上海市徐汇区2013届高三上学期期末教学质量调研数学文试题

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新领航教育特供：上海市徐汇区2013届高三上学期期末教学质量调研数学文试题

小升初中高考高二会考艺考生文化课一对一辅导（教师版）/wxxlhjy QQ:157171090无锡新领航教育特供：2012学年第一学期徐汇区高三年级数学学科学习能力诊断卷（文）（考试时间：120分钟，满分150分） 2013.1一．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填写结果，每个空格填对得4分，否则一律得零分.1．方程组2132x y x y -=⎧⎨+=-⎩的增广矩阵是__________________.【答案】211132-⎛⎫⎪-⎝⎭【解析】根据增广矩阵的定义可知方程组的增广矩阵为211132-⎛⎫⎪-⎝⎭。

2. 已知幂函数()f x 的图像过点18,2⎛⎫⎪⎝⎭，则此幂函数的解析式是()f x =_____________. 【答案】13x -【解析】设幂函数为()f x x α=，则由1(8)2f =得182α=，即3122α-=，所以31α=-，13α=-，所以13()f x x-=。

3．（文）若4cos 5θ=，则=θ2cos ___________.【答案】725【解析】因为4cos 5θ=，所以2247cos 22cos 12()1525θθ=-=⨯-=。

4．若抛物线22(0)y px p =>的焦点与双曲线221610xy-=的右焦点重合，则实数p 的值是 . 【答案】8【解析】抛物线的焦点坐标为(,0)2p ，在双曲线中22610a b ==，，所以22216c a b =+=，所以4c =，即双曲线的右焦点为(4,0)，所以482p p ==，。

上海市徐汇区2013届高三数学一模试卷

上海市徐汇区2013届高三数学一模试卷本试题卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．考生作答时，将答案答在答题卡上（答题注意事项见答题卡），在本试题卷上答题无效．考试结束后，将本试题卷和答题卡一并交回．第Ⅰ卷选择题一、选择题：本大题共12小题，每小题5分．在每小题给出的四个选项中,只有一项是符合题目要求的．1．设全集U是实数集R，集合M＝{x｜≥2x}，N＝{x｜≤0}，则M∩N＝A．{1，2} B．{ 2 } C．{1} D．[1，2]2．i为虚数单位，若复数＝，则｜z｜＝A．1 B．2 C．D．23．双曲线的离心率为A．B．C．D．4．某学生在一门功课的22次考试中，所得分数如下茎叶图所示，则此学生该门功课考试分数的极差与中位数之和为A．117 B．118 C．118．5 D．119．55．在△ABC中，M是AB边所在直线上任意一点，若＝－2 ＋λ，则λ＝A．1 B．2 C．3 D．46．“m＝－1”是“函数f（x）＝ln（mx）在（－∞，0）上单调递减”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7．公差不为0的等差数列{ }的前21项的和等于前8项的和．若，则k＝A．20 B．21 C．22 D．238．在如图所示的程序框图中，若U＝•，V＝，则输出的S＝A．2 B．C．1 D．9．在几何体的三视图如图所示，则该几何体的体积为A．B．2 C．D．10．e，π分别是自然对数的底数和圆周率，则下列不等式中不成立的是A．＞B．＋＞1C．＋＞2 D．－e＞－π11．在△ABC中，a，b，c分别是角A，B，C的对边，若＝2014 ，则的值为A．0 B．1 C．2013 D．201412．四面体ABCD中，AD与BC互相垂直，且AB＋BD＝AC＋CD．则下列结论中错误的是A．若分别作△BAD和△CAD的边AD上的高，则这两条高所在直线异面B．若分别作△BAD和△CAD的边AD上的高，则这两条高长度相等C．AB＝AC且DB＝DCD．∠DAB＝∠DAC第Ⅱ卷非选择题本卷包括必考题和选考题两部分．第13题～第21题为必考题。

上海市黄浦区2013届高三上学期期末教学质量调研数学文试题--含答案

（第9题图）黄浦区2012学年度第一学期高三年级期终考试数学试卷(文科) （一模） 2013年1月17日考生注意：1．每位考生应同时收到试卷和答题卷两份材料，解答必须在答题卷上进行，写在试卷上的解答一律无效；2．答卷前，考生务必将姓名、准考证号等相关信息在答题卷上填写清楚； 3．本试卷共23道试题，满分150分；考试时间120分钟．一、填空题(本大题满分56分) 本大题共有14题，考生应在答题卷的相应编号的空格内直接填写结果，每题填对得4分，否则一律得零分． 1．函数sin 2y x =的最小正周期为． 2．已知集合{|03}A x x =<<，2{|4}B x x =>，则AB = ．3．若(12i)(i)z a =--(i 为虚数单位)为纯虚数，则实数a 的值为． 4．若数列{}n a 的通项公式为3n a n =+(*)N n ∈，则12lim4n n n a a n++∞+=→ ．5．若双曲线2221(0)4x y b b -=>的一条渐近线过点P (1, 2)，则b 的值为_________．6．已知1tan 2α=，1tan()3βα-=-，则tan(2)βα-的值为．7．已知直线1l ：20x ay ++=和2l ：(2)360a x y a -++=，则1l ∥2l 的充要条件是a = ．8．91(x x+的展开式中5x 的系数是（用数字作答）．9．执行右边的程序框图，若10p =，则输出的S = ． 10．盒中装有形状、大小完全相同的7个球，其中红色球4个，黄色球3个．若从中随机取出2个球，则所取出的 2个球颜色不同的概率等于．11．已知⎩⎨⎧=xx x f 3log )(2)0()0(≤>x x ，且函数()()F x f x x a =+-有且仅有两个零点，则实数a 的取值范围是．12．已知函数()x f x a =(0a >且1a ≠)满足(2)(3)f f >，若1()f x -是()f x 的反函数，则关于x 的不等式1(1)1f x -->的解集是．13．已知抛物线22(0)y px p =>上一点(1,)M m (m >0)到其焦点F 的距离为5，该抛物线的顶点在直线MF 上的射影为点P ，则点P 的坐标为． 14．已知命题“若22()f x m x =，2()2g x mx m =-，则集合1{|()(),1}2x f x g x x <≤≤=∅” 是假命题，则实数m 的取值范围是．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，考生应在答题卷的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分．15．在四边形ABCD 中，AB DC =，且AC ·BD ＝0，则四边形ABCD 是（）A ．菱形B ．矩形C ．直角梯形D ．等腰梯形16．已知1z =且z ∈C ，则|22i |z --（i 为虚数单位）的最小值是（）A ．22B ．2C ．122+D ． 122-17．若矩阵12341234a a a a b b b b ⎛⎫⎪⎝⎭满足下列条件：①每行中的四个数所构成的集合均为{1,2,3,4}； ②四列中有且只有两列的上下两数是相同的．则这样的不同矩阵的个数为（） A ．24B ．48C ．144D ．28818．若()f x 是R 上的奇函数，且()f x 在[0,)+∞上单调递增，则下列结论：①|()|y f x =是偶函数；②对任意的x ∈R 都有()|()|0f x f x -+=；③()y f x =-在(,0]-∞上单调递增； ④()()y f x f x =-在(,0]-∞上单调递增．其中正确结论的个数为（） A ．1 B ．2 C ．3 D ．4三、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卷相应的编号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．NPMDCBAFD 1C 1B 1A 1D CBAE如图所示，在棱长为2的正方体1111ABCD A B C D -中，E ,F 分别为线段1DD ,BD 的中点．（1）求三棱锥E ADF -的体积；（2）求异面直线EF 与BC 所成的角．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．在△ABC 中，角A , B , C 的对边分别为a , b , c ，且A , B , C 成等差数列．（1）若3AB BC ⋅=-，且b =a c +的值；（2）若sin cos AM A=，求M 的取值范围．21．（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．如图所示，ABCD 是一个矩形花坛，其中AB = 6米，AD = 4米．现将矩形花坛ABCD 扩建成一个更大的矩形花园AMPN ，要求：B 在AM 上，D 在AN 上，对角线MN 过C 点，且矩形AMPN 的面积小于150平方米．（1）设AN 长为x 米，矩形AMPN 的面积为S 平方米，试用解析式将S 表示成x 的函数，并写出该函数的定义域；（2）当AN 的长度是多少时，矩形AMPN 的面积最小?并求最小面积．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．给定椭圆C ：22221(0)x y a b a b+=>>，称圆心在原点O 的圆为椭圆C 的“准圆”．已知椭圆C 的一个焦点为F ，其短轴的一个端点到点F （1）求椭圆C 和其“准圆”的方程；（2）过椭圆C 的“准圆”与y 轴正半轴的交点P 作直线12,l l ，使得12,l l 与椭圆C 都只有一个交点，求12,l l 的方程；（3）若点A 是椭圆C 的“准圆”与x 轴正半轴的交点，,B D 是椭圆C 上的两相异点，且BD x ⊥轴，求AB AD ⋅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3分，第2小题满分7分，第3小题满分8分．对于函数()y f x =与常数,a b ，若(2)()f x af x b =+恒成立，则称(,)a b 为函数)(x f 的一个“P 数对”．设函数)(x f 的定义域为R +，且(1)3f =．（1）若(1,1)是()f x 的一个“P 数对”，求10(2)f ；（2）若(2,0)-是()f x 的一个“P 数对”，且当[1,2)x ∈时()f x =(2)k x -，求()f x 在区间2[1,2)n(*)N n ∈上的最大值与最小值；（3）若()f x 是增函数，且(2,2)-是()f x 的一个“P 数对”，试比较下列各组中两个式子的大小，并说明理由．①(2)n f -与22n -+(*)N n ∈；②()f x 与22x +1((2,2],*)N n n x n --∈∈．EABCD A 1B 1C 1D 1F黄浦区2012学年度第一学期高三年级期终考试数学试卷（文科）参考答案一、填空题（本大题满分56分）本大题共有14小题，考生应在答题卷相应编号的空格内直接填写结果，每题填对得4分，否则一律得零分．1．π； 2．(2,3)； 3．2； 4．12； 5．4 6．1-； 7．3； 8．36；9．81； 10．47； 11．(,1]-∞ 12．(1,1)a -； 13．6448(,)2525； 14．(7,0)-．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，考生应在答题卷的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分． 15．A 16．D 17．C 18．B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卷相应的编号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．解：（1）在正方体1111ABCD A B C D -中， ∵F 是AC 的中点， ∴112122CDFADC S S ∆∆==⨯=， ………………3分又CE ⊥平面ABCD ，即CE ⊥平面CDF ，故11111333E CDF CDF V S CE -∆=⋅=⋅⋅=，所以三棱锥E ADF -的体积为13．………………6分（2）连1BD ，由E 、F 分别为线段1DD 、BD 的中点，可得EF ∥1BD ，故1D BC ∠即为异面直线EF 与BC 所成的角． ………………… 8分 ∵BC ⊥平面11CDD C ，1CD ⊂≠平面11CDD C ，∴1BC CD ⊥，在Rt △1BCD 中，2BC =，1D C =∴11tan D CD BC BC∠=1D BC ∠= 所以异面直线EF 与BC所成的角为 ………………………… 12分 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．解：（1）A 、B 、C 成等差数列，∴2,B A C =+又A B C π++=，∴3B π=， …………………………2分由3AB BC ⋅=-得，2cos33c a π⋅=-，∴6ac = ① ………………………4分NPM D CB A又由余弦定理得2222cos,3b ac ac π=+-∴2218a c ac =+-，∴2224a c += ② ………………………6分由①、②得，6a c += ……………………………………8分（2）sin sin cos AM A A A==-2sin()3A π=- ……………………………………11分由（1）得3B π=，∴23A C π+=，由203C A π=->且0A >，可得20,3A π<<故333A πππ-<-<，所以2sin()(3A π-∈，即M的取值范围为(． …………………………14分 21．（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．解：（1）由△NDC ∽△NAM ，可得DN DCNA AM=， ∴46x x AM -=，即64x AM x =-，……………………3分故264x S AN AM x =⋅=-， ………………………5分由261504x S x =<-且4x >，可得2251000x x -+<，解得520x <<，故所求函数的解析式为264x S x =-，定义域为(5,20)． …………………………………8分（2）令4x t -=，则由(5,20)x ∈，可得(1,16)t ∈，故2266(4)166(8)x t S t t t +===++ …………………………10分8)96≥=， …………………………12分当且仅当16t t=，即4t =时96S =．又4(1,16)∈，故当4t =时，S 取最小值96．故当AN 的长为8时，矩形AMPN 的面积最小，最小面积为96（平方米）…………14分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．解：（1）由题意知c a =1b =，故椭圆C 的方程为2213x y +=，其“准圆”方程为224x y +=． ………………4分（2）由题意可得P 点坐标为(0,2)，设直线l 过P 且与椭圆C 只有一个交点，则直线l 的方程可设为2y kx =+，将其代入椭圆方程可得 ………………6分223(2)3x kx ++=，即22(31)1290k x kx +++=，由22(12)36(31)0k k ∆=-+=，解得1k =±， ………………8分所以直线1l 的方程为2y x =+，2l 的方程为2y x =-+，或直线1l 的方程为2y x =-+，2l 的方程为2y x =+． ………………10分（3）由题意，可设(,),(,)B m n D m n -(m <，则有2213m n +=，又A 点坐标为(2,0)，故(2,),(2,)AB m n AD m n =-=--， ………………12分故2222(2)44(1)3m AB AD m n m m ⋅=--=-+--2244343()332m m m =-+=-, …………………………14分又m <<，故243()[0,732m -∈+，所以AB AD ⋅的取值范围是[0,7+． …………………………16分23．（本题满分18分）本题共有3个小题，第1小题满分3分，第2小题满分7分，第3小题满分8分．解：（1）由题意知(2)()1f x f x =+恒成立，令2(*)N k x k =∈，可得1(2)(2)1k k f f +=+，∴数列{(2)}k f 是公差为1的等差数列，故100(2)(2)10f f =+，又0(2)3f =，故10(2)13f =． ………………………………3分（2）当[1,2)x ∈时，()(2)f x k x =-，令1x =，可得(1)f k =，由(1)3f =可得3k =，即[1,2)x ∈时，()3(2)f x x =-， …………………………………4分可知()f x 在[1,2)上的取值范围是(0,3]．又(2,0)-是()f x 的一个“P 数对”，故(2)2()f x f x =-恒成立，当1[2,2)k k x -∈(*)N k ∈时，1[1,2)2k x -∈，()2()4()24x x f x f f =-==…11(2)()2k k xf --=-， …………………………………6分故当k 为奇数时，()f x 的取值范围是1(0,32]k -⨯；当k 为偶数时，()f x 的取值范围是1[32,0)k --⨯． ……………………………8分由此可得()f x 在2[1,2)n 上的最大值为2232n -⨯，最小值为2132n --⨯．………………10分（3）由(2,2)-是()f x 的一个“P 数对”，可知(2)2()2f x f x =-恒成立，即1()(2)12f x f x =+恒成立，令12k x =(*)N k ∈，可得1111()()1222k k f f -=+， …………………12分即1111()2[()2]222k k f f --=-(*)N k ∈，又01()2(1)212f f -=-=，∴11{()2}2k f --是一个等比数列，∴11()21()22n n f -=⨯，所以(2)22n n f --=+． …………………………………15分当1(2,2](*)N n n x n --∈∈时，由()f x 是增函数，故11()(2)22n n f x f --≤=+，又12222222n n x --+>⨯+=+，故有()22f x x <+．…………………………………18分。

