-数学｜2023年9月THUSSAT中学生标准学术能力高三上学期诊断性测试数学试卷及答案

符合题目要求的.2．欧拉公式eiθ=cosθ+isinθ把自然对数的底数e、虚数单位i、三角函数联系在一起，充分体现4．已知向量AB.AC=6，线段BC的中点为M，且AM=6，则BC=A．2B．3C．2D．3(3)(1)(2)(4)与平面ABC所成角的正弦值是47．已知三角形ABC中，BC=3，角A的平分线交BC于点D，若=，则三角形ABC面积0.1题目要求．全部选对得5分，部分选对但不全得3分，有错选的得022数x1；最大和最小两个数据的方差为s，平均数x2；原样本数据的方差为S2，平均数x，若x12是函数f(x)是函数f(x)图象的一条对称轴2x2212．已知椭圆x22=1的中心为O，A,B是C上的两个不同的点且满足OA⊥OB，则A．点O在直线AB上投影的轨迹为圆 3B．经AOB的平分线交AB于D 3三、填空题：本题共4小题，每小题5分，共20分.13．已知tanc=2，则sin4c=.x25415．已知四棱锥的各个顶点都在同一个球面上．若该球的体积为36π,则该四棱锥体积的最大值16．已知函数f(x)=ex+msinx−x2−(m+1)x+1，在x=0处取到极小值，则实数m=.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.1710分）已知{an}是各项均为正数的等比数列，设cn=log3an，若数列{cn}的前n项和n2（2）记dn=an.(2n2+6n+5)，求数列1812分）记ΔABC的内角A,B,C的对边分别为a,b,c，已知c=2acosAcosB−bcos2A（2）若D是BC上的一点，且BD:DC=1:2,AD=2，求a的最小值.（2）设随机变量X为3人中得分为100的人数，求随机变量X的数学期望.2012分）已知四棱锥S−ABCD中，底面ABCD是矩形，（2）若SA⊥AD,SA=2，点P是SC上的动点，直线AP与平,F（2）设定点T(t,0)，过点T的直线l交椭圆C于P,Q两点，若在C上存在一点A，使得直线AP的斜率与直线AQ的斜率之和为定值，求t的范围.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)2323222n2nn2n22−6·······························（1）c=2acosAcosB−bcos2A(A<B)，ab+c−bc(c)(c):sinC=2sinAcosAab+c−bc(c)(c):sinC=sin2AcosB−sinBcos2A=sin(2A−B)>0·····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<所以当三角形ABC为等边三角形时a最小，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=0.15（2）A员工得100分的概率为P(A1.AB员工得100分的概率为P(B1.B2)=P(B1).P(B2·····································································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+24t2t2−6λ212λ2+2,12y0−y1x01y0−y2x02(x02−y20(x002x0y0−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)