六五文档>基础教育>试卷>2023年11月诊断性测试数学答案
2023年11月诊断性测试数学答案
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中学生标准学术能力诊断测试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.7.14.131210+215.16.354四、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)2csinBsinC2sinCBCsinsin(1)=,所以由正弦定理得=,bBcossinBBcos1=cosB,得B=············································································3分233+3tanAB+tan4tanCAB=−tan(+)=−=−=−53·················5分1−tanABtan313−4(2)ABC内切圆的面积为,所以内切圆半径r=1,由圆的切线性质得c+a−b=23,b=c+a−23=3+3·····························7分由余弦定理得b2=c2+a2−ac,22(3+3)=c22+a−ac=(a+c)−3ac,将ac+=3+33代入,第1页共8页{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}1=+==+acSac843,sin233··············································10分ABC231(或++=+=++=+abcSrabc643,233()······················10分)ABC218.(12分)(1)过D作AB的垂线交AB于H点,设ACA==Ba,2则BC=2a,BD=2a,HD=HB=BD=2a,AH=BH−BA=2−1a,2()ADAHDHa=+=−22522,由题意得,二面角C−−SAD的平面角为CAD,·········································2分DH234522(+)===cosCAD············································4分DA522−17(2)分别以AB,AC,AS为x轴,y轴,z轴建立直角坐标系,A(0,0,0),B(a,0,0),C(0,a,0),S(0,0,h),则E(a,0,(1−−)h),F(0,a,(1)h),SESFSB=SC且==,SE=SF·····················································6分SBSC又SABSAC,BSA=CSA,那么SAESAF,则AE=AF·····························································································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,111111Ga,a,(1−)h,SG=a,a,−h,AG=a,a,(1−)h,222222平面AEF⊥平面SBC等价于SG⊥AG·················10分aa2222SGAG=+−h(10−h)h=,442又SCB=60,a=h,=························12分319.(12分)1(1)2an++11=an+2,2(an−2)=an−2,即aa−22=(−),nn+12第2页共8页{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}1−a2是公比为的等比数列·······························································2分n211nn−−11·····················································4分aann−==+23,232221nn−1111(2)Saaannnn=+++=+++++=+−122312612222·············································································································6分111nn,−−=SnSnnn266,266−−−=222023即2n6−0169,n取最小值14··································································8分n−12C21(n+)(3)n+1,得,Cnn==,1n233Cnn4即CC,有CCC,又CCC=,==,nn+13451233故Cn中最大项为CC23,·······································································10分27247又bm中最小值为,(bC)−(),即−,mnminmax3337(3−7)(+1)0,又0,·················································12分320.(12分)(1)由题意可知:X的所有可能取值为2.3,0.8,0.5,134PX(=2.3)==0.3······································································1分245X=0.8包含的可能为“高低高”“低高高”“低低高”,114134114PX(=0.8)=++=0.5··········································2分245245245PX(=0.5)=1−0.3−0.5=0.2···································································3分X的分布列为:X2.30.80.5P0.30.50.2第3页共8页{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}··············································································································4分数学期望EX()=1.19················································································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−−−aaa.8,0.5·····················································6分1436b+3PYa(=−=+=2.3)b·························································7分2451013411411456−bPYabbb(=−=−+++−=0.8)·············8分245245245016bb+−3561P(Y=0.5−a)=1−−=························································9分1010563561bb+−EYaaaba()=−+−+−=+−(2.30.80.51.190.9)()()()10105············································································································11分E(Y)E(X)0.9b−a0,110a1即:ba(0,0.4)(备注:不写出不扣分)·······························12分29221.(12分)(1)AB+AF11+BF=42,即AF2+BF2+AF1+BF1=42,又AF1+AF2=BF1+BF2=2a,4a=42,a=2·······························2分c2又e==,cb=1,=1,a2x2故椭圆E的方程是+=y21······································································4分2(2)依题意知直线BC的斜率存在,设直线BC:y=+kxm,22xx2代入+=y21,得+(kx+m)=1,221222即+kx+2kmx+m−1=0①,2第4页共8页{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}212222=−+−=(2412240kmkmmk)−++(),2即2km1022−+②,设Bx(yC1122,x,,y)(),则Ax(y22,−),−2kmm2−1xx12+=xx12=12③,1④·····················································6分+k+k222A,,FB2三点共线,F2(1,0),直线AB不与坐标轴垂直,yy12−=,(x2−1)(kx1+m)=−(x1−1)(kx2+m),xx12−−11++−+−=220kxxmxxkxxm121212()(),221km(−)22km22km−+−20m=111,+k2+k2+k22222km2−2k−2km2+2k2m−m−2k2m=0,mk=−2,直线BC:2y=−k(x),1由②得:2k2−4k2+10,k2,23k2BC=1+kx−x,点F1(−1,0)到直线BC:kx−y−2k=0的距离d=,121+k213S=BCd=kx−x·································································8分F1BC221222224k4(4k−1)x−x=(x+x)−4xx=−12121211++kk2222第5页共8页{#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}422116441kkk−−+()2224−k==22,1122++kk22kk22(24−)SkFBC=302(),1(12+k2)t−1设2kt112+=,则k2=,2tt−−12−+−tt232112()()===−+−SFBC333231221tttt2131=−32−+···························
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