上海市2014届十三校第一次联考试题(理科)(2014.12.12)

高三学科测试数学试题(理科)考试时间 120分钟 满分150分考生注意：1．答卷前，考生务必在答题纸上将学校、姓名及准考证号等填写清楚．答题时客观题用2B 铅笔按要求涂写，主观题用黑色水笔填写．2．本试卷共有23道题，共4页．满分150分，考试时间120分钟． 3．考试后只交答题纸，试卷由考生自己保留．一. 填空题(本大题满分56分)本大题共有14题，考生应在答题纸上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. 已知集合1{|0}{||1|2}1x A x B x x x -=<=-<+，，则B A =ð . 2. 已知tan 2α=-，则sin 3cos cos sin αααα-=+____________． 3. 在复平面中，复数2(1)3i i++(i 是虚数单位)对应的点在第 象限.4. 函数()2|sin |3f x x =+的最小正周期是 .5. 已知函数{2 2()(2) 2x x f x f x x ≥=+<，，，则2(log 3)f = .6. 已知351()log log 2014f x a x b x =++，若12015()20142014f =，则(2014)f = ．7. 满足2arccos()arccos(2)x x >的实数x 的取值范围是 . 8. 设n a是(1(234)n n = ，，，的展开式中x 的一次项的系数，若21(1)n n n n a b a +++=，则nb 的最小值是 .9. 若存在正数x 使221x x m x<成立，则实数m 的取值范围是 . 10. 某班班会准备从甲、乙等7名学生中选4名学生发言，要求甲、乙至少有一人参加，那么不同的发言顺序的种数为 (用数字作答).11. 已知函数()|||2|(0)f x x k x k k =-+->，若当34x ≤≤时，()f x 能取到最小值，则实数k 的取值范围是 . 12. 已知数列{}n a 中，1112 1n n a a a +==-+，，若k 是5的倍数，且2k a =，则k = . 13. 如果一个正整数能表示为两个连续偶数的平方差，那么称这个正整数为“神秘数”．则区间[1 200]，内的所有“神秘数”之和为 ．14. 已知0 m m >≠，1 l y m =：与函数2|log |y x =的图像从左至右相交于点 A B 、，直线241l y m =+：与函数2|log |y x =的图像从左至右相交于点 C D 、.记线段AC 和BD 在x 轴上的投影长度分别为 a b ,，当m 变化时，ba的最小值是 .二. 选择题(本大题满分20分)本大题共有4题，每题只有一个正确答案.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. 设z 表示复数z 的共轭复数，则与“复数z 为实数”不等价的说法是 ( ) A.z z = B.20z ≥ C.0z z += D.Im 0z =(Im z 表示复数z 的虚部)16. 在ABC ∆中，内角 A B C 、、所对的边分别是 a b c 、、，则“cos cos a A b B =”是“ABC ∆是以 A B 、为底角的等腰三角形”的 ( ) A.充分非必要条件 B.必要非充分条件 C.充要条件 D.既非充分也非必要条件 17. 函数()43x g x =⨯的图像可看成将函数()3x f x =的图像 ( ) A.向左平移3log 4个单位得到 B.各点纵坐标不变，横坐标伸长到原来的4倍得到 C.向右平移3log 4个单位得到 D.各点横坐标不变，纵坐标缩短到原来的14得到 18. 如图所示，医用输液瓶可以视为两个圆柱的组合体. 开始输液时，滴管内匀速滴下液体(滴管内液体忽略不计)，设输液开始后x 分钟，瓶内液面与进气管 的距离为h 厘米，已知当0x =时，13h =. 如果瓶内的药液恰 好156分钟滴完. 则函数()h f x =的大致图像为 ( )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)本题共有2个小题，第(1)小题满分5分，第(2)小题满分7分．已知二次函数2()23f x mx x =--，若不等式()0f x <的解集为(1)n -， (1)解关于x 的不等式：224(1)1x x n m x -+>+-；(2)是否存在实数(0 1)a ∈，，使得关于x 的函数1()4x x y f a a +=-([12]x ∈，)的最小值为4-？若存在，求a 的值；若不存在，说明理由.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．某同学用“五点法”画函数()sin()(0 )2f x A x πωϕωϕ=+><，在某一个周期内的图像(1)请写出上表的1、2、3，并直接写出函数的解析式；(2)设()()(1)g x x f x =+-，当[0 4]x ∈，时，求()g x 的单调递增区间.B CDA21. (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．某环境保护部门对某处的环境状况用“污染指数”来监测．据测定，该处的“污染指数”与附近污染源的强度和距离之比成正比，比例系数为常数(0)k k >．现已知相距36km 的A B 、两家化工厂(污染源)的污染强度分别为1和a ，它们连线段上任意一点C 处的污染指数y 等于两化工厂对该处的污染指数之和．设(km)AC x =． (1)试将y 表示为x 的函数，指出其定义域；(2)当6x =时，C 处的“污染指数”最小，试求B 化工厂的污染强度a 的值．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6分，第(3)小题满分7分.已知数列{}n a 的各项均为正数，记12231() ()k k A k a a a B k a a a +=+++=+++ ，，342()k C k a a a +=+++ .(1)若113(5)n n n a =+-，求lim ()n B n →∞； (2)若121 5a a ==，，且对任意*N k ∈，()B k 都是()A k 与()C k 的等差中项，求数列{}n a 的通项公式； (3)已知命题：“若数列{}n a 是公比为q 的等比数列，则对任意*N () () ()k A k B k C k ∈，，，都是公比为q 的等比数列”是真命题.试写出该命题的逆命题，判断真假，并证明.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M 是满足下列条件的集合：①0 1M M ∈∈，；②若 x y M ∈，，则x y M -∈；③若x M ∈且0x ≠，则1M x∈.(1)判断13M ∈是否正确，说明理由；(2)证明：“x Z ∈”是“x M ∈”的充分条件，其中Z 是正整数集；(3)证明：若 x y M ∈，，则xy M ∈.。

