[潍坊期中]潍坊市2014届高三11月期中考试(英语)

听力选择题Where are the speakers?A.At an airport.B.In a hotel.C.In a market.【答案】B【解析】【原文】W: What can I do for you, sir?M: Hi, I'm staying in room 326 and I'd like to order an extra coffee without milk.W: OK! Your coffee will arrive in a minute, sir.听力选择题Why doesn't the woman try to get a PhD now?A.She doesn't need one for her job.B.She can't afford the payment.C.She can't spare the time.【答案】C【解析】【原文】M: Would you like to get a PhD in something?W: Yeah, maybe one day. But it's impossible right now. I have my job and the kids. I'd have to study at midnight!M: Oh, yeah. That would be difficult.听力选择题How does the boy find his parents?A.Strict.B.Tolerant.C.Considerate.【答案】A【解析】【原文】M: My parents never let me out after 7:00 on school nights.W: Well, it probably kept you out of trouble.M: Yeah, and it also stopped me from seeing any movies or shows in the evenings.听力选择题What did the woman mean?A.She failed to see the whole program.B.She stayed awake the whole night.C.She went home very late.【答案】A【解析】【原文】M: I really enjoyed the TV special about Chinese culture last night. Did you get home in time to see it?W: Oh, yes, but I wish I could have stayed awake long enough to see the whole thing.听力选择题What are the speakers mainly talking about?A.How fruits were harvested.B.Why the fruit sales increased.C.What caused the low price of fruits.【答案】B【解析】【原文】M: This year our sales increased by 23%.W: What caused the increase?M: Well, the harvest of oranges, bananas and some other fruits was good. Prices dropped and so sales increased.听力选择题听下面一段较长对话，回答以下小题。

试卷类型：A高一期中调研监测考试地理试题2024.11注意事项：1．答题前，考生先将自己的学校、姓名、班级、座号、考号填涂在相应位置。

2．选择题答案必须使用2B铅笔（按填涂样例）正确填涂；非选择题答案必须使用0.5毫米黑色签字笔书写，绘图时，可用2B铅笔作答，字体工整、笔迹清楚。

3．请按照题号在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。

保持卡面清洁，不折叠、不破损。

一、选择题（本题共15小题，每小题3分，共45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。

）2024年10月30日，神州十八号和神州十九号航天员乘组会师于中国空间站，该站距地面约400千米。

如图示意某摄影师拍到的中国空间站凌日（太阳被一个小的暗星体遮挡）轨迹。

据此完成1～2题。

1．能正确反映中国空间站凌日时太阳、地球与空间站三者位置关系的示意图是（）A．B．C．D．2．与飞船发射中心所处的大气层相比，中国空间站所处的大气层（）A．气压较高B．温度较高C．天气多变D．密度较小2024年5月10日至11日，地球磁场爆发了特大地磁暴，一场极光盛宴如期而至。

据此完成3～4题。

3．下列属于极光现象形成的条件是（）①地球磁场作用②高层大气电离③气候温和干燥④强劲的太阳风A．①②③B．①③④C．①②④D．②③④4．极光观赏宜选择在（）A．晴朗的白天B．阴天的白天C．晴朗的夜晚D．阴天的夜晚对流层顶是对流层与平流层之间的过渡层，研究对流层顶的分布特征与变动情况具有重大的理论意义和应用价值。

如图示意1968～1996年中国部分地区甲、乙、丙三个月份的平均对流层顶高度场分布（单位：m，等值距为500）。

据此完成5～6题。

甲乙丙5．a、b、c的数值依次是（）A．11500 10000 10500B．10500 12000 10500C．11500 10000 11500D．10500 12000 115006．甲、乙、丙分别对应的月份是（）A．7月、1月、4月B．1月、4月、7月C．1月、7月、4月D．7月、4月、1月2024年6月30日，上海中北部普降暴雨，城区多条道路被淹。

2024年潍坊高三期中考试试卷
一、选择题
1.中国古代四大发明包括（） A. 纸张 B. 水车 C. 火药 D. 指南针
2.南宋的开国君主是（） A. 赵匡胤 B. 宋徽宗 C. 赵构 D. 赵昀
3.丝绸之路是连接（）和（）的重要商业通道。

A. 中国；印度 B. 中国；
欧洲 C. 中国；非洲 D. 中国；日本
4.以下哪个不属于唐朝文学代表作家（） A. 李白 B. 白居易 C. 骆宾王 D.
李商隐
5.五岳之一的泰山在（）省。

A. 山东 B. 河北 C. 河南 D. 江苏
二、填空题
1.《三国演义》的作者是（）。

2.中国现代著名小说家巴金的原名是（）。

3.中国著名广场舞《小苹果》的歌手是（）。

4.我国的“国粹”是指（）。

5.论语的作者是（）。

三、简答题
1.请简单概述唐朝盛世的主要特点。

2.《西游记》是中国哪个朝代的文学作品？请简要介绍。

3.什么是世界遗产？请列举一些中国的世界遗产名称。

4.请解释什么是“中华文明的核心价值观”？
5.你理解的旅游安全和文明游的意义是什么？
四、综合应用题
某人行程如下：第1天从潍坊出发，每天步行15公里，第10天抵达目的地，请问行程共多长路程？如果遇到每3天休息1天的情况，那么在此情况下到达目
的地需要多少天？
五、作文题
请从“快乐生活”这个主题出发，写一篇不少于800字的文章，探讨你对快乐生活的看法和实践经验。

以上为2024年潍坊高三期中考试试卷，请同学们认真答题，祝你们顺利。

山东省各地2014届高三上学期期中考试试题分类汇编集合一、选择题1、（德州市2014高三期中）若{}{}{}1,2,3,4,5,6,1,2,4,2,3,6U M N ===，则()U C M N ⋃=A ．{}1,2,3B ．{}5C ．{}1,3,4D ．{}2答案：B2、（菏泽市2014高三期中）．设集合{}{}2,ln ,,A x B x y ==，若{}0A B ⋂=，则y 的值为A ．0B ．1C ．eD ．2答案：A3、（济南外国语学校2014高三期中）已知集合1,={|1},U U R A x y C A x ==-集合则=（ ） A ．}10|{<≤x x B ．}10|{≥<x x x 或 C ．}1|{≥x x D ．}0|{<x x 答案：A4、（济南一中等四校2014高三期中）已知全集U={0，1，2，3，4），集合A={1，2，3），B={2，4}，则()U C A B 为A.{1,2,4)B.{2,3,4)C.{0,2,4)D.{0,2,3,4)答案：C5、（临沂市2014高三期中）已知全集U=R ， 集合{}{}2320,0U A x x x B x x a C B A =-+>=-≤⊆，若，则实数a 的取值范围是 A.()1-∞，B.(]2-∞，C.[)1+∞，D.[)2+∞，答案：D 6、（青岛市2014高三期中）已知全集R U =,{|21}x A x y ==-,则U C A =A ．[0,)+∞B ．(,0)-∞C ．(0,)+∞D ．(,0]-∞答案：B7、（山东师大附中2014高三期中）若集合{}{}02|,822|22>-∈=≤<∈=+x x R x B Z x A x ，则()B C A R 所含的元素个数为（ ）A.0B.1C.2D.3 答案：C 8、（威海市2014高三期中）已知集合{}{}11|,,A B m m x y x A y A =-==+∈∈，，，则集合B 等于（A ）{}2,2- （B ）{}2,0,2- （C ）{}2,0- （D ）{}0答案：B9、（潍坊市2014高三期中）设集合A={1,2,3}，B={4,5}，C={x |x =B b A a a b ∈∈-,,}，则C 中元素的个数是A ．3B ．4C ．5D ． 6答案：B 10、（文登市2014高三期中）已知集合4{|log 1}A x x =<，{|2}B x x =≥，则R A C B =A.(,2)-∞B.(0,2)C.(,2]-∞D.[2,4)答案：B11、（枣庄市2014高三期中）已知全集U ＝{1,2,3,4,5,6)，集合A ＝{2,4,5)，B ＝{1,3,5)，则A. {1}B. {3}C. {1,3,5,6}D. {1,3}答案：D二、填空题1、（济南一中等四校2014高三期中）已知集合{}{}{}22,3,23,21,2,5U U a a A a C A =+-=-=，则实数a 的值为___________.答案：2。

