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吉林市普通中学2020—2021学年度高中毕业班第三次调研测试文科数学参考答案一、选择题:本大题共12题,每小题5分,共60分.二、填空题:本大题共4个小题,每小题5分,共20分.其中第16题的第一个空填对得2分,第二个空填对得3分.13.1−14.,,cb
a15.242516.22143xy+=(2分),①②③④(3分)17.【解析】(1)nm⊥0cos2=+−=Baanm.()0a.........................................3分22cos=B(),0B4=B......
.................................................6分(2)由正弦定理得BbAasinsin=2222sin32=A23sin=A..........................................9分
(),0A3=A或32.........................................12分[注:①只写出一种情形且算对,扣2分;②未说明角范围各扣1分.]18.【解析】(Ⅰ)散点图如右
图....................1分由散点图可知,管理时间y与土地使用面积x线性相关............................2分依题意:3554321=++++=x,又16=y437281)2(0)5()1(
)8()2())((51=++−+−−+−−=−−=yyxxiii..........3分10210)1()2()(22222251=+++−+−=−=xxii,206)(251=−=yyii.................4分则947.04.45435152432061043)
()())((21211==−−−−====niiniiiniiyyxxyyxxr................5分由于75.0947.0,故管理时间y与土地使用面积x线性相关性较强.....................6分(Ⅱ
)22列联表如下:..............8分假设0H:该村村民的性别与参与管理的意愿无关2K的观测值828.1025100200120180)604060140(3002=−=k.............................10分001.0)828.10(2
KP所以有9.99%的把握认为该村的村民的性别与参与管理意愿有关...................12分19.【解析】(1)证明:连.BN连,11HCAAC=连MHMHBCMBAMHCAH//,11==又MH面CMA1,1BC面CMA1//1BC面CMA1.
................2分四边形NBMA1是平行四边形,MABN1//BN面CMA1,MA1面CMA1//BN面CMA1............................4分=BNBCBBNBC,,11面NBC1
面//1CMA面NBC1...................................................5分【注:也可以利用CAPNCMNC11//,//证明】PQ面NBC1//PQ面CMA1............................
..........6分(2)由(1)知,面//1CMA面NBC1.则点B到面CMA1的距离h即为所求.由⊥1AA面ABC得1AA为锥体CMBA−1的高.3422212313111===−CMBBMCASAAVΔ.............................
.8分MCA1中2222=+=AMACMC,222211=+=ACAACA,222211=+=AMAAMA则328431==MCASΔ................................................10分由BMCA
MCABVV−−=11即343231=h得332=h即点Q到面CMA1的距离为332............................12分20.【解析】解:(Ⅰ)由抛物线的定义知,2321=+p,解得1=p..........................2分所以抛物线C的
方程为yx22=................................................3分焦点)21,0(F................................................................
4分(Ⅱ)由(Ⅰ)知焦点)21,0(F,设),(),,(2211yxByxA易知直线l存在斜率,设为k,直线l方程为21+=kxy,联立=+=yxkxy2212,消去x得:041)12(22=++−yky04424+=kkΔ恒成
立,则12221+=+kyy................................5分22||221+=++=kpyyAB................................................6分设原点O到直线l的距
离为1d,12121+=kd所以121121)1(221||2122211+=++==kkkdABS........................7分1S解法二联立=+=yxkxy22
12,消去y得:0122=−−kxx,0442+=kΔ恒成立,则kxx221=+,121−=xx)1(24414)(1||222212212+=++=−++=kkkxxxxkAB设原点O到直线l的距离为1d,12121+=kd所以12
1121)1(221||2122211+=++==kkkdABS1S解法三1214421214)(||21||||212221221211+=+=−+=−=kkxxxxOFxxOFS易知)21,21(−kQ设Q到
直线l的距离为2d,122222++=kkd所以1)2(21122)1(221||212222222++=+++==kkkkkdABS................8分故2111SS−=2121)2()1(
21)2(21222222222++=+++=++−+kkkkkkkk..........9分设112+=km,1122121211221=+=+=−mmmmmmSS................10分当且仅当mm1=,即1=m时取等号.......
...................................11分所以2111SS−的最大值为1...................................................12分21.【解析】解:(Ⅰ)axxexfx−=)(,
则axexfx−+=)1()(...........................1分)(xf有两个极值点,则0)1()(=−+=axexfx有两个不等实根即ay=与)1()(+=xexHx有两个公共点)1()(+=xexH
x)(Rx)2()(+=xexHx令0)(=xH解得2−=x......................................................2分)()(xHxH,变化情况如下表所示:x)2,(−
−2−),2(+−)(xH−0+)(xH单调递减21e−单调递增2min1)2()(eHxH−=−=..............................................
....4分当+→x时,+→)(xH当−→x时,与一次函数相比xey−=呈爆炸增长,01)(→+=−xexxH..............5分故)0,1(2ea−............................................
.......................6分(Ⅱ)当0=a时,)2ln()2ln()()(+−=+−=xexxxfxgx)02(−x...........7分21)(+−=xexgx在)0,2(−单调递增,并且011)1(−=−eg,0531)31(3−=−eg(也可
以取其它点)在)31,1(−−上存在唯一实数根0x使得0)(0=xg...................................8分2100+=xex,即)2ln(00+−=xx①........................................9分02xx−时
,0)(xg,)(xg在),2(0x−上单调递减00xx时,0)(xg,)(xg在)0(0,x上单调递增)2ln()()(00min0+−==xexgxgx②...............................
........10分由①②知,02)1(21)()(020000++=++=xxxxxgxg)311(0−−x即证当0=a时,)0,2(−x,0)(xg..........................................1
2分(Ⅱ)方法二:当0=a时,)2ln()2ln()()(+−=+−=xexxxfxgx)02(−x......7分令)1()(+−=xexx)02(−x01)(−=xex)(x在)0,2(−单调
递减0)0()(min=→x,0)(x即1+xex①..............................9分再令)2ln(1)(+−+=xxxS)02(−x21211)(++=+−=xxxxS
当12−−x时,0)(xS,)(xS在)1,2(−−单调递减当01−x时,0)(xS,)(xS在)0,1(−单调递增0)1()()(min=−=SxSxS即)2ln(1++xx②....................................
..............11分由①②知,02−x时,)2ln(+xex即证当0=a时,)0,2(−x,0)(xg.........................................12分22.【解
析】(1)sin4=sin42=.......................................2分0422=−+yyx即()4222=−+yx.......................................4分(2)将直线l参数方程+=
−=tytx22122(t为参数)代入曲线C()4222=−+yx中得:0322=−−tt.....................................................5分设方程的两根为21
,tt则−==+322121tttt.......................................7分021tt1t与2t异号.............................................
.......8分141222121=+=−=+=+ttttPBPA...............................10分23.【解析】(Ⅰ)()−−=4,5241,31,25xxxxxxf........
.................................1分5)(xf−1525xx或4153x或−4552xx...................3分50x
不等式解集为50xx..............................4分(注:结果不表示成集合或区间扣1分)(Ⅱ)由(Ⅰ)知,()xf在()1,−上单调递减,()+,4上单调递增,()3min=xf3=M
..............................................5分解法1:3=+ba()()612=+++ba1121+++ba()()12112161++++++=baba()3222611221261=+
++++++=baab.................................8分解法2:由柯西不等式得:1121+++ba()()12112161++++++=baba326411
1221612==+++++bbaa......................8分当且仅当=++=+312baba时,即2,1==ba时............................9分1121+++b
a的最小值为32...........................................10分