上海中学高三上周测01和02(图片版)-word文档资料

大同中学2022-2023学年第一学期阶段性质量检测高三数学试卷（第5周）说明：1.本试卷共6页，若试卷有印刷错误，请立刻向监考老师反映. 2.本次考试满分150分，120分钟完成.3.答案请一律写在答题纸上，做在试卷和草稿纸上一律不得分.4.凡是标明A 、B 组的试题均为选做题，平行班同学请选做A 组，实验班同学请选做B 组，评卷时会根据学生的班级号进行分组阅卷，故不要跨组选做，否则后果自负. 一、填空题：（本大题满分54分，前6题每小题满分4分，后6题每小题满分5分）1. 已知集合则_____.{}1,2,3,6,{|23},A B x x =-=-<<=A B 【答案】 {}1,2-【解析】【详解】试题分析：．故答案应填： {}{}{}1,2,3,6|231,2A B x x ⋂=-⋂-<<=-{}1,2-【考点】集合运算【名师点睛】本题重点考查集合的运算，容易出错的地方是审错题意，属于基本题，难度不大.一要注意培养良好的答题习惯，避免出现粗心而出错，二是明确江苏高考对于集合题的考查立足于列举法，强调对集合运算有关概念及法则的理解.2. 复数其中i 为虚数单位，则z 的实部是________________. (12)(3),z i i =+-【答案】5 【解析】【详解】试题分析：．故答案应填：5 (12i)(3i)55i z =+-=+【考点】复数概念【名师点睛】本题重点考查复数的基本运算和复数的概念，属于基本题.首先对于复数的四则运算，要切实掌握其运算技巧和常规思路，如，其次要熟悉复()()()(),,,,a bi c di ac bd ad bc i a b c d R ++=-++∈数的相关概念，如复数的实部为，虚部为，共轭为 (,)a bi a b R +∈a b a bi -3. 在平面直角坐标系中，双曲线的焦距是____________.xOy 22173x y -=【答案】 【解析】【详解】试题分析：．故答案应222227,3,7310,2a b c a b c c ==∴=+=+=∴=∴=填：【考点】双曲线性质【名师点睛】本题重点考查双曲线几何性质，而双曲线的几何性质与双曲线的标准方程息息相关，明确双曲线标准方程中各个量的对应关系是解题的关键，揭示焦点在x 轴，实轴长22221(0,0)x y a b a b-=>>为，虚轴长为，焦距为，离心率为．2a 2b 2c =b y x a =±c a=4. 已知{a n }是等差数列，S n 是其前n 项和.若a 1+a 22=3，S 5=10，则a 9的值是______. -【答案】 20.【解析】【详解】由得，因此510S =32a =2922(2)33,23620.d d d a -+-=-⇒==+⨯=考点：等差数列性质5. 定义在区间[0,3π]上的函数y=sin2x 的图象与y=cosx 的图象的交点个数是______. 【答案】7 【解析】 【详解】由，因为，所以1sin 2cos cos 0sin 2x x x x =⇒==或[0,3]x π∈共7个 3551317,,,,,,,2226666x πππππππ=考点：三角函数图像6. 如图，一个酒杯的内壁的轴截面是抛物线的一部分，杯口宽，杯深，称为抛物线酒杯.在8cm 杯内放入一个小的玻璃球，要使球触及酒杯底部，则玻璃球的半径的取值范围为______.【答案】10,2⎛⎤ ⎥⎝⎦【解析】【分析】根据圆的方程、抛物线方程以及两点的距离公式建立不等式求解.【详解】由题可知，设抛物线方程为，因为，22x py =()B所以，解得，所以抛物线方程为， (216p =12p =2x y =在杯内放入一个小的玻璃球，要使球触及酒杯底部， 设玻璃球截面所在圆的方程为，()222x y r r +-=，r ≥则有恒成立，解得，得， ()22120xxr +-≥120r -≥102r <≤所以玻璃球的半径的取值范围为.10,2⎛⎤ ⎥⎝⎦故答案为：.10,2⎛⎤ ⎥⎝⎦7. 定义在R 上的函数f(x )满足+>1, ,则不等式（其中e 为自然对()f x ()f x '()04f =()3xxe f x e >+数的底数）的解集为_________. 【答案】 (0,)+∞【解析】【分析】构造函数，根据，利用导数研究的单调性，()()(),xxg x e f x e x R =-∈()3xxe f x e >+()g x 结合原函数的性质和函数值，利用单调性转化不等式，从而可得结果. 【详解】设，()()(),xxg x e f x e x R =-∈则()()()()()''1x x x xg x e f x e f x e e f x f x ⎡⎤=+-=+-⎣⎦，()()()()'1,'10f x f x f x f x +>∴+-> ，在定义域上单调递增，()'0g x ∴>()y g x ∴=，()()3,3x x e f x e g x >+∴> 又，()()000413g e f e =-=-= ，()()0,0g x g x ∴>∴>即不等式的解集为，故答案为.()3xxe f x e >+()0,+∞()0,+∞【点睛】本题主要考查抽象函数的单调性以及函数的求导法则，属于难题.求解这类问题一定要耐心读题、读懂题，通过对问题的条件和结论进行类比、联想、抽象、概括，准确构造出符合题意的函数是解题的关键；解这类不等式的关键点也是难点就是构造合适的函数，构造函数时往往从两方面着手：①根据导函数的“形状”变换不等式“形状”；②若是选择题，可根据选项的共性归纳构造恰当的函数.8. 在平面直角坐标系中，已知直线的参数方程为（为参数），椭圆的参数方程为xOyl 112x t y ⎧=+⎪⎪⎨⎪=⎪⎩t C （为参数），设直线与椭圆相交于、两点，则线段的长是________ cos 2sin x y θθ=⎧⎨=⎩θl C A B AB 【答案】167【解析】【分析】将椭圆方程和直线方程化为普通方程，联立可求得两交点的横坐标，利用弦长公式可求得12,x x 结果.【详解】由椭圆参数方程得椭圆的普通方程为：C 2214y x +=由直线参数方程得直线的普通方程为：l y =-联立可得：，即，解得：， ()224314x x +-=27610x x --=11x =217x =-11677AB ∴==故答案为167【点睛】本题考查参数方程中的弦长问题的求解，关键是能够熟练掌握参数方程化普通方程的方法，利用弦长公式求得弦长.9. 设 是定义在R 上且周期为2的函数，在区间[）上， 其中()f x 1,1-,10,(){2,01,5x a x f x x x +-≤<=-≤<.a R ∈若 ，则的值是.(5)f a 【答案】 25-【解析】【详解】， 51911123()()()(22222255f f f f a a -=-==⇒-+=-⇒=因此 32(5)(3)(1)(1)1.55f a f f f ===-=-+=-【考点】分段函数，周期性质【名师点睛】分段函数的考查方向注重对应性，即必须明确不同的自变量所对应的函数解析式是什么.函数周期性质可以将未知区间上的自变量转化到已知区间上.解决此类问题时，要注意区间端点是否可以取到及其所对应的函数值，尤其是分段函数分界点处的函数值.10. 如图，在中，是的中点，是上的两个三等分点，，ABC ∆D BC ,E F ,A D 4⋅=BA CA ，则 的值是_______.1BF CF ⋅=- BE CE ⋅u u r u ur【答案】78【解析】【详解】因为，222211436=42244AD BC FD BC BA CA BC AD BC AD --⋅=-⋅--== （）（），2211114123234FD BC BF CF BC AD BC AD -⋅=-⋅--==- （）（）因此，22513,82FD BC ==2222114167.22448ED BC FD BC BE CE BC ED BC ED --⋅=-⋅--=== （）（）【考点】向量数量积【名师点睛】研究向量的数量积，一般有两个思路，一是建立平面直角坐标系，利用坐标研究向量的数量积；二是利用一组基底表示所有向量，两种思路实质相同，但坐标法更易理解和化简. 对于涉及中线的向量问题，一般利用向量加、减法的平行四边形法则进行求解．11. 若函数的表达式（常数）对于任意两个不同的、，当()y f x =()21f x tx x =++t ∈R 1x 2x 时，均有（为常数，）成立，如果满足条件的最小[]122,2x x ∈-、()()1212f x f x k x x -≤-k k ∈N 自然数为6，则实数的取值范围是______. k t 【答案】或. 5342t -≤<-3524t <≤【解析】【分析】根据已知，利用对任意且恒成立， ()1216x x t ++≤[]122,2x x ∈-、12x x ≠且对且有解，进而得到关于实数t 的不等式组求解.()1215x x t ++>[]122,2x x ∈-、12x x ≠【详解】因为，所以()21f x tx x =++，()()()()()()122211*********f x f tx x t t x x x x x x x +++-=-=++-+⎡⎤⎣⎦因为，所以由有：， 12x x ≠()()1212f x f x k x x -≤-()121t x x k ++≤由题可知，对任意且恒成立， ()1216x x t ++≤[]122,2x x ∈-、12x x ≠且对且有解， ()1215x x t ++>[]122,2x x ∈-、12x x ≠则对任意且恒成立，()1275x t x +-≤≤[]122,2x x ∈-、12x x ≠或对且有解，()124t x x +>()126x x t <-+[]122,2x x ∈-、12x x ≠因为，且，所以， []122,2x x ∈-、12x x ≠1244x x -<+<若对任意且恒成立，()1275x t x +-≤≤[]122,2x x ∈-、12x x ≠则，解得；745745t t -≤-≤⎧⎨-≤≤⎩5544t -≤≤若或对且有解， ()124t x x +>()126x x t <-+[]122,2x x ∈-、12x x ≠则时，或；时，或； 0t >46t -<-44t >0t <44t ->46t <-解得或.32t <-32t >因此，实数的取值范围是或. t 5342t -≤<-3524t <≤故答案为：或. 5342t -≤<-3524t <≤12. 如图所示，在四面体 中，底面 是一个边长为2的等边三角形， 的外心为点-P ABC ABC ABC O ，平面 ，且 ，动点 分别在线段（含端点）上和所在的平面中运PA ⊥ABC 1PA =M N 、PA PBC 动，满足 ．1MN =（1）则的最大值为 __． 2ON （2）则的取值范围为 __．2ON【答案】 ①.; ②. . 193193⎤⎥⎦【解析】【分析】取 的中点E ，连接 ，等边三角形的中心O 在 上，过点O 作 于BC PE AE ，AE OG PE ⊥G ，过点M 作 于I ，由几何关系可得N 在 所在平面中运动，所以N 的轨迹是以I 为圆MI PE ⊥PBC 心， 为半径的圆，据此分别求得的最大值和最小值，即可确定其取值范围． IN 2ON 【详解】取的中点E ，连接，等边三角形的中心O 在上， BC PE AE ，AE 过点O 作于G ，过点M 作于I ，OG PE ⊥MI PE ⊥∵是等边三角形，所以 ， ABC AE BC ⊥∵平面，平面,PA ⊥ABC BC ⊂ABC ∴ ，又平面 ，∴BC ⊥平面， PA BC ⊥,,AP AE A AP AE =⊂ PAE PAE 又平面，∴平面平面，平面平面,BC ⊂PBC PBC⊥PAE PBC ⋂PAE PE =∵ ，平面，∴平面， MI PE ⊥MI ⊂PAE MI ⊥PBC 同理 平面，OG ⊥PBC设 ，∵ 是边长为2的等边三角形，所以，∵ ，故 ，,01PM m m =≤≤ABC AE =1PA =2PE =所以 ， 3060PEA EPA ∠=︒∠=︒，则， 11113232OG GE ====在 中，，在 中，∵ ，∴PMI MI m =Rt MNI 1MN =IN =∵N 在所在平面中运动，所以N 的轨迹是心I 为圆心，为半径的圆， PBC IN又∵平面 ，所以， OG ⊥PBC 22222ON OG GN GN =+=+所以 最小时， 的值最小， GN ON 又的最小值为的值减去圆的半径，GN GI，所以最小值为， 12PI m =GN 11222GN m =--设，11()2122f x x x =---≤≤则单调递增，131()1222x f x x '=-+=-≤≤令，解得，当，递减， ()0f x '=x =x <()0f x '<()f x 当，递增，故，x >()0f x '>()f x min3()2fx f ==-即当时，最小，最小值为， m =GN 32故最小值为2ON 223(2+当的值最大时最大，最大值 GN ON 113122222GN m m =--+=-+由于，（）为m 的减函数，所以 时最大， 3122y m =-+01m ≤≤0m =GN 最大值为， 52GN =此时的最大值为， 2ON 22519()23+=故答案为：，． 193193⎤⎥⎦【点睛】本题主要考查空间距离的计算，涉及到空间线面垂直以及面面垂直的证明和性质的应用，考查了空间想象能力，解答的关键是将线段长度的平方表示为参变量的函数，然后利用函数的最值求解线段的最值.二、选择题：（本大题满分20分，共4小题，每题有且只有一个结论是正确的，必须把正确结论的代号填涂在答题纸相应的空格中.每题选对得5分）13. 若在区间内有定义，且x 0∈，则“”是“x 0是函数的极值点”的()f x (,)a b (,)a b 0()0f x '=()f x （ ）A. 充分非必要条件B. 必要非充分条件C. 充要条件D. 既非充分条件也非必要条件 【答案】D 【解析】【分析】根据极值的概念，导数的几何意义即可求解． 【详解】由不一定能得到x 0是函数的极值点，0()0f x '=()f x 反例，，但并不是的极值点，3()f x x =(0)0f '=0x =()f x 反过来：x 0是函数的极值点也不一定能得到， ()f x 0()0f x '=反例，为的极小值点，但不存在，()||f x x =0x =()f x 0()f x '∴“”是“x 0是函数的极值点”的既非充分条件也非必要条件， 0()0f x '=()f x 故选：D ．14. 对于函数f （x ）=asinx+bx+c(其中，a,b R,c Z),选取a,b,c 的一组值计算f （1）和f （-1），所得出的∈∈正确结果一定不可能是 A. 4和6 B. 3和1 C. 2和4 D. 1和2【答案】D 【解析】【详解】试题分析：求出f （1）和f （﹣1），求出它们的和；由于c ∈Z ，判断出f （1）+f （﹣1）为偶数． 解：f （1）=asin1+b+c ① f （﹣1）=﹣asin1﹣b+c ② ①+②得： f （1）+f （﹣1）=2c ∵c ∈Z∴f （1）+f （﹣1）是偶数 故选D考点：函数的值．15. 在平面直角坐标系中，已知点，，动点满足xOy ()2,1M -()2,1N -P ()22PM PN a a -=∈R ，记点的轨迹为曲线，则下列命题中，可能成立的个数为（ ）P C （I ）曲线上所有的点到点的距离大于2 C 1,4a ⎛⎫⎪⎝⎭（II ）曲线上有两点到点与的距离之和为6 C ())（III ）曲线上有两点到点与的距离之差为2C ())（IV ）曲线上有两点到点的距离与到直线的距离相等 C (),0a x a =-A. 1个 B. 2个C. 3个D. 4个【答案】B 【解析】【分析】根据已知可得出点的轨迹方程为直线：.根据点到直线的距离，即可判断（I ）；P 84x y a -+=根据椭圆、双曲线以及抛物线的定义，可求出（II ）（III ）（IV ）的点的轨迹方程，结合各个图象的性质，即可判断（II ）（III ）（IV ）.【详解】设，则由已知可得，(),P x y ，22PM PN -()()()()22222121x y x y =-++-+--84x y a =-+=所以，点的轨迹方程为直线：. P 84x y a -+=对于（I ），点到直线的距离，故（I ）错误； 1,4a ⎛⎫ ⎪⎝⎭84x y a -+=2d <对于（II ），根据椭圆的定义可知，到点与的距离之和为6的点在椭圆上，设椭圆方程为())， ()2211221110x y a b a b +=>>则，，所以，，126a =1c =13a =2221114b a c =-=所以，椭圆方程为.22194x y +=当时，直线方程为，显然与椭圆有两个交点， 0a =2y x =即曲线上有两点到点与的距离之和为6，故（II）正确；C ())对于（III ），根据双曲线的定义可知，到点与的距离之差为2的点的轨迹为双曲线的一())支.设双曲线的方程为，()2222222210,0x y a b a b -=>>则，，所以，，222a =2c =21a =2222224b c a =-=所以，双曲线的方程为.()22104y x x -=>因为双曲线的渐近线方程为，直线的斜率为2， 2y x =±84x y a -+=所以直线与双曲线的一条渐近线平行或重合， 所以，直线与双曲线最多有一个交点，故（III ）错误；对于（IV ），当时，根据抛物线的定义可知，到点的距离与到直线的距离相等的点的轨迹为抛物0a ≠(),0a x a =-线.由已知可设抛物线的方程为，24y ax =联立可得，，2484y ax x y a⎧=⎨-+=⎩22240y ay a -+=，()22244280a a a ∆=--⨯=>所以，当时，直线与抛物线有两个交点，0a ≠即曲线上有两点到点的距离与到直线的距离相等，故（IV ）正确. C (),0a x a =-综上所述，可能成立的为（II ）（IV ）. 故选：B .16. 老张每天下班回家，通常步行5分钟后乘坐公交车再步行到家，公交车有，两条线路可以17:00A B 选择.乘坐线路所需时间（单位：分钟）服从正态分布，下车后步行到家要5分钟；乘坐线路A ()44,4N 所需时间（单位：分钟）服从正态分布，下车后步行到家要12分钟.下列说法从统计角度认B ()33,16N为不合理的是（ ） （参考数据：，则，，()2,Z N μσ~(),0.6827P Z μσμσ-<+≈()2,20.9545P Z μσμσ-<+≈）()3,30.9973P Z μσμσ-<+≈A. 若乘坐线路，前一定能到家B 18:00B. 乘坐线路和乘坐线路在前到家的可能性一样 A B 17:58C. 乘坐线路比乘坐线路在前到家的可能性更大 B A 17:54D. 若乘坐线路，则在前到家的可能性不超过 A 17:481%【答案】A 【解析】【分析】利用正态分布曲线的对称性及正态分布的概率，对四个选项逐个分析判断即可. 