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宁德市2020-2021学年度第二学期期末高二质量检测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.B2.D3．B4．D5.A6.C7.D8.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.ABD10.BCD11.AD12.ACD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）(本小题第一个空2分，第二个空

3分)13.214.4815.（1）12,（2）31016.24四、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：（1）z是虚数2230mm··············································

········································3分解得：1m且3m·················································································5分（2）复数z表示的点在第四

象限，21>0230mmm······················································································

7分即>1-1<<3mm·······························································································9分得：1<

<3m所以，m的取值范围为1,3········································································10分18.（本小题满分12分）解：法一：（1）∵当2k，()

(21)xfxxe，(0)1f，································1分()(21)xfxxe，(0)1f················································

·························3分yfx在在点0,0f处的切线方程为1yx·········································5分（2）()(1)xfxkxke···························

················································6分∵函数fx在区间2,2内单调递减，∴0fx在区间2,2上恒成立；····································

·························7分即(1)0xkxke，0xe10kxk即10kxk在区间2,2上恒成立···························································8分∴210210kkk

k，··················································································11分解得113k，∴k的取值范围是113k

········································································12分法二：（1）同法一。（2）()(1)xfxkxke········································

···································6分∵函数fx在区间2,2内单调递减，∴0fx在区间2,2上恒成立·······························································

·7分(1)0xkxke，0xe10kxk即10kxk在区间2,2上恒成立·····················································

······8分设()1gxkxk当0k时，()10gx，符合题意·····························································9分当0k

时，()gx在2,2单调递减，(2)10gk，解得1k，得10k······························································

···············10分当0k时，()gx在2,2单调递增，(2)310gk，解得13k，得103k···········································································11分∴

k的取值范围是113k.·······································································12分（备注：第（2）小题得到如下答案：113k或113k或

113k的统一扣1分）19.（本小题满分12分）解：（1）由表格知：2018x，1.6y，·······················································2分∴521410

1410iixx，····························································3分由上，有122515512.50.960.752.61002.5.6

6tiiiiiixxyyrxxyy，(备注：未算出0.96r，直接判断2.50.752.6r的不扣分！）······················································

························································5分则y与x的线性相关.·················································································

·6分(2)依题意，完善表格如下:购置传统燃油车购置新能源车总计男性车主361248女性车主4812总计402060··············································

································································8分22603684127.56.63540204812K

·························································11分（备注：列式得1分；计算正确得1分；比大小得1分）故有99%的把握认为购车车主是否购置新能源乘用车与性别有关.··················

·····12分20.（本小题满分12分）解:法一(1)取PD的中点F,连接,AFEF.E是PC的中点EF是PCD的中位线1//,12EFCDEFCD································

············································1分//,1ABCDAB//,ABEFABEF且四边形ABEF是平行四边形························································

·············2分FEPCABD//BEAF······························································································3分BE平面PAD,AF平面PAD

//BE平面PAD····················································································5分（备注：BE

平面PAD,AF平面PAD，未写全的统一扣一分）（2）由平面几何知识知：DBBC且==2DBBC···········································6分又PC底面ABCD，设ECh，则11111(2233233DBCEEBCDB

CDVVSEChh）解得1h,2PC·······································································

·············7分∴以D为原点，DA为x轴，DC为y轴，建立空间直角坐标系，如图所示，则1,0,0A，(0,2,0)C，0,0,0D，(0,2,2)P1,2,2AP,1,0,0AD,0,0,2CP....................8分

面ADP的一个法向量为0,1,1m...............................9分，面ACP的一个法向量为2,1,0n..................10分10cos,10mn，·········

·········································································11分二面角CAPD的平面角的正弦值为31010··················

·······························12分法二，(1)取CD的中点G,连接,EGBG.PCD中，CEEP，CGDG//EGPDEG平面PAD，PD平面PAD//EG平面PAD，··································

················································2分//,1ABCDABBG四边形ABGD是平行四边形//BGADBG平面PAD，AD平面PAD//BG平面PAD··················

···································································3分BGEGG，,BGEG平面BEGPCABDxyz平面//B

