鞍山市普通高中2023—2024学年度高三第二次质量监测数学参考答案选择题:题号1234567891011答案DCCBBCADADABDBCD412.12π13.14.5e215.(1)解:m=0.012.................................................................................2分设中位数为x0,由0.260.50.56可知,x060,70)且(x0−+=600.030.260.5).....................................4分x0=68即这50名学生分数的中位数为68................................5分(2)解:分数在[70,80)的7人,在[80,90)的3人,[90100],的1人.6分可取值的集合为0,1,23,CC035638,分P(=0)=3=................................................................7C11165CC12842838,分P(====1)3.......................................................8C1116555CC2124838,分P(====2)3........................................................9C1116555CC30138,分P(===3)3................................................................10C11165的分布列为:0123562881P1655555165............................................................................................................11分562881的数学期望E()=0+1+2+3,...............12分16555551651359==...........................................................................13分16511516.(1)证明:在△BCD中,CD==BC7,cos=DCB,7由余弦定理可得BD2=BC2+CD2−2BCCDcosDCB...........1分即BD=2.......................................................................................2分{#{QQABAQgAggAgAJAAARgCUQGCCAOQkBGAAAoGRAAMsAAAiRFABAA=}#}又因为DE=2,E=DB60,所以△BDE为正三角形设BD中点为F,连接EF、PF,故(由△BDE为正三角形可知)BD⊥EF,(由CDB=C可知)BDP⊥F,(EF、PF平面PEF,EFPFF=)所以BD⊥平面PEF,.....................................................................4分(又因为PF平面PEF)故BDP⊥E.....................................5分在△PFD中易得PF=6,在△BDE中易得EF=3,又因为EP=3,由勾股定理可知PEE⊥F,.............................6分又因为EF、BD平面ABDE,EFBDF=,..........................7分所以PE平面ABDE...................................................................8分(2)解:因为AE∥BD,所以AE⊥EF,z再由PE⊥平面ABDE可知,PPE⊥AE,PE⊥EF,故可以E为坐标原点,EA,EF,EP分别为x轴,y轴,z轴正方向,建立如图所示空间直角坐标系.....................................................E9分D在坐标系中,各点坐标分别为:AFxyA(1,0,0),B(130,,),F(030,,),BP(0,0,3).......................................................................................10分设平面PAB的法向量为n1111=(xyz,,),AB=03,0,AP=-1,0,3因为(),(),由nn11==ABAP00,可得,,,平面PAB的一个法向量n1=(301),..........................................12分设平面PDB的法向量为n2222=(xyz,,),同理可得,平面PAB的一个法向量n2=(0,11,),.........................13分nn设所求锐二面角为,则cos=12,..................................14分nn122整理得cos=..........................................................................15分4217.(1)解:f(x)=ax−(a+2)+,.........................................................2分x曲线y=f()x在x=2处的切线与y轴垂直,{#{QQABAQgAggAgAJAAARgCUQGCCAOQkBGAAAoGRAAMsAAAiRFABAA=}#}2故faa(2)2(2)0=−++=,........................................................4分2解得a=1........................................................................................5分2axaxaxx2−++−−(2221)()()(2)解:fxaxa()2=−++==()xxxx+(0,)........................................................................................6分①当a0时,由fx()0得01x,由fx()0得x1........7分2②当a0时,由fx()=0得x=或x=1,................................8分a22ⅰ若1,即02a,则由得或x,aa2由得1x,..........................................................10分a2ⅱ若=1,即a=2,则fx()0当且仅当x=1时取等,.....11分a22ⅲ若1,即a2,则由得0x或x1,aa2由得x1,..........................................................13分a综上,当时,fx()的单调增区间为(01,),单调减区间为(1,+);2当时,的单调增区间为,,+,a2单调减区间为1,;a当a=2时,的单调增区间为(0,+);2当a2时,的单调增区间为0,,(1,+),a2单调减区间为,1.a18.(1)由若OB=OC可知,B,O,C三点共线,再由椭圆的对称性可知,B,C关于原点中心对称,即xx32=−,yy32=−,.................................1分1易知M(−20,),由直线MB和直线MC的斜率之积为−可得,4yy22−1=−对任意x2成立,.....................................................3分xx22+2−+24221即yx22=−(4)对任意x2成立,4{#{QQABAQgAggAgAJAAARgCUQGCCAOQkBGAAAoGRAAMsAAAiRFABAA=}#}2222xy22xx224−再由+=1,消去y2,可知+=1对任意x2成立,4b244b2故b2=1,.................................................................................................4分所以b=1.(注:原题中忘了给出条件b0,所以,将错就错吧☺)...........................................................................................................5分2x1(2)若xx12=,则y=1−,1,2421x12△OAB的面积为1,即x−=211,解得x1=2,2411故xx22==2,yy22==,所以xx22+=4,....................6分212121222yy12+=1若xx12,即直线AB斜率存在,设直线AB的方程为:yk=+xm,2x2代入椭圆方程得++=(kxm)1,4222所以xx12,是关于x的方程(418440kxkmxm+++−=)的两实根,由韦达定理可知,当△=−+−(84kmkm41440)2(22)()即−8km44m2−41km22+时,xx+=,xx=,......7分1241k2+1241k2+22222−−−+8444kmmkm41xxxxx12121−=+−=−=2x()44222414141kkk+++1441km22−+△OAB的面积为1,即m=1241k2+整理得412km22+=.....................................................................9分故2222222−−−−844844kmmkmmxxxxx12121+=+−=−=−2x()22222241412kkmm++1644km22−即xx22+=−=4...........................................10分12mm2222综上,xx12+=4,222222x1x2x1+x2yy12+=1−+1−=2−=1......................11分444x1++x2y1y2(3)显然,D,,22{#{QQABAQgAggAgAJAAARgCUQGCCAOQkBGAAAoGRAAMsAAAiRFABAA=}#}222222由(2)可知,OAOBxyxy+=+++=11225,222222又OAOBOAOBODDAODDAODDA22+=+=++−=+()()2()即OA2+OB2=2(OD2+DA2)4ODDA=2ODAB5所以ODAB,.....................................................................15分21当且仅当ODDAAB==时取等号,2此时△OAB为直角三角形且∠AOB为直角,故OAOBx=+=+++xy12121212yxxkxmkx
辽宁省鞍山市普通高中2024届高三下学期第二次质量监测试题数学答案
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