning a cultural exchange project,” which indicates the topic of the project is a cultural exchange.10.You will hear a short announcement about a school event. Listen and answer the question.Question: When will the event take place?A) This Friday at 2 PMB) Next Tuesday at 3 PMC) This Saturday at 4 PMD) Next Wednesday at 2 PMAnswer: A) This Friday at 2 PMExplanation: The announcement clearly states, “The event will be held this Friday at 2 PM in the school auditoriu m.”11.You will hear a conversation between two students discussing their weekend plans. Listen and answer the following question.What does the student say they will do on Saturday?A)Go shopping with a friend.B)Attend a sports event.C)Stay at home and watch movies.Answer: C) Stay at home and watch movies.Explanation: In the conversation, the student mentions that they have no plans for Saturday and would rather stay at home and watch movies.12.You will hear a short lecture about the effects of climate change on wildlife. Listen and answer the following question.What is one significant effect of climate change mentioned in the lecture?A)Increased food availability for wildlife.B)The migration patterns of animals shifting.C) A decrease in the number of polar bears.Answer: B) The migration patterns of animals shifting.Explanation: The lecture discusses how climate change is causing the migration patterns of animals to change, which can disrupt their natural habitats and food sources.13.You will hear a conversation between two students discussing their weekend plans. Listen carefully and answer the question.What activity does the second student plan to do on Saturday afternoon?A)Go to the movies.B)Visit a friend.C)Go for a hike.Answer: B) Visit a friend.Explanation: In the conversation, the second student mentions that they have plans to visit a friend on Saturday afternoon.14.You will hear a short lecture about the importance of exercise. Listen carefully and answer the question.According to the lecture, what is one of the benefits of regular exercise?A)Improved memory.B)Better sleep.C)Increased appetite.Answer: B) Better sleep.Explanation: The lecturer discusses how regular exercise can lead to better sleep quality, which is one of the benefits mentioned.15.How many students are studying at the university library on a sunny weekend?A. 200B. 300C. 400D. 500Answer: BExplanation: The conversation between two students discussing their weekend plans mentions that there are about 300 students at the library, which is a common spot for studying on sunny weekends. Therefore, the correct answer is B. 300.三、阅读第一节(第1题7.5分,其余每题10分,总37.5分)第一题Reading PassageIn the small coastal town of Porthkerry, the local community has always valued its rich history and natural beauty. The town is known for its stunning cliffs, picturesque beaches, and a vibrant local culture. One of the most iconic landmarks in the town is the Porthkerry Castle, which has stood for over 800 years.The castle was built in the 13th century by the de Braose family, a prominent noble family at the time. Over the centuries, it has been home to many different owners, each leaving their mark on the structure and the surrounding area. The castle has witnessed battles, royal visits, and the daily lives of ordinary people.Today, the Porthkerry Castle is a museum and a popular tourist destination. It houses a collection of historical artifacts, including weaponry, furniture, and paintings. The castle grounds are also home to a variety of plants and animals, making it a haven for nature lovers.1.What is the main purpose of the Porthkerry Castle today?A. It is a residential home for a noble family.B. It is a museum and a tourist destination.C. It is a center for educational workshops.D. It is a military base.2.Who built the Porthkerry Castle in the 13th century?A. The de Braose familyB. The King of EnglandC. The local villagersD. A famous architect3.What has the Porthkerry Castle witnessed over the centuries?A. Only battles and royal visitsB. The daily lives of ordinary peopleC. The construction of new buildingsD. The expansion of the town4.What is one of the attractions for nature lovers at the Porthkerry Castle?A. The historical artifactsB. The paintings in the museumC. The variety of plants and animalsD. The stunning cliffsAnswers:1.B2.A3.B4.C第二题Passage:The following is a story about the importance of teamwork.In a small town, there was a race scheduled between two teams: the Lightning Team and the Thunder Team. Both teams were known for their competitiveness and were equally matched. The race was set to take place on a sunny Saturday morning, and the entire town was buzzing with excitement.The Lightning Team had a star runner, Alex, who was known for his speed and determination. The Thunder Team had a strong team spirit, with each member contributing to the overall effort. The team captain, Jamie, was a master strategist and knew how to maximize the team’s potential.On the day of the race, the weather was perfect. The crowd gathered at the starting line, eager to see who would win. The race began, and the Lightning Team took an early lead. However, the Thunder Team was not to be outdone. Jamie had planned the strategy well, and each member of the team played their part perfectly.As the race progressed, the teams exchanged leads several times. The crowd cheered loudly for both teams, but it was clear that the race was coming down to the wire. With only a few meters to go, the Lightning Team seemed to have the upper hand. But then, disaster struck. Alex tripped over his own shoelaceand fell to the ground.The crowd gasped in shock. The Lightning Team was in disarray, and it looked like the Thunder Team would win. But instead of celebrating, the Thunder Team members rushed to help Alex. They formed a human chain and lifted him to his feet. With renewed determination, Alex and the Thunder Team member who had helped him continued the race.In the end, the Thunder Team won the race by a mere second. The crowd erupted in cheers, and the entire town was filled with a sense of camaraderie and joy. The Lightning Team was gracious in defeat, and the Thunder Team was hailed as heroes.This race taught the town an important lesson: teamwork can overcome individual setbacks and lead to unexpected victories.Questions:1、What was the main theme of the story?A) The importance of speed in a raceB) The power of teamworkC) The importance of individual talentD) The unpredictability of a race2、Who was the star runner on the Lightning Team?A) JamieB) AlexC) The team captainD) A member of the Thunder Team3、What strategy did Jamie use to help the Thunder Team win the race?A) He focused on improving the speed of his team membersB) He encouraged individual efforts from each team memberC) He planned for potential setbacks and had backup strategiesD) He tried to distract the Lightning Team4、How did the Thunder Team’s victory affect the town?A) It caused a lot of conflict and argumentB) It brought the town closer togetherC) It led to increased competition between the teamsD) It resulted in a decrease in town spiritAnswers:1、B2、B3、C4、B第三题Read the following passage and answer the questions that follow.In the small coastal town of Amble, the local library has been a hub of community activity for decades. Known for its vast collection of books and friendly staff, the library has played a significant role in the lives of its residents. However, with the advent of digital technology and e-books, thelibrary’s future seemed