2008年修改-辽宁卷[word版]
2008年高考文科数学辽宁卷试题
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三、解答题:本大题共 6 小题,共 74 分.解答应写出文字说明,证明过程或演算步骤. 17. (本小题满分 12 分) 在 △ABC 中,内角 A,B,C 对边的边长分别是 a,b,c ,已知 c 2 , C (Ⅰ)若 △ABC 的面积等于 3 ,求 a,b ; (Ⅱ)若 sin B 2sin A ,求 △ABC 的面积.
2008年普通高等学校招生全国统一考试英语试题及答案-辽宁卷
2008年普通高等学校招生全国统一考试 (辽宁卷)英语第一卷(三部分,共115分)第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例:How much is the shirt?A.$19.15.B.$9.15.C.$9.18.答案是B。
1. What is the weather like?A. It's raining.B. It’s cloudy.C. It’s sunny.2. Who will go to China next month?A. Lucy.B. Alice.C. Richard..3. What arc the speakers talking about?A. The man’s sister.B. A filmC. An actor,4. Where will the speakers meet?A. In Room 34O.B. In Room 3l4.C. In Room 223.5. Where does the conversation most probably take place?A. In a restaurant.B. In an office.C. At home.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话。
每段对话后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话读两遍。
听第6段材料,回答第6至8题。
2008年普通高等学校招生全国统一考试辽宁卷数学
第1页,总6页…………○…………外…………○…………装…………○…………订…………○…………线…………○…………姓名:____________班级:____________学号:___________…………○…………内…………○…………装…………○…………订…………○…………线…………○…………2008年普通高等学校招生全国统一考试辽宁卷数学考试时间:**分钟 满分:**分姓名:____________班级:____________学号:___________题号 一 二 三 总分 核分人 得分注意事项:1、填写答题卡的内容用2B铅笔填写2、提前 15 分钟收取答题卡第Ⅰ卷 客观题第Ⅰ卷的注释评卷人 得分一、单选题(共10题)1. 若函数为偶函数,则a =()A .B .C .D .2. 圆与直线没有公共点的充要条件是( )A .B .C .D .3. 已知,,,,则下列关系正确的是A .B .C .D .4. 已知四边形的三个顶点,,,且,则顶点的坐标为( ) A .B .C .D .5. 设为曲线上的点,且曲线在点处切线的倾斜角的取值范围为,则点横坐标的取值范围为( )A .B .C .D .6. 将函数的图象按向量平移得到函数的图象,则( )A .B .C .D .答案第2页,总6页………○…………外…………○…………装…………○…………订…………○…………线…………○…………※※请※※不※※要※※在※※装※※订※※线※※内※※答※※题※※………○…………内…………○…………装…………○…………订…………○…………线…………○…………7. 已知变量满足约束条件则的最大值为()A . B. C. D.8. 一生产过程有4道工序,每道工序需要安排一人照看.现从甲、乙、丙等6名工人中安排4人分别照看一道工序,第一道工序只能从甲、乙两工人中安排1人,第四道工序只能从甲、丙两工人中安排1人,则不同的安排方案共有( )A .24种B .36种C .48种D .72种9. 已知双曲线的一个顶点到它的一条渐近线的距离为,则()A .1 B. 2 C. 3 D. 410. 在正方体ABCD A 1B 1C 1D 1中,E ,F 分别为棱AA 1,CC 1的中点,则在空间中与三条直线A 1D 1,EF ,CD 都相交的直线( )A .不存在B .有且只有两条C .有且只有三条D .有无数条第Ⅱ卷 主观题第Ⅱ卷的注释评卷人 得分一、填空题(共4题)1. 设,则函数的最小值为 .2. 函数的反函数是 .3. 体积为的球面上有三点,,,两点的球面距离为,则球心到平面的距离为_______________.4. 展开式中的常数项为 .评卷人 得分二、解答题(共6题)5. (本小题满分12分)在中,内角对边的边长分别是,已知,.(Ⅰ)。
2008年辽宁高考英语试题真题及答案(WORD排版)
2008年普通高等学校招生统一考试(辽宁卷)第二部分:英语知识运用(共两节,满分15分)第一节:单项选择(共15 小题;每小题1分,满分15分)21. —Did you have a good time in Thailand last week?—______, it was too hot.A. No! reallyB. Yeah, why notC. Oh, greatD. You‟re right22. Peter ______ be really difficult at times even though he‟s a nice person in general.A. shallB. shouldC. canD. must23. We first met on a train in 2000. We both felt immediately that we ______ each other for years.A. knewB. have knownC. had knownD. know24. My neighbor asked me to go for ______ walk, but I don‟t think I‟ve got ______ energy.A. a; 不填B. the; theC. 不填;theD. a; the25. You have to be a fairly good speaker to ______ listener s‟ interest for over an hour.A. holdB. makeC. improveD. receive26. —Could you tell me how to get to Victoria Street?—Victoria Street? ______ is where the Grand Theatre is.A. SuchB. ThereC. ThatD. This27. He was busy writing a story, only ______ once in a while to smoke a cigarette.A. to stopB. stoppingC. to have stoppedD. having stopped28. ______ hungry I am, I never seem to be able to finish off this loaf of bread.A. WhateverB. WheneverC. WhereverD. However29. —Have you got any job offers?—No. I ______.A. waitedB. had been waitingC. have waitedD. am waiting30. It looks like the weather is changing for ______. Shall we stick to our plan?A. the worseB. worseC. the worstD. worst31. Please remain ______; the winner of the prize will be announced soon.A. seatingB. seatedC. to seatD. to be seated32. I used to love that film ______ I was a child, but I don‟t feel it that way any more.A. onceB. whenC. sinceD. although33. I like Mr. Miner‟s speech; it was clear and ______ the point.A. atB. onC. toD. of34. —My name is Jonathan. Shall I spell it for you?—______.A. If you don‟t mindB. Not at allC. Take it easyD. Nice to meet you35. Bill wasn‟t happy about the delay of the report by Jason, and ______.A. I was neitherB. neither was IC. I was eitherD. either was IC. mustn‟t shout C. mustn‟t have shouted第二节:完形填空(共20小题;每小题1.5分,满分30分)阅读下面短文,从短文后各题所给的四个选项(A、B、C和D)中,选出可以填入空白处的最佳选项,并在答题卡上将该项涂黑。
2008年高考数学试卷(辽宁.文)含详解
2008年普通高等学校招生全国统一考试(辽宁卷)数 学(供文科考生使用) 第Ⅰ卷(选择题 共60分)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。
参考公式:如果事件A 、B 互斥,那么 球的表面积公式P(A+B)=P(A)+P(B) S =4πR 2如果事件A 、B 相互独立,那么 其中R 表示球的半径P(A ·B)=P(A) ·P(B) 球的体积公式如果事件A 在一次试验中发生的概率是P ,那么 V=43πR3n 次独立重复试验中事件A 恰好发生k 次的概率 其中R 表示球的半径 P n (k )=C k n P k (1-p )n-k (k =0,1,2,…,n )一、选择题:本大题共12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.(1)已知集合M ={x |-3<x <1|,N={x |x ≤-3},则M =⋃N (A)∅ (B) {x|x ≥-3} (C){x|x ≥1}(D){x |x <1|(2)若函数y=(x +1)(x-a )为偶函数,则a = (A)-2 (B) -2 (C)1 (D)2(3)圆x 2+y 2=1与直线y=kx +2没有公共点的充要条件是 (A)2,2(-∈k )(B) 3,3(-∈k )(C)k ),2()2,(+∞⋃--∞∈(D) k ),3()3,(+∞⋃--∞∈(4)已知0<a <1,x =log a 2log a 3,y =,5log 21a z =loga 3,则 (A)x >y >z(B)z >y >x(C)y >x >z(D)z >x >y(5)已知四边形ABCD 的三个顶点A (0,2),B (-1,-2),C (3,1),且AD BC 2=,则顶点D 的坐标为 (A)(2,27) (B)(2,-21) (C)(3,2) (D)(1,3)(6)设P 为曲线C :y =x 2+2x +3上的点,且曲线C 在点P 处切线倾斜角的取值范围为⎥⎦⎤⎢⎣⎡4,0π,则点P 横坐标的取值范围为 (A)⎥⎦⎤⎢⎣⎡--21,1(B)[-1,0] (C)[0,1](D)⎥⎦⎤⎢⎣⎡1,21(7)4张卡片上分别写有数字1,2,3,4从这4张卡片中随机抽取2张,则取出的2张卡片上的数字之和为奇数的概率为 (A)31 (B)21 (C)32 (D)43 (8)将函数y=2x +1的图象按向量a 平移得到函数y =2x +1的图象,则 (A)a =(-1,-1) (B)a =(1,-1) (C)a =(1,1) (D)a=(-1,1)(9)已知变量x 、y 满足约束条件⎪⎩⎪⎨⎧≥+-≤--≤-+,01,013,01x y x y x y 则z =2x+y 的最大值为第Ⅱ卷(非选择题 共90分)二、填空题:本大题共4小题,每小题4分,共16分.(13)函数23()x y ex +=-∞+∞的反函数是 .(14)在体积为的球的表面上有A 、B 、C 三点,AB =1,BCA 、C 两点的球面距离为3π,则球心到平面ABC 的距离为 . (15)3621(1)()x x x++展开式中的常数项为 . (16)设(0,)2x π∈,则函数22sin 1sin 2x y x +=的最小值为 .三、解答题:本大题共6小题,共74分.解答应写出文字说明,证明过程或演算步骤.(17)(本小题满分12分)在△ABC 中,内角A ,B ,C ,对边的边长分别是a ,b ,c .已知2,3c C π==. (Ⅰ)若△ABC,求a ,b ;(Ⅱ)若sin 2sin B A =,求△ABC 的面积. (18)(本小题满分12分)某批发市场对某种商品的周销售量(单位:吨)进行统计,最近100周的统计结果如下表所示:频数205030(Ⅱ)若以上述频率作为概率,且各周的销售量相互独立,求 (i )4周中该种商品至少有一周的销售量为4吨的概率; (ii )该种商品4周的销售量总和至少为15吨的概率. (19)(本小题满分12分)如图,在棱长为1的正方体ABCD -A ′B ′C ′D ′中,AP =BQ =b (0<b <1),截面PQEF ∥A ′D ,截面PQGH ∥AD ′.(Ⅰ)证明:平面PQEF 和平面PQGH 互相垂直;(Ⅱ)证明:截面PQEF 和截面PQGH 面积之和是定值,并求出这个值; (Ⅲ)若12b =,求D ′E 与平面PQEF 所成角的正弦值. (20)(本小题满分12分)已知数列{a n },{b n }是各项均为正数的等比数列,设(N*)nn nb c n a =∈. (Ⅰ)数列{c n }是否为等比数列?证明你的结论;(Ⅱ)设数列{tna n },{lnb n }的前n 项和分别为S n ,T n .若12,,21n n S n a T n ==+求数列{c n }的前n 项和.(21)(本小题满分12分)在平面直角坐标系xOy 中,点P 到两点(0,-3)、(0,3)的距离之和等于4.设点P 的轨迹为C .(Ⅰ)写出C 的方程;(Ⅱ)设直线y =kx +1与C 交于A 、B 两点.k 为何值时?⊥此时||的值是多少?(22)(本小题满分14分)设函数f (x )=ax 3+bx 2-3a 2x +1(a 、b ∈R )在x =x 1,x =x2处取得极值,且|x 1-x 2|=2. (Ⅰ)若a =1,求b 的值,并求f (x )的单调区间; (Ⅱ)若a >0,求b 的取值范围.2008年普通高等学校招生全国统一考试(辽宁卷)数学(供文科考生使用)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(选择题共60分)参考公式:如果事件A B ,互斥,那么球的表面积公式()()()P A B P A P B +=+24πS R =如果事件A B ,相互独立,那么 其中R 表示球的半径()()()P A B P A P B =球的体积公式如果事件A 在一次试验中发生的概率是P ,那么 34π3V R =n 次独立重复试验中事件A 恰好发生k 次的概率()(1)(012)k kn k n n P k C P p k n -=-=,,,,其中R 表示球的半径一、选择题:本大题共12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. 1.已知集合{}31M x x =-<<,{}3N x x =-≤,则M N =( D )A .∅B .{}3x x -≥C .{}1x x ≥D .{}1x x <答案:D解析:本小题主要考查集合的相关运算知识。
2008年普通高等学校招生全国统一考试数学(辽宁卷_文科)(附答案,完全word版)
