真的是2011广州二模理科数学答案
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2011年广州市普通高中毕业班综合测试(二)
数学(理科)试题参考答案及评分
说明:1(参考答案与评分标准指出了每道题要考查的主要知识和能力,并给出了一种或几种
解法供参考,如果考生的解法与参考答案不同,可根据试题主要考查的知识点和能力
比照评分标准给以相应的分数(
2(对解答题中的
,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得分,但所给分数不得超过
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,选择题和填空题不给中间分(
一、选择题:本大题主要考查基本知识和基本运算(共8小题,每小题5分,满分40分(
1 2 3 4 5 6 7 8 题号
D C A B A B C B 答案
二、填空题:本大题主要考查基本知识和基本运算(本大题共7小题,考生作答6小题,每
小题5分,满分30分(其中14,15题是选做题,考生只能选做一题(
323329( 10(fxxx,,, 11(4 12(11 13(?? 14( ,,27
,,4,,,,,,,sin1cos1,,,,sin1,,15(或或或,,,,,,,,,,,,363,,,,,,
3cossin20,,,,,,,
三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程和演算步骤. 16((本小题满分12分)
(本小题主要考查方位角、正弦定理、余弦定理等基础知识,考查运算求解能力等()
,,BAC120AB,12解:(1)依题意,,,AC,,,10220,
北 ,,BCA,(„„„„„„„„„2分
在?ABC中,由余弦定理,得 C
222BCABACABACBAC,,,,,,2cos „„„„„„„„4分
22,,,,,,,122021220cos120784 (
东 西 解得BC,28( „„„„„„„„„„„„„„„„„„„„„6分 ,A 60 BC所以渔船甲的速度为海里/小时( ,14B 2南 14答:渔船甲的速度为海里/小时(„„„„„„„„„„„„„7分
,,BAC120AB,12ABCBC,28(2)方法1:在?中,因为,,,
,,BCA,,
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由正弦定理,得
ABBC(„„„„„„„„„„„„„„„„„„„„„„„„„„9分 ,sinsin120,
3,12ABsin120332,即,,,( sinBC2814
答:的值为sin,
33(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„12分 14
方法2:在?中,因为AB,12,,,, ABCAC,20BC,28,,BCA,
由余弦定理,得
222ACBCAB,,cos,,(„„„„„„„„„„„„„„„„„„„„„„9分 2ACBC,
22220281213,,,即cos,,( 2202814,,
23313,,2因为为锐角,所以,,( sin1cos1,,,,,,,,1414,,
答:的值为sin,
33(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„12分 14
17((本小题满分12分)
(本小题主要考查概率与统计的概念、随机变量的分布列等基础知识,考查运算求解能力等() 解:(1)由表格数据可知,视觉记忆能力恰为中等,且听觉记忆能力为中等或中等以上的学
10,a生共有人(记“视觉记忆能力恰为中等,且听觉记忆能力为中等或中等以上”为,,
A事件,
102,a则,解得PA(),,405
a,6(„„„„„„„„„„„„„„„„„„„„„„„„„„2分
ba,,,,,,40(32)40382所以(
答:的值为6,b的值为a
2(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„3分 (2)由表格数据可知,具有听觉记忆能力或视觉记忆能力超常的学生共有8人(
B方法1:记“至少有一位具有听觉记忆能力或视觉记忆能力超常的学生”为事件,
B则“没有一位具有听觉记忆能力或视觉记忆能力超常的学生”为事件,
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3C12412332所以( PBPB()1()11,,,,,,,3C24724740
答:从这40人中任意抽取3人,其中至少有一位具有听觉记忆能力或视觉记忆能力超常的学生的概率为
123(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„247
„„6分
方法2:记“至少有一位具有听觉记忆能力或视觉记忆能力超常的学生”为事件B,
12213CCCCC,,1238328328所以( PB,,,,3C24740
答:从这40人中任意抽取3人,其中至少有一位具有听觉记忆能力或视觉记忆能力超常的学生的概率为
123(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„247
