????????h?h?n?AD?x,y,z?2,0,?2x?z?0?????233， ··············· 5分 ????????????n?AC??x,y,z???0,6,h??6y?hz?01?2???取z??6，则x?y?h，故n2??h,h,?6?． ·············· 6分
??????n?n??????1267?∴cos??cos?n1,n2????， ············································ 7分 ????＝27n1n21?2h?36解得h?63．
∴ AA1?63． ········································································································· 8分 （Ⅲ）在平面BCC1B1内，分别延长CB、C1D，交于点F，连结AF，则直线AF为平面ADC1与平面ABC的交线． ························································································ 9分
11∵ BD//CC1，BD=BB1=CC1，
33∴
BFBD1??． FCCC13????1????∴ BF?CB，
2????????????????1????1∴ AF?AB?BF?AB?CB??2,0,0???2,?6,0???3,?3,0?． ························ 11分
22?????h?由（Ⅱ）知，h?63，故DE???2,3,???2,3,3，
6??????????????????AF?DE?155∴ cos?AF,DE?????·············································· 12分 ??2． ·??????832?4AFDE??∴ 直线l与DE所成的角的余弦值为?18．（本小题满分13分）
552?2． ········································· 13分 88解：（Ⅰ）设圆C的半径为r（r?0），依题意，圆心坐标为(r,2)． ····················· 1分 ∵ MN?32
225?3?∴ r????22，解得r2?． ·········································································· 3分
4?2?5?252?∴ 圆C的方程为?x????y?2??． ···························································· 5分
24??5?252?（Ⅱ）把y?0代入方程?x????y?2??，解得x?1，或x?4，
2?4?22即点M?1,0?，N?4,0?． ···························································································· 6分
（1）当AB?x轴时，由椭圆对称性可知?ANM??BNM． ································· 7分 （2）当AB与x轴不垂直时，可设直线AB的方程为y?k?x?1?．
?y?k?x?1?联立方程?2，消去y得，k2?2x2?2k2x?k2?8?0． ······················· 8分 22x?y?8???设直线AB交椭圆?于A?x1,y1?、B?x2,y2?两点，则
2k2k2?8，x1?x2?2． ············································································· 9分 x1?x2?2k?2k?2∵ y1?k?x1?2?,y2?k?x2?2?， ∴ kAN?kBN??k?x1?1?k?x2?1?y1y2??? x1?4x2?4x1?4x2?4k?x1?1??x2?4??k?x2?1??x1?4??x1?4??x2?4?．····································································· 10分
2?k2?8?10k2?2?8?0， ∵?x1?1??x2?4???x2?1??x1?4??2x1x2?5?x1?x2??8?k2?2k?2 ··································································································································· 11分 ∴ kAN?kBN?0，?ANM??BNM． ·································································· 12分 综上所述，?ANM??BNM． ················································································ 13分 19．（本小题满分13分）
解：（Ⅰ）当a?2时，f(x)?ln(x?1)?∴f?(x)?2x， x?112x?3??， ·········································································· 1分 x?1(x?1)2(x?1)2∴ f?(0)?3，所以所求的切线的斜率为3. ····························································· 2分 又∵f?0??0，所以切点为?0,0?. ·········································································· 3分 故所求的切线方程为：y?3x. ··············································································· 4分 （Ⅱ）∵f(x)?ln(x?1)?∴f?(x)?ax(x??1)， x?11a(x?1)?axx?1?a??． ································································· 5分 x?1(x?1)2(x?1)2ss5uKu①当a?0时，∵x??1，∴f?(x)?0；KKs55u ································································ 6分
