福建省福州市2021-2022学年高三上学期期末质量抽测数学试卷官方答案解析（高考一检）

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．*21.已知集合A2,1，BxNxx20≤，则ABA．B．2，11，C．2，112，，D．2，1012，，，【答案】C*【解析】BxN1≤x≤212，，所以AB2，112，，，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z34i，则zz+iA．1+3iB．84iC．9+3iD．29+3i【答案】C【解析】因为z34i，所以zz+i5+34ii=9+3i，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1页共21页） 614.3x展开式中的常数项为xA．540B．15C．15D．135【答案】D61rr6r【解析】二项式3x的展开式的第r1项为Tr16C3)x(xx633rr6r62r421展开(1)C36x．令60r，解得r4，所以T5C63135，所以3x2x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．3xx10，＞，a5.已知函数fx()为偶函数，则2+b3axb，x＜0313A．3B．C．D．222【答案】B33【解析】解法一、当x0时，x0，所以f(x)(x)1x+1，因为fx()为偶33a3函数，所以fx()f(x)x+1，又fx()axb，所以ab1,1，所以2+b．2ff(1)1,ab2,a1，解法二、因为fx()为偶函数，所以所以解得经ff(2)2,8ab9,b1，a1，3a检验，符合题意，所以2+b．b12【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弧度，则该纸片扫4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，所形成的几何体为柱体，412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S22rhr8212222222282，故选C．8高三数学参考答案及评分细则（第2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fxsinx的部分图象如图所示，则fx的单调递增区间为1515A．k,,kkZB．2k,2k,kZ66661515C．k,,kkZD．2k,2k,kZ6666【答案】D2mmZ,,3【解析】解法一、由解得42,mmZ2,mmZ33fxsinx2msinx，3315令kxkkZ，解得kxkkZ．故选D．2326614T41335解法二、由图象知1,T2,又x时，fx()取得最大值，23326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．22xy8.已知O为坐标原点，F是双曲线C:1a＞0，b＞0的左焦点，A为C的右22ab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34A．2B．C．3D．23【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，FMb，OFcOM，a，由△FMN∽△FPO，△AQO∽△AMN得高三数学参考答案及评分细则（第3页共21页） 2acaFNMNAOQOFNAO1c1,,以上两式相乘得，所以，2FOPOANMNFOAN2a2cac22aca1ca111所以，即，所以1，解得e2．故选A．acca2c2e2yPMQFNOAx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byx，baaca线方程为yx，则直线FP的方程为yxc，令x0，得yP．由abbyaxcbab2baabcb可得M，，则k，所以直线AM的方程为yxa，AM2ccaacacacab2abac2222令x0得yQ．因为Q为OP中点，所以，整理得acc22bca，acacb222即cac20a，所以ee20，解得e2，或e1（舍去）．故选A．yPMQFNOAx解法三：如图所示，设AM交y轴于Q．由双曲线性质可知，FMb，OFc，OMaOA，所以OMQOAQ．在Rt△OPM中，Q为OP中点，所以MQOQPQ，所以QOMOMQOAQ，所以Rt△OMP≌Rt△AOQ，所以MPOQ，所以△MPQ为正三角形，所以MPQ60，故MFO30，2ab所以ca2,所以e2（也可以利用.ktan30，即e12），故选A．FPba高三数学参考答案及评分细则（第4页共21页） yPMQFOAx【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mn=31，，mn=1，1，则A．mn∥nB．mnnC．mn2D．mn，45【答案】BCD．11【解析】依题意，m=mnmn20，，n=[mnmn]11，，22所以mnn1，111，0，所以mnn，所以选项A错误，B正确．mn22所以mn22，选项C正确；cosmn，，mn222因为0≤mn，≤180，所以mn，45，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11A.当n1时，pB.当n2时，p，6334C.当n3时，pD.当n4时，p105【答案】AC511424【解析】当n1时，p，选项A正确；当n2时，p，选项B6566515333244错误；当n3时，p，选项C正确；当n4时，p，选项D65106515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；高三数学参考答案及评分细则（第5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AB30，，30，，动点C满足CA2CB，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PMQ45时，PQ42D．直线AC的斜率的取值范围为33，【答案】AC．2222【解析】设Cxy，，由CA2CB得xy323xy，整理得的方22程为x5y16，其图象是以D50，为圆心，半径r4的圆．故11S△PABmaxABr6412，选项B错误．因为PA2PB，MA2MB，所以22PAPB，所以PABMAB，又轨迹的图象关于x轴对称，所以QM，关于x轴对MAMB称，选项A正确．当PMQ45时，PDQ45290，则△DPQ为等腰直角三角形，PQ2r42，选项C正确．当直线AC与圆D相切时，CDAC，此时AD82r2CD，切线AC的倾斜角为30和150，结合图象，可得直线AC的斜率33的取值范围为，．选项D错误．故选AC．33y4Q32P1ABD–3–2–1O123456789x–1–2–3–4M【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．xx112.设函数fx()xe1xe，则A．fx()f(1x)高三数学参考答案及评分细则（第6页共21页） B．函数fx()有极大值为eC．若xx1，则xfx()xfx≥e1211221D．若xx＜1，且x＞，则fx()＜f1x122212【答案】ACD1【解析】易验证A是正确的，也即函数fx()关于直线x对称．