福建省泉州市三检2022届高三数学试卷及答案(解析版) PDF版含解析

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高三数学试题第1页(共12页)泉州市2022届高中毕业班质量监测(三)2022.03高三数学参考答案(选择题)本试卷共22题,满分150分,共6页。考试用时120分钟。一、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.若集合

1Axx,2Bxx,则ABRðA.B.RC.2,1D.2,1【命题意图】本小题主要考查集合的交集、补集运算等基础知识;考查运算求解能力;体现基础性,导向对发展数学运算核心素养

的关注.【试题解析】解法一:因为2Bxx,所以[2,+)BðR,所以[2,1)ABðR,故选D.解法二:由2A且2B,得2ABRð,排除A,C;又由1A,得1ABRð,排

除B,故选D.【试题评析】解题关键在于读懂集合相关符号语言的含义.2.已知向量3,1a=,1,3b=,且abab,则的值为A.2B.1C.1D.2【命题意图】本小题主要考查向量的坐标表示、向量的运

算及其几何意义等基础知识;考查运算求解等能力;考查数形结合思想;体现基础性与综合性,导向对发展数学运算、直观想象、逻辑推理等核心素养的关注.【试题解析】解法一(坐标法):(4,4)ab,(3,13)

ab,由0abab得16160,即1,故选C.解法二:利用22||aa,由0abab得:22(1)0abab,所以16160,即1,故选C.保密★使用前高三数学试题第2页(共12页)解

法三(验证法):对A:22223480abababab=,错误;对B:222320abababab=,错误;对C:220ababab,正确,故选C.解法四(数形结合法):如右图知

OACB为菱形,所以OCAB,即abab,所以1,故选C.【试题评析】平面向量问题的处理,从代数运算入手和从几何图形入手两个途径并重,各有利弊,可视具体问题灵活选择.3.已知双曲线2222:10,0xyCabab的焦距为25,点2,1

P在C的一条渐近线上,则C的方程为A.2214yxB.2214xyC.22331205xyD.221164xy【命题意图】本小题主要考查双曲线的方程、渐近线等基础知识;考查逻辑推理、运算求解等能力;渗透函数与方程、化归与转化等数学思想;

体现基础性,导向对发展逻辑推理、数学运算、直观想象等核心素养的关注.【试题解析】解法一:由已知225c,则5c,又12ba,且222bac,所以2,1ab.则C的方程为2214xy,故选B.解法二:由已知225c,则5c

,对于C,222553ab,对于D,22205ba,所以排除C,D;又由点2,1P在C的一条渐近线上,坐标代入方程检验可排除A.故选B.【试题评析】圆锥曲线的方程与基本性质,属于必考的基础知识,应作为必得分试题处理.4.6211xxx

的展开式中7x的系数为A.5B.6C.7D.15高三数学试题第3页(共12页)【命题意图】本小题主要考查二项式定理、计数原理等基础知识;考查逻辑推理能力和运算求解能力;考查化归与转化等数学思想;体现基础性、综合性,导向对发展数学运算、逻辑推理等核心素养的关注.【试题解析】解法一:展开式中含7

x的项为:215067766(61)5xCxxCxxx,所以7x的系数为5,故选A.解法二:因为65231111xxxxx,所以展开式中含7x的项为314755xCxx,所以7x

的系数为5,故选A.5.已知圆锥SO的底面半径为1,若其底面上存在两点,AB,使得90ASB,则该圆锥侧面积的最大值为A.2B.2C.22D.4【命题意图】本题考查圆锥的轴截面、侧面积等有关基础知识;考查运算求解等能力;考查数形结合思

想;体现基础性与综合性,导向对发展数学运算、直观想象、逻辑推理等核心素养的关注.【试题解析】解法一:设圆锥的母线长为(1)ll.如图,作圆锥SO的轴截面SAC,由题意得90ASCASB≥,所

以222242ACSASCl≥,即12l≤,所以ππ2πSrll圆锥侧≤,故选A.解法二:设圆锥的母线长为(1)ll.如图,作出圆锥的轴截面SAC,由题意得90ASC≥,所以4590ASO≤,则2sin12r

ASOl≤,即12l≤,所以ππ2πSrll圆锥侧≤,故选A.解法三:设圆锥的母线长为(1)ll.因为90ASB,所以2ABl,且OAOBAB≥,即22rl≥,所以12l≤,所以ππ2πSrll圆锥侧

≤,故选A.高三数学试题第4页(共12页)解法四:设圆锥的母线长为(1)ll.所以ππSrll圆锥侧取最大值时,母线取最大值,轴截面顶角取最小值.由圆锥底面上存在两点,AB,使得90ASB,知轴截面顶角最小值为90,此时母线长为2,圆锥侧面积为2.6.已知函数s

in04fxx在0,2有且仅有一个零点,则的值可以是A.1B.3C.5D.7【命题意图】本小题主要考查三角函数的图象和性质、三角函数的周期等基础知识;考查逻辑推理;考查函数与方程、转化与化归、数形结合等数学思想;体现基础性与综合性,导向对发展逻

辑推理、数学运算、直观想象等核心素养的关注.【试题解析】解法一:函数πsin04fxx,当π02x时,ππππ4424x,因为fx在π0,2上有且仅有一个零点,所以πππ2

π24≤,解得3722≤,故选B.解法二:πsin04fxx是由sin0yx向左平移8T所得.由已知fx在0,2有且仅有一个零点,则11π112828TT

≤,即32ππ72π828≤,所以3722≤,故选B.解法三:当1时,由sin04x,可得π4xkkZ,所以在区间0,2没有零点,当3时,由sin304x,可得π123

kxkZ,所以在区间0,2有且仅有一个零点,故选B.另外当5≥时,fx的周期2ππ2T,故fx在0,2的零点至少有两个,舍去.【试题评析】本题是检测核心素养的典型试题.体现对弦型曲线特征的整体把握,整体上把握图象特征,结合

