福建省2023-2024学年高三上学期第一次质量检测数学试卷

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2023－2024学年第一学期福州市高中毕业班开门考数学试题（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2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．11.已知复数z满足1i，则在复平面内，z对应的点在zA．第一象限B．第二象限C．第三象限D．第四象限【考查意图】本小题以复数为载体，主要考查复数的基本运算、几何意义等基础知识；考查运算求解能力、推理论证能力；考查数学运算、逻辑推理等数学核心素养，体现基础性．【答案】A．111i【解析】由1i得z，应选A．z1i222.已知集合Axx＜1，Bxx＞0，则ABA．0，1B．0，C．1，D．，【考查意图】本小题以不等式为载体，主要考查集合运算等基础知识；考查运算求解能力、推理论证能力；考查数学运算、逻辑推理等核心素养，体现基础性．【答案】C．【解析】Ax1＜x＜1，Bxx＞0，故AB(1,)，应选C．23.已知点Px，2在抛物线C:y4x上，则P到C的准线的距离为0A．4B．3C．2D．11{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 【考查意图】本小题以抛物线为载体，主要考查抛物线的图象和性质、直线与抛物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．【答案】C．2【解析】抛物线y4x的准线为x1，由PC得x1，故P到准线的距离为2，0应选C．4.“二十四节气”是中国古代劳动人民伟大的智慧结晶，其划分如图所示．小明打算在网上搜集一些与二十四节气有关的古诗．他准备在春季的6个节气与夏季的6个节气中共选出3个节气，若春季的节气和夏季的节气各至少选出1个，则小明选取节气的不同情况的种数是A．90B．180C．270D．360【考查意图】本小题以二十四节气为载体，主要考查排列与组合等基础知识；考查运算求解能力、推理论证能力和应用意识；考查数学运算、逻辑推理等核心素养，体现基础性和应用性．【答案】B．【解析】根据题意可知，小明可以选取1春2夏或2春1夏．其中1春2夏的不同情况1221有：C6C690种；2春1夏的不同情况有：C6C690种，所以小明选取节气的不同情况有：9090180种．应选B．35.一个正四棱台形油槽可以装煤油190000cm，其上、下底面边长分别为60cm和40cm，则该油槽的深度为75A．cmB．25cmC．50cmD．75cm4【考查意图】本小题以正四棱台形油槽为载体，主要考查空间几何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，体现基础性和应用性．【答案】D．h22【解析】设正四棱台的高，即深度为hcm，依题意，得19000060406040，3解得h75，应选D．6.一个袋子中有大小和质地相同的4个球，其中有2个红球，2个黄球，每次从中随机摸出1个球，摸出的球不再放回．则第二次摸到黄球的条件下，第一次摸到红球的概率为2{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 1123A．B．C．D．3234【考查意图】本小题主要考查条件概率、全概率公式等基础知识；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；考查数学建模、逻辑推理、数据分析等核心素养，体现综合性、应用性与创新性．【答案】C．【解析】解法一：记第i次摸到红球为事件A，摸到黄球为事件B（i1，2），则ii12111PB2PA1PB2A1PB1PB2B1，23232221P(AB)2PABPAPBA1212121，故P(A1B2)．应选C．433P(B)32解法二：记第i次摸到红球为事件A，摸到黄球为事件B（i1，2）．由抽签的公平ii21221P(AB)2PB，又12性可知2PA1B2，所以P(A1B2)．应选C．42433P(B)32157.已知a，bln2，cln5，则eA．a＞b＞cB．b＞c＞aC．a＞c＞bD．c＞a＞b【考查意图】本小题以数的大小比较为载体，主要考查函数与导数等基础知识；考查运算求解能力、推理论证能力、应用意识；考查数学建模、数学运算、逻辑推理等核心素养，体现基础性、应用性和综合性．【答案】A．1lneln2ln45ln5lnx【解答】解法一：a，bln2，cln5，令fx，ee245x1lnxfx，当x≥e时，fx≤0，故fx在区间e，上单调递减，所以a＞b＞c．2x10510105解法二：因为2232＞255，所以5ln2＞ln5，即b＞c．x2在同一坐标系中作出函数fx2，gxx的图象，e2如图所示，由图可知，fe＜ge，即2＜e，所以e21111122e＜e2e，即22＜ee，所以ln2＜lne，即b＜a．2eelnx1lnx（令fx，fx，当0xe时，fx0，故fx在区间0，e上2xx1lneln2单调递增，所以aln2b．）ee23{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 综上，a＞b＞c．应选A．8.若定义在R上的函数fxsinxcosx（＞0）的图象在区间0，上恰有5条对称轴，则的取值范围为1721172517253341A．，B．，C．，D．，44444444【考查意图】本小题以三角函数为载体，考查三角函数的图象与性质、三角恒等变换等基础知识；考查抽象概括能力、推理论证能力、应用意识；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，体现基础性和综合性．【答案】A．