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★秘密·2022年12月15日16:00前重庆市2022-2023学年(上)12月月度质量检测高三数学答案及评分标准【命题单位:重庆缙云教育联盟】1.B2.B3.C4.B5.D【详解】解:由()(sinsin)
sinsinacACbBaB+−+=及正弦定理,得2()()acacbab+−+=,即222abcab+−=,由余弦定理得,2221cos22abcCab+−==,∵()0,C,∴3C=.由32CACDCB=−,1233CDCACB=+,两边平方,得22144999CDCACA
CBCB=++即222144cos999CDbaabC=++22142999baab=++()212299baab−=+()221122992baba++−()21212ba=+,当且仅当224baba=+=,即12ab==时取等
号,即2214(2)123CDba+=,∴线段CD长度的最小值为233.故选:D.6.D【详解】如图,连接11,ACAC,取AC中点E,过M作MF⊥面11AC,垂足为F,在正方体1111ABCDABCD−中,1AA⊥平面1111DCBA,且
1AA平面11ACCA,平面11ACCA⊥平面1111DCBA,平面11ACCA平面111111ABCDAC=,且MF平面11ACCA,MF⊥平面1111DCBA,P为1AB的中点,111,,APAEBA
CCACAMAM===APMAEM,故PMEM=,而对固定点M,当11MNBC⊥时,MN最小,此时由MF⊥面1111DCBA,11BCQ面1111DCBA,11MFBC⊥,又11MNBC⊥,MFNFF=,且,MFNF面MNF
,故11BC⊥面MNF,又11BC⊥面11ABA,则面//MNF面11ABA,根据三棱锥特点,可知11MNFABA,而易知11ABA△为等腰直角三角形,可知MFN△为等腰直角三角形,122,2222()22
22MFMNPMMNPMMNEMMFAA=+=+=+=.故选:D.2022.127.D【详解】由题意,圆C:(),2Ca−,半径1r=,C点到直线l的距离()22342134adr+−===+,a为整数,1a=;C到直线'l的距离'2222324
111mmmdmm−+−−==++,考察()2'221611mdm=−+,令221mpm=+,则有()2212,20pmmpmmp+=−+=,22240p=−,21,11pp−,即221mm+的取值范围是1,1−
,当2211mm=−+时,1m=−,()2'd最大;故选:D.8.D【详解】解:构造一个底面半径为2,高为3的圆柱,在圆柱中挖去一个以圆柱下底面圆心为顶点的圆锥,则当截面与顶点距离为()03hh时,小圆锥底面半径为r,则32hr=,23rh=,故截面面积为:2
449h−,把yh=代入22149xy+=,即22149xh+=,解得:2293xh=−,橄榄球形几何体的截面面积为22449xh=−,由祖暅原理可得橄榄球形几何体的体积为:(2VV=圆柱V−圆锥1)24343163
=−=.故选:D.9.AB10.AD11.ABD【详解】取FD中点Q,连接MQBQ、.若A正确,//BM平面AEF,且MQ为三角形AFD中位线,则//MQAF,AF面AFE,则//MQ面AFE,因为,,BMMQMBMMQ=平面,M
QB所以平面//MQB平面AFE,因为面MQB平面,ABCDQB=面AFE平面,ABCDEF=所以//BQFE,显然,EF为三角形ABD中位线,//EFBD,矛盾,故假设不成立,A错误;以A为坐标原点,AD为y轴正半轴,在平面ABCD中作与AD垂直方向为x轴正半轴,z轴垂直平面ABCD
,建立空间坐标系.因为60A=,1,12AEAF==,所以2221cos22AEAFEFAAEAF+−==,所以32EF=,所以222AEEFAF+=,所以AEEF⊥,即AEEF⊥,又因为//EFBD,则AEBD⊥,若B正确,则有BDAC⊥,因为,,AAEA
EACAC=平面AEC,所以BD⊥平面AEC,因为EC平面AEC,则BDEC⊥必定成立.则根据题意,可得31,,044E、31,,022B、35,,022C、()0,2,0D.33,
,022BD=−,39,,044EC=,则327388BDEC=−+=,即BDEC⊥不成立,故矛盾,所以B不成立;当二面角AEFB−−为直二面角时,即平面AEF⊥平
