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高三数学试题答案第1页（共8页）吉林市普通中学2022—2023学年度高三毕业年级第四次调研测试数学试题参考答案一、单项选择题：本大题共8小题，每小题5分，共40分．12345678BACDAABB二、多项选择题：本大题共4小题，共20分．全部选对的得5分，部分选对的得2分，有选错的得0分

．教学提示：4．选项D，可以参考人教A版选择性必修三第82~81页，探究与发现《二项分布的性质》，研读如何利用分布列的表达式研究kP的增减变化及最大值，并掌握当Pn)1(为正整数及非正整数时k的取值情况．5.①针对函数x

xln)xf(，了解函数图象的变化情况（指导学生阅读人教A版选择性必修二第68页例4和第89页例4），体会极值点左侧图象“陡峭”，右侧图象“平缓”，如图所示：)3()3(2eff②利用极值点偏移，对于函数xxlnxf)(，当)()(

21xfxf且21xx时，221exx.设)()3(02xfef且320ex则2203eex)3()3()3()(3200feffxfx补充：程度较好的学生熟练掌握六种经典函数图象（

单调性、极值点、零点）：xxxexy,xlnxy,xey,xxlny,xey,xlnxy7．法一：由杨辉三角中观察得可得2010631.推广，得到3221242322nnCCCCC即322)1(243232221nCnn由

题意，2021层“刍童垛”小球的总个数为)1(22023202254433232024CS法二：指导学生阅读人教A版选择性必修二第42~43页，阅读与思考《中国古代数学家求数列和的方法》.可以结合人教A版选择性必修

三39~42页数学探究《杨辉三角的性质与应用》，探索堆垛问题以及高阶等差数列的求和方法.由已知顶层6个圆球，设为321a，向下每边依次加1个，第n层23)2()1(2nnnnan个.利用分组求和法，得到n层“刍童垛”小球的总个数为91011

12ACADADABC高三数学试题答案第2页（共8页）2223)3)(2)(1(23]65)[1(222)1(236)12)(1(2)21(3)21(23)2232()2131(232222222

nnCnnnnnnnnnnnnnnnnnS即2021层“刍童垛”小球的总个数为2232024CS.11．统计数据来源于国家统计局网站，选项的命制主要根据网站公布数据的解释说明．为使学生深刻体会数学的实用性，结合本题的实

际特点，选择恰当的统计图表对数据进行可视化描述，体会合理使用统计图表的重要性．三、填空题：本大题共4小题，每小题5分，共20分．其中第16题的第一个空填对得2分，第二个空填对得3分．13．314．115．2116．6；]

33643332[,（注：可以用不等关系表示）教学提示：16．过P作PH平面ABC垂足为H,则3060PBH,PAH333BHPHPBHtan,AHPHPAHtanAHBH3H点轨迹是半径

为3的圆（阿波罗尼斯圆），即轨迹长为6易知]3432[3]42[,AHPH,AH，33643332316)8421(31,PHPHV四、解答题17．【解析】（Ⅰ）由

bacAsin26）（，得：acAcosbAsinb3由正弦定理，得AsinBAsinAsinCsinBsinAcosBsinAsin）（3整理得：AsinBcosAsinBsinAsin3因为0Asin，所以13BcosBsin········

···························································3分两边平方得，消Bsin，得0122BcosBcos解得:21Bcos或1Bcos（舍去）又)0(,B，所以3B·······

·················································································5分（也可化简为21)6(Bsin求得B）（Ⅱ）法一：高三数学试题答案第3页（共8页）因为212222

acbcaBcos且332Bsinb所以：acca922··························································································

···7分所以9439322）（）（caacca所以6ca，当且仅当3ca时，取“”号所以△ABC的周长9cba，即当3ca时，△ABC的周长最大值为9···················

···········································10分法二：设△ABC的外接圆半径为R，因为RCsincBsinbAsina2，所以AsinAsinRa322，32BsinR

b，AsinAcosAsinABsinCsinCsinRc3333232322）（）（······7分所以△ABC的周长AsinAcosAsincba333323333AsinAcos366）（Asin因为

）（320π,A，所以）（6566π,πA当3A时，）（6Asin的最大值1，此时△ABC的周长的最大值为9························10分（注：没有写清取等条件扣1分）18．【解析】（Ⅰ）对于有放回抽检，每次抽到混合动力汽车的概率为41，且各次抽检结果是