上海市徐汇区2013年高三上学期教学质量检测(高考一模)英语试题及答案

徐汇区2012-2013学年第一学期教学质量监测高三年级英语学科试卷2013.1 考生注意：1.考试时间120分钟，试卷满分140分。

2.本次考试设试卷和答题纸两部分。

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。

3.答题前，务必在答题纸上填写准考证号和姓名，并将核对后的条形码贴在指定位置上，在答题纸反面清楚地填写姓名。

第一卷I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversation and the question will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. At the booking office. B. In a Hong Kong hotel.C. On a busy street.D. At an airport.2. A. The bag. B. The sofa. C. The table. D. The lamp.3. A. Ask the man for help. B. Have the refrigerator fixed.C. Wait until tomorrow.D. Send the fax right away.4. A. The watch is on the television. B. The program will be over soon.C. The woman should leave the television on.D. The woman shouldn’t stand in the way.5. A. The man lent the tools to other people. B. The man couldn’t find the bookshelf.C. The tools the man borrowed are missing.D. The tools have already been returned.6. A. Because Susan doesn’t like football. B. Because Susan invited him to a movie.C. Because he wanted to watch the movie.D. Because he didn’t know about the game.7. A. Give his ankle a good rest. B. Treat his injury immediately.C. Continue his regular activities.D. Go to the doctor for help.8. A. The man missed the way to the class.B. The woman will bring the man the news everyday.C. The woman will help the man make up the lessons.D. The man is always worried about his lessons.9. A. The woman will go to China this weekend.B. The woman will treat the man dinner tonight.C. The man will join her for dinner this weekend.D. The woman refuses to join the man for dinner.10. A. Spending money puts him in a good mood.B. He has just earned a big sum of money.C. He is pleased with his new purchase.D. He paid a high price for his new cell phone.Section BDirections: In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions11through 13 are based on the following passage.11. A. Planting some trees in the greenhouse.B. Writing a want ad to a local newspaper.C. Putting up a going out of business sign.D. Helping a customer select some purchases.12. A. Opening an office in the new office park.B. Keeping better relations with her company.C. Developing fresh business opportunities.D. Building a big greenhouse of his own.13. A. Owning the greenhouse one day. B. Securing a job at the office park.C. Cultivating more potted plants.D. Finding customers out of town.Questions14through 16are based on the following passage.14. A. She had run a long way. B. She felt weak and tired in the subway.C. She had done a 1ot of work.D. She had given blood the night before.15. A. By lifting her to the platform to get others’ help.B. By moving her with the help of his girlfriend.C. By holding her arm and pulling her along the ground.D. By waking her up and dragging her away from the edge.16. A. Danger in the subway. B. A subway rescue.C. How to save people.D. A traffic accident.Section CDirections: In Section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet.Blanks17through20are based on the following conversation.Job InformationCompany Name: ALBA, the ___(17)____ groupJob: Marketing ____(18)____Qualifications: More than 5 years' ___(19)___ marketing experience Closing date for application: The ___(20)___ of JulyComplete the form. Write ONE WORD for each answer. Blanks 21 through 24 are based on the following conversation.What is Internet addiction（上瘾）? People have relationship problems, orproblems ____(21)___because they are spending so much time on the Net.What are the symptoms of Internet addiction? ① It begins to affect other areas of your life, such asyour work or ___(22)____.② You are beginning to feel ___(23)___ and youreally look forward to going online.What advice did Dr James give? Have some sort of balance in your life. Spend some time on the Internet and then go out and ___(24)___.Complete the form. Write NO MORE THAN THREE WORDS for each answer.II. Grammar and VocabularySection ADirections: Beneath each of the following sentences there are four choices marked A, B. C andD. Choose the one answer that best completes the sentence.25. Books are the most important records we keep man’s thought, ideas and feelings.A. upB. toC. ofD. on26. Knowledge begins to increase as soon as one individual communicates his ideas to _______by means of speech.A. otherB. anotherC. the otherD. some other27. ---I hope you will be ready to leave on time.---Don’t worry. I’ll be ready by the time the taxi _______.A. arrivedB. arrivesC. will arriveD. will have arrived28. ---Are you pleased with what he has done?---It couldn’t be _______. Why didn’t he put more effort into his work?A. any worseB. much betterC. so badD. the best29. Almost every one of the graduates wants to deliver the keynote speech at the graduationceremony, because for the speaker, _______is an honour.A. invitedB. being invitedC. be invitedD. inviting30. ---My e-dictionary is nowhere to be found. Who have taken it?---I don’t know. But keep looking and you will find it.A. shouldB. couldC. needD. shall31. I hope to achieve this objective by calling on the smokers ________ good judgment and showconcern for others rather than by regulation.A. to usingB. usingC. useD. to use32. John was so disappointed when Susan turned down his proposal ________ he decided to staysingle for the rest of his life.A. thatB. asC. whereD. since33. ________ his own boss for such a long time, he found it hard to accept orders from another.A. BeingB. To beC. Having beenD. Been34. ---I don’t think our coach knows the real reason for our losing the match.---Well, surprisingly, he does. Our team leader has been called in and now.A. has been questionedB. is being questionedC. is questioningD. has questioned35. The Nobel Prize in literature has been awarded annually to an author from any country_________ has produced "in the field of literature the most outstanding work in an ideal direction".A. whichB. whoC. whereD. as36. Though ______ of danger, sightseers have been flocking to the site where the world’s biggestterrorist Bin Laden lived.A. warnedB. warningC. being warnedD. having warned37. Owing to the level of the damage to factories and infrastructure, it will be weeks or evenmonths ________ the country’s supply chain returns to normal.A. thatB. beforeC. whenD. since38. ________ she was at the time of the murder was of major concern to the police that areinvestigating the case.A. WhenB. WhyC. WhetherD. Where39. So _________ that even the people in the next room could hear him.A. loudly he spokeB. he spoke loudlyC. loudly did he speakD. loudly spoke did he40. People may forget what you said or what you did, but they will never forget _______ youmade them feel.A. whyB. howC. whatD.thatSection BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.A. considerationB. originallyC. involvementD. finalizedE. sponsoredF. increaseG. balancedH. representI. proceedJ. includedThe Man Booker Prize for Fiction is awarded every year for a novel written by a writer from the Commonwealth or the Republic of Ireland and it aims to ___(41)___ the very best incontemporary fiction. The prize was ___(42)___ called the Booker-McConnell Prize, which was the name of the company that ___(43)___ it, though it was better-known as simply the ‘Booker Prize’. In 2002, the Man Group became the sponsor and they chose the new name, keeping‘Booker’.Publishers can submit（提交） books for ___(44)___ for the prize, but the judges can also ask for books to be submitted they think should be ___(45)___. Firstly, the Advisory Committee give advice if there have been any changes to the rules for the prize and selects the people who will judge the books. The judging panel（评审团）changes every year and usually a person is only a judge once.Great efforts are made to ensure that the judging panel is ___(46)___ in terms of gender and professions within the industry, so that a writer, a critic, an editor and an academic are chosen along with a well-known person from wider society. However, when the panel of judges has been ___(47)___, they are left to make their own decisions without any further ___(48)___ or interference from the prize sponsor.The Man Booker judges include critics, writers and academics to maintain the consistent（始终如一的） quality of the prize and its influence is such that the winner will almost certainly see the sales ___(49)___ considerably, in addition to the £50,000 that comes with the prize.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases markedA, B, C and D. Fill in each blank with the word or phrase that best fits the context.Auditing（旁听）classes at university is an ideal way of learning or trying out new areas of study without committing yourself. You can study alongside fellow students without any___(50)___ to participate in formal assessments or gain credits for your degree.The trend of auditing university classes has ___(51)___ across universities. Recently, Fudan Postgraduate, a publication by Fudan University, even ___(52)___ a map guide on its campus. Altogether, it features 25 courses, including details of lectures, venues, times and recommendations. Students can design their own auditing ___(53)___ based on a handy map. Zhu, editor of the guide, said that she has ___(54)___ from the trend herself – she successfully moved from a bachelor’s degree in biochemistry to her current one after auditing classes. “Visiting classes helped me a lot. The teachers of outstanding courses seldom teach strictly according to textbooks, which is attractive to students. Their tutoring methods have ___(55)___ my mind,” Zhu said.Other students audit purely out of ___(56)___. Shi Shuai, 22, a senior majoring in administration management at Shantou University, attended courses in economics and finance for one year. “It is a great way to explore professional knowledge outside your ___(57)___,” said Shi, who acquired the basics of accounting and finance management.Despite the ___(58)___ of learning in an open environment, university administrators remind us that there are still rules to follow. Not all classes accept outsiders, especially minority language classes and science modules. These are often taught in small groups to ensure the ___(59)___oflearning, or require special equipment and individual instruction, which is only ___(60)___ to registered students who pay tuition fees.Regulations vary across departments and universities. According to Huang Xiaoxiong, a journalism teacher at Fudan University, students need to ensure that auditing a class is permitted. “It is about basic manners. You need to let the teacher know about your ___(61)___,” he explained. “It’s po lite to greet the teacher when attending a course. Participate in discussions, but do not distract (使分心) others,” Huang added. “It’s okay to hand in homework assigned by the teacher, but expect general advice rather than careful correction.”Auditing class es ___(62)___ or beyond one’s ability can be a waste of time, warns Xu Jun, 27, HR manager at Guangzhou Automobile Industry Group. “It’s good to audit some classes. But you have to manage your time well, as you can’t get any academic credit or formal ___(63)___ for these courses,” Xu suggested.