山东省潍坊市高密市第三中学2023届高三上学期9月月考数学试题 word版含答案

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2020级高三9月数学学科模拟考试试题满分150分,考试时间120分钟2022.09.25一、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.已知集合301xAxx−=

+,{|ln(3)}Bxyx==−,则如图中阴影部分表示的集合为()A.[1,3]−B.(3,)+C.(,3]−D.[1,3)−2.已知n,m为正整数,且nm,则在下列各式中错误的是()A.36A120=;B.77712127ACA=;C.111CCC

mmmnnn++++=;D.CCmnmnn−=3.已知1.1log0.9a=,1.10.9b=,0.91.1c=,则a,b,c的大小关系为()A.abcB.acbC.bacD.bca4.若等比数列na的各

项均为正数,且12327aaa=,4283aa−=−,则12naaa的最大值为()A.9B.8C.3D.275.函数22()41xxxfx=−的图象大致为()A.B.C.D.6.函数2()1fxxax=−+在区间1,32上有零点,则实数a的取值范围是()A.(2,)+

B.102,3C.52,2D.[2,)+7.下列命题中,错误的命题有()A.五位同学站成一排合影,张三站在最右边,李四、王五相邻,则不同的站法种数为12B.命题“0[0,1]x,2001xx+”的否定为“[0,1]x,21xx

+”C.设函数22,0()2,0xxxfxx+=,则()fx在R上单调递增D.设x,yR,则“xy”是“2()0xyy−”的必要不充分条件8.设函数()fx的定义域为R,(1)fx+为奇函数,(2)fx+为偶函数,当[1,2]x时,()2xfxab=+.若(0)(3

)6ff+=,则()2log96f的值是()A.-12B.-2C.2D.12二、多项选择题:本题共4小题,每小题5分,共20分。在每小题给出的选项中,有多项符合题目要求的。全部选对的得5分,部分选对的得2分,有选错的得0分.9.若110ab,则下列结论中正确的是()A.

22abB.2abbC.||||||abab++D.33ab10.已知函数()fx是定义在R上的奇函数,当0x时,()(1)xfxex=+,则下列命题正确的是()A.当0x时,()(1)xfxex−=−B.函数()fx有

2个零点C.()0fx的解集为(,1)(0,1)−−D.1x,2xR,都有()()122fxfx−11.已知函数()lnln(2)1fxxx=+−+,则()A.()fx有一个极值点B.()fx没有零

点C.直线yx=是曲线()yfx=的切线D.曲线()yfx=关于直线1x=对称12.英国数学家贝叶斯在概率论研究方面成就显著,根据贝叶斯统计理论,随机事件A、B存在如下关系:()(|)(|)()PAPBAPABPB=.王同学连续两天在某高校的甲、乙两家餐厅就餐,王同学第一天去甲、乙两家餐厅

就餐的概率分别为0.4和0.6.如果他第一天去甲餐厅,那么第二天去甲餐厅的概率为0.6;如果第一天去乙餐厅,那么第二天去甲餐厅的概率为0.5,则王同学()A.第二天去甲餐厅的概率为0.54B.第二天去乙餐厅的概率为0.44C.第二

天去了甲餐厅,则第一天去乙餐厅的概率为59D.第二天去了乙餐厅,则第一天去甲餐厅的概率为49三、填空题:本题共4小题,每小题5分,共20分。13.若函数()fx的定义域为[1,1]−,则(lg)fx的定义域为__

__________.14.若212nxx−的展开式中二项式系数的和为64,则该展开式中的常数项是____________.15.数据:1,2,2,3,4,5,6,6,7,8,中位数为m,60%百分位数为a,则m

=__________,a=__________.16.已知正实数a,b满足abab=+,则2ab+的最小值为____________.四、解答题:本题共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)已知等差数列na的前n项

的和为nS,2320aS+=,514a=.(1)求na的通项公式;(2)求数列11nnaa+的前n项和nT.并证明16nT.18.(本小题满分12分)某校为了缓解高三学子复习压力,举行“趣味数学”闯关活动,规定每人从10道题中至少随机抽3道回答,至少答对

2题即可闯过第一关,某班有5位同学参加闯关活动,假设每位同学都能答对10道題中的6道题,且每位同学能否闯过第一关相互独立.(1)求B同学闯过第一关的概率;(2)求这5位同学闯过第一关的人数X的分布列和数学期望.19.(本小题满分12分)已知奇函数21(

)21xxafx−=+的定义域为[2,]ab−−(1)求实数a,b的值;(2)当[1,2]x时,2()20xmfx++恒成立,求m的取值范围.20.(本小题满分12分)已知数列na的前n项和为nS,且满足11a=,12nnSna+=,*nN.