2023-2024学年山东省潍坊市高三（上）期中数学试卷一、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．已知向量a →=（1，k ），b →=（2，1），若a →∥b →，则实数k ＝（ ） A ．12B ．−12C ．2D ．﹣22．若“∃x ∈R ，sin x ＜a ”为真命题，则实数a 的取值范围为（ ） A ．a ≥1B ．a ＞1C ．a ≥﹣1D ．a ＞﹣13．已知集合A ＝{1，3，a 2}，B ＝{1，a +2}，则满足A ∪B ＝A 的实数a 的个数为（ ） A ．0B ．1C ．2D ．34．北京故宫博物院展示着一件来自2200年前的宝物——秦诏文权（如图1）．此文权下部呈圆台形，上部为鼻钮，被誉为最美、最具文化、最有政治和历史意义的文物之一．某公司仿照该文权制成一纸镇（如图2），已知该纸镇下部的上、下底面半径分别为3，4，高为3，则该纸镇下部的侧面积与体积分别为（ ）A ．21π 37πB ．21π 111πC ．7√10π 37πD ．7√10π 111π5．设等差数列{a n }的前n 项和为S n ，且公差不为0，若a 4，a 5，a 7构成等比数列，S 11＝66，则a 7＝（ ） A ．5B ．6C ．7D ．86．已知a ＝20.5，b ＝log 25，c ＝log 410，则a ，b ，c 的大小关系为（ ） A ．a ＜b ＜cB ．a ＜c ＜bC ．c ＜a ＜bD ．b ＜c ＜a7．设函数f （x ）={x +1，x ≤0√x −1，x ＞0，则方程f （f （x ））＝0的实根个数为（ ）A ．4B ．3C ．2D ．18．已知cos(π4−α)=35，sin(5π4+β)=−1213，其中α∈(π4，3π4)，β∈(0，π4)，则tanαtanβ=（ ）A ．−5663B ．5663C ．﹣17D ．17二、多项选择题：本大题共4小题，每小题5分，共20分，在每个小题给出的选项中，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．在正方体ABCD ﹣A 1B 1C 1D 1中，直线l ⊂平面ABB 1A 1，直线m ⊂平面BCC 1B 1，直线n ⊂平面ABCD ，则直线l ，m ，n 的位置关系可能是（ ）A ．l ，m ，n 两两垂直B ．l ，m ，n 两两平行C ．l ，m ，n 两两相交D ．l ，m ，n 两两异面10．已知函数f(x)=2sin(2x +π3)，把f （x ）的图象向左平移π3个单位长度得到函数g （x ）的图象，则（ ）A ．g （x ）是奇函数B ．g （x ）的图象关于直线x =−π4对称C ．g （x ）在[0，π2]上单调递增D ．不等式g （x ）≤0的解集为[kπ+π2，kπ+π]，k ∈Z11．已知a ，b 为方程2x 2﹣8x +m ＝0（m ＞0）的两个实根，则（ ） A ．a 2+b 2≥8 B ．ab ≥4 C ．√a +√b ≤2√2D ．1a+2+12b≥3+2√21212．已知正项数列{a n }满足：a 1＝1，a n =na n+12na n+1+1，则（ ）A ．a 2=√5−12B ．{a n }是递增数列C ．a n+1−a n ＞1n+1D ．a n+1＜1+∑ n k=11k三、填空题：本大题共4小题，每小题5分，共20分.13．已知点A （2，1），向量OA →绕原点O 顺时针旋转π2得到向量OB →，则点B 的坐标为 ．14．诺沃尔（Knowall ）在1740年发现了一颗彗星，并推算出在1823年、1906年…人类都可以看到这颗彗星，即该彗星每隔83年出现一次．从现在开始到公元3000年，人类可以看到这颗彗星的次数为 ．15．已知函数f（x）是R上的偶函数，f（x+2）为奇函数，若f（0）＝1，则f（1）+f（2）+…+f（2023）＝．16．右图为几何体Ω的一个表面展开图，其中Ω的各面都是边长为1的等边三角形，将Ω放入一个球体中，则该球表面积的最小值为；在Ω中，异面直线AB与DE的距离为．四、解答题：本大题共6道小题，共70分，解答应写出文字说明、证明过程或演算步骤17．（10分）已知函数f(x)=log12x，F（x）＝f（x+1）+f（1﹣x）．（1）判断F（x）的奇偶性，并证明；（2）解不等式|F（x）|≤1．18．（12分）已知函数f（x）＝A sin（ωx+φ）+B（其中A，ω，φ，B均为常数，ω＞0，A＞0，|φ|＜π2)的部分图象如图所示．（1）求f（x）的解析式；（2）求函数y=f(x+5π12)+f(x)在[−π3，π2]上的值域．19．（12分）在四棱柱ABCD﹣A1B1C1D1中，底面ABCD是矩形，AD＝2CD＝2，AA1=A1D=√5，A1C=√6．（1）证明：平面AA1D1D⊥平面ABCD；（2）求二面角A1﹣CD﹣D1的余弦值．20．（12分）为方便居民休闲娱乐，某市计划在一块三角形空地上修建一个口袋公园，如图所示．在公园内部计划修建景观道路CD （道路的宽度忽略不计），已知CD 把三角形空地分成两个区域，△ACD 区域为儿童娱乐区，△BCD 区域为休闲健身区．经测量，AC ＝BC ＝100米，AB =100√3米．若儿童娱乐区每平方米的造价为100元，休闲健身区每平方米的造价为50元，景观道路每米的造价为2500元． （1）若∠ADC =π4，求景观道路CD 的长度；（2）求∠ADC 为何值时，口袋公园的造价最低？21．（12分）设S n 为数列{a n }的前n 项和，s n =3n+1−32．（1）求{a n }的通项公式； （2）若数列{S 2n +15a n}的最小项为第m 项，求m ； （3）设b n =2a n (a n −2)2，数列{b n }的前n 项和为T n ，证明：T n ＜132．22．（12分）已知函数f （x ）＝e x +aln （x +1）（a ∈R ）．（1）当a ＝﹣2时，求曲线y ＝f （x ）在点（0，f （0））处的切线方程； （2）若f （x ）在定义域上存在极值，求a 的取值范围； （3）若f （x ）≥1﹣sin x 恒成立，求a ．2023-2024学年山东省潍坊市高三（上）期中数学试卷参考答案与试题解析一、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．已知向量a →=（1，k ），b →=（2，1），若a →∥b →，则实数k ＝（ ） A ．12B ．−12C ．2D ．﹣2解：因为a →=（1，k ），b →=（2，1），且a →∥b →，所以2k ﹣1＝0，解得k =12．故选：A ．2．若“∃x ∈R ，sin x ＜a ”为真命题，则实数a 的取值范围为（ ） A ．a ≥1B ．a ＞1C ．a ≥﹣1D ．a ＞﹣1解：“∃x ∈R ，sin x ＜a ”，故a ＞（sin x ）min ，a ＞﹣1． 故选：D ．3．已知集合A ＝{1，3，a 2}，B ＝{1，a +2}，则满足A ∪B ＝A 的实数a 的个数为（ ） A ．0B ．1C ．2D ．3解：A ∪B ＝A ，则B ⊆A ，当a +2＝3，即a ＝1时，集合A 不满足元素的互异性，舍去， 当a +2＝a 2，即a ＝2或a ＝﹣1，当a ＝2时，A ＝{1，3，4}，B ＝{1，4}，满足题意， 当a ＝﹣1时，集合B 不满足元素的互异性，舍去， 综上所述，a ＝2，故满足A ∪B ＝A 的实数a 的个数为1． 故选：B ．4．北京故宫博物院展示着一件来自2200年前的宝物——秦诏文权（如图1）．此文权下部呈圆台形，上部为鼻钮，被誉为最美、最具文化、最有政治和历史意义的文物之一．某公司仿照该文权制成一纸镇（如图2），已知该纸镇下部的上、下底面半径分别为3，4，高为3，则该纸镇下部的侧面积与体积分别为（ ）A ．21π 37πB ．21π 111πC ．7√10π 37πD ．7√10π 111π解：由题意得，S 侧＝π（3+4）×√32+(4−3)2=7√10π，V =13π×（42+32+4×3）×3＝37π．故选：C ．5．设等差数列{a n }的前n 项和为S n ，且公差不为0，若a 4，a 5，a 7构成等比数列，S 11＝66，则a 7＝（ ） A ．5B ．6C ．7D ．8解：等差数列{a n }的前n 项和为S n ，且公差d 不为0，若a 4，a 5，a 7构成等比数列，S 11＝66， 故S 11=11(a 1+a 11)2=11a 6=66，解得a 6＝6，故{a 6=6a 52=a 4⋅a 7，整理得{a 1+5d =6(a 1+4d)2=(a 1+3d)(a 1+6d)，解得{a 1=−4d =2，故a 7＝a 1+6d ＝8． 故选：D ．6．已知a ＝20.5，b ＝log 25，c ＝log 410，则a ，b ，c 的大小关系为（ ） A ．a ＜b ＜cB ．a ＜c ＜bC ．c ＜a ＜bD ．b ＜c ＜a解：因为a =20.5=√2，c =log 410=log 2√10＜log 25，所以b ＞c ，c =log 410=log 2√10＞log 22√2=32＞√2，所以 c ＞a ，所以a ＜c ＜b ．故选：B ．7．设函数f （x ）={x +1，x ≤0√x −1，x ＞0，则方程f （f （x ））＝0的实根个数为（ ）A ．4B ．3C ．2D ．1解：令t ＝f （x ），则方程f （f （x ））＝0，即f （t ）＝0， 当t ≤0时，t +1＝0，∴t ＝﹣1； 当t ＞0时，√t −1=0，∴t ＝1；当t ＝﹣1时，若x ≤0，则x +1＝﹣1，∴x ＝﹣2，符合题意； 若x ＞0，则√x −1=−1，∴x ＝0，不合题意； 当t ＝1时，若x ≤0，则x +1＝1，∴x ＝0，符合题意；若x ＞0，则√x −1=1，∴x ＝4，符合题意，即方程f （f （x ））＝0的实根个数为3． 故选：B ．8．已知cos(π4−α)=35，sin(5π4+β)=−1213，其中α∈(π4，3π4)，β∈(0，π4)，则tanαtanβ=（ ）A ．−5663B ．5663C ．﹣17D ．17解：cos(π4−α)=35，∵α∈(π4，3π4)，∴π4−α∈（−π2，0），∴sin （π4−α）=−√1−cos 2(π4−α)=−45，sin （α−π4）=45，cos α＝cos[（α−π4）+π4]＝cos （α−π4）cos π4−sin （α−π4）sin π4=35×√22−45×√22=−√210，则sin α=√1−(√210)2=7√210，则tan α=sinαcosα=−7， sin(5π4+β)=−1213，∵β∈(0，π4)，∴5π4+β∈(5π4，3π2)， ∴cos （5π4+β）=−√1−sin 2(5π4+β)=−513，sin β＝sin [(5π4+β)−5π4]=sin(5π4+β)cos 5π4−cos(5π4+β)sin 5π4=−1213×(−√22)−513×√22=7√226，cos β=√1−(7226)2=17√226，则tan β=sinβcosβ=717，则tanαtanβ=−7717=−17． 故选：C ．二、多项选择题：本大题共4小题，每小题5分，共20分，在每个小题给出的选项中，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．在正方体ABCD ﹣A 1B 1C 1D 1中，直线l ⊂平面ABB 1A 1，直线m ⊂平面BCC 1B 1，直线n ⊂平面ABCD ，则直线l ，m ，n 的位置关系可能是（ ）A ．l ，m ，n 两两垂直B ．l ，m ，n 两两平行C ．l ，m ，n 两两相交D ．l ，m ，n 两两异面解：如图，当l 为BB 1，m 为BC ，n 为CD 时，满足直线l ⊂平面ABB 1A 1，直线m ⊂平面BCC 1B 1，直线n ⊂平面ABCD ，l ，m ，n 两两相交且垂直，当l 为A 1B ，m 为B 1C 1，n 为AC 时，三条直线两两异面，故ACD 正确； 三条直线不可能两两平行，若l ∥n ，则l ∥AB ∥n ，而AB 与平面BCC 1B 1相交，则AB 与M 不平行，故B 错误． 故选：ACD ．10．已知函数f(x)=2sin(2x +π3)，把f （x ）的图象向左平移π3个单位长度得到函数g （x ）的图象，则（ ）A ．g （x ）是奇函数B ．g （x ）的图象关于直线x =−π4对称C ．g （x ）在[0，π2]上单调递增D ．不等式g （x ）≤0的解集为[kπ+π2，kπ+π]，k ∈Z解：由题意g （x ）＝2sin[2（x +π3）+π3]＝2sin （2x +π）＝﹣2sin2x ，A 中，可得g （x ）为奇函数，所以A 正确；B 中，函数g （x ）的对称轴方程满足2x =π2+k π，k ∈Z ， 解得x =π4+k 2π，k ∈Z ，当k ＝﹣1时，x =−π4，所以函数g （x ）的图象关于x =−π4对称，所以B 正确； C 中，x ∈[0，π2]，则2x ∈[0，π]，显然g （x ）不单调，所以C 不正确；D 中，令g （x ）≤0，则2k π≤2x ≤π+2k π，k ∈Z ，解得k π≤x ≤π2+k π，k ∈Z ，即x ∈[k π，π2+k π]，k ∈Z ，所以D 不正确． 故选：AB ．11．已知a ，b 为方程2x 2﹣8x +m ＝0（m ＞0）的两个实根，则（ ） A ．a 2+b 2≥8 B ．ab ≥4 C ．√a +√b ≤2√2D ．1a+2+12b≥3+2√212解：因为已知a ，b 为方程2x 2﹣8x +m ＝0（m ＞0）的两个实根， 所以Δ＝64﹣8m ≥0，即m ≤8，又因为m ＞0，所以0＜m ≤8， 由韦达定理可得：a +b ＝4，ab =m2＞0，所以a ＞0，b ＞0． 对于选项A ，由a+b 2≤√a 2+b 22，当且仅当a ＝b 时等号成立可得：a 2+b 2≥8，当且仅当a ＝b 时等号成立，故A 正确；对于选项B ，由a +b ＝4≥2√ab ，当且仅当a ＝b 时等号成立可得：ab ≤4，当且仅当a ＝b 时等号成立，故B 不正确；对于选项C ，由a+b 2≤√a 2+b 22，当且仅当a ＝b 时等号成立可得：√a+√b2≤√a+b 2，即√a +√b ≤2√2，当且仅当a ＝b 时等号成立，故C 正确；对于选项D ，1a+2+12b =（1a+2+12b）[（2a +4）+2b ]×112=112（2+2b a+2+a+2b +1）≥112（3+2√2b a+2⋅a+2b ）=112（3+2√2），当且仅当2b a+2=a+2b，即a =√2b ﹣2时等号成立，故D 正确． 故选：ACD ．12．已知正项数列{a n }满足：a 1＝1，a n =na n+12na n+1+1，则（ ）A ．a 2=√5−12B ．