浙江省杭州市2020届高三上学期期末教学质量检测

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浙江省杭州市2020届高三年级教学质量检测英语含答案

浙江省杭州市2020届高三年级教学质量检测英语试题注意事项：1.答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

2.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，用橡及擦资干净后，再选涂其他答案标号。

不能答在本试卷上，否则无效。

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，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 planning to do?A.Take a photography course.B.Take a nice photo.C.Make a photo album.2.How did the woman learn to make the cake?A.From a cookbook.B.From the man's wife..C.From a food market.3.How much did the man pay for the jacket?A.40$.B.20$.C.10$.4.Who will go to the movie at last?A.The man.B.The woman.C.The woman's sister.5.Where does the conversation probably take place?A.In a post office.B.In a book store.C.In a library第二节（共15小题：每小题1.5分，共22.5分）听下面可，或独自。

每段对的或班日后有儿个小题，从题中所给的A、B、C三个选项对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各中选出最佳选项。

浙江省杭州市2019-2020学年第一学期高三年级期末教学质量检测(一模)数学试题附答案

杭州市2019-2020学年度高三期末教学质量统一检测卷试题数学一、选择题1. 设集合{}|2A x x =>，()(){}|130B x x x =--<，则A B =I （）A. {}|1x x >B. {}|23x x <<C. {}|13x x <<D. {}|2,1x x x >< 2. 双曲线2214x y -=的离心率等于（） A. 52 B. 5 C. 32 D. 33. 已知非零向量a r ，b r ，则“0a b ⋅>r r ”是“向量a r ，b r 为锐角”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4. 若实数x ，y 满足不等式组010x y x x y +≥⎧⎪≥⎨⎪-≥⎩，则（）A. 1y ≥B. 2x ≥C. 20x y +≥D. 210x y -+≥ 5. 设正实数x ，y 满足()y x y x e e e⋅=，则当x y +取得最小值时，x =（） A. 1 B. 2 C. 3 D. 46. 已知随机变量ξ的取值为()0,1,2i i =.若()105P ξ==，()1E ξ=，则（） A. ()()1P D ξξ=< B. ()()1P D ξξ== C. ()()1P D ξξ=>D. ()()115P D ξξ== 7. 下列不可能．．．是函数()()()22a x x x a Z f x -=+∈的图象的是（） A. B. C. D .8. 若函数()y f x =，()y g x =定义域为R ，且都不恒为零，则（）A. 若()()y f g x =为周期函数，则()y g x =为周期函数B. 若()()y f g x =为偶函数，则()y g x =为偶函数C. 若()y f x =，()y g x =均为单调递增函数，则()()y f x g x =⋅为单调递增函数D. 若()y f x =，()y g x =均为奇函数，则()()y f g x =为奇函数 9. 已知椭圆()222210x y a b a b+=>>的左右焦点分别为1F ，2F ，抛物线()220y px p =>的焦点为2F ，设两曲线的一个交点为P ，若221216PF F F P ⋅=u u u u r u u u u r ，则椭圆的离心率为（） A. 12 B. 22 C. 34 D. 3210. 已知非常数列{}n a 满足()*12n n n a a a n N αβαβ+++=∈+，若0αβ+≠，则（） A. 存在α，β，对任意1a ，2a ，都有{}n a 为等比数列B. 存在α，β，对任意1a ，2a ，都有{}n a 为等差数列C. 存在1a ，2a ，对任意α，β，都有{}n a 为等差数列D. 存在1a ，2a ，对任意α，β，都有{}n a 为等比数列二、填空题11. 设复数z 满足()12i z i +⋅=（i 为虚数单位），则z =______，z =______.12. 已知二项式()60a x a x ⎛⎫+> ⎪⎝⎭的展开式中含2x 的项的系数为15，则a =______，展开式中各项系数和等于______.13. 在ABC ∆中，BAC ∠的平分线与BC 边交于点D ，sin 2sin C B =，则BD CD=______；若1AD AC ==，则BC =______. 14. 已知函数()210cos 0x x f x x x π⎧-≤=⎨>⎩，则()()2019f f =______；若关于x 的方程()0f x a +=在(),0-∞内有唯一实根，则实数a 的取值范围是______.15. 杭州亚运会启动志愿者招募工作，甲、乙等5人报名参加了A 、B 、C 三个项目的志愿者工作，因工。

浙江省杭州市2020-2021学年高三上学期期末(一模)语文试题附答案

2020学年第一学期杭州市高三年级教学质量检测语文试题卷考生须知：1．本试卷分试题卷和答题卷，满分150分，考试时间150分钟。

2．答题前，在答题卷密封区内填写学校、班级和姓名。

3．所有答案必须写在答题卷上，写在试题卷上无效。

一、语言文字运用（共20分）1．下列各句中，没有错别字且加点字的注音全都正确的一项是（3分）A．汉语历史悠久且沿续至今，在世界各种语言中独树一帜，如果简单套用其它语言的相关理论来研究，难免会削．（xiāo）足适履，方枘．（ruì）圆凿。

B．果敢就是跳出自我设定的框．（kuàng）架，将举棋不定和优柔寡断从自己的人生字典中剔．（tī）除，但果敢不等于鲁莽，不贸然行事是其智慧所在。

C．“佛系”流行体现年轻人对锱．（zī）铢必较、非理性争执的反感；“确认过眼神”风靡．（mǐ）一时，反映了人们被海量信息裹挟时渴望斟别真伪的心理。

D．个人隐私被泄露倒．（dào）卖的现象之所以屡禁不止，除不法之徒受暴利驱使之外，公民信息保护意识淡薄、监管追惩．（chéng）乏力也是其中重要原因。

阅读下面的文字，完成2-3题。

【甲】宋朝是一个中华文化绚烂至极的朝代，“华夏民族之文化，历数千载之演进，造极于赵宋之世”（陈寅恪《<宋史职官志考正）序》）。

相比唐朝在科技方面的不温不火，宋朝可以说奇峰突起，鹤立鸡群．．．．。

作为宋代文化巅峰的代表，苏轼在诗词、散文、书法、绘画等方面，都让后人难以望其项背．．．．。

【乙】同时，这个朝代还诞生了另一位可与苏轼比肩．．，在科技领域影响深远的“百科全书式的大师”——沈括。

但沈括的历史评价比苏轼差远了。

【丙】在《宋史》中，沈括的传记并没有单独成篇，而是依附其堂侄沈遴、沈辽的关系才得以位入“列传”的。

该列传中，关于沈括的部分固然．．有近2000字，而对沈括的科技研究，则是寥寥数语——“括博学善文，于天文、方志、律历、音乐、医药、卜算，无所不通，皆有所论著”。

第四关以立体几何为背景的新颖问题为背景的填空题-(原卷版)

压轴填空题第四关以立体几何为背景的新颖问题为背景的填空题【名师综述】以立体几何为背景的新颖问题常见的有折叠问题，与函数图象相结合问题、最值问题，探索性问题等. 对探索、开放、存在型问题的考查，探索性试题使问题具有不确定性、探究性和开放性，对学生的能力要求较高，有利于考查学生的探究能力以及思维的创造性，是新课程下高考命题改革的重要方向之一；开放性问题，一般将平面几何问题类比推广到立体几何的中，不过并非所有平面几何中的性质都可以类比推广到立体几何中，这需要具有较好的基础知识和敏锐的洞察力;对折叠、展开问题的考查，图形的折叠与展开问题（三视图问题可看作是特殊的图形变换）蕴涵了“二维——三维——二维” 的维数升降变化，求解时须对变化前后的图形作“同中求异、异中求同”的思辩，考查空间想象能力和分析辨别能力，是立几解答题的重要题型.类型一几何体在变化过程中体积的最值问题典例1．如图，等腰直角三角形ABE 的斜边AB 为正四面体A BCD -的侧棱，2AB =，直角边AE 绕斜边AB 旋转一周，在旋转的过程中，三棱锥E BCD -体积的取值范围是___________.【来源】山东省菏泽市2021-2022学年高三上学期期末数学试题【举一反三】如果一个棱锥底面为正多边形，且顶点在底面的射影是底面的中心，这样的棱锥称为正棱锥.已知正四棱锥P ABCD -内接于半径为1的球，则当此正四棱锥的体积最大时，其高为_____类型二几何体的外接球或者内切球问题典例2．已知正三棱锥S ABC -的底面边长为32P ，Q ，R 分别是棱SA ，AB ，AC 的中点，若PQR 是等腰直角三角形，则该三棱锥的外接球的表面积为______．【来源】陕西省宝鸡市2022届高三上学期高考模拟检测（一）文科数学试题【举一反三】已知菱形ABCD 中，对角线23BD =，将ABD △沿着BD 折叠，使得二面角A BD C --为120°，AC 33= ，则三棱锥A BCD -的外接球的表面积为________. 【来源】江西宜春市2021届高三上学期数学（理）期末试题类型三立体几何与函数的结合典例3. 已知正方体1111ABCD A B C D -的棱长为1，E 为线段11A D 上的点，过点E 作垂直于1B D 的平面截正方体，其截面图形为M ，下列命题中正确的是______． ①M 在平面ABCD 上投影的面积取值范围是17,28⎡⎤⎢⎥⎣⎦；②M 的面积最大值为334； ③M 的周长为定值．【来源】江西省九江市2022届高三第一次高考模拟统一考试数学（理）试题【举一反三】如图，点C 在以AB 为直径的圆周上运动（C 点与A ，B 不重合)，P 是平面ABC 外一点，且PA ⊥平面ABC ，2PA AB ==，过C 点分别作直线AB ，PB 的垂线，垂足分别为M ，N ，则三棱锥B CMN -体积的最大值为______．【来源】百校联盟2020-2021学年高三教育教学质量监测考试12月全国卷（新高考）数学试题类型四立体几何中的轨迹问题典例4. 已知P 为正方体1111ABCD A B C D -表面上的一动点，且满足2,2PA PB AB ==，则动点P 运动轨迹的周长为__________.【来源】福建省莆田市2022届高三第一次教学质量检测数学试题【举一反三】在棱长为2的正方体1111ABCD A B C D -中，棱1BB ，11B C 的中点分别为E ，F ，点P 在平面11BCC B 内，作PQ ⊥平面1ACD ，垂足为Q ．当点P 在1EFB △内（包含边界）运动时，点Q 的轨迹所组成的图形的面积等于_____________．【来源】浙江省杭州市2020-2021学年高三上学期期末教学质量检测数学试题【精选名校模拟】1．已知在圆柱12O O 内有一个球O ，该球与圆柱的上､下底面及母线均相切.过直线12O O 的平面截圆柱得到四边形ABCD ，其面积为8.若P 为圆柱底面圆弧CD 的中点，则平面PAB 与球O 的交线长为___________. 【来源】江苏省南通市2020-2021高三下学期一模试卷2．已知二面角PAB C 的大小为120°，且90PAB ABC ∠=∠=︒，AB AP =，6AB BC +=.若点P 、A 、B 、C 都在同一个球面上，则该球的表面积的最小值为______.【来源】山东省枣庄市滕州市2020-2021学年高三上学期期中数学试题3．四面体A BCD -中，AB BC ⊥，CD BC ⊥，2BC =，且异面直线AB 和CD 所成的角为60︒，若四面体ABCD 的外接球半径为5，则四面体A BCD -的体积的最大值为_________. 【来源】浙江省宁波市镇海中学2020-2021学年高三上学期11月期中数学试题4．我国古代《九章算术》中将上，下两面为平行矩形的六面体称为刍童，如图的刍童ABCD EFGH -有外接球，且43,4,26,62AB AD EH EF ====，点E 到平面ABCD 距离为4，则该刍童外接球的表面积为__________.【来源】江苏省苏州市张家港市2020-2021学年高三上学期12月阶段性调研测试数学试题5．已知正三棱柱111ABC A B C -的外接球表面积为40π，则正三棱柱111ABC A B C -的所有棱长之和的最大值为______．【来源】河南省中原名校2020-2021学年高三第一学期数学理科质量考评二6．已知体积为72的长方体1111ABCD A B C D -的底面ABCD 为正方形，且13BC BB =，点M 是线段BC 的中点，点N 在矩形11DCC D 内运动（含边界），且满足AND CNM ∠=∠，则点N 的轨迹的长度为______. 【来源】百校联盟2021届普通高中教育教学质量监测考试（全国卷11月）文科数学试卷7．矩形ABCD 中，3,1AB BC ==，现将ACD △沿对角线AC 向上翻折，得到四面体D ABC -，则该四面体外接球的表面积为______；若翻折过程中BD 的长度在710,22⎡⎤⎢⎥⎣⎦范围内变化，则点D 的运动轨迹的长度是______．【来源】江苏省无锡市江阴市青阳中学2020-2021学年高三上学期1月阶段检测数学试题8．如图，在四面体ABCD 中，AB ⊥BC ，CD ⊥BC ，BC =2，AB =CD =23，且异面直线AB 与CD 所成的角为60，则四面体ABCD 的外接球的表面积为_________.【来源】山东省新高考2020-2021学年高三上学期联考数学试题9．已知三棱锥P ABC -外接球的表面积为100π，PB ⊥平面ABC ，8PB =，120BAC ∠=︒，则三棱锥体积的最大值为________.【来源】江苏省徐州市三校联考2020-2021学年高三上学期期末数学试题10．已知直三棱柱111ABC A B C -的底面为直角三角形，且内接于球O ，若此三棱柱111ABC A B C -的高为2，体积是1，则球O 的半径的最小值为___________.【来源】广西普通高中2021届高三高考精准备考原创模拟卷（一）数学（理）试题11．如图，已知长方体1111ABCD A B C D -的底面ABCD 为正方形，P 为棱11A D 的中点，且6PA AB ==，则四棱锥P ABCD -的外接球的体积为______.【来源】2021年届国著名重点中学新高考冲刺数学试题（7）12．如图所示，在三棱锥B ACD -中，3ABC ABD DBC π∠=∠=∠=，3AB =，2BC BD ==，则三棱锥B ACD -的外接球的表面积为______.【来源】江西省南昌市八一中学、洪都中学、十七中三校2021届高三上学期期末联考数学（理）试题13．在三棱锥P ABC -中，平面PAB 垂直平面ABC ，23PA PB AB AC ====120BAC ∠=︒，则三棱锥P ABC -外接球的表面积为_________.【来源】福建省福州市八县（市）一中2021届高三上学期期中联考数学试题14．已知A ，B ，C ，D 205的球体表面上四点，若4AB =，2AC =，23BC =且三棱维A BCD -的体积为23CD 长度的最大值为________．【来源】福建省四地市2022届高三第一次质量检测数学试题15．如图，在四棱锥P ABCD -中，PA ⊥平面ABCD ，底面ABCD 是直角梯形，//AB CD ，AB ⊥AD ，22CD AD AB ===，3PA =，若动点Q 在PAD △内及边上运动，使得CQD BQA ∠=∠，则三棱锥Q ABC -的体积最大值为______.【来源】八省市2021届高三新高考统一适应性考试江苏省无锡市天一中学考前热身模拟数学试题16．已知正三棱锥A BCD -的底面是边长为23其内切球的表面积为π，且和各侧面分别相切于点F 、M 、N 三点，则FMN 的周长为______．【来源】湖南省常德市2021-2022学年高三上学期期末数学试题17．在三棱锥P ABC -中，PA ⊥平面ABC ，AC CB ⊥，4===PA AC BC ．以A 为球心，表面积为36π的球面与侧面PBC 的交线长为______．【来源】山东省威海市2021-2022学年高三上学期期末数学试题18．在棱长为1的正方体1111ABCD A B C D -中，过点A 的平面α分别与棱1BB ，1CC ，1DD 交于点E ，F ，G ，记四边形AEFG 在平面11BCC B 上的正投影的面积为1S ，四边形AEFG 在平面11ABB A 上的正投影的面积为2S .给出下面四个结论：①四边形AEFG 是平行四边形； ②12S S +的最大值为2； ③12S S 的最大值为14；④四边形AEFG 6则其中所有正确结论的序号是___________.【来源】北京西城区2022届高三上学期期末数学试题196，在该圆柱内放置一个棱长为a 的正四面体，并且正四面体在该圆柱内可以任意转动，则a 的最大值为__________．【来源】河南省郑州市2021-2022学年高三上学期高中毕业班第一次质量预测数学（文）试题20．在三棱锥P －ABC 中，P A ＝PB ＝PC ＝2，二面角A －PB －C 为直二面角，∠APB ＝2∠BPC (∠BPC <4π)，M ，N 分别为侧棱P A ，PC 上的动点，设直线MN 与平面P AB 所成的角为α.当tan α的最大值为2532时，则三棱锥P －ABC 的体积为__________.【来源】湖南省长沙市长郡中学2020-2021学年高三上学期入学摸底考试数学试题21．体积为8的四棱锥P ABCD -的底面是边长为22底面ABCD 的中心为1O ，四棱锥P ABCD -的外接球球心O 到底面ABCD 的距离为1，则点P 的轨迹长度为_______________________．22．如图，在ABC 中，2BC AC =，120ACB ∠=︒，CD 是ACB ∠的角平分线，沿CD 将ACD △折起到A CD'△的位置，使得平面A CD '⊥平面BCD ．若63A B '=，则三棱锥A BCD '-外接球的表面积是________．【来源】河南省2021-2022学年高三下学期开学考试数学理科试题23．在三棱锥P ABC -中，4AB BC ==，8PC =，异面直线P A ，BC 所成角为π3，AB PA ⊥，AB BC ⊥，则该三棱锥外接球的表面积为______．【来源】辽宁省营口市2021-2022学年高三上学期期末数学试题24．在棱长为2的正方体1111ABCD A B C D -中，E 是CD 的中点，F 是1CC 上的动点，则三棱锥A DEF -外接球表面积的最小值为_______.【来源】安徽省淮北市2020-2021学年高三上学期第一次模拟考试理科数学试题25．如图，在正方体1111ABCD A B C D -中，点M ，N 分别为棱11,B C CD 上的动点(包含端点)，则下列说法正确的是___________．①当M 为棱11B C 的中点时，则在棱CD 上存在点N 使得MN AC ⊥；②当M ，N 分别为棱11,B C CD 的中点时，则在正方体中存在棱与平面1A MN 平行；③当M ，N 分别为棱11,B C CD 的中点时，则过1A ，M ，N 三点作正方体的截面，所得截面为五边形； ④直线MN 与平面ABCD 2；⑤若正方体的棱长为2，点1D 到平面1A MN 2．【来源】四川省成都市第七中学2021-2022学年高三上学期1月阶段性考试理科数学试题11。

