清华附中高三十月月考

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2024届清华大学附中高三语文上学期10月考试题卷

2024届清华大学附中高三语文上学期10月考试题卷

2024届清华大学附中高三语文上学期10月考试卷2023.10(试卷满分150分;考试时间150分)一、本大题共5小题,共18分。

阅读下面材料,完成1-5 题。

材料一盛世修典兴藏是中华民族的传统,我国历朝历代都非常重视典籍的收藏保存,从而使得中华文明得以生生不息。

早在周朝时,由朝廷设立的守藏室就保存着历算、预测、谋略等方面的典籍,可谓是最早的“国家图书馆”。

史料显示,老子曾在周王朝担任守藏室史,类似后来的图书管理员。

秦朝收藏各类典籍的地方被称为“石室金匮”。

金匮指用金属封缄的柜子,石室即用石头修筑的房子,这种封闭性好的空间用来储藏典籍再合适不过了。

但是,秦朝的焚书坑儒导致不少先秦典籍损毁,实在是遗憾。

位于西安未央区的著名古迹天禄阁是汉朝廷御用收藏典籍和开展学术活动的地方,兼具图书馆和档案馆功能。

西汉学者刘向就曾在这里编校图书,搜集先秦典籍,为保存先秦文化做了大量工作。

《战国策》一书就是在这里辑录完成的。

唐朝廷在门下省设立修文馆保藏典籍,后更名为弘文馆。

据记载,当时的弘文馆有藏书二十余万卷,并有校书郎负责校理图籍,勘正错谬。

宋代的崇文院是中央藏书之所,宋仁宗景祐年间,欧阳修、宋祁等人奉诏整理图籍,编撰了《崇文总目》。

此外,崇文院还主持编纂了四部丛书,即《太平御览》《太平广记》《文苑英华》《册府元龟》,厥功至伟。

蒙古族建立元朝后,积极学习汉文化。

元文宗图帖睦尔有较高的汉文化修养,曾在京师创建奎章阁,以收藏各类经史图书。

他还组织人员编修《经世大典》,整理并保存了大量元代典籍。

遗憾的是,这部图书已经散佚。

明清时期的文渊阁是赫赫有名的皇家藏书场所。

明朝时,一批饱学之士曾在此主持编修《永乐大典》,该书被称为“世界上有史以来最大的百科全书”。

清廷组织人员在此编修的《四库全书》也颇为有名。

因《四库全书》规模宏大,为方便收藏,朝廷先后建造7座藏书楼,因此有“四库七阁”之说。

除了朝廷建造的藏书设施,中国古代还有一些著名的私人藏书楼,比如宁波范氏“天一阁”、苏州顾氏“过云楼”等,它们在保存中华典籍方面也功不可没。

清华大学附中高三第一学期第一次月考试卷

清华大学附中高三第一学期第一次月考试卷

清华大学附中高三第一学期第一次月考试卷(10年10月5日)语文试卷一、本大题共5小题,每小题3分,共15分。

1.下列词语中,字形和加点的字的读音全都正确的一项是A.揣摹耸人听闻连累(lèi)方兴未艾(ài)B.广袤成群结对庇护(bì)岿然不动(guī)C.饿莩铺张浪废拮据(jù)大腹便便(pián)D.果脯食不果腹狙击(jū)以一当十(dāng)选D(A.揣摩,连累lěi B.成群结队岿kuīC.铺张浪费拮据(jū))2.下列句中加点的成语,运用恰当的一项是A.一场春雨过后,校园里的柳树冒出一片新绿,栩栩如生,让人不觉而生盎然之感。

B. 两个孩子相继出世,使老刘家本来就捉襟见肘的生活更加不堪重负。

C. 如今小学也有外教课,虽然外教刻意放慢语速,但英语基础差的学生听在耳中还是不知所云。

D. 读古人文章,困难就在于古人文不加点,今天的人读起来相当吃力。

选B( A栩栩如生:好像活的一样。

形容生动,逼真。

C 不知所云:指语言紊乱或空洞。

这里对象错。

D 文不加点:指文章一气写成,无须修改,形容文思敏捷,写作技艺纯熟。

误为写文章不加标点。

)3.下列句子中,没有语病的一句是A.中国2010年上海世博会会徽,以中国汉字“世”字书法创意为形,塑造出世博会“理解、沟通、欢聚、合作”的理念。

B.日本贸易企业官员称,从21日开始,中国已经停止向日本出口稀土,这对生产超导体、计算机、混合动力汽车和其他高科技产品是非常重要的。

C.经罗氏公司总部和上海药品鉴定机构分别鉴定,确定在上海市第一人民医院造成大量眼疾患者产生不良反应事件的药物为假药。

D.从学术打假到今天的打架,从抄论文到今天的抄起家伙,学术不端的问题,已经完全演变成为一个社会问题。

选D(A塑造理念动宾不搭配;B“这”指代“稀土”这种资源还是“中国已经停止向日本出口稀土”这个事实不清楚;C主语残缺。

)4.下列有关文学常识的表述,有错误的一项是A.余光中在《论朱自清的散文》一文中说“他的节奏慢,调门平,情绪稳,境界是和风细雨,不是苏海韩潮。

北京市清华大学附属中学2022届高三10月月考化学试题 Word版含答案

北京市清华大学附属中学2022届高三10月月考化学试题 Word版含答案

高三月考试题化学一、选择题(共14题,每题3分,共42分。

每小题只有一个答案符合题意)1.下列陈述正确并且有因果关系的是()A.SiO2有导电性,可用于制备光导纤维B.浓硫酸具有氧化性,可用于干燥H2C.Al2O3具有很高的熔点,可用于制造熔融烧碱的坩埚D.FeCl3溶液能与Cu反应,可用于蚀刻印刷电路板【答案】D2.试验室常用NaNO2和NH4Cl反应制取N2,下列有关说法正确的是()A.NaNO2是还原性B.NH4Cl中N元素被还原C.生成1molN2时转移6mol电子D.氧化剂和还原剂的物质的量之比是1:1【答案】D3.将SO2气体通入BaCl2溶液中,未见沉淀生成。

若同时通入另一种气体X则有沉淀生成。

则X气体不行能是()A.H2S B.NH3C.CO2D.NO2【答案】C4.下列解释事实的方程式正确的是()A.将NO2通入水中,红棕色消逝:3NO2+H2O=2HNO3+O2B.漂白粉实现漂白性:CO2+2ClO-+H2O=CO32-+2HClOC.Na与水反应产生气体:Na+H2O=Na++OH-+H2↑D.燃煤时加入石灰石可削减SO2的排放:2CaCO3+O2+2SO 2 2CaSO4+2CO2【答案】D 5.肼(N2H4)是火箭常用的高能燃料,常温下为液体,其球棍模型如下图所示。

肼能与双氧水发生反应:N2H4+2H2O2=N2+4H2O。

用N A表示阿伏加德罗常数的值,下列说法正确的是()A.标准状况下,11.2LN2中含电子总数为5N AB.标准状况下,22.4LN2H4中所含原子总数为6N AC.标准状况下,3.2gN2H4中含有共价键的总数为0.6N AD.若生成3.6gH2O,则上述反应转移电子的数目为0.2N A【答案】D6.聚维酮碘的水溶液是一种常用的碘伏类缓释消毒剂,聚维酮通过氢键与HI3形成聚维酮碘,其结构表示如下:(图中虚线表示氢键)下列说法不正确的是()A.聚维酮的单体是B.聚维酮分子由(m+n)个单体聚合而成C.聚维酮碘是一种水溶性物质D.聚维酮在肯定条件下能发生水解反应【答案】B7.下列有关试验的选项正确的是()A.配制0.10mol/LNaOH溶液B.除去Cl2中的HCl C.CCl4萃取碘水中的I2D.制备纯洁的Fe(OH)2【答案】B8.依据元素周期表和元素周期律,下列叙述不正确的是()A.气态氢化物的稳定性:HF>H2O>NH3B.氢元素与其他元素可形成共价化合物或离子化合物C.右图所示试验可证明元素的非金属性:Cl>C>SiD.第118号元素在周期表中位于第七周期0族【答案】C9.氢氧化铜和碱式碳酸铜均可溶于盐酸转化为氯化铜。