{a n }是递增数列C ．a n+1−a n ＞1n+1D ．a n+1＜1+∑ n k=11k解：由a 1＝1，a n =na n+12na n+1+1，可得a 1=a 22a 2+1=1，解得a 2=1+√52（负的舍去），故A 错误；由a n +1﹣a n =na n+12+a n+1−na n+12na n+1+1=a n+1na n+1+1＞0，即a n +1＞a n ，则{a n }是递增数列，故B 正确；由a n+1na n+1+1−1n+1=a n+1−1(n+1)(na n+1+1)＞0，则a n +1﹣a n ＞1n+1，故C 正确；由a n+1na n+1+1−1n=−1n(na n+1+1)＜0，则a n +1﹣a n ＜1n ，所以a n +1＝a 1+（a 2﹣a 1）+（a 3﹣a 2）+...+（a n +1﹣a n ）＜1+1+12+...+1n，故D 正确．故选：BCD ．三、填空题：本大题共4小题，每小题5分，共20分.13．已知点A （2，1），向量OA →绕原点O 顺时针旋转π2得到向量OB →，则点B 的坐标为 （1，﹣2） ．解：点A （2，1），向量OA →绕原点O 顺时针旋转π2后等于OB →，则OA →=（2，1），OB →=（1，﹣2），则点B 的坐标为（1，﹣2）． 故答案为：（1，﹣2）．14．诺沃尔（Knowall ）在1740年发现了一颗彗星，并推算出在1823年、1906年…人类都可以看到这颗彗星，即该彗星每隔83年出现一次．从现在开始到公元3000年，人类可以看到这颗彗星的次数为 12 ． 解：由题意可知：彗星出现的年份构成一个公差为d ＝83，首项为a 1＝1740的等差数列，所以a n＝a1+（n﹣1）d＝1740+83（n﹣1）＝83n+1657，令2023≤a n≤3000，即2023≤83n+1657≤3000，解得36683≤n≤134383，又n∈N*，所以n＝5、6、 (16)所以从现在开始到公元3000年，人类可以看到这颗彗星的次数为16﹣5+1＝12次．故答案为：12．15．已知函数f（x）是R上的偶函数，f（x+2）为奇函数，若f（0）＝1，则f（1）+f（2）+…+f（2023）＝﹣1．解：f（x+2）是奇函数，故f（x+2）＝﹣f（﹣x+2）且f（2）＝0，因为f（x）为偶函数，故f（x+2）＝﹣f（﹣x+2）＝﹣f（x﹣2），则f（x+4）＝﹣f（x），f（x+8）＝﹣f（x+4）＝f（x），即函数周期为8，因为f（x+2）＝﹣f（﹣x+2），故f（3）+f（1）＝0，f（4）+f（0）＝0，即f（4）＝﹣1，f（5）＝﹣f（1），f（6）＝﹣f（2）＝0，f（7）＝﹣f（3），f（8）＝f（0）＝1，故f（1）+f（2）+…+f（8）＝0，f（1）+f（2）+…+f（2023）＝﹣f（8）＝﹣1．故答案为：﹣1．16．右图为几何体Ω的一个表面展开图，其中Ω的各面都是边长为1的等边三角形，将Ω放入一个球体中，则该球表面积的最小值为2π；在Ω中，异面直线AB与DE的距离为√63．解：把平面展开图还原为空间几何体为正八面体，如图所示：球表面积最小，则正八面体的八个顶点在球面上，∴正八面体外接球的球心为正方形ACFD的中心O，半径R＝OA=12AF=12√12+12=√22，∴S表＝4πR2＝4π×12=2π；∵平面ABC∥平面DEF，∴异面直线AB与DE的距离为平面ABC与平面DEF的距离，又∵O到平面ABC的距离与O到平面DEF的距离相等，∴直线AB与DE的距离为O到平面ABC的距离2倍，∵V O﹣ABC＝V B﹣AOC，∴13S△ABC•h=13S△AOC•OB，∴√34h=12×√22×√22×√22，∴h=√66，∴异面直线AB与DE的距离为√6 3．故答案为：2π；√6 3．四、解答题：本大题共6道小题，共70分，解答应写出文字说明、证明过程或演算步骤17．（10分）已知函数f(x)=log12x，F（x）＝f（x+1）+f（1﹣x）．（1）判断F（x）的奇偶性，并证明；（2）解不等式|F（x）|≤1．解：（1）F（x）为偶函数；证明：∵f(x)=log12x，由{x+1＞01−x＞0，得x∈（﹣1，1），∴F（x）＝f（x+1）+f（1﹣x）=log12(x+1)+log12(1−x)的定义域为（﹣1，1），又F（﹣x）=log12(1−x)+log12(x+1)=F（x），∴F（x）为偶函数；（2）∵F（x）=log12(x+1)+log12(1−x)=log12(1−x2)≥log121=0，∴|F（x）|≤1⇔0≤F（x）=log12(1−x2)≤1，∴1≥1﹣x2≥12，解得−√22≤x≤√22，∴原不等式的解集为[−√22，√22]．18．（12分）已知函数f（x）＝A sin（ωx+φ）+B（其中A，ω，φ，B均为常数，ω＞0，A＞0，|φ|＜π2)的部分图象如图所示．（1）求f （x ）的解析式；（2）求函数y =f(x +5π12)+f(x)在[−π3，π2]上的值域．解：（1）由图知A =3−02=32，B =3+02=32， 且{ω⋅(−π3)+φ=−π2+2kπ，k ∈Z ω⋅π2+φ=π2+2kπ，k ∈Z ，|φ|＜π2，解得ω=65，φ=−π10， 所以f （x ）=32sin （65x −π10）+32； （2）y ＝f （x +5π12）+f （x ）=32sin[65（x +5π12）−π10]+32+32sin （65x −π10）+32=32[sin （65x −π10+π2）+32sin （65x x −π10）+3=32 [cos （65x x −π10）+sin （65x x −π10）]+3=3√22 s in （65x x −π10+π4）+3=3√22 s in （65x x +3π20）+3， 因为x ∈[−π3，π2]，所以65x +3π20∈[−π4，3π4]， 所以sin （65x +3π20）∈[−√22，1]， 所以y ∈[3√22•−√22+3，3√22×1+3]＝[32，3√22+3]． 即函数y 的值域为[32，3√22+3]． 19．（12分）在四棱柱ABCD ﹣A 1B 1C 1D 1中，底面ABCD 是矩形，AD ＝2CD ＝2，AA 1=A 1D =√5，A 1C =√6．（1）证明：平面AA 1D 1D ⊥平面ABCD ；（2）求二面角A 1﹣CD ﹣D 1的余弦值．（1）证明：取AD 的中点O ，连接OC ，因为AA 1=A 1D =√5，得A 1O ⊥AD ，因为A 1D =√5，OD ＝1，所以A 1O ＝2，又OD ＝DC ＝1，所以OC =√2，在△A 1OC 中，OC =√2，A 1C =√6，A 1O ＝2，所以A 1C 2=A 1O 2+OC 2，故△A 1OC 为直角三角形，A 1O ⊥OC ，因为OC ∩AD ＝O ，故A 1O ⊥平面ABCD ，因为A 1O ⊂平面AA 1D 1D ，所以平面AA 1D 1D ⊥平面ABCD ；（2）解：如图，以O 为坐标原点，分别以DC →，OD →，OA 1→的正方向为x 轴，y 轴，z 轴正方向， 建立如图所示空间直角坐标系：故A 1（0，0，2），C （1，1，0），D （0，1，0），D 1（0，2，2），则CD →=(−1，0，0)，A 1C →=（1，1，﹣2），DD 1→=(0，1，2)，设平面A 1CD 的一个法向量为m →=（x 1，y 1，z 1），则{m →⋅CD →=−x 1=0m →⋅A 1C →=x 1+y 1−2z 1=0，令y 1＝2，则m →=（0，2，1），设平面CDD 1C 1的一个法向量为n →=（x 2，y 2，z 2），则{n →⋅CD →=x 2=0n →⋅DD 1→=y 2+2z 2=0，令y 2＝2，则n →=（0，2，﹣1），所以cos ＜m →，n →＞=|m →⋅n →||m →||n →|=3√5×√5=35， 由图可知二面角A 1﹣CD ﹣D 1为锐角，所以二面角A1﹣CD﹣D1的余弦值为3 5．20．（12分）为方便居民休闲娱乐，某市计划在一块三角形空地上修建一个口袋公园，如图所示．在公园内部计划修建景观道路CD（道路的宽度忽略不计），已知CD把三角形空地分成两个区域，△ACD区域为儿童娱乐区，△BCD区域为休闲健身区．经测量，AC＝BC＝100米，AB=100√3米．若儿童娱乐区每平方米的造价为100元，休闲健身区每平方米的造价为50元，景观道路每米的造价为2500元．（1）若∠ADC=π4，求景观道路CD的长度；（2）求∠ADC为何值时，口袋公园的造价最低？解：（1）在△ABC中，AC＝BC＝100，AB＝100√3，所以AC2+AB2﹣BC2＝1002﹣（100√3）2﹣1002＝30000，则cosA=AC2+AB2−BC22AC⋅AB=√32，A∈（0，π），所以A=B=π6，在△ACD中，∠ADC=π4，由正弦定理得ACsin∠ADC=CDsinA，即CD=AC⋅sinAsin∠ADC=10Osinπ6sinπ4=50√2，所以景观道路CD的长度为50√2米．（2）设∠ADC=θ(π6＜θ＜5π6)，在△ACD中，CD=50sinθ，所以S△ADC=12AC⋅CD sin∠ACD=12×100×50sin(5π6−θ)sinθ=2500sin(5π6−θ)sinθ，又S△ABC=12AC⋅AB•sin A=12×100×100√3×12=2500√3，所以S△BCD＝2500√3−2500sin(5π6−θ)sinθ，所以投资总额y＝2500CD+100S△ACD+50S△BCD＝2500×50sinθ+100×2500sin(5π6−θ)sinθ+50[2500√3−2500sin(5π6−θ)sinθ]＝2500×50[√3+1+sin(5π6−θ)sinθ]＝2500×50（3√32+2+cosθ2sinθ），因为2+cosθ2sinθ=3cos2θ2+sin2θ24sinθ2cosθ2=34tanθ2+tanθ24≥2√34tanθ2⋅tanθ24=√34，当且仅当tan θ2=√3，即θ=2π3时取等号， 此时y 取得最小值，即公园造价最低，所以∠ADC =2π3，口袋公园的造价最低． 21．（12分）设S n 为数列{a n }的前n 项和，s n =3n+1−32． （1）求{a n }的通项公式；（2）若数列{S 2n +15a n }的最小项为第m 项，求m ； （3）设b n =2a n (a n −2)2，数列{b n }的前n 项和为T n ，证明：T n ＜132． （1）解：当n ＝1时，a 1＝S 1=32−32=3； 当n ≥2时，a n ＝S n ﹣S n ﹣1=3n+1−32−3n−32=3n ， 因为a 1＝3满足上式，所以a n ＝3n ．（2）解：S 2n +15a n =32n+1−32+153n =32n+1+272⋅3n =32•（3n +93n ）≥32•2√3n ⋅93n =9， 当且仅当3n =93n ，即n ＝1时，等号成立， 所以m ＝1． （3）证明：b n =2a n (a n −2)2=2⋅3n(3n −2)2， 当n ＝1时，b 1=2⋅31(31−2)2=6； 当n ≥2时，b n =2⋅3n 32n −4⋅3n +4＜2⋅3n 32n −4⋅3n +3=2⋅3n (3n −1)(3n −3)=3n 3n −3−3n 3n −1=11−3−n+1−11−3−n ， 所以T n ＝b 1+b 2+b 3+…+b n ＜6+（11−3−1−11−3−2）+（11−3−2−11−3−3）+…+（11−3−n+1−11−3−n ）＝6+11−3−1−11−3−n =152−11−3−n ＜152−1=132，命题得证． 22．（12分）已知函数f （x ）＝e x +aln （x +1）（a ∈R ）．（1）当a ＝﹣2时，求曲线y ＝f （x ）在点（0，f （0））处的切线方程；（2）若f （x ）在定义域上存在极值，求a 的取值范围；（3）若f （x ）≥1﹣sin x 恒成立，求a ．解：（1）当a ＝﹣2时，f （x ）＝e x ﹣2ln （x +1），可得f ′(x)=e x −2x+1，此时f′(0)=e0−21=−1，又f（0）＝e0﹣2ln1＝1，曲线y＝f（x）在点（0，f（0））处的切线方程为y﹣1＝﹣（x﹣0），即x+y﹣1＝0；（2）易知f′(x)=e x+ax+1(x＞−1)，当a≥0时，f′（x）≥0恒成立，此时函数f（x）在（﹣1，+∞）上单调递增，不符合题意；当a＜0时，f′(x)=e x−a(x+1)2＞0，所以当a＜0时，f′（x）在定义域上单调递增，又f′(a2)=e a2+aa2+1，因为aa2+1≥−12，e a2＞1，所以f′（a2）＞0；当a＜﹣1时，易知f′（0）＝1+a＜0，所以函数f（x）在（0，a2）上存在极值点；当a＝﹣1时，f′(x)=e x−1x+1，易知f′（0）＝0，所以x＝0为f（x）的极值点；当﹣1＜a＜0时，f′(a2−1)=e a2−1+1 a ，因为e a2−1＜1，1a＜−1，所以f′（a2﹣1）＜0，则函数f（x）在（a2﹣1，a2）上存在极值点，综上所述，满足条件的a的取值范围为（﹣∞，0）；（3）若f（x）≥1﹣sin x恒成立，即sin x+e x+aln（x+1）≥1恒成立，不妨设g（x）＝sin x+e x+aln（x+1），函数定义域为（﹣1，+∞），可得g′(x)=cosx+e x+ax+1，不妨设h（x）＝cos x+e x+ax+1，函数定义域为（﹣1，+∞），可得h′（x）＝﹣sin x+e x−a(x+1)2，若a＝﹣2，当x∈（﹣1，0]时，cosx+e x≤2，−2x+1≤−2，所以g'（x）≤0，当x∈[0，+∞）时，e x≥1，h′（x）≥0，所以g′（x）≥g′（0）＝cos0+e0﹣2＝0，则x＝0时，函数g（x）在x∈（﹣1，+∞）上取得唯一极小值点，此时g（x）≥g（0）＝1，所以a＝﹣2时，f（x）≥1﹣sin x恒成立；若a＜﹣2，易知e x﹣sin x＞0，−a(x+1)2＞0，所以h′（x）＞0，即函数g'（x）单调递增，又g′(−a)=e−a+cos(−a)+a−a+1＞e2−1−1＞0，因为g'（0）＝2+a＜0，所以存在x1∈（0，﹣a），使得g'（x1）＝0，当0＜x＜x1时，g′（x1）＜0，g（x）单调递减，所以g（x1）＜g（0）＝1，不符合题意；若﹣2＜a＜0，由（2）知g′（x）单调递增，当﹣1＜x＜﹣1−a2＜0时，ax+1＜−2，g′(x)＜1+1+ax+1＜0，又g′（0）＝2+a＞0，所以存在x2∈（﹣1，0），使得g′（x2）＝0，当x2＜x＜0 时，g′（x）＞0，g（x）单调递增，所以g（x2）＜g（0）＝1，不符合题意；若a≥0，易知cos x+e x＞0，ax+1≥0，所以g′（x）＞0，g（x）单调递增，又g（0）＝1，所以当﹣1＜x＜0时，g（x）＜g（0）＝1，不符合题意，综上所述，满足条件的a的值为﹣2．。

