高三9月月考试卷
山西省晋城市2024-2025学年高三上学期9月月考试题 语文 含答案
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山西省晋城市2024-2025学年高三上学期9月月考语文试题考生注意:1.本试卷共150分,考试时间150分钟。
2.请将各题答案填写在答题卡上。
3.本试卷主要考试内容:高考全部内容。
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:随着西方艺术史研究的不断深入,大量理论、研究方法被引入国内,对中国艺术史研究的发展产生了一系列深刻的影响。
一方面,中国艺术史研究借鉴西方经验,引入了许多研究方法和理论,如社会历史学、文化研究、后现代主义等,这些方法和理论帮助中国艺术史研究者更好地分析和解读艺术作品,关注艺术与社会、文化、政治等方面的关系,研究领域不断扩大。
受新艺术史研究的影响,中国艺术史研究者开始关注非传统的艺术领域,如民间艺术、当代艺术、女性艺术等,这种拓展使中国艺术史的研究更加多元化和综合化,进一步丰富了中国艺术史的研究内容。
例如,在分析绘画中的女性形象时,研究者会更多地结合作品的历史背景和女性心理学,分析作品的精神内涵,尝试解释其中的历史、文化、政治因素,而不是仅仅停留在笔触、品质等层面,这显示出我国美术史研究发生的深刻变化。
另一方面,艺术史研究的对象范围逐渐扩大,现代中国艺术史研究的视野早已不再局限于研究内部艺术变化,如风格、样式、语言、技法,而是扩展外向型研究;艺术史的研究方法也不再局限于本学科的理论方法,而是选择跨学科的方法和理论体系,如符号学、社会学、心理学等。
受西方艺术史研究的影响,中国艺术史研究者与国际学术界进行了更加广泛的交流,这种跨文化的对话促进了不同文化间的艺术交流和相互借鉴,拓宽了中国艺术史研究的视野,促使中国艺术史研究者对传统的艺术史观念和叙述进行批判和反思,推动了中国艺术史研究的发展。
总之,西方艺术史研究包括新艺术史研究,对中国艺术领域产生了广泛而深远的影响。
它为中国艺术史研究提供了新的研究方法和理论,拓展了研究领域,激发了中国艺术史研究者的理论创新和批判精神,使中国艺术史研究更加多元化、综合化和国际化。
哈尔滨市第九中学2024年高三上学期9月月考英语试卷
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哈尔滨市第九中学校2023-2024学年度高三上学期九月份考试英语试卷(考试时间:120分钟满分150分共5页)第1卷(满分95分)第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。
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. How long has Adam lived alone?A.30 years.B.22 years.C.15 years.2. How will the speakers go to the concert?A.By car.B. By taxi.C.By subway.3. What did the man probably do just now?A. He did some shopping.B. He bought some coffee.C. He walked a dog.4. Where are the speakers?A. In a bookstore.B. In a library.C. In a classroom.5. What does the man mean?A. He hates to build the bookshelf.B. He has lost the woman's tools.C.He hasn’t finished his work.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
每段对话或独长后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。
听每段对话或独白前你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料,回答第6、76. What is the relationship between the speakers?A. Boss and employee.B.Teacher and student.C. Interviewer and interviewee.7. What is the man mainly talking about?A. His special skills.B.His mother language.C.His dream job.听第7段材料,回答第8、9题。
湖北省沙市2024-2025学年高三上学期9月月考试题 数学含解析
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2024—2025学年度上学期2022级9月月考数学试卷(答案在最后)命题人:考试时间:2024年9月25日一、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,只有一个选项是正确的.请把正确的选项填涂在答题卡相应的位置上.1.集合{}215=∈<N M x x ,若{}05⋃=≤<M N x x ,则集合N 可以为()A.{}4 B.{}45≤<x x C.{}05<<x x D.{}5<x x 2.若复数232022202320241i i i i +i i z =-+-++- ,则z =()A.B.C.1D.23.已知2b a = ,若a 与b 的夹角为60︒,则2a b - 在b 上的投影向量为()A .12br B .12b- C .32b- D .32b4.纯电动汽车是以车载电源为动力,用电机驱动车轮行驶,符合道路交通、安全法规各项要求的车辆,它使用存储在电池中的电来发动.因其对环境影响较小,逐渐成为当今世界的乘用车的发展方向.研究发现电池的容量随放电电流的大小而改变,1898年Peukert 提出铅酸电池的容量C 、放电时间t 和放电电流I 之间关系的经验公式:C I t λ=,其中λ为与蓄电池结构有关的常数(称为Peukert 常数),在电池容量不变的条件下,当放电电流为7.5A 时,放电时间为60h ;当放电电流为25A 时,放电时间为15h ,则该蓄电池的Peukert 常数λ约为(参考数据:lg 20.301≈,lg 30.477≈)()A .1.12B .1.13C .1.14D .1.155.已知,(0,π)αβ∈,且cos 5α=,sin()10αβ+=,则αβ-=()A .4πB .34πC .4π-D .34π-6.已知函数2()()ln 0f x x ax b x =++≥恒成立,则实数a 的最小值为()A .2-B .1-C .1D .27.函数()ln 1f x x =-与函数()πsin 2g x x =的图象交点个数为()A .6B .7C .8D .98.斐波拉契数列因数学家斐波拉契以兔子繁殖为例而引入,又称“兔子数列”.这一数列如下定义:设{}n a 为斐波拉契数列,()*12121,1,3,N n n n a a a a a n n --===+≥∈,其通项公式为1122n nna⎡⎤⎛⎫⎛⎫⎥=-⎪ ⎪⎪ ⎪⎥⎝⎭⎝⎭⎦,设n是2log1(14(xx x⎡⎤⎣-⎦-<+的正整数解,则n的最大值为()A.5B.6C.7D.8二、选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全部选对得6分,部分选对的得部分分,选对但不全的得部分分,有选错的得0分. 9.给出下列命题,其中正确命题为()A.已知数据12310x x x x、、、、,满足:()12210i ix x i--=≤≤,若去掉110x x、后组成一组新数据,则新数据的方差为168B.随机变量X服从正态分布()21,,( 1.5)0.34N P xσ>=,若()0.34P x a<=,则0.5a=C.一组数据()(),1,2,3,4,5,6i ix y i=的线性回归方程为 23y x=+,若6130iix==∑,则6163iiy==∑D.对于独立性检验,随机变量2χ的值越大,则推断“两变量有关系”犯错误的概率越小10.如图,棱长为2的正方体1111ABCD A B C D-中,E为棱1DD的中点,F为正方形11C CDD内一个动点(包括边界),且1//B F平面1A BE,则下列说法正确的有()A.动点FB.1B F与1A B不可能垂直C.三棱锥11B D EF-体积的最小值为13D.当三棱锥11B D DF-的体积最大时,其外接球的表面积为25π211.已知抛物线2:2(0)C y px p=>的焦点为F,准线交x轴于点D,直线l经过F且与C交于,A B 两点,其中点A在第一象限,线段AF的中点M在y轴上的射影为点N.若MN NF=,则()A.lB.ABD△是锐角三角形C.四边形MNDF2D.2||BF FA FD⋅>三、填空题:本题共3小题,每小题5分,共15分.12.若“[]1,4x∃∈使20040x ax-+>”为假命题,则实数a的取值范围为___________.13.在ABC∆中,BC=,∠3Aπ=,D为线段AB靠近点A的三等分点,E为线段CD的中点,若14BF BC=,则AE AF⋅的最大值为________.14.将1,2,3,4,5,6,7这七个数随机地排成一个数列,记第i项为()1,2,,7ia i= ,若47a=,123567a a a a a a++<++,则这样的数列共有个.四、解答题:本题共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.已知ABC △的内角A ,B ,C 的对边分别为a ,b ,c ,若()4sin sin sin -=-A b B c A B .(1)求a 的值;(2)若ABC △的面积为()22234+-b c a ,求ABC △周长的取值范围.16.已知正项数列{}n a 的前n 项和为n S ,且222n n n a a n S +-=.(1)求数列{}n a 的通项公式;(2)设21na nb =-,若数列{}nc 满足11n n n n b c b b ++=⋅,且数列{}n c 的前n 项和为n T ,若()12n T n λ-+≤恒成立,求λ的取值范围.17.如图所示,半圆柱1OO 与四棱锥A BCDE -拼接而成的组合体中,F 是半圆弧BC 上(不含,B C )的动点,FG 为圆柱的一条母线,点A 在半圆柱下底面所在平面内,122,22OB OO AB AC ====.(1)求证:CG BF ⊥;(2)若//DF 平面ABE ,求平面FOD 与平面GOD 夹角的余弦值;(3)求点G 到直线OD 距离的最大值.18.已知双曲线E 的中心为坐标原点,渐近线方程为y =,点(2,1)-在双曲线E 上.互相垂直的两条直线12,l l 均过点()(,0n n P p p >,且)*n ∈N ,直线1l 交E 于,A B 两点,直线2l 交E于,C D 两点,,M N 分别为弦AB 和CD 的中点.(1)求E 的方程;(2)若直线MN 交x 轴于点()()*,0n Q t n ∈N ,设2nn p =.①求n t ;②记n a PQ =,()*21n b n n =-∈N ,求211(1)nkk k k k b b a +=⎡⎤--⎣⎦∑.19.如果函数的导数为()()F x f x '=,可记为()()d f x x F x ⎰=,若()0f x ≥,则()()()baf x dx F b F a =-⎰表示曲线=op ,直线x a x b ==,以及x 轴围成的“曲边梯形”的面积.如:22d x x x C ⎰=+,其中C 为常数;()()2202204xdx C C =+-+=⎰,则表0,1,2x x y x ===及x 轴围成图形面积为4.(1)若()()()e 1d 02xf x x f =⎰+=,,求()f x 的表达式;(2)求曲线2y x =与直线6y x =-+所围成图形的面积;(3)若()[)e 120,xf x mx x ∞=--∈+,,其中Rm ∈,对[)0,a b ∞∀∈+,,若a b >,都满足()()0d d a bf x x f x x >⎰⎰,求m 的取值范围.1.C2.C 【详解】()()32024+1232022022022024241i 1i ()1+1i 1i 1i 11i i iiiii z i =-+----⨯-+====--+-+++C6.B 【详解】∵()0f x ≥恒成立,设2()g x x ax b =++,则当1x >时()0g x ≥,01x <<时()0g x <,∴(1)0g =⎧⎨≤,即101a b a b++=⇒=--⎧⎨≤,∴1a ≥-11.ABD 【详解】由题意可知:抛物线的焦点为,02p F ⎛⎫ ⎪⎝⎭,准线为x 则11,,0,242xy p M N ⎛⎫⎛+ ⎪ ⎝⎭⎝可知MNF 为等边三角形,即且MN ∥x 轴,可知直线则直线:32p l y x ⎛⎫=- ⎪12.【详解】因为“0使00”为假命题,所以“[]1,4x ∀∈,240x ax -+≤”为真命题,其等价于4≥+a x x在[]1,4上恒成立,又因为对勾函数()4f x x x=+在[]1,2上单调递减,在[]2,4上单调递增,而()()145f f ==,所以()max 5f x =,所以5a ≥,即实数a 的取值范围为[5,)+∞.13.11814.360【解析】∵12345621+++++=,∴310S ≤,列举可知:①(1,2,3)……(1,2,6)有4个;②(1,3,4),……,(1,3,6)有3个;③(1,4,5)有1个;④(2,3,4),(2,3,5)有2个;故共有10个组合,∴共计有333310360A A ⨯⨯=个这样的数列。
江苏省南通市名校2025届高三上学期9月月考语文试卷(解析版)
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江苏省南通市名校2024-2025学年高三上学期9月月考语文试卷一、现代文阅读(35分)(一)现代文阅读I(本题共5小题,18分)1.(18分)阅读下面的文字,完成下列各题。
材料一:西北地区是中华文明的发祥地之一。
丝绸之路上各民族往来贸易,交流融合,造就了西北地区多民族共居和多元文化汇聚的独特景观。
先秦时期有戎、羌、氐,秦汉时期有匈奴、月氏、乌孙,后来还有吐谷浑、吐蕃、回鹘、党项等民族。
汉代接收浑邪王十万降众时,采取了“因其故俗为属国”的制度。
“属国”体制保留了匈奴人原有的社会组织和生产生活方式,宽容的民族政策推动西北地区多民族共居局面的形成,和谐的民族关系促进了丝绸之路的繁荣。
汉唐时期,丝绸之路上建立过很多民族政权,但都未影响丝绸之路的畅通与繁荣,丝绸之路的繁荣是沿线民族合作联动的产物,而各民族互相联动的基础首先是沿线民族友好交往、民心相通。
贸易双方的双向共赢也是丝路长期繁荣的重要原因。
丝路贸易不是单向的,而是双向共赢的。
东西方的经济、文化交流是“不断地交流、发展、融合后,再交流、再发展、再融合,从而达到了更高的发展”。
以“佛教的倒流”为例,佛教并非简单地从西向东单向传播,经过中国佛教界的改造和发展后又传回中亚、印度,从而对中亚和印度的佛教作出贡献。
丝路贸易的主体是“转输贸易”,在古代丝绸之路的各个重镇,都有数量不等的胡商或胡人聚落,“胡人尤其是粟特人充当了丝绸之路上的贸易担当者”。
由于转输贸易的需要,丝路上很多重镇是“商胡”入华之路,也是商贸中心。
如新疆的吐鲁番,“一些商胡从西边来到高昌后便不再东行,而是将货物在当地出售,东边来的商人也不再西行到货物的产地去收购商品,而是在高昌购买后,再到其他地方去出售”。
正是因为丝路贸易的这个特点,沿线一些城市成为国际贸易集散中心,如君士坦丁堡、巴格达、敦煌、吐鲁番、兰州等。
丝路贸易的主要承担者,来自中亚的粟特商人成群结队地徙居中国并长期定居,其后裔有的继续经商,有的则在中国入仕,完全融入中华民族大家庭中。
湖北省沙市2024-2025学年高三上学期9月月考试题 地理含解析
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2024—2025学年度上学期2022级9月月考地理试卷(答案在最后)命题人:考试时间:2024年9月25日一、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。
)我国某中学地理教师给地理社团成员布置模拟天文观测作业,给出以下条件。
完成下面小题。
1.春分日或秋分日,天气晴朗。
2.月亮比太阳早6小时升起。
3.太阳与月亮最大高度角相等。
1.这一天,下列现象可信的是()A.日出时,月亮高度角最大B.月出时,当地地方时为6点C.日落时,月亮的亮面朝东D.月落时,天空出现满天星星2.当天可见直立杆的月影范围最有可能的是()A.①B.②C.③D.④2023年1月24日凌晨,山东烟台受寒潮影响,出现罕见的“雷打雪”现象。
