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高三数学参考答案及评分细则（第1页共21页）2021－2022学年第一学期福州市高三期末质量检测数学试题（完卷时间：120分钟；满分:150分）本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．第Ⅰ卷1到3页，第Ⅱ卷3至4页．注意事项：1.答题前，考生务必在

试题卷、答题卡规定的地方填写自己的准考证号、姓名．考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致．2.第Ⅰ卷每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑．如需改动，用橡皮擦干净后，再选涂其

它答案标号．第Ⅱ卷用0.5毫米黑色签字笔在答题卡上书写作答．在试题卷上作答，答案无效．3.考试结束，考生必须将答题卡交回．第Ⅰ卷一、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，只有一项是符合题目要求的．1.已知集合*22,120ABxxxN，≤

，则ABA．B．211，，C．2112，，，D．21012，，，，【答案】C【解析】*1212BxxN≤≤，，所以AB2112，，，，故选C．【考查意图】本小题以集合为载体，主要考查集合的概念和基

本运算等基础知识；考查运算求解能力、推理论证能力；考查数学运算、逻辑推理等数学核心素养，体现基础性．2.已知34iz，则+izzA．1+3iB．84iC．9+3iD．29+3i【答案】C【解析】因为34iz，所以+i5+34ii=9+3izz，故选C．【考查意图】本小题以复

数为载体，主要考查复数的基本概念等基础知识；考查运算求解能力、推理论证能力；考查数学运算、逻辑推理等数学核心素养，体现基础性．3.已知甲、乙、丙、丁、戊五位同学高一入学时年龄的平均数、中位数均为16，方差为0.8，则三年后，下列判断错误的是

A．这五位同学年龄的平均数变为19B．这五位同学年龄的中位数变为19C．这五位同学年龄的方差仍为0.8D．这五位同学年龄的方差变为3.8【答案】D【考查意图】本小题以“五位同学的年龄”为载体，考查平均数、中位数、方差等基础知识；考查应用意识；考查逻辑推理等核心素养；体现基础性与应用性．高三数学

参考答案及评分细则（第2页共21页）4.613xx展开式中的常数项为A．540B．15C．15D．135【答案】D【解析】二项式613xx的展开式的第1r项为616(3)rrrr

TCxx36626(1)C3rrrrx．令3602r，解得4r，所以42651353TC，所以613xx展开式中的常数项为135．故选D.【考查意图】本小题以二项式为载体，主要考查二项

式定理等基础知识；考查运算求解能力、推理论证能力；考查数学运算、逻辑推理等核心素养，体现基础性．5.已知函数3310()0xxfxaxbx，＞，，＜为偶函数，则2+abA．3B．32C．12D．32【答案】B【解析】解法一、当0x时，0

x，所以33()()1+1fxxx，因为()fx为偶函数，所以3()()+1fxfxx，又3()fxaxb，所以1,1ab，所以32+2ab．解法二、因为()fx为偶函数，所以(1)1,(2)2,ffff所以2,89,abab

解得11ab，，经检验，11ab，符合题意，所以32+2ab．【考查意图】本小题以分段函数为载体，主要考查函数的奇偶性的定义等基础知识；考查运算求解能力、推理论证能力；考查数学运

算、逻辑推理等核心素养，体现基础性．6.已知一张边长为2的正方形纸片绕着它的一条边所在的直线旋转4弧度，则该纸片扫过的区域形成的几何体的表面积为A．2B．C．D．【答案】C【解析】因为一个边长为2的正方形纸片绕着一条边

旋转4，所形成的几何体为柱体，该柱体是底面半径r为2，高h为2的圆柱的八分之一，所以其表面积21228Srhr2222122222828，故选C．高三数学参考答案及评分细则（第3页共21页）【考查意图】本小题以旋转体为载体，主要考查旋转体的

体积等基础知识；考查空间想象能力、推理论证能力、运算求解能力；考查数形结合思想、化归与转化思想；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性．7.已知函数sinfxx的部分图象如图所示，则fx的单调递增区间为