理性思维,才能找到问题解决的突破口.高三数学试题第5页(共12页)7.已知函数2()axxfxbc,若3log31bac,则A.()()()fafbfcB.()()()fcfbfaC.()()()fbfafcD

.()()()fbfcfa【命题意图】本小题主要考查基本初等函数的图象、运用函数单调性比较数的大小等基础知识;考查逻辑推理、运算求解;考查数形结合、转化与化归等数学思想;体现基础性、综合性和应用性,导向对发展直观想象、逻辑推理、数学运算等核心素养的关注

.【试题解析】解法一(特值法):令3c得27a,1b,2()axxfxbc的对称轴为1254bax,所以2bbcaa,因为()fx在,2ba单调递增,所以()()()fbfcfa,故选D.解法二:作函数3logyx,3xy

,yx的图像,易得直线1yc与函数的交点的横坐标分别为,,abc,如图所示,由图知01bca,因为3log1a,所以3a,所以2bbcaa,下同解法一.8.1883年,德国数学家康托提出了三分康托集,亦称康托尔集.右图是其构造过程的图示,其详细构造

过程可用文字描述为:第一步,把闭区间01,平均分成三段,去掉中间的一段,剩下两个闭区间103,和213,;第二步,将剩下的两个闭区间分别平均分为三段,各自去掉中间的一段,剩下四段闭区间:109,,2193,,2739,,819

,;如此不断的构造下去,最后剩下的各个区间段就构成了三分康托集.若经历n步构造后,20212022不属于剩下的闭区间,则n的最小值是A.7B.8C.9D.10【命题意图】本小题主要考查集合、数列递推等基础知识;

考查抽象概括与运算求解能力;考查数据处理能力、应用意识等;体现基础性、综合性与应用性,导向对发展数学高三数学试题第6页(共12页)运算、数学建模、数据分析等核心素养的关注.【试题解析】解:由题意可知,构造过程中每一步剩下的最右边的闭区间依次为213,,819

,,,所以第n步构造后,剩下的最右边的区间段为1113n,.若1202111120223n,且202111120223n,,即112021111320223nn≤,即33log2022log20221n≤,因为

*nN,所以7n.经历第7次构造后,最右侧的两个闭区间为67121133,和71113,,且77112320223,7722021111320223,即20212

022不属于剩下的任何闭区间,故选A.二、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,有选错的得0分,部分选对的得2分。9.已知点M在直线:43lykx上,点N在圆22:9Ox

y上,则下列说法正确的是A.点N到l的最大距离为8B.若l被圆O所截得的弦长最大,则43kC.若l为圆O的切线,则k的取值范围为70,24D.若点M也在圆O上,则O到l的距离的最大值为3

【命题意图】本小题主要考查直线与圆的方程、直线与圆的位置关系等基础知识;考查逻辑推理、运算求解;考查函数与方程、数形结合、转化与化归等数学思想;体现基础性、综合性和应用性,导向对发展直观想象、逻辑推理、数学运算等核心素养的关注.【试题解析】解:易知直线l过定点(3,4)

P,如图所示:对A:点O到直线l的距离的最大值为22||345OP,所以点N到l的最大距离为||538OPr,故选项A正确;对B:若l被圆O所截得的弦长最大,则直线l过圆心O,即高三数学试题第7页(共1

2页)0403k,所以43k,故选项B正确;对C:若l为圆O的切线,则圆心O到直线l的距离2|34|31kdk,解得724k,故选项C错误;【或由0k,直线为4y,显然不与圆O相切,故选项C错误.】对D:

若点M也在圆O上,则l与圆O相交或相切,当l与圆O相切时,O到l的距离的最大值为半径3,故选项D正确;故选ABD.10.设1z,2z为复数,则下列命题正确的是A.若12||0zz,则12zzB.若12||||zz,则2212zzC.若120zz,则21zzD.

若120zz,则10z或20z【命题意图】本小题主要考查复数的概念、复数的表示方法、复数的模和复数的复数的运算等基础知识;考查逻辑推理、运算求解;考查数形结合、转化与化归等数学思想;体现基础性和综合性,导向对发展直观想象、逻辑推理、数学运算等核

心素养的关注.【试题解析】解:对A:设1izab,2izcd,则12i(i)()izzabcdacbd,若2212||()()0zzacbd,则ac且bd,所以12zz,故A正确;【另解:若12||0zz,由运算的

几何意义知12zz,对应的点重合,故12zz.】对B:取11z,2iz,则12||||1zz,此时211z,221z,2212zz,故B错误;对C:取11iz,2iz,则12

10zz,此时21zz,故选项C错误;对D:设1izab,2izcd,则12(i)(i)()izzabcdacbdadbc,若120zz,则0,0,acbdadbc得,,a

cbdadbc所以22acdbcd①,若00cd且,则①显然成立,此时20z;若00cd且,则易得0ab,此时10z;高三数学试题第8页(共12页)若00cd且,即得1

0z;若00cd且,得22ab,则0ab,此时10z,故选项D正确.【另解:因为1212||||zzzz||,所以,若120zz,则1||0z或2||0z,故10z或20z,故D正确.】故选AD.11.某校高三1班4

8名物理方向的学生在一次质量检测中,语文成绩、数学成绩与六科总成绩在全年级中的排名情况如下图所示,“”表示的是该班甲、乙、丙三位同学对应的点.从这次考试的成绩看,下列结论正确的是A.该班六科总成绩排名前6的同学语文成绩比数学成绩排名更好B.在语文和数学两个科目中,丙同学

