高考卷 08届 普通高等学校招生全国统一考试数学(全国Ⅱ·文科)(答案,完全word版)

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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34π3VR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.abcB.cabC.bacD.bca6.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.3D.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3610aaa,,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(kkxy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.C5.C6.D7.A8.B9.A10.B11.B12.C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.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,642102aadd,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.·············································12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1AHG是二面角1ADEB的平面角.························································8分223EFCFCE,23CECFCGEF,2233EGCECG.13EGEF,12315EFFDGHDE.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,,,,,,,,,,,.(021)(220)DEDB,,,,,,11(224)(204)ACDA,,,,,.······························3分ABCDEA1B1C1D1yxzABCDEA1B1C1D1FHG(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1DA

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