A．15,,66kkkZB．152,2,66kkkZC．15,,66kkkZD．152,2,66kkkZ【答案】D【解析】解法一、由2342,3mmmmZ,Z解得

2,3mm,Zsin2sin33fxxmx，令232kxkkZ，解得1566kxkkZ．故选D．解法二、由图象知411

,2,233TT又1453326x时，()fx取得最大值，排除A、B、C.故选D．【考查意图】本小题以三角函数的图象为载体，考查三角函数的图象和性质等基础知识；考查抽象概括能力、推理论证能力；考查数形结合思想；考查直观想象、逻辑推理等核心素养，体现基础性和综合性．8.已知O

为坐标原点，F是双曲线2222:100xyCabab＞，＞的左焦点，A为C的右顶点．过F作C的渐近线的垂线，垂足为M，且与y轴交于点P．若直线AM经过OP的中点，则C的离心率为A．2B．32C．3D．43【答案】A【解析】解法

一、如图所示，设AM交y轴于Q，过M作x轴的垂线，垂足为N．由双曲线性质可知，FMb，OFcOMa，，由FMN△∽FPO△，AQOAMN△∽△得高三数学参考答案及评分细则（第4页共21页）FNMNFOPO,AOQOANMN,以上两式

相乘得12FNAOFOAN，所以2212acacacac，所以2212acaacca，即12cac，所以1112e，解得2e．故选A．解法二、如图所示，设AM交y轴于Q，过M作x轴的垂线，垂足为N．不

妨设渐近线方程为byxa，则直线FP的方程为ayxcb，令0x，得Pacyb．由byxaayxcb，可得2aabMcc，，则2AMabbckaacac，所以直线AM的方程为byxaac，令0x得Qabyac

．因为Q为OP中点，所以2abacacb，整理得222222accbca，即2220caca，所以220ee，解得2e，或1e（舍去）．故选A．解法三：如图所示，设AM交y轴于Q．由双曲线性质可知，FMb，OFc

，OMaOA，所以OMQOAQ．在RtOPM△中，Q为OP中点，所以MQOQPQ，所以QOMOMQOAQ，所以RtRtOMPAOQ△≌△，所以MPOQ，所以MPQ△为正三角形，所以60MPQ

，故30MFO，所以2,ca所以2e.（也可以利用tan30FPakb，即212bea），故选A．xyOFAMPQNxyOFAMPQN高三数学参考答案及评分细则（第5页共21页）【考查意图】本小题以双曲线为载体，主要考查双曲线的图象和性质、直线与双曲线

的位置关系等基础知识；考查运算求解能力、推理论证能力；考查数形结合思想、化归与转化思想；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性和综合性．二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，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.已知向量3111，，，mn=mn=，则A．∥mnnB．mnnC．2mnD．45，mn【答案】BCD．【解析】依题意，1202，m=mnmn，1[2n

=mn]11，mn，所以11110，，mnn，所以mnn，所以选项A错误，B正确．所以22mn，选项C正确；22cos222，mnmnmn，因为0180mn≤，≤，所以45，mn，选项D正确．【考查意图】本小题以平面向量为载体，考

查平面向量的坐标表示、平面向量共线与垂直、平面向量模长、夹角等基础知识；考查推理论证能力、运算求解能力；考查数形结合思想、函数与方程思想；考查直观想象、逻辑推理等核心素养，体现基础性和综合性．10.某人有6把钥匙，其中n把能打开门．如果随机地取一把钥匙试着开门，把不能开门的钥匙扔掉，设第

二次才能打开门的概率为p，则下列结论正确的是A.当1n时，16pB.当2n时，13p，C.当3n时，310pD.当4n时，45p【答案】AC【解析】当1n时，511656p，选项A正确；当2n时，4246515p，选项B错误；当3n时，3

