2004年高考试题全国卷2

2004年高考试题全国卷2
2004年高考试题全国卷2

2004年高考试题全国卷2 理科数学(必修+选修Ⅱ)

(四川、吉林、黑龙江、云南等地区)

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,只有一个选项是符合题目要求的. (1)已知集合M ={x|x2<4},N ={x|x2-2x -3<0},则集合M ∩N = (A ){x|x <-2} (B ){x|x >3} (C ){x|-1<x <2} (D ){x|2<x <3}

(2)542lim 2

21-+-+→x x x x n =

(A )21

(B )1

(C )52 (D )41 (3)设复数ω=-21+23

i ,则1+ω=

(A )–ω (B )ω2

(C )

ω1-

(D )21

ω

(4)已知圆C 与圆(x -1)2+y2=1关于直线y =-x 对称,则圆C 的方程为 (A )(x +1)2+y2=1 (B )x2+y2=1 (C )x2+(y +1)2=1 (D )x2+(y -1)2=1

(5)已知函数y =tan(2x +φ)的图象过点(12π

,0),则φ可以是

(A )-6π (B )6π (C )-12π (D )12π

(6)函数y =-ex 的图象

(A )与y =ex 的图象关于y 轴对称 (B )与y =ex 的图象关于坐标原点对称 (C )与y =e -x 的图象关于y 轴对称 (D )与y =e -x 的图象关于坐标原点对称

(7)已知球O 的半径为1,A 、B 、C 三点都在球面上,且每两点间的球面距离为2π

,则球心O 到平面ABC 的距离为

(A )31 (B )33 (C )32 (D )36

(8)在坐标平面内,与点A(1,2)距离为1,且与点B(3,1)距离为2的直线共有

(A )1条 (B )2条 (C )3条 (D )4条

(9)已知平面上直线l 的方向向量)53,54(-=e ,点O(0,0)和A(1,-2)在l 上的射影分别是O1和A1,则11A O =λe ,其中λ

(A )511 (B )-511

(C )2 (D )-2

(10)函数y =xcosx -sinx 在下面哪个区间内是增函数

(A )(2π,23π) (B )(π,2π) (C )(23π,25π

) (D )(2π,3π)

(11)函数y =sin4x +cos2x 的最小正周期为

(A )4π (B )2π

(C )π (D )2π

(12)在由数字1,2,3,4,5组成的所有没有重复数字的5位数中,大于23145且小于43521的数共有 (A )56个 (B )57个 (C )58个 (D )60个

二、填空题:本大题共4小题,每小题4分,共16分.把答案填在题中横线上.

(13)从装有3个红球,2

ξ的概率分布为

(14)设x ,y 满足约束条件

??

?

??≤-≥≥,y x y ,x ,x 120

则z =3x +2y 的最大值是 .

(15)设中心在原点的椭圆与双曲线2x2-2y2=1有公共的焦点,且它们的离心率互为倒数,则该椭圆的方程是 . (16)下面是关于四棱柱的四个命题:

①若有两个侧面垂直于底面,则该四棱柱为直四棱柱

②若两个过相对侧棱的截面都垂直于底面,则该四棱柱为直四棱柱 ③若四个侧面两两全等,则该四棱柱为直四棱柱

④若四棱柱的四条对角线两两相等,则该四棱柱为直四棱柱 其中,真命题的编号是 (写出所有真命题的编号).

解答题:本大题共6个小题,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)

已知锐角三角形ABC 中,sin(A +B)=53,sin(A -B)=51

(Ⅰ)求证:tanA =2tanB ;

(Ⅱ)设AB =3,求AB 边上的高.

(18)(本小题满分12分)

已知8个球队中有3个弱队,以抽签方式将这8个球队分为A 、B 两组,每组4个.求 (Ⅰ)A 、B 两组中有一组恰有两个弱队的概率; (Ⅱ)A 组中至少有两个弱队的概率.

(19)(本小题满分12分)

数列{an}的前n 项和记为Sn ,已知a1=1,an +1=n n 2

+Sn (n =1,2,3,…).证明:

(Ⅰ)数列{n S n

}是等比数列;

(Ⅱ)Sn +1=4an .

