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第1页共4页第2页共4页中学生标准学术能力诊断性测试2024年1月测试数学试卷本试卷共150分，考试时间120分钟。一、单项选择题：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mR，集合2,1

,2,AmBaaA=−=，若CAB=，且C的所有元素和为12，则m=A．3−B．0C．1D．22．已知数列na满足1111,2nnnnnaaaaa++=−=，则na=A．1221n−+B．112n−C．221n+D．12

1n−3．复数z满足()2i1iz+=−（i为虚数单位），则z的共轭复数的虚部是A．3−B．1C．iD．i−4．在直三棱柱111ABCABC−中，所有棱长均为1，则点1A到平面1ABC的距离为A．217B．105C．216D．1045．设()201212nnn

xaaxaxax+=++++，若56aa=，则n=A．6B．7C．8D．96．若不等式22458174xxxx−++−+的解集为,ab，则ab+的值是A．5B．42C．6D．77．已知32ee,ln2,155ln

52abc===−，则A．abcB．bcaC．acbD．bac8．已知3311,0,344xyxyxy+−−=，则13xy+的最大值是A．15B．18C．20D．24二、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

的四个选项中，有多项符合题目要求．全部选对得5分，部分选对但不全得2分，有错选的得0分.9．设,,为互不重合的平面，,mn为互不重合的直线，则下列命题为真命题的是A．若,，则C．若,,mnmn，则B．若,mm=⊥，则,⊥⊥D．若,⊥⊥，则1

0．已知点P为双曲线22:14xCy−=上的任意一点，过点P作渐近线的垂线，垂足分别为,EF，则A．455PEPF+=C．1225PEPF=−B．45PEPF=D．PEFS的最大值为82511．直线1:0laxbyc++=和2:0l

bxcya++=将圆()()22:111Cxy−+−=分成长度相等的四段弧，则()()()222111abc−+−+−的取值可以是A．43B．2C．83D．312．已知()sin2sin22sin22+=+，且kk+Z，则()t

an2tan+++3tan的值可能为A．6−B．5−C．52D．8三、填空题：本题共4小题，每小题5分，共20分．13．设函数()fx的定义域为R，()fx为偶函数，()21fx+−为奇函数，当2,4x时，()fx=

2logaxb+，若()()064ff+=，则2ab+=．14．已知12,FF是椭圆()2222:10xyCabab+=的左、右焦点，P是C上一点，线段2PF的中垂线l过点1F，与椭圆C相交于,AB两点，且95ABa=，则椭圆C的离

心率为．15．已知函数()gx的图象与函数()exfxx=−的图象关于原点对称，动直线()0xaa=与函数()(),fxgx的图象分别交于点,AB，函数()fx的图象在A处的切线1l与函数()gx的图象在B处的切线2l相交于点C，则ABC面积的最小值是．16．对任意的xR，不等式()()

()2222714613817xxmxxxx−+−+−+恒成立，则实数m的取值范围为．{#{QQABDYKAogiAABAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第3页

共4页第4页共4页四、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．（10分）数列na的前n项和为1,1nSa=，当2n时，212nnnSaS=−．（1）求证：数列1nS是等差数列，并求nS的表达式；（2）设221nnnSbn=+，数

列nb的前n项和为nT，不等式23nTmmn−+对所有的n*N恒成立，求正整数m的最小值．18．（12分）如图所示，在ABC中，1,ABD=是BC上的点，12BADDAC=．（1）若2BAC=，求证：213ADAC−=；（2）若14BDDC=，求A

BC面积的最大值．19．（12分）如图所示，一只蚂蚁从正方体1111ABCDABCD−的顶点1A出发沿棱爬行，记蚂蚁从一个顶点到另一个顶点为一次爬行，每次爬行的方向是随机的，蚂蚁沿正方体上、下底面上的棱爬行的概率为16，沿正方体的侧棱爬行

的概率为23．（1）若蚂蚁爬行n次，求蚂蚁在下底面顶点的概率；（2）若蚂蚁爬行5次，记它在顶点C出现的次数为X，求X的分布列与数学期望．20．（12分）如图所示，已知ABC是以BC为斜边的等腰直角三角形，点M是边AB的中点，点N在边BC上，且3

BNNC=．以MN为折痕将BMN折起，使点B到达点D的位置，且平面DMC⊥平面ABC，连接,DADC．（1）若E是线段DM的中点，求证：NE平面DAC；（2）求二面角DACB−−的余弦值．21．（12分）如图所示，已知抛物线()21,0,1yxM=−，A，B是抛物线与x轴的交点，过点M作

斜率不为零的直线l与抛物线交于C，D两点，与x轴交于点Q，直线AC与直线BD交于点P．（1）求CMDMCD的取值范围；（2）问在平面内是否存在一定点T，使得TPTQ为定值？若存在，求出点T的坐标；若不存在，请说明理由．22．（12分）已知函数()2

1lnxfxxax−=+−有两个零点()1212,xxxx．（1）求实数a的取值范围；（2）求证：121xx；（3）求证：22221214xxaxx−−−．（第19题图）（第20题图）（第21题图）（第18题图）{#{QQABDYKAogiAA

BAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第1页共9页中学生标准学术能力诊断性测试2024年1月测试数学参考答案一、单项选择题：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．12345678ADBACCAC二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对但不全的得2分，有错选的得0分．9101112ABBCDCDACD三、填空题：本题共4小题，每小题5分，共20分

．13．414．51315．216．12−，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．（10分）（1）当2n时，数列na的前n项和为nS，满足212nnnSaS

=−，即()22111111222nnnnnnnnnSSSSSSSSS−−−=−−=−−+，整理可得112nnnnSSSS−−=−······················································

··················1分11S=，则21122SSSS=−，即2221SS=−，可得213S=·······························2分由23232SSSS=−，即332133SS=−，可得31,,5S=以此类推可知，对任意的,0nnS*N，在等式112

nnnnSSSS−−=−两边同时除以1nnSS−可得1112nnSS−−=·······················4分{#{QQABDYKAogiAABAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第2页共9页所以数列1nS为等差

数列，且其首项为111S=，公差为2·································5分()112121nnnS=+−=−，因此，121nSn=−·······················

·····················6分（2）解：()()211111112142121482121nnnSbnnnnn==+=+−+−+−+，1114821nnTn=+−+

············································································8分不等式23nTmmn−+对所有的n*N恒成立，则22303mm−+，即9576m+或9576m−·············

························································9分因此，满足条件的正整数m的最小值为3··································

····················10分18．（12分）（1）证明：由1,22BACBADDAC==，知,63BADDAC==，111,sinsin26232ABCABDACDSSSABADADACABAC=++=，即32ADADACA

C+=，两边同除以ADAC，得213ADAC−=······················································5分（2）设BAD=，则2DAC=，ABD中，由正弦

定理，得sinsinABBDBDA=①，ACD中，由正弦定理，得sinsin2ACDCCDA=②，②①，结合sinsin,4BDACDADCBD==，得2cosAC=····················7分321sin33sin4sinsin3

3tan4tansin2coscosABCSABAC−====−{#{QQABDYKAogiAABAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第3页共9页2322tan3tantan3tan4

tan1tan1tan−=−=++····································9分设()tan0,3t=，即求函数()323,0,31ttytt−=+的最大

值，()()()()()()()2222322222332333313211tttttttytt−−++−+−−==++，()20,233t−时，0y，函数单调递增；()2233,3t−时，0y，函数单调递减，当2233t

=−时，函数有最大值，max639y=−，ABC面积的最大值为639−····························································12分19．（12分）（1）记蚂

蚁爬行n次在底面ABCD的概率为nP，由题意可得，()11212,1333nnnPPPP+==+−····················································3分11

111,2322nnnPPP+−=−−−是等比数列，首项为16，公比为13−，11111111,263263nnnnPP−−−=−=+−··············

··········································5分（2）X=0,1,2，X=2时，蚂蚁第3次、第5次都在C处，()11212112122111112222663636636336666

18PX==++++=········································································

······································7分X=1时，蚂蚁第3次在C处或第5次在C处，设蚂蚁第3次在C处的概率为1P，1112121121151521122266363663666663318P

=++++=·····················································································

·························8分设蚂蚁第5次在C处的概率为2P，{#{QQABDYKAogiAABAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第4页共9页设蚂蚁不过点C且

第3次在1D的概率为3P，设蚂蚁不过点C且第3次在1B的概率为4P，设蚂蚁不过点C且第3次在A的概率为5P，由对称性知，34PP=，3111212134366636354P=+=，512122211663633327P=+=，得2351211

7222636654PPP=+=···················································11分()125127PXPP==+=，()()()41011254PXPXPX==−

=−==，X的分布列为：X012P4154527118X的数学期望()()()()800112227EXPXPXPX==+=+==············12分20．（12分）（1）过点E作AM的平行线交AD于点F，过点N作AB的平行线交AC于点G，

连接FG．因为点E是线段DM的中点，3BNNC=，12EFNGAM==，且EFNG，四边形EFGN是平行四边形．由,NEFGNE平面DAC，FG平面DAC，NE平面DAC·························································

·····························5分（2）解法1：以点A为原点，AB，AC所在的直线为x轴、y轴，过点A垂直于平面ABC的直线为z轴，建立空间直角坐标系··································

···································6分设2ABAC==，则()()130,0,0,,1,0,0,,,022AMN，设(),,,Dxyz，{#{QQABDYKAogiA

ABAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第5页共9页因为平面DMC⊥平面ABC，所以点D在平面ABC上的射影落在直线CM上，12yx+=①，由题意可知，()22231,2

,112DMDNxyz==−++=②，222139222xyz−+−+=③，由①②③解得，8221182211,,,,,777777xyzD==−=−···········