“雷打雪”是指因冷暖气团交汇产生强对流天气,并在降雪过程中伴有雷电的天气现象,其发生与风速、风向、温度露点差(温度与露点的差值,温度露点差越大,湿度越小,当温度露点差接近0℃时,表示空气中的水汽达到近似饱和状态)等气象要素的短期变化密切相关,下图示意烟台该次“雷打雪”事件发生时气象要素随时间的变化状况。
据此完成下面小题。
3.烟台发生“雷打雪”时,()A.近地面暖湿气流受热迅速上升B.强冷气团自渤海南下,遇陆地主动爬升C.冷暖气流强烈交汇,暖气团快速抬升D.暖湿气流迅速北推,冷暖气流强烈交汇4.推测图示“雷打雪”现象发生时刻为()A.T0B.T2C.T1D.T35.“雷打雪”发生后一周内,该地的天气特点为()A.冷干B.暖湿C.暖干D.冷湿气候林线是指高海拔山地森林分布的上限,亚热带山地海拔达到2200—3600m时才可能发育气候林线。
但有调查发现,我国亚热带东段山地在海拔2000m以下的山顶部位也出现了林线景观,因其成因与气候林线不同,称之为“假林线”。
我国广东M自然保护区内部分山顶或山脊处,分布有斑块状草地,林线海拔介于700-1200m,呈现出南坡低于北坡的特点。
云南2024-2025学年高三上学期9月月考数学试题含答案
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数学试卷(答案在最后)注意事项:1.答题前,考生务必用黑色碳素笔将自己的姓名、准考证号、考场号、座位号在答题卡上填写清楚.2.每小题选出答案后,用2B 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.在试题卷上作答无效.3.考试结束后,请将本试卷和答题卡一并交回.满分150分,考试用时120分钟.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,只有一项符合题目要求)1.已知集合{13},{(2)(4)0}A xx B x x x =≤≤=--<∣∣,则A B = ()A.(2,3] B.[1,2)C.(,4)-∞ D.[1,4)【答案】A 【解析】【分析】解出集合B ,再利用交集含义即可得到答案.【详解】{(2)(4)0}{24}B xx x x x =--<=<<∣∣,而{|13}A x x =≤≤,则(2,3]A B ⋂=.故选:A.2.已知命题2:,10p z z ∃∈+<C ,则p 的否定是()A.2,10z z ∀∈+<CB.2,10z z ∀∈+≥C C.2,10z z ∃∈+<C D.2,10z z ∃∈+≥C 【答案】B 【解析】【分析】根据存在量词命题的否定形式可得.【详解】由存在量词命题的否定形式可知:2:,10p z z ∃∈+<C 的否定为2,10z z ∀∈+≥C .故选:B3.正项等差数列{}n a 的公差为d ,已知14a =,且135,2,a a a -三项成等比数列,则d =()A.7B.5C.3D.1【答案】C【解析】【分析】由等比中项的性质再结合等差数列性质列方程计算即可;【详解】由题意可得()23152a a a -=,又正项等差数列{}n a 的公差为d ,已知14a =,所以()()2111224a d a a d +-=+,即()()222444d d +=+,解得3d =或1-(舍去),故选:C.4.若sin160m ︒=,则︒=sin 40()A.2m -B.2-C.2-D.2【答案】D 【解析】【分析】利用诱导公式求出sin 20︒,然后结合平方公式和二倍角公式可得.【详解】因为()sin160sin 18020sin 20m ︒=︒-︒=︒=,所以cos 20︒==,所以sin 402sin 20cos 202︒=︒︒=故选:D5.已知向量(1,2),||a a b =+= ,若(2)b b a ⊥- ,则cos ,a b 〈〉=()A.5-B.10-C.10D.5【答案】C 【解析】【分析】联立||a b += 和(2)0b b a ⋅-=求出,b a b ⋅ 即可得解.【详解】因为(1,2)a = ,所以a =,所以222||27a b a b a b +=++⋅=,整理得222b a b +⋅=①,又(2)b b a ⊥- ,所以2(2)20b b a b a b ⋅-=-⋅=②,联立①②求解得11,2b a b =⋅= ,所以12cos ,10a b a b a b⋅〈〉=== .故选:C 6.函数)()ln f x kx =是奇函数且在R 上单调递增,则k 的取值集合为()A.{}1-B.{0}C.{1}D.{1,1}-【答案】C 【解析】【分析】根据奇函数的定义得()))()222()ln lnln 10f x f x kx kx x k x -+=-+=+-=得1k =±,即可验证单调性求解.【详解】)()lnf x kx =+是奇函数,故()))()222()ln ln ln 10f x f x kx kx x k x -+=-+=+-=,则22211x k x +-=,210k -=,解得1k =±,当1k =-时,)()lnf x x ==,由于y x =在0,+∞为单调递增函数,故()lnf x =0,+∞单调递减,不符合题意,当1k =时,)()lnf x x =+,由于y x =在0,+∞为单调递增函数且()00f =,故)()ln f x x =为0,+∞单调递增,根据奇函数的性质可得)()ln f x x =+在上单调递增,符合题意,故1k =,故选:C7.函数π()3sin ,06f x x ωω⎛⎫=+> ⎪⎝⎭,若()(2π)f x f ≤对x ∈R 恒成立,且()f x 在π13π,66⎡⎤⎢⎣⎦上有3条对称轴,则ω=()A.16 B.76C.136D.16或76【答案】B【解析】【分析】根据()2π3,2π2f T T =≤<求解即可.【详解】由题知,当2πx =时()f x 取得最大值,即π(2π)3sin 2π36f ω⎛⎫=+= ⎪⎝⎭,所以ππ2π2π,Z 62k k ω+=+∈,即1,Z 6k k ω=+∈,又()f x 在π13π,66⎡⎤⎢⎥⎣⎦上有3条对称轴,所以13ππ2π266T T ≤-=<,所以2π12T ω≤=<,所以76ω=.故选:B8.设椭圆2222:1(0)x y E a b a b +=>>的右焦点为F ,过坐标原点O 的直线与E 交于A ,B 两点,点C 满足23AF FC = ,若0,0AB OC AC BF ⋅=⋅=,则E 的离心率为()A.9B.7C.5D.3【答案】D 【解析】【分析】设(),A m n ,表示出,,,OA OC AF BF,根据0,0AB OC AC BF ⋅=⋅= 列方程,用c 表示出,m n ,然后代入椭圆方程构造齐次式求解可得.【详解】设(),A m n ,则()(),,,0B m n F c --,则()()(),,,,,OA m n AF c m n BF c m n ==--=+,因为23AF FC = ,所以()555,222n AC AF c m ⎛⎫==-- ⎪⎝⎭,所以()()55533,,,22222n c n OC OA AC m n c m m ⎛⎫⎛⎫=+=+--=-- ⎪ ⎪⎝⎭⎝⎭ ,因为0,0AB OC AC BF ⋅=⋅=,所以222253302220c OA OC m m n AF BF c m n ⎧⎛⎫⋅=--=⎪ ⎪⎝⎭⎨⎪⋅=--=⎩ ,得34,55m c n c ==,又(),A m n 在椭圆上,所以222291625251c ca b+=,即()()222222229162525c a c a c a a c -+=-,整理得4224255090a a c c -+=,即42950250e e -+=,解得259e =或25e =(舍去),所以3e =.故选:D【点睛】关键点睛:根据在于利用向量关系找到点A 坐标与c 的关系,然后代入椭圆方程构造齐次式求解.二、多项选择题(本大题共3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,部分选对的得部分分,有选错的得0分)9.数列{}n a 的前n 项和为n S ,已知22()n S kn n k =-∈R ,则下列结论正确的是()A.{}n a 为等差数列B.{}n a 不可能为常数列C.若{}n a 为递增数列,则0k >D.若{}n S 为递增数列,则1k >【答案】AC 【解析】【分析】根据,n n a S 的关系求出通项n a ,然后根据公差即可判断ABC ;利用数列的函数性,分析对应二次函数的开口方向和对称轴位置即可判断D .【详解】当1n =时,112a S k ==-,当2n ≥时,()()()221212122n n n a S S kn n k n n kn k -⎡⎤=-=-----=-+⎣⎦,显然1n =时,上式也成立,所以()22n a kn k =-+.对A ,因为()()()1222122n n a a kn k k n k k -⎡⎤-=-+---+=⎣⎦,所以是以2k 为公差的等差数列,A 正确;对B ,由上可知,当0k =时,为常数列,B 错误;对C ,若为递增数列,则公差20k >,即0k >,C 正确;对D ,若{}n S 为递增数列,由函数性质可知02322k k >⎧⎪⎨<⎪⎩,解得23k >,D 错误.故选:AC10.甲、乙两班各有50位同学参加某科目考试(满分100分),考后分别以110.820y x =+、220.7525y x =+的方式赋分,其中12,x x 分别表示甲、乙两班原始考分,12,y y 分别表示甲、乙两班考后赋分.已知赋分后两班的平均分均为60分,标准差分别为16分和15分,则()A.甲班原始分数的平均数比乙班原始分数的平均数高B.甲班原始分数的标准差比乙班原始分数的标准差高C.甲班每位同学赋分后的分数不低于原始分数D.若甲班王同学赋分后的分数比乙班李同学赋分后的分数高,则王同学的原始分数比李同学的原始分数高【答案】ACD 【解析】【分析】根据期望和标准差的性质求出赋分前的期望和标准差即可判断AB ;作差比较,结合自变量范围即可判断C ;作出函数0.820,0.7525y x y x =+=+的图象,结合图象可判断D .【详解】对AB ,由题知()()1215E y E y ====,因为110.820y x =+,220.7525y x =+,所以()()120.82060,0.752515E x E x +=+===,解得()()1250,20E x E x =≈==,所以()()12E x E x >=,故A 正确,B 错误;对C ,因为111200.2y x x -=-,[]10,100x ∈,所以10200.220x ≤-≤,即110y x -≥,所以C 正确;对D ,作出函数0.820,0.7525y x y x =+=+的图象,如图所示:由图可知,当12100y y =<时,有21x x <,又因为0.820y x =+单调递增,所以当12y y >时必有12x x >,D 正确.故选:ACD11.已知函数()f x 及其导函数()f x '的定义域为R ,若(1)f x +与()f x '均为偶函数,且(1)(1)2f f -+=,则下列结论正确的是()A.(1)0f '=B.4是()f x '的一个周期C.(2024)0f =D.()f x 的图象关于点(2,1)对称【答案】ABD 【解析】【分析】注意到()f x '为偶函数则()()2f x f x -+=,由()(1)1f x f x -+=+两边求导,令0x =可判断A ;()()11f x f x --='+'结合导函数的奇偶性可判断B ;利用()f x 的周期性和奇偶性可判断C ;根据()()2f x f x -+=和()(1)1f x f x -+=+可判断D .【详解】因为()f x '为偶函数,所以()()f x f x -'=',即()()f x f x c --=+,而(1)(1)2f f -+=,故2c =-,故()()2f x f x +-=,又(1)f x +为偶函数,所以()(1)1f x f x -+=+,即()()2f x f x =-,所以()2()2f x f x -+-=,故()(2)2f x f x ++=即()2(4)2f x f x +++=,()()4f x f x =+,所以4是()f x 的周期,故B 正确.对A ,由()(1)1f x f x -+=+两边求导得()()11f x f x --='+',令0x =得()()11f f -'=',解得()10f '=,A 正确;对C ,由上知()()2f x f x +-=,所以()01f =,所以()()(2024)450601f f f =⨯==,C 错误;对D ,因为()()2f x f x +-=,()()2f x f x =-,故()2(2)2f x f x -++=,故()f x 的图象关于2,1对称,故选:ABD【点睛】关键点睛:本题解答关键在于原函数与导数数的奇偶性关系,以及对()(1)1f x f x -+=+两边求导,通过代换求导函数的周期.三、填空题(本大题共3小题,每小题5分,共15分)12.曲线()e xf x x =-在0x =处的切线方程为______.【答案】1y =##10y -=【解析】【分析】求出函数的导函数,利用导数的几何意义求出切线的斜率,即可求出切线方程.【详解】因为()e xf x x =-,则()01f =,又()e 1xf x '=-,所以()00f '=,所以曲线()e xf x x =-在0x =处的切线方程为1y =.故答案为:1y =13.若复数cos 21sin isin (0π)2z θλθθθ⎛⎫=+-+<< ⎪⎝⎭在复平面内对应的点位于直线y x =上,则λ的最大值为__________.【答案】1-##1-+【解析】【分析】根据复数对应的点cos 21sin ,sin 2θλθθ⎛⎫⎛⎫+- ⎪ ⎪⎝⎭⎝⎭在y x =得212sin 1sin sin 2θλθθ⎛⎫-+-= ⎪⎝⎭,即可利用二倍角公式以及基本不等式求解.【详解】cos 21sin isin (0π)2z θλθθθ⎛⎫=+-+<< ⎪⎝⎭对应的点为cos 21sin ,sin 2θλθθ⎛⎫⎛⎫+- ⎪ ⎪⎝⎭⎝⎭,故cos 21sin sin 2θλθθ⎛⎫+-= ⎪⎝⎭,故212sin 1sin sin 2θλθθ⎛⎫-+-= ⎪⎝⎭,由于()0,πθ∈,故sin 0θ>,则2sin 1111sin sin sin 122sin θλθθθθ==≤++++,当且仅当1sin 2sin θθ=,即2sin 2θ=,解得π3π,44θθ==时等号成立,114.过抛物线2:3C y x =的焦点作直线l 交C 于A ,B 两点,过A ,B 分别作l 的垂线与x 轴交于M ,N 两点,若||12AB =,则||MN =__________.【答案】【解析】【分析】联立直线与抛物线方程,得韦达定理,根据焦点弦的公式可得223332122k AB k +=+=,解得213k =,即可求解()111:AM y x x y k=--+得11M x ky x =+,即可代入求解.【详解】2:3C y x =0,根据题意可知直线l 有斜率,且斜率不为0,根据对称性不设直线方程为34y k x ⎛⎫=-⎪⎝⎭,联立直线34y k x ⎛⎫=-⎪⎝⎭与23y x =可得22223930216k x k x k ⎛⎫-++= ⎪⎝⎭,设()()1122,,,A x y B x y ,故2121223392,16k x x x x k ++==,故21223332122k AB x x p k +=++=+=,解得213k =,直线()111:AM y x x y k=--+,令0y =,则11M x ky x =+,同理可得22N x ky x =+,如下图,故()()()211221212121M N MN x x ky x ky x k y y x x k x x =-=+--=-+-=+-,()()22221212233192141483316k MN k x x x x k ⎛⎫+ ⎪⎛⎫=++-=+-⨯= ⎪ ⎪⎝⎭ ⎪⎝⎭故答案为:83四、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.记ABC V 的内角A ,B ,C 的对边分别为a ,b ,c ,已知22cos 0a b c A -+=.(1)求角C ;(2)若AB 边上的高为1,ABC V 的面积为33,求ABC V 的周长.【答案】(1)π3C =;(2)23.【解析】【分析】(1)利用余弦定理角化边,整理后代入余弦定理即可得解;(2)利用面积公式求出c ,然后由面积公式结合余弦定理联立求解可得a b +,可得周长.【小问1详解】由余弦定理角化边得,2222202b c a a b c bc +--+⨯=,整理得222a b c ab +-=,所以2221cos 222a b c ab C ab ab +-===,因为()0,πC ∈,所以π3C =.【小问2详解】由题知,13123c ⨯=,即233c =,由三角形面积公式得1πsin 233ab =,所以43ab =,由余弦定理得()222π42cos 333a b ab a b ab +-=+-=,所以()2416433a b +=+=,所以3a b +=,所以ABC V 的周长为33a b c ++=+=16.