uncertain.Last year, the library’s director, Sarah Miller, noticed a decline in the number of visitors. She decided to take action and launched a new initiative aimed at attracting young people to the library. The program, called “Read & Win,” offered prizes for students who r ead a certain number of books during the school year. The prizes ranged from small gadgets to gift cards for local bookstores.Sarah’s initiative paid off. The number of young people visiting the library increased significantly. Many students expressed their gratitude for the program, saying it not only encouraged them to read more but also helped them connect with other book lovers in the community.1、What was the main concern of the library director, Sarah Miller?A. The lack of digital technology in the libraryB. The declining number of visitorsC. The outdated collection of booksD. The high cost of maintaining the library2、What was the purpose of the “Read & Win” program?A. To reduce the library’s budgetB. To encourage students to read moreC. To promote the sale of e-booksD. To attract tourists to the town3、How did the “Read & Win” program affect the library?A. It decreased the number of visitorsB. It increased the number of visitorsC. It improved the library’s financial situationD. It led to the closure of the library4、What did many students say about the “Read & Win” program?A. It was too expensive for themB. It did not encourage them to read moreC. It helped them connect with other book loversD. It was not as effective as expectedAnswers:1、B2、B3、B4、CFourth QuestionRead the passage carefully and choose the best answer to each of the questions that follow. You will find the answers immediately after each question.Passage:In an age where technology seems to be taking over every aspect of our lives, there remains one activity that has not been overshadowed by digital advancements: reading traditional books. Despite the popularity of e-books andaudiobooks, many people still prefer the tactile experience of holding a physical book and turning its pages. This preference for traditional books is not only due to nostalgia but also because of the unique benefits they offer.A recent study conducted by the National Reading Agency found that individuals who read physical books tend to have a better memory of the content compared to those who read the same material on digital devices. Furthermore, there is a growing concern about the impact of screen time on sleep patterns, with research suggesting that reading from screens before bedtime can negatively affect the quality of sleep. Physical books, on the other hand, do not emit blue light which can disrupt melatonin production, thus helping readers get a good night’s rest.Moreover, libraries and bookstores continue to thrive as communal spaces where people gather to share ideas and engage in discussions about literature. The atmosphere in these places fosters a sense of community and a love for reading that cannot be replicated online. While digital media offers convenience and access to a vast amount of information, it does not replace the unique experience of visiting a library or bookstore.In conclusion, while technology offers many conveniences, the tradition of reading physical books remains strong. It provides a way to disconnect from the digital world, improve memory retention, and foster social connections within communities.End of PassageQuestions:1、According to the passage, what is one benefit of reading physical books over digital devices?Answer: Better memory retention of the content.2、What potential negative effect on health is associated with reading from screens before bedtime?Answer: Disruption of sleep quality due to blue light affecting melatonin production.3、How do libraries and bookstores contribute to community life according to the text?Answer: They serve as communal spaces where people gather to share ideas and engage in discussions about literature.4、What does the passage suggest about the future of physical books despite technological advances?Answer: Physical books will remain significant as they offer ways to disconnect, improve memory, and foster social connections.四、阅读第二节(12.5分)Title: The Art of MeditationReading Passage:Meditation has been a part of human culture for thousands of years, serving as a means for relaxation, stress reduction, and spiritual growth. Its originscan be traced back to ancient India, where it was practiced by sages and monks to achieve enlightenment. Over time, meditation has spread across the world, gaining popularity in various forms, such as mindfulness, transcendental meditation, and yoga.One of the most common types of meditation is mindfulness meditation, which involves paying attention to the present moment and accepting it without judgment. This form of meditation can be practiced anywhere, at any time, and does not require any special equipment. It is particularly useful for individuals looking to improve their focus, reduce anxiety, and increase their overall sense of well-being.Mindfulness meditation i s based on the concept of being aware of one’s thoughts, feelings, and bodily sensations. It encourages individuals to observe their experiences without getting caught up in them. By practicing mindfulness, one can learn to let go of negative thoughts and emotions, leading to a more peaceful and balanced state of mind.Here is a brief guide on how to practice mindfulness meditation:1.Find a quiet and comfortable place to sit or lie down.2.Close your eyes and take a few deep breaths to relax.3.Focus on your breath, noticing the sensation of air entering and leaving your nostrils.4.When your mind starts to wander, gently bring your attention back to your breath.5.Continue this practice for 5 to 10 minutes, or as long as you feel comfortable.Question:What is the primary purpose of mindfulness meditation?A)To achieve enlightenmentB)To improve focus and reduce anxietyC)To practice yogaD)To increase spiritual growthAnswer: B) To improve focus and reduce anxiety五、语言运用第一节 _ 完形填空(15分)Cloze Test (完形填空)Directions: For this part, you are going to read a passage with 15 blanks. Choose the best word or phrase from the four choices marked A, B, C, and D for each blank. There is only one correct answer for each blank.Passage:The world of the internet has transformed our lives in countless ways. It has made information more accessible than ever before, connecting people across the globe in real-time. However, it also poses new challenges and risks. One of these is the increasing amount of misinformation that can be found online. While the internet is undoubtedly a valuable tool for learning and communication, it requires users to develop critical thinking skills to discern fact fromfiction. This ability is particularly important for young people who have grown up with easy access to technology but may lack the experience necessary to judge the 1 of the information they encounter.In addition to being able to evaluate the credibility of online content, digital literacy involves understanding the importance of privacy and security measures when using the web. Sharing personal details on social media platforms can make individuals vulnerable to identity theft and other forms of cybercrime. Therefore, it is essential to educate children about the risks associated with 2 too much personal information online.Another challenge of the digital age is the potential for addiction to electronic devices and the internet itself. The constant connectivity can lead to a feeling of being overwhelmed by the constant influx of 3. This is why many experts recommend setting boundaries around screen time and encouraging alternative activities that promote mental well-being.Despite these challenges, the internet remains a powerful force for good, providing opportunities for collaboration, innovation, and learning. By teaching students how to use it responsibly, we can help ensure they benefit from its advantages while avoiding its pitfalls.Now choose the best option to complete the passage:1.A. quantity B. quality C. quantity D. availability2.A. sharing B. hiding C. collecting D. deleting3.A. emails B. letters C. information D. stimuliKey:1.B. quality2.A. sharing3.D. stimuliThis exercise tests students’ comprehension of the passage as well as their vocabulary and grammar usage. The context here relates to the impact of the internet on modern life, which is a relevant topic for today’s youth.