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2022—2023学年度上学期常德市高三检测考试数学（试题卷）本试卷满分150分，考试时间120分钟注意事项：1．所有试题的答案请在答题卡的指定区域内作答．2．考试结束后，只交答题卡．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2

{|320}Axxx=−+，{|22}Bxx=−，则AB=A.[02]，B.[12]，C.[22]−，D.2.已知复数zi=−3，则复数zzi−2在复平面内对应的点所在的象限为A．第一象限B.第二象限C.第三象限D.第四象限3.已知向量,ab满足ab2=，且b(3

,4)=−，则向量a在向量b上的投影向量为A.(,)−6855B.(,)−6855C.(,)−682525D.(,)−6825254.沙漏是我国古代的一种计时工具，是用两个完全相同的圆锥顶对顶叠放在一起组成的（如图）．在一个圆锥中装满沙子，放在上方，沙子就从顶

点处漏到另一个圆锥中，假定沙子漏下来的速度是恒定的．已知一个沙漏中沙子全部从一个圆锥中漏到另一个圆锥中需用时80分钟．设经过t分钟沙漏上方圆锥中的沙子的高度与下方圆锥中的沙子的高度恰好相等（假定沙堆的底面是水平的），则t的值为A.10B.20C.60D.705.在平面直角坐标系中，已知点(3,4)

P为角终边上的点，则cos2cos+=A.825B.1325C.2225D.27256.在平面直角坐标系中，已知直线4390xy+−=与圆2220Cxxya−++=：相交的弦长为42，则a=A.8−B.2−C.2D.87.已知lna=22，lnb=33，ec

=22，则A.abcB.bcaC.bacD.acb8.已知双曲线2222:1(0,0)xyCabab−=的左右焦点分别为1F、2F，过1F的直线与曲线C的左右两支分别交于点M、N，且12||:||:||1:2:3FMFNMN=，则曲线C的离心率为A.2B.3

33C.223D.113二、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，部分选对的得2分，有选错的得0分．9.已知抛物线2:4Cyx=，O为坐标原点，点P为

直线2x=−上一点，过点P作抛物线C的两条切线，切点分别为A，B，则A.抛物线的焦点坐标为(0，1)B.抛物线的准线方程为1x=−C.直线AB一定过抛物线的焦点D.OPAB⊥10.已知定义在R上的函数()fx满足0)()27(=++xfxf，且)4

7(−=xfy为奇函数，则下列说法一定正确的是A.函数()fx的周期为27B.函数()fx的图像关于)0,47(−对称C.函数()fx为偶函数D.函数()fx的图像关于47=x对称11．下列说法正确的是A.数据6，5，3，4，2，7，8，9的上四分位数为7B.若2~(,)N

，且函数()(2)fxPxx=+为偶函数，则1=C.若随机事件A，B满足：()()1PABPA+=，则A，B相互独立D.已知采用分层抽样得到的样本数据由两部分组成，第一部分样本数据()1,2,,ixin=的平均数为x，方差为2xs；第二部分样本数据()1,2,,iyin=的

平均数为y，方差为2ys，若总的样本方差为2222xySSS+=，则xy=12.如图，已知正方体1111ABCDABCD−的棱长为2，,EF分别是棱1,BCCC的中点，P是侧面11BCCB内(含边界)的动点，则下列说法正确的是A.若直线1AP与平面AEF平行，则三棱锥P－AEF的体积为23B.若直

线1AP与平面AEF平行，则直线A1B1上存在唯一的点Q，使得DQ与A1P始终垂直C.若15AP=，则EP的最小值为51−D.若15AP=，则11APBC的最大值为42三、填空题：本题共4小题，每小题5分，共20分.13.已知函数()lnxfx

eaxa−=−−1，若曲线()yfx=在点(,())f11处的切线与直线012=−+yx垂直，则切线的方程为_____________．14.1631(1)(2)xyx−−+的展开式中的常数项为_____________．15.若函数

()2sin()(0)6fxx=−在(0,)3内存在唯一极值点，且在2[,]23ππ上单调递减，则的取值范围为_____________.16.已知数列{}na满足首项a=11，nnnanaan++=12，为奇数3，为偶数，则数列{}na的前2n项的和

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)已知数列na的首项a=138，且满足nnnaaa+=+1332．(1)求证：数列{}na−13是等比

数列；(2)若...naaaa++++1231111101，求满足条件的最大整数n的值．18．(本小题满分12分)如图，在梯形ABCD中，AD//BC，且2AD=，4BC=.(1)若3AB=，2CD=，求梯形ABCD的面积；(2)若2DB=，证明：ABC