如图,PC 是圆台12O O 的一条母线,ABC V 是圆2O 的内接三角形,AB 为圆2O 的直径,4,AB AC ==.(1)证明:AB PC ⊥;(2)若圆台12O O 的高为3,体积为7π,求直线AB 与平面PBC 夹角的正弦值.【答案】(1)证明见详解;(2)19.【解析】【分析】(1)转化为证明AB ⊥平面12O O CP ,利用圆台性质即可证明;(2)先利用圆台体积求出上底面的半径,建立空间坐标系,利用空间向量求线面角即可.【小问1详解】由题知,因为AB 为圆2O 的直径,所以AC BC ⊥,又4,AB AC ==AB ==,因为2O 为AB 的中点,所以2O C AB ⊥,由圆台性质可知,12O O ⊥平面ABC ,且12,,,O O P C 四点共面,因为AB ⊂平面ABC ,所以12O O AB ⊥,因为122,O O O C 是平面12O O CP 内的两条相交直线,所以AB ⊥平面12O O CP ,因为PC ⊂平面12O O CP ,所以AB PC ⊥.【小问2详解】圆台12O O的体积(2211ππ237π3V r =⋅+⋅⨯=,其中11r PO =,解得11r =或13r =-(舍去).由(1)知122,,O O AB O C 两两垂直,分别以2221,,O B O C O O 为x 轴、y 轴、z 轴建立空间直角坐标系,如图,则(2,0,0),(2,0,0),(0,2,0),(0,1,3)A B C P -,所以(4,0,0),(2,1,3),(2,2,0)AB BP BC ==-=-.设平面PBC 的一个法向量为(,,)n x y z =,则230,220,n BP x y z n BC x y ⎧⋅=-++=⎪⎨⋅=-+=⎪⎩解得,3,x y x z =⎧⎨=⎩于是可取(3,3,1)n =.设直线AB 与平面PBC 的夹角为θ,则sin cos ,19AB n θ===,故所求正弦值为19.17.已知函数()ln f x x ax =+.(1)若()0f x ≤在(0,)x ∈+∞恒成立,求a 的取值范围;(2)若()1,()e()xa g x f f x ==-,证明:()g x 存在唯一极小值点01,12x⎛⎫∈ ⎪⎝⎭,且()02g x >.【答案】(1)1,e⎛⎤-∞- ⎥⎝⎦;(2)证明见解析.【解析】【分析】(1)参变分离,构造函数()ln xh x x=-,利用导数求最值即可;(2121内,利用零点方程代入()0g x ,使用放缩法即可得证.【小问1详解】()0f x ≤在(0,)x ∈+∞恒成立,等价于ln xa x≤-在(0,)+∞上恒成立,记()ln x h x x =-,则()2ln 1x h x x='-,当0e x <<时,ℎ′<0,当e x >时,ℎ′>0,所以ℎ在()0,e 上单调递减,在()e,∞+上单调递增,所以当e x =时,ℎ取得最小值()ln e 1e e eh =-=-,所以1a e≤-,即a 的取值范围1,e ∞⎛⎤-- ⎥⎝⎦.【小问2详解】当1a =时,()()e()eln ,0xxg x f f x x x =-=->,则1()e x g x x'=-,因为1e ,xy y x==-在(0,)+∞上均为增函数,所以()g x '在(0,)+∞单调递增,又()121e 20,1e 102g g ⎛⎫=-''=- ⎪⎝⎭,1存在0x ,使得当∈0,0时,()0g x '<,当∈0,+∞时,()0g x '>,所以()g x 在()00,x 上单调递减,在()0,x ∞+上单调递增,所以()g x 存在唯一极小值点01,12x ⎛⎫∈⎪⎝⎭.因为01e 0x x -=,即00ln x x =-,所以00000()e ln =e x x g x x x =-+,因为01,12x ⎛⎫∈⎪⎝⎭,且=e x y x+1上单调递增,所以012001()=e e 2x g x x +>+,又9e 4>,所以123e 2>,所以00031()=e 222xg x x +>+=.18.动点(,)M xy 到直线1:l y=与直线2:l y =的距离之积等于34,且|||y x <.记点M 的轨迹方程为Γ.(1)求Γ的方程;(2)过Γ上的点P 作圆22:(4)1Q x y +-=的切线PT ,T 为切点,求||PT 的最小值;(3)已知点40,3G ⎛⎫⎪⎝⎭,直线:2(0)l y kx k =+>交Γ于点A ,B ,Γ上是否存在点C 满足0GA GB GC ++= ?若存在,求出点C 的坐标;若不存在,说明理由.【答案】(1)2213y x -=(2)2(3)3,44C ⎛⎫-- ⎪ ⎪⎝⎭【解析】【分析】(1)根据点到直线距离公式,即可代入化简求解,(2)由相切,利用勾股定理,结合点到点的距离公式可得PT =,即可由二次函数的性质求解,(3)联立直线与双曲线方程得到韦达定理,进而根据向量的坐标关系可得()02201224,3443k x k k y y y k ⎧=-⎪⎪-⎨-⎪=-+=⎪-⎩,将其代入双曲线方程即可求解.【小问1详解】根据(,)M xy 到直线1:l y=与直线2:l y =的距离之积等于3434=,化简得2233x y -=,由于|||y x <,故2233x y -=,即2213y x -=.【小问2详解】设(,)P x y,PT ====故当3y =时,PT 最小值为2【小问3详解】联立:2(0)l y kx k =+>与2233x y -=可得()223470k x kx ---=,设()()()112200,,,,,A x y B x y C x y ,则12122247,33k x x x x k k-+==--,故()212122444,3k y y k x x k+=++=+-设存在点C 满足0GA GB GC ++= ,则1201200433x x x y y y ++=⎧⎪⎨++=⨯⎪⎩,故()02201224,3443k x k k y y y k ⎧=-⎪⎪-⎨-⎪=-+=⎪-⎩,由于()00,C x y 在2233x y -=,故22222443333k k k k ⎛⎫-⎛⎫--= ⎪⎪--⎝⎭⎝⎭,化简得421966270k k -+=,即()()2231990k k --=,解得2919k =或23k =(舍去),由于()22Δ162830k k =+->,解得27k<且23k ≠,故2919k =符合题意,由于0k >,故31919k =,故022024,344334k x k k y k ⎧=-=-⎪⎪-⎨-⎪==-⎪-⎩,故3,44C ⎛⎫-- ⎪ ⎪⎝⎭,故存在3,44C ⎛⎫-- ⎪ ⎪⎝⎭,使得0GA GB GC ++= 19.设n ∈N ,数对(),n n a b 按如下方式生成:()00,(0,0)a b =,抛掷一枚均匀的硬币,当硬币的正面朝上时,若n n a b >,则()()11,1,1n n n n a b a b ++=++,否则()()11,1,n n n n a b a b ++=+;当硬币的反面朝上时,若n n b a >,则()()11,1,1n n n n a b a b ++=++,否则()()11,,1n n n n a b a b ++=+.抛掷n 次硬币后,记n n a b =的概率为n P .(1)写出()22,a b 的所有可能情况,并求12,P P ;(2)证明:13n P ⎧⎫-⎨⎬⎩⎭是等比数列,并求n P ;(3)设抛掷n 次硬币后n a 的期望为n E ,求n E .【答案】(1)答案见详解;(2)证明见详解,1111332n n P -⎛⎫=-⨯- ⎪⎝⎭;(3)21113929nn E n ⎛⎫=+--⎪⎝⎭【解析】【分析】(1)列出所有()11,a b 和()22,a b 的情况,再利用古典概型公式计算即可;(2)构造得1111323n n P P +⎛⎫-=-- ⎪⎝⎭,再利用等比数列公式即可;(3)由(2)得()11111232nn n Q P ⎡⎤⎛⎫=-=--⎢⎥ ⎪⎝⎭⎢⎥⎣⎦,再分n n a b >,n n a b =和n n a b <讨论即可.【小问1详解】当抛掷一次硬币结果为正时,()()11,1,0a b =;当抛掷一次硬币结果为反时,()()11,0,1a b =.当抛掷两次硬币结果为(正,正)时,()()22,2,1a b =;当抛掷两次硬币结果为(正,反)时,()()22,1,1a b =;当抛掷两次硬币结果为(反,正)时,()()22,1,1a b =;当抛掷两次硬币结果为(反,反)时,()()22,1,2a b =.所以,12210,42P P ===.【小问2详解】由题知,1n n a b -≤,当n n a b >,且掷出反面时,有()()11,,1n n n n a b a b ++=+,此时11n n a b ++=,当n n a b <,且掷出正面时,有()()11,1,n n n n a b a b ++=+,此时11n n a b ++=,所以()()()()()1111112222n n n n n n n n n n P P a b P a b P a b P a b P +⎡⎤=>+<=>+<=-⎣⎦,所以1111323n n P P +⎛⎫-=-- ⎪⎝⎭,所以13n P ⎧⎫-⎨⎬⎩⎭是以11133P -=-为首项,12-为公比的等比数列,所以1111332n n P -⎛⎫-=-⨯- ⎪⎝⎭,所以1111332n n P -⎛⎫=-⨯- ⎪⎝⎭.【小问3详解】设n n a b >与n n a b <的概率均为n Q ,由(2)知,()11111232nn n Q P ⎡⎤⎛⎫=-=--⎢⎥⎪⎝⎭⎢⎥⎣⎦显然,111110222E =⨯+⨯=.若n n a b >,则1n n a b =+,当下次投掷硬币为正面朝上时,11n n a a +=+,当下次投掷硬币为反面朝上时,1n n a a +=;若n n a b =,则当下次投掷硬币为正面朝上时,11n n a a +=+,当下次投掷硬币为反面朝上时,1n n a a +=;若n n a b <,则1n n b a =+,当下次投掷硬币为正面朝上时,11n n a a +=+,当下次投掷硬币为反面朝上时,11n n a a +=+.所以1n n a a +=时,期望不变,概率为111122262nn n Q P ⎡⎤⎛⎫+=+-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦;11n n a a +=+时,期望加1,概率为1111111124226262n nn n Q P ⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫-+=-+-=--⎢⎥⎢⎥ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦.所以()11111112144626262nn nn nn n E E E E +⎡⎤⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫=⨯+-++⨯--=+--⎢⎥⎢⎥⎢⎥ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦.故12112111111444626262n n n n n n E E E -----⎡⎤⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫=+--=+--+--⎢⎥⎢⎥⎥ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦=1111111446262n E -⎡⎤⎡⎤⎛⎫⎛⎫=+--++--⎢⎥⎢⎥⎪⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦011111111444626262n -⎡⎤⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫=--+--++--⎢⎥⎢⎥⎢⎥ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦ 111241612n n ⎡⎤⎛⎫--⎢⎥ ⎪⎝⎭⎢⎥=-⎢⎥⎛⎫-- ⎪⎢⎥⎝⎭⎣⎦21113929nn ⎛⎫=+-- ⎪⎝⎭.经检验,当1n =时也成立.21113929nn E n ⎛⎫∴=+-- ⎪⎝⎭.【点睛】关键点点睛:本题第三问的关键是分1n n a a +=和11n n a a +=+时讨论,最后再化简n E 的表达式即可.。
江苏省2024-2025学年高三上学期9月月考化学试题附解答
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化学注意:本试卷分第一部分选择题和第二部分非选择题,共100 分,考试时间75 分钟。
可能用到的相对原子质量:H-1 C-12 O-16 Na-23 S-32 C1-35.5 K-39 Co-59 单项选择题:本题包括13 小题,每小题3 分,共计39 分。
每小题只有一个选项符合题意。
1.2024 年 4 月 24 日是第九个“中国航天日”,主题是“极目楚天 共襄星汉”。
下列有关中国空间站说法不正确的是()A .太阳能电池中的单晶硅——半导体材料B .外层的热控保温材料石墨烯——无机材料C .外表面的高温结构碳化硅陶瓷——硅酸盐材料D .外壳的烧蚀材料之一酚醛树脂——高分子材料2.反应可制含氯消毒剂。
下列说法正确的()A .HCl 和NaCl 所含化学键类型相同B .的空间结构为三角锥形C .是由极性键构成的非极性分子D .中子数为18的Cl 原子:3.实验室制取并探究其性质的实验原理和装置均正确的是()A .制取B .收集C .验证漂白性D .吸收尾气中的4.对金属材料中C 、H 、O 、N 、S 的含量进行定性和定量分析,可以确定金属材料的等级。
下列说法正确的是()A .电离能大小:B .沸点高低:C .酸性强弱:D .半径大小:阅读下列材料,完成5~7题周期表中第ⅤA 族元素及其化合物应用广泛。
以为原料可制得、等产品;32222NaClO 4HCl 2NaCl 2ClO Cl 2H O +=+↑+↑+3ClO -2H O 3518Cl 2SO 2SO 2SO 2SO 2SO ()()11N O I I >22H S H O >233H CO HNO >()()23O N r r -->3NH 3HNO 43NH NO 3PH可以用来防治仓库害虫,次磷酸()是一元酸,具有较强还原性,可由与反应制得;砷化镓可用作半导体材料,其晶胞结构如图所示;锑(Sb )导电性能好,在电池行业有独特应用。
山西省部分学校2024-2025学年高三上学期9月月考化学试题(含解析)
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2025届高三9月质量检测化学全卷满分100分,考试时间75分钟。
注意事项:1.答题前,先将自己的姓名、准考证号填写在试卷和答题卡上,并将条形码粘贴在答题卡上的指定位置。
2.请按题号顺序在答题卡上各题目的答题区域内作答,写在试卷、草稿纸和答题卡上的非答题区域均无效。
3.选择题用2B 铅笔在答题卡上把所选答案的标号涂黑;非选择题用黑色签字笔在答题卡上作答;字体工整,笔迹清楚。
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可能用到的相对原子质量:H1 B11 C12 N14 O16 S32 Cl35.5 Ca40 Fe56一、选择题:本题共14小题,每小题3分,共42分。
在每小题给出的四个选项中,只有一项符合题目要求。
1.化学与科技创新密切相关。
下列说法错误的是( )A.“快舟一号甲”运载火箭利用燃料与氧化剂反应放热并产生大量气体实现助推B.“天目一号”气象星座卫星的光伏发电系统工作时可将化学能转化为电能C “爱达·魔都号”邮轮使用的镁铝合金具有密度低、抗腐蚀性强的特点D.“AG60E ”电动飞机使用的动力型锂电池具有质量轻、比能量高的特点2.下列化学用语表述正确的是( )A.基态Cr 原子的价层电子排布图为B.的化学名称为甲基丁烯C.分子的VSEPR 模型为D.用电子式表示的形成过程为:3.下列生产活动中对应的离子方程式正确的是( )A.铅酸蓄电池充电时的阳极反应:B.向冷的石灰乳中通入制漂白粉:C.用溶液除去锅炉水垢中的:D.用葡萄糖制镜或保温瓶胆:()332CH CH C CH =3-2--3NH 2CaCl 222Pb 2H O 2e PbO 4H +-++-=+2Cl 22Cl 2OH Cl ClO H O---+=++23Na CO 4CaSO 224334CaSO (s)CO (aq)CaCO (s)SO (aq)--++A()2432CH OH(CHOH)CHO 2Ag NH OH ⎡⎤+−−→⎣⎦△24432CH OH(CHOH)COO NH 2Ag 3NH H O-+++↓++4.某化学兴趣小组进行如下实验:实验①:向晶体中滴加浓盐酸,产生黄绿色气体。
贵州省贵阳市第一中学2024-2025学年高三上学期9月月考化学试题(含答案)
