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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.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.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)(2
3)==,,,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(2
2)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aR,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|{−=−=ZxxxNM3、555==d4、)(xf为奇函数5、cabxxe−−0ln1116、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(xxx−=+−,x的系数为414−=−C10、)4sin(2cossin)(−=−=xxxxf)(xf最大值为211、设1||=A
B,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OO2CO1CDBAS为C,则四边形COOO21为矩形,所以OCACACOAOCOO⊥===,1||,2|||,|||
213||||||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4,0421
2====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5cos
13A=−,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1sin2SBCAC
C=1131652365=83=.·····························10分18.解:设数列na的公差为d,则3410aadd=−=−,642102aadd=+=+,1046106
aadd=+=+.·········································································································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()()()P
ABPABPAB=++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()()()()0.0960.0080.104PBPCCPCP
C=+=+=+=.·····································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arcta
n55.··································································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14arccos42.·································································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221
4xy+=,直线ABEF,的方程分别为22xy+=,(0)ykxk=.··················································2分如图,设001122()()()DxkxExkxFxkx,,,,,,其中12xx,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2242560kk−+=,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=
,22ykx=,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