苏北四市2011届高三第一次调研考试数学试题及答案

发布时间:   来源:文档文库   
字号:

苏北四市2011届高三第一次调研考试数学Ⅰ试题


考生在答题前请认真阅读本注意事项及各题答题要求
包含填空题(第1题——第14题)解答题(第15题——第20题)1.本试卷共4页,
考试时间为120分钟。考试结束后,请将本卷和答题卡一并交回。本卷满分160分,
2.答题前,请您务必将自己的姓名、准考证号用0.5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置。
3.请在答题卡上按照顺序在对应的答题区域内作答,在其他位置作答一律无效。作答必须用0.5毫米黑色墨水的签字笔。请注意字体工整,笔迹清楚。4.请保持答题卡卡面清洁,不要折叠、破损。
参考公式:
s
2
1
n
样本数据x1,x2,,xn的方差
n
(xix
2
x
1
n
i1
,其中
n
xi
i1
.
一、填空题:本大题共14小题,每小题5分,共计70分.请把答案填写在答题卡相应位置上.1.若复数z1
1i
z2
24i
,其中i是虚数单位,则复数z1z2的虚部是.
,若A
B
2.已知集合A3.若函数
(,0]
B
m
{1,3,a}
,则实数a的取值范围是.
f(x
221
x
为奇函数,则实数m
(2,0

.
开始
S0,n1
4.若抛物线的焦点坐标为
,则抛物线的标准方程

5.从某项综合能力测试中抽取10人的成绩,统计如下表,则这10人成绩的方差为.
分数人数
53
41
31
23
12
n12

N输出S结束
SS
Yn



6.如图是一个算法的流程图,则最后输出的S
7.已知直线l1ax3y
10

nn2
(第6题图)
.
l22x
(a1y10
,若l1l2,则实数a的值是
8.一个质地均匀的正四面体(侧棱长与底面边长相等的正三棱锥)玩具的四个面上分别标有1

y
yf(x

yf(x


234这四个数字.若连续两次抛掷这个玩具,则两次向下的面上的数字之积为偶数的概率是.9.已知
cos(
π4
35
(
π2
,π

,则cos
f(x


10.已知函数y则曲线y
f(x
及其导函数y
的图象如图所示,
f(x
在点P处的切线方程是
MAMBMC0满足
11.在△ABC中,点M
,若

ABACmAM0
,则实数m的值为
12.mn是两条不同的直线,是三个不同的平面,给出下列命题:①若m



,则m,则,则
n


②若m//m③若④若




m
m//n,则//
上面命题中,真命题的序号是(写出所有真命题的序号).w.w.k.s13.若关于x的不等式
14.已知数列{an}{bn}满足a1
1
(2x1ax
22
的解集中的整数恰有2个,则实数a的取值范围
a2
2010
2
b1
2
且对任意的正整数i,
j,k,l
ij
kl
1
时,都有
aibjakbl
,则
2010

i1
(aibi
的值是
二、解答题:本大题共6小题,共计90分.请在答题卡指定位置内作答,解答时应写出文字说明、证明过程或演算步骤.15.(本小题满分14分)
如图,在△ABC中,已知AB3AC6BC1)求证:DC

2)求ABDC
2BD
7
ADBAC平分线.
A

的值.
B
D(第15题图)
C





16.(本小题满分14分)如图,在四棱锥P.求证:
1EF∥平面PBD2)平面PEF⊥平面PAC.
17.(本小题满分14分)
在各项均为正数的等比数列{an}中,已知a21)求数列{an}的通项公式;2)设bn
log3an
ABCD
中,四边形ABCD是菱形,PB
PD
,且EF分别是BCCD的中
P
A
FD
B
E(第16题图)
C
2a13
,且3a2a45a3成等差数列.
,求数列
anbn的前n项和Sn.

18.(本小题满分16分)
x
2
已知椭圆E
8

y
2
4
1
的左焦点为F,左准线lx轴的交点是圆C的圆心,圆C恰好经过坐
标原点O,设G是圆C上任意一点.1)求圆C的方程;
2)若直线FG与直线l交于点T,且G为线段FT的中点,求直线FG被圆C所截得的弦长;
GF
12
3)在平面上是否存在一点P,使得GP.