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化学试卷注意事项:1.答题前,考生务必用黑色碳素笔将自己的姓名、准考证号、考场号、座位号在答题卡上填写清楚。
2.每小题选出答案后,用2B 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。
在试题卷上作答无效。
3.考试结束后,请将本试卷和答题卡一并交回。
满分100分,考试用时75分钟。
以下数据可供解题时参考。
可能用到的相对原子质量:H-1 C-12 N-14 O-16 Sc-45一、选择题:本题共14小题,每小题3分。
在每小题给出的四个选项中,只有一项是符合题目要求的。
1.下列说法错误的是( )A .纯碱和()3Al OH 都可以用作胃酸中和剂,制胃药B .“青铜器时期”早于“铁器时期”的原因之一是铜比铁稳定C .快餐餐盒使用的聚乳酸材料可由乳酸()缩聚制得D .诗句“司南之杓(勺)投之于地,其柢(勺柄)指南”,“杓”的材质为34Fe O 2.下列说法正确的是( ) A .1-丁醇的键线式:B .基态铜原子的价层电子排布式为923d 4sC .3Fe +的价电子的轨道表示式:D .2SO 的VSEPR 模型:3.A N 为阿伏加德罗常数的值。
下列叙述正确的是( ) A .1mol 苯乙烯中含有碳碳双键数为4A NB .用电解粗铜的方法精炼铜,当电路中通过的电子数为A N 时,阳极应有32gCu 转化为2Cu +C .pH 12=的23Na CO 溶液中OH -数目为0.01A N D .常温常压下,32g 24N H 所含共价键的数目为5A N 4.下列方程式与所给事实相符的一项是( )A .22Na O 吸收2SO 气体:2222322Na O 2SO 2Na SO O +=+B .乙醇与227K Cr O 酸性溶液反应:233232273CH CH OH 2Cr O 16H 3CH COOH 4Cr 11H O -+++→++ C .泡沫灭火器的反应原理:()23223AlO HCO H O Al OH CO --++=↓+↑D .向溶液中通入少量2CO :5.奥司他韦可以用于治疗流行性感冒,其结构如图1所示。
2024-2025学年进才中学高三上9月月考化学试卷
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上海市进才中学2024学年第一学期高三年级化学测试2024.09相对原子质量:H-1;C-12;N-14;O-16;S-32一、元素化合物1(本题共16分)某学习小组发现中学阶段只学习了SO3具有酸性氧化物的性质,为了探究SO3的其他化学性质,设计了如下实验装置验证。
1.写出铜与浓硫酸反应的化学方程式________________________________。
该反应中,浓硫酸体现的性质有________。
(不定项)A.难挥发性B.强氧化性C.脱水性D.强酸性2.由于浓硫酸具有较大的粘稠度,分液漏斗中的浓硫酸很难顺利流入到烧瓶中,实验中可以将分液漏斗更换为________(填写仪器名称)。
3.装置B不具有的作用是________。
A.干燥气体B.混合气体C.与气体反应D.观察通入的气泡速率4.写出H2S的电子式________。
5.下列SO3与H2S反应的方程式中,不可能成立的有________。
(不定项)A.H2S+SO3=H2O+SO2+S B.H2S+2SO3=H2SO4+SO2+SC.H2S+SO3=H2O+O2+2S D.3H2S+SO3=3H2O+4S6.实验进行一段时间后,装置D中有_____________和_____________出现(填写实验现象)。
7.指导教师指出,该装置无法验证SO3与H2S的反应,请问为什么?_______________________________8.E装置中吸收尾气的试剂可以是________。
(不定项)A.KMnO4溶液B.BaCl2溶液C.蒸馏水D.NaOH溶液二、分子结构(本题共16分)分子结构在很大程度上决定了物质的各种性质,所以研究分子结构有很重要的意义。
9.下列分子或离子的空间构型是三角锥形的有________。
(选填编号)①BF3②SO3③PH3④O3⑤ClO3-⑥CO32-⑦H3O+⑧CH3-10.下列分子或离子的中心原子是sp2杂化的有________。
2024-2025学年高三上学期第一次联考(9月月考) 数学试题[含答案]
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2024~2025学年高三第一次联考(月考)试卷数学考生注意:1.本试卷分选择题和非选择题两部分.满分150分,考试时间120分钟.2.答题前,考生务必用直径0.5毫米黑色墨水签字笔将密封线内项目填写清楚.3.考生作答时,请将答案答在答题卡上.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑;非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答,超出答题区域书写的答案无效,在试题卷、草稿纸上作答无效.4.本卷命题范围:集合、常用逻辑用语、不等式、函数、导数及其应用.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,只有一项是符合题目要求的.1.已知集合,,则集合的真子集的个数为(){}4,3,2,0,2,3,4A =---{}2290B x x =-≤A B ⋂A.7B.8C.31D.322.已知,,则“,”是“”的( )0x >0y >4x ≥6y ≥24xy ≥A.充分不必要条件 B.必要不充分条件C.充要条件D.既不充分又不必要条件3.国家速滑馆又称“冰丝带”,是北京冬奥会的标志性场馆,拥有亚洲最大的全冰面设计,但整个系统的碳排放接近于零,做到了真正的智慧场馆、绿色场馆,并且为了倡导绿色可循环的理念,场馆还配备了先进的污水、雨水过滤系统,已知过滤过程中废水的污染物数量与时间(小时)的关系为()mg /L N t (为最初污染物数量,且).如果前4个小时消除了的污染物,那么污染物消0e kt N N -=0N 00N >20%除至最初的还需要( )64%A.3.8小时 B.4小时C.4.4小时D.5小时4.若函数的值域为,则的取值范围是()()()2ln 22f x x mx m =-++R m A.B.()1,2-[]1,2-C.D.()(),12,-∞-⋃+∞(][),12,-∞-⋃+∞5.已知点在幂函数的图象上,设,(),27m ()()2n f x m x =-(4log a f =,,则,,的大小关系为( )()ln 3b f =123c f -⎛⎫= ⎪⎝⎭a b c A.B.c a b <<b a c<<C. D.a c b <<a b c<<6.已知函数若关于的不等式的解集为,则的()()2e ,0,44,0,x ax xf x x a x a x ⎧->⎪=⎨-+-+≤⎪⎩x ()0f x ≥[)4,-+∞a 取值范围为( )A.B. C. D.(2,e ⎤-∞⎦(],e -∞20,e ⎡⎤⎣⎦[]0,e 7.已知函数,的零点分别为,,则( )()41log 4xf x x ⎛⎫=- ⎪⎝⎭()141log 4xg x x ⎛⎫=- ⎪⎝⎭a b A. B.01ab <<1ab =C.D.12ab <<2ab ≥8.已知,,,且,则的最小值为( )0a >0b >0c >30a b c +-≥6b a a b c ++A. B. C. D.29495989二、多选题:本题共3小题,每小题6分,共18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6分,部分选对的得部分分,有选错的得0分.9.下列说法正确的是( )A.函数是相同的函数()f x =()g x =B.函数6()f x =C.若函数在定义域上为奇函数,则()313xx k f x k -=+⋅1k =D.已知函数的定义域为,则函数的定义域为()21f x +[]1,1-()f x []1,3-10.若,且,则下列说法正确的是()0a b <<0a b +>A. B.1a b >-110a b+>C. D.22a b <()()110a b --<11.已知函数,则下列说法正确的是( )()()3233f x x x a x b=-+--A.若在上单调递增,则的取值范围是()f x ()0,+∞a (),0-∞B.点为曲线的对称中心()()1,1f ()y f x =C.若过点可作出曲线的三条切线,则的取值范围是()2,m ()()3y f x a x b =+-+m ()5,4--D.若存在极值点,且,其中,则()f x 0x ()()01f x f x =01x x ≠1023x x +=三、填空题:本题共3小题,每小题5分,共15分.12.__________.22lg 2lg3381527log 5log 210--+⋅+=13.已知函数称为高斯函数,表示不超过的最大整数,如,,则不等式[]y x =x []3.43=[]1.62-=-的解集为__________;当时,的最大值为__________.[][]06x x <-0x >[][]29x x +14.设函数,若,则的最小值为__________.()()()ln ln f x x a x b =++()0f x ≥ab 四、解答题:本题共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.(本小题满分13分)已知全集,集合,.U =R {}231030A x x x =-+≤{}220B x xa =+<(1)若,求和;8a =-A B ⋂A B ⋃(2)若,求的取值范围.()UA B B ⋂= a 16.(本小题满分15分)已知关于的不等式的解集为.x 2280ax x --<{}2x x b-<<(1)求,的值;a b (2)若,,且,求的最小值.0x >2y >-42a bx y +=+2x y +17.(本小题满分15分)已知函数.()()()211e 2x f x x ax a =--∈R (1)讨论的单调性;()f x (2)若对任意的恒成立,求的取值范围.()e x f x x ≥-[)0,x ∈+∞a 18.(本小题满分17分)已知函数是定义在上的奇函数.()22x xf x a -=⋅-R(1)求的值,并证明:在上单调递增;a ()f x R (2)求不等式的解集;()()23540f x x f x -+->(3)若在区间上的最小值为,求的值.()()442x x g x mf x -=+-[)1,-+∞2-m 19.(本小题满分17分)已知函数.()()214ln 32f x x a x x a =---∈R (1)若,求的图像在处的切线方程;1a =()f x 1x =(2)若恰有两个极值点,.()f x 1x ()212x x x <(i )求的取值范围;a (ii )证明:.()()124ln f x f x a+<-数学一参考答案、提示及评分细则1.A 由题意知,又,所以{}2290B x x ⎡=-=⎢⎣∣ {}4,3,2,0,2,3,4A =---,所以的元素个数为3,真子集的个数为.故选.{}2,0,2A B ⋂=-A B ⋂3217-=A 2.A 若,则,所以“”是“”的充分条件;若,满足4,6x y 24xy 4,6x y 24xy 1,25x y ==,但是,所以“”不是“”的必要条件,所以“”是24xy 4x <4,6x y 24xy 4,6x y “”的充分不必要条件.故选A.24xy 3.B 由题意可得,解得,令,可得4004e 5N N -=44e 5k -=20004e 0.645t N N N -⎛⎫== ⎪⎝⎭,解得,所以污染物消除至最初的还需要4小时.故选B.()248e e ek kk---==8t =64%4.D 依题意,函数的值域为,所以,解得()()2ln 22f x x mx m =-++R ()2Δ(2)420m m =--+ 或,即的取值范围是.故选D.2m 1m - m ][(),12,∞∞--⋃+5.C 因为是軍函数,所以,解得,又点在函数的图()()2nf x m x =-21m -=3m =()3,27()n f x x =象上,所以,解得,所以,易得函数在上单调递增,又273n=3n =()3f x x =()f x (),∞∞-+,所以.故选C.1241ln3lne 133log 2log 2->==>=>=>a c b <<6.D 由题意知,当时,;当时,;当时,(),4x ∞∈--()0f x <[]4,0x ∈-()0f x ()0,x ∞∈+.当时,,结合图象知;当时,,当()0f x 0x ()()()4f x x x a =-+-0a 0x >()e 0x f x ax =- 时,显然成立;当时,,令,所以,令,解0a =0a >1e x x a (),0e x x g x x =>()1e xxg x -='()0g x '>得,令0,解得,所以在上单调递增,在上单调递减,所以01x <<()g x '<1x >()g x ()0,1()1,∞+,所以,解得综上,的取值范围为.故选D.()max 1()1e g x g ==11e a0e a < a []0,e 7.A 依题意得,即两式相减得4141log ,41log ,4a b a b ⎧⎛⎫=⎪ ⎪⎝⎭⎪⎨⎛⎫⎪= ⎪⎪⎝⎭⎩441log ,41log ,4a ba b ⎧⎛⎫=⎪ ⎪⎪⎝⎭⎨⎛⎫⎪-= ⎪⎪⎝⎭⎩.在同一直角坐标系中作出的图()44411log log log 44a ba b ab ⎛⎫⎛⎫+==- ⎪ ⎪⎝⎭⎝⎭4141log ,log ,4xy x y x y ⎛⎫=== ⎪⎝⎭象,如图所示:由图象可知,所以,即,所以.故选A.a b >1144ab⎛⎫⎛⎫< ⎪ ⎪⎝⎭⎝⎭()4log 0ab <01ab <<8.C 因为,所以,所以30a b c +- 30a b c +> 11911121519966399939911b a b a b b b b a b c a b a b a a a a ⎛⎫++=+=++--=-= ⎪+++⎝⎭++ ,当且仅当,即时等号成立,所以的最小值为.故选C.1911991b b a a ⎛⎫+= ⎪⎝⎭+29b a =6b aa b c ++599.AD 由解得,所以,由,解得10,10x x +⎧⎨-⎩ 11x - ()f x =[]1,1-210x -,所以的定义域为,又,故函数11x - ()g x =[]1,1-()()f x g x ===与是相同的函数,故A 正确;,()f x ()g x ()6f x ==当且仅当方程无解,等号不成立,故B 错误;函数=2169x +=在定义域上为奇函数,则,即,即()313x x k f x k -=+⋅()()f x f x -=-331313x xx x k k k k ----=-+⋅+⋅,即,整理得,即,()()33313313x x xxxxk k k k ----=-+⋅+⋅313313x x x x k kk k ⋅--=++⋅22919x x k k ⋅-=-()()21910x k -+=所以,解得.当时,,该函数定义域为,满足,210k -=1k =±1k =()1313xx f x -=+R ()()f x f x -=-符合题意;当时,,由可得,此时函数定义域为1k =-()13311331x x xxf x --+==--310x -≠0x ≠,满足,符合题意.综上,,故C 错误;由,得{}0x x ≠∣()()f x f x -=-1k =±[]1,1x ∈-,所以的定义域为,故D 正确.故选AD.[]211,3x +∈-()f x []1,3-10.AC 因为,且,所以,所以,即,故A 正确;0a b <<0a b +>0b a >->01a b <-<10ab -<<因为,所以,故В错误;因为,所以,0,0b a a b >->+>110a ba b ab ++=<0a b <<,a a b b =-=由可得,所以,故C 正确;因为当,此时,故0a b +>b a >22a b <11,32a b =-=()()110a b -->D 错误.故选AC.11.BCD 若在上单调递增,则在上佰成立,所以()f x ()0,∞+()23630f x x x a '=-+- ()0,x ∞∈+,解得,即的取值范围是,故A 错误;因为()min ()13630f x f a '==--'+ 0a a (],0∞-,所以,又()()32333(1)1f x x x a x b x ax b =-+--=---+()11f a b =--+,所以点()()()332(21)21(1)1222f x f x x a x b x ax b a b -+=-----++---+=--+为曲线的对称中心,故B 正确;由题意知,所以()()1,1f ()y f x =()()3233y f x a x b xx =+-+=-,设切点为,所以切线的斜率,所以切线的方程为236y x x =-'()32000,3x x x -20036k x x =-,所以,整理得()()()3220000336y x x x x x x --=--()()()322000003362m xx x x x --=--.