②当a?0时，
?f?(x)?0?f?(x)?0由?，得?1?x??1?a；由?，得x??1?a； ····························· 7分
x??1x??1??综上，当a?0时，函数f(x)在(?1,??)单调递增；
当a?0时，函数f(x)在(?1,?1?a)单调递减，在(?1?a,??)上单调递增． ········· 8分
（Ⅲ）方法一：由（Ⅱ）可知，当a??1时，
f?x??ln?x?1??x在?0,???上单调递增． ························································· 9分 x?1x． ······································· 10分 x?1∴ 当x?0时，f?x??f?0??0，即ln?x?1??111?1?令x?（n?N*），则ln?1???n?． ··················································· 11分
1n?n??1n?1n11111?2，即?另一方面，∵?2,
n?n?1?nnn?1n∴
111································································································ 12分 ??2． ·
n?1nn?1?11∴ ln?1????2（n?N*）． ········································································· 13分
?n?nn方法二：构造函数F(x)?ln(1?x)?x?x2，(0?x?1) ········································· 9分
1x(2x?1)?1?2x?， ··································································· 10分 1?xx?1∴当0?x?1时,F'(x)?0；
∴F'(x)?∴函数F(x)在(0,1]单调递增． ·············································································· 11分 ∴函数F(x)?F(0) ，即F(x)?0
∴?x?(0,1]，ln(1?x)?x?x2?0，即ln(1?x)?x?x2 ··································· 12分 令x?1?1?11（n?N*），则有ln?1????2． ························································ 13分 n?n?nn20．（本小题满分14分）
34解：（Ⅰ）已知?是锐角，根据三角函数的定义，得sin??， ·············· 1分 cos??，55又cos??512，且?是锐角，所以sin??． ························································· 2分 13134531216所以cos(???)?cos?cos??sin?sin???????． ··························· 4分
51351365（Ⅱ）证明：依题意得，MA?sin?，NB?sin?，PC?sin(???) ???因为?，???0，?，所以cos??(0,1)，cos??(0,1)，于是有
?2?sin(???)?sin?cos??cos?sin??sin??sin?，① ·············································· 6分
又∵?+???0,??,??1?cos(?+?)?1，
sin??sin((???)??)?sin(???)?cos??cos(???)?sin??sin(???)?sin?，②
··········································································································································· 7分 同理，sin??sin(???)?sin?，③ 由①，②，③可得，
线段MA、NB、PC能构成一个三角形. ································································· 8分
（III）第（Ⅱ）小题中的三角形的外接圆面积是定值，且定值为
?. 4不妨设?A?B?C?的边长分别为sin?、sin?、sin?????，其中角A?、B?、C?的对边分别为sin?????、sin?、sin?.则由余弦定理，得：
sin2??sin2??sin2(???) ···································································· 9分 cosA??2sin??sin?sin2??sin2??sin2?cos2??cos2?sin2??2sin?cos?cos?sin? ?2sin??sin?sin2??sin2??sin2?sin2??2sin?cos?cos?sin? ?2sin??sin??sin??sin??cos?cos?
??cos(???) ································································································· 11分
???因为?，???0，?，所以????(0,?)，所以sinA??sin(???)，······················· 12分
?2?设?A?B?C?的外接圆半径为R， 由正弦定理，得2R?B?C?sin(???)1
??1，∴R?， ······································· 13分 sinA?sin(???)2
所以?A?B?C?的外接圆的面积为
