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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2

0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=14.3

0.15.222nn−+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...2128n

nnnCCC−+++==·····························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22325(1)()24xy−++=····································11分22325(31)(2)24−++=Q,即(3,0)B满足上述方程,,,AB

CD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2

nan=............................................................................................................

...........5分2nnb=··························································································6分（2）2,2nnnnnc=为

奇数;，为偶数.············································································7分2112342021..Tcc

cccc=+++++···························································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111n

nnnaabb++−=−+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−=++++++()1231111111123

1222222nnnTnn−=++++−+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,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，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++==++=+++=−()212222212122212xxmkmkk

mkkmkxxxxm+−++=+++=−化简得23mk=−，··········································································11分代入直线A

B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)20xymymnnynnxmyn−+=++−+−==+122212202

2121nmnyymnnxxm−+=−+−=+························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=−uuu

r···········································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