记,所以3200029120x x x m -++=()322912h x x x x m =-++()26h x x '=-,令,解得或,当时,取得极大值,当时,1812x +()0h x '=1x =2x =1x =()h x ()15h m =+2x =取得极小值,因为过点可作出曲线的三条切线,所以()h x ()24h m=+()2,m ()()3y f x a x b =+-+解得,即的取值范围是,故C 正确;由题意知()()150,240,h m h m ⎧=+>⎪⎨=+<⎪⎩54m -<<-m ()5,4--,当在上单调递增,不符合题意;当,()223633(1)f x x x a x a =-+-=--'()0,a f x (),∞∞-+0a >令,解得,令,解得在()0f x '>1x <-1x >+()0f x '<11x -<<+()f x 上单调递增,在上单调递堿,在上单调递增,因为,1∞⎛- ⎝1⎛+ ⎝1∞⎛⎫+ ⎪ ⎪⎝⎭存在极值点,所以.由,得,令,所以,()f x 0x 0a >()00f x '=()2031x a-=102x x t+=102x t x =-又,所以,又,()()01f x f x =()()002f x f t x =-()()32333(1)1f x x x a x b x ax b =-+--=---+所以,又,所以()()()330000112121x ax b t x a t x b ---+=-----+()2031x a-=,化简得()()()()()()()322320000000013112121312x x x b x x b t x x t x b----=----=------,又,所以,故D 正确.故选BCD.()()20330t x t --=010,30x x x t ≠-≠103,23t x x =+=12. 由题意知10932232862log 184163381255127log 5log 210log 5log 121027---⎛⎫+⋅+=+⋅-+ ⎪⎝⎭62511411410log 5log 2109339339=-⋅+=-+=13.(2分)(3分) 因为,所以,解得,又函数[)1,616[][]06x x <-[][]()60x x -<[]06x <<称为高斯函数,表示不超过的最大整数,所以,即不等式的解集为.当[]y x =x 16x < [][]06x x <-[)1,6时,,此时;当时,,此时01x <<[]0x =[]2[]9x x =+1x []1x ,当且仅当3时等号成立.综上可得,当时,的[][][]2119[]96x x x x ==++[]x =0x >[]2[]9x x +最大值为.1614. 由题意可知:的定义域为,令,解得令,解21e -()f x (),b ∞-+ln 0x a +=ln ;x a =-()ln 0x b +=得.若,当时,可知,此时,不合题1x b =-ln a b -- (),1x b b ∈--()ln 0,ln 0x a x b +>+<()0f x <意;若,当时,可知,此时,不合ln 1b a b -<-<-()ln ,1x a b ∈--()ln 0,ln 0x a x b +>+<()0f x <题意;若,当时,可知,此时;当ln 1a b -=-(),1x b b ∈--()ln 0,ln 0x a x b +<+<()0f x >时,可知,此时,可知若,符合题意;若[)1,x b ∞∈-+()ln 0,ln 0x a x b ++ ()0f x ln 1a b -=-,当时,可知,此时,不合题意.综上所ln 1a b ->-()1,ln x b a ∈--()ln 0,ln 0x a x b +<+>()0f x <述:,即.所以,令,所以ln 1a b -=-ln 1b a =+()ln 1ab a a =+()()ln 1h x x x =+,令,然得,令,解得,所以在()ln 11ln 2h x x x '=++=+()0h x '<210e x <<()0h x '>21e x >()h x 上单调递堿,在上单调递增,所以,所以的最小值为.210,e ⎛⎫ ⎪⎝⎭21,e ∞⎛⎫+ ⎪⎝⎭min 2211()e e h x h ⎛⎫==- ⎪⎝⎭ab 21e -15.解:(1)由题意知,{}2131030,33A x x x ⎡⎤=-+=⎢⎥⎣⎦∣ 若,则,8a =-{}()22802,2B x x =-<=-∣所以.(]1,2,2,33A B A B ⎡⎫⋂=⋃=-⎪⎢⎣⎭(2)因为,所以,()UA B B ⋂= ()UB A ⊆ 当时,此时,符合题意;B =∅0a 当时,此时,所以,B ≠∅0a <{}220Bx x a ⎛=+<= ⎝∣又,U A ()1,3,3∞∞⎛⎫=-⋃+ ⎪⎝⎭13解得.209a -< 综上,的取值范围是.a 2,9∞⎡⎫-+⎪⎢⎣⎭16.解:(1)因为关于的不等式的解集为,x 2280ax x --<{2}xx b -<<∣所以和是关于的方程的两个实数根,且,所以2-b x 2280ax x --=0a >22,82,b a b a⎧=-⎪⎪⎨⎪-=-⎪⎩解得.1,4a b ==(2)由(1)知,所以1442x y +=+()()()221141422242241844242y xx y x y x y x y y x ⎡⎤+⎛⎫⎡⎤+=++-=+++-=+++-⎢⎥ ⎪⎣⎦++⎝⎭⎣⎦,179444⎡⎢+-=⎢⎣ 当且仅当,即时等号成立,所以.()2242y x y x +=+x y ==2x y +74-17.解:(1)由题意知,()()e e x x f x x ax x a=-=-'若,令.解得,令,解得,所以在上单调递琙,在0a ()0f x '<0x <()0f x '>0x >()f x (),0∞-上单调递增.()0,∞+若,当,即时,,所以在上单调递增;0a >ln 0a =1a =()0f x ' ()f x (),∞∞-+当,即时,令,解得或,令,解得,ln 0a >1a >()0f x '>0x <ln x a >()0f x '<0ln x a <<所以在上单调递增,在上单调递减,在上单调递增;()f x (),0∞-()0,ln a ()ln ,a ∞+当,即时,令,解得或,令,解得,ln 0a <01a <<()0f x '>ln x a <0x >()0f x '<ln 0a x <<所以在上单调递增,在上单调递减,在上单调递增.()f x (),ln a ∞-()ln ,0a ()0,∞+综上,当时,在上单调递减,在上单调递增;当时,在0a ()f x (),0∞-()0,∞+01a <<()f x 上单调递增,在上单调递减,在上单调递增当时,在上(,ln )a ∞-()ln ,0a ()0,∞+1a =()f x (),∞∞-+单调递增;当时,在上单调递增,在上单调递减,在上单调递增.1a >()f x (),0∞-()0,ln a ()ln ,a ∞+(2)若对任意的恒成立,即对任意的恒成立,()e xf x x - [)0,x ∞∈+21e 02xx ax x -- [)0,x ∞∈+即对任意的恒成立.1e 102x ax -- [)0,x ∞∈+令,所以,所以在上单调递增,当()1e 12x g x ax =--()1e 2x g x a=-'()g x '[)0,∞+,即时,,所以在上单调递增,所以()10102g a =-' 2a ()()00g x g '' ()g x [)0,∞+,符合题意;()()00g x g = 当,即时,令,解得,令,解得,所()10102g a =-<'2a >()0g x '>ln 2a x >()0g x '<0ln 2a x < 以在上单调递减,()g x 0,ln 2a ⎡⎫⎪⎢⎣⎭所以当时,,不符合题意.0,ln 2a x ⎛⎫∈ ⎪⎝⎭()()00g x g <=综上,的取值范围是.a (],2∞-18.(1)证明:因为是定义在上的奇函数,所以,()f x R ()010f a =-=解得,所以,1a =()22x xf x -=-此时,满足题意,所以.()()22x x f x f x --=-=-1a =任取,所以12x x <,()()()()211122121211122222122222222122x x x x x x x x x x x x f x f x x x --⎛⎫--=---=--=-+ ⎪++⎝⎭又,所以,即,又,12x x <1222x x <12220x x -<121102x x ++>所以,即,所以在上单调递增.()()120f x f x -<()()12f x f x <()f x R (2)解:因为,所以,()()23540f x x f x -+->()()2354f x x f x ->--又是定义在上的奇函数,所以,()f x R ()()2354f x x f x ->-+又在上单调递增,所以,()f x R 2354x x x ->-+解得或,即不等式的解集为.2x >23x <-()()23540f x x f x -+->()2,2,3∞∞⎛⎫--⋃+ ⎪⎝⎭(3)解:由题意知,令,()()()44244222xxxxxxg x mf x m ---=+-=+--322,,2x x t t ∞-⎡⎫=-∈-+⎪⎢⎣⎭所以,所以.()2222442x xxxt --=-=+-()2322,,2y g x t mt t ∞⎡⎫==-+∈-+⎪⎢⎣⎭当时,在上单调递增,所以32m -222y t mt =-+3,2∞⎡⎫-+⎪⎢⎣⎭,解得,符合题意;2min317()323224g x m m ⎛⎫=-++=+=- ⎪⎝⎭2512m =-当时,在上单调递减,在上单调递增,32m >-222y t mt =-+3,2m ⎛⎫- ⎪⎝⎭(),m ∞+所以,解得或(舍).222min ()2222g x m m m =-+=-=-2m =2m =-综上,的值为或2.m 2512-19.(1)解:若,则,所以,1a =()214ln 32f x x x x =---()14f x x x =--'所以,又,()14112f =--='()1114322f =--=所以的图象在处的切线方程为,即.()f x 1x =()1212y x -=-4230x y --=(2)(i )解:由题意知,()22444a x a x x x af x x x x x '---+=--==-又函数恰有两个极值点,所以在上有两个不等实根,()f x ()1212,x x x x <240x x a -+=()0,∞+令,所以()24h x x x a =-+()()00,240,h a h a ⎧=>⎪⎨=-<⎪⎩解得,即的取值范围是.04a <<a ()0,4(ii )证明:由(i )知,,且,12124,x x x x a +==04a <<所以()()2212111222114ln 34ln 322f x f x x a x x x a x x ⎛⎫⎛⎫+=---+--- ⎪ ⎪⎝⎭⎝⎭()()()2212121214ln ln 62x x a x x x x =+-+-+-,()()()21212121214ln 262x x a x x x x x x ⎡⎤=+--+--⎣⎦()116ln 1626ln 22a a a a a a =----=-+要证,即证,只需证.()()124ln f x f x a+<-ln 24ln a a a a -+<-()1ln 20a a a -+-<令,所以,()()()1ln 2,0,4m a a a a a =-+-∈()11ln 1ln a m a a a a a -=-++=-'令,所以,所以即在上单调递减,()()h a m a ='()2110h a a a =--<'()h a ()m a '()0,4又,所以,使得,即,()()1110,2ln202m m '-'=>=<()01,2a ∃∈()00m a '=001ln a a =所以当时,,当时,,所以在上单调递增,在()00,a a ∈()0m a '>()0,4a a ∈()0m a '<()m a ()00,a 上单调递减,所以.()0,4a ()()()max 00000000011()1ln 2123m a m a a a a a a a a a ==-+-=-+-=+-令,所以,所以在上单调递增,所以()()13,1,2u x x x x =+-∈()2110u x x =->'()u x ()1,2,所以,即,得证.()000111323022u a a a =+-<+-=-<()0m a <()()124ln f x f x a +<-。
湖北省宜城市第一中学2024-2025学年高三上学期9月月考英语试卷(含答案)
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2025届高三9月月考参考答案第一部分听力1—5 BABCA 6—10 CACBA 11—15 BCBCA 16—20 BCBACA: CBDB: DCABC: AACDD: CDBAGABDFABCDC BAACD BDCAB56.themselves 57.creativity 58.imaginative 59.to cross 60.supported 61.has received 62.consisting 63.which 64.as 65.whyDear Chris,With the Teachers’ Day approaching, I’m writing to express my gratitude to you. I present you a handcrafted gift with Chinese characteristics because you have taught me so much and inspired me greatly. The gift is a beautifully painted fan, featuring traditional Chinese landscapes. It symbolizes the elegance and charm of our culture.I sincerely hope it will bring you joy and you will continue to guide us in the future.Yours sincerely,Li Hua【导语】本篇书面表达属于应用文,要求考生给外教Chris一份有中国特色的礼物并写一封信表达对他的感谢。
其内容包括:赠礼原因、礼物简介和表达期待。
【详解】1.词汇积累感激:gratitude→appreciation鼓励:inspire→motivate代表:symbolize→stand for快乐:joy→happiness2.句式拓展简单句变复合句原句:With the Teachers’ Day approaching, I’m writing to express my gratitude to you.拓展句:As the Teachers’ Day is approaching, I’m writin g to express my gratitude to you.【点睛】【高分句型1】The gift is a beautifully painted fan, featuring traditional Chinese landscapes.(运用了现在分词作状语)【高分句型2】I sincerely hope it will bring you joy and you will continue to guide us in the future.(hope后运用了省略that的宾语从句)范文From that day on, I didn’t talk to Alex. Each day at school, Inavigated the hallways with a new route, avoiding the places he frequented. The shared excitement that once connected us was replaced by a painful silence. Alex tried to do something to mend it. One day, he even left me a note, trying to comfort me. But I looked at him coldly, “Leave me alone.” I knew I was hurting him, but the sting of rejection was too fresh, and I couldn’t find the words to bridge the gap.Finally, Mom talked with me, saying Alex wasn’t the one to blame. She reminded me that success and failure were parts of life’s journey. “You should have congratulated your friend on this instead of feeling pitiful for yourself,” she said seriously. It suddenly struck me that how unfairly I treated Alex. Seized by guilt, I followed mom’s advice and left Alex a note, expressing my feelings—not as an apology, but as a step towards healing and rebuilding our bond. It was time to face my fears and embrace the reality that our friendship could weather this storm.【导语】本文以人物为线索展开,讲述了作者和亚历克斯一起申请项目,结果作者被拒绝了,作者感到不公平,拒绝和亚历克斯说话。
