2017届上海市高三八校联合调研考试文科数学试卷及答案精品

2017届上海市高三八校联合调研考试语文试卷及答案

2017届上海市高三八校联合调研考试试卷语文试卷考生注意：1．本考试设试卷和答题纸两部分，试卷包括试题与答题要求，所有答题必须涂（选择题）或写（非选择题）在答题纸上，做在试卷上一律不得分。

2．答题纸与试卷在试题编号上是一一对应的，答题时应特别注意，不能错位。

3．考试时间150分钟。

试卷满分150分。

一阅读80分（一）阅读下文，完成第1─6题。

（17分）①日本的“过劳死”和美国由不公平而流行的饥饿情况使哲学家罗素得出了一个令人毛骨悚然的结论：“人类将比预想的更快地消亡。

”②在人类生存的20世纪，几千万人无辜地死去，这证明了罗素判断的正确性。

而且如果我们只把自己看作一个长长的生物链中的一环，前景就将更悲哀。

在我们这个地球上，现在生存着大约4000万不同种类的植物和动物。

然而，在从前某个时期，地球上曾生存着50亿到400亿个物种。

也就是说，只有千分之一的物种仍然存在，创造了地球上99.9％的物种都已死去的生存记录。

大批物种的出现和消失提示了有关我们自己命运的问题。

人类与那99.9％的物种的不同之处，是因为只有我们才懂得如何运用客观环境来为我们服务——改变环境来适应我们这个种群，而不是像其他所有种群那样去适应环境最终还是因为环境的变化如冰河时代的到来而消失。

因而，人类是独一无二的，他是造物主，而不是自己命运的产物。

③然而我们的智慧却并没有保证我们所使用的方法都是明智的。

我们的技术能够让我们建造出所希望的任何世界，但我们迄今创造了一个值得生活在其中的世界吗？或者更确切的讲，我们是不是在使我们生活的星球像环境学家所警告的那样不适合居住呢?④如果是这样的话，那么我们是不是正在复制一个如星象学家米奇奥·卡库所总结的自我毁灭的银河模型呢?卡库认为在我们银河中的2000亿颗星球当中成千上万的星球上有大量的智慧生物存在。

卡库总结道：“也许其他的文明是自我毁灭的。

当然这已无法知道，但是银河系中我们这个部分的明显贫瘠可能对准了那个方向。

2017年全国高考卷文科数学试题及答案详细解析(选择、填空、解答全解全析) 精品

2017年普通高等学校招生全国统一考试文科数学(必修+选修I)解析版本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。

第Ⅰ卷1至2页。

第Ⅱ卷3至4页。

考试结束后，将本试卷和答题卡一并交回。

第Ⅰ卷注意事项：1．答题前，考生在答题卡上务必用直径0.5毫米黑色墨水签字笔将自己的姓名、准考证号填写清楚，并贴好条形码.请认真核准条形码上的准考证号、姓名和科目.2．每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，如需改动,用橡皮擦干净后，再选涂其他答案标号，在试题卷上作答无效.3．第Ⅰ卷共l2小题，每小题5分，共60分。

在每小题给出的四个选项中，只有一项是符合题目要求的. 一、选择题 (1)设集合U={}1,2,3,4,{}1,2,3,M ={}2,3,4,N =则U =（M N ）Ið（A ）{}12，（B ）{}23，（C ）{}2，4 （D ）{}1，4 【命题意图】本题主要考查集合交并补运算.【解析】{2,3},(){1,4}U M N C M N =∴=【答案】D(2)函数0)y x =≥的反函数为（A ）2()4x y x R =∈ （B ）2(0)4x y x =≥（C ）24y x =()x R ∈ （D ）24(0)y x x =≥ 【命题意图】本题主要考查反函数的求法.【解析】由0)y x =≥反解得24y x =,又原函数的值域为0y ≥,所以函数0)y x =≥的反函数为2(0)4x y x =≥.【答案】B(3)设向量,a b 满足||||1a b == ,12a b ⋅=-r r ,则2a b +=（A（B（C（D【命题意图】本题主要考查平面向量的数量积与长度的计算方法.【解析】2221|2|||44||14()432a b a a b b +=+⋅+=+⨯-+= ,所以2a b +=【答案】B(4)若变量x ，y 满足约束条件63-21x y x y x +≤⎧⎪-≤⎨⎪≥⎩，则=23z x y +的最小值为（A ）17 （B ）14 （C ）5 （D ）3 【命题意图】本题主要考查简单的线性规划.【解析】作出不等式组表示的可行域，从图中不难观察当直线=23z x y +过直线x=1与x-3y=-2的交点（1，1）时取得最小值,所以最小值为5. 【答案】C(5)下面四个条件中，使a b >成立的充分而不必要的条件是（A ）1a b +＞（B ）1a b -＞（C ）22a b ＞（D ）33a b ＞【命题意图】本题主要考查充要条件及不等式的性质.【解析】即寻找命题P ,只需由P a b ⇒>,且由a b >不能推出P ,可采用逐项验证的方法，对A ，由1a b +＞，且1b b +>，所以a b >，但a b >时，并不能得到1a b +＞，故答案为A 。

2017年上海高考数学真题试卷(word解析版)

6 绝密★启用前2017 年普通高等学校招生全国统一考试（上海卷）数学试卷（满分 150 分，考试时间 120 分钟）1、考生注意2、1.本场考试时间 120 分钟，试卷共 4 页，满分 150 分，答题纸共 2 页.3、2.作答前，在答题纸正面填写姓名、准考证号，反面填写姓名，将核对后的条形码贴在答题纸指定位置.4、3.所有作答务必填涂或书写在答题纸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. 5、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 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 A = {1, 2,3, 4}，集合 B = {3, 4,5}，则 AB =2. 若排列数P m= 6 ⨯ 5 ⨯ 4 ，则m =3. 不等式x -1> 1 的解集为 x4. 已知球的体积为36π ，则该球主视图的面积等于5. 已知复数 z 满足 z + 3= 0 ，则| z | =z 6. 设双曲线 x 9- y2 b 2 = 1 (b > 0) 的焦点为 F 1 、 F 2， P 为该双曲线上的一点，若| PF 1 | = 5 ，则| PF 2 | =7. 如图，以长方体 ABCD - A 1B 1C 1D 1 的顶点 D 为坐标原点，过 D 的三条棱所在的直线为坐标轴，建立空间直角坐标系，若 DB 1 的坐标为(4,3, 2) ，则 AC 1 的坐标为- ⎧⎪3x -1, x ≤ 08. 定义在(0, +∞) 上的函数 y = f (x ) 的反函数为 y = f 1(x ) ，若 g (x ) = ⎨ ⎪⎩ f (x ), 为 x > 0奇函数，则 f -1(x ) = 2 的解为119. 已知四个函数：① y = -x ；② y =- ；③ xy = x 3 ；④ y = x 2 . 从中任选 2 个，则事件“所选 2 个函数的图像有且仅有一个公共点”的概率为10. 已知数列{a } 和{b } ，其中a = n 2 ， n ∈ N * ，{b } 的项是互不相等的正整数，若对于nnnn任意n ∈ N * ，{b } 的第a 项等于{a } 的第b 项，则lg(b 1b 4b 9b 16 ) =nnnnlg(b 1b 2b 3b 4 )2⎨2x + 3y = 4 n n 211. 设a 、 a ∈ R ，且1+1= 2 ，则| 10π - α - α |的最小值等于122 + sin α2 + sin(2α ) 121212. 如图，用 35 个单位正方形拼成一个矩形，点 P 1 、 P 2 、 P 3 、 P 4 以及四个标记为“#”的点在正方形的顶点处，设集合Ω = {P 1 , P 2 , P 3 , P 4 }，点P ∈Ω，过 P 作直线l P ，使得不在l P 上的“#”的点分布在l P 的两侧. 用 D 1 (l P ) 和 D 2 (l P ) 分别表示l P 一侧和另一侧的“#”的点到l P 的距离之和. 若过 P 的直线l P 中有且只有一条满足 D 1 (l P ) = D 2 (l P ) ，则Ω 中所有这样的 P 为二. 选择题（本大题共 4 题，每题 5 分，共 20 分）13. 关于 x 、 y 的二元一次方程组⎧x + 5y = 0 ⎩的系数行列式 D 为（）0 5 1 0 A.B. 4 32 4 1 5 6 0 C. D.2 35 414. 在数列{a } 中， a = (- 1)n ， n ∈ N * ，则lim a （）n n2n →∞ n A. 等于- 1 2 B. 等于 0 C. 等于 12D. 不存在15. 已知a 、b 、c 为实常数，数列{x } 的通项 x = an 2+ bn + c ，n ∈ N * ，则“存在k ∈ N * ，使得 x 100+ k 、 x 200+ k 、 x 300+ k 成等差数列”的一个必要条件是（） A. a ≥ 0B. b ≤ 0C. c = 0D. a - 2b + c = 0x 2y 2 16. 在平面直角坐标系 xOy 中，已知椭圆C 1 : 36 + 4= 1 和C : x 2 + y 9 = 1 . P 为C 1 上的动点，Q 为C 2 上的动点， w 是OP ⋅ OQ 的最大值. 记Ω = {(P ,Q ) | P 在C 1 上，Q 在C 2 上，且OP ⋅ OQ = w }，则Ω 中元素个数为（） A. 2 个B. 4 个C. 8 个D. 无穷个三. 解答题（本大题共 5 题，共 14+14+14+16+18=76 分）17. 如图，直三棱柱 ABC - A 1B 1C 1 的底面为直角三角形，两直角边 AB 和 AC 的长分别为 4 和 2，侧棱 AA 1 的长为 5.（1）求三棱柱 ABC - A 1B 1C 1 的体积；（2）设 M 是 BC 中点，求直线 A 1M与平面 ABC 所成角的大小.219 2 ⎪⎩ny18. 已知函数 f (x ) = cos 2 x - sin 2 x + 1， x ∈ (0,π ) .2（1）求 f (x ) 的单调递增区间；（2）设△ABC 为锐角三角形，角 A 所对边a = ，角 B 所对边b = 5 ，若 f ( A ) = 0 ，求△ABC 的面积.19. 根据预测，某地第n (n ∈ N * ) 个月共享单车的投放量和损失量分别为a 和b （单位：辆），nn⎧⎪5n 4 +15, 1 ≤ n ≤ 3其中a n = ⎨-10n + 470, ， b n = n + 5 ，第n 个月底的共享单车的保有量是前n 个月的 n ≥ 4 累计投放量与累计损失量的差.（1）求该地区第 4 个月底的共享单车的保有量；（2）已知该地共享单车停放点第n 个月底的单车容纳量 S = -4(n - 46)2 + 8800（单位：辆）. 设在某月底，共享单车保有量达到最大，问该保有量是否超出了此时停放点的单车容纳量？20. 在平面直角坐标系 xOy 中，已知椭圆Γ : x 2 + 24= 1 ， A 为Γ 的上顶点， P 为Γ 上异于上、下顶点的动点， M 为 x 正半轴上的动点.（1）若 P 在第一象限，且| OP | = ，求 P 的坐标；8 3 P ( , ) 5 5，若以 A 、P 、M 为顶点的三角形是直角三角形，求 M 的横坐标；（3）若| MA | = | MP | ，直线 AQ 与Γ 交于另一点 C ，且 AQ = 2 A C ， PQ = 4PM ，求直线 AQ 的方程.21. 设定义在 R 上的函数 f (x 1) ≤ f (x 2 ) .f (x ) 满足：对于任意的 x 1 、 x 2 ∈ R ，当 x 1 < x 2 时，都有（2）设（1）若f (x) =ax3+1，求a 的取值范围；（2）若f (x) 为周期函数，证明：f (x) 是常值函数；（3）设f (x) 恒大于零，g(x) 是定义在R 上、恒大于零的周期函数，M 是g(x) 的最大值.函数h(x) =f (x)g(x) .证明：“h(x) 是周期函数”的充要条件是“ f (x) 是常值函数”.6 2 2017 年普通高等学校招生全国统一考试上海--数学试卷考生注意1. 本场考试时间 120 分钟，试卷共 4 页，满分 150 分，答题纸共 2 页.2. 作答前，在答题纸正面填写姓名、准考证号，反面填写姓名，将核对后的条形码贴在答题纸指定位置.3. 所有作答务必填涂或书写在答题纸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. 已知集合 A ={1, 2, 3, 4}, B = {3, 4, 5} ,则 AB = .【解析】本题考查集合的运算，交集，属于基础题【答案】{3, 4}2. 若排列数P m = 6⨯ 5⨯ 4 ,则m = . 【解析】本题考查排列的计算，属于基础题【答案】3x -1 3. 不等式x> 1的解集为.【解析】本题考查分式不等式的解法，属于基础题【答案】(-∞,0)4. 已知球的体积为36π ，则该球主视图的面积等于.【解析】本题考查球的体积公式和三视图的概念，4π R 3 = 36π ⇒ R = 3 ,3所以 S = π R 2 = 9π ，属于基础题【答案】9π5. 已知复数 z 满足 z +3 = 0 ，则 z = .z【解析】本题考查复数的四则运算和复数的模， z + 3= 0 ⇒ z 2 = -3 设 z = a + bi ，z则 a 2- b 2+ 2abi = -3 ⇒ a = 0, b = ± 3i ,z =，属于基础题【答案】6. 设双曲线x - y 29 b 2= 1(b > 0) 的焦点为 F 1、F 2 , P 为该双曲线上的一点.若 PF 1= 5 ,则 a 2 + b 2 34 PF 2 = .【解析】本题考查双曲线的定义和性质，PF 1 - PF 2 = 2a = 6 （舍），PF 2 - PF 1 = 2a = 6 ⇒ PF 2 = 11【答案】117. 如图，以长方体 ABCD - A 1B 1C 1D 1 的顶点 D 为坐标原点，过 D 的三条棱所在的直线为坐标轴，建立空间直角坐标系.若 DB 1 的坐标为(4, 3, 2) ,则 AC 1 的坐标是.【解析】本题考查空间向量，可得 A (4，0，0)，C 1(0，3, 2) ⇒ AC 1 = (-4，3，2) ,属于基础题【答案】(-4，3，2)8. 定义在(0, +∞) 上的函数 y =数，则 f -1(x )=2 的解为.⎧3x -1, x ≤ 0, f (x ) 的反函数 y = f -1(x ) .若 g (x ) = ⎨ ⎩ f (x ), x > 0 为奇函【解析】本题考查函数基本性质和互为反函数的两个函数之间的关系，属于中档题x > 0, -x < 0, g (-x ) = 3-x -1 = -g (x ) ⇒ g (x ) = 1- 1 3x,所以 f (x ) = 1- 1,3x 当 x = 2 时， f (x ) = 8,所以 f 9(8) = 29 【答案】 x = 89119. 已知四个函数：① y = - x ；② y =-；③ y = x 3；④ y = x 2.从中任选 2 个，则事件“所x选 2 个函数的图像有且仅有一个公共点”的概率为.【解析】本题考查事件的概率，幂函数的图像画法和特征，属于基础题总的情况有： C 2 = 6 种，符合题意的就两种：①和③，①和④-11 2 3 4 2 π nnnn1⎧ π ⎨ 1 【答案】310. 已知数列{a } 和{b } ,其中 a = n 2 , n ∈ N * ，{b } 的项是互不相等的正整数.若对于任意n ∈ N *，{b } 中的第 a 项等于{a } 中的第b 项，则 lg (b 1b 4b 9b 16 )= .nnn lg (b 1b 2b 3b 4 )【解析】本题考查数列概念的理解，对数的运算，属于中档题由题意可得： b = a ⇒ b = (b )2 ⇒ b = b 2 , b = b 2 , b = b 2 ,b = b 2 ，a nb nn 2n1 1 42 93 16 4lg (b 1b 4b 9b 16 ) lg (b 1b 2b 3b 4 ) lg (bb b b )2lg (b 1b 2b 3b 4 )【答案】211. 设α1，α2 ∈ R ,且12 + sin α+2 + sin(2α = 2 ，则 10π - α)1 - α2的最小值等于. 12【解析】考查三角函数的性质和值域，1∈ ⎡1 ,1⎤，1 ∈ ⎡1 ,1⎤2 + sin α1 ⎢⎣3 ⎥⎦ 2 + sin(2α2 ) ⎢⎣3 ⎥⎦ ，要使 1 + 1 = 2 ⎧ 1 =1 ⎪ 2 + sin α1 则⎨ α1 = - + 2k 1⎪ , k , k ∈ Z 2 + sin α 2 + sin(2α ) 1 π 1 2 1 2 ，⎪ =1 ⎪ α = - + k π ⎪⎩ 2 + sin(2α2 )⎪⎩ 2 4 2 10π -α -α= 10π + 3π - (2k + k )π = π 当2k + k =11时成立 1 2 minπ4 1 2 min4 , 【答案】 412. 如图，用 35 个单位正方形拼成一个矩形，点 P 1, P 2 , P 3 , P 4 以及四个标记为“▲”的点在正方形的顶点处.设集合Ω={P 1, P 2 , P 3 , P 4 } ，点 P ∈Ω .过 P 作直线l P ，使得不在l P 上的“▲” 的点分布在l P 的两侧.用 D 1 (l P ) 和 D 2 (l P ) 分别表示l P 一侧和另一侧的“▲”的点到l P 的距离之和.若过 P 的直线l P 中有且只有一条满足 D 1 (l P )=D 2 (l P ) ，则Ω 中所有这样的 P 为.⇒ n所以 = =21 2⎩ ⎨【解析】本题考查有向距离，以左下角的顶点为原点建立直角坐标系。

