2011 英国高考数学试卷之一
2011年新课标高考数学试题及答案（理科）
2011年新课标高考数学试题及答案（理科）Part IV Translation & WritingTranslationA. Translate the following sentences from Chinese intoEnglish.1) 约翰同时干许多事情。
我觉得他应当休息一下。
(work on, all at once, take a break)John works on many things all at once. I think he should takea break2) 杨教授说的话有着神奇的力量。
许多同学接受他的忠告，开始专注学业了。
(what, magical, advice, focus on)What Prof. Yang said has magical power. On his advice, many students began to focus on their schoolwork.3) 由于星期天晚上汤姆没有提示他将做何种选择，我无法弄清楚他会如何完成这项任务。
(clue, option, fgure out, accomplish) As Tom gave no clue Sunday night about which option he would choose, I can’t figure out how he will accomplish the task4) 我的父亲是极负责任的人。
虽然他总是很忙，但他设法每天都给家庭留出一些时间。
(responsibility, on the go, set aside) My father is a man of great responsibility. Though he is on the go all the time, he manages to set aside some time for the family every day.5) 这个项目的成功与否取决于我们如何确定轻重缓急。
2011年高考数学理(重庆)(含答案解析)
2011年高考数学理（重庆）（含答案解析）高考真题高考模拟高中联考期中试卷期末考试月考试卷学业水平同步练习2011年高考数学理（重庆）（含答案解析）1 复数（A）（B）（C）（D）【答案解析】 C本题主要考查复数单位i及复数的四则基本运算，以及转化的思想．难度较小＝＝－＝－＝－＝－i．2 是的（A）充分而不必要条件（B）必要而不充分条件（C）充要条件（D）既不充分也不必要条件【答案解析】 A本题主要考查一元二次不等式的解法及充要条件的判断．难度较小．解不等式x2－1＞0，得x＜－1或x＞1，因此当x＜－1成立时，x2－1＞0成立，而当x＜－1或x＞1成立时，x＜－1不一定成立，故选A．3 已知，则（A） (B)2 (C)3 (D)6【答案解析】 D本题主要考查函数的极限求法，同时考查转化的思想及简单的方程思想．难度较小．(＋)＝＝＝＝2，即a＝6．4 的展开式中与的系数相等，则=（A）6 （B）7 （C）8 （D）9【答案解析】 B本题主要考查二项定理、组合数的应用，以及考查方程的思想、转化的思想，同时考查逻辑思维能力及运算能力．难度较小．方法1由题意可得C35＝C36，即C＝3C，即＝3·，即＝，解得n＝7．方法2当n＝6时，x5项的系数为C35＝1456，x6项的系数＝C36＝729，显然不成立．当n ＝7时，x5项的系数为C35＝5103，x6项的系数＝C36＝5103，满足条件．5 下列区间中，函数=在其上为增函数的是（A）（-（B）（C）（D）【答案解析】 D本题主要考查对数函数的单调性、图象的作法及应用，同时考查函数的数形结合的思想、转化的思想．难度较小．化f(x)为分段函数f(x)＝，作出函数的图象，如图所示，根据图象知f(x)在[1，2）上为增函数．6 若的内角A、B、C所对的边a、b、c满足，且C=60°，则ab的值为（A）（B） (C)1 (D)【答案解析】 A本题主要考查余弦定理的应用，同时考查逻辑思维能力、简单的运算能力、整体代换的思想与转化的思想．难度较小．由(a＋b) 2－c2＝4，得a2＋b 2－c2＋2ab＝4．由余弦定理得a2＋b 2－c2＝2abcocosC＝ab，所以ab＋2ab＝4，即ab＝．7 已知a＞0，b＞0，a+b=2，则y=的最小值是（A）（B）4(C) (D)5【答案解析】 C本题主要考查均值定理的应用，以及考查对代数式结构的变换能力．逻辑思维能力，同时考查转化的思想．难度一般．＋＝(＋)(a＋b)＝(5＋＋a)≥(5＋2)＝．8 在圆内，过点的最长弦和最短弦分别是和，则四边形的面积为（A）（B）(C)(D)【答案解析】 B本题主要考查圆的方程、直线与圆相交的弦长问题、两点之间的距离、四边形面积等，同时考查数形结合的思想、逻辑思维能力、作图能力．难度中等．将圆方程配方得(x－1)2＋(y－3)2＝10，则设圆心为G，易知G(1，3)．最长弦AC为过E的直径，则|AC|＝2．最短弦BD为与GE垂直的弦，如图所示．易知|BG|＝，|EG|＝＝，|BD|＝2|BE|＝＝2．所以四边形ABCD的面积为|AC||BD|＝10．9 高为的四棱锥S-ABCD的底面是边长为1的正方形，点S、A、B、C、D均在半径为1的同一球面上，则底面ABCD的中心与顶点S之间的距离为（A）（B） (C)1 (D) [来源:学|科|网Z|X|X|K]【答案解析】 C本题主要考查球的性质、棱锥的性质、平面间的距离等基础知识，以及考查转化的思想、构造的思想，同时考查空间想象能力、逻辑思维能力、图形的变换能力、创新解决问题的能力．难度中等偏上．如图所示，设球心为O，正方形的中心为O1，则OB＝1，O1B＝BD＝，所以点O到平面ABCD的距离OO1＝＝，因为四棱锥S－ABCD的底面的高为，可以想到四棱锥的顶点S是与平面ABCD平行且距离为的一个小圆的圆周上，此小圆的圆心O2在OO1的中点上，易知SO为线段OO1的垂直平分线，所以SO1＝SO＝1．10 设m，k为整数，方程在区间（0,1）内有两个不同的根，则m+k的最小值为（A）-8 （B）8 (C)12 (D)13【答案解析】 D本题主要考查不等式的性质、一元二次方程根的分布、一元二次函数的图象、不等式组表示的平面区域及最值问题等，以及考查数形结合的思想、转化的思想、方程与函数思想，同时考查逻辑思维能力、作图能力、运算能力、综合分析问题和解决问题的能力．难度较大．设f(x)＝mx2－kx＋2，由f(0)＝2，易知f(x)的图象恒过定点(0，2)，因此要使已知方程在区间(0，1)内两个不同的根，即f(x)的图象在区间(0，1)内有两个不同的交点，必有1，即，在直角坐标系mok中作出满足不等式平面区域，设z＝m＋k，则直线m＋k－z＝0经过图中的阴影中的整点(6，7)时，z＝m＋k取得最小值，即zmin＝13．11 在等差数列中，，则__________【答案解析】 74本题主要考查等差数列的基本运算及转化的能力．难度较小．方法1：由a3＋a7＝37，得(a1＋2d)＋(a1＋6d)＝37，即2a1＋8d＝37．a2＋a4＋a6＋a8＝(a1＋d)＋(a1＋3d)＋(a1＋5d)＋(a1＋7d)＝2(2a1＋8d)＝74．方法2：∵a3＋a7＝a2＋a8＝a4＋a6＝37，∴a2＋a4＋a6＋a8＝74．12 已知单位向量，的夹角为60°，则__________【答案解析】本题主要考查平面向量的夹角、数量积公式、模的求法等知识，以及考查转化的思想与运算能力．难度较小．方法1：|2e1－e2|2＝4e12－4e1·e2＋e22＝4|e1|2－4|e1||e2|·cos60°＋|e2|2＝4×12－4×1×1×＋12＝3，∴|2e1－e2|＝．方法2：在平面直角坐标系中取e1＝(1，0)，e1＝(，)，则|2e1－e2|＝|(0，)|＝．13 将一枚均匀的硬币投掷6次，则正面出现的次数比反面出现的次数多的概率为__________【答案解析】本题主要考查独立重复试验的概率计算及分类思想．难度较小．方法1：处理为独立重复试验，正面出现的次数比反面出现的次数多的情况就是出现了4次、5次、6次正面，所以所求概率为C()4()2＋C()5()＋C()6＝．方法2：一枚均匀的硬币投掷6次，出现的正反面的情况有26，其中正面出现的次数比反面出现的次数的情况有C＋C＋C，则所求概率有＝．14 已知，且，则的值为__________【答案解析】－本题主要考查同角三角函数的基本关系、两角和与差公式、二倍角公式的三角恒等变换，以及考查转化的能力、逻辑思维能力、运算能力．难度中等．方法1：＝＝＝－(cosa＋sina)，∵sina＝＋cosa，∴cos a－sina＝－，两边平方得1－2sinacosa＝－，所以2sinacosa＝．∵a∈(0，)，∴cosa＋sina＝＝＝，∴＝－．方法2：由条件得cosa－sina＝－，两边平方得1－2sinacosa＝－，所以sin2a＝．所以由a∈(0，)，知cos2a＝＝．