【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()345222cos17522DHCADDA+===−············································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()22221044aaSG
AGhhh=+−−=,又260,,3SCBah===························12分19.(12分)(1)()1122,222nnnnaaaa++=+−=−,即()11222nnaa+−=−,{#{QQABLYQUggCgQBBAAQhCA
wEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第3页共8页2na−是公比为12的等比数列·······························································2分111123,2322nnnna
a−−−==+·····················································4分(2)211211112312612222nnnnSaaann−=+++
=+++++=+−··········································································
···································6分111266,266222023nnnnSnSn−−=−−−=,即126069,nn−取最小值14················
··················································8分(3)()11212,133nnnnnCCnCn−++==,得2n,即1nnCC+,有345CCC,又1234,3
CCC===,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,()1342.30.3
245PX===······································································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()635610.5110105
bbPYa+−=−=−−=························································9分()()()()()635612.30.80.51.190.910105bbEY
aaaba+−=−+−+−=+−········································································································
····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221142A
FBFAFBF+++=,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,{#{QQABLYQU
ggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第5页共8页()()22222124122402kmkmmk=−+−=−++,即22210km−+②,设()()1122,,,BxyCx
y,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,()()()()12211212,1
111yyxkxmxkxmxx−=−+=−−+−−,()()121212220kxxmxxkxxm++−+−=,()222222212220111222kmkmkmmkkk−−+−=+++,2222222220km
kkmkmmkm−−+−−=,2,mk=−直线():2BCykx=−,由②得:22212410,2kkk−+,2121BCkxx=+−,点()11,0F−到直线:20BCkxyk−−=的距离231kdk=+,1121322FB
CSBCdkxx==−·································································8分()()222212121222441441122kkxxxxxxkk−−=+−=−++
{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}第6页共8页()422222221164412421122kkkkkk−−+−==++,()()()1222
2243012FBCkkSkk−=+,设2211kt+=,则212tk−=,()()12222123211333231FBCttttStttt−−−+−===−+−213132
48t=−−+···································································10分所以当134t=时,即22441,21,336tk
k=+==(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{#{QQABLYQUggCgQBBAAQhCAwEQ
CgAQkBACAAoGxAAIMAIBgRFABAA=}#}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12
1x−−,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,
1,1eexxxx=−+−−−−+()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{#{QQABLYQUggCgQBBAAQhCAwEQCgAQ
kBACAAoGxAAIMAIBgRFABAA=}#}获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com