试卷类型：A高三英语2014.01 本试卷分第I卷（选择题）和第Ⅱ卷（非选择题）两部分。

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第一部分英语知识运用（共两节，满分50分）第一节单项填空（共20小题，每小题1分，满分20分）从每题所给的A、B、C、D四个选项中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。

1. We would often go for _______ drive with some friends at that time ,purely for ______ fun.A. the;aB. a; theC.a;不填D.不填; a2. - Do you like jazz?-_____ I prefer classical music.A. Of course I doB. It's hard to sayC. It's not my cup of teaD. Don't mention it3. One of our rules is that no students ______carry mobile phones while taking an exam.A. willB. needC. mayD. shall4. It is reported that his new book ______ by that company next year.A. will publishB. will be publishedC. was publishedD. has been published5. Don't leave home ____you're fully prepared for your trip.A. afterB. untilC. ifD. when6. People all over the world have a dream ______they will always live a peaceful life.A. thatB. whatC. whichD. where7.______more consumers, many online shops provide great discounts on Single's Day.A. AttractedB. To attractC. AttractingD. Having attracted8. Mo Yan, most of_____ novels have been adapted for films , is now a world famous writer.A. whichB. whomC. whoseD. his9. Everything was perfect ______ the bad weather when we visited Hong Kong.A. in case ofB. in place ofC. far fromD. apart from10. Mom often phones me, asking _____I am getting along with my studies.A. howB. whenC. whetherD. why11. Never lose heart______ you'll make a quick advance in your project.A. orB. andC. butD. for12. A good teacher is ______ who teaches not only with mind but also with heart.A. oneB. the oneC. itD. that13. The trees there are extremely tall ,some _____ over 90 meters.A. measureB. measuringC. to measureD. measured14.- How about your trip to Hawaii?- Wonderful! I_____ have enjoyed myself more.A. shouldn 'tB. needn 'tC. couldn'tD. wouldn ' t15. It is our responsibility, so _____ I don't want now is to push it to the students.A. howB. whichC. thatD. what16. He left home last month and no one_ _____him ever since.A. has seenB. sawC. had seenD. will see17. It was several hours ______the traffic returned to normal after the terrible accident.A. beforeB. sinceC. asD. when18. Her progress in speaking ability._____ the truth ,is a big comfort to me.A. tellB. to tellC. toldD. telling19. There are many examples in our life_____ a word of appreciation changes a person completely.A. whichB. thatC. whenD. where20 - I failed my exam again!-_____ lt isn't the end of the world.A. Pardon me?B. Good luck !C. What for?D. Come on!第二节完形填空（共20小题；每小题1.5分，满分30分）阅读下面短文，掌握其大意，从每题所给的A、B、C徊四个选项中，选出最佳选项，并在答题卡上将该项涂黑。

山东省潍坊市2023-2023学年高三上学期期中考试英语试题阅读理解Passage 1In his book How Music Works, David Byrne, founder of the band Talking Heads and a fine artist and musical collaborator in his own right, put his finger on a key moment in the evolution of music. For most of human history, Byrne points out, music mainly took place at home, in small gatherings of amateurs who made music for their own enjoyment and that of their friends and family. But today, music is primarily a professional product, created by a few stars supported by a vast industry.What this means, among other things, is that most of us aren’t encouraged to make music ourselves. Instead, we consume it, just as we consume DVDs and microwave meals. And when we do make music, it is in a far more professional manner than was supported in the past. Fewer people play music for fun these days, in other words.Byrne’s observation helps us get a handle on Shane Snow’s forthcoming book, Smartcuts: How Hackers, Innovators, and Icons Accelerate Success, which draws on a series of author interviews and his own life observations to decode how extraordinary success can happen.问题：1.What does David Byrne claim is different abouttoday’s music in comparison to music in the past?2.According to the passage, what is one result of theprofessional nature of today’s music industry?3.What does the passage mainly focus on?Passage 2The energy industry is undergoing a revolution. Old ways of generating electricity are being rethought and replaced with renewable energy sources. Solar, wind, wave, and tidal power could provide a clean, sustainable solution to the world’s growing demands for power. But these renewable energy projects have been designed for large scale usage, and their potential in providing electricity to rural and remote communities has been largely overlooked.Microgrids, which generate and supply electricity to a local area, are key to providing power in remote areas. They allow small communities to generate, distribute, and consume their own renewable energy. Microgrids can operate independently of the main power grid, or they can be connected to it, functioning as a backup source of electricity during power outages.By implementing microgrids in remote communities, the benefits are numerous. Residents will experience improved reliability and reduced electricity costs. In addition, microgridscan potentially reduce the carbon footprint of these communities by replacing diesel generators with cleaner sources of energy.问题：1.What is the main topic of this passage?2.According to the passage, what are the benefits ofimplementing microgrids in remote communities?3.How are microgrids different from the traditionalpower grid?完形填空Once upon a time, there was a huge elephant that was completely covered in dirt. The elephant had been 1 through many fields and had rolled all over them, 2 its beautiful white skin covered in dirt. The other animals in the jungle saw how dirty the elephant was and began to 3 to themselves, “What a dirty elephant! It has no 4 !One day, the dirty elephant was walking through the jungle when it came across a small monkey. The monkey, 5 of the dirty elephant, asked, “Why are you so dirty?” The elephant answered, “Because I’ve walked through many fields, and it was great 6 so!”The monkey thought for a while a nd then said, “I’ve heard that the peacock is the most beautiful bird in the world. Why 7 you try washing yourself with its feathers?”The dirty elephant thanked the monkey and immediately went to find a peacock. When it found a group of peacocks 8 by a river, the dirty elephant asked the peacocks if it could borrow a few of their feathers to clean its skin. The 9 considered for a moment, and then decided to help the elephant.The peacocks 10 the dirty elephant in the morning sun, which reflected off their beautiful feathers and 11 the elephant to change its ways. After the dirt had gone, the elephant’s beautiful white skin shone brighter than ever before. The other animals in the jungle, who couldn’t 12 recognize the elephant, praised its beauty and said, “What a beautiful elephant! Look at its beautiful white skin!”And so the elephant 13 washed and went to the river every day, not only to clean its skin, but also to do a good 14 for the peacocks. In return for their help, the elephant would bring them fruit and leaves to eat.Moral: Even the dirtiest people can change their ways, if given the 15 to do so.问题：1.What was the initial condition of the elephant?2.How did the elephant become clean?3.What did the elephant do in return for the peacocks’help?写作任务假设你作为学校的一员，为了减少食堂的食品浪费，你决定写一篇英文海报，向同学们宣传节约食物的重要性。

