THUSSAT2023年9月诊断性测试数学答案

项是符合题目要求的.12345678BBBACACD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.9ABDABDABC三、填空题：本题共4小题，每小题5分，共20分.3四、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.（1）Sn=,:Sn−1=，:Sn−Sn−1=n=cn(n>1,neN)··································································2分又c1:cn=n(neN+),:an=3n·······················································3分(n+1)2+1n2n···············:Tn=(22+1)2222n2nn2n22−6·····················（1）c=2acosAcosB−bcos2A(A<B)，ab+c−bc(c)(c):sinC=2sinAcosAab+c−bc(c)(c):sinC=sin2AcosB−sinBcos2A=sin(2A−又0<2A−B<π,则C=2A−B或C+2A−B=π,若C=2A−B，则A=；若C+2A−B=π,则A=2B，又A<B，不符合题意，舍去，综上所述A=······································:AD=,:(AD)2=2····························8分:b2+4c2+2bc=36①,又a2=b2+c2−bc②,(c)2(c)22（b)（b)（b)（b)令b=x，又A<B,:a<b,:a<b,:b+c−bc<b，:c<b,:0<=x<1，:1<f(t)<7,:a2<PPBP·····································································PB.APPAP·····································································2PCCAB2所以三人得分之和为20分的概率为0.048+0.09+0.02=0.15B员工得100分的概率为P(B1.B2)=P(·····································································X~B(3,0.3)······················································································11分E(X)=30.3=0.9············································································12分2012分）MNSA，又SA⊥BD,BD⊥MN························································2分BQ1,BD3NC=a,NQ=同理BD=a,BQ=a，又NQ2+BQ2=a2+a2=a2=BN2···························································5分632BD⊥CN,BD⊥面MNC,MC⊥BD···················································6分22,M，，设平面AMC的法向量n=(x,y,z)，设直线AP与平面AMC所成的角为θ,····················································································则sinθ=—→→cosAP,n—→→ . . :λ=,:=··················································································12分2SC22112分）r，rrr 22222rc222:2a2−c2=9,a2=6,:c2=3,:b2=3，（2）设A(x0,y0),P(x1,y1),Q(x2,y2)，直线l的方程为x=λy+t，2tλt2−6:y1+y2=−2,y1y2=2,x1=λy1+t,x2=λy2+t，λ+2λ+24t2t2−6λ212λ2+4t2t2−6λ212λ2+2,12y0−y1x01y0−y2x02(x02)+(y0(x002)2x0y0−y0(x1+x2)+2λy1y2+(t−x0)(y1+y2)=2)x0+x1x22x0y0λ2+(2tx0−12)λ+4y0(x0−t))λ2+2(x0−t)2=p4y0(x0−t)2y0=2(x0(618)22y0的斜率之和为定值················································x0lnx