钦州市2006年高三毕业班第一次调研测试
理科数学(必修+选修Ⅱ)
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8页.考试结束后,将本试卷和答题卡一并交回.共150分.考试时间120分钟.
第Ⅰ卷
注意事项:
1.答题前,考生在答题卡上务必用黑色签字笔将自己的姓名、准考证号填写清楚,并贴好条形码.请认真核准条形码上的准考证号、姓名和科目.
2.每小题选出答案后,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,再选涂其它答案,不能答在试题卷上.
3.本卷共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.
参考公式:
如果事件A、B互斥,那么 P(A+B)=P(A)+P(B).
如果事件A、B相互独立,那么 P(A·B)=P(A)·P(B).
如果事件A在1次试验中发生的概率是P,那么n次独立重复试验中恰好发生次的概率.
球的表面积公式 S球=4πR2 其中R表示球的半径.
球的体积公式 V球=πR3 其中R表示球的半径.
一、选择题:
1.如图,U是全集,M、P、S是U的三个子集,则阴影部分所表示的集合是
(A)(MP)S
(B)(MP)S
(C)(MP)(US)
(D)(MP)(US)
2.设函数f(x)=sin(πx-),则下列命题中正确的是
(A)f(x)是周期为1的奇函数 (B)f(x)是周期为2的偶函数
(C)f(x)是周期为1的非奇非偶函数 (D)f(x)是周期为2的非奇非偶函数
3.有如下三个命题:
①分别在两个平面内的两条直线一定是异面直线;
②垂直于同一个平面的两条直线是平行直线;
③垂直于同一个平面的两个平面互相平行;
④过平面的一条斜线有一个平面与平面垂直.
其中正确命题的个数为
(A)1 (B)2 (C)3 (D)4
4.函数y=log(x-1)的反函数的图象是
(A) (B) (C) (D)
5.设abc≠0,“ac>0”是“曲线ax2+by2=c为椭圆”的
(A)充分非必要条件 (B)必要非充分条件
(C)充分必要条件 (D)既非充分又非必要条件
6.已知向量、为单位向量,它们的夹角为60°,则+3的值是
(A) (B) (C) (D)4
7.用1,2,3,4,5这五个数字,组成比20 000大,而且百位数字不是3的没有重复数字的五位数共有
(A)64个 (B)72个 (C)78个 (D)96个
8.等差数列{an}中,如a1+a2+a3=6,a10+a11+a12=9,则a1+a2+…+a12=
(A)15 (B)30 (C)45 (D)60
9.椭圆=1,两焦点间距离为6,则t=
(A)16 (B)34 (C)16或34 (D)11
10.已知双曲线的两个焦点为F1(-,0),F2(,0),P是此双曲线上的一点,且PF1⊥PF2,PF1·PF2=2,则该双曲线的方程
(A)=1 (B)=1 (C)=1 (D)=1
11.已知奇函数f(x)的定义域为:{x||x+2-a|<a,a>0},则a的值为
(A)0 (B)1 (C)2 (D)3
12.函数f(x)在上是增函数,A(0,-2)、B(4,2)是其图象上的两点,则不等式
f(x+2)<2的解集是
(A)(-∞,-2)∪(2,+∞) (B)(-2,2)
(C)(-∞,0)∪(4,+∞) (D)(0,4)
钦州市2006年高三毕业班第一次调研测试
理科数学(必修+选修Ⅱ)
第Ⅱ卷
题号 | 一 | 二 | 三 | 总分 | |||||
17 | 18 | 19 | 20 | 21 | 22 | ||||
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注意事项:
1.用钢笔或圆珠笔直接答在试题卷中.
2.答卷前将密封线内的项目填写清楚.
3.本卷共10小题,共90分.
得分 | 评卷人 |
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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.= .
14.圆x2+y2+x-6y+3=0上两点P、Q关于直线kx-y+4=0对称,则k= .
