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2014北京各区高考数学二模试题及答案解析

2014北京各区高考数学二模试题及答案解析
2014北京各区高考数学二模试题及答案解析

2014北京各区高考数学二模

试题及答案解析

2014年北京市各县区的高考二模对于测验高三考生的复习成果和接下来的高考志愿填报具有非常重要的参考价值。本人特将一模试题进行整理汇总,以下是2014年北京各城区高考二模试题及答案汇总,供考生

参考!

北京市西城区2014年高三二模试卷

数 学(理科) 2014.5

第Ⅰ卷(选择题 共40分)

一、选择题:本大题共8小题,每小题5分,共40分.在每小题列出的四个选项中,选出符合

题目要求的一项.

1.已知集合{|20}A x x =-<,{|}B x x a =<,若A B A =,

则实数a 的取值范围是( ) (A )(,2]-∞-

(B )[2,)-+∞

(C )(,2]-∞

(D )[2,)+∞

2.在复平面内,复数2

=(12i)z +对应的点位于( ) (A )第一象限 (B )第二象限 (C )第三象限

(D )第四象限

3.直线2y x =为双曲线22

22 1(0,0)x y C a b a b

-=>>:的一条渐近线,则双曲线C 的离心率是( )

(A (B (C

(D

2014年浙江省高考数学试卷(理科)

2014年浙江省高考数学试卷(理科) 一、选择题(每小题5分,共50分) 2 2 3.(5分)(2014?浙江)某几何体的三视图(单位:cm)如图所示,则此几何体的表面积是() 4.(5分)(2014?浙江)为了得到函数y=sin3x+cos3x的图象,可以将函数y=cos3x的图 向右平移向左平移个单位 向右平移向左平移个单位 5.(5分)(2014?浙江)在(1+x)6(1+y)4的展开式中,记x m y n项的系数为f(m,n), 6.(5分)(2014?浙江)已知函数f(x)=x3+ax2+bx+c,其0<f(﹣1)=f(﹣2)=f(﹣3) 7.(5分)(2014?浙江)在同一直角坐标系中,函数f(x)=x a(x≥0),g(x)=log a x的图象可能是()

B . . D . 8.(5分)(2014?浙江)记max{x ,y}=,min{x ,y}=,设,为 +||﹣min{|||} min{|+﹣|}min{||||} ||﹣||||max{|||﹣|+||9.(5分)(2014?浙江)已知甲盒中仅有1个球且为红球,乙盒中有m 个红球和n 个蓝球(m ≥3,n ≥3),从乙盒中随机抽取i (i=1,2)个球放入甲盒中. (a )放入i 个球后,甲盒中含有红球的个数记为ξi (i=1,2) ; (b )放入i 个球后,从甲盒中取1个球是红球的概率记为p i (i=1,2). 10.(5分)(2014?浙江)设函数f 1(x )=x 2 ,f 2(x )=2(x ﹣x 2 ), , ,i=0,1,2,…,99 .记I k =|f k (a 1)﹣f k (a 0)|+|f k (a 2)﹣f k (a 1)丨+…+|f k (a 99) 二、填空题 11.(4分)(2014?浙江)在某程序框图如图所示,当输入50时,则该程序运算后输出的结果是 .

2014年北京市高考数学试卷(理科)

2014年北京市高考数学试卷(理科) 一、选择题(共8小题,每小题5分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项) 1.(5分)(2014?北京)已知集合A={x|x2﹣2x=0},B={0,1,2},则A∩B=()A.{0}B.{0,1}C.{0,2}D.{0,1,2} 2.(5分)(2014?北京)下列函数中,在区间(0,+∞)上为增函数的是() A.y=B.y=(x﹣1)2 C.y=2﹣x D.y=log0.5(x+1) 3.(5分)(2014?北京)曲线(θ为参数)的对称中心() A.在直线y=2x上B.在直线y=﹣2x上 C.在直线y=x﹣1上D.在直线y=x+1上 4.(5分)(2014?北京)当m=7,n=3时,执行如图所示的程序框图,输出的S的值为() A.7B.42C.210D.840 5.(5分)(2014?北京)设{a n}是公比为q的等比数列,则“q>1”是“{a n}为递增数列” 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.既不充分也不必要条件