浙江省杭州市2020届高三上学期期末教学质量检测语文试卷(含答案).doc

浙江省杭州市2020届高三上学期期末教学质量检测语文试题卷考生须知：1.本试卷分试题卷和答题卷，满分150分，考试时间150分钟。

2.答题前，在答题卷密封区内填写学校、班级和姓名。

3.所有答案必须写在答题卷上，写在试题卷上无效。

一、语言文字运用（共20分）1．下列各句中，没有错别字且加点字的注音全都正确的一项是（3分）A．面对安全生产问题，不可麻痹．（bì）大意、心存侥幸，更不能敷衍（yǎn）搪塞、推卸责任，而应该积极防犯，将隐患消除在萌芽状态。

B．嫉．（jí）恶如仇、自命不凡而又滑稽可笑的堂吉诃德，模仿真正的骑士锄强扶弱，虽然以失败告终，其形象却至今熠．（yì）熠生辉。

C．有着混（hǔn）凝土式防守的意大利队倒在了世界杯大门口，让球迷潸然泪下，但相信有着深厚底蕴的“亚平宁雄鹰”终究会涅槃．（pán）重生。

D．春风和煦，绿树成荫，树上栖着几只雀儿，湖上掠过一群白鹭，年轻恋人湖边徜．（cháng）徉，白发伴侣相偎小憩．（qī），这份悠闲正是西湖迷人之处。

阅读下面的文字，完后2—3题。

【甲】美学的基本概念、词汇很多来自日常语言，不免．．存在着多义性、隐喻性、含混性。

美学和文艺理论中的许多争论，主要就由此引发．．。

【乙】例如前些年十分热闹的关于形象思维的讨论便相当典型，争论了半天，“形象思维”这个词究竟是什么意思？它包含哪几种不同含义？却并没弄清楚。

【丙】但另一方面，也不必因噎废食．．．．，不必因语词概念的多义含混而取消美学的生存；正如并不因为审美艺术领域内突出的个性差异和主观自由，便根本否认研究它的可能一样。

事实上，尽管一直有各种怀疑和反对，迄今为止，并没有一种理论能够严格证实传统意义上的美学不能成立或不存在。

相反，从古到今，关于美、审美和艺术哲学性的探索、讨论和研究始终不绝如缕．．．．，许多时候还相当兴盛。

2．文段中加点的词，运用不正确的一项是（3分）A．不免 B．引发 C．因噎废食 D．不绝如缕3．文段中画线的甲、乙、丙句，标点有误的一项是（2分）A．甲 B．乙 C．丙4．下列各句中，没有语病的一项是（3分）A．号称“全球首家共享书店”的新华书店合肥三孝口店开启共享模式，初衷是降低阅读成本，提高阅读频次，从而提升大众阅读素养。

浙江杭州市-2020学年高三第一学期教学质量检测杭州一模语文试题及答案高三总复习

浙江杭州市-2020学年高三第一学期教学质量检测杭州一模语文试题及答案高三总复习学年第一学期杭州市高三年级教学质量检测语文试题卷考生须知：1.本试卷分试题卷和答题卷，满分150分，考试时间150分钟。

2.答题前，在答题卷密封区内填写学校班级和姓名。

3.所有答案必须写在答题卷上，写在试题卷上无效。

一、语言文字运用（共20分）1.下列各句中，没有错别字且加点字的注音全都正确的一项是（3分）A.书本扉（fěi）页赫然粘贴着一张山水画，画面上山峦起伏，流水淙（cóng）淙，几间茅屋掩映在缥缈的雾霭中，若隐若现，犹如世外桃源。

