2015厦门适应性考试打印版 福建省厦门市2015届高三毕业班适应性考试数学理试题 Word版含答案

2015年厦门市高三适应性考试

数学（理科）试卷

注意事项：

1．本科考试分试题卷和答题卷，考生须在答题卷指定位置上作答，答题前，请在答题卷的密封线内填写学校、班级、考号、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．

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，只有一

项是符合题目要求的.

1． 复数i(1i)（i为虚数单位）的共轭复数是

A. 1i B. 1i C. 1i D. 1i 2． 随机变量~N(0,1)，则P12=

C. 0.1574 D. 0.2718 )0.6，P(22)0.9544，（参考数据：P(Ａ.0.0215 B. 0.1359

P(33)0.9974）

x2y23． 直线y2x2恰好经过椭圆221的右焦点和上顶点，则椭圆的离心率等于 ab15255A. B. C. D. 2552y4． 已知函数fx的图像如图所示，则fx的解析式可能是

1x3 2x11C.fxx3

2x1A.fx

B.fx1x3 2x11D.fxx3

2x1Oxyx2,5．已知实数x,y满足，则zxy的取值范围是

xy20A.[0,6]

6． 命题p:函数yx2在

1B.[,6]

41C.[,0]

4223D.[,6] 42x9；命题q:loga1logaa0. 1,4上的值域为3,11B．D．

下列命题中，真命题的是 A．

pq pq

*C．

pqpq

7． 已知数列{a}满足: 当pq11p,qN,pq时，anpaq2p，则{an}的前10项

和S10

A.31 B.62 C.170 D.1023

8．如图，圆O与x轴的正半轴的交点为A，点B，C在圆O上，点B的坐标为(1,2)，点C位于第一象限，AOC．若BC5， ByCα- 1 - AOx

则sin2cos2

3cos2

A.25 53=

225 B.5C.5 5

D.25 5QC1

9． 如图1，已知正方体ABCD－A1B1ClD1的棱长为a，

动点M、N、Q分别在线段AD1,B1C,C1D1上.

当三棱锥Q-BMN的俯视图如图2所示时， 三棱锥Q-BMN的正视图面积等于

D1A1MB11212A. a B. a

422232C. a D. a

44NCDA正视方向 B图1 图2 10．如图所示，由直线xa,xa1a0,yx2及x轴围成的曲边梯形的面积介于相应小矩形与大矩形的面积之间，即a

2a1ax2dx(a1)2.类比之，

1恒成立, 2n1Oy11nN,n1n2则实数A等于

*

111A2nnn1135 A. B. C.ln2 D.ln

522

aa+1x第Ⅱ卷 （非选择题 共100分）

二、填空题：本大题共5小题，每小题4分，共20分．

11．阅读如图所示的程序，该程序输出的结果是 ▲ ．

12．设1x5a0a1(x1)a2(x1)2a5(x1)5,

则a1a2

13．一个口袋内有5个不同的红球，4个不同的白球.若取一个红球记2分，

取一个白球记1分，从中任取4个球，使总分不少于7分的取法有 ▲ 种.

14．如图，在△ABC中，ADBC0,BC3BD，过点D的直线分别交

直线AB，AC于点M，N.若AMAB,ANAC0,0， 则2的最小值是 ▲ ．

15．十八世纪，法国数学家布丰和勒可莱尔提出投针问题：在平面上画有一

组间距为a的平行线，将一根长度为l的针任意掷在这个平面上，求得此

MBDANCa5 ▲ ．

a=0 S=1 WHILE a<3 S=S*3 a=a+1 WEND PRINT S END

- 2 -

2l（为圆周率）. a针与平行线中任一条相交的概率p已知l3.14,a6，3.14，现随机掷14根相同的针（长度为l）在这个平面上，记这些针与平行线（间距为a）相交的根数为m，其相应的 概率为P(m).当P(m)取得最大值时，m ▲ ．

三、解答题：本大题共6小题，共80分.解答应写出文字说明、证明过程或演算步骤. 16．（本小题满分13分）

如图，平面直角坐标系xOy中，ABC为3.

