福建省漳州市2020-2021学年高一上学期期末教学质量检测英语试题(无答案)

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2020-2021学年度高一上学期期末考试英语试卷及答案两套(附听力录音稿)1

2020-2021学年度高一上学期期末考试英语试卷及答案两套(附听力录音稿)1

2020-2021学年度高一上学期期末考试英语试卷注意事项:1. 答卷前,考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置上。

2. 回答选择题时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动,用橡皮擦干净后,再选涂其他答案标号。

回答非选择题时,将答案写在答题卡上。

写在本试卷上无效。

3. 考试结束后,将本试卷和答题卡一并交回。

第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)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 will the speakers do next?A. Ship the goods.B. Have a meeting.C. Discuss a report.2. How much does a buffet meal cost at 7:00 pm?A. $5.B. $8.C. $10.3. What does the man think of his work?A. Boring.B. Satisfactory.C. Tough.4. Whom has the man sent an invitation to?A. Laura.B. Rosa.C. Maria.5. What is the man doing?A. Looking for a car.B. Visiting a company.C. Picking up the woman.第二节听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

福建省漳州市2020-2021学年高二上学期期末考试化学试题

福建省漳州市2020-2021学年高二上学期期末考试化学试题

漳州市2020—201学年(上)期末高中教学质量检测高二化学(化学反应原理)试题试卷说明:1.本试卷分为第Ⅰ卷和第Ⅱ卷两部分.第Ⅰ卷为选择题,第Ⅱ卷为非选择题.2.本试卷满分100分,考试时间90分钟.3.本试卷可能用到的相对原子质量:H 1 C 12 O 16 S 32 Cu 64 Ag 108 Ba 137第Ⅰ卷(选择题 共42分)一、选择题(本题包括14小题,每小题3分,共42分,每小题只有一个选项符合题目要求)1.化学与生活密切相关,下列说法错误的是( ) A.明矾可用于自来水的杀菌消毒 B.热的纯碱溶液可用于去除餐具的油污 C.燃烧木柴时,采用较细木柴并架空有利于燃烧D.甲烷燃料电池的能量转换效窣比甲烷直接燃烧的能量转换效率高 2.下列化学用语正确的是( ) A.醋酸铵在水溶液中电离:3434CH COONH CH COO NH -++ B.碳酸氢钠在水溶液中电离:233NaHCO Na H CO ++-++ C.2CuCl 溶液显酸性:222Cu 2H O Cu(OH)2H ++++ D.2Na S 溶液显碱性:222S H OH S OH --++3.一定温度下,在密闭容器中可逆反应2X(g)Y(g)3Z(g)+达到平衡状态的标志是( )A.混合气体的密度不再发生变化B.单位时间内,生成3mol Z 同时生成2mol XC.混合气体总物质的量不再发生变化D.X 、Y 、Z 的物质的量之比为2:1:3 4.要增大铁与盐酸反应的速率,下列措施中无效的是( )A.提高反应的温度B.增大盐酸的浓度C.用铁粉代替铁片D.增大压强 5.下列各组离子在常温下一定能大量共存的是( ) A.33K Al HCO ++-、、 B.23H Fe NO ++-、、C.Na Ag I ++-、、 D.24Mg NH Cl ++-、、6.向氯化铁溶液中加入过量氢氧化钠溶液,振荡后静置一段时间.下列关于该体系的说法中错误的是( )A.溶液中不再存在3Fe +B.生成了氢氧化铁沉淀C.体系中存在氢氧化铁的沉淀溶解平衡D.加入少量盐酸,则溶液中3Fe +浓度会上升7.为了保护舰艇(主要是铁合金材料),根据原电池原理可在舰体表面镶嵌金属块R.下列有关说法错误的是( )A.金属块R 可能是锌B.舰艇在海水中易发生吸氧腐蚀C.这种保护舰体的方法叫做牺牲阳极的阴极保护法D.舰体腐蚀时正极的电极反应式为22O 4H 4e 2H O +-++8.工业上用甲醇分解制氢,反应原理为32CH OH(g)CO(g)2H (g)ΔH +.一定温度下,该反应过程中的能量变化如图所示,下列说法正确的是( )A.21ΔH E E =-B.该反应为放热反应C.曲线b 为使用催化剂后的能量变化D.平衡后增大压强,平衡向左移动,ΔH 减小9.3CaCO 与100mL 稀盐酸反应生成2CO 的量与反应时间的关系如图所示,下列结论错误的是( )A.开始阶段反应速率增大主要是受体系温度升高的影响B.24min ~内平均反应速率最大C.24min ~内用盐酸表示平均反应速率为1(HCl)0.1mol (L min)v -=⋅⋅ D.4min 后,反应速率减小的主要原因是()Hc +减小10.32NH H O ⋅在水溶液中存在电离平衡324NH H ONH OH +-⋅+,下列说法正确的是( )A.加入氯化铵晶体后,溶液的pH 减小B.加入碳酸钠固体,平衡向右移动C.加水稀释,溶液中的离子浓度均减小D.降低温度,()b 32NH H O K ⋅增大11.普通电解精炼铜的方法所制备的铜中仍含杂质,利用下图的双膜(阴离子交换膜和过滤膜)电解装置可制备高纯度的铜.下列有关叙述错误的是( )A.当电路中通过1mol 电子时,可生成32g 精铜B.电极a 为粗铜,电极b 为精铜C.甲膜为阴离子交换膜,可阻止杂质阳离子进入阴极区D.乙膜为过滤膜,可阻止阳极泥及漂浮物杂质进入阴极区 12.下列有关实验操作的解释或结论正确的是( )13.已知可逆反应2242NO (g)N O (g)中2NO 、24N O 的消耗速率与其浓度存在如下关系:()()2212NO NO v k c =⋅,()()24224N O N O v k c =⋅(其中1k 、2k 是只与温度有关的常数),一定温度下根据上述关系式建立如图关系.下列说法正确的是( )A.图中A 点表示该反应达到化学平衡状态B.若某温度时12k k =,则该温度下反应的平衡常数0.5K =C.在1L 密闭容器中充入21mol NO ,平衡时:()()1224NO N O 1mol L c c -+=⋅D.在1L 密闭容器中充入21mol NO ,当()()224NO N O c c =时,2NO 的转化率是33.3% 14.已知25℃时23H SO 的 1.85a110K -=、7.19a 210K -=,用10.1mol L NaOH -⋅溶液滴定12320mL 0.1mol L H SO -⋅溶液的滴定曲线如下图所示.下列说法错误的是( )A.a 点,溶液中()()2233H SO SO c c ->B.b 点,溶液中()()()()2233H SO HSO OH c c c c +--+=+C.c 点,所加NaOH 溶液的体积为30mLD.d 点,溶液中()()()233NaSO HSO c c c +-->>第Ⅱ卷(非选择题 共58分)二、非选择题(本题包括6小题,共58分)15.(7分)铝作为一种应用广泛的金属,在电化学领域发挥着举足轻重的作用.请回答下列问题:(1)某同学根据氧化还原反应232Al 3Cu2Al 3Cu ++++设计如图所示的原电池.①电极Y 的材料为__________(填化学式,下同). ②盐桥中的阳离子向________溶液中移动. (2)新型电池中的铝电池类型较多.①2Al Ag O -电池可用作水下动力电源,该电池反应原理为:2222Al 3Ag O 2NaOH 2NaAlO 6Ag H O ++++.放电时,Al 电极附近溶液的pH_________(填“增大”、“减小”或“不变”),当电极上析出1.08g Ag 时,电路中转移的电子为________mol . ②Li Al /FeS -是一种二次电池,可用于车载电源.