B.不管是攀登上巅峰，还是徘徊（huái）于低谷，中国女排始终葆有奋力拼搏、砥砺前行的精神，创造了足以标炳史册的卓（zhu ó）著战绩。

C要克服工作中自由散（sǎn）漫敷衍拖沓（tà）、推诿扯皮这些痼疾，必须制订并实施科学的考核制度，从根本上铲除它们滋生的土壤。

D.毋（wù）庸置疑，功利、僵化的阅读教学容易造成学生阅读的荒芜和贫瘠（jí），当务之急是改变心态，让学生多接受那些经典名著的熏陶。

阅读下面的文字，完成2-3题。

【甲】翻看世界历史，当康熙正在轰轰烈烈地除鳌升、削三藩之时，欧洲已进入科学高速发展的时期，产生了培根、牛顿、笛卡儿等伟大的科学家。

可能有人会说，康熙大帝不是也极关注西方的科学技术吗？是的，他不止学会了天文历算的基础知识，还组织编写了《律历渊源》和《数理精蕴》。

但是，康熙对西方科学文化知识的学习仅停留在个人爱好阶段。

到了乾隆朝，乾隆帝在早期靠着祖宗的阴功，凭着自己的聪明，尚能开疆拓土，但到了中后期就沉湎于“文治武功”的颂扬声中朝政日趋腐败，其间爆发了四次农民起义。

【乙】此时，英国的产业革命正如火如荼，资本主义走向黄金时代；而大清帝国却正处于“盲人骑瞎马，夜半临深渊”的危险境地，两者的发展状况已有了天壤之别。

可以说，“康乾盛世”的统治者们正是在纵向比较中产生了盲目乐观的自恋，夜郎自大。

2020届杭州市采荷中学高三语文期末试卷及参考答案

2020届杭州市采荷中学高三语文期末试卷及参考答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

药砚练建安阳光朗照，河头城浮动飘忽的浓雾渐渐消散。

石钵头赤裸脊背，噔噔踏入石坝码头肉铺摊点，立定，双肩一耸，大块猪肉扇啪嗒一声脆响，平摊在了肉案上。

两个伙计手忙脚乱，将猪肉扇挂上一根铜皮红木大秤。

一个掌挂钩，一个挪秤砣报数：“二百……三十一斤半。

”石钵头斜了他们一眼，操起两把剔骨尖刀，咔咔磨擦，笑骂：“黄疸后生！”墟镇巷道，湿漉漉的，水气淋漓。

此时悠悠然走来一位身穿灰布长衫、手摇折扇的精瘦老人。

他迈着方步在猪肉摊边踱了三二个来回，瞧瞧，点点头，似笑非笑。

石钵头认得此人，是个老童生。

传说是满腹诗书，考到胡子花白，连一个秀才也没捞着。

长衫洗得发白，几块补丁格外刺眼，看着老穷酸装模作样赛百万的架势，石钵头扭头噗地吐出了一口浓痰。

华昌驻足停步，收起折扇，倒转扇柄指点，问：“前蹄，几多钱啊？”石钵头利刀游走剔骨，沙沙响。

“老弟，几多钱？”华昌再问。

石钵头说：“现钱，不赊账。

”华昌说：“你这后生哥啊，好没道理，咋就说俺要赊账呢？”石钵头说：“搞笑嘴！”华昌在衣兜里摸索良久，拍出了一把制钱。

石钵头将制钱收拢、叠好，放在案板前沿，说：“钱你拿走，莫挡俺做生意。

”华昌说：“无怨无仇，做嘛介不卖？”石钵头斫下猪蹄，说：“看好了，可是这副？”华昌点头。

石钵头抓起猪蹄，猛地往后抛入汀江，说：“俺要敬孝龙王爷。

不行么？”华昌拣起制钱，一声不吭地走了。

身后传来阵阵哄笑声。

半个月后，华昌带着几个破蒙童子江岸踏青，歇息于城东风雨亭。

彼时，石钵头正惬意地嚼吃着亭间售卖的糠酥花生。

一扬手，花生壳撒落遍地。

石钵头说：“咦，巧了，今天倒有八副猪蹄，老先生有现钱么？”华昌面无表情，牵着童子匆匆离去。

走不远，就听到石钵头的两个伙计阴阳怪气地高唱一首当地歌谣：“先生教俺一本书，俺教先生打野猪。

野猪逐过河，逐去先生背驼驼……”后来，他们还遇过几次。

浙江省杭州市2023届高三上学期教学质量检测数学试题含解析

2022学年第一学期杭州市高三年级教学质量检测数学试题卷（答案在最后）考生须知：1.本试卷分试题卷和答题卷两部分.满分150分，考试时间120分钟.2.请用黑色字迹的钢笔或签字笔在答题卡指定的区域（黑色边框）内作答，超出答题区域的作答无效!3.考试结束，只需上交答题卡.一､选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，有一项是符合题目要求的.1.已知集合{}{}210,lg 0A x x B x x =-<=≤∣∣，则A B ⋃=（）A.{01}x x <<∣B.{01}xx <≤∣C.{11}xx -<<∣ D.{11}xx -<≤∣【答案】D 【解析】【分析】根据一元二次不等式解法以及对数函数性质可求得集合,A B ,根据集合的并集运算即可求得答案.【详解】由题意可得{}{}210{|11},lg 0{|01}A xx x x B x x x x =-<=-<<=≤=<≤∣∣，故{|11}A B x x ⋃=-<≤，故选：D.2.若复数4i1iz =+（其中i 为虚数单位），则||z =（）A.B.2C. D.4【答案】C 【解析】【分析】由除法运算化简复数，再根据定义求模即可.【详解】因为()()()4i 1i 4i 22i 1i 1i 1i z -===+++-，则||z ==.故选：C ．3.已知1tan 2α=-，则sin 22cos24cos24sin 2αααα+=-（）A.114B.114-C.52D.52-【答案】A 【解析】【分析】由三角恒等变换及齐次式弦化切，即可求值.【详解】2222221sin 22cos22sin cos 2cos 2sin tan 1tan 1474cos24sin 24cos 4sin 8sin cos 22tan 4tan 142αααααααααααααααα++-+-====-----.故选：A ．4.已知二次函数()f x 的图象如图所示，将其向右平移2个单位长度得到函数()g x 的图象，则不等式()2log g x x >的解集是（）A.(),2-∞ B.()2,+∞ C.()0,2 D.()0,1【答案】C 【解析】【分析】作出函数()g x 与2log y x =的图象，数形结合可得出不等式()2log g x x >的解集.【详解】根据图中信息作出函数()g x 、2log y x =的图象如下图所示：因为()01f =，则()21g =，且2log 21=，由图可知，不等式()2log g x x >的解集为()0,2.故选：C.5.已知非零向量,a b的夹角的余弦值为15，且(3)(2)a b a b +⊥- ，则||||a b =（）A.1B.23C.32D.2【答案】A 【解析】【分析】结合向量数量积运算及向量垂直的表示，可得关于||||a b、的齐次方程，即可进一步求得||||a b 的值.【详解】1cos ,5a b <>= 由(3)(2)a b a b +⊥-得2222(3)(2)253230a b a b a ab b a a b b+⋅-=+-=+-=.∴2230a ab b ⎛⎫⎪+-= ⎪⎝⎭，令0at b=> ，∴2230t t +-=，解得1at b ==或32-（舍去）.故选：A.6.冬末春初，人们容易感冒发热，某公司规定：若任意连续7天，每天不超过5人体温高于37.3℃，则称没有发生群体性发热．根据下列连续7天体温高于37.3℃人数的统计量，能判定该公司没有发生群体性发热的为（）①中位数是3，众数为2；②均值小于1，中位数为1；③均值为3，众数为4；④均值为22．A.①③B.③④C.②③D.②④【分析】根据中位数、众数、平均数、标准差等知识确定正确答案.【详解】任意连续7天，每天不超过5人体温高于37.3℃的人数为2，2，2，3，3，4，6，则满足中位数是3，众数为2，但第7天是6人高于5人，故①错误；任意连续7天，每天不超过5人体温高于37.3℃的人数为0，1，2，4，4，4，6，则满足均值是3，众数为4，但第7天是6人高于5人，故③错误；对于②，将7个数据从小到大排列为1234567,,,,,,x x x x x x x ，41x =，123567117x x x x x x ++++++<，所以1235676x x x x x x +++++<，由于123567,,,,,x x x x x x 是自然数，且12356701x x x x x x ≤≤≤≤≤≤≤，所以1234567,,,,,,x x x x x x x 都不超过5，②正确.对于④，将7个数据从小到大排列为1234567,,,,,,x x x x x x x ，123456727x x x x x x x ++++++=，123456714x x x x x x x ++++++=，()()()()()()()22222221234567222222227x x x x x x x -+-+-+-+-+-+-=，()()()()()()()22222221234567222222214x x x x x x x -+-+-+-+-+-+-=，由于123567,,,,,x x x x x x 是自然数，若自然数x 大于5，则()2216x -≥，矛盾，所以123567,,,,,x x x x x x 都不超过5，④正确.综上所述，正确的为②④.故选:D7.已知抛物线2:4C y x =的焦点为F ，直线l 过焦点F 与C 交于A ，B 两点，以AB 为直径的圆与y 轴交于D ，E 两点，且4||||5DE AB =，则直线l 的方程为（）A.10x ±-=B.10x y ±-=C.220x y ±-= D.210x y ±-=【分析】设||2(24),AB r r AB =≥的中点为M ，根据4||||5DE AB =求出r ，进而得到M 点横坐标；再设直线()()1122:(1),,,,l y k x A x y B x y =-，由韦达定理得到k 与M 横坐标的关系，进而求出k ．【详解】设||2(24),AB r r AB =≥的中点为M ，MN y ⊥轴于点N ，过A ，B 作准线=1x -的垂线，垂足分别为11,A B ，如下图：由抛物线的定义知112(||1)||||||2MN AA BB AF BF AB r +=+=+==，故||1MN r =-，所以8||5DE r ==，即21650250r r -+=，解得52r =或58r =（舍去），故M 的横坐标为32，设直线()()1122:(1),,,,l y k x A x y B x y =-，将(1)y k x =-代入24y x =，得()2222240k x k x k -++=，则2122243k x x k++==，解得2k =±，故直线l 的方程为220x y ±-=．故选：C ．【点睛】本题解题的关键是要抓住圆的两要素：圆心和半径，用圆心的横坐标得到斜率的等量关系．8.若过点(,)a b 可以作曲线1(0)y x x x=->的两条切线，则（）A.0b a >> B.1a b a a >>- C.10a b aa<-<< D.10a b a a-<<<【答案】B 【解析】【分析】设切点为()00,x y ，结合导数法有000|x y bk y x a-='=-，则存在两条切线等价于方程有两个不同正解，结合判别式法及韦达定理列不等式组即可化简判断选项.【详解】设切点为()00,x y ，则00x >，∴211(0)y x x '=+>，则0002000111x b y b x k x x ax a---=+==--，化简得：200()20a b x x a --+=①，则44()a a b ∆=--，∵过点(,)a b 可以作曲线的两条切线，∴方程①有两个不同正解，∴()202Δ00a b aa b -⎧->⎪-⎪⎪>⎨⎪⎪>-⎪⎩，∴1a b a a >>-.故选：B ．二､多选题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9.已知函数()f x ．下列命题中正确的是（）A.()f x 的图象是轴对称图形，不是中心对称图形B.()f x 在30,2⎛⎫ ⎪⎝⎭上单调递增，在3,32⎛⎫⎪⎝⎭上单调递减C.()f x，最小值为0D.()f x【答案】ABD 【解析】【分析】利用函数对称的结论，即可验证选项A,由()0f x ≥，知()y f x =和2()y f x =在定义域内的单调性相同，可验证选项B ,通过单调性，即可求得最大值和最小值.【详解】对于选项A ，由(3)()f x f x -=，得()f x 的对称轴为直线32x =，因此()f x 的图象是轴对称图形，不是中心对称图形．对于选项BCD ，因为2()0,()3f x f x ≥=+，函数()y f x =和2()y f x =在定义域内的单调性相同，而2()y f x =在30,2⎛⎫ ⎪⎝⎭上单调递增，在3,32⎛⎫ ⎪⎝⎭上单调递减，所以()f x 在30,2⎛⎫ ⎪⎝⎭上单调递增，在3,32⎛⎫⎪⎝⎭上单调递减；当32x =时，()f x ；当0x =或3时，()f x 取到最小值故选：ABD.10.甲箱中有5个红球，2个白球和3个黑球，乙箱中有4个红球，3个白球和3个黑球．先从甲箱中随机取出一球放入乙箱，分别以12,A A 和3A 表示由甲箱取出的球是红球，白球和黑球的事件；再从乙箱中随机取出一球，以B 表示由乙箱取出的球是红球的事件，则下列结论正确的是（）A.事件B 与事件(1,2,3)i A i =相互独立B.()1522P A B =C.()25P B = D.()245|8P A B =【答案】BD 【解析】【分析】由题设求出()i P A 、(|)i P B A (1,2,3)i =，利用条件概率公式、全概率公式判断B 、C 、D ，根据()(),()i i P A P B P A B 是否相等判断事件的独立性.【详解】由题意11()2P A =，21()5P A =，33()10P A =，先1A 发生，此时乙袋有5个红球，3个白球和3个黑球，则15(|)11P B A =，先2A 发生，此时乙袋有4个红球，4个白球和3个黑球，则24(|)11P B A =，先3A 发生，此时乙袋有4个红球，3个白球和4个黑球，则34(|)11P B A =，所以1115()(|)()22P A B P B A P A ==，B 正确；2224()(|)()55P A B P B A P A ==，3336()(|)()55P A B P B A P A ==，()1122339(|)()(|)()(|)()22P B A P A P B A P A P B B A A P P ++==，C 错误；则11()()()P A P B P A B ≠，22()()()P A P B P A B ≠，33()()()P A P B P A B ≠，A 错误；()2222()(|)()|()(845)P A B P B A P A P A B P B P B ⋂===，D 正确.故选：BD11.若函数()π1sin 2(0)62f x x ωω⎛⎫=+-> ⎪⎝⎭在区间0,24π⎛⎫ ⎪⎝⎭上单调递增，则（）A.存在ω，使得函数()f x 为奇函数B.函数()f x 的最大值为12C.ω的取值范围为(0,4]D.存在4个不同的ω，使得函数()f x 的图象关于直线π2x =对称【答案】BCD 【解析】【分析】对A 选项，计算()f x -，得到其与()f x -的关系即可判断，对B 选项，根据正弦函数的值域即可求出()f x 的最大值，对C 选项，根据()f x 在区间π0,24⎛⎫ ⎪⎝⎭上单调递增，得到不等式组ππ2π62πππ2π1262k k ω⎧-+⎪⎪⎨⎪++⎪⎩ ，解出即可，对D 选项，令πππ2π,Z 262m m ω⨯+=+∈，解出ω，再结合C 选项ω范围则可得到ω的值.【详解】解：()π1sin 262f x x ω⎛⎫=+- ⎪⎝⎭，定义域为R ，()()π1π1sin 2sin 26262f x x x f x ωω⎛⎫⎛⎫-=-+-=-++=- ⎪ ⎪⎝⎭⎝⎭不恒成立，则不存在ω，使得函数()f x 为奇函数，故A 错误;由π1sin 216x ω⎛⎫-+⎪⎝⎭ ，得31()22f x - ，则()f x 的最大值为12，故B 正确;由于()f x 在区间π0,24⎛⎫ ⎪⎝⎭上单调递增，故ππ2π62πππ2π1262k k ω⎧-+⎪⎪⎨⎪++⎪⎩ ,Z k ∈解第一个不等式得13k ≤，Z k ∈ ,故max 0k =，解二式得244k ω≤+，故4ω≤，又0ω>，所以04ω< ，故C 正确;令πππ2π262m ω⨯+=+，m ∈Z ，解得13m ω=+，Z m ∈，由04ω< 知ω的取值为13，43，73，103，共4个值，故D 正确.故选：BCD.【点睛】关键点睛：本题的难点在于C,D 选项的判断，根据()f x 的某个单调增区间，则其整体应该在ππ2π,2π,Z 22k k k ⎡⎤-+∈⎢⎥⎣⎦，即应该是后者的子集，再结合0ω>，从而得到关键的不等式组，解出ω范围，而D 选项我们采取代入法，将π2x =代入则内部整体应等于对称轴通项即ππ2x k =+，Z k ∈再结合ω范围，则得到所有ω取值.12.已知函数3()13xxf x =+，设(1,2,3)i x i =为实数，且1230x x x ++=．下列结论正确的是（）A.函数()f x 的图象关于点10,2⎛⎫ ⎪⎝⎭对称B.不等式1(1)2f x ->的解集为{}1x x >C.若1230x x x ⋅⋅<，则()()()12332f x f x f x ++<D.若1230x x x ⋅⋅<，则()()()12332f x f x f x ++>【答案】ABD 【解析】【分析】对A ，由()()1f x f x -+=可判断；对B ，根据函数单调递增可求解；对CD ，根据()f x 的性质画出函数图象，表示出直线AD 的方程，根据,B C 均在直线AD 上方建立不等关系可得.【详解】对A ，()()3313113133113x x xx x x xf x f x ---+=+=+=++++ ，∴函数()f x 的图象关于点10,2⎛⎫ ⎪⎝⎭对称，故A 正确；对B ，31()11313x x xf x ==-++ 在R 上单调递增，且()102f =，则1(1)2f x ->化为()(1)0f x f ->，则10x ->，解得1x >，故不等式1(1)2f x ->的解集为{}1x x >，故B 正确；对CD ，30x >，则可得101113x<-<+，且()f x 关于点10,2⎛⎫⎪⎝⎭对称，在R 上单调递增，可得()f x 函数图象如下：,B C 均在直线AD 上方，其中直线AD 的方程为()23231122f x x y x x x +-=++，则可得()()2322231122f x x f x x x x +->++，()()2333231122f x x f x x x x +->++，所以()()()()()232323232323231111122222f x x f x x f x f x x x f x x x x x x +-+-+>+++=++++，()()()23111f x x f x f x +=-=- ，()()()231112f x f x f x ∴+>-+，即()()()12332f x f x f x ++>，故C 错误，D 正确.故选：ABD.【点睛】关键点睛：解决本题的关键是判断出函数的对称性和单调性画出函数图象，数形结合求解.三､填空题：本大题共4小题，每小题5分，共20分.13.()241(12)xx ++的展开式中3x 的系数为_______________.【答案】40【解析】【分析】利用4(12)x +的展开式的通项，令x 的指数等于3和1，即得展开式中3x 的系数.【详解】因为4(12)x +的展开式的通项()14422C C rrr r r r T x x +==，令3r =和1r =，可得3x 的系数为331442C 2C 842440+=⨯+⨯=.故答案为：40.14.将函数y =π3sin 24x ⎛⎫+ ⎪⎝⎭的图象向右平移π6个单位长度，则平移后的图象中与y 轴最近的对称轴的方程是____.【答案】524x π=-##5π24-【解析】【分析】先根据图象变换得解析式，再求对称轴方程，最后确定结果.【详解】3sin[2()]3sin(2)6412y x x πππ=-+=-72()()122242k x k k Z x k Z πππππ-=+∈∴=+∈当1k =-时524x π=-故答案为：524x π=-【点睛】本题考查三角函数图象变换、正弦函数对称轴，考查基本分析求解能力，属基础题.15.已知双曲线2221y x a-=，若过点(2,2)能作该双曲线的两条切线，则该双曲线的离心率e 的取值范围为__________．【答案】213⎭【解析】【分析】设出切线方程，联立双曲线方程消元得一元二次方程，则两条切线等价于Δ0=有两个不等实根，即可进一步列判别式不等式求得参数范围，从而求得离心率的取值范围.【详解】设切线方程为(22)y k x -=-代入2221y x a-=得()222224(1)4(1)0a k x k k x k a -+----=，易得k a ≠±，由2203840k k a ∆=⇒-++=，由题意此方程有两个不等的实根，故()22146412403a a ∆=-+>⇒<，则22713c a =+<，所以13c e =<，即13e <<，又k a =±代入223840k k a -++=得1a =±，所以e =，故离心率e 的取值范围为3⎫⎪⎭．故答案为：213⎫⎪⎭．16.已知不等式()()ln ln 10,1xa a a x a a >->≠，对()1,x ∀∈+∞恒成立，则a 的取值范围是__________．【答案】1e e ,∞⎛⎫+ ⎪⎝⎭【解析】【分析】根据已知得出()1ln ln 1x aa x ->-，对()1,x ∀∈+∞恒成立，而在0a >，1x >上10x a ->，()ln 10x ->，可得1a >，将()1ln ln 1x a a x ->-化为()()()()1ln ln 11ln e ln 1e x a x x a x ---⋅>-⋅，令()e x f x x =，根据导数得出其单调性，则()()()()1ln ln 11ln e ln 1e x a x x a x ---⋅>-⋅可化为()()()()1ln ln 1fx a f x ->-，即可根据单调性得出ln(1)ln 1x a x ->-，令()ln(1)1x x g x -=-，根据导数得出()()1e 1eg x g ≤+=，即可得出1ee a >，即可得出答案.【详解】()()ln ln 10,1xa a a x a a >->≠，对()1,x ∀∈+∞恒成立，则()1ln ln 1x aa x ->-，对()1,x ∀∈+∞恒成立，0a > ，1x >，10x a -∴>，()ln 10x ->，则要满足()1ln ln 1x aa x ->-，则ln 0a >，即1a >，()1ln ln 1x a a x ->-化为：()()1ln ln e ln 1x a a x ->-，两边乘1x -得：()()()()()()1ln ln 11ln e1ln 1ln 1ex ax x a x x x ---⋅>--=-⋅，令()e x f x x =，则()e e x xf x x ='+，令()e e 0xxf x x =+>'，解得1x >-，则()e xf x x =在()1,-+∞上单调递增，不等式()()ln ln10,1x a a a x a a >->≠，对()1,x ∀∈+∞恒成立，即1x >时，()()()()1ln ln 11ln e ln 1ex ax x a x ---⋅>-⋅恒成立，则()()()()1ln ln 11ln eln 1ex ax x a x ---⋅>-⋅可化为：()()()()1ln ln 1fx a f x ->-，当1x >，1a >时，(1)ln 0x a ->，()ln 10x ->，则根据单调性可得()()1ln ln 1x a x ->-，则ln(1)ln 1x a x ->-，令()ln(1)1x x g x -=-，则()()()21ln 11x g x x --'=-，令()0g x '>，解得e 1x <+，即()ln(1)1x x g x -=-在()1,e 1+上单调递增，令()0g x '<，解得e 1x >+，即()ln(1)1x x g x -=-在()e 1,∞++上单调递减，则()()1e 1e g x g ≤+=，则1ln ea >，即1e e a >，10ee e 1>= ，综上1ee a >，故答案为1ee ,∞⎛⎫+ ⎪⎝⎭.四､解答题：本题共6小题，共70分.解答应写出文字说明､证明过程或演算步骤.17.已知等比数列{}n a 的前n 项和为n S ，数列{}1n S +是公比为2的等比数列.（1）求数列{}n a 的通项公式；（2）求数列{}n na 的前n 项和n T .【答案】（1）12n n a -=（2）()121nn T n =-⋅+【解析】【分析】(1)由数列{}1n S +是公比为2的等比数列，写出通项公式，用1(2)n n n a s s n -=-≥求得通项公式.(2)考查用错位相减法求和,写出n T ，n qT ，两式错位作差，化简.【小问1详解】()11112n n S a -+=+⋅，①2n ≥时，()211112n n S a --+=+⋅，②①-②()()21222n n a a n -⇒=+⋅≥，∵{}n a 为等比数列∴211111221aa a a a +=⇒=⇒=∴12n n a -=【小问2详解】12n n na n -=⋅，∴()01221122232122n n n T n n --=⋅+⋅+⋅++-⋅+⋅ ，①()()12212122222122n n n n T n n n --=⋅+⋅++-⋅+-⋅+⋅ ，②①-②得，2112222n nn T n --=++++-⋅ ()()112221212112n n n n n n n n ⋅-=-⋅=--⋅=-⋅--∴()121nn T n =-⋅+.18.ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，已知()()2sin 2sin 2sin a c A c a C b B -+-=.（1）求B ；（2）若ABC 为锐角三角形，且2b =，求ABC 周长的取值范围.【答案】（1）3B π=；（2）2,6]+.【解析】【分析】（1）根据给定条件，利用正弦定理角化边，再利用余弦定理计算作答.（2）由（1）的结论，利用正弦定理结合三角恒等变换求解作答.【小问1详解】在ABC 中，()()2sin 2sin 2sin a c A c a C b B -+-=，由正弦定理得：2(2)(2)2a c a c a c b -+-=，整理得222b ac ac =+-，由余弦定理得：2221cos 22a cb B ac +-==，而0B π<<，所以3B π=.【小问2详解】由（1）知，3B π=，由正弦定理得：2sin sin sin sin 3a cb A C B π====，则,a A c C ==，而23A C π+=，令,33A C ππθθ=+=-，在锐角ABC 中，032032ππθππθ⎧<+<⎪⎪⎨⎪<-<⎪⎩，解得66ππθ-<<，cos 12θ<≤，于是得(sin sin 2sin cos 4cos 24333a c πππθθθθ⎤⎛⎫⎛⎫⎤+=++-==∈ ⎪ ⎪⎥⎦⎝⎭⎝⎭⎦，则26a b c +<++≤，所以ABC周长的取值范围是2,6]+.19.已知函数()f x 满足()2()31f x f x x =-+-．（1）求函数()f x 的解析式；（2）若关于x 的方程2()1f x k x x =--恰有四个不同的实根，求实数k 的取值范围．【答案】（1）()1f x x =+（2）10,(1,)5⎛⎫+∞ ⎪⎝⎭【解析】【分析】（1）构造等式()2()31f x f x x -=--，即可解得()f x 的解析式；（2）对k 的符号分类讨论，其中0k >时，由参变分离可得11(1)31x k x =++-+恰有四个不相等的实根，结合对勾函数性质数形结合讨论即可.【小问1详解】由题意得：()2()31f x f x x -=--，∴()2[2()31]31f x f x x x =--+-，解得()1f x x =+；【小问2详解】i.当0k <时，明显无解；ii.当0k =时，|1|0x +=只有一个实根，不符合条件；iii.当0k >时，2111(1)311x x x k x x --==++-++恰有四个不相等的实根.∴11(1)31x x k ++=++与11(1)31x x k++=-+共有四个不相等的实根.∴132132k k ⎧+>⎪⎪⎨⎪->⎪⎩解得15k >或101k <<，∴105k <<或1k >，∴实数k 的取值范围是10,(1,)5⎛⎫+∞ ⎪⎝⎭．20.第24届冬季奥运会将于2022年2月4日在北京开幕，本次冬季奥运会共设7个大项，15个分项，109个小项.为调查学生对冬季奥运会项目的了解情况，某大学进行了一次抽样调查，若被调查的男女生人数均为()10n n *∈N，统计得到以下22⨯列联表，经过计算可得24.040K≈.男生女生合计了解6n不了解5n合计10n 10n（1）求n 的值，并判断有多大的把握认为该校学生对冬季奥运会项目的了解情况与性别有关；（2）①为弄清学生不了解冬季奥运会项目的原因，采用分层抽样的方法从抽取的不理解冬季奥运会项目的学生中随机抽取9人，再从这9人中抽取3人进行面对面交流，“至少抽到一名女生”的概率；②将频率视为概率，用样本估计总体，从该校全体学生中随机抽取10人，记其中对冬季奥运会项目了解的人数为X ，求X 的数学期望.附表：()20P K k ≥0.100.050.0250.0100.0010k 2.7063.8415.0246.63510.828附：()()()()()22n ad bc K a b c d a c b d -=++++.【答案】（1）20n =，有95%的把握；（2）①2021；②()112E X =.【解析】【分析】（1）完善22⨯列联表，根据2K 的计算可得出关于n 的等式，即可解得正整数n 的值，结合临界值表可得出结论；（2）①分析可知这9人中男生的人数为4，女生的人数为5，利用组合计数原理结合古典概型和对立事件的概率公式可求得所求事件的概率；②分析可知11~10,20X B ⎛⎫⎪⎝⎭，利用二项分布的期望公式可求得()E X 的值.【小问1详解】解：22⨯列联表如下表所示：男生女生合计了解6n 5n 11n 不了解4n 5n9n 合计10n10n20n()2220654520 4.040101011999n n n n n n K n n n n⨯⨯-⨯==≈⨯⨯⨯，N n *∈ ，可得20n =，()2 3.8410.05P K ≥= ，因此，有95%的把握认为该校学生对冬季奥运会项目的了解情况与性别有关；【小问2详解】解：①采用分层抽样的方法从抽取的不理解冬季奥运会项目的学生中随机抽取9人，这9人中男生的人数为4，女生的人数为5，再从这9人中抽取3人进行面对面交流，“至少抽到一名女生”的概率为3439C 42011C 8421-=-=；②由题意可知11~10,20X B ⎛⎫ ⎪⎝⎭，故()111110202E X =⨯=.21.已知椭圆2222:1x y C a b +=的离心率为32，上顶点为M ，下顶点为N ，2MN =，设点(,2)(0)T t t ≠在直线2y =上，过点T 的直线,TM TN 分别交椭圆C 于点E 和点F．（1）求椭圆C 的标准方程；（2）求证：直线EF 恒过定点，并求出该定点；（3）若TMN △的面积为TEF 的面积的k 倍，则当t 为何值时，k 取得最大值？【答案】（1）2214x y +=（2）证明见解析，10,2⎛⎫ ⎪⎝⎭（3）t =±【解析】【分析】（1）由短轴长及离心率求得参数，可得标准方程；（2）分别联立直线TM 、直线TN 与椭圆的方程，解出E F 、坐标，即可写出直线EF 方程，判断定点；（3）设EF 交y 轴与P ，由TMN TMNTEF TMN FPN PEMS S S S S S =-+△△△△△△得到关于t 的齐次式函数，结合均值不等式讨论最值即可.【小问1详解】由题意可得2221,MN b b ==⇒=由椭圆的离心率为2可得22223144b e a a =-=⇒=，所以椭圆C 的标准方程为2214x y +=．【小问2详解】由题意知直线TM 的方程为1x y t =+，直线TN 的方程为31x y t=-．由22141x y x y t ⎧+=⎪⎪⎨⎪=+⎪⎩，得22284,44t t E t t ⎛⎫-- ⎪++⎝⎭．同理，2222436,3636t t F t t ⎛⎫- ⎪++⎝⎭．所以()()()()()()2222222243222236814416192436443636424824436364EFt k t t t t t t t t t t t t t t t t t t -===+----+--++-+++++++()()()2222121212161612t t ttt t -+-=-=-+，所以直线EF 的方程为：222241284164t t t y x t t t --⎛⎫-=-+ ⎪++⎝⎭，即21210162t x y t -+-=，所以，直线EF 过定点10,2P ⎛⎫ ⎪⎝⎭．【小问3详解】设EF 交y 轴与P ，则2218||2||||364TEF TMN FPN PEM t t S S S S t t t =-+=-+++△△△△．因为||TMNS t =△，所以424222401441641114424144324TMN TEF S t t k S t t t t++===+≤+++++△．当且仅当22144t t=，即t =±时，等号成立．所以当t =±时，k 取得最大值43．【点睛】方法点睛：（1）直线过顶点问题，一般可写出直线方程，通过方程判断定点，本题直线上的点由其它直线与圆锥曲线相交所得，故可联立直线与圆锥曲线求得交点，即可写出直线方程。