“Even though your interests are important, your ___(64)___ is the first thing to consider at university.”50. A. limitation B. hesitation C. obligation D. tendency51. A. swept B. split C. smashed D. survived52. A. imposed B. inserted C. extended D. released53. A. habit B. route C. data D. rule54. A. benefited B. transferred C. suffered D. managed55. A. kept B. burdened C. expanded D. changed56. A. kindness B. interest C. pressure D. instinct57. A. major B. campus C. control D. potential58. A. instructions B. advantages C. possibilities D. qualifications59. A. effectiveness B. uniqueness C. consciousness D. seriousness60. A. affordable B. portable C. adaptable D. available61. A. preference B. existence C. expectation D. performance62. A. regularly B. purposefully C. moderately D. aimlessly63. A. education B. investigation C. recognition D. comment64. A. decision B. diligence C. devotion D. degreeSection BDirections: 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)Marlon Brando is widely considered the greatest movie actor of alltime. He was born in Omaha Nebraska in 1924. He was named after hisfather, a salesman. His mother, Dorothy, was an actress in the local theater.Marlon Brando moved to New York City when he was 19 years old in 1943. He took acting classes at the New School for Social Research. One ofhis teachers was Stella Adler, who taught the "Method" style of realisticacting. The Method teaches actors how to use their own memories and emotions to identify with the characters they are playing.Marlon Brando learned the Method style quickly and easily. Critics said he was probably the greatest Method actor ever. One famous actress commented on his natural ability for it. She said teaching Marlon Brando the Method was like sending a tiger to jungle school.Marlon Brando appeared in several plays. He got his first major part in a Broadway play in 1947, at the age of 23. He received great praise for his powerful performance at Stanley Kowalski in Tennessee Williams' play, A Streetcar Named Desire. His fame grew when he acted the same part in the movie version, released in 1951.Marlon Brando won the Best Actor Oscar for The God Father, but he rejected it. He sent a woman named Sasheen Littlefeather to speak for him at the Academy Awards ceremony. She said that Brando could not accept the award because of the way the American film industry treated Native Americans. The people at the Academy Awards ceremony did not like the speech. But some experts think the action helped change the way American Indians were shown in movies.Marlon Brando acted in about forty movies. He was nominated for a total of eight Academy Awards. But he earned a "bad boy" reputation for his public outbursts and unusual behaviours. According to Los Angeles magazine, "Brando was rock and roll before anybody knew what rock and roll was". His later life was marked with family tragedies. His son Christian went to prison for killing his daughter Cheyene’s boyfriend. Cheyene later committed suicide. Brando became lonely. He worked occasionally for the money. But, in his prime, Marlon Brando was an actor other actors could only hope to become.65. According to the passage, the "Method" ______.A. requires actors to use their imagination in actingB. made Marlon Brando a great and famous actorC. wasn't very difficult for Marlon BrandoD. can be most effectively learned in a jungle66. The speech at the Academy Awards ceremony ______ at the time.A. made Sasheen Littlefeather well-knownB. was well-received by Native AmericansC. changed people's attitude to American IndiansD. received both positive and negative responses67. Which of the following is NOT true according to the passage?A. Marlon Brando’s later life was troubled and unhappy.B. A Streetcar Named Desire was later adapted into a movie.C. Marlon Brando was the first rock star in USA.D. As an actor, Brando’s talent was unparalleled.68. The purpose of this passage is to ______.A. inform us of Marlon Brando's attitude to civil rights movementB. introduce Marlon Brando as one of the greatest actorsC. help us understand Marlon Brando's secret to successD. instruct us how to become a great actor like Marlon Brando(B)69. From the passage we can conclude that “Learning English Video Project” is most probably________.A. an online language learning courseB. audio documents on language learningC. a series of short video programsD. a set of films on English-speaking countries70. If someone is interested in the comparison between English and other languages, he might beinterested to watch __________.A. Encounters in the UKB. Stories from MoroccoC. Thoughts from BrazilD. Insights from China71. What can we know about English learning in Sao Paulo, Brazil?A. Classroom teaching is more interactive and communicative.B. Homestay arrangement provides positive experience for learners.C. The Internet and games plays a major role in language learning.D. The principle of learning by doing is widely accepted by learners.(C)It is found that American students spend less than 15% of their time in school. While there’s no doubt that school is important, a number of recent studies reminds us that parents are even more so. A study published earlier this month by researchers at North Carolina State University, for example, finds that parental involvement — checking homework, attending school meetings and events, discussing school activities at home —has a more powerful influence on students’ academic performance than anything about the school the students attend. Another study, published in the Review of Economics and Statistics, reports that the effort put forth by parents (reading stories aloud, meeting with teachers) has a bigger impact on their children’s educational achievement than the effort devoted by either teachers or the students themselves. And a third study concludes that schools would have to increase their spending by more than $1,000 per pupil in order to achieve the same results that are gained with parental involvement.So parents matter. But it is also revealed in researches that parents, of all backgrounds, don’t need to buy expensive educational toys or digital devices for their kids in order to give them an advantage. They don’t need to drive their offspring to enrichment classes or test-preparation courses. What they need to do with their children is much simpler: talk.But not just any talk. Recent research has indicated exactly what kinds of talk at home encourage children’s success at school. For example, a study conducted by researchers at the UCLA School of Public Health and published in the journal Pediatrics found that two-wayadult-child conversations were six times as potent in promoting language development as the ones in which the adult did all the talking. Engaging in this reciprocal（双向的) back-and-forth gives children a chance to try out language for themselves, and also gives them the sense that their thoughts and opinions matter.The content of parents’ conversations with kids matters, too. Children who hear talk about counting and numbers at home start school with much more extensive mathematical knowledge, report researchers from the University of Chicago. While the conversations parents have with their children change as kids grow older, the effect of these exchanges on academic achievement remains strong. Research finds that parents play an important role in what is called “academic socialization” — setting expectations and making connections between current behavior andfuture goals. Engaging in these sorts of conversations has a greater impact on educational accomplishment.72. Parents are even more important than schools in that ______.A. parental involvement makes up for what schools are not able to doB. teachers and students themselves do not put in enough effortC. parental involvement saves money for schools and the local governmentD. students may well make greater achievements with parents' attention73. It can be inferred from the 2nd paragraph that ______.A. educational toys are unaffordable nowadaysB. digital devices can give children an advantageC. some parents believe in enrichment classesD. talking with children is a very simple task74. The word "potent" is closest in meaning to ______.A. powerfulB. difficultC. necessaryD. resistant75. Which of the following will more encourage children's success at school according to thepassage?A. Parents order their children to stop playing video games.B. Parents discuss with their children the possible future career.C. Parents lecture their children on getting too low marks on tests.D. Parents introduce colleges around the US to their children.Section CDirections: Read the following text and choose the most suitable heading from A-F for each paragraph. There is one extra heading which you do not need.Start with an idea. This doesn't have to be a brand new invention or new product. In fact, many successful small businesses have found a way to deliver an existing service more efficiently or economically or have customized an existing product or service. Evaluate your competitors - how many competitors, how strong are they, where are they, how will you compete. State what isrequired to enter this market, barriers to entry such as high fixed costs (factories, restaurants) and government regulations that must be met.77.This doesn't mean build a big factory or a fancy office. It simply means keep accurate customer records, a clean set of updated books and a technology foundation, if necessary. One of the downfalls of many small businesses is that they don't know if they're making or losing money (i.e. the need for a clean set of books). Another downfall is when small business owners try to sell their company years later but lack accurate customer history and customer information.78.Once you know you can be profitable take the leap and get started. Besides getting business supplies or advertising, plan ahead by establishing some new business clients ahead of time. If your business is unlike a restaurant, that physically needs to wait for customers to walk into its doors, establish accounts ahead of time. In this way, you will have pre-planned future receivables to look forward to.79.Use every technology available that will give your business a competitive advantage. The internet is a customers’ research tool. Help future c ustomers learn more about you and the details about what you sell and why your products or services are different and better for them than other competitors.80.Carrying cash can be risky. Therefore, most people choose to carry Visa, Mastercard, American Express and Discover. All t hese credit cards are part of our everyday life when it comes to making purchases. Debit cards are especially popular. So, along with having a sales counter cash register, get set up to accept credit cards for your business. To do this, also purchase a new credit machine.Section DDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.If you need another reason to give thanks at the dinner table on Thanksgi ving Day, how’s this: people who maintain an “attitude of gratitude” tend to be happier and healthier than those who don’t, according to an instructive article this week in the Wall Street Journal.The WSJ’s Melinda Beck reports that adults who feel grateful have “more energy, more optimism, more social connections and more happiness than those who do not, according to studies conducted over the past decade. Now a new study conducted by researchers at Hofstra University — the results of which are set to be published in an upcoming issue of the Journal of Happiness Studies — finds similar benefits of gratitude for adolescents as well.Dr. Jeffrey J. Froh, assistant professor of psychology and lead researcher of the new study, surveyed 1,035 students aged from 14 to 19 and found that grateful students reported higher grades, more life satisfaction, better social integration and less envy and depression than theirpeers who were less thankful and more materialistic. Additionally, feelings of gratitude had a more powerful impact on the students’ lives overall than materialism.What the majority of the research suggests is that gratitude should be chronic（长期的） in order to make a lasting difference in well-being. Dr. Robert Emmons, professor of psychology at the University of California, Davis, and a pioneer in gratitude research, told the WSJ that in order to reap（收获）all of its benefits, feeling gratitude must be rooted into your personality, and you must frequently acknowledge and be thankful for the role other people play in your happiness: “The key is not to leave it on the Thanksgiving table,” he said.For older children and adults, one simple way to cultivate gratitude is to literally count your blessings. Keep a journal and regularly record whatever you are grateful for that day. Be specific. Listing “my friends, my school, my dog” day after day means that “gratitude tiredness” has set in, Dr. Froh says. Writing “my dog licked my face when I was sad” keeps it fresher. The real benefit comes in changing how you e xperience the world. Look for things to be grateful for, and you’ll start seeing them.Studies show that using negative, insulting words — even as you talk to yourself — can darken your mood, as well. Fill your head with positive thoughts, express thanks and encouragement aloud and look for something to be grateful for, not criticize, in those around you, especially loved ones.(Note: Answer the questions or complete the statements in NO MORE THAN TEN WORDS.)81. According to the article in the Wall Str eet Journal, people who don’t maintain an “attitude of gratitude” tend to be ______________.82. What are the major findings of the new study at Hofstra University about?83. According to the passage, how can people probably avoid “gratitude tiredness”?84. In order not to darken our moods, we’d better stop ________________.第II卷（共45分）I . TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 每个市民都应自觉遵守交通法规。

