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宁德市2021-2022学年度第一学期期末高二质量检测数学参考答案及评分标准说明:一、本解答指出了每题要考查的主要知识和能力，并给出了一种或几种解法供参考，如果考生的解法与本解法不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则.二、对计算题，

当考生的解答在某一部分解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分.三、解答右端所注分数，表示考生正确做到

这一步应得的累加分数.四、只给整数分数，选择题和填空题不给中间分.一、单项选择题:本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一个选项是符合题目要求的.1.A2．C3．B4．D5.A6.C7．B８．C二、多项选择题：本题共4小题，每

小题5分，共20分.在每小题给出的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2分，有选错的得0分）9．ABC10.BC11.ABC12.BCD三、填空题:（本大题共4小题，每小题5分，共20分.把答案填在答题卡的相应位置）13.3yx=14.30.15.22

2nn−+16.31−（答案为423−不扣分）四、解答题：本大题共6小题，共70分.解答应写出文字说明，证明过程或演算步骤.17.（本小题满分10分）解：选①:012(1)1372nnnnnCCCn−

++=++=..................................................................2分8n=或9n=−（舍去）·····························································

··············4分选②:依题意得2172nnCC=···············································································2

分即(1)722nnn−=得8n=···················································································

·················4分选③:偶数项的二项式系数之和为1351...2128nnnnCCC−+++==·····························2分8n=解得··········

·······················································································4分（1）82()xx−展开式的通项公式为()83821882(2)rrrrrrrTCx

Cxx−−+=−=−...................6分令8312r−=，得2r=·················································································7分展开式中x的项的系数为112·

····································································8分（2）展开式中二项式系数最大的项为422816=1120Cx

x−−··································10分18.（本小题满分12分）解：法一：（1）ABQ边上的中线所在的直线的方程为1x=可设(1,)Cm················

········································································1分ACQ边上的高所在的直线方程为26yx=−+，其斜率为-

2(2)(2)12ACmk−=−=−，····································································3分解得：1m=···········

··············································································4分(1,1)C··························

····································································5分（2）(1,0),(3,0),(1,1),ABC−Q(1,4)D−1,22

ACADkk==−············································································6分1ACADkk=−ACADkk⊥··················

······································································8分1,22BCBDkk=−=BCBDkk⊥········································

················································10分,,,ABCD在以CD为直径的圆上.························································12分法二：（1）同法一.

..........................................................................................................5分（2）(1,0),(3,0),(1,1),ABC−Q(1,4

)D−1,22ACADkk==−·············································································7分1ACADkk=−，ACADkk⊥·······

················································································9分ACDV的外接圆的圆心为3(1,)2−，半径为52ACDV的外接圆的方程为22

325(1)()24xy−++=····································11分22325(31)(2)24−++=Q,即(3,0)B满足上述方程,,,ABCD四点共圆.·············································

···························12分法三：（1）同法一...................................................................................................

........5分（2）设ABC外接圆的方程为：222()()xaxbr−+−=·······························6分将三点(1,0),(3,0),(1,1)ABC−分别代入圆的方程：222222222(1)(0)(3

)(0)(1)(1)abrabrabr−−+−=−+−=−+−=····7分解得2132254abr==−=·····················································

·····························10分(1,0),(3,0),(1,1)ABC−的外接圆方程：22325(1)()24xy−++=···················11分(1,4)D−Q满足上述方程,,,,ABCD四点共圆.··········

···························································12分法四：（1）同法一......................................................

.....................................................5分（2）设ABC外接圆的方程为：22220(40)xyDxEyFDEF++++=+−··········6分将三点(1,0),(3,0),(1,1)ABC−代入圆的方程：10009

0300110DFDFDEF+−++=++++=++++=··············7分解得233DEF=−==−...........................................

.......................................................................10分ABC的外接圆方程：222330xyxy+−+−=····

····································11分(1,4)D−Q满足上述方程,,,ABCD四点共圆.···············································

·························12分19.（本小题满分12分）解：（1）设na的公差为d，设nb的公比为q，.....................................................1分则22

2226324dqdq+=++=+·············································································3分0qQ2,2dq==·················

·····························································4分2nan=........................................................

...............................................................5分2nnb=····················································

······································6分（2）2,2nnnnnc=为奇数;，为偶数.·······························································

·············7分2112342021..Tcccccc=+++++···························································8分246202122322522221=++++++++

L246202[13521](2222)=+++++++++LL·····································9分123102[13521](4444)=+++++++++LL104242(41)3=+−11722433=+········

·················································12分(答案是104242(41)3+−或41950263不扣分）)20.（本小题满分12分）解：（1）设椭圆

C的该方程：22221(0)yxabab+=......................................................１分则222232acaabc===+············

···································································２分21ab==··············································

···················································４分椭圆C的方程：2214xy+=·····························································５分

（2）解法一：1(0,)2P，设1122(,),(,)AxyBxy，··············································６分联立2221422301122xyxxyx+=+−==+

··························································７分解得：1172x−−=，2172x−+=·······················································９分所以，171

7(,)24A−−−，1717(,)24B−++2217171355(0)(),2424PA−−−+=−+−=··································１０分2217171355(0)(),2424PB−++−=−+−=···

·······························１１分35535515448PAPB+−==···················································１２分解法二：1(0,)2P，设1122(,),(,)AxyBxy，··

····················································６分2211115(0)(),22PAxyx=−+−=·······················································７分

2222215(0)(),22PBxyx=−+−=························································８分联立2221422301122xyxxyx+=+−==+···················

········································９分1232xx=−······················································································１０分12515=

48PAPBxx=g...............................................................................................１２分21(本小题满分12分)解：（1）Q111ab−=，

111ab+=,数列nnab−是公比为12的等比数列...................................................................1分112nnnab−−=............................