（Ⅰ）求AB的长；

（Ⅱ）若函数f(x)Msin(x)(M0,0,3, ?ADCp,AC7，BCD的面积62)的图象经过

A,B,C三点，其中A,B为fx的图象与x轴相邻的两个交点，求函数f(x)的解析式.

yC BOD 17．（本小题满分13分）

如图，梯形ABCD中，AB⊥AD，AD∥BC，AD＝6，BC＝2AB＝4，E，F分别在线段BC，AD上，EF∥AB．将四边形ABEF沿EF折起，连接AD，AC.

（Ⅰ）若BE＝3，在线段AD上一点取一点P，使APAx1PD，求证：CP∥平面ABEF； 2（Ⅱ）若平面ABEF⊥平面EFDC，且线段FA,FC,FD的长成等比数列，求二面角E－AC－F的大小． A P

AFDB

FD

BCEEC

18．（本小题满分13分）

某茶厂现有三块茶园，每块茶园的茶叶估值为6万元.根据以往经验：今年5月12日至14日是采茶的最佳时间，在此期间，若遇到下雨，当天茶园的茶叶估值减少为前一天的一半.现有两种采摘方案：

方案①：茶厂不额外聘请工人，一天采摘一块茶园的茶叶；

方案②：茶厂额外聘请工人，在12日采摘完全部茶叶，额外聘请工人的成本为3.2万元.

- 3 -

根据天气预报，该地区5月12日不降雨，13日和14日这两天降雨的概率均为40%.每天是否下雨不相互影响.

（Ⅰ）若采用方案①，求茶厂14日当天采茶的预期收益； （Ⅱ）从统计学的角度分析，茶厂采用哪种方案更合理. 19．（本小题满分13分）

如图，抛物线E:y2pxp0的焦点为F，其准线l与x轴交于点A，2yDP过抛物线E上的动点．．P作PDl于点D. 当DPF2时， PF4. 3（Ⅰ）求抛物线E的方程；

（Ⅱ）过点P作直线mDF，求直线m与抛物线E的交点个数；

（Ⅲ）点C是DPF的外心，是否存在点P，使得CDP的面积最小.若存在，请求出面积的最小值及P的坐标；若不存在，请说明理由. 20．（本小题满分14分）

已知函数f(x)exax（e为自然对数的底数）.

（Ⅰ）讨论f(x)的单调性；

AOFx（Ⅱ）定义：函数F(x)的定义域为D，若x0D，使F(x0)x0成立，则称x0为F(x)的不动点.

当a1时， （ⅰ）证明：函数y1(x0)存在唯一的不动点x0，且x0(ln2,1)； f(x)1(nN*)， f(an)（ⅱ）已知数列an满足a1ln2,an1求证：nN,*f(a2n)f(x0)1f(x0)x01，(x0)（其中x0为ya2nx0f(x)的不动点）.

21．本题有（1）、（2）、（3）三个选答题，每小题7分，请考生任选2题作答，满分14分，

如果多做，则按所做的前两题计分．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

已知矩阵Aa2，A的一个特征值2.

14（Ⅰ）求矩阵A；

（Ⅱ）在平面直角坐标系中，点P(1,1)依次在矩阵A所对应的变换和关于x轴的反射．．

变换的作用下得到点P，写出复合变换的变换公式，并求出点P的坐标.

（2）（本小题满分7分）选修4-4：坐标系与参数方程

已知曲线C的极坐标方程为2cos4sin.以极点为原点，极轴为x轴的正半轴，

- 4 -

建立平面直角坐标系，直线l的参数方程为x1tcos,(t为参数).

y1tsin（Ⅰ）判断直线l与曲线C的位置关系，并说明理由；

（Ⅱ）若直线l和曲线C相交于A,B两点，且AB32，求直线l的斜率.

（3）（本小题满分7分）选修4-5：不等式选讲

已知a0,b0,c0，

1113abc的最小值为m. a3b3c3（Ⅰ）求m的值；

（Ⅱ）解关于x的不等式|x1|2xm.

