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参考答案:1.D2.B3.D4.A5.C6.D7.A8.B9.AD10.BC11.CD12.AC13.2814.1015.15516.3π17.(1)解:若6=,1,3BCCD==,所以,在BCD△中,22232cos42312BDBCCDBCCD=+−=−=所以1BD=;(2)解
:因为BCD=,1,3BCCD==,所以,根据余弦定理得2222cos423cosBDBCCDBCCD=+−=−,因为,1,3,ABBDDABCCD====所以()3423cos4ABDS=−△,3sin2BCDS=,所以()
33sin423cos3sin3243ABDBCDSSS==+−=−++△△四边形,所以,56=时,S四边形取到最大值2318.(1)2nnnaSn−=①2n时,()()11112nnnaSn−−−−−=②则①-②得()11221
nnnanana−−−=−,当3n时可整理得()()12112121212nnaannnnnn−−=−=−−−−−−−,即1221122nnaannnn−−=−−−−−,由①当1n=时,111112aSa−=−=,得12a=,当3n=时,331233332aSaaa++−=−=,得3
8a=,1232222311223322nnnaaaannnnnn−−−=−=−==−=−−−−−−,()31231nann=−+=−,又12a=,25a=,符合31nan=−,31nan=−;
(2)由(1)得()()()()()()22222323333132323223291111nnnbnnnnnn++===+−−+−−+++,1132113nnbn−−+−121313231211111
125572nnnbbbnnn+++−−+−++−=−+−++,1032n+1212nbbbn+++−19.(1)取1BB中点F,连接DF,EF,如图所示:因为,,DEF分别为1AA,1BC,1BB的中点,所以//DFAB,//EFBC,又因为DF平面ABC,
EF平面ABC,AB平面ABC,BC平面ABC,所以//DF平面ABC,//EF平面ABC,又因为DFEFF=,,DEEF平面DEF,所以平面//DEF平面ABC,又因为DE平面DEF,所以//DE平面ABC.(2
)连接,EOAO,如图所示:因为,EO分别为1,BCBC的中点,所以1//EOBB,11=2EOBB,又因为D为1AA的中点,所以1//DABB,112DABB=,所以EODA=,//EODA,即四边形AOED为平行四边形,即//DEAO.因为DE⊥面1CBB,所以AO⊥面1CBB.又因为B
C⊥面1CBB,所以AOBC⊥,即ABAC=.以A为原点,1,,ABACAA分别为,,xyz轴建立空间直角坐标系,设12BBBC==,则2ABAC==,()12,0,2B,()0,2,0C,()10,0,2A,()0,
0,1D,22,,122E,22,,022DE=,()112,0,0AB=,()10,2,2AC=−,设平面11ABC的法向量为(),,nxyz=,则11120220nABxnACy
z===−=,令2y=得()0,2,1n=.则13cos,3112122nDE==++,所以平面11ABC与平面1BBC的夹角的余弦值为33.20.(1)由题意可知,甲、乙乘车超过12公里且不超过22公里的概率分别为14,13则甲、乙两人所付乘车费用相同的概率111111
114323433P=++=,所以甲、乙两人所付乘车费用不相同的概率1121133PP=−=−=.(2)由题意可知,6,7,8,9,10=,则111(6)4312P===,11111(7)43234P==+
=,1111111(8)4343233P==++=,11111(9)23434P==+=,111(10)4312P===,所以的分布列为678910P112141314112则11111()67891081243412E=++++=.21.(
1)设(),Pxy为椭圆上一点,则4226PFPEPAPEAEEF+=+===,所以P点轨迹是以,FE为焦点,长轴长为242a=的椭圆,设椭圆的方程为()222210xyabab+=,所以6,22ca==,则2222bac=−=,所以椭圆方程
为22182xy+=;(2)设()()()112200,,,,,AxyBxyMxy,则220010xy+=,切线MA方程:11182xxyy+=,切线MB方程:22182xxyy+=,两直线都经过点M,所以,得1010182xxyy+=,2020182xxyy+=,从而直线AB的方程是
:00182xyxy+=,由0022182182xyxyxy+=+=,得()2220003101664320yxxxy+−+−=,由韦达定理,得20012122200166432,310310xyxxxxyy−+==++,2012014xABx
xy=+−−2220002220001664321416310310xxyyyy−=+−++()202021032310yy+=+,点M到直线AB的距离220020220013282582xyydxy+−+==+
,()32202023212310MABySABdy+==+,其中2010y,令2032ty=+,则3222,42,8MABttSt=+,令()3228tftt=+,则()()()422222408ttftt+=+,()ft在
2,42t上递增,42t=,即2010y=时,MAB△的面积取到最大值325,此时点()0,10M.22.(1)易知函数()fx的定义域为(),a+.又()()11xxaafxxxaxa−−=−+=−−−.当0a=时,()()21,2fxxxfx=−+在()
0,1上单调递增,()fx在()1,+上单调递减;当0a时,()fx在(),1aa+上单调递增,()fx在()1,a++上单调递减;当10a−时,()fx在()0,1a+上单调递增,()fx在(),0a和()1,a++上单调递减
;当1a=−时,()fx在()1,−+上单调递减;当1a−时,()fx在()1,0a+上单调递增,()fx在(),1aa+和()0,+上单调递减.(2)由()21ln2gxaxxx=−+,0x则()21axxagxxxx−−=−+=−
,由题意知12,xx是方程20xxa−−=的两根,因此,121xx=+,12xxa=−,且104a−,121012xx.所以,()()()()221211122211lnln22fxfxaxaxxaxaxx−=−−+−−−+
()()()()()()()()1122121212122211lnln222xaxaaxxxxaxxxxxaxa−−=−−+−=−−+−−−把121xx=+,12xxa=−代入得()()()()()()()()111212121212221211lnln22xaxxxfxfxaxxx
xxxxaxxx−+−=+−=−+−−+212121222111111111111lnln1ln122211xxxxxxxxxxxx+=−+−=−+−−+++要证
()()120fxfx−,只需证明22111111ln1ln1022xxxx+−−++,即22111111ln1ln122xxxx+−+−,也即2211111111ln11ln1122xxxx+−+
+−+.令1111tx=+,2211tx=+,由121012xx,得212tt.设()1ln2httt=−,要证()()21htht.因为,()112022thttt−=−=,()ht在()2,+上单调递减,所以,()()21ht
ht,即证.获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com