15.一个袋子里装有大小相同的3个红球和2个黄球,从中同时取出两个,则其中含红球个数的数学期望是_________________.
16.如图,正方体ABCD-A1B1C1D1的棱长为a,将该正方体沿对角面BB1D1D切成两块,再将这两块拼接成一个不是正方体的四棱柱,那么所得四棱柱的全面积为 .
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.
得 分 | 评卷人 |
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17.(本小题满分12分)
已知一扇形的周长为c(c>0),当扇形的弧长为何值时,它有最大面积?并求出面积的最大值.
得 分 | 评卷人 |
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18.(本小题满分12分)
从6位女同学和4位男同学中随机选出3位同学进行体能测试,每位女同学能通过测试的概率均为,每位男同学能通过测试的概率均为,试求:
(1)选出的3位同学中,至少有一位男同学的概率;
(2)10位同学中的女同学甲和男同学乙同时被选中且通过测试的概率.
19.(本小题满分12分)
得 分 | 评卷人 |
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已知有三个居民小区A、B、C构成△ABC,AB=700、BC =800、AC=300.现计划在与A、B、C三个小区距离相等处建造一个工厂,为不影响小区居民的正常生活和休息,需在厂房的四周安装隔音窗或建造隔音墙.据测算,从厂房发出的噪音是85分贝,而维持居民正常生活和休息时的噪音不得超过50分贝.每安装一道隔音窗噪音降低3分贝,成本3万元,隔音窗不能超过3道;每建造一堵隔音墙噪音降低15分贝,成本10万元;距离厂房平均每25噪音均匀降低1分贝.
(1)求∠C的大小;
(2)求加工厂与小区A的距离.(精确到1);
(3)为了不影响小区居民的正常生活和休息且花费成本最低,需要安装几道隔音窗,建造几堵隔音墙?
(计算时厂房和小区的大小忽略不计)
得 分 | 评卷人 |
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20.(本小题满分12分)
如图①所示的等腰梯形ABCD中,上底和高均为2,下底边长为2+2,DE⊥AB于E,CF⊥AB于F,将△AED、△BFC分别沿DE、CF折起,使A、B重合于P得图形②.在空间图形②中:
(1)求证:FP⊥平面PDE;
(2)求EF与面PDF所成的角的大小.
21.(本小题满分13分)
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已知等比数列{an}中,a1=64,公比q≠1,a2、a3、a4又分别是某等差数列的第7项、第3项、第1项.
(1)求等比数列{an}的通项公式an;
(2)设bn=log2an,求数列{bn}的前n项和.
22.(本小题满分13分)
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如图,已知过点D(-2,0)的直线l 与椭圆+y2=1交于不同的两点A、B,点M是弦AB的中点.
(1)若=+,求点P的轨迹方程;
(2)求的取值范围.
钦州市2006年高三毕业班第一次调研测试
理科数学参考答案及评分标准
说明:
1、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.
2、对计算题,当考生的解答在某一步出错时,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.
3、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.
4、只给整数分.选择题和填空题不给中间分.
