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12020～2021学年第一学期期中考试高二数学参考答案及评分标准一、单项选择题：共8小题，每小题5分，共40分．1．D2．C3．B4．A5．A6．D7．D8．D二、多项选择题：共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求，全部选对的得5分，部分选对的得3分

，有选错的得0分．9．AC10．AB11．ACD12．BCD三、填空题：共4小题，每小题5分，共20分．13．614．33,3215．23n16．36四、解答题：共6小题，共70分．解答时写出必要的文字说明、证明过程或演算步骤．17．（本小题满分12分

）解：（1）由8ababab≥8+2，整理得280abab≥，(4)(2)0abab≥，所以ab≥4，故ab最小值为16，当4ab时取得最小．·····················6分（2）由8abab，得81bab，由0a，0b得1b．···

··························8分所以89222(1)336211babbbbb≥，·····························10分当且仅当92(1)1bb，

即3212b时，2ab的最小值为362．·······························································12分18．（本小题满分12分）解：（1

）设等比数列na的公比为q，又11a则21aaqq，2231aaqq，·······················································2分由于22a是3a和14a的等差中

项，得23144aaa，即244qq，解得2q.·································4分2所以，1111122nnnnaaq········································

················6分（2）21224nnnbnan012112324446424nnnSbbbbn212221441246214442143nnnnnnnn

·······························································································

····12分19．（本小题满分12分）解：（1）依题意得()()0gxfx对任意[3,3]x恒成立，即26kxx对任意[3,3]x恒成立，则2max(6),3,3kxxx≥，当3x时，2max(6)27xx，所以27k．···

············································································5分（2）因存在1x，2[3,3]x，使12()()fxgx成立则有mi

nmax[()][()]fxgx，·····································································8分因2()24fxxxk22(1)2xk，2()2gxxx2(1)1x，[3,3]x

所以minmax()(1),g()(3)fxfxg················································10分于是1(3)fg，即222(1)4(1)(3)2(3)k，解得17k

．··················································································12分20．（本小题满分12分）解：（1）直线l斜率不为0，(1,0)F:1lxty

设直线112212(,),(,).A0,0AxyBxyxyy设点在轴上方221,4404,xtyytyyx由得12124,4yytyy··································

···································3分31122122(1,y)2(1,)2AFFBxxyyy由1211224,8,2,4,yytytyyyt代入124yy，因10y，所以0t，解得122

t，22220ABxy所在直线方程为·············································6分（2）ABNNNxy设中点为（,），212+==2,=212NNyyytxt

2(21,2)Ntt所以AB中垂线2:2(21)lyttxt2(23,0)Dt2223122DFtt···················································

··············9分222121=()()ABxxyy222121=()()tytyyy2212121()4tyyyy2211616tt244t22442(22ABtDFt定值）··············

···················································12分21．（本小题满分10分）解：选择条件①：由114(2)nnaan≥，得数列21na和2na成公差为4的等差数列

，··········································2分则有21114(1)44naanna，2224(1)44naanna，································

·····························4分当2n时，314aa，又123,,aaa成等差数列，所以其公差为2，则有212aa，所以22214(1)4442naannana，···························

············6分则当n为奇数时，122nana；当n为偶数时，122nana，4所以122nana，()nN.·····························································

8分则12(1)nnaan≥，所以数列na是等差数列.··································10分选择条件②：因为21124(3)nnnnnaaaaan≥，所以2112()()2(3)nnnnnnnaaaaaaan≥，

又114(2)nnaan≥，所以24(3)nnaan≥，24()nnaan≥1，所以112(3)nnnaaan≥，·····························································6分所以数列na从第二项

开始等差，设公差为d，因4242aad，所以322aad，又313221()()4aaaaaa，所以212aa，综上有数列na是公差为2的等差数列.···········

····································10分选择条件③：因为114(2)nnaan≥，所以24()nnaan≥1，又361nnaan（≥），两式相减得3221nnaan（≥），所以数列na从第三项开始等差.·······

················································6分将3n带入114nnaa得424aa，因为432aa，所以42433232()()2()4aaaaaaaa

，所以322aa，同理212aa.综上有数列na是公差为2的等差数列.···············································10分22．（本小题满分12分）解（1）222222ABaa设焦距为2c，22122ceca离心

率2221bac5所以，所求的椭圆方程为2212xy·····················································3分（2）设直线MN方程为1xmy，设11Mxy（

,），22Nxy（,）由221,21,xyxmy22(2)210mymy得12122221+=,=22myyyymm··········································

·················5分11(2)2yAMyxx直线方程是，22(2)2yBNyxx直线方程是1122(2),2(2),2yyxxyyxx由·····································

···································7分21211221212121(2)(12)(12)2=2(2)(12)(12)yxymymyyyxxyxymymyyy可得212212112

12()(12)()(12)2(2)221(12)(2)()(12)2mmymmmymmmmymym21212121(322)(12)(2)(12)(2)[(322)(12)(2)](12)[(12)(2)](12)mmymmymmymmy

2121[(322)(12)(2)](12)(12)(2)mmymmy221212[(12)(2)](12)(12)(2)(12)322mmymmy········

···········································11分2=3+22=22xxx，解得=2.AMBNx因此直线与直线的交点P落在定直线上·················

··················12分