于是＝＝－．15 设圆C位于抛物线与直线3所围成的封闭区域（包含边界）内，则圆C的半径能取到的最大值为__________【答案解析】－1本题主要考查圆与直线相切、圆与抛物线相切问题，以及考查判别式法、转化法等数学方法，同时考查方程思想、逻辑推理能力、运算能力．难度较大．由题意知，半径取得最大值的圆的圆心必在x轴上，高为(a,o)(0＜a＜3)，则半径为3－a，于是圆的方程为(x－a)2＋y2＝(3－a)2，将抛物线方程y2＝2x代入圆方程得(x－a)2＋2x ＝(a－3)2，即x2－2(a－1)x＋6a－9＝0，由△＝4(a－1)2－4(6a－9)＝0，即a2－8a＋10＝0，a＝4±，∵0＜a＜3，∴a＝4－，故圆C的半径能取到的最大值为3－a＝－1．16 设，满足，求函数在上的最大值和最小值.【答案解析】解：由因此当为增函数，当为减函数，所以又因为故上的最小值为17 某市公租房的房源位于A，B，C三个片区，设每位申请人只申请其中一个片区的房源，且申请其中任一个片区的房源是等可能的求该市的任4位申请人中：（Ⅰ）恰有2人申请A片区房源的概率；（Ⅱ）申请的房源所在片区的个数的分布列与期望【答案解析】解：这是等可能性事件的概率计算问题.（I）解法一：所有可能的申请方式有34种，恰有2人申请A片区房源的申请方式种，从而恰有2人申请A片区房源的概率为解法二：设对每位申请人的观察为一次试验，这是4次独立重复试验.记“申请A片区房源”为事件A，则从而，由独立重复试验中事件A恰发生k次的概率计算公式知，恰有2人申请A片区房源的概率为（II）ξ的所有可能值为1，2，3.又综上知，ξ有分布列ξ12 3P从而有18 设的导数满足，其中常数．（Ⅰ）求曲线在点处的切线方程；（Ⅱ）设，求函数的极值．【答案解析】解：（I）因故令由已知又令由已知因此解得因此又因为故曲线处的切线方程为（II）由（I）知，从而有令当上为减函数；当在（0，3）上为增函数；当时，上为减函数；从而函数处取得极小值处取得极大值19 如题（19）图，在四面体中，平面平面，，，．（Ⅰ）若，，求四面体的体积；（Ⅱ）若二面角为，求异面直线与所成角的余弦值．【答案解析】（I）解：如答（19）图1，设F为AC的中点，由于AD=CD，所以DF⊥AC.故由平面ABC⊥平面ACD，知DF⊥平面ABC，即DF是四面体ABCD的面ABC上的高，且DF=ADsin30°=1，AF=ADcos30°=.在Rt△ABC中，因AC=2AF=，AB=2BC，由勾股定理易知故四面体ABCD的体积（II）解法一：如答（19）图1，设G，H 分别为边CD，BD的中点，则FG//AD，GH//BC，从而∠FGH是异面直线AD与BC所成的角或其补角.设E为边AB的中点，则EF//BC，由AB⊥BC，知EF⊥AB.又由（I）有DF⊥平面ABC，故由三垂线定理知DE⊥AB.所以∠DEF为二面角C—AB—D的平面角，由题设知∠DEF=60°设在从而因Rt△ADE≌Rt△BDE，故BD=AD=a，从而，在Rt△BDF中，，又从而在△FGH中，因FG=FH，由余弦定理得因此，异面直线AD与BC所成角的余弦值为解法二：如答（19）图2，过F作FM⊥AC，交AB于M，已知AD=CD，平面ABC⊥平面ACD，易知FC，FD，FM两两垂直，以F为原点，射线FM，FC，FD分别为x 轴，y轴，z轴的正半轴，建立空间直角坐标系F—xyz.不妨设AD=2，由CD=AD，∠CAD=30°，易知点A，C，D的坐标分别为显然向量是平面ABC的法向量.已知二面角C—AB—D为60°，故可取平面ABD的单位法向量，使得设点B的坐标为，有易知与坐标系的建立方式不合，舍去.因此点B的坐标为所以从而故异面直线AD与BC所成的角的余弦值为20 如题（20）图，椭圆的中心为原点，离心率，一条准线的方程为．（Ⅰ）求该椭圆的标准方程；（Ⅱ）设动点满足：，其中是椭圆上的点，直线与的斜率之积为，问：是否存在两个定点，使得为定值？若存在，求的坐标；若不存在，说明理由．【答案解析】解：（I）由解得，故椭圆的标准方程为（II）设，则由得因为点M，N在椭圆上，所以，故设分别为直线OM，ON的斜率，由题设条件知因此所以所以P点是椭圆上的点，设该椭圆的左、右焦点为F1，F2，则由椭圆的定义|PF1|+|PF2|为定值，又因，因此两焦点的坐标为21 设实数数列的前n项和，满足（I）若成等比数列，求和；（II）求证：对【答案解析】（I）解：由题意，由S2是等比中项知由解得（II）证法一：由题设条件有故从而对有①因，由①得要证，由①只要证即证此式明显成立.因此最后证若不然又因矛盾.因此证法二：由题设知，故方程（可能相同）.因此判别式又由因此，解得因此由，得因此。
2011年高考数学(理)真题(Word版)——天津卷(试题+答案解析)
2011年普通高等学校招生全国统一考试（天津卷）数学（理）试题本试卷分为第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共150分，考试用时120分钟。
第Ⅰ卷参考公式：如果事件A ，B 互斥，那么如果事件A ，B 相互独立，那么()()()P A B P A P B =+U()()().P AB P A P B =棱柱的体积公式.V Sh =圆锥的体积公式1.3V Sh =其中S 表示棱柱的底面面积 其中S 表示圆锥的底面面积 h 表示棱柱的高 h 表示圆锥的高一、选择题：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．i 是虚数单位，复数131ii--= A ．2i + B ．2i -C ．12i -+D ．12i --2．设,,x y R ∈则“2x ≥且2y ≥”是“224x y +≥”的A ．充分而不必要条件B ．必要而不充分条件C ．充分必要条件D ．即不充分也不必要条件 3．阅读右边的程序框图，运行相应的程序，则输出i 的值为 A ．3 B ．4 C ．5 D ．6 4．已知{}n a 为等差数列，其公差为-2，且7a 是3a 与9a 的等比中项，n S 为{}n a 的前n 项和，*n N ∈，则10S 的值为A ．-110B ．-90C ．90D ．1105．在62x x ⎛⎫- ⎪ ⎪⎝⎭的二项展开式中，2x 的系数为A ．154-B ．154C ．38-D ．386．如图，在△ABC 中，D 是边AC 上的点，且AB=AD ，2AB=BD ，BC=2BD ，则sin C 的值为A 3B 3C 6D 67．已知324log 0.3log 3.4log 3.615,5,,5a b c ⎛⎫=== ⎪⎝⎭则A ．a b c >>B ．b a c >>C ．a c b >>D ．c a b >>8．对实数a 和b ，定义运算“⊗”：,1,, 1.a a b a b b a b -≤⎧⊗=⎨->⎩设函数()()22()2,.f x x x x x R =-⊗-∈若函数()y f x c =-的图像与x 轴恰有两个公共点，则实数c 的取值范围是A ．(]3,21,2⎛⎫-∞-⋃- ⎪⎝⎭B ．(]3,21,4⎛⎫-∞-⋃--⎪⎝⎭C ．111,,44⎛⎫⎛⎫-⋃+∞ ⎪ ⎪⎝⎭⎝⎭D ．311,,44⎛⎫⎡⎫--⋃+∞ ⎪⎪⎢⎝⎭⎣⎭第II 卷二、填空题：本大题共6小题，每小题5分，共30分．9．一支田径队有男运动员48人，女运动员36人，若用分层抽样的方法从该队的全体运动员中抽取一个容量为21的样本，则抽取男运动员的人数为___________ 10．一个几何体的三视图如右图所示（单位：m ），则该几何体的体积为__________3m11．已知抛物线C 的参数方程为28,8.x t y t ⎧=⎨=⎩（t 为参数）若斜率为1的直线经过抛物线C 的焦点，且与圆()2224(0)x y r r -+=>相切，则r =________.12．如图，已知圆中两条弦AB 与CD 相交于点F ，E 是AB 延长线上一点，且2,::4:2:1.DF CF AF FB BE ===若CE 与圆相切，则线段CE 的长为__________.13．已知集合{}1|349,|46,(0,)A x R x x B x R x t t t⎧⎫=∈++-≤=∈=+-∈+∞⎨⎬⎩⎭，则集合A B ⋂=________.14．已知直角梯形ABCD 中，AD //BC ,090ADC ∠=,2,1AD BC ==,P 是腰DC 上的动点，则3PA PB +u u u r u u u r的最小值为____________.三、解答题：本大题共6小题，共80分．解答应写出文字说明，证明过程或演算步骤． 15．（本小题满分13分）已知函数()tan(2),4f x x π=+（Ⅰ）求()f x 的定义域与最小正周期；（II ）设0,4πα⎛⎫∈ ⎪⎝⎭，若()2cos 2,2f αα=求α的大小．16．（本小题满分13分）学校游园活动有这样一个游戏项目：甲箱子里装有3个白球、2个黑球，乙箱子里装有1个白球、2个黑球，这些球除颜色外完全相同，每次游戏从这两个箱子里各随机摸出2个球，若摸出的白球不少于2个，则获奖．