面EBD.根据上面可知AEEF⊥,所以AEBD⊥,又BDEB⊥,因为AEEBE=,,AEEB平面AEB,所以BD⊥平面AEB,因为AB平面AEB,所以BDAB⊥,故四面体ABDE−为所有面都是直角三角形的四面体,根据外接球性质可知,球
心必为AD中点K,即KB为外接球半径.12AEEB==,3BD=,由勾股定理可知22142ADABBD=+=,则144KB=,外接球面积为2147442=,故C正确.当平面AEF⊥平面ABCD时,直线AC和平面ABCD所成的角的最大,
记此时角为.由上图可知,在EBC中,1120,,22EBCEBBC===,由余弦定理222122cos1201222EC+−=可解得212EC=.此时1212tan21212AEEC===.此时6,故D错.故选:ABD12.CD【详解】因为()()elnln0,0xf
xaxaxa=−+,且()0fx恒成立,所以elnln0xaxa−+,则elnlnlnxxaxaa−=,故1elnxxaa,则elnxxxxaa,当0xa时,e0x,01xa,则ln0xa,故ln0xxaa,则e
lnxxxxaa恒成立,当0ax时,1xa,ln0xa,则ln0xxaa,对elnxxxxaa两边取对数,得lnlnlnlnxxxxaa++,令()()ln0gxxxx=+,则()lnxgxga,又()
110gxx=+,所以()gx在()0,+上单调递增,故lnxxa,即exxa在()0,+上恒成立,令()()0exxhxx=,则()ahx在()0,+上恒成立,即()maxahx,又()1exxhx−=,令()0hx
,得01x;()0hx,得1x;所以()hx在()0,1上单调递增,在()1,+上单调递减,则()()max11ehxh==,故1ea,对于AB,易得112ee,211ee,故AB错误;对于CD,易得11ee=,21ee,故CD正
确.故选:CD.13.2πarccos10−14.-10015.()2211kjjjjnsxxn=+−【详解】解:111,jjnnjjjjjjjjxxxnxn====.∴样本均值11111()()knkjjjjjjxxnxnn=====.又2222
111(),()jjnnjjjjjjjjjjxxxsxnsn===−−=.计算总体22221111()()()2()()()jjjjnnnnjjjjjjjjjjjjjjjxxxxxxxxxxxxnxx====−=−+−=−+−−+−又11()
0jjnnjjjjjjjjjjjxxxnxnxnx==−=−=−=.2221()()jnjjjjjjxxnsnxx=−=+−.222211111()()jnkkjjjjjjjjxxnsnxxnns====−=+−.故答
案为:()2211kjjjjnsxxn=+−16.3【详解】根据题意,为使两交点距离最小,只需两交点在同一周期内;由题意,令3sin3cosxx=,可得sincos0xx−=,则sin04x
−=,所以4xk−=,Zk,即14xk=+;当0k=,14x=,1322y=,当1k=,254x=,2322y=−,如图所示,由勾股定理得()()()222212133y
yxx−+−=,即()()2225323344+−=,即29=,解得:3=.故答案为:317.(1)因为,()0fc=,()()()2111fxaxbxc−−=−
−+−−+()22axabxabc=+−+−+,又()00f=,()()1fxfx=−−,所以有02cabbabcc=−=−+=,解得0abc==,所以()2fxaxax=+,()2fx
axa=+.因为函数()2fxaxax=+与直线22yx=−−相切,设切点为()00,xy,则()02fx=−,()0022fxx=−−,即020002222axaaxaxx+=−+=−−,解得021ax==−,所以,2a=,2b=,0c=,
所以()222fxxx=+.(2)由(1)知,2422nnnSaa=+,即22nnnSaa=+.当1n=时,2111122Saaa=+=,解得11a=或10a=(舍去);当2n时,有22nnnSaa=+,21112nnnSaa−−−=+,所以有()221112nnnnnnSSaaaa−−
−−=+−−,整理可得()()1110nnnnaaaa−−+−−=,因为10nnaa−+,所以110nnaa−−−=,即11nnaa−−=.所以,na是以11a=为首项,1为公差的等差数列.所以,()111naann=+−=,()()1122
