河南省新鄉(xiāng)市2020-2021學(xué)年高二上學(xué)期期末考試數(shù)學(xué)試題

河南省新鄉(xiāng)市2020-2021學(xué)年高二上學(xué)期期末考試數(shù)學(xué)試題xx2x70）2000xx2x70xx2x70．．D．20200000x0,x2x70x0,x2x70．22{x|32x1,Bx2xx20AB（A．(0,2]）．[1,0)．0,1．(0,1]x2y2F,F1|7P4112|（）2或或3log3xx）22,N,．如圖，在四面體OABCMOAOBMN（）111．2221．21．21．2112121212122p:；1p:xR，x020；20p:3）p(p)(p)．p．D．1213(p)p(p)(p)．1323ABCD,1，MNABCD1111CC1ABCD1（）形與DNBM1111710305103．．．．6ba1bababa）1a}na1，a2S，{}S）annn1n1n31713171．．．．22344322344312,0P:y4x,ABPAABC22點(diǎn)A1．32．357．D．33C．ABC,,,,sincos1cosABCabcCC213ABCS()sinABCab）C22．275．75．273．73x2y2b2,FFC:1(a0,b0)F121a2M,N|MF|FF|2CC21211）4353325．．．．2a}aa24a}．n4nABC,,,,sinA3sinB，5ABCabcc5cosC．a(chǎn)6x30x,y0yazxya2xy0．Exy22F(1,0)E:ab(Aa2b2PAPF9EEP．S,S2a2,ba}nnnnnnba,bb10.3226a},nn（2）求數(shù)列ab3)n.TnnnBCBC3ABC2sincosBC2cossin.222A；2ABC.ABCDEABEAB//CD，F(xiàn),G,MBE,BC,AD1，AECDAD2，AB.AB3AE//MFEG:y2px(p0)；FF45C22線C，QPQyPpPQ（2）直線lCAlAABCD4ABCDABCD1111，BE3EBBAD60，BB且.E11BDDB；1ACE1CAEBxyb221ab0FF、M12a221，MFN42．線F211、FllQQPP12S3SlFF11Cxx2x70“2000x0,x2x70“2D20|02xxA{x|32x1{x|1x,Bx2xxAB{x|0xDC2a4PF7,PF3PFca422或，12122PF3故或2A332893∵2。2222232log3xAx21111122222Cp:p11p:xR，x0p120202p率對(duì)于1p33B1x,y,z,DC,DD1為Dxyz，BM則,0,1,以212710114cosBM,DNDN,,2，302214114441Ababa0，ba111ba=123ba1ba1+2ba1,bababaa=1ba3.當(dāng)bba或.Da2Sa2Sa3an2a1a2,aa3由n1nnn1n11221n1,n11,n11a，=11n223,n2an,22nnn23111111101......aa1......2333a12111313171111=122234431813DaS與nnn.BAx,y,Bx,y,B2DE,設(shè)x，112213224xx2yx221PAABx1，又12yy3413xy22122DCCCCCsinCcosC1cos2sincos2cos211cos由22222CCCCcos2cos-2sin10,cos022221234CC12CCC22223CCC70CCCcos得222424413absinCababab2又S22c2a2b22abcosC827c71ABC73DBFM1MF,FMca,MPca,FPca取211NF2MFNF4ca,NFc2aRtNPF11122RtMPF222222222224c2a5ca22cca22，ca02223850ca，caca，5e又e3【.a,ce②a,cea,ce．a(chǎn)a2a24a12,aa7d14,d2a22，.48881na2b2c2b255,sinA3sinB,3,abC2b26b1,a33.x300yaxy0z2xyz2xyA3,a6aa1010.z此.11[,]．54F'0PF'PF2aPF2aPF'設(shè)EPPAPF9,PAPFPA2aPF'9PAPF'9a，AF'PAPF'AF'1PAPF'1，a1921114114ace,.5542）an1n(2n261）T,bnnnn試題分析：分類(lèi)討論n1和n2兩種情況可得數(shù)列的通項(xiàng)公式為2，據(jù)此計(jì)算aannn可得bn1；n結(jié)合數(shù)列的通項(xiàng)公式錯(cuò)位相減可得數(shù)列23的前項(xiàng)和abnnnT2n3216n.n試題解析：（1）當(dāng)1時(shí)，，2na1當(dāng)時(shí)，aSS222a，即aa，2nannn1nn1nn1所以a是以2為首項(xiàng)，2為公比的等比數(shù)列，即2，annn又ba4,bbb10bn1.，所以32264nb32n12（2）因?yàn)閍，nnnn1232522n12所以T，①23nT12322n322n1223nn1，②n由①②得T222222n12，23nn1n2n326.所以Tn1n332A.S3BCBC32sincos2cosBsinCsinBCsinAA．2222，A3BCBC32sincos2cosBsinC，2223，2sinBC2cosBsinC33sinBcosCcosBsinC2cosBsinCsinBCsinA則由，22ABCA.312ABCaBCbACa2b2c2,,2cos74222bcA，ccc22c30c31332bcsinAS.26）.FGMGCDEMGF//CDEMF//CDE；ABCD，（2ABCDABEAEABEGCDE.AExyz以AAF,G,MBE,BC,AD//ABCDFG//CE，MG//CD，又FGMGGMGF//CDE，//CDEMGFMFMF.ABCDABEAEABAEABCD.xyz以AA，53EC0,2,3D0,1,3G,，，則，，221,1,3530,1,0DC，ED，,22y0nDC0nx,y,z設(shè)CDE，nED0xy3z0n3,0,1.3nEG62設(shè)EGCDE所成角為，sin2816nEG6故EGCDE.162y.(1)(2)2）p）x222y21pxPx,yQx,yyy4，12px112212y222yy2p2p1lyxb12xxyy14122xb4xb04x入y222離公式1pF,0，2Px,yQx,yyy4.設(shè)，1122122yy2py21px2p412.112p2pxxxyy1y221222xb4x，y2lyx2b4xb20，得b4b0∵l2，C2A21b.1212，∵PQdA21y22.22x33926）.BDABCDACBBAC，1ACBDDB11ACEBDDBAC與BDOAC與BDOO1111111zOA、OB、OOOxyz1ACEABEAEBCABCDACABCDBB.BD.BBAC.ABCDABCD，∴111111BDBACBDDB，11∵BB1又ACACEACEBDDB.11、OB、OOAC與BDO，AC與BDOOOA111111z23,0,0，AOxyzC23,0,0，E0,2,3，D0,2.4，10,4,1，AE23,2,3，AC43,0,0，ED則1nx,y,z設(shè)則ACE111n23x2y3z0111，n43x01n0,3,2.2取z1AB，取ABFDFDF3,3,0EFABEABE.nm33926n,m∴，nm33926AEBCAEB∴由圖可知，二面角C..14x2）1）.y222442aMFFa21b1c、b出cxmy1，MFNl1Px,yQx,y、l1122y2yml.21FMN42442a2，a1b1，MF的斜率11cx2c12abcb1；222y21yxMF1yxx2x，340x211y224112S3,N1S，333FF1111221NFQFsinQFNMFPFsinPFMQF2PF，即32111111110lxmy1Px,y、Qx,y，l1122y2yy2y，QP21211xmym2y21022，x2y1222m1yyyy，，2m212m21222m2y81m221471m22得2m2，m，y122m27m22y221m21471471142mml．m..14x2）1）.y222442aMFFa21b1c、b出cxmy1，MFNl1Px,yQx,y、l1122y2yml.21FMN42442a2，a1b1，MF的斜率11cx2c12abcb1；222y21yxMF1yxx2x，340x211y224112