【文档说明】中学生标准学术能力诊断性测试2024-2025学年高三上学期12月月考试题 数学 PDF版含答案.pdf，共(6)页，643.146 KB，由envi的店铺上传

转载请保留链接：https://www.doc5u.com/view-5be8de6a3297d3db48ae65f11ba3c7ff.html

以下为本文档部分文字说明：

第1页共4页第2页共4页标准学术能力诊断性测试2024年12月测试数学试卷本试卷共150分一、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．已知集合1,2,3,4

,2,2PQ==−，下列结论成立的是A．QPC．PQQ=B．PQP=D．2PQ=2．已知aR，i为虚数单位，若复数()()2iia++的实部与虚部相等，则a=A．3−B．2−C．2D．33．已知()sin2sinsin,tantan2+==−，则()tan

+=A．43−B．43C．2−D．24．斜率为1的直线l经过()1,0点，且与抛物线24yx=交于,AB两点，则AB=A．4B．42C．8D．825．已知()PA、()PB分别表示随机事件A、B发生的概

率，则()1PAB−是下列哪个事件的概率A．事件A、B同时发生B．事件A、B至少有一个发生C．事件A、B都不发生D．事件A、B至多有一个发生6．已知01a，设2log,log4aman==，则mn+的取值范围为A．(,2−−B．)2,0−C．(0,2D．)2,+7．

设()()23fxxx=−，若方程()()fxkk=R有3个不同的根,,abc，则abc的取值范围为A．()4,0−B．()2,0−C．()0,4D．()0,28．已知双曲线的左、右焦点为()()122,0,2,0FF−，的一条渐近线为yx=

，点P位于第一象限且在双曲线上，点M满足：121,FPMMPFMFMP=⊥，则12MFMF+的最大值为A．26B．42C．210+D．46二、多项选择题：本题共3小题，每小题6分，共18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6分，部分选对但不全得3分，有

错选的得0分.9．已知直线00:lxxyym+=与圆22:1Cxy+=，()00,Pxy，则下列判断正确的是A．若点P在圆C上，且直线l与圆C相切，则1m=B．若点P在圆C内，且1m=，则直线l与圆C相交C．若01,2my==，则直线l与圆C相交D．若0m=，则直线l截圆C所得

弦长为210．我们熟知的五面体有三棱柱、三棱台、四棱锥等．《九章算术》中将有三条棱互相平行且不全相等，有一个面为矩形的五面体称之为“刍甍”，对于“刍甍”下列判断正确的是A．三棱台体不是“刍甍”B．“刍甍”有且仅有

两个面为三角形C．存在有两个面为平行四边形的“刍甍”D．“刍甍”存在两个互相平行的面11．已知等差数列n的公差为，cosnnb=，数列nb的前n项和为nS，*nSSn=N，若存在1，使得,,Sabc=，则可能的取值为A

．3B．2C．23D．三、填空题：本题共3小题，每小题5分，共15分．12．如图所示，在正方形ABCD中，E是AB的中点，F在BC上且2CFFB=，AF与DE交于点M，则cosDMF=．13．已知()sinfxx=，记函数()yfx=在闭区间I上的最大值为IM．若正数k满足0

,,22kkkMM=，则k=．14．已知()()e,lnxfxaxgxxax=−=−，若对任意()10,x+，都存在()20,x+，使得()()1212fxgxxx=，则实数a的取值范围为．MFEDCBA（

第12题图）{#{QQABBQYEogiAABIAABgCEwXCCkEQkgCACSgOhFAAsAAByBNABAA=}#}{#{QQABDQKg5giQgAbACB6qE0XGC0sQkoAjLSgEAUAMKAwDCBNIB

AA=}#}第3页共4页第4页共4页四、解答题：本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15．（13分）设等差数列na的前n项和为nS，已知315S=−．（1）若17a=−，求na的通项公式；（2）若对