(20)(本小题满分12分) .

如图,直三棱柱ABC-A1B1C1中,∠ACB =90o ,AC =1,CB =2,侧棱AA1=1,侧面AA1B1B 的两条对角线交点为D ,B1C1的中点为M .

(Ⅰ)求证:CD ⊥平面BDM ;

(Ⅱ)求面B1BD 与面CBD 所成二面角的大小.

(21)(本小题满分12分)

给定抛物线C :y2=4x ,F 是C 的焦点,过点F 的直线l 与C 相交于A 、B 两点. (Ⅰ)设l 的斜率为1,求与夹角的大小;

(Ⅱ)设FB =AF λ,若λ∈[4,9],求l 在y 轴上截距的变化范围.

(22)(本小题满分14分)

已知函数f(x)=ln(1+x)-x ,g(x)=xlnx . (1)求函数f(x)的最大值;

(2)设0<a <b ,证明:0<g(a)+g(b)-2g(2b

a +)<(

b -a)ln2.

2004年高考试题全国卷2 理科数学(必修+选修Ⅱ)

(四川、吉林、黑龙江、云南等地区) 答案:

一、选择题:本大题共12小题,每小题5分,共60分.

(1)C (2)A (3)C (4)C (5)A (6)D (7)B (8)B (9)D (10)B (11)B (12)C

二、填空题:本大题共4小题,每小题4分,共16分.

(13)0.1,0.6,0.3 (14)5 (15)21

x2+y2=1 (16)②④

17.(I)证明:∵sin(A+B)=53,sin(A-B)=51

∴???

???

?

=-=+51sin cos cos sin 5

3sin cos cos sin B A B A B A B A ???????

=

=?5

1sin cos 52

cos sin B A B A ?

2tan tan =B A

,∴B A tan 2tan =.

(II)解:∵2π

53)sin(=

+B A , ∴54)cos(-=+B A , 43

)tan(-=+B A

即43

tan tan 1tan tan -

=-+B

A B A ,将B A tan 2tan =代入上式并整理得01tan 4tan 22=--B B 解得

262tan ±=

B ,因为B 为锐角,所以26

2tan +=

B ,∴B A tan 2tan =

设AB 上的高为CD ,则AB=AD+DB=623tan tan +=

+CD

B CD A CD ,由AB=3得CD=2+

6

故AB 边上的高为2+6

18.(I) 解:有一组恰有两支弱队的概率

7624

8

2523=C C C (II)解:A 组中至少有两支弱队的概率

21

4

81

533482523=+C C C C C C 19.(I )证: 由a1=1,an+1=n n 2

+Sn(n=1,2,3,…),

知a2=112+S1=3a1,224212==a S , 1

11

=S ,∴21212

=S S

又an+1=Sn+1-Sn(n=1,2,3,…),则Sn+1-Sn=n n 2+Sn(n=1,2,3,…),∴nSn+1=2(n+1)Sn, 2

11

=++n S n S n

n (n=1,2,3,…).故数列{n S n }是首

项为1,公比为2的等比数列

(II )解:由(I )知,)

2(14111≥-?=+-+n n S n S n n ,于是Sn+1=4(n+1)·11

--n S n =4an(n 2≥)

又a2=3S1=3,则S2=a1+a2=4=4a1,因此对于任意正整数n ≥1都有Sn+1=4an.

20.解法一:(I)如图,连结CA1、AC1、CM ,则CA1=2, ∵CB=CA1=2,∴△CBA1为等腰三角形, 又知D 为其底边A1B 的中点,∴CD ⊥A1B , ∵A1C1=1,C1B1=2,∴A1B1=3, 又BB1=1,∴A1B=2,

∵△A1CB 为直角三角形,D 为A1B 的中点,CD=21

A1B=1,

CD=CC1

又DM=21AC1=22

,DM=C1M ,∴△CDN ≌△CC1M ,∠CDM=∠CC1M=90°,即CD ⊥DM ,

因为A1B 、DM 为平面BDM 内两条相交直线,所以CD ⊥平面BDM (II)设F 、G 分别为BC 、BD 的中点,连结B1G 、FG 、B1F ,

则FG ∥CD ,FG=21CD ∴FG=21

,FG ⊥BD.