已知FeS 难溶于水,电池总反应为:22Li FeSLi S Fe ++放电充电.充电时,阳极的电极反应式为__________.16.(9分)二甲醚()33CH OCH 是一种重要的清洁燃料,可替代氟利昂作制冷剂,对臭氧层无破坏作用.工业上可利用水煤气合成二甲醚.请回答下列问题:(1)水煤气是水蒸气通过炽热的焦炭而生成的气体,主要成分是CO 和2H .已知该反应生成1mol CO 气体需要吸收131.3kJ 的热量,请写出该反应的热化学方程式__________. (2)利用水煤气合成二甲醚的三步反应如下: (1)12312H (g)CO(g)CH OH(g)Δ90.8kJ mol H -+=-⋅(2)1333222CH OH(g)CH OCH (g)H O(g)Δ23.5kJ mol H -+=-⋅(3)12223CO(g)H O(g)CO (g)H (g)Δ41.3kJ mol H -++=-⋅由此可知总反应23323H (g)3CO(g)CH OCH (g)CO (g)++的焓变ΔH =_______1kJ mol -⋅.(3)若总反应达到平衡后,要提高CO 的转化率,可以釆取的措施是_______(填字母). a.增大压强 b.增大CO 的浓度 c.加催化剂 d.分离出二甲醚 (4)已知某温度下反应②33322CH OH(g)CH OCH (g)H O(g)+的平衡常数400K =.此温度下,测得某时刻各组分的浓度如下:①比较此时正、逆反应速率的大小:()v 正_______()v 逆(填“>”、“<”或“=”).②若从此刻开始又经过10min 达到平衡,则这段时间内反应速率()3CH OH v =________. 17.(7分)已知25℃时部分弱电解质的电离平衡常数如表所示:回答下列问题:(1)写出23H CO 的第一级电离平衡常数表达式:al K =__________. (2)等物质的量浓度的下列溶液,pH 由大到小的顺序为_______(填字母). a.3CH COONa b.NaCN c.23Na CO(3)一定温度下,体积相同、pH 均为2的3CH COOH 溶液与HX 溶液,加水稀释时pH 的变化如右图所示.稀释相同倍数后,HX 溶液中由水电离出的()Hc +__________3CH COOH 溶液中由水电离出的()H c +(填“大于”、小于”或“等于”).(4)25℃时,测得某3CH COOH 与3CH COONa 的混合溶液pH 6=,则溶液中()()3CH COO Na c c -+-=______1mol L -⋅(填精确值).18.(10分)滴定原理在实际生产生活中应用广泛.用25I O 定量测定CO 的实验步骤如下:步骤Ⅰ:取250mL (标准状况)含有CO 的某气体样品,通过盛有足量25I O 的硬质玻璃管,在170℃下充分反应(已知:Δ25225CO I O 5CO I ++,气体样品中其他成分不与25I O 反应); 步骤Ⅱ:用水—乙醇溶液充分溶解产物2I ,配制成100mL 溶液;步骤Ⅲ:量取步骤Ⅱ中溶液25.00mL 于锥形瓶中,然后用10.0100mol L -⋅的223Na S O 标准溶液滴定,反应原理为:22322462Na S O I 2NaI Na S O ++,消耗223Na S O 标准溶液的体积如下表所示.(1)步骤Ⅱ中配制100mL 待测溶液需要用到的玻璃仪器有烧杯、量筒、玻璃棒、胶头滴管和______. (2)①滴定过程中应选用的指示剂是______. ②滴定过程中,眼睛应始终注视_______变化. ③判断达到滴定终点的现象是__________. (3)气体样品中CO 的体积分数为_________.(4)下列操作会造成所测CO 的体积分数偏大的是__________(填字母). a.配制100mL 待测溶液时,有少量溶液溅出 b.锥形瓶用待测溶液润洗 C.滴定前滴定管尖嘴处有气泡,滴定后没有气泡 d.滴定终点时俯视读数19.(11分)某工厂以重晶石(主要含4BaSO )为原料,生产“电子陶瓷工业支柱”——钛酸钡()3BaTiO 的工艺流程如下:回答下列问题:(1)为提高3BaCO 的酸浸速率,可采取的措施为__________(写出一条即可).常温下,4TiCl 为液体且易水解,配制一定浓度的4TiCl 溶液的方法是___________.(2)用23Na CO 溶液浸泡重晶石(假设杂质不与23Na CO 反应),能将4BaSO 转化为3BaCO ,此反应的平衡常数K =____________(填写计算结果).若不考虑23CO -的水解,要使42.33g BaSO 恰好完全转化为3BaCO ,则至少需要浓度为1231.0mol L Na CO -⋅溶液________mL (已知:()10sp 4BaSO 1.010K -=⨯、()9sp 3BaCO 5.010K -=⨯).(3)流程中“混合”溶液的钛元素以TiO(OH)+、24TiOC O 、()2242TiO C O -三种形式存在,其分布随pH 发生变化.在制备过程中,先用氨水调节混合溶液的pH ,再进行“沉淀”.请写出“沉淀”过程的离子方程式_____________.(4)流程中“滤液”的主要成分为_________(写化学式). 20.(14分)已知2CO 催化加氢合成乙醇的反应原理为:1222522CO (g)6H (g)C H OH(g)3H O(g)Δ173.6kJ mol H -++=-⋅.请回答下列问题:(1)图1、图2分别是2CO 的平衡转化率随投料比()m 、压强()p 及温度()T 的变化关系. ①图1中1m 、2m 、3m 从小到大的顺序为_______.②图2中投料比相同,温度从高到低的顺序为_______,理由是_________.(2)图3表示在总压为5MPa 的恒压条件下,同一投料比在不同温度下反应达到平衡时各物质的物质的量分数()y 与温度的关系.①曲线a 代表的物质为_________(填化学式). ②2图3中P 点时,2H 的转化率为_________.③4T 温度时,投料比m =_________,该反应的平衡常数p K =_________.(用平衡分压代替平衡浓度来计算,某组分平衡分压=总压×该组分的物质的量分数,计算结果保留小数点后三位数字).漳州市2020—2021学年(上)期末高中教学质量检测高二化学(化学反应原理)试题参考答案评分说明:1、考生若写出其它正确答案,可参照评分标准给分.2、化学方程式或离子方程式未能正确配平的均不给分,未正确标注反应条件、“↑”、“↓”等总扣1分.第Ⅰ卷(选择题 共42分)一、选择题(本题包括14小题,每小题3分,共42分.每小题只有一个选项符合题目要求)第Ⅱ卷(非选择题 共58分)二、非选择题(本题包括6小题,共58分)15.(7分)(1)①Al (1分) ②4CuSO (1分)(2)①减小(1分) 0.01(2分) ②2Fe 2e S FeS ---+(2分)16.(9分) (1)122C(s)H O(g)CO(g)H (g)Δ131.3kJ mol H -++=+⋅(2分)(2)246.4-(2分) (3)a 、d (2分)(4)①>(1分) ②10.08mol (L min)-⋅⋅(2分) 17.(7分) (1)()()()323HCO H H CO c c c -+(1分) (2)cha (2分)(3)小于(2分) (4)79.910-⨯(2分) 18.(10分)(1)100mL 容量瓶(1分)(2)①淀粉溶液(1分) ②锥形瓶中溶液颜色(2分)③滴加最后一滴标准液,溶液由蓝色变为无色且半分钟内不恢复原色(2分) (3)17.92%(2分) (4)bc (2分) 19.(11分)(1)适当加热、搅拌、将3BaCO 研成粉末、增大盐酸浓度等任写一点(1分) 将4TiCl 溶于浓盐酸,再加适量水稀释至所需浓度(2分) (2)0.02(2分) 510(2分) (3)()()2224224222BaTiO C O 4H OBaTiO C O 4H O -+++⋅↓(2分)(4)4NH Cl (2分) 20.(14分)(1)①321m m m <<(2分) ②321T T T >>(2分)该反应为放热反应,其它条件相同时,升高温度,平衡逆向移动,二氧化碳的转化率减小(2分) (2)①2H (2分) ②66.7%(2分) ③3(2分) 0.243(2分)。