π【解析】由已知，fx2sinx，4π(4k1)π令x=kπ，kZ，得x，kZ，424(4k1)π依题意知，有5个整数k满足0≤≤π，即0≤4k1≤4，所以k0，1，417212，3，4，则441≤4＜451，故≤＜，应选A．44二、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.某市抽查一周空气质量指数变化情况，得到一组数据：80，76，73，82，86，75，81．以下关于这组数据判断正确的有A．极差为13B．中位数为82C．平均数为79D．方差为124【考查意图】本小题主要考查极差、中位数、平均数、方差等基础知识；考查推理论证能力、运算求解能力；考查化归与转化思想；考查数据分析等核心素养，体现基础性．【答案】AC．2210.已知圆M:xy1，直线l:ykx31，则3A．l恒过定点3，1B．若l平分圆周M，则k3C．当k3时，l与圆M相切D．当3＜k＜3时，l与圆M相交【考查意图】本小题以直线与圆为载体，考查直线的方程、圆的方程、直线与圆的位置关系等基础知识；考查运算求解能力；考查直观想象、逻辑推理等核心素养；体现基础性和综合性．【答案】BC．【解析】依题意，l恒过定点3，1，选项A错误；4{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 3若l平分圆周M，则l经过圆M的圆心0，0，代入直线方程得k，选项B正确；33k1圆心O0，0到l的距离d，当k3时，d1r，l与圆M相切，选项C2k122正确；若l与圆M相交，则d＜1，即3k1＜k1，即0＜k＜3，故选项D错误．综上，应选BC．311.已知函数fxx3ax2有两个极值点．则A．fx的图象关于点0，2对称B．fx的极值之和为4C．aR，使得fx有三个零点D．当0＜a＜1时，f(x)只有一个零点【考查意图】本小题以三次函数为载体，主要考查函数与导数等基础知识；考查运算求解能力、推理论证能力、应用意识；考查数学建模、数学运算、逻辑推理等核心素养，体现基础性、应用性和综合性．【答案】ACD．3【解答】fx的图象可由奇函数gxx3ax的图象向上平移2个单位长度得到，故fx的图象关于点0，2对称，选项A正确．设fx的极值点分别为x，x（x＜x），则由对称性可知xx0，故121212fx1fx2224，即fx的极值之和为4，选项B错误．2依题意，方程fx3x3a0有两异根，则a＞0，xa，xa，fx在区间12，a上单调递增，在区间a，a上单调递减，在区间a，单调递增．由图象可知，当fx＞0＞fx时，fx的图象与x轴有3个交点，即fx有3个零点，选项12C正确．当0＜a＜1时，faaa3aa221aa＞0，此时fx只有一个零点，选项D正确．综上，应选ACD．12.已知正四棱柱ABCDABCD的底面边长为2，球O与正四棱柱的上、下底面及侧棱1111都相切．P为平面CDD上一点，且直线BP与球O相切，则1A．球O的表面积为4πB．直线BD与BP夹角等于451C．该正四棱柱的侧面积为162D．侧面ABBA与球面的交线长为2π11【考查意图】本小题以正四棱柱为载体，主要考查球、直线与平面的位置关系等基础知识；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、应用性和综合性．【答案】BCD．5{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 【解答】如图，设球O与下底面相切于点O，则OO平面ABCD，连接OA，则OAO1111为直线OA与平面ABCD所成的角．因为球O与正四棱柱的侧棱相切，所以其半径ROOOA2，所以S4π28π，四棱柱的侧面积为2422162，故选项11表A错误，C正确．依题意，BB，BP均为球O的切线，BD经过球心O，所以BBDPBD，又11111BD22BB，所以PBDBBD45，选项B正确．111111对于选项D，棱AA的中点F，即球O与棱AA的切点应为11交线上的点，故交线应为过F的圆．截面圆的圆心即为矩形1ABBA的中心E，在Rt△OEF中，OFR2，OEBC1，112所以截面圆半径rEF211，周长为2π，该选项正确．综上，应选BCD．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.已知向量a1，2，b1，2，若ab，则实数的值为．【考查意图】本小题以平面向量为载体，主要考查平面向量的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、直观想象等核心素养，体现基础性．【答案】5．【解析】由ab得1220，解得5．14.将圆周16等分，设每份圆弧所对的圆心角为，则sincos的值为．【考查意图】本小题以圆的等分为载体，考查三角恒等变换等基础知识；考查推理论证能力，抽象概括能力；考查逻辑推理等核心素养；体现基础性与应用性．2【答案】．411π2【解析】依题意，得，所以sincossin2sin．8224415.已知定义域为R的函数fx同时具有下列三个性质，则fx．（写出一个满足条件的函数即可）①fxyfxfy；②fx是偶函数；③当xy＞0时，fxfy＜0．6{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 【考查意图】本小题以函数的性质为载体，考查函数的奇偶性、函数与导数等基础知识；考查推理论证能力；考查逻辑推理等核心素养；体现基础性、综合性与应用性．【答案】x（答案不唯一，kxk＜0均可）．22xy16.已知双曲线C:1（a＞0，b＞0）的左焦点为F，两条渐近线分别为l，l．点2212abA在l上，点B在l上，且点A位于第一象限，原点O与B关于直线AF对称．若12AF2b，则C的离心率为．【考查意图】本小题以双曲线为载体，主要考查双曲线的离心率、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．【答案】2．b【解答】依题意，l的方程为yx，AFl，设垂足为P，则FPb．