【文档说明】山东省烟台市2021-2022学年高二上学期期末考试 数学答案.pdf，共(7)页，367.374 KB，由管理员店铺上传

转载请保留链接：https://www.doc5u.com/view-532b0be5c77ada79b0e93676878fec6f.html

以下为本文档部分文字说明：

高二数学答案（第1页，共6页）2021～2022学年度第一学期期末学业水平诊断高二数学参考答案及评分标准一、选择题CDBABBAC二、选择题9.ACD10.BD11.ACD12.BCD三、填空题13.17714.915.3416.22yx=，2四、解答题17.解：由已知214aad=+=，511

545510302Sadad×=+=+=.················2分联立方程组11426adad+=+=，解得12a=，2d=.··········································4

分所以2nan=.·····················································································5分(22)(1)2nnnSnn+==+，·············

·····················································6分由题意28nSnnnλ=++≥，即27nnλ−≤.········································

·7分令2()7fxxx=−，其图象为开口向上的抛物线，对称轴为72x=，···············8分所以当3n=或4时，()fn取得最小值12−，······································

···9分故12λ−≤.······································································10分18.解：（1）因为离心率2222e15

cabbaaa+===+=，所以224ba=.·······3分又因为点(2,23)M−在双曲线C上，所以224121ab−=.························4分联系上述方程，解得21a=，24b=

，即1a=,2b=.·····························6分（2）设所求双曲线的方程为22(0)4yxλλ−=≠，······································8分高二数学答案（第2页，共6页）由双曲线经过点(3,25)P，得2034λ−=，即

2λ=−.························10分所以双曲线的方程为2224yx−=−，其标准方程为22182yx−=.··················12分19.解：（1）由已知得，11nnSna++=−，当2n≥时，111nnSna−+−=−.·

····················································1分两式相减得，121nnaa+=+.········································

······················2分于是112(1)nnaa++=+，即1121nnaa++=+，············································4分又23a=，214a+=，11

20a+=≠，所以21121aa+=+满足上式，··················5分所以1121nnaa++=+对n∗∀∈N都成立，故数列{1}na+是等比数列.··················6分（2）由（1）得21nna=−，···························

································7分21nnbn=+−，··························································8分[]23(2222)123(1)nnTn=+++

++++++−·····························9分21222nnn+−=−+.··············································

···12分20.解：（1）由抛物线定义||122pPF=+=，解得2p=，···························2分所以抛物线C的方程为24yx=.·············································3分

（2）因为点(1,)Pt在C：24yx=上，且0t>，所以2t=，即(1,2)P.············4分由题意可知0m≠，设11(,)Axy，22(,)Bxy，高二数学答案（第3页，共6页）联立241yxxmy==+，得2440ymy−−=，所以1

24yym+=，124yy=−，·························································5分于是1121112241214PAyykyxy−−===−+−，则直线PA的方程为14

2(1)2yxy−=−+，令0y=，则12yx=−；令0x=，则1122yyy=+，所以C点的坐标为1(,0)2y−，D点的坐标为112(0,)2yy+.······························7分同理M点的坐标为2(,0)2y−，N点的坐标为222(0,

)2yy+.···························8分所以12||||2yyCM−=，1212||4||(2)(2)yyDNyy−=++.···························

··9分121212||11||||4||222(2)(2)CDMNyyyySCMDNyy−−=⋅=××++四边形22121212121212()()4|(2)(2)||2()4|yyyyyyyyyyyy−+−==+++++2216162222||4|8|||||mm

mmmm++===+≥，···········································11分当且仅当1m=±时等号成立，此时四边形CDMN的面积最小值为4，直线l的方程为1yx=−或1yx=−+.·······

·························································12分21.解：（1）由123213142(1)aaaaaa++=+=+得，24a=.································

······1分又14aq=，34aq=，所以44414qq++=，即22520qq−+=，高二数学答案（第4页，共6页）解得2q=或12q=（舍去）.··················································

·············3分所以2nna=（n∗∈N），·································································4分当1n=时，111bS==，当2n≥时，2211111()[(1)(1)]2222

nnnbSSnnnnn−=−=+−−+−=，············5分经检验，1n=时，11b=适合上式，故nbn=（n∗∈N）.·······································································6分（

2）由（1）可知，22,4(2),nnnnnacnnna=−−为奇数为偶数当n为奇数时，2nnnc=，当n为偶数时，222(2)22nnnnnc−−=−，··················

··································7分由题意，有135213521135212222nnnTcccc−−−=++++=++++奇，①357212111352321422222nnnnT−+−−=+++++奇，②①−②得，357212121113122222

112144142222222214nnnnnnT−++−−−=+++++−=+−−奇542156563424664nnnnn−+=−−=−×××，························9分所以110659184nnT−+=−×奇.··········

··········································10分222222222462204264222204264(2)(22)()()()22222222nnnnnTcccc−−=++++=−+−+−+−

偶高二数学答案（第5页，共6页）2222201(2)042244nnnnnn−=−==.························································11分故22211110651018659184491

84nnnnnnnnT−−−+−−=−+=+××.································12分22.解：（1）由已知得1b=，······································

············1分离心率222e12ba=−=，·································································2分所以222

2ab==，·····································································3分故椭圆E的方程为2212xy+=.··········································4分（2）

当直线l的斜率存在时，设l：(1)ykx=+，11(,)Axy，22(,)Bxy，联立方程组2212(1)xyykx+==+得，2222(12)4220kxkxk+++−=，所以2122412kx

xk+=−+，21222212kxxk−=+.··········································5分212||1||ABkxx=+−2212121()4kxxxx=++−2222224881()121

2kkkkk−=+−−++22288112kkk+=+×+2222(1)12kk+=+.···························································7分222211111121||(1)(1)12

2xAFxyxx+=++=++−=，222221222221||(1)(1)122xBFxyxx+=++=++−=.·····························8分高二数学答案（第6页，共6页）所以121122||||22xxAFBF++=×12122()42xxxx+++=22

22228412122kkkk−−+++=22112kk+=+.··································9分所以11||22||||ABAFBF=.·············

··········································10分当直线l的斜率不存在时，l：1x=−，联立方程组22121xyx+==−，得2(1,)2A−，2(1,)2B−−.||2AB=，11221||||222AFBF=×=，··········

······························11分所以11||22||||ABAFBF=.综上，存在实数22t=使得11||||||ABtAFBF=恒成立.··························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