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石家庄市2024届高中毕业年级教学质量摸底检测(数学答案)(一、二)选择题123456789101112BACBCDADABDBDBCDAC(三)填空题13.16014.2315.1816.2211225144xy,229236yx(四)解答题(阅卷过程中发现
的其他解法参照本答案的评分细则,教研组研讨决定)17.【解析】(1)设等差数列的公差为d,由,64583Sa可得6427885211dada…...................………...
....……………………2分解得211da,…...................……….......…………………….4分所以12nan,………………………….......…………………
….5分(2)因为11nnaa=)121121(21)12)(12(1nnnn,所以nT=)12112171515131311(21nn………………………......………………………7分=
)1211(21n………………………………………………....…………………9分因为*Nn,所以21)1211(21n,即nT<21.……………………......………………………10分18.【解析】(1)由余弦定理可得:52150cos32421216
cos22222BaccabAC,则AC132,…………………………..........…………………2分又因为BbAasinsin,即150sin132sin32A……
……………………..........…………………4分{#{QQABKQKEogAoQABAABhCEwVyCgKQkAGACCoOxBAEIAAAARFABCA=}#}所以2639sinBAC.………………………….........…………………5分(2)因为2639sinBAC,
所以26137cosBAC,从而73tanBAC,……………………………...........……………........…7分在RtABD中,343tan477BDABBAC……………………………..…......……...……9分1s
in2BCDSBDBCDBC1433272…...........……………………….......…......…11分BCDS637……................……………………….......…......…12分19.【解析】(1)因为PC平面A
BC,AC平面ABC,所以PCAC,又4PC,26PA,所以22AC.......................................................................……………………………..................
.….......2分在ABC中,因为2ABBC,所以222ABBCAC,所以ABBC因为PC平面ABC,AB平面ABC,所以PCAB,..………........…….........……........4分又因为PCB
CC,所以AB平面PBC..................................……………...................................5分(2)(方法一)由(1)知ABBC,以B为
坐标原点,,BCBA所在直线分别为,xy轴,过点B且与平面ABC垂直的直线为z轴,建立空间直角坐标系,如图所示.zyxPCBAM{#{QQABKQKEogAoQABAABhCEwVyCgKQkAGACCoOxB
AEIAAAARFABCA=}#}则0,0,0,0,2,0,2,0,4,2,0,0BAPC,1,1,2M,所以1,1,2CM,0,2,0BA,2,0,4BP..................…………….....
......................6分设平面PAB的法向量为,,nxyz,则00nBAnBP,即20240yxz,令2x,则1z,所以2,0,1n,.........................…………….....
..............................8分设CM与平面PAB所成角为,则22230cos,1565CMnCMnCMn,....................................................................
...10分230sincos,15CMn,105cos15即CM与平面PAB所成角的余弦值为10515.............................................
.................................12分(2)(方法二)过点C作CNPB,垂足为N,连接MN,............................................
..........6分由(1)知AB平面PBC,AB平面PAB,平面PAB平面PBC,平面PAB平面PBCPB,CN平面PBC,CNPB,CN平面PAB,CMN为CM与平面PAB所成角,.........................................
.8分在RtPAC中,162CMPA,在RtPBC中,4245525PCBCCNPB,...........................................................................10分
在RtCMN中,22224570655MNCMCN,故105cos15MNCMNCM,即CM与平面PAB所成角的余弦值为10515..........................................................
......................12分20.【解析】(1)设A“张某选择甲类问题”,B“张某答对所选问题”,M“张某至少答对一道问题”,{#{QQABKQKEogAoQABAABhCEwVyCgKQkA
GACCoOxBAEIAAAARFABCA=}#}则A“张某选择乙类问题”,B“张某未答对所选问题”M“张某一道问题都没答对”.......................................
.................................................................1分由题意得,𝑃(𝐴)=𝑃(𝐴̅)=0.5,0.9PBA,0.1PBA,0
.7PBA,0.3PBA,.....................................2分由全概率公式,得0.50.10.50.30.2PMPAPBAPAPBA.....................................
....................4分110.20.8PMPM.....................................................................
