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高一数学试题第1页（共6页）2021-2022学年度上学期泉州市高中教学质量监测2022.02高一数学选择填空题参考答案本试卷共22题，满分150分，共6页。考试用时120分钟。一、选择题：本大题共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．如图所示，已

知全集UR，集合{1,3,5,7}A，{4,5,6,7,8}B，则图中阴影部分表示的集合为A．{1,3}B．{5,7}C．{1,3,5}D．{1,3,7}【答案】A．2．函数3()fxxx的零点所在区间为A．(0,1

)B．(1,2)C．(2,3)D．(3,4)【答案】C．3．函数2()22xxxfx的图象大致是A．B．C．D．【答案】D．保密★启用前高一数学试题第2页（共6页）4．将整体一分为二，较大部分与整体部分的比值等于较小部分与较大部分的比值，这样的分割被称为黄金分割．黄金分割蕴藏着丰富的数学知

识和美学价值，被广泛运用于艺术创作、工艺设计等领域．黄金分割的比值为无理数512，该值恰好等于2sin18，则cos36A．52B．154C．154D．512【答案】C．5．下列命题中正确的是A．若11ab，则abB．若a

cbc，则abC．若ab，则abD．若22ab，则ab【答案】C．6．若函数()ln2fxax在(1,)单调递增，则实数a的取值范围为A．(0,)B．(2,)C．(0,2]D．[2,)【答案】D．7．第24届冬季奥林匹克运动会，将于2022年

2月4日～2月20日在北京和张家口联合举行．为了更好地安排志愿者工作，现需要了解每个志愿者掌握的外语情况，已知志愿者小明只会一门外语，甲说，小明不会法语，也不会日语；乙说，小明会英语或法语；丙说，小明会德语．已知三人中只有一人说对了，由此推断小明掌握的外语是A．德语B．法语C．日语D．英语【答案】

B．8．已知ln22a，122b，4log3c，则A．abcB．acbC．bacD．cab【答案】A．二、选择题：本大题共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24loglogmn，则高一数学试题第3页（共6页）A．2nmB．93loglognmC．ln2lnnmD．28loglog()mmn【答案】

BCD．10．已知正实数,ab满足1ab，则A．14abB．221+2abC．2abD．114113ab【答案】BD．11．若定义在R上的奇函数()fx满足()(2)fxfx，当1,0x时，()2

fxx，则A．()fx在3,5上单调递增B．()fx所有零点的集合为2,xxnnZC．()fx的最小正周期T=4D．(1)yfx为偶函数【答案】BCD．12．已知圆O的半径为1米，A为

圆O上一定点，动点,MN均以每秒1米的速度同时从A出发，M沿着OA方向向右运动，N沿着圆周按逆时针运动，当N运动回到A时，M停止运动，连接,MNON，记运动时间为t秒，三角形OMN的面积为1S，扇形AON（阴影部分）的面积为2S，则A

．当1t时，ONM为钝角B．当t时，,MN之间距离最大C．(0,)2t，MN与圆O相切D．(,)2t，12SS【答案】AC．三、填空题：本大题共4小题，每小题5分，共20分。将

答案填在答题卡的相应位置。13．已知13,0,()1,0,xxfxxx则((2))ff________．14．已知点(1,3)P落在角的终边上，则cos()=2_________．15．用()Mx表示

()fx，()gx中的较大者，记为()max{(),()}Mxfxgx，则函数2()max{log,3}Mxxx的最小值为________．高一数学试题第4页（共6页）16．写出一个满足(1)(1)fxfx，且(0)(3)ff的函数()fx的解析式___

______．高一数学试题第1页（共6页）2021-2022学年度上学期泉州市高中教学质量监测2022.02高一数学解答题参考答案本试卷共22题，满分150分，共6页。考试用时120分钟。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）已知集合

|31Axaxa，2|2150Bxxx．（1）当2a时，求()ABRð；（2）若“xA”是“xB”的充分条件，求实数a的取值范围．【命题意图】本小题主要考查集合的概念与基本运算、一元二次不等式、充分条件等基础知识，考查运算求解能力、推理

论证能力，考查化归与转化思想、分类与整合思想、数形结合思想，考查逻辑推理、数学运算等核心素养，体现基础性．解析：（1）依题意，2|2150|35Bxxxxxx或，·············

··················2分|35BxxRð．·········································································3分当2a时，|51Axx，······

··················································4分所以，()|31ABxxRð．····························

···························5分（2）依题意，“xA”是“xB”的充分条件，故AB，···························7分所以13a或35a，·····························

······································9分解得4a或8a，故实数a的取值范围为(,4][8,)．·················10分保密★启用前高一数学试题

第2页（共6页）18．（12分）已知函数2()4fxaxaxb（0a）在03，上的最大值为3，最小值为1．（1）求()fx的解析式；（2）1,x，fxmx，求实数m的取值范围．【命题意图】本小题主要考查二次函数性质、基本不等式、存在量词命题等