四川省绵阳南山中学2024-2025学年高三上学期9月月考语文试题(答案+答题卡)
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绵阳南山中学高2022级9月月考语文试卷语文本试卷共 8 页,满分 150 分,考试时间 150 分钟。
注意事项:1.答题前,务必将自己的姓名、班级、考号填写在答题卡规定的位置上。
2.答选择题时,必须使用 2B 铅笔将答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦擦干净后,再选涂其它答案标号。
3.答非选择题时,必须使用 0.5 毫米黑色签字笔,将答案书写在答题卡规定的位置上。
4.所有题目必须在答题卡上作答,在试题卷上答题无效。
5.考试结束后,只将答题卡交回。
一、现代文阅读(35分)(一)现代文阅读I(本题共5小题,19分)阅读下面的文字,完成1~5题。
材料一:中国社会学自诞生之日起,就树立了注重社会调查的学术传统。
今天,发展中国特色社会主义社会学,需要继承这一学术传统,通过社会调查走进大众的日常生活和精神世界,不断加深对“人”的认识和理解,在将理论创新与现实关怀融为一体的过程中,承担起建构中国社会学自主知识体系的责任。
社会调查是连接理论与实践的枢纽环节,也是对社会的基础与本质予以理解的一种方法体系。
首先,社会调查是具象化的。
社会调查扎根于特定的地域和人群,以一时一地的社会风貌为对象,试图理解和把握在具体情景中展现出来的情感、气质、风俗、社会发展进程以及变迁趋势。
其次,社会调查是历史性的。
要对一个特定资料作出有效解释,就要善于挖掘其背后的历史发展脉络,深入了解那些习以为常却又容易视而不见的背景;就要将材料置于时间维度之下,避免片面解读或过度诠释。
再次,社会调查是整合性的。
人类社会发展既存在一般规律,也充斥着碎片化、非预期和非理性现象,社会调查就是要将这些规律和现象都纳入总体性分析框架,做到对事实本身的把握和尊重。
最后,社会调查长于共情。
在具体场景中所唤起的感性认识,可以使调查者尽可能避免成为置身事外的旁观者。
这种从情感上激发的对“人”的体恤之情,可以使事物“活”的一面被充分挖掘出来,由此提炼出的理论便有了直指人心的力量。
贵州省毕节市七星关区燕子口中学2024-2025学年高三上学期9月月考试卷英语
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毕节市七星关区燕子口中学高三2024年9月月考试卷英语试题第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话,每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例:How much is the shirt?A. £19.15.B. £9.18.C. £9.15.答案是C。
1. What is Kate doing?A. Boarding a flight.B. Arranging a trip.C. Seeing a friend off.2. What are the speakers talking about?A. A pop star.B. An old song.C. A radio program.3. What will the speakers do today?A. Go to an art show.B. Meet the man’s aunt.C. Eat out with Mark.4. What does the man want to do?A. Cancel an order.B. Ask for a receipt.C. Reschedule a delivery.5. When will the next train to Bedford leave?A. At 9:45.B. At 10:15.C. At 11:00.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C 三个选项中选出最佳选项。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
湖北省襄阳市2024-2025学年高三上学期9月月考试题 物理含答案
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襄阳2025届高三上学期9月月考物理试卷(答案在最后)一、单项选择题:本题共7小题,每小题4分,共28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.已知氘()21H 核的质量为D m ,质子()11H 的质量为p m ,中子()10n 的质量为nm ,光速为c ,则氘核的比结合能为()A .()212p n D m m m c +-B .()213p n D m m m c +-C .212D m c D .213D m c 2.北京时间2024年4月25日,神舟十八号载人飞船发射取得成功。
假如“神舟十八号”仅受地球施加的万有引力作用下,绕地球沿如图椭圆形轨道运动,它在A B C 、、三点以下关系正确的是()A .从A 点运动至B 点的过程中,“神舟十八号”的机械能不断减少B .从A 点运动至C 点的时间小于从C 点运动至B 点的时间C .“神舟十八号”在A 点处的加速度最小D .“神舟十八号”在B 点处受到的地球施加的万有引力最大3.如图,在水池底中部放一线状光源,光源平行于水面,则水面观察到的发光区域形状为()A .B .C .D .4.一物体做匀变速直线运动,其运动的位移x 随时间的变化关系图像如图所示,则在4s t =末,物体的速度大小为()A .3m /sB .4m /sC .5m /sD .6m /s5.如图所示,有一个边长为L 的立方体空间ABCD MNPQ -的导体棒沿AP 方向放置。
空间内加上某一方向的匀强磁场(图中未画出),磁感应强度的大小为B 。
在导体棒中通以从A 至P 、大小为I 的电流,则关于导体棒受到的安培力,下列说法中正确的是()A .若磁场沿M 指向AB .若磁场沿M 指向AC .若磁场沿M 指向Q 的方向,安培力的大小为2ILB D .若磁场沿M 指向Q 的方向,安培力的大小为32ILB 6.如图所示,半球形容器内有三块不同长度的滑板AO BO CO '''、、,其下端都固定于容器底部O '点,上端搁在容器侧壁上,与水平面间的夹角分别为304537︒︒︒、、。
湖北省沙市2024-2025学年高三上学期9月月考试题 英语含答案
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2024—2025学年度上学期2022级9月月考英语试卷(答案在最后)命题人:考试时间:2024年9月25日第一部分听力(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 weather like now?A.Hot.B.Windy.C.Cold.2.What does the woman do?A.A dentist.B.A saleswoman.C.A waitress.3.What are the speakers mainly talking about?A.What materials to prepare.B.How many guests to invite.C.When to decorate the house.4.How does the man feel about the final exam?A.Worried.B.Confident.C.Relaxed.5.Where are the speakers?A.At the man’s house.B.At a housing agency.C.At a household service company.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听下面一段对话,回答6-7题。
绵阳南山中学2024-2025学年高三上学期9月月考英语试题(含答案)
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2024年9月绵阳南山中学2024-2025学年高三9月月考英语试卷命题人:审题人:考试时间共120分钟,满分150分注意事项:1.答题前,考生务必在答题卡上将自己的学校、姓名、班级、准考证号用0.5毫米黑色签字笔填写清楚,考生考试条形码由监考老师粘贴在答题卡上的“条形码粘贴处”。
2. 选择题使用2B铅笔填涂在答题卡上对应题目标号的位置上,如需改动,用橡皮擦擦干净后再填涂其它答案;非选择题用0.5毫米黑色签字笔在答题卡的对应区域内作答, 超出答题区域答题的答案无效;在草稿纸上、试卷上答题无效。
3.考试结束后由监考老师将答题卡收回。
第I 卷 (共95分)第一部分听力(共两节,满分30分)第一节(共5小题,每小题1.5分,满分7.5)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳答案,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一个小题。
每段对话仅读一遍。
1. What will the speakers do next?A. Pack bags.B. Gas up their car.C. Get into a taxi.2. What did Alice think of her new job?A. It was just so so.B. It was difficult.C. It was easy.3. What is Ben going to do later?A. Go home from work.B. Have dinner with Sarah.C. Visit his doctor.4. Who is the man?A. A gardener.B. A flower seller.C. A private home chef.5. What are the speakers talking about?A. Their favorite fruit.B. Items on a menu.C. Drink orders.第二节(共15小题,每小题1.5分,满分22.5)听下面5段对话或独白。
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2013-2014学年高三级9月月考英语试题Ⅰ.语言知识及应用(共两节,满分45分)第一节完形填空(共10小题,每小题2分,满分20分)There are about fifteen hundred languages in the world.But only a few of them are very ___1 . English is one of these. Many, many people use it, not only in England and the U. S. A, but in other parts of the world. About 200, 000, 000 speak it as their own language. It is difficult to say how many people are learning it as a __2_ language. Many millions are trying to do so.Is it easy or difficult to learn English? Different people may have different ___3 Have you ever __4 ads of this kind in the newspapers or magazines?“Learn English in six month, or your ___5__ back ...” “Easy and funny? Our records and tapes ___6__ you master your English in a month. From the first day your ___7__ will be excellent. Just send...” Of course, it never ___8__ quite like this.The only language that seems easy to learn is the mother tongue. We should remember that we all learned our own language well when we were ___9__. If we could learn English in the same way, it would not seem so difficult. ___10__ what a small child does. He listens to what people say. He tries what he hears. When he is using the language, talking in it, and ___11_ in it all the time, just imagine how much ___12__ that gets!So it is hard to say that learning English is easy, because a good command of English ___13__ upon a lot of practice. And practice needs great effort and ___14__ much time. Good teachers, records, tapes, books, and dictionaries will ___15 . But they cannot do the student’s work for him.1. A. difficult B. important C. necessary D. easy2. A. native B. foreign C. useful D. mother3. A. questions B. problems C. ideas D. answers4. A. found B. watched C. noticed D. known5. A. knowledge B. time C. money D. English6. A. make B. help C. let D. allow7. A. spelling B. grammar C. English D. pronunciation8. A. happened B. know C. seemed D. felt9. A. students B. children C. babies D. grown-ups10. A. Imagine B. Mind C. Do D. Think of11. using B. thinking C. trying D. practicing12 A. time B. money C. language D. practice13 A. depends B. tries C. has D. takes14 A. uses B. takes C. gets D. costs15A. do B. work C. help D. master第二节:语法填空(共10小题,每小题1.5分,满分15分)Two summers ago I went to