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第1页共9页数学参考答案题号123456789101112答案BACDCABCABDABDABCABD1.【解析】ln01,Axx，20,4Bxx，故1,4AB，选B。2.【解析】22

2211222ziii，故选A。3.【解析】对于A选项，样本相关系数r来刻画成对样本数据的相关程度，当r越大，则成对样本数据的线性相关程度越强，故A正确；对于B选项，经验回归方程ybxa一定经过样本中心点,xy，故B

正确；对于C选项，相关指数2R来刻画模型的拟合效果，若2R越大，则相应模型的拟合效果越好，故错误.对于D选项，残差平方和越小，则相应模型的拟合效果越好，故C正确；故选C。4.【解析】由已知得3log41m

，2log31n，又233332333log2log411log2log42log3log40log2log2mn，所以1nm，故选D。5.【解析】由已知得2222abac，解

得512e，故选C。6.【解析】取,BCBD中点,EG，可知2ABADAG，且BGEG，取BE的中点O，则G为圆O上一点，所以AG的最大值为16AO=，故ABAD的

最大值为12，所以选A。7.【解析】圆台12OO与球O的截面图如图所示，设球O的半径为R，圆台12OO的上、下底面半径分别为,2rr，则母线长为3r，由已知得222Rr。22148SRr，222224914Srr

rr，212284147SrSr，即1247SS，故选B。8.【解析】由gx为偶函数，故gxgx，即11fxfx，所以fx图象关于1x对称；GEOABCDr2r1RRRr2r1OO2O1第2页共9页22g

x为奇函数，故2gx为奇函数，gx图象关于20,对称，fx图象关于10,对称。fx是周期为8T的函数。1370fff，12380ffff20221123660ififffff

，20221237560igifffff。20231123780ififfffff202312381igi

ffff故选C。9.【解析】12xxxxfxeeee，当且仅当10xex即时，等号成立，故A正确；xxfxee，故fx在,单调递增，故B正确；设曲线yfx的切点为00,xy，故000xxfxee，所

以002xxee，解得00x，曲线yfx的切线方程为2yx，故D正确；设曲线yfx的切点为11,xy，故曲线yfx的切线方程为11111xxxxyeeeexx，所以有111111110xxxxx

xeeeeeexee，无解，故C错误.所以选ABD.10.【解析】抛物线2:4Cxy的焦点为0,1F，准线为1y。2,3M在抛物线上方，PMPF的最小值为M到准线1y的距离4，故A正确；由已知0,4PNF，故

2cos,12，所以B正确；由于点3,2在C的下方，故只有一个公共点的直线有三条，所以C不正确；设PQ的中点为G，由已知及抛物线定义知PQPFQF，所以直线PQ过F，故D正确。

故选ABD.11.【解析】如图，1,OO分别是正三棱台的底面中心，因为3OA，1133OA,可求得1263hOO。由棱台体积公式13VSSSSh，可得四面体11ABBC的体积111332623432ABBCV，故A正确

；设四面体11ABBC的外接球的半径为r，因为2211113OAOOOAOA，故3rOA，2412Sr，故B正确；截棱台所得截面为长方形11MNCB，故其面积为2，故C正确；第3页共9页棱台体积1333261323334436V，11132624

32AMNABCV，1111112523::2310AMNABCAMNABCVVV，故D错误。所以选ABC。12.【解析】由123aaa，345aaa，…，21221nnnaaa，故24221

1nnaaaa，故A正确；因为21nnnaaa，2212nnnnaaaa，222222221233434422245511nnnnnnnnnaaaaaaaaaaaaaaaaaaaa故B正确；假设1152n

naa为公比为q等比数列，故211151522nnnnaaqaa，即21151522nnnaqaqa，所以15151,122qq，

矛盾，故C不成立.假设1152nnaa为公比为q的等比数列，故211151522nnnnaaqaa，即21151522nnnaqaqa

，由已知得：15151,122qq，解得152q，所以D正确.故选ABD.13.【解析】008604.8，故进球总数的第60百分位数是169.14.【解析】5sin26fxx等.15.【解析】2ln2xfxaxx有

两个正实数根，即2lnxax有两个正实数根令2lnxgxx，则312lnxgxx，由0gx，得xe。当0,xe时，0gx，gx递增；当,xe时，0gx，gx递减。又10g，

12gee，故10,2ae.第4页共9页16.【解析】设22:11Mxy，则1,0M半径为1；圆:N2224xy，则2,0N，半径为2.（如图）当且仅当AMOB，

BNOA时，OAB面积最大。此时设AOM，22OBNMOD，4ONE。在ONE中，由42得6。此时23,23,3OAOBAOB，故max133

3323222S。17.【解析】（1）由已知，0,1nnSS，+112nnSS，所以111111111111121nnnnnnnSSSSSSS故数列11nS为公差为1等差数列·······

·······························································4分（2）因为12a，不满足条件，此时12S，1111S由（1）知数列11nS

为首项为1公差为1等差数列，所以11nnS，故11nSn当2n时，111111nnnaSSnnnn由11000na，故1111000nn，即11000nn因为nN，所以33n。故满足11000na