···············8分82211816211,,,,,777777ADCD=−=−，设平面ACD的法向量为(),,nxyz=，00ADnCDn==，即411048110xyzxyz−+=−+=，取11,0,4

xyz===−······················11分取平面ABC的法向量()0,0,1m=．设二面角DACB−−的平面角为，则43coscos,9mnmnmn===，所以，二面角DACB−−的余弦值为439···················

································12分解法2：如图，过点B作直线MN的垂线交于点I，交直线CM于点H．由题意知，点D在底面ABC上的射影在直线BI上且在直线MC上，所以点H即点D在底面上的射影

，即DH⊥平面ABC·····················································································6分设2AB=，则31,2,24BMB

NMBN===，由余弦定理，得102MN=，()1031010cos,sin,cos101010BMNBMNMIBMBMN=−==−=，{#{QQABDYKAogiAABAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第6页共9页()1053

102572coscos10510510IMHIMBHMB=−=+=，5cos7MIMHIMH==．过点H作AC的垂线交于点O，连接DO，由三垂线定理知，DOAC⊥，DOH是二面角DACB−−

的平面角········································································9分由AMCMHOCH=，解得228211,77HODHDMMH==−=，11tan4DHDOHHO==，得4

3cos9DOH=，所以，二面角DACB−−的余弦值为439··················································12分21．（12分）（1）设点()()1122,,,CxyDxy，设直线l的方程为()10yk

xk=+，代入抛物线21yx=−，得220xkx−−=（*），222122221211217221,28218CMDMkxkxkCDkkxxk+++===−++−+··········4分（2）()(

)2211221,1,,1,,0CxxDxxQk−−−，设(),Tmn，由（*）式，知1212,2xxkxx+==−·····························································

·5分直线AC的方程为()()111yxx=−+，直线BD的方程为()()211yxx=+−，解得()()121212122121212123,222xxxxxxxxxyxxxxxx+−−−−+===−+−+−+，所以点P的坐标为()1212212123,22xxxxxxx

x−−+−+−+···············································7分()12122121231,,,22xxxxTPmnTQmnxxxxk−−+=−−=−−−−+−+，()()1212

212123122xxxxTPTQmmnnxxkxx−−+=−−−+−−−+−+{#{QQABDYKAogiAABAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第7页共9页()()12221212212121231222x

xxxxxmmnnxxkkxxxx−−++=−−−+−−+−+−+222121121222knmmnnxxkxx−=−−+++−+−+2218xxk−=+，22221282kmnmTPTQmnnkk−+−=++++++··········

·····································10分当10,,2mnTPTQ==为定值54，所以存在定点T的坐标为10,2·····················································

·············12分22．（12分）（1）()()32221ln2ln2xxxfxxxx−+−+=+=····················································1分又因为函

数()()321lngxxx=−+递增，且()10g=，()01fxx，()fx在()0,1递减，在)1,+递增······················································

······2分当()120fa=−，即2a时，2211111lnln0faaaaaaaa=+−−=+，()()()()222211111ln10aaaaafaaaaaaaaaaa−−−+−−=+−−+−−=，()fx在()

1,1,1,aa上各有一个零点·························································3分当2a时，()fx的最小值为()1f，且()120fa=−，()fx在()0,+内至多只有一个零点，综上，实数a的取值范围

是2a··································································4分（2）设()()1,1Fxfxfxx=−，{#{QQABDYKAogiAABAAAB

gCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第8页共9页则()()()()22322111121lnxxFxfxfxxxxxx−−=+=−−+()()3323

21112ln221lnxxxxxxxxxxx+−=−−−=−−+当1x时，ln1xx−，()()()()33222112120xxxxxxxxx−−+−=+−=−++，()()()322111lnxxxxxxx−+−

+，()Fx在()1,+上递增，当1x时，()()10FxF=，即当1x时，()1fxfx······································································6分又因为函

数()fx有两个零点()1212,xxxx，由（1）知，122101,01xxx，()()1221fxfxfx=，又()fx在()0,1递减，121xx，即121xx················································

················································8分（3）设()()211lnxGxfxxaxxxx=−+−=−−，()()()232122212

1ln1ln21ln21xxxxxxxxGxxxxx−+++−−−+=−−==，()110G=，当1x时，()()()2121ln211xxGxxxxx−=+++−，显然()2ln2101xxxx+++−{#{QQABDYKAogiAABAAABgCAQ

FKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}第9页共9页()1Gx在()0,1递减，()1,+递增，()()1110GxG=，即()()11fxxahxx+−=，设()1hx的零点为()2343443,,4xxxxxxa−=−，由图象可知312

4xxxx，2214xxa−−··················································································10分设()22211ln1111lnxfxxaxx

xxxx−−+−=−=−−，设()211lnGxxx=−−，易得()20Gx恒成立，即()()2221fxxahxx+−=，设()2hx的零点为()222565665,,4x

xxxxxa−=−，由图象可知，1562xxxx，22221562xxxx，222214xxa−−，22221214xxaxx−−−·································

····································12分{#{QQABDYKAogiAABAAABgCAQFKCEOQkAGCCAoOxAAEsAAAgRNABCA=}#}获得更多资源请扫码加入享学资

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