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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211
da,…...................……….......…………………….4分所以12nan,………………………….......…………………….5分(2)因为11nnaa=)121121(21)12)(12(1nnnn,所以nT=)1211217
1515131311(21nn………………………......………………………7分=)1211(21n………………………………………………....…………………9分因为*Nn,所以21)121
1(21n,即nT<21.……………………......………………………10分18.【解析】(1)由余弦定理可得:52150cos32421216cos22222BaccabAC,则AC132,……………………
……..........…………………2分又因为BbAasinsin,即150sin132sin32A…………………………..........…………………4分{#{QQABKQKEogAoQAB
AABhCEwVyCgKQkAGACCoOxBAEIAAAARFABCA=}#}所以2639sinBAC.………………………….........…………………5分(2)因为2639sinBAC,所以261
37cosBAC,从而73tanBAC,……………………………...........……………........…7分在RtABD中,343tan477BDABBAC……………………………..…......……...……9分1sin2BCDSBD
BCDBC1433272…...........……………………….......…......…11分BCDS637……................……………………….......…......…12分19.【解析】(1)因为PC平面
ABC,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PCBCC,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{#{QQABKQKEogAoQABAABhCEwVyCgKQkAGACCoOxBAEIAAAARFABCA=}#}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,4245525PCBCCNP
B,...........................................................................10分在RtCMN中,22224570655M
NCMCN,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”,{#{QQABKQKEogAoQABAABhCEwVyCgKQkAGACCoOxBAEIAAAARFABCA=}#}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0.50.10.50.30.2PM
PAPBAPAPBA.........................................................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.
910.70.27PX;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00.3500.0
7800.6353.9EY...................................................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(,)Mx
y22(,)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)()15343424634kkkkkkxkk221560316242
kk,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1ln0a,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01a...........................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21ln1112gxgaa,所以1231313
2lngxgxgxaxxxx3111132lnxxaxxaeae213312ln11axxaxx42lnaa,...........................................
......................................................10分{#{QQABKQKEogAoQABAABhCEwVyCgKQkAGACCoOxBAEIAAAAR
FABCA=}#}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{#{QQABKQKEogAoQABA
ABhCEwVyCgKQkAGACCoOxBAEIAAAARFABCA=}#}获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com