因为12aAF2b2FP，所以点F，A关于直线l对称，FOPAOP，又l，l关于y轴对称，21221bb所以l的倾斜角为18060，故tan603，所以离心率e12．123aa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）已知等比数列a的前n项和为S，且aS2．nnn1n（1）求a的通项公式；n（2）若bloga，求数列b的前n项和T．n22n1nn【命题意图】本小题主要考查等差数列、等比数列、递推数列及数列求和等基础知识，考查运算求解能力、逻辑推理能力和创新能力等，考查化归与转化思想、分类与整合思想、函数与方程思想、特殊与一般思想等，考查逻辑推理、数学运算等核心素养，体现基础性和综合性．满分10分．aS2，【解答】（1）解法一：由aS2得21········································1分n1naS2，32设等比数列a的公比为q，na1q12，所以················································································2分2a1qq12，a2，a2，11解得或（舍去）．··························································4分q2，q07{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} n所以a2．··························································································5分n2n1（2）blogalog22n1，························································7分n22n12故b1，bb2n12(n1)12（n≥2），1nn1所以b是首项为1，公差为2的等差数列，nnb1bnn12n12所以Tn．······················································10分n22解法二：（1）因为aS2，①n1n所以当n≥2时，aS2，②·······························································1分nn1①－②得a2a，················································································2分n1nan1所以等比数列a的公比q2．························································3分nan由①式得aa2，得a2，·································································4分211n所以a2．··························································································5分n（2）Tbbbn12nlogalogaloga212322n1log2a1a3a2n1········································································7分132n1log2212n1nlog2222n．·······················································································10分18.（本小题满分12分）π记△ABC的内角A，B，C所对的边分别为a，b，c，已知b2，B．6（1）若c2，求a；（2）求△ABC面积的最大值．【命题意图】本小题主要考查正弦定理、余弦定理及三角恒等变换等基础知识，考查逻辑推理能力、运算求解能力等，考查化归与转化思想、函数与方程思想、数形结合思想等，考查数学运算、逻辑推理等核心素养，体现基础性和综合性．满分12分．π【解答】解法一：（1）因为b2，c2，B，6222根据余弦定理得bac2accosB，2π22所以2a24acos，···································································3分68{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 2即a23a20，解得a31．······················································································6分222（2）根据余弦定理,得bac2accosB,22π22所以2ac2accosac3ac≥2ac3ac23ac，·················8分6（当且仅当ac31时取等号），···························································9分2即ac≤223，2311π1123所以△ABC面积S△ABCacsinBacsinac≤223，22644223即△ABC面积的最大值为．