2020-2021学年高一英语下学期期末教学质量检测试题(含解析)

2020-2021学年高一英语下学期期末教学质量检测试题(含解析)

第二学期期末教学质量检测高一英语试卷考生注意:1.本试卷共150分,考试时间120分钟。

2.请将各题答案填写在答题卡上。

第一部分听力(共两节,满分30分做题时,先将答案标在试卷上。

录音内容结束后你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题;每小题1.5分满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题每段对话仅读一遍。

1.【此处有音频,请去附件查看】What are the speakers mainly talking about?A. Turning on the light.B. Power failure.C. Buyingair-conditioning.【答案】B【解析】【原文】W:The light has gone out!M: We often have power failure in this city when too many people are using air conditioning.2.【此处有音频,请去附件查看】How does the man usually go to work?A. By bus.B. By car.C. On foot.【答案】A【解析】【原文】W:Do you go to work on foot every day?M: Well. I'm too fat to walk a lot. I usually take a bus. But sometimes when the bus is crowded. I wish someone would give me a ride.3.【此处有音频,请去附件查看】What color does the man prefer to paint the bedroom?A. Light blue.B. Yellow.C. Pink.【答案】A【解析】【原文】W:Do you think we should paint our bedroom yellow or pink?M:Why not light blue?4.【此处有音频,请去附件查看】Where was the man last night?A. In the library.B. At home.C. At the concert. 【答案】B【解析】【原文】W: Hello. Tony. Where were you last night?M: I had to stay at home.W:But you missed a really wonderful concert!5.【此处有音频,请去附件查看】What do we know about the man?A. He likes driving.B. He enjoys traveling by car.C. He used to have a car.【答案】C【解析】【原文】W: You’ve sold your old car. You don’t need a new one?M:Not really. I’ve never liked driving anyway. Now that I’ve moved to a place near the subway entrance. I can go to work quite conveniently.第二节(共15小题每小题1.5分满分22.5分)听下面5段对话或独白。

福建省漳州市2020-2021学年学年高一数学上学期期末考试试题(含解析)

福建省漳州市2020-2021学年学年高一数学上学期期末考试试题(含解析)