2024届贵州省贵阳市清华中学高三上学期10月月考数学试题及答案

2024届贵州省贵阳市清华中学高三上学期10月月考数学试题及答案

贵阳市清华中学2024届高三10月月考试卷数学本试卷分第I 卷(选择题)和第II 卷(非选择题)两部分,共4页.时间120分钟,满分150分.第Ⅰ卷一、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(){}()12,e ,log ⎧⎫====⎨⎬⎩⎭∣∣xA x y yB x y y x ,则A B ⋂的元素个数为( )D. 4C. 3B. 2A. 12. 若2ii 3ia b +=-,其中i 是虚数单位,,a b ∈R 且0b ≠,设i z a b =+,则z 为( )A. 2B. C. 6D. 3. PM2.5的监测值是用来评价环境空气质量的指标之一.划分等级为:PM2.5日均值在335μg /m 以下,空气质量为一级;PM2.5日均值在335~75μg /m ,空气质量为二级;PM2.5日均值超过375μg /m 为超标.如图是某地8月1日至10日PM2.5的日均值(单位:3μg /m )变化的折线图,下列关于PM2.5日均值说法正确的是( )A. 这10天日均值的70百分位数为60B. 前4天的日均值的极差小于后4天的日均值的极差C. 前4天的日均值的方差大于后4天的日均值的方差D. 这10天的日均值的中位数为454. 数列{}n a 的通项公式为2n a n kn =+,则“2k ≥-”是“{}n a 为递增数列”的( )A. 充分不必要条件B. 必要不充分条件C. 既不充分也不必要条件D. 充要条件5. 若方程22121x y t t-=--所表示的曲线为C ,则下列命题错误的是( )A. 若曲线C 为双曲线,则1t <或2t >B. 若曲线C 为椭圆,则12t <<C. 曲线C 可能是圆D. 若曲线C 为焦点在x 轴上的椭圆,则312t <<6. 两个单位向量1e 与2e 满足120e e ⋅=,则向量12e - 与2e 的夹角为( )A. 30︒B. 60︒C. 120︒D. 150︒7. 设1()cos coscos ...cos 242n n x x x f x x -=,则58π3f ⎛⎫= ⎪⎝⎭( )A. B.C. 116-D.8. 若对任意正实数,x y 都有()12ln ln 0e x x y y m⎛⎫---≤ ⎪⎝⎭,则实数m 的取值范围为( )A. (]0,1 B. (]0,e C. ()[),01,-∞⋃+∞ D. ()[),0e,-∞⋃+∞二、多项选择题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分.9. 若0a b <<,且0a b +>,则( )A. a b > B.1ab>-C.110a b+> D. ()()111a b --<10. 如图,点P 是棱长为2的正方体1111ABCD A B C D -的表面上一个动点,则以下说法正确的是( )A. 当P 在平面11BCC B 上运动时,四棱锥11P AA D D -的体积不变B. 当P 在线段AC 上运动时,1D P 与11A C 所成角的取值范围是ππ,32⎡⎤⎢⎥⎣⎦C. 若点P 在底面ABCD 上运动,则使直线1A P 与平面ABCD 所成角为45的点P 的轨迹为椭圆D. 若F 是11A B 的中点,点P 在底面ABCD 上运动时,不存在点P 满足PF 平面11B CD 11. 阿波罗尼斯是古希腊著名数学家,与欧几里得、阿基米德被称为亚历山大时期数学三巨匠,阿波罗尼斯发现:平面内到两个定点,A B 的距离之比为定值(0λλ>,且1)λ≠的点的轨迹是圆,此圆被称为“阿波罗尼斯圆”.在平面直角坐标系xOy 中,()()2,0,4,0A B -,点P 满足12PA PB =.设点P 的轨迹为曲线C ,则下列说法正确的是( )A. C 的方程为22(4)16x y ++=B. 点,A B 都曲线C 内部C. 当,,A B P 三点不共线时,则APO BPO ∠=∠D. 若()2,2D ,则2PB PD +的最小值为12. 定义在[0,)+∞的函数()f x 满足()()6f x f x +=,且()()ln 2,02sin π,23x x f x x x ⎧-≤<⎪=⎨≤<⎪⎩,(]0,3x ∀∈都有(6)()0f x f x -+=,若方程()()f x a a =∈R 的解构成单调递增数列{}n x ,则下列说法中正确的是( )A. (2023)0f =B. 若数列{}n x 为等差数列,则公差为6C. 若12122()3x x x x +=+,则0ln 2a <<D. 若11ln 2a -<<,则()2323116ni i i x x n n--=+=+∑第II 卷三、填空题(本大题共4小题,共20分)13. 已知一个扇形的圆心角为π9,弧长为π3,则该扇形的面积为________.14. ,,a b c 分别为ABC 内角,,A B C 的对边.已知22252a b c +=,则cos C 的最小值为________.的在15. 已知函数()21ln 2f x x ax x =--存在单调递减区间,则实数a 的取值范围是__________.16. 已知椭圆2222:1(0)x y C a b a b+=>>左、右焦点分别为12,F F ,经过2F 的直线交椭圆C 于,P Q 两点,O 为坐标原点,且()2220,2OP OF PQ PF F Q +⋅==,则椭圆C 的离心率为______.四、解答题:本题共6小题,共70分,解答应写出文字说明、证明过程及演算步骤.17. 已知函数()()sin f x A x =+ωϕ(0A >,0ω>,π02ϕ<<)图象相邻两条对称轴间的距离为π2.函数()f x 的最大值为2,且______.请从以下3个条件中任选一个,补充在上面横线上,①π6f x ⎛⎫-⎪⎝⎭为奇函数;②当0x =时()f x =;③π12x =是函数()f x 的一条对称轴.并解答下列问题:(1)求函数()f x 的解析式;(2)在ABC 中,a 、b ,c 分别是角A ,B ,C 的对边,若()f A =3c =,ABC的面积ABC S = ,求a 的值.18. 已知数列{}n a 的前n 项和为n S ,194a =-,且1439n n S S +=-.(1)求数列{}n a 的通项公式;(2)设数列{}n b 满足()*30Nn n b na n +=∈,记数列{}nb 的前n 项和为nT ,若12nn Tb λ≤+对任意*N n ∈恒成立,求实数λ的取值范围.19. 如图,在三棱柱111ABC A B C -中,ABC 为等边三角形,四边形11BCC B 是边长为2的正方形,D 为AB中点,且1A D =.(1)求证:CD ⊥平面11ABB A ;的的(2)若点P 在线段1B C 上,且直线AP 与平面1ACD,求点P 到平面1ACD 的距离.20. 第22届亚运会将于2023年9月23日至10月8日在我国杭州举行,这是我国继北京后第二次举办亚运会,为迎接这场体育盛会,浙江某市决定举办一次亚运会知识竞赛,该市A 社区举办了一场选拔赛,选拔赛分为初赛和决赛,初赛通过后才能参加决赛,决赛通过后将代表A 社区参加市亚运知识竞赛.已知A 社区甲、乙、丙3位选手都参加了初赛且通过初赛的概率依次为12,13,12,通过初赛后再通过决赛的概率均为13,假设他们之间通过与否互不影响.(1)求这3人中至少有1人参加市知识竞赛概率.(2)某品牌商赞助了A 社区的这次知识竞赛,给参加选拔赛的选手提供了两种奖励方案:方案一:参加了选拔赛的选手都可参与抽奖,每人抽奖1次,每次中奖的概率均为13,且每次抽奖互不影响,中奖一次奖励600元:方案二:只参加了初赛的选手奖励100元,参加了决赛的选手奖励400元(包含参加初赛的100元),若品牌商希望给予选手更多的奖励,试从三人奖金总额的数学期望的角度分析,品牌商选择哪种方案更好.21. 已知双曲线()2222:10,0x y C a b a b-=>>的左右焦点分别为12,F F ,点P在双曲线上,若12PF PF -=,且双曲线焦距为4.(1)求双曲线C 的方程;(2)如果Q 为双曲线C 右支上的动点,在x 轴负半轴上是否存在定点M 使得222QF M QMF ∠=∠?若存在,求出点M 的坐标;若不存在,说明理由.22. 已知曲线()322f x x x mx n =+++在0x =处的切线方程为43y x =+.(1)求,m n 的值;(2)已知k 为整数,关于x 的不等式()()()ln 1f x x x f k x +>-在1x >时恒成立,求k 的最大值.的贵阳市清华中学2024届高三10月月考试卷数学本试卷分第I 卷(选择题)和第II 卷(非选择题)两部分,共4页.时间120分钟,满分150分.第Ⅰ卷一、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(){}()12,e ,log ⎧⎫====⎨⎬⎩⎭∣∣xA x y yB x y y x ,则A B ⋂的元素个数为( )A. 1 B. 2 C. 3 D. 4【答案】A 【解析】【分析】令()12e log =-xf x x ,可知A B ⋂的元素个数即为()f x 的零点个数,根据函数单调性结合零点存在性定理运算求解.【详解】由题意可知:A B ⋂的元素个数即为函数e x y =与12log y x =的交点个数,令()12e log =-xf x x ,则函数e x y =与12log y x =的交点个数即为()f x 的零点个数,因为e x y =在R 内单调递增,12log y x =在()0,∞+内单调递减,则()f x 在()0,∞+内单调递增,且()141e 20,1e 04⎛⎫=-<=> ⎪⎝⎭f f ,可知()f x 在()0,∞+内有且仅有1个零点,即函数e x y =与12log y x =有且仅有1个交点,所以A B ⋂的元素个数为1.故选:A.2. 若2ii 3ia b +=-,其中i 是虚数单位,,a b ∈R 且0b ≠,设i z a b =+,则z 为( )A. 2B.C. 6D. 【答案】D 【解析】【分析】化简可得2i 3i a b b +=+,然后根据复数相等的条件列出关系式,求出,a b 的值,根据共轭复数的概念以及复数的求模运算,即可得出答案.【详解】由2ii 3ia b +=-得,()2i i 3i 3i a b b b +=-=+,所以2b =且3b a =,解得6a =,2b =,所以,62i z =+,所以62i z =-==故选:D.3. PM2.5的监测值是用来评价环境空气质量的指标之一.划分等级为:PM2.5日均值在335μg /m 以下,空气质量为一级;PM2.5日均值在335~75μg /m ,空气质量为二级;PM2.5日均值超过375μg /m 为超标.如图是某地8月1日至10日PM2.5的日均值(单位:3μg /m )变化的折线图,下列关于PM2.5日均值说法正确的是( )A. 这10天日均值的70百分位数为60B. 前4天的日均值的极差小于后C. 前4天的日均值的方差大于后4天的日均值的方差D. 这10天的日均值的中位数为45【答案】B 【解析】【详解】解:对于A ,将10天中PM2.5日均值按从小到大排列为30,32,34,40,41,45,48,60,78,80,根据百分位数的定义可得,这10天中PM2.5日均值的70百分位数是4860542+=,故选A 错误;对于B ,前4天的日均值的极差为413011-=,后4天的日均值的极差为784533-=,故选项B 正确;对于C ,由折线图和方差的定义可知,前4天的日均值波动性小,所以前4天的日均值的方差小于后4天日均值的方差,故选项C 错误;的对于D ,这10天中PM2.5日均值的中位数为4145432+=,故选项D 错误.故选:B4. 数列{}n a 的通项公式为2n a n kn =+,则“2k ≥-”是“{}n a 为递增数列”的( )A. 充分不必要条件 B. 必要不充分条件C. 既不充分也不必要条件 D. 充要条件【答案】A 【解析】【分析】根据10n n a a +->以及充分条件和必要条件的定义分别进行判断即可【详解】由题意得数列{}n a 为递增数列等价于对任意*1N ,210n n n a a n k +∈-=++>恒成立,即21>--k n 对任意*N n ∈恒成立,故()max 213k n >--=-,所以“2k ≥-”是“{}n a 为递增数列”的充分不必要条件,故选:A5. 若方程22121x y t t-=--所表示的曲线为C ,则下列命题错误的是( )A. 若曲线C 为双曲线,则1t <或2tB. 若曲线C 椭圆,则12t <<C. 曲线C 可能是圆D. 若曲线C 为焦点在x 轴上的椭圆,则312t <<【答案】B 【解析】【分析】利用方程表示双曲线求解t 的取值范围可判断A ;方程表示椭圆求解t 可判断B ;方程是否表示圆可判断C ;方程表示焦点在x 轴上的椭圆求解t 可判断D.【详解】对于选项A :方程表示双曲线,则()()210t t -->,解得1t <或2t >,故A 正确;对于选项B :方程表示椭圆,则201021t t t t ->⎧⎪->⎨⎪-≠-⎩,解得12t <<且32t ≠,故B 错误;.为对于选项C :当3212t t t -=-⇒=时,方程表示圆,故C 正确;对于选项D :方程表示焦点在x 轴上的椭圆,则210t t ->->,解得312t <<,故D 正确;故选:B.6. 两个单位向量1e 与2e 满足120e e ⋅=,则向量12e - 与2e 的夹角为( )A. 30︒B. 60︒C. 120︒D. 150︒【答案】D 【解析】【分析】由题意可得11e = ,21e =,根据1e -可得12e -,设12e - 与2e的夹角为θ,利用cos θ.【详解】由题意可得11e = ,21e = ,且120e e ⋅=,所以12e == .设12e - 与2e 的夹角为θ,0180θ︒≤≤︒,则cos θ=所以150θ=︒.故选;D.7. 设1()cos coscos ...cos 242n n x x x f x x -=,则58π3f ⎛⎫= ⎪⎝⎭( )A. B.C. 116-D.【答案】A 【解析】【分析】利用倍角公式化简可得()1sin 22sin 2n n n xf x x -=,代入结合诱导公式运算求解.【详解】∵1111cos cos cos ...cos sin 2422()cos cos cos ...cos 242sin 2n n n n n x x x xx x x x f x x x ----==2211cos cos cos ...cos sin sin 22422...2sin 2sin 22n n n n n x x x xx x x x ----===,所以5548π4ππ4ππsin 2sin 4πsin πsin sin 8π33333π18π316161632sin 2sin 623f ⎛⎫⎛⎫⎛⎫⨯++ ⎪ ⎪ ⎪⎛⎫⎝⎭⎝⎭⎝⎭=====-= ⎪⎛⎫⎝⎭⨯ ⎪⎝⎭.