·························································12分2π解法二：（1）因为b2，c2且B,6bc根据正弦定理,得,sinBsinC222所以，即sinC，·····························································1分πsinC2sin6π5π因为cb，所以CB，所以C，66π3π所以C或C，··············································································2分44πππ123262当C时，sinAsinBCsin，46422224ab根据正弦定理,得,sinAsinB622bsinA4所以a31；······················································4分sinBπsin63ππ3π123262当C时，sinAsinBCsin，46422224ab根据正弦定理,得,sinAsinB622bsinA4所以a31；sinBπsin69{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 综上，a31．···················································································6分（2）略，同解法一．π解法三：（1）因为b2，c2且B,6bc根据正弦定理,得,sinBsinC222所以，即sinC，·····························································1分πsinC2sin6π5π因为cb，所以CB，所以C，66π3π所以C或C，··············································································2分44πππ7π当C时，AπBCπ，46412ab根据正弦定理,得,sinAsinB7π2sinbsinA12ππππππ所以a22sin22sincoscossinsinBπ343434sin6ππππ22sincoscossin31；······················································4分34343ππ3ππ当C时，AπBCπ，46412ab根据正弦定理,得,sinAsinBπ2sinbsinA12ππππππ所以a22sin22sincoscossinsinBπ343434sin6ππππ22sincoscossin31；3434综上，a31．···················································································6分acb2（2）根据正弦定理,得22,sinAsinCsinBπsin6所以a22sinA，c22sinC，10{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 25π13即ac22sinAsinC8sinAsinA8sinAcosAsinA62221cos2A2sin2A43sinA2sin2A432sin2A23cos2A23213π4sin2Acos2A234sin2A23，··································8分2235πππ4π因为0A，所以2A，6333ππ5ππ所以当2A，即A时，sin2A取得最大值为1，即ac最大值为423，3212311π1123所以△ABC面积SacsinBacsinac≤423，△ABC22644223即△ABC面积的最大值为．·························································12分219.（本小题满分12分）国际上常采用身体质量指数（BodyMassIndex，缩写BMI）来衡量人体肥瘦程度，其体重（单位：kg）计算公式是BMI．为了解某公司员工的身体肥瘦情况，研究人员从该公22身高（单位：m）司员工体检数据中，采用比例分配的分层随机抽样方法抽取了50名男员工、30名女员工的身高和体重数据．计算得到他们的BMI值，并根据“中国成人的BMI数值标准”简称“指标”整理得到如下结果：指标偏瘦正常偏胖肥胖人数（BMI＜18.5）（18.5≤BMI＜24）（24≤BMI＜28）（BMI≥28）性别男12171110女91173（1）若该公司男员工有1500名，则该公司共有多少名员工？（2）以频率估计概率，分别从该公司男、女员工中各随机抽取2名员工，求抽到的员工中至少有一名是肥胖的概率．【命题意图】本小题主要考查分层抽样、独立事件的概率、互斥事件、对立事件的概率等基础知识；考查数学建模能力，运算求解能力，逻辑推理能力，创新能力以及阅读能力等；考查统计与概率思想、分类与整合思想等；考查数学抽象，数学建模和数学运算等核心素养；体现应用性和创新性．满分12分．【解】（1）设该公司共有x名员工，150050依题意得，·········································································3分x503011{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 解得x2400，所以该公司共有2400名员工．··································································5分404（2）依题意，事件“抽到一名男员工不为肥胖”的概率为，事件“抽到一名女505279员工不为肥胖”的概率为，···································································7分30104416由事件的独立性，得抽到的两个男员工都不存在肥胖的概率为，·········8分55259981抽到的两个女员工都不存在肥胖的概率为，·································9分1010100设事件M为“抽到的员工中至少有一名是肥胖”，则事件M为“抽到的员工都不存在肥胖”，8116324所以PM，···································································10分10025625324301所以PM1，625625301所以抽到的员工中至少有一名是肥胖的概率为．