福建省漳州市2020-2021学年学年高一数学上学期期末考试试题(含解析)本试卷共5页,22题.全卷满分150分.考试用时120分钟.注意事项:1.答题前,考生务必在试题卷、答题卡规定的地方填写自己的准考证号、姓名.考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致.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,再选涂其它答案标号.回答非选择题时,将答案写在答题卡上.写在本试卷上无效.3.考试结束,考生必须将试题卷和答题卡一并交回.一.单项选择题:本大题共8小题,每小题5分,共40分,在每小题给出的四个选项中,只有一项是符合题目要求的. 1.已知集合{|4}A x x =>,{|2}B x x ,则A B =( )A. (2,)+∞B. (4,)+∞C. (2,4)D. (,4)-∞【答案】B 【解析】 【分析】由交集的定义求解即可. 【详解】{|{|2}4}{|4}x A B x x x x x =>>=>故选:B【点睛】本题主要考查了集合间的交集运算,属于基础题. 2.sin(600)-︒的值是( )A.12B. 12-C.2D. 【答案】C 【解析】 【分析】原式中的角度变形后,利用诱导公式及特殊角的三角函数值计算即可得到结果.【详解】解:()()()sin 600sin 720120sin120sin 18060sin60-︒=-︒+︒=︒=︒-︒=︒= 故选C .【点睛】此题考查了运用诱导公式化简求值,熟练掌握诱导公式是解本题的关键. 3.下列各函数的值域与函数y x =的值域相同的是( ) A. 2yxB. 2xy =C. sin y x =D.2log y x =【答案】D 【解析】 【分析】分别求出下列函数的值域,即可判断. 【详解】函数y x =的值域为R20y x =≥,20x y =>则A ,B 错误;函数sin y x =的值域为[]1,1-,则C 错误; 函数2log y x =的值域为R ,则D 正确; 故选:D【点睛】本题主要考查了求具体函数的值域,属于基础题.4.已知函数42,0,()log ,0,x x f x x x ⎧=⎨>⎩则((1))f f -=( )A. 2-B. 12-C.12D. 2【答案】B 【解析】 【分析】分别计算(1)f -,12f ⎛⎫ ⎪⎝⎭即可得出答案.【详解】121(1)2f --==,241211log log 12222f -⎛⎫===- ⎪⎝⎭所以1((1))2f f -=- 故选:B【点睛】本题主要考查了已知自变量求分段函数的函数值,属于基础题. 5.函数log ||()(1)||a x x f x a x =>图象的大致形状是( )A. B.C. D.【答案】A 【解析】 【分析】判断函数函数()f x 为奇函数,排除BD 选项,取特殊值排除C ,即可得出答案. 【详解】log ||log ||()()||||a a x x x x f x f x x x ---==-=--所以函数()f x 为奇函数,故排除BD.log ||()10||a a a f a a ==>,排除C故选:A【点睛】本题主要考查了函数图像的识别,属于基础题.6.已知0.22log 0.2,2,sin 2a b c ===,则( )A. a b c <<B. a c b <<C. c a b <<D.b c a <<【答案】B【解析】 【分析】分别求出a ,b ,c 的大概范围,比较即可.【详解】因为22log 0.2log 10<=,0sin 21<<,0.20221>= 所以a c b <<. 故选:B【点睛】本题主要考查了指数,对数,三角函数的大小关系,找到他们大概的范围再比较是解决本题的关键,属于简单题.7.已知以原点O 为圆心的单位圆上有一质点P ,它从初始位置01(,22P 开始,按逆时针方向以角速度1/rad s 做圆周运动.则点P 的纵坐标y 关于时间t 的函数关系为 A. sin(),03y t t π=+≥ B. sin(),06y t t π=+≥ C. cos(),03y t t π=+≥D. cos(),06y t t π=+≥【答案】A 【解析】当时间为t 时,点P 所在角的终边对应的角等于3t π+, 所以点P 的纵坐标y 关于时间t 的函数关系为sin(),03y t t π=+≥.8.已知函数()f x 为定义在(0,)+∞的增函数,且满足()()()1f x f y f xy +=+.若关于x 的不等式(1sin )(1)(cos )(1sin )f x f f a x f x --<+-+恒成立,则实数a 的取值范围为( ) A. 1a >- B. 14a >-C. 1a >D. 2a >【答案】D 【解析】 【分析】将题设不等式转化为2(cos )(cos )f x f a x <+,根据函数()f x 的单调性解不等式得出2cos cos x a x <+,通过换元法,构造函数2()g x t t =-,[]1,1t ∈-求出最大值,即可得到实数a 的取值范围.【详解】(1sin )(1)(cos )(1sin )f x f f a x f x --<+-+(1sin )(1sin )(cos )(1)f x f x f a x f ∴-++<++因为()()()2(1sin )(1sin )1sin 1sin 1(cos)1f x f x fx x f x -++=-++=+,(cos )(1)(cos )1f a x f f a x ++=++所以2(cos )(cos )f x f a x <+在(0,)x ∈+∞恒成立故2cos cos x a x <+在(0,)x ∈+∞恒成立,即2cos cos x x a -<在(0,)x ∈+∞恒成立 令[]cos ,1,1x t t =∈-,则22()cos cos g x x x t t =-=-所以函数2()g x t t =-在11,2⎡⎤-⎢⎥⎣⎦上单调递减,在1,12⎛⎤ ⎥⎝⎦上单调递增,(1)2(1)0g g -=>= 所以2a > 故选:D【点睛】利用函数的单调性解抽象不等式以及不等式的恒成立问题,属于中档题.二.多项选择题:本大题共4小题,每小题5分,共20分,在每小题给出的四个选项中,有多个选项符合题目要求,全部选对的得5分,选对但不全的得3分,有选错的得0分.9.设11,,1,32α⎧⎫∈-⎨⎬⎩⎭,则使函数y x α=的定义域是R ,且为奇函数的α值可以是( )A. 1-B.12C. 1D. 3【答案】CD 【解析】 【分析】求出对应α值函数y x α=的定义域,利用奇偶性的定义判断即可.【详解】当α的值为11,2-时,函数y x α=的定义域分别为()(),00,-∞+∞,[)0,+∞当1α=时,函数y x =的定义域为R ,令()f x x =,()()f x x f x -=-=-,则函数y x =为R 上的奇函数当3α=时,函数3y x =的定义域为R ,令3()f x x =,3()()f x x f x -=-=-,则函数3y x=为R 上的奇函数故选:CD【点睛】本题主要考查了判断函数的奇偶性,属于基础题. 10.要得到sin 25y x π⎛⎫=- ⎪⎝⎭的图象,可以将函数sin y x =的图象上所有的点( ) A. 向右平行移动5π个单位长度,再把所得各点的横坐标缩短到原来的12倍B. 向右平行移动10π个单位长度,再把所得各点的横坐标缩短到原来的12倍C. 横坐标缩短到原来的12倍,再把所得各点向右平行移动5π个单位长度D. 横坐标缩短到原来的12倍,再把所得各点向右平行移动10π个单位长度【答案】AD 【解析】 【分析】由正弦函数的伸缩变换以及平移变换一一判断选项即可. 【详解】将函数sin y x =的图象上所有的点向右平行移动5π个单位长度,得到函数n 5si y x π⎛⎫=- ⎪⎝⎭的图象,再把所得各点的横坐标缩短到原来的12倍,得到sin 25y x π⎛⎫=- ⎪⎝⎭的图象,故A 正确;将函数sin y x =的图象上所有的点向右平行移动10π个单位长度,得到函数sin 10y x π⎛⎫=- ⎪⎝⎭的图象,再把所得各点的横坐标缩短到原来的12倍,得到sin 210y x π⎛⎫=- ⎪⎝⎭的图象,故B 错误;将函数sin y x =的图象上所有的点横坐标缩短到原来的12倍,得到sin 2y x =的图象,再把所得各点向右平行移动5π个单位长度,得到25sin 2y x π⎛⎫=-⎪⎝⎭的图象,故C 错误; 将函数sin y x =的图象上所有的点横坐标缩短到原来的12倍,得到sin 2y x =的图象,再把所得各点向右平行移动10π个单位长度,得到sin 25y x π⎛⎫=- ⎪⎝⎭的图象,故D 正确;故选:AD【点睛】本题主要考查了正弦函数的伸缩变换以及平移变换,属于基础题.11.