的成绩名次更靠前的科目是语文C.数学成绩与六科总成绩的相关性比语文成绩与六科总成绩的相关性更强D.在甲、乙两人中,其语文成绩名次比其六科总成绩名次靠前的学生是甲【命题意图】本小题主要考查统计图表、相关关系等基础知识;考查读图、识图、用图的能力以及逻辑推理能力;体现基础性、

综合性和创新性,导向对发展数据分析、逻辑推理等核心素养的关注.【试题解析】解:对A:该班总成绩排名前6的同学即为年级前100名的同学,数学成绩都在前200名,而语文成绩比较离散,有2个是前200名,剩下4位同学都在200名之后,且有1位同学的语文成绩

大约是400名,故选项A错误;对B:由右图丙同学六科总成绩是400~500的三位同学中靠前的一位,其语文成绩在250~300名,对应左图找通过六科总成绩找到丙同学,其数学成绩排名大约是400名,所以丙同学语文成绩靠前,故选项B正确;对C:由散点图可知,数学成

绩与总成绩的分布呈左下到右上的趋势,且在一条直线附丙甲乙高三数学试题第9页(共12页)近,语文成绩与总体成绩比较分散,故选项C正确;对D:由左图知甲同学总成绩排名是在100~110名,由总成绩排名可在右图找到甲同学对应的点,其语文成绩大约是50名,所以甲同学语文成绩靠前.同理,由左图知乙

同学的总成绩排名是在240~250名,可在右图找到乙同学对应的点,其语文成绩大约是250名,故选项D正确;故选BCD.12.已知函数()fx的定义域为0,,且满足2210,1,()log31,2,xxfxxx,,当2x

≥时,2fxfx,为非零常数,则下列说法正确的是A.当1时,21log804fB.当0时,()fx在10,11单调递增C.当1时,()fx在*04nnN,的值域为2122,nn

D.当0,且1时,若将函数12()xgx与()fx的图象在*02nnN,的m个交点记为,iixy(123)im,,,,,则211mniiixyn【命题意图】本小题主要考查分段函数、函数的值域、函数的单调性、函数的

周期性、函数图象的变换、等比数列的前n项和等基础知识;考查逻辑推理、运算求解;考查数形结合、转化与化归等数学思想;体现基础性、综合性和创新性,导向对发展直观想象、逻辑推理、数学运算等核心素养的关注.【试题解析】解:对A:当1时,2fxfx,则

42fxfxfx,所以,当2x≥时,()fx满足4fxfx.因为222log64log80log128,即26log807,所以222log80log804

log5fff2log52f25log4f,因为25log0,14,所以2log80f25log4f25log42114,故选项A错误;高三数学试题第10页(共12

页)对B:当0时,()fx在0,1的单调性与()fx在2,21nn*nN的单调性相同,因为()fx在0,1单调递增,所以()fx在10,11单调递增,故选项B正确.对C:由2fxfx得,242()fxfxfx,则2

4()nfxnfx.因为1,如图可知,()fx在01,和*41,41nnnN单调递增,在*43,41nnnN单调递减.当*04xnn,N时,min()fx2(1)(41)(414(

1))nfnfnn22(3)nf21(1)nf21n;max()fx2(1)(43)(434(1))nfnfnn22(1)nf22n.所以()fx在*

04nnN,的值域为2122,nn,故选项C正确.对D:由图像可知,12()xgx与()fx的图象在*02nnN,有n个交点,且21ixi,1iiy,1,2,3

in,因为0,且1,所以数列ix是等差数列,数列iy是等比数列.所以111mnniiiiiiixyxy(121)121nnn211nn,故选项D错误.故选BC.高三数学试题第1页

(共12页)泉州市2022届高中毕业班质量监测(三)2022.03高三数学参考答案(填空题)本试卷共22题,满分150分,共6页。考试用时120分钟。三、填空题:本题共4小题,每小题5分,共20分。13.若sin21cos213,则tan.【命题意图】本小题主要考查二倍角公

式、三角恒等变换等基础知识;体现基础性,导向对发展数学运算、直观想象等核心素养的关注.【试题解析】解:因为22sin22sincos2sincostancos212cos2cos11,所以1tan3.14.写出一个满足1fx为偶函数,

且在0,单调递增的函数fx.【命题意图】本小题主要考查函数奇偶性、函数单调性等基础知识;考查抽象概括;体现基础性、应用性和开放性,导向对发展逻辑推理、数学抽象等核心素养的关注.【试题解析】解:21fxx或1fxx或12xfx或ln1fxx或

11xxfxee(答案不唯一)15.已知抛物线2:4Eyx的焦点为F,准线为l,过F的直线m与E交于,AB两点,AF的垂直平分线分别交l和x轴于,PQ两点.若AFPAFQ,则||AB.【命题意图】本小题主

要考查抛物线的方程、定义、焦点弦长公式、平面几何的性质等基础知识;考查数形结合、转化与化归等数学思想;体现基础性、综合性和创新性,导向对发展直观想象、数学运算等核心素养的关注.【试题解析】解:因为AF的垂直平分线交l于点P,所以||||PFPA,且AFPPAF;又AFPAFQ,

所以PAFAFQ,所以PAx∥轴,即APl,保密★使用前高三数学试题第2页(共12页)由抛物线的定义得||||PAAF,所以PAF△为等边三角形.所以60PAFAFQ.解法一:所以直线m的

方程为31yx.设11,Axy,22,Bxy.由231,4,yxyx消去y整理得231030xx,由韦达定理,得12103xx,所以1216||23ABxx.解法二:设l和x轴交于点G,过B作BCl

