高考全国二卷理科数学题及其答案.docx
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高考全国二卷理科数学题及其答案
2008年普通高等学校招生全国统一考试(全国卷2数学)
理科数学(必修+选修Ⅱ)
第Ⅰ卷
一、选择题
1.设集合M{mZ|3m2},N{nZ|1≤n≤3},则MN()
A.0,1B.1,0,1C.0,1,2D.1,0,1,2
2.设a,bR且b0,若复数
3
(abi)是实数,则()
A.
22
baB.
3
22
abC.
3
22
baD.
9
22
a9b
3.函数
1
f(x)x
x
的图像关于()
A.y轴对称B.直线yx对称
C.坐标原点对称D.直线yx对称
4.若
13
x(e,1),alnx,b2lnx,clnx,则()
A.a
≥
,yx
≤
5.设变量x,y满足约束条件:
x2y2,则zx3y的最小值()
,
≥
x2.
A.2B.4C.6D.8
6.从20名男同学,10名女同学中任选3名参加体能测试,则选到的3名同学中既有男同学又有女
同学的概率为()
A.
9
29
B.
10
29
C.
19
29
D.
20
29
7.
64
(1x)(1x)的展开式中x的系数是()
A.4B.3C.3D.4
8.若动直线xa与函数f(x)sinx和g(x)cosx的图像分别交于M,N两点,则MN的最
大值为()
A.1B.2C.3D.2
22
xy
9.设a1,则双曲线221
的离心率e的取值范围是()a(a1)
A.(2,2)B.(2,5)C.(2,5)D.(2,5)
第1页(共11页)
10.已知正四棱锥SABCD的侧棱长与底面边长都相等,E是SB的中点,则AE,SD所成的角
的余弦值为()
1232
A.B.C.D.
333
3
11.等腰三角形两腰所在直线的方程分别为xy20与x7y40,原点在等腰三角形的底
边上,则底边所在直线的斜率为()
A.3B.2C.
1
3
D.
1
2
12.已知球的半径为2,相互垂直的两个平面分别截球面得两个圆.若两圆的公共弦长为2,则两圆
的圆心距等于()
A.1B.2C.3D.2
第Ⅱ卷
二、填空题:
本大题共4小题,每小题5分,共20分.把答案填在题中横线上.
13.设向量a(1,2),b(2,3),若向量ab与向量c(4,7)共线,则.
14.设曲线
ax
ye在点(0,1)处的切线与直线x2y10垂直,则a.
15.已知F是抛物线
2
C:
yx的焦点,过F且斜率为1的直线交C于A,B两点.设FAFB,
4
则FA与FB的比值等于.
16.平面内的一个四边形为平行四边形的充要条件有多个,如两组对边分别平行,类似地,写出空
间中的一个四棱柱为平行六面体的两个充要条件:
充要条件①;
充要条件②.
(写出你认为正确的两个充要条件)
三、解答题:
本大题共6小题,共70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10分)
在△ABC中,cos
5
B,
13
cos
4
C.
5
(Ⅰ)求sinA的值;
(Ⅱ)设△ABC的面积
33
△,求BC的长.
S
ABC
2
18.(本小题满分12分)
购买某种保险,每个投保人每年度向保险公司交纳保费a元,若投保人在购买保险的一年度内出险,
则可以获得10000元的赔偿金.假定在一年度内有10000人购买了这种保险,且各投保人是否出险
相互独立.已知保险公司在一年度内至少支付赔偿金10000元的概率为
4
10
10.999.
(Ⅰ)求一投保人在一年度内出险的概率p;
(Ⅱ)设保险公司开办该项险种业务除赔偿金外的成本为50000元,为保证盈利的期望不小于0,
求每位投保人应交纳的最低保费(单位:
元).
第2页(共11页)
19.(本小题满分12分)
如图,正四棱柱ABCDABCD中,AA12AB4,点E在CC1上且C1E3EC.
1111
(Ⅰ)证明:
AC平面BED;
1
D1
C1
(Ⅱ)求二面角
ADEB的大小.