对于函数()sin(cos )f x x =,下列结论正确的是( ) A. ()f x 为偶函数B. ()f x 的一个周期为2πC. ()f x 的值域为[sin1,sin1]-D. ()f x 在[]0,π单调递增【答案】ABC 【解析】 【分析】利用奇偶性的定义以及周期的定义判断A ,B 选项;利用换元法以及正弦函数的单调性判断C 选项;利用复合函数的单调性判断方法判断D 选项. 【详解】函数()f x 的定义域为R ,关于原点对称()()()()sin cos sin cos ()f x x x f x -=-==,则函数()f x 偶函数,故A 正确;()()()sin co 22s sin cos ()f x x x f x ππ+=+==⎡⎤⎣⎦,则函数()f x 的一个周期为2π,故B正确;令[]cos ,1,1t x t =∈-,则()sin f x t =,由于函数sin y t=[]1,1-上单调递增,则()sin 1()sin1sin1()sin1f x f x -≤≤⇒-≤≤,故C 正确;当[]0,x π∈时,函数cos t x =为减函数,由于[]cos 0,1t x =∈,则函数sin y t =在0,1上为增函数,所以函数()f x 在[]0,π单调递减,故D 错误; 故选:ABC【点睛】本题主要考查了判断函数的奇偶性,周期性,求函数值域,复合函数的单调性,属于中档题.12.已知()f x 为R 上的奇函数,且当0x >时,()lg f x x =.记()sin ()cos g x x f x x =+⋅,下列结论正确的是( ) A. ()g x 为奇函数B. 若()g x 的一个零点为0x ,且00x <,则()00lg tan 0x x --=C. ()g x 在区间,2ππ⎛⎫-⎪⎝⎭的零点个数为3个 D. 若()g x 大于1的零点从小到大依次为12,,x x ,则1223x x ππ<+<【答案】ABD 【解析】 【分析】根据奇偶性的定义判断A 选项;将()0g x =等价变形为tan ()x f x =-,结合()f x 的奇偶性判断B 选项,再将零点问题转化为两个函数的交点问题,结合函数()g x 的奇偶性判断C 选项,结合图象,得出12,x x 的范围,由不等式的性质得出12x x +的范围. 【详解】由题意可知()g x 的定义域为R ,关于原点对称因为()()()sin ()cos sin ()cos ()g x x f x x x f x x g x -=-+-⋅-=--⋅=-,所以函数()g x 为奇函数,故A 正确; 假设cos 0x =,即,2x k k Z ππ=+∈时,sin ()co cos s sin 02x k x f x k πππ⎛⎫++⋅==≠ ⎪⎝⎭所以当,2x k k Z ππ=+∈时,()0g x ≠当,2x k k Z ππ≠+∈时,sin ()cos 0tan ()x f x x x f x +⋅=⇔=-当00x <,00x ->,则()000()()lg f x f x x =--=--由于()g x 的一个零点为0x , 则()()00000tan ()lg t lg an 0x x f x x x =-=⇒--=-,故B 正确;当0x >时,令12tan ,lg y x y x ==-,则()g x 大于0的零点为12tan ,lg y x y x ==-的交点,由图可知,函数()g x 在区间()0,π的零点有2个,由于函数()g x 为奇函数,则函数()g x 在区间,02π⎛⎫-⎪⎝⎭的零点有1个,并且(0)sin 0(0)cos00g f =+⋅= 所以函数在区间,2ππ⎛⎫-⎪⎝⎭的零点个数为4个,故C 错误;由图可知,()g x 大于1的零点123,222x x ππππ<<<< 所以1223x x ππ<+< 故选:ABD【点睛】本题主要考查了判断函数的奇偶性以及判断函数的零点个数,属于较难题. 三、填空题:本大题共4题,每小题5分,共20分.13.函数()1xf x a =+(0a >且1a ≠)的图象恒过点__________【答案】()0,2 【解析】分析:根据指数函数xy a =过()0,1可得结果.详解:由指数函数的性质可得xy a =过()0,1,所以1xy a =+过()0,2,故答案为()0,2.点睛:本题主要考查指数函数的简单性质,属于简单题. 14.已知扇形的圆心角为12π,面积为6π,则该扇形的弧长为_______; 【答案】6π 【解析】 【分析】由扇形面积公式求出扇形半径,根据扇形弧长公式即可求解.【详解】设扇形的半径为r 由扇形的面积公式得:216212r ππ=⨯,解得2r该扇形的弧长为2126ππ⨯=故答案为:6π 【点睛】本题主要考查了扇形面积公式以及弧长公式,属于基础题. 15.函数()2sin 23f x x π⎛⎫=- ⎪⎝⎭在区间0,2π⎡⎤⎢⎥⎣⎦上的值域为______;【答案】[2] 【解析】 【分析】由x 的范围,确定23x π-的范围,利用换元法以及正弦函数的单调性,即可得出答案.【详解】0,2x π⎡⎤∈⎢⎥⎣⎦,22,333x πππ⎡⎤∴-∈-⎢⎥⎣⎦令22,333t x πππ⎡⎤=-∈-⎢⎥⎣⎦,函数()2sin g t t =在,32ππ⎡⎤-⎢⎥⎣⎦上单调递增,在2,23ππ⎡⎤⎢⎥⎣⎦上单调递减2si ()(n 33)g ππ--==2si 2()2n 2g ππ==, 222sin (3)3g ππ==所以函数()f x 在区间0,2π⎡⎤⎢⎥⎣⎦上的值域为[2]故答案为:[2]【点睛】本题主要考查了正弦型函数的值域,属于中档题. 16.已知函数1()f x x=,()2sin g x x =,则函数()f x 图象的对称中心为_____,函数()y f x =的图象与函数()y g x =的图象所有交点的横坐标与纵坐标之和为____. 【答案】 (1). (0,0) (2). 0 【解析】 【分析】判断函数()f x ,()g x 为奇函数,即可得出函数()f x ,()g x 图象的对称中心都为原点; 根据对称性即可得出所有交点的横坐标与纵坐标之和. 【详解】1()()f x f x x-=-=-,则函数()f x 为奇函数,即函数()f x 图象的对称中心为(0,0) ()()2sin 2sin ()g x x x g x -=-=-=-,则函数()g x 为奇函数,即函数()g x 的对称中心为(0,0)所以函数()y f x =的图象与函数()y g x =的图象所有交点都关于原点对称 即所有交点的横坐标之和为0,纵坐标之和也为0则函数()y f x =的图象与函数()y g x =的图象所有交点的横坐标与纵坐标之和为0 故答案为:(0,0);0【点睛】本题主要考查了函数奇偶性的应用以及对称性的应用,属于中档题.四、解答题:本大题共6小题,共70分,解答应写出文字说明,证明过程或演算步骤. 17.已知α为锐角,且3cos 5α=. (1)求tan 4πα⎛⎫+ ⎪⎝⎭的值;(2)求cos sin(2)2παπα⎛⎫-+-⎪⎝⎭的值. 【答案】(1)-7(2)4425【解析】 【分析】(1)利用平方关系以及商数关系得出tan α,再利用两角和的正切公式求解即可; (2)利用诱导公式以及二倍角的正弦公式求解即可. 【详解】解:(1)因为α为锐角,且3cos 5α=. 所以24sin 1cos 5αα, 所以sin 4tan cos 3ααα==, 所以41tan tan34tan 7441tan tan 1143παπαπα++⎛⎫+===- ⎪⎝⎭--⨯. (2)因为cos sin 2παα⎛⎫-=⎪⎝⎭, sin(2)sin 2παα-=,所以cos sin(2)sin sin 22παπααα⎛⎫-+-=+ ⎪⎝⎭sin 2sin cos ααα=+4432555=+⨯⨯ 4425= 【点睛】本题主要考查了两角和的正切公式,诱导公式,二倍角的正弦公式,属于中档题. 18.已知集合{}|2216xA x =<<,{|sin 0,(0,2)}B x x x π=>∈. (1)求AB ;(2)集合{|1}C x x a =<<()a ∈R ,若AC C =,求a 的取值范围.【答案】(1){|04}A B x x ⋃=<<(2)4a 【解析】 【分析】(1)利用指数函数以及正弦函数的性质化简集合,A B ,再求并集即可;(2)由题设条件得出C A ⊆,分别讨论集合C =∅和C ≠∅的情况,即可得出答案.【详解】解:(1)依题意{|14}A x x =<<,{|0}B x x π=<<,所以{|04}A B x x ⋃=<<. (2)因为AC C =,所以C A ⊆.①当C =∅时,1a ,满足题意;②当C ≠∅时,1a >,因为C A ⊆,得4a ≤,所以14a <; 综上,4a .【点睛】本题主要考查了集合的并集运算以及根据集合间的包含关系求参数范围,属于中档题.19.已知函数()2sin (sin cos )f x x x x =⋅+. (1)求()f x 的最小正周期; (2)求()f x 的单调区间.【答案】(1)最小正周期为π.