nnnaannS++==.则不等式8(1)nnnSa+−对于任意*Nn恒成立,可转化为()()1812nnnn++−,即()()18812122nnnnnn++−=++对于任意*Nn恒成立.①当n为偶数时,即有8122nn++恒成立,因为818192222
22nnnn+++=,当且仅当82nn=,即4n=时等号成立,此时有92;②当n为奇数时,即有8122nn−++恒成立,令()8122xhxx=++,()()()22441822xxhxxx+−=−=,当04x时,()0hx
,()8122xhxx=++单调递减;当4x时,()0hx,()8122xhxx=++单调递增.又()3811432323h=++=,()5812314525253h=++=,所以当n为奇数时,()8122xhxx=++最小值为()235
5h=.所以,235−,即有235−.综上所述,23952−.18.(1)已知cos3sincBbCa+=,根据正弦定理可得:sincos3sinsinsinCBBCA+=,在ABC中,ABC++=,()()sinsinsinABCB
C=−+=+,所以()sincos3sinsinsinsincoscossinCBBCBCBCBC+=+=+,得3sinsinsincosBCBC=,sin0B,3sincosCC=即3tan3C=
,得6C=.(2)由3ACCB=−,得3coscos32CACBCACBCabCab====,即23=ab.根据余弦定理得2222243cos2243abcabCab+−+−===,解得2210ab+=.19.(1)设双曲线C的方程为221(0)mxnymn−
=,将6,12P,(2,2)Q代入上式得:312221mnmn−=−=,解得112mn==,双曲线C的方程为2212yx−=.(2)设()11,Mxy,()22,Nxy,由题意
易得直线l的斜率存在,设直线l的方程为13ykx=−,代入2212yx−=整理得,()22221896180kxkxk−+−−=,21226189kxxk+=−−,212218189kxxk+=−−,21890k−且0
,则1212124433yykkxx−−+=+−−121211443333kxkxxx−−−−=+−−1281124333kkxx=+−+−−()1212126824339xxkkxxxx+−=+−−++()()22222661898
24318189189kkkkkkk−−−=+−−−++−8324334kk=+−−=,故123kk+=为定值.20.(1)底面ABCD是菱形,ACBD⊥,又平面11BDDB⊥平面ABCD,且平面11BDDB
平面ABCDBD=,AC平面ABCD,AC⊥平面11BDDB,又1BB平面11BDDB,1ACBB⊥.(2)解法一:由(1)知AC⊥面11BDDB,又AC平面11ACCA,平面11ACCA⊥平面11BDDB,作BE⊥交线1OO
,垂足为E,因为平面11ACCA平面11BDDB=1OO,BE平面11BDDB,则BE⊥面11ACCA,又1AA平面11ACCA,所以1AABE⊥.再作1BFAA⊥,垂足为F,BE面BEF,BF面BEF,BEBFB=所以1AA⊥面BEF,又面EFBEF则1EFAA⊥,所以B
FE为二面角1BAAC−−的平面角,111111132231,3322AABCDABCDAAAVShhh−====因为11AC//平面ABCD,所以1O到底面ABCD的距离也为32.作1OHOB⊥,因为平面11BDDB⊥平面ABCD,平面11BDDB平面ABCD
=OB,1OH平面11BDDB,所以1OH⊥平面ABCD,所以132OH=,又1OBO为锐角,所以11,60,2BHOBO==又11OBOB==,所以1BOO为等边三角形,故11OO=,所以32BE=,因为//ABCD,所以1121221sinsin77BAABFA
BBAA===,所以3732sin,cos442217BEBFEBFEBF====.所以二面角1BAAC−−的平面角的余弦值为34.解法二:由(1)知AC⊥面11BDDB,又AC平面ABCD,平面ABCD
⊥平面11BDDB,作1OHBD⊥,因为平面11BDDB⊥平面ABCD,平面11BDDB平面ABCD=BD,1OH平面11BDDB,所以1OH⊥平面ABCD,如图,建立直角坐标系:O为原点,,OAOB为,xy轴方向,z轴//1OH.11111132231,3322AABCDABCDAAAV
Shhh−====因为11AC平面ABCD,所以1O到底面ABCD的距离也为32.所以132OH=,又1OBO为锐角,所以111,,60,22BHOHOBO===又11OBOB==,所以1BOO为等