六、语言运用第二节 _ 语法填空(15分)Language Usage Section II: Grammar CompletionReading Passage:When it comes to learning a new language, many people struggle to find the right balance between memorization and practice. One effective method is to immerse yourself in the language by watching movies, listening to music, and reading books. This not only helps you get accustomed to the pronunciation and intonation but also exposes you to a wide range of vocabulary and expressions.However, it is important to note that simply listening and watching is not enough. You need to actively engage with the language by speaking and writing. Try to find a language exchange partner or join a conversation group to practice speaking. This will not only improve your speaking skills but also help you gain confidence in using the language in real-life situations.When it comes to writing, start by keeping a daily journal in the targetlanguage. This will help you practice your writing skills and keep track of your progress. Don’t worry if you make mistakes; they are an essential part of the learning process. The key is to be consistent and not be afraid to make mistakes.In conclusion, the key to mastering a new language is to find the right balance between memorization and practice. By immersing yourself in the language, actively engaging with it, and being patient and persistent, you will eventually achieve your goals.Grammar Completion:Choose the most appropriate word or phrase to complete the following sentences. Write the letter corresponding to your answer in the blanks.1.One effective method is to________yourself in the language by watching movies, listening to music, and reading books.a)immerseb)immersec)immersedd)immersing2.This not only helps you get________to the pronunciation and intonation but also exposes you to a wide range of vocabulary and expressions.a)accustomedb)accustomedc)accustomedd)accustomed3.Try to find a language exchange partner or join a________to practice speaking.a)conversation groupb)conversation groupc)conversation groupd)conversation group4.This will not only improve your speaking skills but also help you gain________in using the language in real-life situations.a)confidenceb)confidencec)confidenced)confidence5.When it comes to writing, start by keeping a daily journal in the target language. This will help you practice your________skills and keep track of your progress.a)readingb)readingc)readingd)reading6.Don’t worry if you make mistakes; they are an essential part of the________process.a)learningb)learningc)learningd)learning7.The key to mastering a new language is to find the right balancebetween________and practice.a)memorizationb)memorizationc)memorizationd)memorization8.By immersing yourself in the language, actively engaging with it, and being patient and ________, you will eventually achieve your goals.a)persistentb)persistentc)persistentd)persistent9.This method is especially useful for those who are________in learning a new language.a)motivatedb)motivatedc)motivatedd)motivated10.It is important to note that simply listening and watching is not enough; you need to actively engage with the language by________and writing.a)speakingb)speakingc)speakingd)speakingAnswers:1.a) immerse2.b) accustomed3.a) conversation group4.b) confidence5.a) reading6.a) learning7.a) memorization8.d) persistent9.b) motivated10.a) speaking七、写作第一节 _ 应用文写作(15分)Writing Part I: Practical WritingDirections: Write an appropriate response to the following situation in no less than 80 words. Your writing should be clear, well-organized, and appropriate to the context.Scenario:Your school has recently organized a cultural exchange program with a sister school in the UK. You have been chosen to host a student from the UK for one week. Write an email to your guest introducing yourself and offering some suggestions for activities during their stay.To:SarahThompson(*****************************.uk)From:。
高二年级第一学期期中考试(物理)试卷含答案解析
高二年级第一学期期中考试(物理)(考试总分:100 分)一、 单选题 (本题共计10小题,总分40分)1.(4分)下列关于点电荷的说法正确的是( )A.点电荷是带电荷量很小的带电体B.带电体体积很大时不能看成点电荷C.点电荷的带电荷量可能是202.5610C -⨯D.大小和形状对作用力影响可以忽略的带电体可看做点电荷2.(4分)关于电流的说法中正确的是( )A.根据qI t=,可知I 与q 成正比 B.电流有方向,电流也有矢量C.电流的单位“安培”是国际单位制中的基本单位D.在金属导体中,自由电子移动方向为电流方向3.(4分)关于电场线的说法,正确的是( )A .电场线的方向,就是电荷受力的方向B .正电荷只在电场力作用下一定沿电场线运动C .静电场的电场线可能是闭合的D .电场线越密的地方,同一电荷所受电场力越大4.(4分)如图所示,两个半径均为r 的金属球放在绝缘支架上,两球面最近距离为r ,带等量异种电荷,电荷量绝对值均为Q ,两球之间的静电力为( )A .等于229Q k rB .大于229Q k rC .小于229Q k rD .等于22Q k r5.(4分)如图所示,将一电荷量为 q 的正试探电荷置于电场中的 A 点,受到的电场力为 F 。
若把该电荷换为电荷量为2q 的负试探电荷,则 A 点的电场强度 E 为( )A.Fq,方向与 F 相反 B.2Fq,方向与 F 相反 C.Fq,方向与 F 相同 D.2Fq,方向与 F 相同 6.(4分)如图所示,A B C 、、为直角三角形的三个顶点,30A ∠=︒,D 为AB 的中点,负点电荷Q 位于D 点。
A B C 、、三点的电势分别用A B C ϕϕϕ、、表示,下列说法正确的是( )A.C ϕ大于A ϕB.A B 、两点电场强度相同C.负检验电荷在BC 连线上各点具有的电势能都相等D.将正检验电荷沿AC 从A 点移到C 点,电场力先做正功后做负功7.(4分)对于电容QC U=,以下说法正确的是( )A.一只电容充电荷量越大,电容就越大B.对于固定电容器,它所充电荷量跟它两极板间所加电压的比值保持不变C.可变电容器的充电荷量跟加在两极间的电压成反比D.如果一个电容器没有电压,就没有充电荷量,也就没有电容8.(4分)如图,平行板电容器经开关S与电源连接,C点所在的空间位置不变且始终处于由带电平板所激发的匀强电场中,选取B板的电势为零,关于C点电势的变化情况,以下说法正确的是( )A.开关S闭合,将A板上移一小段距离,极板间电场强度变弱,C点电势升高B.开关S闭合,将B板上移一小段距离,极板间电场强度变强,C点电势升高C.开关S先闭合后断开,然后将A板上移一小段距离,场强不变,C点电势升高D.开关S先闭合后断开,然后将B板下移一小段距离,场强不变,C点电势升高9.(4分)如图所示,带箭头的线表示某一电场的电场线。
海南省海口市琼山区海南中学2024-2025学年高二上学期11月期中考试地理试题(含答案)
海南中学2024-2025学年度第一学期高二期中考试地理试题注意事项:1.本试卷分选择题和非选择题两部分。
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。
写在本试卷上无效。
3.回答非选择题时,将答案写在答题卡上。
写在本试卷上无效。
Ⅰ选择题本题为单项选择题,15 小题,每小题 3 分,共45 分。
2024年巴黎奥运会于当地时间2024年7月26日开幕,8月11日闭幕。
中国国家跳水队队员海南姑娘陈艺文在巴黎奥运会女子跳水3米板决赛中夺得金牌。
据此完成1-3题。
1.在奥运会举办期间,下列说法正确的是A.晨昏线与地轴的夹角不变 B.巴黎的日出时间变晚C.地球的公转速度逐渐变慢 D.巴黎的自转角速度比海口快2.此次奥运会阶段,巴黎实施夏令时(时间提早1小时)。
海南观众们于北京时间8月9日21:00准时收看奥运会女子跳水3米板决赛,此时巴黎(东一区)的夏令时为A.8月9日15时B.8月9日14时C.8月9日13时D.8月10日4时3.巴黎奥运会闭幕式当天A.海口昼短夜长 B.巴黎的昼长大于海口C.海口的正午太阳高度角为86.5° D.海口的正午太阳高度角小于巴黎图1是“某条河流的剖面图”,箭头为图示区域某日正午太阳来向(遮蔽区范围随季节有变化),该地全年正午太阳都从一方向来。
据图完成4-6题。
4.图示区域可能位于A.北纬10°附近B.北纬40°附近C.南纬10°附近D.南纬40°附近图15.若仅考虑地转偏向力的影响,下列说法正确的是 A .河流自南向北流B .河流自东向西流C .该河南岸坡度陡,适合挖沙D .该河北岸地势平坦,适合修建港口6.图示遮蔽区的范围最小时,正值A .南极地区科考繁忙B .东北地区水稻播种C .华北地区小麦丰收D .江淮地区梅雨纷纷土耳其的卡帕多奇亚地区,数亿年前分别由岩浆冷却形成了玄武岩和由火山灰堆积形成了凝灰岩,后经长期的风化侵蚀作用,形成头戴玄武岩“帽子”的圆锥形尖塔,被当地人称为“精灵烟囱”,当地人在凝灰岩岩面上开凿窑洞作为客栈。
安徽省蚌埠市2023-2024学年高二上学期期中数学试题含解析