山东省潍坊市昌邑市2024-2025学年高三上学期11月期中考试数学试题一、单选题1．已知集合{}|(1)(2)0A x x x =-+=，{}|3213B x x =-<-<，则A B = （）A ．{}2,1-B ．()2,1-C ．{}1D ．()1,2-2．命题“所有能被3整除的整数都是质数”的否定是（）A ．存在一个能被3整除的整数不是质数B ．所有能被3整除的整数都不是质数C ．存在一个能被3整除的整数是质数D ．不能被3整除的整数不是质数3．已知等差数列{}n a 的前n 项和为n S ，若321523,1S S S a =+=，则{}n a 的公差等于（）A ．2-B ．1-C ．1D ．24．为净化水质，向一个游泳池加入某种化学药品，加药后池水中该药品浓度C 随时间t 的变化关系为22040Ct t C -+=，则C 的最大值为（）A ．1B ．2C ．4D ．55．如图，,,A B C 是圆O 上的三点，且60,90AOB BOC ∠∠=︒=︒，则OA =（）A ．1322OB -B ．1322OB +C ．3122OB OC-D ．3122OB OC+6．已知一个圆锥的底面圆半径为1，其侧面展开图是一个圆心角为2π3的扇形，则该圆锥的体积为（）A ．B ．2π3C ．3D ．37．已知定义在R 上的函数()f x 满足()()()()11f x y f x y f x f y -+-++=，且()12f =，则()()()234f f f ++=（）A ．2B ．0C ．2-D ．4-8．已知函数()()πsin 03,2f x x ωϕωϕ⎛⎫=+<<< ⎪⎝⎭，甲、乙、丙、丁四位同学各说出了这个函数的一条结论：甲：函数()f x 的图象关于π,03⎛⎫⎪⎝⎭对称；乙：函数()f x 在5ππ,1212⎡⎤-⎢⎥⎣⎦上单调递增；丙：函数()f x 在区间()0,π上有3个零点；丁：函数()f x 的图象向左平移π2个单位之后与()f x 的图象关于x 轴对称.若这四位同学中恰有一人的结论错误，则该同学是（）A ．甲B ．乙C ．丙D ．丁二、多选题9．已知直线,m n 是平面α外两条不同的直线，则下列命题正确的是（）A ．若m //,n α//α，则m //nB ．若m //,n n //α，则m //αC ．若m //,n m α⊥，则n α⊥D ．若m //,n αα⊥，则m n ⊥10．已知03a b <<<，则（）A ．(1)1b a +>B ．()log 11a b +>C ．()()cos πcos πa b +<+D ．ππcos cos 22a b ⎛⎫⎛⎫-<- ⎪ ⎪⎝⎭⎝⎭11．设函数()2cosπ2xf x x ax =-+，则（）A ．存在实数a ，使得()f x 为偶函数B ．函数()f x 的图象关于()2ax a =∈Z 对称C ．当2a =时，142f x x⎛⎫-< ⎪⎝⎭D ．当4a =时，函数()f x 在()2,3上单调递增三、填空题12．已知向量a ，b满足1a = ，()1,1b = ，()a ab ⊥- ，则,a b = .13．已知点π,08P ⎛⎫ ⎪⎝⎭在函数()sin (03)2f x x ωω=-<<的图象上，则曲线()y f x =在点P 处的切线方程为.14．已知数列{}n a 满足11a =，且对于任意2n ≥，都存在{}1,2,,1k n ∈-L ，使得4n k a a =+，则4a 的所有可能取值构成的集合M =；若{}n a 的各项均不相等，把半径为123,,a a a （单位：cm ）的三个小球放入一个正方体容器（容器壁厚度忽略不计），则该正方体容器的棱长最小值为cm .四、解答题15．记ABC V 的内角,,A B C 的对边分别为,,a b csin cos C c c A -=.(1)求A ；(2)若6a b c +=，求ABC V 的面积.16．已知数列{}n a 的前n 项和为n S ，且3212321212121n n n a a a a ++++=---- .(1)求n S ；(2)设3642n n nS b +=，若数列{}n b 的最小项为m b ，求m .17．如图，已知平行六面体1111ABCD A B C D -的底面是菱形，2AB =，AC BD O = ，11A AB A AD ∠=∠.(1)证明：1AA BD ⊥；(2)若1122AA AO ==，π3BAD ∠=，点P 在平面1AB C 内，且BP ⊥平面1AB C ，求BP 与平面ABCD 所成角的正弦值.18．已知函数()()()2ln 1f x ax x x a =-++∈R .(1)当0a ≥时，讨论()f x 的单调性；(2)()()()1e 21xg x f x x x =++--.（i ）当0a =时，求()g x 的最小值；（ii ）若()0g x ≥在[)0,+∞上恒成立，求a 的取值范围.19．已知()f x 为定义域M 内的连续函数，()f x '为其导函数，常数a M ∈，若各项不相等的数列{}n a 满足n a M ∈，1a a >，()()()1n n n f a f a f a a a+'-=-，则称{}n a 为()f x 的“拉格朗日数列”，简记为“()L a -数列”.(1)若函数()ln g x x =，数列{}n b 是()g x 的“()1L -数列”，且1e b =.（i ）求2b ，3b ；（ii ）证明：{}n b 是递减数列；(2)正项数列{}n c 是函数()36sin h x x x =+的“()L c -数列”，已知()1,n n c c c +∈，记{}n c 的前n 项和为n S ，证明：0c >时，()112n n S c n c c +≥-+.。

山东省潍坊市2024-2025学年高一上学期11月期中考试语文试题注意事项：1．答卷前，考生务必将自己的姓名、考生号和座号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。

一、现代文阅读（35分）（一）现代文阅读Ⅰ（本题共5小题，19分）阅读下面的文字，完成1～5题。

材料一：费孝通曾指出，在中国传统社会里，我们对于社会的认识主要来自书本与经典，所以才有所谓“秀才不出门，能知天下事”“半部《论语》治天下”的认知格局。

强调书本以及经典的重要性，当然很有道理，因为经典记载了过去我们认识世界的经验和总结。

但是，费孝通认为若是只通过书本来认识社会的话，应该有两个逻辑前提：一个是过去出版的著作里包含了人类所有的知识，因此任何知识都可以从书本中得到了解；另一个是过去的书中应该包含了所有解决问题的方法。

实际上，这两个逻辑前提是不成立的。

特别是在一个发展日益迅速、竞争强度日渐增加、社会状况日趋复杂的时代，我们的认知也处在知识爆炸的情境中。

这样的中国，无论是知识的积累，还是解决问题的需求，要面对的要素越来越多且越来越复杂。

所以按照费孝通的说法，认识和治理中国这样的社会，应该以社会调查作为根本途径。

费孝通认为社会调查是认识论的起点。

这是符合历史唯物主义的，是从实践到理论，又从理论回到实践的一个认识过程。

我们要认识社会，必须从实际调查出发，没有调查就没有发言权。

费孝通真正将社会调查变成理论体系，形成社会学调查，与他攻读博士学位阶段严格的学术训练有关，这使他可以通过专业的眼光去看待过去做过的社会调查。

费孝通强调了社会调查与社会学调查的区别。

社会调查是描述性的，告诉我们社会是什么，《江村经济》这本书就是一种描述性的著作，也就是如今所说的民族志。

山东省各地2014届高三上学期期中考试试题分类汇编应用题1、（德州市2014高三期中）统计表明某型号汽车在匀速行驶中每小时的耗油量y （升）关于行驶速度x （千米/小时）的函数为3138(0120)12800080y x x x =-+<<。

（1）当64x =千米/小时时，要行驶100千米耗油量多少升？（2）若油箱有22.5升油，则该型号汽车最多行驶多少千米？解：（1）当64x =千米/小时时，要行驶100千米需要100256416=小时 要耗油(313256464811.95()1280008016⨯-⨯+⨯=）升 （2）设22.5升油该型号汽车可行驶a 千米，由题意得313(822.512800080a x x x -⨯+⨯=） 222.518312800080a x x ∴=+- 设2183()12800080h x x x =+-则当()h x 最小时，a 取最大值， 由332221880()6400064000x h x x x x -'=-=令()080h x x '=⇒= 当(0,80)x ∈时，()0h x '<，当(80,120)x ∈时，()0h x '>故当(0,80)x ∈时，函数()h x 为减函数，当(80,120)x ∈时，函数()h x 为增函数所以当80x =时，()h x 取得最小值，此时a 取最大值为222.5200183801280008080a ∴==⨯+-答：若油箱有22.5升油，则该型号汽车最多行驶200千米。

2、（临沂市2014高三期中）某厂以x 千克/小时的速度匀速生产某种产品（生产条件要求110x ≤≤），每小时可获得的利润是310041x x ⎛⎫+- ⎪⎝⎭元. （I ）要使生产该产品1小时获得的利润不低于1200元，求x 的取值范围；（II ）要使生产120千克该产品获得的利润最大，问：该厂应该选取何种生产速度？并求此最大利润.3、（青岛市2014高三期中）某连锁分店销售某种商品,每件商品的成本为4元,并且每件商品需向总店交(13)a a ≤≤元的管理费,预计当每件商品的售价为(79)x x ≤≤元时,一年的销售量为2(10)x -万件．（Ⅰ）求该连锁分店一年的利润L （万元）与每件商品的售价x 的函数关系式()L x ;（Ⅱ）当每件商品的售价为多少元时,该连锁分店一年的利润L 最大,并求出L 的最大值． 解: （Ⅰ）由题得该连锁分店一年的利润L （万元）与售价x 的函数关系式为2()(4)(10),[7,9]L x x a x x =---∈. ……………………………3分（Ⅱ）2()(10)2(4)(10)L x x x a x '=-----(10)(1823),x a x =-+- …………………………………………6分令'()0L x =，得263x a =+或10x = ……………………………8分 20213,6833a a ≤≤∴≤+≤Q . ①当2673a +≤，即312a ≤≤时， [7,9]x ∴∈时，()0L x '≤，()L x 在[7,9]x ∈上单调递减，故max ()(7)279L x L a ==- ……………10分 ②当2673a +>，即332a <≤时， 2[7,6]3x a ∴∈+时，'()0L x >；2[6,9]3x a ∈+时，()0L x '< ()L x ∴在2[7,6]3x a ∈+上单调递增；在2[6,9]3x a ∈+上单调递减， 故3max 2()(6)4(2)33a L x L a =+=- ……………12分答:当312a ≤≤每件商品的售价为7元时,该连锁分店一年的利润L 最大,最大值为279a -万元； 当332a <≤每件商品的售价为263a +元时,该连锁分店一年的利润L 最大,最大值为34(2)3a -万元. ……………13分4、（潍坊市2014高三期中）如图，某广场要划定一矩形区域ABCD ，并在该区域内开辟出三块形状大小相同的矩形绿化区，这三块绿化区四周和绿化区之间设有1米宽的走道。

试卷类型：A2024—2025学年度第一学期期中质量检测七年级英语试题2024. 11注意事项：1. 本试题满分80分, 考试时间为60分钟；2. 答卷前务必将试题密封线内及答题卡上面的项目填涂清楚。

所有答案都必须涂、写在答题卡相应位置, 答在本试卷上一律无效；3. 非选择题用黑色钢笔或碳素笔作答。

友情提示：相信自己, 祝你成功!一、词汇考查(共10小题；每小题1分, 满分10分)根据音标或首字母提示写出句子中的正确单词。

1. Are you r ______ for your new life?2. I thought about the m ______of the sentence.3. Emma put the notebook on a s________ and left.4. It's p _________to smile back when someone gives you a smile.5. Mr Li gave us two pieces of a __________: make good use of time and try our best.6. Olive was ________ /a'weik/at midnight.7. Remember this: no pain, no _________ /gein/.8. I have two pencils, an ___________/i'reizə/and a ruler in it.9. The sky__________/'sʌdnli/went dark and it started to rain.10. We do morning exercises and ___________ /həʊld/school sports events here.二、句型转换(共5小题；每小题2分, 满分10分)11. It is my pencil. (变为复数句)_________________________________________________________________12. There is some food on the table. (变为一般疑问句)_________________________________________________________________13. are, problems, the, What, students', new (连词成句)_________________________________________________________________14. sea, through, towards, will, the, go, You, storms (连词成句)_________________________________________________________________15. magic, like, Jimmy, stories, doesn't, writing(连词成句)_________________________________________________________________三、完成句子(共5小题；每空1分, 满分10分)16. __________ _______ (事实上), this is my first lesson for all of you.17. Please design a _________ _________ (校服).18. We want to __________ some thoughts _________ (和…. . . 分享. . . . . . )you.19. Dad and Mum always________ our boat _________ (保护. . . 免受)winds.20. It is OK to ________ __________ (指出)the teacher's mistakes.四、选词填空(共5小题；每小题1分, 满分5分)阅读短文, 从方框中选择恰当的词填空, 使短文通顺完整。