.....................................5分(2)根据条件可知:若张某先回答甲类问题,则张某的累计得分𝑋的可能值为0,30,80,...........................................
...................................................6分∵张某能正确回答甲类问题的概率为0.9,能正确回答乙类问题的概率为0.7,010.90.1PX;300.910.70.27P
X;800.90.70.63PX,则𝑋的分布列为X03080P0.10.270.63当张某先回答甲类问题时,累计得分的期望为00.1300.27800.6358.5EX,......................
...................................................................8分若张某先回答乙类问题,则张某的累计得分𝑌的可能值为0,50,80,..................................
....................9分同理可求010.70.3PY;500.710.90.07PY;800.70.90.63PY,则此时累计得分的期望为00.3500.07800.6353.9EY
...................................................11分因为EXEY,所以,以累计得分多为决策依据,张某应选择先回答甲类问题.........................................
....................12分21.【解析】(1)设E的方程为0,0122nmnymx,................................................
..................1分代入3,12A和0,2B两点得41,31nm,................................................................................
........2分所以E的方程为22134xy.............................................................................................
..............................4分{#{QQABKQKEogAoQABAABhCEwVyCgKQkAGACCoOxBAEIAAAARFABCA=}#}(2)设过点C的直线方程为4ykx
,224341ykxxy消去y得220234364kxkx,222443634144420kkk,解得22kk或,.............................
.........5分设11(,)Mxy22(,)Nxy,则1222434kxxk,1223634xxk............................................................
..6分设过点M且斜率为-2的直线为112()yyxx,令1y,所以1121(,1)2xyQ,111(1,2)Hxyy,所以直线NH的斜率为1221121yyxxy,直线NH为12222112()1yyyyxxxxy
,..........................................................................................8分令1y,22112112121221212(1)
(1)1122yxxyyxyxyyxxxxyyyy,①将114ykx,224ykx代入①式,得21212121212121223()1(2)(43)()15112()6kxxxxyykkxxkxxxyykxx
,②........................................................................................................................
..................................10分将1222434kxxk,1224363xxk代入②式,得222223624(2)(43)()15343424634kkkkkk
xkk221560316242kk,所以直线HN与直线1y交点为定点3,12.............................................................................12分22.【解
析】(1)1(1)xfxaea,..........................................................................................
.........1分当01a时,令0fx,得11lnaxa,{#{QQABKQKEogAoQABAABhCEwVyCgKQkAGACCoOxBAEIAAAARFABCA=}#}当1,1lnaxa
时,0fx;当11ln,axa时,0fx.故fx在1,1lnaa上单调递减,在11ln,aa上单调递增..............................
...........3分当1a时,0fx,fx在,上单调递增.............................................................................4分(2)
1ln1xaegxxxx,0,x,121xxaexgxx,显然,1x是方程0gx的一个根.令11111,()1,()0xxfxaexfxaefx的解为1lnxa,当1l
n0a,即ae时,1()0fx,1fx在0,上单调递增,不符合题意,故舍去;当1ln0a,即ae时,1fx在0,1lna单调递减,在1ln,a单调递增.若1fx有两个零点,则ln11ln1lnln0afaaeaa,解得0
1a...........................6分因为100afe,1110fa,221222122111022aafaeaaaaaa
,所以方程10xaex有两个根,设为13,xx,且13201xxa,gx在10,x,31,x上单调递减,在1,1x,3,x上单调递增,故当01a时,gx有三个极值点1x,2x,3x,其中21x......
...................................................8分由111xaex,得11111111ln1lnxaegxxxxxx,同理333lngxxx,又因为21
ln1112gxgaa,所以12313132lngxgxgxaxxxx3111132lnxxaxxaeae213312ln11axxaxx42lnaa,..
...............................................................................................10分{#{QQABKQKEogAoQABAABhCEwVyCgKQkAGACCoOxBAEIAA
AARFABCA=}#}令42lnmaaa,01a,则210maa,所以ma在(0,1)上单调递增,0a时,ma,13m,所以3ma,所以
123gxgxgx的取值范围是,3....................................................................................12分{#{QQABKQKEogAoQA
BAABhCEwVyCgKQkAGACCoOxBAEIAAAARFABCA=}#}获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com