THUSSAT2023年9月诊断性测试数学答案

2023-11-08 · U1 上传 · 6页 · 482.6 K

中学生标准学术能力诊断测试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
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