【详解】对于A ，因为， 11(45)[1(2145)](10.9973)0.0013522P B P Z >=-≤≤=⨯-=即乘坐线路能到家的概率为，18:020.00135所以乘坐线路，前不一定能到家，所以A 错误， B 18:00对于B ，乘坐线路在前到家的概率为A 17:58，11(48)[1(4048)](4048)(10.9545)0.95450.9772522P A P Z P Z <=-≤≤+≤≤≈⨯-+=乘坐线路在前到家的概率为B 17:58，11(41)[1(2541)](2541)(10.9545)0.95450.9772522P B P Z P Z <=-≤≤+≤≤≈⨯-+=所以乘坐线路和乘坐线路在前到家的可能性一样，所以B 正确， A B 17:58对于C ，乘坐线路在前到家的概率为， A 17:541(44)2P A <=乘坐线路在前到家的概率为B 17:54，111(37)[1(2937)](2937)(10.6827)0.68270.84135222P B P Z P Z <=-≤≤+≤≤≈⨯-+=>所以乘坐线路比乘坐线路在前到家的可能性更大， B A 17:54对于D ，乘坐线路，则在前到家的概率为A 17:48，所以D 正确，11(38)[1(3850)](10.9973)0.001350.0122P A P Z <=-≤≤≈⨯-=<故选：A三、解答题：（本大题满分76分，共5小题，解答本大题需要有必要的步骤和过程）17. 尝试使用概率的“可加性”解决下面的问题：（1）设是同一样本空间中的两个事件，探索，，，之间的等量,A B ()P A B ()P A ()P B ()P A B ⋂关系，并说明理由.（2）甲、乙各抛郑枚硬币，证明：“甲得到的正面数比乙得到的正面数少”这一事件的概率小于. n 12【答案】（1） ()()()()P A B P A P B P A B =+- （2）证明见解析 【解析】【分析】（1）通过事件间的关系，表示和事件，表示积事件，即可找出关系得到结果； A B ⋃A B ⋂（2）先设出事件事件“甲得到的正面数比乙得到的正面数少”，事件“甲得到的正面数比乙得到的正面A B 数多”，事件“甲得到的正面数比乙得到的正面数一样多”，再利用硬币的对称生和C 即可得到证明.()()()1P A P B P C ++=【小问1详解】，()()()()P A B P A P B P A B =+- 因为表示事件和事件至少有一个发生，表示事件和事件同时发生，所以A B ⋃A B A B ⋂A B()()()()P A B P A P B P A B =+- 【小问2详解】设事件“甲得到的正面数比乙得到的正面数少”，事件“甲得到的正面数比乙得到的正面数多”，事A B C 件“甲得到的正面数比乙得到的正面数一样多”，由硬币的对称性知，，又，，()()P A P B =()01P C <<()()()1P A P B P C ++=所以，故， ()1()1(0,22P C P A -=∈()12P A <即“甲得到的正面数比乙得到的正面数少”这一事件的概率小于.1218. 现需要设计一个仓库，它由上下两部分组成，上部分的形状是正四棱锥，下部分的形状1111P A B C D -是正四棱柱（如图所示），并要求正四棱柱的高是正四棱锥的高的4倍.1111ABCD A B C D -1OO 1PO（1）若则仓库的容积是多少？16,2,AB m PO m ==（2）若正四棱锥的侧棱长为，则当为多少时，仓库的容积最大？ 6m 1PO【答案】（1）312（2） 1PO =【解析】【详解】试题分析：（1）明确柱体与锥体积公式的区别，分别代入对应公式求解；（2）先根据体积关系建立函数解析式，，然后利用导数求其最值. ()()32636063V V V h h h =+=-<<锥柱试题解析：解：（1）由PO 1=2知OO 1=4PO 1=8. 因为A 1B 1=AB=6，所以正四棱锥P-A 1B 1C 1D 1的体积 ()22311111=6224m ;33V A B PO ⋅⋅=⨯⨯=锥正四棱柱ABCD-A 1B 1C 1D 1的体积 ()2231=68288m .V AB OO ⋅=⨯=柱所以仓库的容积V=V 锥+V 柱=24+288=312（m 3）.（2）设A 1B 1=a （m ），PO 1=h （m ），则0<h<6，OO 1=4h .连结O 1B 1.因为在中， Rt 11PO B 2221111O B PO PB ，+=所以，即 2236h +=()22236.a h =-于是仓库的容积， ()()22231132643606333V V V a h a h a h h h h =+=⋅+⋅==-<<柱锥从而. ()()2226'36326123V h h =-=-令，得 或. '0V =h =h =-当时， ，V 是单调增函数；0h <<'0V >当时，，V 是单调减函数. 6h <<'0V <故时，V 取得极大值，也是最大值. h =因此，当m 时，仓库的容积最大.1PO =【考点】函数的概念、导数的应用、棱柱和棱锥的体积【名师点睛】对应用题的训练，一般从读题、审题、剖析题目、寻找切入点等方面进行强化，注重培养将文字语言转化为数学语言的能力，强化构建数学模型的几种方法.而江苏高考的应用题往往需结合导数知识解决相应的最值问题，因此掌握利用导数求最值方法是一项基本要求，需熟练掌握.19. 2022年冬季奥林匹克运动会主办城市是北京，北京成为第一个举办过夏季奥林匹克运动会和冬季奥林匹克运动会以及亚洲运动会三项国际赛事的城市.为迎接冬奥会的到来，某地很多中小学开展了模拟冬奥会赛事的活动，为了深入了解学生在“自由式滑雪”和“单板滑雪”两项活动的参与情况，在该地随机选取了10所学校进行研究，得到如下数据：（1）在这10所学校中随机选取3所来调查研究，求在抽到学校至少有一个参与“自由式滑雪”超过40人的条件下，“单板滑雪”不超过30人的概率；（2）“单板滑雪”参与人数超过45人的学校可以作为“基地学校”，现在从这10所学校中随机选出3所，记为可选作为“基地学校”的学校个数，求的分布列和数学期望；X X （3）现在有一个“单板滑雪”集训营，对“滑行､转弯､停止”这3个动作技巧进行集训，且在集训中进行了多轮测试.规定：在一轮测试中，这3个动作中至少有2个动作达到“优秀”.则该轮测试记为“优秀”，在集训测试中，小明同学3个动作中每个动作达到“优秀”的概率均为，每个动作互不影响且每轮测试互不13影响.如果小明同学在集训测试中要想获得“优秀”的次数的平均值达到3次，那么理论上至少要进行多少轮测试？ 【答案】（1）125（2）分布列见解析， 6()5E X =（3）12轮 【解析】【分析】（1）根据已知条件结合条件概率的概率公式求解，（2）的可能取值为0，1，2，3，分别求出对应的概率，从而可求得的分布列和数学期望， X X （3）根据题意，结合二项分布的概率公式求解 【小问1详解】由题可知10个学校，参与“自由式滑雪”的人数依次为27，15，43，41，32，26，56，36，49，20，参与“单板滑雪”的人数依次为46，52，26，37，58，18，25，48，32，30，其中参与“自由式滑雪”的人数超过40人的有4个，参与“自由式滑雪”的人数超过40人，且“单板滑雪”的人数超过30人的有2个，设事件为“从这10所学校中抽到学校至少有一个参与“自由式滑雪”的人数超过40人”，事件为“从A B 10所学校中选出的3所学校中参与“单板滑雪”的人数不超过30人”则， ， ()1221346464310C C +C C +C 100C 120P A == ()12212222310C C C C 4C 120P AB +==所以()()()4112010025120P AB P B A P A ===【小问2详解】“单板滑雪”参与人数超过45人的学校有4所，则的可能取值为0，1，2，3，X ,0346310C C 1(0)C 6P X ===,1246310C C 1(1)C 2P X ===,2146310C C 3(2)C 10P X ===,34310C 1(3)C 30P X ===所以的分布列为XX 0 1 2 3P 1612310 130所以 11316()01236210305E X =⨯+⨯+⨯+⨯=【小问3详解】由题意可得小明同学在一轮测试中为“优秀”的概率为， 2323331117C 1C 33327P ⎛⎫⎛⎫⎛⎫=⨯-+=⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭所以小在轮测试中获得“优秀”的次数满足n Y ，7,27Y B n ⎛⎫ ⎪⎝⎭由，得， 7327n ⋅≥8111.67n ≥≈所以理论上至少要进行12轮测试20. 某疾病可分为Ⅰ、Ⅱ两种类型．为了解该疾病类型与性别的关系，在某地区随机抽取了患该疾病的病人进行调查，其中女性是男性的2倍，男性患Ⅰ型病的人数占男性性别病人的，女性患Ⅰ型病的人数56占女性性别病人的． 13（1）若在犯错误的概率不超过0.001的前提下认为“所患疾病类型”与“性别”有关，求男性患者至少有多少人？（2）某药品研发公司欲安排甲乙两个研发团队来研发此疾病的治疗药物．两个团队各至多安排2个接种周期进行试验．甲团队研发的药物每次接种后产生抗体的概率为，每人每次花费()01p p <<()0m m >元，每人每次接种每个周期至多接种3次，第一个周期连续2次出现抗体则终止本接种周期进入第二个接种周期，否则需依次接种至第一周期结束，再进入第二周期：第二接种周期连续2次出现抗体则终止试验，否则需依次接种至至试验结束；乙团队研发的药物每次接种后产生抗体的概率为，每()01q q <<人每次花费元，每个周期接种3次，每个周期必须完成3次接种，若一个周期内至少出现2次()0n n >抗体，则该周期结束后终止试验，否则进入第二个接种周期，假设两个研发团队每次接种后产生抗体与否均相互独立．当，时，从两个团队试验的平均花费考虑，公司应选择哪个团队？ 23n m =p q =（3）乙团队为奖励参与研发的工作人员，特地给参与本次研发的工作人员每人发放价值1000元的购物卡，并推出一档“勇闯关，送大奖”的活动．规则是：在某张方格图上标有第0格、第1格、第2格、…第30格共31个方格．棋子开始在第0格，然后掷一枚均匀的硬币（已知硬币出现正、反面的概率都是，其中）．1201P =若掷出正面，将棋子向前移动一格（从k 到），若掷出反面，则将棋子向前移动两格（从k 到1k +2k +）．重复多次，若这枚棋子最终停在第29格，则认为“闯关成功”，并赠送1000元购物卡；若这枚棋子最终停在第30格，则认为“闯关失败”，不再获得其他奖励，活动结束．设棋子移到第n 格的概率为，若n P 某员工参与这档“闯关游戏”，试比较一名员工闯关成功和失败的概率，并说明理由．附：，()()()()()22n ad bc K a b c d a c b d -=++++()20P K k ≥0.10 0.05 0.01 0.005 0.0010k 2.7063.8416.6357.87910.828【答案】（1）18人 （2）该公司选择乙团队进行药品研发（3）该大学生闯关成功的概率大于闯关失败的概率，理由见解析 【解析】【分析】（1）列出列联表，并计算出，再列不等式即可求得男性患者至少有多少人； 2K （2）分别求得甲、乙两研发团队试验总花费的数学期望，再对二者进行比较即可解决；（3）先找到棋子移到第格，第格，第格的概率间的递推关系，再利用叠加法求得棋子移到n n 1-2n -第格的概率公式，即可求得闯关成功的概率和闯关失败的概率，二者再进行大小比较即可解决. n 【小问1详解】设男性患者有a 人，则女性患者有人，列联表如下：2a Ⅰ型病Ⅱ型病合计 男56a6a a女23a 43a2a 合计32a 32a3a 要使在犯错误的概率不超过0.001的前提下认为“所患疾病类型”与“性别”有关，则，解得．2254232636310.828333222a a a a a a K a a a a ⎛⎫⨯-⨯ ⎪⎝⎭==>⨯⨯⨯16.242a >∵，，∴a 的最小整数值为18，因此．男性患者至少有18人 6a Z ∈3aZ ∈【小问2详解】设甲研发团队试验总花费为元，则的可能取值为、、，X X 4m 5m 6m∵，()2244P X m p p p ==⋅=，()()()()2222245112p P x m p p p p p ⋅==⋅-+-=-，()()22426121P X m p p p ==-=-+∴，()()424422410()62126X mp m p p m p p mp m E =+-+-+=-+函数在递减，∴，()226f p mp m =-+()0,1()4E X m >设乙研发团队试验总花费为Y 元，则Y 的可能取值为、，3n 6n ∴，，()()2233233123Y n C q q P q q q ==-+=-+()326123P Y n q q ==+-∴，()()()3232323236123696E Y n q qn qq nq nq n =⋅-++⋅+-=-+设，，()32696,(01)g q nq nq n q =-+<<()()()23661810g q n q q nq q '=-=-<函数在递减，，()32696g q nq nq n =-+()0,1()64E Y n m <=∴恒成立，所以，该公司选择乙团队进行药品研发 ()()E X E Y >【小问3详解】棋子开始在第0格为必然事件，．01P =第一次掷硬币出现正面，棋子移到第1格，其概率为，． 12112P =棋子移到第格的情况是下列两种，而且也只有两种： ()229n n ≤≤棋子先到第格，又掷出反面，其概率为； 2n -212n P -棋子先到第格，又掷出正面，其概率为， n 1-112n P -所以，即，且， 211122n n n P P P --=+()11212n n n n P P P P ----=--1012P P -=-所以当时，数列是首项，公比为的等比数列． 129n ≤≤{}1n n P P --1012P P -=-12-则，，，…，，1112P -=-12212P P ⎛⎫-=- ⎪⎝⎭33212P P ⎛⎫-=- ⎪⎝⎭112nn n P P -⎛⎫-=- ⎪⎝⎭以上各式相加，得，21111222n nP ⎛⎫⎛⎫⎛⎫-=-+-++- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭所以．()2111121110,1,2,,2922232n n n P n +⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫=+-+-++-=--=⎢⎥ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦所以闯关成功（即这枚棋子最终停在第29格）的概率为， 3030292121113232P ⎡⎤⎡⎤⎛⎫⎛⎫=--=-⎢⎥⎢⎥ ⎪⎪⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦闯关失败（即这枚棋子最终停在第30格）的概率为．2929302811211111223232P P ⎡⎤⎡⎤⎛⎫⎛⎫==⨯--=+⎢⎥⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦30292829302111111110323232P P ⎡⎤⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫-=--+=->⎢⎥⎢⎥⎢⎥ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦所以该大学生闯关成功的概率大于闯关失败的概率． 21. 已知抛物线C :, 过抛物线C 上点M 且与M 处的切线垂直的直线称为抛物线C 在点2742y x x =++M 的法线．（1）若抛物线C 在点M 的法线的斜率为，求点M 的坐标； 12-()00,x y （2）设P 为C 对称轴上的一点，在C 上是否存在点，使得C 在该点的法线通过点P ．若有，求()2,a -出这些点，以及C 在这些点的法线方程；若没有，请说明理由． 【答案】（1）；（2）当时，在上有三点，及，在该点的法线通过点，法线方程分别为，，，当时，在上有一点，在该点的法线通过点，法线方程为．【解析】【详解】试题分析：（1）求导可得点处切线的斜率法线斜率为⇒⇒=点的坐标为；（2）设为上一点，由上点⇒⇒处的切线斜率,法线方程为法线过点；若的法线方程M ⇒⇒⇒为：．再讨论和⇒⇒，即可求得：当时，有三点和三条法线；当时，有一点和一条法线．试题解析：（1）函数的导数，点处切线的斜率过点的法线斜率为=，解得，．故点的坐标为∴．（2）设为上一点，若，则上点处的切线斜率,过点的法线方程为， 法线过点; 若，则过点的法线方程为：．若法线过点，则，即．若，则，从而，代入得，．若，与矛盾，若，则无解．综上，当时，在上有三点，及，在该点的法线通过点，法线方程分别为，，．当时，在上有一点，在该点的法线通过点，法线方程为．考点：1.导数；2.切线；3.法线；4.直线方程.22. 如图，在平面直角坐标系中，已知以为圆心的圆:及其上一点xOy M M 221214600x y x y +--+=A (2，4).（1）设圆N 与x 轴相切，与圆外切，且圆心N 在直线x =6上，求圆N 的标准方程； M （2）设平行于OA 的直线l 与圆相交于B ，C 两点，且BC=OA ，求直线l 的方程；M （3）设点T （t ，0）满足：存在圆上的两点P 和Q ，使得求实数t 的取值范围.M ,TA TP TQ +=【答案】(1)；(2)2x −y +5=0或2x −y −15=0.(3). 22(6)(1)1x y -+-=[2-+【解析】【详解】试题分析：（1）根据直线与x 轴相切确定圆心位置，再根据两圆外切建立等量关系求半径;（2）根据垂径定理确定等量关系，求直线方程；（3）利用向量加法几何意义建立等量关系，根据圆中弦长范围建立不等式，求解即得参数取值范围.试题解析：解：圆M 的标准方程为，所以圆心M （6，7），半径为5，. ()()226725x y -+-=（1）由圆心N 在直线x=6上，可设.因为N 与x 轴相切，与圆M 外切，()06,N y 所以，于是圆N 的半径为，从而，解得.007y <<0y 0075y y -=+01y =因此，圆N 的标准方程为.()()22611x y -+-=（2）因为直线l ∥OA ，所以直线l 的斜率为. 40220-=-设直线l 的方程为y=2x+m ，即2x-y+m=0，则圆心M 到直线l 的距离d因为BC OA ===而 222,2BC MC d =+（）所以，解得m=5或m=-15. ()252555m +=+故直线l 的方程为2x-y+5=0或2x-y-15=0.（3）设()()1122,,,.P x y Q x y 因为，所以……① ()()2,4,,0,A T t TA TP TQ += 因为点Q 在圆M 上，所以…….②()()22226725.x y -+-=将①代入②，得.()()22114325x t y --+-=于是点既在圆M 上，又在圆上， ()11,P x y ()()224325x t y -++-=⎡⎤⎣⎦从而圆与圆有公共点，()()226725x y -+-=()()224325x t y -++-=⎡⎤⎣⎦所以解得. 5555,-≤≤+22t -≤≤+因此，实数t 的取值范围是.22⎡-+⎣【考点】直线方程、圆的方程、直线与直线、直线与圆、圆与圆的位置关系、平面向量的运算【名师点睛】直线与圆中的三个定理：切线的性质定理，切线长定理，垂径定理；两个公式：点到直线距离公式及弦长公式，其核心都是转化到与圆心、半径的关系上，这是解决直线与圆的根本思路.对于多元问题，也可先确定主元，如本题以为主元，揭示在两个圆上运动，从而转化为两个圆有交点这一P P 位置关系，这也是解决直线与圆问题的一个思路，即将问题转化为直线与圆、圆与圆的位置关系问题.。