于点C,作BDAP于点D.在PGFRt△中,||2GF,60PFG,所以||||||4PFPAAF.因为||||||BFBCPD,所以||||||||||4||ADPAPDPABFBF.在ABD

Rt△中,||2||ABAD,即4||24||BFBF,所以4||3BF.故16||||||3ABAFBF.解法三:由二级结论112||||AFBFp得,1114||BF,所以4||3BF.故16||||||3ABAFBF.解法四:由二级结论,得222416||

sinsin603pAB.【注:答案写成153者,也是对的,但写成小数5.333者,不能得分.】16.已知三棱锥ABCD的所有顶点都在球O的球面上,ABACDBDC,24ADBC,则球O的表面积的最小值为.【命题意图】本

小题考查球的有关基础知识;考查空间想象、推理论证、运算求解等能力;考查数形结合、化归与转化等思想;体现基础性与综合性,导向对发展直观想象、逻辑推理、数学运算等核心素养的关注.【试题解析】解:设,EF分别

为BCAD,的中点,连结,,,AEDEBFCF.由已知,ABDB,ACDC,BCBC得ABCDBC△≌△.因为E是BC中点,所以AEDE,又因为AFDF,所以EFAD,高三数学试题第3页(共12页)即直线EF是线段AD的垂直平分线.①同理,ABAC,BDDC,ADAD

得ABDACD△≌△.因为F是AD中点,所以BFCF,又因为BEDE,所以EFBC,即直线EF是线段BC的垂直平分线.②由①②知,球心O在直线EF上,设球O的半径为R,则2,4,OBOCBCOAODAD≥≥即22,24,RR≥

≥即2R≥,当且仅当O是AD中点时,2R,所以球O的表面积的最小值为24π16πSR.高三数学试题第1页(共26页)17.(10分)在平面四边形ABCD中,1AB,3BC,60B,30ACD.(1)若213AD,求ADC;(2)若BDCD,求ACD△的面积.【命

题意图】本小题主要考查正弦定理、余弦定理、三角形面积公式和三角恒等变换等基础知识;考查抽象概括、逻辑推理、运算求解等能力;考查数形结合思想、函数与方程思想;体现基础性,导向对发展逻辑推理、数学运算等核心素养的关注.【试题解析】(1)法一:在ABC△中,由余弦定理

,得2222cosACABBCABBCABC,·······2分即211921372AC,所以7AC.·········································3分在ACD△中,由正弦定理,得sinsinAD

ACACDADC,···························4分即2173sin30sinADC,所以3sin2ADC,所以60ADC或120.····································

································5分法二:在ABC△中,由余弦定理,得2222cosACABBCABBCABC,··2分即211921372AC,所以7AC.································

······3分在ACD△中,由余弦定理,得2222cosADACCDACCDACD,即23321140CDCD,所以213CD或2213CD.当213CD时,CDAD,此时30

ACDDAC,所以1803030120ADC.····················································4分当2213CD时,222CDADAC,此时90DAC,所以180309

060ADC.·····················································5分综上60ADC或120.···········································

······················5分高三数学试题第2页(共26页)法三:在ABC△中,由余弦定理,得2222cosACABBCABBCABC,····2分即211921372AC,所以7AC.································

·······3分在ACD△中,由余弦定理,得2222cosADACCDACCDACD,即23321140CDCD,所以213CD或2213CD.在ACD△中,由余弦定理,得,222cos2ADCDACADCADCD所以777133cos1423ADC

或7287133cos2823ADC·····················4分所以60ADC或120.········································

···························5分【补充说明】本小题答案具有开放性。因调控难度需要,没有在开放性答案的选择上另行设置给分点.若考生没讲理由,直接舍去60答案者,扣1分,直接舍去120

答案者,扣1分;有指出理由(如认为平面四边形一般指的是凸四边形)才舍去60答案者,不论在解题过程中的哪个地方指出,均不扣分.【说理样例】(默认平面四边形一般指的是凸四边形)当60ADC时,90DAC,又由179cos0217BAC,知90BAC,从而180BAD

BACDAC,不符合题意,舍去.故120ADC.(2)法一:设ACB.在ABC△中,由正弦定理,得71sin60sin,所以21sin14.··················6分又ABBC,所

以为锐角,因此257cos1sin14,······················7分所以3121coscos30cossin227BCD.·····························

8分设E为线段BC中点,连接DE.因为BDCD,故DEBC,32CE,所以3212cos2217CECDBCD.·························································9分因此173sin3028

ACDSACCD△.····················································10分法二:设ACB.在ABC△中,由正弦定理,得2222cosABBCAC

BCAC,··············6分数字代入求解得57cos14,·····························································

··7分下同法一.高三数学试题第1页(共26页)18.(12分)体育课程的实施可以有效地促进学生身体的正常发育,提高身体的健康水平.某校对高一年男生进行1000米测试,经对随机抽取的100名学生的成绩数据处理

后,得到如下频率分布直方图:(1)从这100名学生中,任意选取2人,求两人测试成绩都低于60分的概率;(2)从该校所有高一年男生中任意选取3人,记70分以上的人数为,求的分布列和期望;(3)从样本频率分布直方图中发现该校男生的1000米成绩Χ近似服从2,N,

已知样本方差211644s.,高一年男生共有1000人,试预估该校高一年男生1000米成绩在89.2分以上的人数.附:11644108...若2,ΧN,则06826.PX,2209544.