336510p，选项C正确；当4n时，2446515p，选项D错误．故选AC．【考查意图】本小题以“取钥匙开门”为载体，考查随机事件的概率等基础知识；考查推理论证能力、运算求解能力与创新意识；考查化归与转化思想、或然与

必然思想；xyOFAMPQ高三数学参考答案及评分细则（第6页共21页）考查数学建模、逻辑推理、数学运算等核心素养，体现综合性、应用性．11.已知3030AB，，，，动点C满足2CACB，记C的轨迹为．过A的直线与交于PQ，两点，直线BP与的另一个交点

为M，则A．QM，关于x轴对称B．PAB△的面积的最大值为63C．当45PMQ时，42PQD．直线AC的斜率的取值范围为33，【答案】AC．【解析】设Cxy，，由2CACB得223xy2223xy，整理得的方程

为25x2y16，其图象是以50D，为圆心，半径4r的圆．故max11641222PABSABr△，选项B错误．因为2PAPB，2MAMB，所以PAPBMAMB，所以PABMAB，又轨迹的图象关于x轴对称，所以QM，关

于x轴对称，选项A正确．当45PMQ时，452PDQ90，则DPQ△为等腰直角三角形，242PQr，选项C正确．当直线AC与圆D相切时，CDAC，此时822ADrCD，切线AC的倾斜角为30和150，结合图象，可得直线AC的斜率的取值范围为3333

，．选项D错误．故选AC．【考查意图】本小题以圆为载体，主要考查直线与圆的位置关系、弦长、图象的对称性等基础知识；考查运算求解能力、推理论证能力；考查数形结合思想、函数与方程思想；考查直观想象、逻辑推

理、数学运算等核心素养，体现基础性和综合性．12.设函数1()e1exxfxxx，则A．()(1)fxfxxy–1–2–3123456789–1–2–3–41234POABDQM高三数学参考答案及评分细则（第7页共21页）B．函数()fx有极大值为eC．若121xx

，则1122()exfxxfx≥D．若121xx＜，且212x＞，则21()1fxfx＜【答案】ACD【解析】易验证A是正确的，也即函数()fx关于直线12x对称．故选项A正确；因为11()ee2eexxxxfxxx1+1e2exxxx，所

以1()02f，当12x时，113()+1e2e(ee)2xxxxfxxx，此时1xx，所以()0fx，故函数()fx在1(,)2上单调递增；由于函数()fx关于直线12x对称，

所以函数()fx在1(,)2上单调递减．所以函数()fx在12x处有极小值，也是最小值，1()e2f．故选项B错误；若121xx，且212x，则21112xx，由()fx在1(,)2上

单调递增得21()1fxfx．故选项D正确；由于函数()fx的最小值为e，所以1()efx，若121xx，则211xx，所以211fxfx，又因为111fxfx，所以12fxfx，故112211211211()()()()()exfx

xfxxfxxfxxxfxfx，故选项C正确．故选ACD.【考查意图】本小题以函数为载体，考查函数与导数、函数的基本性质、函数的极值等基础知识；考查抽象概括能力、推理论证能力、创新意识；考查数形结合思想、化归

与转化思想；考查数学抽象、逻辑推理、数学运算、直观想象等核心素养；体现综合性与创新性．第Ⅱ卷注意事项：用0.5毫米黑色签字笔在答题卡上书写作答．在试题卷上作答，答案无效．三、填空题：本大题共4小题，每小题5分，

共20分．13.曲线3()2fxxx在0x处的切线方程是．【答案】20xy高三数学参考答案及评分细则（第8页共21页）【解析】依题意得，(0)0f，232fxx，所以02f，所以

所求的切线方程为2yx，即20xy．【考查意图】本小题以三次函数为载体、考查导数的几何意义等基础知识；考查抽象概括能力、运算求解能力，考查化归与转化思想、数形结合思想；考查数学抽象、直观想象、数学运算等核心素养，体现基础性．14.在正三棱柱