2014年普通高等学校招生全国统一考试（上海）卷数学（理科）一．填空题：共14小题，每小题4分，共56分。

1．函数()212cos 2y x =-的最小正周期是______________。

2．若复数12z i =+，其中i 是虚数单位，则1z z z ⎛⎫+⋅= ⎪⎝⎭______________。

3．若抛物线22y px =的焦点与椭圆22195x y +=的右焦点重合，则该抛物线的准线方程为_______________。

4．设()()()2x x a f x x x a <⎧⎪=⎨≥⎪⎩，若()24f =，则a 的取值范围是_______________。

5．若实数,x y 满足1xy =，则222x y +的最小值为______________。

6．若圆锥的侧面积是底面积的3倍，则其母线与底面夹角的大小为____________（结果用反三角函数表示）。

7．已知曲线C 的极坐标方程为()3cos 4sin 1ρθθ-=，则C 与极轴的交点到极点的距离为___________。

8．设无穷等比数列{}n a 的公比为q ，若()134lim n n a a a a →∞=+++，则q =________。

9．若()2132f x x x-=-，则满足()0f x <的x 的取值范围为________________。

10．为强化安全意识，某商场拟在未来的连续10天中随机选择3天进行紧急疏散演练，则选择的3天恰好为连续3天的概率是______________（结果用最简分数表示）。

11．已知互异的实数,a b 满足0ab ≠，集合{}{}22,,a b a b =，则a b +=________。

12．设常数a使方程sin x x a =在闭区间[]0,2π上恰有三个解123,,x x x ，则123x x x ++=___________。

13．某游戏的得分为1,2,3,4,5，随机变量ξ表示小白玩该游戏的得分。

2014年全国普通高等学校招生统一考试上海数学试卷（理工农医类）考生注意：1、本试卷共4页，23道试题，满分150分。

考试时间120分钟。

2、本考试分设试卷和答题纸。

试卷包括试题与答题要求。

作答必须涂（选择题）或写（非选择题）在答题纸上。

在试卷上作答一律不得分。

3、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，在答题纸正面清楚地填写姓名。

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(cos 212的最小正周期是x y -=【答案】2π【解析】2π4π2∴4cos -)2(cos 2-12====T x x y 周期 2、若复数z=1+2i ，其中i 是虚数单位，则⎪⎪⎭⎫ ⎝⎛_z 1 +z z ⋅=___________.【答案】6【解析】61)41(11(∴21=++=+=•++=z z z zz i z 3、若抛物线y 2=2px 的焦点与椭圆15922=+y x 的右焦点重合，则该抛物线的准线方程为___________.【答案】x=-2【解析】2-2-)0,2(2)0,2(159222==∴=∴=+x x px y y x 所以，是其准线方程为焦点为右焦点为 4、设⎩⎨⎧+∞∈-∞∈=),,[,),,(,)(2a x x a x x x f 若4)2(=f ，则a 的取值范围为_____________.【答案】]2,∞-(【解析】]2,∞-(.2≤),∞,[∈2∴4)2(所以，是解得a a f += 5、若实数x,y 满足xy=1,则2x +22y 的最小值为______________.【答案】22【解析】22,2222≥22y ∴1222222所以，是=•+=+=xx x x x xy 6.若圆锥的侧面积是底面积的3倍，则其母线与底面所成角的大小为（结果用反三角函数值表示）。

