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．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−+···························