一、选择题:(每小题5分,共60分)
题号 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
答案 | D | B | B | C | B | C | C | B | C | D | C | B |
二、填空题:(每小题4分,共16分)
13. 14.2 15. 16.(4+2)a2
三、解答题:
17.解:设扇形的半径为R,弧长为l,面积为S,
∵c=2R+l,∴R=,(l<c).·················································································· 2分
则S=Rl=×·l=(cl-l2)·········································································· 6分
=-(l2-cl)=-+,································································ 8分
∴当l=时,Smax=.··································································································· 11分
答:当扇形的弧长为时,扇形有最大面积,面积的最大值是.············· 12分
18.解:(1)易知选出的3位同学中,没有选到一位男同学的概率是,·············· 4分
∴选出的3位同学中,至少有一位男同学的概率P=1-=;···· 6分
(2)除女同学甲和男同学乙同时被选中外还有另外的8个同学中有一位同学的概率是, 10分
∵每位女同学能通过测试的概率均为,
每位男同学能通过测试的概率均为,
∴10位同学中的女同学甲和男同学乙同时被选中且通过测试的概率
P=××=.····················································································· 12分
19.解:(1)由余弦定理得cos∠C=,∠C=60º;······························································· 3分
(2)由题设知,所求距离为△ABC外接圆半径R,················································· 4分
由正弦定理得R==404.·········································································· 6分
答:加工厂与小区A的距离约为404;···························································· 7分
(3)设需要安装x道隔音窗,建造y堵隔音墙,总成本为S万元,由题意得:
即····················································· 9分
其中S=3x+10y,当x=2,y=1时,S最小值为16万元.····················· 11分
答:需安装2道隔音窗,建造1堵隔音墙即可.··········································· 12分
20.解法一:
(1)在等腰梯形ABCD中,EF=DC=2,AE=BF=,·········································· 1分
而AE=PE,BF=PF,∴PE2+FP2=EF2,∴PF⊥EP.···································· 2分
又∵AE⊥DE(即PE⊥DE),EF⊥DE,∴DE⊥面PEF,·································· 4分
∴DE⊥FP,∴FP⊥面PDE;···························································································· 6分
(2)由(1)得FP⊥面PDE,∴面FPD⊥面PDE,························· 7分
作EM⊥DP于M,则EM⊥面PDF,················································ 8分
连结FM,则∠EFM为EF与面PDF所成的角.························ 9分
在△PED中,=,可得,EM==,························ 10分
∴sin∠EFM===,∴∠EFM=arcsin.························ 11分
故EF与面PDF所成的角的大小为arcsin.····················································· 12分
解法二:
(1)以E为坐标原点,建立如图所示的空间直角坐标系E-xyz,························· 1分
则依题意得D(0,0,2),C(0,2,2),F(0,2,0),P(1,1,0),· 2分
∴=(1,-1,0),=(0,0,2),=(-1,-1,2).·········· 3分
∵·=(1,-1,0)·(0,0,2)=0··························· 4分
·=(1,-1,0)·(-1,-1,2)
=-1+1+0=0,································································ 5分
∴FP⊥ED,FP⊥PD,∴FP⊥平面PDE;······························ 6分
(2)设面PDF的法向量=(x,y,z),则
即∴············································· 8分
取y=2,则=(2,2,2).······························································································ 9分
设EF与面PDF所成的角为,则=·sin,················· 10分
∴sin==,=arcsin.····························· 11分
故EF与面PDF所成的角的大小为arcsin.······················································· 12分
21.解:(1)∵=,∴=,·············································· 2分
2q2-3q+1=0,∴q=,··························································································· 4分
∴an=2;························································································································· 6分
(2)bn=log2an=log22=7-n,····················································································· 8分
∴当1≤n≤7时,
b1+b2+…+bn=b1+b2+…+bn=(13n-n2);······················· 10分
∴当n≥8时,
b1+b2+…+bn=(b1+b2+…+b7)-(b8+b9+…+bn)····· 11分
=2(b1+b2+…+b7)-(b1+b2+…+bn)····· 12分
=42-(13n-n2).····················································· 13分
22.解:(1)设直线l的方程为y=k(x+2),P(x,y),A(x1,y1),B(x2,y2), 1分
由得:(1+2k2)x2+8k2x+8k2-2=0,··································· 3分
∴△=64k4-4(1+2k2)(8k2-2)>0,∴0≤k2<.······························ 4分
∵=+,
∴x=x1+x2=-,y=y1+y2=k(x1+x2+4)=,········ 5分
消去k得:x2+2y2+4x=0,························································································ 6分
又x=-=-4+∈,
∴点P的轨迹方程为:x2+2y2+4x=0,(-2<x≤0);······························ 7分
(2)===····································· 9分
==··············································· 11分
∵0≤k2<,∴∈.································································· 13分