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2008年普通高等学校招生全国统一考试文科数学(必修+选修I)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10页．考试结束后，将本试卷和答题卡一并交回．第Ⅰ卷注意事项：1．答第Ⅰ卷前，考生务必将自己的姓名、准考证号、考试科目涂写在答题卡上．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，再选涂其它答案标号．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AB，互斥，那么球的表面积公式()()()PABPAPB+=+24πSR=如果事件AB，相互独立，那么其中R表示球的半径()()()PABPAPB=球的体积公式如果事件A在一次试验中发生的概率是p，那么34π3V

R=n次独立重复试验中事件A恰好发生k次的概率其中R表示球的半径()(1)(012)kknkknPkCppkn−=−=，，，，一、选择题1．若sin0且tan0是，则是（）A．第一象限角B．第二象限角C．

第三象限角D．第四象限角2．设集合{|32}Mmm=−Z，{|13}NnnMN=−=Z则，≤≤（）A．01，B．101−，，C．012，，D．1012−，，，3．原点到直线052=−

+yx的距离为（）A．1B．3C．2D．54．函数1()fxxx=−的图像关于（）A．y轴对称B．直线xy−=对称C．坐标原点对称D．直线xy=对称5．若13(1)ln2lnlnxeaxbxcx−===，，，，，则（）A．a<b<cB．c<a

<bC．b<a<cD．b<c<a6．设变量xy，满足约束条件：222yxxyx+−，，．≥≤≥，则yxz3−=的最小值为（）A．2−B．4−C．6−D．8−7．设曲线2axy=在点（1，a）处的切线与直线062=−−yx平行，则=a（）A．1B．12C．12−D．1−8．正四棱

锥的侧棱长为32，侧棱与底面所成的角为60，则该棱锥的体积为（）A．3B．6C．9D．189．44)1()1(xx+−的展开式中x的系数是（）A．4−B．3−C．3D．410．函数xxxfcossin)(−=的最大值为（）A．1B．2C．3

D．211．设ABC△是等腰三角形，120ABC=，则以AB，为焦点且过点C的双曲线的离心率为（）A．221+B．231+C．21+D．31+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．

22008年普通高等学校招生全国统一考试文科数学(必修+选修I)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(12)(23)==，，，ab，若向量+ab与向量(47)=−−，c共线，则=．14．从10名男同学，6名女同学中选

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不同选法共有种（用数字作答）15．已知F是抛物线24Cyx=：的焦点，AB，是C上的两个点，线段AB的中点为(22)M，，则ABF△的面积等于．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，5cos13A=−，3cos5B=．（Ⅰ）求sinC的值；（Ⅱ）设5BC=，求ABC△的面积．18．（本小题满分12分）等差数列na中，410a=且3610

aaa，，成等比数列，求数列na前20项的和20S．19．（本小题满分12分）甲、乙两人进行射击比赛，在一轮比赛中，甲、乙各射击一发子弹．根据以往资料知，甲击中8环，9环，10环的概率分别为0.6，0.3，0.1

，乙击中8环，9环，10环的概率分别为0.4，0.4，0.2．设甲、乙的射击相互独立．（Ⅰ）求在一轮比赛中甲击中的环数多于乙击中环数的概率；（Ⅱ）求在独立的三轮比赛中，至少有两轮甲击中的环数多于乙击中环数的概

率．20．（本小题满分12分）如图，正四棱柱1111ABCDABCD−中，124AAAB==，点E在1CC上且ECEC31=．（Ⅰ）证明：1AC⊥平面BED；（Ⅱ）求二面角1ADEB−−的大小．21．（本小题满分12分）设aR

，函数233)(xaxxf−=．（Ⅰ）若2=x是函数)(xfy=的极值点，求a的值；（Ⅱ）若函数()()()[02]gxfxfxx=+，，，在0=x处取得最大值，求a的取值范围．22．（本小题满分12分）设椭圆中心在坐标原点，(20)(01)AB，，，是它的两个顶点，直线)0(=kkx

y与AB相交于点D，与椭圆相交于E、F两点．（Ⅰ）若6EDDF=，求k的值；（Ⅱ）求四边形AEBF面积的最大值．ABCDEA1B1C1D12008年普通高等学校招生全国统一考试文科数学试题（必修+选修