上海市八校2017高三联考英语试题

2017届高三年级八校联合调研英语试卷2016年11月(满分140分，考试时间120分钟)第I卷（共90分）II. Grammar and VocabularySection A(10×1=10分)Directions: Read the following passage. Fill in the blanks to make the passage coherent. For the blanks with a given word, fill in each blank with the proper form of the given word. For the other blanks, fill in each blank with one proper word. Make sure that your answers are grammatically correct.Have you ever seen an old movie called Three Coins in the Fountain? It is about three young American women (21) _______(search) for permanent romance in Rome and they all find it. Far-fetched Hollywood? Well, from the world history point of view, romance did, in fact, set down its roots in Rome.The word romance evolved in Latin from Roma to Romanicus of the Roman language, to the Old French romanz escrive, (22) _______means “to write in a Romance language,” and on to the English romance.The Romance languages (23) ____________(compose) of seven groups of languages that all have Latin (24) ______their basis. These languages include French, Italian, Spanish and Portuguese. The common people in ancient Rome spoke (25)________ is referred to as Vulgar Latin, an informal speech, as opposed to the classical Latin of the more educated. Most language experts agree that Vulgar Latin is the chief source of the Romance languages.Medieval Romances were tales (26) __________(write)primary in French verse about brave heroes. The notion of having a romance with another person is thought (27) __________(develop) sometime during the Middle Ages. In the late 18th century and on through the 19th, a romance was not a love story (28) _________ a work of prose fiction that contained far-fetched, mysterious events. Romances of this period (29) _________(include) English Gothic novels like The Castle of Otranto by Horace Walpole.What exactly is a twentieth-century romance ? Does it have any relationship with the lively, popular novels written today, with their fantastic plots of love affairs? Or did the playwright Oscar Wilde have it right in The Picture of Dorian Gray: “ When one is in love, one always begins by deceiving (30) _________, and one always ends by deceiving others. That is what the world calls a romance.”Section B（10×1=10分）Directions: Complete the following passage by using the words in the box. Each word can be used only once. Note that there is one word more than you need.A.astonishinglyB. surroundingC. collapseD. unnoticedE. interruptedF. previouslyG. congratulateH. predictionsI. potential J. producing K. propertiesIn the wake of the historic announcement of the discovery of gravitational waves on February 11, 2016 by the Laser Interferometer Gravitational-Wave Observatory (LIGO), British physicist and black hole theorist Stephen Hawking was quick to ____31___ the US-led collaboration, sharing his excitement for the historic news.According to Hawking, these results confirm several very important ____32___ of Einstein’s theory of general relativity and it also confirms the existence of gravitational waves directly.As is becoming clear, the direct detection of these ripples in space time not only confirms Einstein’s famous theory of general theory but it also opens our eyes to a(n) _33________ “dark” universe. Astronomers employ the electromagnetic spectrum（电磁光谱）to study the universe, but objects that do not radiate in the electromagnetic spectrum will go ___34____. But now we know how to detect gravitational waves, which can help us detect and study some of the most energetic cosmic phenomena.“Gravitat ional waves provide a completely new way of looking at the universe and the ability to detect them has the ___35___ to revolutionize astronomy” said Hawking. “The discovery is the first observation of black holes merging. The observed __36____ of this system are consistent with predictions about black holes that I made in 1970 in Cambridge.”However, this discovery also presents a puzzle for astrophysicists. The mass of each of the black holes are larger than expected for those formed by the gravitational __37_____ of a star---so how did both of these black holes become so massive?This question touches on one of the biggest mysteries ___38___ black hole evolution. Currently, astronomers are having a hard time understanding how black holes grow to be so massive. On the one end of the scale, there are “stellar mass（恒星质量）” black holes that form immediately after a massive star explodes, ___39____ an extremely bright light. And we also have an abundance of evidence for the existence of the super-massive that live in the centers of most galaxies. There is a disconnect, however. If black holes grow by merging and consuming stellar matter, there should be evidence of black holes of all sizes, but “intermediate mass” black holes and black holes of a few dozen solar masses are ____40____ rare, throwing some black holes evolution theories into doubt.One thing is clear, however. This is the first time that we’ve acquired direct evidence of a black hole merger. So it’s good to know we’re on the right track.III. Reading ComprehensionSection A（15×1=15分）Directions: 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.Cowboy or spaceman ? A dilemma for a children’s party, perhaps. But also a question for economists, argued Kenneth Boulding, in an essay published in 1966. We have run our 41 , he warned, like cowboys on the open grassland: taking and using the world’s resources, 42 ____ that more lies over the horizon. But the Earth is 43 a grassland than a spaceship---a closed system, alone in space, carrying exhaustible supplies. We need, said Boulding, an economics that takes seriously the idea of environmental 44 . In the half century since his essay, a new movement has respo nded to his challenge. “Ecological economists,” as they call themselves, want to 45 __ its aims and assumptions. What do they say -- and will their ideas take off?To its 46 , ecological economics is neither ecology nor economics, but a mix of both. Their starting point is to recognize that the human economy is part of the natural world. Our environment, theynote, is both a source of resources and a sink for wastes. But it is 47 in traditional textbooks, where neat diagrams trace the flows between firms, households and the government as though nature did not exist. That is a huge mistake.There are two ways our economies can grow, ecological economists point out: through technological change, or through maximum use of resources. Only the 48 , they say, is worth having. They are suspicious of GDP (gross domestic product), a simple 49 which does not take into account resource exhaustion, unpaid work and countless other factors. 50 , they advocate more holistic approaches, such as GPI (genuine progress indicator)，a composite（复合的）index that include things like the cost of pollution, deforestation and car accidents. While GDP has kept growing, global GPI per person 51 in 1978: by destroying our environment, we are making ourselves poorer, not richer. The solution, according to experts, lies in a “steady-state” economy, where the use of materials and energy is held 52 .Mainstream economists are not 53 . GPI, they point out, is a subjective standard. And talk of limits to growth has had a bad press since the days of Thomas Malthus, who predicted in the 18th century, wrongly, that overpopulation would lead to famine. Human beings find solutions to some of the most annoying problems. But ecological economists 54 self-satisfaction. In 2009, a paper in Nature argued that human activity is already 55 safe planetary boundaries on issues such as biodiversity and climate change. That suggests ecologist economists are at least asking some important questions, even if their answers turn out to be wrong.41. A. grassland B. nation C. economy D. spaceship42. A. ignorant B. confident C. astonished D. anxious43. A. less B. smaller C. more D. larger44. A. movements B. influences C. limits D. threats45. A. reject B. realize C. resemble D. revolutionize46. A. challengers B. learners C. advocates D. professors47. A. addressed B. ignored C. opposed D. reflected48. A. advanced B. former C. latter D. scientific49. A. number B. product C. idea D. measure50. A. In addition B. For example C. In other words D. In its place51.A. peaked B. plunged C. persisted D. paused52.A. sufficient B. efficient C. constant D. adequate53.A. impressed B. involved C. concerned D. appointed54.A. call for B. contribute to C. warn against D. refer to55.A. setting B. overstepping C. extending D. redrawingSection B（11×2=22分）Directions:Read the following three 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)Every April I am troubled by the same concern -- that spring might not occur this year. The landscape looks dull, with hills, sky and forest appearing gray. My spirits ebb, as they did during an April snowfallwhen I first came to Maine 15 years ago. "Just wait," a neighbor advised. "You'll wake up one morning and spring will just be here."And look, on May 3 that year I awoke to a green so amazing as to be almost electric, as if spring were simply a matter of flipping a switch. Hills, sky and forest revealed their purples, blues and green. Leaves had unfolded and daffodils were fighting their way heavenward.Then there was the old apple tree. It sits on an undeveloped lot in my neighborhood. It belongs to no one and therefore to everyone. The tree's dark twisted branches stretch out in unpruned（未经修剪的）abandon. Each spring it blossoms so freely that the air becomes filled with the scent of apple.Until last year, I thought I was the only one aware of this tree. And then one day, in a bit of spring madness, I set out with pruner to remove a few disorderly branches. No sooner had I arrived under the tree than neighbors opened their windows and stepped onto their porches. These were people I barely knew and seldom spoke to, but it was as if I had come uninvited into their personal gardens.My mobile-home neighbor was the first to speak."You're not cutting it down, are you?" she asked anxiously. Another neighbor frowned as I cut off a branch. "Don't kill it, now," he warned. Soon half the neighborhood had joined me under the apple tree. It struck me that I had lived there for five years and only now was learning these people's names, what they did for a living and how they passed the winter. It was as if the old apple tree was gathering us under its branches for the purpose of both acquaintanceship and shared wonder. I couldn't help recalling Robert Frost's words:The trees that have it in their pent-up budsTo darken nature and be summer woodsOne thaw led to another. Just the other day I saw one of my neighbors at the local store. He remarked how this recent winter had been especially long and complained of not having seen or spoken at length to anyone in our neighborhood. And then, he looked at me and said, "We need to prune that apple tree again."56. By saying that “my spirits ebb” (Para. 1), the author means that _________________.A.he feels relievedB. he feels blueC. he is surprisedD. he is tired57. The apple tree mentioned in the passage is most likely to _________________.A. be regarded as a delight in the neighborhoodB. have been abandoned by its original ownerC. have been neglected by everyone in the communityD. be appealing only to the author58. In Para. 4, “neighbors opened their windows and stepped onto their porches” probably because ___________________________.A.they were surprised that someone unknown was pruning the treeB.they wanted to prevent the author from pruning the treeC.they were concerned about the safety of the treeD.they wanted to get to know the author59. It can be inferred that the author’s neighbor mentioned in the last paragraph most cared about _______________.A.when spring would arriveB. how to pass the long winterC. the neighborhood gatheringD. the pruning of the apple tree(B)Mount Cook National Park is home of the highest mountains and the longest glaciers. It is alpine(高山) in the purest sense---with skyscraping peaks, glaciers and permanent snow fields, all set under a star-studded sky.Key HighlightsAlthough it includes 23 peaks over 3,000 metres high, this park is very accessible. State Highway 80 leads to Mt Cook Village which is situated beside scenic Lake Pukaki and provides a comfortable base for alpine activities. Far from city lights, the stargazing here is magnificent—Aoraki Mount Cook National Park forms the majority of New Zealand's only International Dark Sky Reserve.Mountaineers regard the area to be the best climbing region, while less skilled adventurers find plenty of satisfaction with the mountain walks that lead to alpine tarns, herb fields and spectacular glacier views. Encounters with cheeky kea (mountain parrots) are part of the fun.Key ActivitiesMountain walksThere are 10 short walks beginning near the village. All tracks are formed and well marked. The Red Tarns Track, Kea Point and the Hooker Valley Track each take around two hours return. For more experienced alpine hikers, there are three mountain pass routes—over the Mueller, Copland and Ball passes.Glacier viewing and skiingHelicopters and ski－planes provide access to the park's fabulous glaciers. The Tasman Glacier is an excellent choice for intermediate skiers, while the Murchison, Darwin and Bonney glaciers promise excitement for advanced skiers. From October until May, you can explore the Tasman Glacier's terminal lake by boat.MountaineeringClimbing Mount Cook remains the ultimate challenge, but there are many other peaks to tempt experienced climbers. Tasman, Malte Brun, Elie de Beaumont, Sefton and La Perouse are quite popular..Key Tips●Climbers don't require permits, but are requested to complete a trip intentions form.●Local guides are available for climbing, walking and glacier skiing.●Winter climbing is an extreme sport—only recommended for well-prepared, experienced mountaineers.●The weather can change very suddenly—be prepared for heavy rainfall, snow and/or high winds.●The park has an airport serving domestic commercial flights and scenic flight operators.60. Which is one of the characteristics of Mount Cook National Park?A. It is alpine in the purest sense and hard to reach.B. It provides star-shining night skies for visitors.C. It attracts less skilled climbers to all alpine activities.D. It guarantees visitors a sight of cheeky kea.61. Mike is an experienced adventurer and may find ________ the most exciting.A. Mountaineering on Elie de BeaumontB. Mountain walks via Hooker Valley TrackC. Skiing on Tasman GlacierD. Climbing Mount Cook62. If you are a visitor to the park, you should ________．A. properly evaluate your own experience and skillB. get your permit prepared before you start to climbC. hire local guides to help you to train for climbingD. avoid exploring glaciers in winter(C)How many really suffer as a result of labor market problems? This is one of the most critical yet debatable social policy questions.In many ways, our social statistics overstate the degree of hardship. Unemployment does not have the same horrible consequences today as it did in the 1930’s when most of the unemployed were primary breadwinners, when income and earnings were usually much closer to the margin of survival, and when there were fewer effective social programs for those failing in the labor market. Increasing wealth, the rise of families with more than one wage earner, the growing dominance of secondary earners among the unemployed and improved social welfare protection have unquestionably relieved the consequences of joblessness. Earnings and income data also overestimate the scale of hardship. Among the millions with hourly earnings at or below the minimum wage level, the majority are from multiple-earner, relatively well-off families. Most of those counted by the poverty statistics are elderly or handicapped or have family responsibilities which keep them out of the labor force, so the poverty statistics are by no means an accurate indicator of labor market problems.Yet there are also many ways our social statistics underestimate the degree of labor-market-related hardship. The unemployment counts exclude the millions of fully employed workers whose wages are so low that their families remain in poverty. Low wages and repeated or long-time unemployment frequently interact to weaken the capacity for self-support. Since the number experiencing joblessness at some time during the year is several times that unemployed in any month, those who suffer as a result of forced idleness can equal or exceed average annual unemployment, even though only a minority of the jobless in any month really suffer. For every person counted in the monthly unemployment totals, there is another working part-time because of the inability to find full-time work, or else outside the labor force but wanting a job. Finally, income transfers in our country have always focused on the elderly, disabled, and dependent, neglecting the needs of the working poor, so that the dramatic expansion of cash and non-cash transfers does not necessarily mean that those failing in the labor market are adequately protected.As a result of such conflicting evidence, it is uncertain whether those suffering seriously as a result of labor market problems number in the hundreds of thousands or the tens of millions, and, hence, whether high levels of joblessness can be tolerated or must be counteracted(抵消）by job creation and economic stimulation. There is only one area of agreement in this debate—that the existing poverty, employment, and earnings statistics are inadequate for one of their primary applications, measuring the consequences of labormarket problems.63. In Paragraph 2, the author contrasts the 1930’s with the present in order to show that_____________.A. more people were unemployed in the 1930’sB. unemployment is more intolerable todayC. social programs are more in need nowD. i ncome level has increased since the 1930’s64.Which of the following is true according to the passage?A.A majority of low-wage workers receive earnings from more than one job.B.Repetition of short-term unemployment mainly contributes to people’s loss of workin g capacity.C. Many unemployed people are from families where other members are working.D. Labor market hardship is understated because fewer individuals are jobless than counted.65.It can be inferred from the passage that the effect of income transfers is often not felt by _________________.A. those doing a low-paid, part-time jobB. children in single-earner familiesC. workers who have just retiredD. full-time workers who become unemployed66. Which of the following is the principal topic of the passage?A. What causes labor market problems that result in suffering.B. Why income statistics are imprecise in measuring degrees of poverty.C. When poverty, employment, and earnings figures agree with each other.D. How statistics give an unclear picture of the labor-market-related suffering.Section C（4×2=8分）Directions: Read the following passage. 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. Even being good at getting others to fight most efficiently is not being civilized.B. Most people believe those who have conquered the most nations are the greatest.C. However, every year conflicts between countries and nations still claim thousands of lives.D. And not only has it won, but also because it has won, it has been in the right.E. So there has been little time to learn in, but there will be oceans of time in which to learn better.F. People don’t fight and kill each other in the streets, but nations still behave like savages.Most of the people who appear most often and most gloriously in the history books are great conquerors and generals, while the people who really helped civilization forward are often never mentioned. We do not know who first set a broken leg, or launched a seagoing boat, or calculated the length of the year but we know all about the killers and destroyers. People think so much of them that on all the highestpillars in the great cities of the world you will find the figure of a conqueror or a general. ____67_________It is possible they are, but they are not the most civilized. Animals fight, so do savages; so to be good at fighting is to be good in the way an animal or a savage is good, but it is not to be civilized. ____68_______. People fight to settle quarrels. Fighting means killing, and civilized peoples ought to be able to find some ways of settling their disputes other than by seeing which side can kill off greater number of the other side, and then saying that the side which has killed most has won. ___69______. For that is what going to war means; it means power is right.This is what the story of mankind has been like. But we must not expect too much. After all, the race of men has only just started. From the point of view of evolution, human beings are very young indeed, babies of a few months old. Scientists assume that there has been life of some sort on the earth for about twelve hundred million years; but there have been men for only one million years, and there has been civilized men for about eight thousand years.____70_____. Taking man’s civilized past at about seven or eight hours, we may estimate his future at about one hundred thousand years. Thus mankind is only at the beginning of its whole a pretty beastly business, a business of fighting and killing. We must not expect even civilized peoples not to have done these things. All we can ask is that they will sometimes have done something else.第II卷(共50分）I.Summary (10分)Directions: Reading the following passage. Summarize the main idea and the main point(s) of the passage with no more than 60 words. Use your own words as far as possible.It's not piano lessons or dance classes. Nowadays, the biggest extra-curricular activity is going to a tutor. "I spend about 800 Canadian dollars a month on tutors. It's costly," says Pat, a mother in Canada. However, she adds, "after finding out half my daughter's class had tutors, I felt like my child was going to fall behind because everyone else seemed to be ahead"Shelley, a mother of three, also has tutors constantly coming in and out of her home. "When I used to sit down with my children, it was hard to get them focused. I was always yelling. When I got a tutor once a week, they became focused for one entire hour and could get most of their homework done."Tutoring isn't simply a private school phenomenon. Nor is it geared only toward lower-achieving students. In Canada alone, seven percent of high school students reported using a tutor in 2010. That increased to 15 percent last year.Overall, parents hire tutors because they are worried schools are not meeting their expectations, but there is also a cultural shift. A special value is placed on education in Asia, where tutoring is viewed as an extension of the school day. As a large number of Asians emigrated to the West over the recent years, their attitudes towards education have had an impact.Another reason for the growth in business is parental frustration and their packed schedules. "A lot of parents just don’t have time to help their children with homework," says Julie Diamond, president of an American tutoring company. "Others couldn't help their children after Grade 3."There has been a shift in the attitudes, too. "Children used to get bullied (欺侮) for having a tutor,"Diamond says. "Now it's becoming the norm to have one."Children don't seem to mind that they have a tutor. One parent feels surprised that so many of her child’s classmates have tutors. "For the amount we pay in tuition, they should have as much extra help as they need," she says. Still, she’s now thinking of getting a tutor. Why? Her daughter has actually asked for one.II. Translation（3+3+4+5=15分）Directions: Translate the following sentences into English, using the words given in the brackets.1. 没过多久，失主就来认领他的行李了。

2017年高考数学上海卷含答案

数学试卷第 6页（共 14页）
【解析】an n2 ，n N* ，若对于一切 n N* ，bn 中的第 an 项恒等于 an 中的第 bn 项，
可得 ban abn (bn )2 ．于是 b1 a1 1 ， (b2 )2 b4 ， (b3 )2 b9 ， (b4 )2 b16 .即可得出.
C2 : {(
x2 P,
Q
y2 1 9 )} ，P
．P 为 C1 上的
在 C1 上，Q 在（）
A.2 个
B.4 个
C.8 个
D.无穷个
数学试卷第 2页（共 14页）
三、解答题：本大题共 5 题，共 14+14+14+16+18=76 分. 17.如图，直三棱柱 ABC A1B1C1 的底面为直角三角形，两直角边 AB 和 AC 的长分别为
故答案为： 9π ．
【考点】球的体积公式，以及主视图的形状和面积求法。 5.【答案】 3 【解析】设 z a bi(a,b R) ，代入 z2 3 ，由复数相等的条件列式求得 a，b 的值得
答案．
数学试卷第 5页（共 14页）
【考点】复数代数形式的乘除运算。 6.【答案】11 【解析】根据题意，由双曲线的方程可得 a 的值，结合双曲线的定义可得 || PF1 | | PF2 || 6 ，解可得 | PF2 | 的值，即可得答案．【考点】双曲线的几何性质。 7.【答案】 (4,3, 2) 【解析】解：如图，以长方体 ABCD﹣A1B1C1D1 的顶点 D 为坐标原点，过 D的三条棱所在的直线为坐标轴，建立空间直角坐标系， ∵ DB1 的坐标为 (4,3, 2) ，∴ A(4,0,0) ， C1(0,3, 2) ， ∴ AC1 (4,3, 2) . 故答案为： (4,3, 2) ．

上海市高三数学八校联合调研考试试题文沪教版

2014届高三年级八校联合调研考试试卷数学（文科）一、填空题（本题满分56分）本大题共有14题，要求在答题纸相应题序的空格内直接填写结果，每个空格填对得4分，否则一律得零分． 1. 在复平面上，复数()232i -对应的点到原点的距离为．2. 已知函数()44sin cos f x x x ωω=-()0>ω的最小正周期是π，则=ω ．3. 向量在向量方向上的投影为．4. 直线220x y -+=过椭圆22221x y a b+=的左焦点1F 和一个顶点B ，则椭圆的方程为．5. 已知直线l 的法向量为()1,2=，则该直线的倾斜角为．（用反三角函数值表示）6. 已知正数,a b 满足2a b +=，则行列式111111ab++的最小值为．7. 阅读右边的程序框图，如果输出的函数值y 在区间⎥⎦⎤⎢⎣⎡141,内，则输入的实数x 的取值范围是． 8. 设αβ、是一元二次方程022=+-m x x 的两个虚根.若||4αβ=，则实数=m ．9. 在△ABC 中，A B C 、、所对边分别为a b c 、、.若tan 21tan A cB b+=，则A = ．10. 已知数列{}n a 的首项12a =，其前n 项和为n S .若121n n S S +=+，则n S = ． 11. 某地球仪上北纬30︒纬线长度为12πcm ，该地球仪的表面积为 cm 2．12. 已知直线()2y k x =-与抛物线x y C 8:2=相交于A 、B 两点，F 为抛物线C 的焦点．若||2||FA FB =u u u r u u u r，则实数=k ．13. 已知“,,,c d e f ”是从1,3,4,5,7中取出4个元素的一个排列．设x 是实数，若“(2)(6)0x x --<”可推出“()()0x c x d --<或()()0x e x f --<”，则满足条件的排列“,,,c d e f ”共有_________个． 14. 将()22xxaf x =-的图像向右平移2个单位后得曲线1C ，将函数()y g x =的图像向下平移2个单位后得曲线2C ，1C 与2C 关于x 轴对称．若()()()f x F x g x a=+的最小值为m 且m >2+7，则实数a 的取值范围为．二．选择题（本题满分20分）本大题共有4题，每题都给出四个结论，其中有且只有一个结论是正确的，必须把答题纸上相应题序内的正确结论代号涂黑，选对得 5分，否则一律得零分．15. 已知关于x 的不等式21<++ax x 的解集为P . 若P ∉1，则实数a 的取值范围为（）（A ）(][)+∞∞-,,10Y . （B ）[]01,-. （C ）()()+∞-∞-,,01Y . （D ）(]01,-. 16. 函数()()21212-<+=x x x f 的反函数是（）（A ）3)y x =≤<. （B ） 3)y x =>.（C ）3)y x =≤<. （D ）3)y x =>. 17. 已知A 、B 、C 是单位圆上三个互不相同的点.若，则的最小值是（）（A ）0．（B ）14-．（C ）12-．（D ）34-．18. 已知公比为q 的等比数列{}n a 的前n 项和为,*n S n N ∈，则下列结论中：（1）232,,n n n n n S S S S S --成等比数列；（2）2232()()n n n n n S S S S S -=-；（3）322()n n n n n S S q S S -=-正确的结论为（）（A ）（1）（2）．（B ）（1）（3）．（C ）（2）（3）．（D ）（1）（2）（3）．三．解答题：（本题满分74分）本大题共有5题，解答下列各题必须在答题纸的规定区域（对应的题号）内写出必要的步骤．19. （本题满分12分；第（1）小题满分6分，第（2）小题满分6分）在直三棱柱111ABC -A B C 中,90 ABC =∠︒ ,11,2AB =BC =BB =，求: （1）异面直线11B C 与1A C 所成角的大小；（2）四棱锥111A B BCC -的体积．20. (本题满分14分；第（1）小题满分6分，第（2）小题满分8分)已知()()x b xx f 24lg2++=，其中常数0>b ．求证：（1）当1b =时，()x f 是奇函数；（2）当4b =时，()x f y =的图像上不存在两点A 、B ，使得直线AB 平行于x 轴．21. （本题满分14分；第（1）小题6分，第（2）小题8分）已知点1F 、2F 为双曲线C ：()01222>=-b by x 的左、右焦点，过2F 作垂直于x 轴的直线，在x 轴上方交双曲线C 于点M ，1230MF F ∠=︒．（1）求双曲线C 的方程；（2）过双曲线C 上任意一点P 作该双曲线两条渐近线的垂线，垂足分别为1P 、2P ，求21PP PP ⋅的值．22. （本题满分16分；第（1）小题满分8分，第（2）小题满分8分）如图，制图工程师用两个同中心的边长均为4的正方形合成一个八角形图形．由对称性，图中8个三角形都是全等的三角形，设α=∠11H AA ．（1）试用α表示11H AA ∆的面积；（2）求八角形所覆盖面积的最大值，并指出此时α的大小．23. （本题满分18分；第（1）小题６分，第（2）小题6分，第（3）小题６分）在等差数列{}n a 和等比数列{}n b 中，112a b ==，222a b b ==+，n S 是{}n b 前n 项和．（1）若lim 3n n S b →∞=-，求实数b 的值；（2）是否存在正整数b ，使得数列{}n b 的所有项都在数列{}n a 中？若存在，求出所有的b ，若不存在，说明理由；（3）是否存在实数b ，使得数列{}n b 中至少有三项在数列{}n a 中，但{}n b 中的项不都在数列{}n a 中？若存在，求出一个可能的b 的值，若不存在，说明理由．2014届高三年级八校联合调研考试试卷数学（文科）一、填空题（本题满分56分）本大题共有14题，要求在答题纸相应题序的空格内直接填写结果，每个空格填对得4分，否则一律得零分．答案 351 22- 2215x y += 2arctan -π 3 []02,-题号 8 910111213 14答案 43π 321n⋅+ 192π 22±481(,2)2二．选择题（本题满分20分）本大题共有4题，每题都给出四个结论，其中有且只有一个结论是正确的，必须把答题纸上相应题序内的正确结论代号涂黑，选对得 5分，否则一律得零分．题号 15 16 17 18 答案DBCC三．解答题：（本题满分74分）本大题共有5题，解答下列各题必须在答题纸的规定区域（对应的题号）内写出必要的步骤．19.（本题满分12分；第（1）小题满分6分，第（2）小题满分6分）在直三棱柱111ABC -A B C 中,90 ABC =∠︒ ,11,2AB =BC =BB =，求: （1）异面直线11B C 与1A C 所成角的大小；（2）四棱锥111A B BCC -的体积．解：（1）因为11//B C BC ，所以1A CB ∠（或其补角）是异面直线11B C 与1A C 所成角. ………………1分因为BC ^AB ,BC ^BB 1,所以BC ⊥平面1ABB ，所以1BC A B ⊥. (3)分在Rt D A 1BC 中,11tan 5A BACB BC∠==所以15ACB ∠=5分所以异面直线11B C 与1A C 所成角的大小为5． ………………6分（2）因为A 1B 1^B 1C 1,A 1B 1^BB 1所以11A B ⊥平面11B BCC ……………9分则11111111233A B BCC B BCC V S A B -=⨯= ……………12分20.(本题满分14分；第（1）小题满分6分，第（2）小题满分8分)函数()()x b xx f 24lg2++=，其中常数0>b ．求证：（1）当1b =时， ()y f x =是奇函数；（2）当4b =时，()x f y =的图像上不存在两点A,B ，使得直线AB 平行于x 轴．证明：（1）由题意，函数定义域R ， ……………1分对定义域任意x ，有：()))lg2lg2f x x x -===- ……4分所以()()f x f x -=-，即y f x =是奇函数. ……………6分（2）假设存在不同的B A ,两点，使得AB 平行x 轴，则))lg2lg2A B x x = ……………9分A B x x =-化简得：2220A B A B x x x x +-=，即A B x x =，与A B 、不同矛盾。