?. ········································································ 14分 40?································································· 2分 ?． ·2?21．（1）（本小题满分7分）选修4－2：矩阵与变换 ?1解：（Ⅰ）由条件得矩阵M???0?1（Ⅱ）因为矩阵M???0??100?的特征多项式为f(?)??(??1)(??2)， ?2?0??2令f(?)?0，解得特征值为?1?1，?2?2，····························································· 4分 ???x????x??x?设属于特征值?1的矩阵M的一个特征向量为e1???，则Me1??????，解得
?2y??y??y????1?y?0，取x?1，得e1???， ··························································································· 5分
?0?????0?同理，对于特征值?2，解得x?0，取y?1，得e2???， ····································· 6分
1?????1?????0?所以e1???是矩阵M属于特征值?1?1的一个特征向量，e2???是矩阵M属于特
?0??1?征值?2?2 的一个特征向量．··························································································· 7分
（2）（本小题满分7分) 选修4—4：极坐标与参数方程 解：（Ⅰ）∵点A、B的极坐标分别为(1,∴点A、B的直角坐标分别为(?3)、(3,2?)， 313333······································· 2分 )， ,)、(?,2222∴直线AB的直角坐标方程为23x?4y?33?0． ············································ 4分
?x?rcos?,（Ⅱ）由曲线C的参数方程?化为普通方程为x2?y2?r2，(?为参数）?y?rsin? ············································································································································· 5分
∵直线AB和曲线C只有一个交点， ∴半径r?····································································· 7分 ?321． ·
2214(23)?433（3）（本小题满分7分) 选修4—5：不等式选讲
解：（Ⅰ）∵关于x的不等式2?x?x?1?m对于任意的x?[?1,2]恒成立
························································································· 1分 ?m?(2?x?x?1)max ·根据柯西不等式，有
(2?x?x?1)2?(1?2?x?1?x?1)2?[12?12]?[(2?x)2?(x?1)2]?6 所以2?x?x?1?6，当且仅当x?（Ⅱ）由（Ⅰ）得m?2?0，则f?m??m?∴f?m??331时等号成立，故m?6． ···················· 3分 21111?(m?2)?(m?2)??2
(m?2)222(m?2)211133(m?2)?(m?2)??2?2?2 ············································ 5分 22(m?2)2211当且仅当(m?2)?，即m?32?2?6时取等号， ································ 6分 22(m?2)所以函数f?m??m?
13ss5uKu的最小值为32?2．KKs55u ··········································· 7分 2(m?2)2
2012年福州市高中毕业班综合练习
数学(理科)试卷
（完卷时间：120分钟；满分：150分）
注意事项：
1．本科考试分试题卷和答题卷，考生须在答题卷上作答，答题前，请在答题卷的密封线内填写学校、班级、准考证号、姓名；
2．本试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分，全卷满分150分，考试时间120分钟．
参考公式：
样本数据x1，x2，?，xn的标准差 锥体体积公式： s?1?222x1?x???x2?x?????xn?x?? ??n?1V?Sh 其中S为底面面积，h为高 3球的表面积、体积公式 其中x为样本平均数 柱体体积公式 V?Sh 其中S为底面面积，h为高 S?4?R2，V?43?R 3其中R为球的半径 第Ⅰ卷 （选择题 共50分）
一、选择题（本大题共10小题，每小题5分，共50分．在每小题给出的四个选项中有且只有一项是符合题目要求的，把答案填在答题卡的相应位置．）
1．已知全集U?R，集合M?{xx?x?0}，则eUM? A．{x|0?x?1} C．{x|x?0或x?1} 数z1?z2所对应的点位于
A．第一象限
C．第三象限
ss5uKuB．第二象限KKs55u
D．第四象限
2B．{x|0?x?1}
yAD．{x|x?0或x?1}
????????2．如图，在复平面内，若复数z1,z2对应的向量分别是OA,OB，则复
OBx3．设等比数列{an}的前n项和为Sn，则“a1?0”是“S3?S2”的 A．充分而不必要条件 B．必要而不充分条件 C．充要条件 D．既不充分也不必要条件 4．若一个几何体的三视图，其正视图和侧视图均为矩形、俯视图为正三角形，尺寸如图所示，则该几何体的体积为
第2题图 23 3C．3
A．
33 2 D．23
B．第4题图
5．如图，执行程序框图后，输出的结果为
A．8 B．10 C．12 D．32
????6．下列函数中，周期为?，且在?,?上单调递增的奇函数是
?42????A．y?sin?x??
2???2?
???B．y?cos?2x??
2?????D．y?cos?2x??
2????C．y?sin??2x??
????????????????????????????7．已知AB?BC?0，AB?1，BC?2，AD?DC?0，则BD
的最大值为 A.
第5题图
25 B. 2 C. 5 D. 25 58．若从区间(0,e)内随机取两个数，则这两个数之积不小于．．．e的概率为
2121A．1? B. 1? C. D.
