则
uuuruuur1uuuruuur?uuur1uuur?1uuur21uuur21uuuruuur111BE?BF?(BA?BD)??BD?BA??BD?BA?BA?BD??12??4?23?263263??2?23?322?, 23故答案为:
22. 3三、解答题(共5个小题,每小题15分,共75分) 16.解:(Ⅰ)在VABC中,由余弦定理得:
a2?b2?c2?2bccosA,
又a?4,c?3,cosA??1, 42b2?3b?14?0,
解得b?2; (Ⅱ)由cosA??1, 4所以sinA?15, 4由正弦定理得:
ab?, sinAsinB得sinB?15, 8又0?B??2,
所以cosB?7, 8715172,cos2B?2cosB?1?, 3232- 9 -
所以sin2B?2sinBcosB?
所以sin?2B?????371517117?215, ??????6?23232264故答案为:
17?215.
6417.【解答】证明:(Ⅰ)Q四边形ABCD是正方形,?AB//CD,
Q四边形ADPQ是梯形,PD//QA,AB?QA?A,CD?PD?D, ?平面ABP//平面DCP,
QQB?平面ABQ,?QB//平面PDC.
解:(Ⅱ)以D为原点,DA为x轴,DC为y轴,DP为z轴,建立空间直角坐标系, 则C(0,2,0),P(0,0,2),B(2,2,0),Q(2,0,1),
uuuruuuruuurPB?(2,2,?2),PC?(0,2,?2),PQ?(2,0,?1),
r设平面PBC的法向量n?(x,y,z),
ruuurr??n?PB?2x?2y?2z?0则?ruuu,取y?1,得n?(0,1,1), r??n?PC?2y?2z?0ur设平面PBQ的法向量??(x,y,z),
uruuurur??m?PB?2x?2y?2z?0则?u,取x?1,得??(1,1,2), ruuur??m?PQ?2x?z?0设二面角C?PB?Q的大小为?,由图形得?为钝角,
urr|m?n|33rr???则cos???u,
2|m|?|n|2?6???5?, 65?. 673, 15?二面角C?PB?Q的大小为
(Ⅲ)点H在棱PD上,且异面直线AH与PB所成角的余弦值为
uuuruuur设DH?t,则H(0,0,t),A(2,0,0),AH?(?2,0,t),PB?(2,2,?2),
- 10 -
uuuruuuruuuruuur|AH?PB|4?2t73r??|cos?AH,PB?|?uuuruuu?,
215|AH|?|PB|4?t?12解得t?33,?线段DH的长为.
22
cb2118.解:(Ⅰ)由椭圆的离心率e??1?2?,则a?2c,b?3c,
aa2VB1F1F2的面积的面积S?解得:a?2,b?1?2c?b?3,则bc?3, 23,c?1,
x2y2?椭圆的标准方程:??1;
43(Ⅱ)由(Ⅰ)可知:F(?1,0),由函数的对称性,直线的斜率存在且不为0, 设直线AC:y?k(x?1),且k??3,A?x1,y1?,C?x2,y2?,
?y?k(x?1)?22222,整理得:?3?4k?x?8kx?4k?12?0, ?xy2??1?3?48k24k2?12x1?x2??,x1x2?,
3?4k23?4k2则|AC|?1?k2??x1?x2?2?4x1x2?212?1?k2?3?4k2,
12?1?k?1将?代入上式可得|BD|?, 2k4?3k
- 11 -
|AC|3k2?43712则,k?3, ?2???2|BD|4k?3444k?3由k?0,则
233714???2?,k2?3 4444k?33?|AC|?313??134?的取值范围?,???,?. |BD|?415??153?19.解:(1)等比数列?an?的各项均为正数,设公比为q,q?0,
2由2a5,a4,4a6成等差数列,可得2a4?2a5?4a6,即a4?a4q?2a4q,即2q?q?1?0,
2解得q?1(?1舍去), 2232由a4?4a3,可得a1q?4a1q??2,即a1?n1812a1, 411?1?解得a1?,则an????22?2?数列?bn?的前n项和Sn?n?1?1????; ?2?(n?1)bn,n?N*,且b1?1, 2可得n…2时,Sn?1?n(n?1)n?1nbn?1,又Sn?bn,两式相减可得bn?bn?bn?1, 2222化为
bnnbbb23n??n, ,则bn?b1?2?3?n?1???bn?1n?1b1b2bn?112n?1上式对n?1也成立,则bn?n,n?N*;
?n,n为奇数 ?bn,n为奇数????1?n(2)cn??,
?an,n为偶数???,n为偶数??2?当
n为偶数时,前n项和
1?1?11Pn?(1?3?5?L?n?1)??????n??(1?n?1)2?2?416 - 12 -
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