2017届上海市嘉定区高三1月第一次质量调研文科数学试题及答案精品

上海市嘉定区2017届高三上学期期末质量调研（一模）数学（文）试卷2017年1月考生注意：1．每位考生应同时收到试卷和答题纸两份材料，解答必须写在答题纸上，写在试卷或草稿纸上的解答一律无效．2．答卷前，考生务必在答题纸上将姓名、学校、班级等相关信息填写清楚，并在规定的区域内贴上条形码．答题纸不能折叠．3．本试卷共有23道试题，满分150分；考试时间120分钟．一．填空题（本大题共有14题，满分56分）考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得4分，否则一律得零分．1．函数)2(log 2-=x y 的定义域是_____________．2．已知i 是虚数单位，复数z 满足1)31(=+⋅i z ，则=||z _______． 3．已知函数)(x f y =存在反函数)(1x f y -=，若函数)1(-=x f y 的图像经过点)1,3(，则)1(1-f的值是___________．4．已知数列}{n a 的前n 项和2n S n =（*N ∈n ），则8a 的值是__________．5．已知圆锥的母线长为5cm ，侧面积为π202cm ，则此圆锥的体积为________3cm ．6．已知θ为第二象限角，54sin =θ，则=⎪⎭⎫ ⎝⎛+4tan πθ____________． 7．已知双曲线12222=-by a x （0>a ，0>b ）满足021=b a ，且双曲线的右焦点与抛物线x y 342=的焦点重合，则该双曲线的方程为______________．8．分别从集合}4,3,2,1{=A 和集合}8,7,6,5{=B 中各任取一个数，则这两数之积为偶数的概率是_________．9．在边长为1的正方形ABCD 中，M 为BC 的中点，点E 在线段AB 上运动，则⋅的最大值为___________．10．函数xa y =（0>a ,1≠a ）的图像经过点⎪⎭⎫⎝⎛41,2P ，则=+++∞→)(lim 2n n a a a ______． 11．设等比数列}{n a 的前n 项和为n S ，且55S a =，则=2014S ________．12．在平面直角坐标系中，动点P 到两条直线03=-y x 与03=+y x 的距离之积等于4，则P 到原点距离的最小值为_________．13．设集合}1)4(),{(22=+-=y x y x A ，}1)2()(),{(22=+-+-=at y t x y x B ，若存在实数t ，使得∅≠B A ，则实数a 的取值范围是___________．14．已知函数⎪⎩⎪⎨⎧<+-≥+=0,,0,2)(22x bx x x x ax x f 是偶函数，直线t y =与函数)(x f 的图像自左至右依次交于四个不同点A 、B 、C 、D ，若||||BC AB =，则实数t 的值为_______．二．选择题（本大题共有4题，满分20分）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，每题选对得5分，否则一律得零分．15．设向量)1,1(-=x a，)1,3(+=x b ，则“a ∥b ”是“2=x ”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 ．既非充分又非必要条件16．若nx x ⎪⎭⎫ ⎝⎛+22展开式中只有第六项的二项式系数最大，则展开式中的常数项是（）A ．180B ．120C ．90D ．4517．若将函数x x y cos 3sin +=（R ∈x ）的图像向左平移m （0>m ）个单位后，所得图像关于原点对称，则m 的最小值是……………………………………………（） A ．6π B ．3π C ．32π D ．65π18．设函数)(x f 的定义域为D ，若存在闭区间D b a ⊆],[，使得函数)(x f 满足：①)(x f在],[b a 上是单调函数；②)(x f 在],[b a 上的值域是]2,2[b a ，则称区间],[b a 是函数)(x f 的“和谐区间”．下列结论错误的是…………………………………………（）A ．函数2)(x x f =（0≥x ）存在“和谐区间”B ．函数xx f 2)(=（R ∈x ）不存在“和谐区间” C ．函数14)(2+=x xx f (0≥x ）存在“和谐区间” D ．函数x x f 2log )(=（0>x ）不存在“和谐区间”三．解答题（本大题共有5题，满分74分）解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤． 19．（本题满分12分）本题共有2个小题，第1小题满分6分，第2小题满分6分．如图，正三棱锥BCD A -的底面边长为2，侧棱长为3，E 为棱BC 的中点．（1）求该三棱锥的表面积S ；（2）求异面直线AE 与CD 所成角的大小（结果用反三角函数值表示）．20．（本题满分14分）本题共有2个小题，第1小题满分8分，第2小题满分6分．设R ∈x ，函数x x x f sin cos )(+=，x x x g sin cos )(-=．（1）求函数)()()()(2x f x g x f x F +⋅=的最小正周期和单调递增区间；（2）若)(2)(x g x f =，求xx x xcos sin cos sin 122-+的值． 21．（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．已知椭圆C 的中心在原点，焦点在x 轴上，长轴长为4，且点⎪⎪⎭⎫⎝⎛23,1在椭圆C 上．（1）求椭圆C 的方程；（2）设P 是椭圆C 长轴上的一个动点，过P 作方向向量)1,2(=d的直线l 交椭圆C 于A 、BB AE D两点，求证：22||||PB PA +为定值．22．（本题满分16分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分6分．已知函数)(x f 和)(x g 的图像关于原点对称，且x x x g 2)(2+-=．（1）求函数)(x f 的解析式；（2）解不等式|1|)()(-+≤x x g x f ；（3）若函数1)()()(+⋅+=x g x f x h λ在区间]1,1[-上是增函数，求实数λ的取值范围． 23．（本题满分18分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分8分．已知数列}{n a 满足121++=+n a a n n （*N ∈n ）．（1）若数列}{n a 是等差数列，求它的首项和公差；（2）证明：数列}{n a 不可能是等比数列；（3）若11-=a ，b kn a c n n ++=（*N ∈n ），试求实数k 和b 的值，使得数列}{n c 为等比数列；并求此时数列}{n a 的通项公式．上海市嘉定区2013—2017学年高三年级第一次质量调研（文）参考答案与评分标准一．填空题（每小题4分，满分56分）1．),2(∞+ 2．21 3．2 4．15 5．π16 6．71- 7．1222=-y x 8．43 9．23 10．1 11．0 12．22 13．⎥⎦⎤⎢⎣⎡34,0 14．43二．选择题（每小题5分，满分20分）15．B 16．A 17．C 18．B三．解答题 19．（本题满分12分，第1小题满分6分，第2小题满分6分）（1）正三棱锥的斜高221322=-='h ， ……………………（2分）所以，2632222132432+=⨯⨯⨯+⨯=S ． ……………………（4分）（2）取BD 中点F ，连结AF 、EF ，因为EF ∥CD ，所以AEF ∠就是异面直线AE 与CD 所成的角（或其补角）． ……………………………………（2分）在△AEF 中，22==AF AE ，1=EF ， ………………（1分）所以822221cos ==∠AEF ． ………………………………（2分）所以，异面直线AE 与CD 所成的角的大小为82arccos． ………………（1分） 20．（本题满分14分，第1小题满分8分，第2小题满分6分）（1）2)sin (cos )sin )(cos sin (cos )(x x x x x x x F ++-+= …………（1分）142sin 212cos 2sin cos sin 21sin cos 22+⎪⎭⎫ ⎝⎛+=++=++-=πx x x x x x x ，…（2分）所以，函数)(x F 的最小正周期为π． ……………………（2分）由224222πππππ+≤+≤-k x k （Z ∈k ），得883ππππ+≤≤-k x k （Z ∈k ），（2分）所以函数)(x F 的单调递增区间是⎥⎦⎤⎢⎣⎡+-8,83ππππk k （Z ∈k ）． ………………（1分）（2）由题意，)sin (cos 2sin cos x x x x -=+，x x cos sin 3=，…………（1分）所以，31tan =x ． ………………………………（1分）所以，x x x x cos sin cos sin 122-+611tan 1tan 21cos sin cos sin 2cos 2222=-+=-+=x x xx x x x ． ……（4分）（中间步骤每步1分，答案2分）21．（本题满分14分，第1小题满分6分，第2小题满分8分）（1）因为C 的焦点在x 轴上且长轴为4，故可设椭圆C 的方程为14222=+by x （0>>b a ）， ……………………（1分）因为点⎪⎪⎭⎫⎝⎛23,1在椭圆C 上，所以143412=+b ， ………………（2分）解得12=b ， …………（1分）所以，椭圆C 的方程为1422=+y x ． ………………（2分）（2）设)0,(m P （22≤≤-m ），由已知，直线l 的方程是2mx y -=， ……（1分）由⎪⎪⎩⎪⎪⎨⎧=+-=,14,)(2122y x m x y ⇒ 042222=-+-m mx x （*） ………………（2分）设),(11y x A ，),(22y x B ，则1x 、2x 是方程（*）的两个根，所以有，,22m x x =+24221-=m x x ， ……………………（1分）所以，2222212122)()(||||y m x y m x PB PA +-++-=+])()[(45)(41)()(41)(222122222121m x m x m x m x m x m x -+-=-+-+-+-=]22)(2)[(45]2)(2[45221212212212221m x x x x m x x m x x m x x +-+-+=++-+= 5]2)4(2[452222=+---=m m m m （定值）． ………………（3分）所以，22||||PB PA +为定值． …………（1分）（写到倒数第2行，最后1分可不扣）22．（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）（1）设),(y x P 是函数)(x f 图像上任一点，则P 关于原点对称的点),(y x Q --在函数)(x g 的图像上， …………………………（1分）所以)(2)(2x x y -+--=-，故x x y 22+=． …………（2分）所以，函数)(x f 的解析式是x x x f 2)(2+=． …………（1分）（2）由|1|)()(-+≤x x g x f ，得|1|2222-++-≤+x x x x x ， …………（1分）即0|1|22≤--x x ． ………………（1分）当1≥x 时，有0122≤+-x x ，△0781<-=-=，不等式无解； ……（1分）当1<x 时，有0122≤-+x x ，0)1)(12(≤+-x x ，解得211≤≤-x ．……（2分）综上，不等式|1|)()(-+≤x x g x f 的解集为⎥⎦⎤⎢⎣⎡-21,1． ……………………（1分）（3）1)1(2)1(1)2(2)(222+++-=++-++=x x x x x x x h λλλ．…………（1分） ①当1=λ时，14)(+=x x h 在区间]1,1[-上是增函数，符合题意． …………（1分） ②当1≠λ时，函数)(x h 图像的对称轴是直线11-+=λλx ． …………（1分）因为)(x h 在区间]1,1[-上是增函数，所以，1）当1<λ时，01>-λ，函数)(x h 图像开口向上，故111-≤-+λλ，解得10<≤λ； ……………………………………………………（1分） 2）当1>λ时，01<-λ，函数)(x h 图像开口向下，故111≥-+λλ，解得1>λ．…（1分）综上，λ的取值范围是),0[∞+． ……………………（1分）23．（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）（1）解法一：由已知2212+=a a ，7432123+=+=a a a ， ……（1分）若}{n a 是等差数列，则3122a a a +=，即754411+=+a a ， ……（1分）得31-=a ，42-=a ，故1-=d ． ……………………（1分）所以，数列}{n a 的首项为3-，公差为1-． ………………（1分）解法二：因为数列}{n a 是等差数列，设公差为d ，则d a a n n +=+1，故12++=+n a d a n n ， ……（1分）1-+-=d n a n ，又d n a a n )1(1-+=，所以有1-=d ， …………（1分）又11-=-d d a ，从而31-=a ． …………（1分）所以，数列}{n a 的首项为3-，公差为1-． …………（1分）（2）假设数列}{n a 是等比数列，则有3122a a a =，即)74()1(41121+=+a a a ， ………………（1分）解得41-=a ，从而62-=a ，93-=a ， …………（1分）又144234-=+=a a ． …………（2分）因为1a ，2a ，3a ，4a 不成等比数列，与假设矛盾，所以数列}{n a 不是等比数列． ………………（2分）（3）由题意，对任意*N ∈n ，有q c c nn =+1（q 为定值且0≠q ），即q bkn a bn k a n n =++++++)1(1． ………………（2分）即q bkn a b k n k a b kn a b n k n a n n n n =+++++++=+++++++1)1(2)1(12， …………（1分）于是，qb kqn qa b k n k a n n ++=+++++1)1(2， …………（1分）所以，⎪⎩⎪⎨⎧=++=+=,1,1,2qb b k kq k q ⎪⎩⎪⎨⎧===⇒.2,1,2b k q …………（2分）所以，当1=k ，2=b 时，数列}{n c 为等比数列． …………（1分）此数列的首项为2211=++a ，公比为2=q ，所以n n n a 22=++．因此，}{n a 的通项公式为22--=n a n n ． ………………（1分）。

上海市八校2017届高三联考英语试题

2017届高三年级八校联合调研英语试卷2016年11月(满分140分，考试时间120分钟)第I卷（共90分）II. Grammar and VocabularySection A(10×1=10分)Directions: Read the following passage. Fill in the blanks to make the passage coherent. For the blanks with a given word, fill in each blank with the proper form of the given word. For the other blanks, fill in each blank with one proper word. Make sure that your answers are grammatically correct.Have you ever seen an old movie called Three Coins in the Fountain? It is about three young American women (21) _______(search) for permanent romance in Rome and they all find it. Far-fetched Hollywood? Well, from the world history point of view, romance did, in fact, set down its roots in Rome.The word romance evolved in Latin from Roma to Romanicus of the Roman language, to the Old French romanz escrive, (22) _______means “to write in a Romance language,” and on to the English romance.The Romance languages (23) ____________(compose) of seven groups of languages that all have Latin (24) ______their basis. These languages include French, Italian, Spanish and Portuguese. The common people in ancient Rome spoke (25)________ is referred to as Vulgar Latin, an informal speech, as opposed to the classical Latin of the more educated. Most language experts agree that Vulgar Latin is the chief source of the Romance languages.Medieval Romances were tales (26) __________(write)primary in French verse about brave heroes. The notion of having a romance with another person is thought (27) __________(develop) sometime during the Middle Ages. In the late 18th century and on through the 19th, a romance was not a love story (28) _________ a work of prose fiction that contained far-fetched, mysterious events. Romances of this period (29) _________(include) English Gothic novels like The Castle of Otranto by Horace Walpole.What exactly is a twentieth-century romance ? Does it have any relationship with the lively, popular novels written today, with their fantastic plots of love affairs? Or did the playwright Oscar Wilde have it right in The Picture of Dorian Gray: “ When one is in love, one always begins by deceiving (30) _________, and one always ends by deceiving others. That is what the world calls a romance.”Section B（10×1=10分）Directions: Complete the following passage by using the words in the box. Each word can be used only once. Note that there is one word more than you need.In the wake of the historic announcement of the discovery of gravitational waves on February 11, 2016 by the Laser Interferometer Gravitational-Wave Observatory (LIGO), British physicist and black hole theorist Stephen Hawking was quick to ____31___ the US-led collaboration, sharing his excitement for the historic news.According to Hawking, these results confirm several very important ____32___ of Einstein’s theory of general relativity and it also confirms the existence of gravitational waves directly.As is becoming clear, the direct detection of these ripples in space time not only confirms Einstein’s famous theory of general theory but it also opens our eyes to a(n) _33________ “dark” universe. Astronomers employ the electromagnetic spectrum（电磁光谱）to study the universe, but objects that do not radiate in the electromagnetic spectrum will go ___34____. But now we know how to detect gravitational waves, which can help us detect and study some of the most energetic cosmic phenomena.“Gravitat ional waves provide a completely new way of looking at the universe and the ability to detect them has the ___35___ to revolutionize astronomy” said Hawking. “The discovery is the first observation of black holes merging. The observed __36____ of this system are consistent with predictions about black holes that I made in 1970 in Cambridge.”However, this discovery also presents a puzzle for astrophysicists. The mass of each of the black holes are larger than expected for those formed by the gravitational __37_____ of a star---so how did both of these black holes become so massive?This question touches on one of the biggest mysteries ___38___ black hole evolution. Currently, astronomers are having a hard time understanding how black holes grow to be so massive. On the one end of the scale, there are “stellar mass（恒星质量）” black holes that form immediately after a massive star explodes, ___39____ an extremely bright light. And we also have an abundance of evidence for the existence of the super-massive that live in the centers of most galaxies. There is a disconnect, however. If black holes grow by merging and consuming stellar matter, there should be evidence of black holes of all sizes, but “intermediate mass” black holes and black holes of a few dozen solar masses are ____40____ rare, throwing some black holes evolution theories into doubt.One thing is clear, however. This is the first time that we’ve acquired direct evidence of a black hole merger. So it’s good to know we’re on the right track.III. Reading ComprehensionSection A（15×1=15分）Directions: 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.Cowboy or spaceman ? A dilemma for a children’s party, perhaps. But also a question for economists, argued Kenneth Boulding, in an essay published in 1966. We have run our 41 , he warned, like cowboys on the open grassland: taking and using the world’s resources, 42 ____ that more lies over the horizon. But the Earth is 43 a grassland than a spaceship---a closed system, alone in space, carrying exhaustible supplies. We need, said Boulding, an economics that takes seriously the idea of environmental 44 . In the half century since his essay, a new movement has respo nded to his challenge. “Ecological economists,” as they call themselves, want to 45 __ its aims and assumptions. What do they say -- and will their ideas take off?To its 46 , ecological economics is neither ecology nor economics, but a mix of both. Their starting point is to recognize that the human economy is part of the natural world. Our environment, theynote, is both a source of resources and a sink for wastes. But it is 47 in traditional textbooks, where neat diagrams trace the flows between firms, households and the government as though nature did not exist. That is a huge mistake.There are two ways our economies can grow, ecological economists point out: through technological change, or through maximum use of resources. Only the 48 , they say, is worth having. They are suspicious of GDP (gross domestic product), a simple 49 which does not take into account resource exhaustion, unpaid work and countless other factors. 50 , they advocate more holistic approaches, such as GPI (genuine progress indicator)，a composite（复合的）index that include things like the cost of pollution, deforestation and car accidents. While GDP has kept growing, global GPI per person 51 in 1978: by destroying our environment, we are making ourselves poorer, not richer. The solution, according to experts, lies in a “steady-state” economy, where the use of materials and energy is held 52 .Mainstream economists are not 53 . GPI, they point out, is a subjective standard. And talk of limits to growth has had a bad press since the days of Thomas Malthus, who predicted in the 18th century, wrongly, that overpopulation would lead to famine. Human beings find solutions to some of the most annoying problems. But ecological economists 54 self-satisfaction. In 2009, a paper in Nature argued that human activity is already 55 safe planetary boundaries on issues such as biodiversity and climate change. That suggests ecologist economists are at least asking some important questions, even if their answers turn out to be wrong.41. A. grassland B. nation C. economy D. spaceship42. A. ignorant B. confident C. astonished D. anxious43. A. less B. smaller C. more D. larger44. A. movements B. influences C. limits D. threats45. A. reject B. realize C. resemble D. revolutionize46. A. challengers B. learners C. advocates D. professors47. A. addressed B. ignored C. opposed D. reflected48. A. advanced B. former C. latter D. scientific49. A. number B. product C. idea D. measure50. A. In addition B. For example C. In other words D. In its place51.A. peaked B. plunged C. persisted D. paused52.A. sufficient B. efficient C. constant D. adequate53.A. impressed B. involved C. concerned D. appointed54.A. call for B. contribute to C. warn against D. refer to55.A. setting B. overstepping C. extending D. redrawingSection B（11×2=22分）Directions:Read the following three 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)Every April I am troubled by the same concern -- that spring might not occur this year. The landscape looks dull, with hills, sky and forest appearing gray. My spirits ebb, as they did during an April snowfallwhen I first came to Maine 15 years ago. "Just wait," a neighbor advised. "You'll wake up one morning and spring will just be here."And look, on May 3 that year I awoke to a green so amazing as to be almost electric, as if spring were simply a matter of flipping a switch. Hills, sky and forest revealed their purples, blues and green. Leaves had unfolded and daffodils were fighting their way heavenward.Then there was the old apple tree. It sits on an undeveloped lot in my neighborhood. It belongs to no one and therefore to everyone. The tree's dark twisted branches stretch out in unpruned（未经修剪的）abandon. Each spring it blossoms so freely that the air becomes filled with the scent of apple.Until last year, I thought I was the only one aware of this tree. And then one day, in a bit of spring madness, I set out with pruner to remove a few disorderly branches. No sooner had I arrived under the tree than neighbors opened their windows and stepped onto their porches. These were people I barely knew and seldom spoke to, but it was as if I had come uninvited into their personal gardens.My mobile-home neighbor was the first to speak."You're not cutting it down, are you?" she asked anxiously. Another neighbor frowned as I cut off a branch. "Don't kill it, now," he warned. Soon half the neighborhood had joined me under the apple tree. It struck me that I had lived there for five years and only now was learning these people's names, what they did for a living and how they passed the winter. It was as if the old apple tree was gathering us under its branches for the purpose of both acquaintanceship and shared wonder. I couldn't help recalling Robert Frost's words:The trees that have it in their pent-up budsTo darken nature and be summer woodsOne thaw led to another. Just the other day I saw one of my neighbors at the local store. He remarked how this recent winter had been especially long and complained of not having seen or spoken at length to anyone in our neighborhood. And then, he looked at me and said, "We need to prune that apple tree again."56. By saying that “my spirits ebb” (Para. 1), the author means that _________________.A.he feels relievedB. he feels blueC. he is surprisedD. he is tired57. The apple tree mentioned in the passage is most likely to _________________.A. be regarded as a delight in the neighborhoodB. have been abandoned by its original ownerC. have been neglected by everyone in the communityD. be appealing only to the author58. In Para. 4, “neighbors opened their windows and stepped onto their porches” probably because ___________________________.A.they were surprised that someone unknown was pruning the treeB.they wanted to prevent the author from pruning the treeC.they were concerned about the safety of the treeD.they wanted to get to know the author59. It can be inferred that the author’s neighbor mentioned in the last paragraph most cared about _______________.A.when spring would arriveB. how to pass the long winterC. the neighborhood gatheringD. the pruning of the apple tree(B)60. Which is one of the characteristics of Mount Cook National Park?A. It is alpine in the purest sense and hard to reach.B. It provides star-shining night skies for visitors.C. It attracts less skilled climbers to all alpine activities.D. It guarantees visitors a sight of cheeky kea.61. Mike is an experienced adventurer and may find ________ the most exciting.A. Mountaineering on Elie de BeaumontB. Mountain walks via Hooker Valley TrackC. Skiing on Tasman GlacierD. Climbing Mount Cook62. If you are a visitor to the park, you should ________．A. properly evaluate your own experience and skillB. get your permit prepared before you start to climbC. hire local guides to help you to train for climbingD. avoid exploring glaciers in winter(C)How many really suffer as a result of labor market problems? This is one of the most critical yet debatable social policy questions.In many ways, our social statistics overstate the degree of hardship. Unemployment does not have the same horrible consequences today as it did in the 1930’s when most of the unemployed were primary breadwinners, when income and earnings were usually much closer to the margin of survival, and when there were fewer effective social programs for those failing in the labor market. Increasing wealth, the rise of families with more than one wage earner, the growing dominance of secondary earners among the unemployed and improved social welfare protection have unquestionably relieved the consequences of joblessness. Earnings and income data also overestimate the scale of hardship. Among the millions with hourly earnings at or below the minimum wage level, the majority are from multiple-earner, relatively well-off families. Most of those counted by the poverty statistics are elderly or handicapped or have family responsibilities which keep them out of the labor force, so the poverty statistics are by no means an accurate indicator of labor market problems.Yet there are also many ways our social statistics underestimate the degree of labor-market-related hardship. The unemployment counts exclude the millions of fully employed workers whose wages are so low that their families remain in poverty. Low wages and repeated or long-time unemployment frequently interact to weaken the capacity for self-support. Since the number experiencing joblessness at some time during the year is several times that unemployed in any month, those who suffer as a result of forced idleness can equal or exceed average annual unemployment, even though only a minority of the jobless in any month really suffer. For every person counted in the monthly unemployment totals, there is another working part-time because of the inability to find full-time work, or else outside the labor force but wanting a job. Finally, income transfers in our country have always focused on the elderly, disabled, and dependent, neglecting the needs of the working poor, so that the dramatic expansion of cash and non-cash transfers does not necessarily mean that those failing in the labor market are adequately protected.As a result of such conflicting evidence, it is uncertain whether those suffering seriously as a result of labor market problems number in the hundreds of thousands or the tens of millions, and, hence, whether high levels of joblessness can be tolerated or must be counteracted(抵消）by job creation and economic stimulation. There is only one area of agreement in this debate—that the existing poverty, employment, and earnings statistics are inadequate for one of their primary applications, measuring the consequences of labormarket problems.63. In Paragraph 2, the author contrasts the 1930’s with the present in order to show that_____________.A. more people were unemployed in the 1930’sB. unemployment is more intolerable todayC. social programs are more in need nowD. i ncome level has increased since the 1930’s64.Which of the following is true according to the passage?A.A majority of low-wage workers receive earnings from more than one job.B.Repetition of short-term unemployment mainly contributes to people’s loss of workin g capacity.C. Many unemployed people are from families where other members are working.D. Labor market hardship is understated because fewer individuals are jobless than counted.65.It can be inferred from the passage that the effect of income transfers is often not felt by _________________.A. those doing a low-paid, part-time jobB. children in single-earner familiesC. workers who have just retiredD. full-time workers who become unemployed66. Which of the following is the principal topic of the passage?A. What causes labor market problems that result in suffering.B. Why income statistics are imprecise in measuring degrees of poverty.C. When poverty, employment, and earnings figures agree with each other.D. How statistics give an unclear picture of the labor-market-related suffering.Section C（4×2=8分）Directions: Read the following passage. 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.Most of the people who appear most often and most gloriously in the history books are great conquerors and generals, while the people who really helped civilization forward are often never mentioned. We do not know who first set a broken leg, or launched a seagoing boat, or calculated the length of the year but we know all about the killers and destroyers. People think so much of them that on all the highestpillars in the great cities of the world you will find the figure of a conqueror or a general. ____67_________It is possible they are, but they are not the most civilized. Animals fight, so do savages; so to be good at fighting is to be good in the way an animal or a savage is good, but it is not to be civilized. ____68_______. People fight to settle quarrels. Fighting means killing, and civilized peoples ought to be able to find some ways of settling their disputes other than by seeing which side can kill off greater number of the other side, and then saying that the side which has killed most has won. ___69______. For that is what going to war means; it means power is right.This is what the story of mankind has been like. But we must not expect too much. After all, the race of men has only just started. From the point of view of evolution, human beings are very young indeed, babies of a few months old. Scientists assume that there has been life of some sort on the earth for about twelve hundred million years; but there have been men for only one million years, and there has been civilized men for about eight thousand years.____70_____. Taking man’s civilized past at about seven or eight hours, we may estimate his future at about one hundred thousand years. Thus mankind is only at the beginning of its whole a pretty beastly business, a business of fighting and killing. We must not expect even civilized peoples not to have done these things. All we can ask is that they will sometimes have done something else.第II卷(共50分）I.Summary (10分)Directions: Reading the following passage. Summarize the main idea and the main point(s) of the passage with no more than 60 words. Use your own words as far as possible.It's not piano lessons or dance classes. Nowadays, the biggest extra-curricular activity is going to a tutor. "I spend about 800 Canadian dollars a month on tutors. It's costly," says Pat, a mother in Canada. However, she adds, "after finding out half my daughter's class had tutors, I felt like my child was going to fall behind because everyone else seemed to be ahead"Shelley, a mother of three, also has tutors constantly coming in and out of her home. "When I used to sit down with my children, it was hard to get them focused. I was always yelling. When I got a tutor once a week, they became focused for one entire hour and could get most of their homework done."Tutoring isn't simply a private school phenomenon. Nor is it geared only toward lower-achieving students. In Canada alone, seven percent of high school students reported using a tutor in 2010. That increased to 15 percent last year.Overall, parents hire tutors because they are worried schools are not meeting their expectations, but there is also a cultural shift. A special value is placed on education in Asia, where tutoring is viewed as an extension of the school day. As a large number of Asians emigrated to the West over the recent years, their attitudes towards education have had an impact.Another reason for the growth in business is parental frustration and their packed schedules. "A lot of parents just don’t have time to help their children with homework," says Julie Diamond, president of an American tutoring company. "Others couldn't help their children after Grade 3."There has been a shift in the attitudes, too. "Children used to get bullied (欺侮) for having a tutor,"Diamond says. "Now it's becoming the norm to have one."Children don't seem to mind that they have a tutor. One parent feels surprised that so many of her child’s classmates have tutors. "For the amount we pay in tuition, they should have as much extra help as they need," she says. Still, she’s now thinking of getting a tutor. Why? Her daughter has actually asked for one.II. Translation（3+3+4+5=15分）Directions: Translate the following sentences into English, using the words given in the brackets.1. 没过多久，失主就来认领他的行李了。