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、化归与转化思想、分类与整合思想、数形结合思想，考查逻辑推理、数学运算等核心素养，体现基础性、综合性．解析一（1）因为224(2)4fxaxax

baxab．所以二次函数fx图象的对称轴为2x．··············································1分因为0a，所以fx在[0,2]单调递减，在[2,3]单调递减．··············

········2分所以maxmin()(0)3,()(2)1,fxffxf·····································································3分即3,481,baab

解得3,1.ba······························································5分故243fxxx．········································

·································6分（2）由（1）得243fxxx．1,x，fxmx成立，等价于243xxmx在1,有解．等价于34mxx在1,有解，

·······················································7分令3(),(1,)gxxxx，则min4()mgx．·············

···························8分又33()223gxxxxx，···························································9分当且仅当3xx，即3x时等号成立，··

··············································10分因为31,+x（），·······································································11分所以423m

，即234m．······················································12分所以实数m的取值范围为234m．·················

································12分高一数学试题第3页（共6页）解析二（1）同解析一．（2）由（1）得243fxxx．因为1,x，fxmx成立，

所以243xxmx在1,有解．即2(4)30xmx在1,有解．····················································7分令2()(4)3,(1,)hxxmxx．·················

····································8分则函数()hx图象的对称轴为42mx．①当412m，即2m时，(1)0h，即0m，得0m（不合题意，舍去）；··································9分②当412m

，即2m时，4()02mh，········································10分即224(4)()3022mm，解得234m或234m（不合题意，舍去）.·················

············11分综上所述，实数m的取值范围为234m．··················12分【整合的分数】高一数学试题第4页（共6页）19．（12分）已知sin2cos0．（1）求2sin2cos；（2）若0，且1cos()4

，求cos．【命题意图】本小题主要考查两角和差的正余弦公式、同角三角基本关系、二倍角公式等基础知识，考查推理论证能力、运算求解能力，考查函数与方程思想、化归与转化思想，考查数学运算等核心素养，体现基础性、综合性．解析：（1

）法一：依题意，sin2cos0，得sintan2cos，·························2分2222222sin2cos2sincoscossin2cossincossincos

···························································4分【1的代换与二倍角公式各1分】22tan1413415tan1

．········6分【转化正切计算结果各1分】法二：sin2cos0，联立22sincos1，可得1cos5或1cos5．························································

··2分①当1cos5时，2sin5，························································3分223sin2cos2sincoscos5

；·········································4分②当1cos5时，2sin5，223sin2cos2sincoscos5．·······································

·····5分综上所述，23sin2cos5．··························································6分（2）sin2cos0，联立22sincos1，因为0

2p-<，故2sin5，1cos5．··············7分【结果对就给分】因为02p-<p，1cos()04，故2p<p，···············8分215sin()

1cos()4，·····················································9分coscos[()]cos()cossin()sin

························11分111525103()()442055．································12分高一数学试题第5页（共6页）20．（12分）已知函数2()2xxmfxn为定义在R

上的奇函数．（1）求实数m，n的值；（2）解关于x的不等式2(26)(3)(0)fxxfaaxf．【命题意图】本小题主要考查函数的奇偶性、单调性、二次不等式等基础知识，考查推理论证能力、运算求解能力，考查函数与方程思想、化归与转化

思想、分类与整合思想，考查逻辑推理、数学运算等核心素养，体现基础性、综合性．解析：（1）因为()fx为R上的奇函数，所以(0)=0f，即02+0m，即1m．···········1分所以21()2xxfxn，根据(

)fx为R上的奇函数可得()+()=0fxfx，············2分所以2121+022xxxxnn，即2(21)(1)0(21)(2)xxxnnn对任意xR上恒成

立，····3分所以1n．························································································4分（2）2()121

xfx，设任意12,xxR，且12xx，则12121212222(22)()()(1)(1)02121(21)(21)xxxxxxfxfx，所以()fx在R上单调递增．··········································

························6分又()fx为R上的奇函数，不等式22630fxxfaaxf可化为2263fxxfaxa·······7分由()fx在R

上单调递增，可得2263xxaxa，即230xax，方程230xax的两根分别为3，2a．······································

·····8分当6a时，32a，所以x；···························································9分当6a时，32a，所以32ax；························

·····························10分当6a时，32a，所以32ax．·····················································11分综上所述，当6a时，不等式的解集为；当6a时，不等式的

解集为32axx；当6a时，不等式的解集为32axx．··················12分高一数学试题第6页（共6页）21．（12分）函数()sin()fxAx（0A，0，02）在一个周期内的图象如图所示，,,MNP为

该图象上三个点，其中,MN为相邻的最高点与最低点，1(,0)2P．且172OM，25MN．（1）求()fx的解析式；（2）()fx的图象向左平移1个单位后得到()gx的图象，分析函数()()()Fxfxgx在1

5[,]43的单调性及最值．【命题意图】本小题主要考查三角函数的图象与性质等基础知识，考查抽象概括能力、推理论证能力、运算求解能力，考查函数与方程思想、化归与转化思想、数形结合思想，考查数学抽象、直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性．解析

一（1）过M作'MMx轴于'M，连接MN与x轴交于B，则5MB．·············1分设'OMt，则1'2BMt，由22222'''MMMOOMMBBM，即222171()5()22tt，可得1'2OM．进而可得'2MMA，'1MB．········