·····································12分62520.（本小题满分12分）如图，在底面为菱形的四棱锥MABCD中，ADBDMB2，MAMD2．（1）求证：平面MAD平面ABCD；（2）已知MN2NB，求直线BN与平面ACN所成角的正弦值．【命题意图】本小题主要考查直线与直线、直线与平面、平面与平面的位置关系，直线与平面所成角等基础知识；考查空间想象能力，逻辑推理能力，运算求解能力等；考查化归与转化思想，数形结合思想，函数与方程思想等；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．满分12分．【解答】（1）取AD的中点为O，连结OM,OB，因为四边形ABCD是为菱形，且ADBD2，所以△ABD为正三角形，所以BOAD，且BO3．因为MAMD2，所以MOAD，·····················································2分12{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 2222所以MOMAAO211，222又因为MB2，所以MOBOMB，所以MOBO，···················································································4分因为ADBOO，AD平面ABCD，BO平面ABCD所以MO平面ABCD，········································································5分又因为MO平面MAD，所以平面MAD平面ABCD．································································6分（2）由（1）知，OA,OB,OM两两垂直，故以O为坐标原点，分别以OA,OB,OM为x，y，z轴的正方向建立如图所示的空间直角坐标系Oxyz．231则A1,0,0,B0,3,0,C2,3,0,M0,0,1,N0,,,·························7分3331所以CA3,3,0,CN2,,,CB2,0,0，33设平面ACN的法向量为nx,y,z，3x3y0，nCA0，则即31nCN0，2xyz0，33取x1，则n1,3,3．·····································································9分因为BM0,3,1,BMn33313则cosBM，n，··············································11分BMn21313313所以直线BN与平面ACN所成角的正弦值为．···································12分1321.（本小题满分12分）13{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 22xy已知椭圆E:1的右焦点为F，左、右顶点分别为A，B．点C在E上，43P4，yP，Q4，yQ分别为直线AC，BC上的点．（1）求yy的值；PQ（2）设直线BP与E的另一个交点为D，求证：直线CD经过F．【命题意图】本小题主要考查椭圆的标准方程及简单几何性质，直线与圆、椭圆的位置关系，平面向量等基础知识；考查运算求解能力，逻辑推理能力，直观想象能力和创新能力等；考查数形结合思想，函数与方程思想，化归与转化思想等；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．【解答】（1）依题意，A2，0，B2，0．···············································1分22xy11设Cx，y，则1，1143y6y11直线AC方程为yx2，令x4得y，·····························2分Px2x211y2y11直线BC方程为yx2，令x4得y，·····························3分Qx2x211212y1所以yyPQ2x412x112314········································································4分2x419，即yy的值为9．·············································································5分PQ（2）设Dx，y，P4，t，则22tt直线AP方程为yx2，直线BP的方程为yx2，62tyx2，2222由6得t27x4tx4t1080，···································6分3x24y212224t108542tt18t所以2x，即x，故yx2．················7分1212112t2727t627ttyx2，2222由2得t3x4tx4t120，3x24y212224t122t6t6t所以2x，即x，故yx2．·····················8分2222222t3t32t314{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 所以x1yx1y122122273t6tt918t222227tt3t327t226t273t3t270，···························································10分22t327t又F1，0，所以向量FCx1，y与FDx1，y共线，···················11分1122所以直线CD经过F．