A1eeee9．如图，在正方体ABCD?A1BC11D1中，若平面A1BCD1上一动点 D1B1C1P到AB1和BC的距离相等，则点P的轨迹为
DPCABA．椭圆的一部分 B．圆的一部分
第9题图 C．一条线段 D．抛物线的一部分
10．将方程x?tanx?0的正根从小到大地依次排列为a1,a2,?,an,?，给出以下不
等式：
①0?an?1?an??2③2an?1?an?2?an；
；
②?an?1?an??； 2④2an?1?an?2?an；
?其中，正确的判断是
A. ①③ B. ①④ C. ②③ D. ②④
二、填空题：本大题共5小题，每小题4分，共20分．把答案填在答题卡的相应位置．
??x,x?011．已知函数f(x)??x，则f?f??1??? .
??2,x?0x2y2??1(m?0,n?0)的离心率为2，12．已知双曲线有一个焦点与抛物线y2?16xmn的焦点重合，则n?__________.
13．已知等差数列?an?的公差不为零，且a1、a1?a2?a5?13，a2、
a5成等比数列，则a1的取值范围为 .
14．已知三次函数f(x)?ax3?bx2?cx?d的图象如图所示， 则
f?(?3)? ★★★ ． f?(1)第14题图
15．假定平面内的一条直线将该平面内的一个区域分成面积相等的两个区域，则称这条直线平分这个区域．如图，?是平面?内的任意一个封闭区域.现给出如下结论：
① 过平面内的任意一点至少存在一条直线平分区域?；
② 过平面内的任意一点至多存在一条直线平分区域?； ③ 区域?内的任意一点至少存在两条直线平分区域?； ④ 平面内存在互相垂直的两条直线平分区域?成四份． 其中正确结论的序号是 ． 证明过程或演算步骤．
16．（本小题满分13分）
招聘会上，某公司决定先试用后再聘用小强，该公司的甲、乙两个部门各有4个不同岗位．
（Ⅰ）公司随机安排小强在这两个部门中的3个岗位上进行试用，求小强试用的3个岗位中恰有2个在甲部门的概率；
（Ⅱ）经试用，甲、乙两个部门都愿意聘用他．据估计，小强可能获得的岗位月工资及相应概率如下表所示：
甲部门不同岗位月工资X1（元） 2200 2400 2600 2800
获得相应岗位的概率P0.4 0.3 0.2 0.1 1 乙部门不同岗位月工资X2（元） 2000 2400 2800 3200
获得相应岗位的概率P2 0.4 0.3 0.2 0.1
求甲、乙两部门月岗位工资的期望与方差，据此请帮助小强选择一个部门，并说明理由． 17．（本小题满分13分）
如图，三棱柱ABC?A1B1C1中，AA1?平面ABC，?BAC?90?，
1AB?2,AC?6, 点D在线段BB1上，且BD?BB1，AC?AC1?E． 13（Ⅰ）求证：直线DE与平面ABC不平行；
（Ⅱ）设平面ADC1与平面ABC所成的锐二面角为?，若cos??7，求AA1的长； 7（Ⅲ）在（Ⅱ）的条件下，设平面ADC1?平面ABC?l，求直线
y第15题图
三、解答题：本大题共6小题，共80分．解答写在答题卡相位置，应写出文字说明、
l与DE所成的角的余弦值．
18．（本小题满分13分）
第17题图 如图，圆C与y轴相切于点T?0,2?，与x轴正半轴相交于两点 M,N（点M在点N的左侧），且MN?3．
TAC（Ⅰ）求圆C的方程；
xy（Ⅱ）过点M任作一条直线与椭圆?:??1相交于两点
48A、B，连接AN、BN，求证：?ANM??BNM．
B22OMNx第18题图
19．（本小题满分13分） 已知函数f(x)?ln(x?1)?ax?a?R?． x?1（Ⅰ）当a?2时，求函数y?f?x?的图象在x?0处的切线方程； （Ⅱ）判断函数f(x)的单调性；
?1?11（Ⅲ）求证：ln?1????2（n?N*）．
?n?nn20．（本小题满分14分）
如图，在平面直角坐标系中，锐角?、?的终边分别与单位圆交于A，B两点． （Ⅰ）如果tan??35，B点的横坐标为，求c os?????的值；413（Ⅱ）若角???的终边与单位圆交于C点，设角?、?、
???的正弦线分别为MA、NB、PC，求证：线段MA、NB、PC
能构成一个三角形；
（III）探究第（Ⅱ）小题中的三角形的外接圆面积是否为定值？
若是，求出该定值；若不是，请说明理由. 第20题图