2014年普通高等学校招生全国统一考试（上海卷）数学（理科）第Ⅰ卷（选择题 共50分）一、填空题（本大题共14小题，共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）【2014年上海，理1，4分】函数212cos (2)y x =-的最小正周期是 ． 【答案】2π【解析】原式＝cos4x -，242T ππ==． （2）【2014年上海，理2，4分】若复数12i z =+，其中i 是虚数单位，则1z z z ⎛⎫+⋅= ⎪⎝⎭ ．【答案】6【解析】原式＝211516z z z ⋅+=+=+=．（3）【2014年上海，理3，4分】若抛物线22y px =的焦点与椭圆22195x y +=的右焦点重合，则该抛物线的准线方程为 ． 【答案】2x =-【解析】椭圆右焦点为(2,0)，即抛物线焦点，所以准线方程2x =-．（4）【2014年上海，理4，4分】设2(,)()[,)xx a f x xx a ∈-∞⎧=⎨∈+∞⎩，若(2)4f =，则a 的取值范围为 ．【答案】2a ≤【解析】根据题意，2[,)a ∈+∞，∴2a ≤．（5）【2014年上海，理5，4分】若实数x ，y 满足1xy =，则222x y +的最小值为 ．【答案】【解析】2222x y x +≥⋅=． （6）【2014年上海，理6，4分】若圆锥的侧面积是底面积的3倍，则其母线与底面夹角的大小为 ．（结果用反三角函数值表示）【答案】1arccos 3【解析】设圆锥母线长为R ，底面圆半径为r ，∵3S S =侧底，∴23r R r ππ⋅⋅=⋅，即3R r =，∴1cos 3θ=，即母线与底面夹角大小为1arccos 3．（7）【2014年上海，理7，4分】已知曲线C 的极坐标方程为(3cos 4sin )1ρθθ-=，则C 与极轴的交点到极点的距离是 ．【答案】13【解析】曲线C 的直角坐标方程为341x y -=，与x 轴的交点为1(,0)3，到原点距离为13．（8）【2014年上海，理8，4分】设无穷等比数列{}n a 的公比为q ，若()134lim n n a a a a →∞=+++L ，则q = ．【解析】223111011a a q a q q q q q ==⇒+-=⇒=--，∵01q <<，∴q =P2P 5P6P 7P 8P 4P 3P 1B A （9）【2014年上海，理9，4分】若2132()f x x x -=-，则满足()0f x <的x 的取值范围是 ． 【答案】(0,1)【解析】2132()0f x x x -<⇒<，结合幂函数图像，如下图，可得x 的取值范围是(0,1)． （10）【2014年上海，理10，4分】为强化安全意识，某商场拟在未来的连续10天中随机选择3天进行紧急疏散演练，则选择的3天恰好为连续3天的概率是 ．（结果用最简分数表示） 【答案】115【解析】3108115P C ==．（11）【2014年上海，理11，4分】已知互异的复数,a b 满足0ab ≠，集合{}{}22,,a b a b =，则a b += ． 【答案】1-【解析】第一种情况：22,a a b b ==，∵0ab ≠，∴1a b ==，与已知条件矛盾，不符；第二种情况：22,a b b a ==，∴431a a a =⇒=，∴210a a ++=，即1a b +=-．（12）【2014年上海，理12，4分】设常数a使方程sin x x a +=在闭区间[0,2]π上恰有三个解123,,x x x ，则123x x x ++= ． 【答案】73π【解析】化简得2sin()3x a π+=，根据下图，当且仅当a =恰有三个交点，即12370233x x x πππ++=++=．（13）【2014年上海，理13，4分】某游戏的得分为1,2,3,4,5，随机变量ξ表示小白玩该游戏的得分．若() 4.2E ξ=，则小白得5分的概率至少为 ．【答案】0.2【解析】设得i 分的概率为i p ，∴123452345 4.2p p p p p ++++=，且123451p p p p p ++++=，∴12345444444p p p p p ++++=，与前式相减得：1235320.2p p p p ---+=， ∵0i p ≥，∴1235532p p p p p ---+≤，即50.2p ≥．（14）【2014年上海，理14，4分】已知曲线:C x =，直线:6l x =． 若对于点(,0)A m ，存在C 上的点P 和l 上的Q 使得0AP AQ +=u r r，则m 的取值范围为 ．【答案】1615-【解析】根据题意，A 是PQ 中点，即622P Q P x x x m ++==，∵20P x -≤≤，∴[2,3]m ∈． 二、选择题（本大题共有4题，满分20分）考生应在答题纸相应编号位置填涂，每题只有一个正确选项，选对得5分，否则一律得零分． （15）【2014年上海，理15，5分】设,a b ∈R ，则“4a b +>”是“2a >且2b >”的（ ）（A ）充分条件 （B ）必要条件 （C ）充要条件 （D ）既非充分也非必要条件 【答案】B【解析】充分性不成立，如5a =，1b =；必要性成立，故选B ．（16）【2014年上海，理16，5分】如图，四个棱长为1的正方体排成一个正四棱柱，AB 是一条侧棱，(1,2,,8)i P i =L 是上底面上其余的八个点，则(1, 2, , 8)i AB AP i ⋅=u u u r u u u r K 的 不同值的个数为（ ）（A ）1 （B ）2 （C ）4 （D ）8【答案】AA βC BαD【解析】根据向量数量积的几何意义，i AB AP ⋅u u u r u u u r 等于AB u u u r乘以i AP u u u r 在AB u u u r 方向上的投影，而i AP u u u r 在AB u u u r 方向上的投影是定值，AB u u u r也是定值，∴i AB AP ⋅u u u r u u u r 为定值1，故选A ．（17）【2014年上海，理17，5分】已知111(,)P a b 与222(,)P a b 是直线1y kx =+（k 为常数）上两个不同的点，则关于x 和y 的方程组112211a x b y a x b y +=⎧⎨+=⎩的解的情况是（ ）（A ）无论12,,k P P 如何，总是无解 （B ）无论12,,k P P 如何，总有唯一解（C ）存在12,,k P P ，使之恰有两解 （D ）存在12,,k P P ，使之有无穷多解 【答案】B【解析】由已知条件111b ka =+，221b ka =+，11122122a b D a b a b a b ==-122112(1)(1)0a ka a ka a a =+-+=-≠， ∴有唯一解，故选B ．（18）【2014年上海，理18，5分】设2(),0,()1,0.x a x f x x a x x ⎧-≤⎪=⎨++>⎪⎩若(0)f 是()f x 的最小值，则a 的取值范围为（ ） （A ）[1,2]- （B ）[1,0]- （C ）[1,2] （D ）[0,2]【答案】D【解析】先分析0x ≤的情况，是一个对称轴为x a =的二次函数，当0a <时，min ()()(0)f x f a f =≠，不符合题意，排除AB 选项；当0a =时，根据图像min ()(0)f x f =，即0a =符合题意，排除C 选项，故选D ．三、解答题（本题共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）【2014年上海，理19，12分】底面边长为2的正三棱锥P ABC -，其表面展开图是三角形123PP P ，如图．求123PP P ∆的各边长及此三棱锥的体积V ． 解：根据题意可得12,,P B P 共线，∵112ABP BAP CBP ∠=∠=∠，60ABC ∠=︒，∴11260ABP BAP CBP ∠=∠=∠=︒，∴160P ∠=︒，同理2360P P ∠=∠=︒，∴123PP P ∆是等 边三角形，P ABC -是正四面体，所以123PP P ∆边长为4；∴3123V AB =⨯=． （20）【2014年上海，理20，14分】设常数0a ≥，函数2()2xx af x a+=-．（1）若4a =，求函数()y f x =的反函数1()y f x -=；（2）根据a 的不同取值，讨论函数()y f x =的奇偶性，并说明理由．解：（1）∵4a =，∴24()24x x f x y +==-，∴4421x y y +=-，∴244log 1y x y +=-， ∴1244()log 1x y f x x -+==-，(,1)(1,)x ∈-∞-+∞U ． ……6分（2）若()f x 为偶函数，则()()f x f x =-，∴2222x x x xa aa a --++=--，整理得(22)0x x a --=，∴0a =，此时为偶函， 若()f x 为奇函数，则()()f x f x =--，∴2222x x x xa aa a--++=---，整理得210a -=，∵0a ≥，∴1a =，此时 为奇函数，当(0,1)(1,)a ∈⋃+∞时，此时()f x 既非奇函数也非偶函数． ……14分（21）【2014年上海，理21，14分】如图，某公司要在A B 、两地连线上的定点C 处建造广告牌CD ，其中D 为顶端，AC 长35米，CB 长80米． 设点A B 、在同一水平面上，从A 和B 看D 的仰角分别为α和β．（1）设计中CD 是铅垂方向． 若要求2αβ≥，问CD 的长至多为多少（结 果精确到0.01米）？（2）施工完成后，CD 与铅垂方向有偏差．现在实测得38.12α=︒，18.45β=︒，求CD 的长（结果精确到0.01米）．P 2解：（1）设CD 的长为x 米，则tan ,tan 3580x x αβ==，∵202παβ>≥>， ∴tan tan 2αβ≥，∴22tan tan 1tan βαβ≥-， ∴2221608035640016400xx x x x ≥=--，解得028.28x <≤≈，∴CD 的长至多为28.28米． ……6分 （2）设,,DB a DA b DC m ===，180123.43ADB αβ∠=︒--=︒，则sin sin a ABADBα=∠， 解得115sin38.1285.06sin123.43a ︒=≈︒∴26.93m =≈∴CD 的长为26.93米． ……14分（22）【2014年上海，理22，16分】在平面直角坐标系xOy 中，对于直线:0l ax by c ++=和点111222(,),(,)P x y P x y ，记1122()()ax by c ax by c η=++++． 若0η<，则称点12,P P 被直线l 分割． 若曲线C 与直线l 没有公共点，且曲线C 上存在点12,P P 被直线l 分割，则称直线l 为曲线C 的一条分割线． （1）求证：点(1,2),(1,0)A B -被直线10x y +-=分割；（2）若直线y kx =是曲线2241x y -=的分割线，求实数k 的取值范围；（3）动点M 到点(0,2)Q 的距离与到y 轴的距离之积为1，设点M 的轨迹为曲线E ．求证：通过原点的直线中，有且仅有一条直线是E 的分割线．解：（1）将(1,2),(1,0)A B -分别代入1x y +-，得(121)(11)40+-⨯--=-<，∴点(1,2),(1,0)A B -被直线10x y +-=分割． ……3分 （2）联立2241x y y kx⎧-=⎨=⎩，得22(14)1k x -=，依题意，方程无解∴2140k -≤，∴12k ≤-或12k ≥．……8分（3）设(,)M x y1=，∴曲线E 的方程为222[(2)]1x y x +-= ① 当斜率不存在时，直线0x =，显然与方程①联立无解，又12(1,2),(1,2)P P -为E 上两点，且代入0x =，有10η=-<，∴0x =是一条分割线；当斜率存在时，设直线为y kx =，代入方程得：2432(1)4410k x kx x +-+-=， 令2432()(1)441f x k x kx x =+-+-，则(0)1f =-，22(1)143(2)f k k k =+-+=-，22(1)143(2)f k k k -=+++=+，当2k ≠时，(1)0f >，∴(0)(1)0f f <，即()0f x =在(0,1)之间存在实根，∴y kx =与曲线E 有公共点当2k =时，(0)(1)0f f -<，即()0f x =在(1,0)-之间存在实根，∴y kx =与曲线E 有公共点， ∴直线y kx =与曲线E 始终有公共点，∴不是分割线，综上，所有通过原点的直线中，有且仅有一条直线0x =是E 的分割线． ……16分（23）【2014年上海，理23，18分】已知数列{}n a 满足1133n n n a a a +≤≤，*n ∈N ，11a =．（1）若2342,,9a a x a ===，求x 的取值范围；（2）设{}n a 是公比为q 的等比数列，12n n S a a a =+++L ． 若1133n n n S S S +≤≤，*n ∈N ，求q 的取值范围；（3）若12,,,k a a a L 成等差数列，且121000k a a a +++=L ，求正整数k 的最大值，以及k 取最大值时相应数列12,,,k a a a L 的公差．解：（1）依题意，232133a a a ≤≤，∴263x ≤≤，又343133a a a ≤≤，∴327x ≤≤，综上可得36x ≤≤．……3分（2）由已知得1n n a q -=，又121133a a a ≤≤，∴133q ≤≤，当1q =时，n S n =，1133n n n S S S +≤≤，即133nn n ≤+≤，成立；当13q <≤时，11n n q S q -=-，1133n n n S S S +≤≤，即1111133111n n n q q q q q q +---≤≤---， ∴111331n nq q +-≤≤-， 此不等式即11320320n n n nq q q q ++⎧--≥⎨-+≤⎩，∵1q >，∴132(31)2220n n n n q q q q q +--=-->->，对于不等式1320n n q q +-+≤，令1n =，得2320q q -+≤，解得12q ≤≤，又当12q <≤时，30q -<，∴132(3)2(3)2(1)(2)0n n n q q q q q q q q +-+=-+≤-+=--≤成立，∴12q <≤，当113q ≤<时，11n n q S q -=-，1133n n n S S S +≤≤，即1111133111n n nq q q q q q +---≤≤---，即11320320n n n nq q q q ++⎧--≤⎨-+≥⎩，310,30q q ->-<， ∵132(31)2220n n n n q q q q q +--=--<-<，132(3)2(3)2(1)(2)0n n n q q q q q q q q +-+=-+≥-+=--> ∴113q ≤<时，不等式恒成立，综上，q 的取值范围为123q ≤≤． ……10分 （3）设公差为d ，显然，当1000,0k d ==时，是一组符合题意的解，∴max 1000k ≥，则由已知得1(2)1(1)3[1(2)]3k dk d k d +-≤+-≤+-，∴(21)2(25)2k d k d -≥-⎧⎨-≥-⎩， 当1000k ≥时，不等式即22,2125d d k k ≥-≥---，∴221d k ≥--，12(1) (10002)k k k da a a k -+++=+=，∴1000k ≥时，200022(1)21k d k k k -=≥---，解得10001000k ≤+1999k ≤， ∴k 的最大值为1999，此时公差2000219981(1)199919981999k d k k -==-=--⨯． ……18分。