故选:A .8. 若对任意正实数,x y 都有()12ln ln 0e x x y y m ⎛⎫---≤ ⎪⎝⎭,则实数m 的取值范围为( )A. (]0,1 B. (]0,e C. ()[),01,-∞⋃+∞ D. ()[),0e,-∞⋃+∞【答案】A 【解析】【分析】运用分离参数求最值,即将原不等式化为e(2e )(ln )x x y y m-≤,再构造函数()(2e )ln h t t t =-(0t >),求其最大值,进而求得结果.【详解】化简不等式可得1(2)(ln )e x x y y m -≤,即:e(2e )x x y y m-≤,令x t y =(0t >),则对任意的0t >,e(2e )ln t t m-≤,所以max e[(2e )ln ]t t m -≤,设()(2e )ln h t t t =-,0t >,则2e ()ln 1h t t t'=-+-,令()()2eln 10g t t t t =-+->,所以()212e0g t t t=--<',所以()g t 在(0,)+∞上单调递减,又因为2e(e)ln e 10eg =-+-=,所以()00e g t t >⇒<<,()0e g t t ⇒,所以当0e t <<时,()0h t '>,当t e >时,()0h t '<,所以()h t 在(0,e)上单调递增,在(e,)+∞上单调递减,所以max ()(e)e h t h ==,所以ee m≤,解得:01m <≤,即:m 的取值范围为(0,1].故选:A .二、多项选择题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分.9. 若0a b <<,且0a b +>,则( )A. a b > B.1ab>-C.110a b+> D. ()()111a b --<【答案】BD 【解析】【分析】利用不等式的性质结合作差法逐一判断即可.【详解】因为0a b <<,且0a b +>,则0ab <,所以b a >-,即b a >,故A 错误;则1a b >-,所以1ab >-,故B 正确;则110a ba b ab++=<,故C 错误;因为()()()1110a b ab a b ---=-+<,所以()()111a b --<,故D 正确.故选:BD.10. 如图,点P 是棱长为2的正方体1111ABCD A B C D -的表面上一个动点,则以下说法正确的是( )A. 当P 在平面11BCC B 上运动时,四棱锥11P AA D D -的体积不变B. 当P 在线段AC 上运动时,1D P 与11A C 所成角的取值范围是ππ,32⎡⎤⎢⎥⎣⎦C. 若点P 在底面ABCD 上运动,则使直线1A P 与平面ABCD 所成的角为45的点P 的轨迹为椭圆D. 若F 是11A B 的中点,点P 在底面ABCD 上运动时,不存在点P 满足PF 平面11B CD 【答案】AB 【解析】【分析】根据体积公式,即可判断A ,建立空间直角坐标系,根据线线角,线面角,以及利用法向量判断线面关系,即可判断BCD.【详解】A. 当P 在平面11BCC B 上运动时,点P 到平面11AA D D 的距离为2,所以四棱锥11P AA D D -的体积1822233V =⨯⨯⨯=,故A 正确;B .如图,建立空间直角坐标系,()10,0,2D ,(),2,0P x x -,()12,0,2A ,()10,2,2C ,()1,2,2D P x x =-- ,()112,2,0A C =-,[]0,2x ∈,设1D P 与11A C 所成角为θ,则111cos cos ,D P A C θ===当1x ≠时,cos θ=,[)(]0,11,2x ∈ ,则1cos 0,2θ⎛⎤∈ ⎥⎝⎦,当1x =时,cos 0θ=,所以1cos 0,2θ⎡⎤∈⎢⎥⎣⎦,ππ,32θ⎡⎤∈⎢⎥⎣⎦,故B 正确;C. 若点P 在底面ABCD 上运动,设(),,0P x y ,()12,0,2A ,()12,,2P x A y =--,平面ABCD 的法向量为()0,0,1m =,线直则1A P 面平与ABCD 为角的成所45 ,时111sin 45cos ,A P m A P m A P m⋅====,化简为()2224x y -+=,则点P 的轨迹为圆,故C 错误;D.如图,()10,0,2D ,()0,2,0C ,()12,2,2B ,()2,0,0A ,()10,2,2C ,()10,2,2D C =- ,()12,0,2CB = ,()12,2,2AC =-,110AC D C ⋅= ,110CB AC ⋅=,且11D C CB C = ,且11,D C CB ⊂平面11CB D ,所以1AC ⊥平面11CB D ,即向量1AC u u u r是平面11CB D 的法向量,()2,1,2F ,(),,0P x y ,()2,1,2FP x y =---,若//FP 平面11CB D ,则()()1222140FP AC x y ⋅=--+--=,即10x y -+=,直线10x y -+=与底面ABCD 有公共点,即存在点P 满足PF 平面11B CD ,故D 错误.故选:AB11. 阿波罗尼斯是古希腊著名数学家,与欧几里得、阿基米德被称为亚历山大时期数学三巨匠,阿波罗尼斯发现:平面内到两个定点,A B 的距离之比为定值(0λλ>,且1)λ≠的点的轨迹是圆,此圆被称为“阿波罗尼斯圆”.在平面直角坐标系xOy 中,()()2,0,4,0A B -,点P 满足12PA PB=.设点P 的轨迹为曲线C ,则下列说法正确的是( )A. C 的方程为22(4)16x y ++=B. 点,A B 都在曲线C 内部C. 当,,A B P 三点不共线时,则APO BPO ∠=∠D. 若()2,2D ,则2PB PD +的最小值为【答案】ACD 【解析】【分析】对于A ,通过直接法求出点P 的轨迹方程即可判断;对于B ,利用点到圆心的距离,判断点与圆的位置关系;对于C ,由题意,结合三角形内角平分线定理进行判断即可;对于D ,将||2||PB PD +转化为2||2||PA PD +进行判断即可.【详解】设(,)P x y ,(P 不与A ,B 重合),由(2,0)A -,(4,0)B,有||PA =,||PB =||1||2PA PB =12=,化简得22(4)16x y ++=,所以点P 的轨迹曲线C 是以(4,0)C -为圆心,半径4r =的圆,如图所示,对于A 选项,由曲线C 方程为22(4)16x y ++=,选项A 正确;对于B 选项,由8BC =,点B 在曲线C 外,选项B 错误;对于C 选项,由||2OA =,||4OB =,有||1||||2||OA PA OB PB ==,则当A ,B ,P 三点不共线时,由三角形内角平分线定理知,PO 是APB 内角APB ∠的角平分线,所以APO BPO ∠=∠,选项C 正确;对于D 选项,由||1||2PA PB =,得||2||PB PA =,则()||2||2||2||2||||2||2PB PD PA PD PA PD AD +=+=+≥==,当且仅当P 在线段AD 上时,等号成立,的则2PB PD +的最小值为D 正确.故选:ACD .12. 定义在[0,)+∞的函数()f x 满足()()6f x f x +=,且()()ln 2,02sin π,23x x f x x x ⎧-≤<⎪=⎨≤<⎪⎩,(]0,3x ∀∈都有(6)()0f x f x -+=,若方程()()f x a a =∈R 的解构成单调递增数列{}n x ,则下列说法中正确的是( )A. (2023)0f =B. 若数列{}n x 为等差数列,则公差为6C. 若12122()3x x x x +=+,则0ln 2a <<D. 若11ln 2a -<<,则()2323116ni i i x x n n--=+=+∑【答案】ABD 【解析】【分析】对于A :根据题意结合周期性运算求解;对于B :根据题意结合图象分析判断;对于B :整理可得()()12ln 2ln 2x x -=-,结合图象分析判断;对于D :根据图象结合对称性分析可得数列{}323i 1i x x ---是以首项为7,公差为12的等差数列,进而利用等差数列知识运算求解.【详解】因为(]0,3x ∀∈都有(6)()0f x f x -+=,即()f x 的图象关于()3,0对称,令3x =,则(3)(3)0f f +=,即(3)0f =,可知()f x 在()0,6内的图象关于点()3,0对称,根据题意作出()f x 在[)0,6内的图象,如图所示:对于选项A :因为定义在[0,)+∞的函数()f x 满足()()6f x f x +=,则()()(2023)6337110f f f =⨯+==,故A 正确;对于选项B :由图象可知:若数列{}n x 为等差数列,则()(),11,a ∈-∞+∞U ,此时()y f x =与y a =在[)0,6内有且仅有一个交点,因为()()6f x f x +=,则16n n x x +-=,所以公差为6,故B 正确;对于选项C :若12122()3x x x x +=+,则()()12221x x --=,可得()()()()1212ln 22ln 2ln 20x x x x ⎡⎤--=-+-=⎣⎦,则()()12ln 2ln 2x x -=-,即()y f x =与y a =在[)0,2内有且仅有2个交点,结合图象可得0ln 2a <≤,故C 错误;对于选项D :若11lnln 22a -<<=-,则()y f x =与y a =在[)0,6内有且仅有3个交点,且127x x +=,因为()()6f x f x +=,则()()()()()31323231323132316612i i i i i i i i x x x x x x x x ++---------=+-+--=⎡⎤⎣⎦,所以数列{}3231i i x x ---是以首项为7,公差为12的等差数列,可得()32317121125i i x x n n ---=+-=-,所以()()232311712562ni i i n n x x n n --=+-+==+∑,故D 正确;故选:ABD.【点睛】方法点睛:应用函数思想确定方程解的个数的两种方法(1)转化为两熟悉的函数图象的交点个数问题、数形结合、构建不等式(方程)求解;(2)分离参数、转化为求函数的值域问题求解.第II 卷三、填空题(本大题共4小题,共20分)13. 已知一个扇形的圆心角为π9,弧长为π3,则该扇形的面积为________.【答案】π2【解析】【分析】利用扇形面积公式计算即可.【详解】设扇形半径为r ,由题意可知π33π3r ==,所以该扇形的面积为1ππ3232S =⨯⨯=.故答案为:π2.14. ,,a b c 分别为ABC 内角,,A B C 的对边.已知22252a b c +=,则cos C 的最小值为________.【答案】35##0.6【解析】【分析】因为222cos 2a b c C ab +-=,所以代入()22225c a b =+,得到()2235cos 2a b C ab+=,并结合基本不等式,得到cos C 的最小值.详解】由余弦定理得()()22222222223323555cos 22225a b a b a b ab a b cC abab ab ab +-++⨯+-===≥=.当且仅当a b =时,取等号.所以cos C 的最小值为35故答案为:3515. 已知函数()21ln 2f x x ax x =--存在单调递减区间,则实数a 的取值范围是__________.【答案】1,4⎛⎫-+∞ ⎪⎝⎭【解析】【【分析】利用函数单调性与导数的关系,列出不等式即可求解.【详解】函数()21ln 2f x x ax x =--的定义域为()0,∞+,求导得()11f x ax x '--=,依题意,不等式()0f x '<在()0,∞+上有解,等价于211a x x>->在()0,∞+上有解,而22111111244x x x ⎛⎫-=--≥- ⎪⎝⎭,当且仅当2x =时取等号,则14a >-,所以实数a 的取值范围是1,4⎛⎫-+∞ ⎪⎝⎭.故答案为:1,4⎛⎫-+∞ ⎪⎝⎭.16. 已知椭圆2222:1(0)x y C a b a b+=>>的左、右焦点分别为12,F F ,经过2F 的直线交椭圆C 于,P Q 两点,O 为坐标原点,且()2220,2OP OF PQ PF F Q +⋅==,则椭圆C 的离心率为______.【解析】.【详解】因为()2220,2OP OF PQ PF F Q +⋅== ,所以()22302OP OF PF +⋅=,即()()22302OP OF OF OP +⋅-=,所以21OP OF OF c === ,所以12π2F PF ∠=.设2F Q x =,则22PF x =,所以1122,2PF a x QF a x =-=-,由22211||PF PQ QF +=得222(22)(3)(2)a x x a x -+=-,所以3a x =,所以2124,33a PF a PF ==,在12Rt PFF △中,由2221212PFPF F F +=,得22224(2)33a a c ⎛⎫⎛⎫+= ⎪ ⎪⎝⎭⎝⎭,所以c e a ==.故答案为:四、解答题:本题共6小题,共70分,解答应写出文字说明、证明过程及演算步骤.17. 已知函数()()sin f x A x =+ωϕ(0A >,0ω>,π02ϕ<<)的图象相邻两条对称轴间的距离为π2.函数()f x 的最大值为2,且______.请从以下3个条件中任选一个,补充在上面横线上,①π6f x ⎛⎫-⎪⎝⎭为奇函数;②当0x =时()f x =;③π12x =是函数()f x 的一条对称轴.并解答下列问题:(1)求函数()f x 的解析式;(2)在ABC 中,a 、b ,c 分别是角A ,B ,C 的对边,若()f A =3c =,ABC 的面积ABC S = ,求a 的值.【答案】(1)()π2sin 23f x x ⎛⎫=+ ⎪⎝⎭(2【解析】【分析】(1)由最大值确定A ,根据相邻两条对称轴间的距离为π2确定最小正周期,从而确定ω,选①,可得=0 6f π⎛⎫-⎪⎝⎭,求解即可;选②,()0f ,求解即可;选③,整体思想ππ2π(Z)122k x ϕ⨯+=+∈,求解即可.(2)利用面积公式求出b ,结合余弦定理即可求解.【小问1详解】由题意得π2,22T A ==,∴最小正周期πT =,则2π2Tω==,∴()()2sin 2f x x ϕ=+.