2017年上海市高考数学试卷及解析

2017年上海市高考数学试卷一、填空题（本大题共12题，满分54分，第1～6题每题4分，第7～12题每题5分）1、（4分）已知集合A={1，2，3，4}，集合B={3，4，5}，则A∩B=、2、（4分）若排列数=6×5×4，则m=、3、（4分）不等式＞1的解集为、4、（4分）已知球的体积为36π，则该球主视图的面积等于、5、（4分）已知复数z满足z+=0，则|z|=、6、（4分）设双曲线﹣=1（b＞0）的焦点为F1、F2，P为该双曲线上的一点，若|PF1|=5，则|PF2|=、7、（5分）如图，以长方体ABCD﹣A1B1C1D1的顶点D为坐标原点，过D的三条棱所在的直线为坐标轴，建立空间直角坐标系，若的坐标为（4，3，2），则的坐标是、8、（5分）定义在（0，+∞）上的函数y=f（x）的反函数为y=f﹣1（x），若g（x）=为奇函数，则f﹣1（x）=2的解为、9、（5分）已知四个函数：①y=﹣x，②y=﹣，③y=x3，④y=x，从中任选2个，则事件“所选2个函数的图象有且仅有一个公共点”的概率为、10、（5分）已知数列{a n}和{b n}，其中a n=n2，n∈N*，{b n}的项是互不相等的正整数，若对于任意n∈N*，{b n}的第a n项等于{a n}的第b n项，则=、11、（5分）设a1、a2∈R，且，则|10π﹣a1﹣a2|的最小值等于、12、（5分）如图，用35个单位正方形拼成一个矩形，点P1、P2、P3、P4以及四个标记为“▲”的点在正方形的顶点处，设集合Ω={P1，P2，P3，P4}，点P∈Ω，过P作直线l P，使得不在l P上的“▲”的点分布在l P的两侧、用D1（l P）和D2（l P）分别表示l P一侧和另一侧的“▲”的点到l P的距离之和、若过P的直线l P中有且只有一条满足D1（l P）=D2（l P），则Ω中所有这样的P为、二、选择题（本大题共4题，每题5分，共20分）13、（5分）关于x、y的二元一次方程组的系数行列式D为（）A、B、C、D、14、（5分）在数列{a n}中，a n=（﹣）n，n∈N*，则a n（）A、等于B、等于0C、等于D、不存在15、（5分）已知a、b、c为实常数，数列{x n}的通项x n=an2+bn+c，n∈N*，则“存、x200+k、x300+k成等差数列”的一个必要条件是（）在k∈N*，使得x100+kA、a≥0B、b≤0C、c=0D、a﹣2b+c=016、（5分）在平面直角坐标系xOy中，已知椭圆C1：=1和C2：x2+=1、P为C1上的动点，Q为C2上的动点，w是的最大值、记Ω={（P，Q）|P 在C1上，Q在C2上，且=w}，则Ω中元素个数为（）A、2个B、4个C、8个D、无穷个三、解答题（本大题共5题，共14+14+14+16+18=76分）17、（14分）如图，直三棱柱ABC﹣A1B1C1的底面为直角三角形，两直角边AB 和AC的长分别为4和2，侧棱AA1的长为5、（1）求三棱柱ABC﹣A1B1C1的体积；（2）设M是BC中点，求直线A1M与平面ABC所成角的大小、18、（14分）已知函数f（x）=cos2x﹣sin2x+，x∈（0，π）、（1）求f（x）的单调递增区间；（2）设△ABC为锐角三角形，角A所对边a=，角B所对边b=5，若f（A）=0，求△ABC的面积、19、（14分）根据预测，某地第n（n∈N*）个月共享单车的投放量和损失量分别为a n和b n（单位：辆），其中a n=，b n=n+5，第n个月底的共享单车的保有量是前n个月的累计投放量与累计损失量的差、（1）求该地区第4个月底的共享单车的保有量；（2）已知该地共享单车停放点第n个月底的单车容纳量S n=﹣4（n﹣46）2+8800（单位：辆）、设在某月底，共享单车保有量达到最大，问该保有量是否超出了此时停放点的单车容纳量？20、（16分）在平面直角坐标系xOy中，已知椭圆Γ：=1，A为Γ的上顶点，P为Γ上异于上、下顶点的动点，M为x正半轴上的动点、（1）若P在第一象限，且|OP|=，求P的坐标；（2）设P（），若以A、P、M为顶点的三角形是直角三角形，求M的横坐标；（3）若|MA|=|MP|，直线AQ与Γ交于另一点C，且，，求直线AQ的方程、21、（18分）设定义在R上的函数f（x）满足：对于任意的x1、x2∈R，当x1＜x2时，都有f（x1）≤f（x2）、（1）若f（x）=ax3+1，求a的取值范围；（2）若f（x）是周期函数，证明：f（x）是常值函数；（3）设f（x）恒大于零，g（x）是定义在R上的、恒大于零的周期函数，M是g（x）的最大值、函数h（x）=f（x）g（x）、证明：“h（x）是周期函数”的充要条件是“f（x）是常值函数”、参考答案与试题解析一、填空题（本大题共12题，满分54分，第1～6题每题4分，第7～12题每题5分）1、（4分）已知集合A={1，2，3，4}，集合B={3，4，5}，则A∩B={3，4} 、题目分析：利用交集定义直接求解、试题解答：解：∵集合A={1，2，3，4}，集合B={3，4，5}，∴A∩B={3，4}、故答案为：{3，4}、点评：本题考查交集的求法，是基础题，解题时要认真审题，注意交集定义的合理运用、2、（4分）若排列数=6×5×4，则m=3、题目分析：利用排列数公式直接求解、试题解答：解：∵排列数=6×5×4，∴由排列数公式得，∴m=3、故答案为：m=3、点评：本题考查实数值的求法，是基础题，解题时要认真审题，注意排列数公式的合理运用、3、（4分）不等式＞1的解集为（﹣∞，0）、题目分析：根据分式不等式的解法求出不等式的解集即可、试题解答：解：由＞1得：，故不等式的解集为：（﹣∞，0），故答案为：（﹣∞，0）、点评：本题考查了解分式不等式，考查转化思想，是一道基础题、4、（4分）已知球的体积为36π，则该球主视图的面积等于9π、题目分析：由球的体积公式，可得半径R=3，再由主视图为圆，可得面积、试题解答：解：球的体积为36π，设球的半径为R，可得πR3=36π，可得R=3，该球主视图为半径为3的圆，可得面积为πR2=9π、故答案为：9π、点评：本题考查球的体积公式，以及主视图的形状和面积求法，考查运算能力，属于基础题、5、（4分）已知复数z满足z+=0，则|z|=、题目分析：设z=a+bi（a，b∈R），代入z2=﹣3，由复数相等的条件列式求得a，b的值得答案、试题解答：解：由z+=0，得z2=﹣3，设z=a+bi（a，b∈R），由z2=﹣3，得（a+bi）2=a2﹣b2+2abi=﹣3，即，解得：、∴、则|z|=、故答案为：、点评：本题考查复数代数形式的乘除运算，考查了复数相等的条件以及复数模的求法，是基础题、6、（4分）设双曲线﹣=1（b＞0）的焦点为F1、F2，P为该双曲线上的一点，若|PF1|=5，则|PF2|=11、题目分析：根据题意，由双曲线的方程可得a的值，结合双曲线的定义可得||PF1|﹣|PF2||=6，解可得|PF2|的值，即可得答案、试题解答：解：根据题意，双曲线的方程为：﹣=1，其中a==3，则有||PF1|﹣|PF2||=6，又由|PF1|=5，解可得|PF2|=11或﹣1（舍）故|PF2|=11，故答案为：11、点评：本题考查双曲线的几何性质，关键是掌握双曲线的定义、7、（5分）如图，以长方体ABCD﹣A1B1C1D1的顶点D为坐标原点，过D的三条棱所在的直线为坐标轴，建立空间直角坐标系，若的坐标为（4，3，2），则的坐标是（﹣4，3，2）、题目分析：由的坐标为（4，3，2），分别求出A和C1的坐标，由此能求出结果、试题解答：解：如图，以长方体ABCD﹣A1B1C1D1的顶点D为坐标原点，过D的三条棱所在的直线为坐标轴，建立空间直角坐标系，∵的坐标为（4，3，2），∴A（4，0，0），C1（0，3，2），∴、故答案为：（﹣4，3，2）、点评：本题考查空间向量的坐标的求法，考查空间直角坐标系等基础知识，考查运算求解能力，考查数形结合思想，是基础题、8、（5分）定义在（0，+∞）上的函数y=f（x）的反函数为y=f﹣1（x），若g（x）=为奇函数，则f﹣1（x）=2的解为、题目分析：由奇函数的定义，当x＞0时，﹣x＜0，代入已知解析式，即可得到所求x＞0的解析式，再由互为反函数的两函数的自变量和函数值相反，即可得到所求值、试题解答：解：若g（x）=为奇函数，可得当x＞0时，﹣x＜0，即有g（﹣x）=3﹣x﹣1，由g（x）为奇函数，可得g（﹣x）=﹣g（x），则g（x）=f（x）=1﹣3﹣x，x＞0，由定义在（0，+∞）上的函数y=f（x）的反函数为y=f﹣1（x），且f﹣1（x）=2，可由f（2）=1﹣3﹣2=，可得f﹣1（x）=2的解为x=、故答案为：、点评：本题考查函数的奇偶性和运用，考查互为反函数的自变量和函数值的关系，考查运算能力，属于基础题、9、（5分）已知四个函数：①y=﹣x，②y=﹣，③y=x3，④y=x，从中任选2个，则事件“所选2个函数的图象有且仅有一个公共点”的概率为、题目分析：从四个函数中任选2个，基本事件总数n=，再利用列举法求出事件A：“所选2个函数的图象有且只有一个公共点”包含的基本事件的个数，由此能求出事件A：“所选2个函数的图象有且只有一个公共点”的概率、试题解答：解：给出四个函数：①y=﹣x，②y=﹣，③y=x3，④y=x，从四个函数中任选2个，基本事件总数n=，③④有两个公共点（0，0），（1，1）、事件A：“所选2个函数的图象有且只有一个公共点”包含的基本事件有：①③，①④共2个，∴事件A：“所选2个函数的图象有且只有一个公共点”的概率为P（A）==、故答案为：、点评：本题考查概率的求法，是基础题，解题时要认真审题，注意列举法的合理运用、10、（5分）已知数列{a n}和{b n}，其中a n=n2，n∈N*，{b n}的项是互不相等的正整数，若对于任意n∈N*，{b n}的第a n项等于{a n}的第b n项，则=2、题目分析：a n=n2，n∈N*，若对于一切n∈N*，{b n}中的第a n项恒等于{a n}中的第b n项，可得==、于是b1=a1=1，=b4，=b9，=b16、即可得出、试题解答：解：∵a n=n2，n∈N*，若对于一切n∈N*，{b n}中的第a n项恒等于{a n}中的第b n项，∴==、∴b1=a1=1，=b4，=b9，=b16、∴b1b4b9b16=、∴=2、故答案为：2、点评：本题考查了数列递推关系、对数的运算性质，考查了推理能力与计算能力，属于中档题、11、（5分）设a1、a2∈R，且，则|10π﹣a1﹣a2|的最小值等于、题目分析：由题意，要使+=2，可得sinα1=﹣1，sin2α2=﹣1、求出α1和α2，即可求出|10π﹣α1﹣α2|的最小值试题解答：解：根据三角函数的性质，可知sinα1，sin2α2的范围在[﹣1，1]，要使+=2，∴sinα1=﹣1，sin2α2=﹣1、则：，k1∈Z、，即，k2∈Z、那么：α1+α2=（2k1+k2）π，k1、k2∈Z、∴|10π﹣α1﹣α2|=|10π﹣（2k1+k2）π|的最小值为、故答案为：、点评：本题主要考察三角函数性质，有界限的范围的灵活应用，属于基本知识的考查、12、（5分）如图，用35个单位正方形拼成一个矩形，点P1、P2、P3、P4以及四个标记为“▲”的点在正方形的顶点处，设集合Ω={P1，P2，P3，P4}，点P∈Ω，过P作直线l P，使得不在l P上的“▲”的点分布在l P的两侧、用D1（l P）和D2（l P）分别表示l P一侧和另一侧的“▲”的点到l P的距离之和、若过P的直线l P中有且只有一条满足D1（l P）=D2（l P），则Ω中所有这样的P为P1、P3、P4、题目分析：根据任意四边形ABCD两组对边中点的连线交于一点，过此点作直线，使四边形的四个顶点不在该直线的同一侧，则该直线两侧的四边形的顶点到直线的距离之和相等；由此得出结论、试题解答：解：设记为“▲”的四个点是A，B，C，D，线段AB，BC，CD，DA的中点分别为E，F，G，H，易知EFGH为平行四边形，如图所示；又平行四边形EFGH的对角线交于点P2，则符合条件的直线l P一定经过点P2，且过点P2的直线有无数条；由过点P1和P2的直线有且仅有1条，过点P3和P2的直线有且仅有1条，过点P4和P2的直线有且仅有1条，所以符合条件的点是P1、P3、P4、故答案为：P1、P3、P4、点评：本题考查了数学理解力与转化力的应用问题，也考查了对基本问题的阅读理解和应用转化能力、二、选择题（本大题共4题，每题5分，共20分）13、（5分）关于x、y的二元一次方程组的系数行列式D为（）A、B、C、D、题目分析：利用线性方程组的系数行列式的定义直接求解、试题解答：解：关于x、y的二元一次方程组的系数行列式：D=、故选：C、点评：本题考查线性方程组的系数行列式的求法，是基础题，解题时要认真审题，注意线性方程组的系数行列式的定义的合理运用、14、（5分）在数列{a n}中，a n=（﹣）n，n∈N*，则a n（）A、等于B、等于0C、等于D、不存在题目分析：根据极限的定义，求出a n=的值、试题解答：解：数列{a n}中，a n=（﹣）n，n∈N*，则a n==0、故选：B、点评：本题考查了极限的定义与应用问题，是基础题、15、（5分）已知a、b、c为实常数，数列{x n}的通项x n=an2+bn+c，n∈N*，则“存、x200+k、x300+k成等差数列”的一个必要条件是（）在k∈N*，使得x100+kA、a≥0B、b≤0C、c=0D、a﹣2b+c=0，x200+k，x300+k成等差数列，可得：2x200+k=x100+k x300+k，代入化题目分析：由x100+k简即可得出、、x200+k、x300+k成等差数列，可得：2[a（200+k）试题解答：解：存在k∈N*，使得x100+k2+b（200+k）+c]=a（100+k）2+b（100+k）+c+a（300+k）2+b（300+k）+c，化为：a=0、，x200+k，x300+k成等差数列的必要条件是a≥0、∴使得x100+k故选：A、点评：本题考查了等差数列的通项公式、简易逻辑的判定方法，考查了推理能力与计算能力，属于基础题、16、（5分）在平面直角坐标系xOy中，已知椭圆C1：=1和C2：x2+=1、P为C1上的动点，Q为C2上的动点，w是的最大值、记Ω={（P，Q）|P 在C1上，Q在C2上，且=w}，则Ω中元素个数为（）A、2个B、4个C、8个D、无穷个题目分析：设出P（6cosα，2sinα），Q（cosβ，3sinβ），0≤α\β＜2π，由向量数量积的坐标表示和两角差的余弦公式和余弦函数的值域，可得最大值及取得的条件，即可判断所求元素的个数、试题解答：解：椭圆C1：=1和C2：x2+=1、P为C1上的动点，Q为C2上的动点，可设P（6cosα，2sinα），Q（cosβ，3sinβ），0≤α\β＜2π，则=6cosαcosβ+6sinαsinβ=6cos（α﹣β），当α﹣β=2kπ，k∈Z时，w取得最大值6，则Ω={（P，Q）|P在C1上，Q在C2上，且=w}中的元素有无穷多对、另解：令P（m，n），Q（u，v），则m2+9n2=36，9u2+v2=9，由柯西不等式（m2+9n2）（9u2+v2）=324≥（3mu+3nv）2，当且仅当mv=nu，即O、P、Q共线时，取得最大值6，显然，满足条件的P、Q有无穷多对，D项正确、故选：D、点评：本题考查椭圆的参数方程的运用，以及向量数量积的坐标表示和余弦函数的值域，考查集合的几何意义，属于中档题、三、解答题（本大题共5题，共14+14+14+16+18=76分）17、（14分）如图，直三棱柱ABC﹣A1B1C1的底面为直角三角形，两直角边AB 和AC的长分别为4和2，侧棱AA1的长为5、（1）求三棱柱ABC﹣A1B1C1的体积；（2）设M是BC中点，求直线A1M与平面ABC所成角的大小、题目分析：（1）三棱柱ABC﹣A1B1C1的体积V=S△ABC×AA1=，由此能求出结果、（2）连结AM，∠A1MA是直线A1M与平面ABC所成角，由此能求出直线A1M 与平面ABC所成角的大小、试题解答：解：（1）∵直三棱柱ABC﹣A1B1C1的底面为直角三角形，两直角边AB和AC的长分别为4和2，侧棱AA1的长为5、∴三棱柱ABC﹣A1B1C1的体积：V=S△ABC×AA1===20、（2）连结AM，∵直三棱柱ABC﹣A1B1C1的底面为直角三角形，两直角边AB和AC的长分别为4和2，侧棱AA1的长为5，M是BC中点，∴AA1⊥底面ABC，AM==，∴∠A1MA是直线A1M与平面ABC所成角，tan∠A1MA===，∴直线A1M与平面ABC所成角的大小为arctan、点评：本题考查三棱柱的体积的求法，考查线面角的大小的求法，考查空间中线线、线面、面面间的位置关系等基础知识，考查推理论证能力、运算求解能力、空间想象能力，考查化归与转化思想、函数与方程思想、数形结合思想，是中档题、18、（14分）已知函数f（x）=cos2x﹣sin2x+，x∈（0，π）、（1）求f（x）的单调递增区间；（2）设△ABC为锐角三角形，角A所对边a=，角B所对边b=5，若f（A）=0，求△ABC的面积、题目分析：（1）由二倍角的余弦公式和余弦函数的递增区间，解不等式可得所求增区间；（2）由f（A）=0，解得A，再由余弦定理解方程可得c，再由三角形的面积公式，计算即可得到所求值、试题解答：解：（1）函数f（x）=cos2x﹣sin2x+=cos2x+，x∈（0，π），由2kπ﹣π≤2x≤2kπ，解得kπ﹣π≤x≤kπ，k∈Z，k=1时，π≤x≤π，可得f（x）的增区间为[，π）；（2）设△ABC为锐角三角形，角A所对边a=，角B所对边b=5，若f（A）=0，即有cos2A+=0，解得2A=π，即A=π，由余弦定理可得a2=b2+c2﹣2bccosA，化为c2﹣5c+6=0，解得c=2或3，若c=2，则cosB=＜0，即有B为钝角，c=2不成立，则c=3，△ABC的面积为S=bcsinA=×5×3×=、点评：本题考查二倍角公式和余弦函数的图象和性质，考查解三角形的余弦定理和面积公式的运用，考查运算能力，属于中档题、19、（14分）根据预测，某地第n（n∈N*）个月共享单车的投放量和损失量分别为a n和b n（单位：辆），其中a n=，b n=n+5，第n个月底的共享单车的保有量是前n个月的累计投放量与累计损失量的差、（1）求该地区第4个月底的共享单车的保有量；（2）已知该地共享单车停放点第n个月底的单车容纳量S n=﹣4（n﹣46）2+8800（单位：辆）、设在某月底，共享单车保有量达到最大，问该保有量是否超出了此时停放点的单车容纳量？题目分析：（1）计算出{a n}和{b n}的前4项和的差即可得出答案；（2）令a n≥b n得出n≤42，再计算第42个月底的保有量和容纳量即可得出结论、试题解答：解：（1）∵a n=，b n=n+5∴a1=5×14+15=20a2=5×24+15=95a3=5×34+15=420a4=﹣10×4+470=430b1=1+5=6b2=2+5=7b3=3+5=8b4=4+5=9∴前4个月共投放单车为a1+a2+a3+a4=20+95+420+430=965，前4个月共损失单车为b1+b2+b3+b4=6+7+8+9=30，∴该地区第4个月底的共享单车的保有量为965﹣30=935、（2）令a n≥b n，显然n≤3时恒成立，当n≥4时，有﹣10n+470≥n+5，解得n≤，∴第42个月底，保有量达到最大、当n≥4，{a n}为公差为﹣10等差数列，而{b n}为等差为1的等差数列，∴到第42个月底，单车保有量为×39+535﹣×42=×39+535﹣×42=8782、S42=﹣4×16+8800=8736、∵8782＞8736，∴第42个月底单车保有量超过了容纳量、点评：本题考查了数列模型的应用，等差数列的求和公式，属于中档题、20、（16分）在平面直角坐标系xOy中，已知椭圆Γ：=1，A为Γ的上顶点，P为Γ上异于上、下顶点的动点，M为x正半轴上的动点、（1）若P在第一象限，且|OP|=，求P的坐标；（2）设P（），若以A、P、M为顶点的三角形是直角三角形，求M的横坐标；（3）若|MA|=|MP|，直线AQ与Γ交于另一点C，且，，求直线AQ的方程、题目分析：（1）设P（x，y）（x＞0，y＞0），联立，能求出P点坐标、（2）设M（x0，0），A（0，1），P（），由∠P=90°，求出x0=；由∠M=90°，求出x0=1或x0=；由∠A=90°，则M点在x轴负半轴，不合题意、由此能求出点M的横坐标、（3）设C（2cosα，sinα），推导出Q（4cosα，2sinα﹣1），设P（2cosβ，sinβ），M（x0，0）推导出x0=cosβ，从而4cosα﹣2cosβ=﹣5cosβ，且2sinα﹣sinβ﹣1=﹣4sinβ，cosβ=﹣cosα，且sinα=（1﹣2sinα），由此能求出直线AQ、试题解答：解：（1）设P（x，y）（x＞0，y＞0），∵椭圆Γ：=1，A为Γ的上顶点，P为Γ上异于上、下顶点的动点，P在第一象限，且|OP|=，∴联立，解得P（，）、（2）设M（x0，0），A（0，1），P（），若∠P=90°，则•，即（x0﹣，﹣）•（﹣，）=0，∴（﹣）x0+﹣=0，解得x0=、如图，若∠M=90°，则•=0，即（﹣x0，1）•（﹣x0，）=0，∴=0，解得x0=1或x0=，若∠A=90°，则M点在x轴负半轴，不合题意、∴点M的横坐标为，或1，或、（3）设C（2cosα，sinα），∵，A（0，1），∴Q（4cosα，2sinα﹣1），又设P（2cosβ，sinβ），M（x0，0），∵|MA|=|MP|，∴x02+1=（2cosβ﹣x0）2+（sinβ）2，整理得：x0=cosβ，∵=（4cosα﹣2cosβ，2sinα﹣sinβ﹣1），=（﹣cosβ，﹣sinβ），，∴4cosα﹣2cosβ=﹣5cosβ，且2sinα﹣sinβ﹣1=﹣4sinβ，∴cosβ=﹣cosα，且sinα=（1﹣2sinα），以上两式平方相加，整理得3（sinα）2+sinα﹣2=0，∴sinα=，或sinα=﹣1（舍去），此时，直线AC的斜率k AC=﹣=（负值已舍去），如图、∴直线AQ为y=x+1、点评：本题考查点的坐标的求法，考查直线方程的求法，考查椭圆、直线方程、三角函数等基础知识，考查推理论证能力、运算求解能力，考查化归与转化思想、函数与方思想，是中档题、21、（18分）设定义在R上的函数f（x）满足：对于任意的x1、x2∈R，当x1＜x2时，都有f（x1）≤f（x2）、（1）若f（x）=ax3+1，求a的取值范围；（2）若f（x）是周期函数，证明：f（x）是常值函数；（3）设f（x）恒大于零，g（x）是定义在R上的、恒大于零的周期函数，M是g（x）的最大值、函数h（x）=f（x）g（x）、证明：“h（x）是周期函数”的充要条件是“f（x）是常值函数”、题目分析：（1）直接由f（x1）﹣f（x2）≤0求得a的取值范围；（2）若f（x）是周期函数，记其周期为T k，任取x0∈R，则有f（x0）=f（x0+T k），证明对任意x∈[x0，x0+T k]，f（x0）≤f（x）≤f（x0+T k），可得f（x0）=f（x0+nT k），n∈Z，再由…∪[x0﹣3T k，x0﹣2T k]∪[x0﹣2T k，x0﹣T k]∪[x0﹣T k，x0]∪[x0，x0+T k]∪[x0+T k，x0+2T k]∪…=R，可得对任意x∈R，f（x）=f（x0）=C，为常数；（3）分充分性及必要性证明、类似（2）证明充分性；再证必要性，然后分类证明试题解答：（1）解：由f（x1）≤f（x2），得f（x1）﹣f（x2）=a（x13﹣x23）≤0，∵x1＜x2，∴x13﹣x23＜0，得a≥0、故a的范围是[0，+∞）；（2）证明：若f（x）是周期函数，记其周期为T k，任取x0∈R，则有f（x0）=f（x0+T k），由题意，对任意x∈[x0，x0+T k]，f（x0）≤f（x）≤f（x0+T k），∴f（x0）=f（x）=f（x0+T k）、又∵f（x0）=f（x0+nT k），n∈Z，并且…∪[x0﹣3T k，x0﹣2T k]∪[x0﹣2T k，x0﹣T k]∪[x0﹣T k，x0]∪[x0，x0+T k]∪[x0+T k，x0+2T k]∪…=R，∴对任意x∈R，f（x）=f（x0）=C，为常数；（3）证明：充分性：若f（x）是常值函数，记f（x）=c1，设g（x）的一个周期为T g，则h（x）=c1•g（x），则对任意x0∈R，h（x0+T g）=c1•g（x0+T g）=c1•g（x0）=h（x0），故h（x）是周期函数；必要性：若h（x）是周期函数，记其一个周期为T h、若存在x1，x2，使得f（x1）＞0，且f（x2）＜0，则由题意可知，x1＞x2，那么必然存在正整数N1，使得x2+N1T k＞x1，∴f（x2+N1T k）＞f（x1）＞0，且h（x2+N1T k）=h（x2）、又h（x2）=g（x2）f（x2）＜0，而h（x2+N1T k）=g（x2+N1T k）f（x2+N1T k）＞0≠h（x2），矛盾、综上，f（x）＞0恒成立、由f（x）＞0恒成立，任取x0∈A，则必存在N2∈N，使得x0﹣N2T h≤x0﹣T g，即[x0﹣T g，x0]⊆[x0﹣N2T h，x0]，∵…∪[x0﹣3T k，x0﹣2T k]∪[x0﹣2T k，x0﹣T k]∪[x0﹣T k，x0]∪[x0，x0+T k]∪[x0+T k，x0+2T k]∪…=R，∴…∪[x0﹣2N2T h，x0﹣N2T h]∪[x0﹣N2T h，x0]∪[x0，x0+N2T h]∪[x0+N2T h，x0+2N2T h]∪…=Rh（x0）=g（x0）•f（x0）=h（x0﹣N2T h）=g（x0﹣N2T h）•f（x0﹣N2T h），∵g（x0）=M≥g（x0﹣N2T h）＞0，f（x0）≥f（x0﹣N2T h）＞0、因此若h（x0）=h（x0﹣N2T h），必有g（x0）=M=g（x0﹣N2T h），且f（x0）=f（x0﹣N2T h）=c而由（2）证明可知，对任意x∈R，f（x）=f（x0）=C，为常数21/ 21。