高三学科测试英语试卷 2013.12.13考试时间：120分钟满分：150分第一卷(共103 分)I. Listening Comprehension （30分）Section 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 conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. A student. B. A secretary. C. A teacher. D. A boss.2. A. In the department store. B. At the airport.C. At the railway station.D. At the furniture store.3. A. By car. B. By bike. C. By bus. D. On foot.4. A. The English test took place a week ago.B. The woman was confident about the English test.C. Tomorrow’s English test will be easy.D. There won’t be an English test tomorrow.5. A. She doesn’t like English poetry. B. Poetry class is very popular.C. Dr. Robinson is easy to get along with.D. The course will be difficult.6. A. Five feet tall. B. Only one foot tall. C. Six feet tall. D. Seven feet tall.7. A. Go to the library. B. Go to see a film.C. Get some exercise.D. Do homework in school.8. A. It is 4: 10 now. B. It is 4: 20 now. C. It is 4: 50 now. D. It is 5:00 now.9. A. He is not very enthusiastic about his English lessons.B. He has made great progress in his English.C. He is a student of the music department.D. He is not very interested in English songs.10. A. He has always been so productive. B. He got his job done within a minute.C. He finished the paper just in time.D. He hasn’t finished his paper yet.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.Questions 11 through 13 are based on the following passage.11. A. When directions are long. B. When directions are short.C. When homework is given.D. When your mother talks.12. A. Your pen and paper. B. A few words. C. Your mind and ears. D. Some pictures.13. A Topics or page numbers. B. Key words or a picture in mind.C. Some details.D. School assignments.Questions 14 through 16 are based on the following passage.14. A. It is based on individual needs, personal goals and different life styles.B. It is decided by the healthiest lifestyle.C. It is based on some tests and comparison with standards.D. It is based more on individual needs than personal goals.15. A. It is more accurate. B. It is more flexible.C. It is less enjoyable.D. It is less effective.16. A. An accountant who can be as physically fit as an athlete.B. The importance of three basic factors concerning fitness.C. New concept of fitness and its essential factors.D. Some sports with significant training effect.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.Complete the form. Write ONE WORD for each answer.II. Grammar and Vocabulary （26分）Section ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)Many kinds of music can stir the imagination and produce strong feeling. For some people, romantic composers such as Chopin and Tchaikovsky enhance feelings of love and sympathy. Religious and spiritual music 25 help some people feel peace or lessen their pain. But one musician seems to have a unique ability of healing(治愈) the human body – Wolfgang Amadeus Mozart. Scientists have found Mozart’s music to be remarkable in its abil ity 26 (calm) its listeners. It can also increase their perceptions, and help them express themselves more clearly.Many amazing cases have been documented using Mozart 27 a healing aid. For example, a tiny premature baby 28 (name) Krissy, who weighed just 1.5 pounds at birth, was on total life support. Doctors thought she had little chance of survival. Her mother insisted on playing Mozart for Krissy, and thought 29 saved her daughter’s life. Krissy lived, 30 she was very small for her age and slower than the average child. At the age of four, she showeda n i n t e r e s t i n m u s i c a n d h e r p a r e n t s g a v e h e r v i o l i n l e s s o n s.31 their astonishment, Krissy was able to play musical pieces from memory that were far beyond the ability of an average four-year-old. 32 (play) music helped her improve in all areas of her life.(B)Touch is the first tool we turn to when we face pain. We react similarly to 33 injured toe and a broken heart, tenderly 34 (grasp) the affected area. We are all born with the power to heal ourselves and to heal others because healing energy does not come from within but from outside ourselves. The energy is there for anyone to use, and no formal training 35 (require). When you have the intent to heal, and love is your only motive, you become a channel for healing energy. Through the simple touch of hands, you can use that universal healing energy to comfort those who 36 (experience) pain or distress.In performing healing energy work, it is necessary that you let your intuition (直觉) guide you to the affected area. Just imagine a healing light being drawn in through the top of your head and flowing through your hands. The energy will begin to flow once you have made a physical connection, and your touch will help awaken the body’s capacity for self-healing. The affected areas of the body, 37 were previously tense or tight, will relax with enough healing energy.38 (try) not to feel like you aren’t helping if you don’t feel the flow. The work you ar e doing is indeed helping.Understanding 39 energy works is less important than consciously making use of it. Performing a loving healing treatment on your loved ones can be a wonderful experience that brings you closer together. And as the healing energy passes through you, it can awaken a feeling within you that helps you 40 (well) understand the interactions between the spiritual, physical and mental selves.Section 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.Genes are found in every plant or animal cell. They are the basic units of life that are capable of passing specific 41 from one generation to another. For example, whether you are tall or short depends upon the genes that you 42 from your parents. Cloning plants or animals is a process that 43 the production of a new organism that is genetically identical to the organism from which cells were taken. Although this 44 has aroused great interest around the world over the last ten years or so, cloning is not something new in nature. Most mothers give birth to just one child at a time; however, on rare 45 a mother may give birth to a set of twins. If these two children are identical twins, then they are in fact naturally 46 clones of each other, although not of the parents. The reason for them being clones is that the two children originated from one 47 egg and consequently they are genetically identical. Twins thatd e v e l o p f r o m t w o48 eggs are not clones of each other.Cloning also occurs naturally in plants. Strawberry plants reproduce themselves by sending out runners, or modified stems, and these runners take root a short distance from the 49 plant and start to grow, and to produce fruit. In time, this plant sends out runners of its own that also take root. This is known as asexual reproduction; in other words, reproduction that does noti n v o l v e t h e f u s i o n(融合)o f m a l e a n d f e m a l e c e l l s.T h e n e w l y50 plants are clones of the original. In addition to strawberry plants, certain types of flowers and grasses reproduce themselves in the same way. Throughout history, people have also facilitated cloning by taking small cuttings from plants and then growing them into new plants. This form of cloning is called vegetative propagation.III. Reading Comprehension（47分）Section ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Robert Frank, an economist at Cornell, believes that his profession is restricting cooperation and generosity. In the U.S., economics professors give __51__ money to charity than professors in other fields. Economics students in Germany are more likely than those from other majors to recommend an overpriced plumber (水管工) when they are __52__ to do it. Economics majors tend to rate __53__ as “generally good,” “correct,” and “moral” more than their peers.Does studying economics change people? Maybe not. It could be self-selection: students who already believe in self-interest are __54__ to economics. But this doesn't exclude the possibility that studying economics pushes people further toward the selfish extreme. By spending time with like-minded people, economics students may become __55__ that selfishness is widespread and reasonable -- or at least that giving is rare and foolish.“As a business school professor, these effects worry me, as economics, __56__ every aspect of our lives, is taught widely in business schools, providing a __57__ for courses in management, finance, and accounting.” says Frank.If economics can __58__ pro-social behavior, which is central to the well-being of people or society, what should we do about it? A change in economics and business __59__ is suggested. Courses in behavioral economics, which considers the role of “social preferences” like __60__, fairness and cooperation, are required for students of economics major. In fact, economics courses not involving some behavioral economics are considered both an inadequate education and a poorpreparation to be a practising economist. Also, __61__ width, economics majors are required to take courses in social sciences like sociology and psychology, which place considerable emphasis on how people are __62__ about others, not only themselves. __63__, within economics courses, we should do a better job __64__ the principle of self-interest, which involves anything a person values -- including helping others.Not until then may the prophecy (预言) by Nobel Prize-winning economist and philosopher Amartya Sen be __65__. Calling economists “rational fools,” he observed: “The purely economic man is indeed close to being a social fool.”51. A. fewer B. less C. smaller D. more52. A. encouraged B. requested C. assigned D. paid53. A. teamwork B. greed C. desire D. economics54. A. opposed B. entitled C. drawn D. attached55. A. convinced B. depressed C. relaxed D. doubtful56. A. depending on B. adapting to C. differing from D. relating to57. A. potential B. judgment C. foundation D. reason58. A. assess B. research C. discourage D. cause59. A. education B. standard C. approach D. application60. A. competition B. evaluation C. community D. generosity61. A. in case of B. in terms of C. in relation to D. in need of62. A. concerned B. anxious C. curious D. enthusiastic63. A. However B. Therefore C. Furthermore D. Otherwise64. A. claiming B. defining C. overlooking D. recalling65. A. broken B. predicted C. challenged D. fulfilledSection 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)I used to think of myself as a fairly cosmopolitan（全球的）sort of person, but my bookshelves told a different story as my literature collection mainly consisted of British and American titles. Worse still, I hardly ever read anything in translation. My reading was limited to stories by English-speaking authors.So, at the start of 2012, I set myself the challenge of trying to read a book from every country in a year to find out what I was missing.With no idea how to go about this, thinking that I was unlikely to find books from nearly 200 nations from my local bookshop, I decided to ask the planet’s readers for help. I created a blog called A Year of Reading the World, appealing for suggestions of titles that