111ABCABC中，12ABAA，F是线段11AB上的动点，则1AFFC的最小值为．【答案】62【解析】将正三棱柱111ABCABC上底面沿11AB展开至平面11ABBA上，如图所示，因为1112AAAC，且1190+60=150AAC，所以1111

2sin2AACACAA22sin7562，所以1AFFC的最小值为62．【考查意图】本小题以正三棱柱为载体，主要考查空间中动点到两定点的距离和最小值等基础知识；考查空间想象能力、推理论证能力；考查化归与转化思想；考查直观想象、逻辑推理等核心素

养，体现基础性．15.抛物线2:2(0)Eypxp的焦点为F，点A是E的准线与坐标轴的交点，点P在E上，若30PAF，则sinPFA．【答案】33【解析】过P作准线的垂线，垂足为B，所以30BPAPAF，PBPF．在RtBPA△中，cos30PBPA，即在PA

F△中，cos30PFPA，xyOAPFB高三数学参考答案及评分细则（第9页共21页）又由正弦定理sinsinPAPFPFAPAF，所以sin303sinsincos303PAPFAPAFPF．【考查意图】本小题以抛物线为载体，主要考查抛物线的方程与定义、解三角

形等基础知识；考查运算求解能力、推理论证能力；考查数形结合思想；考查数学运算、逻辑推理等数学核心素养，体现基础性．16.函数yx称为高斯函数，x表示不超过x的最大整数，如0.90，lg991．已知数列na满足3

3a，且1nnnanaa，若lgnnba，则数列nb的前2022项和为．【答案】4959【解析】利用累乘法(或13113nnaaann),得nan．记nb的前n项和为nT，当19n时，0l

g1na时，lg0nnba；当1099n时，1lg2na时，1nb；当100999n时，2lg3na时，2nb；当10002022n时，3lg4na时，3nb；所以2022122022lglglg9019002102334959Ta

aa．【考查意图】本小题以数列为载体，考查数列求通项、递推数列、数列前n项的和等基础知识；考查推理论证能力、运算求解能力、创新意识；考查数形结合思想、化归与转化思想；考查数学抽象、逻辑推理、直观想象、数学运算等核心素养；体现综合性与创新性．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.（10分）设数列na是首项为1的等差数列，若2a是1a，5a的等比中项，且23aa.（1）求na的通项公式；（2）设11nnnb

aa，求数列nb的前n项的和nS．【考查意图】本小题主要考查等比中项、等差数列的通项公式、裂项相消求和等基础知识；考查推理论证能力、运算求解能力；考查化归与转化思想、函数与方程思想；考查逻辑推理、数学运算等核心素养，体现基础

性、综合性．满分10分．【解】（1）设等差数列na的公差为d，因为11a，且2a是1a，5a的等比中项，所以2215aaa，···································································

················1分高三数学参考答案及评分细则（第10页共21页）所以21114adaad，·················································

·················2分又因为11a，所以2114dd解得2d，或0d，··········································································3分

又因为23aa，所以0d，所以2d，·······································································4分所以111212

1nandnan．···············································5分（2）由（1）知，21nan，因为11nnnaba，所以1111()(21)(21)22121nbnnnn

，············································7分所以11111111(1)()()()2335572121nSnn··················

·········8分=11(1)221n．·········································································9分.21nn··································

···············································10分18.（12分）为让人民享受到更优质的教育服务，我国逐年加大对教育的投入．下图是我国2001年至2019年间每年普通本科招生数y（单位：万人）的条形图．普通高等学校本科招生数（万人）

（数据来源：国家统计局网站）为了预测2022年全国普通本科招生数，建立了y与时间变量t的三个回归模型．其中根据2001年至2019年的数据（时间变量t的值依次为12319，，，，）建立模型①：高三数学参考答案及评分细则（第11页共21页）0.058ˆ166.9ety，相关指数210