2017届上海市奉贤区高三下学期二模文科数学试题及答案

2017学年奉贤区调研测试高三数学试卷(文科)（考试时间：120分钟，满分150分）一. 填空题 (本大题满分56分）本大题共有14题，考生应在答题纸相应编号的空格内直接写结果，1-14题每个空格填对得4分)1、函数()()42lg -=x x f 的定义域为________.2、设z a i =+（a R +∈，i 是虚数单位），满足2z=，则a =________. 3、如果函数x x f a log )(=的图像过点⎪⎭⎫⎝⎛121，P ，则2lim()n n a a a →∞+++⋅⋅⋅=________.4、执行如图所示的程序框图，输出的S 的值为________.5、若圆C 的半径为1，圆心在第一象限，且与直线034=-y x 和x 轴相切，则该圆的标准方程是________.6、在(1)n x +的二项展开式中，按x 的降幂排列，只有第5项的系数最大，则各项的二项式系数之和为________（答案用数值表示）.7、将外形和质地一样的4个红球和6个白球放入同一个袋中，将它们充分混合后，现从中取出4个球，取出一个红球记2分，取出一个白球记1分，若取出4个球总分不少于5分，则有________种不同的取法.第4题图8、若一个圆锥的侧面展开图是面积为π2的半圆面，则该圆锥的体积为________.9、设实数,x y 满足⎪⎩⎪⎨⎧≤≤-≥+,4,42,2y y x y x 则2x y -的最大值等于________.10、将函数cos ()sin xf x x=的图像向左平移m 个单位(0)m >，若所得图像对应的函数为偶函数，则m 的最小值是________.11、已知抛物线220y x =焦点F 恰好是双曲线22221x y a b-=的右焦点，且双曲线过点15(,3)4，则该双曲线的渐近线方程为________.12、定义在(0,)+∞上的函数()f x 满足：①当[1,3)x ∈时，1,12,()3,23,x x f x x x -≤≤⎧=⎨-<<⎩②(3)3()f x f x =，设关于x 的函数()()1F x f x =-的零点从小到大依次记为123,,,x x x ⋅⋅⋅，则123x x x ++=________.13、已知{}n a 是首项为a ，公差为1的等差数列，1nn na b a +=，若对任意的*n N ∈，都有8n b b ≥成立，则实数a 的取值范围是________.14、以()m ,0间的整数()N m m ∈>,1为分子，以m 为分母组成分数集合1A ，其所有元素和为1a ；以()2,0m 间的整数()N m m ∈>,1为分子，以2m 为分母组成不属于集合1A 的分数集合2A ，其所有元素和为2a ；……，依次类推以()n m ,0间的整数()N m m ∈>,1为分子，以n m 为分母组成不属于121,,,n A A A -⋅⋅⋅的分数集合n A ，其所有元素和为n a ；则12n a a a ⋅⋅⋅+++=________.二．选择题（本大题满分20分）本大题共有4题，每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分.15、三角形ABC 中，设,AB a BC b == ，若()0a a b ⋅+<，则三角形ABC 的形状是( )A ．锐角三角形B ．钝角三角形C ．直角三角形D ．无法确定16、设数列{}n a ，以下说法正确的是（）A ．若2=4n n a ，*n N ∈，则{}n a 为等比数列B ．若221n n n a a a ++⋅=，*n N ∈，则{}n a 为等比数列 C ．若2m n m n a a +⋅=，*,m n N ∈，则{}n a 为等比数列 D ．若312n n n n a a a a +++⋅=⋅，*n N ∈，则{}n a 为等比数列17、下列命题正确的是( )A ．若Z k k x ∈≠,π,则4sin 4sin 22≥+x x B ．若,0<a 则44-≥+aa C ．若0,0>>b a ,则b a b a lg lg 2lg lg ⋅≥+ D ．若0,0<<b a ,则2≥+baa b18、已知R ∈βα,,且设βα>:p ，设:sin cos sin cos q ααβββα+>+⋅，则p 是q 的（）A ．充分必要条件B ．充分不必要条件C ．必要不充分条件D ．既不充分又不必要条件三．解答题（本大题满分74分）本大题共有5题，解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.19、如图,在直三棱柱111ABC A B C -中, 3AC =,4BC =,5AB =,点D 是AB 的中点.四面体BCD B -1的体积是2，求异面直线1DB 与1CC 所成的角.A B1BC(文19题图）20、已知函数9()||f x x a a x=--+，[1,6]x ∈，a R ∈．（1）若1a =，试判断并用定义证明函数()f x 的单调性；（2）当()3,1∈a 时，求函数()f x 的最大值的表达式()M a ．21、某人沿一条折线段组成的小路前进，从A 到B ，方位角（从正北方向顺时针转到AB 方向所成的角）是050，距离是3km ；从B 到C ，方位角是110°，距离是3km ；从C 到D ，方位角是140°，距离是（339+）km.试画出大致示意图，并计算出从A 到D 的方位角和距离（结果保留根号）.22、如图，已知平面内一动点A 到两个定点1F 、2F 的距离之和为4，线段12F F的长为（1）求动点A 的轨迹Γ的方程；（2）过点1F 作直线l 与轨迹Γ交于A 、C 两点，且点A 在线段12F F 的上方，线段AC 的垂直平分线为m . ①求12AF F ∆的面积的最大值；②轨迹Γ上是否存在除A 、C 外的两点S 、T关于直线m 对称，请说明理由.2F23、若函数()f x 满足：集合*{()|}A f n n =∈N 中至少存在三个不同的数构成等比数列，则称函数()f x 是等比源函数.（1）判断下列函数：①2x y =；②lg y x =中，哪些是等比源函数？（不需证明）（2）证明：函数()23g x x =+是等比源函数；（3）判断函数()21x f x =+是否为等比源函数，并证明你的结论.2017—2017学年奉贤区调研测试高三数学试卷(文科)参考答案 4一、填空题（每小题4分，共56分）二、选择题（每小题5分，共20分）三、解答题19、（文）【解】直三棱柱111ABC A B C -中11//CC BB所以1DB B ∠为异面直线1DB 与1CC 所成的角（或其补角） 3分直三棱柱111ABC A B C -中1111113423322B BCD BCD V S B B B B -∆=⋅=⨯⨯⨯=得12B B = 7分由点D 是AB 的中点得52DB =直三棱柱111ABC A B C -中1B B BD ⊥1Rt B BD ∆中11552tan 24BD DB B B B ∠=== 所以15arctan 4DB B ∠=（或1DB B ∠=）所以异面直线1DB 与1BC 所成的角为5arctan 4（或 12分 20、【解】 (1)判断：若1a =，函数()f x 在[1,6]上是增函数. 证明：当1a =时，9()f x x x=-， ()f x 在[1,6]上是增函数.2分在区间[1,6]上任取12,x x ，设12x x <，12121212121212129999()()()()()()()(9)0f x f x x x x x x x x x x x x x x x -=---=----+=<所以12()()f x f x <，即()f x 在[1,6]上是增函数. 6分A B 1BC(2) （理）因为13a <≤，所以92(),1,()9,6,a x x a xf x x a x x ⎧-+≤≤⎪⎪=⎨⎪-<≤⎪⎩8分当13a <≤时，()f x 在[1,]a 上是增函数， 9分证明：当13a <≤时，()f x 在[1,]a 上是增函数（过程略）11分()f x 在在[,6]a 上也是增函数当13a <≤时，()x f y =[]6,1∈x 上是增函数 12分所以任意一个[]6,1∈x ，均能找到唯一的y 和它对应，所以()x f y =[]6,1∈x 时，()f x 存在反函数14分(2) （文）因为13a <≤，所以92(),1,()9,6,a x x a xf x x a x x ⎧-+≤≤⎪⎪=⎨⎪-<≤⎪⎩8分当13a <≤时，()f x 在[1,]a 上是增函数，9分证明：当13a <≤时，()f x 在[1,]a 上是增函数（过程略）11分()f x 在在[,6]a 上也是增函数当13a <≤时，()f x 在[1,6]上是增函数12分证明：当13a <≤时，()f x 在[1,]a 上是增函数（过程略） 13分所以当6x =时，()f x 取得最大值为92；14分21、【解】示意图，如图所示，4分连接AC ，在△ABC 中，∠ABC=50°+(180°-110°)=120°, 又AB=BC=3，∴∠BAC=∠BCA=30° 由余弦定理可得33120cos 222=︒⋅-+=BC AB BC AB AC7分在△ACD 中，∠ACD=360°-140°-(70°+30°)=120°,CD=33+9. 由余弦定理得AD=︒⋅-+120cos 222CD AC CD AC =)21()933(332)933(272-⨯+⨯⨯-++=2)62(9+(km). 10分由正弦定理得sin ∠CAD=()()22262923933sin sin =+⨯+=∠⋅=∠AD ACDCD CAD12分∴∠CAD=45°,于是AD 的方位角为50°+30°+45°=125°, 13分所以，从A 到D 的方位角是125°，距离为2)62(9+km. 14分22、（文）【解】（1）因为4>，轨迹是以1F 、2F 为焦点的椭圆，3分（2）以线段21F F 的中点为坐标原点，以21F F 所在直线为x 轴建立平面直角坐标系，可得轨迹Γ的方程为22+14x y = 7分max 23θπ=12F AF S ∆最大值为23tan tan 22πθ==（3）同理23、（文）【解】（1）①②都是等比源函数； 4分（2）证明： (1)5g =，(6)15g =，(21)45g =因为5,15,45成等比数列所以函数()23g x x =+是等比源函数；10分其他的数据也可以（3）函数()21x f x =+不是等比源函数. 证明如下：假设存在正整数,,m n k 且m n k <<，使得(),(),()f m f n f k 成等比数列， 2(21)(21)(21)n m k +=++，整理得2122222n n m k m k +++=++，等式两边同除以2,m 得2122221n m n m k k m --+-+=++. 因为1,2n m k m -≥-≥，所以等式左边为偶数，等式右边为奇数，所以等式2122221n m n m k k m --+-+=++不可能成立，所以假设不成立，说明函数()21x f x =+不是等比源函数. 18分。

