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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.111CCCmmmnnn++++=；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：()(|)(|)()PAPBA

PABPB=.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，2320a

S+=，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()EX.参考公式：22()()()()()nadbcabcd

acbd−=++++（其中nabcd=+++为样本容量）参考数据：()20Pk0.500.400.250.150.1000.0500.0250k0.4550.7081.3232.0722.7063.8415.02422.（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）.2

020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.ACD11.AD12.AC13.[10,100]14.423−15.4016.1017.解：（1）设na的公差为d，由题意得11

4420414adad+=+=，解得123ad==.所以1(1)31naandn=+−=−.·················································

··········································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)C33243

PX===，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()533E

X==.·····································································································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21212121xxxxaa−−−−=−++，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)nnSna−=−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22b=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220.05200(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()()10.20.50.9P

CPAPB=−=−=.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−==（0

k=，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11

0n=，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===.···························································1

2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2211111111ln1ln1ln1144444434nnn+++++++++=−即211111ln111144434nn+++−·························

············································10分因为*nN，所以111114343n−即21111ln1114443n+++，所以13231

11111114444ne++++.····································································12分获得更多资源请扫码加入

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