2013年高考文科数学上海卷及答案

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2013年普通高等学校招生全国统一考试（上海卷）数学试卷(文史类)考生注意：1.答卷前,务必用钢笔或圆珠笔在答题纸正面清楚地填写姓名、准考证号,并将核对后的条形码贴在指定位置上,在答题纸反面清楚地填写姓名.2.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.不等式021xx <-的解为 . 2.在等差数列{}n a 中,若123430a a a a +++=,则23a a += .3.设m ∈R ,222(1)i m m m +-+-是纯虚数,其中i 是虚数单位,则m = .4.若2011x =,111xy=,则y = . 5.已知ABC △的内角A 、B 、C 所对的边分别是a 、b 、c .若2220a ab b c ++-=,则角C 的大小是 .6.某学校高一年级男生人数占该年级学生人数的40%.在一次考试中,男、女生平均分数分别为75、80,则这次考试该年级学生平均分数为 .7.设常数a ∈R .若25()ax x+的二项展开式中7x 项的系数为﹣10,则a = .8.方程91331x x +=-的实数解为 .9.若1cos cos sin sin 3x y x y +=,则2cos(2)x y =－ .10.已知圆柱Ω的母线长为l ,底面半径为r ,O 是上底面圆心,A 、B 是下底面圆周上两个不同的点,BC 是母线,如图.若直线OA 与 BC 所成角的大小为π6,则1r = .11.盒子中装有编号为1,2,3,4,5,6,7的七个球,从中任意取出两个,则这两个球的编号之积为偶数的概率是（结果用最简分数表示）. 12.设AB 是椭圆Γ的长轴,点C 在Γ上,且π4CBA ∠=.若4AB =,BC =则Γ的两个焦点之间的距离为 .13.设常数0a >.若291a x a x ++≥对一切正实数x 成立,则a 的取值范围为 .14.已知正方形ABCD 的边长为1.记以A 为起点,其余顶点为终点的向量分别为1a 、2a 、 3a ；以C 为起点,其余顶点为终点的向量分别为1c 、2c 、3c .若{},,,1,2,3i j k l ∈且 i j ≠,k l ≠,则()()i j k l a a c c ++的最小值是 .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. 15.函数2()=1(0)f x x x -≥的反函数为1()f x -,则1(2)f -的值是( ) AB. C.1D.116.设常数a ∈R ,集合{}|(1)()0A x x x a =--≥,{}|1B x x a =-≥.若A B =R ,则a的取值范围为( ) A .(,2)-∞ B .(,2]-∞ C .(2,)+∞D .[2,)+∞17.钱大姐常说“好货不便宜”,她这句话的意思是：“好货”是“不便宜”的( ) A .充分条件B .必要条件C .充分必要条件D .既非充分又非必要条件18.记椭圆221441x ny n +=+围成的区域（含边界）为(1,2,)n n Ω=,当点(),x y 分别在12,,ΩΩ上时,x y +的最大值分别是12,,M M ,则lim n n M →∞=( )--------在--------------------此--------------------卷--------------------上--------------------答--------------------题--------------------无--------------------效----------------姓名________________ 准考证号_____________数学试卷第3页（共14页）数学试卷第4页（共14页）A .0B .14C .2D.三、解答题（本大题共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）如图,正三棱锥O ABC -的底面边长为2,高为1,求该三棱锥的体积及表面积.第19题图20.（本题满分14分）本题共有2个小题,第1小题满分5分,第2小题满分9分. 甲厂以x 千克/小时的速度匀速生产某种产品（生产条件要求110x ≤≤）,每一小时可获得的利润是3100(51)x x+-元.（Ⅰ）求证：生产a 千克该产品所获得的利润为213100(5)a x x+-元；（Ⅱ）要使生产900千克该产品获得的利润最大,问：甲厂应该选取何种生产速度？并求此最大利润.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.已知函数()2sin()f x x ω=,其中常数0ω>.（Ⅰ）令1ω=,判断函数π()()()2F x f x f x =++的奇偶性,并说明理由；（Ⅱ）令2ω=,将函数()y f x =的图象向左平移π6个单位,再向上平移1个单位,得到函数()y g x =的图象.对任意a ∈R ,求()y g x =在区间[,10π]a a +上零点个数的所有可能值.22.（本题满分16分）本题共有3个小题,第1小题满分3分,第2小题满分5分,第3小题满分8分.已知函数()2||f x x =-,无穷数列{}n a 满足1()n n a f a +=,*n ∈N .（Ⅰ）若10a =,求2a ,3a ,4a ；（Ⅱ）若10a >,且1a ,2a ,3a 成等比数列,求1a 的值；（Ⅲ）是否存在1a ,使得1a ,2a ,3a ,…,n a ,…成等差数列？若存在,求出所有这样的1a ；若不存在,说明理由.23.（本题满分18分）本题共有3个小题,第1小题满分3分,第2小题满分6分,第3小题满分9分.如图,已知双曲线22112x C y -=:,曲线2:||||1C y x =+.P 是平面内一点,若存在过点P 的直线与1C 、2C 都有公共点,则称P 为“12C C -型点”.（Ⅰ）在正确证明1C 的左焦点是“12C C -型点”时,要使用一条过该焦点的直线,试写出一条这样的直线的方程（不要求验证）；（Ⅱ）设直线y kx =与2C 有公共点,求证：||1k >,进而证明原点不是“12C C -型点”；（Ⅲ）求证：圆2212x y +=内的点都不是“12C C -型点”.（文史类）答案解析数学试卷第5页（共14页）数学试卷第6页（共14页）11 4315C=种取法，利用古典概型的概率计算公式即可【考点】古典概型及其概率计算公式．2．CBA∠=，24 3b∴=【解析】根据对称性，当向量()()j k la a c c++与互为相反向量，且它们的模最大时()()i j k la a c c++最小．这时a A C=，a AD=，c CA=，c CB=，2()()|)|5a a c c a a++=-+=-．【提示】如图建立直角坐标系．为起点，其余顶点为终点的向量1a，2a，3a分别为AB，AC，AD，以C为起点，其余顶点为终点的向量c，c，3c分别为CD，CA，CB再分类讨论当，k，l取不同的值时，利用向量的坐标运算计()()k la a c c++的值，从而得出()()k la a c c++的最小值．【考点】平面向量数量积的运算．[1,)(,1][),2)B A A B=+∞=-∞+∞∴=R，，符合题意．综上，选标准解法：[1,)(,1)B a A B A a=-+∞=∴⊇-∞-R，数学试卷第7页（共14页）数学试卷第8页（共14页）数学试卷第9页（共14页）数学试卷第10页（共14页）,1][,)a +∞，解得12a <≤；当][1,)+∞⇒B时，代入解集中的不等式中，确定出A ，求出满足两集合的并集为111333ABC S ∆=的中点为E ，则1OQ =，QE =332BCOE +=，表面积33O ABC S -= 10051a x x ⎛+ ⎝千克该产品所获得的利润为21311100900590000513x x x x ⎡⎤⎛⎫⎛⎫+-=+- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦次函数的知识可知，当1x =16，即6x =时，19000036y ⎤⎫≤⎪⎥⎭⎦数学试卷第11页（共14页）数学试卷第12页（共14页）T =周期∴图像左移）12a a ，，211||)a a -=1)(1,)∞．=，A Bϕ所以原点不是“（3）设直线l数学试卷第13页（共14页）数学试卷第14页（共14页）。