2017年上海市高考数学试卷及参考答案与试题解析

2017年上海市高考数学试卷及参考答案与试题解析一、填空题(本大题共12题,满分54分,第1～6题每题4分,第7～12题每题5分) 1.(4分)已知集合A ＝{1,2,3,4},集合B ＝{3,4,5},则A ∩B ＝ . 2.(4分)若排列数＝6×5×4,则m ＝ .3.(4分)不等式＞1的解集为 .4.(4分)已知球的体积为36π,则该球主视图的面积等于 .5.(4分)已知复数z 满足z ＋＝0,则|z|＝ .6.(4分)设双曲线－＝1(b ＞0)的焦点为F 1、F 2,P 为该双曲线上的一点,若|PF 1|＝5,则|PF 2|＝ .7.(5分)如图,以长方体ABCD －A 1B 1C 1D 1的顶点D 为坐标原点,过D 的三条棱所在的直线为坐标轴,建立空间直角坐标系,若的坐标为(4,3,2),则的坐标是 .8.(5分)定义在(0,＋∞)上的函数y ＝f(x)的反函数为y ＝f －1(x),若g(x)＝为奇函数,则f －1(x)＝2的解为 .9.(5分)已知四个函数：①y ＝－x,②y ＝－,③y ＝x 3,④y ＝x,从中任选2个,则事件“所选2个函数的图象有且仅有一个公共点”的概率为 .10.(5分)已知数列{a n }和{b n },其中a n ＝n 2,n ∈N *,{b n }的项是互不相等的正整数,若对于任意n∈N *,{b n }的第a n 项等于{a n }的第b n 项,则＝ .11.(5分)设a 1、a 2∈R,且,则|10π－a 1－a 2|的最小值等于 .12.(5分)如图,用35个单位正方形拼成一个矩形,点P 1、P 2、P 3、P 4以及四个标记为“▲”的点在正方形的顶点处,设集合Ω＝{P 1,P 2,P 3,P 4},点P ∈Ω,过P 作直线l P ,使得不在l P 上的“▲”的点分布在l P 的两侧.用D 1(l P )和D 2(l P )分别表示l P 一侧和另一侧的“▲”的点到l P 的距离之和.若过P 的直线l P 中有且只有一条满足D 1(l P )＝D 2(l P ),则Ω中所有这样的P 为 .二、选择题(本大题共4题,每题5分,共20分)13.(5分)关于x、y的二元一次方程组的系数行列式D为( )A. B. C. D.14.(5分)在数列{an }中,an＝(－)n,n∈N*,则an( )A.等于B.等于0C.等于D.不存在15.(5分)已知a、b、c为实常数,数列{xn }的通项xn＝an2＋bn＋c,n∈N*,则“存在k∈N*,使得x100＋k 、x200＋k、x300＋k成等差数列”的一个必要条件是( )A.a≥0B.b≤0C.c＝0D.a－2b＋c＝016.(5分)在平面直角坐标系xOy中,已知椭圆C1：＝1和C2：x2＋＝1.P为C1上的动点,Q为C2上的动点,w是的最大值.记Ω＝{(P,Q)|P在C1上,Q在C2上,且＝w},则Ω中元素个数为( )A.2个B.4个C.8个D.无穷个三、解答题(本大题共5题,共14＋14＋14＋16＋18＝76分)17.(14分)如图,直三棱柱ABC－A1B1C1的底面为直角三角形,两直角边AB和AC的长分别为4和2,侧棱AA1的长为5.(1)求三棱柱ABC－A1B1C1的体积；(2)设M是BC中点,求直线A1M与平面ABC所成角的大小.18.(14分)已知函数f(x)＝cos2x－sin2x＋,x∈(0,π).(1)求f(x)的单调递增区间；(2)设△ABC为锐角三角形,角A所对边a＝,角B所对边b＝5,若f(A)＝0,求△ABC的面积.19.(14分)根据预测,某地第n(n∈N*)个月共享单车的投放量和损失量分别为an 和bn(单位：辆),其中an ＝,bn＝n＋5,第n个月底的共享单车的保有量是前n个月的累计投放量与累计损失量的差.(1)求该地区第4个月底的共享单车的保有量；(2)已知该地共享单车停放点第n个月底的单车容纳量Sn＝－4(n－46)2＋8800(单位：辆).设在某月底,共享单车保有量达到最大,问该保有量是否超出了此时停放点的单车容纳量？20.(16分)在平面直角坐标系xOy中,已知椭圆Γ：＝1,A为Γ的上顶点,P为Γ上异于上、下顶点的动点,M为x正半轴上的动点.(1)若P在第一象限,且|OP|＝,求P的坐标；(2)设P(),若以A、P、M为顶点的三角形是直角三角形,求M的横坐标；(3)若|MA|＝|MP|,直线AQ与Γ交于另一点C,且,,求直线AQ的方程.21.(18分)设定义在R上的函数f(x)满足：对于任意的x1、x2∈R,当x1＜x2时,都有f(x1)≤f(x2).(1)若f(x)＝ax3＋1,求a的取值范围；(2)若f(x)是周期函数,证明：f(x)是常值函数；(3)设f(x)恒大于零,g(x)是定义在R上的、恒大于零的周期函数,M是g(x)的最大值.函数h(x)＝f(x)g(x).证明：“h(x)是周期函数”的充要条件是“f(x)是常值函数”.2017年上海市高考数学试卷参考答案与试题解析一、填空题(本大题共12题,满分54分,第1～6题每题4分,第7～12题每题5分)1.(4分)已知集合A＝{1,2,3,4},集合B＝{3,4,5},则A∩B＝{3,4} .【分析】利用交集定义直接求解.【解答】解：∵集合A＝{1,2,3,4},集合B＝{3,4,5},∴A∩B＝{3,4}.故答案为：{3,4}.【点评】本题考查交集的求法,是基础题,解题时要认真审题,注意交集定义的合理运用.2.(4分)若排列数＝6×5×4,则m＝ 3 .【分析】利用排列数公式直接求解.【解答】解：∵排列数＝6×5×4,∴由排列数公式得,∴m＝3.故答案为：m＝3.【点评】本题考查实数值的求法,是基础题,解题时要认真审题,注意排列数公式的合理运用.3.(4分)不等式＞1的解集为(－∞,0) .【分析】根据分式不等式的解法求出不等式的解集即可.【解答】解：由＞1得：,故不等式的解集为：(－∞,0),故答案为：(－∞,0).【点评】本题考查了解分式不等式,考查转化思想,是一道基础题.4.(4分)已知球的体积为36π,则该球主视图的面积等于9π.【分析】由球的体积公式,可得半径R＝3,再由主视图为圆,可得面积.【解答】解：球的体积为36π,设球的半径为R,可得πR3＝36π,可得R＝3,该球主视图为半径为3的圆,可得面积为πR2＝9π.故答案为：9π.【点评】本题考查球的体积公式,以及主视图的形状和面积求法,考查运算能力,属于基础题.5.(4分)已知复数z满足z＋＝0,则|z|＝.【分析】设z＝a＋bi(a,b∈R),代入z2＝－3,由复数相等的条件列式求得a,b的值得答案.【解答】解：由z＋＝0,得z2＝－3,设z＝a＋bi(a,b∈R),由z2＝－3,得(a＋bi)2＝a2－b2＋2abi＝－3,即,解得：.∴.则|z|＝.故答案为：.【点评】本题考查复数代数形式的乘除运算,考查了复数相等的条件以及复数模的求法,是基础题.6.(4分)设双曲线－＝1(b＞0)的焦点为F1、F2,P为该双曲线上的一点,若|PF1|＝5,则|PF2|＝11 .【分析】根据题意,由双曲线的方程可得a的值,结合双曲线的定义可得||PF1|－|PF2||＝6,解可得|PF2|的值,即可得答案.【解答】解：根据题意,双曲线的方程为：－＝1, 其中a＝＝3,则有||PF1|－|PF2||＝6,又由|PF1|＝5,解可得|PF2|＝11或－1(舍)故|PF2|＝11,故答案为：11.【点评】本题考查双曲线的几何性质,关键是掌握双曲线的定义.7.(5分)如图,以长方体ABCD－A1B1C1D1的顶点D为坐标原点,过D的三条棱所在的直线为坐标轴,建立空间直角坐标系,若的坐标为(4,3,2),则的坐标是(－4,3,2) .【分析】由的坐标为(4,3,2),分别求出A和C1的坐标,由此能求出结果.【解答】解：如图,以长方体ABCD－A1B1C1D1的顶点D为坐标原点,过D的三条棱所在的直线为坐标轴,建立空间直角坐标系,∵的坐标为(4,3,2),∴A(4,0,0),C1(0,3,2),∴.故答案为：(－4,3,2).【点评】本题考查空间向量的坐标的求法,考查空间直角坐标系等基础知识,考查运算求解能力,考查数形结合思想,是基础题.8.(5分)定义在(0,＋∞)上的函数y＝f(x)的反函数为y＝f－1(x),若g(x)＝为奇函数,则f－1(x)＝2的解为.【分析】由奇函数的定义,当x＞0时,－x＜0,代入已知解析式,即可得到所求x＞0的解析式,再由互为反函数的两函数的自变量和函数值相反,即可得到所求值.【解答】解：若g(x)＝为奇函数,可得当x＞0时,－x＜0,即有g(－x)＝3－x－1,由g(x)为奇函数,可得g(－x)＝－g(x),则g(x)＝f(x)＝1－3－x,x＞0,由定义在(0,＋∞)上的函数y＝f(x)的反函数为y＝f－1(x),且f－1(x)＝2,可由f(2)＝1－3－2＝,可得f－1(x)＝2的解为x＝.故答案为：.【点评】本题考查函数的奇偶性和运用,考查互为反函数的自变量和函数值的关系,考查运算能力,属于基础题.9.(5分)已知四个函数：①y＝－x,②y＝－,③y＝x3,④y＝x,从中任选2个,则事件“所选2个函数的图象有且仅有一个公共点”的概率为.【分析】从四个函数中任选2个,基本事件总数n＝,再利用列举法求出事件A：“所选2个函数的图象有且只有一个公共点”包含的基本事件的个数,由此能求出事件A：“所选2个函数的图象有且只有一个公共点”的概率.【解答】解：给出四个函数：①y＝－x,②y＝－,③y＝x3,④y＝x,从四个函数中任选2个,基本事件总数n＝,③④有两个公共点(0,0),(1,1).事件A：“所选2个函数的图象有且只有一个公共点”包含的基本事件有：①③,①④共2个,∴事件A：“所选2个函数的图象有且只有一个公共点”的概率为P(A)＝＝.故答案为：.【点评】本题考查概率的求法,是基础题,解题时要认真审题,注意列举法的合理运用.10.(5分)已知数列{an }和{bn},其中an＝n2,n∈N*,{bn}的项是互不相等的正整数,若对于任意n∈N*,{bn }的第an项等于{an}的第bn项,则＝ 2 .【分析】an ＝n2,n∈N*,若对于一切n∈N*,{bn}中的第an项恒等于{an}中的第bn项,可得＝＝.于是b1＝a1＝1,＝b4,＝b9,＝b16.即可得出.【解答】解：∵an ＝n2,n∈N*,若对于一切n∈N*,{bn}中的第an项恒等于{an}中的第bn项,∴＝＝.∴b1＝a1＝1,＝b4,＝b9,＝b16.∴b1b4b9b16＝.∴＝2.故答案为：2.【点评】本题考查了数列递推关系、对数的运算性质,考查了推理能力与计算能力,属于中档题.11.(5分)设a 1、a 2∈R,且,则|10π－a 1－a 2|的最小值等于.【分析】由题意,要使＋＝2,可得sinα1＝－1,sin2α2＝－1.求出α1和α2,即可求出|10π－α1－α2|的最小值【解答】解：根据三角函数的性质,可知sinα1,sin2α2的范围在[－1,1],要使＋＝2,∴sinα1＝－1,sin2α2＝－1.则：,k 1∈Z.,即,k 2∈Z. 那么：α1＋α2＝(2k 1＋k 2)π,k 1、k 2∈Z.∴|10π－α1－α2|＝|10π－(2k 1＋k 2)π|的最小值为.故答案为：.【点评】本题主要考察三角函数性质,有界限的范围的灵活应用,属于基本知识的考查.12.(5分)如图,用35个单位正方形拼成一个矩形,点P 1、P 2、P 3、P 4以及四个标记为“▲”的点在正方形的顶点处,设集合Ω＝{P 1,P 2,P 3,P 4},点P ∈Ω,过P 作直线l P ,使得不在l P 上的“▲”的点分布在l P 的两侧.用D 1(l P )和D 2(l P )分别表示l P 一侧和另一侧的“▲”的点到l P的距离之和.若过P 的直线l P 中有且只有一条满足D 1(l P )＝D 2(l P ),则Ω中所有这样的P 为 P 1、P 3、P 4 .【分析】根据任意四边形ABCD 两组对边中点的连线交于一点, 过此点作直线,使四边形的四个顶点不在该直线的同一侧,则该直线两侧的四边形的顶点到直线的距离之和相等；由此得出结论.【解答】解：设记为“▲”的四个点是A,B,C,D, 线段AB,BC,CD,DA的中点分别为E,F,G,H,易知EFGH为平行四边形,如图所示；又平行四边形EFGH的对角线交于点P2,则符合条件的直线lP 一定经过点P2,且过点P2的直线有无数条；由过点P1和P2的直线有且仅有1条,过点P3和P2的直线有且仅有1条,过点P4和P2的直线有且仅有1条,所以符合条件的点是P1、P3、P4.故答案为：P1、P3、P4.【点评】本题考查了数学理解力与转化力的应用问题,也考查了对基本问题的阅读理解和应用转化能力.二、选择题(本大题共4题,每题5分,共20分)13.(5分)关于x、y的二元一次方程组的系数行列式D为( )A. B. C. D.【分析】利用线性方程组的系数行列式的定义直接求解.【解答】解：关于x、y的二元一次方程组的系数行列式：D＝.故选：C.【点评】本题考查线性方程组的系数行列式的求法,是基础题,解题时要认真审题,注意线性方程组的系数行列式的定义的合理运用.14.(5分)在数列{an }中,an＝(－)n,n∈N*,则an( )A.等于B.等于0C.等于D.不存在【分析】根据极限的定义,求出an＝的值.【解答】解：数列{an }中,an＝(－)n,n∈N*,则an＝＝0.故选：B.【点评】本题考查了极限的定义与应用问题,是基础题.15.(5分)已知a、b、c为实常数,数列{xn }的通项xn＝an2＋bn＋c,n∈N*,则“存在k∈N*,使得x100＋k 、x200＋k、x300＋k成等差数列”的一个必要条件是( )A.a≥0B.b≤0C.c＝0D.a－2b＋c＝0【分析】由x100＋k ,x200＋k,x300＋k成等差数列,可得：2x200＋k＝x100＋kx300＋k,代入化简即可得出.【解答】解：存在k∈N*,使得x100＋k 、x200＋k、x300＋k成等差数列,可得：2[a(200＋k)2＋b(200＋k)＋c]＝a(100＋k)2＋b(100＋k)＋c＋a(300＋k)2＋b(300＋k)＋c,化为：a＝0.∴使得x100＋k ,x200＋k,x300＋k成等差数列的必要条件是a≥0.故选：A.【点评】本题考查了等差数列的通项公式、简易逻辑的判定方法,考查了推理能力与计算能力,属于基础题.16.(5分)在平面直角坐标系xOy中,已知椭圆C1：＝1和C2：x2＋＝1.P为C1上的动点,Q为C2上的动点,w是的最大值.记Ω＝{(P,Q)|P在C1上,Q在C2上,且＝w},则Ω中元素个数为( )A.2个B.4个C.8个D.无穷个【分析】设出P(6cosα,2sinα),Q(cosβ,3sinβ),0≤α\β＜2π,由向量数量积的坐标表示和两角差的余弦公式和余弦函数的值域,可得最大值及取得的条件,即可判断所求元素的个数.【解答】解：椭圆C1：＝1和C2：x2＋＝1.P为C1上的动点,Q为C2上的动点,可设P(6cosα,2sinα),Q(cosβ,3sinβ),0≤α\β＜2π,则＝6cosαcosβ＋6sinαsinβ＝6cos(α－β),当α－β＝2kπ,k∈Z时,w取得最大值6,则Ω＝{(P,Q)|P在C1上,Q在C2上,且＝w}中的元素有无穷多对.另解：令P(m,n),Q(u,v),则m2＋9n2＝36,9u2＋v2＝9,由柯西不等式(m2＋9n2)(9u2＋v2)＝324≥(3mu＋3nv)2,当且仅当mv＝nu,即O、P、Q共线时,取得最大值6,显然,满足条件的P、Q有无穷多对,D项正确.故选：D.【点评】本题考查椭圆的参数方程的运用,以及向量数量积的坐标表示和余弦函数的值域,考查集合的几何意义,属于中档题.三、解答题(本大题共5题,共14＋14＋14＋16＋18＝76分)17.(14分)如图,直三棱柱ABC－A1B1C1的底面为直角三角形,两直角边AB和AC的长分别为4和2,侧棱AA1的长为5.(1)求三棱柱ABC－A1B1C1的体积；(2)设M是BC中点,求直线A1M与平面ABC所成角的大小.【分析】(1)三棱柱ABC－A1B1C1的体积V＝S△ABC×AA1＝,由此能求出结果.(2)连结AM,∠A1MA是直线A1M与平面ABC所成角,由此能求出直线A1M与平面ABC所成角的大小.【解答】解：(1)∵直三棱柱ABC－A1B1C1的底面为直角三角形,两直角边AB和AC的长分别为4和2,侧棱AA1的长为5.∴三棱柱ABC－A1B1C1的体积：V＝S△ABC ×AA1＝＝＝20.(2)连结AM,∵直三棱柱ABC－A1B1C1的底面为直角三角形,两直角边AB和AC的长分别为4和2,侧棱AA1的长为5,M是BC中点,∴AA1⊥底面ABC,AM＝＝,∴∠A1MA是直线A1M与平面ABC所成角,tan∠A1MA＝＝＝,∴直线A1M与平面ABC所成角的大小为arctan.【点评】本题考查三棱柱的体积的求法,考查线面角的大小的求法,考查空间中线线、线面、面面间的位置关系等基础知识,考查推理论证能力、运算求解能力、空间想象能力,考查化归与转化思想、函数与方程思想、数形结合思想,是中档题.18.(14分)已知函数f(x)＝cos2x－sin2x＋,x∈(0,π).(1)求f(x)的单调递增区间；(2)设△ABC为锐角三角形,角A所对边a＝,角B所对边b＝5,若f(A)＝0,求△ABC的面积.【分析】(1)由二倍角的余弦公式和余弦函数的递增区间,解不等式可得所求增区间；(2)由f(A)＝0,解得A,再由余弦定理解方程可得c,再由三角形的面积公式,计算即可得到所求值.【解答】解：(1)函数f(x)＝cos2x－sin2x＋＝cos2x＋,x∈(0,π),由2kπ－π≤2x≤2kπ,解得kπ－π≤x≤kπ,k∈Z,k＝1时,π≤x≤π,可得f(x)的增区间为[,π)；(2)设△ABC为锐角三角形,角A所对边a＝,角B所对边b＝5,若f(A)＝0,即有cos2A＋＝0,解得2A＝π,即A＝π,由余弦定理可得a2＝b2＋c2－2bccosA,化为c2－5c＋6＝0,解得c＝2或3,若c＝2,则cosB＝＜0,即有B为钝角,c＝2不成立,则c＝3,△ABC的面积为S＝bcsinA＝×5×3×＝.【点评】本题考查二倍角公式和余弦函数的图象和性质,考查解三角形的余弦定理和面积公式的运用,考查运算能力,属于中档题.19.(14分)根据预测,某地第n(n∈N*)个月共享单车的投放量和损失量分别为an 和bn(单位：辆),其中an ＝,bn＝n＋5,第n个月底的共享单车的保有量是前n个月的累计投放量与累计损失量的差.(1)求该地区第4个月底的共享单车的保有量；(2)已知该地共享单车停放点第n个月底的单车容纳量Sn＝－4(n－46)2＋8800(单位：辆).设在某月底,共享单车保有量达到最大,问该保有量是否超出了此时停放点的单车容纳量？【分析】(1)计算出{an }和{bn}的前4项和的差即可得出答案；(2)令an ≥bn得出n≤42,再计算第42个月底的保有量和容纳量即可得出结论.【解答】解：(1)∵an ＝,bn＝n＋5∴a1＝5×14＋15＝20a2＝5×24＋15＝95a3＝5×34＋15＝420a4＝－10×4＋470＝430b1＝1＋5＝6b2＝2＋5＝7b3＝3＋5＝8b4＝4＋5＝9∴前4个月共投放单车为a1＋a2＋a3＋a4＝20＋95＋420＋430＝965,前4个月共损失单车为b1＋b2＋b3＋b4＝6＋7＋8＋9＝30,∴该地区第4个月底的共享单车的保有量为965－30＝935.(2)令an ≥bn,显然n≤3时恒成立,当n≥4时,有－10n＋470≥n＋5,解得n≤, ∴第42个月底,保有量达到最大.当n≥4,{an }为公差为－10等差数列,而{bn}为等差为1的等差数列,∴到第42个月底,单车保有量为×39＋535－×42＝×39＋535－×42＝8782.S42＝－4×16＋8800＝8736.∵8782＞8736,∴第42个月底单车保有量超过了容纳量.【点评】本题考查了数列模型的应用,等差数列的求和公式,属于中档题.20.(16分)在平面直角坐标系xOy中,已知椭圆Γ：＝1,A为Γ的上顶点,P为Γ上异于上、下顶点的动点,M为x正半轴上的动点.(1)若P在第一象限,且|OP|＝,求P的坐标；(2)设P(),若以A、P、M为顶点的三角形是直角三角形,求M的横坐标；(3)若|MA|＝|MP|,直线AQ与Γ交于另一点C,且,,求直线AQ的方程.【分析】(1)设P(x,y)(x＞0,y＞0),联立,能求出P点坐标.(2)设M(x0,0),A(0,1),P(),由∠P＝90°,求出x＝；由∠M＝90°,求出x＝1或x＝；由∠A＝90°,则M点在x轴负半轴,不合题意.由此能求出点M的横坐标.(3)设C(2cosα,sinα),推导出Q(4cosα,2sinα－1),设P(2cosβ,sinβ),M(x,0)推导出x＝cosβ,从而4cosα－2cosβ＝－5cosβ,且2sinα－sinβ－1＝－4sinβ,cosβ＝－cosα,且sinα＝(1－2sinα),由此能求出直线AQ.【解答】解：(1)设P(x,y)(x＞0,y＞0),∵椭圆Γ：＝1,A为Γ的上顶点,P为Γ上异于上、下顶点的动点,P在第一象限,且|OP|＝,∴联立,解得P(,).(2)设M(x,0),A(0,1),P(),若∠P＝90°,则•,即(x－,－)•(－,)＝0,∴(－)x0＋－＝0,解得x＝.如图,若∠M＝90°,则•＝0,即(－x0,1)•(－x,)＝0,∴＝0,解得x0＝1或x＝,若∠A＝90°,则M点在x轴负半轴,不合题意.∴点M的横坐标为,或1,或.(3)设C(2cosα,sinα),∵,A(0,1),∴Q(4cosα,2sinα－1),又设P(2cosβ,sinβ),M(x,0),∵|MA|＝|MP|,∴x02＋1＝(2cosβ－x)2＋(sinβ)2,整理得：x＝cosβ,∵＝(4cosα－2cosβ,2sinα－sinβ－1),＝(－cosβ,－sinβ),,∴4cosα－2cosβ＝－5cosβ,且2sinα－sinβ－1＝－4sinβ,∴cosβ＝－cosα,且sinα＝(1－2sinα),以上两式平方相加,整理得3(sinα)2＋sinα－2＝0,∴sinα＝,或sinα＝－1(舍去),此时,直线AC的斜率kAC＝－＝ (负值已舍去),如图.∴直线AQ为y＝x＋1.【点评】本题考查点的坐标的求法,考查直线方程的求法,考查椭圆、直线方程、三角函数等基础知识,考查推理论证能力、运算求解能力,考查化归与转化思想、函数与方思想,是中档题.21.(18分)设定义在R上的函数f(x)满足：对于任意的x1、x2∈R,当x1＜x2时,都有f(x1)≤f(x2).(1)若f(x)＝ax3＋1,求a的取值范围；(2)若f(x)是周期函数,证明：f(x)是常值函数；(3)设f(x)恒大于零,g(x)是定义在R上的、恒大于零的周期函数,M是g(x)的最大值.函数h(x)＝f(x)g(x).证明：“h (x)是周期函数”的充要条件是“f (x)是常值函数”. 【分析】(1)直接由f(x 1)－f(x 2)≤0求得a 的取值范围；(2)若f(x)是周期函数,记其周期为T k ,任取x 0∈R,则有f(x 0)＝f(x 0＋T k ),证明对任意x ∈[x 0,x 0＋T k ],f(x 0)≤f(x)≤f(x 0＋T k ),可得f(x 0)＝f(x 0＋nT k ),n ∈Z,再由…∪[x 0－3T k ,x 0－2T k ]∪[x 0－2T k ,x 0－T k ]∪[x 0－T k ,x 0]∪[x 0,x 0＋T k ]∪[x 0＋T k ,x 0＋2T k ]∪…＝R,可得对任意x ∈R,f(x)＝f(x 0)＝C,为常数；(3)分充分性及必要性证明.类似(2)证明充分性；再证必要性,然后分类证明. 【解答】(1)解：由f(x 1)≤f(x 2),得f(x 1)－f(x 2)＝a(x 13－x 23)≤0, ∵x 1＜x 2,∴x 13－x 23＜0,得a ≥0. 故a 的范围是[0,＋∞)；(2)证明：若f(x)是周期函数,记其周期为T k ,任取x 0∈R,则有 f(x 0)＝f(x 0＋T k ),由题意,对任意x ∈[x 0,x 0＋T k ],f(x 0)≤f(x)≤f(x 0＋T k ), ∴f(x 0)＝f(x)＝f(x 0＋T k ).又∵f(x 0)＝f(x 0＋nT k ),n ∈Z,并且…∪[x 0－3T k ,x 0－2T k ]∪[x 0－2T k ,x 0－T k ]∪[x 0－T k ,x 0]∪[x 0,x 0＋T k ]∪[x 0＋T k ,x 0＋2T k ]∪…＝R,∴对任意x ∈R,f(x)＝f(x 0)＝C,为常数；(3)证明：充分性：若f(x)是常值函数,记f(x)＝c 1,设g(x)的一个周期为T g ,则 h(x)＝c 1•g(x),则对任意x 0∈R,h(x 0＋T g )＝c 1•g(x 0＋T g )＝c 1•g(x 0)＝h(x 0),故h(x)是周期函数；必要性：若h(x)是周期函数,记其一个周期为T h .若存在x 1,x 2,使得f(x 1)＞0,且f(x 2)＜0,则由题意可知, x 1＞x 2,那么必然存在正整数N 1,使得x 2＋N 1T k ＞x 1, ∴f(x 2＋N 1T k )＞f(x 1)＞0,且h(x 2＋N 1T k )＝h(x 2). 又h(x 2)＝g(x 2)f(x 2)＜0,而h(x 2＋N 1T k )＝g(x 2＋N 1T k )f(x 2＋N 1T k )＞0≠h(x 2),矛盾. 综上,f(x)＞0恒成立. 由f(x)＞0恒成立,任取x 0∈A,则必存在N 2∈N,使得x 0－N 2T h ≤x 0－T g , 即[x 0－T g ,x 0]⊆[x 0－N 2T h ,x 0],∵…∪[x 0－3T k ,x 0－2T k ]∪[x 0－2T k ,x 0－T k ]∪[x 0－T k ,x 0]∪[x 0,x 0＋T k ]∪[x 0＋T k ,x 0＋2T k ]∪…＝R,∴…∪[x 0－2N 2T h ,x 0－N 2T h ]∪[x 0－N 2T h ,x 0]∪[x 0,x 0＋N 2T h ]∪[x 0＋N 2T h ,x 0＋2N 2T h ]∪…＝R. h(x 0)＝g(x 0)•f(x 0)＝h(x 0－N 2T h )＝g(x 0－N 2T h )•f(x 0－N 2T h ),∵g(x 0)＝M ≥g(x 0－N 2T h )＞0,f(x 0)≥f(x 0－N 2T h )＞0.因此若h(x 0)＝h(x 0－N 2T h ),必有g(x 0)＝M ＝g(x 0－N 2T h ),且f(x 0)＝f(x 0－N 2T h )＝c. 而由(2)证明可知,对任意x ∈R,f(x)＝f(x 0)＝C,为常数. 综上,必要性得证. 【点评】本题考查抽象函数及其应用,考查逻辑思维能力与理论运算能力考查分类讨论的数学思想方法,题目设置难度过大.。