若选①,6f x π⎛⎫-⎪⎝⎭为奇函数,则=0 6f π⎛⎫-⎪⎝⎭,∴π2sin 03ϕ⎛⎫-+= ⎪⎝⎭,即πsin 03ϕ⎛⎫-+= ⎪⎝⎭∵02πϕ<<,即πππ336ϕ-<-+<,∴π03ϕ-+=即π3ϕ=,∴()π2sin 23f x x ⎛⎫=+⎪⎝⎭.若选②,当0x =时()f x =,∴2sin ϕ=即sin ϕ=∵02πϕ<<,∴π3ϕ=,∴()π2sin 23f x x ⎛⎫=+ ⎪⎝⎭.若选③,π12x =是函数()f x 的一条对称轴,∴ππ2π(Z)122k x ϕ⨯+=+∈即ππ(Z)3k x ϕ=+∈∵02πϕ<<,∴π3ϕ=,∴()π2sin 23f x x ⎛⎫=+ ⎪⎝⎭.【小问2详解】∵()f A =,∴π2sin 23A ⎛⎫+= ⎪⎝⎭即πsin 23A ⎛⎫+= ⎪⎝⎭∵(0,π)A ∈即ππ7π2,333A ⎛⎫⎛⎫+∈ ⎪ ⎪⎝⎭⎝⎭,∴π2π233A +=,即π6A =,又∵3c =,ABC 的面积ABC S = ,∴1sin 2bc A =得b =在ABC 中,由余弦定理得:2222cos a b c bc A =+-,解得a =18. 已知数列{}n a 的前n 项和为n S ,194a =-,且1439n n S S +=-.(1)求数列{}n a 的通项公式;(2)设数列{}n b 满足()*30Nn n b na n +=∈,记数列{}nb 的前n 项和为nT ,若12nn Tb λ≤+对任意*N n ∈恒成立,求实数λ的取值范围.【答案】(1)()()*33N 4nn a n ⎛⎫=-⨯∈ ⎪⎝⎭(2)[)3,∞-+【解析】【分析】(1)由n a 与n S 的关系即可求解;(2)先用错位相减法求出n T ,再由不等式恒成立问题的解法即可求解.【小问1详解】解:当2n ≥时,∵1439n n S S +=-,∴1439n n S S -=-,两式相减得143n n a a +=,2n ≥.∵194a =-,21439S S =-,所以1214()39a a a +=-,∴22716=-a ,∵2134a a =,∴()*13N 4n n a n a +=∈,∴数列{}n a 是以首项94-,公比为34的等比数列.∴()()1*9333N 444n nn a n -⎛⎫⎛⎫⎛⎫=-⨯=-⨯∈ ⎪ ⎪⎪⎝⎭⎝⎭⎝⎭【小问2详解】∵30n n b na +=,∴34nn b n ⎛⎫= ⎪⎝⎭,∴12333331234444nn T n ⎛⎫⎛⎫⎛⎫⎛⎫=⨯+⨯+⨯⋅⋅⋅+⨯ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭,∴()23413333331231444444nn n T n n +⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫=⨯+⨯+⨯⋅⋅⋅+-⨯+⨯ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭,∴1231133333444444n n n T n +⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫=++⋅⋅⋅+-⨯ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭113314433331344414n n n n n n ++⎡⎤⎛⎫⨯-⎢⎥ ⎪⎡⎤⎝⎭⎢⎥⎛⎫⎛⎫⎛⎫⎣⎦=-⨯=⨯--⨯⎢⎥ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦-∴()1333121412312444n n nn T n n +⎡⎤⎛⎫⎛⎫⎛⎫=⨯--⨯=-+⎢⎥ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦,∵12n n T b λ≤+对任意N n ∈恒成立,∴()33123121244n nn n λ⎛⎫⎛⎫-+≤+ ⎪ ⎪⎝⎭⎝⎭,∴()312n n λ-+≤,∴()()*34431N n n n n λ+⎛⎫≥-=-+∈ ⎪⎝⎭恒成立,∵411n+>,∴3λ≥-,∴λ的取值范围是[)3,∞-+.19. 如图,在三棱柱111ABC A B C -中,ABC 为等边三角形,四边形11BCC B 是边长为2的正方形,D 为AB 中点,且1A D =.(1)求证:CD ⊥平面11ABB A ;(2)若点P 在线段1B C 上,且直线AP 与平面1ACD ,求点P 到平面1ACD 的距离.【答案】(1)证明见解析(2【解析】【分析】(1)由勾股定理证明1⊥A A AD ,再由1A A BC ⊥,可证1A A ⊥平面ABC ,即得1CD AA ⊥,由CD AB ⊥,可证CD ⊥平面11ABB A ;(2)由题意证明得,,OA OB OQ 两两垂直,建立空间直角坐标系,写出对应点的坐标与向量的坐标,求解平面1ACD 的法向量,设()[]12,2,0,0,1CP CB λλλλ==∈ ,再由向量夹角的公式代入计算得()1,1,0CP =,根据点到平面的距离公式代入计算,可得答案.【小问1详解】证明:由题知112,1,AA AD A D ===,222115AD A A A D ∴+==1A A AD ⇒⊥,又111,B B BC B B A A ⊥∥,所以1A A BC ⊥,又AD BC B = ,,AD BC ⊂平面ABC ,所以1A A ⊥平面ABC ,又CD ⊂平面ABC ,所以1CD AA ⊥,在正ABC 中,D 为AB 中点,于是CD AB ⊥,又1AB AA A ⋂=,1,AB AA ⊂平面11ABB A ,所以CD ⊥平面11ABB A【小问2详解】取BC 中点为11,O B C 中点为Q ,则,OA BC OQ BC ⊥⊥,由(1)知,1A A ⊥平面ABC ,且OA ⊂平面ABC ,所以1OA AA ⊥,又11B B A A ∥,所以11,OA BB BB BC B ⊥⋂=,1,BB BC ⊂平面11BCC B 所以OA ⊥平面11BCC B ,于是,,OA OB OQ 两两垂直.如图,以O 为坐标原点,,,OB OQ OA的方向为x 轴、y 轴、z 轴的正方向,建立空间直角坐标系,则()((()10,0,0,,0,,1,0,0O A A C -,()11,1,2,02D B ⎛ ⎝,所以(13,1,2CD CA ⎛==⎝,()(12,2,0,1,0,CB AC ==-.设平面1ACD 的法向量为(),,n x y z =,则100n CD n CA ⎧⋅=⎪⎨⋅=⎪⎩,即30220x x y ⎧=⎪⎨⎪++=⎩,令1x =,则1z y ==,于是(1,1,n =.设()[]12,2,0,0,1CP CB λλλλ==∈,则(121,2,AP AC CP AC CB λλλ=+=+=-.由于直线AP 与平面1ACD,cos ,AP n ∴==即21λ+=24830λλ-+=,由于[]0,1λ∈,所以1,2λ=于是()11,1,0CP CB λ==.设点P 到平面1ACD 的距离为d,则CP n d n⋅=== ,所以点P 到平面1ACD.【点睛】方法点睛:对于立体几何中角的计算问题,往往可以利用空间向量法,通过求解平面的法向量,利用向量的夹角公式求解.20. 第22届亚运会将于2023年9月23日至10月8日在我国杭州举行,这是我国继北京后第二次举办亚运会,为迎接这场体育盛会,浙江某市决定举办一次亚运会知识竞赛,该市A 社区举办了一场选拔赛,选拔赛分为初赛和决赛,初赛通过后才能参加决赛,决赛通过后将代表A 社区参加市亚运知识竞赛.已知A 社区甲、乙、丙3位选手都参加了初赛且通过初赛的概率依次为12,13,12,通过初赛后再通过决赛的概率均为13,假设他们之间通过与否互不影响.(1)求这3人中至少有1人参加市知识竞赛的概率.(2)某品牌商赞助了A 方案一:参加了选拔赛的选手都可参与抽奖,每人抽奖1次,每次中奖的概率均为13,且每次抽奖互不影响,中奖一次奖励600元:方案二:只参加了初赛的选手奖励100元,参加了决赛的选手奖励400元(包含参加初赛的100元),若品牌商希望给予选手更多的奖励,试从三人奖金总额的数学期望的角度分析,品牌商选择哪种方案更好.【答案】(1)3181(2)从三人奖金总额的数学期望的角度分析,品牌商选择方案二更好【解析】【分析】(1)根据独立事件的概率,先分别求出甲乙丙三人参加市赛的概率,即可求出至少有1人参加市知识竞赛的概率.(2)分别求出两个方案的奖励期望,比较大小即可.【小问1详解】甲参加市赛的概率为111236⨯=,乙参加市赛的概率为131139⨯=,丙参加市赛的概率为111236⨯=,至少1人参加市赛的概率为:211311116981⎛⎫⎛⎫--⨯-= ⎪ ⎪⎝⎭⎝⎭.【小问2详解】方案一:设三人中奖人数为X ,所获奖金总额为Y 元,则600Y X =,且13,3X B ⎛⎫⎪⎝⎭ .所以()()160060036003E Y E X ==⨯⨯=元,方案二:记甲、乙、丙三人获得奖金之和为Z 元,则Z 的所有可能取值为300、600、900、1200,则()11113001112326P Z ⎛⎫⎛⎫⎛⎫==-⨯-⨯-= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,()212111115600C 1112233212P Z ⎛⎫⎛⎫⎛⎫==⋅--+-= ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭,()2121111119001C 1232233P Z ⎛⎫⎛⎫⎛⎫==⨯-+⋅-⨯= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,()1111120023212P Z ==⨯⨯=,所以,()11130060090012007006312E Z =⨯+⨯+⨯=.所以,()()E Y E Z <,所以从三人奖金总额的数学期望的角度分析,品牌商选择方案二更好.21. 已知双曲线()2222:10,0x y C a b a b-=>>的左右焦点分别为12,F F ,点P 在双曲线上,若12PF PF -=,且双曲线焦距为4.(1)求双曲线C 的方程;(2)如果Q 为双曲线C 右支上的动点,在x 轴负半轴上是否存在定点M 使得222QF M QMF ∠=∠若存在,求出点M 的坐标;若不存在,说明理由.【答案】(1)2213y x -=(2)存在,坐标为()1,0-【解析】【分析】(1)利用双曲线的定义求解即可;(2)在x 轴负半轴上假设存在点M 满足题意,当2QF 垂直于x 轴时,易得()1,0M -,当2QF 不垂直于x 轴时,由斜率公式和二倍角正切公式也可解得()1,0M -.【小问1详解】因为点P 在双曲线上,所以由双曲线的定义可得122PF PF a -==①,又双曲线焦距即24c =,且222+=a b c ③,①②③联立解得1a b =⎧⎪⎨=⎪⎩,所以双曲线C 的方程为2213y x -=.【小问2详解】假设存在点()(),00M t t <满足题设条件,由题目可知()22,0F ,设()()000,1Q x y x ≥为双曲线C 右支上一点,当02x =时,03y =,因为22290QF M QMF ∠=∠=︒,所以245QMF ∠=︒,于是23MF QF ==,所以1t =-,即()1,0M -,当02x ≠时,2020tan 2QF y QF M k x ∠=-=--,020tan QM y QMF k x t∠==-,因为222QF M QMF ∠=∠,所以002000221y y x tx y x t ⨯--=-⎛⎫- ⎪-⎝⎭,将220033=-y x 代入并整理得()22200002424223x t x t x tx t -++-=--++,所以242243t tt t +=-⎧⎨-=+⎩,解得1t =-,即()1,0M -,综上,满足条件的点M 存在,其坐标为()1,0M -.【点睛】方法点睛:(1)解答直线与双曲线的题目时,时常把两个曲线方程联立,消去x (或y )建立一元二次方程,然后借助根与系数的关系,并结合题设条件建立有关参变量的等量关系;(2)涉及到直线方程的设法时,务必考虑全面,不要忽略直线斜率为0或不存在等特殊情形.22. 已知曲线()322f x x x mx n =+++在0x =处的切线方程为43y x =+.(1)求,m n 的值;(2)已知k 为整数,关于x 的不等式()()()ln 1f x x x f k x +>-在1x >时恒成立,求k 的最大值.【答案】(1)4,3m n == (2)3【解析】【分析】(1)根据导数的几何意义求解即可;(2)求导分析单调性可得()f x 在R 上是增函数,进而将不等式转化为()1ln 1x x k x +<-在1x >时恒成立,再构造函数()()1ln (1)1x x g x x x +=>-,再求导分析单调性,结合零点存在性定理分析函数的极值,进而可得k 的最大值.【小问1详解】由题知,()()234,0f x x x m f n =++=',()0f m ∴'=,()f x 在0x =处的切线方程为43y x =+,4,3m n ∴==.【小问2详解】由(1)知()32243f x x x x =+++,()22283443033f x x x x ⎛⎫∴=++=++> ⎪⎝⎭',()f x \在R 上是增函数,关于x 的不等式()()()ln 1f x x x f k x +>-在1x >时恒成立,∴不等式()ln 1x x x k x +>-即()1ln 1x x k x +<-在1x >时恒成立.设()()1ln (1)1x x g x x x +=>-,则()()()()222ln 1ln ln 2(1)(1)x x x x x x x g x x x +--+--==--'.设()ln 2(1)h x x x x =-->,则()1110(1)x h x x x x-=-=>>',()h x ∴在区间()1,+∞是增函数,()()31ln30,422ln20h h =-<=-> ,∴存在()03,4x ∈,使()000ln 20h x x x =--=,当01x x <<时,()()0,0h x g x '<<,当0x x >时,()()0,0h x g x '>>,()g x ∴在区间()01,x 单调递减,在区间()0,x +∞单调递增,()()()0000min 00001ln 12()11x x x x g x g x x x x ++-∴====--,()00k g x x ∴<=,又k 为整数,()03,4,x k ∈∴的最大值为3.【点睛】方法点睛:(1)恒成立问题可考虑参变分离;(2)构造函数分析单调性,极值点求不出的可设极值点,根据零点存在性定理确定极值点的范围;(3)根据极值点满足的关系式,代入极值,继续构造函数分析极值范围.。