„„6分
3(3)由于从40位学生中任意抽取3位的结果数为,其中具有听觉记忆能力或视觉记忆能C40
力偏高或超常的学生共24人,从40位学生中任意抽取3位,其中恰有位具有听觉记忆k能力或视觉记忆能力偏高或超常的结果数为
kk3,,„„„„„„„„„„„„„„„„„„„„„„„„„„„7分 CC2416
所以从40位学生中任意抽取3位,其中恰有位具有听觉记忆能力或视觉记忆能力偏高k
kk3,CC2416或超常的概率为,Pk,,,()3C40k,0,1,2,3„„„„„„„„„„„„„„„„„„„„„„„„8分 ,,
,的可能取值为0,1,2,
3,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„9分
0312CCCC6因为, , ,,,,,,,,P(0)P(1)33C247C2474040
2130CCCC55225324162416,, ,,,,,,,,P(2)P(3)33C1235C12354040
,所以的分布列为
, 0 1 2 3
„„„„„„„„10分 1472552253 P 24724712351235
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E,,,0所以( ,,1,,2,,3,,5
,答:随机变量的数学期望为9(„„„„„„„„„„„„„„„„„„„„„„„„„„„12分 5
,若将抽取的3人理解为可重复抽取~而采用二项分布求解~可酌情给分,
18((本小题满分14分)
(本小题主要考查空间线线、线面关系,二面角,三视图等知识,考查化归与转化数学思想方法,以及空间想象能力、推理论证能力、运算求解能力()
EAABC,平面AABCC,平面1)证明:因为,,所以,即( 方法1:(EAAC,EDAC,
ABEDA,又因为,,所以平面EBD( ACAB,AC,
BDEBD,平面因为,所以
(„„„„„„„„„„„„„„„„„„„„„„„„4分 ACBD,
AB(2)解:因为点、、在圆的圆周上,且,所以为圆的直径( COABAC,BCO
设圆的半径为,圆柱高为,根据正(主)视图、侧(左)视图的面积可得, Ohr
E 1,2210,rhr,,,,,2„„„„„„„„„„„„„„„„6分 ,C 1,22212.rhr,,,,,A ,2A1O
1 B r,2,,解得 ,DD 1h,2.,
1 所以,BC,4
(„„„„„„„„„„„„„„„„„„„„„„„„„„„7分 ABAC,,22
AH过点HC作CHBD,于点,连接,
ACCHC,BD,由(1)知,ACBD,,,所以平面ACH(
AH,BDAH,因为平面ACH,所以(
所以,AHC为二面角ABDC,,的平面角(„„„„„„„„„„„„„„„„„„„„„„9分
ABDAH,ABD由(1)知,AC,平面,平面,
所以ACAH,,即?CAH为直角三角形(
22BADAD,2在Rt?中,,,则BDABAD,,,23( AB,22
26AH,ABADBDAH,,,由,解得( 3
因为
AC(„„„„„„„„„„„„„„„„„„„„„„„„„„„„13tan3,,,AHCAH
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,60所以( ,AHC
所以二面角的平面角大小为ABDC,,
60(„„„„„„„„„„„„„„„„„„„„„14分
方法2:(1)证明:因为点A、B、在圆的圆周上,且,所以为圆的COABAC,BCO
直径(
设圆的半径为,圆柱高为,根据正(主)视图、侧(左)视图的面积可得, Ohr
E 1,2210,rhr,,,,,2„„„„„„„„„„„„„„„„2分 ,C 1,22212.rhr,,,,,A ,2A1O
1 B r,2,,解得 ,DD 1h,2.,
1 所以,BC,4
(„„„„„„„„„„„„„„„„„„„„„„„„„„„3分 ABAC,,22
DDE以点为原点,、所在的射线分别为轴、轴建立如图的空间直角坐标系DDzx1
Dxyz,D0,0,0D4,0,0A0,0,2B2,2,2C2,2,2,,则,,,,,,,,,,,,,,,1
AC,,2,2,0DB,2,2,2,( ,,,,
„„„„„„„„„5分 z
E ACDB,,,2,2,02,2,20因为, ,,,,
C 所以( ACDB,A A1O 所以ACBD,(„„„„„„„„„„„„„„„„„„„9分 1 B
BC,,0,4,0n,xyz,,BCD(2)解:设是平面的法向量,因为, ,,,,D x D1
,nBC,0,,,40,y,,1 y所以即 ,,2220.xyz,,,nDB,0.,,,
z,,1n,,1,0,1取,则是平面BCD的一个法向,,
量(„„„„„„„„„„„„„„„„„11分
ABBDB,ABDACBD,ACAB,AC,由(1)知,,又,,所以平面(
AC,,2,2,0ABD所以是平面的一个法向,,
量(„„„„„„„„„„„„„„„„„„„„12分
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n,AC21因为, cos,nAC,,,2222,n,AC
所以( n,60AC,
而等于二面角的平面角, ABDC,,n,AC
所以二面角的平面角大小为ABDC,,