EG平面PAD·············································································4分//BE平面PAD········································

··········································5分（说明：直接由线线平行推出面面平行，扣2分）（2）由平面几何知识知：DBBC且==2DBBC···········································6分又PC

底面ABCD，设ECh，则11111(2233233DBCEEBCDBCDVVSEChh）解得1h，∴2PC·········································································

········7分∴以C为原点，CD为x轴，CP为z轴建立空间直角坐标系Cxyz，如图所示，则2,1,0A，2,0,0,(0,0,2)DP，2,1,2AP,0,1,0AD,0,0

,2CP····················································8分面ADP的一个法向量为1,0,1m，.......................9分面ACP的一个

法向量为1,2,0n......................10分10cos,10mn.....................................................11分二面角CA

PD的正弦值为31010..............................12分法三：（1）同法一（2）（几何法）作CMPD，垂足为M，作MNPA，垂足为N，连接CN.·······················

·······················································································6分可证得CM平面ADP.PACM,PA平面CNM.PACN.CNM是二面角CAP

D的平面角.···························································9分在RtCMN中，2CM,22525325CN.310sin10CMCNMCN二面角CAPD的正弦值为31010

·····························································12分xPCABDzyMPCABDN21(本小题满分12分)解：（1）用X表示4个疑似病例中化验呈阳性的人数，则14,4XB·····

···········1分由题意可知，4个疑似病例中至少有1例呈阳性的概率为4117510114256PX··········································································

····································4分(2)方案一：逐个检验，检验次数为4.··················································

··········5分方案二：混合在一起检测，记检测次数为X，则随机变量X的可能取值为1，5，所以4181114256PX，8117551256256PX...................................

............................................................6分所以随机变量X的分布列为：X15P所以方案二检测次数X的数学期望为8117523

91525625664EX·····················7分方案三：每组两个样本检测时，呈阴性的概率为2191416，设方案三的检测次数为随机变量Y，则Y的可能取值为2，4，6，所以2981216256PY

，129712641616256PYC，2749616256PY，···········································································10分所以随机变量Y的分布列为：

Y24681256175256P所以方案三检测次数Y的期望为8112649960152462562562562564EY，······11分所以4EXEY，所以选择方案二最优.

·················································································12分22.(本小题满分12分)解：（1）2)(ln1)(xxxxf的定义域为),0(···

·········································1分xxxxfln211)(2············································································

··2分xxxxfxxgln21)()(，则0)1(12211)(22222xxxxxxxxg，所以)(xg在),0(上单调递增，且0)1(g，...............................

.....................3分当)1,0(x时，,0)(,0)(xfxg，则)(xf在)1,0(上单调递减；当)1(，x时，0)(,0)(xfxg，则)(xf在)1(，上单调递增；············4分所以函数2)1()(minfxf.··

····························································5分（2）)11ln(11)11ln()()11(nnananenan，·············

··············7分令]2,1(11nt，则ttaln111，令)21(ln111)(xxxxh，·················8分则222)(ln)1(2)(ln1)(xxxxxxh，···················

················································9分]2,1(x，再由（1）知2)(ln12xxx，即0)(xh，则)(xh在]2,1(上单调递增·········································

····································································11分所以2ln11)2()(maxhxha，2ln11)ln111(maxtt81256

12625649256所以a的取值范围为),2ln11[.·····························································12分法二：(1)解答同法一。（2）)11ln(11)11ln

()()11(nnananenan，···························7分令]2,1(11nt，则ttaln111，令)21(ln111)(xxxxh，·················8分2222)(ln

)1(2)(ln1)(xxxxxxxh··········································································9分设22

()1(ln)2xxxxx，2()2(ln)2ln2xxxx2()2(ln1)xxx，22()lnxxx12()0xx()x在1,2上单调递增，()(1)0x

()x在1,2上单调递增，()(1)0x()x在1,2上单调递增，()(1)0x，即()0hx()hx在1,2上单调递增·························

··················································11分所以2ln11)2()(maxhxha，即a的取值范围为),2ln11[.······················12分