（每次游戏结束后将球放回原箱） （Ⅰ）求在1次游戏中，（i ）摸出3个白球的概率； （ii ）获奖的概率；（Ⅱ）求在2次游戏中获奖次数X 的分布列及数学期望()E X . 17．（本小题满分13分）如图，在三棱柱111ABC A B C -中，H 是正方形11AA B B 的中心，122AA =，1C H ⊥平面11AA B B ，且1 5.C H =（Ⅰ）求异面直线AC 与A 1B 1所成角的余弦值； （Ⅱ）求二面角111A AC B --的正弦值；（Ⅲ）设N 为棱11B C 的中点，点M 在平面11AA B B 内，且MN ⊥平面11A B C ，求线段BM 的长．18．（本小题满分13分）在平面直角坐标系xOy 中，点(,)P a b (0)a b >>为动点，12,F F 分别为椭圆22221x y a b+=的左右焦点．已知△12F PF 为等腰三角形．（Ⅰ）求椭圆的离心率e ；（Ⅱ）设直线2PF 与椭圆相交于,A B 两点，M 是直线2PF 上的点，满足2AM BM ⋅=-u u u u r u u u u r，求点M 的轨迹方程．19．（本小题满分14分）已知0a >，函数2()ln ,0.f x x ax x =->（()f x 的图像连续不断） （Ⅰ）求()f x 的单调区间；（Ⅱ）当18a =时，证明：存在0(2,)x ∈+∞，使03()()2f x f =； （Ⅲ）若存在均属于区间[]1,3的,αβ，且1βα-≥，使()()f f αβ=，证明ln 3ln 2ln 253a -≤≤．20．（本小题满分14分）已知数列{}n a 与{}n b 满足：1123(1)0,2n n n n n n n b a a b a b ++++-++==， *n ∈N ，且122,4a a ==．（Ⅰ）求345,,a a a 的值；（Ⅱ）设*2121,n n n c a a n N -+=+∈，证明：{}n c 是等比数列；（III ）设*242,,k k S a a a k N =++⋅⋅⋅+∈证明：4*17()6nk k kS n N a =<∈∑．答案解析一、选择题( 本大题共8 题, 共计40 分)1、(5分) B.2、(5分) A∵x≥2，∴x2≥4.∵y≥2，∴y2≥4.∴x2＋y2≥8≥4.∴x≥2且y≥2?x2＋y2≥4，反之令x2＋y2＝5≥4，可取x＝2，y＝1，无法推出y≥2.故选A项．3、(5分) B第一次运算：i＝1，a＝2，a＜50；第二次运算：i＝2，a＝5，a＜50；第三次运算：i＝3，a ＝16，a＜50；第四次运算：i＝4，a＝65，a＞50.所以输出i＝4.4、(5分) D设等差数列{a n}的首项为a，公差d＝－2.则a7＝a1＋6d＝a1－12；a3＝a1＋2d＝a1－4；a9＝a1＋8d＝a1－16.∵a7是a3与a9的等比中项，∴＝a3·a9，∴(a1－12)2＝(a1－4)(a1－16)，∴a1＝20.∴S10＝10a1＋d＝110.5、(5分) C设含x2的项是二项展开式中第r＋1项，则.令3－r＝2，得r＝1.∴x2的系数为.6、(5分) D设BD＝a，则BC＝2a，AB＝AD＝A．在△ABD中，由余弦定理，得cos A＝＝＝.又∵A为△ABC的内角，∴sin A＝.在△ABC中，由正弦定理得，.∴sin C＝·sin A＝·＝.7、(5分) C，log23.4＞log22＝1，log43.6＜log44＝1，＞log3 3＝1，又log23.4＞＞，∴log23.4＞＞log43.6.又∵y＝5x是增函数，∴a＞c＞b．8、(5分) B由题意得，即在同一坐标系内画出函数y＝f(x)与y＝c的图象如图所示，结合图象可知，当c∈(－∞，－2]∪(－1，)时两个函数的图象有两个公共点，从而方程f(x)－c＝0有两个不同的根，即y＝f(x)－c与x轴有两个不同交点．二、填空题( 本大题共6 题, 共计30 分)9、(5分) 12解析：设抽取男运动员人数为n，则女运动员人数21－n.由分层抽样知：，∴n＝12.10、(5分) 6＋π解析：由几何体的三视图可知，原几何体是一个长方体和一个圆锥的组合体．下面的长方体的长、宽、高分别是3 m，2 m，1 m，∴体积为3×2×1＝6(m3)．上面的圆锥底面圆半径为1 m，高为3 m，∴圆锥的体积为π×12×3＝π(m3)．∴该几何体的体积为(6＋π) m3.11、(5分)解析：消去参数t，得抛物线标准方程y2＝8x，其焦点F(2，0)，∴过抛物线焦点斜率为1的直线方程：x－y－2＝0，∵直线与圆(x－4)2＋y2＝r2相切，∴r＝d＝.12、(5分)解析：设BE＝m，则BF＝2m，AF＝4m.∵AB与CD是圆的两条相交弦，交点为F，∴由相交弦定理，得AF·FB＝CF·FD＝＝2，∴4m·2m＝2，∴m2＝.又∵CE是圆的切线，根据切割弦定理，得CE2＝EB·EA＝m(m＋2m＋4m)＝7m2＝，∴CE＝.13、(5分) {x|－2≤x≤5}解析：解不等式|x＋3|＋|x－4|≤9.(1)当x＜－3时，|x＋3|＋|x－4|＝－x－3＋4－x≤9，∴x≥－4，即－4≤x＜－3；(2)当－3≤x≤4时，|x＋3|＋|x－4|＝x＋3＋4－x≤9恒成立，∴－3≤x≤4；(3)当x＞4时，|x＋3|＋|x－4|＝x＋3＋x－4≤9，∴x≤5，即4＜x≤5.综上所述，A＝{x∈R|－4≤x≤5}．∵t∈(0，＋∞)，∴x＝4t＋－6≥2－6＝－2，当且仅当t＝时等号成立．∴B＝{x∈R|x≥－2}．∴A∩B＝{x∈R|－4≤x≤5}∩{x∈R|x≥－2}＝{x∈R|－2≤x≤5}．14、(5分) 5解析：根据题意，以AD为x轴，DC为y轴，建立平面直角坐标系，如图所示．∵AD＝2，∴A(－2，0)．∵BC＝1，∴可设B(－1，n)．∴点P在DC上运动，∴可设P(0，y)(0≤y≤n)．∴＝(－2，－y)，＝(－1，n－y)，∴＝(－5，3n－4y)．∴.∴当3n＝4y，即y＝n时，取得最小值5.三、解答题( 本大题共6 题, 共计80 分)15..解：（1）由，得，所以f(x)的定义域为．f(x)的最小正周期为（2）由得，，整理得．因为，所以.因此由，得．所以16.解：（1）①设“在1次游戏中摸出i个白球”为事件A i(i＝0，1，2，3)，则②设“在1次游戏中获奖”为事件B，则B＝A2∪A3.又且A2，A3互斥，所以.（2）由题意可知X的所有可能取值为0，1，2.所以X的分布列是X0 1 2PX的数学期望17. 解：(方法1)如图所示，建立空间直角坐标系，点B为坐标原点.依题意得（1）解：易得，于是所以异面直线AC与A1B1所成角的余弦值为（2）解：易知设平面AA1C1的法向量，则即不妨令可得，同样地，设平面A1B1C1的法向量，则即不妨令，可得于是从而所以二面角A—A1C1—B的正弦值为（3）解：由N为棱B1C1的中点，得设M（a，b，0），则由平面A1B1C1，得即解得故因此，所以线段BM的长为(方法2)（1）解：由于AC//A1C1，故是异面直线AC与A1B1所成的角. 因为平面AA1B1B，又H为正方形AA1B1B的中心，可得因此所以异面直线AC与A1B1所成角的余弦值为（2）解：连接AC1，易知AC1=B1C1，又由于AA1=B1A1，A1C1=A1=C1，所以≌，过点A作于点R，连接B1R，于是，故为二面角A—A1C1—B1的平面角. 在中，连接AB1，在中，，从而所以二面角A—A1C1—B1的正弦值为（3）解：因为平面A1B1C1，所以取HB1中点D，连接ND，由于N是棱B1C1中点，所以ND//C1H且.又平面AA1B1B，所以平面AA1B1B，故又所以平面MND，连接MD并延长交A1B1于点E，则由得，延长EM交AB于点F，可得连接NE.在中，所以可得连接BM，在中，18. 解：（1）设F1(－c，0)，F2(c，0)(c＞0)．由题意，可得|PF2|＝|F1F2|，即整理得(舍)，或，所以.（2）由(1)知，可得椭圆方程为3x2＋4y2＝12c2，直线PF2方程为．A，B两点的坐标满足方程组，消去y并整理，得5x2－8cx＝0.解得得方程组的解.不妨设．设点M的坐标为．由于是由，即，化简得将，所以x＞0.因此，点M的轨迹方程是18x2－16xy－15＝0(x＞0)．19. 解：（1）．令.xf′(x) ＋0 －f(x) ↗极大值↘所以，f(x)的单调递增区间是的单调递减区间是．（2）证明：当由(1)知f(x)在(0，2)内单调递增，在(2，＋∞)内单调递减．令由于f(x)在(0，2)内单调递增，故.取所以存在x0∈(2，x′)，使g(x0)＝0，即存在(说明：x′的取法不唯一，只要满足x′＞2，且g(x′)＜0即可．)（3）证明：由f(α)＝f(β)及(1)的结论知，从而f(x)在[α，β]上的最小值为f(α)．又由β－α≥1，α，β∈[1，3]，知1≤α≤2≤β≤3.故从而20. 解：（1）由可得又b n a n＋a n＋1＋b n＋1a n＋2＝0，当n＝1时，a1＋a2＋2a3＝0，由a1＝2，a2＝4，可得a3＝－3；当n＝2时，2a2＋a3＋a4＝0，可得a4＝－5；当n＝3时，a3＋a4＋2a5＝0，可得a5＝4.（2）证明：对任意n∈N*，a2n－1＋a2n＋2a2n＋1＝0，①2a2n＋a2n＋1＋a2n＋2＝0，②a2n＋1＋a2n＋2＋2a2n＋3＝0.③②－③，得a2n＝a2n＋3.④将④代入①，可得a2n＋1＋a2n＋3＝－(a2n－1＋a2n＋1)．即c n＋1＝－c n(n∈N*)．又c1＝a1＋a3＝－1，故c n≠0，因此是等比数列．（3）证明：由(2)可得a2k－1＋a2k＋1＝(－1)k，于是，对任意k∈N*且k≥2，有a1＋a3＝－1，－(a3＋a5)＝－1，a5＋a7＝－1，(－1)k(a2k－3＋a2k－1)＝－1.将以上各式相加，得a1＋(－1)k a2k－1＝－(k－1)，即a2k－1＝(－1)k＋1(k＋1)，此式当k＝1时也成立．由④式得a2k＝(－1)k＋1·(k＋3)．从而S2k＝(a2＋a4)＋(a6＋a8)＋…＋(a4k－2＋a4k)＝－k，S2k－1＝S2k－a4k＝k＋3所以，对任意n∈N*，n≥2，对于n＝1，不等式显然成立．。