一般初等黉舍招生天下一致测验〔辽宁卷〕数学〔供理科考生应用〕本试卷分第一卷〔选择题〕跟第二卷〔非选择题〕两局部.第一卷 1至2页,第二卷3至4页,测验完毕后,将本试卷跟答题卡一并交回.第一卷〔选择题共60分〕参考公式:假如事情A ,B 互斥,那么球的外表积公式S4πR 2此中R 表现球的半径 球的体积公式4 P(AB)P(A)P(B)假如事情A ,B 相互独破,那么P(AB)P(A)P(B)A 在一次实验中发作的概率是P ,那么VπR 3 3假如事情 n 次独破反复实验中事情A 恰恰发作k 次的概率 k knkP n (k)CP(1p)(k01,,2,,n)此中R 表现球的半径n一、选择题:本年夜题共 12小题,每题5分,共60分,在每题给出的四个选项中,只有一项为哪一项契合标题请求的.1.曾经明白聚集M x3x1,N xx ≤3,那么MN 〔〕xx ≥3xx ≥1xx1D .A .B .C . 2.假定函数 y (x1)(xa)为偶函数,那么a=〔 C .1〕212A .B . D . 223.圆xy1与直线ykx2不年夜众点的充要前提是〔 〕k(2,2) k(3,3) A . B . D .k(∞,2)(2,∞)k(∞,3)(3,∞)C . 10a1xlog2log3,y log5zlog21log3,那么〔 4.曾经明白, , 〕a aa aa 2xyz zyxyxzzxy D .A .B .C . ABCD 的三个极点A(02)B(12)C(31)BC2AD ,那么极点,,且5.曾经明白四边形D 的坐标为〔,, , ,〕A .2,72B .2,12C .(3,2)D .(1,3)2yx2x3上的点,且曲线C 在点P 处切线倾歪角的取值范畴为6.设P 为曲线C :0,,那么点P 横坐标的取值范畴为〔 4〕,1 2D .1,12B .10,C .01,A .17.4张卡片上分不写有数字 1,2,3,4,从这4张卡片中随机抽取2张,那么掏出的2张卡 片上的数字之跟为奇数的概率为〔 〕 1 31 22 33 4A .B .C .D .x8.将函数y21的图象按向量 a 平移失掉函数y2x1的图象,那么〔〕A .a (1,1)B .a (1,1)C .a (11),D .a (11),yx1≤0,x ,y 满意束缚前提 y3x1≤0,那么z2xy 的最年夜值为〔 yx1≥0,〕9.曾经明白变量4 2 C .1 10.一消费进程有4道工序,每道工序需求布置一人照看.现从甲、乙、丙等 排4人分不照看一道工序,第一道工序只能从甲、乙两工人中布置1人,第四道工序只能从1人,那么差别的布置计划共有〔B .36种C .48种D .72种4A .B . D . 6名工人中安甲、丙两工人中布置 〕A .24种15 22 211.曾经明白双曲线9ymx1(m0)的一个极点到它的一条渐近线的间隔为 m,那么〔 〕A .1B .2C .3D .4ABCDABCD ,的中点,那么在空间中与三E ,F 分不为棱AACC1112.在正方体中,1 11 1 条直线AD EFCD 都订交的直线〔 , , 〕1 1 A .不存在B .有且只要两条C .有且只要三条D .有有数条第二卷〔非选择题共90分〕二、填空题:本年夜题共 4小题,每题4分,共16分. 2x113.函数ye(∞x ∞)的反函数是.14.在体积为43的球的外表上有A 、B ,C 三点,AB=1,BC=2,A ,C 两点的球面距3 离为ABC 的间隔为_________.,那么球心到破体 361 315.(1x)x开展式中的常数项为 .x 222sinx1 16.设x0 ,,那么函数y 的最小值为 .2sin2x三、解答题:本年夜题共 6小题,共74分.解容许写出笔墨阐明,证实进程或演算步调. 17.〔本小题总分值12分〕在△ABC 中,内角A ,B ,C 对边的边长分不是a ,b ,c ,曾经明白c2,C .3〔Ⅰ〕假定 △ABC 的面积即是3,求a ,b ;〔Ⅱ〕假定sinB2sinA ,求△ABC 的面积.18.〔本小题总分值12分〕某零售市场对某种商品的周贩卖量〔单元:吨〕进展统计,近来 示:100周的统计后果如下表所 2 3 4 周贩卖量 频数205030〔Ⅰ〕依照下面统计后果,求周贩卖量分不为2吨,3吨跟4吨的频率;〔Ⅱ〕假定以上述频率作为概率,且各周的贩卖量相互独破,求〔ⅰ〕4周中该种商品至多有一周的贩卖量为 〔ⅱ〕该种商品4周的贩卖量总跟至多为4吨的概率; 15吨的概率.19.〔本小题总分值12分〕如图,在棱长为1的正方体ABCDABCD 中,AP=BQ=b 〔0<b<1〕,截面PQEF ∥AD , 截面PQGH ∥AD .D〔Ⅰ〕证实:破体PQEF 跟破体PQGH 相互垂直; CHGB〔Ⅱ〕证实:截面PQEF 跟截面PQGH 面积之跟是定值, A并求出那个值; 1 PQ 〔Ⅲ〕假定bDE 与破体PQEF 所成角的正弦值.,求 DC2FE A B20.〔本小题总分值12分〕b n *(n N ).在数列|a||b|是各项均为负数的等比数列,设, c nn n a n〔Ⅰ〕数列|c|能否为等比数列?证实你的论断;nSTa 12,S n .假定nn|lna||lnb| 〔Ⅱ〕设数列n的前项跟分不为,求数,,n nn T n 2n1列|c|的前项跟. n n21.〔本小题总分值12分〕在破体直角坐标系xOy 中,点P 到两点(0,3),(0,3)的间隔之跟即是4,设点P 的轨 迹为C .〔Ⅰ〕写出C 的方程;〔Ⅱ〕设直线ykx1与C 交于A ,B 两点.k 为何值时 OAOB ?如今AB 的值是多少?22.〔本小题总分值14分〕322设函数f(x)axbx3ax1(a ,b R )xxxx 处获得极值,且 在,1 2x 1x2.2〔Ⅰ〕假定a1,求b 的值,并求f(x)的枯燥区间; 〔Ⅱ〕假定a0,求b 的取值范畴.一般初等黉舍招生天下一致测验〔辽宁卷〕数学〔供理科考生应用〕试题参考谜底跟评分参考一、选择题:此题考察根本常识跟根本运算.每题5分,共60分.1.D 7.C 2.C8.A3.B9.B4.C 5.A 6.A10.B 11.D 12.D二、填空题:此题考察根本常识跟根本运算.每题4分,总分值16分.1 2 3 213.y (lnx1)(x0) 14.15.35 16. 3三、解答题17.本小题要紧考察三角形的边角关联等根底常识,考察综算盘算才能.总分值12分.2 2解:〔Ⅰ〕由余弦定理得,abab4,1又由于△ABC的面积即是 3 ,因而absinC 3,得ab4 .·······················4分22 2abab4,解得a2,b2.··············································6分ab4,联破方程组〔Ⅱ〕由正弦定理,曾经明白前提化为b2a,·························································8分2 2abab4,233 43 3联破方程组解得a ,b .b2a,1 2 23 3因而△ABC的面积S absinC .····················································12分18.本小题要紧考察频率、概率等根底常识,考察应用概率常识处理实践咨询题的才能.总分值12分.解:〔Ⅰ〕周贩卖量为2吨,3吨跟4吨的频率分不为0.2,0.5跟0.3.······················4分〔Ⅱ〕由题意知一周的贩卖量为概率为2吨,3吨跟4吨的频率分不为0.2,0.5跟0.3,故所求的4〔ⅰ〕P10.70.7599.···································································8分13 3 4〔ⅱ〕PC0.50.30.30.0621.···············································12分2 419.本小题要紧考察空间中的线面关联跟面面关联,解三角形等根底常识,考察空间设想能力与逻辑思想才能.总分值解法一:12分.〔Ⅰ〕证实:在正方体中,又由曾经明白可得AD AD,AD AB,PF∥ADPH∥AD,PQAB,,∥因而PHPF ,PHPQ , 因而PH破体PQEF .因而破体PQEF 跟破体PQGH 相互垂直.·························································4分 〔Ⅱ〕证实:由〔Ⅰ〕知PF 2AP ,PH 2PA ,又截面PQEF 跟截面PQGH 基本上矩形,且PQ=1,因而截面PQEF 跟截面PQGH 面积之跟是(2AP2PA)PQ 2,是定值.···························································8分〔Ⅲ〕解:设AD 交PF 于点N ,贯穿连接EN , AD破体PQEF ,由于 D CC 因而∠DEN 为DE 与破体PQEF 所成的角. HB GQ A1 由于b,P ,Q ,E ,F 分不为AA ,BB ,BCAD 的中点.,因而D 2PNFE BA 3243 可知DNDE 32, .22 43 因而sin ∠DEN.···································································12分22解法二:以D 为原点,射线DA ,DC ,DD ′分不为x ,y ,z 轴的正半轴树破如图的空间直角坐标系 DF1b ,故 D -xyz .由曾经明白得A(1,0,0),A(1,0,1),D(0,0,0),D(0,0,1),P(1,0,b),Q(11,,b),E(1b ,1,0), zDCHGABB F(1b ,0,0)G(b ,11),H(b ,0,1)., , C PQ 〔Ⅰ〕证实:在所树破的坐标系中,可得DFyEA PQ(010),,,PF(b ,0,b), xPH(b101,,b),AD(101),,,AD(10,,1).ADPQ0ADPF0,由于AD 是破体PQEF 的法向量.,因而由于ADPQ0ADPH0,因而,AD 是破体PQGH 的法向量.由于ADAD0,因而ADAD ,因而破体PQEF 跟破体PQGH 相互垂直.···························································4分 〔Ⅱ〕证实:由于EF(0,10),,因而EF ∥PQ ,EF=PQ ,又PFPQ ,因而PQEF 为矩形,同理PQGH 为矩形. 在所树破的坐标系中可求得 PH 2(1b),PF 2b ,因而PHPF 2,又PQ1,因而截面PQEF 跟截面PQGH 面积之跟为2,是定值.·······································8分 〔Ⅲ〕解:由〔Ⅰ〕知AD(101),,是破体PQEF 的法向量. PAA 中点可知,Q ,E ,F 分不为BB ,BCAD 的中点. 由 为 ,112因而E ,1,0,DE,1,1,因而DE 与破体PQEF 所成角的正弦值即是 2|cosAD ,DE|2.·············································································12分 220.本小题要紧考察等差数列,等比数列,对数等根底常识,考察综合应用数学常识处理咨询 题的才能.总分值12分. c n 解:〔Ⅰ〕是等比数列.··············································································2分证实:设a n 的公比为q 1(q0)b q 2(q0),那么2,的公比为1nc n1b n1a n b n1a n q 20,故c 为等比数列.····································5分nc na n1b nba n1q 1n〔Ⅱ〕数列lna nlnb nlnqlnq 的等差数列. 跟 分不是公役为 跟 1 2n(n1)lnq 1nlna 12 2 由前提得,即n(n1)lnq 22n1nlnb 122lna(n1)lnq 1 n1 .·········································································7分2lnb(n1)lnq 22n11故对n1,2,⋯,2(2lnqlnq)n(4lnalnq2lnblnq)n(2lnalnq)0.1 2 1 1 1 2 1 1因而2lnqlnq0, 12 4lnalnq2lnblnq 20, 1 1 1 2lnalnq0. 11将a2代入得q 14q16b8.·······················································10分 , , 12 1 816n1 24n1n从而有c n4.因而数列c nn的前项跟为4 244⋯4nn(41).·········································································12分 321.本小题要紧考察破体向量,椭圆的界说、规范方程及直线与椭圆地位关联等根底常识, 考察综合应用剖析多少何常识处理咨询题的才能.总分值 解:12分. 〔Ⅰ〕设P 〔x ,y 〕,由椭圆界说可知,点 P 的轨迹C 是以(0,3),(0,3)为核心,长半22(3)21,轴为2的椭圆.它的短半轴by 2 故曲线C 的方程为x 21 .······································································4分4〔Ⅱ〕设A(x ,y),B(x ,y),其坐标满意 1 1 2 2y 24x 21,ykx1.消去y 并收拾得(k4)x2kx30,2k 2 2 3 故xx 21,xx12.····························································6分2k42k4OAOB ,即xxyy0. 121 22而yykxxk(xx)1, 1 2 1212233k 22k 24k1. 因而xxyy 2112 12222k4k4k4k412因而kx 1x 2yy0,故OAOB .···············································8分12 时,1 24 12172当kx 1x 2,xx12时, .17(xx)(yy)222 AB(1k)(xx),2 12 12 122而(xx)(xx)4xx 2 2 12 114217243413 34,17172因而AB465.····················································································12分 1722.本小题要紧考察函数的导数,枯燥性、极值,最值等根底常识,考察综合应用导数研讨 函数的有关性子的才能.总分值 解:f(x)3ax2bx3a2 〔Ⅰ〕当a1时,14分2 .①·····································································2分2f(x)3x2bx3;2由题意知x ,x3x2bx30的两根,因而为方程1 2 24b36 3x 1x 2.由xx2,得b0.···············································································分 41 2 22从而f(x)x3x1f(x)3x33(x1)(x1).,当x(11),时,f(x)0;当x(∞,1)(1,∞)时,f(x)0.故f(x)在(11),枯燥递加,在(∞,1),(1,∞)枯燥递增.······························6分 223x2bx3a0的两根,〔Ⅱ〕由①式及题意知x ,x1为方程 24b36a 32因而xx 21.3a22从而xx2b9a(1a), 1 2由上式及题设知0a ≤1.············································································8分2思索g(a)9a9a 3,2 g(a)18a27a 227aa.······························································10分32 23 234.3故g(a)在0,枯燥递增,在,1枯燥递加,从而g(a)在01,的极年夜值为g 32 3 4 3又g(a)在g(1)0.因而b2 01,上只要一个极值,因而g 为在g(a)01,上的最年夜值,且最小值为4 2323,.········································14分30,,即b的取值范畴为3 3。
2008年普通高等学校招生全国统一考试数学(辽宁卷·理科)(附答案)
⑵证明: 1 + 1 + + 1 5 .
a1 + b1 a2 + b2
an + bn 12
22.设函数 f (x) = ln x − ln x + ln(x +1) . 1+ x
⑴求 f (x) 的单调区间和极值;
⑵是否存在实数 a ,使得关于 x 的不等式 f (x) a 的解集为 (0, +) ?若存在,求 a 的取值范围;若
不存在,试说明理由.