PX.【命题意图】本小题主要考查样本数字特征、古典概型、二项分布、正态分布等基础知识;考查逻辑推理、运算求解;考查样本估计总体的统计思想;体现基础性、综合性和应用性,导向对发展数据分析、数学建模、逻辑推理、数学运算等核心素养的关注.【试题解析】解:(1)由频率分布直方图知,测试成绩

低于60分的频率为(0.0010.002)100.03,··1分共有0.03100=3人,设事件A表示“从这100名学生中,任意选取2人,两人测试成绩都低于60分”,则232100311009916502C

PAC.································································3分【说明:列式232100CPAC1分,计算得116501分,未交代事件A,回扣1分.】(2)由频率分布直方图知,测试

成绩在70分以上的频率为4(0.0400.0220.018)105,高三数学试题第2页(共26页)用样本的频率估计总体的概率,·······························································4分【说明:未写“用样本

的频率估计总体的概率”,不得此分.】则可判定,,Bnp(其中3n,45p),···········································5分所以随机变量的分布列为:003

3411(0)()()55125PC,11234112(1)()()55125PC,22134148(2)()()55125PC,33034164(3)()()55125PC,···················

·7分···································7分【说明:2个或3个正确得1分,4个正确得2分.】随机变量的分布列也可列表表示为:0123P1125121254812564125······

··································································································7分所以随机变量的数学期望

为412355Enp.··································8分(3)由样本的频率分布直方图得样本平均数为450.01550.02650.17750.40850.22950.18x··········

·········9分45100.02200.17300.40400.22500.18450.23.4128.8978.4,以样本估计总体,得78.4.且由211644s.且11644108..,可得10.8.因为78.410.889.2

,·····························································10分所以10.682689.20.15872PΧ,·············

····································11分所以10000.1587158.7159,··············································

············12分即预估跑步成绩在89.2分以上的人数为159人.······································12分高三数学试题第1页(共26页)19.(12分)已

知数列na满足12121111nnnaaaaaaa.(1)求na的通项公式;(2)在ka和*1kakN中插入k个相同的数11kk,构成一个新数列nb:1234,1,,2,2,,3,3,3,,aaaa,求nb的前100项和100

S.【命题意图】本小题主要考查等差数列的定义、等差数列前n项和公式及递推数列等基础知识;考查抽象概括与运算求解能力;考查化归与转化思想思想.体现基础性与综合性,导向对发展数学运算等核心素养的关注.【试题解析】解:(1)因为12121111n

nnaaaaaaa......①,当2n时,11212111111nnnaaaaaaa......②,·······································

····1分①÷②,得11nnnnaaaa,所以11nnaa,···············································3分又1n时,11111aaa,即12a,······················

······································4分所以数列na是以2为首项,1为公差的等差数列,所以2111nann.················································

···················5分(2)法一:设插入的所有数构成数列nc.因为1231278,781391100,123121391,9114105100,所以,12100,,

,bbb中包含na的前13项及nc的前87项,····························8分所以10012131287()()Taaaccc································

··············9分2222223141234111213913214123121392·························11分【各1分】138136139

143.·························································12分高三数学试题第2页(共26页)(2)法二:根据数列nb的构

造规律,可知nb中等于ka的项之前,有插入的123(1)k项和原数列na中的前1k项,故nb中的ka,即nb的第(1)2kk项,即(1)2kkkba.···································7分因为13

14141510022,所以,12100,,,bbb中包含na的前13项及nc的前87项,····························8分所以10012131287()()Taaaccc·

·············································9分2222223141234111213913214123121392

·········································11分138136139143.............12分(2)法三:设ka和插入的k个数11kk构成一组数,则前k组数共有

1322kkkkk个,···············································6分令31002kk,解得12k,当12k时有312159010022kk,··························

···················7分所以nb的前100项中包含前12组数和第13组数的前10个,·······················8分所以22210012111213(1)(2)(11)(

12)(139)Saaaaa22222231412341112139················10分13214123121392·······················

·············11分138136139143.12分高三数学试题第1页(共26页)20.(12分)如图,多面体ABCEF中,ABAC,BFCE,D为BC的中点,四边形ADEF为矩形.(

1)证明:BECE;(2)若2AB,120BAC,当三棱锥EBCF的体积最大时,求二面角ABFE的余弦值.【命题意图】本小题主要考查线面垂直的判定与性质、二面角的求解及空间向量的运算与应用等基础

知识;考查空间想象能力、推理论证及运算求解能力;考查数形结合思想、化归与转化思想等;体现基础性、综合性与应用性,导向对发展逻辑推理、数学运算、直观想象等核心素养的关注.【试题解析】解法一:(1)因为四边形ADEF为矩形,所以

ADDE.因为ABAC,D为BC的中点,所以ADBC.因为DEBCD,,DEBC平面BCE,所以AD平面BCE.···············································

··································1分又四边形ADEF为矩形,所以ADFE∥,所以FE平面BCE,····································································

············2分又EC平面BCE,所以FECE.······························································3分又BFCE,BFFEF,所以

CE平面BEF.·················································································4分又BE平面BEF,所以BECE

.·······························································5分(2)因为2AB,120BAC,在ADBRt△中,112ADAB,3BD,所以23BC.由(1)知FE平面BCE,

所以EBCFFBCEVV13BCESEF△,....................6分【体积公式】EBCFFBCEVV16BECE22112CEBE≤21112BC,高三数学试题第2页(共26页)当且仅当6BECE时

,等号成立,即三棱锥EBCF的体积最大.........7分【基本不等式】又DBDC,所以DEBC.又ADBC,ADDE,所以,以D为坐标原点,分别以,,DADBDE为,,xyz轴的正方向建立空间直角坐标系,····················

·······················································································8分则所以(1,0,0)A,(0,3,0

)B,C(0,3,0),(0,0,3)E,(1,0,3)F,所以(1,3,0)AB,(0,0,3)AF,(1,0,0)EF,(0,3,3)EB,···9分

设平面ABF的法向量为111(,,)xyzu,平面BEF的法向量为222(,,)xyzv,由0,0,ABAFuu得11130,30,xyz令11y,得(3,1,0)u,················