2011年Alevel数学试卷真题
*2011*UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONSInternational General Certificate of Secondary EducationMATHEMATICS 0580/21 Paper 2 (Extended) May/June 20111 hour 30 minutesCandidates answer on the Question Paper.Additional Materials: E lectronic calculator Geometrical instrumentsMathematical tables (optional) Tracing paper (optional) READ THESE INSTRUCTIONS FIRSTWrite your Centre number, candidate number and name on all the work you hand in.Write in dark blue or black pen.You may use a pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or correction fluid.DO NOT WRITE IN ANY BARCODES.Answer all questions.If working is needed for any question it must be shown below that question.Electronic calculators should be used.If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.For π , use either your calculator value or 3.142.At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.The total of the marks for this paper is 70.This document consists of 12 printed pages.IB11 06_0580_21/7RP© UCLES 2011 [Turn over *9523289317*© UCLES 20110580/21/M/J/116 ÷ ABAB1A concert hall has 1540 seats.Calculate the number of people in the hall when 55% of the seats are occupied.Answer[1]For Examiner's Use2Shade the required region on each Venn diagram.A ⋃ B' (A ⋂B )'[2]0.25−23Calculate 814 .Answer [2]4(a) Find m when 4m × 42 = 412 .Answer(a) m =[1](b) Find p when 6 p ÷ 6 5 = .Answer(b) p = [1][Turn o ver© UCLES 2011 0580/21/M/J/115A hummingbird beats its wings 24 times per second.(a) Calculate the number of times the hummingbird beats its wings in one hour.Answer(a)[1](b) Write your answer to part (a) in standard form.Answer(b)[1]For Examiner's Use6NOT TO SCALEA company makes solid chocolate eggs and their shapes are mathematically similar. The diagram shows eggs of height 2 cm and 6 cm. The mass of the small egg is 4 g. Calculate the mass of the large egg.Answerg [2]7 Find the length of the straight line from Q (−8, 1) to R (4 , 6).Answer QR =[3]6 cm2 cm© UCLES 2011 0580/21/M/J/11338Calculate the radius of a sphere with volume 1260 cm .For [The volume, V , of a sphere with radius r is V =4 πr .]3Examiner's UseAnswer cm [3]9ANOT TOSCALEDAB is parallel to CD .Calculate the value of x .Answer x =[3]10 Solve the simultaneous equations.3x + y = 30 2x – 3y = 53Answer x =y = [3]2x °BC5x ° x °[Turn o ver© UCLES 2011 0580/21/M/J/11211 A rectangular photograph measures 23.3 cm by 19.7 cm, each correct to 1 decimal place.Calculate the lower bound for(a) the perimeter,Answer(a)cm [2](b) the area.Answer(b)cm [1]For Examiner's Use12 A train leaves Barcelona at 21 28 and takes 10 hours and 33 minutes to reach Paris.(a) Calculate the time the next day when the train arrives in Paris.Answer(a)[1](b) The distance from Barcelona to Paris is 827 km.Calculate the average speed of the train in kilometres per hour.Answer(b)km/h [3]© UCLES 2011 0580/21/M/J/1122213 The scale on a map is 1: 20 000.(a) Calculate the actual distance between two points which are 2.7 cm apart on the map.Give your answer in kilometres.Answer(a)km [2]For Examiner's Use(b) A field has an area of 64 400 m . 2Calculate the area of the field on the map in cm .Answer(b) cm [2]14 Solve the equation 2x + 3x – 6 = 0.Show all your working and give your answers correct to 2 decimal places.Answer x =or x =[4][Turn o ver© UCLES 2011 0580/21/M/J/1115 A teacher asks 36 students which musical instruments they play.P = {students who play the piano} G = {students who play the guitar} D = {students who play the drums}The Venn diagram shows the results.