由侧面矩形BB1A1A 的对角线的交点为D,知BD=B1D=21

A1B=1,

所以△BB1D 是边长为1的正三角形,于是B1G ⊥BD ,B1G=23,

∴∠B1GF 是所求二面角的平面角

又B1F2=B1B2+BF2=1+(22

)2=23.

∴cos ∠B1GF=

33

21

23223)21()23(

22212

1221-=?

?-+=

?-+FG

G B F

B FG G B

即所求二面角的大小为π-arccos 33

解法二:如图以C 为原点建立坐标系

(I):B(2,0,0),B1(2,1,0),A1(0,1,1),D(22,21,21

), M(22,1,0),=CD (22,21,21

),=B A 1(2,-1,-1), =DM (0,21,-21

),,0,01=?=?A

∴CD ⊥A1B,CD ⊥DM.

因为A1B 、DM 为平面BDM 内两条相交直线, 所以CD ⊥平面BDM

(II):设BD 中点为G ,连结B1G ,则G ),41,41,423(

=BD (-22,21,21),

=B 1),41

,43,42(--∴01=?B ,∴BD ⊥B1G ,又CD ⊥BD ,∴CD 与B 1的夹角θ等于所求二面角的平面角,

cos .3

3

11-

==

θ

所以所求二面角的大小为π-arccos 33

A'

C'

21.解:(I )C 的焦点为F(1,0),直线l 的斜率为1,所以l 的方程为y=x-1. 将y=x-1代入方程y2=4x ,并整理得x2-6x+1=0. 设A(x1,y1),B(x2,y2),则有x1+x2=6,x1x2=1,

?=(x1,y1)·(x2,y2)=x1x2+y1y2=2x1x2-(x1+x2)+1=-3. 41

]16)(4[||||2121212

2222121=+++=

+?+=

?x x x x x x y x y x OB OA

cos<,>=.41

41

3-

=

所以OA 与OB 夹角的大小为π-arccos 4141

3.

解:(II)由题设知AF

FB λ=得:(x2-1,y2)=λ(1-x1,-y1),即??

?-=-=-)

2()1()1(11212 y y x x λλ

由 (2)得y22=λ2y12, ∵y12=4x1,y22=4x2,∴x2=λ2x1 (3)

联立(1)(3)解得x2=λ.依题意有λ>0. ∴B(λ,2λ)或B(λ,-2λ),又F(1,0),

得直线l 的方程为(λ-1)y=2λ(x-1)或(λ-1)y=-2λ(x-1)

当λ∈[4,9]时,l 在y 轴上的截距为12-λλ或-12-λλ

由12-λλ=1212-+

+λλ,可知1

2-λλ在[4,9]上是递减的,

∴≤4312-λλ34≤,-≤34-1

2-λλ

43-

≤ 直线l 在y 轴上截距的变化范围是]

34,43[]4

3,34[ -- 22.(I)解:函数f(x)的定义域是(-1,∞),'f (x)=1

11

-+x .令'f (x)=0,解得x=0,当-10,当x>0时,'f (x)<0,又f(0)=0,

故当且仅当x=0时,f(x)取得最大值,最大值是0

(II)证法一:g(a)+g(b)-2g(2b a +)=alna+blnb-(a+b)ln 2b a +=a b a b

b b

a a +++2ln

2ln . 由(I)的结论知

ln(1+x)-x<0(x>-1,且x ≠0),由题设0

021,02<-<->-b

b

a a a

b ,因此

a a

b a a b b a a 2)21l n (2ln

-->-+-=+,b b

a b b a b a b 2)21ln(2ln -->-+-=+.

所以a

b a b b b a a +++2ln 2ln

>-022=---b

a a

b .

又,22b b a b a a +<+ a b a b b b a a +++2ln 2ln

a +)<(b-a)ln2.

(II)证法二:g(x)=xlnx,1ln )('+=x x g ,设F(x)= g(a)+g(x)-2g(2x

a +),

.2ln ln )]'2(

[2)(')('x

a x x a g x g x F +==+-=当0a 时,0)('>x F 因此F(x)在(a,+∞)

上为增函数从而,当x=a 时,F(x)有极小值F(a)因为F(a)=0,b>a,所以F(b)>0,即0

a +).