蚌埠2023-2024学年第一学期期中检测试卷高二数学(答案在最后)一、单选题(本大题共8小题,共40.0分.在每小题列出的选项中,选出符合题目的一项)1.若直线l 的一个方向向量为(-,求直线的倾斜角()A.π3B.π6C.2π3D.5π6【答案】C 【解析】【分析】求出直线斜率,进而求出直线倾斜角即得.【详解】直线l 的一个方向向量为(-,则直线l 斜率为,所以直线l 的倾斜角为2π3.故选:C2.如图,在四棱锥P ABCD -中,底面ABCD 是平行四边形,已知PA a = ,PB b = ,PC c = ,12PE PD = ,则BE = ()A.131222a b c -+B.111222a b c-+C.131222a b c ++D.113222a b c -+【答案】A 【解析】【分析】利用空间向量加法法则直接求解.【详解】连接BD ,如图,则()()()1111122222BE BP BD PB BA BC PB PA PB PC PB =+=-++=-+-+-()11131131222222222PB PA PB PC PA PB PC a b c=-+-+=-+=-+故选:A .3.已知点A 与点(1,2)B 关于直线30x y ++=对称,则点A 的坐标为A.(3,4) B.(4,5)C.(4,3)-- D.(5,4)--【答案】D 【解析】【分析】根据对称列式求解.【详解】设(),A x y ,则123052224(1)11x y x y y x ++⎧++=⎪=-⎧⎪∴⎨⎨-=-⎩⎪⋅-=-⎪-⎩,选D.【点睛】本题考查关于直线对称点问题,考查基本分析求解能力,属基础题.4.在一平面直角坐标系中,已知()1,6A -,()2,6B -,现沿x 轴将坐标平面折成60°的二面角,则折叠后A ,B 两点间的距离为()A.27 B.41C.17 D.35【答案】D 【解析】【分析】平面直角坐标系中已知()1,6A -,()2,6B -,现沿x 轴将坐标平面折成60°的二面角后,通过向量的数量积转化求解距离即可.【详解】解:平面直角坐标系中已知()1,6A -,()2,6B -,沿x 轴将坐标平面折成60°的二面角后,作AC ⊥x 轴,交x 轴于C 点,作BD ⊥x 轴,交x 轴于D 点,则6,3,6,AC CD DB === ,AC CD CD DB ⊥⊥ ,,AC DB的夹角为120°∴AB AC CD DB =++ ,222222212+2+2=6+3+6266452AB AC CD DB AC CD CD DB AC DB =+++⋅⋅⋅-⨯⨯⨯= 35AB ∴=,即折叠后A ,B 两点间的距离为35.故选:D .【点睛】本题考查与二面角有关的立体几何综合题,解题时要认真审题,注意数形结合思想的合理运用.5.如果实数x ,y 满足()2222x y -+=,则yx的范围是()A.()1,1- B.[]1,1- C.()(),11,-∞-⋃+∞ D.(][),11,-∞-⋃+∞【答案】B 【解析】【分析】设yk x =,求y x的范围救等价于求同时经过原点和圆上的点(),x y 的直线中斜率的范围,结合图象,易得取值范围.【详解】解:设yk x=,则y kx =表示经过原点的直线,k 为直线的斜率.如果实数x ,y 满足22(2)2x y -+=和yk x=,即直线y kx =同时经过原点和圆上的点(),x y .其中圆心()2,0C ,半径2r =从图中可知,斜率取最大值时对应的直线斜率为正且刚好与圆相切,设此时切点为E则直线的斜率就是其倾斜角EOC ∠的正切值,易得2OC =,CE r ==可由勾股定理求得OE ==,于是可得到tan 1CEk EOC OE =∠==为y x的最大值;同理,yx的最小值为-1.则yx的范围是[]1,1-.故选:B.6.抛物线214x y =的焦点到双曲线22221(0,0)x y a b a b -=>>的渐近线的距离是2,则该双曲线的离心率为()A.B.C.2D.233【答案】A 【解析】【分析】先求得抛物线的焦点,根据点到直线的距离公式列方程,求得22b a =,由此求得双曲线的离心率.【详解】抛物线214x y =即24y x =的焦点坐标为()1,0,双曲线22221(0,0)x y a b a b-=>>的渐近线方程为b y x a =±,即0bx ay ±=,所以点()1,0到直线0bx ay ±=的距离为22=,则22b a =,则双曲线的离心率为c e a =====故选:A7.直线()2200ax by a b a b +--=+≠与圆2220x y +-=的位置关系为()A.相离 B.相切C.相交或相切D.相交【答案】C 【解析】【分析】利用几何法,判断圆心到直线的距离与半径的关系,判断直线与圆的位置关系即可.【详解】由已知得,圆2220x y +-=的圆心为(0,0),所以圆心到直线()2200ax by a b a b +--=+≠.因为222ab a b ≤+,所以()()2222a b a b+≤+≤,所以直线与圆相交或相切;故选:C .8.在正方体1111ABCD A B C D -中,点P 在1AC 上运动(包括端点),则BP 与1AD 所成角的取值范围是()A.ππ,43⎡⎤⎢⎥⎣⎦ B.π0,2⎡⎤⎢⎥⎣⎦C.ππ,62⎡⎤⎢⎥⎣⎦D.ππ,63⎡⎤⎢⎥⎣⎦【答案】B 【解析】【分析】建立空间直角坐标系,设1AB =,则,01λ≤≤,利用1c s o BC BP =,,即可得出答案.【详解】设BP 与1AD 所成角为θ,如图所示,不妨设1AB =,则()0,0,0B ,()0,1,0A ,()10,1,1A ,()11,0,1C ,()111,0,1AD BC == ,()1,0,0BC = ,()11,1,1AC =-.设1AP AC λ= ,则()1,1,BP BA AC λλλλ=+=-,01λ≤≤.所以111c ·o s BC BPBC BP BC BP==⋅,当0λ=时,10cos BC BP = ,,此时BP 与1AD 所成角为π2,当0λ≠时,1c os BC BP =,,此时10cos 1BC BP <≤,,当且仅当1λ=时等号成立,因为cos y x =在π02x ⎡⎤∈⎢⎥⎣⎦,上单调递减,所以1π0,2BC BP ⎡⎫∈⎪⎢⎣⎭ ,,综上,π0,2θ⎡⎤∈⎢⎥⎣⎦.故选:B .二、多选题(本大题共4小题,共20.0分.在每小题有多项符合题目要求)9.下列说法正确的有()A.若直线y kx b =+经过第一、二、四象限,则()k b ,在第二象限B.直线32y ax a =-+过定点()32,C.过点()21-,斜率为的点斜式方程为)12y x +=-D.斜率为2-,在y 轴截距为3的直线方程为23y x =-±.【答案】ABC 【解析】【分析】由直线y kx b =+过一、二、四象限,得到斜率0k <,截距0b >,可判定A 正确;由把直线方程化简为()()320a x y -+-+=,得到点()32,都满足方程,可判定B 正确;由点斜式方程,可判定C 正确;由斜截式直线方程可判定D 错误.【详解】对于A 中,由直线y kx b =+过一、二、四象限,所以直线的斜率0k <,截距0b >,故点()k b ,在第二象限,所以A 正确;对于B 中,由直线方程32y ax a =-+,整理得()()320a x y -+-+=,所以无论a 取何值点()32,都满足方程,所以B 正确;对于C 中,由点斜式方程,可知过点()21-,斜率为的点斜式方程为)12y x +=-,所以C 正确;由斜截式直线方程得到斜率为2-,在y 轴上的截距为3的直线方程为23y x =-+,所以D 错误.故选:ABC .【点睛】本题主要考查了直线的方程的形式,以及直线方程的应用,其中解答中熟记直线的点斜式的概念及形式,以及直线的斜率与截距的概念是解答的关键,着重考查推理与运算能力,属于基础题.10.关于空间向量,以下说法正确的是()A.若直线l 的方向向量为()1,0,3e = ,平面α的法向量为22,0,3n ⎛⎫=- ⎪⎝⎭ ,则直线l α∥B.已知{},,a b c 为空间的一个基底,若m a c =+,则{},,a b m 也是空间的基底C.若对空间中任意一点O ,有111632OP OA OB OC =++,则P ,A ,B ,C 四点共面D.两个非零向量与任何一个向量都不能构成空间的一个基底,则这两个向量共线【答案】BCD 【解析】【分析】计算得到e n ⊥,l α∥或l ⊂α,A 错误,若,,a b a c +r r r r 共面,则,,a b c 共面,不成立,故B 正确,化简得到23PA PB PC =--,C 正确,若这两个向量不共线,则存在向量与其构成空间的一个基底,故D 正确,得到答案.【详解】()22,0,22031,0,3e n ⎛⎫=-=-+= ⎪⎝⎭⋅⋅ ,故e n ⊥ ,故l α∥或l ⊂α,A 错误;若,,a b a c +r r r r共面,设()()b a a c a c λμλμμ=++=++ ,则,,a b c 共面,不成立,故{},,a b m 也是空间的基底,B 正确;111632OP OA OB OC =++ ,则()()()111632OA OP OB OP OC OP -+-+- 1110632PA PB PC =++=,即23PA PB PC =--,故P ,A ,B ,C 四点共面,C 正确;若这两个向量不共线,则存在向量与其构成空间的一个基底,故D 正确.故选:BCD.11.已知平面α的法向量为()1,2,2n =-- ,点()2,21,2A x x +为α内一点,若点()0,1,2P 到平面α的距离为4,则x 的值为()A.2 B.1C.3- D.6-【答案】AD【解析】【分析】利用向量法可知,点P 到平面α的距离公式为||||AP n d n →→→⋅=,代入相关数值,通过解方程即可求解.【详解】解:由向量法可知,点P 到平面α的距离公式为||||AP n d n →→→⋅=,又 ()()22,(,20,2,0)122,1,x x AP x x →+--==-,()1,2,2n =--24AP n x x →→∴⋅=+,||3n ==由点()0,1,2P 到平面α的距离为4,有2443x x+=解得2x =或6x =-故选:AD【点睛】本题考查的是点面距离的计算问题,核心是会利用向量法中点到平面的距离公式,考查运算求解能力,属于基础题.12.已知双曲线C 经过点6,12⎛⎫ ⎪ ⎪⎝⎭,且与椭圆22Γ:12x y +=有公共的焦点12,F F ,点M 为椭圆Γ的上顶点,点P 为C 上一动点,则()A.双曲线CB.sin 3MOP ∠>C.当P 为C 与Γ的交点时,121cos 3F PF ∠= D.||PM 的最小值为1【答案】ACD 【解析】【分析】根据题意中的点求出双曲线方程,结合离心率的定义即可判断A ;根据双曲线的渐近线,结合图形即可判断B ;根据椭圆与双曲线的定义,结合余弦定理计算即可判断C ;由两点距离公式,结合二次函数的性质即可判断D.【详解】A :由题意,12(1,0),(1,0)F F -,设双曲线的标准方程为222221,11x y a a a-=<-,将点,1)2代入得212a =,所以双曲线方程为2211122x y -=,得其离心率为22c e a ===,故A 正确;B :由A 选项的分析知,双曲线的渐近线方程为y x =±,如图,π4MON ∠=,所以π3π44MOP <∠<,得sin 12MOP <∠≤,故B 错误;C :当P为双曲线和椭圆在第一象限的交点时,由椭圆和双曲线的定义知,1212PF PF PF PF +=-=12,22PF PF ==,又122F F =,在12F PF △中,由余弦定理得222121212121cos 23PF PF F F F PF PF PF +-∠==⋅,故C 正确;D :设00(,)P x y ,则22001,(0,1)2x y M -=,所以PM ==,当012y =时,min1PM =,故D 正确.故选:ACD.三、填空题(本大题共4小题,共20.0分)13.若空间向量(,2,2)a x =和(1,1,1)b = 的夹角为锐角,则x 的取值范围是________【答案】4x >-且2x ≠【解析】【分析】结合向量夹角公式、向量共线列不等式来求得x 的取值范围.【详解】依题意04211a b a bx x ⎧⋅=>⎪⋅⎪⇒>-⎨⎪≠⎪⎩ 且2x ≠.故答案为:4x >-且2x ≠14.已知0a >,0b >,直线1l :()110a x y -+-=,2l :210x by ++=,且12l l ⊥,则21a b+的最小值为__________.【答案】8【解析】【分析】根据两条直线的一般式方程及垂直关系,求出a ,b 满足的条件,再由基本不等式求出最小值即可.【详解】因为12l l ⊥,所以()11120a b -⨯+⨯=,即21a b +=,因为0a >,0b >,所以()2121422248b a a b a b a b a b ⎛⎫+=++=+++≥+ ⎪⎝⎭,当且仅当4b a a b =,即12a =,14b =时等号成立,所以21a b+的最小值为8.故答案为:8.15.直线30x y ++=分别与x 轴,y 轴交于,A B 两点,点P 在圆()2232x y -+=上,则ABP 面积的取值范围______.【答案】[]6,12【解析】【分析】由题意求得所以()30A -,,()0,3B -,从而求得AB =,再根据直线与圆的位置关系可求得点P 到直线30x y ++=距离h ⎡∈⎣,再结合面积公式即可求解.【详解】因为直线30x y ++=分别与x 轴,y 轴交于A ,B 两点,所以()30A -,,()0,3B -,因此AB =.因为圆()2232x y -+=的圆心为()3,0,半径r =,设圆心()3,0到直线30x y ++=的距离为d ,则3033222d ++==>,因此直线30x y ++=与圆()2232x y -+=相离.又因为点P 在圆()2232x y -+=上,所以点P 到直线30x y ++=距离h 的最小值为32222d r -=-=,最大值为32242d r +=+=,即22,42h ⎡⎤∈⎣⎦,又因为ABP 面积为13222AB h h ⨯⨯=,所以ABC 面积的取值范围为[]6,12.故答案为:[]6,1216.瑞士数学家欧拉(LeonhardEuler )1765年在其所著的《三角形的几何学》一书中提出:任意三角形的外心、重心、垂心在同一条直线上,后人称这条直线为欧拉线.已知ABC 的顶点()4,0-A ,()0,4B ,其欧拉线方程为20x y -+=,则顶点C 的坐标可以是_________【答案】()2,0或()0,2-【解析】【分析】设(,)C x y ,依题意可确定ABC ∆的外心为(0,2)M ,可得出,x y 一个关系式,求出ABC ∆重心坐标,代入欧拉直线方程,又可得出,x y 另一个关系式,解方程组,即可得出结论.【详解】设(,),C x y AB 的垂直平分线为y x =-,ABC 的外心为欧拉线方程为20x y -+=与直线y x =-的交点为(1,1)M -,∴22||||10,(1)(1)10MC MA x y ==++-=①由()4,0-A ,()0,4B ,ABC 重心为44(,)33x y -+,代入欧拉线方程20x y -+=,得20x y --=②由①②可得2,0x y ==或0,2x y ==-.故答案为:()2,0或()0,2-.【点睛】本题以数学文化为背景,考查圆的性质和三角形的外心与重心,考查逻辑思维能力和计算能力,属于较难题.四、解答题(本大题共6小题,共70.0分.解答应写出文字说明,证明过程或演算步骤)17.已知圆M 的圆心为()2,3,且经过点()5,1C -.(1)求圆M 的标准方程;(2)已知直线:34160l x y -+=与圆M 相交于,A B 两点,求AB .【答案】(1)()()222325x y -+-=(2)AB =【解析】【分析】(1)根据条件求出圆M 的半径,再结合圆心坐标求出标准方程即可;(2)求出圆心M 到直线l 的距离,再由垂径定理求出||AB .【小问1详解】因为圆M 的圆心为(2,3),且经过点(5,1)C -,所以圆M 的半径5r MC ===,所以圆M 的标准方程为()()222325x y -+-=.【小问2详解】由(1)知,圆M 的圆心为()2,3,半径=5r ,所以圆心M 到直线l 的距离2d =,所以由垂径定理,得AB ===.18.已知ABC 的顶点()3,2A ,边AB 上的中线所在直线方程为380x y -+=,边AC 上的高所在直线方程为290x y --=.(1)求顶点,B C 的坐标;(2)求ABC 的面积.【答案】(1)B 的坐标为()8,7,C 的坐标为()1,3(2)152【解析】【分析】(1)设(),B a b ,(),C m n ,由题意列方程求解即可得出答案.(2)先求出AB 和直线AB 所在的方程,再由点到直线的距离公式求出边AB 上的高,即可求出ABC 的面积.【小问1详解】设(),B a b ,因为边AB 上的中线所在直线方程为380x y -+=,边AC 上的高所在直线方程为290x y --=,所以2903238022a b a b --=⎧⎪⎨++-⨯+=⎪⎩,解得87a b =⎧⎨=⎩,即B 的坐标为()8,7.设(),C m n ,因为边AB 上的中线所在直线方程为380x y -+=,边AC 上的高所在直线方程为290x y --=,所以3802132m n n m -+=⎧⎪-⎨=-⎪-⎩,解得13m n =⎧⎨=⎩,即C 的坐标为()1,3.【小问2详解】因为()()3,2,8,7A B,所以AB ==因为边AB 所在直线的方程为237283y x --=--,即10x y --=,所以点()1,3C 到边AB的距离为2=,即边AB上的高为2,故ABC的面积为115222⨯=.19.已知直三棱柱111ABC A B C -,侧面11AA C C 是正方形,点F 在线段1AC 上,且13AF =,点E 为1BB 的中点,1AA =,1AB BC ==.(1)求异面直线CE 与BF 所成的角;(2)求平面CEF 与平面11ACC A 夹角的余弦值.【答案】(1)90(2)21【解析】【分析】(1)利用直棱柱的结构特征,结合线面垂直的性质,建立空间直角坐标系,利用直线与直线所成角的向量求法,计算得结论;(2)分别求出两个平面的法向量,利用平面与平面所成角的向量求法,即可得到结果.【小问1详解】因为侧面11AA C C 是正方形,1AA =,1AB BC ==,所以BA BC ⊥,因为三棱柱111ABC A B C -直三棱柱,所以1BB ⊥面ABC ,而BC ,BA ⊂平面ABC ,因此1BB BC ⊥,1BB BA ⊥,所以BC ,BA ,1BB 两两垂直.