···········10分由0,0,EBEFvv得222330,0,yzx令21y,得(0,1,1)v,···························11分设二面角ABFE的大小为)2(,则||2cos||

||4uvuv,·······································································12分即当三棱锥EBCF的体积最大时,二面

角ABFE的余弦值为24.············12分解法二:(1)同解法一.··········································································

·····················5分(2)因为2AB,120BAC,在ADBRt△中,112ADAB,3BD,所以23BC.由(1)知FE平面BCE,所以EBCFFBCEVV1

3BCESEF△,....................6分【体积公式】EBCFFBCEVV16BECE22112CEBE≤21112BC,当且仅当6BECE时,等号成立,即三棱锥EBCF的体积最大.......

..7分【基本不等式】以E为坐标原点,以,,EFECEB分别为,,xyz轴的正方向建立空间直角坐标系,.......8分所以66(1,,)22A,(0,06)B,,(0,0,0)E,(1,0,0

)F,所以66(1,,)22AB,66(0,,)22FA,············································9分高三数学试题第3页(共26页)易知平面BEF的法向量为(0,1,0)v,·····················

··································10分设平面ABF的法向量为(,,)xyzu,由00ABAFuu得2660,0,xyzyz,令1y得(6,1,1)u,···················11分设

二面角ABFE的大小为)2(,则||2cos||||4uvuv,··········12分即当三棱锥EBCF的体积最大时,二面角ABFE的余弦值为24.···········12分解法三:(1)同解法一.··············

··················································································5分(2)因为2AB,120BAC,

在ADBRt△中,112ADAB,3BD,所以23BC.由(1)知FE平面BCE,所以EBCFFBCEVV13BCESEF△,....................6分【体积公式】EBC

FFBCEVV16BECE22112CEBE≤21112BC,当且仅当6BECE时,等号成立,即三棱锥EBCF的体积最大.........7分【基本不等式】设二面角ABFE的大小为)2(.由EABFBAEFV

V得点E到平面ABF的距离为13322AEFABFSBDEFBDdSAB△△,···············································································8分又点E到直线BF的距离/6142

77BEEFdBF,······················································9分所以/3142sin4427dd,·····························

··································································10分高三数学试题第4页(共26页)所以22cos|cos|1sin4

,即2cos4,故当三棱锥EBCF的体积最大时,二面角ABFE的余弦值为24.···············12分【又点E到直线BF的距离/614277BEEFdBF,···························

·······················9分所以/3142sin4427dd,···················································································

············10分所以22cos|cos|1sin4,即2cos4,故当三棱锥EBCF的体积最大时,二面角ABFE的余弦值为24.············12分】高三数学试题第1页(共26页)21.(12分)已知点

110F,,210F,,M为圆O:224xy上的动点,延长1FM至N,使得1||||MNMF,1FN的垂直平分线与2FN交于点P,记P的轨迹为.(1)求的方程;(2)过2F的直线l与交于AB,两点,纵坐标不为0的点E在直线4x上,线段OE分别与线段AB,交于C,D两点,

且2||||||ODOCOE,证明:||||ACBC.【命题意图】本小题主要考查椭圆的定义、标准方程,直线与椭圆的位置关系等基础知识;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.【试题解析】解法一:(1)连结MO,1PF.因为1FN的垂直平分线交2FN于点P,所以1||||PFPN.所以1222||||||||||PFPFPNPFNF

.···················································1分在12NFF△中,1||||MNMF,12||||FOOF,所以2||2||4NFOM,即1212||||4||PFPFFF.···················

·················2分所以点P的轨迹是以1F,2F为焦点,长轴长为4的椭圆.···························3分由已知11,0F,21,0F,故的方程为22341xy.···················

···················································4分(2)因为2||||||ODOCOE,即||||||||ODOEOCOD,即||||||||DECDxxxx,·······················

·5分由已知,显然0Cx,且4Ex,所以24DCxx.·······································6分(i)当直线l的斜率不存在时,即直线l:1x.这时1Cx,2Dx,24DCxx显然不

成立.高三数学试题第2页(共26页)(ii)当直线l的斜率存在时,设直线l的方程为1ykx,11,Axy,22,Bxy.由221,1,43ykxxy消去y,整理得22224384120kxkxk

,········7分由韦达定理,得2122843kxxk.所以21224243xxkk,即线段AB的中点横坐标为22443kk.················8分设直线OE的方程为0ykxk

.由1,,ykxykx解得kxkk,即Ckxkk.································9分由22,1,43ykxxy解得221243xk,即22124

3Dxk.························10分由24DCxx,得212443kkkk,化简,得34kk,34kk.··11分所以2243434Ckkx

kkk,即点C为线段AB的中点,故||||ACBC.········································································12分【若(ii)得

满分,但漏掉(i)的说明,扣1分;若(ii)未得满分,对漏掉(i)的说明者,不补扣1分】解法二:(1)设(,)Pxy,00(,)Mxy.··························································

···············1分则00(21,2)Nxy,2(1,)FPxy,2002,2)FNxy(,00(,)MPxxyy,100(2+2,2)FNxy.因为1FN的

垂直平分线交2FN于点P,所以10MPFN,22FPFN∥,因此0000()(1)()0xxxyyy①,00(1)yxxy②.······························2分又22004xy③,联立①

②③解得022022(1)(4),(1)(4).(1)xxxxyxyyxy·································3分代入22004xy并化简整理得