.87R；模型②：ˆ152.416.3yt，相关系数20.97r，相关指数220.95R．根据2014年至2019年的数据（时间变量t的值依次为1236，，，，）建立模型③：ˆ372.89.8yt

，相关系数30.99r，相关指数230.99R．（1）可以根据模型①得到2022年全国普通本科招生数的预测值为597.88万人，请你分别利用模型②、③，求2022年全国普通本科招生数的预测值；（2）你认为用哪个

模型得到的预测值更可靠？并说明理由．【考查意图】本小题主要考查回归分析、相关指数、相关系数等基础知识；考查数据处理能力、推理论证能力、运算求解能力与创新意识；考查函数与方程思想、化归与转化思想；考查数学建模、逻辑推理、数学运算等核心素养，体现综合性、应用性与创新性．满分12分．【解

】（1）利用模型②，2022年全国普通本科招生数的预测值为ˆ152.416.322511y（万人）；·······················································3分利用模型③，2022年全国普通本科招生数的预测值为ˆ372.89.8946

1y（万人）．·························································6分（2）利用模型③得到的预测值更可靠．················································7分理由如下：

（ⅰ）从条形图可以看出，2001年至2010年，2011年至2019年两个区间增长率有显著区别，2014年至2019年招生数增长速度趋于稳定，线性关系更为明显，故模型③比模型①、②能更好地描述时间变量与招生数的变化趋势．·································

··9分（ⅱ）从计算结果可以看出，模型③的相关指数230.99R最高，说明其拟合效果最好．模型③的相关系数30.99r比模型②的相关系数20.97r高，说明模型③的两变量的相关性比模型②更强，因此利用模型③得到的预测值更可靠．

···························12分19.（12分）记ABC△的内角A，B，C的对边分别为a，b，c．已知coscoscaBcA．（1）试判断ABC△的形状，并说明理由；（2）设点D在边AC上，若ADBD，sinsinADBABC，求ab

的值.【考查意图】本小题主要考查解三角形等基础知识；考查推理论证能力、运算求解能力；考查函数与方程思想、数形结合思想；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性．满分12分．【解】解法一、（1）ABC△为直角三角形或AB

C△为等腰三角形.·················1分理由如下：在ABC△中，因为coscoscaBcA，根据正弦定理得，sinsincossincosCABCA，···················

··········································2分高三数学参考答案及评分细则（第12页共21页）ABCD又因为π()CAB，所以sin()sincossincosABABCA，

即sincoscossinsincossincosABABABCA，即cossinsincosABCA，···································································3分所以cos0A或sins

inBC，······························································4分若cos0A，则π2A，故此时ABC△为直角三角形.·········

······················５分若sinsinBC，则由正弦定理得，bc.故此时ABC△为等腰三角形.综上，ABC△为直角三角形或ABC△为等腰三角形.···················

··············6分（2）由（1）知，π2A或bc，若π2A，则ADBD，这与已知条件ADBD相矛盾，所以π2A；·······················································

································7分所以bc，所以ABCC.又因为sinsinADBABC，所以sin(π)sinADBC，即sinsinBDCC，故BDCC，·······························

·················································8分所以ADBC.在DBC△中，sinsinBCDBCBDCDC，在ABC△中，sinsinBCABCAAC，················

···········································9分两式相乘得2BCACDC，·······················································

·········10分(也可通过等腰ABC△BDC△得到2BCACDC.)又ACADACBDACBCCD，所以2()abba，··········································································

···11分解得512ab或512ab（舍去）．所以ab的值为512．········································································12分解法二、（1）ABC△为直角三角

形或ABC△为等腰三角形.··························1分理由如下：因为coscoscaBcA，高三数学参考答案及评分细则（第13页共21页）ABCD根据余弦定理，得22222222acbbcacacacbc，·····

··························２分22222222acbbcacb，所以22222222acbbcaccb，·············································

···········３分即2222222222cacbbcacb，所以222222()()bcbacbca，······················································４分所以2