上海市上海中学2019届高三上学期摸底考试英语试题Word版含答案2018-2019学年上海中学高三第一学期摸底考试II.Grammar and Vocabulary温馨提示：多少汗水曾洒下，多少期待曾播种，终是在高考交卷的一刹尘埃落地，多少记忆梦中惦记，多少青春付与流水，人生，总有一次这样的成败，才算长大。

高考保持心平气和，不要紧张，像对待平时考试一样去做题，做完检查一下题目，不要直接交卷，检查下有没有错的地方，然后耐心等待考试结束。

Section ADirection: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the other answer that best completes the sentence.25.While I was waiting to enter ________ university, I saw advertised in a local newspaper ateaching post at a school in ________ suburb of London.A. /, aB. an, aC. a, theD. the ,the26.In most cases, ________ a passenger has his ticket and managers to catch his train, he canreach his destination more comfortably than ________ he had to drive himself.A. once, ifB. that ,ifC. when, whileD. where, when27.The invention of the modern computer is one of the greatcontributions ________ to man’sefficiency.A. having ever been madeB. ever been madeC. ever madeD. having ever made28.I was not able to work out the problem ________ my teacher explained it.A. asB. unlessC. untilD. when29.For him to be re-elected, what is essential is not that his policy works, but ________ thepublic believe that it does.A. /B. whetherC. thatD. if30.What struck the audience most was ________ the blind girl could accomplish with her ownhands.A. thatB. whatC. whoD. so31.The pressure ________ causes Americans to be energetic, but it also puts them under aconstant emotional strain.A. to completeB. completingC. to be completedD. to have completed32.Though ________ money, his parents managed to send him to university.A. lackedB. lacking ofC. lackingD. being lack of33.________ Japanese is certainly complex, it is by no means impossible to learn.A. WhereasB. WhileC. SinceD. As34.To the students________, the new teacher felt very nervous to say anything, with handsslightly________.A. concerned with, shakenB. concerned, shakingC. concerned with, shakingD. concerned, shaken35.–I can’t find Ms. Miller. Where did you meet her this morning?–It was in the hotel ________ he was staying.A. thatB. whichC. the oneD. where36.________ your opinion was worth considering, they won’tplace too much importance on it.A. AsB. SinceC. UnlessD. If only37.We shall meet at the same place ________ we met for the first place.A. thatB. whereC. asD. which38.The monitor suggested ________ to the Sea World in the summer vocation.A. to me visitingB. their visitingC. to me their visitD. they visit39.He often wrote to the writer ________ the thought would help him to become a writer, too.A. whomB. whoC. becauseD. when40.In the past decade, geologists have come closer than ever to ________ the age of the earth.A. calculateB. calculatingC. be calculatingD. have calculatedSection BDirection: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.Traffic science is one of those ____41____ seems permanently poised on the verge of a breakthrough. Professional journals regularly publish promising research, and the ____42____ trumpets their importance. However, it turns out that traffic is a deceptively complicated problem. It ____43____ molecular physics, in fact, because it's a system of individual particles ____44____ in complex ways. Except, with traffic, the particles have minds of their own.There are two kinds of traffic flow. In uncongested stable flows, cars can move at or near the speed limit, and the "unstable regime," what laypeople (外行) call a stop-and-go traffic. What scientists have figured out over the past decade or so is when and why traffic ____45____ between the two.“We see in our models that traffic becomes unstable when the number of cars(passing a specific spot) per lane per hour reaches between 2,000 and 2,500. At that nominal capacity level, traffic is very likely to become unstable,” says Hani Mahmassani, a traffic scientist at Northwestern University in Chicago.Consider a ____46____ case. A slow-moving car shifts into the left lane to pass an even slower-moving car. The car ____47____ behind the lane-changer has to decelerate ____48____ - not just to the speed of the car in front of him, but slow enough to create a safe driving distance between them. The next car back has to slow down even more, again to give itself a ____49____. This slowdown ripples back through the lane and eventually spreads into the other lanes as nearby drivers notice the sea of brakelights and reflexively slow down. Traffic researchers ____50____ to this as a shock wave, and it can travel back for miles.III. Reading ComprehensionSection ADirection: For each blank in the following passages there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.(A)There are many things parents can do to help children with autism (自闭症) overcome their challenges.Learning all you can about autism and getting (51)________ in treatment will go a long way towar d helping yourchild. Additionally, the following tips will make daily home life easier for both you and your autist ic child:●Be consistent (一致的). Children with autism have a hard time (52)________ what they've learned if there is a change of setting. For example, your child may use sign language at school to communicate, but never thinkto do so at home. Creating (53)________ in your child's environment is the best way to reinfo rce learning.Find out what your child's therapists are doing and continue their techniques at home. Explore the(54)________ of having therapy take place in more than one place in order to encourage yourchild to(55)________ what he or she has learned from one environment to another. It's also important to be consistentin the way you (56)________ with your child and deal with challenging behaviors.●(57)________ a schedule. Children with autism tend to do best when they have a highly-structured schedule or routine. Again, this goes back to the consistency they both need and crave. Set up a schedule for your child,with (58)________ times for meals, therapy, school, and bedtime. Try to keep disturbance to t his routine to a(59)________. If there is an unavoidable schedule change, prepare your child for it (60)________.●(61)________ good behavior. Positive reinforc ement can go a long way with children with autism, so makean effort to 'catch them doing something good.' Praise them when they act appropriately or learn a new skill, being very (62)________ about what behavior they?r e being praised for.●Pay attentio n to your child's sensory sensitivities. Many children with autism are hypersensitive to light,sound, touch, taste, and smell. Other children with autism are 'under-sensitive' to sensory stimuli.(63)________ what sights, sounds, smells and movements cause your kid's 'bad' or disruptivebehaviors andwhat brings about a(n) (64)________ response. If you understand what affects your child, you'll be better atsolving problems, preventing situations that cause difficulties, and creating (65)________ experiences.51. A. interested B. balanced C. absorbed D. involved52. A. applying B. devoting C. communicating D. appealing53. A. attraction B. comfort C. steadiness D. attention54. A. possibility B. goal C. process D. solution55. A. transplant B. transfer C. transport D. transform56. A. meet B. interact C. negotiate D. associate57. A. Draw up B. Arrange for C. Work out D. Stick to58. A. regular B. flexible C. appropriate D. normal59. A. decrease B. mystery C. minimum D. degree60. A. without doubt B. in private C. without notice D. in advance61. A. Admire B. Stick C. Reward D. Maintain62. A. curious B. specific C. particular D. anxious63. A. Figure out B. Account for C. Put up D. Take on。