- 5 -

2015年厦门市高中毕业班适应性考试 数学（理科）试题参及评分标准

一、选择题： 1 题号 C 答案 二、填空题：

2 B 3 A 4 A 5 B 6 D 7 B 8 D 9 B 10 C 题号 11 12 13 14 答案 27 31 45 15 8 4或5 3三、解答题：

16．本题考查解三角形和三角函数图象及性质等知识，考查学生运算求解能力、数据处理能

力及推理论证能力，考查学生数形结合思想、函数与方程思想及转化与化归思想，属于中档偏易题.本题满分13分.

pp2p，∴?BCD，?CBD， BC=BD ···················· 1分 66312p3BCsin又∵BCD的面积为3，∴SDBCD=BD鬃······························ 2分 =BC2=3， ·234∴BC=2. ············································································································································· 3分

p在DABC中，AC7，?ABC，

3p222由余弦定理得：AC=AB+BC-2AB BCcos， ································································· 4分

3122AB，整理得AB2-2AB-3=0， 即7=AB+4-2创2∴AB=3，或AB=-1(舍去)，∴AB的长为3. ············································································· 6分 （Ⅱ）由（Ⅰ）知，A(2,0),B(-1,0),C(0,3)， ···················································································· 7分

解：（Ⅰ）∵?ABC∵函数f(x)Msin(x)(M0,0,为fx的图象与x轴相邻的两个交点， ∴函数f(x)的半个周期∴T=6=p，?ADC32)的图象经过A,B,C三点，其中A,BT1=3，对称轴为x=， ·········································································· 9分 222p， wp， ························································································································ 10分 31pppj=+kp,k Z，∴j=+kp,k Z， ∴?2323p又∵，∴j=， ·················································································································· 11分

23∵0，∴w=

- 6 -

ppx+)， 33p3又∵f(0)=Msin=M=32∴f(x)=Msin(······································································ 12分 3，∴M=2， ·

∴函数f(x)的解析式是f(x)=2sin(ppx+). ············································································· 13分 3317．本题以翻折的图形为载体，考查空间点、线、面位置关系、线面平行证明及求二面角大

小等有关基础知识，同时结合考查数列知识．本题考查空间想象能力、运算求解能力和推理论证能力，考查数形结合和化归与转化等数学思想方法. 本题满分13分.

A解：（Ⅰ）证法一：在梯形ABCD中，

AD∥BC， EF∥AB ，BE＝3，∴AF=3，

PQ又AD＝6，BC＝4，∴EC=1，FD=3， ································································································· 1分

1在线段AF上取点Q，使AQQF，连接PQ,QE, ······································································· 2分

2FD11∵APPD，∴PQ//DF，

231EC∵CE//DF，∴CE//PQ， ············································································································· 3分 3A∴四边形ECPQ为平行四边形，∴CP//EQ， ·················································································· 4分

P∵CP平面ABEF，EQ平面ABEF，∴CP∥平面ABEF. ··························································· 5分

B证法二：同证法一，EC=1，FD=3， ···································································································· 1分

B延长DC交FE的延长线于点M，连接AM，则MC∵AP1CD， ······················································ 2分 FD21PD，∴CP//MA， ·········································································································· 4分 2EC∵CP平面ABEF，MA平面ABEF，∴CP∥平面ABEF. ··························································· 5分

证法三：同证法一，EC=1，FD=3， ···································································································· 1分 M在线段DF上取点R，使FR∵AP1RD，连接PR，CR， 2AP1BPD，∴PR//AF，

2∵PR平面ABEF，AF平面ABEF，∴PR∥平面ABEF; ···························································· 2分 RFD1∵FRRD， ∴ECFR1,

2∵EC//FR， ∴四边形ECRF为平行四边形，∴CR//EF， EC∵CR平面ABEF，EF平面ABEF，∴CR∥平面ABEF; ···························································· 3分 ∵PRCRR，∴平面PRC//平面ABEF， ·················································································· 4分 ∵CP平面PRC，∴CP∥平面ABEF. ································································································· 5分