于任意*nN，都有7nSS，求公差d的取值范围．16．（15分）如图所示，在四棱锥PABCD−中，底面ABCD为平行四边形，点E为棱PD的中点．（1）设平面ABE与直线PC相交于点F，求证：EF平面ABCD；（2）若PD⊥平

面ABCD，2,60,22BCCDBCDPD====，求直线BE与平面PCD所成角的大小．17．（15分）某校高一学生共有500人，年级组长利用数字化学习软件记录每位学生每日课后作业完成的时长，期中考试之后统计得到了如下平均作业时长n

与学业成绩m的数据表：平均作业时长n（单位：小时）)1,1.5)1.5,2)2,2.5)2.5,3)3,3.5学业成绩优秀：90100m11437435学业成绩不优秀：090m136

137102187（1）试判断：是否有95%的把握认为学业成绩优秀与日均作业时长不小于2小时且小于3小时有关？（2）常用()()()PBALBAPBA=表示在事件A发生的条件下事件B发生的优势，在统计中称为似然比．已知该校高一学生

女生中成绩优秀的学生占比25%，现从所有高一学生中任选一人，A表示“选到的是男生”，B表示“选到的学生成绩优秀”，若()0.2LBA=，求()PA．附：()()()()()()222,3.8410.05nadbcPabcdacbd−=++++．18．（17分）设()1lnfxaxx=+．（

1）当1a=，求函数()yfx=的递减区间；（2）求证：函数()()()1ln2gxfxaxx=−−−的图象关于()1,0对称；（3）若当且仅当()0,1x时，()fxx，求实数a的取值范围．19．（17分）在直角坐标平面xOy内，对于向量(),mxy=，记mxy=+．设,,a

bc为直角坐标平面xOy内的向量，()1,1a=．（1）若()1,2b=−，求ab−；（2）设()1,1b=−−，若4cacb−+−=，求c的最大值；（3）若2,2bcbc===，求证：3332623bca−+−+．（第16题图）{#{QQ

ABBQYEogiAABIAABgCEwXCCkEQkgCACSgOhFAAsAAByBNABAA=}#}{#{QQABDQKg5giQgAbACB6qE0XGC0sQkoAjLSgEAUAMKAwDCBNIBAA=}#}第1页共4页标准学术能力诊断性测试2024年12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．12345678DDACCACA二、多项选择题：本题共3小题，每小题6分，共18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6分，部分选对但不全的得3分，有错选的得0分．91011CDABABC三、填空题：本题共3小题，每小题5分，共15分．12．210−13．5612或14．(0−，四、解答题：本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15．（13分）解：（

1）32315Sa==−，所以25a=−······································································2分因为17a=−，所以公差2d=················

······················································4分得()72129nann=−+−=−·········································

······························6分（2）因为对任意*nN，都有7nSS，所以8767,SSSS，得870,0aa····························

······························2分由（1）知25a=−，所以8276560,550aaddad=+=−+=−+·················5分得516d··································

····························································7分16．（15分）解：（1）因为ABCD，AB不在平面PCD内，所以AB平面PCD··························2分因

为平面ABE与平面PCD相交于EF，所以ABEF····································4分因为EF不在平面ABCD内，所以EF平面ABCD·····················

··················6分（2）取CD中点H，因为2,60BCCDBCD===，所以,3BHCDBH⊥=·······································································

·····1分{#{QQABBQYEogiAABIAABgCEwXCCkEQkgCACSgOhFAAsAAByBNABAA=}#}{#{QQABDQKg5giQgAbACB6qE0XGC0sQkoAjLSgEAUAMKAwDCBNIBAA=}#}

第2页共4页因为PD⊥平面ABCD，所以PDBH⊥··························3分得BH⊥平面PCD·······················································5分所以BEH即是直线BE与平面PCD所

成角·····················7分因为12,12DEPDDH===，所以3EH=，所以tan1BHBEHEH==，得45BEH=，所以直线BE与平面PCD所成角的大小为45············································

·····9分17．（15分）解：（1）22列联表数据如下：时长n23n其他总计优秀8020100不优秀120280400总计200300500········································