上海市第一中学2024-2025学年高三上学期期中考试数学试题2024.11一、填空题(本大题共12题，第1~6题每题4分，第7~12题每题5分.满分54分)1.已知集合，则______________2.设复数，则______________3.函数的最小正周期为______________4.角的始边与轴的正半轴重合，终边过点，则______________5.若实数x 、y 满足，则的最小值为______________6.已知，则在方向上的投影为______________7.方程的解集为______________8.若函数在区间[0，a ]上是严格减函数，则实数的最大值为______________9.法国数学家拉格朗日于1797年在其著作《解析几何函数论》中给出一个定理，如果函数满足条件：①在闭区间[a ，b ]上是连续不断的；②在区间（a ，b ）上都有导数.则在区间上至少存在一个实数，使得，其中称为“拉格朗日”中值.函数在区间的“拉格朗日”中值______________10.如图，正六边形的边长为2，圆的圆心为正六边形的中心，半径为1，若点在正六边形的边上运动，动点A 、B 在圆上运动且关于圆心对称，则的取值范围是______________11.如图，互不相同的点和分别在角的两条边上，所有相互平行，且所有梯形的面积均相等.设，若，则数列的通项公式______________(0,4),[2,5]A B ==A B ⋂=(1i)2i z -=||z =π()tan 23f x x ⎛⎫=+⎪⎝⎭αx (3,4)-sin(π)α+=1xy =223x y +(2,3),(1,0)a b =-= a b|21||22|3x x ++-=cos sin y x x =-a ()y f x =(,)a b t ()()()()f b f a f t b a '-=-t sin y x =π0,2⎡⎤⎢⎥⎣⎦t =O M O O MA MB ⋅12n A A A 、、、、12n B B B 、、、、O n n A B 11n n n n A B B A ++n n OA a =121,2a a =={}n a n a =12.设函数是奇函数，当时，.若对任意的，不等式都成立，则实数的取值范围为______________二、选择题（本大题满分20分）本大题共有4题，每题5分.13.已知，则“”是“”的（ ）条件A.充分不必要B.必要不充分C.充要D.既不充分也不必要14.若函数在处的导数等于，则的值为（ ）A.0B.C. D.2a15.已知函数，实数，下列选项中正确的是（ ）A.若，函数关于直线对称B.若，函数在上是增函数C.若函数在上最大值为1，则D.若，则函数的最小正周期是16.已知，集合，.关于下列两个命题的判断，说法正确的是（ ）命题①：集合表示的平面图形是中心对称图形；命题②：集合表示的平面图形的面积不大于.（ ）()y f x =0x ≥()2221()232f x x a x a a =-+--x ∈R (1)()f x f x -≤a x ∈R 1x >21x >()y f x =0x x =a ()()0002limx f x x f x x∆→+∆-∆12a aπ(),()2sin 6y f x f x x ω⎛⎫==+⎪⎝⎭0ω>2ω=()y f x =5π12x =12ω=()y f x =[0,π]()y f x =[π,0]-43ω≤1ω=|()|y f x =2π()sin f x x =ππ,,{(,)2()()0,,}22D x y f x f y x y D ⎡⎤=-Γ=+=∈⎢⎥⎣⎦∣{(,)2()()0,,}x y f x f y x y D Ω=+≥∈∣ΓΩ25π12A.①真命题，②假命题B.①假命题，②真命题C.①真命题，②真命题D.①假命题，②假命题三、解答题（本大题满分76分）17.已知，且.（1）求向量与的夹角大小；（2）求.18.设常数.（1）若是奇函数，求实数的值；（2）设中，内角的对边分别为若，求的面积.19.已知递增的等差数列的首项，且成等比数列.（1）求数列的通项公式；（2）设数列满足为数列的前项和，求.20.为了助力企业发展，某地政府决定向当地企业发放补助款，其中对纳税额在3万元至6万元（包括3万元和6万元）的小微企业做统一方案.方案要求同时具备下列两个条件：①补助款（万元）随企业原纳税额（万元）的增加而增加；②补助款不低于原纳税额（万元）的，经测算政府决定采用函数模型（其中为参数）作为补助款发放方案.（1）已知某企业纳税额为4万元，计算该企业将获得的补助款；（2）判断使用参数是否满足条件，并说明理由；（3）求同时满足条件①、②的参数的取值范围.21.已知.（1）若，求曲线在点处的切线方程；（2）若函数存在两个不同的极值点，求证：；（3）若，数列满足.求证：当时，.||1,||2a b == ()(2)6a b a b +⋅-=-a b|2|a b +2,()cos cos ,k f x k x x x x ∈=+∈R R ()f x k 1.k ABC = A B C 、、a b c 、、，()1,f A a ==3b =ABCS {}n a 11a =124a a a 、、{}n a n a {}n b 2(1),n a n n n n b a T =+-{}n b n 2n T ()f x x x 50%()44x bf x x=-+b 12b =b ()ln 1f x a x ax =---0a =()y f x =(1,1)P ()y f x =12x x 、()()120f x f x +>1,()()a g x f x x ==+{}n a ()11(0,1),n n a a g a +∈=2n ≥212n n n a a a +++>2024学年第一学期高三年级数学期中考试参考答案一、填空题（本大题共12题，第题每题4分，第题每题5分.满分54分）1.3.4. 5.6. 7. 8.9. 10.[2，3]12.二、选择题（本大题满分20分）本大题共有4题，每题5分.13.A14.D15.C16.A三、解答题（本大题满分76分）17.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分.解（1）；（218.（本题满分14分）第（1）小题6分，第（2）小题8分.解（1）；（2）.19.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分.解（1）由题可知，且，即，可得（2）.20.（本题满分16分）本题共有3小题，第（1）题4分，第（2）题4分，第（3）题8分.解（1）（2）因为当时，，所以当时不满足条件②.（3）由条件①可知，在[3，6]上单调递增，在恒成立，在恒成立，所以1~67~12[2,4)π245(2,0)1,12⎡⎤-⎢⎥⎣⎦3π42arccos π⎡⎢⎣2π30k =S =10,1d a >=2142a a a ⋅=()()21113a a d a d ⋅+=+2*111,1,(1),n a d d a d a a n d n n N ===∴=+-⋅=∈()12222(1),222[1234(21)2]nnnn n b n T n n =+-=++++-+-+---+ ()2212122212n n n n +-=+=+--(4)54bf =-12b =33(3)42f =<12b =()44x bf x x=-+22214()044b x b f x x x '+⇒=+=≥[3,6]x ∈24x b ⇒≥-[3,6]x ∈94b ≥-由条件②可知，，即不等式在[3，6]上恒成立，等价于，当时，取最小值，所以综上，参数的取值范围是.21.（本题满分18分，第1小题满分4分，第2小题满分6分，第3小题满分8分）解（1）当时，所以曲线在点处的切线方程为…………………………………………4分（2）由，令，则原方程可化为：①，则是方程①的两个不同的根所以，解得………………………………………………………3分所以因为，所以，所以 (6)分（3）由题意，，所以当时，，所以函数在区间上严格减，当时，，所以函数在区间上严格增，………………3分因为，所以，以此类推，当时，，………………………………………………4分()2x f x ≥44x bx+≤22114(8)1644b x x x ≤-+=--+3x =21(8)164y x =--+394394b ≤b 939,44⎡⎤-⎢⎥⎣⎦0a =()(1)1f x f ''==()y f x =(1,1)P y x =()0f x '=0aa x--=t=0t >20at t a -+=12t t ==214010a a⎧∆=->⎪⎨>⎪⎩102a <<()()()()1212122ln ln 2f x f x a x x a x x +=+-+-+-()()()222212121212ln 222t t a t t a t t a a=+--+-=+-102a <<12220a a+->->()()120f x f x +>()ln 1g x x =--()g x '=(0,1)x ∈()0g x '<()y g x =(0,1)(1,)x ∈+∞()0g x '>()y g x =(1,)+∞101a <<()()2132(1)1,(1)1a g a g a g a g =>==>=2n ≥()1(1)1n n a g a g +=>=又，所以函数在区间上严格减，当时，，所以，.....................................7分所以，即，故. (8)分2131124()2102f x x x'⎫---⎪⎝⎭=⨯--=<()y f x =(0,)+∞2n ≥()()(1)0n n n f a g a a f =-<=1n n a a +<()()1n n f a f a +>211n n n n a a a a +++->-212n n n a a a +++>。

2020届高三第一学期英语测试卷考生注意：考试时间120分钟，试卷满分150分。

Ⅰ. Listeng Comprehension (30%)Section A (10%)Directions: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. plumber. B. A dentistC. A physician.D. A programmer.2. A. 40 pence. B. 70 pence.C. 1.4 pounds.D. 1.6 pounds.3. A. She was apologetic. B. She was optimistic.C. She was generous.D. She was unforgiving.4. A. The woman is just in time. B. The plane has taken off.C. He will call the plane back.D. The flight has been cancelled.5. A. The man looks sick in blue. B. The man looks very angry.C. The mans roommate is ill.D. The man’s roommate is back.6. A. At the court. B. At a stationery shopC. Near Robinson street.D. At the emergency room.7. A. She appreciates what Jeff did. B. She feels really sorry for Jeff.C. She thinks Jeff deserves his failure.D. She considers Jeff an upright boy.8. A. Apply at the personnel office. B. Talk to the administrator first.C. Go to the office himself.D. Call the administrative building.9. A. He is preparing for his term paper on economic crisis.B. He doesn’t know the paper well enough to comment yet.C. He finds it difficult to form opinions about economic paper.D. He is not sure whether he should read through the paper.10. A. Mike is beginning to drink heavily again.B. Mike has been having a hard time since he started drinking.C. Mike is taking up his new happy as a long-distance runner.D. Mike will eventually benefit from giving up heavy drinking.Section B (20%) . Directions: In Section B, you will hear two short passages and one longer conversation, and youwill be asked to questions on each of the passages. The passages will be read twice, but thequestions will be spoken only once. When you hear a question, read the four possible answers onyour paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. In her friend’s car. B. In the baby sitter’s.C. In the nearby hospital.D. In the local bar.12. A. There was blood in her eyes.B. The girl was sleeping soundly.C. The girl was looking at the sky.D The girl got seriously burnt in the back.13. A. 2-year-old girl murdered by babysitter.B. Mother of two children using drugs.C. Baby-girl killed due to neglect.D. Couple arrested for selling drugs.Questions 14 through 16 are based on the following passage.14. A. 9.2 billion tons B. 6.9 billion tonsC. 5.3 million tonsD. 14 million tons15. A. Plastics are cheap. B. Plastics are durable.C. Plastics are easily processed.D. Plastics are the cleanest items.16. A. Because hospital plastics will be the only food for ocean animals.B. Because hospital plastics ending up in oceans pose great risk to marine life.C. Because hospital plastics aren’t recycled and take more time to biodegrade.D. Because hospital plastics will ultimately be washed into the Earth’s sink.Questions 17 through 20 are based on the following conversation.17. A. To throw a birthday party. B. To post his photos online..C. To take photos for his birthday.D. To share his friends’ photos online.18. A. It is attractive and more efficient.B. It is a great way to make friends.C. It can make people feel uncomfortable.D. It encourages personal information sharing.19. A. Conservative and cautious. B. Indifferent and stubborn.C Narrow-minded and rude. D. Optimistic and outgoing.20. A. The Best way to celebrate a birthday party.B. Different voices about social media sharing.C. Various ways to interact with people.D. The benefits of privacy protection.Ⅱ. Grammar and vocabulary (25%)Section A (10%)Directions: Read the following two passages. Fill in the blanks to make the passage coherent. For the blanks with a given word, fill in each blank with the proper from of the given word. For the other blanks, fill in each blank with one proper word. Make sure that your answers are grammatically correct.Facebook to buy tech start-up working on mind control technologyTECH giant Facebook announced it has made a deal with a start-up company working on technology that will allow users to make computer commands using their mind ___1___ of more traditional methods like taps and keystrokes.CTRL-Labs will become part of Facebook Reality Labs with an aim of perfecting the technology and ____2____ (build) it into consumer products, according to Andrew Bosworth, vice president of augmented reality and virtual reality at the social media company. “We spend a lot of time trying to get our technology ___3___ (do) what we want rather than enjoying the people around us. We know there are more natural, intuitive ways to interact with devices and technology,” he said in ___4___ online post.The wristband will decode electrical impulses ___5___ (send) to hand muscles, telling them to move in certain ways. The vision for this work is a wristband ___6___ lets people control their devices as a natural extension of movement. Here’s how it ___7___ (work): You have neurons in your spinal cord that send electrical signals to yourhand muscles telling them to move in specific ways such as to click a mouse or press a button.The wristband will decode those signals and translate them into a digital signal your device ___8___ understand, empowering you with control over your digital life. It captures your intention so you can share a photo with a friend using an unperceivable movement or just by, well, intending to. Bosworth also explains how thought-commanded interactions might dramatically alter ___9___ people experience virtual reality scenarios, which currently feature hand-held controls. _______10_______ Facebook hasn’t disclosed financial terms of the deal to buy the NewYork-based CTRL-labs so far, unconfirmed media reports said it paid more than $500 million.【答案】1. instead2. building3. to do4. an5. sent6. that/which7. will work/works8. can9. how10. While/ Although【解析】【分析】本文是一篇新闻报道。