以B 为坐标原点,BC ,BA ,1BB 所在直线分别为x ,y ,z 轴,建立空间直角坐标系,如下图:因此()100C ,,,()000,,B ,()010A ,,,(1102C ,,而点E 为1BB 的中点,所以2002E ⎛⎫ ⎪ ⎪⎝⎭,,,因为F 在线段1AC 上,所以设()()1,201AF AC λλλλλ==-≤≤ ,因此(),12BF BA AF λλλ=+=- ,因为13AF = ()()222123λλλ+-+=解得16λ=,因此152,,666BF ⎛⎫= ⎪ ⎪⎝⎭ ,即152,,666F ⎛⎫ ⎪ ⎪⎝⎭,因为21,0,2CE ⎛⎫=- ⎪ ⎪⎝⎭,所以11066CE BF ⋅=-+= ,因此异面直线CE 与BF 所成的角为90 .【小问2详解】设平面CEF 的法向量为()1n x y z = ,,,而552,,666CF ⎛⎫=- ⎪ ⎪⎝⎭,因此由1100n CE n CF ⎧⋅=⎪⎨⋅=⎪⎩ 得2025520666x z x y z ⎧-+=⎪⎪⎨⎪-++=⎪⎩,取2z =得1x =,35y =,所以13125n ⎛= ⎝ ,,是平面CEF 的一个法向量,设平面11ACC A 的法向量为()2222n x y z = ,,,()110AC =- ,,,(112AC =- ,,,因此由22100n AC n AC ⎧⋅=⎪⎨⋅=⎪⎩ 得020x y x y z -=⎧⎪⎨-+=⎪⎩,取1x =得1y =,0z =,所以()2110n = ,,是平面11ACC A 的一个法向量.设平面CEF 与平面11ACC A 夹角为θ,则02πθ≤≤,因此121212cos cos ,n n n n n n θ⋅==31521+==,所以平面CEF 与平面11ACC A 夹角的余弦值为24221.20.已知双曲线C的焦点坐标为()1F,)2F ,实轴长为4,(1)求双曲线C 的标准方程;(2)若双曲线C 上存在一点P 使得12PF PF ⊥,求12PF F △的面积.【答案】(1)2214x y -=;(2)1.【解析】【分析】(1)由题可知,c a 的值即可求出双曲线C 的标准方程;(2)由双曲线的定义及面积公式即可求出.【详解】(1)设双曲线方程为22221(0,0)x y a b a b-=>>,由条件知c =,24a =,∴2,1a b ==,∴双曲线C 的方程为2214x y -=.(2)由双曲线的定义可知,124PF PF -=±.∵12PF PF ⊥,∴22212420PF PF c +==,即21212()220PF PF PF PF ⨯-+=∴122PF PF ⋅=,∴12PF F △的面积12112122S PF PF =⋅=⨯=.21.在四棱锥P ABCD -中,底面ABCD 为直角梯形,//AD BC ,AB BC ⊥,侧面PAB ⊥底面ABCD ,2PA PB AD ===,4BC =.(1)若PB 的中点为E ,求证://AE 平面PCD ;(2)若PB 与底面ABCD 所成的角为60︒,求PC 与平面PBD 的所成角的余弦值.【答案】(1)证明见解析(2)80535【解析】【分析】(1)取PC 的中点F ,连接,EF DF .先证明四边形ADFE 是平行四边形,即可得出//DF AE ,然后即可证明线面平行;(2)先证明PO ⊥平面ABCD ,即可得出60PBA ∠=︒.然后建立空间直角坐标系,得出点以及向量的坐标,求出平面PBD 的法向量,根据向量求得PC 与平面PBD 的所成角的正弦值,进而求得余弦值.【小问1详解】如图1,取PC 的中点F ,连接,EF DF ,,E F 分别为,PB PC 的中点,∴//EF BC ,且122EF BC ==.//AD BC 且2AD =,//EF AD ∴且2EF AD ==,∴四边形ADFE 是平行四边形,//DF AE ∴.AE ⊄ 平面PCD ,DF ⊂平面PCD ,∴//AE 平面PCD .【小问2详解】若O 是AB 中点,取CD 中点为G ,连结OG .,O G 分别是,AB CD 的中点,∴//OG BC .AB BC ⊥,∴OG AB ⊥.由底面ABCD 为直角梯形且//AD BC ,2PA PB AD ===,4BC =.PA PB =,∴PO AB ⊥.由侧面PAB ⊥底面ABCD ,平面PAB ⋂平面ABCD AB =,PO ⊂面PAB ,∴PO ⊥平面ABCD ,P ∴在平面ABCD 的投影在直线AB 上.又PB 与底面ABCD 所成的角为60︒,PB ∴与底面ABCD 所成角的平面角60PBA ∠=︒,∴PAB 为等边三角形,2AB PA ==.以O 为原点,分别以,,OB OG OP 所在的直线为,,x y z 轴,如图2建空间直角坐标系,则()1,0,0B ,()1,4,0C ,()1,2,0D -,(3P ,则(3BP =- ,(1,2,3PD =- ,(1,4,3PC = .设平面PBD 的法向量(),,n x y z =r,则00n BP n PD ⎧⋅=⎪⎨⋅=⎪⎩,即020x x y ⎧-+=⎪⎨-+-=⎪⎩,取x =,得)n = ,∴cos ,35n PC n PC n PC ⋅==r uu u r r uu u r r uu u r .设PC 与平面PBD 的所成角为θ,则sin cos ,35n PC θ== . π0,2θ⎡⎤∈⎢⎥⎣⎦,∴cos 0θ≥∴cos 35θ==,PC ∴与平面PBD的夹角的余弦值为35.22.已知抛物线C :()220y px p =>的焦点为F ,斜率为1的直线l 经过F ,且与抛物线C 交于A ,B 两点,8AB =.(1)求抛物线C 的方程;(2)过抛物线C 上一点(),2P a -作两条互相垂直的直线与抛物线C 相交于MN 两点(异于点P ),证明:直线MN 恒过定点,并求出该定点坐标.【答案】(1)24y x=(2)证明见解析【解析】【分析】(1)根据条件,得到直线l 方程为2p y x =-,设1122(,),(,)A x y B x y ,联立抛物线方程,根据抛物线的弦长求得p ,即得答案;(2)求得a 的值,设直线MN 的方程为x my n =+,联立抛物线方程,得根与系数的关系,利用PM PN ⊥,得到32(1)n m -=-或32(1)n m -=--,代入直线方程,分离参数,求得定点坐标,证明结论.【小问1详解】设1122(,),(,)A x y B x y ,由题意知(,0)2p F ,则直线l 方程为2p y x =-,代入()220y px p =>,得22304p x px -+=,280p ∆=>,∴123x x p +=,由抛物线定义,知1||2p AF x =+,2||2p BF x =+,∴12348AB AF BF x x p p p p =+=++=+==,∴2p =,∴抛物线的方程为24y x =.【小问2详解】证明: (),2P a -在抛物线24y x =上,∴242),1(a a =∴=-,由题意,直线MN 的斜率不为0,设直线MN 的方程为x my n =+,设3344(,),(,)M x y N x y ,由24y x x my n⎧=⎨=+⎩,得2440y my n --=,则216160m n '∆=+>,且34344,4y y m y y n +==-,又23434)242(x x m y y n m n +=++=+,22234344334()()()x x my n my n m y y mn y y n n =++=+++=,由题意,可知PM PN ⊥,PM PN ∴⊥,故3434(1)(1)(2)(2)0PM PN x x y y +⋅=+--+= ,故()3434343412()40x x x x y y y y -++++++=,整理得2246850n m n m --++=,即22(3)4)(1n m -=-,∴32(1)n m -=-或32(1)n m -=--,即21n m =+或25n m =-+.若21n m =+,则21(2)1x my n my m m y =+=++=++,此时直线MN 过定点(1,2)-,不合题意;若25n m =-+,则()2525x my n my m m y =+=-+=-+,此时直线MN 过定点(5,2),符合题意,综上,直线MN 过异于P 点的定点(5,2).【点睛】方法点睛:直线和抛物线的位置关系中,证明直线过定点问题,一般是设出直线方程,利用根与系数的关系化简,求得参数之间的关系式,再对直线分离参数,求得定点坐标,进而证明直线过定点.。
广东省广州市第二中学2024-2025学年高二上学期期中考试物理试卷(含答案)
广州市第二中学2024学年第一学期期中考试高二物理命题:王**、邱** 审校:2024.11.07本试卷共6页,15小题,满分为100分。
考试用时75分钟。
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一、单项选择题(共7题,每题4分,共28分。
每题所给的四个选项中,只有一个正确答案,选错或多选均不得分)1、下列说法中,正确的是( )A.由可知电场中某点的电势φ与q成反比B.由可知单位时间内通过导体横截面的电荷量越多,导体中的电流越大C.试探电荷在电场中所受的电场力F与其电量q无关D.电容器电容越大,电容器所带的电荷量就越多2、锂离子电池主要依靠锂离子(Li+)在电池的正极和负极之间移动来工作。
图为常用手机锂电池的内部结构,某过程中Li+从负极脱嵌通过膈膜嵌入正极。
此锂电池的电动势为3.7V ,则此过程( )A.电池内部电流方向从正极到负极B.电池内部是静电力使Li+移动到正极C.电源内部每搬运一个Li+非静电力做功3.7eVD.将锂电池接入电路,与用电器形成闭合回路,锂电池两端电压为3.7V3、欧姆不仅发现了欧姆定律,还对导体的电阻进行了研究。
如图所示,一块均匀长方体的金属样品,边长分别为a、b、c,且a>b>c。
当样品两侧面加上相同的电压U时,样品中电流最大的是( )A.B.C.D.4、在与纸面平行的匀强电场中,建立如图甲所示的直角坐标系,a、b、c、d是该坐标系中的4个点,已知、、;现有一电子以某一初速度从O点沿Od方向射入,则图乙中能大致反映电子运动轨迹的是( )A.①B.②C.③D.④5、如图是某一沿x方向电场的电场强度E与位置坐标x的关系图像。
最新高二年级第一学期历史期中考试试卷(含答案)
最新高二年级第一学期历史期中考试试卷(含答案)一、单项选择题(每题2 分,共50 分)1. 中国古代史学名著《左传》中说:“多行不义必自毙。
”下列哪一历史人物的结局可以作为该论断的有力证据?()A. 夏桀B. 商纣C. 周厉王D. 周幽王2. 西周实行分封制的根本目的是()A. 奖赏功臣B. 拱卫王室C. 发展经济D. 安抚宗亲3. 春秋战国时期,社会经济发展比较落后,但当时的诸子百家,群星灿烂,是中国文化史上的一个黄金时代。
《孙子兵法》至今为兵家经典,甚至被应用于当代企业管理。
这一事实主要说明()A. 文化是经济的集中表现B. 文化对社会生产方式产生重大影响C. 文化具有相对独立性D. 一定的文化由一定的经济、政治所决定4. 秦始皇统一六国后,采取了一系列措施加强中央集权,其中对后世政治制度影响最深远的是()A. 建立皇帝制度B. 推行郡县制C. 统一度量衡D. 统一文字5. 汉武帝时期,为加强中央集权采取的主要措施有()①颁布“推恩令” ②设立中朝③实行察举制④罢黜百家,独尊儒术A. ①②③B. ①②④C. ①③④D. ②③④6. 魏晋南北朝时期,江南地区得以开发的最主要原因是()A. 北方人民南迁带来先进技术B. 南方的自然条件优越C. 江南统治者施行仁政D. 南方社会较为稳定7. 科举制是中国古代重要的选官制度,它正式形成于()A. 隋文帝时期B. 隋炀帝时期C. 唐太宗时期D. 唐玄宗时期8. 唐太宗统治时期,政治清明,经济发展,国力强盛,史称()A. “文景之治”B. “光武中兴”C. “贞观之治”D. “开元盛世”9. 唐朝时期,对外交往频繁,其中为中印文化交流作出杰出贡献的是()A. 鉴真B. 玄奘C. 晁衡D. 崔致远10. 两宋时期,我国经济重心南移完成,其主要表现不包括()A. “苏湖熟,天下足”B. 中央的财政收入主要来自南方C. 北方人口大量南迁D. 景德镇成为著名的瓷都11. 元朝在地方实行行省制度,行省长官由中央任命。
河南省洛阳市2023-2024学年高二上学期期中考试语文试题(含答案)
洛阳市2023—2024 学年第一学期期中考试高二语文试卷(本试卷共10页,23 小题,满分150分。
考试用时150 分钟。
)注意事项:Ⅰ、答卷前,考生务必用黑色字迹钢笔或签字笔将自己的姓名、考生号、考场号和座位号填写在答题卡上。
将条形码贴在答题卡右上角“条形码粘贴处”。
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考试结束后,将试卷和答题卡一并交回。
一、现代文阅读(35 分)(一)现代文阅读Ⅰ(本题共5 小题,19分)阅读下面的文字,完成1~2题。
习近平总书记在党的二十大报告中指出,“弘扬以伟大建党精神为源头的中国共产党人精神谱系,用好红色资源,深入开展社会主义核心价值观宣传教育,深化爱国主义、集体主义、社会主义教育,着力培养担当民族复兴大任的时代新人”。
包括伟大建党精神、长征精神、延安精神、抗战精神、抗美援朝精神、焦裕禄精神、抗洪精神、抗疫精神等在内的中国共产党人精神谱系,是我们党的宝贵精神财富,是上好“大思政课”的鲜活素材、融通课堂教学与实践教学的有效媒介。
依托中国共产党人精神谱系相关学术成果和实践教学基地,通过课堂叙事式教学、平台情景式教学、基地体验式教学、网络延展式教学的“四位一体”立体化实践教学模式,将中国共产党人精神谱系全面融入高校思政课实践教学,能够使广大青年学生深刻领悟中国共产党人精神谱系的丰富内涵和时代意义,激励他们继承优良传统、赓续红色血脉,将志气、骨气、底气固化为信仰、转化为信念、强化为信心。
弘扬中国共产党人精神谱系重在实效性,实现课堂叙事式教学、平台情景式教学、基地体验式教学、网络延展式教学的相互渗透、有机融合、功能互补,有效整合校内校外、课内课外、线上线下等教育教学资源,不断增强思政课的思想性、理论性和亲和力、针对性。
北京市大兴区2023-2024学年高二上学期期中语文试题含答案
大兴区2023~2024学年度第一学期期中检测高二语文(答案在最后)2023.11考生须知:1.本试卷共8页,共五道大题,23道小题,满分150分。
考试时间150分钟。
2.试题答案一律涂或写在答题卡上,选择题用2B铅笔作答,其他试题用黑色签字笔作答,在试卷上作答无效。
3.考试结束,只需上交答题卡。
一、本大题共5小题,共18分。
阅读下面材料,完成小题。
材料一民政部养老服务司副司长李邦华介绍,截至2021年年底,全国60岁及以上老年人口达2.67亿,占总人口的18.9%。
预计“十四五”时期,60岁及以上老年人口总量将突破3亿,占比将超过20%,我国将进入中度老龄化阶段。
养老服务已经成为积极应对人口老龄化的重要内容。
“养老”一词,最早见于《礼记·王制》:“凡养老,有虞氏以燕礼,夏后氏以飨礼,殷人以食礼,周人修而兼用之。
”燕、飨、食等礼仪都是借祭祀鬼神之日,以宴会的形式编排长幼序列,演示敬老之礼。
这里的“养老”还并不是常规意义上的养老行为。
周代养老的仪式除了设置公宴外,还给国老颁发上顶端镶有木雕鸠鸟形状的黑色木制拐杖——鸠杖(同王杖、玉杖)。
鸠鸟食道宽,吞咽顺利,意在祝福老年人吃好吃饱;鸠杖象征着一种权利和荣誉,持杖老人凭杖就可以享受一定的待遇。
汉代至南北朝时期,国家实行了一系列的养老优抚政策,除给予老人一些荣誉之外,还向社会颁布养老的法令,明确养老范围,建立了具体的保障监督措施,比如汉代就明确规定“子孙为国而死的父祖”等四类人归社会养老。
唐宋时期,敬老和崇文并举,国家建立了“文学馆”等文史研究机构,组织老年学士修史编志,起草皇帝诏书,协助科举考试。
《唐书》中还有国家为高龄老人配备家庭服务人员的记载。
北宋出现了最先用财政资金救助“老疾孤穷丐”的机构——“福田院”。
明代除参照汉代做法外,还积极组织老年人参加政权建设,并在多地设立了养济院。
清代沿用了明朝的养济制度。
我国古代养老文化的核心是“孝”,而以“孝”为核心的古代养老文化,从诞生之日起,就具有了强大的生命力。
山东省青岛地区2023-2024学年高二上学期期中考试语文试题(含答案)
2023-2024学年度第一学期期中学业水平检测高二语文2023.11注意事项:1.答卷前,考生务必将自己的姓名和考号填涂在答题卡上。
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,将答题卡交回。
一、现代文阅读(35分)(一)现代文阅读Ⅰ(本题共5小题,19分)阅读下面的文字,完成1~5题。
材料一:近几年,短视频作为媒体融合的一种表达形式,成为手机网民中绝对的“宠爱”。
从《中国互联网络发展状况统计报告》中用户规模和网民使用率两项指标看,短视频早在2018年就已成为互联网发展的“黑马”,超过了网络购物、网络视频、网络音乐和网络游戏。
这个结果远远超出了专业机构对短视频行业发展前景的预期。
艾媒咨询发布的两份关于中国短视频行业专题调查分析报告中,对2018年中国短视频用户规模的预判分别为3.53亿和5.01亿。
实际上,根据《中国互联网络发展状况统计报告》,当年短视频用户规模已达6.48亿。