2013年上海市高考文科数学试卷及参考答案与试题解析

2013年上海市高考文科数学试卷及参考答案与试题解析一、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.(4分)不等式＜0的解为.2.(4分)在等差数列{an }中,若a1＋a2＋a3＋a4＝30,则a2＋a3＝.3.(4分)设m∈R,m2＋m－2＋(m2－1)i是纯虚数,其中i是虚数单位,则m＝.4.(4分)已知,,则y＝.5.(4分)已知△ABC的内角A,B,C所对的边分别是a,b,c,若a2＋ab＋b2－c2＝0,则角C的大小是.6.(4分)某学校高一年级男生人数占该年级学生人数的40%,在一次考试中,男,女平均分数分别为75、80,则这次考试该年级学生平均分数为.7.(4分)设常数 a∈R,若(x2＋)5的二项展开式中x7项的系数为－10,则 a＝.8.(4分)方程的实数解为.9.(4分)若cosxcosy＋sinxsiny＝,则cos(2x－2y)＝.10.(4分)已知圆柱Ω的母线长为l,底面半径为r,O是上底面圆心,A,B是下底面圆周上两个不同的点,BC是母线,如图,若直线OA与BC所成角的大小为,则＝.11.(4分)盒子中装有编号为1,2,3,4,5,6,7的七个球,从中任意抽取两个,则这两个球的编号之积为偶数的概率是(结果用最简分数表示)12.(4分)设AB是椭圆Γ的长轴,点C在Γ上,且∠CBA＝,若AB＝4,BC＝,则Γ的两个焦点之间的距离为.13.(4分)设常数a＞0,若9x＋对一切正实数x成立,则a的取值范围为.14.(4分)已知正方形ABCD的边长为1,记以A为起点,其余顶点为终点的向量分别为；以C为起点,其余顶点为终点的向量分别为,若i,j,k,l∈{1,2,3},且i≠j,k≠l,则的最小值是.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15.(5分)函数f(x)＝x2－1(x≥0)的反函数为f－1(x),则f－1(2)的值是( )A. B. C.1＋ D.1－16.(5分)设常数a∈R,集合A＝{x|(x－1)(x－a)≥0},B＝{x|x≥a－1},若A∪B＝R,则a的取值范围为( )A.(－∞,2)B.(－∞,2]C.(2,＋∞)D.[2,＋∞)17.(5分)钱大姐常说“好货不便宜”,她这句话的意思是：“好货”是“不便宜”的( )A.充分条件B.必要条件C.充分必要条件D.既非充分又非必要条件18.(5分)记椭圆围成的区域(含边界)为Ωn(n＝1,2,…),当点(x,y)分别在Ω1,Ω2,…上时,x＋y的最大值分别是M1,M2,…,则Mn＝( )A.0B.C.2D.2三、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.(12分)如图,正三棱锥O－ABC的底面边长为2,高为1,求该三棱锥的体积及表面积.20.(14分)甲厂以x千克/小时的速度匀速生产某种产品(生产条件要求1≤x≤10),每一小时可获得的利润是100(5x＋1－)元.(1)求证：生产a千克该产品所获得的利润为100a(5＋)元；(2)要使生产900千克该产品获得的利润最大,问：甲厂应该选取何种生产速度？并求此最大利润.21.(14分)已知函数f(x)＝2sin(ωx),其中常数ω＞0.(Ⅰ)令ω＝1,判断函数的奇偶性,并说明理由.(Ⅱ) 令ω＝2,将函数y＝f(x)的图象向左平移个单位,再向上平移1个单位,得到函数y ＝g(x)的图象.对任意a∈R,求y＝g(x)在区间[a,a＋10π]上的零点个数的所有可能.22.(16分)已知函数f(x)＝2－|x|,无穷数列{an }满足an＋1＝f(an),n∈N*(1)若a1＝0,求a2,a3,a4；(2)若a1＞0,且a1,a2,a3成等比数列,求a1的值(3)是否存在a1,使得a1,a2,…,an,…成等差数列？若存在,求出所有这样的a1,若不存在,说明理由.23.(18分)如图,已知双曲线C1：,曲线C2：|y|＝|x|＋1,P是平面内一点,若存在过点P的直线与C1,C2都有公共点,则称P为“C1－C2型点”(1)在正确证明C1的左焦点是“C1－C2型点“时,要使用一条过该焦点的直线,试写出一条这样的直线的方程(不要求验证)；(2)设直线y＝kx与C2有公共点,求证|k|＞1,进而证明原点不是“C1－C2型点”；(3)求证：圆x2＋y2＝内的点都不是“C1－C2型点”2013年上海市高考数学试卷(文科)参考答案与试题解析一、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.(4分)不等式＜0的解为0＜x＜.【分析】根据两数相除商为负,得到x与2x－1异号,将原不等式化为两个一元一次不等式组,求出不等式组的解集即可得到原不等式的解集.【解答】解：原不等式化为或,解得：0＜x＜,故答案为：0＜x＜【点评】此题考查了其他不等式的解法,利用了转化的思想,是一道基本试题.2.(4分)在等差数列{an }中,若a1＋a2＋a3＋a4＝30,则a2＋a3＝15 .【分析】根据给出的数列是等差数列,由等差数列的性质可得a1＋a4＝a2＋a3,结合已知条件可求a2＋a3.【解答】解：因为数列{an }是等差数列,根据等差数列的性质有：a1＋a4＝a2＋a3,由a1＋a2＋a3＋a4＝30,所以,2(a2＋a3)＝30,则a2＋a3＝15.故答案为：15.【点评】本题考查了等差中项概念,在等差数列中,若m,n,p,q,t∈N*,且m＋n＝p＋q＝2t,则a m ＋an＝ap＋aq＝2at,此题是基础题.3.(4分)设m∈R,m2＋m－2＋(m2－1)i是纯虚数,其中i是虚数单位,则m＝－2 .【分析】根据纯虚数的定义可得m2－1＝0,m2－1≠0,由此解得实数m的值.【解答】解：∵复数z＝(m2＋m－2)＋(m－1)i为纯虚数,∴m2＋m－2＝0,m2－1≠0,解得m＝－2,故答案为：－2.【点评】本题主要考查复数的基本概念,得到 m2＋m－2＝0,m2－1≠0,是解题的关键,属于基础题.4.(4分)已知,,则y＝ 1 .【分析】利用二阶行列式的运算法则,由写出的式子化简后列出方程,直接求解y即可.【解答】解：由已知,,所以x－2＝0,x－y＝1所以x＝2,y＝1.故答案为：1.【点评】本题考查了二阶行列式的展开式,考查了方程思想,是基础题.5.(4分)已知△ABC的内角A,B,C所对的边分别是a,b,c,若a2＋ab＋b2－c2＝0,则角C的大小是.【分析】利用余弦定理表示出cosC,将已知等式变形后代入求出cosC的值,由C为三角形的内角,利用特殊角的三角函数值即可求出C的度数.【解答】解：∵a2＋ab＋b2－c2＝0,即a2＋b2－c2＝－ab,∴cosC＝＝＝－,∵C为三角形的内角,∴C＝.故答案为：【点评】此题考查了余弦定理,以及特殊角的三角函数值,熟练掌握余弦定理是解本题的关键.6.(4分)某学校高一年级男生人数占该年级学生人数的40%,在一次考试中,男,女平均分数分别为75、80,则这次考试该年级学生平均分数为78 .【分析】设该年级男生有x人,女生有y人,这次考试该年级学生平均分数为a,根据“平均成绩×人数＝总成绩”分别求出男生的总成绩和女生的总成绩以及全班的总成绩,进而根据“男生的总成绩＋女生的总成绩＝全班的总成绩”列出方程,结合高一年级男生人数占该年级学生人数的40%,即可求出这次考试该年级学生平均分数.【解答】解：设该班男生有x人,女生有y人,这次考试该年级学生平均分数为a.根据题意可知：75x＋80y＝(x＋y)×a,且＝40%.所以a＝78,则这次考试该年级学生平均分数为78.故答案为：78.【点评】本题主要考查了平均数.解答此题的关键：设该班男生有x人,女生有y人,根据平均数的意义即平均成绩、人数和总成绩三者之间的关系列出方程解决问题.7.(4分)设常数 a∈R,若(x2＋)5的二项展开式中x7项的系数为－10,则 a＝－2 .【分析】利用二项展开式的通项公式求得二项展开式中的第r＋1项,令x的指数为7求得x7的系数,列出方程求解即可.【解答】解：的展开式的通项为Tr＋1＝C5r x10－2r()r＝C5r x10－3r a r令10－3r＝7得r＝1, ∴x7的系数是aC51∵x7的系数是－10,∴aC51＝－10,解得a＝－2.故答案为：－2.【点评】本题主要考查了二项式系数的性质.二项展开式的通项公式是解决二项展开式的特定项问题的工具.4 .8.(4分)方程的实数解为log3【分析】用换元法,可将方程转化为一个二次方程,然后利用一元二次方程根,即可得到实数x 的取值.【解答】解：令t＝3x(t＞0)则原方程可化为：(t－1)2＝9(t＞0)4可满足条件∴t－1＝3,t＝4,即x＝log3即方程的实数解为 log4.34.故答案为：log3【点评】本题考查的知识点是根的存在性,利用换元法将方程转化为一个一元二次方程是解答本题的关键,但在换元过程中,要注意对中间元取值范围的判断.9.(4分)若cosxcosy＋sinxsiny＝,则cos(2x－2y)＝－.【分析】已知等式左边利用两角和与差的余弦函数公式化简,求出cos(x－y)的值,所求式子利用二倍角的余弦函数公式化简后,将cos(x－y)的值代入计算即可求出值.【解答】解：∵cosxcosy＋sinxsiny＝cos(x－y)＝,∴cos(2x－2y)＝cos2(x－y)＝2cos2(x－y)－1＝－.故答案为：－.【点评】此题考查了两角和与差的余弦函数公式,二倍角的余弦函数公式,熟练掌握公式是解本题的关键.10.(4分)已知圆柱Ω的母线长为l,底面半径为r,O是上底面圆心,A,B是下底面圆周上两个不同的点,BC是母线,如图,若直线OA与BC所成角的大小为,则＝.【分析】过A作与BC平行的母线AD,由异面直线所成角的概念得到∠OAD为.在直角三角形ODA中,直接由得到答案.【解答】解：如图,过A作与BC平行的母线AD,连接OD,则∠OAD为直线OA与BC所成的角,大小为.在直角三角形ODA中,因为,所以.则.故答案为【点评】本题考查了异面直线所成的角,考查了直角三角形的解法,是基础题.11.(4分)盒子中装有编号为1,2,3,4,5,6,7的七个球,从中任意抽取两个,则这两个球的编号之积为偶数的概率是(结果用最简分数表示)【分析】从7个球中任取2个球共有＝21种,两球编号之积为偶数包括均为偶数、一奇一偶两种情况,有＝15种取法,利用古典概型的概率计算公式即可求得答案.【解答】解：从7个球中任取2个球共有＝21种,所取两球编号之积为偶数包括均为偶数、一奇一偶两种情况,共有＝15种取法,所以两球编号之积为偶数的概率为：＝.故答案为：.【点评】本题考查古典概型的概率计算公式,属基础题,其计算公式为：P(A)＝,其中n(A)为事件A所包含的基本事件数,m为基本事件总数.12.(4分)设AB是椭圆Γ的长轴,点C在Γ上,且∠CBA＝,若AB＝4,BC＝,则Γ的两个焦点之间的距离为.【分析】由题意画出图形,设椭圆的标准方程为,由条件结合等腰直角三角形的边角关系解出C的坐标,再根据点C在椭圆上求得b值,最后利用椭圆的几何性质计算可得答案.【解答】解：如图,设椭圆的标准方程为,由题意知,2a＝4,a＝2.∵∠CBA＝,BC＝,∴点C的坐标为C(－1,1),因点C在椭圆上,∴,∴b2＝,∴c2＝a2－b2＝4－＝,c＝,则Γ的两个焦点之间的距离为.故答案为：.【点评】本题考查椭圆的定义、解三角形,以及椭圆的简单性质的应用.13.(4分)设常数a＞0,若9x＋对一切正实数x成立,则a的取值范围为[,＋∞) .≥a＋1,【分析】由题设数a＞0,若9x＋对一切正实数x成立可转化为(9x＋)min利用基本不等式判断出9x＋≥6a,由此可得到关于a的不等式,解之即可得到所求的范围≥a＋1, 【解答】解：常数a＞0,若9x＋≥a＋1对一切正实数x成立,故(9x＋)min又9x＋≥6a,当且仅当9x＝,即x＝时,等号成立故必有6a≥a＋1,解得a≥故答案为[,＋∞)【点评】本题考查函数的最值及利用基本不等式求最值,本题是基本不等式应用的一个很典型的例子14.(4分)已知正方形ABCD的边长为1,记以A为起点,其余顶点为终点的向量分别为；以C为起点,其余顶点为终点的向量分别为,若i,j,k,l∈{1,2,3},且i≠j,k≠l,则的最小值是－5 .【分析】如图建立直角坐标系.不妨记以A为起点,其余顶点为终点的向量分别为,,,以C为起点,其余顶点为终点的向量分别为,,.再分类讨论当i,j,k,l取不同的值时,利用向量的坐标运算计算的值,从而得出的最小值.