(a) Find the value of x .Answer(a) x =[1](b) A student is chosen at random.Find the probability that this student(i) plays the drums but not the guitar,Answer(b)(i)[1](ii) plays only 2 different instruments.Answer(b)(ii)[1](c) A student is chosen at random from those who play the guitar.Find the probability that this student plays no other instrument.Answer(c)[1]For Examiner's UseD 7541 58x 2 GP© UCLES 2011 0580/21/M/J/112222 216For Examiner's Usek cmThe diagram shows a square of side k cm.The circle inside the square touches all four sides of the square.(a) The shaded area is A cm .Show that4A = 4k – πk .Answer (a)[2](b) Make k the subject of the formula 4A = 4k – πk .Answer(b) k =[3]17CNOT TOSCALEFor Examiner'sUseOB24°TAA, B and C are points on a circle, centre O.TA is a tangent to the circle at A and OBT is a straight line.AC is a diameter and angle OTA = 24°.Calculate(a)angle AOT,Answer(a) Angle AOT = [2](b)angle ACB,Answer(b) Angle ACB = [1](c)angle ABT.Answer(c) Angle ABT = [2][Turn o ver © UCLES 2011 0580/21/M/J/11© UCLES 2011 0580/21/M/J/1118For SExaminer's UseNOT TO SCALEPRIn the diagram, PQS , PMR , MXS and QXR are straight lines.PQ = 2 QS .M is the midpoint of PR . QX : XR = 1 : 3.= q and= r .(a) Find, in terms of q and r ,(i),Answer(a)(i)= [1](ii).Answer(a)(ii)=[1](b) By finding, show that X is the midpoint of MS .Answer (b)[3]QXM[Turn o ver© UCLES 20110580/21/M/J/112191510Speed (metres per second)5For Examiner's Use2 4 6 8 10 12 14Time (minutes)The diagram shows the speed-time graph of a train journey between two stations.The train accelerates for two minutes, travels at a constant maximum speed, then slows to a stop.(a) Write down the number of seconds t hat the train travels at its constant maximum speed.Answer(a)s [1](b) Calculate the distance between the two stations in metres .Answer(b)m [3](c) Find the acceleration of the train in the first two minutes .2Give your answer in m/s .Answer(c)m/s [2]Question 20 is printed on the next page.© UCLES 2011 0580/21/M/J/113–120f(x ) = xg(x ) = 2x − 3(a) Find(i) g(6),For Examiner's UseAnswer(a)(i)[1](ii) f(2x ).Answer(a)(ii)[1](b) Solve fg(x ) = 125.Answer(b) x =[3]−1(c) Find the inverse function g (x ) .Answer(c) g (x ) =[2]Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity.University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.。
2011全国一高考数学(理)word版、可编辑、高清无水印
2011年普通高等学校招生全国统一考试理科数学(必修+选修II)第Ⅰ卷第Ⅰ卷共l2小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
一、选择题（1）复数1z i =+，z 为z 的共轭复数，则1zz z --= （A ）2i - （B ）i - （C ）i （D ）2i（2）函数2(0)y x x =≥的反函数为（A ）2()4x y x R =∈ （B ）2(0)4x y x =≥（C ）24y x =()x R ∈ （D ）24(0)y x x =≥（3）下面四个条件中，使a b ＞成立的充分而不必要的条件是 （A ）1a b +＞ （B ）1a b -＞ （C ）22a b ＞ （D ）33a b ＞（4）设n S 为等差数列{}n a 的前n 项和，若11a =，公差2d =，224A n S S +-=，则k =（A ）8 （B ）7 （C ）6 （D ）5（5）设函数()cos (0)f x x ωω=＞，将()y f x =的图像向右平移3π个单位长度后，所得的图像与原图像重合，则ω的最小值等于（A）13（B）3（C）6（D）9(6)已知直二面角α− ι−β，点A∈α，AC⊥ι，C为垂足，B∈β，BD⊥ι，D 为垂足．若AB=2，AC=BD=1，则D到平面ABC的距离等于(A)23(B)33(C)63(D) 1(7)某同学有同样的画册2本，同样的集邮册3本，从中取出4本赠送给4位朋友每位朋友1本，则不同的赠送方法共有(A)4种(B)10种(C)18种(D)20种(8)曲线y=2xe-+1在点(0，2)处的切线与直线y=0和y=x围成的三角形的面积为(A)13(B)12(C)23(D)1(9)设()f x是周期为2的奇函数，当0≤x≤1时，()f x=2(1)x x-，则5 ()2f-=(A) -12(B)14-(C)14(D)12(10)已知抛物线C：24y x=的焦点为F，直线24y x=-与C交于A，B两点．则cos AFB∠=(A)45 (B)35 (C)35- (D)45-(11)已知平面α截一球面得圆M ，过圆心M 且与α成060二面角的平面β截该球面得圆N.若该球面的半径为4，圆M 的面积为4π，则圆N 的面积为 (A)7π (B)9π (C)11π (D)13π(12)设向量a ，b ，c 满足a =b =1，a b =12-，,a c b c --=060，则c 的最大值等于(A)2 (B)3 (c)2 (D)1绝密★启用前2011年普通高等学校招生全国统一考试 理科数学(必修+选修II) 第Ⅱ卷 注意事项：1答题前，考生先在答题卡上用直径0．5毫米黑色墨水签字笔将自己的姓名、准考 证号填写清楚，然后贴好条形码。
STEP 3 final 2011
9475 Jun11
5
5 6 The definite integrals T , U , V and X are defined by T =
1 3 1 2
artanh t dt , t
1 2 1 3
ln 3
U=
ln 2
1 2 1 2
u du , 2 sinh u ln(coth x) dx .