北京市清华大学附属中学2019-2020学年高三年级第一学期10月考数学试卷

北京市清华大学附属中学2019-2020学年高三年级第一学期10月考数学试卷

清华附中高三2019年10月月考试卷数学一、选择题1.已知集合{}2A x x =>,()(){}130B x x x =--<,则A B =( )A .{}1x x >B .{}23x x <<C .{}13x x <<D .{}21x x x ><或2.若角θ的终边过点()3,4P -,则()tan θπ+=( ) A .34B .34-C .43 D .43-3.已知函数a y x =,log b y x =的图象如图所示,则( )A .1b a >>B .1b a >>C .1a b >>D .1a b >>4.设函数()y f x =的定义域为R ,则“()00f =”是“函数()f x 为奇函数”的( ) A .充分而不必要条件 B .必要而不充分条件 C .充分必要条件 D .既不充分也不必要条件5.已知3cos 4α=,,02πα⎛⎫∈- ⎪⎝⎭,则sin 2α的值为( )A .36 B .38- C D .6.我国古代数学名著《算法统宗》中有如下问题:“远望巍巍塔七层,红光点点倍加增,共灯三百八十一,请问尖头几盏灯?”意思是:一座7层塔共挂了381盏灯,且相邻两层中的下一层灯数是上一层灯数的2倍,则塔的顶层共有灯( ) A .1盏 B .3盏 C .5盏 D .9盏 7.某校象棋社团组织中国象棋比赛,采用单循环赛制,即要求每个参赛选手必须且只须和其他选手各比赛一场,胜者得2分,负者得0分,平局两人各得1分.若冠军获得者得分比其他人都多,且获胜场次比其他人都少,则本次比赛的参赛人数至少为( ) A .4 B .5 C .6 D .78.已知定义在R 上的函数()()2,0ln ,0xa x f x x a x ⎧+≤⎪=⎨+>⎪⎩,若方程()12f x =有两个不相等的实数根,则a 的取值范围是( ) A .1122a -≤≤B .102a ≤<C .01a ≤<D .102a -<≤二、填空题9.已知函数()y f x =的导函数有且仅有两个零点,其图象如图所示,则函数()y f x =在x =___________处取得极值.10.32-,123,2log 5三个数中最大的数是_____________. 11.在ABC △中,13cos 14A =,73a b =,则B =____________. 12.去年某地的月平均气温y (℃)与月份x (月)近似地满足函数sin 6y a b x πϕ⎛⎫=++ ⎪⎝⎭(a 、b为常数,0πϕ<<),其中三个月份的月平均气温如表所示:则该地2月份的月平均气温约为_______℃,ϕ=__________.13.在等腰梯形ABCD 中,已知AB DC ∥,2AB =,1BC =,60ABC =︒∠,点E 和点F 分别在线段BC 和CD 上,且23BE BC =,16DF DC =,则AE AF ⋅的值为_____________.14.如图,线段8AB =,点C 在线段AB 上,且2AC =,P 为线段CB 上一动点,点A 绕点C 旋转后与点B 绕点P 旋转后重合于点D .设CP x =,CPD △的面积为()f x ,则()f x 的定义域为_________,()'f x 的零点是__________.三、解答题15.已知函数()()cos f x A x ωϕ=+0,0,02A πωϕ⎛⎫>><< ⎪⎝⎭的图象过点10,2⎛⎫⎪⎝⎭,最小正周期为23π,且最小值为1-. (1)求函数()f x 的解析式;(2)若,6x m π⎡⎤∈⎢⎥⎣⎦,()f x 的值域是1,⎡-⎢⎣⎦,求m 的取值范围.16.数列{}n a 的前n 项和记为n S ,若数列n S n ⎧⎫⎨⎬⎩⎭是首项为9,公差为1-的等差数列.(1)求数列{}n a 的通项公式n a ;(2)若n n b a =,且数列{}n b 的前n 项和记为n T ,求415T T +的值.17.已知ABC △的内角,,A B C 所对的边分别为,,a b c ,()8sin 17A C +=,且角B 为锐角. (1)求cos B 的值;(2)若6a c +=,ABC △的面积为2,求边长b .18.已知函数()1xax f x e-=. (1)当1a =时,求函数()f x 的单调区间;(2)当0a <时,求函数()f x 在区间[]0,1上的最小值.19.已知函数()39f x x x =-,函数()23g x x a =+.(1)若曲线()y f x =与曲线()y g x =在它们的交点处且有公共切线,求a 的值; (2)若存在实数b 使不等式()()f x g x <的解集为(),b -∞,求实数a 的取值范围.20.设满足以下两个条件的有穷数列12,,,n a a a …为()2,3,4,n n =…阶“期待数列”: ①1230n a a a a ++++=…; ②1231n a a a a ++++=…;(1) 分别写出一个单调递增的3阶和4阶“期待数列”;(2) 若某2013阶“期待数列”是等差数列,求该数列的通项公式; (3) 记n 阶“期待数列”的前k 项和为()1,2,3,,k S k n =…,试证:12k S ≤.。

清华大学附属中学蒋台分校22024-2025学年高三上学期10月月考 化学试题(含答案)

清华大学附属中学蒋台分校22024-2025学年高三上学期10月月考 化学试题(含答案)

蒋台分校22024-2025学年高三上学期10月月考化学(清华附中奥森、将台路校区 高22级)2024.10满分:100分 考试时长:90分钟可能用到的相对原子质量:H 1 C 12 O 16 Na 23 Mg 24 Ca 40 Ti 48 Mn 55 Fe 56 Zr 91 Ba 137一.选择题(共14小题,每题3分,共42分)1.2020年12月17日凌晨、嫦娥五号携带月壤等样本成功返回地球,完成中国探月工程的收官之战。

下列说法不正确的是( )A.发射时使用液氢和液氧作推进剂,是利用了燃烧反应提供能量B.制造探测器中的瞄准镜时使用光导纤维,其主要成分是SiC.月壤中含有珍贵的3He ,3He 与4He 互为同位素D.留在月球的国旗长时间不褪色、不分解,是利用了材料的稳定性2.下列物质是强电解质的是( )A. B.HIC.H 2SiO 3D.3.下列化学用语或图示表达不正确的是( )A.SO 3的VSEPR 模型:B.羟基的电子式:C.基态24Cr 原子的价层电子轨道表示式为:D.原子核内有8个中子的碳原子:4.N A 为阿伏加德罗常数的值。

下列叙述正确的是( )A.0.50mol分子中共价键的数目为2N AB.标准状况下,2.24LSO 3中电子的数目为4.00N AC.pH=2的H 2SO 4溶液中H +的数目为0.02N AD.常温常压下,28gCO 和N 2的混合气体原子数目为2N A5.物质的性质决定用途,下列两者对应关系不正确的是( )A.铝有强还原性,可用于制作门窗框架()3Fe SCN ()3Al OH 146CB.氧化钙易吸水,可用作干燥剂C.维生素C 具有还原性,可用作食品抗氧化剂D.过氧化钠能与二氧化碳反应生成氧气,可作潜水艇中的供氧剂6.下列各组离子中,能大量共存的是( )A.、、、B.、、、C.、、、D.、、、7.宏观辨识与微观探析是化学学科核心素养之一。