短视频为何发展如此迅猛?对此,复旦大学传播与国家治理研究中心主任李良荣在一次主题演讲中表示:短视频是依托移动互联网的发展而产生和壮大的,由于短视频只有几十秒或者三五分钟,非常符合当前碎片化阅读场景;其次,短视频主题鲜明,叙事结构紧凑,在一个视频中就能将一件事情的前因后果交代清楚,十分符合当前受众高效获取信息的习惯;再次,短视频直接而又直观的表达方式能让“90后”“95后”感觉到这就是他们的生活,符合他们在手机、动漫等包围的成长环境中使用媒介的习惯。
“这一轮短视频风口颠覆了我们以往的内容逻辑。
我们可能看到,短视频平台上有不够专业的东西,但是另一方面,这些东西是人民群众想知道的、想看到的。
”中国人民大学新闻学院宋建武教授曾这样表示。
他认为:“短视频的出现,让媒体传播手段、信息交互方式等都发生了本质性变化,主流媒体不把握这个机会就会没有未来。
天津市2023-2024学年高二上学期期中考试英语试题含解析
天津市2023~2024学年度第一学期高二年级期中检测试卷(2023.11)英语(答案在最后)第I卷选择题一、听力第一节(共5小题;每小题1分,满分5分)1.Where does the woman find the match?A.Beside the telephone.B.In the desk drawer.C.On the kitchen table.2.What is Tina Marks doing?A.Having a meeting.B.Making a phone call.C.Lining up.3.When will the lecture be given?A.On June10th.B.On June11th.C.On June18th.4.What are the two speakers mainly talking about?A.A rainforest.B.A report.C.A book party.5.Why doesn’t the man want to go to the beach?A.He can’t bear the hot weather.B.He has no interest in the beach.C.He is waiting for the football match.第二节(共10小题;每小题1分,满分10分)听下面一段对话,回答第6至第8小题。
6.When will the race be held?A.This afternoon.B.Tomorrow morning.C.Tomorrow afternoon.7.What might the weather be like on the weekend?A.Rainy.B.Sunny.C.Cloudy.8.What does Mike probably do?A.A news reporter.B.A weatherman.C.A sports reporter.听下面一段对话,回答第9至第11小题。
洛阳市2023—2024学年第一学期期中考试高二英语试卷含答案解析
洛阳市2023—2024学年第一学期期中考试高二英语试卷一、短对话(共5 分)1.What is the woman’s presentation about?A.Watches.B.Channels.C.Animals.【答案】C【原文】W: Hi, Bob. Can I switch to another channel? I need to make a presentation on animals in the English class tomorrow.M: Of course! Go ahead.2.What is the woman doing?A.Preparing a party.B.Planting some trees.C.Making a cake.【答案】A【原文】M: All right, Sara, I know you’re planning something big for my birthday. Could you tell me just what you have in your mind?W: I’m afraid I can’t. It’s supposed to be a surprise.3.Where are the speakers?A.In a school.B.At a shop.C.At a restaurant.【答案】B【原文】M: Good morning. How can I help you, madam?W: Yesterday I took this dress for my daughter here, but it’s a bit small. Can I change a larger one? M: Let me check. Yes, there is a larger one.4.How is the man’s flat?A.Expensive.B.Big.C.North-facing.【答案】B【原文】W: Kevin, how is your new flat?M: It’s great, large and south-facing. And the rent is $345 a month with the property management fee and Internet included.W: Wow! That’s a wonderful price.5.What was the man’s major in the past?A.BusinessB.Management.C.Science.【答案】A【原文】W: John, I heard you have changed your major. Is that true?M: Yes. After studying Business for half a year, I found I didn’t really like it.W: So what are you studying now?M: Wild Life Science.二、长对话(共11 分)听下面一段较长对话,回答以下小题。
福建省宁德市五校联考2024-2025学年高二上学期11月期中物理试题(含答案)
福宁古五校教学联合体2024-2025学年第一学期期中质量监测高二物理试题(考试时间:75分钟 试卷总分:100分)注意:1.请在答题卡各题指定的答题区域内作答,本试卷上作答无效2.本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)第Ⅰ卷(选择题 共40分)一、单项选择题:本题共4小题,每小题4分,共16分。
在每小题给出的四个选项中,只有一项符合题目要求,选对得4分,选错得0分。
1.两个完全相同的带电金属小球,相距为R (R 远大于小球半径),其中一个球的电荷量是另一个的5倍,它们间的吸引力大小是F ,现将两球接触后再把它们固定在距离为2R 处,它们间库仑力的大小是( )A.B .C .D .2.一段粗细均匀的金属导体的横截面积为S .导体单位体积内的自由电子数为n ,导体内的自由电子电荷量为e ,导体中通过的电流为I ,以下说法中正确的是( )A .t 时间内通过导体某个横截面的电子数B .自由电子定向移动的速率C .自由电子热运动的速率D .自由电子定向移动的速率为真空中的光速c3.静电喷涂被广泛用于各种表面处理技术中,相比传统的喷涂技术,其具备生产效率高劳动条件好,易于实现半自动化或自动化,适于大规模流水线作业,其原理如图所示。
涂料雾化装置为负电极,接电源负高压,被涂物为正电极,通常接地。
下列说法正确的是()A .图中喷枪与被涂物之间的实线代表电场线B .涂料颗粒在电场中运动时加速度恒定C .涂料颗粒在电场中运动时电势能逐渐增大D .被涂物上的尖端处,涂料附着较多4.空间中存在沿x 轴方向的静电场,各点电势的变化规律如图中图像所示,电子以一定的初速度,仅受电场力作用,沿x 轴从O 点运动到处的过程中,下列说法正确的是()95F920F 4F 5F It N e=I v ne =0Iv neS=x ϕ-4xA .电子在处电势能最小B .电子在处受电场力沿x 轴负方向C .电子在处速度最大D .处电势为零,电场强度也为零二、双项选择题:本题共4小题,每小题6分,共24分。
高二上学期期中考试数学试卷含答案(共5套)
高二上学期期中考试数学试题本卷分Ⅰ(选择题)、Ⅱ卷(非选择题)两部分,其中Ⅰ卷1至2页,第二卷2至4页,共150分,考试时间120分钟。
第Ⅰ卷(选择题,共60分)一、单选题:本题共12个小题,每小题5分1.“”是“”的()A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件2.有下列四个命题:(1)“若,则,互为倒数”的逆命题;(2)“面积相等的三角形全等”的否命题;(3)“若,则有实数解”的逆否命题;(4)“若,则”的逆否命题.其中真命题为()A.(1)(2)B.(2)(3)C.(4)D.(1)(2)(3)3.若则为()A.等边三角形 B.等腰直角三角形C.有一个内角为30°的直角三角形 D.有一个内角为30°的等腰三角形4.已知.若“”是真命题,则实数a的取值范围是A.(1,+∞)B.(-∞,3)C.(1,3)D.5.的内角,,的对边分别为,,,若,,,则的面积为A.B.C.D.6.已知中,,则等于()A.B.或C.D.或7.等差数列的前项和为,若,则等于()A.58B.54C.56D.528.已知等比数列中,,,则()A.2B.C.D.49.已知,则z=22x+y的最小值是A.1 B.16 C.8 D.410.若关于的不等式的解集为,则的取值范围是()A.B.C.D.11.当a>0,关于代数式,下列说法正确的是()A.有最小值无最大值B.有最大值无最小值C.有最小值也有最大值D.无最小值也无最大值12.在△ABC中,AB=2,C=,则AC+BC的最大值为A.B.3C.4D.2第Ⅱ卷(非选择题,共90分)二、填空题:共4个小题,每小题5分,共20分13.命题的否定是______________.14.已知的三边长构成公差为2的等差数列,且最大角的正弦值为,则这个三角形的周长为________.15.已知数列{a n}的前n项和为S n,a1=1,当n≥2时,a n+2S n-1=n,则S2 017的值____ ___ 16.已知变量满足约束条件若目标函数的最小值为2,则的最小值为__________.三、解答题:共6题,共70分,解答应写出必要的文字说明、证明过程或演算步骤。
2023-2024学年江苏省苏州市高二(上)期中数学试卷【答案版】
2023-2024学年江苏省苏州市高二(上)期中数学试卷一、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,请把正确的选项填涂在答题卡相应的位置上. 1.直线3x +y ﹣2=0的方向向量为( ) A .(﹣1,3)B .(1,3)C .(﹣3,1)D .(3,1)2.等差数列{a n }中,若2a 3+a 9=18,则a 2+3a 6的值为( ) A .36B .24C .18D .93.与直线3x ﹣4y +5=0关于y 轴对称的直线方程是( ) A .3x +4y ﹣5=0B .3x +4y +5=0C .3x ﹣4y +5=0D .3x ﹣4y ﹣5=04.经过原点和点(3,﹣1)且圆心在直线3x +y ﹣5=0上的圆的方程为( ) A .(x ﹣5)2+(y +10)2=125 B .(x +1)2+(y ﹣2)2=5C .(x ﹣1)2+(y ﹣2)2=5D .(x −53)2+y 2=2595.设{a n }是公差不为0的无穷等差数列,则“{a n }为递减数列”是“存在正整数N 0,当n >N 0时,a n <0”的( )A .充分而不必要条件B .必要而不充分条件C .充分必要条件D .既不充分也不必要条件6.已知点P (4,3),点Q 在x 2+y 2=4的圆周上运动,点M 满足PM →=MQ →,则点M 的运动轨迹围成图形的面积为( ) A .πB .2πC .3πD .4π7.等比数列{a n },满足a 1+a 2+a 3+a 4+a 5=3,a 12+a 22+a 32+a 42+a 52=15,则a 1﹣a 2+a 3﹣a 4+a 5的值是( )A .3B .√5C .−√5D .58.过点P (2,0)作圆x 2+y 2﹣4y =1的两条切线,设切点分别为A ,B ,则△P AB 的面积为( ) A .3√158B .√152C .5√158D .√15二、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、全部选对得5分,选对但不全得2分,选错或不答得0分,请把正确的选项填涂在答题卡相应的位置上. 9.已知直线l :x +my +m =0,若直线l 与连接A (﹣3,2),B (2,1)两点的线段总有公共点,则直线l 的倾斜角可以是( ) A .2π3B .π2C .π4D .π610.设S n ,T n 分别是等差数列{a n }和等比数列{b n }的前n (n ∈N *)项和,下列说法正确的是( )A .若a 15+a 16>0,a 15+a 17<0,则使S n >0的最大正整数n 的值为15B .若T n =5n +c (c 为常数),则必有c =﹣1C .S 5,S 10﹣S 5,S 15﹣S 10必为等差数列D .T 5,T 10﹣T 5,T 15﹣T 10必为等比数列11.已知等比数列{a n }的公比为q ,前n (n ∈N *)项和为S n ,前n (n ∈N *)项积为T n ,若a 1=132,T 5=T 6,则( ) A .q =2B .当且仅当n =6时,T n 取得最小值C .T n =T 11﹣n (n ∈N *,n <11)D .S n >T n 的正整数n 的最大值为1112.已知圆C :x 2+y 2=4,圆M :x 2+y 2﹣8x ﹣6y +m =0( ) A .若m =8,则圆C 与圆M 相交且交线长为165B .若m =9,则圆C 与圆M 有两条公切线且它们的交点为(﹣3,﹣4) C .若圆C 与圆M 恰有4条公切线,则m >16D .若圆M 恰好平分圆C 的周长,则m =﹣4三、填空题:本题共4小题,每小题5分,共20分,请把答案写在答题卡相应的位置上.13.若{a n }是公差不为0的等差数列,a 2,a 4,a 8成等比数列,a 1=1,S n 为{a n }的前n (n ∈N *)项和,则1S 1+1S 2+⋯+1S 10的值为 .14.平面直角坐标系xOy 中,过直线l 1:7x ﹣3y +1=0与l 2:x +4y ﹣3=0的交点,且在y 轴上截距为1的直线l 的方程为 .(写成一般式)15.如图,第一个正六边形A 1B 1C 1D 1E 1F 1的面积是1,取正六边形A 1B 1C 1D 1E 1F 1各边的中点A 2,B 2,C 2,D 2,E 2,F 2,作第二个正六边形A 2B 2C 2D 2E 2F 2,然后取正六边形A 2B 2C 2D 2E 2F 2各边的中点A 3,B 3,C 3,D 3,E 3,F 3,作第三个正六边形,依此方法一直继续下去,则前n 个正六边形的面积之和为 .16.已知实数a ,b ,c 成等差数列,在平面直角坐标系xOy 中,点A (4,1),O 是坐标原点,直线l :ax +2by +3c =0.若直线OM 垂直于直线l ,垂足为M ,则线段|AM |的最小值为 .四、解答题:本题共6小题,共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.(10分)已知直线l 1:2x ﹣(a ﹣1)y ﹣2=0,l 2:(a +2)x +(2a +1)y +3=0(a ∈R ). (1)若l 1⊥l 2,求实数a 的值; (2)若l 1∥l 2,求l 1,l 2之间的距离.18.(12分)已知等差数列{a n },前n (n ∈N *)项和为S n ,又a 2=4,S 9=90. (1)求数列{a n }的通项公式a n ;(2)设b n =|9﹣a n |,求数列{b n }的前n 项和T n . 19.(12分)已知数列{a n }的首项a 1=23,且满足a n−1=2a na n +1. (1)求证:数列{1a n−1}为等比数列;(2)设b n =(−1)n−1a n,求数列{b n }的前2n 项和S 2n .20.(12分)如图,等腰梯形ABCD 中,AB ∥CD ,AB =2CD =8,AB ,CD 间的距离为4,以线段AB 的中点为坐标原点O ,建立如图所示的平面直角坐标系xOy ,记经过A ,B ,C ,D 四点的圆为圆M . (1)求圆M 的标准方程;(2)若点E 是线段AO 的中点,P 是圆M 上一动点,满足PO →•PE →≥24,求动点P 横坐标的取值范围.21.(12分)平面直角坐标系xOy 中,直线l :3x +2y ﹣13=0,圆M :x 2+y 2﹣12x ﹣8y +48=0,圆C 与圆M 关于直线l 对称,P 是直线l 上的动点. (1)求圆C 的标准方程;(2)过点P 引圆C 的两条切线,切点分别为A ,B ,设线段AB 的中点是Q ,是否存在定点H ,使得|QH |为定值,若存在,求出该定点H 的坐标;若不存在,请说明理由. 22.(12分)记首项为1的递增数列为“W ﹣数列”.(1)已知正项等比数列{a n },前n (n ∈N *)项和为S n ,且满足:a n +2=2S n +2. 