2022学年静安区高三第一学期期末数学学科练习卷考生注意：1.试卷共4页，另有答题纸2页．2.所有作答必须在答题纸上与试卷题号对应的区域完成，不得错位，在试卷或者草稿纸上作答一律无效．一、填空题（本大题共有12题，满分54分，第1～6题每题4分，第7～12题每题5分）考生应在答题纸的相应编号位置直接填写结果．1. 函数πtan34y x的定义域是____________．2. 已知复数12iiaza（i为虚数单位）在复平面内对应的点位于第二象限，则实数a的取值范围是____________．3. 若直线230x y与直线2100x my平行，则这两条直线间的距离是____________．4. 16-17岁未成年人的体重的主要百分位数表（单位：kg）．小王同学今年17岁，她的体重50kg，她所在城市女性同龄人约有4.2万人．估计小王同学所在的城市有________ 万女性同龄人的体重一定高于她的体重．（单位：万人，结果保留一位小数）5. 已知函数22xf x xe cos e，则函数f x的导数f x____________．6. 现有5根细木棍，长度分别为1、3、5、7、9（单位：cm），从中任取3根，能搭成一个三角形的概率是____________．7. 有一种空心钢球，质量为140.2g，测得球的外直径等于 5.0cm，若球壁厚度均匀，则它的内直径为__________cm．（钢的密度是7.9g/cm3，结果保留一位小数）．8. A、B分别是事件A、B的对立事件，如果A、B两个事件独立，那么以下四个概率等式一定成立的是____________．（填写所有成立的等式序号）①P A B P A P B②P A B P A P B③11P A B P A P B④P A B P A P B9. 2022年11月27日上午7点，时隔两年再度回归的上海马拉松赛在外滩金牛广场鸣枪开跑，途径黄浦、静安和徐汇三区.数千名志愿者为1.8万名跑者提供了良好的志愿服务.现将5名志愿者分配到防疫组、检录组、起点管理组、路线垃圾回收组4个组，每组至少分配1名志愿者，则不同的分配方法共有__________种.（结果用数值表示）10. 已知全集为实数集R ，集合21|225616x M x ，N = 25|log (4)1x x x ，则 M N =____________．11. 在空间直角坐标系O xyz 中，点 7,4,6P 关于坐标平面xOy 的对称点P 在第_______卦限；若点Q 的坐标为 8,1,5 ，则向量PQ 与向量PP 夹角的余弦值是____________．12. 已知函数3232f x ax x -+，若函数f x 只有一个零点0x ，则实数a 的取值范围为________.二、选择题（本大题共有4题，满分18分，第13～14题每题4分，第15～16题每题5分）每题有且仅有一个正确选项，考生应在答题纸的相应编号位置将代表正确选项的小方格涂黑．13. 已知数列 n a 是等差数列，11548a a ，，则38133a a a （ ）A. 120B. 96C. 72D. 4814. 若实数x y 满足2243x y xy ，则（ ）成立． A. 1xy B. 2244x yC. 2x yD. 2x y.15. 在233nx x的二项展开式中，533r n r n r nx C 称为二项展开式的第1r 项，其中r =0，1，2，3，……，n ．下列关于233nx x的命题中，不正确的一项是（ ）A. 若8n ，则二项展开式中系数最大的项是1426383x C ．B. 已知0x ，若9n ，则二项展开式中第2项不大于第3项的实数x 的取值范围是35403x．C. 若10n ，则二项展开式中常数项是44103C ．D. 若27n ，则二项展开式中x 的幂指数是负数的项一共有12项．16. “阳马”，是底面为矩形，且有一条侧棱与底面垂直的四棱锥．《九章算术》总结了先秦时期数学成就，是我国古代内容极为丰富的数学巨著，对后世数学研究产生了广泛而深远的影响．书中有如下问题：“今有阳马，广五尺，袤七尺，高八尺．问积几何？” 其意思为：“今有底面为矩形，一条侧棱垂直于底面的四棱锥，它的底面长、宽分别为7尺和5尺，高为8尺，问它的体积是多少？”若以上的条件不变，则这个四棱锥的外接球的表面积为（ ）平方尺.,的A. 142B. 140C. 138D. 128三、解答题（本大题共有5题，满分78分）解答下列各题必须在答题纸的相应位置写出必要的步骤．17. 已知数列 n a 满足：112a，21a ，2145n n n a a a ，对一切正整数n 成立． （1）证明：数列{1n n a a }是等比数列； （2）求数列 n a 的前n 项之和．18.平面向量2(3sin ,cos ),(cos ,m x x n x，函数()2y f x m n. （1）求函数y =()f x 的最小正周期； （2）若[0,]2x ，求y =()f x 的值域；（3）在△ABC 中，内角A B C 、、的对边分别为a b c 、、，已知()f B，2,a b ABC 的面积.19. 如图所示，在矩形ABCD 中，4AB ，2AD ，E 是CD 的中点，O 为AE 的中点，以AE 为折痕将ADE V 向上折起，使D 点折到P 点，且PC PB ．（1）求证：PO 面ABCE ；（2）求AC 与面P AB 所成角 的正弦值．20. 已知椭圆Γ：22221x y a b （0a b）的离心率为3，它的上顶点为A ，左、右焦点分别为 1,0F c ，2,0F c （常数0c ），直线1AF ，2AF 分别交椭圆Γ于点B ，C ．O 为坐标原点．（1）求证：直线BO 平分线段AC ；（2）如图，设椭圆Γ外一点P 在直线BO 上，点P 横坐标为常数m （m a ），过P 的动直线l 与椭圆Γ交于两个不同点M 、N ，在线段MN 上取点Q ，满足MP MQPN QN，试证明点Q在直线2260mx c 上．的21已知函数f(x)＝－2a ln x－2x，g(x)＝ax－(2a＋1)ln x－2x，其中a∈R.（1）若x＝2是函数f(x)的驻点，求实数a的值；（2）当a >0时，求函数g(x)的单调区间；（3）若存在x [1e，e2 ]（e为自然对数的底），使得不等式f(x) g (x)成立，求实数a的取值范围．.2022学年静安区高三第一学期期末数学学科练习卷考生注意：1.试卷共4页，另有答题纸2页．2.所有作答必须在答题纸上与试卷题号对应的区域完成，不得错位，在试卷或者草稿纸上作答一律无效．一、填空题（本大题共有12题，满分54分，第1～6题每题4分，第7～12题每题5分）考生应在答题纸的相应编号位置直接填写结果．1. 函数πtan 34y x的定义域是____________． 【答案】ππZ 43k x x k， 分析】由342πππ，Z x k k可得答案. 【详解】342πππ，Z x k k，则43ππk x，Z k . 故答案为：ππZ 43k x x k， 2. 已知复数12iia z a（i 为虚数单位）在复平面内对应的点位于第二象限，则实数a 的取值范围是____________．【答案】,2【分析】先由复数的除法运算计算出z ，再由复数的几何意义得出相应点的坐标，列方程组求解即可.【详解】22221i 2i 12i i 12i i i i 1a a a a a a z a a a a 222321i 11a a a a ， ∴复数z 在复平面内对应的点为222321,11a a a a， 由已知，222321,11a a a a在第二象限， ∴2223012101a a a a，解得2a . 综上所述，实数a的取值范围是,2.【故答案为：,2. 3. 若直线230x y 与直线2100x my 平行，则这两条直线间的距离是____________．【答案】5【分析】由两直线平行，可求得m 的值，再利用平行线间距离公式求解. 【详解】由直线230x y 与直线2100x my 平行， 可知220m ，即4m ， 故直线为24100x y ，直线230x y 变形得2460x y ，故5d，. 4. 16-17岁未成年人的体重的主要百分位数表（单位：kg ）．小王同学今年17岁，她的体重50kg ，她所在城市女性同龄人约有4.2万人．估计小王同学所在的城市有________ 万女性同龄人的体重一定高于她的体重．（单位：万人，结果保留一位小数） 【答案】2.1【分析】根据题意，由图表可知，该城市女性同龄人高于小王的50百分位数，由百分位数的定义计算可得答案. 【详解】根据题意，小王同学今年17岁，她的体重，50kg 由图表可知，小王体重的百分位数是50， 所以体重一定高于她的体重的人数有504.2 2.1100(万) 故答案为：2.15. 已知函数 22x f x x e cos e ，则函数 f x 的导数 f x ____________．【答案】e cos 22e sin 2x x x x【分析】根据求导公式和四则运算法则计算即可. 【详解】 cos 22sin 2xxf x x x e e .故答案：cos 22in 2e e s x x x x .6. 现有5根细木棍，长度分别为1、3、5、7、9（单位：cm ），从中任取3根，能搭成一个三角形的概率是____________．为【答案】0.3##310【分析】根据古典概型，先求出样本空间，再求出条件空间即可.【详解】从5根木棍中任取3个共有3255C C 10 种，符合条件有 3,5,7,3,7,9,5,7,9 3种，能搭成一个三角形的概率310P； 故答案为：310. 7. 有一种空心钢球，质量为140.2g ，测得球的外直径等于 5.0cm ，若球壁厚度均匀，则它的内直径为__________cm ．（钢的密度是7.9g/cm 3，结果保留一位小数）． 【答案】4.5【分析】设空心钢球的内直径为2r ，表示空心钢球的体积，由条件列方程求r 即可.【详解】设空心钢球的内直径为2cm r ，则空心钢球的体积为333454ππcm 323r，因为空心钢球的质量为140.2g ，钢的密度是7.9g/cm 3，所以33454ππ7.9140.2323r，所以335140.2327.94πr，解得 2.25r ，所以2 4.5r ， 故答案为：4.5.8. A 、B 分别是事件A 、B 的对立事件，如果A 、B 两个事件独立，那么以下四个概率等式一定成立的是____________．（填写所有成立的等式序号） ① P A B P A P B ②P A B P A P B③11P A B P A P B ④P A B P A P B 【答案】②③【分析】根据事件的独立性定义判断即可.【详解】① P A B P A P B P AB ，故①不一定成立；②③由事件的独立性定义可得A 与B ，A 与B 相互独立，所以P A B P A P B ，11P A B P A P B P A P B ，故②③正确；④ P A B P A P B P AB ，故④不一定成立.故答案为：②③.9. 2022年11月27日上午7点，时隔两年再度回归的上海马拉松赛在外滩金牛广场鸣枪开跑，途径黄浦、静安和徐汇三区.数千名志愿者为1.8万名跑者提供了良好的志愿服务.现将5名志愿者分配到防疫组、检录组、起点管理组、路线垃圾回收组4个组，每组至少分配1名志愿者，则不同的分配方法共有__________种.（结果用数值表示） 【答案】240【分析】先将5名志愿者分成四组，然后再分配到四个地方即可.【详解】将5名志愿者分成四组，且每组至少1名志愿者有25C 种情况，所以不同的分配方法有2454C A 240 .故答案为：240.10. 已知全集为实数集R ，集合21|225616x M x，N = 25|log (4)1x x x ，则 M N =____________． 【答案】 ,25,【分析】根据指数函数和对数函数的单调性解不等式得到M ，M ，N ，然后求交集即可. 【详解】不等式21225616x 可整理为428222x ，所以428x ，解得24x ，所以 24M x x ， 2M x x 或 4x ，不等式25log 41x x 可整理为255log 4log 5x x ，所以245x x ，即 510x x ，解得1x 或5x ，所以1N x x 或 5x ， ,25,M N . 故答案为： ,25, .11. 在空间直角坐标系O xyz 中，点 7,4,6P 关于坐标平面xOy 的对称点P 在第_______卦限；若点Q 的坐标为 8,1,5 ，则向量PQ 与向量PP夹角的余弦值是____________．【答案】 ①. 五 ②.9【分析】根据坐标平面对称先求出P 的坐标，根据卦限在空间中的位置可以得出结果； 利用空间坐标直接求出夹角的余弦值即可得出答案.【详解】点 7,4,6P 关于坐标平面xOy 的对称点P 为(7,4,6) ，根据卦限在空间中的位置，所以点P 在第五卦限.由已知可得(1,5,1)PQ ，(0,0,12)PP，所以cos ,9PQ PP故答案为：五；912. 已知函数 3232f x ax x -+，若函数 f x 只有一个零点0x ，则实数a 的取值范围为________.【答案】,【分析】对a 分类讨论：0a ，0a 和a<0，分别求出对应情况下的实根情况列不等式，即可求解. 【详解】函数 3232f x ax x -+导函数为 263f x ax x -.当0a 时，令 0f x，解得：3x，所以函数 f x 有两个零点，不符合题意. 的当0a 时，要使函数 f x 只有一个零点0x ，只需 f x 的极大值小于0或 f x 的极小值大于0. 令 2036f x ax x -，解得：0x 或20x a. 列表：所以极大值 32030200f a -+不符合题意.所以极小值32243220222f a a a a a-+，解得：a ；当a<0时，要使函数 f x 只有一个零点0x ，只需 f x 极大值小于0或 f x 的极小值大于0. .令 2036f x ax x -，解得：0x 或20x a. 列表：所以极大值 32030200f a -+不符合题意.所以极小值32243220222f a a a a a-+，解得：a .综上所述：实数a 的取值范围为, .故答案为：,.二、选择题（本大题共有4题，满分18分，第13～14题每题4分，第15～16题每题5分）每题有且仅有一个正确选项，考生应在答题纸的相应编号位置将代表正确选项的小方格涂黑．13. 已知数列 n a 是等差数列，11548a a ，，则38133a a a （ ） A. 120 B. 96C. 72D. 48【答案】A【分析】根据等差数列的下标性质计算可得结果. 【详解】因为 n a 是等差数列，11548a a ， 所以8248a ，即824a ，所以38133a a a 888235524120a a a . 故选：A14. 若实数x ,y 满足2243x y xy ，则（ ）成立． A. 1xy B. 2244x yC. 2x yD. 2x y.【答案】B【分析】运用基本不等式，对条件代数式变形，逐项求解.【详解】由2243x y xy 和基本不等式2244x y xy （当224x y 时等号成立）， 22434x y xy xy xy ，当0xy 时，有1xy ，当0xy 时，35xy ，A 错误；由xy xy （当,x y 同号时等号成立）得：2222222243444,,43444x y xy xy x y xy x y xy x y ，2244x y ，B 正确；2224253x y xy x y xy ， 2235358x y xy （当224x y 时等号成立） ，2x y ，C ，D 错误；故选：B.15. 在233nx x 的二项展开式中，533r n r n r nx C 称为二项展开式的第1r 项，其中r =0，1，2，3，……，n ．下列关于233n x x的命题中，不正确的一项是（ ）A. 若8n ，则二项展开式中系数最大的项是1426383x C ．B. 已知0x ，若9n ，则二项展开式中第2项不大于第3项的实数x 的取值范围是35403x．C. 若10n ，则二项展开式中的常数项是44103C ．D. 若27n ，则二项展开式中x 的幂指数是负数的项一共有12项． 【答案】D【分析】A 选项：根据系数最大列不等式，解不等式即可；B 选项：根据题意列不等式，然后分01x 和1x 两种情况解不等式即可；C 选项：令51003r，解方程即可；D 选项：令52703r，解不等式即可.【详解】A 选项：令8178881988C 3C 3C 3C 3r r r r r r r r，解得5944r ，所以2r ，所以A 正确； B 选项：22171827339933xx C C ，整理可得5343x，当01x 时，不等式恒成立；当1x 时，解得35413x ，所以35403x，故B 正确；C 选项：令51003r ，解得6r ，所以常数项为610644101033 C C ，故C 正确； D 选项：令52703r ，解得815r ，所以r 可取17,18,27 ，共11项，故D 错. 故选：D.16. “阳马”，是底面为矩形，且有一条侧棱与底面垂直的四棱锥．《九章算术》总结了先秦时期数学成就，是我国古代内容极为丰富的数学巨著，对后世数学研究产生了广泛而深远的影响．书中有如下问题：“今有阳马，广五尺，袤七尺，高八尺．问积几何？” 其意思为：“今有底面为矩形，一条侧棱垂直于底面的四棱锥，它的底面长、宽分别为7尺和5尺，高为8尺，问它的体积是多少？”若以上的条件不变，则这个四棱锥的外接球的表面积为（ ）平方尺.A. 142B. 140C. 138D. 128【答案】C【分析】将四棱锥的外接球转化为长方体的外接球，然后求外接球表面积即可.【详解】如图所示，这个四棱锥的外接球和长方体的外接球相同，所以外接球的半径为22R，外接球的表面积24138S R . 故选：C.三、解答题（本大题共有5题，满分78分）解答下列各题必须在答题纸的相应位置写出必要的步骤．17. 已知数列 n a 满足：112a，21a ，2145n n n a a a ，对一切正整数n 成立． （1）证明：数列{1n n a a }是等比数列； （2）求数列 n a 的前n 项之和． 【答案】（1）证明见解析（2）4118318n n n S【分析】(1)结合递推公式利用等比数列的定义证明即可；(2)结合(1)中结论，利用累加法和等比数列求和公式求解出数列的通项公式，再利用分组求和即可得到结果. 【小问1详解】证明：∵12112a a ，∴2112a a ，∵2145n n n a a a ，对一切正整数n 成立，∴ *2114n n n n a a a a n N ，，即2114n n n na a a a ． ∴数列{1n n a a }是以12为首项，4为公比的等比数列． 小问2详解】 由(1)知，12311422n-n n n a a， ∴112211()()()+n n n n n a a a a a a a a 2527291122222n n n1231114(21)633n n, 当n =1时，1111(21)32a 满足上式，综上所述，n a23*1213n nN . 设数列 n a 的前n 项之和为n S ，则1146314n n n S =144131818318n nn n ．18.平面向量2(3sin ,cos ),(cos ,m x x n x，函数()2y f x m n. （1）求函数y =()f x 的最小正周期； （2）若[0,]2x ，求y =()f x 的值域；（3）在△ABC 中，内角A B C 、、的对边分别为a b c 、、，已知()f B，2,a b ABC 的面积.【答案】（1）π （2）2【（3）2【分析】（1）利用数量积、二倍角公式和辅助角公式化简得到26xf x，然后求最小正周期即可；（2）利用换元法和三角函数单调性求值域即可；（3）利用余弦定理得到c，然后利用三角形面积公式求面积即可. 【小问1详解】233sin cos sin2cos2222262m n x x x x x x，所以26xf x，最小正周期为 .【小问2详解】设26x，0,2x，566，在,62上严格增，在5,26上严格减，1sin62，51sin62，sin12，所以y= ()f x的值域为2．【小问3详解】()f B ，即sin216B，因为B为三角形内角，所以3B．2471cos222cBc，即2230c c，解得3c .所以△ABC的面积为1sin3222ac B .19. 如图所示，在矩形ABCD中，4AB ，2AD ，E是CD的中点，O为AE的中点，以AE为折痕将ADEV向上折起，使D点折到P点，且PC PB．（1）求证：PO 面ABCE；（2）求AC与面P AB所成角 的正弦值．【答案】（1）证明见解析； （2）15． 【分析】（1）取BC 的中点F ，连OF ，PF ，证明OF BC ，BC PF ，得到BC 面POF ，从而证明BC PO ，然后可得PO 面ABCE ；（2）作//OG BC 交AB 于G ，则OG OF ，然后以点O 为原点建立空间直角坐标系，然后利用向量求解即可. 【小问1详解】由题意，可得PA PE ，OA OE ，则PO AE ，取BC 的中点F ，连OF ，PF ，可得OF AB ∥，所以OF BC ， 因为PB PC ，BC PF ，且PF OF F ，所以BC 平面POF ， 又因为PO 平面POF ，所以BC PO ．又由BC 与AE 相交直线，所以PO 平面ABCE ． 【小问2详解】作//OG BC 交AB 于G ，则OG OF 如图建立空间直角坐标系，则(1,1,0),(1,3,0),(1,3,0),(2,4,0),((0,4,0)A B C P AC AP AB，设平面PAB 的法向量为(,,)n x y z，则040n AP x y n AB y，所以可取n ， 所以AC 与面PAB 所成角的正弦值sin cos ,15n AC. 20. 已知椭圆Γ：22221x y a b （0a b）的离心率为3，它的上顶点为A ，左、右焦点分别为 1,0F c ，2,0F c （常数0c ），直线1AF ，2AF 分别交椭圆Γ于点B ，C ．O 为坐标原点．为（1）求证：直线BO 平分线段AC ；（2）如图，设椭圆Γ外一点P 在直线BO 上，点P 的横坐标为常数m （m a ），过P 的动直线l 与椭圆Γ交于两个不同点M 、N ，在线段MN 上取点Q ，满足MP MQPN QN，试证明点Q在直线2260mx c 上． 【答案】（1）证明见解析 （2）证明见解析 【分析】（1）由离心率3c e a，将a ，b 均用c 表示，求出直线1AF 的方程，与椭圆方程联立求得点B 坐标，即可得到直线BO 的方程，根据椭圆的对称性，求出点C 的坐标，再证出AC 中点在直线BO 上即可； （2）设 11,M x y ， 22,N x y 和MP MQPN QN，用线段定比分点坐标公式将P ，Q 坐标表示出来，并代入2260mx c ，结合P 的横坐标为m 和M ，N 在椭圆上，进行运算证明即可.【小问1详解】由题意，3c e a，则a，b，∴ A ， ∴椭圆Γ方程为2222132x y c c，即222236x y c ，∴直线1AF的斜率100AF k c，直线1AF的方程为 y x c ，联立 222236,,x y c y x c 消去y ，化简得2230x cx ，解得0A x ，32B x c ，即点B 的横坐标为32B x c ，代入直线1AF的方程，得3,22B c c， ∴直线BO的斜率023302OBc k c，直线BO的方程为3y x ，∵3,22B c c，∴由椭圆的对称性知3,22C c c，又∵ A ，∴线段AC 的中点坐标为3,44c c，∵3344c c ，∴线段AC 的中点3,44c c在直线BO ：3y x 上， 即直线BO 平分线段AC . 【小问2详解】设过点P 的直线l 与椭圆Γ交于两个不同点的坐标为 11,M x y ， 22,N x y ，∵M ，N 在椭圆Γ上，∴22211236x y c ，22222236x y c ∵MP MQ PN QN ，∴设MP MQPN QN，易知0 ，且1 ， 则由已知，有MP PN ，MQ QN，∴由线段定比分点坐标公式，有1212,11x x y y P，1212,11x x y y Q，∵点P 的横坐标为常数m （m a ）， ∴121x x m，又∵点P 在直线BO ：3y x 上，∴1213y y m 1231y y，将1212,11x x y y Q代入2mx ，得121212122311121mx y y y x x x x y y2222221212222311x x y y222221122223231x y x y2222661c c222611c26c ，即点1212,11x x y y Q在直线2260mx c 上. 【点睛】本题的两问证明，实质上都是证明点在直线上，第（1）问证明AC 中点在直线BO 上，即可证明直线BO 平分线段AC ，第（2）问设MP MQPN QN，由线段定比分点坐标公式求得Q 的坐标，即可结合M ，N ，P 的坐标，证明点Q 在直线2260mx c 上. 21. 已知函数f (x )＝－2a ln x －2x ，g (x )＝ax －(2a ＋1)ln x －2x，其中a ∈R. （1）若x ＝2是函数f (x )的驻点，求实数a 的值； （2）当a >0时，求函数g (x )的单调区间； （3）若存在x [1e，e 2 ]（e 为自然对数的底），使得不等式f (x ) g (x )成立，求实数a 的取值范围． 【答案】（1）12（2）答案见解析 （3） e,【分析】（1）根据2x 是函数 f x 的驻点得到 20f ，然后列方程求a 即可； （2）求导，分12a、12a 和102a 三种情况讨论单调性即可； （3）将存在21,e e x，使得不等式 f x g x 成立转化为 min a h x ，然后利用单调性求最值即可. 【小问1详解】若2x 是函数 f x 的驻点，则 20f ，可得2122022a ，即得12a ． 【小问2详解】函数 g x 的定义域为 0, ，222221212212ax a x ax x a g x a x x x x， 当0a 时，令 0g x ，可得10x a或2x ， ①当12a ，即12a 时，对任意的0x ， 0g x ， 此时，函数 g x 的单调递增区间为 0, ． ②当102a，即12a 时， 令 0g x ，得10x a或2x ， 令 0g x，得12x a， 此时，函数 g x 的单调递增区间为10,a和2, ，单调递减区间为1,2a. ③当12a ，即102a 时，令 0g x ，得02x 或1x a ；令 0g x ，得12x a，此时，函数 g x 的单调递增区间为 0,2和1,a，单调递减区间为12,a .【小问3详解】由 f x g x ，可得ln 0ax x ，即ln x a x ，其中21,e e x， 令 ln x h x x，21,e e x ，若存在21,e e x，使得不等式 f x g x 成立，则 min a h x ，21,e e x， 21ln x h x x ，令 0h x ，得e x ， 当1e ex 时， 0h x ，当2e e x 时， 0h x ， ∴函数 h x 在1,e e上严格递增，在 2e,e上严格递减， ∴函数 h x 在端点1ex 或2e x 处取得最小值． ∵1e e h，222ee h∴ 1h he e ， ∴ min 1h x he e ，∴e a ≥，因此，实数a 的取值范围是 ,e【点睛】对于存在问题，常用到以下两个结论： （1）存在 min a f x a f x ； （2）存在maxa f x a f x .。