的方程为22341xy.·································4分高三数学试题第3页(共26页)(2)由已知直线OE的斜率存在,设为k,又E的纵坐标不为0,则0k,则直线OE的方程为0ykxk.则22222||1DDDODxyk

x,222||1||CCCOCxykx,222||1||EEEOExykx.·························································

·····5分由已知4Ex,0Cx.且2||||||ODOCOE,所以24DCxx.·····················6分由22,1,43ykxxy解得221243xk,即221243Dxk.·

···································7分由已知,直线l过点2F,且l不能与x轴重合,则设直线l的方程为1myx,因为直线l与直线OE相交,故1mk.设11,Axy,22,Bxy.由1,,myxykx

解得11xmk,即11Cxmk.··········································8分由24DCxx,得2124431kmk,即34km,故243

4Cxm.··················9分由221,1,43myxxy消去y,整理得2223484120mxxm,··················10分由韦达定理,得1

22834xxm.·························································11分所以122Cxxx,即点C为线段AB的中点,故||||ACBC.···················12分解法三:(1

)连结MO,1PF.因为1FN的垂直平分线交2FN于点P,所以1||||PFPN.所以1222||||||||||PFPFPNPFNF.················································1分

在12NFF△中,1||||MNMF,12||||FOOF,所以2||2||4NFOM,即12||||4PFPF.···········································2分设(,)Pxy.由已知11,0F,21,

0F,高三数学试题第4页(共26页)所以2222114xyxy()().···················································3分化简,得22341

xy,故的方程为22341xy.·····································4分(2)由已知,OC与OE方向相同,且2||||||ODOCOE,所以2ODOCOE

,·5分即2DCExxx,因为4Ex,所以24DCxx.··········································6分(i)当直线l的斜率不存在时,即直线l:1x.这时1Cx,2Dx,24DCxx,不符合题意.(ii)当直线l的

斜率存在时,设直线l的方程为1ykx,11,Axy,22,Bxy,线段AB的中点00,Mxy.221122221,431,43xyxy相减得,21212121043xxxxyyyy,·······7分即02

121034yyyxxx,即34OMkk.···············································8分设直线OE的方程为0ykxk.由1,,ykxykx解得kxkk,即Ckx

kk.································9分由22,1,43ykxxy解得221243xk,即221243Dxk.························10分由24DCxx,得212443kkkk

,化简,得34kk,················11分又因为34OMkk,所以OMkk,即直线OE与直线OM重合,即点C与点M重合,则点C为线段AB的中点,故||||ACBC.···························

·········12分高三数学试题第1页(共26页)22.(12分)已知函数()()sincosfxxmxx,5π04x,.(1)当π2m≤时,讨论()fx的单调性;(2)若0m,()1(π)fxax

≤,求a.【命题意图】本小题主要考查运用导数判断函数的单调性、求函数的最值、零点存在定理等基础知识;考查抽象概括、推理论证、运算求解等能力;考查函数与方程、化归与转化、分类与整合、数形结合等数学思想;体现综合性、应用性与创新性,导向对发展逻辑推理、数学运算、数学抽象、数学建模等核心素养的关注

.【试题解析】解法一:(1)由已知得()sin()cossinfxxxmxx()cosxmx.························1分①当π2m时,()fxπcos2xx,若π02x

,,π02x,cos0x,则()0fx,()fx在π02,单调递减;若π5π,24x,π02x,cos0x,则()0fx,()fx在π5π,24单调递减.由于()fx的图象

在π2x处连续,故()fx在5π04,单调递减.···············2分②当π02m时,若π2mx,()0fx,故()fx在π2m,单调递增;若0xm≤或π5π24x≤,()0fx,故()fx在0,m和π

5π,24单调递减.·················································3分③当0m≤时,对于5π04x,,恒有0xm.若π02x,,cos0x,()0fx,故()

fx在π02,单调递增;若π5π,24x,cos0x,()0fx,故()fx在π5π,24单调递减.········4分由于()fx的图象在前述所讨论的各区间端点处均

连续,故综上所述,可得:①当0m≤时,()fx在π02,单调递增,在π5π,24单调递减;②当π02m时,()fx在π2m,单调递增,在0,m和π5π,24单调递减;③当π2m时,()fx在5π04

,单调递减.···············································5分【说明:基于本小题的思维量和分类讨论的复杂性,对于图象连续性的交代和单调区间是否包括区间端点高三数学试题第2页(共26页)的问题,不做严格要求,不做扣分处理.】(2

)当0m时,()1(π)fxax≤,即sincosπ10xxxaxa≤,令()sincosπ1gxxxxaxa,5π04x,,则只需max()0gx≤.···········6分()sincossingxx

xxxacosxxa,令()()hxgx,则()cossinhxxxx.当π,π2x时,()0hx,所以()()hxgx单调递减.①当0a≥时,π()02gxga

≤,()gx在π,π2单调递减,又(π)0g,当π,π2x,()0gx,故max()0gx≤不成立;··································

7分【①的另解:当0a≥时,由于ππ()(1)1022ga,故max()0gx≤不成立;】②当π<0a时,因为(π)π0ga,π02ga,由零点存在定理得,0π,π2x,0()0gx,且当0,πxx时,()0gx,()

gx在0,πx单调递减,当0,πxx,均有()(π)0gxg,故max()0gx≤不成立;····················8分③当πa时,()cosπgxxx,(π)π+π0g.(i)当π0,2x

时,()0gx,所以()gx在π0,2单调递增;当π,π2x时,因为()gx单调递减,所以当π,π2x,()(π)0gxg,所以()gx在π,π2单调递增.由()g

x在π0,2单调递增,且在π,π2单调递增,可知,()gx在0,π单调递增.·················································9分(ii)当5ππ,4x时,令()()coss

inFxhxxxx,()cos2sinFxxxx,则()0Fx,()Fx在5ππ,4单调递增,(π)10F,5π25π()(1)0424F,由零点存在定理得,15ππ,4x