【解答】解：不妨记以A为起点,其余顶点为终点的向量分别为,,,以C为起点,其余顶点为终点的向量分别为,,.如图建立坐标系.(1)当i＝1,j＝2,k＝1,l＝2时,则＝[(1,0)＋(1,1)]•[((－1,0)＋(－1,－1)]＝－5；(2)当i＝1,j＝2,k＝1,l＝3时,则＝[(1,0)＋(1,1)]•[((－1,0)＋(0,－1)]＝－3；(3)当i＝1,j＝2,k＝2,l＝3时,则＝[(1,0)＋(1,1)]•[((－1,－1)＋(0,－1)]＝－4；(4)当i＝1,j＝3,k＝1,l＝2时,则＝[(1,0)＋(0,1)]•[((－1,0)＋(－1,－1)]＝－3；同样地,当i,j,k,l取其它值时,＝－5,－4,或－3.则的最小值是－5.故答案为：－5.【点评】本小题主要考查平面向量坐标表示、平面向量数量积的运算等基本知识,考查考查分类讨论、化归以及数形结合等数学思想方法,考查分析问题、解决问题的能力.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15.(5分)函数f(x)＝x2－1(x≥0)的反函数为f－1(x),则f－1(2)的值是( )A. B. C.1＋ D.1－【分析】根据反函数的性质,求f－1(2)的问题可以变为解方程2＝x2－1(x≥0).【解答】解：由题意令2＝x2－1(x≥0),解得x＝所以f－1(2)＝.故选：A.【点评】本题考查反函数的定义,解题的关键是把求函数值的问题变为解反函数的方程问题.16.(5分)设常数a∈R,集合A＝{x|(x－1)(x－a)≥0},B＝{x|x≥a－1},若A∪B＝R,则a的取值范围为( )A.(－∞,2)B.(－∞,2]C.(2,＋∞)D.[2,＋∞)【分析】当a＞1时,代入解集中的不等式中,确定出A,求出满足两集合的并集为R时的a的范围；当a＝1时,易得A＝R,符合题意；当a＜1时,同样求出集合A,列出关于a的不等式,求出不等式的解集得到a的范围.综上,得到满足题意的a范围.【解答】解：当a＞1时,A＝(－∞,1]∪[a,＋∞),B＝[a－1,＋∞),若A∪B＝R,则a－1≤1,∴1＜a≤2；当a＝1时,易得A＝R,此时A∪B＝R；当a＜1时,A＝(－∞,a]∪[1,＋∞),B＝[a－1,＋∞),若A∪B＝R,则a－1≤a,显然成立,∴a＜1；综上,a的取值范围是(－∞,2].故选：B.【点评】此题考查了并集及其运算,二次不等式,以及不等式恒成立的条件,熟练掌握并集的定义是解本题的关键.17.(5分)钱大姐常说“好货不便宜”,她这句话的意思是：“好货”是“不便宜”的( )A.充分条件B.必要条件C.充分必要条件D.既非充分又非必要条件【分析】“好货不便宜”,其条件是：此货是好货,结论是此货不便宜,根据充要条件的定义进行判断即可,【解答】解：若p⇒q为真命题,则命题p是命题q的充分条件；“好货不便宜”,其条件是：此货是好货,结论是此货不便宜,由条件⇒结论.故“好货”是“不便宜”的充分条件.故选：A.【点评】本题考查了必要条件、充分条件与充要条件的判断,属于基础题.18.(5分)记椭圆围成的区域(含边界)为Ωn(n＝1,2,…),当点(x,y)分别在Ω1,Ω2,…上时,x＋y的最大值分别是M1,M2,…,则Mn＝( )A.0B.C.2D.2【分析】先由椭圆得到这个椭圆的参数方程为：(θ为参数),再由三角函数知识求x＋y的最大值,从而求出极限的值.【解答】解：把椭圆得,椭圆的参数方程为：(θ为参数),∴x＋y＝2cosθ＋sinθ,∴(x＋y)max＝＝.∴Mn＝＝2.故选：D.【点评】本题考查数列的极限,椭圆的参数方程和最大值的求法,解题时要认真审题,注意三角函数知识的灵活运用.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.(12分)如图,正三棱锥O－ABC的底面边长为2,高为1,求该三棱锥的体积及表面积.【分析】根据题意画出图形,结合正三棱锥O－ABC的底面边长为2,高为1,由此入手,能够求出此三棱锥的体积及表面积.【解答】解：∵O－ABC是正三棱锥,其底面三角形ABC是边长为2的正三角形,其面积为,∴该三棱锥的体积＝＝；设O′是正三角形ABC的中心,则OO′⊥平面ABC,延长AO′交BC于D.则AD＝,O′D＝,又OO′＝1,∴三棱锥的斜高OD＝,∴三棱锥的侧面积为×＝2,∴该三棱锥的表面积为.【点评】本题考查三棱锥的体积、表面积的求法,解题时要认真审题,注意合理地化立体问题为平面问题.20.(14分)甲厂以x千克/小时的速度匀速生产某种产品(生产条件要求1≤x≤10),每一小时可获得的利润是100(5x＋1－)元.(1)求证：生产a千克该产品所获得的利润为100a(5＋)元；(2)要使生产900千克该产品获得的利润最大,问：甲厂应该选取何种生产速度？并求此最大利润.【分析】(1)由题意可得生产a千克该产品所用的时间是小时,由于每一小时可获得的利润是100(5x＋1－)元,即可得到生产a千克该产品所获得的利润；(2)利用(1)的结论可得生产1千克所获得的利润为90000(5＋),1≤x≤10.进而得到生产900千克该产品获得的利润,利用二次函数的单调性即可得出.【解答】解：(1)生产a千克该产品所用的时间是小时,∵每一小时可获得的利润是100(5x＋1－)元,∴获得的利润为100(5x＋1－)×元.因此生产a千克该产品所获得的利润为100a(5＋)元.(2)生产900千克该产品获得的利润为90000(5＋),1≤x≤10.设f(x)＝,1≤x≤10.则f(x)＝,当且仅当x＝6取得最大值.故获得最大利润为＝457500元.因此甲厂应以6千克/小时的速度生产,可获得最大利润457500元.【点评】正确理解题意和熟练掌握二次函数的单调性是解题的关键.21.(14分)已知函数f(x)＝2sin(ωx),其中常数ω＞0.(Ⅰ)令ω＝1,判断函数的奇偶性,并说明理由.(Ⅱ) 令ω＝2,将函数y＝f(x)的图象向左平移个单位,再向上平移1个单位,得到函数y ＝g(x)的图象.对任意a∈R,求y＝g(x)在区间[a,a＋10π]上的零点个数的所有可能.【分析】(1)特值法：ω＝1时,写出f(x)、F(x),求出F()、F(－),结合函数奇偶性的定义可作出正确判断；(2)根据图象平移变换求出g(x),令g(x)＝0可得g(x)可能的零点,而[a,a＋10π]恰含10个周期,分a是零点,a不是零点两种情况讨论,结合图象可得g(x)在[a,a＋10π]上零点个数的所有可能值；【解答】解：(1)f(x)＝2sinx,F(x)＝f(x)＋f(x＋)＝2sinx＋2sin(x＋)＝2(sinx＋cosx),F()＝2,F(－)＝0,F(－)≠F(),F(－)≠－F(),所以,F(x)既不是奇函数,也不是偶函数.(2)f(x)＝2sin2x,将y＝f(x)的图象向左平移个单位,再向上平移1个单位后得到y＝2sin2(x＋)＋1的图象,所以g(x)＝2sin2(x＋)＋1.令g(x)＝0,得x＝kπ＋或x＝kπ＋(k∈z),因为[a,a＋10π]恰含10个周期,所以,当a是零点时,在[a,a＋10π]上零点个数21,当a不是零点时,a＋kπ(k∈z)也都不是零点,区间[a＋kπ,a＋(k＋1)π]上恰有两个零点,故在[a,a＋10π]上有20个零点.综上,y＝g(x)在[a,a＋10π]上零点个数的所有可能值为21或20.【点评】本题考查函数y＝Asin(ωx＋φ)的图象变换、函数的奇偶性、根的存在性及根的个数的判断,考查数形结合思想,结合图象分析是解决(2)问的关键22.(16分)已知函数f(x)＝2－|x|,无穷数列{an }满足an＋1＝f(an),n∈N*(1)若a1＝0,求a2,a3,a4；(2)若a1＞0,且a1,a2,a3成等比数列,求a1的值(3)是否存在a1,使得a1,a2,…,an,…成等差数列？若存在,求出所有这样的a1,若不存在,说明理由.【分析】(1)由题意代入式子计算即可；(2)把a2,a3表示为a1的式子,通过对a1的范围进行讨论去掉绝对值符号,根据a1,a2,a3成等比数列可得关于a1的方程,解出即可；(3)假设这样的等差数列存在,则a1,a2,a3成等差数列,即2a2＝a1＋a3,亦即2－a1＋|2－|a1||＝2|a1|(*),分情况①当a1＞2时②当0＜a1≤2时③当a1≤0时讨论,由(*)式可求得a1进行判断；③当a1≤0时,由公差d＞2可得矛盾；【解答】解：(1)由题意,代入计算得a2＝2,a3＝0,a4＝2；(2)a2＝2－|a1|＝2－a1,a3＝2－|a2|＝2－|2－a1|,①当0＜a1≤2时,a3＝2－(2－a1)＝a1,所以,得a1＝1；②当a1＞2时,a3＝2－(a1－2)＝4－a1,所以,得(舍去)或.综合①②得a1＝1或.(3)假设这样的等差数列存在,那么a2＝2－|a1|,a 3＝2－|2－|a1||,由2a2＝a1＋a3得2－a1＋|2－|a1||＝2|a1|(*),以下分情况讨论：①当a1＞2时,由(*)得a1＝0,与a1＞2矛盾；②当0＜a1≤2时,由(*)得a1＝1,从而an＝1(n＝1,2,…),所以{an}是一个等差数列；③当a1≤0时,则公差d＝a2－a1＝(a1＋2)－a1＝2＞0,因此存在m≥2使得am ＝a1＋2(m－1)＞2,此时d＝am＋1－am＝2－|am|－am＜0,矛盾.综合①②③可知,当且仅当a1＝1时,a1,a2,…,an,…成等差数列.【点评】本题考查数列的函数特性、等差关系等比关系的确定,考查分类讨论思想,考查学生逻辑推理能力、分析解决问题的能力,综合性强,难度较大.23.(18分)如图,已知双曲线C1：,曲线C2：|y|＝|x|＋1,P是平面内一点,若存在过点P的直线与C1,C2都有公共点,则称P为“C1－C2型点”(1)在正确证明C1的左焦点是“C1－C2型点“时,要使用一条过该焦点的直线,试写出一条这样的直线的方程(不要求验证)；(2)设直线y＝kx与C2有公共点,求证|k|＞1,进而证明原点不是“C1－C2型点”；(3)求证：圆x2＋y2＝内的点都不是“C1－C2型点”【分析】(1)由双曲线方程可知,双曲线的左焦点为(),当过左焦点的直线的斜率不存在时满足左焦点是“C1－C2型点”,当斜率存在时,要保证斜率的绝对值大于等于该焦点与(0,1)连线的斜率；(2)由直线y＝kx与C2有公共点联立方程组有实数解得到|k|＞1,分过原点的直线斜率不存在和斜率存在两种情况说明过远点的直线不可能同时与C1和C2有公共点；(3)由给出的圆的方程得到圆的图形夹在直线y＝x±1与y＝－x±1之间,进而说明当|k|≤1时过圆内的点且斜率为k的直线与C2无公共点,当|k|＞1时,过圆内的点且斜率为k的直线与C2有公共点,再由圆心到直线的距离小于半径列式得出k的范围,结果与|k|＞1矛盾.从而证明了结论.【解答】(1)解：C1的左焦点为(),写出的直线方程可以是以下形式：或,其中.(2)证明：因为直线y＝kx与C2有公共点,所以方程组有实数解,因此|kx|＝|x|＋1,得.若原点是“C1－C2型点”,则存在过原点的直线与C1、C2都有公共点.考虑过原点与C2有公共点的直线x＝0或y＝kx(|k|＞1).显然直线x＝0与C1无公共点.如果直线为y＝kx(|k|＞1),则由方程组,得,矛盾.所以直线y＝kx(|k|＞1)与C1也无公共点.因此原点不是“C1－C2型点”.(3)证明：记圆O：,取圆O内的一点Q,设有经过Q的直线l与C1,C2都有公共点,显然l不与x轴垂直,故可设l：y＝kx＋b.若|k|≤1,由于圆O夹在两组平行线y＝x±1与y＝－x±1之间,因此圆O也夹在直线y＝kx ±1与y＝－kx±1之间,从而过Q且以k为斜率的直线l与C2无公共点,矛盾,所以|k|＞1.因为l与C1由公共点,所以方程组有实数解,得(1－2k2)x2－4kbx－2b2－2＝0.因为|k|＞1,所以1－2k2≠0,因此△＝(4kb)2－4(1－2k2)(－2b2－2)＝8(b2＋1－2k2)≥0,即b2≥2k2－1.因为圆O的圆心(0,0)到直线l的距离,所以,从而,得k2＜1,与|k|＞1矛盾.因此,圆内的点不是“C1－C2型点”.【点评】本题考查了双曲线的简单几何性质,考查了点到直线的距离公式,考查了直线与圆锥曲线的关系,直线与圆锥曲线联系在一起的综合题在高考中多以高档题、压轴题出现,主要涉及位置关系的判定,弦长问题、最值问题、对称问题、轨迹问题等.突出考查了数形结合、分类讨论、函数与方程、等价转化等数学思想方法.属难题.。

