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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】222211222ziii,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233332333log2log41
1log2log42log3log40log2log2mn,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,222224914Srrrr,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22gx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,12380ffff20221123660ififffff,20221237560igifffff。
20231123780ififfffff202312381igiffff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111111110xxxxxxeeeeeexee
,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11ABB
C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,11132
62432AMNABCV,1111112523::2310AMNABCAMNABCVVV,故D错误。所以选ABC。12.【解析】由123aaa,345aaa,…,21221nnnaaa
,故242211nnaaaa,故A正确;因为21nnnaaa,2212nnnnaaaa,222222221233434422245511nnnnnnnnnaaaaaaaaaaaaaaaaaaaa故B正确;假
设1152nnaa为公比为q等比数列,故211151522nnnnaaqaa,即21151522nnnaqaqa,所以15151,122
qq,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】5sin26fx
x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:N2224xy,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max1333323222S。
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,111111nn
naSSnnnn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222448
335CCPABC。··········································································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:sinsin6ABBD
ADB,所以12262sin33ADB。·······································6分故6sin3ADC,3cos3ADC,所以3sincos3CADC。323si
nsinsincoscossin6666BADCADCADC··············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3c
os3ADC,所以6sin3ADC。··········································6分所以6sinsin3ADBADC。后续同①选③:在ABD中,由正弦定理得:si
nsin6ABBDADB;在ACD中,由正弦定理得:sinsin2ACCDADC。········································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P
E平面PBC,故PEBC.又BEBC,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,2
2220,,,1,,2222EPCPzyxOEBDCAP第7页共9页0,0mEPmCP,故11111022022yzxyz,取111221,,2
2xyz,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ykxm0k,故,2mNmk·······································6分与双曲线方程联立得:2223163120kxkmxm,由已知得
231,0k,设1122,,,PxyQxy,则21212226312,3131kmmxxxxkk①··················································8分由,PMP
NMQQN得:11mxxk,22mxxk消去得:2112mmxxxxkk,即121220mxxxxk②···········
·············································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33322xxxe
gxexx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00Fx00,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
2··························································12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com