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2008 年普通高等学校招生全国统一考试(辽宁卷)
数学(供理科考生使用)试题参考答案和评分参考
说明:
一、 解答指出了每题要考查的主要知识和能力,并给出了一种或几种解决供参考,如果考生
的解法与本解答不同,可根据试题的主要考查内容比照评分标准制订相应的评分细则.
A. 17
B. 3
2
C. 5 D. 9 2
11.在正方体 ABCD − A1B1C1D1 中, E, F 分别为棱 AA1,CC1 的中点,则在空间中与三条直线
A1D1, EF,CD 都相交的直线( )
A.不存在 B.有且只有两条 C.有且只有三条 D.有无数条
12.设 f (x) 是连续的偶函数,且当 x 0 时 f (x) 是单调函数,则满足 f (x) = f ( x + 3) 的所有 x 之 x+4
4
2
3.圆 x2 + y2 = 1与直线 y = kx + 2 没有公共点的充要条件是( )
A. k (− 2, 2) B. k (−, − 2) ( 2, +)
2008年普通高等学校招生全国统一考试辽宁数学文科试题及答案
2008年普通高等学校招生全国统一考试(辽宁卷)数学(供文科考生使用)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(选择题共60分)参考公式:如果事件互斥,那么球的表面积公式如果事件相互独立,那么其中表示球的半径球的体积公式如果事件在一次试验中发生的概率是,那么次独立重复试验中事件恰好发生次的概率其中表示球的半径一、选择题:本大题共12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.1.已知集合,,则()A.B.C.D.2.若函数为偶函数,则a=()A.B.C.D.3.圆与直线没有..公共点的充要条件是()A.B.C.D.4.已知,,,,则()A.B.C.D.5.已知四边形的三个顶点,,,且,则顶点的坐标为()A.B.C.D.6.设P为曲线C:上的点,且曲线C在点P处切线倾斜角的取值范围为,则点P横坐标的取值范围为()A.B.C.D.7.4张卡片上分别写有数字1,2,3,4,从这4张卡片中随机抽取2张,则取出的2张卡片上的数字之和为奇数的概率为()A.B.C.D.8.将函数的图象按向量平移得到函数的图象,则()A.B.C.D.9.已知变量满足约束条件则的最大值为()A.B.C.D.10.一生产过程有4道工序,每道工序需要安排一人照看.现从甲、乙、丙等6名工人中安排4人分别照看一道工序,第一道工序只能从甲、乙两工人中安排1人,第四道工序只能从甲、丙两工人中安排1人,则不同的安排方案共有()A.24种B.36种C.48种D.72种11.已知双曲线的一个顶点到它的一条渐近线的距离为,则()A.1 B.2 C.3 D.412.在正方体中,分别为棱,的中点,则在空间中与三条直线,,都相交的直线()A.不存在B.有且只有两条C.有且只有三条D.有无数条第Ⅱ卷(非选择题共90分)二、填空题:本大题共4小题,每小题4分,共16分.13.函数的反函数是.14.在体积为的球的表面上有A、B,C三点,AB=1,BC=,A,C两点的球面距离为,则球心到平面ABC 的距离为_________.15.展开式中的常数项为 .16.设,则函数的最小值为 .三、解答题:本大题共6小题,共74分.解答应写出文字说明,证明过程或演算步骤. 17.(本小题满分12分) 在中,内角对边的边长分别是,已知,.(Ⅰ)若的面积等于,求;(Ⅱ)若,求的面积.18.(本小题满分12分)某批发市场对某种商品的周销售量(单位:吨)进行统计,最近100周的统计结果如下表所示:周销售量 2 3 4 频数205030(Ⅰ)根据上面统计结果,求周销售量分别为2吨,3吨和4吨的频率; (Ⅱ)若以上述频率作为概率,且各周的销售量相互独立,求(ⅰ)4周中该种商品至少有一周的销售量为4吨的概率; (ⅱ)该种商品4周的销售量总和至少为15吨的概率.19.(本小题满分12分) 如图,在棱长为1的正方体中,AP=BQ=b (0<b <1),截面PQEF ∥,截面PQGH ∥.(Ⅰ)证明:平面PQEF 和平面PQGH 互相垂直; (Ⅱ)证明:截面PQEF 和截面PQGH 面积之和是定值,并求出这个值; (Ⅲ)若,求与平面PQEF 所成角的正弦值.A B CD E FP Q HG20.(本小题满分12分)在数列,是各项均为正数的等比数列,设.(Ⅰ)数列是否为等比数列?证明你的结论;(Ⅱ)设数列,的前项和分别为,.若,,求数列的前项和.21.(本小题满分12分)在平面直角坐标系中,点P到两点,的距离之和等于4,设点P的轨迹为.(Ⅰ)写出C的方程;(Ⅱ)设直线与C交于A,B两点.k为何值时?此时的值是多少?22.(本小题满分14分)设函数在,处取得极值,且.(Ⅰ)若,求的值,并求的单调区间;(Ⅱ)若,求的取值范围.2008年普通高等学校招生全国统一考试(辽宁卷)数学(供文科考生使用)试题参考答案和评分参考一、选择题:本题考查基本知识和基本运算.每小题5分,共60分.1.D 2.C 3.B 4.C 5.A 6.A7.C 8.A 9.B 10.B 11.D 12.D二、填空题:本题考查基本知识和基本运算.每小题4分,满分16分.13.14.15.16.三、解答题17.本小题主要考查三角形的边角关系等基础知识,考查综合计算能力.满分12分.解:(Ⅰ)由余弦定理得,,又因为的面积等于,所以,得. ·························4分联立方程组解得,. ···············································6分(Ⅱ)由正弦定理,已知条件化为,··························································8分联立方程组解得,.所以的面积. ······················································ 12分18.本小题主要考查频率、概率等基础知识,考查运用概率知识解决实际问题的能力.满分12分.解:(Ⅰ)周销售量为2吨,3吨和4吨的频率分别为0.2,0.5和0.3.······················4分(Ⅱ)由题意知一周的销售量为2吨,3吨和4吨的频率分别为0.2,0.5和0.3,故所求的概率为(ⅰ).···································································8分(ⅱ).··············································· 12分19.本小题主要考查空间中的线面关系和面面关系,解三角形等基础知识,考查空间想象能力与逻辑思维能力.满分12分.解法一:(Ⅰ)证明:在正方体中,,,又由已知可得,,,所以,,所以平面.所以平面和平面互相垂直. ························································· 4分(Ⅱ)证明:由(Ⅰ)知,又截面PQEF 和截面PQGH 都是矩形,且PQ =1,所以截面PQEF 和截面PQGH 面积之和是,是定值. ···························································· 8分(Ⅲ)解:设交于点,连结,因为平面, 所以为与平面所成的角. 因为,所以分别为,,,的中点. 可知,.所以. ···································································· 12分解法二:以D 为原点,射线DA ,DC ,DD ′分别为x ,y ,z 轴的正半轴建立如图的空间直角坐标系D -xyz .由已知得,故,,,,,,, ,,.(Ⅰ)证明:在所建立的坐标系中,可得,,.因为,所以是平面PQEF 的法向量.A B C D EFP QHy xz G A BC DE FP Q HG N因为,所以是平面PQGH的法向量.因为,所以,所以平面PQEF和平面PQGH互相垂直. ···························································4分(Ⅱ)证明:因为,所以,又,所以PQEF 为矩形,同理PQGH为矩形.在所建立的坐标系中可求得,,所以,又,所以截面PQEF和截面PQGH面积之和为,是定值.·······································8分(Ⅲ)解:由(Ⅰ)知是平面的法向量.由为中点可知,分别为,,的中点.所以,,因此与平面所成角的正弦值等于.············································································· 12分20.本小题主要考查等差数列,等比数列,对数等基础知识,考查综合运用数学知识解决问题的能力.满分12分.解:(Ⅰ)是等比数列.··············································································2分证明:设的公比为,的公比为,则,故为等比数列. ·····································5分(Ⅱ)数列和分别是公差为和的等差数列.由条件得,即.·········································································7分故对,,…,.于是将代入得,,. ······················································ 10分从而有.所以数列的前项和为.········································································· 12分21.本小题主要考查平面向量,椭圆的定义、标准方程及直线与椭圆位置关系等基础知识,考查综合运用解析几何知识解决问题的能力.满分12分.解:(Ⅰ)设P(x,y),由椭圆定义可知,点P的轨迹C是以为焦点,长半轴为2的椭圆.它的短半轴,故曲线C的方程为. ·······································································4分(Ⅱ)设,其坐标满足消去y并整理得,故. ·····························································6分,即.而,于是.所以时,,故. ················································8分当时,,.,而,所以. ····················································································· 12分22.本小题主要考查函数的导数,单调性、极值,最值等基础知识,考查综合利用导数研究函数的有关性质的能力.满分14分解:.①·····································································2分(Ⅰ)当时,;由题意知为方程的两根,所以.由,得.···············································································4分从而,.当时,;当时,.故在单调递减,在,单调递增. ·······························6分(Ⅱ)由①式及题意知为方程的两根,所以.从而,由上式及题设知. ·············································································8分考虑,.······························································ 10分故在单调递增,在单调递减,从而在的极大值为.又在上只有一个极值,所以为在上的最大值,且最小值为.所以,即的取值范围为.········································ 14分。
2008年普通高等学校招生全国统一考试数学卷(辽宁.理)含答案
2008年普通高等学校招生全国统一考试(辽宁卷)数 学(供理科考生使用)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(选择题共60分) 参考公式:如果事件A 、B 互斥,那么 球的表面积公式P(A+B)=P(A)+P(B) S=42R π如果事件A 、B 相互独立,那么 其中R 表示球的半径 P(A ·B)=P(A)·P(B) 球的体和只公式 如果事件A 在一次试验中发生的概率是p ,那么n 次独立重复试验中事件A 恰好发生k 次的概率 V =243R π()(1)(0,1,2,k k n kn nP k C P p k n -=-= 其中R 表示球的半径一、选择题:本大题共12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.(1)已知集合3||0|,||3|1x M x x N x x x +==<=≤--,则集合||1|x x ≥= (A )M N ⋂ (B )M N ⋃ (C )R (M N ⋂) (D ) R (M N ⋃)(2)135(21)lim(21)x n n n →∞++++-=+(A )14 (B )12(C )1 (D )2 (3)圆221x y +=与直线2y kx =+没有..公共点的充要条件是()2)A k ∈ ()(,)B k ∈-∞⋃+∞()3)C k ∈ ()(,)D k ∈-∞⋃+∞ (4)复数11212i i +-+-的虚部是 1()5A i 1()5B 1()5C i - 1()5D -(5)已知O 、A 、B 是平面上的三个点,直线AB 上有一点C ,满足20AC CB +=,则OC - ()2A O A O B - ()2B O A O B-+ 21()33C OA OB - 12()33D OA OB -- (6)设P 为曲线C :223y x x =++上的点,且曲线C 在点P 处切线倾斜角的取值范围为[0,4π],则点P 横坐标的取值范围为 1()[1,]2A -- ()[1,0]B - ()[0,1]C 1()[,1]2D (7)4张卡片上分别写有数字1,2,3,4,从这4张卡片中随机抽取2张,则取出的2张卡片上的数学之和为奇数的概率为 1()3A 1()2B 2()3C 3()4D (8)将函数21212a y a y +=+=的图象按向量平移得到函数的图象,则()(1,1)A a =-- ()(1,1)B a =- ()(1,1)C a = ()(1,1)D a =-(9)一生产过程有4道工序,每道工序需要安排一人照看,现从甲、乙、丙等6名工人中安排4人分别照看一道工序,第一道工序只能从甲、乙两工人中安排1人,第四道工序只能从甲、丙两工人中安排1人,则不同的安排方案共有(A )24种 (B )36种 (C )48种 (D )72种(10)已知点P 是抛物线22y x =上的一个动点,则点P 到点(0,2)的距离与P 到该抛物线准线的距离之和的最小值为(A ()3B (C 9()2D(11)在正方体ABCD -A 1B 1C 1D 1中,E 、F 分别为棱AA 1,CC 1的中点,则在空间中与三条直线A 1D 1、EF 、CD 都相交的直线 ()A 不存在 (B )有且只有两条 (C )有且只有三条 (D )有无数条 (12)设f(x)是连续的偶函数,且当x >0时f(x)是单调函数,则满足f(x)=f 3()4x x ++的所有x 之和为(A )-3 (B )3 (C )-8 (D )8第Ⅰ卷(选择题共60分)二、填空题:本大题共4小题,每小题4分,共16分.(13)函数1,0,,0x x x y e x +<⎧=⎨≥⎩的反函数是__________.(14)在体积为的球的表面上有A 、B 、C 三点,AB =1,BC A 、C 两点的球,则球心到平面ABC 的距离为_________. (15)已知21(1)()n y x x x x+++的展开式中没有..常数项,*n N ∈,且2≤n ≤8,则n =______.(16)已知()sin()(0),()()363f x x f f πππωω=+>=,且()f x 在区间(,)63ππ有最小值,无最大值,则ω=__________.三、解答题:本大题共6小题,共74分.解答应写出文字说明,证明过程或演算步骤. (17)(本小题满分12分) 在ABC ∆中,内角A ,B,C 对边的边长分别是a,b,c ,已知c =2,C =3π.(Ⅰ)若ABC ∆a,b ;(Ⅱ)若sin sin()2sin 2C B A A +-=,求ABC ∆的面积.(18)(本小题满分12分)某批发市场对某种商品的周销售量(单位:吨)进行统计,最近100周的统计结果如下表所示:(Ⅰ)根据上面统计结果,求周销售量分别为2吨,3吨和4吨的频率; (Ⅱ)已知每吨该商品的销售利润为2千元,ξ表示该种商品两周销售利润的和(单位:千元).若以上述频率作为概率,且各周的销售量相互独立,求ξ的分布列和数学期望.(19)(本小题满分12分)如图,在棱长为1的正方体ABCD AB C D ''''-中,AP=BP=b (0<b <1),截面PQEF ∥A D ',截面PQGH ∥A D '.(Ⅰ)证明:平面PQEF 和平面PQGH 互相垂直;(Ⅱ)证明:截面PQEF 和截面PQGH 面积之和是定值,并求出这个值;(Ⅲ)若D E '与平面PQEF 所成的角为45°,求D E '与平面PQGH 所成角的正弦值.(20)本小题主要考查平面向量,椭圆的定义、标准方程及直线与椭圆位置关系等基础知识,考查综合运用解析几何知识解决问题的能力.满分12分. 解:(Ⅰ)设P (x ,y ),由椭圆定义可知,点P 的轨迹C 是以(0,为焦长,长半轴为2的椭圆.