（Ⅱ）解法一：在梯形ABCD中，AB⊥AD，AD∥BC，∴EF⊥AF， EF⊥FD， ∵平面ABEF⊥平面EFDC，平面ABEF平面EFDC=EF，AF平面EFDC， ∴AF⊥平面EFDC， ······························································································································· 6分 设AFx(0x4)，

∵EF＝BA＝2，∴FD6x,EC4x， ∴FC4(4x)2， ······················································································································· 7分

2∵线段AF，FC，FD的长成等比数列，∴FCAFFD，

- 7 -

4(4x)2x(6x)，化简得x27x100， ∴x2或x5（舍）， ························································································································ 9分 以F为原点，FE，FD，FA分别为x,y,z轴建立空间直角坐标系，如图， 则F(0,0,0)，E(2,0,0)，C(2,2,0)，D(0,4,0)，A(0,0,2)， ··················································· 10分

∴EC(0,2,0)，EA(2,0,2)，

设n1(x1,y1,z1)是平面ACE的一个法向量，

BAzP2y10,n1EC0,则，即，

F2x12z10Dn1EA0y取z11，则x11,y10，∴n1(1,0,1)； ···················································································· 11分

又FC(2,2,0)，FA(0,0,2)，

设n2(x2,y2,z2)是平面ACF的一个法向量，

xEC2x22y20,n2FC0,则，即，

2z20n2FA0A取x21，则y21,z20,∴n2(1,1,0)； ··············································································· 12分

∴cosn1n2n1n211，

|n1||n2|2220PB∵ 二面角E-AC-F为锐角， ∴二面角E-AC-F为60. ······································································· 13分

F解法二：同解法一得x2或x5（舍）， ························································································· 9分 DHAF2,EC2，

G∵EF2，∴EFEC，

EC设点G为FC的中点，连接EG，则EGFC， ··········································································· 10分

∵AF⊥平面EFDC，AF平面AFC，∴平面AFC平面EFDC， ∵平面AFC平面EFDC=FC，∴EG⊥平面AFC， ∵AC平面AFC ， ∴EG⊥AC， ···································································································· 11分 过G作GH⊥AC交AC于H，连接EH， ∵EGGH=G， ∴AC⊥平面EGH， ∵EH平面EGH，∴AC⊥EH，

∴EHG是二面角E-AC-F的平面角， ····························································································· 12分

在RtCEF中，EG2， 在RtACE中，AE2∴sinEHG2,EC，2∴AC23，

EH22226，323EG23， EH262300∵EHG为锐角， ∴EHG60，即二面角E－AC－F为60. ················································ 13分 18．本题考查概率概念及其意义，事件，对立事件和互斥事件的概率，随机变量的分布

列及数学期望等相关知识；考查运算求解能力，数据处理能力和应用意识，考查转化与化归和必然与或然等数学思想方法. 本题满分13分. 解：(Ⅰ)设茶厂14日当天采茶的预期收益为万元，则的可能取值为6，3，1.5 ································· 1分

339322312224 ······························· 4分 P(6),P(3),P(1.5)55255555255525

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所以的分布列为

······························································································································································ 5分 所以的数学期望为E()6 6 3 1.5 9124P 25 25 25 912431.53.84， 252525即茶厂14日当天采茶的预期收益为3.84万元. ················································································ 6分

(Ⅱ)茶厂若采用方案①，设茶厂第二天采茶的预期收益为万元，则的可能取值为6和3，

因为P(6)32,P(3)， ····································································································· 7分 55所以的分布列为

 6 3 ································································································································································ 8分

41P 5 5

茶厂若采用方案②则其采茶总收益为y2633.214.8， ····················································· 12分

因为14.<14.8，所以茶厂应该采用方案②收益高，风险小，和谐社会，提供就业岗位. ······································································································································································ 13分 19．本题考查抛物线定义，直线方程，直线方程与抛物线、圆的位置关系等知识，考查学生

运算求解能力、推理论能力、抽象概括能力，考查数形结合、函数与方程、化归与转化等数学思想. 本题满分13分. 解：(Ⅰ)过点P作PQx轴于点Q，

当DPF3234.8， ··············································································· 9分 55