高三生物2013.11 注意事项：1．本试题分第I卷和第Ⅱ卷两部分。

第I卷为选择题，共45分；第Ⅱ卷为非选择题，共55分。

满分100分，考试时间为90分钟。

2．答第I卷前务必将自己的姓名、考号、考试科目涂写在答题卡上。

考试结束，试题和答题卡一并收回。

3．第I卷每题选出答案后，都必须用2B铅笔把答题卡上对应题目的答案标号(ABCD)涂黑，如需改动，必须先用橡皮擦干净，再改涂其它答案。

第I卷(选择题共45分)本卷共30小题．每小题1．5分，共45分。

每小题只有一个选项最符合题意。

1．下列各组化合物中，所含元素种类相同的是A．纤维素和甲状腺激素B．三磷酸腺苷和磷脂C．叶绿素和核糖核酸D．血红蛋白和胰岛素2．下列有关细胞物质组成和结构的叙述，正确的是A．在人体活细胞中碳元素的含量最高B．线粒体可在有氧条件下将葡萄糖氧化分解为CO2和水C．消化腺细胞与口腔上皮细胞相比，前者细胞核上核孔数目更多D．放线菌比根霉菌具有更为发达的生物膜系统3．某研究人员分析了小白兔组织、小麦组织、大肠杆菌、T2噬菌体四种样品的化学成分，以下结论错误的是A．含有糖原的样品是小白兔组织B．只含有蛋白质和DNA的样品是T2噬菌体C．含有纤维素的样品是小麦组织和大肠杆菌D．既含有DNA，又含有RNA的样品是小麦组织、小白兔组织、大肠杆菌4．电子显微镜下观察小麦根尖细胞，不可能观察到A．线粒体B．内质网C．叶绿体D．核糖体5．下图表示某分泌蛋白的合成、加工和运输过程，①②③表示细胞器，a、b、c、d表示过程。

下列说法不正确的是A．物质X可能是细胞呼吸酶B．磷脂总量最多的细胞器是①C．③的内膜中蛋白质种类和含量比外膜多D．d过程体现了细胞膜具有流动性的特点6．下图中四条曲线表示物质进入人体成熟红细胞的方式，研究发现物质M与曲线②和③相符，则物质M可能是A．甘油B．氨基酸C．葡萄糖D．Fe2+7．下列生理过程的反应式错误的是8．下列有关酶的叙述，正确的是A．基本组成单位是氨基酸或核糖核酸B．温度过高或过低时都会失活C．通过为底物供能加快化学反应速度D．在生物体内仅具有催化功能9．将某种植物不同部位的细胞(初始状态都如右图所示)置于相同浓度的蔗糖溶液中，一段时间后四个部位的细胞状态如下图，则细胞液浓度最高的是10．下列有关ATP的叙述，不正确的是A．ATP直接为生命活动供能的过程消耗水B．人体在饥饿状态下，细胞中ATP的含量仍可保持动态平衡C．ATP脱去两个磷酸基团后可以参与合成RNAD．ATP的合成与水解所需的酶相同11．下列有关细胞生命历程的说法，错误的是A．伴随细胞的生长，与外界进行物质交换的效率逐渐提高B．人体内绝大多数高度分化细胞的DNA相同但遗传信息执行情况不同C．癌细胞的表面蛋白质减少，黏着性降低D．细胞凋亡过程中有新蛋白质合成，体现了基因的选择性表达12．下列关于高等绿色植物细胞代谢的叙述，正确的是A．光能经过光反应转化为ATP中活跃的化学能和热能B．与有氧呼吸有关的酶只分布在线粒体内膜和线粒体基质中C．细胞呼吸可为光合作用暗反应提供ATP和CO2D．光合作用产生的O2被细胞呼吸利用至少要穿过8层磷脂分子13．某同学用玻璃槽做成如图所示的实验装置，甲、乙两槽间的隔板中央圆圈处为玻璃纸(允许单糖和水分子自由通过)。

潍坊市高三上学期期中考试试题2019.11第二部分阅读理解（共两节，满分50分）第一节（共16小题;每小题2.5分，满分40分）阅读下列短文，从每题所给的A、B、C和D四个选项中，选岀最佳选项。

AGot 48 hours to relax yourself? You'll maximize your time and fun at "these amazing destinations.Alabama:Gulf ShoresFor relaxing weekend getaways in Alabama, head south to the white-sand beaches of Gulf Shores. Get on the water for some deep-sea fishing. History lovers can get a lesson in the Civil War along a 90-mile path with more than a dozen historic sites.Alaska:HomerHow does fresh seafood sound? You'll find it in Homer, located in south-central Alaska, which is either a short flight or a great four-hour scenic drive on the Seward Highway from Anchorage. Get your fill of fishing in Kachemak Bay, explore local trails and enjoy exceptional birding in this remote natural setting.Arizona:SedonaAdventurous types can get their fill in the magical red rocks of Sedona, which is said to be the location of healing centers of energy. People flood to the area to meditate, practice yoga and more. Sedona also offers fantastic hiking, off-roading, biking, and even climbing.Arkansas:Hot SpringsRelax and renew in Arkansashot waters in Hot Springs, an art community with Art Deco architecture that is also the home of Hot Springs National Park. Make time for a visit to Bathhouse Row, a National Historic Landmark of eight bathhouses, and Superior Bathhouse Brewery, which produces beer made of the hot spring water inside the National Park. Don't miss these National Parks that are off the beaten path.21. Which activity can be found in both Gulf Shores and Homer?A. Going fishing.B. Taking a flight.C. Enjoying beautiful birds.D. Visiting historic sites.22. Which destination is popular with hikers?A. Gulf Shores.B. Homer.C. Sedona.D. Hot Springs.23. What is Superior Bathhouse Brewery known for?A. Its self-made beer.B.Its architecture.C. Its bathhouses.D.Its paths.24. Where is this text most likely from?A. A diary.B. A survey.C. A novel.D. A magazine.BI was excited to take my rescue Beagle（比格犬）to the river the other day, as it’s within walking distance of our new home. Having spent her first six years in a cage, Georgie had never seen a body of water, and I wanted to get there before sunset to watch her experience it.I was growing increasingly impatient about all the stops her little Beagle nose required. She inspected the grass, dirt, and trees, and licked（舔）whatever was stuck to the road. These were all new discoveries for her, and she took her time studying them.When I accepted that it was fruitless to hurry her along, I whipped out my cell phone and began texting. I thought that I needed something to do while Georgie was slowing us down. Then, for some reason, I heard the cicadas（蝉）,and I remembered that the sound of cicadas is my favorite sound in the world. That awakened something buried within me that longed for the simple pleasures that had been replaced by technology.I made a conscious decision to be present, and to enjoy the journey to the river Just like Georgie. Now, I admired the flowers and the winding ivy on our path. I smelled the grass and the flowers, and the dirt and the air. I treasured each one equally, as if discovering them for the first time.As we neared our destination, I realized something even more important: It didn't matter if we even reached the river. Why must there always be a destination? Georgie had no idea that we had a destination. She was present for the journey, and she savored every bit of that sweet experience. There was no race and no finish line.Now I'm not sure who rescued whom.25. Why did the author take out the cellphone?A. To kill time.B. To awaken the simple pleasures.C. To hurry the dog along.D. To record the sound of cicadas.26. What is the fourth paragraph mainly about?A. How I enjoyed the journey.B. What I saw on the way.C. Why I traveled to the river.D. Whether I reached my destination.27. What does the underlined word“savored” in the fifth paragraph mean?A. Smelt.B. Expected.C. Enjoyed.D. Deserved.28. What is the best title for the text?A. Say Goodbye to Cell PhonesB. Think like a DogC. Learn to Get along with PetsD. Travel toward DestinationsCMosquitos can infect you with Malaria, yellow fever, and dengue. All can kill you. So we spray large areas with pesticides（杀虫剂）. But this negatively affects the whole ecosystem and your health. Now there's a new device, which claims to keep you and your loved one’s mosquito free.The device was designed bytwo German engineers who werefed up with being bitten whilecamping. They found solutionscontaining ' DEET' — a damagingchemical which could be bothharmful to us and the environment—worked best, but still only for45 minutes maximum and that thetraditional anti-pest devices didn'twork well! So, they designed thebrand new type of bug killer, Moskinator.Instead of trying to keep the mosquitos away, it actually attracts them, then catches and kills them! It lures（引诱）the flies and mosquitos near using a safe, completely harmless LED light. The flying pests（害虫）can't resist the UV-PT light! Next to the LED light is a powerful, yet efficient reverse fan that sucks（吸）the insects through a one-way trapdoor. Once trapped in the drying basket they are dried to death by the fan a matter of 2 minutes. You can simply empty the dead bugs into the trash, without even touching them.Does Moskinator really work in real life? We had some of our customers try it out in their homes. The results speak for themselves:“My wife would complain she had a headache in the mornings whenever we used chemical pesticides. Now we use our Moskinator, and we wake up bite and headache free!”—Shane Maguire “This Moskinator really works and you don't have to worry abo ut breathing any harmful chemicals.”—Jake Shearer29. What is the name of the new device?A. DEET.B. Maskinator.C. Malaria.D. UV-PT.30. What is the function of the LED light?A. It gives off natural light.B. It attracts flying pests.C. It sucks pests inside.D. It dries pests to death.31.Which of the following best describes the new device?A. Innovative and effective.B. Traditional and cheap.C. Power-saving and portable.D. Chemical free and soundless.32. What is the purpose of writing the text?A. To educate the readers.B. To entertain the readers.C. To introduce a new device.D. To evaluate a new device.DStep into Moving to Mars, an exhibition of Mars mission and colony design at London's Design Museum, and immediately you have good reasons not to move there.Frightening glowing wall-texts announce that Mars wasn’t made for you；that there is no life and precious little water；that, dressed in a spacesuit, you will never touch, taste or smell the planet you now call “home”. As Lisa Grossman wrote for New Scientist a couple of years ago, “What's different about Mars is that there is nothing to do there except try not to die”.It is an odd beginning for such a celebratory exhibition, but it provides a valuable, dark background against which the rest of the show can sparkle（闪耀）一a show that is, as its chief manager Justin remarks，“not about Mars ；this is an exhibition about people”.Moving along, there is a quick yet clear flash through what the science-fiction writer Robinson calls “the history of Mars in the human mind”. A Babylonian clay tablet and a Greek vase speak to early ideas about the planet. A poster for the original Total Recall film reminds us of Mars's psychological threat.The main part of the show is our current plans for the Red Planet. There are real spacesuits and models of 3D-printed Martian settlements and suitable clothing and furniture. Mission architectures and engineering sketches line the walls. Real hammers meant for the International Space Station are wall-mounted beside a low-gravity table that has yet to leave, and may indeed never leave, Earth.This, of course, is the great strength of approaching science through design: reality and assumption can be given equal visual weight, drawing us into an informed conversation about what it is that we actually want from a future on Mars.33. What is the text mainly about?A. How to move to Mars.B. How to survive on Mars.C. What preparations we made for Mars.D. What the exhibition of Mars truly tells us.34. What can we learn from Lisa Grossman?A. It’s impossible to live on Mars.B. It's no good settling on Mars.C. You have nothing to do living on Mars.D. You can live on Mars in spacesuit.35. What does the exhibition focus on?A. The current plans for Mars.B. The advantages of living on Mars.C. The early ideas about Mars.D. The history of Mars in the human mind.36. What does the author want to tell us in the last paragraph?A. An experience.B. An opinion.C. A fantasy.D. A solution.第二节（共5小题;每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