上海市风华中学2024学年度第一学期高三年级英语阶段测试（2024.9）（满分140分考试时间：120分钟）第Ⅰ卷I. Listening ComprehensionSection A 10%Directions: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. She has no appetite at all. B. She wants to dine out.C. She is too tired to go out.D. She prefers to cook at home.2. A. 6 pounds B. 7 pounds. C. 8 pounds. D. 9 pounds.3. A. At the professor's office. B. In the bookstore.C. In the library.D. In the laboratory.4. A. Because something went wrong with his car.B. Because his car was broken in an accident.C. Because he wanted to take a walk for a rest.D. Because he was stuck in a traffic jam.5. A. The morning flight. B. The afternoon flight.C. The evening flight.D. The midnight flight.6. A. She is not interested in going camping with him.B. She wants the man to stay at home with her.C. She thinks the man needs to have a good rest.D. She thinks the man should prepare for the exams.7. A. Some major revisions are needed. B. It should be revised by a tutor.C. Only a few changes should be made.D. The draft needs no revision at all.8 A. He is going away for a while. B. He worked hard to earn money.C. He did very well in the exam.D. He can't wait to have a rest.9. A. He forgot to bring his own camera. B. He is not good at taking pictures.C. He cannot take a photo with the camera.D. He doesn't know how to use the camera.10. A. She was interrupted by a visiting friend. B. She didn't come back until midnight.C. She stayed up late for the final exam.D. She visited her friend instead of studying.Section BDirections: In Section B, you will hear two short passages, and you will be asked several questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. They can maintain their body temperature stable.B. They conserve enough energy before the long sleep.C. They can keep their heart beat at a regular rate.D. They have their weight increased to the maximum.12. A. By staying in bidıng places and eating little.B. By seeking extra food and warm shelter.C. By growing thicker hair to stay warm.D. By storing enough food in advance.13. A To stay safe. B. To save energy. C. To get more food. D. To protect the young. Questions 14 through 16 are based on the following passage.14. A. Four to six hours. B. Six to nine hours.C. Around eight hours.D. More than eight hours.15. A. They may not be able to concentrate well.B. They may get the feeling of being drunk.C. They may suffer from high blood pressure.D. They may lose weight easily in a short period of time.16. A. Military people are used to being deprived of sleep.B. Training can make people sleep less and suffer less.C. People can bank sleep by sleeping more beforehand.D. Sleeping earlier than usual makes people sleep less.Questions 17 through 20 are based on the following passage.17. A. Double Eleven sales in 2021.B. Unreliable factors of online shopping.C. Key points of Taobao's success ı n sales.D. Advantages and disadvantages of online shopping.18. A. People who are good at doing business.B. People who work seven days a week.C. People who have very busy schedules.D. People who dislike telephone shopping.19. A. Consumers can save a lot of time.B. It provides round- the- clock service.C. People can buy things without leaving their homes or offices.D. The quality of the product is the same as what is described online.20. A. Inferior quality. B. Various retailers.C. Efficient sales return.D. Convenient delivery.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blankAI Weather Forecasting Can't Replace Humans- YetAs Hurricane Lee was curving（呈曲线）northward to the west of Bermuda in mid- September of last year, forecasters were busily consulting weather models and data from hurricane- hunter aircraft to calculate （21）________ the dangerous storm was likely to make landfall （着陆）: New England or farther east, in Canada. The sooner the meteorologists（气象学家）could do so, the earlier they could warn those in the path of damaging wind gusts and fierce storm surges By six days ahead of landfall, it was clear that Lee （22）________（follow）the eastward path, and warnings were issued, accordingly. But （mother tool- an experimental AI model called GraphCast-（23）________（mate）that outcome accurately three whole days before the forecasters' traditional models.GraphCast's prediction is a window into AI's potential （24）________（improve）weather forecasts. But whether it is a forecaster of a true sea change in the field or will simply become one of many tools （25）________ human forecasters consult to determine which way the winds will blow is still up in the air.GraphCast, developed by Google DeepMind, is the latest of several AI weather models （26）________（release）in recent years. Google's Metnet, first introduced in2020, is already being used in products such as the company's “now cast” in its weather app. All are advertised as having an accuracy that is comparable with or higher than（27）________ on the best non-AI forecasting computer models and have caused a sensation in meteorology, with GraphCast （28）________（cause）the most significant stir so far.The DeepMind research team had put GraphCast through its paces by feeding it historical weather data to see if it could accurately “predict” what happened. The study showed the AI performed equal to or even better than the gold standard.Yet （29）________ GraphCast becomes probabilistic-- and even if the model's resolution improves and the AI becomes more accurate in its forecasts of rain and storm intensity - modeling remains just a single component of the weather- prediction pipeline, says Hendrik Tolman, senior adviser for advanced modeling systems at the NWS. However, every expert described GraphCast and other Al models as additional devices in their tool kit. If AI （30）________produce accurate forecasts quickly and cheaply, there's no reason not to begin using it together with existing methods.But will there be a world where AI models replace physics- based models— and people -- in the future? Forecasts suggest there's little chance.Section BDirections: Complete the following passage by using the words or phrases in the box. Each word or phrase can only be used once. Note that there is one word or phrase more than you need.A. desperatelyB. functionalC. green- lightingD. impactE. independenceF longstanding G. progressive H reasonably I. reversing J. swallowK. targetedAlzheimer's Drug Approved Despite Fierce DebateThe -U. S. Food and Drug Administration （FDA）recently approved the drug Aduhelm, produced by American biotechnology company Biogen with Japan's Eisai Co., to treat patients with Alzheimer's（老年痴呆症）disease. The approval was based on study results showing that the drug seemed“（31）________ likely” to benefit Alzheimer's patients, the FDA said.The decision, which could （32）________ millions of Alzheimer's patients and their families, has sparked disagreements among medical researchers. While the drug was shown to be effective in slowing the mental decline in patients' suffering from the disease, it was not proven to be effective in （33）________ its effects, the Associated Press reported. citing a study. The rate of mental decline in patients that had been administered Aduhelm was slowed by 22 percent when compared to patients who had received a placebo（安慰剂）. But even given these results, on a test that is conducted to evaluate the cognitive and （34）________ abilities of a patient, patients who were administered Adubelm only showed an increase of 0.39 in their- scores. And it's unclear how such metrics （度量标准）translate into practical benefits, like greater（35）________ or the ability to recall important details.The FDA's review of the drug has become a flashpoint in （36）________ debates over standards used to evaluate therapies for hard- to- treat conditions. On one side, groups representing Alzheimer's patients and their families say any new therapy - even one of small benefit -deserves approval. But many experts ward that（37）________ the drug could set a dangerous example by opening the door to treatments of questionable benefit.Alzheimer's is an irreversible, （38）________ brain disorder that slowly attacks areas of the brain that are essential to memory, reasoning, communication, and basic daily tasks. In the final stages of the disease, the patients will lose the ability to（39）________Science doesn't fully understand what causes Alzheimer's, but there's broad agreement that the brain plaque（斑点）that is being （40）________ by Aduhelm is one of the contributing factors. Evidence suggests family history, education, and chronic conditions like heart disease may all play a role. “This is a sign of hope but not the final answer,” said Dr. Richard Hodes, director of the U. S. National Institute on Aging.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Some people like to read the instructions from start to finish before they take action while others study the diagrams and then jump right in. This 41 for one approach over another when learning new information is not uncommon. Indeed, the notion that people learn in different ways is such a universal belief in American culture that there is a thriving industry dedicated to 42 learning styles and training teachers to meet the needs of different learners.Just because a notion is popular, 43 , doesn't make it true. A recent review of learning styles found evidence to clearly support the idea that outcomes are 44 when instructional techniques align with （匹配）individuals' learning styles. Most previous investigations on learning styles focused on classroom learning, and assessed whether instructional style 45 outcomes for different types of learners. But is the 46 really where most of the serious learning occurs? Some might argue that, in this era of flipped classrooms and online course materials, students 47 more of the information on their own. That might explain why instructional style in theclassroom matters little. It also 48 the possibility that learning styles do matter. Perhaps a 49 between students' individual learning styles and their study strategies is the key to ideal outcomes.To explore this 50 , researchers asked students enrolled in an anatomy class （解剖课）to complete an online learning styles assessment, answer questions about their study strategies and report details about the 51 they used outside of class（e. g. flash cards, review of lecture notes, anatomy coloring books）.Scores suggested that most students used multiple learning styles, but that no particular style 52 better outcomes than another. The focus in this study, however, was not on whether a particular learning style was more53 . Despite knowing their own, self- reported learning preferences, nearly 70% of students 54 to employ study techniques that supported those preferences. Given the popular belief that learning styles matter, and the fact that many students 55 poor academic performance on the lack of a match between their learning style and teachers' instructional methods, one might expect students to rely on techniques that support their personal learning preferences when working on their own.41. A. preference B. tendency C. phenomenon D. practice42. A. identifying B. exposing C. revealing D. establishing43. A. therefore B. moreover C. however D. instead44. A. best B. acceptable C. disappointing D. undesirable45. A. impacted B challenged C. confirmed D. supported46. A. network B. classroom C. school D. lecture47. A. require B. collect C. master D. demand48. A. limits B eliminates C examines D. raises49. A. comparison B. link C. balance D. match50. A. issue B. possibility C. field D. proposal51. A. equipment B. techniques C. notebooks D. assistance52. A. originated in B. resulted from C. resulted in D. took over53. A. important B. advantageous C meaningful D popular54. A. failed B. managed C. struggled D. attempted55. A. count B. concentrate C. blame D. conductSection BDirections: Read the following four passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.（A）I was sure that I was to be killed. I became terribly nervous. I fumbled （摸索）in my pockets to see if there were any cigarettes, which had escaped their search. I found one and because of my shaking hands, I could barely get it to my lips. But I had no matches; they had taken those. I looked through the bars at my jailer. He did not make eye contact with me. I called out to him, “Have you got a light?” He looked at me, shrugged and came over to light my cigarette. As he came close and lit the match, his eyes unconsciously locked with mine. At that moment, I smiled. I don't know why I did that. Perhaps it was nervousness, perhaps it was because, when you get very close, one toanother, it is very hard not to smile. In any case, I smiled. In that instant, it was as though a spark jumped across the gap between our two hearts, our two human souls. I know he didn't want to, but my smile leaped through the bars and generated a smile on his lips, too. He lit my cigarette but stayed near, looking at me directly in the eyes and continuing to smile.I kept smiling at him, now aware of him as a person and not just a jailer. And his looking at me seemed to have a new dimension too.“Do you have kids?” he asked.“Yes, here, here.”I took out my wallet and nervously fumbled for the pictures of my family. He, too, took out the pictures of his family and began to talk about his plans and hopes for them. My eyes filled with tears. I said that I feared that I'd never see my family again, never have the chance to see them grow up. Tears came to his eyes, too. Suddenly, without another word, he unlocked my cell and silently led me out. Out of the jail, quietly and by back routes, out of the town. There, at the edge of town, he released me. And without another word, he turned back toward the town.My life was saved by a smile, yes, the smile- the unaffected, unplanned, natural connection between people. I really believe that if that part of you and that part of me could recognize each other, we wouldn't be enemies. We couldn't have hate or envy or fear.56. The underlined sentence indicates that the author and the jailer started to have a ________ conversation.A. less impersonalB. more intenseC. less formalD. more friendly57. Which is true based on the first paragraph?A. My hands were shaking because of fear.B. The jailer was going to shoot me.C. I smile8 because I had to beg for life.D. He smiled to me because he wanted to.58. Their eyes were filled with tears because they both ________.A. took out the pictures of their familiesB. missed their families far awayC. had plans and hopes for futureD. feared that they would die59. How does a smile succeed in saving the author's life?A. By asking for the jailer to light a cigarette.B. By planning for an exchange of family pictures.C. By establishing natural connection between people.D. By hiding the human feelings of hate, envy or fear.（B）CareersHome The UN Workforce Pay and Benefits Career Journeys Career Options How to Apply Apply Now Home>> How To ApplyFAQs on preparing your ApplicationQ: Should I target my Application to a specific Job Opening（JO）?A: Yes. Naturally, a customized cover note will also help you focus on the key aspects of your Application that relate to the JO, but it is also in your interest to target the Application according to the responsibilities and competencies of the position.Q: What's the difference between duties and achievements?A: Duties describe the specific responsibilities of your job. They accurately reflect what you are doing or have done in each of your previous jobs. In other words, it is the “what you do” of your job. Achievements describe in specific terms “how well” you did in your job.Q: Many of my achievements are team- based, how do I draft them in my Application?A: You should include your team- based achievements in your Application. Indicate that you were part of a team, and describe your specific role in reaching the goal.FAQs on general Application guidelinesQ: Can I save my Application?A: Yes. You should save your Application when you make changes and/ or update it. It is recommended that you save different versions of your Application in Word format and then edit the Application online according to the post for which you are applying.Q: Can I update my Application to apply for a new JO?A: Yes. Each time you apply for a new JO, we recommend that you review your Application and update it , if appropriate, or target it to better reflect your suitability for the new JO. Your updates will not affect the content of Applications previously submitted against other JOs.Q: Must I use up all the available characters in each section of my Application?A: No. In fact, doing so may result in an unnecessary lengthy Application. Unless you have an enormous range of experiences, there is no reason to use up all the space given. Applicants are encouraged to list their duties and achievements in a clear and brief manner.60. Which of the following descriptions best shows your achievements?A. I've developed various interests, ranging from oil painting to designing model.B. I'm good at creating proposals for new product ideas aimed at a specific market.C. I'm in charge of the clearance, production and distribution of information material.D. I succeeded in directing a video presentation, assisting our group to win the first prize.61. If you want to apply for another JO, you'd better __________.A. target your focus on your interest in the JOB. save your latest application in Word formatC. Serape one application with all your competenciesD. update your application to match new requirements62. Applicants are expected to __________ in their applications.A. introduce what JOs they have previously applied toB. list the greatest achievements they have made in detailC. give key information about their experiences and achievementsD provide the results of their tests, assessments and examinations（C）Atlantis is the legendary island that sank beneath the waves in the distant past, taking down with it an advanced civilization. Is it possible that we will ever find it? Or, more importantly did it even exist?The short answer to both: No. All available evidence indicates that the philosopher Plato, sometime around 360 B. C., invented the island nation to illustrate a point about the dangers of aggressive imperialism（势力扩张）. In Plato's telling, Atlantis was no utopia. Rather, it was a centrist to an idealized version of Athens from long before Plato's time. This ancient Athens was very similar to Plato's notion of the ideal state. Plato laid out the details for what such a state would look like in his famous work, I he Republic. It should be small and virtuous. The residents of Atlantis, on the other hand, were eventually “filled with an unjust lust for possessions and power," according to Plato's character who described the island.In Plato's texts, Atlantis was “larger than Libya and Asia combined,”（which, in Plato's time. would have referred to modern- day northern Africa and over half of Turkey）. It was situated in the Atlantic Ocean, somewhere outward from the Strait of Gibraltar. It's a landmass large enough that, if it really existed somewhere underwater in the Atlantic, it would certainly appear on sonar maps of the ocean floor.So how did Atlantis come to represent a lost utopic civilization? For that, you can mostly blame （or thank）Ignatius Donnelly. In 1882, the former U. S. Congressman published Atlantis. The Antediluvian World. The book laid out 13 hypotheses, centered on the idea that Atlantis had truly existed, and indeed represented a place “where early mankind dwelt for ages in peace and happiness. According to Donnelly, Atlantis was the original source of many ancient civilizations around the world. If one followed the clues in Plato's writing, Donnelly believed, Atlantis could be found. He was inspired by a remarkable discovery in the early 1870s. An amateur archaeologist claimed to have unearthed the legendary city of Troy based on Homer's The Iliad. If Troy, long thought to be fictional, was real, why shouldn't Atlantis be, too?Donnelly was certain of his theory, predicting that hard evidence of the sunken city would soon be found, and that museums around the world would one day be filled with artifacts from Atlantis. Yet about 140 years have passed without a trace of evidence. The Atlantis legend has been kept alive, fueled by the public's imagination and fascination with the idea of a hidden. long- lost utopia. Yet the “lost city of Atlantis was never lost; it is where it always was: in Plato's books.63. What can we learn about Plato?A. He predicted that Atlantis would be' destroyed by aggressive imperialism.B. He was inspired by utopia to gradually form the notion of the ideal state.C. He created the setting in which residents of Atlantis were not virtuous.D. He witnessed Atlanteans' pursuit of an unjust lust for possessions and power.64. Homer's The Iliad is mentioned ___________.A. to demonstrate the actual existence of the legendary city of Troy.B. as indirect evidence of the credibility of Plato's account of Atlantis.C. because it is a great piece of fictional writing about an ancient legend.D. because it contains many clues about the legendary city of Troy.65. According to the passage, Atlantis was ___________.A. a long- lost small utopia with many virtuous residents.B. a large landmass situated in the Atlantic Ocean.C. the original source of many ancient civilizations.D. Plato's invention against which to highlight his ideal.66. Which of the following is the best title for this passage?A. Plato, Atlantis and How the City Collapsed and Finally Got LostB Plato Told a Lie, and Ignatius Donnelly was to Blame for ItC. The History, Legends, and Evidence of the Lost City of AtlantisD. Where Is the Lost City of Atlantis — and Does It Even Exist?Section CDirections: Read the passage carefully. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.A. In addition to painting on canvas, Crisco also paints on musical instruments.B. However, no matter what the scene is, none of the paintings focus on the darkness.C. His paintings take you on an unknown adventure into an unknown space, reminding you to be alert and brave.D. Through the use of glowing paint. he brings a sense of magic and energy to the scenes that he creates.E. Studies have shown that expressing themselves. through art can help people with depression and anxietyF. As a self- taught artist s be found his passion for art at a dark time in his life when he found himself suddenly out or work.Bringing Light to the Darkness with Crisco ArtMost paintings are best enjoyed in galleries with good lighting. But an Italian artist who goes by the name Crisco is changing the way we look at paintings with a new approach: glow （发光）- in- the- dark paint.Crisco's paintings are beautiful in normal lighting, but it is when the lights go down that they really come alive. （67）_________ His art mostly shows landscapes. Trees, horizons, and especially starry skies come alive with the glow of his paints. At the center of most of his work, there is often a human or animal figure. The figure may be just a shadow surrounded by the glowing colors, but it often appears to be the source of the light. （68）_________ Instead, they are all bright pictures of hope, life, wonder, and growth. They are Crisco's way o t adding a little light to the world.Crisco's full name is Cristoforo Scorpiniti.（69）_________ Instead of letting a negative experience get the best of him, he threw himself into a new pursuit: art. According to Crisco, he paints with glowing colors to inspire hope. Though his paintings often show night scenes that look good in the dark, Crisco does not focus on the darkness.Instead, he uses his paintings to express positivity by creating light in the darkness.A lot of his best work has come out of just painting what he felt at the time without any plan or structure.（70）_________ With over half a million followers on Instagram, Crisco is already popular on social media for his unique paintings. He'll surely only get more famous in the future for his inspiring paintings that beautifully mix darkness and light.IV. Summary Writing71. Directions: Read the following passage. Summarize the main idea and the main point（s）of the passage in no more than 60 words. Use your own words as far as possible.Are EV Really Environmentally Friendly?Many consumers are opting for an electric vehicle （EV）or plug- in hybrid electric vehicle （PHEV）to replace their polluting gas- powered cars. These electrified vehicles are rising to popularity on the premise of environmental conservation and eliminating the need for harmful emissions. There are a couple of things. however, to consider before concluding that EVs are the most environmentally friendly option for consumers.Where do electric cars get their energy? Although EVs create no emissions on board, they typically draw power from lithium- ion batteries. These batteries require charging, either at home or via a publicly accessible charging station. Since EV charging infrastructure is mainly reliant on the power grid - specifically, the grid draws power from plants like coal plants - although your EV does not produce any harmful emissions as you drive it, burning fossil fuels is involved in fueling it. Moreover, temperature extremes like excessive coldness or heat can dramatically reduce lithium- ion battery life. Carnegie Mellon University's Department of Engineering and Technology says that the most extreme cases of coldness will compromise efficiency by as much as 40%. The decreased efficiency is an issue if the power stored in the battery packs of EVs is sourced from fossil fuel- burning.Besides the power source, metal s such as lithium and cobalt are wrapped up in environmentally and socially questionable processes, too. One of the first environmental issues lithium batteries pose is how to dispose of them properly. In an average battery recycling plant, all parts of the battery are shredded into a powder using a mechanical shredder and then either melted or dissolved into acid — recycling lithium batteries isn't as simple. Lithium batteries are typically made up of a mix of different elements including cobalt, nickel, manganese and iron —cobalt especially known to be a hazardous substance. In addition, most studies associate lithium mining in South America from salt brine with salinization（盐化）of freshwater that the locals need to survive. Since the mineral contains dangerous substances, the mining process also contaminates the local water basins. So, lithium extraction exposes the local ecosystems to poisoning and other related health problems.V. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.72.他那种急于求成的心态让他无缘冠军宝座。