它的短半轴1,b ==故曲线C 的方程为224; 1.yx - ……3分(Ⅱ)设1122(,),(,)A x y B x y ,其坐标满足221,41.y x y kx ⎧⎪+=⎨⎪=+⎩ 消去y 并整理得22(4)2k x kx ++ 3.0, 故12122223,.44k x x x x k k +==-++ ……5分 若,OA OB ⊥即12120.x x y y +=面22121222233210,444k k x x y y k k k +----+=+++ 化简得2410,k -+=所以1.2k =± ……8分(Ⅲ)2222221122;()OA OB x y x y -=++=22221222()4(11)x x x x -+--+=12123()()x x x x --+ =1226().4k x x k -+ 因为A 在第一象限,故x 1>0.由12234x x k =+知20,x 从而120.x x -又0,k故220,OA OB-即在题设条件下,恒有.OAOB ……12分(21)本小题主要考查等差数列,等比数例,数学归纳法,不等式等基础知识,考查综合运用数学知识进行归纳、总结、推理、论证等能力.满分12分. 解:(Ⅰ)由条件得21112,.n n n a n n b a a a b b +++=+-由此可得2223446,9,12,16,20,25.a b a b a b ====== ……2分猜测2(1),(1).n n a n n b n =+=+ ……4分 用数学归纳法证明:①当n =1时,由上可得结论成立. ②假设当n =k 时,结论成立,即2(1),(1),k k a k k b k =+=+那么当n =k +1时,22221122(1)(1)(1)(2),(2)bk k k k ka ab a k k k k k b k b +++=-=+-+=++==+所以当n =k +1时,结论也成立.由①②,可知2(1),(1)n n a n n b n =++对一切正整数都成立. ……7分 (Ⅱ)12115.612a b =+ n ≥2时,由(Ⅰ)知(1)(21)2(1).n n a b n n n n +=+++ ……9分故112211111111()622334(1)n na b a b a b n n ++++++++++⨯⨯+ =11111111()6223341n n +-+-++-+ =1111115().62216412n +-+=+ 综上,原不等式成立.……12分(20)(本小题满分12分)在直角坐标系xoy 中,点P 到两点(0,-)、(0,4,设点P 的轨迹为l 、直线y=kx+1与C 交于A 、B 两点. (Ⅰ)写出C 的方程; (Ⅱ)若OA ⊥OB ,求k 的值;(Ⅲ)若点A 在第一象限,证明:当k >0时,恒有|OA |>|OB |. (21)(本小题满分12分)在数列|a n |,|b n |中,a 1=2, b 2=4,且1,,n n n a b a +成等差数列,11,,n n n b a b ++成等比数列(*n N ∈) (Ⅰ)求a 2, a 3, a 4及b 2, b 3, b 4,由此猜测{a n },{b n }的通项公式,并证明你的结论; (Ⅱ)证明:1122111512n n a b a b a b +++<+++.(22)(本小题满分14分) 设函数f (x )=ln ln ln(1).1xx x x-+++ (Ⅰ)求f (x )的单调区间和极值;(Ⅱ)是否存在实数a,使得关于x的不等式f(x)≥a的解集为(0,+∞)?若存在,求a的取值范围;若不存在,试说明理由.2008年普通高等学校招生全国统一考试(辽宁卷)数学(供理科考生使用)试题参考答案和评分参考说明:一、解答指出了每题要考查的主要知识和能力,并给出了一种或几种解决供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制订相应的评分细则.二、对解答题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应得分数的一半;如果后继部分的解答有较严重的错误,就不再给分.三、答右端所注分数,表示考生正确做到这一步应得的累加分数.四、只给整数分数,选择题和填空题不给中间分.一、选择题:本题考查基本知识和基本运算,每小题5分,共60分.(1)D (2)B (3)C (4)B (5)A (6)A(7)C (8)A (9)B (10)A (11)D (12)C(18)本小题主要考查频率、概率、数学期望等基础知识,考查运用概率知识解决实际问题的能力.满分12分。
2008高考辽宁数学文科试卷含详细解答(全word版)080708
2008年普通高等学校招生全国统一考试(辽宁卷)数 学(文科)一、选择题:本大题共12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.1.已知集合{}31M x x =-<<,{}3N x x =-≤,则M N = ( D ) A .∅B .{}3x x -≥C .{}1x x ≥D .{}1x x <解:依题意{}31,M x x =-<< {}3N x x =-…,∴{|1}M N x x ⋃=<. 2.若函数(1)()y x x a =+-为偶函数,则a =( C ) A .2-B .1-C .1D .2解: (1)2(1),f a =-(1)0(1),f f -== 1.a ∴=3.圆221x y +=与直线2y kx =+没有..公共点的充要条件是(B ) A.(k ∈B .(k ∈C .()k ∈-+D.()k ∈-+解:依题圆221x y +=与直线2y kx =+没有公共点1d ⇔=>⇔(k ∈4.已知01a <<,log log a a x =1log 52a y =,log log a a z = C )A .x y z >>B.z y x >>C .y x z >>D .z x y >>解: log ax = log a y =log a z =由01a <<知其为减函数, y x z ∴>>5.已知四边形ABCD 的三个顶点(02)A ,,(12)B --,,(31)C ,,且2BC AD =,则顶点D 的坐标为( A ) A .722⎛⎫ ⎪⎝⎭,B .122⎛⎫-⎪⎝⎭, C .(32),D .(13), 解:(4,3),BC = (,2),AD x y =- 且2BC AD = ,22472432x x y y =⎧=⎧⎪∴⇒⎨⎨-==⎩⎪⎩6.设P 为曲线C :223y x x =++上的点,且曲线C 在点P 处切线倾斜角的取值范围为04π⎡⎤⎢⎥⎣⎦,,则点P 横坐标的取值范围为( A ) A .112⎡⎤--⎢⎥⎣⎦,B .[]10-,C .[]01,D .112⎡⎤⎢⎥⎣⎦,解:设切点P 的横坐标为0x , 且0'22tan y x α=+=(α为点P 处切线的倾斜角), 又∵[0,]4πα∈,∴00221x ≤+≤,∴01[1,].2x ∈-- 7.4张卡片上分别写有数字1,2,3,4,从这4张卡片中随机抽取2张,则取出的2张卡片上的数字之和为奇数的概率为( C ) A .13B .12C .23D .34解:依题要使取出的2张卡片上的数字之和为奇数,则取出的2张卡片上的数字必须一奇一偶,∴取出的2张卡片上的数字之和为奇数的概率11222342.63C C P C ⋅=== 8.将函数21x y =+的图象按向量a 平移得到函数12x y +=的图象,则( A )A .(11)=--,a B .(11)=-,aC .(11)=,a D .(11)=-,a 解:依题由函数21xy =+的图象得到函数12x y +=的图象,需将函数21xy =+的图象向左平移1个单位,向下平移1个单位;故(11).=--,a9.已知变量x y ,满足约束条件1031010y x y x y x +-⎧⎪--⎨⎪-+⎩≤,≤,≥,则2z x y =+的最大值为( B ) A .4B .2C .1D .4-解:作图易知可行域为一个三角形,其三个顶点为 (01),,(10),,(12),--, 验证知在点(10),时取得最大值2. 10.一生产过程有4道工序,每道工序需要安排一人照看.现从甲、乙、丙等6名工人中安排4人分别照看一道工序,第一道工序只能从甲、乙两工人中安排1人,第四道工序 只能从甲、丙两工人中安排1人,则不同的安排方案共有( B ) A .24种B .36种C .48种D .72种解:依题若第一道工序由甲来完成,则第四道工序必由丙来完成,故完成方案共有2412A =种; 若第一道工序由乙来完成,则第四道工序必由甲、丙二人之一来完成,故完成方案共12A ⋅2424A =种;∴则不同的安排方案共有21242436A A A +⋅=种。
2008年高考辽宁数学理(含答案)
2008年普通高等学校招生全国统一考试(辽宁卷)数学(供理科考生使用)第Ⅰ卷(选择题共60分)参考公式:如果事件A B ,互斥,那么球的表面积公式()()()P A B P A P B +=+24πS R =如果事件A B ,相互独立,那么 其中R 表示球的半径()()()P A B P A P B =球的体积公式如果事件A 在一次试验中发生的概率是P ,那么 34π3V R =n 次独立重复试验中事件A 恰好发生k 次的概率()(1)(012)k k n kn n P k C P p k n -=-= ,,,,其中R 表示球的半径一、选择题:本大题共12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. 1.已知集合{}3|0|31x M x x N x x x +⎧⎫==<=-⎨⎬-⎩⎭,≤,则集合{}|1x x ≥=( ) A .M N B .M NC .()M M N ðD .()M M N ð2.135(21)lim(21)x n n n →∞++++-=+ ( )A .14B .12C .1D .23.圆221x y +=与直线2y kx =+没有..公共点的充要条件是( )A .(k ∈B .()k ∈-+C .(k ∈D .()k ∈-+4.复数11212i i +-+-的虚部是( ) A .15i B .15 C .15i - D .15-5.已知O ,A ,B 是平面上的三个点,直线AB 上有一点C ,满足20AC CB += ,则OC =( )A .2OA OB - B .2OA OB -+C .2133OA OB -D .1233OA OB -+6.设P 为曲线C :223y x x =++上的点,且曲线C 在点P 处切线倾斜角的取值范围为04π⎡⎤⎢⎥⎣⎦,,则点P 横坐标的取值范围为( ) A .112⎡⎤--⎢⎥⎣⎦,B .[]10-,C .[]01,D .112⎡⎤⎢⎥⎣⎦,7.4张卡片上分别写有数字1,2,3,4,从这4张卡片中随机抽取2张,则取出的2张卡片上的数字之和为奇数的概率为( )A .13 B .12C .23 D .34 8.将函数21x y =+的图象按向量a 平移得到函数12x y +=的图象,则( )A .(11)=--,a B .(11)=-,aC .(11)=,a D .(11)=-,a 9.一生产过程有4道工序,每道工序需要安排一人照看.现从甲、乙、丙等6名工人中安排4人分别照看一道工序,第一道工序只能从甲、乙两工人中安排1人,第四道工序只能从甲、丙两工人中安排1人,则不同的安排方案共有( ) A .24种 B .36种 C .48种 D .72种 10.已知点P 是抛物线22y x =上的一个动点,则点P 到点(0,2)的距离与P 到该抛物线准线的距离之和的最小值为( ) AB .3CD .9211.在正方体ABCD -A 1B 1C 1D 1中,E ,F 分别为棱AA 1,CC 1的中点,则在空间中与三条直线A 1D 1,EF ,CD 都相交的直线( ) A .不存在 B .有且只有两条 C .有且只有三条 D .有无数条 12.设()f x 是连续的偶函数,且当x >0时()f x 是单调函数,则满足3()4x f x f x +⎛⎫= ⎪+⎝⎭的所有x 之和为( ) A .3- B .3C .8-D .8第Ⅱ卷(非选择题共90分)二、填空题:本大题共4小题,每小题4分,共16分. 13.函数100xx x y e x +<⎧=⎨⎩,,,≥的反函数是__________.14.在体积为的球的表面上有A ,B ,C 三点,AB =1,BC A ,C 两点的球面距离为3π,则球心到平面ABC 的距离为_________.15.已知231(1)nx x x x ⎛⎫+++ ⎪⎝⎭的展开式中没有..常数项,n ∈*N ,且2≤n ≤8,则n =______. 16.已知()sin (0)363f x x f f ωωπππ⎛⎫⎛⎫⎛⎫=+>= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,,且()f x 在区间63ππ⎛⎫⎪⎝⎭,有最小值,无最大值,则ω=__________.三、解答题:本大题共6小题,共74分.解答应写出文字说明,证明过程或演算步骤. 17.(本小题满分12分) 在ABC △中,内角A B C ,,对边的边长分别是a b c ,,,已知2c =,3C π=.(Ⅰ)若ABC △a b ,;(Ⅱ)若sin sin()2sin 2C B A A +-=,求ABC △的面积.18.(本小题满分12分)某批发市场对某种商品的周销售量(单位:吨)进行统计,最近100周的统计结果如下表所示:周销售量 2 3 4 频数205030(Ⅰ)根据上面统计结果,求周销售量分别为2吨,3吨和4吨的频率;(Ⅱ)已知每吨该商品的销售利润为2千元,ξ表示该种商品两周销售利润的和(单位:千元).若以上述频率作为概率,且各周的销售量相互独立,求ξ的分布列和数学期望.19.(本小题满分12分)如图,在棱长为1的正方体ABCD A B C D ''''-中,AP=BQ=b (0<b <1),截面PQEF ∥A D ',截面PQGH ∥AD '.(Ⅰ)证明:平面PQEF 和平面PQGH 互相垂直; (Ⅱ)证明:截面PQEF 和截面PQGH 面积之和是定值,并求出这个值;(Ⅲ)若D E '与平面PQEF 所成的角为45,求D E '与平面PQGH 所成角的正弦值. 20.(本小题满分12分)在直角坐标系xOy 中,点P 到两点(03),(03),的距离之和等于4,设点P 的轨迹为C ,直线1y kx =+与C 交于A ,B 两点. (Ⅰ)写出C 的方程;(Ⅱ)若OA ⊥OB ,求k 的值;(Ⅲ)若点A 在第一象限,证明:当k >0时,恒有|OA |>|OB |.A BCDE FPQ H A ' B 'C 'D ' G21.(本小题满分12分)在数列||n a ,||n b 中,a 1=2,b 1=4,且1n n na b a +,,成等差数列,11n n n b a b ++,,成等比数列(n ∈*N ) (Ⅰ)求a 2,a 3,a 4及b 2,b 3,b 4,由此猜测||n a ,||n b 的通项公式,并证明你的结论; (Ⅱ)证明:1122111512n n a b a b a b +++<+++….22.(本小题满分14分) 设函数ln ()ln ln(1)1xf x x x x=-+++. (Ⅰ)求f (x )的单调区间和极值;(Ⅱ)是否存在实数a ,使得关于x 的不等式()f x a ≥的解集为(0,+∞)?若存在,求a 的取值范围;若不存在,试说明理由.2008年普通高等学校招生全国统一考试(辽宁卷) 数学(供理科考生使用)试题参考答案和评分参考说明:一、本解答指出了每题要考查的主要知识和能力,并给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制订相应的评分细则.二、对解答题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应得分数的一半;如果后继部分的解答有较严重的错误,就不再给分.三、解答右端所注分数,表示考生正确做到这一步应得的累加分数. 四、只给整数分数,选择题和填空题不给中间分.一、选择题:本题考查基本知识和基本运算.每小题5分,共60分. 1.D 2.B 3.C 4.B 5.A 6.A 7.C 8.A 9.B 10.A 11.D 12.C 二、填空题:本题考查基本知识和基本运算.每小题4分,满分16分. 13.11ln 1.x x y x x -<⎧=⎨⎩,,, ≥14.3215.516.143三、解答题17.本小题主要考查三角形的边角关系,三角函数公式等基础知识,考查综合应用三角函数有关知识的能力.满分12分.解:(Ⅰ)由余弦定理及已知条件得,224a b ab +-=, 又因为ABC △31sin 32ab C =4ab =. ······························· 4分 联立方程组2244a b ab ab ⎧+-=⎨=⎩,,解得2a =,2b =. ··························································· 6分(Ⅱ)由题意得sin()sin()4sin cos B A B A A A ++-=,即sin cos 2sin cos B A A A =, ···························································································· 8分 当cos 0A =时,2A π=,6B π=,3a =3b =, 当cos 0A ≠时,得sin 2sin B A =,由正弦定理得2b a =,联立方程组2242a b ab b a ⎧+-=⎨=⎩,,解得a =b =.所以ABC △的面积1sin 2S ab C ==. ···································································· 12分 18.本小题主要考查频率、概率、数学期望等基础知识,考查运用概率知识解决实际问题的能力.满分12分. 解:(Ⅰ)周销售量为2吨,3吨和4吨的频率分别为0.2,0.5和0.3. ···························· 3分 (Ⅱ)ξ的可能值为8,10,12,14,16,且 P (ξ=8)=0.22=0.04, P (ξ=10)=2×0.2×0.5=0.2, P (ξ=12)=0.52+2×0.2×0.3=0.37, P (ξ=14)=2×0.5×0.3=0.3, P (ξ=16)=0.32=0.09.ξ的分布列为········································································································· 9分E ξ=8×0.04+10×0.2+12×0.37+14×0.3+16×0.09=12.4(千元) ···································· 12分19.本小题主要考查空间中的线面关系,面面关系,解三角形等基础知识,考查空间想象能力与逻辑思维能力。