19.(本小题满分16分)

?若存在,求出点P坐标;若不存在,请说明理


如图1OAOB是某地一个湖泊的两条垂直的湖堤,线段CD和曲线EF分别是湖泊中的一条栈桥和防波堤.为观光旅游需要,拟过栈桥CD上某点M分别修建与OAOB平行的栈桥MGMK,且MGMK为边建一个跨越水面的三角形观光平台MGK建立如图2所示的直角坐标系,测得CD的方程是x
2y20(0x20
曲线EF的方程是xy
200(x0
设点M的坐标为(s,t(题
中所涉及长度单位均为米,栈桥及防波堤都不计宽度)
1)求三角形观光平台MGK面积的最小值;
2)若要使MGK的面积不小于320平方米,求t的范围.1
20.(本小题满分16分)已知函数
f(xeax1aR
(f(x
x
2
,且a为常数
1)求函数2)当a
0
的单调区间;
f(x0
时,若方程只有一解,求a的值;
,求a的取值范围.
3)若对所有x0都有
f(xf(x

数学Ⅱ(附加题)
21.【选做题】本题包括ABCD四小题,请选定其中两题,并在相应的答题区域内作答.若多做,则按作答的前两题评分.解答时应写出文字说明、证明过程或演算步骤.A.选修4-1:几何证明选讲



(本小题满分10分)
如图,AB是⊙O的直径,弦BDCA的延长线相交于点EEF垂直BA的延长线于点F.求证:1AED2AB
2
AFD


E
D
BEBDAEAC

B.选修4-2:矩阵与变换(本小题满分10分)
2
F
A
C
·O
B
(第21A题图)
1M
0对应的变换作用下得到的曲线方程,其中
0
2
求曲线
2x2xy10
01
在矩阵MN
1N
1

C.选修4-4:坐标系与参数方程(本小题满分10分)
以直角坐标系的原点为极点,x轴的正半轴为极轴,并在两种坐标系中取相同的单位长度.已知
x4cos,
(
y2sin
的参数方程为
直线l的极坐标方程为cos2sin0,曲线C又直线l与曲线C交于AB两点,求线段AB的长.
D.选修4-5:不等式选讲(本小题满分10分)若存在实数x使
3x6
14xa

成立,求常数a的取值范围.

【必做题】第22题、第23题,每题10分,共计20分.请在答题卡指定区域内作答,解答时应写出文字说明、证明过程或演算步骤.22.(本小题满分10分)如图,在长方体ABCD的点,且EB
A1B1C1D1
中,已知AB
4
AD
3
AA12EF分别是棱ABBC
FB1
1)求异面直线EC1FD1所成角的余弦值;
2)试在面A1B1C1D1上确定一点G,使DG平面D1EF

D1
G
A1
D
FB1
C1
C



23.(本小题满分10分)设二项展开式
Cn(31
2n1
(nN
*
的整数部分为
An
,小数部分为
Bn
.
1)计算C1B1,C2B2的值;2)求
CnBn
.

宿迁市2011届高三第一次调研试卷数学1答案一填空题:
2
12
3
1.22.a0,3.1,4.10.
xy20
y8x
,5.
[
5
,6.36,7.3,8.4,9.,14.2012.
BDsinBAD

210
,
,11.3,12.,13.
925,49
二、解答题
AB
15.1)在ABD中,由正弦定理得sinADB
AC

①,
ACD中,由正弦定理得sinADCAD平分BAC所以BAD
CAD

DCsinCAD
②,„„„„„„„„„2
sinBAD
sinCAD

sinADBsin(ADCsinADC
BD
ABAC
36

.„„„„„„„„„„„„„„„„„„6
由①②得DC

,所以DC
DC
2BD
23
2)因为DC2BD
BC
,所以
2
.
AC
2
在△ABC中,因为
cosB
ABBC
2
2ABBC

376237
222

11
21„„„„10
22
ABDCAB(BC|AB||BC|cos(B
33所以




23
37(
1121

22
3.„„„„„„„„„„„„„„„„„„„„„14
P
161)因为EF分别是BCCD的中点,所以EFBD,„„„„„„„„„„„2因为EF平面PBDBD平面PBD
所以EF∥平面PBD.„„„„„„„„„62)设BDAC于点O,连结PO
因为ABCD是菱形,所以BDACOBD中点,PBPD,所以BDPO
EFBD,所以EFACEFPO.„„„„„„„„„10AC
POO
AB
OE(第16题图)
FC
D
AC