山东省潍坊市北约联盟2023-2024学年高二上学期11月期中英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读理解Stauer Urban BluePacked with high-end watch performance and style, without the high-end price label.It’s everything a high-end watch should be: Solid stainless steel (不锈钢) and real leather construction. What is the most attracting part is its precision timing that’s accurate to four seconds a day — that’s more precise than a 27 jewel (宝石) automatic watch priced at over $ 6,000. And, good looking — with simple, clean lines and a striking metallic blue face. Your great escape from the over-priced watch.At Stauer, we go directly to the source (cutting out the middleman), and engineer our own watch designs. This means we can offer a top quality timepiece that happens to only cost the same as two well-made cocktails at your favorite bar. So, while we’re busy innovating the watch industry to bring you more real value, you can avoid over-priced watches with the Urban Blue.Your satisfaction is 100%promised.Wear the Urban Blue for 30 days. If you’re not convinced that you achieved excellence for less, send it back for a refund of the item price. The Urban Blue is one of our fastest sellers. It takes six months to engineer this watch, so don’t wait. Take a stand against overpriced watches in perfect style.Limited to the first 1900 responders to this ad only.Don’t miss out…call today!Stauer Urban Blue Watch $199Offer Code Price $29+S&P Save $1701-800-333-2045Your Insider Offer Code: UBW598-02You must use this insider offer code to get our special price.Special price only for customers using the offer code and the price on without your offer code.1．What’s the best thing about Stauer Urban Blue?A．Its raw materials.B．Its accurate timekeeping.C．Its advanced design.D．Its attractive appearance.2．Why is the Urban Blue priced so low?A．It is promoted at a local bar.B．It is produced at a very low cost.C．It is machine-made in large quantities.D．It is self-designed from its own factory. 3．How much should you pay for an Urban Blue Watch if you are the 2000th customer?A．$29.B．$141.C．$170.D．$199.The 97-year-old botanist, Margaret Bradshaw, is the chief caretaker of Teesdale’s rarest flowers. “Everything about Teesdale is unique,” says Bradshaw with pride — and the authority of someone who has just written a 288-page book on the subject. Bradshaw has been recording rare plants here since the early 1950s and has witnessed great decreases.Her data was the first to prove that — and the need to do something about it.The main reason for the decrease of these plants is an unusual one. The number of sheep in Teesdale had been reduced by half by 2000, as the uplands were generally believed there were too many sheep. Bradshaw says while sheep are hunted in some upland areas, reducing herding in Teesdale has been destructive. Longer grass overshadows the delicate (娇嫩) flowers, taking away the light they need to grow. As a result of her findings and her work with farmers who herd the land — as well as Natural England, which manages it — sheep numbers are increasing and the timing of herding is being carefully managed. This has led to the partial recovery of some plants.At 93, she set up Teesdale Special Flora Research and Conservation Trust to record rare plants. A keen horse rider, at 95, she did a 55-mile horse journey across Teesdale, raising almost $ 10,000 for the Trust. When asked about the secret to longevity, “Just keep going,” she says.“Keep at it.Don’t sit down and just watch TV.”“I recognise I’m getting older and I’ve been trying to get more people to take over and do the records. They don’t believe I won’t be here forever,” Bradshaw says. Despite Bradshaw’s guardianship of this land, and the love and energy she has put into saving it, the future here is unknown. The last words of her book speak to this endless loss. “This is our heritage, this unique plant species, mine and yours,” she writes. “In spite of trying, I have failed to prevent its decrease, now it is up to you.”4．What is Margaret Bradshaw’s main work with the rare plants?A．Writing a book.B．Preventing their decrease.C．Collecting their data.D．Directing farmers to grow plants. 5．What contributed to the reduction of rare plants in Teesdale?A．Not enough sheep.B．Lack of fund.C．Poor management.D．Too many sheep.6．What can we learn from Bradshaw’s words in the last paragraph?A．Unity brings strength.B．Many hands make light work.C．Youthful at heart, regardless of age.D．Advance what the forerunners began. 7．Which might be a suitable title for the text?A．Teesdale:Home to Rare Plants B．Battling for Teesdale’s Wildflowers C．Bradshaw:A97-Year-Old Gardener D．Exploring Teesdale’s Unique Plants“Mek thisen at ‘ome, lass,” Dimbleby said in the Yorkshire dialect to the group of nearly two-dozen mostly grey-haired students, when the newspaper reporter arrived at his class. This class is part of an increasingly popular movement to revive the local dialect, once spoken by millions but now has almost died out.“We’ve got to get more young people involved,” says Dimbleby. The feedback so far has been positive, but the class is mostly made up of retirees. He is hoping to run another course on Saturday mornings, when younger people are available. “Most people’s experience of dialect is what they heard from their grandparents. They’ve never read it or written it, so it’s been a new experience for them.”There are some challenges in making it accessible though. As a spoken dialect with many variations (变体), the written aspect is various. How to correctly spell is a problem, because different people may pronounce it in different ways. For example, “nothing” in Yorkshire is “nowt”, but some people would read “out”, while in most of Yorkshire it has more of an “ohwt” sound that is hard to get across without resorting to the rules of pronunciation.“I’ve always been fascinated by the Yorkshire dialect,” says Joanna Kaye, a retired host. “When we were children, we learned all Queen’s English. My mother, although she taught me to speak properly, she could speak in the dialect.”Jean Crampton, here for the first time, has a notebook to write down words used by herparents, who died more than 10 years ago. “When I dressed up for the first time to go to grammar school, my dad said ‘dun’t she look grand’,” she said. “And that’s why I’ve come, to be reminded of them. After they died, that immediacy fades, and language is one of the things that can help bring it back into your mind, which is brilliant.”8．What is Dimbleby’s suggestion for the dialect class?A．More activities to revive the dialect.B．More courses provided for retirees.C．Greater participation of young people.D．Greater enthusiasm in elderly learners. 9．What does the underlined phrase ”resorting to“in paragraph 3 mean?A．Turning to B．Writing down.C．Summing up.D．Speaking of. 10．What makes Joanna Kaye interested in the Yorkshire dialect?A．The Queen’s charm.B．His mother’s influence.C．His childhood dream.D．The requirement for a host.11．Why did Jean Crampton go to the dialect class?A．To pass on the Yorkshire dialect.B．To share her father’s experiences.C．To restore memories from her parents.D．To learn the Yorkshire dialect grammar.Scientists once considered sleep to be like a shadow getting drawn over a window between the brain and the outside world: when the shadow is closed, the brain stops reacting to outside stimuli (刺激).A study published on 12 October in Nature Neuroscience suggests that there might be periods during sleep when that shade is partially open.A few years ago, Oudiette, a cognitive scientist in France, began questioning the definition of sleep after she and her team conducted an experiment in which they were able to communicate with people who are aware that they are dreaming while they sleep — otherwise known as clear dreamers. During these people’s dreams, experimenters were able to ask questions and get responses through eye and facial-muscle movements.Karen Konkoly, who was a co-author on that study, says “it was a big open question in our minds whether communication would be possible with non-clear dreamers.”So Oudiette continued with the work. In her latest study, she and her colleagues observed 27 people with narcolepsy (嗜睡症) — characterized by daytime sleepiness and a high frequency of clear dreams — and 22 people without the condition. While they were sleeping, participants were repeatedly asked to frown or smile. All of them respondedaccurately to at least 70% of these tips. Overall response rates were higher for all participants during REM (rapid eye movement) sleep, when the deepest sleep occurs but the brain remains quite active, than during other sleep stages.This study is part of a larger advance in the field of sleep research, says Mélanie Strauss,a cognitive scientist in Brussels. Researchers are moving away from monitoring sleep mainly with EEG (脑电图) and towards “fine grained” approaches that combine EEG with various tasks and stimuli — a strategy that could help to cast light on specific diseases or conditions, she says.12．What did scientists think of sleep in the past?A．It happens during different periods.B．It is a shadow formed within the brain.C．It suggests a weakening in the brain’s functions.D．It means a stop in brain’s response to the outside.13．How did clear dreamers react to the experimenters?A．By moving their face and eyes.B．By repeating the questions.C．By providing accurate answers.D．By talking about their dream. 14．What can we learn about the participants in the latest study?A．They like sleeping in the daytime.B．They remember their dreams clearly.C．They do exactly as they’re told during sleep.D．They show higher response in the deepest sleep.15．What does Mélanie Strauss mainly tell us in the last paragraph?A．The limitation of using EEG.B．A cure for specific diseases.C．The progress of the sleep research.D．A method of detecting sleep二、七选五Play is part of a parent-child friendshipPlay is a crucial element of friendship with children. With babies and very young children, we parents play with them,engaging in back-and-forth activities like hide-and-seek and tower building. 16 They pull us into play, and we feel pleasure.But delight hits a setback when our child learns how to say “no”. From this point onward—into adolescence —we focus on getting a child to behave. 17 It’s parenting belief, not specifically defined, but often translated into “because I said so”.While parents must help children learn to meet adult demands, when we limit our parenting focus to “command and control” we can crowd out “delight”. 18 If we can leave space for child-driven activities —without adult demands —we can return to delight. These interactions expand our relationship and help our child learn self-regulation and communication. It’s not the same as peer friendship, but we are being a friend.19 The key is following your child’s lead. As they get older and they are deep into other things, friendly play might include teaching you to play their favorite video game or share their YouTube. Your child won’t want you to be their primary friend in these spaces, but being curious and kind is a way to be a friend and a parent. You will be an uncool friend who will never fully get it, but you will be a friend.20 Especially as they get older, they will set more limits on it. But if you make a habit of friendly play with your child, that space will always be there between you, and it will vastly expand your parenting toolbox.A．They often start the game, and we join in.B．Kids won’t always want you to play with them.C．We are told in many ways to be the parent, not the friend.D．Highlighting our role as taskmasters limits our parenting tools.E．It can be a powerful tool for developing parent-child relationships.F．This looks different from babyhood to youth but it’s always essential.G．Friendly play with children can range from building blocks to hide-and-seek.三、完形填空I can remember the first time it happened. Class was getting out of hand. Our teacher,gave up and informed we could take a rest. Instantly, the classroom erupted with studentsAs parents started to arrive, our teacher . 25 the chattering children with a half-hearted smile. But when my mother 26 ,she commented, “Your daughter is such a nice girl in class. She’s a 27 .” And there it was: my first “nice”.Whether or not I knew it, a 28 had been set for my future behavior, though my teacher probably didn’t mean it.As years passed, I became 29 in the art of nice. But as much as I loved praise, I was too 30 to dare be anything but nice.People rarely see the Real Me, because the Real Me is 31 .She doesn’t clean her room as often as she should. She’s 32 at making decisions. Nice became a shelter I 33 my real feelings. “It’s like living a. 34 life,” I thought. I am 35 with being nice. I want to be the best of myself, not the nicest.21．A．sight B．focus C．spirit D．preference 22．A．moving B．changing C．adjusting D．leaving 23．A．excitedly B．hopelessly C．quietly D．nervously 24．A．positive B．normal C．complex D．strange 25．A．picked out B．handed over C．showed off D．drove away 26．A．arrived B．demanded C．followed D．insisted 27．A．treasure B．legend C．star D．champion 28．A．plan B．style C．standard D．goal 29．A．poorly-adapted B．well-trained C．annoyed D．confused 30．A．active B．honest C．eager D．scared 31．A．shy B．perfect C．messy D．unique 32．A．terrible B．quick C．wise D．careful 33．A．managed B．shared C．expressed D．hid 34．A．simple B．comfortable C．double D．busy 35．A．familiar B．tired C．confident D．pleased四、用单词的适当形式完成短文阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