Changdao, a small island near Y antai. During the night, I heard strange baby-like 16 (cry). Islanders told me 17 was stray (迷路的;离群的) cats that were calling in the wild. Well, it was strange to me to see so many stray cats on such a small island. So I asked the locals 18 more details. Then they told me 19 surprising and sad story. The island is isolated in the sea and originally had no cats. Ten years ago, when the island _____20_______(develop) into a national park, tourists began to swarm in. These visitors from all parts of the country brought in money, but the rubbish ___21______(leave) by them drew more and more rats onto the island. ____22_________, islanders introduced the cats in order to get rid of the rats. ___23 when the number of rats decreased, the number of cats increased, and the cats began to prey on birds, chickens, and fish. For more than a month after that, my mind was haunted (回响) by the plaintive crying. So now, whenever I travel, I always hear in mind ____24____ I should bring in, because I don’t want to bring anything that is strange, unneeded or even ____25_________(danger) to my destination.Ⅲ.阅读第一节阅读理解(阅读下列短文,从每题所给的A、B、C和D项中,选出最佳选项。
)AMen have always believed that they are smarter than women. Now, a study has found that while this is certainly true, men also have to deal with the fact that they are also more stupid than the fairer sex.In the study, scientists measured the IQ of 2500 brothers and sisters and they found an uneven number of men not only in the top two percent, but also in the bottom two percent. The study's participants were tested on science, maths, English and mechanical abilities.Though there were twice as many men as women in the smartest group, there were also twice as many men among the dolts.The aggregate scores of men and women were similar.One of the study's authors, psychology professor Timothy Bates, said that the phenomenon may be because men have always been expected to be high achievers and women have been restricted to spend more time taking care of their home."The female developmental program may be tilted more towards ensuring survival and the safety of the middle ground.," the Daily Mail quoted Professor Bates, of Edinburgh University, as saying.The research tallies with past results that men were more likely than women to receive first class University degrees or thirds and women secured the seconds.It has been said that men are more ready to take risk when it comes to academics. Women have always found to be steadier in their learning.A past study has shown that women are securing more firsts and seconds, while men are continuing to receive more thirds.The argument for the change is that the increase of coursework at the cost of exams favors women's steady approach.26. The purpose of the passage is to tell us that ________.A. man are smarter then womenB. man are more stupid the womenC. a new fact about the IQ of men and women has been foundD. men are more likely to receive first class university degrees27. According to Timothy Bates, less women are in the smartest group because _________.A. they are born stupidB. they have to spend more time to take care of their home than menC. they don’t like to take riskD. they are not expected to be high achievers28. The underlined word tallies with in the eighth paragraph means________.A. agree withB. deal withC. go againstD. go with29. It can be inferred from the passage that______.A. Women are steadier in their learning.B. men are more ready to take risk in everythingC. women are securing more firsts and secondsD. women are doing much better in academy30. Which of the following questions has NOT been discussed in the passage?A. Why are men smarter than women?B. Why are men more stupid than women?C. How does the result go along with the past research?D. How can we help the men in the bottom?BA “blogger” is a person who writes on an Internet computer Web site called a “blog”. The word “blog” is a short way of saying Web log, or personal Web site. Anyone can start a blog, and they can write about anything they like.There are millions of blogs on the Internet today. They provide news, information and ideas in many people who read them. They contain links to other Web sites. And they provide a place for people to write about their ideas and react to the ideas of others.A research company called Perscus has studied more than 300 Web logs. It says that blogs are most popular with teenage girls. They use them to let their friends know what is happening in their lives. The study also says that more than 100,000 bloggers stopped taking part in the activity after a year.However, some people develop serious blogs to present political and other ideas. For example, the Republican and Democratic parties in the southern state of Kentucky recently started their own blogs . And American companies are beginning to use blogs to advertise their products.At the same time, some long-standing blogs have ended last week, blogging leader Dave Winer closed his free blog service “weblogs. com”. He says the site became too costly to continue. He started the blog four years ago. And thousands of people had written on it. They are now upset because they did not know that the site was closing.One blog that is still going strong is called Rebecca’s Pocket. Rebecca Blood created the Web-site in 1999. She wrote about the history of blogs on the site. That article led to a book called “The Weblog Handbook”. It has been translated into four languages so far.Ms. Blood says Rebecea’s Pocket gets about 30,000 visitors a month. She writes about anything and everything--politics, culture and movies. She recently provided medical advice. And she wrote about how to prevent people from stealing money from on -line bank accounts.31. The author wrote this passage mainly to _________.A. introduce an Inter net computer Web site called “blog”B. introduce a short way of saying WeblogC. tell readers about blogsD. tell readers how to write blogs32. From the passage we can learn that blogs cover almost everything EXCEPT ______.A. different ideasB. medical adviceC. advertisementsD. account passwords33. Which of the following statements is NOT true according to the passage?A. Politicians don' t use blogs at all.B. A lot of bloggers no longer write or read blogs.C. Among school children, fewer boys like to use blogs.D. People have 5 language versions of “The Weblog Handbook” to read.34. Dave Winer closed his “weblogs. Com” because __________.A. more than 100,000 bloggers stopped taking part in the activityB. American companies used blogs to advertise their productsC. people stole money from on -line bank accountsD. He couldn’t afford the increasing money needed to run the blogs35. The reason why Rebecca’s Pocket is still going strong is that ______.A. it was created by a womanB. it is about the history of blogsC. it provides useful information and adviceD. it has editions in at least four different languagesCAbout ten years ago, a young and very successful businessman named Josh was traveling down a Chicago neighborhood street. He was going a bit too fast in his shiny, black, 12 cylinder Jaguar XKE, which was only two months old.He was watching for kids rushing out from between parked cars and slowed down when he thought he saw something. As his car passed, no child came out, but a brick sailed out and—WHUMP! —it