平面PACPO

平面PAC
所以EF⊥平面PAC.„„„„„„„„„„„„„„„„„„„„„„„„„„12因为EF平面PEF,所以平面PEF⊥平面PAC.„„„„„„„„„„„„„„„14171)设
{an}
公比为q,由题意得q0
a22a13,a1(q23,23a5a2a,2q5q30,234„„„„„„„„„„„„„„„„„2
6
a1,5a13,
q1
q3,2(舍去)解之得,„„„„„„„„„„„„„„„„„„„4
所以数列
an
的通项公式为
an33
n1
3
n
nN.„„„„„„„„„„„„„6
n

2)由(1)可得所以所以
bnlog3ann
2
3
,所以
n
anbnn3
.„„„„„„„„„„„„„8
Sn132333n3
2
3
4

n1
3Sn132333n3
2
3

n
n1
两式相减得,
2
2Sn3(333n3
3
n
n1
„„„„„„„„„„„„„10
(3333n3
3(1313
n

n1
n3
n1

3(2n13
2

3(2n13
4
n1
所以数列
anbn
的前n项和为
Sn
.„„„„„„„„„„„„14



x
2
181)由椭圆E8

y
2
4
1
,得lx4C(4,0F(2,0
(x4y16
2
2
2
2
又圆C过原点,所以圆C的方程为2)由题意,得
G(3,yG
.„„„„„„„„„„„„4,得
yG15
,代入
(x4y16

所以FG的斜率为k15FG的方程为y15(x2„„„„„„„8(注意:若点GFG方程只写一种情况扣1分)
d
152
216(
152
2
所以
C(4,0
FG的距离为,直线FG被圆C截得弦长为
7

故直线FG被圆C截得弦长为7.„„„„„„„„„„„„„„„„„„„„„„10
GF
12,得
2
(x02y0
2
22
3)设整理得
P(s,t
2

2
G(x0,y0
,则由GP

(x0s(y0t
2
2

12

3(x0y0(162sx02ty016st0
①,„„„„„„„„„„12
2
G(x0,y0
在圆C
(x4y16
2
22
上,所以
2
x0y08x00
2
②,
②代入①得
(2s8x02ty016st0
„„„„„„„„„„14
2s80,

2t0,
16s2t20,
G(x0,y0又由为圆C上任意一点可知,解得s4,t0
所以在平面上存在一点P,其坐标为(4,0„„„„„„„„„„16
G(100t,t
K(s,
100s
191)由题意,得

(s0,t0

又因为M(s,t在线段CDx2y20(0x20上,所以s2t20(0s20
SMGK
12
MGMK
1200200140000
(s(t(st4002ts2st„„„„„4
20s2t2s2t22u,得0st50,当且仅当s10t5时等号成立.„„„„„„„„„„„„„„6



stu,则
f(u
12
f(uSMGK
12
(u
40000u
400
u(0,50].
(1
10000u
2
0

,故f(u(0,50]上单调递减,
(注意:若f(u(0,50]上单调递减未证明扣1分)所以
f(uminf(50225
,此时s10t5.
所以三角形MGK面积的最小值为225平方米.„„„„„„„„„„„„„„102)由题意得f(u320
1(u
40000u
400320
2,解得u40u1000(舍去)
由(1)知st40„„„„„„„„„„„„„„14(202tt40,解之得5所以t的范围是[5201
5,5
5]
5t5
5.
.„„„„„„„„„„„„„„„„„„„„„16
x
f(xea
,„„„„„„„„„„„„„„„„„„„„„„„„1

a0时,f(x0f(x(,上是单调增函数.„„„„„„„3
a0时,

f(x0,得xln(af(x(ln(a,上是单调增函数;
f(x0,得xln(af(x(,ln(a上是单调减函数.
综上,a0时,f(x的单调增区间是(,
a0时,f(x的单调增区间是(ln(a,,单调减区间是(,ln(a.„6
f(xminf(ln(a
2)由(1)知,当a0xln(a时,f(x最小,即
由方程f(x0只有一解,得f(ln(a0,又考虑到f(00
所以ln(a0,解得a1.„„„„„„„„„„„„„„„„„„„103)当x0时,
f(xf(x
恒成立,



即得eaxe
h(xee
xx
xx
ax恒成立,即得ee2ax2a
xx
2ax0恒成立,
x
x0,即当x0时,h(x0恒成立.,且
h(x2ee
x
x
h(xee
x
2a22a
,当x0时等号成立.
„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„12

①当a1时,h(x0
所以h(x[0,上是增函数,故h(xh(00恒成立.