2020届杭州市大关中学高三英语上学期期末试卷及参考答案

2020届杭州市大关中学高三英语上学期期末试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AFour Best Hikes in the WorldThere's nothing like getting out and getting some fresh air on a hike. No matter whether your idea of a hike is a leisure walk or climbing the highest mountain on Earth, we've got you covered. Below are four best hikes inthe world.Torres del Paine W CircuitLocation (位置): Patagonia. ChileDistance: 37 + milesTime: 5~6 daysBest time to go: October to JanuaryThe W Circuit is one of the most recommended hikes you'll find. Not only will you appreciate the diverse landscapes and striking granite pillars (花岗岩柱子), but you'll probably meet some new friends along the way.Grand Canyon Rim - to - Rim HikeLocation: Arizona, the United StatesDistance: 48 milesTime: 1~3 daysBest time to go: May to June, September to OctoberThere's no better way to experience one of the greatest wonders in the world. Located in one of the USA's most beautiful parks, the views are ly appealing. Just make sure you're prepared for the challenge.Trek to PetraLocation: JordanDistance: 47 milesTime: 5~ 6 daysBest time to go: October to AprilTake the road less traveled through the Kingdom of Jordan and experience one of the seven wonders of the world. Hike through canyons, gorges and ridges, and see tombs and temples along the way all while avoidingcrowds of tourists.Yosemite Grand TraverseLocation: California, the United StatesDistance: 60 milesTime: 6~7 daysBest time to go: July to SeptemberKnown for some of the best hiking in the world, Yosemite National Park is famous for its views and huge sequoia (红杉) trees. Praised byNational Geographic, the Yosemite Grand Traverse will take you through waterfalls and green mountaintops.1.Which of the following is the best time for the hike in Patagonia, Chile?A.AprilB.MayC.AugustD.December2.Where should you go for a less crowded hike?A.JordanB.Patagonia, ChileC.Arizona, the United StatesD.California, the United States3.What can you do along the Yosemite Grand Traverse?A.Plant sequoia treesB.Appreciate waterfallsC.Visit local templesD.Climb granite pillarsBA maverick describes a person who thinks independently. A maverick refuses to follow the customs or rules of a group to which he or she belongs. In the US, a maverick is often admired for his or her free spirit, although others who belong to the maverick’s group may not like the maverick’s independent ways.But where did the word “maverick” come from?Early in the 1800s, a man named Samuel Augustus Maverick settled down in Texas, which was a place of wide-open land, rich soil, cattle ranches(牛场) and cowboys. As the years passed, Mr. Maverick increased his property(财产) in Texas. Before long, he owned huge pieces of land that were good for raising cattle. But he had no cattle. He wasn’t a rancher.One day, a man came to Samuel Maverick to pay him an old debt. But the man didn’t have enough money. So he offered Mr. Maverick 400 head of cattle. Mr. Maverick accepted them, but he didn’t really want them. He simply put the cattle on his land to eat and care for themselves.It was not long before the cows reproduced(繁殖). The calves grew and had more calves. Soon, hundreds ofcows and calves moved freely across Samuel Maverick’s land. They also moved across the land of nearby ranch owners.It was a tradition among ranchers in the West to put a mark of ownership on newborn calves. They burned the name of their ranch into the animal’s skin with a hot iron. The iron made a clear mark called a “brand”. Brands allowed ranchers to easily see who owned which cattle.Samuel Maverick refused to brand his calves. “Why should I?” he asked. If all the other cattle owners branded theirs, then those without a brand belonged to him.And this is how the word “maverick” entered the American language. It meant a calf without a brand. As time passed, the word “maverick” took on a wider meaning. It came to mean a person who was too independent to follow even his or her own group.4. Why did the man give Samuel Maverick 400 head of cattle?A. To get some money.B. To return what he owed him.C. To buy some of his land.D. To ask him to raise them.5. How could the ranchers easily know who the cattle belonged to?A. Through the brand on the cattle.B. Through the name of the cattle.C. Through the appearance of the cattle.D. Through the land on which the cattle stayed.6. What can we learn about Samuel Augustus Maverick from the text?A. He was born in Texas.B. He took good care of all his cattle.C. He didn’t really want to accept the cattle.D. He followed the tradition of ranchers in the West.7. What is the text mainly about?A. How to become an independent thinker.B. “Maverick” means a calf without a brand.C. The life story of Samuel Augustus Maverick.D. How the word “maverick” got into American English.CChinese paleontologists (古生物学家) have determined that, about 47 million years ago, subtropical forests once existed on the high-altitude Qinghai-Tibet Plateau.The conclusion, which appears in a paper published on Tuesday, was drawn based on the large number of fossils found in theBaingoinBasinat an altitude of nearly 5,000 meters during the second comprehensive scientific expedition to the plateau.A joint team from theXishuangbannaTropicalBotanical Gardenconducted the research on the fossils. Bycombining the findings and models, the team recreated the climate and altitude that existed 47 million years ago, showing that the central plateau had an altitude of just 1,500 meters and an annual average temperature of 19℃, says Su Tao, a researcher from the tropical botanical garden and first author of the paper.“It was covered by thick forest and was rich in water and grass. It is fair tocall it the ‘ShangriLa’ of ancient times,” Su adds.The researchers have also found over 70 plant fossils, the majority of which are most closely related to plant life in today's subtropical or tropical regions.“This is enough to show that the central part of the now high-altitude, freezing Qinghai-Tibet Plateau had flourishing subtropical plants 47 million years ago,” Su says.The findings provide new evidence for the study of the evolutionary history of biodiversity and the evolution of the plateau's landscape, according to Zhou Zhekun, the paper's corresponding author and a researcher at the tropical botanical garden.Chinalaunched the second comprehensive scientific expedition to the Qinghai-Tibet Plateau in June 2017, 40 years after the first. Lasting up to 10 years, the expedition will conduct a series of studies focusing on the plateau's glaciers, its biodiversity and ecological changes, and will also monitor the changes in climate.8. How did the paper come to the conclusion?A. Through the observation of the Baingoin basin.B. Through the fossils found in scientific expedition.C. Through the drawing of a large number of fossils.D. Through the adventure on the Qinghai-Tibetan Plateau.9. What can be inferred according to Su Tao?A. The average altitude of the plateau was 1,500 meters.B. “Shangrila”means a place with abundant water and grass.C. The flourishing subtropical plants have covered the plateau.D. The fossils found by researchers are tropical or subtropical plants now.10. Where might the passage come from?A. The Times.B. The Wall Street Journal.C. Chinese National Geography.D. The Economist.11. What is the purpose of the passage?A. To instruct.B. To educate.C. To persuade.D. To inform.DSimply being quiet is a growing appeal. Lots of business have appeared to meet a rising demand for quiet time, from silent weekend getaways to silent dining, silent reading parties and even silent dating. Silence can mean different things to different people. We are usually silent only with those closest to us. So there is something almost radical(不同凡响的)about the recent trend towards enjoying silence with strangers.Mariel started a regular silent reading party inDundeejust under a year ago. Readers bring their books and meet in a bar, where they read together in silence for an hour or sometimes two and then put their books away to chat and have a drink. “When the reading party starts, everything goes quiet,” says Mariel, “ It’s a little bit surreal (超现实的), especially in what is usually a noisy bar. However, there is something special about sharing the silence with others. It offers a chance to escape from reality; everyone is so busy with work and with technology being ever present. An event like thisgives people the opportunity to escape these things for a while.”Honi Ryan is an artist based inBerlinwho began hosting silent dinner back in 2006. The rules of the dinner are: no talking, no using your voice, no reading or writing, trying to make as little noise as possible, not connecting with technology, and staying for at least two hours. So far she has taken her silent dinner project toMexico, theUS,AustraliaandChina. “It’s evident that the age-old connections we make over food do not depend on the words around it. Silence creates the space for the people and places involved to fill with whatever is needed;itis quite different from our usual social behaviors.”12. Why have lots of silent businesses appeared?A. To satisfy people’s demand for silence.B. To make people get close to each other.C. To appeal to young people.D. To change people’s old way of life.13. What can we learn about Mariel’s silent reading parties?A. Readers can use their voice while reading.B. Readers can be busy with their work.C. Readers can connect with technology.D. Readers can chat and drink after reading.14. The underlined word “it” in the last paragraph refers to .A. noiseB. spaceC. silenceD. food15. What can be a suitable title for the text?A. Escape from Reality.B. Enjoying Being Quiet.C. Silent Reading Parties.D. Silent Dining Projects.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020届杭州市春蕾中学高三生物上学期期末考试试卷及答案解析