21．本题有（1）、（2）、（3）三个选答题，每题7分，请考生任选2题作答，满分14分．如果多做，则按所做的前两题记分．作答时，先用2B铅笔在答题卡上把所选题目对应的题号涂黑，并将所选题号填入括号中．
（1）（本小题满分7分）选修4－2：矩阵与变换
设矩阵M是把坐标平面上的点的纵坐标伸长到原来的2倍，横坐标保持不变的伸缩变换．
（Ⅰ）求矩阵M；
（Ⅱ）求矩阵M的特征值以及属于每个特征值的一个特征向量． （2）（本小题满分7分) 选修4—4：极坐标与参数方程
在直角坐标平面内，以坐标原点O为极点，x轴的非负半轴为极轴建立极坐标系. 已知点A、B的极坐标分别为(1,参数）．
（Ⅰ）求直线AB的直角坐标方程；KKKsss555uuu
（Ⅱ）若直线AB和曲线C只有一个交点，求r的值． （3）（本小题满分7分) 选修4—5：不等式选讲
已知关于x的不等式2?x?x?1?m对于任意的x?[?1,2]恒成立 （Ⅰ）求m的取值范围；
（Ⅱ）在（Ⅰ）的条件下求函数f?m??m?1的最小值．
(m?2)2?3)、(3,2??x?rcos?,)，曲线C的参数方程为?(?为3?y?rsin?
2012年福州市高中毕业班综合练习 理科数学试卷参考答案及评分参考
一、选择题(本大题共10小题,每小题5分,共50分)
1.B 2.A 3.C 4.D 5.B 6.D 7. C 8. B 9.D 10. D 二、填空题(本大题共5小题,每小题4分,共20分) 11. 2 12. 12 13. (1,??) 14. ?5 15. ①④ 三、解答题(本大题共6小题，共80分) 16．（本小题满分13分）
解：（Ⅰ）记事件“小强试用的3个岗位中恰有2个在甲部门的概率”为A，则
21C4?C43P?A???． ···································································· 6分
C837（Ⅱ）E甲?2200?0.4?2400?0.3?2600?0.2?2800?0.1?2400（元）， ··············· 7分 E乙?2000?0.4?2400?0.3?2800?0.2?3200?0.1?2400（元）． ·························· 8分
D?X甲???2200?2400??0.4??2400?2400??0.3??2600?2400??0.2??2800?2400??0.12222 ?40000， ········································································································ 9分 D?X乙???2000?2400??0.4??2400?2400??0.3??2800?2400??0.2??3200?2400??0.12222 ?160000． ····································································································· 10分 选择甲部门：因为X甲?X乙，D?X甲??D?X乙?，说明甲部门各岗位的工资待遇波动
ss5uK5u比乙部门小，竞争压力没有乙部门大，比较安稳．KKs5u ··················································· 13分
选择乙部门：因为X甲?X乙，D?X甲??D?X乙?，说明乙部门各岗位的工资待遇波动比甲部门大，岗位工资拉的比较开，工作比较有挑战性，能更好地体现工作价值． ·· 13分
17．（本小题满分13分）
解：依题意，可建立如图所示的空间直角坐标系A?xyz，设AA1?h，则 h?h???B?2,0,0?,C?0,6,0?,D?2,0,?,A1?0,0,h?,C1?0,6,h?,E?0,3,?．2分
3?2??????（Ⅰ）证明：由AA1?平面ABC可知n1??0,0,1?为平面ABC的
一个法向量．
????????h?h∴ DE?n1???2,3,???0,0,1???0． ························· 3分
6?6?∴ 直线DE与平面ABC不平行． ···························· 4分
???（Ⅱ）设平面ADC1的法向量为n2??x,y,z?，则