V =−
(iii) Find the general solution of the differential equation (x + 1) d2 y dy − y = (x + 1)2 . +x dx2 dx
2
The polynomial f(x) is defined by f(x) = xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 , where n 2 and the coefficients a0 , . . . , an−1 are integers, with a0 = 0. Suppose that the equation f(x) = 0 has a rational root p/q , where p and q are integers with no common factor greater than 1, and q > 0. By considering q n−1 f(p/q ), find the value of q and deduce that any rational root of the equation f(x) = 0 must be an integer. (i) (ii) Show that the nth root of 2 is irrational for n Show that the cubic equation x3 − x + 1 = 0 has no rational roots. (iii) Show that the polynomial equation xn − 5x + 7 = 0 has no rational roots for n 2. 2.
2011年普通高等学校招生全国统一考试数学理试题(全国卷,含答案).doc
2011 年普通高等学校招生全国统一考试数学理试题（全国卷，含答案）本试卷分第Ⅰ卷 ( 选择题 ) 和第Ⅱ卷 ( 非选择题 ) 两部分。
第Ⅰ卷 1 至 2 页。
第Ⅱ卷 3 至 4 页。
考试结束后，将本试卷和答题卡一并交回。
第Ⅰ卷注意事项：1．答题前， 考生在答题卡上务必用直径0.5 毫米黑色墨水签字笔将自己的姓名、准考证号填写清楚，并贴好条形码。
请认真核准条形码上的准考证号、姓名和科目。
2．每小题选出答案后，用 2B 铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用 橡皮擦干净后，再选涂其他答案标号，在试题卷上作答无效。
．．．．．．．．．．3．第Ⅰ卷共 l2 小题，每小题 5 分，共 60 分。
在每小题给出的四个选项中，只有一项是 符合题目要求的。
一、选择题(1) 复数 z 1i ， z 为 z 的共轭复数，则 zz z 1（ A ） 2i（ B ） i（ C ） i（ D ） 2i【答案】 B(2) 函数 y 2 x( x 0) 的反函数为（ A ） yx 2( x R)（ B ）4（ C ）y 4x 2( x R)（ ）Dyx 2( x 0)4y 4x 2 ( x 0) 【答案】 B(3) 下面四个条件中，使 a b 成立的充分而不必要的条件是（ A ） a ＞b 1（ B ） a ＞b 1（C ） a 2＞ b 2（ D ） a 3＞ b 3【答案】 A(4) 设 S n 为等差数列a n 的前 n 项和，若 a 1 1，公差 d2 ， S k 2 S k 24 ，则 k（ A ） 8 （B ） 7（ C ） 6（ D ） 5【答案】 D(5) 设函数 f ( x) cos x(0) ，将 yf ( x) 的图像向右平移个单位长度后，所得的图3像与原图像重合，则的最小值等于（ A ）1（B ） 3（C ） 6（ D ） 93【答案】 C(6) 已知直二面角l , 点 A ， AC l , C 为垂足 , B ， BD l , D 为垂足．若 AB2, AC BD 1，则 D 到平面 ABC 的距离等于2 (B) 36 (D) 1(A)3 (C)33【答案】 CA(7) 某同学有同样的画册 2 本，同样的集邮册 3 本，从中取出 4 本赠送给 4 位朋友每位朋友 1 本，则不同的赠送方法共有(A) 4 种(B)10 种(C)18 种(D)20 种lD【答案】 BCB E(8) 曲线 y e 2 x1在点 (0,2) 处的切线与直线 y 0 和 y x 围 成的三角形的面积为(A)1(B)1 (C)2 (D)1323【答案】 A(9) 设 f ( x) 是周期为 2 的奇函数，当 0x 1 时， f (x)2x(1 x) , 则 f (5 )11112(A) -(B)(C)(D)2442【答案】 A(10) 已知抛物线C ： y 24x 的焦点为 F ，直线 y2x 4 与 C 交于 A ， B 两点．则cos AFB(A)4(B)3 (C)3 (D)4 5555【答案】 D(11) 已知平面 α截一球面得圆 M ，过圆心 M 且与 α 成 600 二面角的平面 β 截该球面得圆 N .若该球面的半径为 4，圆 M 的面积为 4 ，则圆 N 的面积为(A) 7 (B) 9(C)11(D)13【答案】 D(12) r r rr rr r 1 rr r rr设向量 a ， b ， c 满足 | a | | b |1， agb， ac,bc60 ，则 | c | 的最大值2等于(A) 2 (B)3(c)2(D) 1【答案】 AB绝密★启用前2011 年普通高等学校招生全国统一考试ACD理科数学 ( 必修 +选修 II)第Ⅱ卷注意事项：1 答题前，考生先在答题卡上用直径0． 5 毫米黑色墨水签字笔将自己的姓名、准考证号填写清楚，然后贴好条形码。
英国留学中学高考数学试卷
1. 下列各数中，绝对值最小的是（）A. -2B. -1C. 0D. 12. 已知函数f(x) = 2x - 3，则f(-1)的值为（）A. -5B. -1C. 1D. 53. 下列方程中，无解的是（）A. 2x + 3 = 7B. 3x - 4 = 5C. 5x - 6 = 7D. 4x + 5 = 94. 下列各数中，是二次根式的是（）A. √9B. √16C. √25D. √365. 已知等差数列{an}的首项为2，公差为3，则第10项的值为（）A. 29B. 31D. 356. 下列函数中，有最小值的是（）A. y = x^2 + 2x + 1B. y = x^2 - 2x + 1C. y = x^2 + 1D. y = x^2 - 17. 已知圆的方程为(x - 2)^2 + (y - 3)^2 = 9，则圆心坐标为（）A. (2, 3)B. (2, -3)C. (-2, 3)D. (-2, -3)8. 下列三角形中，一定是直角三角形的是（）A. a^2 + b^2 = c^2B. a^2 + b^2 = c^2 + d^2C. a^2 + b^2 + c^2 = d^2D. a^2 + b^2 + c^2 = d^2 + e^29. 下列函数中，单调递增的是（）A. y = 2x - 3B. y = 2x + 3C. y = -2x - 3D. y = -2x + 310. 已知平行四边形ABCD的对角线AC和BD相交于点O，若OA = 5，OC = 10，则OB的长度为（）B. 10C. 15D. 20二、填空题（每题5分，共50分）11. 若x + y = 5，xy = 6，则x^2 + y^2的值为______。
12. 已知函数f(x) = x^2 - 4x + 3，则f(2)的值为______。
13. 若等差数列{an}的首项为a1，公差为d，则第n项an =______。
14. 已知函数f(x) = 2x - 3，则f(-x)的值为______。
英国高考数学试卷中文
第一部分：选择题（共30题，每题2分，共60分）1. 如果一个等差数列的第一项是3，公差是2，那么第10项是多少？A. 19B. 21C. 23D. 252. 下列哪个函数是奇函数？A. f(x) = x^2B. f(x) = x^3C. f(x) = |x|D. f(x) = x + 13. 一个正方体的边长为a，那么它的体积是：A. a^2B. a^3C. 2a^2D. 2a^34. 在直角三角形中，如果两个锐角的正弦值分别是√2/2和√3/2，那么这个三角形的第三个角的余弦值是多少？A. √2/2B. √3/2C. 1/2D. √6/25. 一个圆的半径增加了50%，那么它的面积增加了：A. 50%B. 75%C. 100%D. 125%6. 下列哪个数是负数？A. (-1)^3B. (-1)^4C. (-1)^5D. (-1)^67. 一个二次方程x^2 - 5x + 6 = 0的解是：A. x = 2B. x = 3C. x = 2, x = 3D. x = 68. 如果sinθ = 3/5，且θ在第二象限，那么cosθ是多少？A. -4/5B. 4/5C. -3/5D. 3/59. 一个班级有40名学生，其中20名女生，那么随机抽取一名学生是女生的概率是多少？A. 1/2B. 1/4C. 3/4D. 110. 下列哪个图形是圆？A. 矩形B. 正方形C. 椭圆D. 圆第二部分：解答题（共10题，每题10分，共100分）11. 解方程组：\[\begin{cases}2x + 3y = 8 \\4x - y = 6\end{cases}\]12. 一个长方体的长、宽、高分别是4cm、3cm和2cm，求它的表面积。
13. 一个三角形的两边长分别是3cm和4cm，夹角是45度，求第三边的长度。
14. 求函数f(x) = x^2 - 4x + 4的顶点坐标。
15. 一个圆的半径是5cm，求它的面积。