(2)单调递增区间为3,()88k k k ππππ⎡⎤-++∈⎢⎥⎣⎦Z ,()f x 的单调递减区间为37,()88k k k ππππ⎡⎤++∈⎢⎥⎣⎦Z .【解析】 【分析】利用倍角公式以及辅助角公式化简函数()f x ,根据周期公式得出第一问;根据正弦函数的单调增区间和减区间求()f x 的单调区间,即可得出第二问. 【详解】解:因为2()2sin 2sin cos f x x x x =+⋅22sin sin 2x x =+1cos2sin2x x =-+ sin2cos21x x =-+214x π⎛⎫=-+ ⎪⎝⎭(1)所以函数()f x 的最小正周期为22T ππ==.(2)由222,242k x k k πππππ-+-+∈Z ,得3222,44k x k k ππππ-++∈Z , 即3,88k xk k ππππ-++∈Z , 所以()f x 的单调递增区间为3,()88k k k ππππ⎡⎤-++∈⎢⎥⎣⎦Z ,同理可得,()f x 的单调递减区间为37,()88k k k ππππ⎡⎤++∈⎢⎥⎣⎦Z .【点睛】本题主要考查了求正弦型函数的最小正周期以及单调区间,属于中档题. 20.已知2()1x af x x bx +=++是定义在[1,1]-上的奇函数. (1)求a 与b 的值;(2)判断()f x 的单调性,并用单调性定义加以证明; (3)若[0,2)απ∈时,试比较(sin )f α与(cos )f α的大小.【答案】(1)0a =. 0b =.(2)()f x 在[1,1]-单调递增.见解析 (3)见解析 【解析】 【分析】(1)根据奇函数的性质得出(0)0f =,(1)(1)f f -=-,求解方程,即可得出a 与b 的值; (2)利用函数单调性的定义证明即可;(3)分别讨论α的取值使得sin cos αα=,sin cos αα<,sin cos αα>,结合函数()f x 的单调性,即可得出(sin )f α与(cos )f α的大小.【详解】解:(1)因为()f x 是定义在[1,1]-上的奇函数,所以(0)0f =,得0a =.又由(1)(1)f f -=-,得到1122b b -=--+,解得0b =. (2)由(1)可知2()1xf x x =+,()f x 在[1,1]-上为增函数.证明如下:任取12,[1,1]x x ∈-且设12x x <, 所以()()1212221211x x f x f x x x -=-++()()22121212221211x x x x x x x x +--=++ ()()()()122112221211x x x x x x x x -+-=++()()()()21122212111x x x x xx --=++由于12x x <且12,[1,1]x x ∈-,所以210x x ->,且2110x x -<,又2110x +>,2210x +>,所以()()()()211222121011x x x x xx --<++,所以()()12f x f x <,从而()f x 在[1,1]-单调递增. (3)当4πα=或54πα=时,sin cos αα=,所以(sin )(cos )f f αα=;当04πα<或524παπ<<时,sin cos αα<, 又因为sin [1,1]α∈-,cos [1,1]α∈-,且()f x 在[1,1]-上为增函数,所以(sin )(cos )f f αα<当544ππα<<时,sin cos αα>,同理可得(sin )(cos )f f αα>; 综上,当4πα=或54πα=时,(sin )(cos )f f αα=;当50,,244ππαπ⎡⎫⎛⎫∈⋃⎪ ⎪⎢⎣⎭⎝⎭时,(sin )(cos )f f αα<;当5,44ππα⎛⎫∈ ⎪⎝⎭时,(sin )(cos )f f αα>.【点睛】本题主要考查由函数的奇偶性求参数,判断函数的单调性以及利用单调性比较函数值大小,属于中档题.21.海水受日月的引力,在一定的时候发生涨落的现象叫潮.一般地,早潮叫潮,晚潮叫汐.在通常情况下,船在涨潮时驶进航道,靠近码头;卸货后,在落潮时返回海洋.下面是某港口在某季节每天的时间与水深关系表: .(1)设港口在x 时刻的水深为y 米,现给出两个函数模型:sin()(0,0,)y A x h A ωϕωπϕπ=++>>-<<和2(0)y ax bx c a =++≠.请你从两个模型中选择更为合适的函数模型来建立这个港口的水深与时间的函数关系式(直接选择模型,无需说明理由);并求出7x =时,港口的水深.(2)一条货船的吃水深度(船底与水面的距离)为4米,安全条例规定至少要有1.5米的安全间隙(船底与洋底的距离),问该船何时能进入港口,何时应离开港口?一天内货船可以在港口呆多长时间?【答案】(1)选择函数模型Asin()y x h ωϕ=++更适合. 水深为3米 (2)货船可以在1时进入港口,在5时出港;或者在13时进港,17时出港.一天内货船可以在港口呆的时间为8小时. 【解析】 【分析】(1)观察表格中水深的变化具有周期性,则选择函数模型Asin()y x h ωϕ=++更适合,由表格数据得出,,,A h ωϕ的值,将7x =代入解析式求解即可; (2)由题意 5.5y 时,船可以进港,解不等式2.5sin4.255.56x π+,得出x 的范围,由x的范围即可确定进港,出港,一天内在港口呆的时间. 【详解】解:(1)选择函数模型Asin()y x h ωϕ=++更适合因为港口在0:00时刻的水深为4.25米,结合数据和图象可知 4.25h =6.75 1.752.52A -==因为12T =,所以22126T πππω===, 所以 2.5sin 4.256y x πϕ⎛⎫=++⎪⎝⎭, 因为0x =时, 4.25y =,代入上式得sin 0ϕ=,因为πϕπ-<<,所以0ϕ=, 所以 2.5sin4.256y x π=+.当7x =时,712.5sin4.25 2.5 4.25362y π⎛⎫=+=⨯-+= ⎪⎝⎭, 所以在7x =时,港口的水深为3米(2)因为货船需要的安全水深是4 1.5 5.5+=米, 所以 5.5y 时,船可以进港, 令2.5sin4.255.56x π+,则1sin62xπ, 因为024x <,解得15x 或1317x ,所以货船可以在1时进入港口,在5时出港;或者在13时进港,17时出港. 因为(51)(173)8-+-=,一天内货船可以在港口呆的时间为8小时. 【点睛】本题主要考查了三角函数在生活中的应用,属于中档题. 22.已知函数3(1)log (1)f x a x +=+,且(2)1f =. (1)求()f x 的解析式;(2)已知()f x 的定义域为[2,)+∞. (ⅰ)求()41xf +的定义域;(ⅱ)若方程()()412xxf f k k x +-⋅+=有唯一实根,求实数k 取值范围.【答案】(1)2()log f x x =(2)(ⅰ)[0,)+∞.(ⅱ)1k = 【解析】 【分析】(1)利用换元法以及(2)1f =,即可求解()f x 的解析式;(2)(ⅰ)解不等式412x +≥,即可得出()41xf +的定义域;(ⅱ)根据()41xf +,()2x f k k ⋅+的定义域得出1k ,结合函数()f x 的解析式将方程化为()2(1)2210x x k k -⋅+⋅-=,利用换元法得出2()(1)1,[1,)g t k t k t t =-+⋅-∈+∞,讨论k的值,结合二次函数的性质即可得出实数k 的取值范围.【详解】解:(1)令1(0)t x t =+>,则3()log f t a t =,所以3()log f x a x =, 因为3(2)log 21f a ==,所以231log 3log 2a ==, 所以3232()log log 3log log f x a x x x ==⨯= (2)(ⅰ)因为()f x 的定义域为[2,)+∞, 所以412x +≥,解得0x , 所以()41xf +的定义域为[0,)+∞.(ⅱ)因为0,22,x x k k ⎧⎨⋅+⎩,所以221xk +在[0,)+∞恒成立, 因为221x y =+在[0,)+∞单调递减,所以221x y =+最大值为1,所以1k .又因为()()412xxf f k k x +-⋅+=,所以()()22log 41log 2xxk k x +-⋅+=, 化简得()2(1)2210xx k k -⋅+⋅-=,令2(1)xt t =,则2(1)10k t k t -⋅+⋅-=在[1,)+∞有唯一实数根, 令2()(1)1,[1,)g t k t k t t =-+⋅-∈+∞,当1k =时,令()0g t =,则1t =,所以21x =,得0x =符合题意,所以1k =; 当1k >时,2440k k ∆=+->,所以只需(1)220g k =-,解得1k ,因为1k >,所以此时无解; 综上,1k =.【点睛】本题主要考查了利用换元法求函数解析式以及根据函数的零点确定参数的范围,属于较难题.。

