因此，?的分布列如下：
? p 0 14 551 28 552 12 553 1 55……10分
?E??0?1428121?1??2??3??1． ……12分 5555555518．解：（Ⅰ）证明:∵该几何体的正视图为矩形,左视图为等腰直角三角形,俯视图为直角梯形, ∴
BA,BC,BB1两两垂直.
以BA,BB1,BC分别为x,y,z轴建立空间直角坐标系如图．--------------2分 则B?0,0,0?,N?4,4,0?,B1?0,8,0?,C1?0,8,4?,C?0,0,4?．
?????????∴BN?NB1??4,4,0????4,4,0???16?16?0， ?????????BN?B1C1??4,4,0???0,0,4??0．------------4分
zCC1BB1Ny∴NB?NB1,BN?B1C1. 又NB1与B1C1相交于B1，
∴BN⊥平面C1NB1. -------------------6分 （Ⅱ）∵BN⊥平面C1NB1,
MAx???????∴BN是平面C1B1N的一个法向量n1??4,4,0?, ------------8分
设n2??x,y,z?为平面NCB1的一个法向量,
??????????n2?CN?0??x,y,z???4,4,?4??0?x?y?z?0??????????则???,
x?y?0x,y,z?4,?4,0?0????????n2?NB1?0????所以可取n2??1,1,2?． ------------10分
???????????????n?n24?413?1??????则cos?n1,n2????．
3|n1|?|n2|16?16?1?1?43∴所求二面角C－NB1－C1的余弦值为3． ------------12分 319．解：（Ⅰ）由a2a9?232与a4?a7?a2?a9?37
解得:??a2?8?a2?29或?（由于an?1?an，舍去）
?a9?29?a9?8?a2?a1?d?8?a1?5 ,解得?
a?a?8d?29d?31??9设公差为d，则?所以数列?an?的通项公式为an?3n?2(n?N?)……………………………………4分 （Ⅱ）由题意得:
bn?a2n?1?a2n?1?1?a2n?1?2???a2n?1?2n?1?1
?(3?2n?1?2)?(3?2n?1?5)?(3?2n?1?8)???[3?2n?1?(3?2n?1?1)]
?2n?1?3?2n?1?[2?5?8???(3?2n?1?4)?(3?2n?1?1)]…………………………6分
而2?5?8???(3?2n?1?4)?(3?2n?1?1)是首项为2,公差为3的等差数列的前2n?1项的和,?4)?(3?2n?1?1)
所以2?5?8???(3?2n?1n?1?22n?1(2n?1?1)1?2??3?3?22n?3??2n
24所以b?3?22n?2?3?22n?3?1?2n?9?22n?1?2n………………………………10分
n484所以b?1?2n?9?22n
n48n994(1?4)3n2n所以T?(4?16?64???2)???(4?1)……………………12分 n881?42??????????20．解（Ⅰ）由已知M(a,0)，N(0,b)， F2(c,0)，MN?MF2?(?a,b)?(c?a,0)?a2?ac??1，
∵?NMF2?120?，则?NMF1?60?，∴b?3a，∴c?a2?c2?2a，
y2解得a?1，b?3，∴双曲线的方程为x?········································ 4分 ?`1． ·
3（Ⅱ）直线l的斜率存在且不为0，设直线l:y?kx?2，设A(x1,y1)、B(x2,y2)，
2?3?k2?0,?22??16k?28(3?k)?0,??y?kx?2,??由?2y2得(3?k2)x2?4kx?7?0，则?x?x?4k?0,
122?`1?x?k?3?3??7?x1x2?2?0,k?3?解得3?k?7． ①······················································································ 6分
????????∵点H(7,0)在以线段AB为直径的圆的外部，则HA?HB?0， ????????HA?HB?(x1?7,y1)?(x2?7,y2)?(x1?7)?(x2?7)?y1y2?(1?k2)x1x2?(7?2k)(x1?x2)?53
74k7k2?7?8k2?28k?53k2?159?(1?k)?2?(7?2k)?2?53??0，解得k?2． ②
k2?3k?3k?3由①、②得实数k的范围是2?k?7，······································································· 8分
????????S?AQH|AQ|由已知??，∵B在A、Q之间，则QA??QB，且??1， ?S?BQH|BQ|24k?(1??)x?,22??k?3∴(x1,y1?2)??(x2,y2?2)，则x1??x2，∴? ??x2?7,?2k2?3?(1??)216k2163则································································ 10分 ??2?(1?2)， ·
?7k?37k?3(1??)2641∵2?k?7，∴4?，解得???7，又??1，∴1???7． ??77故λ的取值范围是(1,7)． ························································································ 13分
21．解 （Ⅰ）f?(x)??e?x?(?x)??e?x，函数h(x)?f?(x)?g(x)?xe?x，h?(x)?(1?x)?e?x，当x?1时，h?(x)?0；当x?1时，h?(x)?0，故该函数在(??,1)上单调递增，在(1,??)上单调递减．∴
1函数h(x)在x?1处取得极大值h(1)?． ······································································ 4分
exx（Ⅱ）由题1?e?x?在[0,??)上恒成立，∵x?0，1?e?x?[0,1)，∴?0，
ax?1ax?11若x?0，则a?R，若x?0，则a??恒成立，则a?0．
xx不等式1?e?x?恒成立等价于(ax?1)(1?e?x)?x?0在[0,??)上恒成立，······ 6分
ax?1令u(x)?(ax?1)(1?e?x)?x，则u?(x)?a(1?e?x)?(ax?1)e?x?1，
又令?(x)?a(1?e?x)?(ax?1)e?x?1，则??(x)?e?x(2a?ax?1)，∵x?0，a?0．
①当a?0时，??(x)??e?x?0，则?(x)在[0,??)上单调递减，∴?(x)?u?(x)??(0)?0， ∴u(x)在[0,??)上单减，∴u(x)?u(0)?0，即f(x)?g(x)在[0,??)上恒成立； ·· 7分
2a?1②当a?0时，??(x)??a?e?x(x?)．
a1ⅰ）若2a?1，即0?a?时，??(x)?0，则?(x)在[0,??)上单调递减，?02∴?(x)?u?(x)??(0)?0，∴u(x)在[0,??)上单调递减，∴u(x)?u(0)?0，此时f(x)?g(x)在
··················································································································· 8分 [0,??)上恒成立； ·
12a?12a?1ⅱ）若2a?1?0，即a?时，若0?x?时，??(x)?0，则?(x)在(0,)上单调递
aa22a?1增，∴?(x)?u?(x)??(0)?0，∴u(x)在(0,)上也单调递增，
a∴u(x)?u(0)?0，即f(x)?g(x)，不满足条件． ··················································· 9分
1综上，不等式f(x)?g(x)在[0,??)上恒成立时，实数a的取值范围是[0,]． ·· 10分
2x2?x1（Ⅲ）由（Ⅱ）知，当a?时，则1?e?x?， ?e?x?12?x2x?122?x2?x2?x2n?24当x?[0,2)时，e?x?，令， ?x?ln?n，则x??2?2?x2?x2?xn?1n?1nnn444*∴lnn?2?，∴ln(n!)?2n??， ··· 12分 (n?N)，∴?lnk?2n??k?1k?1n?1k?1k?1k?111又由（Ⅰ）得h(x)?h(1)，即xe?x?，当x＞0时，ln(xe?x)?ln??1，∴lnx?x?1，
een(n?1)， ln(n!)?ln2?ln3???lnn?1?2???(n?1)?24n?n综上得2n??，即e?ln(n!)?k?12k?1n22n??k?1k?1n4?n!?en(n?1)2． ····························· 14分