,1()0Fx.当1π,xx时,()0Fx,即()0hx,()()hxgx单调递减;高三数学试题第3页(共26页)当15π,4xx时,()0Fx,即()0hx,(

)()hxgx单调递增,且(π)0g,5π52()π1048g,由零点存在定理得,215π,4xx,当2π,xx时,()0gx,()gx单调递减;当25π,4xx时,()0gx,()gx单调递增.·······

··················10分综合(i)(ii)的结论,又(π)0g,25π25ππ()+1104244g,可得,max()0gx,符合题意.·····························

·····················11分④当πa时,(π)π0ga.当5ππ,4x时,由(ii)知,()gx在1π,x单调递减,在15π,4x单调递增,且111()cosgxxxa.若1()0gx

≥,则()gx在1π,x单调递增,对于1π,xx,均有()(π)0gxg,不符合题意;若1()0gx,则31π,xx,3()0gx,当3π,xx时,()0gx,()gx在3π,x单调递增,对于3π,xx,均有()

(π)0gxg,不符合题意.故当πa时,不符合题意.···················································12分综上所述:πa.·········

······························································12分解法二:(1)同解法一.················································

·······································5分(2)当0m时,()1(π)fxax≤,即sincosπ10xxxaxa≤,令()sincosπ1gxxxxaxa

,5π04x,,只需max()0gx≤.··············6分考察函数式()sincosπ1gxxxxaxa,发现(π)0g,所以,若sincosπ10xxxaxa≤(5π04x,),则π必为()gx的最大值点

,································································7分从而必为()gx的极大值点,必有(π)0g.高三数学试题第4页(共26页)由()gxcosxxa,得(π)π0ga

,解得πa.·························8分【注:前述证得“πa”是“sincosπ10xxxaxa≤(5π04x,)”的必要条件.】下面证明π

a符合题意.【即证明充分性】当πa时,()cosπgxxx,令()()hxgx,则()cossinhxxxx.(i)当π0,2x时,()0gx,所以()gx在π0,2单调递增;当π,π2x时,()0hx,

所以()()hxgx单调递减,所以当π,π2x,()(π)0gxg,所以()gx在π,π2单调递增.由()gx在π0,2单调递增,且在π,π2单调递增,可知,()gx在0,π单调递

增.····················································9分(ii)当5ππ,4x时,令()()cossinFxhxxxx,由()cos2sinFxxxx,得()0Fx,()Fx在5ππ,4单

调递增,因为(π)10F,5π25π()(1)0424F,所以由零点存在定理知,15ππ,4x,1()0Fx.当1π,xx时,()0Fx,即()0hx,()hx单调递减,即(

)gx单调递减;当15π,4xx时,()0Fx,即()0hx,()hx单调递增,即()gx单调递增,··························································

································10分因为(π)0g,5π52()π1048g,所以由零点存在定理知,215π,4xx,2()=0gx,当2π,xx时,()0gx,()gx单调递减;当

25π,4xx时,()0gx,()gx单调递增.·····························11分综合(i)(ii)的结论,又(π)0g,25π25ππ()+1104244g,可得,max()0

gx,符合题意.··················································12分综上所述:πa.········································

·····························12分高三数学试题第5页(共26页)解法三:(1)同解法一.··················································································

·····5分(2)当0m时,()1(π)fxax≤,即sincos1(π)xxxax≤,令()sincos1mxxxx,()(π)nxax,5π04x,.sincos1(π)xxxax≤(5π04x,)等价于曲线段()sincos1

mxxxx(5π04x,)恒在直线()(π)nxax的下方(含在直线上),·····································6分由于直线()(π)nxax过曲线段()sincos

1mxxxx(5π04x,)上的点(π,0),故直线()(π)nxax必与曲线段()sincos1mxxxx相切于点(π,0),·····7分所以()(cos)|xamxx.································

·························8分以下证明当πa时,sincos1(π)xxxax≤(5π04x,):【即证明充分性】令2()sincos+π1gxxxxx,5π04x

,,则()cosπgxxx,令()()hxgx,则()cossinhxxxx.(i)当π0,2x时,()0gx,所以()gx在π0,2单调递增;当π,π2x时,()0

hx,所以()()hxgx单调递减,所以当π,π2x,()(π)0gxg,所以()gx在π,π2单调递增.由()gx在π0,2单调递增,且在π,π2单调递增,可知,(

)gx在0,π单调递增.····················································9分(ii)当5ππ,4x时,令()()cossinFxhxxxx

,由()cos2sinFxxxx,得()0Fx,()Fx在5ππ,4单调递增,因为(π)10F,5π25π()(1)0424F,所以由零点存在定理知,15ππ,4x,1(

)0Fx.当1π,xx时,()0Fx,即()0hx,()hx单调递减,即()gx单调递减;当15π,4xx时,()0Fx,即()0hx,()hx单调递增,即()gx单调递增,···································

·······················································10分因为(π)0g,5π52()π1048g,高三数学试题第6页(共26页)所以由零点存在定理知,215π,4xx

,2()=0gx,当2π,xx时,()0gx,()gx单调递减;当25π,4xx时,()0gx,()gx单调递增.·····························11分综合(i)(i

i)的结论,又(π)0g,25π25ππ()+1104244g,故证得,对于5π04x,,2()sincos+π1()0gxxxxxg≤.综上所述:πa.····

·································································12分【关于25π25ππ()+1104244g的估算:225π25ππ()+1102+45+44244g

(),因22+45+419.7()19.4,,故25π25ππ()+1104244g.】获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com

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