【文档说明】四川省宜宾市南溪第一中学2023-2024学年高三上学期第13周周练数学（理）试题答案.docx，共(8)页，530.158 KB，由小赞的店铺上传

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南溪一中高2021级高三上期第13周周练考试数学（理科）一．选择题：题号123456789101112答案AD.BCDBADBABC二、填空题：13.9014.36515.526+16.212.C【解析】取AE的中点记为H，连DH，FH，由题意知AC⊥平面BD

EF，故ACDF⊥.由AE⊥平面DHF，可得AEDF⊥，所以DF⊥平面AEC，故①正确.由题意可知，DF的中点O为外接球的球心，故半径32R=，所以243SR==，故②正确.把ADF△沿AF折成与BAF△共面，连BD，则min()22BGBD+=+，

所以BGD△的周长的最小值为222++，故③错误.由对称性可知，BE⊥平面AFC，记BE平面AFCI=，连IG，则EGI为EG与平面AFC所成的角，sinEIEGIEG=，因为22333EIBE==，所以当G为中点时，EGI的正弦值最大为223的正弦值最大为2

23，故④正确.三．解答题：17.解：（1）因为()3sincosacbAA+=+，由正弦定理可得()sinsinsin3sincosACBAA+=+即()()sinsinsin3sincosAABBAA++=+即sinsincos3sinsi

nAABAB+=又因为sin0A所以3sincos1BB−=.........................................2分所以1sin62B−=.·················

·······················································································3分又因为()0,B所以5,666B

−−所以66B−=所以3B=··················································································

····································6分（2）因为3ABCS=△所以1sin32acB=得4ac=由余弦定理得：2222cos13acbacB+=+=.又()12BDBABC=+··························

··············································································8分所以()22221117||()2cos444BDBABCcaacB=+=++=···············

········································11分得2||17BD=|故BD的长为172·······························································

··········12分18.解：（1）列联如下表：则2600(24050160150)1200126.63539021040020091k−==所以能在犯错误的概率不超过0.01的前提

下认为性别与对新能源汽车的关注有关．（2）已知600人中男性与女性的比为2:1，故：所抽男性人数为2643=人，所抽女性人数为1623=人由题意知，X服从超几何分布，()()342361,2,3kkCCPX

kkC−===()124236115CCPXC===()214236325CCPXC===()304236135CCPXC===X的分布列为：X123P1535151311232555EX=++=比较关注不太关注总计男性2401

60400女性15050200总计39021060019.解：（1）因为四边形ABCD为正方形，所以ABCD∥，因为AB平面PCD，CD平面PCD，所以AB∥平面PCD，又因为AB平面PAB，平面PAB

平面PCDl=，所以ABl∥.···········································.4分（2）记BC，AD的中点分别为E，F，连接PE，EF，PF，因为ADEF⊥，ADPF⊥，EF，PF平面PEF，EFPFF=，所以AD⊥平面PEF

，所以ADPE⊥.因为3cos4PAB=，所以2PB=，又BCPE⊥，所以1PE=，所以222PEPFEF+=，所以PEPF⊥，PF，AD平面APD，ADPFF=，所以PE⊥平面PAD.以P为坐标原点，以PF，PE所在的直线为

x，z轴，过P与AD平行的直线为y轴，建立如图所示的空间直角坐标系.·········································································

··········································6分则()0,0,0P，()3,1,0A，()0,1,1B，()0,1,1C−，()0,1,1PB=，()3,1,0PA=，()0,1,1PC=−，·······························7分

设平面PAB的法向量为()111,,mxyz=，则11110,30,mPByzmPAxy=+==+=令11x=，得(1,3,3)m=−·····························

····························8分设平面PBC的法向量为()222,,nxyz=，n·PB=y2+22=0，则22220,0,nPByznPCyz=+==−+=令21x=，得(1,0,0)n=······

···························································8分（或者说明PF⊥平面PBC，可取平面PBC的法向量为()1,0,0n=）7cos,|

|||7mnmnmn==，······························································································11分由图可知，二面角ABPC−−为钝角，所以二面角ABPC−−的余弦值

