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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

.15.222nn−+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)rrrrrrrTCxCxx−−+=−

=−...................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=1120Cxx−−··········

························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−1,22ACADkk==−············································································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A

CD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22325(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：100090300110DFDFDEF+−++=++++=++++=··············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222226324dqdq+=++=+·············································································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，1717(,)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−=++++++()12311111111231

222222nnnTnn−=++++−+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1

2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,OMON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uuruur1221122(,)11kTAmkk=−++uurQ，22222

22(,)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,B

xy，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,)xxPM

yy=−uuur,22222244(2,)xxPNyy=−uuur············································9分2212122221124444(2)2xx

xxyyyy−=−12122yyxx=−·····················································································10分()()2122121212

12121211()()2kxmkxmyymmmkkkkmxxxxxxxxxx++==++=+++=−()212222212122212xxmkmkkmkkmkxxxxm+−++=+++=−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,Axy，()22,Bxy，联立()2222211(1)(22)20xym

ymnnynnxmyn−+=++−+−==+1222122022121nmnyymnnxxm−+=−+−=+························5分设,OMO

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

···6分2222244yxxyyyxyx===，所以22222244(,)xxNyy，···········································7分因为21121

144(,)xxMyy，22222244(,)xxNyy三点共线//PMPNuuuruuur21121144(2,)xxPMyy=−uuur,22222244(2,)xxPNyy=−uuur···········································9分221

2122221124444(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