下列物质性质实验对应的反应方程式书写正确的是( )A.Na 2O 2放入水中:B.通过灼热铁粉:C.铜丝插入热的浓硫酸中:D.SO 2通入酸性KMnO 4溶液中: 8.下列指定反应的离子方程式正确的是( )A.碘化亚铁溶液通入少量的氯气:B.用H 2O 2从酸化的海带灰浸出液中制取碘:C.胶体的制备:D.等物质的量的NaHCO 3溶液与溶液混合: 9.KIO 3常用作食盐中的补碘剂,可用“氯酸钾氧化法”制备,该方法的第一步反应为。

2024-2025学年北京市朝阳区清华大学附属中学朝阳分校高三上学期10月月考数学试题(含答案)

2024-2025学年北京市朝阳区清华大学附属中学朝阳分校高三上学期10月月考数学试题(含答案)

2024-2025学年北京市朝阳区清华大学附属中学朝阳分校高三上学期10月月考数学试题一、单选题:本题共10小题,每小题5分,共50分。

在每小题给出的选项中,只有一项是符合题目要求的。

1.已知全集U={x|x>0},集合A={x|2≤x≤3},则∁U A=( )A. (0,2]∪[3,+∞)B. (0,2)∪(3,+∞)C. (−∞,2]∪[3,+∞)D. (−∞,2)∪(3,+∞)2.若等差数列{a n}和等比数列{b n}满足a1=b1,a2=b2=2,a4=8,则{b n}的公比为( )A. 2B. −2C. 4D. −43.在平面直角坐标系xOy中,角α与角β均以Ox为始边,它们的终边关于直线y=x对称.若sinα=35,则cosβ=( )A. −45B. 45C. −35D. 354.若点M(1,1)为圆C: x2+y2−4x=0的弦AB的中点,则直线AB的方程是( )A. x−y−2=0B. x+y−2=0C. x−y=0D. x+y=05.已知D是边长为2的正△ABC边BC上的动点,则AB⋅AD的取值范围是( )A. [3,4]B. [3,2]C. [0,2]D. [2,4]6.若a>b>0,则①1b >1a;②ab>a+1b+1;③a+1−b+1>a−b.上述结论中,所有正确结论的序号是( )A. ①②B. ①③C. ②③D. ①②③7.若命题“∃x∈R,x2+2x+m≤0”是真命题,则实数m的取值范围是( )A. m<1B. m≤1C. m>1D. m≥18.“a=1”是“函数f(x)=2x+a2x−a具有奇偶性”的( )A. 充分而不必要条件B. 必要而不充分条件C. 充分必要条件D. 既不充分也不必要条件9.已知函数f(x)=3x−2x,则( )A. f(x)在R上单调递增B. 对∀x∈R,f(x)>−1恒成立C. 不存在正实数a,使得函数y=f(x)为奇函数a xD. 方程f(x)=x只有一个解10.如图为某无人机飞行时,从某时刻开始15分钟内的速度V(x)(单位:米/分钟)与时间x(单位:分钟)的关系.若定义“速度差函数”v(x)为无人机在时间段[0,x]内的最大速度与最小速度的差,则v(x)的图像为( )A. B.C. D.二、填空题:本题共5小题,每小题5分,共25分。