.......................................................................2分111nnnnaabb++−=−+Q11()1nnnnabab+++−+=

·································································4分nnab+是1为首项，1为公差等差数列············································5分nnabn+=·

···············································································６分（2）由（1）知112nnnab−−=，nnabn+=1221()()2nnnnnnnabababn−−=+

−=..........................................................................７分所以01211111123.........2222nnTn−=++++++

()12311111111231222222nnnTnn−=++++−+L，所以0234111111112222222nnnTn−=+++++−

L······················８分即11112222nnnTn−=−−，得2111422nnnTn−−=−−...............................

............................................................１０分所以21114422nnnTn−−=−−·········

·············································１１分所以4t······················································

································１２分22.（本小题满分12分）解：法一（1）设(),Pxy，Q动点P到直线2x=−的距离比到点()1,0F的距离大1PF等于P到直线1x=−的距离····

····················································１分根据抛物线的定义知，曲线E是以()1,0F为焦点，直线1x=−为准线的抛物线．······························

······························2分故曲线E的方程为24yx=．·······························································４分（2）显然,OM

ON斜率存在，分别设,OMON的直线方程为12,ykxykx==·········５分联立21144yxykykx===，所以21144(,)Mkk，···········································６分同理得：22244(

,)Nkk··········································································７分因为2211224444(,),(,),(2,0)MNQkkkk三点共线//QMQNuuuruuur21144(2,)

QMkk=−uuur，22244(2,)QNkk=−uuur2221124444(2)(2)kkkk−=−122kk=−············································

·······································８分()222111121xyxkykx−+==+=，所以1221122(,)11kAkk++，同理得：2222222(,)11kBkk++·································

··························９分若存在满足题设的定点，由抛物线与圆的对称性知定点必在x轴上，设定点为(,0)Tm由1221122(,)11kAkk++，2222222(,)11kBkk++，(,0)Tm三点共线//TATBuuruur

1221122(,)11kTAmkk=−++uurQ，2222222(,)11kTBmkk=−++uur21222212212222()1111kkmmkkkk−=−++++，·····································

·····１０分122kk=−Q得()()122323mkmk−=−，即4()1223()0mkk−−=即恒成立，···················11分120kk−Q23m=，直线AB过定点2

(,0)3T······················································12分（备注：能写对定点坐标的给1分）法二：（1）设动点(,)Pxy依题意有：2221(1)xxy+−=−+····························

··························2分由几何直观知0x所以，221(1)xxy+=−+化简得：24yx=...........................................................................

........................4分（2）1o当直线AB垂直于x轴时，由对称性可知MN也垂直于x轴，所以(2,22)M，(2,22)N−，,OMON的直线方程为22yx=，22yx=−，()221124(,2)3322xyAyx−+==,同理得：24(,2)33B−，

所以直线AB的方程为23x=···································································································5分2o当直线AB不垂直于x轴

时，设AB的方程为ykxm=+，()11,Axy，()22,Bxy，联立()2222211(1)(22)0xykxkmxmykxm−+=++−+==+122212202211kmxxkmxxk

−+=−+=+·································6分设,OMON的直线方程为11yyxx=，22yyxx=,则2111144yxxyyyxyx===，所以21121144(,

)xxMyy，2222244yxxyyyxyx===，所以22222244(,)xxNyy，··············································7分因为21121144

(,)xxMyy，22222244(,)xxNyy三点共线//PMPNuuuruuur21121144(2,)xxPMyy=−uuur,22222244(2,)xxPNyy=−uuur····················

························9分2212122221124444(2)2xxxxyyyy−=−12122yyxx=−·······································································

··············10分()()212212121212121211()()2kxmkxmyymmmkkkkmxxxxxxxxxx++==++=+++=−()212222212122212xx

mkmkkmkkmkxxxxm+−++=+++=−化简得23mk=−，··········································································11分代

入直线AB的方程为2()3ykx=−，所以直线AB恒过2,03综合1o，2o得直线AB恒过2,03·······················································12分法三：（1）设动点(,)Pxy依题意有

：2221(1)xxy+−=−+······················································2分当2x−时，得221(1)xxy+=−+化简得：24yx=(0)x当2x−时，得223(1)xxy−−=−+化简得：2880yx=+，

舍去所以，24yx=··············································································4分（注：若未舍去288yx=+，扣1分）(2)显然直线AB不平行于x轴，可设AB的方程为xmyn=+，()11,Ax

y，()22,Bxy，联立()2222211(1)(22)20xymymnnynnxmyn−+=++−+−==+1222122022121nmnyymnnxxm−+=−+−=

+························5分设,OMON的直线方程为11yyxx=，22yyxx=2111144yxxyyyxyx===，所以21121144(,)xxMyy，

············································6分2222244yxxyyyxyx===，所以22222244(,)xxNyy，·········································

··7分因为21121144(,)xxMyy，22222244(,)xxNyy三点共线//PMPNuuuruuur21121144(2,)xxPMyy=−uuur,22222244(2,)xxPNyy=−uuur·································

··········9分2212122221124444(2)2xxxxyyyy−=−12122yyxx=−··········································································

··········10分()()()12121222121212122yyyyyyxxmynmynmyymnyyn===−+++++化简得23n=，······································

········································11分代入直线AB的方程为23xmy=+，所以直线AB恒过2,03···················12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com