独立的，设1X为有放回抽检的混合动力汽车的台数，则)41,2(~1BX，1X可取2,1,0，169)43()0(21XP；834143)1(121CXP；161)41()2(21XP.1X的分布列如下：1X012P16983161则211

6128311690)(1XE········································································4分对于不放回抽检，各次抽检的结果不独立，设2X为不放回抽检

的混合动力汽车的台数，则2X服从超几何分布，2X可取2,1,0，476267)0(21202902CCXP；476180)1(21201901302CCCXP；47629)1(21202302CCXP.2X的分布列如下：2X012P4762674761804762

9则2147629247618014762670)(2XE.·······························································8分注：也可按照下面步骤作答.高三数学试题答案第4页（共8页）1X的分布列为2,1,0,)41(

)43()(221kCkXPkkk，21412)(1XE.2X的分布列为2,1,0,)(2120290302kCCCkXPkk，21120302)(2XE.（Ⅱ）样本中混合动力汽车的比例1010Yf是一个随机变量，根据参考数据，有放回抽取：86556

0146000250280281570187710)41()150250(10.....YP.|.f|P不放回抽取：881400147010261340290510182540)41()150250(10.....YP.|.f|P··········

···········································································································10分因为88

140.086556.0，所以，在相同的误差限制下，采用不放回抽取估计的结果更可靠.······························12分（注：（Ⅱ）问，可以参考人教A版选择性必修三第79页例6，分别就放

回抽样和不放回抽样，用样本中的某类品的比例估计总体中这类品的比例，定量地比较估计效果，用概率的方法解释直观常识。对用同一抽取模型，两个分布的均值相同，从两种分布的概率分布看，或者从方差的大小比较（超几何分布的方差较小），都反应超几何分布更集中于均值附近．）

19.解析：（Ⅰ）CEAB||四边形ABCE是平行四边形BC//AE，AE平面PAE，BC平面PAE//BC平面PAE···································································

·······························2分BC平面PBC，平面PAE平面lPBC．l//BC·········································4分（Ⅱ）法一：1|E

B||EC||EP||EA|，323EA,EC,EA,EP设EC,EP，)3(,点T到直线EB的距离22)(EBETETd····················

·······································7分22)](2[)2(ECEAECEPECEP2]1)21(112111[)211(21ECEPECEP2)1()22(21ECEPE

CEP2)(121ECEP2121cossin21·····················································10分)3(,]10(,sin]210(,d即点T到直线EB的距离的取值范围是]

210(,······························································12分法二：取AE中点O，连接OB，OPABE是等边三角形OEOB以O为原点，OB,OE所在直线为

x轴，y轴，过O作平面ABCE的垂线为z轴，建立空间直角坐标系xyzO，如图所示设)0(POB，则)23230(sin,cos,P，)0231(,,C，)0021(,,E，)0230(,,B高三数学试题答案第5页（共8页）)43)1(

4321(sin,cos,T)43)1(43(0sin,cos,ET，)02321(,,EB)1(83)1(2343coscosEBET·············································

················································································6分点T到直线EB的距离22)(EBETETd··········································

·················7分222)]1(83[)43()]1(43[cossincos22)1(4143sincos523832coscos···································

···········10分11cos当31cos时，21maxd；当1cos时，0d即点T到直线EB的距离的取值范围是]210(,·························

·····································12分法三：提示：也可设)430(2t,t,P整理得16154343212ttd当63t时，21maxd；当23t时，0d（命题意图及教学建议：关注投影向量概念及意义的教学，体

会课标中“能用向量方法解决点到直线、点到平面、相互平行直线、相互平行的平面的距离问题和简单的夹角问题”的这一要求．）20．【解析】（Ⅰ）令1n时2221212aaaS01a22a································

····································································1分221nnSS2n时221nnSS，相减得nnaa21·····················

·················3分,n,n,,ann21201·································································································4

分（Ⅱ）由（Ⅰ）知2212)(nalogbnn······································································5分令nnnnnb

ac221，则1223222232221232122)1(222122232221nnnnnnnnTnccccT相减得：1232122)12(252321

nnnnnT············································7分令1323212)12(2)32(232122)12(252321

nnnnnnnGnG相减得：xOylz高三数学试题答案第6页（共8页）62)32(2)12(21)21(222)12()222(222)12(2222222111131321321

nnnnnnnnnnnnG62)32(1nnnG········································································