I could read in English.The response was amazing. Before I knew it, people all over the world were getting in touch with ideas and offers of help. Some posted me books from their home countries. Others did hours of research on my behalf. In addition, several writers sent me unpublished translations of their novels, giving me a rare opportunity to read works otherwise unavailable to the 62% of British people who only speak English. Even with such support, however, sourcing books was no easy task.Tracking down stories in some unfamiliar places even took as much time as the reading and blogging. It was hard to fit it all in around work and many were the nights when I sat bleary-eyed (睡眼惺忪的) into the small hours to make sure I stuck to my target of reading one book every 1.87 days.Still，one by one, the country names on the list made at the start of the year transformed into vital, exciting places filled with laughter, love, anger, hope and fear. Lands that had once seemed exotic became familiar to me – places I could identify with. At its best, I learned, fiction makes the world real.66.What help did the writer receive to achieve his goal?A.Some writers sent him their original works.B.The local bookstore provided books from 200 nations even if it was unlikely.C.Some writers did research work on what books are popular.D.Some people mailed books from their own countries.67.In order to accomplish his goal, the writer did the following except ______.A.asking for help from all over the worldB.making full use of his leisure time after workC.reading until early next morningD.appealing to writers to send him new books68.What does the word "exotic" in the last paragraph most probably mean?A.Unrealistic.B. Uncivilized.C. Strange.D. Unimaginable.69.Which of the following best describe the writer’s experience in the year?A. Original but tiringB. Hard but relaxingC. Busy but fruitfulD. Helpful but challenging(B)Villa d'Este, Tivoli (Italy) - Official Site Useful InformationCall Center 199766166Number to dial from all of Italy for pre-sales and reservations for: tickets, guided tours, school groups, instructional visits.Bookings from abroad:email: villadestetivoli@fax: 0039 0412770747telephone: 0039 0412719036Visiting Hours:Opening 8.30 – closed one hour before sunset.The ticket office closes one hour before the closing of the monument.The hydraulic organ of the Organ Fountain is active daily, from 10.30 am, every two hours.The Fontana della Civetta functions daily, from 10.00 am, every two hours.Ticket Prices:(from May 17 to October 20, 2013)Full ticket (exhibition + villa and gardens, not divisible): € 11.Reduced ticket: € 7.These prices will be valid during the daytime openings of the Villa until the closure of the exhibition, due on the 20th of October, 2013(From the 22nd of October, 2013)Full ticket: € 8Reduced ticket: € 4These fares may vary in conjunction with exhibitions set inside the Villa.The right to purchase reduced price tickets belongs to all citizens of the European Union between the ages of 18 and 24 as well as permanent teachers of state schools (upon presentation of identity documents).School Visits:Reservations are required. The management of Villa d'Este, in the aim of preserving the monument and better distributing the flow of students, has limited the number of students allowed into the Villa to 100 students per hour. Should any school group arrive at the Villa without having made a reservation, it will be admitted to the Villa according to space availability at a particular time and asked to wait until such space becomes available. Right of Reservation cost: € 1,00.Notices:Certain areas of the villa may be closed for restoration: for information inquire at the ticket office.Please pay particular attention to the areas marked with signs indicating danger (in Italian: pericolo).70. How can a visiting Chinese professor of architecture in Rome make a booking?A. By dialing 199766166.B. By writing an email to villadestetivoli@.C. By calling 0039 0412719036.D. By sending a fax to 0039 0412770747.71. The receptionist at the ticket office may recommend you to see ______ first, if you arrive at10.25 am.A. the exhibition inside the villaB. the Organ FountainC. the gardensD. the Fontana della Civetta72. Why are reservations essential for school visits?A. Reservations are more economical.B. Reservations enable as many students as possible to visit the monument.C. Reservations ensure a pleasant visit for students and a manageable one for the Villa.D. Reservation fees can help preserve the site.73. Which of following statements is NOT TRUE?A. Villa d'Este is closed at night.B. Ticket prices are usually higher than usual when there happens to be an exhibition.C. Tourists are n ot allowed to enter areas marked with signs “pericolo”.D. Visitors can buy reduced tickets as long as they make a reservation.(C)Have you ever seen harmful robots refuse to die,no matter how fiercely people fight back? As is known,it is always the case in some of the scariest sciencefiction.Now, science fiction has changed into sciencefact. For the first time, researchers have created arobotic machine that can stand a beating and keep onwalking. Developed by scientists from CornellUniversity and the University of Vermont, the newrobot looks like a spider with four legs. Until now, even the most advanced robot was almost certain to break down when damaged. That’sbecause its inside computer simply doesn’t know how to operate the machine aft er its shape has changed.To get around this problem, the spidery robot’s developers equipped their invention with eight motors and two sensors that read how the machine is tilting(倾斜). The motors and sensors all provide electrical signals to the machine ’s software.Using this information, the system follows a new procedure to figure out the machine’s shape at any given moment. The program chooses from among 100,000 possible arrangements of parts. From there, the computer considers a wide variety of possible next steps, and it calculates how best to move the robot forward the longest possible distance, before trying to move again. “The new strategy is a major advance in robotics”, Professor Skehan, a scientist in Cornell University says, “and it’s far from scary”. The technology may someday help researchers create better artificial arms and legs that give new freedom to people who lack them. The new knowledge might also help scientists understand how people and animals figure out their own sense of place in space.“Designing robots that can adapt to changing environments and can compensate (补偿) for damage has been a difficult problem,” says neuroscientist Olaf Sporns of Indiana University in Bloomington. “This work provides a new way toward solving this important problem.”74. What’s the most important characteristic of the new robot?A. Its shape is like a spider with four legs.B. It can repair itself and move again after damaged.C. It is operated by its inside computer.D. It is certain to change the direction when damaged.75. it can be learnt from the passage that ________.A. animals have no sense of place in spaceB. people have difficulty in figuring out how to get to his destinationC. scientists don ’t understand people ’s ability of locationD. scientists want to know where people and animals are going76. We can infer from the view of Professor Skehan that ________.A. the new technology will help people have a better sense of spaceB. the new technology will help people grow new arms and legsC. designing robots can help to be adaptable to the changing of the environmentD. the new technology will have some medical benefits77. The purpose of the author to write this article is to ________.A. introduce a new robot that will not dieB. discuss science fiction and new inventionC. recommend the latest inventionD. explain the relationship between robot and medicineSection CDirections: Read the passage carefully. Then answer the questions or complete the statements in NO MORE THAN TEN WORDS.With 26 British prime ministers, 58 Nobel prize winners, over 150 Olympic medals and a "notable alumni" list that reads like a historical hall of fame, the institutions of Oxford and Cambridge – collectively known as Oxbridge – are seen, by many, as the dream alma mater(母校).However, as admission season dawns, and the October 15 deadline for Oxbridge inches ever closer, many students may wonder what makes the UK's top two institutions so special. Why does Oxbridge symbolize the academic dream for so many? Do the universities deserve the reputation they hold in the minds of so many teenagers, parents and academics?There is no doubt that both Oxford and Cambridge have a long standing history of academia. Yet, what is often overlooked is the strong reputation of countless other universities in the UK.A simple click on Google can tell students that the London School of Economics and Political Sciences, for example, has over 18 Nobel Prize winners and 50 world leaders. University College London attracts students from 150 countries and has exchange and research links with nearly 300 overseas universities. Manchester University has the largest student union of all UK universities and boasts alumni such as Niels Bohr and James Chadwick.I suddenly realized the Oxbridge effect when I was out in London with friends. I met some new people and was surprised by their reaction when I told them I was going to study at Cambridge. One person even got down on one knee and kissed my hand.From then on, I told people I was going on a gap year. Although I knew that Oxbridge provides a good education, I had never expected to get such extreme reactions. It was this that made me start to feel uncomfortable about the way that Oxbridge is viewed.The Cambridge Tab, the university's student newspaper, has argued that what makes Cambridge special is the fact that the students are told that they are unique and a cut above the rest, so they believe it. Ron Weasley never took the Felix Felicis potion that Harry Potter pretended to give him – all he had to do to win the Quidditch match was to believe he was great.For students applying now, yes, you will receive a world class education at Oxbridge, but you will find the same at numerous other universities throughout the country. Every time you hearabout the brilliance of Oxbridge, consider that just because many believe, it does not have to be so.78. The examples of the London School of Economics and Political Sciences, University College London and Manchester University serve to illustrate the point that __________________________________in the UK.79. According to the passage, we can find that people ___________________ the students at the Oxbridge.80. How did the writer avoid over-reaction from people to her admission into Cambridge?______________________________________________________________________81. What is the writer’s suggestion to students applying to universities?_______________________________________________________________________第II卷（共47分）I．TranslationDirections: Translate the following sentences into English, using the word given in the brackets.1.这个失败给了我们一个教训:今日事，今日毕。