2019-2020学年北京市清华附中高三(上)10月月考数学试卷试题及答案

2019-2020学年北京市清华附中高三(上)10月月考数学试卷试题及答案

2019-2020学年北京市清华附中高三(上)10月月考数学试卷一、选择题1.已知集合{|2}A x x =>,{|(1)(3)0}B x x x =--<,则(A B = )A .{|1}x x >B .{|23}x x <<C .{|13}x x <<D .{|2x x >或1}x <2.若角θ的终边过点(3,4)P -,则tan()(θπ+= ) A .34B .34-C .43 D .43-3.已知函数a y x =,log b y x =的图象如图所示,则( )A .1b a >>B .1b a >>C .1a b >>D .1a b >>4.设函数()y f x =的定义域为R ,则“(0)0f =”是“函数()f x 为奇函数”的( ) A .充分而不必要条件 B .必要而不充分条件C .充分必要条件D .既不充分也不必要条件5.已知3cos 4α=,(2πα∈-,0),则sin 2α的值为( )A .38B .38-C D .6.我国古代数学名著《算法统宗》中有如下问题:“远望巍巍塔七层,红光点点倍加增,共灯三百八十一,请问尖头几盏灯?”意思是:一座7层塔共挂了381盏灯,且相邻两层中的下一层灯数是上一层灯数的2倍,则塔的顶层共有灯( ) A .1盏B .3盏C .5盏D .9盏7.某校象棋社团组织中国象棋比赛,采用单循环赛制,即要求每个参赛选手必须且只须和其他选手各比赛一场,胜者得2分,负者得0分,平局两人各得1分.若冠军获得者得分比其他人都多,且获胜场次比其他人都少,则本次比赛的参赛人数至少为( ) A .4B .5C .6D .78.已知定义在R 上的函数2,0()(),0x a x f x ln x a x ⎧+=⎨+>⎩…,若方程1()2f x =有两个不相等的实数根,则a 的取值范围是( ) A .1122a -<…B .102a <… C .01a <…D .102a -<…二、填空题9.已知函数()y f x =的导函数有且仅有两个零点,其图象如图所示,则函数()y f x =在 x = 处取得极值.10.32-,123,2log 5三个数中最大数的是 . 11.在ABC ∆中,13cos 14A =,73a b =,则B = . 12.去年某地的月平均气温(C)y ︒与月份x (月)近似地满足函数sin()(6y a b x a πϕ=++,b为常数,0)2πϕ<<.其中三个月份的月平均气温如表所示:则该地2月份的月平均气温约为 C ︒,ϕ= .13.在等腰梯形ABCD 中,已知//AB DC ,2AB =,1BC =,60ABC ∠=︒,点E 和F 分别在线段BC 和DC 上,且23BE BC =,16DF DC =,则AE AF 的值为 . 14.如图,线段8AB =,点C 在线段AB 上,且2AC =,P 为线段CB 上一动点,点A 绕点C 旋转后与点B 绕点P 旋转后重合于点D .设CP x =,CPD ∆的面积为()f x .则()f x 的定义域为 ;()0f x '=的解是 .三、解答题15.已知函数()cos()(0f x A x A ωϕ=+>,0ω>,0)2πϕ<< 的图象过点1(0,)2,最小正周期为23π,且最小值为1-. (1)求函数()f x 的解析式.(2)若[6x π∈,]m ,()f x 的值域是[1-,,求m 的取值范围.16.数列{}n a 的前项n 和记为n S ,若数列{}nS n是首项为9,公差为1-的等差数列. (1)求数列{}n a 通项公式n a .(2)若||n n b a =,且数列{}n b 的前项n 和记为n T ,求415T T +的值.17.已知ABC ∆的内角A 、B 、C 所对应的边分别为a ,b ,c ,8sin()17A C +=,且角B 为锐角.(1)求cos B 的值;(2)若6a c +=,ABC ∆的面积为2,求边长b .18.已知函数1()xax f x e -=. (Ⅰ)当1a =时,求函数()f x 的单调区间;(Ⅱ)当0a <时,求函数()f x 在区间[0,1]上的最小值.19.已知函数3()9f x x x =-,函数2()3g x x a =+.(1)若曲线()y f x =与曲线()y g x =在它们的交点处且有公共切线,求a 的值; (2)若存在实数b 使不等式()()f x g x <的解集为(,)b -∞,求实数a 的取值范围.20.设满足以下两个条件的有穷数列1a ,2a ,⋯,n a 为(2n n =,3,4,⋯,)阶“期待数列”:①1230n a a a a +++⋯+=; ②123||||||||1n a a a a +++⋯+=.(1)分别写出一个单调递增的3阶和4阶“期待数列”;(2)若某2013阶“期待数列”是等差数列,求该数列的通项公式;(3)记n 阶“期待数列”的前k 项和为(1k S k =,2,3,⋯,)n ,试证:1||2k S ….2019-2020学年北京市清华附中高三(上)10月月考数学试卷参考答案与试题解析一、选择题1.已知集合{|2}A x x =>,{|(1)(3)0}B x x x =--<,则(A B = )A .{|1}x x >B .{|23}x x <<C .{|13}x x <<D .{|2x x >或1}x <【解答】解:集合{|2}A x x =>, {|(1)(3)0}{|13}B x x x x x =--<=<<,则{|23}A B x x =<<.故选:B .2.若角θ的终边过点(3,4)P -,则tan()(θπ+= ) A .34B .34-C .43 D .43-【解答】解:角θ的终边过点(3,4)P -,则44tan()tan 33y x θπθ-+=-=-=-=, 故选:D .3.已知函数a y x =,log b y x =的图象如图所示,则( )A .1b a >>B .1b a >>C .1a b >>D .1a b >>【解答】解:由图象可知,01a <<,1b >, 故选:A .4.设函数()y f x =的定义域为R ,则“(0)0f =”是“函数()f x 为奇函数”的( ) A .充分而不必要条件B .必要而不充分条件C .充分必要条件D .既不充分也不必要条件【解答】解:函数()y f x =的定义域为R ,若函数()f x 为奇函数,则(0)0f =,反之不成立,例如2()f x x =.∴ “(0)0f =”是“函数()f x 为奇函数”的必要不充分条件.故选:B . 5.已知3cos 4α=,(2πα∈-,0),则sin 2α的值为( )A .38B .38-C D .【解答】解:3cos 4α=,(2πα∈-,0),sin α∴===,3sin 22sin cos 2(4ααα∴==⨯⨯= 故选:D .6.我国古代数学名著《算法统宗》中有如下问题:“远望巍巍塔七层,红光点点倍加增,共灯三百八十一,请问尖头几盏灯?”意思是:一座7层塔共挂了381盏灯,且相邻两层中的下一层灯数是上一层灯数的2倍,则塔的顶层共有灯( ) A .1盏B .3盏C .5盏D .9盏【解答】解:设塔的顶层共有1a 盏灯, 则数列{}n a 公比为2的等比数列, 717(12)38112a S -∴==-,解得13a =. 故选:B .7.某校象棋社团组织中国象棋比赛,采用单循环赛制,即要求每个参赛选手必须且只须和其他选手各比赛一场,胜者得2分,负者得0分,平局两人各得1分.若冠军获得者得分比其他人都多,且获胜场次比其他人都少,则本次比赛的参赛人数至少为( ) A .4B .5C .6D .7【解答】解:由题意可得,冠军得分比其他参赛人员高,且获胜场次比其他人都少,所以冠军与其他匹配场次中,平均至少为3场,A 选项:若最少4人,当冠军3次平局时,得3分,其他人至少1胜1平局,最低得3分,故A 不成立,B 选项:若最少5人,当冠军1负3平局时,得3分,其他人至少1胜1平,最低得3分,不成立,当冠军1胜3平局时,得5分,其他人至少2胜1平,最低得5分,不成立,故B 不成立, C 选项:若最少6人,当冠军2负3平局时,得3分,其他人至少1胜1平,最低得3分,不成立,当冠军1胜4平局时,得6分,其他人至少2胜1平,最低得5分,成立,故C 成立, D 选项:76>,故不为最少人数,故不成立,故选:C .8.已知定义在R 上的函数2,0()(),0x a x f x ln x a x ⎧+=⎨+>⎩…,若方程1()2f x =有两个不相等的实数根,则a 的取值范围是( ) A .1122a -<…B .102a <… C .01a <…D .102a -<…【解答】解:由题意知当0x >时,()()f x ln x a =+,则0a …, 当0x …时,()1a f x a <+…,若0a …,当0x >时,()()f x ln x a lna =+…,若方程1()2f x =有两个不相等的实数根, 则11212a a lna ⎧<+⎪⎪⎨⎪<⎪⎩…,即1212a a a ⎧<⎪⎪⎪-⎨⎪⎪<⎪⎩…,得1122a -<…,0a …,102a ∴<…, 故选:B .二、填空题9.已知函数()y f x =的导函数有且仅有两个零点,其图象如图所示,则函数()y f x =在x = 1- 处取得极值.【解答】解:函数()y f x =的导函数有且仅有两个零点,其图象如图所示, 1x <-时,()0f x '<,1x >-时,()0f x '…, 所以函数只有在1x =-时取得极值. 故答案为:1-.10.32-,123,2log 5三个数中最大数的是 2log 5 . 【解答】解:由于3021-<<,12132<<,22log 5log 42>=,则三个数中最大的数为2log 5. 故答案为:2log 5. 11.在ABC ∆中,13cos 14A =,73a b =,则B 3或3. 【解答】解:在ABC ∆中,13cos 14A =,sin A ∴== 73a b =,sin 7sin 3b A B a ∴===(0,)B π∈, 3B π∴=或23π. 故答案为:3π或23π. 12.去年某地的月平均气温(C)y ︒与月份x (月)近似地满足函数sin()(6y a b x a πϕ=++,b为常数,0)2πϕ<<.其中三个月份的月平均气温如表所示:则该地2月份的月平均气温约为 5- C ︒,ϕ= .【解答】解:函数sin()(6y a b x a πϕ=++,b 为常数),∴当51182x +==时,sin()6x πϕ+取得最大或最小值, ∴862k ππϕπ⨯+=+,k Z ∈,解得56k πϕπ=-,k Z ∈, 又02πϕ<<,6πϕ∴=;31a b ∴-=,且sin 13a b π+=,解得13a =,18b =-;1318sin()66y x ππ∴=-+,当2x =时,1318sin(2)5()66y C ππ=-⨯+=-︒.故答案为:5-,6π.13.在等腰梯形ABCD 中,已知//AB DC ,2AB =,1BC =,60ABC ∠=︒,点E 和F 分别在线段BC 和DC 上,且23BE BC =,16DF DC =,则AE AF 的值为18. 【解答】解:2AB =,1BC =,60ABC ∠=︒,1122BG BC ∴==,211CD =-=,120BCD ∠=︒, 23BE BC =,16DF DC =, ∴21()()()()36AE AF AB BE AD DF AB BC AD DC =++=++ 12216336AB AD AB DC BC AD BC DC =+++ 122121cos6021cos011cos6011cos1206336=⨯⨯︒+⨯⨯⨯︒+⨯⨯⨯︒+⨯⨯⨯⨯︒111291331818=++-=, 故答案为:291814.如图,线段8AB =,点C 在线段AB 上,且2AC =,P 为线段CB 上一动点,点A 绕点C 旋转后与点B 绕点P 旋转后重合于点D .设CP x =,CPD ∆的面积为()f x .则()f x 的定义域为 (2,4) ;()0f x '=的解是 .【解答】解:由题意,2DC =,CP x =,6DP x =- CPD ∆,∴262662x xx x x x +>-⎧⎪+->⎨⎪+->⎩,解得(2,4)x ∈如图,三角形的周长是一个定值8,故其面积可用海伦公式表示出来即()f x==()f x∴'=,令()0f x'=,解得3x=,故答案为:(2,4),3.三、解答题15.已知函数()cos()(0f x A x Aωϕ=+>,0ω>,0)2πϕ<<的图象过点1(0,)2,最小正周期为23π,且最小值为1-.(1)求函数()f x的解析式.(2)若[6xπ∈,]m,()f x的值域是[1-,,求m的取值范围.【解答】解:(1)由函数的最小值为1-,0A>,得1A=,最小正周期为23π,2323πωπ∴==,()cos(3)f x xϕ∴=+,又函数的图象过点1(0,)2,1cos2ϕ∴=,而02πϕ<<,3πϕ∴=,()cos(3)3f x xπ∴=+,(2)由[6xπ∈,]m,可知533633x mπππ++剟,5()cos66fππ==cos1π=-,7cos6π=,由余弦定理的性质得:7336mπππ+剟,∴25918mππ剟,即2[9mπ∈,5]18π.16.数列{}n a 的前项n 和记为n S ,若数列{}nS n是首项为9,公差为1-的等差数列. (1)求数列{}n a 通项公式n a .(2)若||n n b a =,且数列{}n b 的前项n 和记为n T ,求415T T +的值. 【解答】解:(1)数列{}nS n是首项为9,公差为1-的等差数列, ∴9(1)(1)10nS n n n=+-⨯-=-,即210n S n n =-+,① 2n ∴…时,21(1)10(1)n S n n -=--+-,②①-②可得1211n n n a S S n -=-=-+, 又当1n =时,119a S ==,满足上式, 211n a n ∴=-+;(2)由题意,|||112|n n b a n ==-,∴当15n 剟时,212(9112)102n n n nT a a a n n +-=++⋯+===-+;6n …时,2(5)(1211)2510502n n n T n n -+-=+=-+.41524125149T T ∴+=+=.17.已知ABC ∆的内角A 、B 、C 所对应的边分别为a ,b ,c ,8sin()17A C +=,且角B 为锐角.(1)求cos B 的值;(2)若6a c +=,ABC ∆的面积为2,求边长b . 【解答】解:(1)8sin()17A C +=, 8sin sin[()sin()17B AC A C π∴=-+=+=, 角B 为锐角, cos 0B ∴>,即15cos 17B ===.(2)ABC ∆的面积为2,118sin 22217S ac B ac ∴==⨯=, 则172ac =, 6a c +=,2222151717152cos ()2236223617154172217b ac ac B a c ac ac∴=+-=+--=-⨯-⨯⨯=--=, 则2b =.18.已知函数1()xax f x e -=. (Ⅰ)当1a =时,求函数()f x 的单调区间;(Ⅱ)当0a <时,求函数()f x 在区间[0,1]上的最小值. 【解答】解:(Ⅰ)1a =时,1()xx f x e -=,x R ∈, 2()xx f x e -+∴'=, 令()0f x '>,解得:2x <, 令()0f x '<,解得:2x >,()f x ∴在(,2)-∞递增,在(2,)+∞递减;(Ⅱ)由1()xax f x e -=得: 1()xax a f x e -++'=,[0x ∈,1], 令()0f x '=,0a <,解得:111x a=+<, ①110a+…时,即10a -<…时,()0f x '…对[0x ∈,1]恒成立,()f x ∴在[0,1]递增,()(0)1min f x f ==-;②当1011a<+<时,即1a <-时, x ,()f x ',()f x 在[0,1]上的情况如下:111()(1)aaf x min f ae +∴=+=;综上,10a -<…时,()1min f x =-,1a <-时,11()min aa f x e+=.19.已知函数3()9f x x x =-,函数2()3g x x a =+.(1)若曲线()y f x =与曲线()y g x =在它们的交点处且有公共切线,求a 的值; (2)若存在实数b 使不等式()()f x g x <的解集为(,)b -∞,求实数a 的取值范围. 【解答】解:(1)2()39f x x '=-,()6g x x '=,设()f x 与()g x 的交点坐标为0(x ,0)y ,则3200020093396x x x a x x ⎧-=+⎪⎨-=⎪⎩,解得:015x a =-⎧⎨=⎩或0327x a =⎧⎨=-⎩,a ∴的值为5或27-;(2)令32()39h x x x x =--,则()y h x =的图象在直线y a =的下方的部分对应点的横坐标(,)x b ∈-∞,2()3693(1)(3)h x x x x x '=--=+-,∴令()0h x '=,得:1x =-或3,列表:()h x ∴的极大值为(1)5h -=,极小值为h (3)27=-,又当x →+∞时,()h x →+∞,当x →-∞时,()h x →-∞, 如图所示:∴当5a >或27a -…时,满足题意,∴实数a 的取值范围为:(-∞,27](5,)-+∞.20.设满足以下两个条件的有穷数列1a ,2a ,⋯,n a 为(2n n =,3,4,⋯,)阶“期待数列”:①1230n a a a a +++⋯+=; ②123||||||||1n a a a a +++⋯+=.(1)分别写出一个单调递增的3阶和4阶“期待数列”;(2)若某2013阶“期待数列”是等差数列,求该数列的通项公式;(3)记n 阶“期待数列”的前k 项和为(1k S k =,2,3,⋯,)n ,试证:1||2k S …. 【解答】解:(1)数列12-,0,12为三阶期待数列,数列38-,18-,18,38为四阶期待数列.(Ⅱ)设该2013阶“期待数列”的公差为d , 1220130a a a ++⋯+=,∴120132013()02a a +=,120130a a ∴+=,即10070a =, 1008a d ∴=,当0d =时,与期待数列的条件①②矛盾,当0d >时,据期待数列的条件①②可得10081009201312a a a ++⋯+=, 100610051100622d d ⨯∴+=,即110061007d =⨯, *10071007(1007)(10061007n n a a n d n N -∴=+-=∈⨯,2013)n …,当0d <时,同理可得100710061007n n a -+=⨯,*(n N ∈,2013)n ….(Ⅲ)当k n =时,显然1||02n S =…成立; 当k n <时,根据条件①得:1212()k k k k n S a a a a a a ++=++⋯+=-++⋯+, 即1212||||||k k k k n S a a a a a a ++=++⋯+=++⋯+,12121212||||||||||||||||1k k k k n k k n S a a a a a a a a a a a +++∴=++⋯++++⋯+++⋯+++⋯+=…,1||(12k S k ∴=…,2,⋯,)n .。