2008年高考英语辽宁卷试题及答案
2008年普通高等学校招生全国统一考试(辽宁卷)英语本试卷分第I卷(选择题)和第Ⅱ卷(非选择题)两部分。
共150分,考试时间l20分钟。
第工卷(选择题共ll5分)第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例:How much is the shirt?A.$19.15.B.$9.15.C.$9.18.答案是B。
1. What is the weather like?A. It's raining.B. It’s cloudy.C. It’s sunny.2. Who will go to China next month?A. Lucy.B. Alice.C. Richard..3. What arc the speakers talking about?A. The man’s sister.B. A film,C. An actor,4. Where will the speakers meet?A. In Room 34O. B, In Room 3l4.C. In Room 223.5. Where does the conversation most probably take place?A. In a restaurant.B. In an office.C. At home.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话。
每段对话后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
2008年普通高等学校招生全国统一考试辽宁卷
2008年普通高等学校招生全国统一考试(辽宁卷)语文本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
第Ⅰ卷1至4页。
第Ⅱ卷5至8页。
考试结束后,将本试卷和答题卡一并交回。
第Ⅰ卷(选择题共30分)一、(12分,每小题3分)1.下列词语中加点的字,读音全都正确的一组是A. 金钗.(chāi)三缄.其口(jiān)颔.联(hàn)纨绔.子弟(kù)B. 给.养(gěi)危如雷.卵(lěi)分娩.(miǎn)妩.媚多姿(wǔ)C. 俟.机(sì)沆.瀣一气(hàng)奶酪.(luò)不容置喙.(huì)D. 奇葩.(pā)不谙.水性(ān)城垣.(huán)风流倜傥.(tǎng)2.下列各句中,加点的词语使用不恰当的一项是A.瑞士国土面积不大,但民族众多,语言也多,法语、德语、意大利语等都是日常生活通行的语言,不少人都能随心所欲....地使用几种语言。
B.牡丹居社区餐厅明天将开始营业,消息传出。
社区居民口耳相传....。
以前他们到最近的餐厅都要步行半个多小时,现在出门走几步就能吃上饭了。
C.岭南的书法艺术历史悠久,但由于气候潮湿等原因,唐代以前的书法作品鲜有传世,即使是宋元墨迹,今天能见到的也是寥若晨星....。
D.灾情就是命令,地震救援人员们冒着大雨,跋山涉水....,克服重重困难,终于按规定时间抵达四川震中灾区,并立即投入了救援工作。
3.下列各句中,没有语病的一句是A.中国皮影戏的艺术魅力曾经倾倒和征服了无数热爱它的人民,它的传播对中国近代电影艺术也有着不可忽视的启示作用。
B.这篇文章集中分析了形式,辨证地回答了在大开放、大交往、大融合大世界里,我们迫切需要一种全新的观念来协调各种关系。
C.交通台日前播报说,有的人在小轿车内开着空调过夜,由此发生窒息死亡事件每年都有发生,应该引起司机朋友们的高度重视。
D.大学毕业后去农村应聘村官的人当中,多数人希望能在建设新农村这一大环境中找到施展才华、创立事业、实现理想的有效途径。
2008年普通高等学校招生全国统一考试辽宁卷
2008年普通高校招生统一考试(辽宁卷)英语第一节(共5小题;每小题l5分,满分7.5分)听下面5段对话.每段对话后有一个小题.从题中所给的A、B、C三个选项中选出最佳选项.并标在试卷的相应位置。
听完每段对话后.你都有l0秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍.例:How much is me shirt?A. £I915B.£915C. £918答案是B.1. What is the weather like?A. it's rainingB. It’s cloudyC. It's sunny.2. Who will go to China next month?A. LucyB. AliceC. Richard.3. What is the speaker talking about?A. The man’s sisterB. A filmC. An actor4. Where ill the speakers meet?A. In Room 340B. In Room 314C. In Room 2235. Where does the conversation most probably take place?A. In a restaurantB. In an officeC. At home第二节(共l5小题;每小题l.5分,满分22.5分)听下面5段对话或独白,每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项.并标在试卷的相应位置。
听每段对话或独白前你将有时间阅读各个小题,每小题5秒钟;听完后,各个小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段村料.回答第6至8题。
6. Why did the woman go to New York?A. To spend some time with the baby.B. To look after her sister.C. To find a new job7. How old was the baby when the woman left New York?A Two monthsB Five monthsC seven months8. What did me women like doing most with the baby?A Holding himB Playing with himC Feeding him听第7段材料,回答第9至11题。
2008年普通高等学校招生全国统一考试数学(辽宁卷·理科)试卷与答案
实用文档2008年普通高等学校招生全国统一考试(辽宁卷)数 学(供理科考生使用)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(选择题 共60分)参考公式:如果事件A B 、互斥,那么 球的表面积公式()()()P A B P A P B +=+ 24S R π=如果事件A B 、互相独立,那么 其中R 表示球的半径()()()P A B P A P B = 球的体积公式如果事件A 在一次试验中发生的概率是p ,那么 343V R π=n 次独立重复试验中事件A 恰好发生k 次的概率 其中R 表示球的半径()(1)(0,1,2,,)k kn k n n P k C p p k n -=-=实用文档一、选择题:本大题共12小题,每小题5分,共60分. 在每小题给出的四个选项中,只有一项最符合题目要求的.(1)已知集合3{|0}1x M x x +=<-,{|3}N x x =≤-,则集合{|1}x x ≥= (A )MN (B )MN (C )()R C MN (D )()R C MN(2)135(21)lim(21)n n n n →∞++++-=+(A )14 (B )12(C )1 (D )2(3)圆221x y +=与直线2y kx=+没有..公共点的充要条件是 (A)(k ∈ (B )(,(2,)k ∈-∞+∞(C)(k ∈ (D )(,(3,)k ∈-∞+∞(4)复数11212i i+-+-的虚部是 (A )15i (B )15 (C )15i - (D )15-实用文档(5)已知O A B 、、是平面上的三个点,直线AB 上有一点C ,满足20AC CB +=,则OC =(A )2OA OB - (B )2OA OB -+ (C )2133OA OB - (D )1233OA OB -+ (6)设P 为曲线2:23C y x x =++上的点,且曲线C 在点P 处切线倾斜角的取值范围为[0,]4π,则点P 横坐标的取值范围为(A )1[1,]2-- (B )[1,0]- (C )[0,1] (D )1[,1]2(7)4张卡片上分别写有数字1,2,3,4,从这4张卡片中随机抽取2张,则取出的2张卡片上的数字之和为奇数的概率为(A )13 (B )12 (C )23(D )34(8)将函数21x y =+的图象按向量a 平移得到函数12x y +=的图象,则实用文档(A )(1,1)a =-- (B )(1,1)a =- (C )(1,1)a = (D )(1,1)a =-(9)一生产过程有4道工序,每道工序需要安排一人照看. 现从甲、乙、丙等6名工人中安排4人分别照看一道工序,第一道工序只能从甲、乙两工人中安排1人,第四道工序只能从甲、丙两工人中安排1人,则不同的安排方案共有(A )24种 (B )36种 (C )48种 (D )72种(10)已知点P 是抛物线22y x =上的一个动点,则点P 到点(0,2)的距离与P 到该抛物线准线的距离之和的最小值为(A )2(B )3 (C(D )92(11)在正方体1111ABCD A B C D -中,E 、F 分别为棱1AA 、1CC 的中点,则在空间中与三条直线11A D 、EF 、CD 都相交的直线(A )不存在 (B )有且只有两条 (C )有且只有三条 (D )有无数条实用文档(12)设()f x 是连续的偶函数,且当0x >时()f x 是单调函数,则满足3()()4x f x f x +=+的所有x 之和为 (A )3- (B )3 (C )8- (D )82008年普通高等学校招生全国统一考试(辽宁卷)数 学(供理科考生使用)第Ⅱ卷(非选择题 共90分)二.填空题:本大题共4小题,每小题4分,共16分.(13)函数1,0,0x x x y e x +<⎧=⎨≥⎩ ,的反函数是______________________.(14)在体积为的球的表面上有A 、B 、C 三点,1AB =,BC =A 、C,则球心到平面ABC 的距离为___________________.实用文档(15)已知231(1)()nx x x x+++的展开式中没有..常数项,*n N ∈且28n ≤≤,则n =____________.(16)已知()sin()(0)3f x x πωω=+>,()()63f f ππ=,且()f x 在区间(,)63ππ有最小值,无最大值,则ω=________________.三、解答题:本大题共6小题,共74分.解答应写出文字说明,证明过程或演算步骤.(17)(本小题满分12分)在ABC 中,内角A ,B ,C 对边的边长分别是a ,b ,c .已知2c =,3C π=.(Ⅰ)若ABCa ,b ;实用文档ABCD EF PQ'A'B'C'D H G(Ⅱ)若sin sin()2sin 2C B A A +-=,求ABC 的面积.(18)(本小题满分12分)某批发市场对某种产品的周销售量(单位:吨)进行统计,最近100周的统计结果如下表所示:(Ⅰ)根据上面的统计结果,求周销售量分别为2吨,3吨和4吨的频率;(Ⅱ)已知每吨该产品的销售利润为2千元,ξ表示该种商品两周销售利润的和(单位:千元).若以上述频率作为概率,且每周的销售量互相独立,求ξ的分布列和数学期望.(19)(本小题满分12分)如图,在棱长为1的正方体''''ABCD A B C D -中,(01)AP BQ b b ==<<,截面'PQEF A D ,截面'PQGH AD .实用文档(Ⅰ)证明:平面PQEF 和平面PQGH 互相垂直;(Ⅱ)证明:截面PQEF 和截面PQGH 面积之积是定值,并求出这个值;(Ⅲ)若'D E 与平面PQEF 所成的角为45︒,求'D E 与平面PQGH 所成角的正弦值.(20)(本小题满分12分)在直角坐标系xOy 中,点P到两点(0,、的距离之和等于4,设点P 的轨迹为C ,直线1y kx =+与C 交于A 、B 两点.(Ⅰ)写出C 的方程;(Ⅱ)若OA OB ⊥,求k 的值;(Ⅲ)若点A 在第一象限,证明:当0k >时,恒有||||OA OB > . (21)(本小题满分12分)在数列{}n a ,{}n b 中,12a =,14b =,且n a ,n b ,1n a +成等差数列,n b ,1n a +,1n b +成等比数列(*n N ∈).(Ⅰ)求2a ,3a ,4a 及2b ,3b ,4b ,由此猜测{}n a ,{}n b 的通项公式,并实用文档证明你的结论;(Ⅱ)证明:1122111512n n a b a b a b +++<+++.(22)(本小题满分14分)设函数ln ()ln ln(1)1xf x x x x=-+++. (Ⅰ)求()f x 的单调区间和极值;(Ⅱ)是否存在实数a ,使得关于x 的不等式()f x a ≥的解集为(0,)+∞?若存在,求a 的取值范围;若不存在,试说明理由.2008年普通高等学校招生全国统一考试(辽宁卷)数学(供理科考生使用)试题参考答案和评分参考说明:一. 解答指出了每题要考查的主要知识和能力,并给出了一种或几种解决供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制订相应的评分细则.二. 对解答题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应得分数的一半;如果后继部分的解答有较严重的错误,就不再给分.三.解答右端所注分数,表示考生正确做到这一步应得的累加分数.四.只给整数分数. 选择题和填空题不给中间分.一. 选择题:本题考查基本知识和基本运算. 每小题5分,共60分.(1)D (2)B (3)C (4)B (5)A (6)A(7)C (8)A (9)B (10)A (11)D (12)C 二. 填空题:本题考查基本知识和基本运算. 每小题4分,共16分.(13)1,1,ln, 1.x xyx x-<⎧=⎨≥⎩(14)32(15)5 (16)143三. 解答题(17)本小题主要考查三角形的边角关系,三角函数公式等基础知识,考查实用文档实用文档综合应用三角函数有关知识的能力. 满分12分.解:(Ⅰ)由余弦定理及已知条件得,224a b ab +-=,又因为ABC 的面积等于,所以1sin 2ab C =,得4ab =. ……4分联立方程组224,4,a b ab ab ⎧+-=⎨=⎩ 解得2, 2.a b == ……6分(Ⅱ)由题意得sin()sin()4sin cos ,B A B A A A ++-=即sin cos 2sin cos .B A A A =……8分当cos 0A =时,2A π=,6B π=,3a =3b =. 所以ABC 的面积1sin 23S ab C ==. ……12分 (18)本小题主要考查频率、概率、数学期望等基础知识,考查运用概率知识解决实际问题的能力.满分12分.解:(Ⅰ)周销售量为2吨,3吨和4吨的频率分别为0.2,0.5和0.3. ……3分(Ⅱ)ξ的可能值为8,10,12,14,16,且P(ξ=8)=0.22=0.04,P(ξ=10)=2×0.2×0.5=0.2,P(ξ=12)=0.52+2×0.2×0.3=0.37,P(ξ=14)=2×0.5×0.3=0.3,P(ξ=16)=0.32=0.09.ξ的分布列为……9分实用文档实用文档ABCDEF PQ'A'B'C 'D H GE ξ=8×0.04+10×0.2+12×0.37+14×0.3+16×0.09=12.4(千元). ……12分(19)本小题主要考查空间中的线面关系,面面关系,解三角形等基础知识,考查空间想象能力与逻辑思维能力. 满分12分.解法一:(Ⅰ)证明:在正方体中,''AD A D ⊥,'AD AB ⊥,又由已知可得'PF A D ,'PH AD ,PQ AB ,所以PH PF ⊥,PH PQ ⊥,所以PH ⊥平面PQEF ,所以平面PQEF 和平面PQGH 互相垂直.……4分(Ⅱ)证明:由(Ⅰ)知PF =,'PH =. 又截面PQEF 和截面PQGH 都是矩形,且1PQ =,所以截 面PQEF 和截面PQGH 面积之和是实用文档')PQ ⨯=. ……8分(Ⅲ)解:连结'BC 交EQ 于点M .因为'PH AD ,PQ AB ,所以平面''ABC D 和平面PQGH 互相平行,因此'D E 与平面PQGH 所成角与'D E 与平面''ABC D 所成角相等.与(Ⅰ)同理可证EQ ⊥平面PQGH ,可知EM ⊥平面''ABC D ,因此EM 与'D E 的比值就是所求的正弦值.设'AD 交PF 于点N ,连结EN ,由1FD b =-知'D E ='ND=(1)22b +-. 因为'AD ⊥平面PQEF ,又已知'D E 与平面PQEF 成45︒角,所以''D E =2[(1)]22b +-= 解得12b =,可知E 为BC 中点.所以4EM =,又3'2D E ==,实用文档故'D E 与平面PQGH 所成角的正弦值为'6EM D E =. ……12分 解法二:以D 为原点,射线DA 、DC ,DD ′分别为x,y,z 轴的正半轴建立如图的空间直角坐标系D-xyz 由已知得DF=l-b ,故A (1,0,0),A ′(1,0,1),D (0,0,0),D ′(0,0,1),P (1,0,b ),Q(1,1,b),E(1,-b,1,0),F(1-b,0,0),G(b,1,1),H(b,0,1).