2013学年第一学期十二校联考高三数学（理）考试试卷一、填空题 （本大题满分56分，每题4分）1．已知全集U {}5,4,3,2,1=，A {}3,1=，B {}4,3,2=，那么=⋃)(B C A U __． 2．函数)12arcsin(-=x y 的定义域为 ．3．若数列{}n a 满足：111,2()n n a a a n N *+==∈，则前6项的和6S = .（用数字作答） 4. 计算：2(1)(13)lim(2)(1)n n n n n n →∞+-=-++________．5．集合{}12-<<=x x A ，{}0<-=a x x B ，若B A ⊆，则实数a 的取值范围是 ． 6. 设()887872x a x a x -=++…10a x a +，则87a a ++…0a +＝ ．7. 已知函数)(x f 有反函数)(1x f -，且[),,0,24)(1+∞∈-=+x x f x x则=-)0(1f ． 8. 已知袋中有大小相同的红球和白球若干个，其中红、白球个数的比为4:3．假设从袋中任取2个球，取到的都是红球的概率为413．那么袋中的红球有 __个． 9. 已知函数32tansin )(x xx x f ++=，)1,1(-∈x ，则满足不等式0)12()1(<-+-a f a f 的实数a 的取值范围是 ．10. 已知x 是7,6,5,,3,2,1x 这7个数据的中位数，且y x -,,2,12这四个数据的平均数为1，则xy 1-的最小值为 ． 11．设ω>0，若函数)(x f = sin 2x ω cos 2x ω 在区间[－3π，4π]上单调递增，则ω的范围是_____________．12. 设正项数列}{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1a =_______________.13．函数)(x f y =的图像与直线b x a x ==,及x 轴所围成图形的面积称为函数)(x f 在[]b a ,上的面积，已知函数nx y sin =在⎥⎦⎤⎢⎣⎡n π,0上的面积为)(2*∈N n n ，则函数1)3sin(+-=πx y 在⎥⎦⎤⎢⎣⎡34,3ππ上的面积为 ．14．（理）函数)(x f 的定义域为A ，若A x x ∈21,且)()(21x f x f =时总有21x x =，则称)(x f 为单函数，例如，函数)(12)(R x x x f ∈+=是单函数．下列命题： ①函数)()(2R x x x f ∈=是单函数；②指数函数)(2)(R x x f x ∈=是单函数；③若)(x f 为单函数，A x x ∈21,且21x x ≠，则)()(21x f x f ≠； ④在定义域上具有单调性的函数一定是单函数；⑤若)(x f 为单函数，则函数)(x f 在定义域上具有单调性。