60(„„„„„„„„„„„„„„„„„„„„„14分
EAABC,平面AABCC,平面方法3:(1)证明:因为,,所以,即( EAAC,EDAC,
ABEDA,又因为,,所以平面EBD( ACAB,AC,
BDEBD,平面因为,
所以
(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„4ACBD,
分
AB(2)解:因为点、、在圆的圆周上,且,所以为圆的直径( COABAC,BCO设圆的半径为,圆柱高为,根据正(主)视图、侧(左)视图的面积可得, Ohr
E 1,2210,rhr,,,,,2„„„„„„„„„„„„„„„„6分 ,C 1,22212.rhr,,,,,A ,2A1O
1 B r,2,,解得 ,DD 1h,2.,
1 所以BC,4,
(„„„„„„„„„„„„„„„„„„„„„„„„„„„7分 ABAC,,22
DDE以点为原点,、所在的射线分别为轴、轴建立如图的空间直角坐标系DDzx1
Dxyz,D0,0,0D4,0,0A0,0,2B2,2,2C2,2,2,,则,,,,,,,,,,,,,,,1
BC,,0,4,0DB,2,2,2,( ,,,,
„„„„„„„„„„9分 z
E n,xyz,,设是平面BCD的法向量, ,,
C ,nBC,0,,,40,y,,则即 ,,A A12220.xyz,,,O nDB,0.,,,1 B z,,1n,,1,0,1BCD取,则是平面的一个法向量(„„„11分 ,,D x D1
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ABBDB,由(1)知,,又,, ACBD,ACAB,
所以平面( ABDAC,
AC,,2,2,0所以是平面ABD的一个法向,,
量(„„„„„„„„„„„„„„„„„„„„12分
n,AC21因为, cos,nAC,,,2222,n,AC
所以( n,60AC,
而等于二面角的平面角, ABDC,,n,AC
所以二面角的平面角大小为ABDC,,
60(„„„„„„„„„„„„„„„„„„„„„14分
19((本小题满分14分)
(本小题主要考查等差数列、等比数列和不等式等基础知识,考查运算求解能力、推理论证
能力,以及
与方程、化归与转化等数学思想() (1)解法1:当时,n,2
na,1,,nann,1aSS,,,,,„„„„„„„„„„„„„„„„2分 nnn,122
即
aann,1n,2(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„,,,1,nn
„„„4分
aa,,n1所以数列是首项为的常数,1,,n1,,
列(„„„„„„„„„„„„„„„„„„„„„„„5分
a*nn,N所以,即( an,1,,,nn
a的通项
为所以数列,,n
*n,N(„„„„„„„„„„„„„„„„„„„„„7分 an,,,n
解法2:当n,2时,
na,1,,nann,1aSS,,,,,„„„„„„„„„„„„„„„„2分 nnn,122
即
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ann(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„,n,2,,1an,n,1
„4分
所以
aaaann,132nn,132(„„„„„„„„„5aan,,,,,,,,,,,,,1n1aaaann,,1221nn,,1221
分
因为,符合的表达a,1a1n式(„„„„„„„„„„„„„„„„„„„„„„„„„„„„6分
所以数列的通项公式为a,,n
*n,N(„„„„„„„„„„„„„„„„„„„„„7分 an,,,n
*kmk,,2,,N(2)假设存在,使得、、成等比数列, kbbb,,kk,1k,2
则
2(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„bbb,,kk,2k1
8分
因为(n?2), ban,,lnlnnn
所以
222,,ln2kk,,,lnln2kk,,,,,, „„„„„„„„„„„„112),,bbkk,,,,,lnln(kk,,,222,,,,,,
分
22,,ln1k,2,,2(„„„„„„„„„ln1kb,,,,,,,,,,k,1,,2,,,,
„„„„13分
2这与b矛盾( bb,,kk,2k1
,故不存在(),使得、、成等比数kbkk,,2,Nbbkk,1k,2列(„„„„„„„„„„„„„14分
20((本小题满分14分)
(本小题主要考查圆、双曲线、直线方程和不等式等基础知识,考查运算求解能力和推理论
证能力,以及数形结合、分类讨论思想和创新意识等()
bab,,0解:(1)因为,所以,所以,1a
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222cabb,,,(„„„„„„„1分 e,,,,1,2,,aaa,,
,,APB90OPb,2由及圆的性质,可知四边形是正方形,所以( PAOB
b2OPba,,2,因为,所以,所以a2
2226cabb,,,,(„„„„„3分 e,,,,1,,2aaa,,
故双曲线离心率的取值范围为e