Ⅰ）参考答案和评分参考评分说明：1．本解答给出了一种或几种解法供参考，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参考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现错误时，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．只给整数分数．选择题不给中间分．一、选择题1．C2．B3．D4．C

5．C6．D7．A8．B9．A10．B11．B12．C提示：1、,0sin在第三或四象限，0tan，在第一或三象限为第三象限角2、}1,0,1{},21|{−=−=ZxxxNM3、555==d4、)(xf为奇函数5、cabxxe−

−0ln1116、当=−=22yx时，83min−=−=yxZ7、axy2'=，当1=x时，122,2'===aaay8、如图，,60,32oSAOSA==则6,3,360sin====ABAOSASOo636312==V9、444)1()1()1(xx

x−=+−，x的系数为414−=−C10、)4sin(2cossin)(−=−=xxxxf)(xf最大值为211、设1||=AB，则3=AC，13||||2−=−=CBACa，1||2==ABC，21322+==ace12、1O与2O的公共弦为AB，球心为Ｏ，AB中点ＯＯ２ＣＯ

１CDBAS为C，则四边形COOO21为矩形，所以OCACACOAOCOO⊥===,1||,2|||,|||213||||||22=−=ACOAOC二、填空题13．214．42015．216．两组相对侧面分

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．13、20)2(7)32(4)32,2(==+−+++=+ba；14、42036310316=−−CCC；15、设),(),(2211yxByxA，),(

444122122121222xxyyxyxy−=−==14121212=+=−−yyxxyyAB所在直线方程为22−=−xy即xy=，又4,04212====xxxyxy，22||||211||24||2||12==

==−=OFABSOFxxABABF;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5cos13A=−，得12sin13A=，由3cos5B=，得4sin5B=．···································

································································2分所以16sinsin()sincoscossin65CABABAB=+=+=．·····

············································5分（Ⅱ）由正弦定理得45sin13512sin313BCBACA===．···················································

······8分所以ABC△的面积1sin2SBCACC=1131652365=83=．·····························10分18．解：设数列na的公差为d，则3410aadd=−=−，642102a

add=+=+，1046106aadd=+=+．·········································································································3分由3610aaa

，，成等比数列得23106aaa=，即2(10)(106)(102)ddd−+=+，整理得210100dd−=，解得0d=或1d=．·······································································

···········································7分当0d=时，20420200Sa==．·····················································································

········9分当1d=时，14310317aad=−=−=，于是2012019202Sad=+207190330=+=．·····························································1

2分19．解：记12AA，分别表示甲击中9环，10环，12BB，分别表示乙击中8环，9环，A表示在一轮比赛中甲击中的环数多于乙击中的环数，B表示在三轮比赛中至少有两轮甲击中的环数多于乙击中的环数，12CC，

分别表示三轮中恰有两轮，三轮甲击中环数多于乙击中的环数．（Ⅰ）112122AABABAB=++，·················································································

·······2分112122()()PAPABABAB=++112122()()()PABPABPAB=++112122()()()()()()PAPBPAPBPAPB=++0.30.40.10.40.10.40.2=++=．··························

·······················································6分（Ⅱ）12BCC=+，·······································································

···········································8分22213()[()][1()]30.2(10.2)0.096PCCPAPA=−=−=，332()[()]0.20.008PCPA==

=，1212()()()()0.0960.0080.104PBPCCPCPC=+=+=+=．·····································12分20．解法一：依题设，2AB

=，1CE=．（Ⅰ）连结AC交BD于点F，则BDAC⊥．由三垂线定理知，1BDAC⊥．································································································3分在平面1ACA内