1
A1B
1
E
DC
AB
20.(本小题满分12分)
设数列
a的前n项和为S.已知
nn
aa,
1
n
a1S3,
nn
*
nN.
n
(Ⅰ)设bS3,求数列
nn
b的通项公式;
n
(Ⅱ)若
a≥a,
n1n
*
nN,求a的取值范围.
21.(本小题满分12分)
设椭圆中心在坐标原点,A(2,0),B(0,1)是它的两个顶点,直线ykx(k0)与AB相交于点D,
与椭圆相交于E、F两点.
(Ⅰ)若ED6DF,求k的值;
(Ⅱ)求四边形AEBF面积的最大值.
22.(本小题满分12分)
sinx
设函数f(x)
.2cosx
(Ⅰ)求f(x)的单调区间;
(Ⅱ)如果对任何x≥0,都有f(x)≤ax,求a的取值范围.
第3页(共11页)
2008年参考答案和评分参考
一、选择题
1.B2.A3.C4.C5.D6.D
7.B8.B9.B10.C11.A12.C
部分题解析:
2.设a,bR且b0,若复数
3
(abi)是实数,则()
A.
22
baB.
3
22
abC.
3
22
baD.
9
22
ab,
9
解:
33223
(abi)a3abi3a(bi)(bi)(←考查和的立方公式,或二项式定理)
3223
(a3ab)(3abb)i(←考查虚数单位i的运算性质)
R(←题设条件)
∵a,bR且b0
∴
23
3abb0(←考查复数与实数的概念)
∴
22
ba.
3
故选A.
6.从20名男同学,10名女同学中任选3名参加体能测试,则选到的3名同学中既有男同学又
有女同学的概率为()
A.
9
29
B.
10
29
C.
19
29
D.
20
29
思路1:
设事件A:
“选到的3名同学中既有男同学又有女同学”,其概率为:
P(A)
2112
CCCC
20102010
3
C
30
(←考查组合应用及概率计算公式)
2019109
1020
2121
302928
(←考查组合数公式)
321
10191010109
(←考查运算技能)102914
20
29
故选D.
思路2:
设事件A:
“选到的3名同学中既有男同学又有女同学”,
事件A的对立事件为A:
“选到的3名同学中要么全男同学要么全女同学”
其概率为:
P(A)1P(A)(←考查对立事件概率计算公式)
1
33
CC
2010
3
C
30
(←考查组合应用及概率计算公式)
第4页(共11页)
201981098
1
321321
302928
(←考查组合数公式)
321
2019181098
(←考查运算技能)302928
20
29
故选D.
7.已知球的半径为2,相互垂直的两个平面分别截球面得两个圆.若两圆的公共弦长为2,则
两圆的圆心距等于()
A.1B.2C.3D.2
分析:
如果把公共弦长为2的相互垂直的两个截球面圆,想成一般情况,问题解决起来就比较
麻烦,许多考生就是因为这样思考的,所以浪费了很多时间才得道答案;但是,如果把公共弦长为
2的相互垂直的两个截球面圆,想成其中一个恰好是大圆,那么两圆的圆心距就是球心到另一个小
圆的距离3,问题解决起来就很容易了.
二、填空题
13.214.25.322
16.两组相对侧面分别平行;一组相对侧面平行且全等;对角线交于一点;底面是平行四边形.
注:
上面给出了四个充要条件.如果考生写出其他正确答案,同样给分.
三、解答题
17.解:
(Ⅰ)由
cos
5
B,得
13
sin
12
B,
13
由cos
4
C,得
5
sin
3
C.
5
所以
33
sinAsin(BC)sinBcosCcosBsinC.···········································5分
65
(Ⅱ)由
33
S
△得
ABC
2
133
ABACsinA,
22
由(Ⅰ)知sin
33
A,
65
故ABAC65,·······································································································8分
又
ABsinB20
ACAB
sinC13
,
故
20
13
2
AB65,
13
AB.
2
所以BC
ABsinA11
sinC2
.·····················································································10分
18.解:
各投保人是否出险互相独立,且出险的概率都是p,记投保的10000人中出险的人数为,
第5页(共11页)
4
则~B(10,p).