2020届杭州市春蕾中学高三生物上学期期末考试试卷及答案解析一、选择题：本题共15小题，每小题2分，共30分。

每小题只有一个选项符合题目要求。

1. 黑藻是一种叶片薄且叶绿体较大的水生植物，分布广泛、易于取材，可用作生物学实验材料。

下列说法错误的是（）A. 在高倍光学显微镜下，观察不到黑藻叶绿体的双层膜结构B. 观察植物细胞的有丝分裂不宜选用黑藻成熟叶片C. 质壁分离过程中，黑藻细胞绿色加深、吸水能力减小D. 探究黑藻叶片中光合色素的种类时，可用无水乙醇作提取液2. 下列关于生长素及其作用的叙述，正确的是A. 植物的生长是由单侧光引起的B. 生长素在细胞内可由色氨酸合成C. 生长素由苗尖端产生并促进苗尖端的伸长D. 不同浓度的生长素对植物同一器官的作用效果一定不同3. 下列关于遗传信息和遗传密码在核酸中的位置和碱基构成的叙述中，正确的是()A. 遗传信息位于mRNA上，遗传密码位于DNA上，碱基构成相同B. 遗传信息位于DNA上，遗传密码位于mRNA、tRNA或rRNA上，碱基构成相同C. 遗传信息和遗传密码都位于DNA上，碱基构成相同D. 遗传信息位于DNA上，遗传密码位于mRNA上，碱基构成不同4. 有关显微镜的知识正确的是A. 物像若被显微镜放大50倍，这里“被放大50倍”是指放大该标本的面积B. 显微镜目镜为10×、物镜为10×时，在视野直径范围内看到一行相连的8个细胞，若只把物镜换成40×时，则在视野直径范围内可看到一行相连细胞2个C. 如果观察的物像位于视野的左上方，应向右下方移动玻片，才能将要观察的物像移到视野的中央D. 在低倍镜下观察到物像时，可以直接使用高倍物镜观察5. 细胞生命大厦的基本框架是由生物大分子构成。

下列关于细胞中生物大分子的叙述，错误的是（）A. 生物大分子都是以碳链为骨架的B. 生物大分子都是生命活动的产物C. 生物大分子都可以为生命活动提供能量D. 生物大分子都是由单体通过脱水缩合形成的6. 哺乳动物的成熟红细胞做为获取细胞膜的实验研究材料，其依据是（）A.哺乳动物的成熟红细胞容易获取B.哺乳动物的成熟红细胞中没有任何物质C.哺乳动物的成熟红细胞的细胞膜在光镜下容易观察到D.哺乳动物的成熟红细胞中，没有膜结构的细胞器和细胞核7. 科学家用显微技术除去变形虫的细胞核后，发现其细胞代谢减弱，运动停止；当重新植入细胞核后，细胞代谢恢复，运动恢复。

浙江省杭州市2020届高三数学上学期期末教学质量检测试题(含解析)

浙江省杭州市2020届高三上学期期末教课质量检测数学试题一、选择题（本大题共10小题，共分）1.设会合A={1，2}，B={x∈Z||x|＜2}，则A∩B=（）A.B.C.D.2. 椭圆+ =1的离心率等于（）A.B.C.D.3.设∈，则“＞2”是“||＞2”的（）RA.充足不用要条件B.必需不充足条件C.充要条件D.既不充足也不用要条件4.若复数z知足（1-2i）z=2+i，则|z|=（）A.C.D.5.函数y=的图象大概为（）A.B.C.D.6.已知正三角形ABC的边长为2，设=2，=，则（（）A.B.C.D.7.已知函数f （）（x∈）的周期为（＞0），且在（0，）上单一，则（）x R TT TA.是周期函数，且在上单一B.不是周期函数，且在上单一C.是周期函数，且在上单一D.不是周期函数，且在上单一8. 设θ∈[，]，随机变量ξ的散布列如表所示，则Eξ（）ξ123P sin2θcos2θA.有最大值，最小值B.有最大值，最小值C.有最大值，无最小值D.无最大值，有最小值9.设a ＜0，不等式（3x2+）（2+）≥0，在（a，）上恒建立，则-的最大值为（）a xb b baB.C.D.10.设函数则（A.存在B.存在C.存在f（x）=sin）（2x+φ）+cos，使得，使得，使得2x．记f（x）的最大值为M（φ），最小值为m（φ），D.存在，使得二、填空题（本大题共7小题，共分）11.设a=log23，b=log3a8，则2=______，ab=______．12.设a ，，c分别为△的三边长，若=3，=5，=7，则cos=______，△的外接b ABC ab c C ABC圆半径等于______．13.若双曲线M：x2-=1的离心率小于，则m的取值范围是______；若m=2，双曲线M的渐近线方程为______．14.某几何体的三视图如下图（单位：cm），则该几体的体积32是______cm；表面积是______cm．15.若实数x 、y知足不等式组，则2+3y的最小值是______．x16.若函数f （）=+-（≠0）存在零点，则a的取值范围是______．x aa17.设为△的外接圆圆心．若存在正实数k ，使得=+k，则k的取值范围为______．O ABC三、解答题（本大题共5小题，共分）18.已知f（x）=sin2x+cos2x（x∈R）．（Ⅰ）求f（）的值．（Ⅱ）若x∈[0，]，求函数f（x）的取值范围．19.设函数f（x）=-k（x-1）2．（Ⅰ）若k=1，解方程f（x）=0．（Ⅱ）若对于x的方程f（x）=0有四个不一样的解，求k的取值范围．如图，在△ABC中，AB=8，AC=6，AD⊥BC，M，N分别为AB，AC的中点．（Ⅰ）若?=-6，求|BC|．（Ⅱ）若+=5，BAC的大小．求∠设公差不为0的等差数列{a n}的前n项和为S n，若S6=60，且a6为a1和a21的等比中项．（Ⅰ）求a n和S n．（Ⅱ）设数列{b n}知足b n+1-b n=a n，若b1=3，求数列{}的前n项和T n（n∈N*）．已知函数f（x）=x2+ax+ln x，a∈R．（Ⅰ）若函数f（x）存在两个极值，i）求a的取值范围；ii）证明：函数f（x）存在独一零点．（Ⅱ）若存在实数x1，x2，使f′（x1）+f′（x2）=0，且x2＜x1＜2x2，求f（x1）-f（x2）取值范围．答案和分析【答案】B【分析】解：B={-1，0，1}，A={1，2}；∴A∩B={1}．应选：B．可求出会合B，而后进行交集的运算即可．考察描绘法、列举法表示会合的定义，以及交集的运算．【答案】B【分析】解：椭圆+=1，可得a=，b=2，则c=1，因此椭圆的离心率等于=．应选：B．利用椭圆的标准方程，求解椭圆的离心率即可．此题考察椭圆的简单性质的应用，是基本知识的考察．【答案】A【分析】解：由|x|＞2得x＞2或x＜-2，即“x＞2”是“|x|＞2”充足不用要条件．应选：A．依据绝对值不等式的性质联合充足条件和必需条件的定义进行判断即可．此题主要考察充足条件和必需条件的判断，联合不等式的性质是解决此题的重点．【答案】B【分析】解：∵（1-2i）z=2+i，∴z=，则|z|=||=．应选：B．把已知等式变形，再由商的模等于模的商求解．此题考察复数模的求法，是基础的计算题．【分析】分析：函数存心义，需使e x-e-x≠0，其定义域为{x|x≠0}，清除C，D，又由于，因此当x＞0时函数为减函数，应选A应选：A．欲判断图象大概图象，可从函数的定义域{x|x≠0}方面考虑，还可从函数的单一性（在函数当x＞0时函数为减函数）方面进行考虑即可．此题考察了函数的图象以及函数的定义域、值域、单一性等性质．此题的难点在于给出的函数比较复杂，需要对其先变形，再在定义域内对其进行考察其他的性质．【答案】D【分析】解：如图，令D为AB中点，设==．且AD=BD=BE=1，∠EBC=120°．∴作平行四边形BEFC，∴||=||≠1．故A错；不垂直，故B错；，故C错；应选：D．画出图形，利用向量的运算性质求解．此题考察了向量的运算性质，属于中档题．【分析】解：函数f（x）（x∈R）的周期为T（T＞0），可是x2≥0，因此函数的定义域变小，故f（x2）不是周期函数．且：在（0，T）上单一，故：0＜x2＜T，解得：，故：在（0，）上单一．应选：B．直接利用函数的性质单一性和周期性的应用求出结果．此题考察的知识重点：函数的性质周期性和单一性的应用，主要考察学生的运算能力和转变能力，属于基础题型．【答案】B【分析】解：∵θ∈[，]，随机变量ξ的散布列如表所示，ξ123Psin2θcos2θ∴Eξ=+2×+cos2θ+cos2θ，∵θ∈[，]，∴，，∴∈[]，cos2θ∈[，]，由随机变量ξ的散布列的性质得：cos2θ∈[，]，∴Eξ=∈[]．故Eξ有最大值，最小值．应选：B．推导出Eξ=+cos2θ，θ∈[，]，联合随机变量ξ的散布列的性质得：cos2θ∈[，]，由此能求出Eξ的最大值和最小值．此题考察失散型随机变量的数学希望的取值范围的求法，考察失散型随机变量的数学希望的性质、三角函数的性质等基础知识，考察运算求解能力，是中档题．【答案】C【分析】解：∵（3x2+a）（2x+b）≥0在（a，b）上恒建立，3x2+a≥0，2x+b≥0或3x2+a≤0，2x+b≤0，①若2x+b≥0在（a，b）上恒建立，则2a+b≥0，即b≥-2a＞0，此时当x=0时，3x2+a=a≥0不建立，②若2x+b≤0在（a，b）上恒建立，则2b+b≤0，即b≤0，若3x2+a≤0在（a，b）上恒建立，则3a2+a≤0，即-≤a≤0，故b-a的最大值为，应选：C．若（3x2+a）（2x+b）≥0在（a，b）上恒建立，则3x2+a≥0，2x+b≥0或3x2+a≤0，2x+b≤0，联合一次函数和二次函数的图象和性质，可得a，b的范围，从而获得答案．此题考察的知识点是恒建立问题，二次函数的图象和性质，分类议论思想，难度中档．【答案】D【分析】解：由f（x）=sin（2x+φ）+cos2x=sin（2x+φ）+ cos2x=sin2xcosφ+cos2xsinφ+ cos2x=cosφsin2x+（sinφ+）cos2x+ =sin（2x+θ），则M（φ）=，m（φ）=-，对于选项A，M（φ）+m（φ）=+（-）=1，即不存在φ∈R，使得M（φ）+m（φ）=π，故A错误，对于选项B，M（φ）-m（φ）=-（-）=2∈[1，3]，即不存在φ∈R，使得M（φ）-m（φ）=π，故B错误，对于选项C，M（φ）?m（φ）=（）?（-）=-1-sinφ∈[-2，0]，即不存在φ∈R，使得|M（φ）?m（φ）|=π，故C错误，对于选项D，||=||=||∈[2，+∞），即存在φ∈R，使得||=π，故D正确，应选：D．由三角函数的协助角公式及三角函数求最值逐个查验即可得解．此题考察了三角函数的协助角公式及三角函数求最值，属中档题．11.【答案】33【分析】解：∵a=log23；∴2a=3；又b=log38；∴．故答案为：3，3．由a=log23即可得出2a=3，利用换底公式可得出，从而可求出ab=3．考察对数式和指数式的互化，对数的定义，对数的换底公式．【答案】-【分析】解：∵a=3，b=5，c=7，∴cosC===-．∴sinC==，∴设△ABC的外接圆半径为R，则由2R==，解得：R=．故答案为：-，．由已知利用余弦定理可求cosC的值，依据同角三角函数基本关系式可求sinC的值，利用正弦定理即可求解．此题主要考察了余弦定理，同角三角函数基本关系式，正弦定理在解三角形中的综合应用，考察了计算能力和转变思想，属于基础题．13.【答案】（0，1）y=±x【分析】解：双曲线M：x2-=1的离心率小于，可得：，解得m∈（0，1）．则m的取值范围是：（0，1）．m=2，双曲线M化为：x2-=1，双曲线的渐近线方程：y=x．故答案为：（0，1）；y=x．利用双曲线的离心率的范围列出不等式，求解可得m的范围，经过m的值，求解双曲线的渐近线方程．此题考察双曲线的简单性质的应用，考察转变思想以及计算能力．14.【答案】288-24π264+12π【分析】解：依据三视图知该几何体是一长方体，挖去两个对极点的圆锥，且圆锥的底面圆内切与长方体，画出图形，如下图；则该几何体的体积为V=8×6×6-2××π×32×4=288-24π；表面积为S=4×6×8+6×6-2×π×32+2×π×3×=264+12π．故答案为：288-24π，264+12π．依据三视图还原几何体的形状，联合图中数据求出几何体的体积和表面积．此题考察了利用三视图求几何体的体积和表面积的应用问题，也考察了空间想象能力和计算能力，是基础题．【答案】4【分析】解：依题意作出可行性地区如图，目标函数z=2x+3y在界限点（2，0）处取到最小值z=2×2+3×0=4．故答案为：4此题考察的知识点是简单线性规划的应用，我们要先画出知足拘束条件的平面区域，而后剖析平面地区里各个角点，而后将其代入2x+3y中，求出2x+3y的最小值．在解决线性规划的小题时，常用“角点法”，其步骤为：①由拘束条件画出可行域?②求出可行域各个角点的坐标?③将坐标逐个代入目标函数?④考证，求出最优解．16.【答案】[2，4]【分析】解：要使函数存心义，则，即，即-a≤x≤a，则（a＞0），由f（x）=+-a=0得+=a，平方得a-x+a+x+2=a2，即2=a2-2a，即=，设y=，则y=的图象是以原点为圆心半径为a的上半圆，要使=有解，则知足0≤≤a，即，即，得，得2≤a≤4或a=0（舍），即实数a的取值范围是[2，4]，故答案为：[2，4]先求出函数的定义域，依据函数与方程之间的关系，进行整理，获得=有解，借助y=的几何意义，利用数形联合进行求解即可．此题主要考察函数与方程的应用，利用转变法，转变为两个函数交点问题，以及利用数形结合是解决此题的重点．【答案】k＞【分析】解：由三角形外心的定义，联合向量的投影的几何意义可得：2，即（+k）=2，化简得：k=-2＜0，又k＞0，可得＜0，同理：=2，即（+k）?=2，化简得：=2，又＜0，即2＜0，即1-2k＜0，即k，故答案为：k由三角形外心的定义即外心为各边中垂线的交点，联合向量的投影的几何意义可得：=2，即＜0，同理：=2，即=2，又＜0，即2＜0，即1-2k＜0，即k，故得解此题考察了三角形外心的定义即外心为各边中垂线的交点、向量的投影，属中档题．18.【答案】解：（Ⅰ）f（）=sin+ cos =- +=0，（Ⅱ）f（x）=sin2x+cos2x=2sin（2x+），当x∈[0，]时，2x+∈[，]，sin（2x+）∈[，1]，∴函数f（x）的取值范围为[1，2]【分析】（Ⅰ）直接代值计算即可，（Ⅱ）先化简，再依据三角函数的性质即可求出．此题考察了三角函数值的求法和三角函数的性质，属于基础题19.【答案】解：（Ⅰ）当k=1时，-k（x-1）2=0，∴|x-1|?=0，∴|x-1|?=0，∴|x-1|?=0，|x-1|=0或1-|x-1|（x-2）=0，∴x=1或x=．（Ⅱ）∵|x-1|?（）即|x-1|=0或，当x-1=0时，x=1，此时k∈R，-k|x-1|=0有三个不等于1的解，依据函数y=|x-1|?（x-2）的图象，得-，解得k＜-4，∴k的取值范围是（-∞，-4）．【分析】（Ⅰ）当k=1时，-k（x-1）2=0，推导出|x-1|=0程f（x）=0的解．（Ⅱ）|x-1|?（），得|x-1|=0或或1-|x-1|（x-2，从而）=0，由此能求出方-k|x-1|=0有三个不等于1的解，由此能求出k的取值范围．此题考察方程的解法，考察实数的取值范围的求法，考察函数性质等基础知识，考察运算求解能力，考察化归与转变思想、数形联合思想，是中档题．【答案】解：（Ⅰ）由AD⊥BC可知，|DM|=|AM|，|DN|=|AN|，因此∠MDN=∠MAN，由于=12cos∠MAN=-6，因此cos∠MAN=-，因此|BC|=|AB|+|AC|-2|AB||AC|cos∠MAN=148，因此|BC|=2，故答案为：2（Ⅱ）由于+=（|DB|+|DC|）=5，因此|BC|=10，因此∠BAC=90°，故答案为：90°．【分析】（Ⅰ）由平面向量的数目运算及余弦定理得：cos∠MAN=-，|BC|2=|AB|2+|AC|2-2|AB||AC|cos∠MAN=148，（Ⅱ）由平面向量的数目运算得：+=（|DB|+|DC|）=5，即|BC|=10，因此∠BAC=90°，得解本考了平面向量的数目运算及余弦定理，属．【答案】解：（Ⅰ）等差数列的公差d，，解得a 1=5，=2，da n=2n+3，∴S n==n（n+4），（Ⅱ）∵b n+1-b n=a n，b n-b n-1=a n-1，n≥2，n∈N*，当n≥2，b n=（b n-b n-1）+（b n-1-b n-2）+）+（b2-b1）+b1=a n-1+a n-2+⋯+a1+b1，=（n-1）（n-1+4）+3=n（n+2），于b1=3也合适，b n=n（n+2），∴==（-），∴T n=（1-+-+⋯+-+-）=（--）=【分析】（Ⅰ）由意可得等差数列的公差d，，算即可求出a1，d 的，即可求出a n和S n．（Ⅱ）先依据迭代法求出数列的通项公式，再依据裂项乞降即可求出．此题考察了数列的通项公式和递推公式以及裂项乞降，考察了运算能力，属于中档题．22.【答案】解：（Ⅰ）（i）依据题意，f′（x）=，（x＞0）方程2x2+ax+1=0有2个正根m，n，（不如设m＜n），故，解得：a＜-2；（ii）证明：易知f （x）在x=m时取极大值，在x=n时取极小值，2由（i）知2m+am+1=0，2故f（m）=-m+ln m-1，令g（x）=-x2+ln x-1，故g′（x）=-2x，由-2x=0，解得：x=，故g（x）≤g（）=ln-＜0，=故f（m）＜0，f（x）至多只有1个零点，又f（-a）=ln（-a）＞0，故f（x）存在独一零点；（Ⅱ）由题意知：2x1+a++2x2+a+=0，即a=-（x1+x2）-，故f（x1）-f（x2）-+a（x1-x2）+ln=-（-）+ln，设t=∈（1，2），记h（t）=-++ln t，则h′（t）=-≤0，故h（t）递加，故h（t）∈（h（2），h（1）），即h（t）∈（-+ln2，0），即f （1）-f（2）取值范围是（-+ln2，0）．x x【分析】（Ⅰ）（i）求出函数的导数，联合二次函数的性质获得对于a的不等式组，解出即可；（ii）令g（x）=-x2+lnx-1，求出g（x）≤g（）=ln-＜0，获得f（x）至多只有1个零点，从而证明结论；（Ⅱ）求出a=-（x1+x2）-，以及f（x1）-f（x2）=-（-）+ln，设t=∈（1，2），记h（t）=-++lnt，依据函数的单一性求出其范围即可．此题考察了函数的单一性，最值问题，考察导数的应用以及换元思想，转变思想，是一道综合题．。