英国高中数学 C1 - 2011(试卷)
Paper Reference(s)6663/01Edexcel GCECore Mathematics C1Advanced SubsidiaryMonday 10 January 2011 MorningTime: 1 hour 30 minutesMaterials required for examination Items included with question papersMathematical Formulae (Pink)NilCalculators may NOT be used in this examination.Instructions to CandidatesWrite the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Core Mathematics C1), the paper reference (6663), your surname, initials and signature.Information for CandidatesA booklet ‘Mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.The marks for the parts of questions are shown in round brackets, e.g. (2).There are 11 questions in this question paper. The total mark for this paper is 75.Advice to CandidatesYou must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the Examiner.Answers without working may not gain full credit.H35402A This publication may only be reproduced in accordance with Edexcel Limited copyright policy.©2011 Edexcel Limited.H35402A21. (a ) Find the value of 4116-.(2)(b ) Simplify 4412⎪⎪⎭⎫ ⎝⎛-xx . (2)2.Find⎜⎠⎛+-x x x x d )4312(3125,giving each term in its simplest form.(5)3.Simplify13325-√√-,giving your answer in the form p + q √3, where p and q are rational numbers.(4)4.A sequence 1a , 2a , 3a , ... is defined by1a = 2,1+n a = n a 3– cwhere c is a constant.(a ) Find an expression for 2a in terms of c .(1)Given that ∑=31i i a = 0,(b ) find the value of c .(4)H35402A3 Turn over5.Figure 1Figure 1 shows a sketch of the curve with equation y = f(x ) wheref(x ) = 2-x x, x ≠ 2.The curve passes through the origin and has two asymptotes, with equations y = 1 and x = 2, as shown in Figure 1.(a ) In the space below, sketch the curve with equation y = f(x − 1) and state the equations of theasymptotes of this curve.(3) (b ) Find the coordinates of the points where the curve with equation y = f(x − 1) crosses thecoordinate axes.(4)6.An arithmetic sequence has first term a and common difference d . The sum of the first 10 terms of the sequence is 162.(a ) Show that 10a + 45d = 162.(2) Given also that the sixth term of the sequence is 17,(b ) write down a second equation in a and d ,(1)(c ) find the value of a and the value of d .(4)H35402A47. The curve with equation y = f(x ) passes through the point (−1, 0).Given thatf ′(x ) = 12x 2 − 8x + 1,find f(x ).(5)8.The equation x 2 + (k − 3)x + (3 − 2k ) = 0, where k is a constant, has two distinct real roots.(a ) Show that k satisfiesk 2 + 2k − 3 > 0.(3)(b ) Find the set of possible values of k .(4)9. The line 1L has equation 2y − 3x − k = 0, where k is a constant.Given that the point A (1, 4) lies on 1L , find(a ) the value of k ,(1)(b ) the gradient of 1L .(2)The line 2L passes through A and is perpendicular to 1L .(c ) Find an equation of 2L giving your answer in the form ax + by + c = 0, where a , b and c areintegers.(4)The line 2L crosses the x -axis at the point B .(d ) Find the coordinates of B .(2) (e ) Find the exact length of AB .(2)H35402A510. (a ) Sketch the graphs of(i) y = x (x + 2)(3 − x ),(ii) y = –x2.showing clearly the coordinates of all the points where the curves cross the coordinate axes.(6)(b ) Using your sketch state, giving a reason, the number of real solutions to the equationx (x + 2)(3 – x ) + x2= 0.(2)11. The curve C has equationy = 321x – 239x + x 8+ 30, x > 0.(a ) Find xyd d .(4)(b ) Show that the point P (4, –8) lies on C .(2)(c ) Find an equation of the normal to C at the point P , giving your answer in the form ax + by + c= 0 , where a, b and c are integers.(6)TOTAL FOR PAPER: 75 MARKSEND。
2011 November igcse 英文数学试卷
Answer(b)
[1]
© UCLES 2011
0580/23/O/N/11
[Turn over
8 18 Write as a single fraction, in its simplest form.
For Examiner's Use
1− x 2 + x − x 1 − 2x
Answer
[4]
[1]
Answer(b)
[1]
6
Pedro invested $800 at a rate of 5% per year compound interest. Calculate the total amount he has after 2 years.
Answer $
[2]
7
Show that
. 36 Write down all the steps of your working.
°C [1]
Answer(b)
°C [2]
10 Maria decides to increase her homework time of 8 hours per week by 15%. Calculate her new homework time. Give your answer in hours and minutes.