求证:数列{a n }为“W ﹣数列”;(2)设数列{b n }(n ∈N ∗)为“W ﹣数列”,前n (n ∈N *)项和为S n ,且满足∑b i 3=S n 2(n ∈N ∗)ni=1.(注:∑b i 3=b 13+b 23+⋯+b n 3ni=1) ①求数列{b n }的通项公式b n ; ②数列{c n }(n ∈N ∗)满足c n =b n33bn,数列{c n }是否存在最大项?若存在,请求出最大项的值,若不存在,请说明理由.(参考数据:√2≈1.41,√33≈1.44)2023-2024学年江苏省苏州市高二(上)期中数学试卷参考答案与试题解析一、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,请把正确的选项填涂在答题卡相应的位置上.1.直线3x+y﹣2=0的方向向量为()A.(﹣1,3)B.(1,3)C.(﹣3,1)D.(3,1)解:根据直线方程3x+y﹣2=0,可得直线的斜率为﹣3,所以直线的一个方向向量为(1,﹣3),又(1,﹣3)=﹣(﹣1,3),所以(﹣1,3)也是直线的一个方向向量.故选:A.2.等差数列{a n}中,若2a3+a9=18,则a2+3a6的值为()A.36B.24C.18D.9解:设等差数列{a n}的公差为d,2a3+a9=18,则2(a1+2d)+a1+8d=3a1+12d=18,即a1+4d=6,a2+3a6=a1+d+3(a1+5d)=4a1+16d=4(a1+4d)=4×6=24.故选:B.3.与直线3x﹣4y+5=0关于y轴对称的直线方程是()A.3x+4y﹣5=0B.3x+4y+5=0C.3x﹣4y+5=0D.3x﹣4y﹣5=0解:令x=0,则y=54,可得直线3x﹣4y+5=0与y轴的交点(0,54).令y=0,可得x=−53,可得直线3x﹣4y+5=0与x轴的交点(−53,0),此点关于y轴的对称点为(53,0).∴与直线3x﹣4y+5=0关于y轴对称的直线经过两点:(0,54),(53,0).其方程为:x53+y54=1,化为:3x+4y﹣5=0.故选:A.4.经过原点和点(3,﹣1)且圆心在直线3x+y﹣5=0上的圆的方程为()A.(x﹣5)2+(y+10)2=125B.(x+1)2+(y﹣2)2=5C.(x﹣1)2+(y﹣2)2=5D.(x−53)2+y2=259解:设圆心C(a,5﹣3a),则由所求的圆经过原点和点(3,﹣1),即√a 2+(5−3a)2=√(a −3)2+(5−3a +1)2,求得a =53,可得圆心为(53,0),半径为√a 2+(5−3a)2=53,故圆的方程为(x −53)2+y 2=259. 故选:D .5.设{a n }是公差不为0的无穷等差数列,则“{a n }为递减数列”是“存在正整数N 0,当n >N 0时,a n <0”的( )A .充分而不必要条件B .必要而不充分条件C .充分必要条件D .既不充分也不必要条件解:因为{a n }是公差不为0的无穷等差数列,若“{a n }为递减数列”,可得{a n }的通项公式为一次函数且一次项系数小于0,一定有a n <0,即“{a n }为递减数列”是“存在正整数N 0,当n >N 0时,a n <0”的充分条件;若“存在正整数N 0,当n >N 0时,a n <0”,设通项公式为a n =pn +q ,则p <0,n ∈N +, 即{a n }为递减数列,所以“{a n }为递减数列”是“存在正整数N 0,当n >N 0时,a n <0”的必要条件, 综上所述:“{a n }为递减数列”是“存在正整数N 0,当n >N 0时,a n <0”的充要条件. 故选:C .6.已知点P (4,3),点Q 在x 2+y 2=4的圆周上运动,点M 满足PM →=MQ →,则点M 的运动轨迹围成图形的面积为( ) A .πB .2πC .3πD .4π解:设M (x ,y ),点P (4,3),点M 满足PM →=MQ →, 可得Q (2x ﹣4,2y ﹣3), 点Q 在x 2+y 2=4的圆周上运动, 可得(2x ﹣4)2+(2y ﹣3)2=4, 即(x ﹣2)2+(y −32)2=1,点M 的运动轨迹是以(2,32)为圆心,1为半径的圆,点M 的运动轨迹围成图形的面积为π. 故选:A .7.等比数列{a n },满足a 1+a 2+a 3+a 4+a 5=3,a 12+a 22+a 32+a 42+a 52=15,则a 1﹣a 2+a 3﹣a 4+a 5的值是( )A .3B .√5C .−√5D .5解:设数列{a n }的公比为q ,且q ≠1,则a 1+a 2+a 3+a 4+a 5=a 1(1−q 5)1−q=3①,a 12+a 22+a 32+a 42+a 52=a 12(1−q 10)1−q 2=15②∴②÷①得a 12(1−q 10)1−q 2÷a 1(1−q 5)1−q=a 1(1+q 5)1+q=5,∴a 1﹣a 2+a 3﹣a 4+a 5=a 1(1+q 5)1+q=5. 故选:D .8.过点P (2,0)作圆x 2+y 2﹣4y =1的两条切线,设切点分别为A ,B ,则△P AB 的面积为( ) A .3√158B .√152C .5√158D .√15解:由题设,圆的标准方程为x 2+(y ﹣2)2=5, 圆心为C (0,2),半径r =√5,所以|CP|=2√2,如图所示,切点分别为A ,B ,则|BP|=|AP|=√8−5=√3, 所以sin ∠BPC =|BC||CP|=√52√2,cos ∠BPC =|BP||CP|=32√2,又∠BP A =2∠BPC ,所以sin ∠BP A =sin2∠BPC =2sin ∠BPC cos ∠BPC =2√52√2×32√2=√154,所以S △PAB =12|BP||AP|sin∠BPA =12×√3×√3×√154=3√158. 故选:A .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、全部选对得5分,选对但不全得2分,选错或不答得0分,请把正确的选项填涂在答题卡相应的位置上. 9.已知直线l :x +my +m =0,若直线l 与连接A (﹣3,2),B (2,1)两点的线段总有公共点,则直线l 的倾斜角可以是( ) A .2π3B .π2C .π4D .π6解:直线l :x +my +m =0,即x +(y +1)m =0,故直线l 过定点C (0,﹣1), A (﹣3,2),B (2,1), 则k AC =2−(−1)−3−0=−1,k BC =1−(−1)2−0=1, 直线AC 的倾斜角为3π4,直线BC 的倾斜角为π4,直线l 与连接A (﹣3,2),B (2,1)两点的线段总有公共点, 则直线l 的倾斜角范围为[π4,3π4].故选:ABC .10.设S n ,T n 分别是等差数列{a n }和等比数列{b n }的前n (n ∈N *)项和,下列说法正确的是( ) A .若a 15+a 16>0,a 15+a 17<0,则使S n >0的最大正整数n 的值为15 B .若T n =5n +c (c 为常数),则必有c =﹣1 C .S 5,S 10﹣S 5,S 15﹣S 10必为等差数列D .T 5,T 10﹣T 5,T 15﹣T 10必为等比数列解:令{a n }的公差为d ,则a n =a 1+(n ﹣1)d =dn +(a 1﹣d ), 所以{a 15+a 16=2a 1+29d >0a 15+a 17=2a 1+30d <0,故−292d <a 1<−15d ,且d <0,使S n =na 1+n(n−1)2d =d 2n 2+(a 1−d2)n >0, 则0<n <1−2a1d , 而29<−2a 1d<30, 即1−2a 1d∈(30,31),故0<n ≤30, 所以使S n >0的最大正整数n 的值为30,故A 错;令{b n }的公比为q 且q ≠0,则T n =b 1(1−q n )1−q =b 11−q −b 1⋅q n1−q =5n +c (公比不能为1),所以{q =5b 11−q=−1,即c =﹣1,故B 对;根据等差、等比数列片段和的性质知:S 5 S 10﹣S 5,S 15﹣S 10必为等差数列,T 5,T 10﹣T 5,T 15﹣T 10必为等比数列,C 、D 对. 故选:BCD .11.已知等比数列{a n }的公比为q ,前n (n ∈N *)项和为S n ,前n (n ∈N *)项积为T n ,若a 1=132,T 5=T 6,则( )A .q =2B .当且仅当n =6时,T n 取得最小值C .T n =T 11﹣n (n ∈N *,n <11)D .S n >T n 的正整数n 的最大值为11 解:根据题意,依次分析选项:对于A ,若T 5=T 6,则a 6=T6T 5=1,又由a 1=132,则q 5=a6a 1=32,则q =2,A 正确;对于B ,由A 的结论,当1≤n ≤5时,a n <1,a 6=1,当n >6时,a n >1,故当n =5或6时,T n 取得最小值,B 错误;对于C ,由A 的结论,a 6=1,则有a n a 12﹣n =(a 6)2=1, 当n <6时,11﹣n >n ,则有T 11−n T n =a n +1a n +2……a 10﹣n a 11﹣n =1,即T n =T 11﹣n ,同理:当6≤n <11时,也有T n =T 11﹣n , 故T n =T 11﹣n (n ∈N *,n <11)成立,C 正确; 对于D ,若S n >T n ,即a 1(1−q n )1−q>a 1a 2a 3……a n ,即2n −125>2n 2−11n 2,当n =12时,S 12=212−125=27−132,T 12=26,此时S n >T n ,D 错误.故选:AC .12.已知圆C :x 2+y 2=4,圆M :x 2+y 2﹣8x ﹣6y +m =0( ) A .若m =8,则圆C 与圆M 相交且交线长为165B .若m =9,则圆C 与圆M 有两条公切线且它们的交点为(﹣3,﹣4) C .若圆C 与圆M 恰有4条公切线,则m >16D .若圆M 恰好平分圆C 的周长,则m =﹣4解:对于A ,m =8时,圆M :(x ﹣4)2+(y ﹣3)2=17,则M (4,3),半径r =√17. 而圆C :x 2+y 2=4中C (0,0),半径r =2,所以|CM |=√42+32=5, 故√17−2<|CM|<√17+2,即两圆相交,此时相交弦方程为4x +3y ﹣6=0, 所以C (0,0)到4x +3y ﹣6=0的距离d =65,故相交弦长为2×√22−(65)2=165,故A 正确; 对于B ,当m =9时,圆M :(x ﹣4)2+(y ﹣3)2=16,则M (4,3),半径r =4, 类似于A 的分析,可得4﹣2<|CM |<4+2,故两圆相交,故B 错误;对于C ,若圆C 与圆M 恰有4条公切线,则两圆相离,可得|CM |>r +r ′=2+r , 而圆M :(x ﹣4)2+(y ﹣3)2=25﹣m ,即r =√25−m ,所以{25−m >02+√25−m <5,解得16<m <25,故C 错误;对于D ,若圆M 恰好平分圆C 的周长,则相交弦所在直线必过C (0,0),两圆方程相减,可得相交弦方程为8x +6y ﹣m ﹣4=0,将点代入可得m =﹣4,故D 正确. 故选:AD .三、填空题:本题共4小题,每小题5分,共20分,请把答案写在答题卡相应的位置上.13.若{a n }是公差不为0的等差数列,a 2,a 4,a 8成等比数列,a 1=1,S n 为{a n }的前n (n ∈N *)项和,则1S 1+1S 2+⋯+1S 10的值为2011.解:设数列{a n }是公差d 不为0的等差数列,a 2,a 4,a 8成等比数列,a 1=1, 故(a 1+3d)2=(a 1+d)(a 1+7d),整理得(1+3d )2=(1+d )(1+7d ),解得d =1; 故a n =1+(n ﹣1)=n , 所以S n =1+2+3+...+n =n(n+1)2, 故1S n=2n(n+1)=2(1n −1n+1);所以1S 1+1S 2+⋯+1S 10=2(1−12+12−13+...+110−111)=2×1011=2011.故答案为:2011.14.平面直角坐标系xOy 中,过直线l 1:7x ﹣3y +1=0与l 2:x +4y ﹣3=0的交点,且在y 轴上截距为1的直线l 的方程为 9x +5y ﹣5=0 .(写成一般式)解:联立{7x −3y +1=0x +4y −3=0,解得x =531,y =2231,即直线l 1,l 2的交点(531,2231),由题意设l 的方程为:y =kx +1,即2231=531k +1,即k =−95,所以直线l 的方程为y =−95x +1, 即9x +5y ﹣5=0. 故答案为:9x +5y ﹣5=0.15.如图,第一个正六边形A 1B 1C 1D 1E 1F 1的面积是1,取正六边形A 1B 1C 1D 1E 1F 1各边的中点A 2,B 2,C 2,D 2,E 2,F 2,作第二个正六边形A 2B 2C 2D 2E 2F 2,然后取正六边形A 2B 2C 2D 2E 2F 2各边的中点A 3,B 3,C 3,D 3,E 3,F 3,作第三个正六边形,依此方法一直继续下去,则前n 个正六边形的面积之和为 4[1−(34)n ] .解:由题设知:后一个正六边形与前一个正六边形的边长比值为√32, 故它们面积比为34, 所以前n 个正六边形的面积是首项为1,公比为34的等比数列, 所以前n 个正六边形的面积之和S =1−(34)n 1−34=4[1﹣(34)n ]. 故答案为:4[1﹣(34)n ]. 16.已知实数a ,b ,c 成等差数列,在平面直角坐标系xOy 中,点A (4,1),O 是坐标原点,直线l :ax +2by +3c =0.若直线OM 垂直于直线l ,垂足为M ,则线段|AM |的最小值为 √2 .解:因为实数a ,b ,c 成等差数列,所以2b =a +c ,所以直线l :ax +2by +3c =0为ax +(a +c )y +3c =0,整理得a (x +y )+c (y +3)=0,令{x +y =0y +3=0,解得x =3,y =﹣3, 即直线l 过定点(3,﹣3),设该点为点P ,如图所示,因为OM ⊥l ,所以点M 在以OP 为直径的圆上,该圆的圆心为Q (32,−32),半径为r =12|OP |=3√22, 所以|AM |≥|AQ |﹣r =√(4−32)2+(1+32)2−3√22=√2,当且仅当A ,M ,Q 三点共线时,等号成立, 所以线段|AM |的最小值为√2.故答案为:√2.四、解答题:本题共6小题,共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.(10分)已知直线l1:2x﹣(a﹣1)y﹣2=0,l2:(a+2)x+(2a+1)y+3=0(a∈R).(1)若l1⊥l2,求实数a的值;(2)若l1∥l2,求l1,l2之间的距离.解:(1)因为l1⊥l2,可得2(a+2)﹣(a﹣1)(2a+1)=0,即2a2﹣3a﹣5=0,解得a=﹣1或a=−5 2;(2)因为l1∥l2,则2(2a+1)=(a+2)[﹣(a﹣1)],且﹣2(2a+1)=﹣(a﹣1)×3=0,解得:a=0或a=﹣5(舍),所以直线l1的方程为:2x+y﹣2=0,直线l2的方程:2x+y+3=0.所以l1,l2之间的距离d=|−2−3|√2+1=√5.18.(12分)已知等差数列{a n},前n(n∈N*)项和为S n,又a2=4,S9=90.(1)求数列{a n}的通项公式a n;(2)设b n=|9﹣a n|,求数列{b n}的前n项和T n.解:(1)等差数列{a n},前n(n∈N*)项和为S n,又a2=4,S9=90.设首项为a1,公差为d,所以{a1+d=49a1+9×82d=90,解得{a1=2d=2.故a n=2n;(2)由(1)得:b n=|9﹣a n|=|9﹣2n|;当n≤4时,T n=7+9−2n2⋅n=8n−n2,当n≥5时,T n=(b1+b2+b3+b4)﹣(b5+b6+...+b n)=32﹣(8n﹣n2)=n2﹣8n+32.故T n={8n−n2(n≤4的正整数)n2−8n+32(n≥5的正整数).19.(12分)已知数列{a n}的首项a1=23,且满足a n−1=2a na n+1.(1)求证:数列{1a n−1}为等比数列;(2)设b n=(−1)n−1a n,求数列{b n}的前2n项和S2n.证明:(1)因为a n+1=2a n a n +1,a 1=23,所以a n ≠0, 所以1a n+1=a n +12a n =12a n +12,所以1a n+1−1=12a n −12, 因为a 1=23,1a 1−1=12≠0,1a n+1−11a n −1=12, 所以{1a n −1}是以12为首项,12为公比的等比数列; (2)S 2n =1a 1−1a 2+1a 3−1a 4+⋯+1a 2n−1−1a 2n=(1a 1−1)−(1a 2−1)+(1a 3−1)−(1a 4−1)+⋯+(1a 2n−1−1)−(1a 2n−1). 又{1a n −1}是以12为首项,−12为公比的等比数列, 所以S 2n =12[1−(−12)2n ]1−(−12)=1−(12)2n 3=4n−13×4n . 20.(12分)如图,等腰梯形ABCD 中,AB ∥CD ,AB =2CD =8,AB ,CD 间的距离为4,以线段AB 的中点为坐标原点O ,建立如图所示的平面直角坐标系xOy ,记经过A ,B ,C ,D 四点的圆为圆M .