22()()()()0bcbcbcbbccabc，所以222()()0bcbca，···············································

·················５分即222abc或bc，所以ABC△为直角三角形或ABC△为等腰三角形.····································6分（2）由（1）知，π

2A或bc，若π2A，则ADBD，这与已知条件ADBD相矛盾，所以π2A；··········7分所以bc，所以ABCC.又因为sinsinADBABC，所以sin(π)sinADBC，即

sinsinBDCC，故BDCC，················································································8分所以ADBC.又因为ADBD，所以ADBA，即BD平分ABC

，··············································································9分所以BADABCDC，·······································

········································10分所以caaba，所以baaba，······································

··········································11分解得512ab或512ab（舍去）．所以ab的值为512．·····························

···········································12分高三数学参考答案及评分细则（第14页共21页）20.（12分）如图，在三棱锥DABC中，DA底面ABC，1ACBCDA

，2AB，E是CD的中点，点F在DB上，且EFDB．（1）证明:DB平面AEF；（2）求二面角ADBC的大小．【考查意图】本小题主要考查空间直线与直线、直线与平面的位置关系，二面角等基础知识；考查推理论证能力、运算求解能力与空间想象能力；考查数形结合思想；考查直观想象、逻辑推理、数

学运算等核心素养，体现基础性、综合性．满分12分．【解】解法一、（1）DA平面ABC，且BC平面ABC，DABC，12ACBCAB，，222ACBCAB，ACBC，·······················································

··1分DAACA，BC平面DAC，······················································2分AE平面DAC，BCAE．又DAAC，E是C

D的中点，DCAE，······················································································3分又BCDCC，AE平面DBC．DB平面DBC，DBAE，···············

···········································4分,EFDBEFAEE，DB平面AEF．········································

·····································5分（2）过点A作AGBC∥，由（1）知BC平面DAC，所以AG平面DAC．以点A为原点，分别以向量ACAGAD，，为x轴，y轴，z轴的正方向，建立如图所示的空间直角坐标系Axyz，·······

······································································6分则0,0,0A，1,0,0C，1,1,0B，0,0,1D，则0,0,1,AD1,1,11,0,1BDCD

，，设平面ADB的法向量111,,mxyz，则0,0,mADmBD所以11110,0,zxyz令11y，DEFABC高三数学参考答案及评分细则（第15页共21页）则1,1,0.m··

···················································································8分设平面DBC的法向量为222,,nxyz，则0,0,nCDnBD所以222220,0

,xzxyz令21x，则1,0,1n．·····································································10分

所以11cos,222mn，··························································11分又因为二面角ADBC的平面角为锐角，

所以二面角ADBC的大小为3．·····················································12分解法二、（1）DA平面ABC，且BC平面ABC，DABC，12ACBCAB，，2

22ACBCAB，ACBC，·························································1分DAACA，BC平面DAC，····················································

·························2分过点A作AGBC∥，所以AG平面DAC.以点A为原点，分别以向量ACAGAD，，为x轴，y轴，z轴的正方向，建立如图所示的空间直角坐标系Axyz

，则110,0,0,1,1,0,0,0,10,,22ABDE，，所以1,1,1DB，110,,22AE，11101(1)022DBAE，·································

·······································································4分DBAE，DBAE，DBEF，且AEEFF，DB平

面AEF.················································································5分（2）因为EFDB，由（1）得DBAF，所以AFE为二面角ADBC的

平面角，··············································6分因为1,0,0C，由（1）知，1,1,1DB.设,,Fxyz，则,,z1DFxy.·

····························································7分高三数学参考答案及评分细则（第16页共21页）因为点F在DB上，所以存在实数k，使得DFkDB，所以,,1xkykzk

，即,,1Fkkk，因为AFDB，所以0AFDB，··························································8分所以110kkk，解

得13k，················································9分所以点112,,333F，所以111,,366FE，112,,333FA，·············