,,6,2(„„„„„„„„„„„„„„„„„„„„„„4分 ,,,2,,
222222(2)方法1:因为, PAOPOAxyb,,,,,00
PPAP所以以点为圆心,为半径的圆的方程为
22222xxyyxyb,,,,,,(„„„5分 ,,,,0000
P因为圆O与圆两圆的公共弦所在的直线即为直线AB,„„„„„„„„„„„„„„„„„6分
所以联立方程组
222,xyb,,,,„„„„„„„„„„„„„„„„„„7分 ,22222xxyyxyb,,,,,,.,,,,,0000,
22xAB消去,,即得直线的方程为y
2(„„„„„„„„„„„„„„„„„„8分 xxyyb,,00
Axy,Bxy,Pxy,方法2:设,已知点, ,,,,,,112200
yy,y011k,其中xxx,,,0则,( k,,,101OAPAxx,x011
PAOA,因为,所以,即kk,,1PAOAyy,y011,,,1(„„„„„„„„„„„„„„„„5分 xxx,011
22xxyyxy,,,整理得( 010111
222因为xyb,,,所以11
2xxyyb,,(„„„„„„„„„„„„„„„„„„„„„„„6分 0101
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因为,PAPB,,根据平面几何知识可知,( OAOB,ABOP,
y0因为,所以k,OPx0
x0(„„„„„„„„„„„„„„„„„„„„„„„„„„„7分 k,,ABy0
x0所以直线AB方程为( yyxx,,,,,,11y0
( 即xxyyxxyy,,,000101
AB所以直线的方程为
2(„„„„„„„„„„„„„„„„„„„„„„„„8分 xxyyb,,00
方法3:设AxyBxy,,,,已知点Pxy,, ,,,,,,112200
yy,y011则k,其中xxx,,,0,( k,,,101OAPAxx,x011
因为,所以,即PAOA,kk,,1PAOA
yy,y011,,,1(„„„„„„„„„„„„„„„„5分 xxx,011
22整理得( xxyyxy,,,010111y P 2222因为,所以(„„6分 xyb,,xxyyb,,010111
A 2A这说明点在直线上( „„„„7分 xxyyb,,00
2B同理点也在直线上( xxyyb,,00x O
2B AB所以xxyyb,,就是直线的方程( „„8分 00
2AB(3)由(2)知,直线的方程为, xxyyb,,00
2bAB所以点O到直线的距离为d,( 22xy,00
22242bxyb,,b20022因为, ABOAdb,,,,,222222xy,xy,0000所以三角形OAB的面积
3222bxyb,,100(„„„„„„„„„„„„„„10分 SABd,,,,222xy,00
OABS以下给出求三角形的面积的三种方法:
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22xy方法1:因为点在双曲线,,1上, Pxy,,,0022ab
222222xybxab,222000xa,所以,,1,即( y,,,00222aba
2,,b2222222txybxbab,,,,,,,,12设, ,,0002a,,
所以
3btS,(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„1122tb,
分
3,,,btbtb,,,,,因为, S,222tb,,,
,,所以当时,,当时,( 0,,tbS,0tb,S,0
3btS,0,bb,,,所以在上单调递增,在上单调递,,,,22tb,
减(„„„„„„„„„„„„„„12分
22当,即时,abb,,bab,,2
3bb,12Sb,,,„„„„„„„„„„„„„13分 最大值22bb,2
322322babbab,,,22当abb,,,即时,( S,,ab,2最大值22a222abb,,,,
12综上可知,当时,;当时,Sb,bab,,2ab,2最大值2
322bab,S,(„„„14分 最大值2a
222方法2:设,则txyb,,,00
33btb(„„„„„„„„„„„„„„„„11分 S,,222btb,t,t
2222xyxy00,,1,,1Pxy,因为点在双曲线上,即,即,,002222abab
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2222bxab,2220xa,( y,,,002a
2,,b2222222txybxbab,,,,,,,,12所以( ,,0002a,,
22tbtb,,,,,,bb,gt,,,1令,则( gtt,,,,,,22ttt
,,所以当时,,当时,( gt,0gt,00,,tbtb,,,,,
2b所以gtt,,在0,b上单调递减,在b,,,上单调递,,,,,,t
增(„„„„„„„„„„„„„12分
22当,即时,abb,,bab,,2
3b12,„„„„„„„„„„„„„„13分 Sb,,最大值2b2b,b
3322bbab,22当,即时,( abb,,S,,ab,2最大值22ba22ab,,22ab,
12综上可知,当时,;当时,Sb,bab,,2ab,2最大值2
322bab,S,(„„„14分 最大值2a