6.(5分)(2014?北京)若x,y满足,且z=y﹣x的最小值为﹣4,则k的值为() A.2B.﹣2C.D.﹣ 7.(5分)(2014?北京)在空间直角坐标系Oxyz中,已知A(2,0,0),B(2,2,0),C (0,2,0),D(1,1,),若S1,S2,S3分别表示三棱锥D﹣ABC在xOy,yOz,zOx 坐标平面上的正投影图形的面积,则() A.S1=S2=S3B.S2=S1且S2≠S3 C.S3=S1且S3≠S2D.S3=S2且S3≠S1 8.(5分)(2014?北京)学生的语文、数学成绩均被评定为三个等级,依次为“优秀”“合格”“不合格”.若学生甲的语文、数学成绩都不低于学生乙,且其中至少有一门成绩高于乙,则称“学生甲比学生乙成绩好”.如果一组学生中没有哪位学生比另一位学生成绩好,并且不存在语文成绩相同、数学成绩也相同的两位学生,则这一组学生最多有()A.2人B.3人C.4人D.5人 二、填空题(共6小题,每小题5分,共30分) 9.(5分)(2014?北京)复数()2=. 10.(5分)(2014?北京)已知向量,满足||=1,=(2,1),且+=(λ∈R),则|λ|=. 11.(5分)(2014?北京)设双曲线C经过点(2,2),且与﹣x2=1具有相同渐近线,则 C的方程为;渐近线方程为. 12.(5分)(2014?北京)若等差数列{a n}满足a7+a8+a9>0,a7+a10<0,则当n=时,{a n}的前n项和最大. 13.(5分)(2014?北京)把5件不同产品摆成一排,若产品A与产品B相邻,且产品A与产品C不相邻,则不同的摆法有种. 14.(5分)(2014?北京)设函数f(x)=A sin(ωx+φ)(A,ω,φ是常数,A>0,ω>0) 若f(x)在区间[,]上具有单调性,且f()=f()=﹣f(),则f(x)的最小正周期为.

2013年北京高考理科数学试题及标准答案

绝密★启封前 机密★使用完毕前 2013年普通高等学校招生全国统一考试 数 学(理)(北京卷) 本试卷共5页,150分,考试时长120分钟,考生务必将答案答在答题卡上,在试卷上作答无效.考试结束后,将本试卷和答题卡一并交回. 第一部分(选择题 共40分) 一、选择题共8小题,每小题5分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项. (1)已知集合{}101A =-, ,,{}|11B x x =-<≤,则A B = A.{}0 B.{}10-, ? C.{}01,?D.{}101-,, (2)在复平面内,复数()2 2i -对应的点位于( ) A.第一象限?B.第二象限?C .第三象限 D.第四象限 (3)“π?=”是“曲线()sin 2y x ?=+过坐标原点”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.既不充分也不必要条件 (4)执行如图所示的程序框图,输出的S 值为 A .1? B . 23??C.1321 D.610987 (5)函数()f x 的图象向右平移1个单位长度,所得图象与曲线e x y =关于y 轴对称,则()f x = A .1e x +????B.1e x - C.1e x -+? D.1e x -- (6)若双曲线22 221x y a b -= 则其渐近线方程为 A .2y x =± ?? B.y = C .1 2 y x =± D .y = (7)直线l 过抛物线2:4C x y =的焦点且与y 轴垂直,则l 与C 所围成的图形的面积等于 A.43 ? ?B .2 C.8 3 ?

2014年北京高考英语试卷及答案(word)完美版可直接打印

2014 年普通高等学校招生全国统一考试 英语(北京卷) 本试卷共16页,共150分。考试时间为120分钟。考生务必将答案答在答题卡上,在试卷上作答无效。考试结束后,将本试卷和答题卡一并交回。 第一部分:听力理解(共三节:30 分) 第一节(共5 小题;每小题 1.5 分,共7.5 分) 听下面 5 段对话,每段对话有一道小题,从每题所给的A、B、C 三个选项中选出最佳选项,听完每段对话后,你将有10 秒钟的时间来回答有关小题和阅读下一小题。每段对话你将听一遍。 例:What is the man going to rend? A. A newspaper B. A magazine C. A book 答案是A 1. What juice does the man order? A. Lemon B. Apple C. Orange 2. What subject does the man like best? A. History. B. Biology. C. Chemistry. 3. Where is the woman from? A. Britain. B. Russia. C. America. 4. What kind of student bus pass does the woman want? A. Weekly. B. Monthly. C. Yearly. 5. What are the two speakers going to buy for Mary’s birthday? A. A bicycle. B. A pen. C. A book. 第二节(共10 小题;每小题 1.5 分,共15 分) 听下面4 段对话或独白。每段对话或独白后有几道小题,从每题所给的A、B、C 三个选项中选出最佳选项。听每段对话或独白前,你将有5 秒钟的时间阅读每小题。听完后,每小题将给出5 秒钟的作答时间。每段对话或独 白你将听两遍。 听第6 段材料,回答第6 至7 题。 6. What’s wrong with the woman ? A. She has a cough. B. She has a headache. C. She has a fever. 7. How long is the medicine for? A. One day. B. Two days. C. Three days. 听第7 段材料,回答第8 至9 题。 8. What does the woman need? A. Some ink. B. A printer. C. Some paper. 9. What problem does the man have? A. He can’t send a text message..