边三角形,故11OO=,在空间直角坐标系中:()()()3,0,0,0,1,0,3,0,0ABC−,设113,,22Aa,则()1133,,,3,1,022AAaDCAB=−==−()11211332732cos,.7213(3)244aDCAADCAAaDCAAa
−−+====−++则()()1313,,,3,1,0,23,0,0222AAABAC=−=−=−,设平面1ABA的法向量为(),,mxyz=,1313022230mAAxyzmABxy=−++==−+=,取(1,3,0)m=设平面1ACA的法
向量为(),,nxyz=,13130222230nAAxyznACx=−++==−=,取()0,3,1n=−所以3cos,4mnmnmn==−,由题知二面角为锐角,故二面角1BAAC−−的平
面角的余弦值为34.21.(1)()fx的定义域为()0,+.∵()()221ln12ln(ln1)0fxxxxxx=++=+,仅当1ex=时取等号,∴()fx的单调递增区间为()0,+.
(2)由题可得211112lneeeeef=+=,若211xxe,则必有()()2112fxfxfee=,则()()124efxfx+;若2110exx,则必有()()1212eefxfxf=,则()()12
4efxfx+.∴若()()124efxfx+=,则1210exx.要证()12lnln21xx+−,只需证122exx+,只需证212exx−,即证()212efxfx−
,又()()214efxfx=−,故只需证()1142eefxfx−−.令()()241,0,eeegxfxfxx=+−−.则()()2222(ln1)ln1eegxfxfxxx=−−=+−−+22
lnlnlnln2eexxxx=−−+−+.∵10ex,∴21xxee−,∴2lnln0exx−−,且2222elnln2ln2ln20ee2xxxxxx+−+−+=−++=
,∴()22lnlnlnln20eegxxxxx=−−+−+,故()gx在10,e上单调递增.∵10ge=
,∴()10egxg=,∴()11240fxfxee+−−,∴()1142eefxfx−−,得证.22.(1)由Shapley值的评判标准知:利用边界贡献()()()iSvSivS=−计算
出员工的Shapley值,使员工所得与员工的贡献率相等,相对比较公平,也可以促进员工之间工作的积极性.(2)由题意知:加入的顺序有6种,①按ABC−−的顺序:()110000A=,()()()111,2700010
00017000BABA=−=−=,()()()111,,,500002700023000CABCAB=−=−=;②按ACB−−的顺序:()210000A=,()()()222,,,500003750012500BABCAC
=−=−=,()()()222,375001000027500CACA=−=−=;③按BAC−−的顺序:()()()333,270001250014500AABB=−=−=,()312500B=,()()()333,,,50000270002300
0CABCAB=−=−=;④按BCA−−的顺序:()()()444,,,500003500015000AABCBC=−=−=,()412500B=,()()()444,350001250022500CBCB
=−=−=;⑤按CAB−−的顺序:()()()555,1000050005000AACC=−=−=,()()()555,,,500003750012500BABCAC=−=−=,()55000C=;⑥按CBA−−的顺序:()()()666,,,5
00003500015000AABCBC=−=−=,()()()666,35000500030000BBCC=−=−=,()65000C=;A的Shapley值为:10000100001450015000325001500016166.76+++++=,B的Shapley值
为:17000125001250012500125003000016166.76+++++=,C的Shapley值为:230002750023000225005000500017666.76+++++=;故A分得奖金的16166.732.3%50000=;
故B分得奖金的16166.732.3%50000=;故C分得奖金的17666.735.4%50000=.获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com