THUSSAT2023年9月诊断性测试数学答案

中学生标准学术能力诊断性测试2023年9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．12345678BBBACACD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3分，有错选的得0分．9101112ABDABDBCDABC三、填空题：本题共4小题，每小题5分，共20分．2413．−2514．−466415．316．1四、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）2nn2+(nn−11)+−（1）SS=,=，nn22−1Sn−Sn−1=n=cn(n1,nN)··································································2分+n又c11=S=1,cnn=n(nN),a=3·······················································3分2（）d=3n2n2++=++−+6n5n113n+1n213n分2n()()()·····························72222322Tn=(213+−++)(113313)(+−)(213+++)n2+−−+13nn12213n−+1+++n113n1−+n213n()()()()=−121+13+n+12+13n+()()=(nn21+2+2)3n+−6·····································································10分18．（12分）（1）c=2acosAcosB−bcos2A(AB)，第1页共6页{#{QQABCQqQogAgABIAAQhCQQVCCkCQkBEACAoGAFAAIAABwQFABAA=}#}sinCAABBA=2sincoscos−sincos2····················································2分sinCABBAAB=sin2cos−sincos2=sin(2−)0···································4分又02−AB，则CA=−B2或CA+B−2=，若CA=−B2，则A=；3若CA+B−2=，则AB=2，又AB，不符合题意，舍去，综上所述A=·························································································6分3222AB++AC2ABAC（2）2BD=DC,AD=,(AD)=····························8分33b+22c+b=c4236①，又ab222c=b+c−②，2cc224++213642c+b+bcbb①②得：2==222········································9分ab+−cbccc−+1bbc令=x，又ABaba,,,2b2bcbcb2+2−2，bccb,0=x1，b42x2+x+−16x3令f(x)=(0x1,)f(x)=4+······························10分xx22−xx+1−+1t+3令6x−3=t,x=，636t36f(t)=+4(−3t3),f(t)=+4(−3t3)227，t+27t+t27273667又t+12或t+−12,1f(t)7,7,a，tta2767所以当三角形ABC为等边三角形时a最小，最小值为·····························12分719．（12分）（1）设事件A1为A员工答对甲类问题；设事件A2为A员工答对乙类问题；设事件B1为B员工答对甲类问题；设事件B2为B员工答对乙类问题；第2页共6页{#{QQABCQqQogAgABIAAQhCQQVCCkCQkBEACAoGAFAAIAABwQFABAA=}#}设事件C1为C员工答对甲类问题；设事件C2为C员工答对乙类问题；三人得分之和为20分的情况有：①A员工答对甲类题，答错乙类题；B与C员工均答错甲类题，则PAABCPAPAPBPC(1211)=(1)(2)(1)(1)=0.50.40.40.6=0.048··············································································································2分②B员工答对甲类题，答错乙类题；A与C员工均答错甲类题，PBBACPBPBPAPC(1211)=(1)(2)(1)(1)=0.60.50.50.6=0.09··············································································································4分③C员工答对甲类题，答错乙类题；A与B员工均答错甲类题，，PCCABPCPCPAPB(1211)=(1)(2)(1)(1)=0.40.250.50.4=0.02所以三人得分之和为20分的概率为0.048+0.09+0.02=0.158··································6分（2）A员工得100分的概率为PAAPAPA(12)=(1)(2)=0.3，B员工得100分的概率为PBBPBPB(12)=(1)(2)=0.3，C员工得100分的概率为PCCPCPC(12)=(1)(2)=0.3，··············································································································9分XB~(3,0.3)······················································································11分EX()=30.3=0.9············································································12分20．（12分）（1）取AB的中点N，连接MN，NC，则线段MN为三角形SAB的中位线，MNSA，又SA⊥BD,BD⊥MN························································2分设直线CN与直线BD交于Q点，NQBQ1则BNQCDQ,==，NCBD366设AD=a,CD=2a,NC=a,NQ=a，263同理BD==3,aBQa，3a2a2a2又NQ2+BQ2=+==BN2···························································5分632BD⊥CN,BD⊥面MNC,⊥MCBD···················································6分第3页共6页{#{QQABCQqQogAgABIAAQhCQQVCCkCQkBEACAoGAFAAIAABwQFABAA=}#}（2）分别以直线AD，AB，AS为x轴，y轴，z轴建立直角坐标系，则ASCBM(0,0,0,)(0,0,2,)(2,22,0,)(0,22,0,)(0,2,1)，设SPS=C，P(2,22,21(−)),AP=(2,22,21(−))·································8分又AM==(0,2,1,)AC(2,22,0)，设平面AMC的法向量nx=yz(,,)，nAM=20y+z=则,n=(−2,1,−2)··········································10分nAC=2x+22y=0设直线AP与平面AMC所成的角为，22(1−)10则sin=cosAP,n==，5162−8+41011SP=,=··················································································12分22SC21．（12分）（1）设MF1==r1,MF2r2，在MF12F中，设=F12MF，2222F1F2=r1+r2−2rr12cos=4c，12rrcos=r2+r2−4c2，又MC=+MFMF，12122(12)2221122rr22122，=MCMF1++=++MF22MF1MF2(r1r22r1r2cos)=+−c4()4222rr22(r+−r)2rrMC2=12+−c2=1212−c2=2a2−c2−5=4·························3分2222a2−c2=9,a2=6,c2=3,b2=3，xy22所以椭圆C的方程为：+=1·······························································4分63（2）设A(x0,,,,,y0)P(x1y1)Q(x2y2)，直线l的方程为x=+yt，第4页共6页{#{QQABCQqQogAgABIAAQhCQQVCCkCQkBEACAoGAFAAIAABwQFABAA=}#}xy22+=163(2+2)y2+2ty+t2−6=0，x=+yt26tt2−y+y=−,,,yy=x=y+tx=y+t，1222++221211224tt222−6x+x=,xx=································································7分1222++2212y−−yyy(y−y)(x−x)+(y−y)(x−x)设01+=0201020201x0−x1x0−x2(x0−x1)(x0−x2)22xyyxx00−0(1+2)+yy12+(txyy−0)(1+2)=2x0−(x1+x2)x0+x1x22xy2+(2tx−12)+4y(x−t)==00000p222(x00−62)+(x−t)若p为常数，则2tx0−=120·····································································10分22x0y04y00(x−t)y0即6=tx，而此时==，022xt−(x0−6)2(xt0−)06又−6x6,−66，即t6或t−6，0t618综上所述，或，存在点A,3−，使得直线AP的斜率与直线AQt6t−62tt2y的斜率之和为定值0············································································12分xt0−22．（12分）lnx(1−lnx)1−+lnxx2（1）g(x)=+x,1g(x)=+=······································1分xx22x212令h(x)=1−ln