··············································2分记()fx的最小正周期为T，则'142TMB，·········3分【周期公式1分】得2．·································

························································4分故()2sin()2fxx，又1()02f，且02，得4．即()2sin()24fx

x．······································································5分（2）依题意，()(1)2sin()24gxfxx，··································

············6分()()()4sin()sin()2(sincos)(cossin)24242222Fxfxgxxxxxxx222(cossin)2cos

22xxx．······················································8分高一数学试题第7页（共6页）由22kxk，可得()Fx单调减区间为[2,21],kkkZ；由22kxk，可得()Fx单调增区

间为[21,2],kkkZ．故()Fx在1[,1]4单调递减；在5[1,]3单调递增，·········································10分则min()(1)2FxF；··········

····························································11分max151()max{(),()}max{2,1}()2434FxFFF．····························12分解析

二：（1）过M作'MMx轴于'M，连接MN与x轴交于B，则5MB．·············1分'MMA，'4TBM，则1'42TOM，由22222'''MMMOOMMBBM，可得142T，得2．··································

······························2分进而可得2A．····································································

·············4分故()2sin()2fxx，又1()02f，且02，得4．即()2sin()24fxx．·······························································

·······5分（2）依题意，()(1)2sin()24gxfxx，··············································6分()()()4sin()sin()4sin()sin[()]242424224Fxfxgxxx

xx4sin()cos()2sin()2cos24242xxxx．········8分由15[,]43x，得5[,]43x，由余弦函数图象可知，[,]4x时，余弦函数单调递

减；[]3x时，余弦函数单调递增．即()Fx在1[,1]4单调递减；在5[1,]3单调递增，·········································10分则min()(1)2FxF；····································

··································11分max151()max{(),()}max{2,1}()2434FxFFF．····························12分高一数学试题第8页（

共6页）22．（12分）我们知道，声音由物体的振动产生，以波的形式在一定的介质(如固体、液体、气体)中进行传播．在物理学中，声波在单位时间内作用在与其传递方向垂直的单位面积上的能量称为声强I（2W/cm）．但在实际生

活中，常用声音的声强级D（分贝dB）来度量．为了描述声强级D（dB）与声强I（2W/cm）之间的函数关系，经过多次测定，得到数据如下表：组别1234567声强I（2W/cm）11101121011310114101010①7910声

强级D（dB）1013.0114.7716.022040②现有以下三种模型供选择：DkIb，2DaIc，lgDmIn．（1）试根据第1—5组的数据选出你认为符合实际的函数模型，简单叙述理由，并根据第1组和第5组数据求出相应的解析式；（2）根据（1）中所求解析式，结合表中已知数据，求

出表格中①、②数据的值；（3）已知烟花的噪声分贝一般在(90,100)，其声强为1I；鞭炮的噪声分贝一般在(100,110)，其声强为2I；飞机起飞时发动机的噪声分别一般在(135,145)，其声强为3I，试判断13II与22I的大小关系

，并说明理由．【命题意图】本小题主要考查函数的基本性质、对数运算、不等式基本性质等基础知识，考查抽象概括能力、推理论证能力、运算求解能力、数据处理能力、应用意识与创新意识，考查函数与方程思想、数形结合思想，考查数学抽象、直观

想象、数学运算、数学建模、数据分析等核心素养，体现基础性、综合性、应用性与创新性．解析一：（1）选择lgDmIn．·······························································

··········1分由表格中的前四组数据可知，当自变量增加量为1110时，函数值的增加量不是同一个常数，所以不应该选择一次函数；··············································2分同时当自变量增加量为1110时，函数值的增加量从3.

01变为1.76，后又缩小为高一数学试题第9页（共6页）1.25，函数值的增加量越来越小，也不应该选择二次函数．故应选择lgDmIn．···································

································3分由已知可得：111010lg1020lg10mnmn，即10112010mnmn，解之得10

120mn，所以解析式为10lg120DI．··························································5分（2）由（1）知10lg120D

I，令10lg12040I，可得lg8I，810I，故①处应填810．·············6分由已知可得11310I时，10lg311012010lg31014.77D，所以lg30.477，·················

···························································7分又当7910I时，10lg95020lg350200.4775059.5

4D，故②处应填59.54．··········································································8分（3）设烟花噪声、鞭

炮噪声和飞机起飞时发动机噪声的声强级分别为123,,DDD，由已知190100D，2100110D，3135145D，故有1322DDD，································································

·······10分所以13210lg12010lg1202(10lg120)III，·······························11分因此132lglg2lgIII，即2132l

g()lgIII，所以2132III．··············12分解析二：（1）（2）同解法一．（3）设烟花噪声、鞭炮噪声和飞机起飞时发动机噪声的声强级分别为123,,DDD，由已知190100D，21

00110D，3135145D，因此19010lg120100I，所以3211010I，··································9分同理210010lg12011

0I，所以2121010I，·······························10分313510lg120145I，所以352231010I，·······································11分因此31221310

10II，42221010I，所以2132III．······················12分