故选项A正确；2xx11xxxx11因为fx()xee2exexx+1e2e，所以f()0，21x11x3xx当x时，fx()x+1ex2e(ee)，此时xx1，所以fx()0，221故函数fx()在(,)上单调递增；211由于函数fx()关于直线x对称，所以函数fx()在(,)上单调递减．2211所以函数fx()在x处有极小值，也是最小值，f()e．故选项B错误；22111若xx1，且x，则xx1，由fx()在(,)上单调递增得12221222fx()21f1x．故选项D正确；由于函数fx()的最小值为e，所以fx()e，1若xx1，则xx1，所以fxf1x，122121又因为fxf1x，所以fxfx，1112故xfx()xfxxfx()xfx()xxfx()fx()e，故选项C正确．112211211211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．313.曲线fx()x2x在x0处的切线方程是．【答案】20xy高三数学参考答案及评分细则（第7页共21页） 【解析】依题意得，f(0)0，fx32x2，所以f02，所以所求的切线方程为yx2，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ABCABC中，ABAA2，F是线段AB上的动点，则AFFC的1111111最小值为．【答案】62【解析】将正三棱柱ABCABC上底面沿AB展开至平面ABBA上，如图所示，因1111111AAC11为AAAC2，且AAC90+60=150，所以AC2AAsin22sin751111111262，所以AFFC的最小值为62．1【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．215.抛物线Ey:2pxp(0)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PAF30，则sinPFA．3【答案】3y【解析】过P作准线的垂线，垂足为B，所以BPAPAF30，PBPF．BPPB在Rt△BPA中，cos30，AOFxPAPF即在△PAF中，cos30，PA高三数学参考答案及评分细则（第8页共21页） PAPF又由正弦定理，sinPFAsinPAFPAsin303所以sinPFAsinPAF．PFcos303【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0.90，lg991．已知数列a满足a3，且anaa，若balg，则数列b的前2022项n3nn1nnnn和为．【答案】4959aaann13【解析】利用累乘法(或1),得an．记b的前n项和为T，nnnnn13当19n时，0lga1时，balg0；nnn当10n99时，1lga2时，b1；nn当100n999时，2lga3时，b2；nn当1000n2022时，3lga4时，b3；nn所以Tlgalgalga9019002102334959．2022122022【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a是首项为1的等差数列，若a是a，a的等比中项，且aa.n21523（1）求a的通项公式；n1（2）设bn，求数列bn的前n项的和Sn．aann1【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a的公差为d，n因为a1，且a是a，a的等比中项，12152所以aaa，···················································································1分215高三数学参考答案及评分细则（第9页共21页） 2所以adaa4d，··································································2分111又因为a1，12所以1dd14解得d2，或d0，··········································································3分又因为aa，23所以d0，所以d2，·······································································4分所以aan1d12n12n1．···············································5分n1（2）由（1）知，a21n，n1因为b，naann11111所以b()，············································7分n(2n1)(2n1)22n12n111111111所以S(1)()()()···························8分n2335572nn12111=(1)．·········································································9分22n1n.·················································································10分21n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123，，，，19）建立模型①：高三数学参考答案及评分细则（第10页共21页） 0.058t2yˆ166.9e，相关指数R0.87；模型②：ytˆ152.416.3，相关系数r0.97，相关122指数R0.95．根据2014年至2019年的数据（时间变量t的值依次为123，，，，6）建22立模型③：ytˆ372.89.8，相关系数r0.99，相关指数R0.99．33（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yˆ152.416.322511（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yˆ372.89.89461（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2（ⅱ）从计算结果可以看出，模型③的相关指数R0.99最高，说明其拟合效果最3好．模型③的相关系数r0.99比模型②的相关系数r0.97高，说明模型③的两变量的32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cacosBccosA．（1）试判断△ABC的形状，并说明理由；a（2）设点D在边AC上，若ADBD，sinADBsinABC，求的值.b【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△ABC为等腰三角形.