························2分()22500802802012083.3100400200300−=·············································

············4分因为23.841，所以有95%的把握认为学业成绩优秀与日均作业时长不小于2小时且小于3小时有关·······································

·····················································6分（2）设()PAx=，则()1PAx=−，因为()0.25PBA=，所以()()()0.250.251PBAPAx==−····················

····2分因为()()()0.2PBPBAPBA=+=，所以()0.250.05PABx=−···········4分因为()()()0.2PBALBAPBA==，所以()()0.2PBAPBA=，得()()0.2PBA

PBA=········································································5分因为()()()PAPBAPBAx=+=，所以()6xPAB=··················

··········7分由0.250.056xx−=，得0.6x=，所以()0.6PA=··········································9分18．（17分）解：（1）函数的定义域为()0,+········

······································································1分因为()211fxxx=−·····················

······························································3分H{#{QQABBQYEogiAABIAABgCEwXCCkEQkgCACSgOhFAAsAAByBNABAA=}#}{#{QQABDQKg5giQgA

bACB6qE0XGC0sQkoAjLSgEAUAMKAwDCBNIBAA=}#}第3页共4页当()0,1x时，()0fx；当()1,x+时，()0fx，所以函数()yfx=在区间(0,1上递减·····················

······································5分（2）由已知()()lnln2gxaxax=−−，定义域为()0,2·····································

·····1分设()0,2x，则()()()2ln2lngxaxaxgx−=−−=−，所以函数()ygx=的图象关于()1,0对称························································3分（3）设()1lnhxa

xxx=+−，则()211ahxxx=−−，①当2a时，因为10,2xxx+，所以()110hxaxxx=−+···············2分得()yhx=在()0,1上递减，因为()10h=，所以当且仅当()0,1

x时，()0hx，即()fxx····················4分②当()2,0,1ax=时，()1120hxxxx=−+，所以()yhx=在()0,1上递减，

因为()10h=，所以当且仅当()0,1x时，()0hx，即()fxx····················6分③当2a时，令()0hx=，得242aax−=，()240,12aa−−，当24,12aax−−时，()0hx，函数()yhx=在24,12aa

−−上递增，所以当204,12aax−−时，()()010hxh=，即()00fxx，得2a不满足题意·································································

····················8分综上所述，满足题意的实数a的范围为(,2−·················································9分19．（17分）解：（1）()2,1ab−=−，所以3ab−=····················

···············································3分（2）设(),cxy=，则11114xxyy−+++−++=············································2分因为()()11112xxxx−++−−+

=，当11x−时取等，因为()()11112yyyy−++−−+=，当11y−时取等，11114xxyy−+++−++=等价于11x−且11y−··························4分得2222cxy=+，c的最大值为2，当1

x=且1y=时取得···················6分{#{QQABBQYEogiAABIAABgCEwXCCkEQkgCACSgOhFAAsAAByBNABAA=}#}{#{QQABDQKg5giQgAbACB

6qE0XGC0sQkoAjLSgEAUAMKAwDCBNIBAA=}#}第4页共4页（3）由2,2bcbc===知，可设2cos,2sin,2cos,2sin6666bc=++=−−，11323c

os,sin22bca+−=−−，11323cossin22bca+−=−+−··················································2分设()11co

ssin22g=−+−，则()yg=以2为周期，当36−时，()31cossin2cos,2422g=−=+−，当63时，()31cossin12s

in1,21422g=+−=+−−−，当36时，()31sincos2sin,2422g=−=−−，当63时，()311cossin12sin,21422g=−−=

−+++，综上所述，当5,33−时，()31,2122g−+································6分因为()yg=以2为周期，所以当R时，()31,2122g−+

，得()32333,2623bcag+−=−+，所以3332623bca−+−+························································8分{#{QQABBQYEogiAABIAA

BgCEwXCCkEQkgCACSgOhFAAsAAByBNABAA=}#}{#{QQABDQKg5giQgAbACB6qE0XGC0sQkoAjLSgEAUAMKAwDCBNIBAA=}#}