福建省2020-2021学年高一英语上学期期末考试试题

福建省2020-2021学年高一英语上学期期末考试试题

福建省2020-2021学年高一英语上学期期末考试试题高一英语上学期期末考试试题(考试时间:120分钟满分150分)注意:请将全部答案填写在答题卡上。

第一部分听力(共两节,共30分)第一节(共5小题;每小题1.5分,共7.5分)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答案,并标在试卷的相应位置。

听完每段对话后,你将有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 are the speakers doing now?A. Having a walk.B. Visiting a friend.C. Choosing a gift.2. Where does the conversation probably take place?A. In a post office.B. In a restaurant.C. In a supermarket.3. Why are the speakers going to the library?A. To find a partner.B. To meet Mr. Smith.C. To work on a project.4. How many eggs does the woman need today?A. Ten.B. Twenty.C. Forty.5. What does the woman do?A. A tour guide.B. A waitress.C. A student.第二节(共15小题;每小题1.5分,共22.5分)听下面5段对话或独白,每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳答案,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,每小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料,回答第6、7两个小题。

04 七选五专练(解析版)-2020-2021学年上学期高一英语期末专项训练

04 七选五专练(解析版)-2020-2021学年上学期高一英语期末专项训练

高一上学期期末专项训练4 七选五01Things I Wish I'd Known in High SchoolSometimes high school seems like a necessary evil. Now at the end of my college journey, I know that there are still more things I could have done to make my high school experience more beneficial.1Take a class that scares you during high school. The experience will definitely prepare you for future academic challenges.Don't worry about your grades too much. Getting a B is not the end of the world. 2I have started college with all A's since seventh grade, and it was hard to deal with not being able to repeat that. I stayed away from taking some harder courses I might have done well in because I was afraid of doing poorly. Now, I can fully accept that if I try my best in at least one harder class and get a B+, that is still something to be proud of.Find your own de-stressing(解压) routine. Another essential skill to begin building in high school is how to manage work and stress. Maybe you deal with stress by going for a run, or watching a silly movie. Talk with a friend for a while or play with your dog.3Spend your summers wisely. Don't take on a full-time job the first summer you decide to work.4Do find something to do that will make the ten weeks worthwhile. Look for programs that take you to interesting places, or a job that will give you insights into something you might be interested in.Finally, remember to have fun. Your life does not depend on high school. 5You will have plenty of time to work hard in college and beyond. Be prepared, but don't stress out too much about it. If you can make your high school experience fun, you will have no problem doing the same for your life in college and beyond.A.In fact, it's a kind of good thing.B.Take full advantage of your brain.C.You do have to be active in making friends.D.Do your best, but leave time for fun with friends.E.They will be improved if you make great efforts.F.Find out what relaxes you, and you will feel less stressed.G.Forty hours a week will be a huge shock and you'll have plenty of time for that in the future.1. 2. 3. 4. 5.答案[语篇解读]作者讲述了上高中应该做的几件事。

2020-2021学年高一第一学期期末考试(一)英语试卷含答案

2020-2021学年高一第一学期期末考试(一)英语试卷含答案

汪清四中2020-2021学年度高一年级第一学期期末考试英语试卷(考试时间:120分钟试题总分:120分)第一部分听力(共两节,满分20分)第一节(共5小题;每小题1分,满分5分)1.What are the speakers doing?A.Seeing a doctor.B. Enjoying a concert.C. Buying some tickets.2.What will the the man do next Thursday?A.Start his new job.B. Have an interview.C. Meet Professor Green.3.What is the woman’s first lesson this morning?A.ScienceB. GeographyC. Math.4.What day is it probably today?A.MondayB. Wednesday.C. Friday5.What kind of music does the woman like best?A.Country music. B. Rock music. C. Pop music第二节(共15小题:每小题1分,满分15分)听第六段材料,回答第6、7题。

6.How old is the woman now?A.Eight years old. B.Ten years old. C. Eighteen years old.7.Where does the woman plan to go next year ?A.To Canada. B. To America. C. To France.听第七段材料,回答第8、9题。

8.How does the man feel when he sees the sandwiches?A.DisappointedB. ExcitedC. Annoyed9.What will the man order ?A.Steak B. Seafood C. Fried chicken听第八段材料,回答第10至12题。

漳州市2020_2021学年高一地理上学期期末考试试题

漳州市2020_2021学年高一地理上学期期末考试试题

福建省漳州市2020—2021学年高一地理上学期期末考试试题本试卷分第I卷(选择题)和第II卷(非选择题)两部分.满分100分,考试时间90分钟。

请将所有答案写在答题纸上。

第I卷(选择题50分)一、选择题(25小题,每小题2分,共50分.每小题四个选项中只有一项是最符合要求的)嫦城五号探测器开启我国首次地外天体采样返回之旅.12月1日(农历十月十七日)22时57分,嫦城五号探测器在月面风暴洋安全着陆.图1示意太阳、地球与月球的相对位置。

据此完成1-3题。

1.嫦娥五号探测器在月面安全着陆时,月球最接近于图中的A.①位置B。

②位置C.③位置D.④位置2.嫦娥五号探测器着陆月面时,地球上的人们观测到的月相最有可能是3。

嫦娥五号探测器选择北京时间23时许着陆的原因是A.月球处于近地点地月距离短信号传输衰减小B.中国地面测控站面向月球利于信号实时传输C。

此时风暴洋地区风力较弱信号传输受干扰小D。

此时风暴洋地区云量较少信号传输受干扰小2018年1月15日,我国科研人员发现了一种乌鸦大小、像鸟一样的恐龙化石,它有色彩艳丽的羽毛,生活在1。

61亿年前的中国东北,科研人员将它命名为“彩虹”.据此完成4-5题。

4.彩虹恐龙生活的地质年代及生物发展阶段,对应正确的是A。

中生代侏罗纪裸子植物繁盛B。

中生代白垩纪哺乳动物繁盛C.古生代寒武纪蕨类植物繁盛D。

新生代第四纪爬行动物繁盛5。

彩虹恐龙生活的时期A.原始海洋初步形成B。

大量的金属矿藏出现C.是主要的成煤时期D。

无脊椎动物全部灭绝图2示意大气垂直分层,据此完成6—7题。

6。

I层高度最大的地区是A。

冬季的低纬度地区B。

夏季的低纬度地区C.冬季的高纬度地区D。

夏季的高纬度地区7。

对电离层的叙述,下列说法正确的是A.电离层位于II层B.对无线电短波具有反射作用C.受地磁干扰而形成D.是大气层中闪电的能量来源图3某时刻亚洲部分地区海平面气压分布,据此完成8-9题.8。

风力最小的城市是A.北京B.上海C。

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漳州市2020-2021学年(上)期末高中教学质量检测高一英语试题(共150分,考试时间120分钟)第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)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 is the man looking for?A. His car.B. His key.C. The woman's house.2. How does the man sound?A. Surprised.B. Unhappy.C. Fine.3. Who owns the company Google?A. LarryB. GloriaC. Carl4. What does Jason plan to do?A. Work for Bruce.B. Look for a new job.C. Sell things in a market.5. What are the speakers discussing?A. A study.B. A new theater.C. A trip to a museum.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听下面一段对话,回答第6和第7两个小题。