(Ⅰ)记A表示事件:
保险公司为该险种至少支付10000元赔偿金,则A发生当且仅当0,
·····································································································································2分
P(A)1P(A)
1P(0)
4
10
1(1p),
又
4
10
P(A)10.999,
故p0.001.··············································································································5分
(Ⅱ)该险种总收入为10000a元,支出是赔偿金总额与成本的和.
支出10000500,0
盈利1000a0(1000050,0
盈利的期望为E1000a0100E005,0·················································9分
由
43
~B(10,10)知,
3
E1000010,
444
E10a10E510
44434
10a101010510.
E≥
0
444
10a1010510≥0
a≥
1050
a≥(元).
15
故每位投保人应交纳的最低保费为15元.·································································12分
19.解法一:
D1
依题设知AB2,CE1.C1
(Ⅰ)连结AC交BD于点F,则BDAC.
A1B
1
由三垂线定理知,
BDAC.
1
···················································································3分
HE
在平面ACA内,连结EF交A1C于点G,
1
GD
AAACC
122
由于,AB
FFCCE
第6页(共11页)
故
Rt△AAC∽Rt△FCE,
1
AACCFE,
1
CFE与
FCA互余.
1
于是
ACEF.
1
AC与平面BED内两条相交直线BD,EF都垂直,
1
所以
AC平面BED.·······························································································6分
1
(Ⅱ)作GHDE,垂足为H,连结AH.由三垂线定理知AHDE,
11
故
AHG是二面角
1
ADEB的平面角.
1
·······························································8分
22
EFCFCE3,
CG
CECF
EF
2
3
,
223
EGCECG.
3
EG11EFFD2
,GH.EF33DE15
又
22
A1CAA1AC26,
56
AGACCG.
11
3
AG
1
tanAHG55
1
HG
.
所以二面角
ADEB的大小为arctan55.
1
························································12分
z
解法二:
以D为坐标原点,射线DA为x轴的正半轴,
D1
C1
建立如图所示直角坐标系Dxyz.
A1B1
依题设,
B(2,2,0),C(0,2,0),E(0,2,1),A(2,0,4).
1
E
DE(0,2,1),DB(2,2,0),
x
D
AB
C
y
A1C(2,2,4),DA1(2,0,4).················································································3分
(Ⅰ)因为
A1CDB0,A1CDE0,
故
ACBD,A1CDE.
1
又DBDED,
第7页(共11页)
所以
AC平面DBE.·······························································································6分
1
(Ⅱ)设向量n(x,y,z)是平面
DAE的法向量,则
1
nDE,
nDA.
1
故2yz0,2x4z0.
令y1,则z2,x4,n(4,1,2).······························································9分
n等于二面角
,AC
1
ADEB的平面角,
1
cos
nAC
,
1
n
n
AC
1
AC
1
14
42
.
所以二面角
ADEB的大小为arccos
1
14
42
.·························································12分
20.解:
(Ⅰ)依题意,
n
S1Sa1S3,即
nnnn
n
S12S3,
nn
由此得
n1n
SS.···················································································4分
132(3)
nn
因此,所求通项公式为
nn1
bS3(a3)2,
nn
*
nN.①········································································6分
(Ⅱ)由①知
nn1
S3(a3)2,
n
*
nN,
于是,当n≥2时,
aSS
nnn
1
nn1n1n23(a3)23(a3)2
n1n2
23(a3)2,
n1n2
a1a43(a3)2
nn
n2
n
23
212a3,
2
当n≥2时,
n23
a≥a12a3≥0
n1n
2
第8页(共11页)
a≥.
9
又
a2a13a1.
综上,所求的a的取值范围是9,.·································································12分
21.(Ⅰ)解:
依题设得椭圆的方程为
2
x
4
21
y,
直线AB,EF的方程分别为x2y2,ykx(k0).··········································2分
如图,设
D(x,kx),E(x,kx),F(x,kx),其中
001122
xx,
12
且
x,x满足方程
12
22
(14k)x4,
y
B
F
故
xx
21
2
14k
2
.①
E
O
D
A
x
由ED6DF知
x0x16(x2x0),得
1510
x(6xx)x
0212
77714
k
2
;