·················1分理由如下：在△ABC中，因为cacosBccosA，根据正弦定理得，sinCsincosABsincosCA，·····························································2分高三数学参考答案及评分细则（第11页共21页） 又因为Cπ()AB，所以sin(AB)sincosABsincosCA，即sincosABcossinABsincosABsincosCA，即cossinABsinCcosA，···································································3分所以cosA0或sinBCsin，······························································4分π若cosA0，则A，故此时△ABC为直角三角形.·······························５分2若sinBCsin，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π（2）由（1）知，A或bc，2π若A，则ADBD，这与已知条件ADBD相矛盾，2π所以A；·······················································································7分2所以bc，A所以ABCC.又因为sinADBsinABC，D所以sin(πADB)sinC，即sinBDCsinC，BC故BDCC，················································································8分所以ADBC.DCBC在△DBC中，，sinDBCsinBDCACBC在△ABC中，，···························································9分sinABCsinA2两式相乘得ACDCBC，································································10分2(也可通过等腰△ABC△BDC得到ACDCBC.)又CDACADACBDACBC，2所以bba()a，·············································································11分a51a51解得或（舍去）．b2b2a51所以的值为．········································································12分b2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cacosBccosA，高三数学参考答案及评分细则（第12页共21页） 222222acbbca根据余弦定理，得cac，·······························２分22acbc222222acbbca，22cb222222acbbca所以c，························································３分22cb22222222cacbbca即，22cb222222所以bc(ba)cb(ca)，······················································４分222所以bcbc()(bcb)(bcc)abc()0，222所以(bcb)(ca)0，································································５分222即abc或bc，所以△ABC为直角三角形或△ABC为等腰三角形.····································6分π（2）由（1）知，A或bc，2ππ若A，则ADBD，这与已知条件ADBD相矛盾，所以A；··········7分22所以bc，A所以ABCC.又因为sinADBsinABC，D所以sin(πADB)sinC，即sinBDCsinC，BC故BDCC，················································································8分所以ADBC.又因为ADBD，所以ADBA，即BD平分ABC，··············································································9分BADA所以，···············································································10分BCDCca所以，ababa所以，················································································11分abaa51a51解得或（舍去）．b2b2a51所以的值为．········································································12分b2高三数学参考答案及评分细则（第13页共21页） 20.（12分）如图，在三棱锥DABC中，DA底面ABC，DACBCDA1，AB2，E是CD的中点，点F在DB上，且EFDB．（1）证明:DB平面AEF；F（2）求二面角ADBC的大小．E【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A推理、数学运算等核心素养，体现基础性、综合性．满分12分．CB【解】解法一、（1）DA平面ABC，且BC平面ABC，DABC，ACBC12，AB，222ACBCAB，ACBC，·························································1分DAACA，BC平面DAC，······················································2分AE平面DAC，BCAE．又DAAC，E是CD的中点，DCAE，······················································································3分又BCDCC，AE平面DBC．DB平面DBC，DBAE，··························································4分EFDBEF,AEE，DB平面AEF．·············································································5分（2）过点A作AG∥BC，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Axyz，·············································································6分则A0,0,0，C1,0,0，B1,1,0，D0,0,1，则AD0,0,1,BD1,1,1，CD1,0,1，设平面ADB的法向量mxyz,,，111mAD0,则mBD0,z0,1所以xyz0,111令y11，高三数学参考答案及评分细则（第14页共21页） 则m1,1,0.