则:(1)
P1?P(A?B?C)?P(A?B?C)?P(A?B?C)?P(A?B?C)
21221221212?()2?(1?)??(1?)??(1?)???()2??32332332323
(2)?可能取0, 3000, 4000, 6000, 7000, 10000
? P 0 3000 4000 6000 7000 10000 118 29 118 29 29 29 E?=6000
f?(x)?3ax2?2x?18.解:(1)由已知
1x?1
f?(1)?3a?2?同时
3213?kl?22 ∴a?1
213x3?(x?1)2f(x)?x?x?ln(x?1),f?(x)?3x?2x???0x?1x?1∴
[f(x)]min?f(0)?0
∴f(x)在[0,??)上? ∴
x? (2)令
11111f()?0??ln(1?)?032n. 则:nnnn ∴
n?11?ln(1?)3n 即:n1121?ln(1?),?ln(1?)......33233∴2
12n?1111??...??ln(1?)?ln(1?)?...ln(1?)33323n23n∴
34n?1n?1?ln(?...)?ln?ln(n?1)23n2
∴不等式成立.
19.法一:(1)证:E,F为CC1,BC中点?EF//BC1?EF//面BC1M F,M为BC, A1D1中点?AF//C1M?AF//面BC1M ?面AEF//面BC1M?AE//面BC1M
(2)分别取AE,ED中点O,O′.连结FO, CO′,OO′
1//?2则OO′AD?FC ∴平行四边形FCO′O ∴FO//CO′
//∵EC=CE ∴CO′⊥ED
?CO′⊥面AED
AD⊥面CD D1C1?AD⊥CO′ ?FO⊥面AED ∵OO′⊥ED. 连结O′F. 则O′F⊥ED ∴∠OO′F为二面角F—ED—A平面角, 不妨设AB=1 AA′=2
A1 B1 C1 M D1 115在Rt△FOO′中,OO′=2AD=2, AF=AE=2,
E O A B F C O′ D
2 AE=3 ∴FO=2
FO?2?tanOO∴∠OO′F= ∠OO′F=arctan2 ∴二面角为arctan2.
???? 法二:建立如图坐标系,不妨设AA1=2AB=4.则AE=(2,2,2)
??? (1)设平面BC1M的法向量为a,则:
?????????????BC1?a?0?a?(2,?1,?1)???????????C1M?a?0 ??????????????∵AE?a?0 ∴AE?a ????∴AE//面BC1M
(2)同理,可解得面ADE的法向量
z A1 B1 C1 M D1 E O A y B F C D x ??b?(0,1,?1)
??面FED的法向量c?(?2,?1,1)
??????????c?d?23??cos(c,a)???????3 |c||d|2?6∴
显然二面角F—ED—A为锐二面角
arccos∴二面角F—ED—A为
33.
xx12k?f?(x1)?1A(x1,)p 2p,A点处切线斜率20.(1)设
x12x1y??(x?x1)2pp∴直线方程为: x12x12x?0?y??c(0,?)2p 即2p 令
∴AC中点M(x,y)满足
y?yA?yC?02
又∵A,B为l与抛物线交点 ∴
x1?0 ∴
xM?x1?02
∴M点轨迹方程为y?0(x?0) (2)假设存在符合题意的点E.
由已知
l1:y?p33p?xx2?2p(x?)23 联立抛物线方程有:32
x2?∴
23px?p2?03
3p3p3,x2?3pA(?p,),B(3p,p)3362 故
kAE??33
∴
x1??∵△ABE为正△ ∴
y?∴AE:
p333p??(x?p)y??x?63336 即
ppE(?,3p)2 ∴2
准线
l2:y??欲使△ABE为正△,则
kBE不存在. 即xB?xE不符合
∴不存在符合条件的点E.
21.(数学归纳法)先证:
xn?2.∵当n?1时,x1?a?2成立
2xk1[(xk?1)?1]2xk?1???x?22(x?1)2xk?1k 假设n?k时,k.则:
111?[(xk?1)??2]??4?22xk?12
∴
xn?2
2xnx(2?xn)xn?1?xn??xn?n?0x?xn?1 2(x?1)2(x?1)nn 又: ∴n 综上知:
2?xn?1?xn
2xn(xn?2)2x?2xn?1?2??2??n?(xn?2)2(xn?1)2(xn?1)2(xn?1) (2)
xn?211111?[1?]??(1?)?2?xn?x1?3 ∴2(xn?1)2xn?1224 ∵
xn?1?2?111(xn?2)xn?2?()n?1?(x1?2)?()n?1?(a?2)444 ∴
nn∴
114Sn?2n??(xi?2)??()i?1?(a?2)?(a?2)??(a?2)13i?1i?141?4∴ 4a5a?164a(a?2)???0(a?2)?262 ∵3 ∴3Sn?2n?aaSn?2n?2 即 2
∴