，连结EF交1AC于点G，由于122AAACFCCE==，故1RtRtAACFCE△∽△，1AACCFE=，CFE与1FCA互余．于是1ACEF⊥．1AC与平面BED内两条相交直线BDEF，都垂直，所以1AC⊥平面BED．··············

······························································································6分（Ⅱ）作GHDE⊥，垂足为H，连结1AH．由三垂线定理知1AHDE⊥，故1

AHG是二面角1ADEB−−的平面角．·········································································8分223EFCFCE=+=，23CECFCGEF==，2233EGCECG=−=．13EGEF=，1231

5EFFDGHDE==．又221126ACAAAC=+=，11563AGACCG=−=．11tan55AGAHGHG==．所以二面角1ADEB−−的大小为arctan55．·························

·········································12分解法二：以D为坐标原点，射线DA为x轴的正半轴，建立如图所示直角坐标系Dxyz−．依题设，1(220)(020)(021)(204)BCEA，，，，，，，，，，，

．ABCDEA1B1C1D1yxzABCDEA1B1C1D1FHG(021)(220)DEDB==，，，，，，11(224)(204)ACDA=−−=，，，，，．·······································3分（Ⅰ）因为

10ACDB=，10ACDE=，故1ACBD⊥，1ACDE⊥．又DBDED=，所以1AC⊥平面DBE．···················································

·························································6分（Ⅱ）设向量()xyz=，，n是平面1DAE的法向量，则DE⊥n，1DA⊥n．故20yz+=，240xz+=．令1y=，则2z=−，4x=，(412)=−，，n．·····

·································································9分1AC，n等于二面角1ADEB−−的平面角，11114cos42ACACAC==，nnn．所以二面角1ADEB−−的大小为14

arccos42．·································································12分21．解：（Ⅰ）2()363(2)fxaxxxax=−=−．因为2x=是函数()yfx=的极值点，所以(2)0f=

，即6(22)0a−=，因此1a=．经验证，当1a=时，2x=是函数()yfx=的极值点．······················································4分（Ⅱ）

由题设，3222()336(3)3(2)gxaxxaxxaxxxx=−+−=+−+．当()gx在区间[02]，上的最大值为(0)g时，(0)(2)gg≥，即02024a−≥．故得65a≤．·········

·····················································································································9分反之，当65a≤时，对任意[02

]x，，26()(3)3(2)5gxxxxx+−+≤23(210)5xxx=+−3(25)(2)5xxx=+−0≤，而(0)0g=，故()gx在区间[02]，上的最大值为(0)g．综上，a的取值范围为65−，．························

··································································12分22．（Ⅰ）解：依题设得椭圆的方程为2214xy+=，直线ABEF，

的方程分别为22xy+=，(0)ykxk=．··················································2分如图，设001122()()()DxkxExkxFxkx，，，，，，其中12

xx，且12xx，满足方程22(14)4kx+=，故212214xxk=−=+．①由6EDDF=知01206()xxxx−=−，得021221510(6)77714xxxxk=+==+；由D在AB上知0022xkx+=，得0212xk=+．所以221012714kk=++，化简得224

2560kk−+=，解得23k=或38k=．·················································································································6分

（Ⅱ）解法一：根据点到直线的距离公式和①式知，点EF，到AB的距离分别为21112222(1214)55(14)xkxkkhk+−+++==+，DFByxAOE22222222(1214)55(14)xkxkkhk+

−+−+==+．········································································9分又2215AB=+=，所以四边形AEBF的面积为121()2SABhh=+214(12)525(14)kk+=+22(12)14kk

+=+22144214kkk++=+22≤，当21k=，即当12k=时，上式取等号．所以S的最大值为22．·································12分解法二：由题设，1BO=，2AO=．设11ykx=，22yk

x=，由①得20x，210yy=−，故四边形AEBF的面积为BEFAEFSSS=+△△222xy=+··························································

··········································································9分222(2)xy=+22222244xyxy=++22222(4)xy+≤22=，当222xy=时，上式取等号．所以S的最大值为22．·····

···············································12分