浙江省杭州市2023届高三上学期11月份教学质量检测化学试题(含解析)

浙江省杭州市2023届高三上学期11月份教学质量检测化学试题学校:___________姓名：___________班级：___________考号：___________一、单选题1．下列物质的主要化学成分不正确的是A ．软锰矿 2MnO B ．不锈钢 La Ni -合金C ．漂粉精 ()2Ca ClO D ．重晶石 4BaSO 2．北京冬奥会“冰丝带”场馆安装发电玻璃实现零碳供电，原理是在玻璃表面涂抹一层碲化镉(CdTe)，使其具有光电转换功能。

下列说法正确的是A ．CdTe 属于盐B ．12852Te 原子核内中子数为128C ．碲元素在元素周期表中位于s 区D ．发电玻璃发电过程是化学能转化为电能3．下列化学用语表示正确的是A ．的名称：4-甲基戊烷B ．乙烷分子中碳碳σ键电子云的轮廓图：C ．2CO 的空间填充模型：D ．NaClO 的电子式：Na :O :Cl :4．下列物质的相互关系描述正确的是A ．12H 、22H 互为同位素B ．和互为同系物C ．金刚石、碳纳米管互为同素异形体D ．新戊烷和2，2-二甲基丙烷互为同分异构体5．下列说法不正确的是A ．如果不慎将酸液滴到皮肤，应该立刻用大量清水冲洗，然后涂上1%的硼酸B ．强氧化剂高锰酸钾、氯酸钾、过氧化钠等固体不能随意丢弃，可配成溶液或者通过化学方法将其转化成一般化学品后，再进行常规处理C ．轻微烫伤或者烧伤，可先用洁净的冷水处理，然后涂上烫伤膏药SOA．图1：亚硫酸钠粉末与浓硫酸反应制备收集2C HB．图2：电石与氯化钠、硫酸铜混和液反应制备收集22NHC．图3：氢氧化钠固体与氯化铵固体反应制备收集3ClD．图4：二氧化锰与浓盐酸反应制备收集212．下列事实可用键能数据解释的是A．HCl热稳定性强于HBr B．AsH3的沸点高于PH3C．乙醇在水中的溶解度大于二甲醚D．NH₃分子的极性大于BF313．分子TCCA(结构式如图)是一种高效消毒漂白剂。

浙江省杭州市2020届高三上学期期末教学质量检测(一模)数学试题 Word版含答案

杭州市2019-2020学年高三上学期期末教学质量检测数学试题一、选择题：每小题4分，共40分1. 设集合{}2A x x =>，()(){}130B x x x =--<，则A B =I （）A ．{}1x x >B ．{}23x x <<C ．{}13x x <<D ．{}21x x x ><或2. 双曲线2214x y -=的离心率等于（）A ．52B ．5C ．32D ．33. 已知非零向量a ，b ，则“0⋅>a b ”是“向量a ，b 夹角为锐角”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4. 若实数x ，y 满足不等式组010x y x x y +≥⎧⎪≥⎨⎪-≥⎩，则（）A ．1y ≥B ．2x ≥C ．20x y +≥D ．210x y -+≥5. 设正实数x ，y 满足()e e e yx y x ⋅=，则当x y +取得最小值时，x =（） A ．1 B ．2 C ．3 D ．46. 已知随机变量ξ的取值为i （0,1,2i =）．若()105P ξ==，()1E ξ=，则（）A ．()()1P D ξξ=<B ．()()1P D ξξ==C ．()()1PD ξξ=>D ．()()115P D ξξ==7. 下列不可能．．．是函数()()()22a x x f x x a -=+∈Z 的图象的是（） D.C.B.A.xxxyyyyxOOOO8. 若函数()y f x =，()y g x =定义域为R ，且都不恒为零，则（） A ．若()()y f g x =为周期函数，则()y g x =为周期函数 B ．若()()y f g x =为偶函数，则()y g x =为偶函数C ．若()y f x =，()y g x =均为单调递增函数，则()()y f x g x =⋅为单调递增函数D ．若()y f x =，()y g x =均为奇函数，则()()y f g x =为奇函数9. 已知椭圆22221x y a b+=（0a b >>）的左右焦点分别为1F ，2F ，抛物线22y px =（0p >）的焦点为2F ．设两曲线的一个交点为P ，若221216PF F F p ⋅=u u u u r u u u u r ，则椭圆的离心率为（）A ．12B ．22C ．34D ．3210. 已知非常数数列{}n a 满足12n nn a a a αβαβ+++=+（*n ∈N ，α，β为非零常数）．若0αβ+≠，则（） A ．存在α，β，对任意1a ，2a ，都有数列{}n a 为等比数列 B ．存在α，β，对任意1a ，2a ，都有数列{}n a 为等差数列 C ．存在1a ，2a ，对任意α，β，都有数列{}n a 为等差数列D ．存在1a ，2a ，对任意α，β，都有数列{}n a 为等比数列二、填空题：单空题每题4分，多空题每题6分，共36分11. 设复数z 满足()1i 2i z +⋅=（i 为虚数单位），则z = ，z = ．12. 已知二项式()60a x a x ⎛⎫+> ⎪⎝⎭的展开式中含2x 的项的系数为15，则a = ，展开式中各项系数和等于．13. 在ABC △中，BAC ∠的平分线与BC 边交于点D ，sin 2sin C B =，则BDCD= ；若1AD AC ==，则BC = ．14. 已知函数()()()210cos 0x x f x x x π⎧-≤⎪=⎨>⎪⎩，则()2019f f =⎡⎤⎣⎦ ；若关于x 的方程()0f x a +=在(),0-∞内有唯一实根，则实数a 的取值范围是．15. 杭州亚运会启动志愿者招募工作，甲、乙等5人报名参加了A ，B ，C 三个项目的志愿者工作，因工作需要，每个项目仅需1名志愿者．若甲不能参加A ，B 项目，乙不能参加B ，C 项目，那么共有种不同的选拔志愿者的方案．（用数字作答）16. 已知函数()39f x x x =-，()()23g x x a a =+∈R ．若方程()()f x g x =有三个不同的实数解1x ，2x ，3x ，且它们可以构成等差数列，则a = ．17. 在平面凸四边形ABCD 中，2AB =，点M ，N 分别是边AD ，BC 的中点，且32MN =，若()32MN AD BC ⋅-=u u u u r u u u r u u u r ，则AB CD ⋅=u u u r u u u r ．三、解答题：5小题，共74分18. （本题满分14分）已知函数()22sin cos 3f x x x π⎛⎫=-+ ⎪⎝⎭（x ∈R ）．（1）求()f x 的最小正周期；（2）求()f x 在区间,34ππ⎡⎤-⎢⎥⎣⎦上的值域．19. （本题满分15分）已知函数()212f x x k x =+--．（1）当1k =时，求函数()f x 的单调递增区间．（2）若2k ≤-，试判断方程()1f x =-的根的个数．20. （本题满分15分）如图，在ABC △中，23BAC π=∠，3AD DB =u u u r u u u r ，P 为CD 上一点，且满足12AP mAC AB =+u u u r u u u r u u u r，若ABC △的面积为23．（1）求m 的值；（2）求AP u u u r的最小值．PBD AC21. （本题满分15分）设公差不为0的等差数列{}n a 的前n 项和为n S ，等比数列{}n b 的前n 项和为n T ，若2a 是1a 与4a 的等比中项，612a =，11221a b a b ==．（1）求n a ，n S 与n T ；（2）若n n n c S T =⋅，求证：12(2)2n n n c c c ++++<L ．22. （本题满分15分）设函数()e x f x ax =+，a ∈R ．（1）若()f x 有两个零点，求a 的取值范围；（2）若对任意[)0,x ∈+∞均有()2223f x x a +≥+，求a 的取值范围．。