Answer
kg [3]
© UCLES 2011
0580/23/O/N/11
[Turn over
6 14
For Examiner's Use
NOT TO SCALE r 2r
The sphere of radius r fits exactly inside the cylinder of radius r and height 2r. Calculate the percentage of the cylinder occupied by the sphere. [The volume, V, of a sphere with radius r is V =
英国高考数学试卷
试卷时长：2小时30分钟总分：100分一、选择题（每题2分，共20分）1. 已知函数f(x) = 3x - 2，若f(x + 1) = 9，则x的值为：A. 2B. 3C. 4D. 52. 若a、b、c是等差数列的前三项，且a + b + c = 12，a + c = 8，则b的值为：A. 2B. 4C. 6D. 83. 在直角坐标系中，点A(1, 2)关于直线y = x的对称点为：A. (2, 1)B. (1, -2)C. (-2, 1)D. (-1, 2)4. 已知等比数列{an}的首项a1 = 2，公比q = 3，则第n项an的值为：A. 2 3^(n-1)B. 2 3^nC. 6 3^(n-1)D. 6 3^n5. 若复数z = 3 + 4i，则|z|的值为：A. 5B. 7C. 9D. 116. 在△ABC中，角A、B、C的对边分别为a、b、c，若a = 5，b = 7，c = 8，则sinB的值为：A. 5/7B. 7/8C. 8/9D. 9/77. 若函数f(x) = x^2 - 4x + 3，则f(x)的对称轴方程为：A. x = 2B. x = -2C. y = 2D. y = -28. 在△ABC中，角A、B、C的对边分别为a、b、c，若a^2 + b^2 = c^2，则△ABC是：A. 直角三角形B. 锐角三角形C. 钝角三角形D. 等腰三角形9. 若复数z = 1 - i，则z的共轭复数为：A. 1 + iB. -1 + iC. 1 - iD. -1 - i10. 在△ABC中，角A、B、C的对边分别为a、b、c，若a = 3，b = 4，c = 5，则cosA的值为：A. 3/5B. 4/5C. 5/3D. 5/4二、填空题（每题5分，共20分）11. 若函数f(x) = x^2 - 4x + 3，则f(2)的值为______。
12. 在△ABC中，角A、B、C的对边分别为a、b、c，若a = 5，b = 7，c = 8，则△ABC的面积S为______。
英国高考数学试卷中文版
试卷名称：A-Level数学试卷考试时长：2小时30分钟总分：100分一、选择题（共20题，每题2分，共40分）1. 下列哪个数是平方根？A. √4B. √-4C. √0D. √92. 如果一个等差数列的第一项是2，公差是3，那么第五项是多少？A. 7B. 8C. 9D. 103. 已知函数f(x) = 2x - 3，求f(5)的值。
A. 7B. 8C. 9D. 104. 一个三角形的三个内角分别是30°、60°和90°，那么这个三角形是？A. 等腰三角形B. 等边三角形C. 直角三角形D. 钝角三角形5. 已知一个正方形的对角线长为10cm，求这个正方形的面积。
A. 50cm²B. 100cm²C. 50πcm²D. 100πcm²6. 如果一个数的平方是16，那么这个数是？A. 4B. -4C. 4 或 -4D. 07. 下列哪个不等式是正确的？A. 2x + 3 > 5B. 2x + 3 < 5C. 2x + 3 = 5D. 2x + 3 ≠ 58. 一个圆的半径增加了50%，那么这个圆的面积增加了多少？A. 25%B. 50%C. 100%D. 200%9. 如果一个二次方程的解是x = 2和x = 3，那么这个方程是？A. x² - 5x + 6 = 0B. x² - 6x + 8 = 0C. x² - 5x - 6 = 0D. x² - 6x - 8 = 010. 下列哪个数是立方根？A. ∛27B. ∛-27C. ∛0D. ∛811. 如果一个数的立方是64，那么这个数是？A. 4B. -4C. 4 或 -4D. 012. 下列哪个函数是反比例函数？A. y = x + 2B. y = 2xC. y = 2/xD. y = x²13. 一个长方体的长、宽、高分别是4cm、3cm和2cm，求这个长方体的体积。
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Centre Number Candidate NumberSurnameOther NamesCandidate SignatureGeneral Certificate of EducationAdvanced Level ExaminationJanuary2011Mathematics MPC4Unit Pure Core4Monday24January20119.00am to10.30amFor this paper you must have:*the blue AQA booklet of formulae and statistical tables.You may use a graphics calculator.Time allowed*1hour30minutesInstructions*Use black ink or black ball-point pen.Pencil should only be used for drawing.*Fill in the boxes at the top of this page.*Answer all questions.*Write the question part reference(eg(a),(b)(i)etc)in the left-hand margin.*You must answer the questions in the spaces provided.Do not write outside the box around each page.*Show all necessary working;otherwise marks for method may be lost.*Do all rough work in this book.Cross through any work that you do not want to be marked.Information*The marks for questions are shown in brackets.*The maximum mark for this paper is75.Advice*Unless stated otherwise,you may quote formulae,without proof, from the booklet.For Examiner’s UseExaminer’s InitialsQuestion Mark12345678TOTALP38267/Jan11/MPC46/6/6/MPC4 (JAN11MPC401)Answer all questions in the spaces provided.1(a)Express 2sin x þ5cos x in the form R sin ðx þa Þ,where R >0and 0°<a <90°.Give your value of a to the nearest 0.1°.(3marks)(b)(i)Write down the maximum value of 2sin x þ5cos x .(1mark)(ii)Find the value of x in the interval 0°4x 4360°at which this maximum occurs,giving the value of x to the nearest 0.1°.(2marks)...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................P38267/Jan11/MPC4QUESTIONPARTREFERENCE(02)..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................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overs(03)QUESTIONPARTREFERENCE2(a)The polynomial f ðx Þis defined by f ðx Þ¼9x 3þ18x 2Àx À2.(i)Use the Factor Theorem to show that 3x þ1is a factor of f ðx Þ.(2marks)(ii)Express f ðx Þas a product of three linear factors.(3marks)(iii)Simplify 9x 3þ21x 2þ6x f ðx Þ.(3marks)(b)When the polynomial 9x 3þpx 2Àx À2is divided by 3x À2,the remainder is À4.Find the value of the constant p .(2marks)..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................P38267/Jan11/MPC4QUESTIONPARTREFERENCE(04)...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................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overs(05)QUESTIONPARTREFERENCE(06)P38267/Jan11/MPC4.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................P38267/Jan11/MPC4Turn overs(07)QUESTIONPARTREFERENCE(08)P38267/Jan11/MPC4.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................P38267/Jan11/MPC4Turn overs(09)QUESTIONPARTREFERENCE5A model for the radioactive decay of a form of iodine is given bym ¼m 02À18t The mass of the iodine after t days is m grams.Its initial mass is m 0grams.(a)Use the given model to find the mass that remains after 10grams of this form ofiodine have decayed for 14days,giving your answer to the nearest gram.(2marks)(b)A mass of m 0grams of this form of iodine decays to m 016grams in d days.Find the value of d .(2marks)(c)After n days,a mass of this form of iodine has decayed to less than 1%of its initialmass.Find the minimum integer value of n .(3marks)................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................P38267/Jan11/MPC4QUESTIONPARTREFERENCE(10)QUESTIONPARTREFERENCE 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