6. Where does the man work now?A. In a hospital.B. In a school.C. In a bank.7. What relation is Claire to the woman?A. Her friend.B. Her sister.C. Her workmate.听下面一段对话,回答第8和第9两个小题。

8. What is the weather like today?A. Sunny.B. Cloudy.C. Rainy.9. What does the man suggest doing?A. Making some plans for tomorrow.B. Going to the woman's hometown.C. Watching the weather report online.听下面一段对话,回答第10至第12三个小题。

10. Why does the man look forward to his birthday?A. He can hold a party.B. He expects a birthday cake.C. He enjoys spending time with his family.11. Where does the woman usually hold a party?A. In a restaurant.B. At home.C. At school.12. What does the woman enjoy doing on her birthday?A. Going out with her family.B. Playing loud music to celebrate it.C. Talking and eating with her friends.听下面一段对话,回答第13至第16四个小题。

13. Where did Mark work last summer?A. In a clothes shop.B. In a shoe shop.C. In a bookshop.14. What will Mark do with his money he made?A. Pay for college.B. Take a holiday.C. Buy a car.15. How long did the woman work every day?A. 10 hours.B. 8 hours.C. 6 hours.16. What does the woman think of her summer job?A. Tiring.B. Easy.C. Interesting.听下面一段独白,回答第17至第20三个小题。

17. Where does the speaker usually play soccer with Tom?A. In a park.B. In Tom's yard.C. On a playground.18. With whom does the speaker go shopping on Saturdays?A. His friend.B. His brother.C. His mother.19. Why does the speaker like Saturday evening best?A. He can play computer games.B. He can stay up late watching TV.C. He can have fun with his friends.20. What does the speaker often do on Sundays?A. Clean his room.B. Read some books.C. Go to the countryside.第二部分阅读(共两节,满分40分)第一节(共11小题;每小题2.5分,满分27.5分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项。

AHow can you apply to be a firefighter at Clayton County Fire & Emergency Services? First, you need to know more about your work choice and what it takes to get there. Each and every emergency places physical demands on our firefighters, from medical emergencies to apartment fires and everything in between. Do not wait to get in shape; start right now. Our physical agility (敏捷性) test is designed to simulate (模拟) not only actions you would take at a fire but also the tiredness you will go through. Our “Firefighters Combat Challenge” course serves as the physical agility test that applicants are expected to complete and our firefighters complete it every year.Selection Process and Requirements:◇ Applicants must be at least 18 years old◇ Must have got a High School Diploma◇ Have a Driver's License◇ Written Exam◇ Physical Agility Test / Firefighter's Combat Challenge CourseOnce these steps are completed, all application information will be turned over to the Fire Chietand you'll be notified whether you are chosen for an interview.To see if we are hiring, click https: / selsenice. claytoncountyga. gov / MSS. aspz and Lock under “Fire & Emergency Science”. (This page is renewed weekly, so check back regularly.) If we are not hiring at present, click to be placed on a notification list. You will be contacted by mail when the application process opens. To learn about our courses and how to sign up, visit the “Train Here” pages under the Join / Train menu at the top.( ) 21. What do we know about the physical agility test?A. Those who fail it can still apply for the job.B. It better prepares firefighters for the job.C. It works in the form of a written exam.D. Applicants under 18 needn't take it( ) 22. What does the underlined word “notified” mean?A. Trained.B. Accepted.C. Told.D. Examined( ) 23. Why is the text written?A. To stress firefighters’ physical agility.B. To show the requirements for firefighters.C. To explain the process of job application.D. To tell how to join the Services as a firefighter.( ) 24. Where is the text probably from?A. A textbook.B. A magazine.C. A note.D. A website.BWith intelligent systems and new-age networks, life in the big cities will likely be happier and more efficient (有效率的). After all, more than 60 percent of the world's population is expected tore in cities by 2050, according to a UN report. The answer to making these cities more livable for so many people lies in creating “smart” cities. They will use 5G networks and the “internet of things” (IoT) to make everyday life safer and more convenient. Some cities are already using smart technology to improve the lives of their people.But what exactly does a smart city do? Let's look at a few examples. In the United States cities of Boston and Baltimore, sensors (传感器) in smart rubbish cans can tell cleaning workers how full they are and when they need to be emptied. In Amsterdam, the Netherlands, traffic low and energy usage are monitored and adjusted (监控和调整) according to real-time data (数据) collected from sensors around the city. And in Copenhagen, Denmark, a smart bike system allows riders to check on air quality and traffic jam as they ride.Smart cities will be interactive (互动的), allowing their people to feel like they're truly shaping their environment, instead of only living in it It's very important that we can actually communicate with our environment in a way that we never had in the past Smart cities will also allow us to save resources. By using sensors and 5G networks to monitor the use of water, gas and electricity, city managers can figure out how to share and save these resources more efficiently.Of course, it will take time and money to turn our present cities into the smart cities of the future. But as we've already seen, more cities around the world are already using smart technology in small ways. It won't be long before more cities start to develop their own smart systems.( ) 25. What makes it necessary to create smart cities?A. The development of present cities.B. The increasing population in cities.C. The wide use of 5G networks and IoT.D. The growing number of present cities( ) 26. What can we learn from Paragraph 2?A. The sensors can control the traffic low.B. Smart bike systems can improve air quality.C. Smart technology can make city life convenient.D. Smart technology has been used throughout the world.( ) 27. What is Paragraph 3 mainly about?A. How smart cities work.B. How smart cities are created.C. How smart cities save resources.D. How smart cities communicate with us.( ) 28. What is the author's attitude towards smart cities?A. Supportive.B. Worried.C. Uninterested.D. Doubtful.CI have always believed this to be true: people we meet will have a role in our life, whether it's big or small. Some will help us grow, some will hurt us, and some will inspire us to do better. We don't meet people by accident; they' re to cross our path for a reason. We may not realize it at the time of our coming together, but I always do at some point in the future and find I've learnt a lot from them.I come to know the best teachers are those who don't tell you how to pet there but show the way. For them there is no better joy than helping you see a bright future, and seeing you go to levels higher than you would ever have imagined on your own. But that doesn't mean they'll try to fix you or enable you; instead, they guide you to the source of your own power. They believe in you and offer you support.I also learn the importance of never looking down on (瞧不起) someone unless you are helping them up. We like to think of life as a meritocracy (英才管理), so it's easy to look down on someone who isn't as successful or well educated as you are But you've no idea how far he has gone to be here he is now. Be sure we treat everyone with dignity (尊严), for they're worthy to be treated the way you want to be treated.People we meet are like passengers who get on and off the car of our life journey. As they are there to share a certain part of our life, why not take advantage of the chances and learn from them when dealing with them? That way, life is sure to get better.( ) 29. What would the best teachers like to see most?A. You share your joy with them.B. You believe in their support.C. You develop your ability to the full.D. You show them the way to power.( ) 30. What can we infer from the text?A. The writer understands why he meets someone at first.B. Only those who help us make a difference to our life.C. We should treat people with dignity for their success.D. Those who work hard in life should be respected.( ) 31. What can be a suitable title for the text?A. We Don't Meet People by AccidentB. We Need Help And Support in LifeC. What Role Learning Plays in Our LifeD. What the Best Teachers Are Like第二节(共5小题;每小题2.5分,满分12.5分)根据下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

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