(I )证明:在所建立的坐标系中,可得(0,1,0),(,0,),(1,0,1).PQ PF b b PH b b ==--=--'(1,0,1),(1,0,1).AD AD =-=--因为''0,'0,AD PQ AD PF AD ==所以是平面PQEF 的法向量.因为'0,'0,'A D PQ A D PH A D ==所以是平面PQGH 的法向量.实用文档因为''0,''AD A D A D AD =⊥所以,所以平面PQEF 和平面PQGH 互相垂直. ……4分(II )证明:因为(0,1,0)EF -,所以,||||.EF PQ EF PQ PF PQ =⊥又,所以PQEF 为矩形,同理PQGH 为矩形.在所建立的坐标系中可求得||2(1),||2,PH b PF b =-=所以||||2,||1PH PF PQ +==又,所以截面PQEF和截面PQGH 面积之和为,是定值. ……8分(III )解:由已知得''D E AD 与成45角,又'(1,1,1),'(1,0,1),D E b AD =--=-可得''|'||'|D E AD D E AD ==即11,.2b ==解得所以1'(,1,1),'(1,0,1).2D E A D =-=--又所以D ′E 与平面PQGH 所成角的正弦值为实用文档|cos ','|36D E A D -<>== ……12分(20)本小题主要考查平面向量,椭圆的定义、标准方程及直线与椭圆位置关系等基础知识,考查综合运用解析几何知识解决问题的能力. 满分12分.解:(Ⅰ)设P(x ,y ),由椭圆定义可知,点P 的轨迹C 是以(0,为焦点,长半轴为2的椭圆.它的短半轴1,b ==故曲线C 的方程为221.4y x += (3)分(Ⅱ)设1122(,),(,)A x y B x y ,其坐标满足221,41.y x y kx ⎧⎪+=⎨⎪=+⎩ 消去y 并整理得22(4)230k x kx ++-=,实用文档故12122223,.44k x x x x k k +=-=-++ ……5分若,OA OB ⊥即12120.x x y y +=而 2121212()1y y k x x k x x =+++,于是 22121222233210,444k k x x y y k k k +=---+=+++ 化简得2410,k -+=所以1.2k =± ……8分(Ⅲ)2222221122()OA OB x y x y -=+-+ =22221222()4(11)x x x x -+--+ =12123()()x x x x --+=1226().4k x x k -+因为A 在第一象限,故x 1>0.由12234x x k =+知20,x <从而120.x x -> 又0,k >实用文档故220,OA OB ->即在题设条件下,恒有.OA OB > ……12分(21)本小题主要考查等差数列,等比数例,数学归纳法,不等式等基础知识,考查综合运用数学知识进行归纳、总结、推理、论证等能力. 满分12分.解:(Ⅰ)由条件得21112,.n n n a n n b a a a b b +++=+=由此可得2233446,9,12,16,20,25.a b a b a b ======……2分猜测2(1),(1).n n a n n b n =+=+……4分用数学归纳法证明:实用文档①当n =1时,由上可得结论成立.②假设当n =k 时,结论成立,即2(1),(1),k k a k k b k =+=+那么当n =k +1时,22211122(1)(1)(1)(2),(2)kk k k k ka ab a k k k k k b k b +++=-=+-+=++==+所以当n =k +1时,结论也成立.由①②,可知2(1),(1)n n a n n b n =++对一切正整数都成立. ……7分(Ⅱ)11115.612a b =<+n ≥2时,由(Ⅰ)知(1)(21)2(1).n n a b n n n n +=++>+ ……9分故112211111111()622334(1)n n a b a b a b n n +++<+++++++⨯⨯+=11111111()6223341n n +-+-++-+实用文档 =1111115().62216412n +-<+=+ 综上,原不等式成立. ……12分(22)本小题主要考查函数的导数,单调性,极值,不等式等基础知识,考查综合利用数学知识分析问题、解决问题的能力. 满分14分.解:(Ⅰ)221ln 11ln '()(1)(1)1(1)x x f x x x x x x x =--+=-++++. ……2分故当(0,1)x ∈时,'()0f x >,(1,)x ∈+∞时,'()0f x <.所以,()f x 在(0,1)单调递增,在(1,)+∞单调递减. ……4分由此知 ()f x 在(0,)+∞的极大值为(1)ln 2f =,没有极小值. ……6分实用文档(Ⅱ)(ⅰ)当0a ≤时,由于(1)ln(1)ln ln(1)(ln(1)ln )()011x x x x x x x x f x x x++-+++-==>++, 故关于x 的不等式()f x a ≥的解集为(0,).+∞ ……10分(ⅱ)当0a >时,由ln 1()ln(1)1x f x x x=+++ 知(2)n f =ln 21ln(1)122n n n +++,其中n 为正整数,且有22211ln(1)1log (1)222a a n n a e n e +<⇔<-⇔>--.……12分又2n ≥时,ln 2ln 2ln 22ln 2(1)121(11)12n n n n n n n n =<=-+++-. 且2ln 24ln 2112a n a<⇔+-. 取整数0n 满足202log (1)an e >--,04ln 21n a>+,且02n ≥, 则0000ln 21(2)ln(1)12222n n n n a a f a =++<+=+, 即当0a >时,关于x 的不等式()f x a ≥的解集不是(0,)+∞.综合(ⅰ)(ⅱ)知,存在a,使得关于x的不等式()+∞,f x a≥的解集为(0,)且a的取值范围为(,0]-∞.……14分实用文档。
2008年高考试题——数学文(辽宁卷)
2008年普通高等学校招生全国统一考试(辽宁卷)数 学(供文科考生使用)第Ⅰ卷(选择题 共60分)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。
参考公式:如果事件A 、B 互斥,那么 球的表面积公式P(A+B)=P(A)+P(B) S =4πR 2如果事件A 、B 相互独立,那么 其中R 表示球的半径P(A ·B)=P(A) ·P(B) 球的体积公式如果事件A 在一次试验中发生的概率是P ,那么 V=43πR 3n 次独立重复试验中事件A 恰好发生k 次的概率 其中R 表示球的半径 P n (k )=C k n P k (1-p )n-k (k =0,1,2,…,n )一、选择题:本大题共12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.(1)已知集合M ={x |-3<x <1|,N={x |x ≤-3},则M =⋃N (A)∅ (B) {x|x ≥-3} (C){x|x ≥1}(D){x |x <1|(2)若函数()()1y x x a =+-为偶函数,则a = (A)2-(B) 1-(C)1(D)2(3)圆221x y +=与直线2y kx =+没有公共点的充要条件是 (A)2,2(-∈k )(B) 3,3(-∈k ) (C)k ),2()2,(+∞⋃--∞∈(D) k ),3()3,(+∞⋃--∞∈(4)已知0<a <1,log 2log 3a a x =+,1log 5,log 21log 32a a a y z ==-,则(A)x >y >z(B)z >y >x(C)y >x >z(D)z >x >y(5)已知四边形ABCD 的三个顶点A (0,2),B (-1,-2),C (3,1),且AD BC 2=,则顶点D 的坐标为 (A)(2,27) (B)(2,-21) (C)(3,2) (D)(1,3)(6)设P 为曲线2:23C y x x =++上的点,且曲线C 在点P 处切线倾斜角的取值范围为⎥⎦⎤⎢⎣⎡4,0π,则点P 横坐标的取值范围为 (A)⎥⎦⎤⎢⎣⎡--21,1(B)[-1,0] (C)[0,1](D)⎥⎦⎤⎢⎣⎡1,21(7)4张卡片上分别写有数字1,2,3,4从这4张卡片中随机抽取2张,则取出的2张卡片上的数字之和为奇数的概率为 (A)31 (B)21 (C)32 (D)43 (8)将函数y=2x +1的图象按向量a 平移得到函数y =2x +1的图象,则 (A)a =(-1,-1) (B)a =(1,-1) (C)a =(1,1) (D)a=(-1,1)(9)已知变量,x y 满足约束条件⎪⎩⎪⎨⎧≥+-≤--≤-+,01,013,01x y x y x y 则2z x y =+的最大值为(A )4 (B )2 (C )1 (D )4-(10)一生产过程有4道工序,每道工序需要安排一人照看,现从甲、乙、丙等6名工人中安排4人分别照看一道工序,第一道工序只能从甲、乙两工人中安排1人,第四道工序只能从甲、丙两工人中安排1人,则不同的安排方案共有(A )24种 (B )36种 (C )48种 (D )72种 (11)已知双曲线()222910y m x m -=>的一个顶点到它的一条渐近线的距离为15,则m = (A )1 (B )2 (C )3 (D )4(12)在正方体1111ABCD A BC D -中,E F 、分别为棱11,AA CC 的中点,则在空间中与三条直线A 1D 1、EF 、CD 都相交的直线()A 不存在 (B )有且只有两条 (C )有且只有三条 (D )有无数条第Ⅱ卷(非选择题 共90分)二、填空题:本大题共4小题,每小题4分,共16分.(13)函数23()x y ex +=-∞<<+∞的反函数是 .(14)在体积为43π的球的表面上有A 、B 、C 三点,AB =1,BC =2,A 、C 两点的球面距离为33π,则球心到平面ABC 的距离为 . (15)3621(1)()x x x ++展开式中的常数项为 . (16)设(0,)2x π∈,则函数22sin 1sin 2x y x+=的最小值为 .(17)(本小题满分12分)在△ABC 中,内角,,A B C ,对边的边长分别是,,a b c .已知2,3c C π==. (Ⅰ)若△ABC 的面积等于3,求,a b ;(Ⅱ)若sin 2sin B A =,求△ABC 的面积.(18)(本小题满分12分)某批发市场对某种商品的周销售量(单位:吨)进行统计,最近100周的统计结果如下表所示:周销售量 2 3 4 频数205030(Ⅰ)根据上面统计结果,求周销售量分别为2吨,3吨和4吨的频率; (Ⅱ)若以上述频率作为概率,且各周的销售量相互独立,求 (i )4周中该种商品至少有一周的销售量为4吨的概率; (ii )该种商品4周的销售量总和至少为15吨的概率. (19)(本小题满分12分)如图,在棱长为1的正方体ABCD A B C D ''''-中,AP =BQ =b (0<b <1),截面PQEF ∥A ′D ,截面PQGH ∥AD ′.(Ⅰ)证明:平面PQEF 和平面PQGH 互相垂直;(Ⅱ)证明:截面PQEF 和截面PQGH 面积之和是定值,并求出这个值; (Ⅲ)若12b =,求D ′E 与平面PQEF 所成角的正弦值.(20)(本小题满分12分)已知数列{a n },{b n }是各项均为正数的等比数列,设(N*)nn nb c n a =∈. (Ⅰ)数列{c n }是否为等比数列?证明你的结论;(Ⅱ)设数列{}{}ln ,ln n n a b 的前n 项和分别为,n n S T .若12,,21n n S n a T n ==+求数列{c n }的前n 项和.(21)(本小题满分12分)在平面直角坐标系xOy 中,点P 到两点(0,-3)、(0,3)的距离之和等于4.设点P 的轨迹为C .(Ⅰ)写出C 的方程;(Ⅱ)设直线y =kx +1与C 交于A 、B 两点.k 为何值时?OB OA ⊥此时|AB |的值是多少?(22)(本小题满分14分)设函数()322()31f x ax bx a x a b R =+-+∈、在12,x x x x ==处取得极值,且122x x -=.(Ⅰ)若a =1,求b 的值,并求f (x )的单调区间; (Ⅱ)若a >0,求b 的取值范围.答案1. 答案:D解析:本小题主要考查集合的相关运算知识。
2008高考辽宁数学理科试卷含详细解答(全word版)
2008年普通高等学校招生全国统一考试(辽宁卷)数学(供理科考生使用)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(选择题共60分) 参考公式:如果事件A 、B 互斥,那么 球的表面积公式P(A+B)=P(A)+P(B) S=42R π如果事件A 、B 相互独立,那么 其中R 表示球的半径 P(A ·B)=P(A)·P(B) 球的体和只公式 如果事件A 在一次试验中发生的概率是p ,那么n 次独立重复试验中事件A 恰好发生k 次的概率 V =243R π()(1)(0,1,2,,)k k n kn n P k C P p k n -=-= 其中R 表示球的半径一、选择题 1.已知集合{}30,31x M x N x xx ⎧+⎫=<=-⎨⎬-⎩⎭,则集合{}1x x为( )A.M NB.M NC.()RMN D.()RMN答案:C解析:本小题主要考查集合的相关运算知识。
依题{}{}31,3M x x N x x=-<<=-,∴{|1}M N x x ⋃=<,()RMN ={}1.x x2.135(21)lim(21)n n n n →∞++++-+等于( )A.14 B.12C.1D.2 答案:B解析:本小题主要考查对数列极限的求解。
依题22135(21)1limlim .(21)22n n n n n n n n →∞→∞++++-==++ 3.圆221x y +=与直线2y kx =+没有公共点的充要条件是( )A.(k ∈B.(,(2,)k ∈-∞+∞C.(k ∈D.(,(3,)k ∈-∞+∞答案:C解析:本小题主要考查直线和圆的位置关系。
依题圆221x y +=与直线2y kx =+没有公共点1d ⇔=>⇔(k ∈4.复数11212i i +-+-的虚部是( ) A.15i B.15 C.15i - D.15-答案:B解析:本小题主要考查复数的相关运算及虚部概念。
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2008年(辽宁卷)21. —Did you have a good time in Thailand last week?—______, it was too hot.A. No! reallyB. Yeah, why notC. Oh, greatD. You‟re right22. Peter ______ be really difficult at times even though he‟sa nice person in general.A. shallB. shouldC. canD. must23. We first met on a train in 2000. We both felt immediately that we ______ each other for years.A. knewB. have knownC. had knownD. know24. My neighbor asked me to go for ______ walk, but I don‟tthink I‟ve got ______ energy.A. a; 不填B. the; theC. 不填;theD. a; the25. You have to be a fairly good speaker to ______ listener s‟interest for over an hour.A. holdB. makeC. improveD. receive26. —Could you tell me how to get to Victoria Street?—Victoria Street? ______ is where the Grand Theatre is.A. SuchB. ThereC. ThatD. This27. He was busy writing a story, only ______ once in a whileto smoke a cigarette.A. to stopB. stoppingC. to have stoppedD. having stopped28. ______ hungry I am, I never seem to be able to finish off this loaf of bread.A. WhateverB. WheneverC. WhereverD. However29. —Have you got any job offers?—No. I ______.A. waitedB. had been waitingC. have waitedD. am waiting30. It looks like the weather is changing for ______. Shallwe stick to our plan?A. the worseB. worseC. the worstD. worst31. Please remain ______; the winner of the prize will be announced soon.A. seatingB. seatedC. to seatD. to be seated32. I used to love that film ______ I was a child, but I don‟t feel it that way any more.A. onceB. whenC. sinceD. although33. I like Mr. Miner‟s speech; it was clear and ______ the point.A. atB. onC. toD. of34. —My name is Jonathan. Shall I spell it for you?