【文档说明】吉林省吉林市2021届高三下学期第三次调研测试(3月) 数学(理) 答案.docx,共(5)页,387.511 KB,由管理员店铺上传
转载请保留链接:https://www.doc5u.com/view-40fb5408bff29a86de5b740a67c6bdbd.html
以下为本文档部分文字说明:
吉林市普通中学2020—2021学年度高中毕业班第三次调研测试理科数学参考答案一、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.,,cba15.3816.
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()(22222
251=+++−+−=−=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)由题知调查的300名村名中不愿意参与管理的女性村民人数为60)604
0140(300=++−该贫困县中任选一人,取到不愿意参与管理的女性村民的概率5130060==p.............7分则X可取3,2,1,0.............................................................8分
12564)54()0(303===CXP12548)54(51)1(213===CXP1251254)51()2(223===CXP1251)51()3(333===CXP...............10分也可以写如下形式:)51,3(~BX3,
2,1,0,)54()51()(33===−kCkXPkkkX的分布列为..............11分5312513125122125481125640)(=+++=XE............................
..12分(或53513)(===nPXE)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N
BMA1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)以A为原点,1,,AAABAC所在直线分别为x轴、y轴、z轴,建立如图所示的空间直角坐标系..........................................7分设),0,0(1
hA)0(h)0,4,0(),0,0,2(),0,2,0(BCM所以),2,0(1hMA−=,),0,2(1hCA−=设平面CMA1的法向量()zyxn,,=则=−==−=020211hzxCA
nhzyMAn即===2zhyhx)2,,(hhn=平面ACM的法向量)1,0,0(0=n...............................................9分由二面角AC
MA−−1的余弦值是33则331422|||||)1,0,0()2,,(||||||||,cos|20000=+===hnnhhnnnnnn又0h,解得2=h)2,2,2(=n......
........................................................11分又()0,2,0=MB,()()332322,2,20,2,0===nnMBd即点B到平面CMA1的距离为33
2.............................................12分[方法二:利用传统方法求出𝐀𝐀𝟏=𝟐.............................................9分再利用等积法
求出点B到平面CMA1的距离为332............................12分][方法三:利用传统方法求出𝑨𝑨𝟏=𝟐.....................................
....9分再利用点B到平面CMA1的距离即为点A到平面CMA1的距离.由直接法求出距离为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=+=yxkxy2212,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121121)1(221||2122211+=++==kkkdABS1S解法三1214421214)(||21||||2
12221221211+=+=−+=−=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】(Ⅰ)xxexfxsin2)(+−=,xexfxcos2)(+−=,且0)0(=f...................1分①当0x
时,1xe,1cosx02cos−+xex即0)(xf解集为)0,(−)(xf在)0,(−上是减函数...........................2分②当0x时,设xexhxcos2)(+−=则xexhx
sin)(−=而1xe,1sinx,所以0)(xh因此)(xh在),0(+上为增函数,且0)0(=h所以0)(xh在),0(+上恒成立)(xf在),0(+上是增函数.......................3分综上:)(xf的减区间为)0,(−,增
区间为),0(+.....................................4分(2)由(1)知,函数min()(0)1fxf==]2,0[,21xx,使得不等式)()(21xfxg成立等价于不等式(sinc
os)1xexxa−++在[0,]2x时有解即不等式sincosxaxxe−−+在[0,]2x时有解................................5分设()sincosxFxxxe−=−+,()sincosxFxxxe−=+−[
0,]2x时,(sincos)[1,2]xx+,而1xe−所以()0Fx恒成立.............6分即()Fx在[0,]2上是增函数,则min()(0)0FxF==...............................7分因此a的取值范围是[0,)+...
..................................................8分(3)(1,)x+.xxmxexlncos2−+−恒成立等价于()minlncos2xxxemx−+−........................
...................9分令()()1lncos2H−+−=xxxxexx()1lnsinxH−−−=xxex()xxexx1cosH−−=1xeex1cos−−x11−−x()02−exH()xH在
()+,1递增()()01111sin1H−−−−=eeHx....................................10分()xH在()+,1上递增()()1cos21+−=
eHxH1cos2+−em...........................................................11分()2,11cos2+−e且zm因此整数m的最大值为1.......................
...............................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[方法二:也可用|𝑷𝑨|+|𝑷𝑩|=|𝑨𝑩|=𝟐√𝒓𝟐−𝒅𝟐14
=]23.【解析】(Ⅰ)()−−=4,5241,31,25xxxxxxf...........................................1分5)(xf−1525xx或4153x或−4
552xx.....................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++++++=bab
a()3222611221261=+++++++=baab.................................8分解法2:由柯西不等式得:1121+++ba()()12112161++++++=baba3264111221612==
+++++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