2013年高考文科数学上海卷试题与答案word解析版

22．已知函数f(x)＝2－|x|，无穷数列{ n}满足 n＋1＝f( n)，n N*.
(1)若 1＝0，求 2， 3， 4；
(2)若 1＞0，且 1， 2， 3成等比数列，求 1的值；
(3)是否存在 1，使得 1， 2，…， n，…成等差数列？若存在，求出所有这样的 1；若不存在，说明理由．
【考点】本题主要考查等比数列的性质、等差数列的通项公式及等差数列求和，考查考生分类讨论的思想，运算求解能力。
【解析】：(1) 2＝2， 3＝0， 4＝2.
(2) 2＝2－| 1|＝2－ 1， 3＝2－| 2|＝2－|2－ 1|.
①当0＜ 1≤2时， 3＝2－(2－ 1)＝ 1，所以＝(2－ 1)2，得 1＝1.
②当 1＞2时， 3＝2－( 1－2)＝4－ 1，所以 1(4－ 1)＝(2－ 1)2，
得 1＝ (舍去)或 1＝ .
10．已知圆柱Ω的母线长为l，底面半径为r，O是上底面圆心，A、B是下底面圆周上两个不同的点，BC是母线，如图．若直线OA与BC所成角的大小为，则＝______.
【答案】
【考点】本题主要考查圆柱中直线与直线成角的问题。
【解析】由题知， .
11．盒子中装有编号为1,2,3,4,5,6,7的七个球，从中任意取出两个，则这两个球的编号之积为偶数的概率是______(结果用最简分数表示)．
获得最大利润90 000× ＝457 500元．
因此甲厂应以6千克/小时的速度生产，可获得最大利润为457 500元．
21．已知函数f(x)＝2sin(ωx)，其中常数ω＞0.
(1)令ω＝1，判断函数F(x)＝f(x)＋的奇偶性，并说明理由；
(2)令ω＝2，将函数y＝f(x)的图像向左平移个单位，再向上平移1个单位，得到函数y＝g(x)的图像．对任意 R，求y＝g(x)在区间[ ，＋10π]上零点个数的所有可能值．

上海市十三校2013年高三调研考数学试卷(文科)

上海市十三校2013年高三调研考数学试卷（文科）2013.12一、填空题（本大题满分56分，每小题4分）1．函数()f x =的定义域是___________．2．幂函数)(x f y =的图像经过点)21,4(，则1()4f 的值为 . 3．方程tan 2cos()2x x π=+在区间()0,π内的解为 .4．计算：21lim 1n n n n →∞⎡⎤⎛⎫-⎪⎢⎥+⎝⎭⎣⎦=_________． 5．已知二元一次方程组的增广矩阵是421m m mm +⎛⎫⎪⎝⎭，若该方程组无解，则实数m 的值为___________．6．已知流程图如图所示，为使输出的b 值为16，则判断框内①处可以填数字．（填入一个满足要求的数字即可）7．等差数列{}n a 中，1102,15a S ==，记2462n n B a a a a =+++ ，则当n =____时，n B 取得最大值. 8．已知x y R +∈、，且41x y +=，求19x y+的最小值.某同学做如下解答：因为 x y R +∈、，所以14x y =+≥19x y +≥ ①⨯②得1924x y +≥=，所以 19x y +的最小值为24。

判断该同学解答是否正确，若不正确，请在以下空格内填写正确的最小值；若正确，请在以下空格内填写取得最小值时x 、y 的值. . 9．若4mx x+≥在[]3,4x ∈内恒成立，则实数m 的取值范围是 . 10．函数()()x x y 2arccos1arcsin +-=的值域是 . 11．已知函数()(2318,343x tx x f x t x ⎧-+<⎪=⎨-≥⎪⎩在R 递减，则实数t 的取值范围是_________.12．设正数数列{}n a 的前n 项和是n S ,若{}n a 和{n S }都是等差数列,且公差相等,则=+d a 1__ _.13．函数()()g x x R ∈的图像如图所示，关于x 的方程 2[()]()230g x m g x m +⋅++=有三个不同的实数解，则m 的取值范围是_______________．14．已知无穷数列{}n a 具有如下性质:①1a 为正整数；②对于任意的正整数n ,当n a 为偶数时,12nn a a +=;当n a 为奇数时,112n n a a ++=.在数列{}n a 中，若当n k ≥时，1n a =，当1n k ≤<时，1n a >（2k ≥，*k N ∈），则首项1a 可取数值的个数为（用k 表示）二、选择题（本大题满分20分，每小题5分） 15．函数22log x y x =+的零点在区间（）内．（A ）11(,)43 （B ）12(,)35 （C ）21(,)52 （D ）12(,)2316．如果a b c 、、满足c b a <<，且0ac <，那么下列选项不恒成立的是（）．（A ）ab ac > （B ）22cb ab <（C ）()0c b a -> （D ）()0ac a c -<17．如图，点P 在边长为1的正方形的边上运动，M 是CD 的中点，则当P 沿A B C M ---运动时，点P 经过的路程x 与APM ∆的面积y 的函数()y f x =的图像的形状大致是下图中的（）．（A ）（B ）（C ）（D ）18．已知x y R ∈、，命题p 为x y >，命题q 为sin cos sin cos x y x y x y +>+.则命题p 成立是命题q 成立的 ( )．（A ）充分非必要条件（B ）必要非充分条件（C ）充要条件（D ）非充分非必要条件三、解答题（本大题满分74分） 19．（本题满分12分，第一小题满分4分，第二小题满分8分）已知集合21|1,1x A x x R x -⎧⎫=≤∈⎨⎬+⎩⎭，集合{}1,B x x a x R =-≤∈. （1）求集合A ；（2）若R B A B = ð，求实数a 的取值范围.20．（本题满分14分，第一小题满分7分，第二小题满分7分）P AB行列式c o s 2s i n 01c o sA A x A x x ()0A >按第一列展开得1121312M M -+，记函数()1121fx M M =+，且()f x 的最大值是4. （1）求A ；（2）将函数()y f x =的图像向左平移12π个单位，再将所得图像上各点的横坐标扩大为原来的2倍，纵坐标不变，得到函数()y g x =的图像，求()g x 在11,1212ππ⎛⎫-⎪⎝⎭上的值域．21．（本题满分14分，第一小题满分6分，第二小题满分8分）钓鱼岛及其附属岛屿是中国固有领土，如图：点A 、B 、C 分别表示钓鱼岛、南小岛、黄尾屿，点C 在点A 的北偏东47°方向，点B 在点C 的南偏西36°方向，点B 在点A 的南偏东79°方向，且A 、B 两点的距离约为3海里。