所以茶厂若采用方案①则其采茶总收益为y164.83.8414.， ····································· 11分

所以的数学期望为E62 时，PF4，PFPD4，3yDPFPQ6·································································································································· 1分 ， ·

RtPQF中，QFPFsin62， ··························································································· 2分

AOQFxDPPF.AFDPQF6，即 p6， ····································································· 3分 抛物线E的方程：y212x， ········································································································· 4分

（也可由余弦定理求得DF43，在RtDAF中，AF6，即p6） (Ⅱ) 解法一：当点P为原点O时，直线m的方程：x0与抛物线E切于点O； 设Px0,y0，则D3,y0，F3,0，kDF6y0k··········································· 5分 ，， ·my·y06DP6xx0，化简得：6xy0yy026x0， 直线m:yy0y022······················································································ 6分 代入y212x得y2y0yy06x0， ·CnOFx···················································································· 7分 y2y0yy00，yy0（0），

直线m与抛物线E有且只有一个交点P.························································································· 8分 2··········································· 5分 解法二：由(Ⅰ)得A3,0,F3,0，设P3t,6t，则D3,6t， ·22A

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11kDFt，km，直线m:y6tx3t2，即xty3t2， ············································ 6分

tt····················································································· 7分 代入y212x中，得y212ty36t20， ·

y6t，直线m与抛物线E有且只有一个交点P. ····································································· 8分

x32（Ⅲ）解法一：由已知得DP的中垂线：x0，与直线m:6xy0yy06x0联立，

2y023x029························································································ 9分 得到圆心C的纵坐标yc， ·

y0y023x029y0236BCy0ycy0，

y04y0又DPx03，则SCDP不妨设fy0y072y03y0361y372y1296 ·1BCDP················ 10分 ，·00296y096y0221296（y00）， y012963y0232fy03y0722y0当y023时，函数fy0有最小值；

y20232y2036y0········································ 11分 ， ·

由fy00得0y023，由fy00得y023，

当点P的坐标为1,23或1,23时， ··················································································· 12分

··················································································································· 13分 SCDP43取得最小值. ·

3t23解法二：由(Ⅱ)得DP的中垂线：x，又直线m:xty3t2垂直平分DF， 29t23圆心C的纵坐标：yC·································································································· 9分 ， ·y2tPD9t233t232BC6t，又DP3t3，

2t2t则SCDP1BCDP21t3t234t2Cn931At2t， ·································································· 10分 x·OF4t3不妨设ftt2t（t0），

3323ttt1， ·22·································· 11分 3t1t13312ft3t222ttt23330,,ft在递减，在递增；当时，函数ft有最小值； t333当点P的坐标为1,23或1,23时， ··················································································· 12分



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931t2t43取得最小值. ························································································ 13分 4t解法三：设DPF外接的圆C半径为R，DFP，不妨设t0，

6tt, ···················································· 9分 DFPPDFAFD，sin2236t36t1SCDP由正弦定理得：

t1， 3t232R3sinsintDP232DP92，又t1，

24232R29t4221t2BC3t11，则SCDPBCDP22t3t234t2931t2t. ········································· 10分 4t以下解法同上.

20．本小题主要考查函数的零点、函数的单调性、函数的最值、导数及其应用等基础知识，

考查归纳猜想、推理论证能力、运算求解能力、创新意识等，考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等．本题满分14分．

解：（Ⅰ）f'(x)exa， ·························································································································· 1分

当a0时，f'(x)0，f(x)在R上单调递增； ············································································ 2分 当a0时，令f'(x)0，得xlna；令f'(x)0，得xlna；

所以f(x)在(,lna)单调递减，在(lna,)单调递增. ································································ 4分 （Ⅱ）（ⅰ）证法一：依题意，只需研究关于x的方程