··10分所以116cos26663FAFEAFEFAFE，·············································11分又因为0,AFE，所以3AFE.所以二面角ADBC的大小为.3···············

·········································12分解法三、（1）略，同解法一；（2）因为EFDB，由（1）得DBAF，所以AFE为二面角ADBC的平面角，···············

·······························6分DA底面ABC，,,DAACDAAB因为1ACDA，所以1222AEDC，···············································7

分因为2AB，所以1122DABSDAABDBAF△，所以126312DAABAFDB，················································

·········9分由（1）知，AE平面DBC，因为EF平面DBC，所以AEEF，················································································10分所以232sin

263AEAFEAF，·························································11分因为AFE为锐角，所以3AFE，所以二面角ADBC的大小为.3·······················

·································12分21.（12分）高三数学参考答案及评分细则（第17页共21页）定义：若点00(,)xy，00(,)xy在椭圆2222:1xy

Mab（0ab＞＞）上，且满足0000220xxyyab，则称这两点是关于M的一对共轭点，或称00(,)xy关于M的一个共轭点为00(,)xy．已知点A(3,1)在椭圆22:1124x

yM，O为坐标原点．（1）求点A关于M的所有共轭点的坐标；（2）设点P，Q在M上，且PQOA∥，求点A关于M的所有共轭点和点P，Q所围成封闭图形面积的最大值．【考查意图】本小题主要考查新定义、点与椭圆的位

置关系、平面向量共线、四边形的面积等基础知识；考查推理论证能力、运算求解能力；考查函数与方程思想、数形结合思想、化归与转化思想；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性与创新性．满分12分．【解】解法一、（1）设点A关于

M的共轭点的坐标为11(,)xy，依题意得，11221130,1241124xyxy，····································································2分解得113,3xy，或11

3,3xy；·····························································4分即A关于M的两个共轭点1B，2B的坐标分别为3,3，3,3．········5分（2）由（1）知，1B3,3，2B3,3

，所以12BB=22333326．··········································6分设点,ppPxy，,qqQxy．则22221,1241124ppqqxyxy，两式相减得()()()()0124pq

pqpqpqxxxxyyyy，········7分又PQ∥OA，所以13pqpqyyxx，····························································

8分故()pqpqyyxx，则22pqpqyyxx，高三数学参考答案及评分细则（第18页共21页）所以线段PQ的中点在直线0xy上，即线段PQ被12BB平分．···············

····9分设点,ppPxy到0xy的距离为d，则点1B，P，2B，Q所围成四边形面积121121=22262PBBBPBQSSBBdd△四边形，······················································

······················································10分设过P与直线0xy平行的直线l为xym，当l与M相切时，d取得最大值．由22,1124xymxy

，消去y得2246340xmxm，令22364840mm，解得4m，············································11分所以4222maxd，故点1B，P，2B，Q所围成四边形面积的最大值为83

．·························12分解法二、（1）略，同解法一.（2）由（1）知，1B3,3，2B3,3，设点,ppPxy，,qqQxy．由PQOA∥，得直线PQ的斜率为13，·····················

································6分故设直线PQ的方程为13yxt，联立221,31124yxtxy，消去y得，22469360xtxt，········

························7分由22236436(4)36(163)0ttt，解得443333t,且32pqxxt，21(936)4pqxxt，···

··················································8分所以2211()3pqPQxx221031()4936924tt210163

2t．······································································9分设1B3,3，2B3,3到直线PQ的距离分别为1d,2d．高三数学参考答案及评分细则（第19页共21页）则134310td

，243310td，由443333t，得1243343383101010ttdd．······················10分所以四边形1BP2BQ面积等于12121122BPBQSPQdPQd121()2PQdd2110831632210t

223163t···········································································11分故当0t时，12BPBQS取得最大值83，即点1B，P，2B，Q所围成封闭图形面积的