高三数学（文）本试卷共5页，分第I 卷（选择题）和第II 卷（非选择题）两部分，共150分.考试时间120分钟.第I 卷（选择题 共60分）注意事项：1.答第I 卷前，考生务必将自己的姓名、准考证号、考试科目用铅笔涂写在答题卡上.2.每题选出答案后，用2B 铅笔把答题卡对应题目的答案标号涂黑.如需改动，用橡皮擦干净后，再改涂其它答案标号.一、选择题：本大题共12小题，每小题5分，共60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.设x Z ∈，集合A 为偶数集，若命题:,2p x Z x A p ∀∈∈⌝，则为 A.,2x Z x A ∀∈∉ B.,2x Z x A ∀∉∈ C.,2x Z x A ∃∈∈D.,2x Z x A ∃∈∉2.设集合{}{}{}1,2,3,4,5,,,A B C x x b a a A b B ====-∈∈，则C 中元素的个数是 A.3B.4C.5D.63.已知幂函数()y f x =的图象过点()21,log 222f ⎛ ⎝⎭，则的值为A.12B.12-C.1-D.14.在ABC ∆中，内角A 、B 的对边分别是a 、b ，若cos cos A bB a=，则ABC ∆为 A.等腰三角形 B.直角三角形C.等腰三角形或直角三角形D.等腰直角三角形5.若当x R ∈时，函数()()()011xf x a a a f x =>≠≤且满足，则函数()log 1a y x =+的图象大致为6.已知110a b<<，给出下列四个结论； ①a b <；②a b ab +<；③a b >；④2ab b <.其中正确结论的序号是 A.①② B.②④C.②③D.③④7.等差数列{}n a 的前20项和为300，则468131517a a a a a a +++++等于A.60B.80C.90D.1208.已知函数()()2,021,0x a x f x a R x x ⎧-≤=∈⎨->⎩，若函数()f x 在R 上有两个零点，则a 的取值范围是A.(),1-∞-B.(],1-∞C.[)1,0-D.(]0,19.已知数列{}n a 的前n 项和为()*2n n n S S a n n N +=∈，且，则下列数列中一定是等比数列的是 A.{}n aB.{}1n a -C.{}2n a -D.{}2n a +10.已知函数()()sin 03f x x πωω⎛⎫=+> ⎪⎝⎭的最小正周期为π，将函数()y f x =的图象向右平移()0m m >个单位长度后，所得到的图象关于原点对称，则m 的最小值为 A.6π B.3π C.512π D.56π11.已知函数()2sin f x x x x =+，,22x ππ⎛⎫∈-⎪⎝⎭，则下列式子成立的是 A.()13122f f f ⎛⎫⎛⎫-<<⎪ ⎪⎝⎭⎝⎭B.()13122f f f ⎛⎫⎛⎫<-<⎪ ⎪⎝⎭⎝⎭ C.()13122f f f ⎛⎫⎛⎫<<-⎪ ⎪⎝⎭⎝⎭D.()31122f f f ⎛⎫⎛⎫<-<⎪ ⎪⎝⎭⎝⎭12.不等式2220x axy y -+≤对于任意[][]1,21,3x y ∈∈及恒成立，则实数a 的取值范围是A.a ≤B.a ≥C.113a ≥D.92a ≥第II 卷 （非选择题 共90分）注意事项：1.将第II 卷答案用0.5mm 的黑色签字笔答在答题纸的相应位置上.2.答卷前将密封线内的项目填写清楚.二、填空题：本大题共4小题，每小题4分，共16分13.计算121lg lg 251004-⎛⎫-÷= ⎪⎝⎭____________.14.若1sin 2cos tan 424sin cos πθθθθθ+⎛⎫-==⎪-⎝⎭，则__________. 15.已知一元二次不等式()0f x <的解集为()12202x xx f ⎛⎫<<> ⎪⎝⎭，则的解集为_______.16.给出下列命题①若()y f x =是奇函数，则()y f x =的图象关于y 轴对称；②若函数()f x 对任意()()41x R f x f x ∈⋅+=满足，则8是函数()f x 的一个周期； ③若log 3log 3001m n m n <<<<<，则； ④若()[)1x af x e-=+∞在，上是增函数，则1a ≤.其中正确命题的序号是_____.三、解答题：本大题共6小题，共74分.解答应写出文字说明，证明过程或演算步骤. 17.（本小题满分12分）已知全集U=R ，集合[]{231,0,2,2A y y x x x B x y ⎧⎫==-+∈==⎨⎬⎩⎭. （I ）求()U C A UB ； （II ）若集合212C x x m ⎧⎫=+≥⎨⎬⎩⎭，命题:p x A ∈，命题:q x C ∈，且命题p 是命题q 的充分条件，求实数m 的取值范围.18.（本小题满分12分）已知函数()()2sin sin sin .2f x x x x x π⎛⎫=+-+⎪⎝⎭（I ）求函数()f x 的最大值和单调增区间；（II ）△ABC 的内角A 、B 、C 的对边分别为a 、b 、c ，已知2,2sin 3sin 2C f c B A ⎛⎫===⎪⎝⎭且，求△ABC 的面积.19.（本小题满分12分）如图，某广场要划定一矩形区域ABCD ，并在该区域内开辟出三块形状大小相同的小矩形绿化区，这三块绿化区四周和绿化区之间设有1米宽的走道.已知三块绿化区的总面积为800平方米，求该矩形区域ABCD 占地面积的最小值.20.（本小题满分12分） 解关于x 的不等式()11ax a a R x+≥+∈.21.（本小题满分12分）已知公比为q 的等比数列{}n a 是递减数列，且满足123123131,.927a a a a a a ++== （I ）求数列{}n a 的通项公式； （II ）求数列(){}21nn a - 的前n 项和nT；22.（本小题满分14分）已知()()()()()()2ln 1,,1f x a x g x x bx F x f x g x =-=+=+-，其中,a b R ∈.（I ）若()()y f x y g x ==与的图象在交点（2，k ）处的切线互相垂直，求a,b 的值； （II ）当2,0b a a =->时，求()F x 的最大值；（III ）若x=2是函数F （x ）的一个极值点，0x 和1是F （x ）的两个零点，且()0,1,x n n n N ∈+∈，求n ；。

2023年最新的2023潍坊高三11月考2023潍坊高三11月考潍坊市2023-2023学年上学期期末考高三英语2023.1本试卷分第I卷(选择题)和第Ⅱ卷(非选择题)两部分。

第I卷1至10页，第Ⅱ卷11至12页。

满分为150分。

考试用时为120分钟。

第I卷(共100分)注意事项：1.答第I卷前，考生务必将自己的姓名、准考证号、考试科目涂写在答题卡上。

2.每小题选出答案后，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。

不能答在试卷上。

第一部分听力(共两节，满分30分)该部分分为第一、第二两节。

注意：回答听力部分时，请先将答案标在试卷上。

听力部分结束前，你将有两分钟的时间将你的答案转涂到客观题答题卡上。

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。

每段对话后面有一个小题，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秒钟时向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 man do firstA.Fix the bike.B.Have a coffee.C.Clean the windows.2.What is the date todayA.May lst.B.May 2nd.C.May 3rd.3.Where will the speakers probably buy something to eatA.At a cafe.B.At a service station.C.At the airport.4.How does the woman feelA.Proud.B.Pleased.C.Angry.5.What is the conversation mainly aboutA.A film.B.A book.C.A bookcase.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。

试卷类型：A高三阶段性调研监测考试数学试题2024.11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.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.考试结束后，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个小题，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{}{}|(1)(2)0,|3213A x x x B x x =-+==-<-<，则A B ⋂=（ ）A.{}2,1-B.()2,1-C.{}1D.()1,2-2.命题“所有能被3整除的整数都是质数”的否定是（ ）A.存在一个能被3整除的整数不是质数B.所有能被3整除的整数都不是质数C.存在一个能被3整除的整数是质数D.不能被3整除的整数不是质数3.已知等差数列{}n a 的前n 项和为n S ，若321523,1S S S a =+=，则{}n a 的公差等于（ ）A.2-B.1-C.1D.24.为净化水质，向一个游泳池加入某种化学药品，加药后池水中该药品浓度C 随时间t 的变化关系为22040Ct t C -+=，则C 的最大值为（ ）A.1B.2C.4D.55.如图，,,A B C 是圆O 上的三点，且60,90AOB BOC ∠∠==，则OA =（ ）A.1322OB OC -B.1322OB OC +312OC - 312OC + 6.已知一个圆锥的底面圆半径为1，其侧面展开图是一个圆心角为2π3的扇形，则该圆锥的体积为（ ） A.22π B.2π3 C.2π3 D.22π37.已知定义在R 上的函数()f x 满足()()()()11f x y f x y f x f y -+-++=，且()12f =，则()()()234f f f ++=（ ）A.2B.0C.2-D.4-8.已知函数()()πsin 03,2f x x ωϕωϕ⎛⎫=+<<<⎪⎝⎭，甲､乙､丙､丁四位同学各说出了这个函数的一条结论： 甲：函数()f x 的图象关于π,03⎛⎫ ⎪⎝⎭对称； 乙：函数()f x 在5ππ,1212⎡⎤-⎢⎥⎣⎦上单调递增； 丙：函数()f x 在区间()0,π上有3个零点；丁：函数()f x 的图象向左平移π2个单位之后与()f x 的图象关于x 轴对称. 若这四位同学中恰有一人的结论错误，则该同学是（ ）A.甲B.乙C.丙D.丁二､多项选择题：本大题共3个小题，每小题6分，共18分.在每小题给出的四个选项中，有多项符合题目要求，全部选对的得6分，选对但不全的得部分分，有选错的得0分. 9.已知直线,m n 是平面α外两条不同的直线，则下列命题正确的是（ ）A.若m ∥,n α∥α，则m ∥nB.若m ∥,n n ∥α，则m ∥αC.若m ∥,n m α⊥，则n α⊥D.若m ∥,n αα⊥，则m n ⊥10.已知03a b <<<，则（ ）A.(1)1b a +>B.()log 11a b +>C.()()cos πcos πa b +<+D.ππcos cos 22a b ⎛⎫⎛⎫-<- ⎪ ⎪⎝⎭⎝⎭11.设函数()2cos π2x f x x ax =-+，则（ ） A.存在实数a ，使得()f x 为偶函数B.函数()f x 的图象关于()2a x a =∈Z 对称 C.当2a =时，142f x x ⎛⎫-< ⎪⎝⎭D.当4a =时，函数()f x 在()2,3上单调递增三､填空题：本大题共3个小题，每小题5分，共15分.12.已知向量,a b 满足()()1,1,1,a b a a b ==⊥-，则,a b <>=__________.13.已知点π,08P ⎛⎫ ⎪⎝⎭在函数()2sin 3)f x x ωω=<<的图象上，则曲线()y f x =在点P 处的切线方程为__________.14.已知数列{}n a 满足11a =，且对于任意2n ，都存在{}1,2,,1k n ∈-，使得4n k a a =+，则4a 的所有可能取值构成的集合M =__________；若{}n a 的各项均不相等，把半径为123,,a a a （单位：cm ）的三个小球放入一个正方体容器（容器壁厚度忽略不计），则该正方体容器的棱长最小值为__________cm . 四､解答题：本大题共5小题，共77分.解答应写出文字说明､证明过程或演算步骤. 15.（13分）记ABC 的内角,,A B C 的对边分别为,,a b c 3sin cos a C c c A -=.（1）求A ；（2）若21,6a b c =+=，求ABC 的面积.16.（15分）已知数列{}n a 的前n 项和为n S ，且3212321212121n n n a a a a ++++=----. （1）求n S ；（2）设3642n n n S b +=，若数列{}n b 的最小项为m b ，求m . 17.（15分）如图，已知平行六面体1111ABCD A BC D -的底面是菱形，112,,AB AC BD O A AB A AD ∠∠=⋂==.（1）证明：1AA BD ⊥；（2）若11π22,3AA A O BAD ∠===，点P 在平面1ABC 内，且BP ⊥平面1ABC ，求BP 与平面ABCD 所成角的正弦值.18.（17分）已知函数()()()2ln 1f x ax x x a =-++∈R . （1）当0a 时，讨论()f x 的单调性；（2）()()()1e 21xg x f x x x =++--. （i ）当0a =时，求()g x 的最小值；（ii ）若()0g x 在[)0,∞+上恒成立，求a 的取值范围.19.（17分）已知()f x 为定义域M 内的连续函数，()f x '为其导函数，常数a M ∈，若各项不相等的数列{}n a 满足()()()11,,n n n n f a f a a M a a f a a a +>'-∈=-，则称{}n a 为()f x 的“拉格朗日数列”，简记为“()L a -数列”. （1）若函数()ln g x x =，数列{}n b 是()g x 的“()1L -数列”，且1e b =.（i ）求23,b b ；（ii ）证明：{}n b 是递减数列；（2）正项数列{}n c 是函数()36sin h x x x =+的“()L c -数列”，已知()1,n n c c c +∈，记{}n c 的前n 项和为n S ，证明：0c >时，()112n n S c n c c +-+.。