hit the Jag’s shiny black side door! Immediately Josh stopped the car, jumped out, seized the kid and pushed him up against a parked car. He shouted at the kid, “What was that all about and who are you? That’s my new Jag, that brick you threw is gonna cost you a lot of money. Why did you throw it? ”“Please, mister, please. . . I’m sorry! I didn’t know what else to do! ” begged the youngster. “I threw the brick because no one else would stop! ” Tears were streaming down the boy’s face as he pointed around the parked car. “It’s my brother, mister, ” he said. “He rolled of the curb (路沿) and fell out of his wheelcha ir and I can’t lift him up. ”Sobbing, the boy asked the businessman, “ Would you please help me get him back into his wheelchair? He’s hurt and he’s too heavy for me. ”Moved beyond words, the young businessman tried hard to swallow the rapidly swelling lump in his throat. Straining, he lifted the young man back into the wheelchair and took out his handkerchief and wiped the scrapers and cuts, checking to see that everything was going to be OK. He then watched the younger brother push him down the sidewalk toward their home.It was a long walk back to the black, shining 12 cylinder Jaguar XKE—a long and slow walk. Josh never did fix the side door of his Jaguar. He kept the dent (凹痕) to remind him not to go through life so fast that someone has to throw a brick at him to get his attention. Feel for the bricks of life coming at you.36.what did the driver reflect firstly when he found the boy throwing a brick onhis car?A.surprisedB.angryC. Strange D .sad37. The boy threw a brick at the businessman’s car because _____ .A. the businessman drove at a high speedB. he envied the brand-new car very muchC. he wanted to ask for some moneyD. he wanted to get help from the driver38. Which of the following is the right order of the story?a. The younger brother threw a brick at Josh’s car.b. The elder brother fell out of his wheelchair.c. The younger brother begged Josh for help.d. Josh lifted the elder brother back into his wheelchair.e. Josh shouted at the younger brother.A. b, a, e, c, dB. a, c, d, b, eC. b, a, c, e, dD. a, c, b, e, d39. What can we learn from the passage?A. Josh would accept the money from the kids.B. The two kids were Josh’s neighbors.C. Josh was a kind-hearted man.D. Josh’s new car broke down easily.40. According to the passage, the last sentence means _____ .A. trying to get ready for the trouble in your future lifeB. driving fast in a neighborhood street is dangerousC. trying to be more understanding seeing others in troubleD. protecting oneself from being hurtDHow to eat healthfully can be especially complex for working women who often have neither the desire nor the time to cook for themselves (or for anyone else). Registered dietitian(营养专家) Barhara Morrissey suggests that a few simple rules can help.“Go for nutrient dense foods,” she suggests, “foods that contain a multiple of nutrients. For example, select whole wheat bread as a breakfast food, rather than coffee cake. Or drink orange juice rather than orange drink, which contains only a small percentage of real juice-the rest is largely colored sugar water. You just can’t compare the value of these foods, the nutrient dense ones are so superior,” she emphasizes.Morrissey believes that variety is not only the spice of life —it’s the foundation of a healthful diet. Diets which are based on one or two foods are not only virtually impossible to keep up the strength, they can be very harmful, she say s, because nutrients aren’t supplied in sufficient amounts or balance.According to Morrissey, trying to find a diet that can cure your illnesses, or make you superwoman is a fruitless search. As women, many of us are too concerned with staying thin, she says, and we believe that vitamins are some kind of magic cure to replace food.“We need carbohydrates, protein and fat — they are like the wood in the fireplace. The vitamins and minerals are like the match, the spark, for the fuel,” she explains. “We need them all, but in a very different proportions. And if the fuel isn’t there, the spark is useless.”41. From the paragraph we know that working women .A.think cooking is especially complicatedB.do not share the same views with registered dietitiansC.are busy and not interested in cookingD. are likely to eat healthfully42. Orange juice is different from orange drink in that .A. it contains only a small percentage of real juiceB. it is natural, nutritious and prepared from real orangesC. it is largely orange-colored sugar waterD. it produces nothing but calories43. In Paragraph 4, “a fruitless search” means .A. an effort with no resultsB. a search for a diet without fruitsC. a research on fruitless dietD. a diet serving as medicine44. Many women take it for granted, according to passage that .A. a balanced diet can result in being fatB. staying thin and healthy are both possibleC. lack of variety in diets leads to staying thinD. vitamins are some kind of substitutions (替代品) for food45. By “if the fuel isn’t there, the spark is useless”, the author means .A. carbohydrates, protein and fat are enough to support a human lifeB. vitamins and minerals are virtually of no valueC. carbohydrates, protein and fat are as important as vitamins and mineralsD. without carbohydrates, protein and fat, vitamins and minerals are of no use第二节信息匹配(共5小题;每小题2分,满分10分)以下是个人求职广告信息:A. My name is Cheney. I have a genuine interest in providing technical interpretation services (English and Chinese) on a part time-basis. Currently I am working at one of fortune 500 companies in Shanghai on a full-time basis. However I would be really interested if there are opportunities for me to work on the weekends or part-time positions available.B. Hello everyone, currently I am looking for part time job in Shanghai especially in Changning District, I am Indonesian, fluent in English and I can speak Chinese, learn Chinese for 2 years, and learn English for my whole life.C. Hello. I'm Marcos from Washington,D.C and am currently in Shanghai for work. I am offering tutoring services in order to help you improve your oral English. I have over 2 years of experience teaching at the university level and am also TEFL certified. I have bachelor degrees in Economics, Finance, and German. I currently am working on my master's degree in field of Management Information Systems. I enjoy teaching especially to those who want to learn. I'm patient, friendly, and a very humorous guy! I love sports, movies, politics, business and most topics of choice.D. Well, I am Canadian who is currently studying in Shanghai, and I would like to find part time jobs during my spare time. I am fluent in Mandarin and English, and had experience in teaching primary pupils before.E. Experienced teacher / lecturer teaching French (as native speaker) and business English in a prestigious university in Shanghai looking to tutor students or doing translation jobs. I am fluent in French, English and Mandarin, conversant in German and Cantonese. Available to teach on the evenings or week ends.F. Your Nice Personal Tour Guide & Shopping Assistant & Interpreter with 3 years' experience. I can formulate a gorgeous tour plan for you according to your interests and arrange everything well. Besides, I can help you bargaining and getting the lowest prices以下是一些读者的信息,请匹配与它们所对应的广告。