②当a1时,若x0h(x0
x0h(x0
所以h(x[0,上是增函数,故h(xh(00恒成立.„„„„„„„14
xln(a
③当a1时,方程h(x0的正根为1
a1
2

此时,若所以,
x(0x1

,则h(x0,故h(x在该区间为减函数.
x(0x1
时,h(xh(00,与x0时,h(x0恒成立矛盾.
综上,满足条件的a的取值范围是[1,.„„„„„„„„„„„„„„16数学Ⅱ(附加题)参考答案21【选做题】
A.选修4-1:几何证明选讲证明:(1连结AD
因为AB为圆的直径,所以ADB90EFABEFA90ADEF四点共圆,
DEADFA.„„„„„„„„„„„5(2(1知,BDBEBABF连结BC,显然ABCAEF
AB
AC
AF,即ABAFAEAC
E
D
F
A
C
·O
B
(第21A题图)
AE




BEBDAEACBABFABAFAB(BFAFAB
2
„„„„„„„„„„
10
B.选修4-2:矩阵与变换
10
解:MN=
0121
0112=02
„„„„„„„„„„„„„„4

Px,y
是曲线
2x2xy10
2
上任意一点,点P在矩阵MN对应的变换下变为点
x,yP

0
2
x'x

y'2x2yy
2.„„„„„„„„„„„„„„8
x1
y2
则有
于是xx代入
yx
2x2xy10
2
2
xy1
MN对应的变换作用下得到的曲线方程为xy1.„„„„„10
所以曲线
2x2xy10

C.选修4-4:坐标系与参数方程
x
2
解:直线l的直角坐标方程为6
(2
x2y0
曲线C的普通方程为16

y
2
4
1
„„„„„„„

2,

2
AB
(22,2

„„„„„„„„„„„„„„„8

AB
(42(22
2
2
210.
„„„„„„„„„„„„„
10
D.选修4-5:不等式选讲
3x

„„„„„„„„„„„„„2
6
x

(3

x21
2
西
14x(31(x214x64
,„„„„„„„„„8



所以3x614x8,当且仅当x10时取“=
(,8.

a

„„„„„„„„„„„„„„„„„„„„„„„„10
【必做题】

DD1
22.解:1)以D为原点,DADC分别为x轴,y轴,z轴的正向建立空间直角坐
标系,
则有D(0,0,0于是
EC1
D1(0,0,2

C1(0,4,2
E(3,3,0F(2,4,0
.„„„„„„„„„„„„„„„„„„„„„3

EC1(3,1,2


FD1(2,4,2

FD1
所成角为,则
(3(21(422(312
2
2
2
EC1FD1
cos
|EC1||FD1|
(2(42
222

2114

zD1
G
C1B1
CFE
B
y




21
线
EC1

FD1

14.„„„„„„„„„„„„„„„„„„„„5
2)因点G在平面A1B1C1D1上,故可设G(x,y,2
DG(x,y,2
A1
D


FD1(2,4,2


EF(1,1,0

Ax
„„„„„„„„„„„„„„„„„„„„7
2
x,3FD10,2x4y40,
y2.
EF0xy0,3解得
2

DGDG
故当点G在面A1B1C1D1上,且到A1D1C1D1距离均为3时,DG平面D1EF„„„„„„„„„„„„…1023.解:1)因为所以C1

31
Cn(31
A12
2n1

31

B1
,所以C1B12„„„„„„„„„„„2
20
C2(311063
3
,其整数部分A2,小数部分B2
6310




所以C2B28.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„42)因为
Cn(31
2n1
0
2n1
C2n1(3
2n1
1
02n1
C2n1(3
12n2
C2n13C2n1
2n1
2n2
3C2n1,
2n1

(31
C2n1(3C2n1(3
2n2
C2n1
2n2

①—②得:
(31
2n1
—(
2n1
31
2n1
=2
C2n1(3
12n2
C2n1(3
32n4
C2n1N*
„„8
2n1
2n1
0(31所以
2n12n1
B(31311,所以An(31—(n

CnBn(31
2n1
(31
2n1
2n1
2
2n1
.„„„„„„„„„„„„„„„„„„10
(注:若猜想出
CnBn2
而未给出证明只给2


本文来源:https://www.2haoxitong.net/k/doc/83228e0df12d2af90242e6e5.html

《苏北四市2011届高三第一次调研考试数学试题及答案.doc》
将本文的Word文档下载到电脑,方便收藏和打印
推荐度:
点击下载文档

文档为doc格式