(1)求数列na的通项公式;(2)设数列nb满足11b=,12nnnbb+=,*nN,按照如下规律构造新数列nc:1a,2b,3a,4b,5a,6b,7a,8b,…,求数列nc的前2n项和.21.(本小题

满分12分)为了检测某种抗病毒疫苗的免疫效果,需要进行动物与人体试验.研究人员将疫苗注射到200只小白鼠体内,一段时间后测量小白鼠的某项指标值,按[0,20),[20,40),[40,60),[60,80)

分组,绘制频率分布直方图如图所示.试验发现小白鼠体内产生抗体的共有160只,其中该项指标值不小于60的有110只.假设小白鼠注射痕苗后是否产生抗体相互独立.(1)填写下面的22列联表,并根据列联表及0.05=的独立性检验,判断能否认为注射疫苗后小白鼠产生抗体与指标值不小于60

有关.单位:只抗体指标值合计小于60不小于60有抗体没有抗体合计(2)为检验疫苗二次接种的免疫抗体性,对第一次注射疫苗后没有产生抗体的40只小白鼠进行第二次注射疫苗,结果又有20只小白鼠产生抗体.(ⅰ)用频率估计概率,求一只小白鼠注射2次疫苗后产生抗体的概率p;(ⅱ)

以(ⅰ)中确定的概率p作为人体注射2次疫苗后产生抗体的概率,进行人体接种试验,记n个人注射2次疫苗后产生抗体的数量为随机变量X.试验后统计数据显示,当99X=时,()PX取最大值,求参加人体接种试验的人数n及()E

X.参考公式:22()()()()()nadbcabcdacbd−=++++(其中nabcd=+++为样本容量)参考数据:()20Pk0.500.400.250.150.1000.0500.0250k0.45

50.7081.3232.0722.7063.8415.02422.(本小题满分12分)已知函数()(2)ln(2)fxxx=++,2g()(3)2(1)xxaxa=+−+−(aR).(1)求函数()fx的极值;(2)若不等式()g()fxx在(2,)x−+上恒成立,求的取值

范围;(3)证明不等式:1323111111114444ne++++(*nN).2020级高三9月数学学科模拟考试试题答案及评分标准1.C2.C3.A4.D5.A6.B7.C8.B9.AC10.A

CD11.AD12.AC13.[10,100]14.423−15.4016.1017.解:(1)设na的公差为d,由题意得114420414adad+=+=,解得123ad==.所以1(1)31naan

dn=+−=−.···························································································5分(2)令11nnnbaa+=,则

1111(31)(32)33132nbnnnn==−−+−+所以1231111111111132558313232326nnTbbbbnnn=++++=−+−++−=−−++·······10分18.解:(1)B同学闯过第一关的情况有答对2题和答

对3题,故B同学闯过第一关的概率321664310CCC2C3P+==································································································

·········2分(2)由题意可知X的所有可能取值为0,1,2,3,4,5,且X服从二项分布,即2~5,3XB.·····3分511(0)3243PX===,4152110(1)C33243PX===,23252140(

2)C33243PX===,32352180(3)C33243PX===,4452180(4)C33243PX===,5232(5)3243PX===.········

············································9分故X的分布列为X012345P12431024340243802438024332243··························································

·········································································11分所以1104080803210()0123452432432432432432433EX=++++

+=或210()533EX==.·····················································································

················12分(1)因为函数f(x)-4-2*-1是奇函数,所以f(-x)=-f(x),19.解:(1)因为函数21()21xxafx−=+是奇函数,所以()()fxfx−=−,即21212121xxx

xaa−−−−=−++,即2212121xxxxaa−−+=++,即221xxaa−=−+,整理得()(1)210xa−+=,所以10a−=,即1a=,·····························································4分则23a

−−=−,因为定义域为[2,]ab−−关于原点对称,所以3b=;·······································6分(2)因为[1,2]x,所以21()021xxfx−=+,又当[1,2]x

时,2()20xmfx++恒成立,所以()()222121xxxm++−−,[1,2]x时恒成立,令21xt=−,则2(3)(2)5665ttttmtttt++++−==++,[1,3]t时恒成立