曹杨二中2024学年第一学期高三年级数学月考2024.09一、填空题（本大题共有12题,满分54分,第1-6题每题4分,第7-12题每题5分） 1.已知集合()()3,2A ,B ,=−∞=+∞,则A B ⋂= . 2.已知复数z 满足15i z =−(i 为虚数单位),则z = . 3.已知向量()()102,210a ,,b ,,==,则a ,b <>= .4.523x ⎫⎪⎭的二项展开式中的常数项为 .（结果用数值表示）5.设()y f x =是以1为周期的周期函数.若当01x <≤时,()2f x log x =,则32f ⎛⎫= ⎪⎝⎭.6.设m 为正实数.若直线0x y m −+=被圆()()22113x y −+−=所截得的弦长为m ，则m = .7.从一副去掉大小王的52张扑克牌中无放回地任意抽取两次。

在第一次抽到A 的条件下，第二次也抽到A 的概率为 .（结果用最简分数表示）8.设数列{}n a 前n 项和为n S 。

若()21n n S a n ,n N +=≥∈,则5S = . 9.已知,x y 为正实数,且1x y +=,则当21x y+取最小值时,x = . 10.设(),1a R f x lnx ax ∈=−+.若函数()y f x =的图像都在x 轴下方（不含x 轴），则a 的取值范围是 .11.已知{}n a 是严格增数列,且点()()1n n P n,a n ,n N ≥∈均在双曲线2231x y −=上。

设M R ∈，若对任意正整数n ，都有1n n P P M +>，则M 的最大值为 .12.设(){}2,235a R f x min x ,x ax a ∈=−−+−，其中{}min u,v 表示,u v 中的较小值.若函数()y f x =至少有3个零点,则a 的取值范围是 .二、选择题（本大题共4题，满分18分，第13-14题每题4分，第15-16题每题5分）13.已知a R ∈，则"1a >"是"11a<"的（ ）. A.充分非必要条件 B.必要非充分条件 C.充要条件 D.既非充分又非必要条件14.为研究某药品的疗效,选取若干名志愿者进行临床试验,所有志愿者的舒张压（单位：kPa ）的分组区间为[)[)[)[)1213,1314,1415,1516,,,,，[]1617,.将其按从左到右的顺序分别编号为第一组，第二组，，第五组，下图是根据试验数据制成的频率分布直方图。