—______.A. If you don‟t mindB. Not at allC. Take it easyD. Nice to meet you35. Bill wasn‟t happy about the delay of the report by Jason, and ______.A. I was neitherB. neither was IC. I was eitherD. either was I第二节:完形填空(共20小题;每小题1.5分I was a single parent of four small children, working ata low-paid job. Money was always tight, but we had a 36 over our heads, food on the table, clothes on our backs, and if not a lot, always 37 . Not knowing we were poor, my kids(孩子们) just thought I was 38 . I‟ve always been glad about that.It was Christmas time, and although there wasn‟t 39 for a lot of gifts, we planned to celebrate with a family party. But the big 40 for the kids was the fun of Christmas 41 .They planned weeks ahead of time, asking 42 what they wanted for Christmas. Fortunately, I had saved $120 for 43 to share by all five of us.The big 44 arrived. I gave each kid a twenty-dollar bill and 45 them to look for gifts of about four dollars each. Then everyone scattered(散开). We had two hours to shop; then we would 46 back at the “Santa‟s Workshop”.Driving home, everyone was in high Christmas spirits,47 my younger daughter, Ginger, who was unusually48 . She had only one small, flat bag with a few candies —fifty-cent candies! I was so angry, but I didn‟t say anything 49 we got home. I called her into my bedroom and closed the door, 50 to be angry again. This is what she told me.“I was looking 51 thinking of what to buy, and I 52 to read the little cards on the …Giving Trees.‟ One was for a little girl, four years old, and all she 53 for Christmas was a doll(玩具娃娃). So I took the card off the tree and 54 the doll for her. We have so much and she doesn‟t have anything.”I never felt so 55 as I did that day.36. A. roof B. hat C. sky D. star37. A. little B. less C. enough D. more38. A. busy B. seriousC. strict D. kind39. A. effort B. room C. time D. money40. A. improvement B. problemC. surpriseD. excitement41. A. shopping B. travelling C. parties D. greetings42. A. the other B. each otherC. one by oneD. every other one43. A. toys B. clothes C. presents D. bills44. A. day B. chance C. cheque D. tree45. A. forced B. reminded C. invited D. begged46. A. draw B. stay C. move D. meet47. A. including B. besidesC. except D. regarding48. A. quiet B. excited C. happy D. ashamed49. A. since B. after C. while D. until50. A. waiting B. ready C. hoping D. afraid51. A. out B. over C. forward D. around52. A. forgot B. stopped C. failed D. hated53. A. wanted B. did C. got D. played54. A. made B. searched C. bought D. fetched55. A. angry B. rich C. patient D. bitter第三部分:阅读理解(共20小题;每小题2分,满分40分)AI travel a lot, and I find out different “styles”(风格) of directions every time I ask “How can I get to the post office?”Foreign tourists are often confused(困惑) in Japan because most streets there don‟t have names; in Japan, people use landmarks(地标) in their directions instead of street names. For example, the Japanese will say to travelers, “Go straight down to the corner. Turn left at the big hotel and go past a fruit market. The post office is across from the bus stop.”In the countryside of the American Midwest, there are not usually many landmarks. There are no mountains, so the land is very flat; in many places there are no towns or buildings within miles. Instead of landmarks, people will tell you directions and distances. In Kansas or Iowa, for example, people will say, “Go north two miles. Turn east, and then go another mile.”People in Los Angeles, California, have no idea of distance on the map; they measure distance in time, not miles. “How far away is the post office?”you ask. “Oh,”they answer, “it‟s about five minutes from here.” You say, “Yes, but how many miles away is it?” They don‟t know.It‟s true that a person doesn‟t know the answer to your question sometimes. What happens in such a situation? A new Yorker might say, “Sorry, I have no idea.”But in Yucatan, Mexico, no one answers “I don‟t know.” People in Yucatan believe that “I don‟t know” is impolite. They usually give an answer, often a wrong one. A tourist can get very, very lost in Yucatan!56. When a tourist asks the Japanese the way to a certain place, they usually ______.A. describe the place carefullyB. show him a map of the placeC. tell him the names of the streetsD. refer to recognizable buildings and places57. What is the place where people measure distance in time?A. New York.B. Los Angeles.C. Kansas.D. Iowa.58. People in Yucatan may give a tourist a wrong answer ______.A. in order to save timeB. Los Angeles.C. so as to be politeD. for fun59. What can we infer from the text?A. It‟s important for travelers to understand cultural differences.B. It‟s useful for travelers to know how to ask the way properly.C. People have similar understandings of politeness.D. New Yorkers are generally friendly to visitors.BHeroes of Our TimeA good heartDikembe Mutombo grew up in Africa among great poverty and disease. He came to Georgetown University on a scholarship(奖学金) to study medicine —but Coach(教练) John Thompson got a look at Dikembe and had a different idea. Dikembe became a star in the NBA, and a citizen of the United States. But he never forgot the land of his birth, or the duty to share his fortune with others. He built a new hospital in his old hometown in the Congo. Af friend has said of this good-hearted man: “Mutombo believes that God has given him this chance to do great things.”Success and kindnessAfter her daughter was born, Julie Aigner-Clark searched for ways to share her love of music and art with her child. So she borrowed some equipment, and began filming children‟s videos(录像) in her own house. The Baby Einstein Company was born, and in just five years her business grew to more than $20 million in sales. And she is using her success to help others —producing child safety videos with John Walsh of the National Center for Missing and Exploited Children. Julie says of her new program: “I believe it‟s the most important thing that I have ever done. I believe thatchildren have the right to live in a world that is safe.”Bravery and courageA few weeks ago, Wesley Autrey was waiting at a Harlem subway station with his two little girls when he saw a man fall into the path of a train. With seconds to act, Wesley jumped onto the tracks, pulled the man into the space between the rails(铁轨), and held him as the train passed right above their heads. He insists he‟s not a hero. He says: “We have got to show each other some love.”60. What was Mutombo praised for?A. Being a star in the NBA.B. Being a student of medicine.C. His work in the church.D. His willingness to help the needy.61. Mutombo believes that building the new hospital is ______.A. helpful to his personal developmentB. something he should do for his homelandC. a chance for his friends to share his moneyD. a way of showing his respect to the NBA62. What did the Baby Einstein Company do at its beginning?A. Produce safety equipment for children.B. Make videos to help protect children.C. Sell children‟s music and artwork.D. Look for missing and exploited children.63. Why was Wesley Autrey praised as a hero?A. He helped a man get across the rails.B. He stopped a man from destroying the rails.C. He protected two little girls from getting hurt.D. He saved a person without considering his own safety.CTom was one of the brightest boys in the year, with supportive parents. But when he was 15 he suddenly stopped trying. He left school at 16 with only two scores for secondary school subjects. One of the reasons that made it cool for him not to care was the power of his peer(同龄人) group.The lack of right male(男性的) role models in many of their