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清华附中高三十月月考第三部分:阅读理解(共两小节;40分)ALife is all about making choices. At any stage along the way, you may find yourself asking: Am I headed in the right direction What would I really enjoy doing What do I want to do nextThe six-week summer Career Discovery program at the Harvard University Graduate School of Design (GSD) welcomes people—from recent high school and college graduates to seasoned professionals—who are grappling with questions like these. And not just people considering a career in design or planning, but people with a broad spectrum of interests and remarkably diverse plans and goals. What they have in common is the drive and desire to seek the answers to those questions.Participants in our program commit themselves fully to a path of intensive studio work, lectures, workshops, and field trips. Deeply immersed in a culture that is both challenging and rewarding, they experience what education and work are like in the design and planning professions. They emerge—many of them exhilarated—with a more profound understanding of the possibilities ahead and the choices they will make.We invite you to explore the possibilities in Career Discovery: If you would like to be placed on our mailing list to receive a copy of this information in brochure format, please complete an online request form. You may also view a PDF of the brochure here.Career Discovery is open to anyone graduating from high school in 2016 or older with an interest in design or planning. A high level of academic skill is necessary to take full advantage of our rigorous program.The online application deadlines are as follows:For international applicants: on or before April 1, 2016For all other applicants: on or before April 30, 2016We encourage you to submit your application early as applications are reviewed and accepted in the order that they are received.Harvard UniversityGraduate School of DesignCareer Discovery*Underrepresented minority college students are encouraged to apply. Need-based financial aid is available.56. Where is the passage probably fromA. A website brochure magazine poster57. To make full use of the Career Discovery Program, the applicants need to .A. have a wide range of interestsB. consider a future career in designC. have a high level of academic skillD. graduate from high school in 201658. What can be inferred from the passageA. The program offers mainly lectures and workshops.B. A Chinese applicant needs to apply before April 30, 2016.C. Scholarships are offered to outstanding international students.D. Applicants need to fill in a request form to get more details.BIt took me a long time to understand the difference between a present and a gift.I grew up in a household where presents marked special occasions. There was always something for each of us at Christmas or on our birthdays. So when I married a man who did not give presents on regular basis, it was an adjustment. Gary didnot wholly shun gift-giving, though. Sometimes he would return from sea with something he found that reminded of me – a meat cleaver (切肉机) on our first Christmas, a paring (削皮) knife on our fifth. But mostly, he ignored holidays, refusing to shop for a thing to present to me as a sign of his affection.I could not accept this presentless marriage with the one I had grown up observing. I tried to change him by example. I knitted him sweaters, socks, hats and gloves for Christmas; made him shirts; and bought books for his birthdays. He appreciated the caring these gifts represented, but refused to follow my example.I began to tell him what I wanted, giving specific instruction. When Gary left for the local auction one Saturday (my birthday as it happened), I asked him to find me a necklace or diamond earring, as a birthday gift. He came home with a road scraper (刮板). I was stunned that he had missed the mark by so much. He attached it to the back of the ancient tractor, and then enthusiastically showed me how to use it, not seeing that I was not grateful.But when a snowstorm hit later that year and he was at sea, I used the road scraper to plow out both our drive and our neighbor’s. Gary had wisely chosen not the thing that I wanted, but the thing that he knew I would need.I finally realized that he had been giving me gifts all along. The gestures, large and small, born of his caring and concern for me, for our children and for our lives together were the gifts that he gave daily. We struggle to teach others how to love us. In that struggle, we often forget how to appreciate the love they already give us as only they can give it. There are two parts to a gift –the giving and the accepting.I finally began to understand the difference between a present and a gift. A present is a thing. But a gift is a broader and often deeper. It is a small act of kindne ss, the willingness to bend to another’s needs. Love is a gift. Any expression of it, freely given, is an offering from the heart that is immeasurably better thana present.59. Why didn’t Gary give the author presents as often as she expected him toA. Because his family didn’t have this tradition.B. Because he thought gifts ought to be useful.C. Because he was too busy to think of preparing a gift.D. Because he knew the difference a present and a gift.60. Which of the following word has th e closest meaning to “shun” in Paragraph 2A. preventB. acceptC. hideD. avoid61. Which of the following is TRUE according to the passageA. The author finally realized that love means more than giving presents.B. Gary pretended tha t he didn’t know what his wife expected from him.C. The author never expected to receive any jewelry from her husband.D. The couple accepted each other and changed their views on love.62. What is the passage mainly aboutA. Marriage needs compromises.B. A gift is different from a present.C. A small kindness goes a long way.D. Family traditions must be cherished.CThe $ 11 billion self – help industry is built on the idea that you should turn negative th oughts like “ I never do anything right” into positive ones like “I can succeed.” But was positive thinking advocated by Norman Vincent Peale rightIs there power in positive thinkingResearchers in Canada just published a study in the journal Psychological Science that says trying to get people to think more positively can actually havethe opposite effect: it can simply highlight how unhappy they are.The study's authors, Joanne Wood and John Lee of the University of Waterloo and Elaine Perunovic of the University of New Brunswick, begin by citing older research showing that when people get feedback which they believe is overly positive, they actually feel worse, not better. If you tell your dim friend that he has the potential of an Einstein, you're just underlining his faults. In one 1990s experiment, a team including psychologist Joel Cooper of Princeton asked participants to write essays opposing funding for the disabled. When the essayists were later praised for their sympathy, they felt even worse about what they had written.In this experiment, Wood, Lee and Perunovic measured 68 students' self-esteem. The participants were then asked to write down their thoughts and feelings for four minutes. Every 15 seconds, one group of students heard a bell. When it rang, they were supposed to tell themselves, "I am lovable."Those with low self-esteem didn't feel better after the forcedself-affirmation. In fact, their moods turned significantly darker than those of members of the control group, who weren't urged to think positive thoughts.The paper provides support for newer forms of psychotherapy (心理治疗) that urge people to accept their negative thoughts and feelings rather than fight them. In the fighting, we not only often fail but can make things worse. Meditation (静思) techniques, in contrast, can teach people to put their shortcomings into a larger, more realistic perspective. Call it the power of negative thinking.63. According to the first paragraph, the self-help industry_____A.has been declining sharply.B.has produced positive results.C.was founded by Norman Vincent Peale.D.is based on the concept of positive thinking.64. What is the finding of the Canadian researchersA.Unhappy people cannot think positively.B.The power of positive thinking is limited.C.Encouraging positive thinking many do more harm than good.D.There can be no simple treatment for psychological problems.65. The underlined word “dim” in the third paragraph probablymeans______A.not hard-working B.not intelligent C.not happy D. not right 66. We can learn from the last paragraph that _______A.Meditation may prove to be a good form of treatment.B.People can avoid making mistakes through meditation.C.Different people tend to have different ways of thinking.D.The effect of positive thinking vary from person to person.DLots of people assume that the virtual will replace the physical with something cheaper, faster and more efficient. In education, however, the virtual will create a very different type of disruption. We should not aim to replace the physical classroom. Instead we have an opportunity to blend the virtual with the physical and reimagine education entirely.Today students in most classrooms sit, listen and take notes while a professor lectures. Despite there being anywhere from 20 to 300 human beings in the room, there is little to no human interaction(互动). Exams often offer the first opportunity for the professor to get realinformation on how well the students digested the knowledge. If the test identifies gaps in students' understanding of a basic concept, the class still moves on to a more advanced concept.Virtual tools are providing an opportunity to rethink this methodology. If a lecture is available online, class time can be freed for discussion, peer tutoring or professor-led exploration. If a lecture is removed from class time and we have on-demand adaptive exercises and diagnostics, there is no need to continue the Prussian education model—where students are pushed together at a set pace. Instead students can progress at their own pace and continue to prove their knowledge long after the formal course is over.Perhaps the most powerful effect of this reality is what it does to the quality of lectures and other learning material in general. Traditional lecturers and textbook publishers have little to no information on how their content is being used or whether it is even effective. By coupling rich physical experiences with online tools, content creators and professors can finally have granular, up-to-date data on the efficacy of the experiences they create.In this “blended learning” reality, the professors' role is moved up the value chain. Rather than spending the bulk of their time lecturing, writing exams and grading them, they can now interact with their students. Rather than enforcing a sit-and-listen passivity, teachers will mentor and challenge their students to take control of their own learning—the most important skill of all. Yes, for a motivated student in an underdeveloped part of the world, these virtual tools—supposing we can overcome the difficulty in equipment—may facilitate most of their learning. In the developed world, the best experience will be to make useof the online tools so that the physical time can be less passive and more, well, human.67. In today’s class, teachers get real information of students’ learning by______A. carrying out examsB. checking students’ notesC. interacting with studentsD. introducing advanced concept68. Professors can benefit from virtual tools because they can_______A. find high-quality lectures on the Internet.B. identify the gaps of students’ understanding.C. get effective information about their teaching.D. lead students to progress together at the same pace.69. What can we learn from the passageA. The author holds a negative attitude towards virtual tools.B. Virtual class will eventually replace the present class model.C. Students usually spend a lot of time managing their learning.D. Teachers become more important in a “ blended learning” class.70. The main purpose of the passage is to ______A. introduce and evaluateB. examine and assessC. compare and argueD. inform and explain第二节七选五(共5小题; 每小题2分,共10分)About five percent of people are left-handed; they tend to use left hand more often, for more purposes, than the right hand. Some parents of the left-handed children worry about the condition. 71 But most authorities agree that left-handed children should be allowed to perform naturally. Many of the most talented people in history have been left-handed––the greatest artist Leonardo da Vinci and Michelangelo among them.Left-handed persons find out early in life that they are living in a"right-handed society." Most of the objects they encounter are made for the convenience of right-handed people––locks, screws, doorknobs, golf clubs, and even automobiles. Left-handed people must adjust to this "right-handed." 72 They are able to do anything that right-handed people can do.73 But it is known that the body is not exactly the same on both sides. The left side of the face is a little different from the right. One leg is usually somewhat stronger than the other, as is the arm.Scientists even have found that the right half and the left half of the brain do not function in the same way. This is still being studied. It is believed that in most people the left half of the brain predominates, or "rules", over the right half. The nerves from the brain cross over at the level of neck, going opposite sides of the body and making most people right-handed. 74Whatever the cause, the research indicates that ever since prehistoric times few humans have been left-handed. 75 Ancient tools and weapons that have been unearthed were clearly made for right-handed, not left-handed, people.A. We can even see the evidence.B. They think about trying to change it.C. Most of them soon learn to manage quite well.D. Our hands are controlled by the motor center of the brain.E. Most left-handed people face the right side when writing or drawing.F. But no one has explained why the left side of the brain dominates in most people.G. No one is sure of what makes most people right-handed and a minority left-handed.第二节完形填空(共20小题;每小题分,共30分)The requirements for high school graduation have just changed in my community. As a result, all students must ______36__ sixty hours ofservice learning, or they will not receive a ____37___. Service learning is academic learning that also helps the of service learning include cleaning up a polluted river, working in a soup kitchen, or tutoring a student. _____39__ a service experience, students must keep a journal and then write a ___40______ about what they have learned.____41___ claim that there are many benefits of service learning. Perhaps most importantly, students are forced to think __42___ their own interests and become aware of the needs of others. Students are also able to learn real-life ___43_____ that include responsibility, problem-solving, and working as part of the team. _____44___ students can explore possible careers ___45______ service learning. For example, if a student wonders what teaching is like, he or she can choose to work in a elementary school classroom a few afternoon each month.____46____ there are many benefits, opponents ____47____ problems with the new requirements. First, they ___48______ that the main reason students go to school is to learn core subjects and skills. Because service learning is time-consuming, students spend ____49____ time studying the core subjects. Second, they believe that forcing students to work without ___50____ goes against the law. By requiring service, the school takes away an individual’s freedom to choose.In my view, service learning is a great way to ___51_____ to the community, learn new skills, and explore different careers,___52_____ I don’t believe schools should force people to help others –the ____53___ to help must come from the heart. I think the best ____54____ is one that gives students choice: a student should be able to choose sixty hours of independent study or sixty hours of service. Choice encourages both freedom and responsibility, and as young adults, they must learn to___55_____ both wisely.. spend B. gain C. complete D. save. certificate B. diploma C. degree D. grade. Subjects B. Ideas C. Procedures D. Examples 39. A. With B. Before C. During D. Since. report B. diary C. note D. notice. Teachers B. Supporters C. Parents D. Students . beyond B. about C. over D. of. tools B. secrets C. services D. skills. Gradually B. Finally C. Luckily D. Quickly . through B. across C. of D. on46. A. So B. Thus C. But D. While. deal with B. look into C. point out D. take down . argue B. doubt C. overlook D. admit. much B. full C. less D. more. cost B. pay C. care D. praise. contribute B. appeal C. attend D. belong 52. A. Therefore B. Otherwise C. Besides D. However . courage B. desire C. emotion D. spirit. decision B. purpose C. solution D. result. serve B. choose C. explore D. handle第四部分:书面表达(共两节,35分)第一节:应用文写作(15分)你是红星中学的高三学生李华。

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