又∵ MN∥CD， ∴ △MNG∽△DCG． ∴
MNDC?GHGF，即MN?2[3?1?x]3． ························ 4分
故△EMN的面积S＝1?2[3?1?x]?x
23＝?3x2?(1?333)x；
············································· 6分
综合可得：
?x，?0＜x≤1??S??32?3??x?1?x．1＜x＜1??????33????3? ···························································7分
（2）①当MN在矩形区域滑动时，S?x，所以有0②当MN在三角形区域滑动时，S=?33x?(1?2·······································8?S?1； ·
)x.
分
331因而，当x?∵
12?331?23（米）时，S得到最大值，最大值S=2?33（平方米）.
?1，
133∴ S有最大值，最大值为?2平方米. ··································································· 12分
（文）解：（1）bn?an?n?3an?1?2n?3?n?3an?1?3n?3?3(an?1?(n?1))?3bn?1，
n?2
又b1?a1?1?1，所以bn?0（n?N*），
bnbn?1?3(n?2)
所以，数列{bn}是以1为首项3为公比的等比数列． ··················································6分
n?1（2）bn?3，an?bn?n······················································································8分
所以数列{an}的前 n项和Sn?(b1?b2???bn)?(1?2???n)=
3?n?n?12n2．
·································································································································· 14分
2
20（理）解：（1）∵a、b、c成等比数列，∴b=ac ························································1分 则cosB=
2
2
a2?c2?b22ac=
aa22?c22?ac2ac?c?ac ·················································································3分
ac2ac?12而a+c≥2ac∴cosB=
2ac≥，等号当且仅当a=c时取得，即
12≤cosB＜1，得到
- 9 -静安区高三第一学期数学
0?B??3．·················································································································7分
π4?x2（2）cos2x?4sin()sin(
π4?x232)=cos2x?4sin(?4πx2)cos(?4πx2)
=2cosx2?2cosx?1=2(cosx?
∵x=B ∴2(cosx?
∴
12122)? ·················································································· 11分
≤cosx＜1
32122)?
32≥?
3232
··················································································· 14分
则由题意有：?m＜?即m＞
（说明：这样分离变量m?2cosx?cos2x??2cos2x?2cosx?1参照评分） （文）解：（1）由正弦定理
asinA?bsinB?csinC得sinAcosB?sinBcosA?25sinAcosB?8535sinC，2分
又sinC?sin(A?B)?sinAcosB?cosAsinB，所以可得
tanAtanB?sinAcosBsinBcosAsinBcosA，5分
?4． ···················································································7分
（2）若A?600，则sinA?cosB?4191932，cosA?12，tanA?3，得tanB?34，可得
，sinB?3?1919． ··········································································· 10分
53?1938sinC?sin(A?B)?sinAcosB?cosAsinB?asinAbsinBcsinC，
由正弦定理
a?csinC??得
c?sinB?2 ························································ 14分
3?sinA?19，b?sinC2321（理）解：（1）当n?1时，a1?a1，由a1?0得a1?1．····································1分
2当n?2时，(1?a2)?1?a2，由a2?0得a2?2或a2??1．当n?3时，
(1?a2?a3)?1?a2?a3，若a2?2得a3?3或a3??2；若a2??1得a3?1； 5分
233综上讨论，满足条件的数列有三个： 1，2，3或1，2，?2或1，?1，1． ·············································································6分 （2）令Sn?a1?a2???an，则Sn?a1?a2???an（n?N*）．
- 10 -静安区高三第一学期数学
2333从而(Sn?an?1)2?a1?a2???an?an?1． ·······················································7分 两式相减，结合an?1?0，得2Sn?an?1?an?1． ······················································8分 当n?1时，由（1）知a1?1；当n?2时，2an?2(Sn?Sn?1)=(an?1?an?1)?(an?an)，即(an?1?an)(an?1?an?1)?0，所以an?1??an或an?1?an?1． ·························· 12分 又a1?1，a2013??2012，所以无穷数列?an?的前2012项组成首项和公差均为1的等差数列，从第2013项开始组成首项为?2012，公比为?1的等比数列．故
(1?n?2012)?nan??． ··········································································· 14分 n(n?2012)?2012?(?1)（说明：本题用余弦定理，或者正弦定理余弦定理共同使用也可解得，请参照评分）
2223333（文）解：（1）
①如图1所示，当MN在正方形区域滑动， 即0＜x≤2时，
△EMN的面积S=?2?x=x； ··························· 2分
21G ②如图2所示，当MN在三角形区域滑动， 即2＜x＜2?3时，
D M C N
如图，连接EG，交CD于点F，交MN于点H， ∵ E为AB中点，
∴ F为CD中点，GF⊥CD，且FG＝3. 又∵ MN∥CD， ∴ △MNG∽△DCG． ∴
MNDC?GHGFA E
B 图1
G ，即MN?2(3?2?x)3． ················· 5分
D M H F N C
故△EMN的面积S＝
12(3?2?x)??x 23A B
＝?3x2?(1?23)x； ··········································· 7分
33E 综合可得： 图2
?x,0?x?2?S?? ····························································8分 3223x?(1?)x,2?x?2?3??33?说明：讨论的分段点x=2写在下半段也可．
- 11 -静安区高三第一学期数学
（2）①当MN在正方形区域滑动时，S?x，所以有0?②当MN在三角形区域滑动时，S=?因而，当x?1?3233x?(1?2S?2； ·································· 10分
233)x.
，S在(2,2??2（米）
3)上递减，无最大值，0?S?2．
所以当x?2时，S有最大值，最大值为2平方米. ···················································· 14分 22．解：（1）在椭圆中，由已知得c?a?b?过点A(0,?b)和B(a,0)的直线方程为
32xay?b222a?b222 ···········································1分
??1，即bx?ay?ab?0，该直线与原点的距离
为，由点到直线的距离公式得：
aba?b22?32 ···················································3分
解得：a?3,b?1；所以椭圆方程为
22x23?y21···················································4分 ?1 ·
（2）（理）F1(?2,0)，直线F1A1的方程为y?2x?2，F1A1?6，当椭圆上的点P到直
线F1A1距离最大时，△F1PA1面积取得最大值·······························································6分
x2设与直线F1A1平行的直线方程为y?73x?2d22x?d，将其代入椭圆方程
3?y21?1得：
2x?d2?1?0，??0，即8d2?283d2?283?0，解得d2当d??7时，?7，
椭圆上的点P到直线F1A1距离最大为
122?3722?222?37，此时△F1PA1面积为
6?14··························································································9分
2（文）设M(x,y)，则x?3(1?y)，AM2?x?(y?1)??2y?2y?4，其中?1?y?1222 ·····································································································································6分 当y?12时，AM2取得最大值
92，所以AM长度的最大值为
322····························9分
（3）将y?kx?t代入椭圆方程，得(1?3k)x?6ktx?3t?3?0，由直线与椭圆有两个交
- 12 -静安区高三第一学期数学
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