而

1x在R+上根的个数， f(x)1xxf(x)10， f(x)记g(x)xf(x)1xexx21，即g(x)xexx21(x0)， ················································ 5分

g'(x)(x1)ex2x，

x由（Ⅰ）知，当a1时，f(x)f(0)，即ex10，

g'(x)(x1)ex2x(x1)22xx210，

·················································································································· 7分 g(x)在R+上单调递增； ·22又g(ln2)2ln2(ln2)1(ln21)0,g(1)e20, ···················································· 8分 函数g(x)的图像在R+上连续不断，存在唯一x0(ln2,1)，使得g(x0)0，

1(x0)在R+上存在唯一的不动点x0，且x0(ln2,1). ········································· 9分 函数yf(x)1证法二：依题意，只需研究关于x的方程x在R+上根的个数，

f(x)1xxf(x)10， 而

f(x)x2x2记g(x)xf(x)1xex1，即g(x)xex1(x0)， ················································ 5分 g'(x)(x1)ex2x，g''(x)(x2)ex2， x0，(x2)ex2e02，g''(x)0， ·········································································· 6分 g'(x)(x1)ex2x在R+上单调递增，g'(x)g'(0)10，

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·················································································································· 7分 g(x)在R+上单调递增； ·

以下同证法一. ········································································································································ 9分 （ⅱ）证法一：要证不等式

f(a2n)f(x0)f(x0)x01成立，

a2nx0(ea2na2n)(ex0x0)ea2nex0x0只需证·············· 10分 (ex0)x01成立，即证ex0.…（*） ·

a2nx0a2nx0下面用数学归纳法证明：nN*,a2nx0.

①当n1时，由（Ⅰ）得，f(x)在(0,)上单调递增，

由（ⅰ）得，0ln2x0，f(ln2)f(x0)，又a21，

f(a1)a211x0，

f(ln2)f(x0)即a2x0，n1时，结论成立. ····································································································· 11分

*② 假设nkkN时，结论成立，即a2kx0，

f(x)在(0,)上单调递增，且a2kx00，f(a2k)f(x0)f(0)1，

011， f(a2k)f(x0)又a2k1a2k211x0，所以0a2k1x0，f(0)f(a2k1)f(x0)，

f(a2k)f(x0)11x0，即a2k2x0. nk1时，结论成立. f(a2k1)f(x0)根据①，②可得，nN*,a2nx0. ································································································ 12分 所以要证不等式（*）成立，只需证ea2nex0ex0(a2nx0)成立， 即证ea2na2nex0ex0x0ex0. …………（**） 记h(x)exxex0(xx0)，

······················ 13分 h'(x)exex00，当且仅当xx0取到等号， ·

h(x)在[x0,)上单调递增，

nN*,a2nx0，h(a2n)h(x0)，即（**）式成立. f(a2n)f(x0)··············································································· 14分 f(x0)x01，命题得证. ·

a2nx0f(a2n)f(x0)证法二：要证不等式f(x0)x01成立，

a2nx0f(a2n)f(x0)f(a2n)f(x0)只需证············· 10分 ex01成立，只需证f'(x0). ………（*） ·

a2nx0a2nx0下面用数学归纳法证明：nN*,a2nx0.（过程、得分同证法一） ············ （此步骤共2分）12分 所以要证不等式（*）成立，只需证f(a2n)f(x0)f'(x0)(a2nx0)成立， 即证f(a2n)f'(x0)a2nf(x0)f'(x0)x0. …………（**）

记h(x)f(x)f'(x0)x(xx0)，h'(x)f'(x)f'(x0)， ·················································· 13分

f'(x)ex1在[x0,)上单调递增，xx0，

当且仅当xx0取到等号， h(x)在[x0,)上单调递增，h'(x)f'(x)f'(x0)0，

又

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nN*,a2nx0，h(a2n)h(x0)，即（**）式成立. f(a2n)f(x0)··············································································· 14分 f(x0)x01，命题得证. ·

a2nx0f(a2n)f(x0)证法三：要证不等式f(x0)x01成立，

a2nx0f(a2n)f(x0)1只需证····················································································· 10分 x01成立， ·

a2nx0x0下面用数学归纳法证明：nN*,a2nx0.（过程、得分同证法一） ············ （此步骤共2分）12分 所以要证不等式（*）成立，只需证f(a2n)f(x0)(即证f(a2n)(1x01)(a2nx0)成立， x011x01)a2nf(x0)(x01)x0. …………（**） x0x0111记h(x)f(x)(x01)x(xx0)，h'(x)f'(x)(x01)exx0，

x0x0x0 ······································································································································································ 13分