2014年高考浙江理科数学试题及答案(word解析版)

2014年普通高等学校招生全国统一考试(浙江卷) 数学(理科) 第Ⅰ卷(选择题 共50分) 一、选择题:本大题共10小题,每小题5分,共50分,在每小题给出的四个选项中,只有一项符合题目要求. (1)【2014年浙江,理1,5分】设全集{|2}U x N x =∈≥,集合2{|5}A x N x =∈≥,则U A =e( ) (A )? (B ){2} (C ){5} (D ){2,5} 【答案】B 【解析】2{|5}{|A x N x x N x =∈≥=∈,{|2{2}U C A x N x =∈≤=,故选B . 【点评】本题主要考查全集、补集的定义,求集合的补集,属于基础题. (2)【2014年浙江,理2,5分】已知i 是虚数单位,,a b R ∈,则“1a b ==”是“2(i)2i a b +=”的( ) (A )充分不必要条件 (B )必要不充分条件 (C )充分必要条件 (D )既不充分也不必要条件 【答案】A 【解析】当1a b ==时,22(i)(1i)2i a b +=+=,反之,2 (i)2i a b +=,即222i 2i a b ab -+=,则22022 a b ab ?-=?=?, 解得11a b =??=? 或11a b =-??=-?,故选A . 【点评】本题考查的知识点是充要条件的定义,复数的运算,难度不大,属于基础题. (3)【2014年浙江,理3,5分】某几何体的三视图(单位:cm )如图所示,则此几何体的表 面积是( ) (A )902cm (B )1292cm (C )1322cm (D )1382cm 【答案】D 【解析】由三视图可知直观图左边一个横放的三棱柱右侧一个长方体,故几何体的表面积为: 1 246234363334352341382 S =??+??+?+?+?+?+???=,故选D . 【点评】本题考查了由三视图求几何体的表面积,根据三视图判断几何体的形状及数据所对应的几何量是解题的 关键. (4)【2014年浙江,理4,5分】为了得到函数sin 3cos3y x x =+的图像,可以将函数y x 的图像( ) (A )向右平移4π个单位 (B )向左平移4 π个单位 (C )向右平移12π个单位 (D )向左平移12π 个单位 【答案】C 【解析】sin3cos3))]412y x x x x ππ=+=+=+,而)2y x x π=+)]6x π +, 由3()3()612x x ππ+→+,即12x x π→-,故只需将y x =的图象向右平移12 π 个单位,故选C . 【点评】本题考查两角和与差的三角函数以及三角函数的平移变换的应用,基本知识的考查. (5)【2014年浙江,理5,5分】在64(1)(1)x y ++的展开式中,记m n x y 项的系数(,)f m n ,则 (3,0)(2,1)(1,2)f f f f +++=( ) (A )45 (B )60 (C )120 (D )210 【答案】C 【解析】令x y =,由题意知(3,0)(2,1)(1,2)(0,3)f f f f +++即为10 (1)x +展开式中3x 的系数, 故(3,0)(2,1)(1,2)(0,3)f f f f +++=7 10120C =,故选C . 【点评】本题考查二项式定理系数的性质,二项式定理的应用,考查计算能力. (6)【2014年浙江,理6,5分】已知函数32()f x x ax bx c =+++ ,且0(1)(2)(3)3f f f <-=-=-≤( ) (A )3c ≤ (B )36c <≤ (C )69c <≤ (D )9c >

2014年北京市高考数学试卷(理科)答案与解析

2014年北京市高考数学试卷(理科) 参考答案与试题解析 一、选择题(共8小题,每小题5分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项) 2 y= 3.(5分)(2014?北京)曲线(θ为参数)的对称中心() ( (

4.(5分)(2014?北京)当m=7,n=3时,执行如图所示的程序框图,输出的S的值为() 1>

6.(5分)(2014?北京)若x,y满足且z=y﹣x的最小值为﹣4,则k的值为 作出可行域如图, (﹣ (﹣ ﹣

7.(5分)(2014?北京)在空间直角坐标系Oxyz中,已知A(2,0,0),B(2,2,0),C (0,2,0),D(1,1,),若S1,S2,S3分别表示三棱锥D﹣ABC在xOy,yOz,zOx , = 8.(5分)(2014?北京)学生的语文、数学成绩均被评定为三个等级,依次为“优秀”“合格”“不合格”.若学生甲的语文、数学成绩都不低于学生乙,且其中至少有一门成绩高于乙,则称“学生甲比学生乙成绩好”.如果一组学生中没有哪位学生比另一位学生成绩好,并且不存在语

二、填空题(共6小题,每小题5分,共30分) 9.(5分)(2014?北京)复数()2=﹣1. ) 10.(5分)(2014?北京)已知向量,满足||=1,=(2,1),且+=(λ∈R),则|λ|= . =.由于向量,|,且+( = ,满足||=1=+=( 故答案为:

11.(5分)(2014?北京)设双曲线C经过点(2,2),且与﹣x2=1具有相同渐近线,则 C的方程为;渐近线方程为y=±2x. ﹣具有相同渐近线的双曲线方程可设为 , ﹣, 故答案为:, 12.(5分)(2014?北京)若等差数列{a n}满足a7+a8+a9>0,a7+a10<0,则当n=8时,{a n}的前n项和最大. 13.(5分)(2014?北京)把5件不同产品摆成一排,若产品A与产品B相邻,且产品A与产品C不相邻,则不同的摆法有36种.

(完整word版)2014年北京高考英语试题及答案详解,推荐文档

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