lives —at home and particularly in the school environment(环境) —means that their peers are the only people they have to judge themselves against.They don‟t see men succeeding in society so it doesn‟t occur to them that they could make something of themselves. Without male teachers as a role model, the effect of peer actions and street culture(文化) is all-powerful. Boys want to be part of a club. However, schools can provide the environment for change, and provide the right role models for them. Teachers need to be trained to stop that but not in front of a child‟s peers. You have to do it one to one, because that is when you see the real child.It‟s pointless sending a child home if he or she has done wrong. They see it as a welcome day off to watch television or play computer games. Instead, schools should have a special unit where a child who has done wrong goes for the day and gets advice about his problems —somewhere he can work away from his peers and go home after the other children.64. Why did Tom give up studying?A. He disliked his teachers.B. His parents no longer supported him.C. It‟s cool for boys of his age not to care about studies.D. There were too many subjects in his secondary school.65. What seems to have a bad effect on students like Tom?A. Peer groups.B. A special unit.C. The student judges.D. The home environment.66. What should schools do to help the problem schoolboys?A. Wait for their change patiently.B. Train leaders of their peer groups.C. Stop the development of street culture.D. Give them lessons in a separate area.67. A teacher‟s work is most effective with a schoolboy when he ______.A. is with the boy aloneB. teaches the boy a lessonC. sends the boy home as punishmentD. works together with another teacherDFar from the land of Antarctica(南极洲), a huge shelf of ice meets the ocean. At the underside of the shelf there lives a small fish, the Antarctic cod.For forty years scientists have been curious about that fish. How does it live where most fish would freeze to death? It must have some secret. The Antarctic is not a comfortable place to work and research has been slow. Now it seems we have an answer.Research was begun by cutting holes in the ice and catching the fish. Scientists studied the fish‟s blood and measured its freezing point.The fish were taken from seawater that had a temperature of -1.88℃and many tiny pieces of ice floating in it. The blood of the fish did not begin to freeze until itstemperature was lowered to -2.05℃. That small difference is enough for the fish to live at the freezing temperature of the ice-salt mixture.The scientists‟ next research job was clear: Find out what in the fish‟s blood kept it from freezing. Their search led to some really strange thing made up of a protein(蛋白质) never before seen in the blood of a fish. When it was removed, the blood froze at seawater temperature. When it was put back, the blood again had its antifreeze quality and a lowered freezing point.Study showed that it is an unusual kind of protein. It has many small sugar molecules(分子) held in special positions within each big protein molecule. Because of its sugar content. It is called a glycoprotein. So it has come to be called the antifreeze fish glycoprotein. Or AFGP.68. What is the text mainly about?A. The terrible conditions in the Antarctic.B. A special fish living in freezing waters.C. The ice shelf around Antarctica.D. Protection of the Antarctic cod.69. Why can the Antarctic cod live at the freezing temperature?A. The seawater has a temperature of -1.88℃.B. It loves to live in the ice-salt mixture.C. A special protein keeps it from freezing.D. Its blood has a temperature lower than -2.05℃.70. What does the underlined word “it” in Paragraph 5 refer to?A. A type of ice-salt mixture.B. A newly found protein.C. Fish blood.D. Sugar molecule.71. What does “glycol-” in the underlined word “glycoprotein” in the last paragraph mean?A. sugarB. iceC. bloodD. moleculeEIf you boss asks you to work in Moscow this year, he‟d better offer you more money to do so —or even double that depending on where you live now. That‟s because Moscow has just been found to be the world‟s most expensive city for the second year in a row by Mercer Human Resources Consulting.Using the cost of living in New York as a base, Mercer determined Moscow is 34.4 percent more expensive including the cost of housing, transportation, food, clothing, household goods and entertainment(娱乐).A two-bedroom flat in Moscow now costs $4,000 a month; a CD $24.83, and an international newspaper $6.30, according to Mercer. By comparison, a fast food meal with a hamburger(汉堡包) is a steal at $4.80.London takes the No.2 place, up from No.5 a year ago, thanks to higher cost of housing and a stronger British pound relative to the dollar. Mercer estimates(估算) London is 26 percent more expensive than New York these days. Following London closely are Seoul and Tokyo, both of which are 22 percent more expensive than New York, while No.5 Hong Kong is 19 percent more costly.Among North American cities, New York and Los Angeles are the most expensive and are the only two listed in the top 50 of the world‟s most expensive cities. But both have fallen since last year‟s study —New York came in15th, down from 10th place, while Los Angeles fell to 42nd from 29th place a year ago. San Francisco came in a distant third at No. 54, down 20 places from a year earlier.Toronto, meanwhile, is Canada‟s most expensive city but fell 35 places to take 82nd place worldwide. In Australia, Sydney is the priciest place to live in and No. 21 worldwide.72. What do the underlined words “a steal” in Paragraph 3 mean?A. an act of stealingB. something deliciousC. something very cheapD. an act of buying73. London has become the second most expensive city because of ______.A. the high cost of clothingB. the stronger pound against the dollarC. its expensive transportationD. the high prices of fast food meals74. Which city is the third most expensive on the list?A. Tokyo.B. Hong Kong.C. Moscow.D. Sydney.75. Which city has dropped most on the list in North America?A. New York.B. Los Angeles.C. San Francisco.D. Toronto.参考答案第一部分1. B2. A3. C4. A5. B6. A7.C 8. C 9. A 10. A11.C 12. C 13. B 14. B 15. C 16. A17. B 18. B 19. A 20.C第二部分21. A 22. C 23. C 24. D 25. A 26. C27. B 28. D 29. D 30. A31. B 32. B 33. C 34. A 35. B 36. A37. C 38. C 39. D 40. D41. A 42. B 43. C 44. A 45. B 46. D47. C 48. A 49. D 50. B51. D 52. B 53. A 54. C 55. B第三部分56. D 57. B 58. C 59. A 60. D 61. B62. C 63. D 64. C 65. A66. D 67. A 68. B 69. C 70. B 71. A72. C 73. B 74. A 75. D第四部分第一节短文改错It is five years now since I graduate from No.3 High76. graduatedSchool. Last Saturday, the class that I was on held a77. inget-together, which took us a long time ∧ prepare. It 78.towas indeed not easy to get in the touch with everybody79.去掉theand set a well time for all of us. We all enjoyed80. goodthis precious day greatly, remember the time we spent 81. rememberingtogether and the people they were familiar with. It was 82. wea pity which some of us were not present as they had 83. thatgone abroad for further studies, but they called back84. √or sent greeting card from different places.85. cards第二节写作One possible versionWe all want to grow up happily and healthily, and for this goal we must do several things.Firstly, we should develop a good attitude to life. Life consists of not only sunshine but also hard times. We should be brave in front of difficulties. Secondly, we must study hard because knowledge is power. If we have the power. We can help to build our country and enjoy life better. In order to study well, we need to do sports so that we can keep fit. We can go running, play ball games or simply take a walk after a day‟s study. If we do those things well, we will be able to grow up happily and healthily.。