1x0在[x0,)上单调递增，xx0， x0111即h( 'x)0当且仅当xx0取到等号，h'(x)h'(x0)(ex0x0)0，

x0x0x0h(x)在[x0,)上单调递增，

又

h'(x)exnN*,a2nx0，h(a2n)h(x0)，即（**）式成立.

f(a2n)f(x0)···················································································· 14分 f(x0)x01，命题得证. ·

a2nx021．（1）本题考查矩阵与变换、矩阵的乘法等知识；考查运算求解能力；函数与方程数学思

想．满分7分.

解：（Ⅰ）解法一：矩阵A的特征多项式f()a2························· 2分 (a)(4)2，

1412··················································· 3分  ·

14x2，(2a)(24)20，解得a1，A0解法二：设属于矩阵A的特征值2的一个特征向量为，

y0x0a2x02······························································································· 2分 则A2，即， ·14y0y0ax0x0ax02y02x0x0即，整理得，为非零向量，则x00，所以a1，

x4y2yx2yy00000012 ····································································································································· 3分 A·14（Ⅱ）设关于x轴的反射变换对应的矩阵为B，则B复合变换对应的矩阵为BA1

10·························································· 4分 ， ·

0101212， ··························································· 5分 011414- 13 -

xx2y， ······················································································· 6分 复合变换的变换公式为yx4yx1x3将 代入的变换公式，得，P的坐标为(3,3). ········································· 7分

y1y3（2）本小题主要考查直线的参数方程及其几何意义、圆的极坐标方程、直线与圆的位置关系

等基础知识；考查运算求解能力；数形结合思想．满分7分．

解：(Ⅰ)

···················································· 1分 2cos4sin，22cos4sin， ·

································· 3分 曲线C的直角坐标方程为x2y22x4y，即(x1)2(y2)25， ·

直线l过点(1,-1)，且该点到圆心的距离为(11)2(12)25，\\直线l与曲线C相交. ····························································································································································· 4分 (Ⅱ)解法一：当直线l的斜率不存在时，直线l过圆心，AB2532， ········································· 5分

则直线l必有斜率，设其方程为y1k(x1)，即kxyk10， 圆心到直线l的距离d1k21解得k1，直线l的斜率为1. ····································································································· 7分

解

法

二

：

将

(5)2(3222， ····························································· 6分 )22x1tcosy1tsin代入

(x1)2(y2)25，得

(tcos)2(1tsin)25，

整理得，t22(sin)t40， ········································································································· 5分 设A,B两点对应的参数分别为t1,t2，则t1t22sin,t1t24，

AB|t1t2|(t1t2)24t1t24sin21632， ···························································· 6分

不妨设为直线的l的倾斜角，则sin32，则或，直线l的斜率为1. ·············· 7分

442（3）本小题主要考查利用二元和三元基本不等式求最值、绝对值不等式的解法等基础知识；

考查运算求解能力；化归与转化、分类与整合的思想．满分7分．

1113 ·3解：（Ⅰ）a,b,cR，13131························································· 1分 33333abcabcabc1113 3333abc3abc ①

abcabc而33abc233abc6 ② ···································································································· 2分 abcabc3 ··················································································································· 3分 a3b3c36 ③abc当且仅当abc时， ①式等号成立；当且仅当33abc时，②式等号成立；

abc333则当且仅当abc1时，③式等号成立，即abc3取得最小值m6. ·················· 4分 abc(Ⅱ)由（Ⅰ）知m6，则|x1|2x6，即|x1|62x，

62xx162x ， ·············································································································· 5分

762xx1x 解得··································································································· 6分 3 ·x162xx5

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·········································································································· 7分 原不等式的解集为(,). ·

73

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