最大值为83．······················12分22.（12分）设函数2()lnfxaxxx．（1）当1a时，判断()fx的单调性；（2）若函数()fx的图象与x轴没有公共点，求a的取值范围．【考查意图】本小题主要考查函数的单调性、方程有解、

导数的应用等基础知识；考查抽象概括能力、推理论证能力、运算求解能力与创新意识，考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想、有限与无限思想；考查数学抽象、直观想象、逻辑推理、数学运算等核心素养，体现综合性、应用性与创新性．满分12分．【解】解法一、（1）当1a

时，2()ln0fxxxxx，·····················1分所以2212112121()21.xxxxxxfxxxxxx··············2

分当01x时，()0fx，所以()fx在(0,1)单调递增；·······························3分当1x时，()0fx，所以()fx在(1,)单调递减.··································

·4分（2）因为函数()fx的图象与x轴没有公共点，所以()0fx恒成立．若()0fx有解，则关于x的方程2ln0axxxx有解，等价于关于x的方程2ln0xxaxx有解.······

·······································6分令2ln()0xxpxxx，则312ln()xxpxx，·······································7分高三数学参考答案及评分细

则（第20页共21页）令()12ln0qxxxx，则2()10qxx恒成立，()qx在(0,)上单调递减，又(1)0,q当01x时，(1)0qxq；当1x时，(1)0qxq；·····

···················8分所以，当01x时，0px；当1x时，0px；()px在(0,1)单调递增，在(1,)上单调递减，max()(1)1pxp，·······························

··············································9分当0x且0x时，px；当1x时，0px；·························1

0分若直线ya与函数()px的图象有交点，则1a，即1a，········································································11分要使函数()fx的图象与x轴没有公共点，则a的取值

范围为,1．··········12分解法二、(1)略，同解法一；（2）因为2()ln0fxaxxxx，所以2121()21axxfxaxxx．···································

··················5分①当0a时，()0fx，所以()fx在0,单调递增，且11()10,(e)1e0,eeff所以函数()fx有唯一的一个零点，这与()fx的图象与x轴没有公共点相矛盾，所以0a；··························

····6分②当0a时，()0fx，所以()fx在(0,)单调递增；（ⅰ）当1a时，22111111()lnln1eeeeeefaaaaaaa21110ee≤，又因为(1)10,fa所以函数()fx有唯一的一个零点，这与()fx的

图象与x轴没有公共点相矛盾，所以1a不成立；······················7分（ⅱ）当01a时，22()lnlnfxaxxxxxx，令2()ln.hxxxx21111()()10eeeefh，(1)10fa，所

以函数()fx有唯一的一个零点，高三数学参考答案及评分细则（第21页共21页）这与()fx的图象与x轴没有公共点相矛盾，所以01a不成立；·················8分（ⅲ）当0a时，方程2210axx有两根1x，2x，不妨设12xx，因为12121

0,210,2xxaxxa所以120xx，所以，当20xx时，()0fx；当2xx时，()0fx；所以()fx在2(0,)x上单调递增，在2(,)x上单调递减，所以()fx在2xx处取到唯一

的极大值，即函数()fx的最大值为2()fx，且222210.axx·····················································································

···9分222222222211()lnlnln.222xxfxaxxxxxx设1()ln22xuxx，则()ux在(0,)单调递增，且(1)0,u当(0,1)x时，()0ux；当(1,)x时，()0ux；

即2(0,1)x时，2()0fx；当2(1,)x时，2()0fx；由于221111()ln10eeeeeeaaf，所以2(1,)x时，()fx的图象与x轴有公共点，这与()fx的图象与x轴没有公共点相矛盾，所以2(1,)x；·····

····································································10分故2(0,1)x时，()fx的最大值2()fx小于0，又2222210(01),axxx即2222112(0

1)axxx，因为2222211111()224xxx，····················································11分所以22a，即1.a要使函数(

)fx的图象与x轴没有公共点，则a的取值范围为,1．··········12分