【文档说明】中学生标准学术能力诊断性测试2023-2024学年高三上学期11月测试 数学答案.pdf，共(9)页，538.327 KB，由管理员店铺上传

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第1页共8页中学生标准学术能力诊断性测试2023年11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．12345678CABBABDD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．9101112BDACABACD三、填空题：本题共4小题，每小题5分，共20分．13．114．7215．1024+16．35四、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）2sinsincoscBCbB=，所以由正弦定理得2sinsinsinsincosCBCBB=，1cos2B=，得3B=·············

·······························································3分()33tantan4tantan531tantan3134ABCABAB++=−+=−=−=−−−··············

···5分（2）ABC内切圆的面积为，所以内切圆半径1r=，由圆的切线性质得23,2333cabbca+−==+−=+·····························7分由余弦定理得222bcaac=+−，()()2222333caacacac+=+−=+−，将333ac+

=+代入，{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第2页共8页1843,sin23323ABCacSac=+==+·····················

·························10分（或()1643,2332ABCabcSrabc++=+=++=+······················10分）18．（12分）（1）过D作AB的垂线交AB于H点，设ACABa

==，则()22,2,2,212BCaBDaHDHBBDaAHBHBAa======−=−，22522ADAHDHa=+=−，由题意得，二面角CSAD−−的平面角为CAD，·········································2分()345222

cos17522DHCADDA+===−············································4分（2）分别以AB，AC，AS为x轴，y轴，z轴建立直角坐标系，()()0,0,0,,0,0ABa，()()0,,

0,0,0,CaSh，则()()()(),0,1,0,,1EahFah−−，SBSC=且,SESFSESFSBSC===·····················································6分又,SABSACBSACSA=，

那么SAESAF，则AEAF=···························································································

··8分故SEF与AEF都是等腰三角形，取EF的中点G，则SG与AG均垂直于EF，()()111111,,1,,,,,,1222222GaahSGaahAGaah−=−=−，平面AEF⊥平面

SBC等价于SGAG⊥·················10分()22221044aaSGAGhhh=+−−=，又260,,3SCBah===························12分19．（12分）（1）()1122,22

2nnnnaaaa++=+−=−，即()11222nnaa+−=−，{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第3页共8页2na−是公比为12的等比数列···············

················································2分111123,2322nnnnaa−−−==+······································

···············4分（2）211211112312612222nnnnSaaann−=+++=+++++=+−·················································

····························································6分111266,266222023nnnnSnSn−−=−−−=，

即126069,nn−取最小值14··································································8分（3）()11212,133nnnnnCCn

Cn−++==，得2n，即1nnCC+，有345CCC，又1234,3CCC===，故nC中最大项为23,CC·············································

··························10分又mb中最小值为()()2minmax7,3mnbC−，即24733−，()()3710−+，又70,3················

·································12分20．（12分）（1）由题意可知：X的所有可能取值为2.3，0.8，0.5，()1342.30.3245PX===····································

··································1分0.8X=包含的可能为“高低高”“低高高”“低低高”，()1141341140.80.5245245245PX==++=·················

·························2分()0.510.30.50.2PX==−−=·······················································

············3分X的分布列为：X2.30.80.5P0.30.50.2{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第4页共8页·······························

···············································································4分数学期望()1.19EX=································

················································5分（2）设升级后一件产品的利润为Y，Y的所有可能取值为2.3,0.8,0.5aaa−−−···············

······································6分()36142.3245103PYabb=−=+=+·························································7分

()1341141140.82452452450561PYabbbb=−=−+++−=−·············8分()635610.5110105bbPYa+−=−=−−=···········

·············································9分()()()()()635612.30.80.51.190.910105bbEYaaaba+−=−+−

+−=+−·····························································································

···············11分()()0.90EYEXba−，即：()1100,0.429aba（备注：12不写出不扣分）·······························12分21．（12分）（1

）1142ABAFBF++=，即221142AFBFAFBF+++=，又12122,442,2AFAFBFBFaaa+=+===·······························2分又2e=,1,12ccba===，故椭圆E的方程是2212xy+=···············

·······················································4分（2）依题意知直线BC的斜率存在，设直线:BCykxm=+，代入2212xy+=，得()2212xkxm++=，即22212102kxkmxm

+++−=①，{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第5页共8页()()22222124122402kmkmmk=−+−=−++，即22210km−+②，设()(

)1122,,,BxyCxy，则()22,Axy−，122212kmxxk−+=+③，2122112mxxk−=+④·····················································6分2,,AFB三点共

线，()21,0F，直线AB不与坐标轴垂直，()()()()12211212,1111yyxkxmxkxmxx−=−+=−−+−−，()()121212220kxxmxxkxxm++−+−=，()22222221222

0111222kmkmkmmkkk−−+−=+++，2222222220kmkkmkmmkm−−+−−=，2,mk=−直线():2BCykx=−，由②得：22212410,2kkk−+，2121BCkx

x=+−，点()11,0F−到直线:20BCkxyk−−=的距离231kdk=+，1121322FBCSBCdkxx==−·································································8分(

)()222212121222441441122kkxxxxxxkk−−=+−=−++{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第6页共8页()422222221164412421

122kkkkkk−−+−==++，()()()12222243012FBCkkSkk−=+，设2211kt+=，则212tk−=，()()12222123211333231FBCttttS

tttt−−−+−===−+−21313248t=−−+···································································10分所以当13

4t=时，即22441,21,336tkk=+==（符合题意），1FBCS的最大值为324，所以当66k=时，1FBC的面积取最大值为324·······································································

·····································12分22．（12分）（1）定义域为()(),11,−−+，由题意知()()2ln1xxxax−−=，则()()1ln1xxa−−=有三个不同的实数根，当1x时，

令()()()1ln1gxxx=−−，()()()ln11,gxxgx=−+在()1,x+上单调递增，又110eg+=，()gx在11,1ex+上单调递减，在11,ex++上单调递增，()111eegxg+=−··

············································································2分{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAA

oGxAAIMAIBgRFABAA=}#}第7页共8页当1x−时，令()()()1ln1hxxx=−−−，()()()()()221123ln1,1111xxhxxhxxxxx−+=−−+=+=++++，又()30h−=，()hx在(),3−

−上单调递减，在()3,1−−上单调递增，()()32ln20hxh−=+，()hx在(),1−−上单调递增·····································································5

分又()20h−=，当1x−→−时，()hx→+，当x→−时，()hx→−，当1x+→时，()0gx−→，当x→+时，()gx→+，10ea−···········································

················································6分（2）由已知可知()()1ln1axxx−−=有3个零点123,,xxx，不妨设123xxx，显然121x−−，由（

1）中函数()gx性质且11e11e11eea−−++，231112exx+，只需证2322exx++，即3222exx+−····················································8分又

()()()222222122211ln11ln10eeegxgxxxxx+−−=+−+−−−−，上面不等式证明如下：令()()()2211ln11ln1,1,1eeexxxxxx

=+−+−−−−+，{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第8页共8页()()21ln1ln12,1,1

eexxxx=−+−−−−+()2ln1120exx=+−−−，()x在11,1ex+上单调递增，()110ex+=····································

·······································10分又221211,21eeexx−−−+−+，()22121egxgx+−，又显然有()()23gxgx，23121exx+−

，2322exx++，1232exxx++··················································································12分{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAo

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