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2012学年第一学期徐汇区高三年级数学学科学习能力诊断卷（文）（考试时间：120分钟，满分150分） 2013.1一．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填写结果，每个空格填对得4分，否则一律得零分.1．方程组2132x y x y -=⎧⎨+=-⎩的增广矩阵是__________________.2. 已知幂函数()f x 的图像过点18,2⎛⎫⎪⎝⎭，则此幂函数的解析式是()f x =_____________. 3．（文）若4cos 5θ=，则=θ2cos ___________.4．若抛物线22(0)y px p =>的焦点与双曲线221610xy-=的右焦点重合，则实数p 的值是 .5．函数()sin()(0,0,||)2f x A x A πωϕωϕ=+>><的部分图像如右图所示，则()f x = _________.6．（文）若(1,2)n = 是直线l 的一个方向向量，则直线l 的倾斜角的大小为_________________.(结果用反三角函数值表示) 7．（文）不等式210xx+≥ 1 2 2的解为 .8．高三（1）班班委会由4名男生和3名女生组成，现从中任选3人参加上海市某社区敬老服务工作，则选出的人中至少有一名女生的概率是 .(结果用最简分数表示)9．如图所示的程序框图，输出b 的结果是_________.10．（文）数列{}n a 的通项公式*1 , 1()1, 2(1)n n a n N n n n =⎧⎪=∈⎨≥⎪+⎩，前n 项和为n S ，则lim n n S →∞=_____________.11.（文）边长为1的正方形ABCD 中，M 为BC 的中点，E 在线段AB 上运动，则EC EM⋅的取值范围是____________.12．（文）函数{}()min 2f x =-，其中{},min ,,a a ba b b a b≤⎧=⎨>⎩，若动直线y m =与函数()y f x =的图像有三个不同的交点，则实数m 的取值范围是______________.13．（文）若平面向量i a 满足 1(1,2,3,4)i a i ==且10(1,2,3)i i a a i +⋅== ,则1234a a a a +++的最大值为 .14．已知线段010A A 的长度为10，点129,,,A A A 依次将线段010A A 十等分.在0A 处标0，往右数1点标1，再往右数2点标2，再往右数3点标3……(如图)，遇到最右端或最左端返回，按照0A →10A →0A →10A → 的方向顺序，不断标下去，（文）那么标到10这个数时，所在点上的最小数为_____________.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．下列排列数中，等于*(5)(6)(12)(13,)n n n n n N ---≥∈ 的是（） (A)712n P - (B) 75n P - (C) 85n P - (D) 812n P -16．在ABC ∆中，“cos sin cos sin A A B B +=+”是“090C ∠=”的（）(A) 充分非必要条件 (B) 必要非充分条件(C) 充要条件 (D) 既不充分也不必要条件17．若函数21()ax f x x-=在()0,+∞上单调递增，那么实数a 的取值范围是（）(A)0a ≥(B)0a >(C)0a ≤(D) 0a <18．（文）对于直角坐标平面xOy 内的点(,)A x y （不是原点），A 的“对偶点”B 是指：满足1OA OB =且在射线OA 上的那个点. 则圆心在原点的圆的对偶图形（）(A) 一定为圆 (B) 一定为椭圆 (C) 可能为圆，也可能为椭圆 (D) 既不是圆，也不是椭圆三．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．(本题满分12分)已知集合3{|0}4x A x x -=<-，实数a 使得集合{}|()(5)0B x x a x =-->满足A B ⊆，求a 的取值范围.20．(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.已知函数)(x f =21log 1x x +-.(1)判断函数)(x f 的奇偶性，并证明； (2)求)(x f 的反函数)(1x f-，并求使得函数12()()log g x f x k -=-有零点的实数k 的取值范围.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.（文）某种型号汽车的四个轮胎半径相同，均为40R cm =，该车的底盘与轮胎中心在同一水平面上. 该车的涉水安全要求．．．．．．是：水面不能超过它的底盘高度. 如图所示：某处有一“坑形”地面，其中坑ABC 形成顶角为0120的等腰三角形，且60AB BC cm ==，如果地面上有()h cm (40h <)高的积水(此时坑内全是水，其它因素忽略不计).(1) 当轮胎与AB 、BC 同时接触时，求证：此轮胎露在水面外的高度(从轮胎最上部到水面的距离)为10d h =+；(2) 假定该汽车能顺利通过这个坑(指汽车在过此坑时，符合涉水安全要求．．．．．．)，求h 的最大值.(精确到1cm).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.（文）已知椭圆2222:1(0)x y C a b a b+=>>的一个焦点为(1,0)F ，点(-在椭圆C上，点T满足OT OF =（其中O 为坐标原点）, 过点F 作一斜率为(0)k k >的直线交椭圆于P 、Q 两点(其中P 点在x 轴上方，Q 点在x 轴下方) .(1)求椭圆C 的方程；(2)若1k =，求PQT ∆的面积；(3)设点P '为点P 关于x 轴的对称点，判断P Q ' 与QT的位置关系，并说明理由.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.（文）对于数列{}n x ，从中选取若干项，不改变它们在原来数列中的先后次序，得到的数列称为是原来数列的一个子数列. 某同学在学习了这一个概念之后，打算研究首项为1a ,公差为d 的无穷等差数列{}n a 的子数列问题，为此，他取了其中第一项1a ，第三项3a 和第五项5a .(1) 若135,,a a a 成等比数列,求d 的值；(2) 在11a =, 3d =的无穷等差数列{}n a 中，是否存在无穷子数列{}n b ，使得数列{}n b 为等比数列？若存在，请给出数列{}n b 的通项公式并证明；若不存在，说明理由；(3) 他在研究过程中猜想了一个命题：“对于首项为正整数a ，公比为正整数q (1q >)的无穷等比数列{}n c ,总可以找到一个子数列{}n d ,使得{}n d 构成等差数列”. 于是，他在数列{}n c 中任取三项,,()k m n c c c k m n <<，由k n c c +与2m c 的大小关系去判断该命题是否正确.他将得到什么结论？参考答案一、填空题：（每题4分）1. 2111-⎛⎫ ⎪⎝⎭ 3　-2 2. 13x - 3.（文）725 4. 8 5. 2sin 4x π6.（文） arctan27.文）x ≥08.31359. 1 10.（文）3211. （文）13,22⎡⎤⎢⎥⎣⎦12.文）-2 13.（文） 14.（文）5二、选择题：（每题5分）15. C 16. B 17.A 18.文）A三、解答题19. 解：A=(3,4)………………………………………………………………………………..2分a ≥5时，B=(,)(,5)a +∞⋃-∞，满足A ⊆B ；…………………………………..6分 a<5时，B=(5,)(,)a +∞⋃-∞，由A ⊆B ，得a ≥4，故4≤a<5，……………..10分综上，得实数a 的取值范围为a ≥4. ……………………………………………..12分20. 解：(1)f(x)的定义域为(,1)(1,)-∞-⋃+∞……………………………………………..2分 f(-x)=log 211x x -+--=log 211x x -+=-f(x)，所以，f(x)为奇函数. ………………………………………..6分 (2)由y=21log 1x x +-，得x=2121yy+-,所以，f -1(x)=2121xx+-,x ≠0. ……………………………………..9分因为函数12()()log g x fx k -=-有零点，所以，2log k 应在)(1x f-的值域内.所以，log 2k=2121xx +-=1+221x-(,1)(1,)∈-∞-⋃+∞, ………………….13分从而，k 1(2,)(0,)2∈+∞⋃. ……………………………………………..14分21．（文）解：(1) 当轮胎与AB 、BC 同时接触时，设轮胎与AB 边的切点为T ，轮胎中心为O ，则|OT|=40，由∠ABC=1200，知∠OBT=600， …………………………………..2分故|OB|=. .…………………………………………………………………..4分所以，从B+40， …………………………..6分此轮胎露在水面外的高度为060cos 60⋅10h +-，得证. …..8分(2)只要d ≥40， …………………………………………………………..12分即10h -≥40，解得h ≤16cm.，所以h 的最大值为16cm. …..14分22．（文）解:(1)由222211112a b ab ⎧-=⎪⎨+=⎪⎩，得 ……………………………………………………………..2分 a 2=2,b 2=1, 所以，椭圆方程为2212xy +=. …………………………………………………..4分(2)设PQ:y=x-1，由22112x y x y =+⎧⎪⎨+=⎪⎩得3y 2+2y-1=0, …………………..6分解得： P(41,33),Q(0,-1)，由条件可知点(2,0)T , PQT S ∆=12|FT||y 1-y 2|=23. ….. ……………………………………10分(3) 判断：P Q ' 与QT共线. ….. …….. …….. ………………………………………11分设1122(,),(,)P x y Q x y则P '(x 1,-y 1)，P Q ' =(x 2-x 1,y 2+y 1)，TQ=(x 2-2,y 2), ……………………………..12分由22(1)12y k x x y =-⎧⎪⎨+=⎪⎩得2222(12)4220k x k x k +-+-=. ………………………..13分(x 2-x 1)y 2-(x 2-2)(y 1+y 2)=(x 2-x 1)k(x 2-1)-(x 2-2)(kx 1-k+kx 2-k) =3k(x 1+x 2)-2kx 1x 2-4k=3k22412kk+-2k222212k k-+-4k=k(2222124441212kk kk---++)=0. …………………………..15分所以，P Q ' 与QT共线. ………………………………………………………..16分23．（文）解:(1)由a 32=a 1a 5， ………………………………………………………………………..2分即(a 1+2d)2=a 1(a 1+4d)，得d=0. ……………………………………………..4分(2) 解：a n =1+3(n-1)，如b n =4n-1便为符合条件的一个子数列. ……………………..7分因为b n =4n-1=(1+3)n-1=1+11n C -3+21n C -32+…+11n n C --3n-1=1+3M, …………………..9分这里M=11n C -+21n C -3+…+11n n C --3n-2为正整数，所以,b n =1+3M =1+3 [(M+1)-1]是{a n }中的第M+1项，得证. ……………….11分(注：b n 的通项公式不唯一)(3) 该命题为假命题. …………………………………………………….12分由已知可得111,,k m n k m n c aq c aqc aq---===, 因此,11k n k n c c aqaq--+=+,又122m m c aq-=,故 1111()22(12)k n m k n km kk n m c c c aqaq aqaq qq------+-=+-=+-, …………..15分由于,,k m n 是正整数，且n m >，则1,1n m n k m k ≥+-≥-+, 又q 是满足1q >的正整数，则2q ≥, 112121212210n km km k m km km km km kqqqqqqqqq---+-----+-≥+-=+-≥+-=>,所以，k n c c +>2m c ,从而原命题为假命题. …………………………………………..18分。