2022学年第一学期徐汇区学习能力诊断卷高三数学 试卷2022．12考生注意：1．本场考试时间120分钟，试卷共4页，满分150分.2．每位考生应同时收到试卷和答题卷两份材料，答卷前，在答题卷上填写姓名、考号等相关信息.3．所有作答务必填涂在答题卷上与试卷题号对应的区域，不得错位，在试卷上作答一律不得分.4. 用2B 铅笔作答选择题，用黑色字迹钢笔、水笔或圆珠笔作答非选择题.一、填空题（本大题共有12题，满分54分，第1-6题每题4分，第7-12题每题5分）考生应在答题纸的相应位置直接填写结果．1．已知全集U =R ，集合{|0}A x x =>，则A = ．2．在复平面内，复数z 所对应的点的坐标为(1,1)-，则z z ⋅= ． 3．不等式x x x 2+5≥1+2+3的解集为 ． 4． 函数tan ,y x ππ3⎛⎫=⎪22⎝⎭在区间上的零点是 ． 5．已知()f x 是定义域为R 的奇函数，且0x ≤时，()1xf x e =-，则()f x 的值域是 ．6．在92()x x-的二项展开式中，3x 项的系数是 ．7．已知圆锥的侧面积为2π，且侧面展开图为半圆，则该圆锥底面半径为 ． 8．在数列{}n a 中，12a =，且1lg(2)1n n na a n n -=+≥-，则100a = ． 9．某中学从甲、乙两个班中各选出15名学生参加知识竞赛，将他们的成绩（满分100分）进行统计分析，绘制成如图所示的茎叶图.设成绩在88分以上（含88分）的学生为优秀学生，现从甲、乙两班的优秀学生中各取1人，记甲班选取的学生成绩不低于乙班选取得学生成绩记为事件A ，则事件A 发生的概率()P A = ．10．在△ABC 中，4AC =，且AC AB 在方向上的数量投影是2-，则(R)BC BA λλ-∈的最小值为 ．11．设R k ∈，函数243y x x =-+的图像与直线1y kx =+有四个交点，且这些交点的横坐标分别为()12341234,,,x x x x x x x x <<<，则22221234x x x x k+++的取值范围为 ．12．已知正实数a b 、满足326a b +=，则b 的最小值为 ． 二、选择题（本大题共有4题，满分18分，第13-14题每题4分，第15-16题每题5分）每题有且只有一个正确选项.考生应在答题纸的相应位置，将代表正确选项的小方格涂黑． 13．设0ab >，则 a b >“”是11a b<“” 的（ ） A ．充分非必要条件 B ．必要非充分条件 C ．充分必要条件 D ．既非充分也非必要条件14．已知圆1C 的半径为3，圆2C 的半径为7，若两圆相交，则两圆的圆心距可能是（ ） A ．0B ．4C ．8D ．1215．已知平面α、β、γ两两垂直，直线a 、b 、c 满足：,a b c αβγ⊂⊂⊂，，则直线a 、b 、c 位置关系不可能是（ ）A ．两两垂直B ．两两平行C ．两两相交D ．两两异面16．设数列{}n a 为：1，12，12，14，14，14，14，18，18，18，18，18，18，18，18，…，其中第1项为11，接下来2项均为12，再接下来4项均为14，再接下来8项均为18，…，以此类推，记1nn i i S a ==∑，现有如下命题：①存在正整数k ，使得1k a k <； ②数列n S n ⎧⎫⎨⎬⎩⎭是严格减数列． 下列判断正确的是( )A ．①和②均为真命题B ．①和②均为假命题C ．①为真命题，②为假命题D ．①为假命题，②为真命题三、解答题（本大题共有5题，满分78分）解答下列各题必须在答题纸的相应位置写出必要的步骤.17.（本题满分14分，第1小题满分6分，第2小题满分8分） 如图，在直三棱柱111ABC A B C -中，2AB AC ==，14AA =，AB AC ⊥，1BE AB ⊥交1AA 于点E ，D 为1CC 的中点．（1）求证：BE ⊥平面1AB C ；（2）求直线1B D 与平面1AB C 所成角的大小．18. （本题满分14分，第1小题满分6分，第2小题满分8分） 已知21()ln (1)(R)2f x x a x ax a =-++∈． （1）当0a =时，求函数()y f x =在点(1(1))f ，处的切线方程； （2）当(]1a ∈0，时，求函数()y f x =的单调区间．19. （本题满分16分，第1小题满分8分，第2小题满分8分）近年来，为“加大城市公园绿地建设力度，形成布局合理的公园体系”，许多城市陆续建起众多“口袋公园”． 现计划在一块边长为200米的正方形的空地上按以下要求建造“口袋公园”. 如图所示，以EF 中点A 为圆心，FG 为半径的扇形草坪区ABC ，点P 在弧BC 上（不与端点重合），AB BC CA 、弧、、PQ PR RQ 、、为步行道，其中PQ 与AB 垂直，PR 与AC 垂直．设PAB θ∠=． （1）如果点P 位于弧BC 的中点，求三条步行道PQ PR RQ 、、的总长度；（2）“地摊经济”对于“拉动灵活就业、增加多源收入、便利居民生活”等都有积极作用．为此街道允许在步行道PQ PR RQ 、、开辟临时摊点，积极推进“地摊经济”发展，预计每年能产生的经济效益分别为每米5万元、5万元及5.9万元．则这三条步行道每年能产生的经济总效益最高为多少？（精确到1万元）RQBC EP20.（本题满分16分，第1小题满分4分，第2小题满分6分，第3小题满分6分） 已知曲线i C 的方程为()221R,1,2,3i i x y i λλ+=∈=，直线():1R l y k x k =+∈，.（1）若曲线1C 是焦点在x 轴上且离心率为2的椭圆，求1λ的值；（2）若211k λ=≠-，时，直线l 与曲线2C 相交于两点M ，N ，且MN =2C 的方程；（3）若直线l 与曲线i C 在第一象限交于点(),i i i P x y ，是否存在不全相等123,λλλ,满足1322λλλ+=，且使得2213x x x =成立．若存在，求出2x 的值；若不存在，请说明理由．21.（本题满分18分，第1小题满分4分，第2小题满分6分，第3小题满分8分） 对于数列{}{}n n x y ，，其中n y Z ∈，对任意正整数n 都有12n n x y -<，则称数列{}n y 为数列{}n x 的“接近数列”.已知{}n b 为数列{}n a 的“接近数列”，且11,n nn inii i A a B b ====∑∑．（1）若14n a n =+(n 是正整数)，求1234,,,b b b b 的值； （2）若139210n n a +⎛⎫=+- ⎪⎝⎭(n 是正整数)，是否存在k (k 是正整数)，使得k k A B <，如果存在，请求出k 的最小值，如果不存在，请说明理由；（3）若{}n a 为无穷等差数列，公差为d ，求证：数列{}n b 为等差数列的充要条件是d Z ∈．参考答案及评分标准 2022.12一． 填空题：（本大题共有12题，满分54分，第1-6题每题4分，第7-12题每题5分）1． {}0 2． 2 3． []2,1- 4． π 5． (1,1)- 6． 672- 7． 1 8． 4 9．29 10．．182,3⎛⎫-∞- ⎪⎝⎭ 12．2913二．选择题：（本大题共有4题，满分18分，第13、14题每题4分，第15、16题每题5分）13． C 14． C 15． B 16． D三、解答题（本大题共有5题，满分78分）解答下列各题必须在答题纸的相应位置写出必要的步骤.17. （本题满分14分，第（1）小题6分，第（2）小题8分）解：（1）因为三棱柱111ABC A B C -为直三棱柱，所以1AA ⊥平面ABC ，所以1AA AC ⊥．因为AC AB ⊥，1ABAA A =，所以AC ⊥平面11AA B B ．因为BE ⊂平面11AA B B ，所以AC BE ⊥． 因为1BE AB ⊥，1ACAB A =，因此，BE ⊥平面1AB C ．（2）由（1）知1,,AB AC AA 两两垂直， 如图建立空间直角坐标系A xyz -则(000)A ，，,1(2,0,4)B ,(0,2,2)D ,(2,0,0)B ，设(0,0,)E a ，所以， 因为，所以，即．所以平面的一个法向量为． 1(2,2,2)B D =--又设直线1B D 与平面1AB C 所成角的大小为(0)2πθθ≤≤，则1115sin cos 215BE B DBE B Dπθθ⋅⎛⎫=-== ⎪⎝⎭．因此，直线1B D 与平面1AB C 所成角的大小为arcsin15． 18. （本题满分14分，第（1）小题6分，第（2）小题8分）解：（1）当0a =时，()ln f x x x =-，1()1f x x'=-， 1=(02,2)=(2,0,4)=(20,)AD AB BE a -,，，,1AB BE ⊥440a -=1a =1AB C =(20,1)BE -,所以(1)1f =-，(1)0f '=．所以函数()y f x =在点(1,(1))f 处的切线方程为10y +=．（2）因为21l )n 1)2((x a a f x x x =-++，定义域为(0,)+∞，所以()2(1)1(1)1((1)1)x ax ax f x a a a x x x x x ---=-++=='++． ①当01a <<时，()f x 与()f x '在(0,)+∞上的变化情况如下：所以函数()y f x =在(0,1)及(,)a +∞内严格增，在(1,)a内严格减；②当1a =时，令0()f x '=，解得1x =，所以函数在1x =时有唯一的驻点.但是，由于永远成立0()f x '≥，驻点两侧导数值的正负没有发生变化，因此该驻点不是单调区间的分界点，所以此函数的单调增区间为(0,)+∞．因此，当01a <<时，函数()y f x =的单调增区间为(0,1)及1(,)a +∞，单调减区间为1(1,)a；当1a=时，函数()yf x =单调增区间为(0,)+∞．19. （本题满分16分，第（1）小题8分，第（2）小题8分）解：（1）易得BAC π∠=3，由于点P 位于弧BC 的中点，所以点P 位于BAC ∠的角平分线上， 则sin sinPQ PRPA PAB π==⋅∠=200⨯=1006，cos AQ AP PAB =∠=200⨯=2因为BAC π∠=3，AQ AR =所以ARQ ∆为等边三角形，则RQ AQ =，因此三条街道的总长度为l PQ PR RQ =++=100+100=200BC（2）||||sin 200sin PQ AP θθ==，||||sin 200sin 100sin 33PR AP ππθθθθ⎛⎫⎛⎫=-=-=- ⎪ ⎪⎝⎭⎝⎭，||||cos 200cos AQ AP θθ==，||||cos 200cos 100cos 33AR AP ππθθθθ⎛⎫⎛⎫=-=-=+ ⎪ ⎪⎝⎭⎝⎭， 由余弦定理可知:222||||||2||cos3RQ AQ AR AQ AR π=+-‖,22(200cos )(100cos )2200cos (100cos )cos30000,3πθθθθθθ=++-⨯+=则||RQ =设三条步行道每年能产生的经济总效益(||||)5|| 5.9W W PQ PR RQ =+⨯+⨯，则(200sin 100sin )5θθθ=+-⨯+1000sin()3πθ=++ 当sin()13πθ+=即6πθ=时W 取最大值,最大值为10002022+≈.答：三条步行道每年能产生的经济总效益最高约为2022万元．20. （本题满分16分，第（1）小题4分，第（2）小题6分，第（3）小题6分）解：（1）由题得10λ>，曲线1C 为：22111y x λ+=，又c a =离心率为，1a =，则2c =又因为21112λ⎛⎫⎛⎫=+ ⎪ ⎪ ⎪⎝⎭⎝⎭，因此，12λ=． （2）设()(),,,M M N N M x y N x y ，联立方程()()2222222112101x y x x y x λλλλ⎧+=+++-=⎨=+⎩，得， 因为21λ≠-，222244(1)(1)4λλλ∆=-+-= 则222221,11M N M N x x x x λλλλ--+==++，所以，MN ==213λ=-或. 因此，曲线2C 的方程为：221x y +=或2231x y -=．（3）联立()()222221111i i x y k x x y k x λλ⎧+=⎪+=-⎨=+⎪⎩，得，又0i x >，得()2221111i i i i k k x x x kλλλ-+=-=+，解得，假设存在1322λλλ+=（123,,λλλ不全相等）,使得2132x x x =成立.故()()()()2222132222213111111k k k kk k λλλλλλ--⎛⎫-= ⎪+++⎝⎭， ()()4224213132242242131322112112k k k k k k k kλλλλλλλλλλλλ+-++-=+++++有， ()()22132422421313222411112k k k k k k λλλλλλλλλ+-=-+++++进一步有，()22224224213132244112k k k k k k λλλλλλλλ=+++++化简得，由(),i i i P x y 在第一象限，0ix >且()10i i y k x =+>，得0k > . （i ）20λ=，则13=0λλ+，2131,1x x x ==；（ii ）20λ≠，则4242132132k k λλλλλλ==，得，又因为1322λλλ+=， 则123λλλ==与已知矛盾．综上所述：存在1322λλλ+=（123,λλλ,不全相等），使得2213x x x =成立，此时21x =．21.（本题满分18分，第（1）小题4分，第（2）小题6分，第（3）小题8分）解：（1）12341,2,3,4b b b b ====．（2）当n 为奇数时，139210n n a +⎛⎫=+ ⎪⎝⎭，由函数139210x y +⎛⎫=+ ⎪⎝⎭的单调性可知132n a a <≤， 即3231131,2,221002100n n n a a b ⎛⎤⎛⎤∈-∈-=⎥⎥⎝⎦⎝⎦，得，进一步有， 当n 为偶数时，139210n n a +⎛⎫=- ⎪⎝⎭，由函数139210x y +⎛⎫=- ⎪⎝⎭的单调性可知232n a a ≤<，即77132291,1,11000210002n n n a a b ⎡⎫⎡⎫∈-∈-=⎪⎪⎢⎢⎣⎭⎣⎭，得，进一步有，综上所述：()*21n n b n N n ⎧=∈⎨⎩为奇数为偶数．()*31381921,32190102nn n n n A n B n N n n +⎧⎪⎡⎤⎪⎛⎫=+--=∈⎢⎥⎨⎪⎝⎭⎢⎥⎪⎣⎦⎪⎩为奇数，为偶数， 当k 为偶数时，令819901011901010kkk k B A ⎡⎤⎛⎫⎛⎫->⇒-->⇒>⎢⎥ ⎪⎪⎝⎭⎝⎭⎢⎥⎣⎦无解；当k 为奇数时，令18199140102190101081k kk k B A ⎡⎤⎛⎫⎛⎫->⇒-+>⇒<⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦，91014lg 16.6681k >≈所以，，即min 17k =． 因此，存在k (k 是正整数)，使得k k A B <，且min 17k =． （3）充要条件为：d Z ∈．①若d Z ∈时，由题意对于任意正整数n 均有12n n a b -<恒成立， 且n b Z ∈， 则1122n n n a b a -<<+，1111122n n n a b a +++-<<+，从而11111n n n n n n a a b b a a +++--<-<-+，即111n n b b d +-<--<. 因为,n b d ∈∈Z Z ， 所以10n n b b d +--=，即1n n b b d +-=.因此{}n b 为等差数列，且公差也为()d d Z ∈； ②若{}n b 为等差数列，设公差为()d d Z ''∈，11111111111111|||||||||'|1n n n n n n n a a a b b b b a a b b b b a n d +++++++-=-+-+-≤-+-+-<+，又11111111|||||'|1n n n n a a b b a b b a n d ++++-≥---+->-， 即|'|1|||'|1n d n d n d -≤≤+，亦即11(|||'|)d d n n-≤-≤对任意正整数n 都成立， 所以，|||'|d d =，又d Z d Z '∈∈，得．因此，所求充要条件为Z d ∈．。

1上海中学2024学年第一学期高三年级数学期中2024.11一、填空题（本大题共有12题,满分54分,第1-6题每题4分,第7-12题每题5分）1.若集合{}1,0,1,5,10,20,{|1}A B x lgx =-=<,则A B ⋂=.2.已知全集U R =,集合{}{}|0,||1|3A x x a ,x R B x x ,x R =+≥∈=-≤∈.若()[]24U C A B ,⋂=-,则实数a 的取值范围是.3.已知幂函数()f x 的图像过点222⎛ ⎪⎝⎭,则()f x 的定义域为.4.若函数()(f x xln x =+是偶函数,则a =.5.已知0a >,则()()141a a a--的最小值为.6.已知函数()22f x x log x =+:则不等式()()120f x f +-<的解集为.7.设,,a b c 都是正实数,则"1abc ="是"a b c ++≥"的条件.8.已知函数()()212f x lg x ax =-+在[]13,-上是减函数,则实数a 的取值范围是.9.已知当[]11a ,∈-时,不等式()24420x a x a +-+->恒成立,则x 的取值范围为.10.已知函数()2f x +=,当(]01x ,∈时，()2f x x =,若在区间(]11,-内()()()1g x f x t x =-+有两个不同的零点,则实数t 的取值范围是.11.设,b c 均为实数,关于x 的方程0bx c x++=在区间[)1,+∞上有解,则22b c +的取值范围是.12.设[],01a b ,∈,记()()1111a b S a b b a=++--++,则它的最大值和最小值的差为.2二、选择题（本大题共4题，满分18分，第13-14题4分，第15-16题5分）13.若0a b <<,则下列不等式恒成立的是（）.A.11a b> B.a b-> C.22a b > D.33a b <14.已知函数()()22,01,0x x ax a x f x e ln x x ⎧---<⎪=⎨++≥⎪⎩是R 上的严格增函数,则实数a 的取值范围是（）.A.(]0,-∞B.[]10,-C.[]11,- D.[)0,+∞15.若12,x x 是方程280x ax ++=的两相异实根,则有（）.A.122,2x x >> B.123,3x x >>C.12x x -≤D.12x x +>16.已知定义在R 上的函数()(),f x g x 的导数满足()()''f x g x ≤,给出两个命题:（1）对任意12,x x R ∈,都有()()()()1212f x f x g x g x -≤-;（2）若()g x 的值域为[]()(),1,1m,M f m f M -==,则对任意x R ∈都有()()f x g x =.则下列判断正确的是（）.A.(1)(2)都是假命题B.(1)(2)都是真命题C.(1)是假命题,(2)是真命题D.(1)是真命题，(2)是假命题三、解答题（本大题共有5题，满分78分）17.(本题满分14分)已知三个集合:(){}22|581A x R log x x =∈-+=,{}22822{|21},|190xx B x R C x R x ax a +-=∈==∈-+->.（1）求A B ⋃;（2）已知,A C B C ⋂≠∅⋂=∅,求实数a 的取值范围.318.(本题满分14分)记函数()f x =的定义域为()()(),21A g x lg x b ax ⎡⎤=-+⎣⎦(0,)b a R >∈的定义域为B .（1）求集合A ;（2）若A B ⊆,求,a b 的取值范围.19.（本题满分14分）某个体户计划经销,A B 两种商品，据调查统计，当投资额为()0x x ≥万元时，在经销,A B 商品中所获得的收益分别为()f x 万元与()g x 万元，其中()()()()12(0),6(0)f x a x a g x ln x b b =-+>=+>.已知投资额为零时,收益为零.（1）试求出,a b 的值;（2）如果该个体户准备投入5万元经营这两种商品，请你帮他制定一个资金投入方案，使他能获得最大收益，并求出其收入的最大值.（精确到0.1万元）加入高中数学资料QQ 群734924357，获取更多精品资料！420.(本题满分18分)已知函数()()()(),,0;b f x ln ax g x ln x a b R ==>∈.（1）若,1a e b ==-,求()()f x g x ⋅的最大值;（2）若2a =，求关于x 的不等式()()0g x f x ≤的解集；（3）记()()()F x f x g x =+，对于给定的实数b ，若存在x 满足()1F x ≤，求a 的取值范围.加入高中数学资料QQ 群734924357，获取更多精品资料！521.（本题满分18分）若定义在R 上的函数()y f x =和()y g x =分别存在导函数()'f x 和()'g x .且对任意x 均有()()''f x g x ≥，则称函数()y f x =是函数()y g x =的"导控函数".我们将满足方程()()''f x g x =的0x 称为"导控点"（1）试问函数y x =是否为函数y sinx =的"导控函数"?（2）若函数32813y x x =++是函数3213y x bx cx =++的"导控函数"，且函数3213y x bx cx =++是函数24y x =的"导控函数",求出所有的"导控点";（3）若()x x p x e ke -=+，函数()y q x =为偶函数，函数()y p x =是函数()y q x =的"导控函数"，求证："1"k =的充要条件是"存在常数c 使得()()p x q x c -=恒成立。