22方法3:设,则txy,,00
232btb,11,,32Sbb,,,,(„„„„„„„„„„„„„11分 ,,ttt,,
2222xyxy00,,1Pxy,,,1因为点在双曲线上,即,即,,002222abab2222bxab,2220xa,y,( ,,002a
2,,b22222所以( txyxba,,,,,,1,,0002a,,
211,,222令, gubuubu,,,,,,,,,,,2224bb,,
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11,,,,,,,,,,所以在上单调递增,在上单调递gu,,,,,,222b2b,,,,
减(„„„„„„„„„„„„12分
11,,u,,0,因为,所以, ta,,2,ta,,
1111,,,,gug,,当,即时,,此时,bab,,2,,,,22,,22max24bb2ba,,
1132( Sbb,,,最大值22b
„„„„„„„„„„„„13分
221ab,11,,gug,,,,当,即时,,此时,ab,2,,,,24,,22maxaa2ba,,
322bab,S,( 最大值2a
12综上可知,当时,;当时,Sb,bab,,2ab,2最大值2
322bab,S,(„„„14分 最大值2a
21((本小题满分14分)
(本小题主要考查函数的值域、导数、不等式等基础知识,考查运算求解能力和推理论证能力,以及创新意识()
fxaxxx,,ln(1)解:因为,所以,,
,fxax,,,ln1(„„„„„„„„„„„„„„„„1分 ,,
fxaxxx,,ln因为函数的图像在点处的切线斜率为3, x,e,,
,fe3,a,,,lne13所以,即( ,,
所以
a,1(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„
„„„„2分
fxxxx,,ln(2)解:由(1)知,, ,,
fx,,xxx,lnk,x,1x,1所以对任意恒成立,即对任意恒成k,x,1x,1
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立(„„„„„„„„„3分
xxx,ln令, gx,,,x,1
则
xx,,ln2,,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„gx,,,2x,1,,
„4分
令hxxx,,,ln2, x,1,,,,
11x,,则, hx,,,,10,,xx
hx在1,,,上单调递所以函数,,,,增(„„„„„„„„„„„„„„„„„„„„„„„„„5分
hh31ln30,422ln20,,,,,,因为, ,,,,
hx,01,,,x,3,4所以方程在上存在唯一实根,且满足( x,,,,,,00
,gx()0,当,即,当,即1()0,,,xxhx时,xxhx,,时,()000,gx()0,,„„„„„„6分
xxx,ln1,xx,,,所以函数在上单调递减,在上单调递增( gx,,,,,,,00x,1
所以
xxxx1ln12,,,,,,,0000(„„„„„„„„„„„7gxgxx,,,,,3,4,,,,,,,,00,,minxx,,1100
分
kgxx,,,3,4,,所以( ,,,,0,,min
故整数的最大值是k
3(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„8分
xxx,ln4,,,(3)证明1:由(2)知,是上的增函gx,,,,,x,1数,„„„„„„„„„„„„„„9分
所以当nm,,4时,nnnmmm,,lnln(„„„„„„„„„„„„„„„„„„„„10分 ,nm,,11
nmnmnm,,,,,11ln11ln即( ,,,,,,,,
整理,得
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mnnmmmnmnnnmlnlnlnln,,,,,(„„„„„„„„„„„„„„„„„11分 ,,
因为, 所以nm,
(„„„„„„„„„„„„„„„„„12分 mnnmmmnmnnlnlnlnln,,,
mnmmnnlnlnlnlnnmmn,,,即(
即
mnmmnnlnlnnmmn,(„„„„„„„„„„„„„„„„„„„„„„„„„„„„,,,,
,fxmmmmmmm,,,,,,,,,1ln1ln1ln0因为xm,,4,所以( ,,,,
fxm,,,所以函数在上单调递,,,,增(„„„„„„„„„„„„„„„„„„„„„„„„11分
fnfm,因为, 所以( nm,,,,,
所以
22mmmmmmmmlnlnlnln0,,,,mnnmmlnlnl,,,,(„„„„„12mnmnnnln
分
即mnnmmmnmnnlnlnlnln,,,(
mnmmnnlnlnlnlnnmmn,,,即(
即
mnmmnnlnlnnmmn,(„„„„„„„„„„„„„„„„„„„„„„„„„„„„,,,,
13分
所以
mnnmmnnm,(„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„14,,,,
分
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