················10分121262)32(nnnnnT62)32(12nnnnT·······················································

···························12分21．【解析】（Ⅰ）11222byx,a故当直线l过)02(,与双曲线C有且仅有一个公共点时，l应与C的渐近线平行设直线)2(xbyl：，即02by

bx，则点B到直线l的距离为2212bb，1b即双曲线C的标准方程为：122yx·······································································4分（或设2myxl：，由点B到直线l的

距离为22112md，得1m1b）（Ⅱ）（ⅰ）由题可知，直线l斜率不为0设直线)()(22211y,xB,y,xA,myxl：由2122myxyx得：034)1(22myym)01(2m01242

m成立1314221221myy,mmyy)(432121yyymy···························································································6分11222111

xyk,xyk121221211221121122123)1()3()1()1(11yymyyymymyymyyxyxyxyxykk343414943)(433)(432121121221yyyyyyyyyy所以存在实数3

λ，使得12kk成立.···································································9分（注：请老师阅卷时注意非对称问题方法不唯一）（ⅱ）直线)1(1xkyAP：，直线

)1(:2xkyBQ联立得：2131112xkkxx所以直线AP和BQ交点E的轨迹方程为：21x························································12分22．【解析】（Ⅰ）

法一：0)()(xmmxlnxxf在)0(，上恒成立高三数学试题答案第7页（共8页）0xmmxln在)0(，上恒成立设221)()(xmxxmxxg,xmmxlnxg①当0m时，0)(xg恒成立)(xg在)0(,上单调递增，且0)1(g

)10(,x时，0)(xg不符合题意，舍去②当0m时，令0)(xg，则mx；令0)(xg，则mx0.)(xg在)0(m,上单调递减，在)(,m上单调递增01)()(mmlnmgxgmin························

··················································2分设1)(xxlnxh，xxxh1)(令0)(xh，则10x；令0)(xh，则1x.)(xh在)10(,上单调递增，在)1(,上单调递减0)1()(

hxhmax，即当0)(mh时，1mm的取值范围是：1m························································································4分法二：0)(0)1(xff，在)0(，上

恒成立)1(f是)(xf上最小值，也是极小值01)1(1)(mfmxlnxf，即1m···················································2分当1m时，xl

nxfxxlnxxf)(1)(，令0)(xf，则1x；令0)(xf，则10x)(xf在)10(，上单调递减，在)1(，上单调递增即0)1()(fxfmin，满足：0)(xf在)0(，上恒成立1

m·················································································································4分法三：①当1x时，00)(xf恒成立

，Rm···················································1分②当1x时，1xxlnxm恒成立，设)(xu1xxlnx，)(xu2)1(1xxlnx设)(xvxlnx1，)(xv011x)(x

v在)1(,上单调递增，0)1()(vxv，0)(xu，)(xu在)1(,上单调递增当1x时，0xlnx，01x，)(xu1xxlnx为“00”型由洛必达法则得)(1xulimx1xli

m1xxlnx1xlim111xln当1x时，1)(xu，即1m····························································

·················2分③当10x时，1xxlnxm恒成立，设)(xu1xxlnx，)(xu2)1(1xxlnx设)(xvxlnx1，)(xv011x)(xv在)10(,上单调递减，0)1()(vxv

，0)(xu，)(xu在)10(,上单调递增当1x时，0xlnx，01x，)(xu1xxlnx为“00”型由洛必达法则得)(1xulimx1xlim1xxlnx1xlim111xln高三数学试题答案第8页（共8

页）当10x时，1)(xu，即1m综上，m的取值范围是：1m·················································································4分（

Ⅱ）由（Ⅰ）知，0)(xh，即1xxln在)0(，上恒成立（当且仅当1x时取等）令nx311，则nnln31)311(·········································

·····································6分21)311(21311)311(31313131)311()311()311(2121nnn

nlnlnln即eken，)311()311)(311(21又13111且Zk，k的最小值为2·········································

·······················8分（注：重新构造新函数得出nnln31)311(也可以）（Ⅲ）不等式1xxln在)0(，上恒成立（当且仅当1x时取等）令202311x，则20231)202311(ln，即e1)20242023(2023··········

···························10分令202411x，则20241)202411(ln，即e1)20242023(2024故20232024)20242023(1)20242023(e······

···········································································12分获得更多资源请扫码加入享学资源网微信公众号www.

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