2017年高考数学上海卷含答案

数学试卷第1页（共14页）数学试卷第2页（共14页）绝密★启用前上海市2017年普通高等学校招生全国统一考试数学本试卷共150分.考试时长120分钟.一、填空题：本大题共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{1,2,3,4}A =，{3,4,5}B =，那么A B = . 2.若排列数6654m P =⨯⨯，则m = .3.不等式11x x-＞的解集为 .4.已知球的体积为36π，则该球主视图的面积等于 .5.已知复数z 满足30z z+=的定义域为 .6.设双曲线2221(0)9x yb b-=＞的焦点为1F 、2F ，P 为该双曲线上的一点，若1||5PF =，则2||PF = .7.如图，以长方体1111ABCD A B C D -的顶点D 为坐标原点，过D 的三条棱所在的直线为坐标轴，建立空间直角坐标系，若1DB 的坐标为(4,3,2)，则1AC 的坐标是 .8.定义在(0,)+∞上的函数()y f x =反函数为1()y f x -=，若31,0()(),0x x g x f x x ⎧-=⎨⎩≤＞为奇函数，则1()2f x -=的解为 .9.已知四个函数：①y x =-，②1y x=-，③3y x =，④12y x =，从中任选2个，则事件“所选2个函数的图象有且仅有一个公共点”的概率为 .10.已知数列{}n a 和{}n b ，其中2na n =，n ∈*N ，{}n b 的项是互不相等的正整数，若对于任意n ∈*N ，{}n b 的第na 项等于{}n a 的第nb 项，则149161234lg()lg()b b b b b b b b == . 11.设1a 、2a ∈R ，且121122sin 2sin(2)a a +=++，则12|10π|a a --的最小值等于 .12.如图，用35个单位正方形拼成一个矩形，点1P 、2P 、3P 、4P 以及四个标记为“▲”的点在正方形的顶点处，设集合1234{P ,P ,P ,P }Ω=，点P ∈Ω，过P 作直线P l ，使得不在P l 上的“▲”的点分布在P l 的两侧．用1D （P l ）和2D （P l ）分别表示P l 一侧和另一侧的“▲”的点到P l 的距离之和．若过P 的直线P l 中有且只有一条满足1D （P l ）2D =（P l ），则Ω中所有这样的P 为 .二、选择题：本大题共4小题，每题5分，共20分.13.关于x 、y 的二元一次方程组50234x y x y +=⎧⎨+=⎩的系数行列式D 为（）A .0543B .1024C .1523D .605414.在数列{}n a 中，12nn a ⎛⎫=- ⎪⎝⎭，n ∈*N ，则lim n n a →∞ （）A .等于12-B .等于0C .等于12D .不存在15.已知a 、b 、c 为实常数，数列{}n x 的通项2n x an bn c =++，n ∈*N ，则“存在k ∈*N ，使得100k x +、200k x +、300k x +成等差数列”的一个必要条件是（）A .0a ≥B .0b ≤C .0c =D .20a b c -+=16.在平面直角坐标系xOy 中，已知椭圆221:1364x y C +=和222:19y C x +=．P 为1C 上的动点，Q 为2C 上的动点，w 是OP OQ 的最大值．记{(,)}P Q Ω=，P 在1C 上，Q 在2C 上，且OP OQ w =，则Ω中元素个数为（）A .2个B .4个C .8个D .无穷个毕业学校_____________ 姓名________________ 考生号________________________________ _____________-------------在--------------------此--------------------卷--------------------上--------------------答--------------------题--------------------无--------------------效----------------数学试卷第3页（共14页）数学试卷第4页（共14页）三、解答题：本大题共5题，共14+14+14+16+18=76分.17.如图，直三棱柱111ABC A B C -的底面为直角三角形，两直角边AB 和AC 的长分别为4和2，侧棱1AA 的长为5.（1）求三棱柱111ABC A B C -的体积；（2）设M 是BC 中点，求直线1A M 与平面ABC 所成角的大小.18.已知函数221()cos sin 2f x x x =-+，(0,π)x ∈. （1）求()f x 的单调递增区间；（2）设ABC △为锐角三角形，角A所对边a =角B 所对边5b =，若()0f A =，求ABC △的面积.19.根据预测，某地第()n n ∈*N 个月共享单车的投放量和损失量分别为n a 和n b （单位：辆），其中4515,1310470,4n n n a n n ⎧+=⎨-+⎩≤≤≥，5n b n =+，第n 个月底的共享单车的保有量是前n 个月的累计投放量与累计损失量的差.（1）求该地区第4个月底的共享单车的保有量；（2）已知该地共享单车停放点第n 个月底的单车容纳量24(46)8800n S n =--+（单位：辆）.设在某月底，共享单车保有量达到最大，问该保有量是否超出了此时停放点的单车容纳量？20.在平面直角坐标系xOy 中，已知椭圆22:14x y Γ+=，A 为Γ的上顶点，P 为Γ上异于上、下顶点的动点，M 为x 正半轴上的动点.（1）若P在第一象限，且||OP =P 的坐标；（2）设83,55P ⎛⎫⎪⎝⎭，若以A 、P 、M 为顶点的三角形是直角三角形，求M 的横坐标；（3）若||||MA MP =，直线AQ 与Γ交于另一点C ，且2AQ AC =，4PQ PM =，求直线AQ 的方程.21.设定义在R 上的函数()f x 满足：对于任意的1x 、2x ∈R ，当12x x ＜时，都有12()()f x f x ≤.（1）若3()1f x ax =+，求a 的取值范围；（2）若()f x 是周期函数，证明：()f x 是常值函数；（3）设()f x 恒大于零，g()x 是定义在R 上的、恒大于零的周期函数，M 是g()x 的最大值.函数()()()h x f x g x =.证明：“()h x 是周期函数”的充要条件是“()f x 是常值函数”.数学试卷第5页（共14页）数学试卷第6页（共14页）上海市2017年普通高等学校招生全国统一考试数学答案解析一、填空题 1.【答案】{3,4}解析：利用交集定义直接求解。

上海市八校2017届高三联考英语试题

2017届高三年级八校联合调研英语试卷2016年11月(满分140分，考试时间120分钟)第I卷（共90分）II. Grammar and VocabularySection A(10×1=10分)Directions: Read the following passage. Fill in the blanks to make the passage coherent. For the blanks with a given word, fill in each blank with the proper form of the given word. For the other blanks, fill in each blank with one proper word. Make sure that your answers are grammatically correct.Have you ever seen an old movie called Three Coins in the Fountain? It is about three young American women (21) _______(search) for permanent romance in Rome and they all find it. Far-fetched Hollywood? Well, from the world history point of view, romance did, in fact, set down its roots in Rome.The word romance evolved in Latin from Roma to Romanicus of the Roman language, to the Old French romanz escrive, (22) _______means “to write in a Romance language,” and on to the English romance.The Romance languages (23) ____________(compose) of seven groups of languages that all have Latin (24) ______their basis. These languages include French, Italian, Spanish and Portuguese. The common people in ancient Rome spoke (25)________ is referred to as Vulgar Latin, an informal speech, as opposed to the classical Latin of the more educated. Most language experts agree that Vulgar Latin is the chief source of the Romance languages.Medieval Romances were tales (26) __________(write)primary in French verse about brave heroes. The notion of having a romance with another person is thought (27) __________(develop) sometime during the Middle Ages. In the late 18th century and on through the 19th, a romance was not a love story (28) _________ a work of prose fiction that contained far-fetched, mysterious events. Romances of this period (29) _________(include) English Gothic novels like The Castle of Otranto by Horace Walpole.What exactly is a twentieth-century romance ? Does it have any relationship with the lively, popular novels written today, with their fantastic plots of love affairs? Or did the playwright Oscar Wilde have it right in The Picture of Dorian Gray: “ When one is in love, one always begins by deceiving (30) _________, and one always ends by deceiving others. That is what the world calls a romance.”Section B（10×1=10分）Directions: Complete the following passage by using the words in the box. Each word can be used only once. Note that there is one word more than you need.A.astonishinglyB. surroundingC. collapseD. unnoticedE. interruptedF. previouslyG. congratulateH. predictionsI. potential J. producing K. propertiesIn the wake of the historic announcement of the discovery of gravitational waves on February 11, 2016 by the Laser Interferometer Gravitational-Wave Observatory (LIGO), British physicist and black hole theorist Stephen Hawking was quick to ____31___ the US-led collaboration, sharing his excitement for the historic news.According to Hawking, these results confirm several very important ____32___ of Einstein’s theory of general relativity and it also confirms the existence of gravitational waves directly.As is becoming clear, the direct detection of these ripples in space time not only confirms Einstein’s famous theory of general theory but it also opens our eyes to a(n) _33________ “dark” universe. Astronomers employ the electromagnetic spectrum（电磁光谱）to study the universe, but objects that do not radiate in the electromagnetic spectrum will go ___34____. But now we know how to detect gravitational waves, which can help us detect and study some of the most energetic cosmic phenomena.“Gravitat ional waves provide a completely new way of looking at the universe and the ability to detect them has the ___35___ to revolutionize astronomy” said Hawking. “The discovery is the first observation of black holes merging. The observed __36____ of this system are consistent with predictions about black holes that I made in 1970 in Cambridge.”However, this discovery also presents a puzzle for astrophysicists. The mass of each of the black holes are larger than expected for those formed by the gravitational __37_____ of a star---so how did both of these black holes become so massive?This question touches on one of the biggest mysteries ___38___ black hole evolution. Currently, astronomers are having a hard time understanding how black holes grow to be so massive. On the one end of the scale, there are “stellar mass（恒星质量）” black holes that form immediately after a massive star explodes, ___39____ an extremely bright light. And we also have an abundance of evidence for the existence of the super-massive that live in the centers of most galaxies. There is a disconnect, however. If black holes grow by merging and consuming stellar matter, there should be evidence of black holes of all sizes, but “intermediate mass” black holes and black holes of a few dozen solar masses are ____40____ rare, throwing some black holes evolution theories into doubt.One thing is clear, however. This is the first time that we’ve acquired direct evidence of a black hole merger. So it’s good to know we’re on the right track.III. Reading ComprehensionSection A（15×1=15分）Directions: 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.Cowboy or spaceman ? A dilemma for a children’s party, perhaps. But also a question for economists, argued Kenneth Boulding, in an essay published in 1966. We have run our 41 , he warned, like cowboys on the open grassland: taking and using the world’s resources, 42 ____ that more lies over the horizon. But the Earth is 43 a grassland than a spaceship---a closed system, alone in space, carrying exhaustible supplies. We need, said Boulding, an economics that takes seriously the idea of environmental 44 . In the half century since his essay, a new movement has respo nded to his challenge. “Ecological economists,” as they call themselves, want to 45 __ its aims and assumptions. What do they say -- and will their ideas take off?To its 46 , ecological economics is neither ecology nor economics, but a mix of both. Their starting point is to recognize that the human economy is part of the natural world. Our environment, theynote, is both a source of resources and a sink for wastes. But it is 47 in traditional textbooks, where neat diagrams trace the flows between firms, households and the government as though nature did not exist. That is a huge mistake.There are two ways our economies can grow, ecological economists point out: through technological change, or through maximum use of resources. Only the 48 , they say, is worth having. They are suspicious of GDP (gross domestic product), a simple 49 which does not take into account resource exhaustion, unpaid work and countless other factors. 50 , they advocate more holistic approaches, such as GPI (genuine progress indicator)，a composite（复合的）index that include things like the cost of pollution, deforestation and car accidents. While GDP has kept growing, global GPI per person 51 in 1978: by destroying our environment, we are making ourselves poorer, not richer. The solution, according to experts, lies in a “steady-state” economy, where the use of materials and energy is held 52 .Mainstream economists are not 53 . GPI, they point out, is a subjective standard. And talk of limits to growth has had a bad press since the days of Thomas Malthus, who predicted in the 18th century, wrongly, that overpopulation would lead to famine. Human beings find solutions to some of the most annoying problems. But ecological economists 54 self-satisfaction. In 2009, a paper in Nature argued that human activity is already 55 safe planetary boundaries on issues such as biodiversity and climate change. That suggests ecologist economists are at least asking some important questions, even if their answers turn out to be wrong.41. A. grassland B. nation C. economy D. spaceship42. A. ignorant B. confident C. astonished D. anxious43. A. less B. smaller C. more D. larger44. A. movements B. influences C. limits D. threats45. A. reject B. realize C. resemble D. revolutionize46. A. challengers B. learners C. advocates D. professors47. A. addressed B. ignored C. opposed D. reflected48. A. advanced B. former C. latter D. scientific49. A. number B. product C. idea D. measure50. A. In addition B. For example C. In other words D. In its place51.A. peaked B. plunged C. persisted D. paused52.A. sufficient B. efficient C. constant D. adequate53.A. impressed B. involved C. concerned D. appointed54.A. call for B. contribute to C. warn against D. refer to55.A. setting B. overstepping C. extending D. redrawingSection B（11×2=22分）Directions:Read the following three 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)Every April I am troubled by the same concern -- that spring might not occur this year. The landscape looks dull, with hills, sky and forest appearing gray. My spirits ebb, as they did during an April snowfallwhen I first came to Maine 15 years ago. "Just wait," a neighbor advised. "You'll wake up one morning and spring will just be here."And look, on May 3 that year I awoke to a green so amazing as to be almost electric, as if spring were simply a matter of flipping a switch. Hills, sky and forest revealed their purples, blues and green. Leaves had unfolded and daffodils were fighting their way heavenward.Then there was the old apple tree. It sits on an undeveloped lot in my neighborhood. It belongs to no one and therefore to everyone. The tree's dark twisted branches stretch out in unpruned（未经修剪的）abandon. Each spring it blossoms so freely that the air becomes filled with the scent of apple.Until last year, I thought I was the only one aware of this tree. And then one day, in a bit of spring madness, I set out with pruner to remove a few disorderly branches. No sooner had I arrived under the tree than neighbors opened their windows and stepped onto their porches. These were people I barely knew and seldom spoke to, but it was as if I had come uninvited into their personal gardens.My mobile-home neighbor was the first to speak."You're not cutting it down, are you?" she asked anxiously. Another neighbor frowned as I cut off a branch. "Don't kill it, now," he warned. Soon half the neighborhood had joined me under the apple tree. It struck me that I had lived there for five years and only now was learning these people's names, what they did for a living and how they passed the winter. It was as if the old apple tree was gathering us under its branches for the purpose of both acquaintanceship and shared wonder. I couldn't help recalling Robert Frost's words:The trees that have it in their pent-up budsTo darken nature and be summer woodsOne thaw led to another. Just the other day I saw one of my neighbors at the local store. He remarked how this recent winter had been especially long and complained of not having seen or spoken at length to anyone in our neighborhood. And then, he looked at me and said, "We need to prune that apple tree again."56. By saying that “my spirits ebb” (Para. 1), the author means that _________________.A.he feels relievedB. he feels blueC. he is surprisedD. he is tired57. The apple tree mentioned in the passage is most likely to _________________.A. be regarded as a delight in the neighborhoodB. have been abandoned by its original ownerC. have been neglected by everyone in the communityD. be appealing only to the author58. In Para. 4, “neighbors opened their windows and stepped onto their porches” probably because ___________________________.A.they were surprised that someone unknown was pruning the treeB.they wanted to prevent the author from pruning the treeC.they were concerned about the safety of the treeD.they wanted to get to know the author59. It can be inferred that the author’s neighbor mentioned in the last paragraph most cared about _______________.A.when spring would arriveB. how to pass the long winterC. the neighborhood gatheringD. the pruning of the apple tree(B)Mount Cook National Park is home of the highest mountains and the longest glaciers. It is alpine(高山) in the purest sense---with skyscraping peaks, glaciers and permanent snow fields, all set under a star-studded sky.Key HighlightsAlthough it includes 23 peaks over 3,000 metres high, this park is very accessible. State Highway 80 leads to Mt Cook Village which is situated beside scenic Lake Pukaki and provides a comfortable base for alpine activities. Far from city lights, the stargazing here is magnificent—Aoraki Mount Cook National Park forms the majority of New Zealand's only International Dark Sky Reserve.Mountaineers regard the area to be the best climbing region, while less skilled adventurers find plenty of satisfaction with the mountain walks that lead to alpine tarns, herb fields and spectacular glacier views. Encounters with cheeky kea (mountain parrots) are part of the fun.Key ActivitiesMountain walksThere are 10 short walks beginning near the village. All tracks are formed and well marked. The Red Tarns Track, Kea Point and the Hooker Valley Track each take around two hours return. For more experienced alpine hikers, there are three mountain pass routes—over the Mueller, Copland and Ball passes.Glacier viewing and skiingHelicopters and ski－planes provide access to the park's fabulous glaciers. The Tasman Glacier is an excellent choice for intermediate skiers, while the Murchison, Darwin and Bonney glaciers promise excitement for advanced skiers. From October until May, you can explore the Tasman Glacier's terminal lake by boat.MountaineeringClimbing Mount Cook remains the ultimate challenge, but there are many other peaks to tempt experienced climbers. Tasman, Malte Brun, Elie de Beaumont, Sefton and La Perouse are quite popular..Key Tips●Climbers don't require permits, but are requested to complete a trip intentions form.●Local guides are available for climbing, walking and glacier skiing.●Winter climbing is an extreme sport—only recommended for well-prepared, experienced mountaineers.●The weather can change very suddenly—be prepared for heavy rainfall, snow and/or high winds.●The park has an airport serving domestic commercial flights and scenic flight operators.60. Which is one of the characteristics of Mount Cook National Park?A. It is alpine in the purest sense and hard to reach.B. It provides star-shining night skies for visitors.C. It attracts less skilled climbers to all alpine activities.D. It guarantees visitors a sight of cheeky kea.61. Mike is an experienced adventurer and may find ________ the most exciting.A. Mountaineering on Elie de BeaumontB. Mountain walks via Hooker Valley TrackC. Skiing on Tasman GlacierD. Climbing Mount Cook62. If you are a visitor to the park, you should ________．A. properly evaluate your own experience and skillB. get your permit prepared before you start to climbC. hire local guides to help you to train for climbingD. avoid exploring glaciers in winter(C)How many really suffer as a result of labor market problems? This is one of the most critical yet debatable social policy questions.In many ways, our social statistics overstate the degree of hardship. Unemployment does not have the same horrible consequences today as it did in the 1930’s when most of the unemployed were primary breadwinners, when income and earnings were usually much closer to the margin of survival, and when there were fewer effective social programs for those failing in the labor market. Increasing wealth, the rise of families with more than one wage earner, the growing dominance of secondary earners among the unemployed and improved social welfare protection have unquestionably relieved the consequences of joblessness. Earnings and income data also overestimate the scale of hardship. Among the millions with hourly earnings at or below the minimum wage level, the majority are from multiple-earner, relatively well-off families. Most of those counted by the poverty statistics are elderly or handicapped or have family responsibilities which keep them out of the labor force, so the poverty statistics are by no means an accurate indicator of labor market problems.Yet there are also many ways our social statistics underestimate the degree of labor-market-related hardship. The unemployment counts exclude the millions of fully employed workers whose wages are so low that their families remain in poverty. Low wages and repeated or long-time unemployment frequently interact to weaken the capacity for self-support. Since the number experiencing joblessness at some time during the year is several times that unemployed in any month, those who suffer as a result of forced idleness can equal or exceed average annual unemployment, even though only a minority of the jobless in any month really suffer. For every person counted in the monthly unemployment totals, there is another working part-time because of the inability to find full-time work, or else outside the labor force but wanting a job. Finally, income transfers in our country have always focused on the elderly, disabled, and dependent, neglecting the needs of the working poor, so that the dramatic expansion of cash and non-cash transfers does not necessarily mean that those failing in the labor market are adequately protected.As a result of such conflicting evidence, it is uncertain whether those suffering seriously as a result of labor market problems number in the hundreds of thousands or the tens of millions, and, hence, whether high levels of joblessness can be tolerated or must be counteracted(抵消）by job creation and economic stimulation. There is only one area of agreement in this debate—that the existing poverty, employment, and earnings statistics are inadequate for one of their primary applications, measuring the consequences of labormarket problems.63. In Paragraph 2, the author contrasts the 1930’s with the present in order to show that_____________.A. more people were unemployed in the 1930’sB. unemployment is more intolerable todayC. social programs are more in need nowD. i ncome level has increased since the 1930’s64.Which of the following is true according to the passage?A.A majority of low-wage workers receive earnings from more than one job.B.Repetition of short-term unemployment mainly contributes to people’s loss of workin g capacity.C. Many unemployed people are from families where other members are working.D. Labor market hardship is understated because fewer individuals are jobless than counted.65.It can be inferred from the passage that the effect of income transfers is often not felt by _________________.A. those doing a low-paid, part-time jobB. children in single-earner familiesC. workers who have just retiredD. full-time workers who become unemployed66. Which of the following is the principal topic of the passage?A. What causes labor market problems that result in suffering.B. Why income statistics are imprecise in measuring degrees of poverty.C. When poverty, employment, and earnings figures agree with each other.D. How statistics give an unclear picture of the labor-market-related suffering.Section C（4×2=8分）Directions: Read the following passage. 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. Even being good at getting others to fight most efficiently is not being civilized.B. Most people believe those who have conquered the most nations are the greatest.C. However, every year conflicts between countries and nations still claim thousands of lives.D. And not only has it won, but also because it has won, it has been in the right.E. So there has been little time to learn in, but there will be oceans of time in which to learn better.F. People don’t fight and kill each other in the streets, but nations still behave like savages.Most of the people who appear most often and most gloriously in the history books are great conquerors and generals, while the people who really helped civilization forward are often never mentioned. We do not know who first set a broken leg, or launched a seagoing boat, or calculated the length of the year but we know all about the killers and destroyers. People think so much of them that on all the highestpillars in the great cities of the world you will find the figure of a conqueror or a general. ____67_________It is possible they are, but they are not the most civilized. Animals fight, so do savages; so to be good at fighting is to be good in the way an animal or a savage is good, but it is not to be civilized. ____68_______. People fight to settle quarrels. Fighting means killing, and civilized peoples ought to be able to find some ways of settling their disputes other than by seeing which side can kill off greater number of the other side, and then saying that the side which has killed most has won. ___69______. For that is what going to war means; it means power is right.This is what the story of mankind has been like. But we must not expect too much. After all, the race of men has only just started. From the point of view of evolution, human beings are very young indeed, babies of a few months old. Scientists assume that there has been life of some sort on the earth for about twelve hundred million years; but there have been men for only one million years, and there has been civilized men for about eight thousand years.____70_____. Taking man’s civilized past at about seven or eight hours, we may estimate his future at about one hundred thousand years. Thus mankind is only at the beginning of its whole a pretty beastly business, a business of fighting and killing. We must not expect even civilized peoples not to have done these things. All we can ask is that they will sometimes have done something else.第II卷(共50分）I.Summary (10分)Directions: Reading the following passage. Summarize the main idea and the main point(s) of the passage with no more than 60 words. Use your own words as far as possible.It's not piano lessons or dance classes. Nowadays, the biggest extra-curricular activity is going to a tutor. "I spend about 800 Canadian dollars a month on tutors. It's costly," says Pat, a mother in Canada. However, she adds, "after finding out half my daughter's class had tutors, I felt like my child was going to fall behind because everyone else seemed to be ahead"Shelley, a mother of three, also has tutors constantly coming in and out of her home. "When I used to sit down with my children, it was hard to get them focused. I was always yelling. When I got a tutor once a week, they became focused for one entire hour and could get most of their homework done."Tutoring isn't simply a private school phenomenon. Nor is it geared only toward lower-achieving students. In Canada alone, seven percent of high school students reported using a tutor in 2010. That increased to 15 percent last year.Overall, parents hire tutors because they are worried schools are not meeting their expectations, but there is also a cultural shift. A special value is placed on education in Asia, where tutoring is viewed as an extension of the school day. As a large number of Asians emigrated to the West over the recent years, their attitudes towards education have had an impact.Another reason for the growth in business is parental frustration and their packed schedules. "A lot of parents just don’t have time to help their children with homework," says Julie Diamond, president of an American tutoring company. "Others couldn't help their children after Grade 3."There has been a shift in the attitudes, too. "Children used to get bullied (欺侮) for having a tutor,"Diamond says. "Now it's becoming the norm to have one."Children don't seem to mind that they have a tutor. One parent feels surprised that so many of her child’s classmates have tutors. "For the amount we pay in tuition, they should have as much extra help as they need," she says. Still, she’s now thinking of getting a tutor. Why? Her daughter has actually asked for one.II. Translation（3+3+4+5=15分）Directions: Translate the following sentences into English, using the words given in the brackets.1. 没过多久，失主就来认领他的行李了。