2020届杭州市第九中学高三英语期末试卷及答案

2020届杭州市第九中学高三英语期末试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AThis is a list of places to spend the 2020 Olympics if you aren't going to Japan.AustraliaThe fact that the summer Olympics fall during Australia's winter season is worth keeping in mind for two reasons: it makes traveling there cheaper, and it's still not too cold. As far as the Olympics are concerned, there will be no shortage of opportunities to watch the Games in between your various explorations of the country and its limitless activities.ThailandThailand is a breathtaking place 10 spend some time during any summer, and provides a perfect setting for Olympie viewing: You can go scuba diving one morning before watching sports all afternoon, or simply duck out for world-famous street food in between events. But Thai sports fans may also be taking extra interest in the Olympics these days.United StatesThe U.S. is likely to be as interested in the Olympics as any other nation, given is collection of athletes who will be taking center stage. You can rest assured it won't be difficult to watch the Games no matter where you're visiting specifically. This leaves you with all sorts of fun options. You can visit a luxury ski resort town like Aspen, which turns into a gorgeous hiking destination in the summer.Great BritainNot unlike the U, s. Britain will make for a fun place to spend the 2020 Olympics because there's a great deal of national interest in a number of different sports, and the local athletes are expected to be competitive. Summer can also be a good time to be in Britain in general," with ly mild temperatures allowing for full exploration of the country. That means you can stay in the beautiful Lake District if you like, hiking or kayaking when you're not watching the Games.1.What is special about Australia as a destination in an Olympic summer?A.Its season.B.Its activities.C.Its sports.D.Its scenery,2.Which country can you go to if you are a street food lover?A.Australia.B.Thailand.C.United States.D.Great Britain.3.What do the U. S. and Britain have in common as fun places to spend the 2020 Olympics?A.They both have a pleasant temperature.B.They are both interested in the Games.CThey both own lots of sports centers. D.They are both English-speaking countries.BCoke was introduced by the Coca Cola company in 1886, making it a rather true andtested favorite of generations of people in over 200 countries. This list should give you some ideas on how to get more from your coke than usual.. Coca Cola is an excellent rust buster (除锈剂). If you have a bunch of small rusty objects, put them in coke overnight and give them a goodscrubin the morning. Coke helps to break down the rust, making cleaning much easier. Be sure to throw out the used coke when you are done with it or you might be taking a trip to the doctor.. Like the previous item, the citric acid (柠檬酸) in coke makes for an excellent window cleaner. This is especially useful for car windows. Pour a can of coke over the window and rub the window, then wipe it off with a wet cloth to remove any sugary matter from the sugar in the drink. As coke is fullof sugar, you should clean the sticky matter off the window glasses, or it will be not a cleaner but a dirt.. For those of you who live in areas where skunk (臭鼬) smells can be an issue from time to time, one can of coke added to water with detergent (清洁剂) really helps to break the smell down. If you have been sprayed, stand in the shower and cover yourself from head to toe with coke — wait for a few minutes, then wash yourself with a shower. Coke is an excellent hair treatment so you get two tips for the price of one with this item!. Pots can sometimes get black on the bottom. The black is almost impossible to remove; this is caused by over-cooking. To remove the black and renew your pot, pour in a can of coke (or as much as you need to cover the blackened area by an inch) and put it on the stove on a low heat. After an hour or so, wash the pot as normal.4. What does the underlined word “scrub”in Paragraph 2 probably mean?A. Start.B. Cleaning.C. Shake.D. Example.5. What is important while using coke to clean car windows?A. Use a dry cloth.B. Rub the window lightly.C. Don’t pour too much coke.D. Clean the sugary matter thoroughly.6. For which purpose does coke have to be mixed with other material?A. To get rid of the black on the pot.B. To breakdown the rust,C. To remove smells.D. To clean windows.7. What type of writing is this text?A. An advertisement.B. A review.C. A news report.D. A practical guide.CAdvertisers tend to think big and perhaps this is why they're always coming in for criticism. Their critics(批评家)seem to hate them because they have so much money to throw around. Why don’t they stop advertising and reduce the price of their goods? After all, it’s the consumer who pays.The poor old consumer! He'd have to pay a great deal more if advertising didn't create mass markets for products. It is precisely because of the heavy advertising that consumer goods are so cheap. But we get the wrong idea if we think the only purpose of advertising is to sell goods. Another equally important function is to inform. A great deal of the knowledge we have about household goods is largely from the advertisements we read. Advertisements introduce us to new products or remind us of the existence of ones we already know about. Supposing you wanted to buy a washing machine, it is more than likely you would obtain details regarding performance, price, etc., from an advertisement.Lots of people pretend that they never read advertisements, but this claim may be seriously doubted. It is hardly possible not to read advertisements these days. And what fun they often are, too! Just think what a railway station or a newspaper would be like without advertisements. Would you enjoy gazing at a blank wall or reading railway byelaws while waiting for a train? A cheerful, witty advertisement makes such a difference to a dull wall or a newspaper full of the incidents and disasters.We must not forget, either, that advertising makes a positive contribution to our pockets. The fact that we pay so little for our daily paper, or can enjoy so many broadcast programmers is due entirely to the money spent by advertisers. Just think what a newspaper would cost if we had to pay its full price!Another thing we mustn't forget is the “small ads.” What a tremendously useful service they perform for thecommunity! Just about anything can be accomplished through these columns. For instance, you can find a job, buy or sell a house, announce a birth, marriage or death in what used to be called the “hatch, match and dispatch” column(栏目) but by far the most fascinating section is the personal or “agony” column. No other item in a newspaper provides such entertaining reading or offers such a deep insight into human nature. It's the best advertisement for advertising there is!8. What is the main idea of this passage?A. Advertisements steal money from our pocketsB. The critics get the wrong idea of advertisements.C. Advertisers perform a useful service to communities.D. Advertisements are everywhere.9. What is the attitude of the author toward advertisements?A. He appreciates the role of advertisements.B. He doubts the effect of advertisements.C. He believes what is said in advertisements.D. He complains too many advertisements in daily life.10. Which of the following is Not True?A. The personal or “agony” column makes us know more about human nature.B. The only purpose of advertising is to sell goods.C. A newspaper will cost us more if there is no advertisement on it.D. Advertisement makes our life color1 ful.11. Whicof the following shows the structure of the passage?( ①=" Paragraph" 1, ②=" paragraph" 2, ③=" paragraph" 3, ④=" paragraph" 4 ⑤=" paragraph" 5)A B.C. D.DFor centuries , tea has been used for far more than quenching thirst. Around the world people drink it to relax, reinvigorate and relieve, and it's something we need now more than ever.Even in the United States, a long coffee-dominated country, tea drinking is growing in popularity, with the country consuming 0.4 kilograms (14 ounces) of tea leaves per person a year compared with 0. 36 kilograms(12. 7 ounces) in 2007 according to the United Nations, as people switch away from soda，milk and fruit drinks.Scientists are beginning to look into just how tea might affect mood and cognition. Specifically, they're investigating whether its relaxing and refreshing effects are a direct biological outcome of the compounds in tea or whetherthey come from the context in which the drink is consumed—preparing your tea, choosing your favorite cup and sitting down for a brief break from the world. Or both.Drinking green tea has been found to improve brain function in healthy people, said Stefan Borgward, chair and director of the department of psychiatry and psychotherapy at the University of Lubeck, Germany.In a 2014 study, he gave one or two cups of green tea to 12 healthy volunteers and imaged their brains to analyze changes in connectivity inside certain brain regions.“We noticed an increased connectivity in regions of the brain associated with working memory,” he said via email.And a 2017 review of more than 100 studies he coauthored found that green tea can impact the brain in three ways: It can influence psychopathological state such as reducing anxiety; cognition by benefiting memory and attention; and brain function, specifically memory.That review concluded that "it would be desirable" for more Westerners to consume at least 100 milliliters(3.3 fluid ounces) of green tea each day “to protect neurocognitive function.”However, Borgward, cautioned that the effects aren't large, and the evidence is mainly provided by small-scale studies.12. How does the author introduce the tendency of drinking in the U.S.?A. By showing examples.B. By explaining reasons.C. By making a comparison.D. By checking existing facts.13. What is paragraph 3 mainly about?A. Scientists are looking into why tea makes people relaxed and refreshed.B. The context in which the drink is consumed is very important.C. People's relaxation is the direct biological outcome of the ingredients in tea.D. Preparing tea, choosing a cup and sitting down for a brief rest is amazing.14. What can we conclude from Stefan's research?A. Tea is a cure for a neurocognitive disease.B. Green tea is particularly beneficial to our memory.C. Drinking tea every day can prevent us from getting ill.D. Westerners drink less than 100 milliliters of green tea.15. What is the researcher'sattitude towards the finding?A. SupportiveB. Doubtful.C. Unclear.D. Objective.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020届杭州市西湖第一实验学校高三英语上学期期末考试试卷及答案

2020届杭州市西湖第一实验学校高三英语上学期期末考试试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AWhat to Eat—and What to Skip—When It Comes to Takeout FoodIf the burden on your wallet doesn't bother you much, the effect your takeout habit can have on your waistline just might arouse your attention. Here's the best and worst of the lot for your belly.Steamed Vegetable Dumplings: Order This.When she orders Chinese, registered dietitian nutritionist Elisa Zied gets an order of steamed vegetable dumplings. "I often pair them with either chicken and broccoli in brown sauce（I ask for a little saucemade without sugar）or steamed shrimp dumplings," she tells us.Crab Wontons: Not That!When you deconstruct crab wontons, it's easy to see why they're a "Not That!" The inside is filled with crabmeat and cream cheese（which is just a fancy, spreadable fat）.The wonton is made of refined flour, egg and salt and the crispy（脆的）coating is a result of a deep oil bath.Peking Duck: Order This.Most of the fat from the skin flows out of the duck over the course of cooking, making this a healthier choice than most of the stir-fry dishes available. Order a side of steamed vegetables and serve it with a small scoop of brown rice. Done and done!Sweet and Sour Anything: Not That!Anything with “sweet and sour” in its title is a powerful cue that something has been deep-fried and covered in a sickly-sweet pink sauce. If you pair your selection with a side of rice, you're looking at a 1,000-calorie meal.Summer Roll: Order This.Summer rolls are steamed instead of fried—and typically filled with lean proteins and vegetables, making them a winning appetizer in our book. Pair them with an order of edamame（毛豆）and a broth-based soup for a satisfying, filling meal.Spring Roll: Not That!Spring=deep-fried, which is why we say to skip them! They're filled with fat and calories your belly doesn't need.1.What kind of cooking method should be skipped according to the text?A.Steaming.B.Stir-frying.C.Deep-frying.D.Boiling.2.Which of the following suits as a good starter for a meal?A.Chicken and broccoli.B.Steamed vegetable dumplings.C.Peking duck.D.Summer rolls.3.Where can the text be found?A.In a recipe.B.In a guidebook.C.In a science fiction.D.In a health magazine.BIt’s a little before8 a.m. when Mathias Schergen pushes open the side door at Chicago’s Jenner Elementary Academy for the Arts. He walks down the hall toward the office to sign in. It’s the same routine he’s had as Jenner’s art teacher for nearly a quarter century. “It’s going to be a good day,” a colleague calls out. “It’s a good day.” They hug. It seems like a typical Friday. Except it’s not. After 23 years at Jenner Elementary, Schergen is retiring. Even on his last day, there are still art projects to finish.Schergen leaves behind a richlegacyat this school. He’s won grants (拨款) for art projects. He turned an empty classroom into a museum. He’s pushed his students to make art about their lives. And he was awarded a Golden Apple — the most honorable teaching award in Chicago. But it wasn’t always easy. For years, Schergen taught in one of the city’s toughest neighborhoods. “When I first got my room, I noticed there were bullet holes in the window. That made me nervous,” he says. So he stuffed Beanie Babies in the holes to make it “look kind of funny”. “I didn’t even tell my wife for a whole year,” he says. “I didn’t want her to know.”With one hour to go, Schergen piles the chairs and sweeps the floor. He cleans out the sink for the last time. Fifth-grader Deontae Barnes, one of his best helpers, has watched him say goodbye all day. He wanders in the doorway. “Ah, come here, son,” Schergen says, signaling him over. He bends down for a hug. “Thank you for making these last days special and being a help to me.”When Deontae leaves, a reporter asks Schergen: When your kids ask why you’re retiring, what do you tell them? “I just tell them that grown people have dreams too,” he says. “I have other things in my life I have to do. It’s time. It’s just time.”4. Why is it a special Friday for Schergen?A. He was retiring on that day.B. He won an honor for his school.C. He was interviewed by a reporter.D. He received a Golden Apple award.5. What does the underlined word “legacy” in Paragraph 2 refer to?A. Art projects.B. Great achievements.C. Respect from students.D. Change in teaching.6. What made Schergen nervous when he first got to the school?A. Safety concerns in the school.B. The poorly-equipped classroom.C. Being misunderstood by his family.D. Students’ poor academic performance.7. What is the best title for the text?A.A Typical Day for an Art TeacherB. Time for Art ProjectsC. A Teacher’s Final Day at SchoolD. The Last Art ClassCPaper is one of our oldest, simplest and most important inventions. But it also presents a danger to the world in two important ways. First, the making of paper requires the loss of many millions of trees each year. And worldwide use of paper is expected to double in the next 40 years. Clearly, the planet cannot stand such a high rate of forest loss. The second great problem with paper is what happens once it is no longer useful. A large amount of wastepaper ends up in landfills, where it can produce harmful gases and finally contribute to global climate change.One simple solution can greatly reduce both of these problems: paper recycling. Instead of cutting down trees, recycle existing paper to feed the paper-making process.Paper is mainly made from cellulose (纤维素)，which can be used repeatedly in papermaking. Unfortunately, it also means that paper waste takes a surprisingly long time to break down in landfills. So far, trees are the only source(来源) of cellulose that can fill the great demand for paper products. Therefore, recycling paper is simply one of the best ways to save trees. Thanks to advances in processing, recycled paper need not be the dark-color1 edstuff many of us are familiar with. It now can offer the same print performance as non-recycled paper.Effective recycling requires a continuous effort from everyone at all levels of society. The way to begin is with education and understanding. Once enough people realize the need for recycling, more effective recycling systems can be developed.The need is real. The massive loss of trees affects everybody on earth. Everyone should do their part to recycle paper and encourage government and industry to do the same. The world will be a better place for it!8. What can we infer from the text?A. The use of recycled paper will double in 40 years.B. Recycling paper helps relieve global climate change.C. Wastepaper can easily break down in landfills.D. There are not enough landfills for wastepaper.9. What makes recycled paper more acceptable?A. The great demand of trees.B. The low processing cost.C. Its dark-color1 ed feature.D. Its improved print performance.10. What does the author propose?A. Punishing the act of cutting trees.B. Recycling paper.C. Improving recycling system.D. Promoting paper industry.11. How is the text mainly developed?A. By analyzing causes and effects.B. By offering research plans and data.C. By discussing problems and solutions.D. By comparing strengths and weaknesses.DRecently, I experienced a wonderful lesson in how little things still meana lot. My brother, mother and I live in Hawaii. Our farm is at least a dozen miles from even the most basic of services. Therefore, I take weekly trips to the shop to gel supplies. About a month ago, I finished loading up the car and was about to leave when a piece of paper on the ground caught my eye. I picked it up and read it carefully.Immediately, I was grateful that I had done that___4___It was a receipt (收据) from the State Motor Vehicle Division, recording the owner's payment of her Vehicle's Registration fees. I put myself in his or her shoes and thought: no one would throw this away. I looked over the receipt for any personal data, perhaps a license plate (车牌) or telephone number, but failed. How could I find the owner in the busy, crowded parking lot? Had it been lying there for a few minutes or a week? So I checked the date, the fees paid and the name of the owner, who must live in our town. I decided that the best and easiest step to take was to put the receipt in an envelope and send it to the owner first the next morning.By the end of the week, I received a beautiful “thank you” letter from a woman including a handwritten message and a card. In the letter, the woman explained how the wind took her receipt from a pocket in her car's passenger door. She had searched everywhere for quite some time before giving up.It felt great to know I had helped someone avoid a loss by doing something that seemed little and unimportant.12. What does the underlined sentence in paragraph 1 mean?A. He was lucky to learn the lesson.B. It was a good idea to do shopping that day.C. He was right to pick up the paper.D. It turned out the paper belonged to the writer.13. What information did the writer get from the paper?A The woman's license plate number.B. The woman's phone number.C. The woman's name.D. The woman's address.14. How did the woman lose the receipt?A. She forgot where she had put it.B. A strong wind blew it away.C. It fell onto the floor.D. She left it in the parking lot.15. What can be the best title for the text?A. A Lesson I Will Never ForgetB. Never Lose Heart or Give upC. Little Things Still Mean a LotD. Think Carefully Before You Act第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020届杭州市文晖中学高三英语期末考试试卷及参考答案