,而66525265tttt+++=+,当且仅当6tt=,即6t=时,等号成立,所以265m−−,即m的取值范围是(265,)−−+.·····························

··························12分20..解:(1)当1n=时,由11a=且12nnSna+=得22a=·························································1分当2n时,由12(1)nn

Sna−=−得12(1)nnnanana+=−−,所以11nnaann+=+(2n).·················2分所以212naan==,nan=(2n),························

··························································3分又当1n=时,11a=,适合上式.·····································

····················································4分所以.nan=.*Nn··········································

································································5分(2)因为12nnnbb+=,1122nnnbb+++=,所以22nnbb+=(*nN),·································

··········6分又122bb=,所以22b=.····································································································7分所以数列nb的偶数项构成以2

2b=为首项、2为公比的等比数列.·············································8分故数列nc的前2n项的和()()21321242nnnTaaabbb−=+++++++,(

)122212(121)22212nnnnnTn+−+−=+=+−−······································································11分所以数列nc的前2n项和为1222nn++−.···

·······································································12分21.(1)由频率分布直方图,知200只小白鼠按指标值分布为:在[0,20)内有0.00252020010=(只);在[20,40)内有0

.006252020025=(只);在[40,60)内有0.008752020035=(只);在[60,80)内有0.02520200100=(只);在[80,100]内有0.00752020030=(只).························

··············································1分由题意,有抗体且指标值小于60的有50只:而指标值小于60的小白鼠共有10253570++=只,所以指标值小于60且没有抗体的小白鼠有20只,同理,指

标值不小于60且没有抗体的小白鼠有20只,故列联表如下:抗体指标值合计小于60不小于60有抗体50110160没有抗体202040合计70130200·················································

····················································································3分零假设为0H:注射疫苗后小白鼠产生抗体与指标值不小60无关联.根据列联表中数据,得22

0.05200(502020110)4.9453.8411604070130x−==.···························5分根据0.05=的独立性检验,推断0H不成立,即认为注射疫苗后小白鼠产生抗体与指标值不小于60

有关,此推断犯错误的概率不大于0.05.··························································································6分(2)(ⅰ)令事件A=“小白鼠第一次注射疫苗产生抗体”,事件B=

“小白鼠第二次注射疫苗产生抗体”,事件C=“小白鼠注射2次疫苗后产生抗体”.记事件A,B,C发生的概率分别为()PA,()PB,()PC,则160()0.8200PA==,20()0.540PB==,()1()()10.20.50.9PCPAPB=−=−=.所以一只小白鼠

注射2次疫苗后产生抗体的概率为0.9.·····························································8分(ⅱ)由题意,知随机变量~(,0.9)XBn,()C0.90.1kknknPXk−

==(0k=,1,2,…,n).因为(99)PX=最大,所以999999989898999999100100100C0.90.1C0.90.1C0.90.1C0.90.1nnnnnnnn−−−−,解得11091109n,因n是整数,故109n=或110

n=,所以接受接种试验的人数为109或110.···············································································10分①当接种人数为109时,()1090.998

.1EXnp===;②当接种人数为110时,()1100.999EXnp===.···························································12分22.(1)解:∵()(2)ln(2)fxxx=++(2)x

−),()ln(2)1fxx=++,由()0fx可得12,ex−+,此时()fx是增函数,由()0fx可得1,2ex−−,此时()fx是减函数,·············

···········································2分所以当12ex=−时()fx有极小值,极小值为1e−,无极大值··················································

··3分(2)解:由不等式()g()fxx在(2,)x−+上恒成立得2(2)ln(2)(3)2(1)xxxaxa+++−+−即(2)ln(2)(2)(1)xxxxa++++−,因为(2,)x−+,所以1ln(2)axx+−+在(2

,)x−+上恒成立····························································································································

·········5分设()1ln(2)hxxx=+−+,(2,)x−+,由1()02xhxx+==+得1x=−,所以()hx在(2,1)−−上递减,在(1,)−+上递增,所以min()(1)0hxh=−=即0a,所以a的取值范围为(

,0]−·······························································································7分(3)证明:由(2)得1ln(2)xx++在(1,)−+上恒成立,令

114nx=−,则有11ln144nn+,················································································8分所以22

11111111ln1ln1ln1144444434nnn+++++++++=−即211111ln111144434nn+++−

·····································································10分因为*nN,所以111114343n−即21111l

n1114443n+++,所以1323111111114444ne++++.······················································

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