2017届高三数学4月等级考调研测试题(上海市黄浦区附答案)

2017届高三数学4月等级考调研测试题（上海市黄浦区附答案）黄浦区2017年高考模拟考数学试卷 2017年4月（完卷时间：120分钟满分：150分）一、填空题（本大题共有12题，满分54分. 其中第1~6题每题满分4分，第7~12题每题满分5分）考生应在答题纸相应编号的空格内直接填写结果.[ 1．函数的定义域是． 2．若关于的方程组有无数多组解，则实数 _________． 3．若“ ”是“ ”的必要不充分条件，则的最大值为． 4．已知复数， (其中i为虚数单位)，且是实数，则实数t等于． 5．若函数 (a>0,且a≠1)是R上的减函数，则a的取值范围是． 6．设变量满足约束条件则目标函数的最小值为． 7. 已知圆和两点，若圆上至少存在一点，使得，则的取值范围是． 8. 已知向量，，如果∥ ，那么的值为． 9．若从正八边形的8个顶点中随机选取3个顶点，则以它们作为顶点的三角形是直角三角形的概率是． 10．若将函数的图像向左平移个单位后，所得图像对应的函数为偶函数，则的最小值是． 11．三棱锥满足：，，，，则该三棱锥的体积V 的取值范围是． 12．对于数列，若存在正整数，对于任意正整数都有成立，则称数列是以为周期的周期数列．设，对任意正整数n都有若数列是以5为周期的周期数列，则的值可以是．(只要求填写满足条件的一个m值即可) 二、选择题（本大题共有4题，满分20分．）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分． 13．下列函数中，周期为π，且在上为减函数的是（） A．y = sin(2x+ B．y = cos(2x+ C．y = sin(x+ D．y = cos(x+ 14．如图是一个几何体的三视图，根据图中数据，可得该几何体的表面积是（） A． B． C． D． 15．已知双曲线的右焦点到左顶点的距离等于它到渐近线距离的2倍，则其渐近线方程为（） A． B． C． D． 16．如图所示，，圆与分别相切于点，，点是圆及其内部任意一点，且，则的取值范围是（）A． B． C． D．三、解答题（本大题共有5题，满分76分．）解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤． 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．如图，在直棱柱中，，，分别是的中点．（1）求证：；（2）求与平面所成角的大小及点到平面的距离．18．（本题满分14分）本题共有2小题，第小题满分6分，第小题满分8分．在中，角的对边分别为，且成等差数列．（1）求角的大小；（2）若，，求的值．19．（本题满分14分）本题共有2个小题，第1小题6分，第2小题8分．如果一条信息有n 种可能的情形（各种情形之间互不相容），且这些情形发生的概率分别为，则称（其中）为该条信息的信息熵．已知．（1）若某班共有32名学生，通过随机抽签的方式选一名学生参加某项活动，试求“谁被选中”的信息熵的大小；（2）某次比赛共有n位选手（分别记为）参加，若当时，选手获得冠军的概率为，求“谁获得冠军”的信息熵关于n的表达式．20．（本题满分16分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分6分．设椭圆M: 的左顶点为、中心为，若椭圆M过点，且．（1）求椭圆M的方程；（2）若△APQ的顶点Q也在椭圆M上，试求△APQ面积的最大值；（3）过点作两条斜率分别为的直线交椭圆M于两点，且，求证：直线恒过一个定点．21．（本题满分18分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分8分．若函数满足:对于任意正数，都有，且，则称函数为“L函数”．（1）试判断函数与是否是“L函数”；（2）若函数为“L函数”，求实数a的取值范围；（3）若函数为“L函数”，且，求证：对任意，都有．高三数学参考答案与评分标准一、填空题：（1~6题每题4分；7~12题每题5分） 1. ； 2. ；3. ；4. ；5. ；6. ；7. ；8. ；9. ； 10. ；11. ； 12. （或，或）．二、选择题：（每题5分） 13.A 14.D 15.C 16. B 三、解答题：（共76分） 17．解：（1）以A为坐标原点、AB为x轴、为y轴、为z轴建立如图的空间直角坐标系．由题意可知，故，.....................4分由，可知，即．.....................6分（2）设是平面的一个法向量，又，故由解得故．............9分设与平面所成角为，则，............12分所以与平面所成角为，点到平面的距离为．.....................14分 18．解：（1）由成等差数列，可得，.....................2分故，所以，.........4分又，所以，故，又由，可知，故，所以．.....................6分（另法：利用求解）（2）在△ABC中，由余弦定理得， (8)分即，故，又，故，..................10分所以 (12)分，故． (14)分 19．解：（1）由，可得，解之得 . …………………2分由32种情形等可能，故，……………………4分所以，答：“谁被选中”的信息熵为．……………………6分（2）获得冠军的概率为，……………8分当时，，又，故，……………………11分，以上两式相减，可得，故，答：“谁获得冠军”的信息熵为．……………………14分20．解：（1）由，可知，又点坐标为故，可得，……………………………2分因为椭圆M过点，故，可得，所以椭圆M的方程为．.................................4分（2）AP的方程为，即，由于是椭圆M上的点，故可设，.................................6分所以 (8)分当，即时，取最大值．故的最大值为．.................................10分法二：由图形可知，若取得最大值，则椭圆在点处的切线必平行于，且在直线的下方．..............................6分设方程为，代入椭圆M方程可得，由，可得，又，故．..............................8分所以的最大值．.................................10分（3）直线方程为，代入，可得，，又故，，..................12分同理可得，，又且，可得且，所以，，，直线的方程为，..................14分令，可得．故直线过定点．..................16分（法二）若垂直于轴，则，此时与题设矛盾．若不垂直于轴，可设的方程为，将其代入，可得，可得，.........12分又，可得，..................14分故，可得或，又不过点，即，故．所以的方程为，故直线过定点．..................16分21．解：（1）对于函数，当时，，又，所以，故是“L函数”. ..................2分对于函数，当时，，故不是“L函数”. ..................4分（2）当时，由是“L函数”，可知，即对一切正数恒成立，又，可得对一切正数恒成立，所以．..................6分由，可得，故，又，故，由对一切正数恒成立，可得，即．..................9分综上可知，a的取值范围是．...........................10分（3）由函数为“L函数”，可知对于任意正数，都有，且，令，可知，即，...........................12分故对于正整数k与正数，都有，....................................14分对任意，可得，又，所以，.....................16分同理，故． (18)分。

高三联合调研考试数学试卷(文科)八校联考

20XX 届高三联合调研考试数学试卷(文科)(本试卷满分150分,测试时间120分钟)参加学校：华师大一附中、曹杨二中、市西、市三女子、控江、格致、市北、(育才、晋元高中)一、填空题（本大题共56分，每小题4分）1.计算：1i2i-=+____________ (其中i 为虚数单位). 2.已知向量(5,3)a =-，(2,)b x =，若向量a 、b 互相平行，则x =____________. 3.已知向量a 与b 的夹角为3π,||1a =,||2b =,若b a λ-与a 垂直，则实数λ=_________. 4.在二项式81()ax x-的展开式中，常数项为70，则实数a =_____________. 5.已知θ是第三象限角，若3cos 5θ=-，则cos sin 1θθ-的值为_______________.6.若133log (|5|9)4M a =-+,[4,17]a ∈,则M 的取值范围是_________________.7.关于x 的方程组(1)21y q x y qx =-+⎧⎪⎨=-⎪⎩有唯一的一组实数解,则实数q 的值为_____________. 8.把编号为1、2、3、4、5的5位运动员排在编号为1、2、3、4、5的5条跑道中，若有且只有两位运动员的编号与其所在跑道编号相同，则不同的排法种数共有___________种.9.过点1(,1)2M 的直线l 与圆C ：22(1)4x y -+=交于A 、B 两点，C 为圆心，当ACB ∠最小时，直线l 的方程为_________________．10.在平面直角坐标系xOy 中，函数()()1f x k x =-（1k >）的图像与x 轴交于点A ，它的反函数()1y f x -=的图像与y 轴交于点B ，并且这两个函数的图像交于点P ．若四边形OAPB 的面积是3，则k =___________.11.已知Z k ∈,向量(,1)A B k =,(2,4)AC =,若||10AB ≤,则ABC ∆为直角三角形的概率是_______________.12.已知ABO ∆中, (OA =-, (1,0)OB =,D 为AB 上的点，若2A D D B =,则ODB ∠=____________(结果用反三角表示).13.设直线:l 220x y ++=关于原点对称的直线为l ',若l '与椭圆2214y x +=的交点为A ,B 两点,点P 是椭圆上的动点,则使PAB ∆的面积为12的点P 的个数为_____________.14.如图所示的程序框图中, ,函数int()x 表示不超过x 的最大整数,则由框图给出的计算结果是____________.二、选择题（本大题满分20分，每小题5分）15.若函数21y a x =⋅,22x y c =⋅,33y b x =⋅,则由表中数据确定()f x 、()g x 、()h x 依次对应（）. (A) 1y 、2y 、3y (B) 2y 、1y 、3y (C ) 3y 、2y 、1y (D) 1y 、3y 、2y16.在证券交易过程中，常用到两种曲线，即时价格曲线()y f x =及平均价格曲线()y g x = (如(2)3f =是指开始买卖后二个小时的即时价格为3元；(2)3g =表示二个小时内的平均价格为3元)，在下图给出的四个图像中实线表示()y f x =，虚线表示()y g x =其中可能正确的是（）.（A ）（B ）（C ）（D ）17. 正四面体ABCD 的表面积为S ，其中四个面的中心分别是E 、F 、G 、H .设四面体EFGH 的表面积为T ,则TS 等于 ( ). (A) 49 (B) 19 (C)14 (D)1318.函数()y f x =的定义域为R ，若对于任意的正数a ，函数()()()g x f x a f x =+-都是其定义域上的增x()f x函数，则函数()y f x =的图像可能是 ( ).（A ） (B) (C) (D)三、解答题(本大题满分74分)19．（本题满分12分）第1小题满分6分，第2小题满分6分．已知函数()()22sin cos 2cos 2f x x x x =++-.（1）求函数()f x 的最小正周期; (2 )当3,44x ππ⎡⎤∈⎢⎥⎣⎦时,求函数()f x 的最大值,最小值.20．（本题满分12分）第1小题满分6分，第2小题满分6分．在一个棱长为2+的正方体的八个顶角上分别截去一个三棱锥，使截掉棱锥后的多面体有六个面为正八边形，八个面为正三角形（如图所示），（1）求异面直线AB 与GH 所成角的大小; （2）求此多面体的体积(结果用最简根式表示).21.（本题满分12分）第1小题满分5分，第2小题满分7分．已知O 为坐标原点，点(2,1),(1,2)A B ，对于k N *∈有向量k OP kOB OA =+，（1）试证明k P 都在同一条直线23y x =-上;（2）是否在存在k N *∈使k P 在圆22(2)5x y +-=上或其内部,若存在求出k ,若不存在说明理由.22.（本题满分19分）第1小题满分5分，第2小题满分5分，第3小题满分9分．已知函数()y f x =的图像(如图所示)过点(0,2)、(1.5,2)和点(2,0)，且函数图像关于点(2,0)对称;直线1x =和3x =及0y =是它的渐近线.现要求根据给出的函数图像研究函数1()()g x f x =的相关性质与图像,(1)写出函数()y g x =的定义域、值域及单调递增区间; （2）作函数()y g x =的大致图像（要充分反映由图像及条件给出的信息）;（3）试写出()y f x =的一个解析式,并简述选择这个式子的理由(按给出理由的完整性及表达式的合理、简洁程度分层给分).23．（本题满分19分）第1小题满分5分，第2小题满分8分，第3小题满分6分．由下面四个图形中的点数分别给出了四个数列的前四项,将每个图形的层数增加可得到这四个数列的后继项.按图中多边形的边数依次称这些数列为“三角形数列”、“四边形数列”，将构图边数增加到n 可得到“n 边形数列”，记它的第r 项为(,)P n r ，1，3，6，10 1，4，9，16 1，5，12，22 1，6，15，28 （1）求使得(3,)36P r >的最小r 的取值；（2）3725是否为“五边形数列”中的项，若是，为第几项；若不是，说明理由； ( 3) 试推导(,)P n r 关于n 、r 的解析式.参考答案: 1.13i 55- 2. 65- 3. 1 4. 1± 5. 136. 3[2log 2,2]--- (或3[log 18,2]--等7.12或 1 8.20 9.2430x y -+= (或11()(1)022x y ---=等) 10.32 11. 41912.13.2 14.1 15.D 16.C 17.B 18.D19. 解: (1)()sin 2cos 224f x x x x π⎛⎫=+=+ ⎪⎝⎭. 4分∴()f x 的最小正周期为π. 6分(2).337,,244444x x πππππ⎡⎤∈∴≤+≤⎢⎥⎣⎦, 8分,1sin 242x π⎛⎫∴-≤+≤⎪⎝⎭ 10分∴()1f x ≤≤. 12分∴当3,44x ππ⎡⎤∈⎢⎥⎣⎦时,函数()f x 的最大值为1,最小值.20. 解： (1) 易知//FE AB ,//GH EC ,所以FEC ∠就是异面直线AB 与GH 所成的余角). 3分经计算得: 351)48FEC ππ∠=-=(也可以直接用4522.567.5+=做)所以异面直线AB 与GH 所成的角的大小为38π1),arc . 6分（2，则由题意得：2x x +=,所以, 9分设多面体的体积为V ,则311(2832V =+-⨯⨯=563+ 12分21.解:(1)点k P 在同一条直线上,直线方程为23y x =-. 2分证明如下:设点(,)k k k P x y ,则(,)(1,2)(2,1)k k x y k =+ 即2,21,k k x k y k =+⎧⎨=+⎩所以23k k y x =-.所以,点k P 在直线23y x =-上. 5分（文科）按证明情况酌情给分(2)由圆22(2)5x y +-=的圆心(0,2)到直线23y x =-=可知直线与圆相切, 所以直线与圆及内部最多只有一个公共点 10分而切点的坐标为:(2,1),此时0k =不满足题意,所以不存在k N *∈满足题意. 12分22.解: (1) 定义域为:{|1,2,3,}x x x x x R ≠≠≠∈ 2分值域为: (,0)(0,)-∞⋃+∞ 3分函数的单调递增区间为: (1,2)和(2,3) 5分 (2)图像要求能反映出零点((1,0)和(3,0),渐近线2x =,过定点,单调性正确. 5分 (3) 结论可能各异如:3(2)()|1||3|x f x x x -=--,2222(3)()22(1)xx x f x x x x -⎧>⎪-⎪=⎨-⎪<⎪-⎩211()2tan()132233x x f x x x x x π-⎧<⎪-⎪⎪=-<<⎨⎪-⎪>⎪-⎩,等层次一:函数图像能满足题意, 但没有说明理由 4分层次二: 函数图像能满足题意,能简述理由(渐近线、定点等部分内容) 6分层次三: 函数图像能满足题意,能说明过定点、渐近线、单调性及对称性 9分解: (1)(1)(3,)2r r P r +=, 3分由题意得(1)362r r +>,所以,最小的9r =. 5分(2)设n 边形数列所对应的图形中第r 层的点数为r a ,则12(,)r P n r a a a =++⋅⋅⋅+ 从图中可以得出:后一层的点在2n -条边上增加了一点,两条边上的点数不变, 所以12r r a a n +-=-,11a =所以{}r a 是首项为1公差为2n -的等差数列, 所以(,)[2(1)(2)]2r P n r r n =+--.(或(2)(1)2n r r r --+等) 13分 (3)2(,1)(,)(2)21P n r P n r n r r ++=-++ 16分显然3n =满足题意, 17分而结论要对于任意的正整数r 都成立,则2(2)21n r r -++的判别式必须为零, 所以,44(2)0n --=,3n = 19分所以,满足题意的数列为“三角形数列”.（文科）（2）为第50项，（3）同理科（2）.。

上海市八校高三数学下学期联合调研考试试题文沪教版

上海市八校2013届高三数学下学期联合调研考试试题文沪教版一、填空题（本题满分56分）本大题共有14题，要求在答题纸相应题序的空格内直接填写结果，每个空格填对得4分，否则一律得零分。

1．若z C ∈，且1)3(=+i z ，则z =________________。

2．函数0.5log y x =的定义域为。

3．已知22)(-=xx f ，那么1(2)f-的值是。

4．已知{}n a 为等差数列，其前n 项和为n S ，若36a =，312S =，则公差d = 。

5．函数xxx x y cos cos sin cos 3=的最小正周期。

6．{}n a 是无穷数列，已知n a 是二项式(12)(*)nx n N +∈的展开式各项系数的和，记12111n nP a a a =+++L ，则lim n n P →∞=__________。

7．若双曲线的渐近线方程为x y 3±=，它的一个焦点与抛物线x y 1042=的焦点重合，则双曲线的标准方程为。

8．已知正方形ABCD 的边长为1，点E 是AB 边上的动点，DC DE •的最大值为。

9．△ABC 中，三内角A 、B 、C 所对边的长分别为a 、b 、c ，已知ο60=∠B ，不等式2680x x -+->的解集为{|}x a x c <<，则b =______。

10．如图为一几何体的的展开图，其中ABCD 是边长为6的正方形，SD=PD ＝6，CR=SC ，AQ=AP ，点S,D,A,Q 及P,D,C,R 共线，沿图中虚线将它们折叠，使P ，Q ，R ，S 四点重合，则这样的几何体的体积为________。

11．从1,2,3,4,5这5个数中任意抽取三个数,其中仅有两个数是连续整数的概率是。

12．已知数列{}n a ,{}n b 满足11a =,22a =,12b =,且对任意的正整数,,,i j k l ,当i j k l +=+时,都有i j k l a b a b =,则1010a b +的值是。

数学上海八校联考试卷(2017.03)

上海八校联考高三数学试卷2017.3一. 填空题1. 已知设集合2{|20170,}S x x x x R =+=∈，2{|20170,}T x x x x R =-=∈，则 S T =2. 不等式13x x+<的解为 3. 已知两平行直线1:210l x y +-=，2:260l x y +-=，则1l 与2l 的距离是 4. 若行列式5413879xx 中，元素4的代数余子式大于0，则x 满足的条件是5. 二项式(51)n x -的展开式的二项式系数和为W ，各项系数和为P ，且62128W P +=，则n 的值是6. 2，四个顶点在同一球面上，则该球的体积为7. 学校要求每位学生从6门选修课程中选修3门，甲、乙两位同学选的三门课程中恰有两门是相同的概率为（以数字作答）8. 已知1F 、2F 是双曲线2222:1x y C a b-=（0a >，0b >）的左、右焦点，过1F 的直线l 与 C 的左、右两支分别交于点A 、B ，若2ABF ∆为等边三角形，则双曲线的渐近线方程是9. 已知函数log (5)6()(4)462a x x f x a x x -≥⎧⎪=⎨--<⎪⎩，数列{}nb 满足()n b f n =（*n N ∈），且{}n b 是单调递增数列，则实数a 的取值范围是10. 已知0a >，1a ≠，若函数()2|sin 2|2x f x a x π=+-，0x >，若函数()f x 至少有五个零点，则实数a 的取值范围是11. 已知集合211{|,1}k M x x kt t kt k==+<<，若对大于1的正整数k ，所有集合k M 的交集为12. 对于定义域和值域均为[0,1]的函数()f x ，定义1()()f x f x =，21()(())f x f f x =， …，1()(())n n f x f f x -=，1,2,3n =，…，满足()n f x x =的点称为f 的n 阶周期点，设1202()12212x x f x x x ⎧<≤⎪⎪=⎨⎪-<≤⎪⎩，则f 的2阶周期点的个数是二. 选择题13. 条件甲：0a b >>，条件乙：11a b<，则甲是乙成立的（） A. 充分不必要条件 B. 必要不充分条件C. 充要条件D. 既不充分也不必要条件14. m 、n 是不重合的两直线，α、β是不重合的两平面，则下列命题正确的是（）A. 若m ∥α，n α⊂，则m ∥nB. 若m ∥α，m ∥β，则α∥βC. 若m α⊥，m β⊥，则α∥βD. 若n αβ=，m ∥n ，则m ∥α且m ∥β15. 下列命题：① 命题“若22am bm >，则a b >”的逆命题是真命题；② 若(4,3)a =，(2,1)b =-，则b 在a 上的投影是5-③ 在164(x x的二项展开式中，有理项共有4项； ④ 已知一组正数1x 、2x 、3x 、4x 的方差为2222212340.25(16)s x x x x =+++-，则数据 12x +、22x +、32x +、42x +的平均数为4；⑤ 复数32i i+的共轭复数是a bi +（,a b R ∈），则6ab =-，其中真命题的个数为（）A. 0B. 1C. 2D. 316. 已知圆221x y +=与x 轴的两个交点为A 、B ，若圆内的动点P 使2PA 、2PO 、2PB 成等比数列（O 为坐标原点），则PA PB ⋅的取值范围为（）A. 1(0,]2B. 1[,0)2-C. 1(,0)2- D. [1,0)-三. 解答题17. 已知()lg(1)f x x =+；（1）若0(12)()1f x f x <--<，求x 的取值范围；（2）若()g x 是以2为周期的偶函数，且当01x ≤≤时，()()g x f x =，求函数()y g x =（[1,2]x ∈）的反函数；18. ABC ∆中，角A 、B 、C 对应边长分别为a 、b 、c ，且221(cos )2c a B b a b -=-；（1）求角A ；（2）求sin sin B C +的最大值；19. 如图所示，正方体1111ABCD A B C D -的棱长为1，点M 、N 分别是面对角线1A B 和 11B D 的中点；（1）求证：MN AB ⊥；（2）求三棱锥N MBC -的体积；20. 已知圆1C 的圆心在坐标原点O ，且恰好与直线1:220l x y --=相切；（1）求圆的标准方程；（2）设点A 为圆上一动点，AN x ⊥轴于N ，若动点Q 满足：(1)OQ mOA m ON =+-，（其中m 为非零常数），试求动点Q 的轨迹方程2C ；（3）在（2）的结论下，当3m =时，得到曲线C ，与1l 垂直的直线l 与曲线C 交于B 、 D 两点，求△OBD 面积的最大值；21. 设无穷数列{}n a 的前n 项和为n S ，且(3)23n n p S pa p -+=+*()n N ∈，p 为常数， 3p <-；（1）求证：{}n a 是等比数列，写出{}n a 的通项公式；（2）若数列{}n a 的公比()q f p =，无穷数列{}n b 满足：11b a =，13()2n n b f b -=(2)n ≥，求证：1{}nb 是等差数列，并写出{}n b 的通项公式；（3）设11n n n c a a +=-，在（2）的条件下，有lim(lg )lg 27n n n b a →∞=，求数列{}n c 各项和；参考答案一. 填空题1. {0}2. {|0x x <或1}2x >3.5 4. 83x < 5.6 6. 3 7. 920 8. 6y x =± 9. 32(,8)5 10. (1,2) 11. 5[2,)2 12. 4二. 选择题13. A 14. C 15. B 16. B三. 解答题17.（1）2133x -<<；（2）1()310x f x -=-，[0,lg 2]x ∈； 18.（1）3π；（2）3(3]； 19.（1）略；（2）124； 20.（1）224x y +=；（2）2224y x m+=；（33 21.（1）12()3n n p a p -=+；（2）32n b n =+；（3）34-；。

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