【文档说明】江苏省连云港市东海县2020-2021学年高二上学期期中考试数学评分标准及答案.pdf，共(5)页，283.117 KB，由小赞的店铺上传

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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，0

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

bb，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212221441246214442143nnn

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

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2

7k．···············································································5分（2）因存在1x，2[3,3]x，使12()()fxgx成立则有minmax[()][()]fxgx，

·····································································8分因2()24fxxxk22(1)2xk，2()2gxxx2(1)1x，[3,3]x所以min

max()(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122t，

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）2222

22ABaa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21221211212()(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