·····················································································8分设平面DBC的法向量为nxyz,,，222nCD0,xz0,22则所以nBD0,x2y2z20,令x1，则n1,0,1．·····································································10分211所以cosmn,，··························································11分222又因为二面角ADBC的平面角为锐角，所以二面角ADBC的大小为．·····················································12分3解法二、（1）DA平面ABC，且BC平面ABC，DABC，ACBC12，AB，222ACBCAB，ACBC，·························································1分DAACA，BC平面DAC，·············································································2分过点A作AG∥BC，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Axyz，11则A0,0,0,B1,1,0,D0,0,1，E0,,，2211所以DB1,1,1，AE0,,，2211DBAE101(1)0，22········································································································4分DBAE，DBAE，DBEF，且AEEFF，DB平面AEF.················································································5分（2）因为EFDB，由（1）得DBAF，所以AFE为二面角ADBC的平面角，··············································6分因为C1,0,0，由（1）知，DB1,1,1.设Fxyz,,，则DFxy,,z1.·····························································7分高三数学参考答案及评分细则（第15页共21页） 因为点F在DB上，所以存在实数k，使得DFkDB，所以xky,kz,1k，即Fkk,,1k，因为AFDB，所以AFDB0，··························································8分1所以kk1k10，解得k，················································9分3112111112所以点F,,，所以FE,,，FA,,，···············10分3333663331FAFE61所以cosAFE，·············································11分FAFE66263又因为AFE0,，所以AFE.3所以二面角ADBC的大小为.························································12分3解法三、（1）略，同解法一；（2）因为EFDB，由（1）得DBAF，所以AFE为二面角ADBC的平面角，··············································6分DA底面ABC，DAACDA,,AB12因为ACDA1，所以AEDC，···············································7分2211因为AB2，所以SDAABDBAF，△DAB22DAAB126所以AF，·························································9分DB123由（1）知，AE平面DBC，因为EF平面DBC，所以AEEF，················································································10分2AE23所以sinAFE，·························································11分AF623因为AFE为锐角，所以AFE，3所以二面角ADBC的大小为.························································12分321.（12分）高三数学参考答案及评分细则（第16页共21页） 22xy定义：若点(,xy)，(xy,)在椭圆M:1（ab＞＞0）上，且满足000022abxxyy00000，则称这两点是关于M的一对共轭点，或称(,xy)关于M的一个共轭点2200ab为(xy,)．0022xy已知点A(3,1)在椭圆M:1，O为坐标原点．124（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PQ∥OA，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(,)xy，113xy110,124依题意得，，····································································2分22xy111124x3,x3,11解得，或；·····························································4分y3y311即A关于M的两个共轭点B，B的坐标分别为3,3，3,3．········5分12（2）由（1）知，B3,3，B3,3，1222所以BB=333326．··········································6分12设点Pxy,，Qxy,．ppqq22xypp1,124(xpxq)(xpxq)(ypyq)(ypyq)则，两式相减得0，········7分22xyqq1241124yypq1又PQ∥OA，所以，····························································8分xx3pqyyxxpqpq故yy()xx，则，pqpq22高三数学参考答案及评分细则（第17页共21页） 所以线段PQ的中点在直线xy0上，即线段PQ被BB平分．···················9分12设点Pxy,到xy0的距离为d，pp1则点B，P，B，Q所围成四边形面积S=2S2BBd26d，12四边形BPBQ1△PBB12212············································································································10分设过P与直线xy0平行的直线l为xym，当l与M相切时，d取得最大值．