··········································································12分解法二：（1）依题意，A2，0，B2，0．···············································1分22xy11设Cx，y，则1，1143yy11所以kk·····································································2分ACBCx2x2112y12x412x1314·······································································3分2x413．············································································4分43yPyQ即kk，故yy的值为9．·····································5分APBQPQ44242（2）设Cx，y，Dx，y，P4，t．1122要证直线CD经过F1，0，只需证向量FCx1，y与FDx1，y共线，··································6分1122即证x1yx1y．（*）·······························································7分1221222202xyy3x2y1111P因为1，所以k，AC4343x24y611y3x2y22P同理可得k，···················································9分BDx24y222kACx22y11所以，即xy3xy6y2y0，①122112kBDx12y23同理可得3xyxy2y6y0，②·················································10分122112①－②得4xy4xy4y4y0，即x1yx1y．···················11分1221121221所以（*）式成立，命题得证．·······························································12分15{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} 22.（本小题满分12分）已知函数fxlnxa，记曲线yfx在点x，fx处的切线为l，l在x轴上的11截距为x2（x2＞0）．（1）当xe，a1时，求切线方程；1aa（2）证明：xe≥xe．12【命题意图】本小题主要考查导数，函数的单调性、零点、不等式等基础知识；考查逻辑推理能力，直观想象能力，运算求解能力和创新能力等；考查函数与方程思想，化归与转化思想，分类与整合思想等；考查逻辑推理，直观想象，数学运算等核心素养；体现基础性、综合性和创新性．满分12分．1【解答】（1）f(x)，··········································································1分x当xe，a1时，fxlne10，即切点为e，0，································2分111所以所求切线斜率kfe，································································3分e11所以所求的切线方程为y(xe)，即yx1．········································4分ee（2）由于fxlnxa，111所以切线l的方程为y(lnxa)(xx)．··············································5分11x11令y0，得(lnx1a)(xx1)，解得x2x1x1(lnx1a)．（*）·················6分x1a1由x20，得x1e．·············································································7分构造函数gxxx(lnxa)，所以gxalnx，aa所以当0＜x＜e时，gx＞0，gx单调递增；当x＞e时，gx＜0，gx单aa调递减．故gxgee．maxa所以x2e．···························································································8分a若x1e，由（*）式知x1x2，a所以x1x2e，aa故x1ex2e．··············································································10分aaaaaa若x1>e，则x1ex2e(x1e)(ex2)(x1x2)2e，16{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#} aaa所以x1ex2e2x1x1(lnx1a)2e．aaa1构造函数(x)2xx(lnxa)2e（exe)，所以(x)(1a)lnx0，aa1故(x)在区间(e，e）上单调递增，a所以(x)(e)0，a所以2x1x1(lnx1a)2e0，即aaaa所以x1ex2e0，即x1ex2e．aaa综上，不等式成立x1ex2e成立（当且仅当x1e时取等号）．············12分17{#{QQABKYaAggCgAAAAARgCEQUgCEGQkAGAAAgOwFAEoAIBCBNABAA=}#}