为直角三角形．19．(本小题满分12分)如图所示的几何体是由等高的直三棱柱和半个圆柱组合而成，点G为DE的中点，DE为半个圆柱上底面的直径，且90BCF=，2CDCBCF===．H为BC的中点．(1)证明：平面DEH//平面GCF；(2)若Q是线段HE上一动点，求直线AQ

与平面GCF所成角的正弦的最大值．20．(本小题满分12分)常益长高铁的试运营，标志着我省迈入“市市通高铁”的新时代.常益长高铁全线长157公里，共设有常德站、汉寿站、益阳南站、宁乡西站、长沙西站5个车站.在试运营期间，铁路公司随机选取了乘坐常德开往长沙G65

75次复兴号列车的200名乘客，记录了他们的乘车情况，得到下表（单位：人）：上车站下车站汉寿站益阳南站宁乡西站长沙西站总计常德站1020104080汉寿站10102040益阳南站104050H第19题图ABCD第18题图宁乡西站3030总计103030130200(用频率代替概率)(1)从

这200名乘客中任选一人，求该乘客仅乘坐一站的概率；(2)在试营运期间，从常德上车的乘客中任选3人，设这3人到长沙下车的人数为X，求X的分布列，及其期望；(3)已知德山经开区的居民到常德站乘车的概率为0.6，到汉寿站乘车的概率为0.4，若经过益阳南站后高铁上有一位来自德山经开区

的乘客，求该乘客到长沙下车的概率.21．(本小题满分12分)已知点(2,1)P为椭圆:()xyCabab+=222210上的一点，椭圆C的离心率为32.(1)求椭圆C的方程；(2)如图，过点P作直线l1、l2，分

别交椭圆于另一点M、R，直线l1，l2交直线l：x=3于N，S，设直线l1，l2的斜率分别为k1，k2，且k1+k2=0，若PMS面积是PRN面积的2倍，求直线l1的方程.22．(本小题满分12分)已知函数()lnfxxaxx=−+12.(1)讨论

函数()fx的零点个数；(2)证明：当*nN，...ln()()nnnn+++++++35721122436421.xyOPMNRSl1l2l第21题图2022-2023学年度上学期常德市高三检测考试数学（参考答案）一、

选择题：本题共8小题，每小题5分，共40分．题号12345678答案BADDAACB二、选择题：本题共4小题，每小题5分，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.题号9101112答案BDBCBCDABC三、填空题：本题共4小题，每小

题5分，共20分．13.20xy−=14.9615.5(2,]216.4344nn−−三、解答题：本大题共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)解：(1)nnnaaa+=+1332Q,nnaa+=+11213························

·············································2分()nnaa+−=−1121333，{}na−13是以−13为首项，23为公比的等比数列········································

················5分(2)由(1)可得：()nna=−+112333，(())...()nnnnnaaaa−−++++=+=+−−12312111112333311012313即：()nn+−2310203·····················

·····································································.8分令()()nfnn=+−231023，当n1时，()()()nfnfn+−=−1213033，()fn单

调递增；又()f330，()f340，满足不等式的最大整数n=33··············································································10分18．（本小题满分

12分）解：(1)在ABC中，由余弦定理得22227cos28ACBCABACBCAACBCAC+−+==····························1分在ACD中，由余弦定理得2222cos24ACADCDACCADACADAC+−==

·········································2分由BCACAD=有，22784ACACACAC+=，解得7AC=······················································3分277cos84ACBCAAC+

==，又(0,)BCA，3sin4BCA=············································4分梯形ABCD的面积1197sinsin224ABCACDSSSBCACBCAADACCAD=+=+=···········

·······························6分法二：(1)取BC的中点E，连AE，则BE=CE=2·························································1分AD=EC，AD//EC，四边形AECD为平行四边

形····················································2分AE=DC=2··············································

··························································3分在ABE中，2223cos24ABBCAEBABBC+−==又(0,)BCA，7sin4B=··················

·······························································4分梯形ABCD的面积3397sin224ABEAECDABESSSSABBEB=+===··············

··················6分(2)设B=，ACB=，则2D=，BAC=−−，2ACD=−−在ABC中，由正弦定理得sinsinACBCBBAC=，即4sinsin()AC=+①在ACD中，由正弦定理得sinsin

ACADDACD=，即2sin2sin(2)AC=+②·······························8分由①②得：sin22sin(2)sinsin()+=+······

·········································································9分化简得，cossin()sin(2)+=+又sin(2)sin[()]sin

cos()cossin()+=++=+++所以sincos()0+=·····························································

·······························11分又(0,),(0,)+所以2+=，2BAC=，ABC为直角三角形··························