2013年高三调研考数学试卷（理科）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．已知4cos 5α=，则cos()2sin()22tan()cot()2παπαππαα-+-+++=______________.4．计算：2211lim[()]12n n 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．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. .8．等差数列{}n a 中，1102,15a S ==，记2482nn B a a a a =++++，则当n =____时，n B取得最大值.9．函数()()x x y 2arccos 1arcsin +-=的值域是 .10．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__ _.11．已知函数()(2318,3133x tx x f x t x ⎧-+≤⎪=⎨->⎪⎩，记()()*n a f n n N =∈，若{}n a 是递减数列，则实数t 的取值范围是______________.12．已知()sin 2cos 2f x a x b x =+（a ，b 为常数），若对于任意x R ∈都有()5()12f x f π≥，则方程()0f x =在区间[]0,π内的解为 . 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,12n n 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 xy x =+的零点在区间（ ）内（A ）11(,)43（B ）12(,)35（C ）21(,)52（D ）12(,)2316．已知a b 、为实数，命题甲：2ab b >，命题乙：110b a<<，则甲是乙的（ ）条件（A ）充分不必要 （B ）必要不充分 （C ）充要 （D ）非充分非必要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．集合()*{,,S x y z x y z N =∈、、，且x y z <<、y z x <<、z x y <<恰有一个成立}，若(),,x y z S ∈且(),,z w x S ∈,则下列选项正确的是( ) （A ）(),,y z w S ∈,(),,x y w S ∉ （B ）(),,y z w S ∈,(),,x y w S ∈ （C ）(),,y z w S ∉,(),,x y w S ∈ （D ）(),,y z w S ∉,(),,x y w S ∉P AB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）行列式cos 2sin 01cos AA x A x x()0A >按第一列展开得1121312M M -+，记函数()1121f x 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。