的n最小值为33·································10分18.【解析】（1）由已知得：设“选出的4名运动员中恰有2名种子选手”为事件A，22444836187035CCPAC····································

···································3分设“选出的4名运动员中恰有2名种子选手来自团体A”为事件B，xyαEDNMOAB第5页共9页222448335CCPABC。·······························

···········································4分故选出的4名运动员中恰有2名种子选手，这2名种子选手来自团体A的概率为16PABPBAPA························

···························································6分（2）种子选手人数为2m，非种子选手人数为6m，设选出的种子选手的人数为i:42648iimmCCPXiC

，1(1)(3)(2)(4)PXiiimPXimii易得当12mi时，(1)()PXiPXi·····················································10分故122m即3m时，

事件X=2的概率最大。·······················································12分19.【解析】（1）由已知及正弦定理得：ababbcc，即222bcabc………

…………………………………2分由余弦定理得：2221cos22bcaAbc，又0,A，所以23A.···············································································4分（2）选①

：在ABD中，6BAD，由正弦定理得：sinsin6ABBDADB，所以12262sin33ADB。·······································6分故6sin3ADC，3cos3ADC，所以3sincos3CADC。323s

insinsincoscossin6666BADCADCADC··············8分在ABC中，sinsinABACCB，故sin232sinABBACC.·

···································10分所以ABC的面积12sin32323SABAC。···········································12分BCAD第6页共9页选②：因为3cos3ADC，所以6sin3AD

C。··········································6分所以6sinsin3ADBADC。后续同①选③：在ABD中，由正弦定理得：sinsin6ABBDADB；在ACD中，由正弦定理得：sinsin2AC

CDADC。········································8分615ABBDACCD，故232AC···················································10

分所以ABC的面积12sin32323SABAC。··········································12分20.【解析】（1）连接BE，由已知可得：2BECD又1PEDE

，1PB在PBC中，222PBBCPC，故PEPB又PEPC，且PBPCP，PE平面PBC…………………2分因为PE平面PAE，所以平面PAE平面PBC………………4分（2）取BE的中点O，连接OP。由（1）知PE平面PBC，故PEBC.又B

EBC，所以BC平面PBE，故BCOP又1PBPE，所以OPBE，所以OP平面ABCE。……………………………6分以,OBOP所在直线分别为,yz轴，建立空间直角坐标系。则222221,,0,0,,0,0,,0,1,,0,0,0,22222ABECP

，设(01)AQAP，则221,2,22BQ设平面PEC的法向量为111,,mxyz，22220,,,1,,2222EPC

PzyxOEBDCAP第7页共9页0,0mEPmCP，故11111022022yzxyz，取111221,,22xyz，则221,,22m。…………

………………………9分故设直线BQ与平面PEC夹角为，则3sincos,6BQmBQmBQm，则13()22或舍去，即Q为线段AP的中点,此时22AQ………………………………

………………………12分21.【解析】（1）由已知2222:10,0yxCabab，4c，焦点120,4,0,4FF，222126864aAFAF···································2分所以2a，22212bca

，故22:1412yxC···············································4分（2）设l的方程为21ymm，则0,2Dm，故0,Mm设直线PQ的方程为ykxm0k，故,2mNmk

·······································6分与双曲线方程联立得：2223163120kxkmxm，由已知得231,0k，设1122,,,PxyQxy，则21212226312,3131kmmxxx

xkk①··················································8分由,PMPNMQQN得：11mxxk，22mxxk消去得：2112mmxxxxk

k，即121220mxxxxk②························································10分由①②得：220km，由已知2m第8页共9页

故存在定直线:22ly满足条件································································12分22.【解析】（1）函数fx的定义域为,00,。21xfxex，记gxfx，则

33322xxxegxexx当0,x时，0gx，故gx在0,上单调递增当,0x时，记32xxxe，23xxxxe所以,3x时，

0x，x递减；3,0x时，0x，x递增x的极小值为33320e，故0x故0gx，所以gx在,0上单调递减。综上：故fx

在0,上单调递增，在,0上单调递减···································4分（2）①当0a时，因为0x，故0fx，此时不满足条件；当0a时，令Fxfxax，所以Fxfxa由（1）可知fx在0,

上单调递增，故Fx在0,上单调递增。由指数函数性质可知：0,;,xFxxFx故存在00x，使得00Fx00,xx，00Fx，Fx单调递

减；0,xx，00Fx，Fx单调递增①若1ae，则110Fea，不符合题意；②若01ae，110Fea当01x时，0,1x，0Fx，不符合题意当01x时，1,x，0Fx

，不符合题意第9页共9页③若1ae，则110Fea，110Fea，所以01x又0,;,xFxxFx，故存在1201xx，使得120FxFx，满足题意；综上：实数a的取值范围是1,e·······

·················································8分②因为1221,(0)xxxx是fxax的两个根，所以12121211,xxeaxeaxxx，由知120,1xx，于是111111113xaxexxx，（其中用到

了1(0)xexx），故113ax；又222222222111(2)132xxaxeexxexxxx，故21136ax，（其中用到了(1)xeexx）；将上面两个结果相加即可得：1112136axx

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