为77−.············································································12分20．解：（1）由题意知()()12,0,,0AaAa−，又31,2M

，则12331,,1,22MAaMAa=−−−=−−()()2331124aa−−−+−=−，解得2a=由31,2M在椭圆C上及2a=得219144b+=，解得23b=．

椭圆C的方程为22143xy+=（2）由（1）知，右焦点为()1,0F据题意设直线l的方程为()()()112210,1,,1,xmymPmyyQmyy=+++则12121211223323232

2,22yyyykkmymymymy−−−−====于是由120kk+=得12122323022yymymy−−+=，化简得()121243yyyy=+（*）由221,34120xmyxy=++−=消去x整理得()2234690

mymy++−=()()222(6)363414410mmm=++=+△由根与系数的关系得：12122269,3434myyyymm+=−=−++代入（*）式得：2218363434mmm−=−++，解得2m=直线l的方程为210xy−−=方法一()21212391442172

0,,416yyyy=+=+=−=−△由求根公式与弦长公式得：21257201512164PQyy=+−==设点M到直线l的距离为d，则2312135251(2)d−−==+−1115359522458MPQSPQd===△．21．解：（1）()()22111(1)(1)1(1)x

xaaafxxxxx−−+=+−=−+++①当11a−−，即0a时，由()0fx得0x，由()0fx得10x−则()fx在()1,0−上为减函数，在()0,+上为增函数．②当110a−−即01a时由()0fx得11xa−−或0x，由()0fx

得10ax−则()fx在()()1,1,0,a−−+上为增函数，在()1,0a−上为减函数．③当10a−=即1a=时，()0fx恒成立，故()fx在()1,−+上为增函数．④当10a−即1a时由()0fx

得10x−或1xa−，由()0fx得01xa−则()fx在()()1,0,1,a−−+上为增函数，在()0,1a−上为减函数．综上：当0a时，()fx在()1,0−上为减函数，在()0,+上为增函数

；当01a时，()fx在()()1,1,0,a−−+上为增函数，在()1,0a−上为减函数；当1a=时，()fx在()1,−+上为增函数；当1a时，()fx在()()1,0,1,a−−+上单增，在

()0,1a−上单减．（2）由已知得()()()11ln1gxxax=+−++，故()11(1)11axagxxxx+−=−=−++①当1a−时，()()0,gxgx在()1,−+上单调递增，不存在两个零点．②当1a−时，令()0gx得xa，令()

0gx得1xa−故()gx在()1,a−上为减函数，在(),a+上为增函数．()()()min()11ln1gxgaaaa==+−++由()gx有两个零点得：即()()11ln10aaa+−++又1a−，故()ln11a+，解得1ae−又()010g=，且当x→+时，()g

x→+当1ae−时，函数()gx有两个零点．综上可知：a的取值范围为()1,e−+22．解：（1）由已知，圆C的标准方程为：22(2)(2)4xy−+−=．可将其化为：224440xyxy+−−+=将cos,sin.xy

==代入以上方程可得：圆C的极坐标方程为24cos4sin40−−+=（2）由已知，直线l的极坐标方程为()R=，则：2,4cos4sin40.=−−+=整理得24cos4sin40−−+=由0△得02

设()()12,,,MN，则12124sin4cos,4+=+=则()22222121212216OMON+=+=+−=216(sincos)816+−=，化简得：1sin22=由02知02得：26=，或526=．12

=，或512．23．解：（1）当2a=时，()222fxxx=+−①()2,2222xxx−−++无解②()202222xxx−++解得102x−③()0,2222xxx+−解得0x综上：原不等

式的解集为1,2−+（2）当()1,1x−时，()24fxxax=+−原不等式等价于：241xaxx+−+即3axx+，则()33xaxx−++()()130,130.axax++−−故()()()()130

,130,130,130.aaaa−++++−−−−−解得2a的取值范围为()2,2−．获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com