设G(x)=F(x)-(x-a)ln2,则

).ln(ln 2ln 2ln

ln )('x a x x

a x x G +-=-+-=当x>0时,0)('

G(a)=0,b>a,所以G(b)<0.即g(a)+g(b)-2g(2b

a +)<(b-a)ln2.

2018年高考英语真题(新课标全国一卷)有答案

绝密★启用前 2018年普通高等学校招生全国统一考试(新课标全国I卷) 英语 (考试时间:120分钟试卷满分:150分) 注意事项: 1. 答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。 2. 回答选择题时,选出每小题答案后,用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。回答非选择题时,将答案写在答题卡上,写在本试卷上无效。 3. 考试结束后,将本试卷和答题卡一并交回。 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。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. 18. C. £ 9. 15. 答案是C。 1.what will James do tomorrow ? A.Watch a TV program. B.Give a talk. C.Write a report. 2.What can we say about the woman? A.She's generour. B.She's curious. C.She's helpful. 3.When does the traif leave?https://www.360docs.net/doc/e32105731.html, A.At 6:30. B.At8:30. C.At 10:30. 4.How does the wonar sRwr?m A.By car. B.On foot. C.By bike 5.What is the probable relationship between the speakers? A.Classmates. B.Teacher and student. C.Doctor and patient. 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。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段材料,回答第6、7题。 6.What does the woman regret? A.Giving up her research. B.Dropping out of college. C.Changiny her major.

2019全国1卷高考英语试题及答案

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2019年全国I卷英语高考真题

英语 注意事项: 1.答卷前,考生务必将自己的姓名、准考证号填写在答题卡和试卷指定位置上。 2.回答选择题时,选出每小题答案后,用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。回答非选择题时,将答案写在答题卡上,写在本试卷上无效。 3.考试结束后,将本试卷和答题卡一并交回。 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。 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。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:Ho w much i s the sh i r t? B. Buy a pa i r o f gym shoes. C. Change h i s work schedu le. C.When to l eave. A. She migh t wan t a t i cke t. B. She i s look ing f or the man. C. She has an ex t ra t i cke t. 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

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普通高等学校招生全国统一考试(全国卷1) 英语 (考试时间:120分钟试卷满分:150分) 注意事项: 1.本试卷由四个部分组成。其中,第一、二部分和第三部分的第一节为选择题。第三部分的第二节和第四部分为非选择题。 2.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。 3.回答选择题时,选出每小题答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑;回答非选择题时,将答案写在答题卡上,写在本试卷上无效。 4.考试结束后,将本试卷和答题卡一并交回。 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. 18. C. £ 9. 15. 答案是C。 1.What does the woman think of the movie? A.It’s amusing B.It’s exciting C.It’s disappointing 2.How will Susan spend most of her time in France? A. Traveling around B.Studying at a school C.Looking after her aunt 3.What are the speakers talking about? A. Going out B.Ordering drinks C.Preparing for a party 4.Where are the speakers? A.In a classroom B.In a library C.In a bookstore 5.What is the man going to do ?

2017全国卷Ⅰ英语高考试题word版

绝密★启封前 2017年普通高等学校招生全国统一考试(新课标I) 英语 (考试时间:120分钟试卷满分:150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) 1.What does the woman think of the movie? A.It’s amusing B.It’s exciting C.It’s disappointing 2.How will Susan spend most of her time in France? A.Traveling around B.Studying at a school C.Looking after her aunt 3.What are the speakers talking about? A.Going out B.Ordering drinks C.Preparing for a party 4.Where are the speakers? A.In a classroom B.In a library C.In a bookstore 5.What is the man going to do? A.Go on the Internet B.Make a phone call C.Take a train trip 第二节(共15小题;每小题1.5分,满分22.5分) 听第6段材料,回答6、7题。 6.What is the woman looking for? A.An information office B.A police station C.A shoe repair shop 7.What is the Town Guide according to the man? A.A brochure B.A newspaper

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