······························12分法二：取BC的中点E，连AE，则BE=CE=2AD=EC，AD//EC，四边形AECD为平行四边形········································

············8分2AECDB==BBAE=··············································································

······················10分2AEBEEC===ECAEAC=2EACEABBECA+=+=2BAC=，ABC为直角三角形·················································

····························12分19．（本小题满分12分）(1)证明：取DE的中点M，连MG、MH···············································

····················1分////1MGADHCMGHC==且MHCG四边行为平行四边形················································································2分//MHCG，又MH

平面CGF//MHCGF平面································································································3分//DECF又，又DE平面CGF//DECG

F平面·································································································4分,,DEMHMMHDHED

EDHE=又平面平面//DHECGF平面平面·························································································5分(2)如图，以C为原点，CB为x轴，C

F为y轴，CD为z轴建立空间直角坐标系，则A(2,0,2)，C(0,0,0)，F(0,2,0)，G(―1,1,2)，H(1,0,0)，E(0,2,2)······································6分则(0,2,0),(1,1,2)CFCG==−,设面CGF的法向量(

,,)nxyz=2020yxyz=−++=令1z=得2,0xy==,即(2,0,1)n=···········································8分(,2,2),HQHE==−(1,2,22)AQAHHQ=+=−−−

·············9分设所求线面角为，则2222|2222|4sin5(1)4(22)5965−−+−==−−++−−+···········································11分所以当1

3=时，sin取得最大值为255·································································12分20．（本小题满分12分）解：(1)仅乘坐一站的乘客有10+10+10+30

=60人该乘客仅乘坐一站的概率600.3200p==······2分(2)从常德上车的乘客到长沙下车的概率401802p==·····················································3分故这3

人到长沙下车的人数1(3,)2XB，331()2kPxkC==·········································5分X0123P18383818········

················································································································7分13()322Ex==···················

················································································8分(3)记事件A：该乘客在过益阳南站后

到长沙站下车，记事件B1:该乘客在常德站上车，记事件B2:该乘客在汉寿站上车.12()0.6()0.4PBPB==1404(|)505pAB==，2202(|)303pAB==·····································

·································10分1122()()(|)()(|)PAPBPABPBPAB=+4256()0.60.45375PA=+=·····················································

······························12分(阅卷说明：①直接4256()0.60.45375PA=+=得结果的不扣分；②11()0.60.40.522PA=+=的本问给1分)21．（本小题满分12分）解：(1)由题可知abcaabc+

===+2222241132·················································································2分

解得：28a=，22b=椭圆C的方程为xy+=22182···················································································4分(2)记1k

k=，设11(,)Mxy，22(,)Rxy则直线l1：(12)ykxk=+−；直线l2：(12)ykxk=−++联立()xyykxk+==+−2218212消y得：222(41)8(12)(16164)0kxkkxkk++

−+−−=则211216164241Pkkxxxk−−==+，即21288241kkxk−−=+·······················································6分221284||1|2|141kPMkxkk+=+−=++又22||

1|2|1NPNkxk=+−=+2||84||41PMkPNk+=+·································································································

8分同理2||84||41PRkPSk−+=+······························································································9分1||||sin2PMSSP

MPSMPS=；1||||sin2PNRSPNPRNPR=||||2||||PMSPNRSPMPSSPNPR==；即||||2||||PMPRPNPS=················································

··············10分22848424141kkkk+−+=++，解得：16k=直线l1的方程为：11(1)63yx=+−即640xy−+=·······················································12分22

．（本小题满分12分）解：(1)由题意()lnfxxaxx=−+12的零点即为方程()fx=0的实数解，即：lnxaxx=+221·························································1分令ln()xgxx

x=+221，则(ln)()xxxgxx−−=321························································································2分令()lnh

xxxx=−−1，()lnhxx=−；当(,)x01时，()hx0，()hx单调递增；当(,)x+1时，()hx0，()hx单调递减.()()hxh=10··························································

·········································3分()gx0，()gx在(,)+0单调递减，又Qx→+，()gx→0；······························

······················································4分所以，当0a，ya=与函数()gx有一个交点，()fx有一个零点；当a0，ya=与函数()gx没有交点，()fx无零点

····················································5分(2)令()ln,[,)Fxxxxx=−++121，()()xFxxxx−−=−−=22221110··························

····················································6分()Fx在[,)+1单调递减，()()FxF=10ln()xxx−112················

·················································································8分ln()nnnnn+++12121················

·············································································9分...lnlnln...ln()nnnnn++++

+++++++35721234122436421123即...ln()()nnnn+++++++35721122436421·································

···························12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com