-數(shù)學｜2023年9月THUSSAT中學生標準學術能力高三上學期診斷性測試數(shù)學試卷及答案

符合題目要求的.2．歐拉公式eiθ=cosθ+isinθ把自然對數(shù)的底數(shù)e、虛數(shù)單位i、三角函數(shù)聯(lián)系在一起，充分體現(xiàn)4．已知向量AB.AC=6，線段BC的中點為M，且AM=6，則BC=A．2B．3C．2D．3(3)(1)(2)(4)與平面ABC所成角的正弦值是47．已知三角形ABC中，BC=3，角A的平分線交BC于點D，若=，則三角形ABC面積0.1題目要求．全部選對得5分，部分選對但不全得3分，有錯選的得022數(shù)x1；最大和最小兩個數(shù)據(jù)的方差為s，平均數(shù)x2；原樣本數(shù)據(jù)的方差為S2，平均數(shù)x，若x12是函數(shù)f(x)是函數(shù)f(x)圖象的一條對稱軸2x2212．已知橢圓x22=1的中心為O，A,B是C上的兩個不同的點且滿足OA⊥OB，則A．點O在直線AB上投影的軌跡為圓 3B．經(jīng)AOB的平分線交AB于D 3三、填空題：本題共4小題，每小題5分，共20分.13．已知tanc=2，則sin4c=.x25415．已知四棱錐的各個頂點都在同一個球面上．若該球的體積為36π,則該四棱錐體積的最大值16．已知函數(shù)f(x)=ex+msinx?x2?(m+1)x+1，在x=0處取到極小值，則實數(shù)m=.四、解答題：本題共6小題，共70分．解答應寫出文字說明、證明過程或演算步驟.1710分）已知{an}是各項均為正數(shù)的等比數(shù)列，設cn=log3an，若數(shù)列{cn}的前n項和n2（2）記dn=an.(2n2+6n+5)，求數(shù)列1812分）記ΔABC的內(nèi)角A,B,C的對邊分別為a,b,c，已知c=2acosAcosB?bcos2A（2）若D是BC上的一點，且BD:DC=1:2,AD=2，求a的最小值.（2）設隨機變量X為3人中得分為100的人數(shù)，求隨機變量X的數(shù)學期望.2012分）已知四棱錐S?ABCD中，底面ABCD是矩形，（2）若SA⊥AD,SA=2，點P是SC上的動點，直線AP與平,F（2）設定點T(t,0)，過點T的直線l交橢圓C于P,Q兩點，若在C上存在一點A，使得直線AP的斜率與直線AQ的斜率之和為定值，求t的范圍.項是符合題目要求的.12345678BBBACACD符合題目要求．全部選對的得5分，部分選對但不全的得3分，有錯選的得0分.9ABDABDABC三、填空題：本題共4小題，每小題5分，共20分.3四、解答題：本題共6小題，共70分．解答應寫出文字說明、證明過程或演算步驟.（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,則C=2A?B或C+2A?B=π,若C=2A?B，則A=；若C+2A?B=π,則A=2B，又A<B，不符合題意，舍去，綜上所述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<所以當三角形ABC為等邊三角形時a最小，PPBP·····································································PB.APPAP·····································································2PCCAB2所以三人得分之和為20分的概率為0.048+0.09+0.02=0.15（2）A員工得100分的概率為P(A1.AB員工得100分的概率為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，，設平面AMC的法向量n=(x,y,z)，設直線AP與平面AMC所成的角為θ,····················································································則sinθ=—→→cosAP,n—→→ . . :λ=,:=··················································································12分2SC22112分）r，rrr 22222rc222:2a2?c2=9,a2=6,:c2=3,:b2=3，（2）設A(x0,y0),P(x1,y1),Q(x2,y2)，直線l的方程為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)