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?a≥15（元）．

故每位投保人应交纳的最低保费为15元． ············································································ 12分

19．解法一：

依题设知AB?2，CE?1．

（Ⅰ）连结AC交BD于点F，则BD?AC．

由三垂线定理知，BD?AC······························································································· 3分 1． ·在平面ACA内，连结EF交AC11于点G，

AA1AC??22， 由于

FCCE故Rt△A??CFE， 1AC∽Rt△FCE，?AAC1D1 A1

B1

C1

?CFE与?FCA1互余．

D 于是AC?EF． 1A F H E

G B C BED内两条相交直线BD，EF都垂直， AC1与平面

?平面BED． ·所以AC··········································································································· 6分 1（Ⅱ）作GH?DE，垂足为H，连结A1H．由三垂线定理知A1H?DE，

故?A········································································ 8分 1HG是二面角A1?DE?B的平面角． ·

EF?CF2?CE2?3， CG?3CE?CF222，EG?CE?CG?． ?3EF3EG11EF?FD2?，GH??． ?EF33DE15?又AC1AA12?AC2?26，AG?AC11?CG?56． 3tan?A1HG?A1G?55． HGz D1 A1 B1 所以二面角A1?DE?B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，

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建立如图所示直角坐标系D?xyz．

依题设，B(2，2，，0)C(0，2，，0)E(0，21)，，A1(2，0，4)．

????????DE?(0，21)，，DB?(2，2，0)，

?????????·························································································· 3分 AC?(?2，2，?4)，DA1?(2，0，4)． ·1（Ⅰ）因为ACDB?0，ACDE?0， 1?1?故AC?BD，AC?DE． 11又DB?DE?D，

所以AC··········································································································· 6分 ?平面DBE． ·1（Ⅱ）设向量n?(x，y，z)是平面DA1E的法向量，则

?????????????????????????n?DE，n?DA1．

故2y?z?0，2x?4z?0．

1，?2)． ·令y?1，则z??2，x?4，n?(4，····································································· 9分

????n，AC等于二面角A1?DE?B的平面角， 1????????n?AC141． cosn，AC??????142nAC1所以二面角A1?DE?B的大小为arccos20．解：

（Ⅰ）依题意，Sn?1?Sn?an?1?Sn?3n，即Sn?1?2Sn?3n， 由此得Sn?1?3n?114． ································································· 12分 42····························································································· 4分 ?2(Sn?3n)． ·

因此，所求通项公式为

················································································· 6分 bn?Sn?3n?(a?3)2n?1，n?N*．① ·（Ⅱ）由①知Sn?3n?(a?3)2n?1，n?N， 于是，当n≥2时，

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an?Sn?Sn?1

?3n?(a?3)?2n?1?3n?1?(a?3)?2n?2 ?2?3n?1?(a?3)2n?2，

an?1?an?4?3n?1?(a?3)2n?2

?2n?2??3?n?2??12????a?3?， ??2????当n≥2时，

?3?an?1≥an?12????2??a≥?9．

又a2?a1?3?a1．

n?2?a?3≥0

综上，所求的a的取值范围是??9，··········································································· 12分 ???． ·

x2?y2?1， 21．（Ⅰ）解：依题设得椭圆的方程为4直线AB，EF的方程分别为x?2y?2，y?kx(k?0)． ·················································· 2分 如图，设D(x0，kx0)，E(x1，kx1)，F(x2，kx2)，其中x1?x2， 且x1，x2满足方程(1?4k)x?4， 故x2??x1?22y B O E F D A x 21?4k2．①

????????1510由ED?6DF知x0?x1?6(x2?x0)，得x0?(6x2?x1)?x2?；

27771?4k由D在AB上知x0?2kx0?2，得x0?所以

2． 1?2k210?，

21?2k71?4k2化简得24k?25k?6?0，

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23或k?． ················································································································· 6分 38（Ⅱ）解法一：根据点到直线的距离公式和①式知，点E，F到AB的距离分别为

解得k?h1?x1?2kx1?25x2?2kx2?25?2(1?2k?1?4k2)5(1?4k)2，

h2??2(1?2k?1?4k2)5(1?4k)2． ········································································ 9分

又AB?22?1?5，所以四边形AEBF的面积为

S?1AB(h1?h2) 214(1?2k) ??5?225(1?4k)?2(1?2k)1?4k2

1?4k2?4k≤22， ?221?4k当2k?1，即当k?1时，上式取等号．所以S的最大值为22． ································· 12分 2解法二：由题设，BO?1，AO?2．

设y1?kx1，y2?kx2，由①得x2?0，y2??y1?0， 故四边形AEBF的面积为

S?S△BEF?S△AEF

··································································································································· 9分 ?x2?2y2 ·

?(x2?2y2)2 22?x2?4y2?4x2y2 22≤2(x2?4y2)

?22，

当x2?2y2时，上式取等号．所以S的最大值为22． ···················································· 12分 22．解：

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（Ⅰ）f?(x)?(2?cosx)cosx?sinx(?sinx)2cosx?1． ······································· 2分 ?22(2?cosx)(2?cosx)2π2π1?x?2kπ?（k?Z）时，cosx??，即f?(x)?0； 3322π4π1?x?2kπ?当2kπ?（k?Z）时，cosx??，即f?(x)?0． 332当2kπ?因此f(x)在每一个区间?2kπ???2π2π?，2kπ??（k?Z）是增函数， 33?2π4π??f(x)在每一个区间?2kπ?，····································· 6分 2kπ??（k?Z）是减函数． ·

33??（Ⅱ）令g(x)?ax?f(x)，则

g?(x)?a?2cosx?1

(2?cosx)2?a?23? 22?cosx(2?cosx)211?1??3????a?．

3?2?cosx3?故当a≥1时，g?(x)≥0． 3又g(0)?0，所以当x≥0时，g(x)≥g(0)?0，即f(x)≤ax． ······························· 9分 当0?a?1时，令h(x)?sinx?3ax，则h?(x)?cosx?3a． 3故当x??0，arccos3a?时，h?(x)?0． 因此h(x)在?0，arccos3a?上单调增加．

arccos3a)时，h(x)?h(0)?0， 故当x?(0，即sinx?3ax．

arccos3a)时，f(x)?于是，当x?(0，当a≤0时，有f?sinxsinx??ax．

2?cosx3π?π?1． ??0≥a??222???1?3??因此，a的取值范围是?，························································································ 12分 ???． ·

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