(1)求圆M 的标准方程;(2)若点E 是线段AO 的中点,P 是圆M 上一动点,满足PO →•PE →≥24,求动点P 横坐标的取值范围.解:(1)如图,因为AB =2CD =8,AB ,CD 间的距离为4,所以A (﹣4,0),B (4,0),C (2,4),D (﹣2,4),则经过A ,B ,C ,D 四点的圆即经过A ,B ,C 三点的圆,又AB 中垂线方程为x =0,BC 中点为(3,2),k BC =0−44−2=−2, 所以BC 的中垂线方程为y −2=12(x −3),即y =12x +12,联立{x =0y =12x +12,得圆心坐标M(0,12), 则MB =√(4−0)2+(0−12)2=√652,所以圆M 的标准方程为x 2+(y −12)2=654;(2)由已知可得E (﹣2,0),设圆M 上一点P (x ,y ),则PO →=(−x ,−y),PE →=(−2−x ,−y),因为PO →⋅PE →≥24,所以﹣x (﹣2﹣x )+(﹣y )(﹣y )≥24,即x 2+y 2+2x ﹣24≥0,所以P 点在圆(x +1)2+y 2=25上及其外部,联立{x 2+y 2−y −16=0x 2+y 2+2x −24=0, 解得x 1=2,x 2=4,所以两圆交点恰为B (4,0),C (2,4),结合图形,当圆M 上一点纵坐标为12时,横坐标为x 3=√652>4,所以点P 横坐标的取值范围是[2,√652].21.(12分)平面直角坐标系xOy 中,直线l :3x +2y ﹣13=0,圆M :x 2+y 2﹣12x ﹣8y +48=0,圆C 与圆M 关于直线l 对称,P 是直线l 上的动点.(1)求圆C 的标准方程;(2)过点P 引圆C 的两条切线,切点分别为A ,B ,设线段AB 的中点是Q ,是否存在定点H ,使得|QH |为定值,若存在,求出该定点H 的坐标;若不存在,请说明理由.解:(1)圆M :(x ﹣6)2+(y ﹣4)2=4,圆心M (6,4),设圆心C (x 0,y 0),由圆C 与圆M 关于直线l :3x +2y ﹣13=0对称,所以{y 0−4x 0−6=233×x 0+62+2×y 0+42−13=0,即{3y 0=2x 03x 02+y 0=0, 解得{x 0=0y 0=0,所以C (0,0),又r =2, 故圆C 的方程为x 2+y 2=4;(2)因为P 是直线l 上的动点,设P(2t ,132−3t), P A ,PB 分别与圆C 切于A ,B 两点,所以CA ⊥P A ,CB ⊥PB , 所以A ,B 在以PC 为直径的圆N 上,圆N 的方程x(x −2t)+y[y −(132−3t)]=0, 即x 2+y 2−2tx +(3t −132)y =0,又AB 为圆C 与圆N 的公共弦,由{x 2+y 2−4=0x 2+y 2−2tx +(3t −132)y =0, 作差可得AB 的方程为2tx −(3t −132)y −4=0,即t(2x −3y)+132y −4=0, 令{2x −3y =0132y −4=0,得{x =1213y =813, 设T(1213,813),则直线AB 过定点T(1213,813), 又Q 是AB 中点,所以CQ ⊥AB ,所以Q 点是在以CT 为直径的圆上,所以存在点H (613,413)是CT 的中点,使得QH 为定值.22.(12分)记首项为1的递增数列为“W ﹣数列”.(1)已知正项等比数列{a n },前n (n ∈N *)项和为S n ,且满足:a n +2=2S n +2. 求证:数列{a n }为“W ﹣数列”;(2)设数列{b n }(n ∈N ∗)为“W ﹣数列”,前n (n ∈N *)项和为S n ,且满足∑b i 3=S n 2(n ∈N ∗)ni=1.(注:∑b i 3=b 13+b 23+⋯+b n 3n i=1) ①求数列{b n }的通项公式b n ;②数列{c n }(n ∈N ∗)满足c n =b n 33b n ,数列{c n }是否存在最大项?若存在,请求出最大项的值,若不存在,请说明理由.(参考数据:√2≈1.41,√33≈1.44)证明:(1)设正项等比数列{a n }的公比为q (q >0),因为a n +2=2S n +2,则a n +3=2S n +1+2,两式相减得a n +3﹣a n +2=2a n +1, 即a n+1(q 2−q −2)=a n+1(q −2)(q +1)=0因为a n >0,q >0,所以q =2,a n +2=2S n +2中,当n =1时,有a 3=2a 1+2,即4a 1=2a 1+2,解得a 1=1, 因此数列{a n }为“W ﹣数列”;解:(2)①因为∑b i 3=S n 2(n ∈N ∗)ni=1所以b 13=b 12,得又{b n }为“W ﹣数列”, 所以b 1=1,且b n +1>b n ,所以{b n }各项为正数,当n ≥2,∑b i 3=S n 2n i=1①,∑b i 3=S n−12n−1i=1②,①一②得:b n 3=S n 2−S n−12,即b n 3=(S n −S n−1)(S n +S n−1),所以b n 2=S n +S n−1③,从而b n+12=S n+1+S n ④,④﹣③得:b n+12−b n 2=b n+1+b n , 由于{b n }为“W ﹣数列”,必有b n +1+b n >0,所以b n +1﹣b n =1,(n ≥2),又由③知b 22=S 2+S 1,即b 22=2b 1+b 2,解得b 2=2或b 2=﹣1(舍);所以b 2﹣b 1=1,故b n+1−b n =1(n ∈N ∗),所以{b n }是以1为首项,公差是1的等差数列,所以b n =n ;②c n =n 33n >0,所以c n+1c n =13(n+1n)3<1, 整理得n √33−1≈2.27,所以当n ≥3时,c n +1<c n ,即c 3>c 4>c 5>⋯,又c 1=13,c 2=89,c 3=1,所以{c n }中存在最大项,为c 3=1.。
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07—08第一学期高二第二次月考第一节单项选择(共25小题;每小题1分,满分25分)从A、B、C、D四个选项中,选出可以填入空白处的最佳选项。
1. The film _______ him _______ what he had seen in China.A. reminded; toB. remembered; ofC. recalled; withD. reminded; of2. I sight of an empty seat at the back of the bus and went directly there.A. lost B caught C. looked D. took3. He asked us to him ______ carrying through their plan.A. assist; withB. help; toC. assist; inD. help; with4. His report was so exciting that it was interrupted by applause(掌声).A. constantlyB. constantC. seldomD. never5. _______ in a friendly way, their quarrel came to an end.A. Being settledB. SettledC. SettlingD. Having settled6. A driver should __ the road when .A. concentrate on; driveB. concentrate in; drivingC. concentrate to; droveD. concentrate on; driving7. W e must work hard to a good knowledge of English.A. takeB. acquireC. catchD. hold8. Tom kept quiet about the accident ________ lose his job.A. so not as toB. so as not toC. so as to notD. not so as to9. Not until the early years of the 19th century ________ what heat was.A. man did knowB. did man knewC. didn't man knowD. did manknow10. The stone bridge __ _ last year is very beautiful.A . built B. was built C . being built D .to be built11.The poor man , ________,ran out of the dark cave.A. tiring and frightenedB. tired and frightenedC. tired and frighteningD. tiring and frightening12.Don‟t be discouraged. ______ things as they are and you will enjoy every day ofyour life.A. TakingB. To takeC. TakeD. Taken13.These articles are written in simple language, _____ makes it easy to read.A. thatB. thisC. whichD. it14.After the war, a new school building was put up __________there used to be a theatre.A. thatB. whereC. whichD. when15. Tom‟s mother kept telling him that he should work harder, but ______didn‟t help.A. heB. whichC. sheD.it16.Along with the letter was his promise _______ he would visit me this coming Christmas.A. whichB. thatC. whatD. whether17. Victor apologized for __________ to inform me of the change in the plan.A. his being not ableB. him not to be ableC. his not being ableD. him to be not able18. The thief admitted over 10 motorbikes and now he has been arrestedA. to stealB. to have stolenC. stealingD. having stolen19.---The novel gone with wind is said _____ into several language.A. to translateB. being translatedC. having been translatedD. to have been translated20. is no needA. It ; complainingB. That; complainingC. There ; to complainD. This; to complain21.Did he find hard to learn a foreign language ?A.thatB. himC. itD. himself22.The population of China is larger than ______ of the United States.A. thisB. thatC. theseD. those23.Y ou may send me an e-mail or just give me a call. will do.A.Neither B.Each C.Any D. Either24. He‟s got himself into a dangerous situation he is likely to be accused of meeting someone.A. whereB. whichC. whileD. why25. The water ______ cool when I joined into the pool for morning exercise.A. was feltB. is feltC. feelsD. felt第二节完形填空(共20小题;每小题1分,满分20分)阅读下面短文,掌握其大意,然后从所给的四个选项(A、B、C和D)中,选出最佳选项。
Michel is a young girl who works for the police 26 a handwriting expert (专家). She has helped 27 many criminals (罪犯) by using her special talents (天才).When she was fourteen, Michel was already 28 interested in the differences in her friends' 29 that she would spend hours 30 them. After 31 college she went to France for a 32 two-year class in handwriting at the School of Police Science.Michel says that it is 33 for people to hide their handwriting. She can discover 34 of what she needs to know simply 35 looking at the writing with her own eyes, 36 she also has machines 37 help her make 38 different kinds of paper and ink. This knowledge is often 39 great help to the police.Michel believes that handwriting is a good 40 of what kind of person the 41 is. "I wouldn't go out with a fellow 42 I didn't like his handwriting. " She says. But she 43 she fell in love with her future husband, a young policeman 44 she studied his handwriting. It is later proved to be 45 , however.26 A. with B. by C. like D. as27 A. search B. follow C. catch D. judge28 A. so B. too C. quite D. extra29 A. books B. letter C. tongues D. handwriting30 A.writing B. studying C. settling D. uncovering31 A.attending B. finishing C. starting D. stepping into32 A. powerful B. natural C. special D. common33 A.main B. safe C. easy D.impossible34 A. most B. nothing C. little D. sight35 A. with B. by C. of D. about36 A. so B. for C. thus D. but37 A. they B. in which C. that D. those38 A. up B. out C. for D. into39 A. of B. to C. with D. for40 A. test B. sign(标记) C. means D. habit41 A. thief B. criminal C. writer D. policeman42 A. whether B. unless C. if D. after43 A. adds B. tells C. repeats D. cries44 A. before B. after C. so D. and45 A.necessary B. all right C.important D. quite easy第三部分:阅读理解(共20小题,每题2分,满分40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。