2014年高考上海卷数学（理）试卷及答案解析考生注意：1、本试卷共4页，23道试题，满分150分。

考试时间120分钟。

2、本考试分设试卷和答题纸。

试卷包括试题与答题要求。

作答必须涂（选择题）或写（非选择题）在答题纸上。

在试卷上作答一律不得分。

3、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，在答题纸正面清楚地填写姓名。

(1) 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(cos 212的最小正周期是x y -=【答案】 2π【解析】2π4π2∴4cos -)2(cos 2-12====T x x y 周期Θ2、若复数z=1+2i ，其中i 是虚数单位，则}1{zz +z ⋅=___________. 【答案】 6 【解析】61)41(1)1(∴21=++=+=•++=z z z zz i z Θ3、若抛物线y 2=2px 的焦点与椭圆15922=+y x 的右焦点重合，则该抛物线的准线方程为___________. 【答案】 x=-2【解析】2-2-)0,2(2)0,2(159222==∴=∴=+x x px y y x 所以，是其准线方程为焦点为右焦点为ΘΘ4、设⎩⎨⎧+∞∈-∞∈=],,[,),,(,)(2a x x a x x x f 若4)2(=f ，则a 的取值范围为_____________.【答案】 ]2,∞-( 【解析】]2,∞-(.2≤),∞,[∈2∴4)2(所以，是解得a a f +=Θ5、若实数x,y 满足xy=1,则2x +22y 的最小值为______________. 【答案】 22 【解析】22,2222≥22y ∴1222222所以，是=•+=+=x x x x x xy Θ6. 若圆锥的侧面积是底面积的3倍，则其母线与底面角的大小为 （结果用反三角函数值表示）。

2014年全国普通高等学校招生统一考试理科（上海卷）数学试题1、【题文】函数的最小正周期是.2、【题文】若复数z=1+2i，其中i是虚数单位，则=___________.3、【题文】若抛物线y2=2px的焦点与椭圆的右焦点重合，则该抛物线的准线方程为___________.4、【题文】设若，则的取值范围为_____________.5、【题文】若实数x,y满足xy=1,则+的最小值为______________.6、【题文】若圆锥的侧面积是底面积的3倍，则其母线与底面角的大小为（结果用反三角函数值表示）.7、【题文】已知曲线C的极坐标方程为，则C与极轴的交点到极点的距离是 .8、【题文】设无穷等比数列{}的公比为q，若，则q= .9、【题文】若，则满足的取值范围是 .10、【题文】为强化安全意识，某商场拟在未来的连续10天中随机选择3天进行紧急疏散演练，则选择的3天恰好为连续3天的概率是（结构用最简分数表示）.11、【题文】已知互异的复数a,b满足ab≠0，集合{a,b}={,},则= .12、【题文】设常数a使方程在闭区间[0,2]上恰有三个解，则 .13、【题文】某游戏的得分为1,2,3,4,5，随机变量表示小白玩游戏的得分.若=4.2，则小白得5分的概率至少为 .14、【题文】已知曲线C：，直线l：x=6.若对于点A（m，0）,存在C上的点P和l上的点Q使得，则m的取值范围为 .15、【题文】设,则“”是“”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16、【题文】如图，四个棱长为1的正方体排成一个正四棱柱，AB是一条侧棱，是上底面上其余的八个点，则的不同值的个数为（）A．1 B．2 C．4 D．817、【题文】已知与是直线y=kx+1（k为常数）上两个不同的点，则关于x和y的方程组的解的情况是（）A．无论k，如何，总是无解B．无论k，如何，总有唯一解C．存在k，，使之恰有两解D．存在k，，使之有无穷多解18、【题文】若是的最小值，则的取值范围为（）. A．[-1,2] B．[-1，0] C．[1，2] D．19、【题文】（本题满分12分）底面边长为2的正三棱锥,其表面展开图是三角形，如图，求△的各边长及此三棱锥的体积.20、【题文】（本题满分14分）本题有2个小题，第一小题满分6分，第二小题满分1分.设常数，函数（1）若=4，求函数的反函数；（2）根据的不同取值，讨论函数的奇偶性，并说明理由.21、【题文】（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，某公司要在两地连线上的定点处建造广告牌，其中为顶端，长35米，长80米，设在同一水平面上，从和看的仰角分别为.（1）设计中是铅垂方向，若要求，问的长至多为多少（结果精确到0.01米）？（2）施工完成后.与铅垂方向有偏差，现在实测得求的长（结果精确到0.01米）？22、【题文】（本题满分16分）本题共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.在平面直角坐标系中，对于直线：和点记若<0，则称点被直线分隔.若曲线C与直线没有公共点，且曲线C上存在点被直线分隔，则称直线为曲线C的一条分隔线.⑴求证：点被直线分隔；⑵若直线是曲线的分隔线，求实数的取值范围；⑶动点M到点的距离与到轴的距离之积为1，设点M的轨迹为E，求证：通过原点的直线中，有且仅有一条直线是E的分割线.23、【题文】（本题满分18分）本题共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.已知数列满足.（1）若，求的取值范围；（2）若是公比为等比数列，，求的取值范围；（3）若成等差数列，且，求正整数的最大值，以及取最大值时相应数列的公差.。