2020届杭州市文晖中学高三英语期末考试试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AWhen the sun shines brightly, it provides a great chance to get outdoor things done. Like making hay! At least, that is what farmers from the past would say. ―Make hay while the sun shines.This idiom is very old, dating back to Medieval times. Rain would often ruin the process of making hay. So, farmers had no choice but to make hay when the sun was shining.Today, we all use this expression, not just farmers. When conditions are perfect to get something done, we can say, ―It’s a good idea to make hay while the sun shines.In other words, you are taking advantage of a good situation or of good conditions. You are making the most of your opportunities. These all mean ―making hay while the sun shines.And sometimes we use this expression to mean we beat someone to the punch, or we got ahead of someone else. And other times you make hay while the sun shines to make good use of the chance to do something while it lasts. You are being opportunistic – taking advantage of a good opportunity. For example, my friend Ozzy was sick for a week and could not go to work. So, his co-worker Sarah -- who doesn’t like him -- took advantage of his illness and stole his project! Talk about making hay while the sun shines.Sometimes when you make hay while the sun shines you are staying ahead of a problem – like in this example:Hey, do you want to go hiking with me and my friends this weekend? The weather is going to be beautiful! I wish I could. But I have to finish my taxes. It’s the last weekend before they’re due.Oh, that’s too bad.Wait. What about your taxes?My taxes are done. I was off from work a couple of weeks ago and made hay while the sun shined. I got all of it done!I wish I would have taken advantage of my time off last week___1___All I did was lay around thehouse.And that’s all the time we have for these Words and Their Stories. But join us again next week. You can listen while you’re making dinner or riding to work. Yeah, make hay while the sun shines.1.Which of the following best matches ―make hay whilethe sun shines in paragraph 2?A.Sow nothing, reap nothing.B.Sharp tools make good work.C.Strike while the iron is hot.D.One swallow doesn’t make a summer.2.According to the underlined sentence, what feeling does the speaker express?A.AdmirableB.RegretfulC.AnnoyedD.Indifferent3.Where is the passage probably taken from?A.A radio programB.A magazineC.A brochureD.A novelBWe use what is known as inner speech, where we talk to ourselves, to evaluate situations and make decisions. Now, a robot has been trained to speak aloud its inner decision-making process, giving us a view of how it responds to contradictory demands.Arianna Pipitone and Antonio Chella at the University of Palermo, Italy, programmed a humanoid robot named Pepper, with software that models human cognitive(认知的)processes, which allowed Pepper to retrieve (检索)relevant information from its memory and find the correct way to act based on human commands, as well as a text — to — speech processor. It allowed Pepper to voice its decision-making process while completing a task, "With inner speech, we can better understand what the robot wants to do and what its plan is," says Chella.The researchers asked Pepper to set a dinner table according to etiquette (礼仪)rules they had programmed into the robot. Inner speech was either enabled or disabled to see how it affected Pepper's ability to do what was instructed.When instructed to place a napkin on a fork with its inner speech enabled, Pepper asked itself what the etiquette required and concluded that this request went against the rules it had been given. It then asked the researchers if putting the napkin on the fork was the correct action. When told it was, Pepper said, "OK, I prefer to follow your desire," and explained how it was going to place the napkin on the fork.When asked to do the same task with inner speech disabled, Pepper knew this contradicted etiquette rules, so it didn't perform the task or explain why.With the potential for robots to become more common in the future, this type of programming could help the public understand their abilities and limitations, says Sarah Sebo at theUniversityofChicago. "It maintains people's trust and enables cooperation and interactions between humans and robots," she says. However, this experiment only used a single human participant, says Sebo. "It's unclear how their approach would compare across a widerange of human participants," she says.4. Why does the author mention how people make decisions in the first paragraph?A. To introduce the topic.B. To make comparisons.C. To provide an example.D. To support his argument.5. How did Pepper react to the contradictory instruction with its inner speech enabled?A. It failed to complete the task.B. It followed the etiquette rules.C. It made a random decision.D. It communicated with the researchers.6. What did Sarah Sebo think of the research?A. It was creative but worthless.B. It was a good try but the result was a failure.C. It was inspiring but needed further evidence.D. It was carefully designed but poorly performed.7. Which of the following is the best title for the text?A. Robot Taught To Be PoliteB. Robot Can Explain Its DecisionC. Robot Making Decisions: No Longer A DreamD. Robot-Human Communication: No Longer A ProblemCBrown cows may not actually make chocolate milk, but pink silkworms（蚕）do produce pink silk, a team of scientists has discovered. To see if they could produce pre-dyed silk-silk that comes color1 ed, straight from the source-the team fed ordinary silkworms mulberry（桑树）leaves that had been sprayed（喷洒）with fabric（织物）dyes（染色剂）. Out of seven tested dyes, only one worked, producing a thread that reminded me of pink-dyed hair.And yes, the worms themselves take on some color1 before they produce silk. Their color1 ful diets did not affect their growth, the team, which included engineers and biologists from the CSIR-National Chemical Laboratory in India, reports in the journalACS Sustainable Chemistry & Engineering. (The researchers didn't look too deeply into how the dyes affected the silkworms' health. After all, silkworms die when people harvest their silk.)The team made dyeing silk this way because color1 ing fabric normally uses large amounts of fresh water. The water gets polluted with dangerous chemicals in the process, requiring costly treatment before factories can send it back into waterways. Dyeing silk directly by feeding silkworms would avoid those water-washing steps. Scientists are just starting to study this idea. However, it remains to be seen if it's commercially successful. In this experiment, the Indian team tested seven dyes, which are cheap and popular in the industry.The scientists found different dyes moved through silkworms' bodies differently. Some never made it into the worms' silk at all. Others color1 ed the worms and their silk but the color1 disappears before the silk is turned into fabric. Only one dye, named "direct acid fast red", showed up in the final, washed silk threads. By the time it made it there, it was a pleasant, light pink.8. The text is most probably a(n) ________.A. science reportB. tourist guideC. animal experimentD. fashion advertisement9. Silkworms can produce pink silk because ________.A. they are born pinkB. they are dyed pinkC. they grow in pink waterD. they are fed dyed food10. Where is the experiment carried out?A. In America.B. In India.C. In Israel.D. In China.11. How many dyes have been proved successful in the experiment?A. One.B. Three.C. Five.D. Seven.DThis is Scientific America's 60-Second Science. I am Christopher Intagliata.The Apollo missions brought back 842 pounds of rock and soil from the moon, that's nearly 2200 different samples. But the most interesting one, according to a scientist Meenakshi Wadhwa, is a sample named "Apollo 1-0-0-8-5collected by Neil Armstrong on Apollo 11.“He was about to step back into the lunar module(登月舱) when he turned around and saw there were little spaces in the rock box. He knew that geologists on earth would be just so excited to study these materials, so he just scooped up nine scoops(勺) of soil and put it into the box." Wadhwa explained.It was one of the most well studied samples of the Apollo missions. And a geologist named John Wood noticed white flecks(微粒) of rock in the soil, which inspired him to dig deeper into the moon's ancient past.“This was quite a leap of imagination — he proposed that the whole of the moon had been almost coveredwith a magma(岩浆) ocean nearly 4.5 billion years ago. This was a revolutionary idea at the time, because people had thought the moon had formed cold, so it completely changed our idea how the moon formed.”But Wadhwa has a more personal reason to appreciate this sample. She met her husband Scott Parazynski also because of this rock sample. Scott, a mountaineer at that time, wanted to climbMount Everestwith a moon rock while Wadhwa was the chairman of the NASA committee that gives access to the samples for scientific purposes.Neil Armstrong's last-minute scoop of moon dust brought two people together here on Earth and upturned our understanding of how the moon — and the Earth itself-got here.Thank you for listening for Scientific American's 60-Second Science.12. It can be learned from Paragraph 3 that ________ .A. Neil Armstrong was excited to find the soilB. the spaceship was about to land on the moonC. Sample "Apollo1-008-5" was collected at the last minuteD. scientists were not satisfied with the samples brought back by Neil13. Which of the following statements is TRUE according to the text?A. Scott made a new proposal about the moon's origin.B. The Apollo missions brought back 842 rock samples.C. "Apollo 1-0-0-8-5" brought new evidence to the moon's formation.D. Wadhwa and her husband climbedMount Everestwith a moon rock.14. What is this text?A. A short interview.B. An introduction to a scientist.C. An inspiring speech.D. A broadcast story of a program.15. What is the text mainly about?A. A romantic story of a moon rock.B. A big leap made by Neil Armstrong.C. An unusual task for Apollo missions.D. An unexpected discovery in moon exploration.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020届杭州市采荷中学高三语文上学期期末试题及参考答案

2020届杭州市采荷中学高三语文上学期期末试题及参考答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面文章，完成下面小题。

北京渐远的叫卖声黄开发上午在家看书，楼下传来叫卖声：“磨剪子磨刀——”声音低沉，有些苍老的感觉。

“磨”字发音很轻，“刀”字音有点像“都”，与以往所听到的不同。

我走到窗前，见一个老师傅推着自行车走过。

他看起来六十来岁，光头，戴眼镜，上下一身咖啡色的衣裳，连自行车的挡泥板和支架都是这个颜色，——后者应是生锈的缘故。

他不时抖动着一串金闺，发出哗啦哗啦的铁片碰击声。

回到书桌前，我顺手从身边的书架上取下一本书—— 布面仿线装的《一岁货声》。

此书高踞书架已两年有余，之所以放在近前，是为了取读的方便，然而一直忙于读其他的书，无暇顾及。

时令交替，应时叫卖，最能反映出农业文明时代鲜明的季节感。

下面从《一岁货声》中略抄几则，可见一斑。

暮春四月，胡同口的菜摊传来：“杏儿来，熟又烂来，酸来还又管换来呀，烂杏儿巴达来，小葱儿来，莴苣菜呀，嫩水萝葡来，白菜呀，蒿子杆来，蒜苗来，豌豆角儿来，黄瓜来，勾葱辣秦椒来，卖粉皮儿一大钱。

”听到这声音，脑子里就会浮现出杏儿黄、水萝卜红、蔬菜绿的画面，丰富多彩。

“巴达杏”是一种出自西域的著名品种，小贩以此招徕顾客。

“嗳……十朵，花啊晚香啊，晚香的玉来，一个大钱十五朵。

”这是叫卖晚香玉的。

农历五月，初夏的黄昏，悠长的青灰色胡同，篮筐中盛开的白花，阵阵馥郁的芬芳……这动人的情景，宛如一幅有声的风俗画，一首有味的乡土诗。

其中的点号不是省略号，原编者在序中解释，这是表示长声与余韵的。

初冬十月，天气转冷，卖蒸白薯的来了：“栗子味的白糖来，是栗子味的白薯来……烫手来，蒸化了，锅底儿，赛过糖了，喝了蜜了，蒸透了白薯啊，真热活呀！”蒸白薯又香又甜又热乎，听了就使人流哈喇子。

时至腊月，临近年根，各种吆喝声、响器声密集起来。

有跑旱船、耍猴儿的，有卖供花、关东糖的，有卖年画、对联的，有熬粥、卖豆豉豆腐的，有卖红头绳、绫绢花的，有卖砂锅、装灯带的……各种吆喝声伴随着锣声、鼓声、唢呐声，在冰冷的空气里酿出了温煦、热闹、欢庆的年节气氛。

2020届杭州市源清中学高三生物期末试卷及答案

2020届杭州市源清中学高三生物期末试卷及答案一、选择题：本题共15小题，每小题2分，共30分。

每小题只有一个选项符合题目要求。

1. 不能用杂交育种方法来培育新品种的生物是（）A. 水稻B. 牛C. 小麦D. 大肠杆菌2. 下图为酥梨成熟过程中有机物的含量变化曲线图，相关分析错误的是（）A. 在图中所示酶作用下，酥梨细胞内液的渗透压逐渐变大B. 图中的五种有机物质中最可能含有S元素的是酶C. 用斐林试剂检测，10月底的酥梨样液砖红色最深D. 图中蔗糖和果糖含量逐渐升高，是由淀粉分解而来3. 不同供氮水平下的四种处理方式（1：不打顶；2：打顶；3：打顶+羊毛脂；4：打顶+含生长素的羊毛脂）对棉株叶片细胞分裂素和脱落酸含量影响的实验结果如图所示。

下列关于打顶处理与叶片早衰间关系的推测，正确的是（）A.低氮水平下，打顶后涂抹生长素提高了脱落酸含量，会促进叶片早衰B. 在高氮水平下，打顶后涂抹生长素降低了脱落酸含量，会延缓叶片衰老C. 在低氮水平下，与不打顶相比打顶处理细胞分裂素含量降低，会促进叶片早衰D. 在高氮水平下，与不打顶相比打顶处理细胞分裂素含量升高，会延缓叶片衰老4. 某亲本DNA分子双链均以白色表示，以灰色表示第一次复制出的DNA子链，以黑色表示第二次复制出的DNA子链，该亲本双链DNA分子连续复制两次后的产物是A. B.C. D.5. 下面甲→戊是用显微镜观察的几个操作步骤，如图在显微镜下要把视野中的物像从图l转为图2，其正确的操作步骤是甲：转动粗准焦螺旋乙：调节光圈丙：转动细准焦螺旋丁：转动转换器戊：移动装片A.甲→乙→丙→丁B.戊→丁→乙→丙C.乙→甲→戊→丁D.丁→戊→丙→乙6. 下列有关细胞膜的叙述，错误的是A. 细胞膜蛋白质分子具有物质运输功能B. 细胞的生长现象不支持细胞膜的静态结构模型C. 水分子因为磷脂双分子层内部疏水而不能通过细胞膜D. 细胞膜的脂质结构使溶于脂质的物质更容易通过细胞膜7. 如图所示，甲同学用小球做遗传规律模拟实验，甲同学每次分别从Ⅰ、Ⅰ小桶中随机抓取一个小球并记录字母组合，他每次都将抓取的小球分别放回原来小桶后再多次重复。

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