xym,22由xy22，消去y得4x6mx3m40，112422令36mm4840，解得m4，············································11分4所以d22，max2故点B，P，B，Q所围成四边形面积的最大值为83．·························12分12解法二、（1）略，同解法一.（2）由（1）知，B3,3，B3,3，12设点Pxy,，Qxy,．ppqq1由PQ∥OA，得直线PQ的斜率为，·····················································6分31故设直线PQ的方程为yxt，31yxt,322联立，消去y得，4x6tx9t360，································7分22xy112444222由36t436(t4)36(163)t0，解得33t,33312且xxt，xx(9t36)，·····················································8分pqpq24122103221所以PQ1()xx(tt)4936pq3924102163t．······································································9分2设B3,3，B3,3到直线PQ的距离分别为d,d．1212高三数学参考答案及评分细则（第18页共21页） 3t43433t则d，d，12101044433tt43383由33t，得dd12．······················10分33101010所以四边形BPBQ面积等于1211SPQdPQdBPBQ1212221PQd()d122110283163t2210223163t···········································································11分故当t0时，S取得最大值83，BPBQ12即点B，P，B，Q所围成封闭图形面积的最大值为83．······················12分1222.（12分）2设函数fx()axlnxx．（1）当a1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．2【解】解法一、（1）当a1时，fx()xlnxxx0，·····················1分12x22x12xx1xx121所以fx()2x1.··············2分xxxx当01x时，fx()0，所以fx()在(0,1)单调递增；·······························3分当x1时，fx()0，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fx()0恒成立．2若fx()0有解，则关于x的方程axlnxxx0有解，lnxx等价于关于x的方程ax0有解.·············································6分2xlnxx1xx2ln令px()x0，则px()，·······································7分23xx高三数学参考答案及评分细则（第19页共21页） 2令qx()1x2lnxx0，则qx()10恒成立，xqx()在(0,)上单调递减，又q(1)0,当01x时，qxq(1)0；当x1时，qxq(1)0；························8分所以，当01x时，px0；当x1时，px0；px()在(0,1)单调递增，在(1,)上单调递减，px()p(1)1，·············································································9分max当x0且x0时，px；当x1时，px0；·························10分若直线ya与函数px()的图象有交点，则a1，即a1，········································································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（2）因为fx()axlnxxx0，212axx1所以fx()2ax1．·····················································5分xx①当a0时，fx()0，11所以fx()在0,单调递增，且ff()10,(e)1e0,ee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a0；······························6分②当a0时，fx()0，所以fx()在(0,)单调递增；111111（ⅰ）当a1时，f()alnlna122aeaeeaeaeae11≤10，e2e又因为fa(1)10,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a1不成立；······················7分22（ⅱ）当01a时，fx()axlnxxxlnxx，2令hx()xlnxx.1111fh()()10，fa(1)10，eee2e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20页共21页） 这与fx()的图象与x轴没有公共点相矛盾，所以01a不成立；·················8分2（ⅲ）当a0时，方程2axx10有两根x1，x2，不妨设xx12，1xx120,2a因为所以xx120，1xx120,2a所以，当0xx2时，fx()0；当xx2时，fx()0；所以fx()在(0,x2)上单调递增，在(,x2)上单调递减，所以fx()在xx2处取到唯一的极大值，即函数fx()的最大值为fx()2，且22ax22x10.························································································9分2xx2211fx()2ax2lnx2x2lnx2x2lnx2.222x1设ux()lnx，则ux()在(0,)单调递增，且u(1)0,22当x(0,1)时，ux()0；当x(1,)时，ux()0；即x2(0,1)时，fx()02；当x2(1,)时，fx()02；1aa111由于f()ln10，eee22eee所以x2(1,)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x2(1,)；·········································································10分故x2(0,1)时，fx()的最大值fx()2小于0，211又2ax2x210(0x21),即2ax2(021)，x2x2111121因为()2，····················································11分xxx224222所以22a，即a1.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分高三数学参考答案及评分细则（第21页共21页）