2024年1月10日九省联考高三数学真题及答案解析.pdf

该【2024年1月10日九省联考高三数学真题及答案解析 】是由【小屁孩】上传分享，文档一共【10】页，该文档可以免费在线阅读，需要了解更多关于【2024年1月10日九省联考高三数学真题及答案解析 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:..2024年1月10日九省联考高三数学真题及答案解析一、选择题:本题共8小题,每小题5分,,,24,14,10,20,30,12,14,40的中位数为()?y2?1?a?1?,则a?()?a?的前n项和为S,a?a?6a?17S?(),,?,?是两个平面,m,l是两条直线,则下列命题为真命题的是()???,m∥?,l∥?,则m???,l??,m∥l,则?∥?????m,l∥?,l∥?,则m∥??,l??,m∥l,则???、乙、丙等5人站成一排,且甲不在两端,乙和丙之间恰有2人,则不同排法共有():x?2y?1?0PQP??1,?3?,点满足,记的轨迹为,则()?3?????1?2sin2????,??,tan2???4tan????,则?()?4??4?2cos2??sin2?:..x2y2C:??1?a?0,b?0?的左、右焦点分别为F,F,,BFB?2FA,FA?FB?4a2,则C的离心率为()线与交于两点,、选择题:本题共3小题,每小题6分,,,部分选对的得部分分,有选错的得0分.?3???3??f?x??sin2x??cos2x?,则()?????4??4??????f?x???fx的对称轴为x?k?,k??4?????f?x?,f?x?????32?,w均不为0,则()zz2z2?.??w?z?wD.?ww?1?????????,且f???0,若fx?y?fxfy?4xy,则()?2??1??1?????????2?2??2??1??1??x???x??是减函数?2??2?三、填空题:本题共3小题,每小题5分,共15分.??A???2,0,2,4?B?xx?3?mA?B?,,若,?的高与球O的直径相等,则圆锥MM?的体积与球O的体积的比值是,圆锥MM?:..?a?b?c?1,已知b?2a或a?b?1,则max?b?a,c?b,1?c?、解答题:本题共5小题,、?x??lnx?x2?ax?2?2,f?2??2x?3y?015.(13分)已知函数在点处的切线与直线垂直.(1)求a;f?x?(2).(15分)盒中有标记数字1,2,3,4的小球各2个,随机一次取出3个小球.(1)求取出的3个小球上的数字两两不同的概率;XXE?X?(2)记取出的3个小球上的最小数字为,求的分布列及数学期望17.(15分)如图,平行六面体ABCD?ABCD中,底面ABCD是边长为2的正方形,1111O为AC与BD的交点,AA?2,?CCB??CCD,?CCO?45?.1111(1)证明:CO⊥平面ABCD;1(2)求二面角B?AA?:..18.(17分)已知抛物线C:y2?4x的焦点为F,过F的直线l交C于A,B两点,过F与l垂直的直线交C于D,E两点,其中B,D在x轴上方,M,N分别为AB,DE的中点.(1)证明:直线MN过定点;(2)设G为直线AE与直线BD的交点,求???1,2,?p?1?19.(17分),集合,若u,v?X,m?N,记u?v为uv除以p的余数,um,?为um除以p的余数;设a?X,1,a,a2,?,?,ap?2,?两两不同,若an,??bn??0,1,?,p?2?b(),则称n是以a为底n?log?p?,记为a(1)若p?11,a?2,求ap?1,?;m,m??0,1,?,p?2?m?m为m?m除以p?1的余数(当m?m(2)对,记12121212p?1m?m?0log?p??b?c??log?p?b?log?p?c,能被整除时,).证明:12aaa其中b,c?X;????(3)已知n??X,k?1,2,?,p?2,令y?ak,?,y?x?bk,?.a12xyyn?p?2?,?.证明:??214:..答案解析一、选择题,:将数据从小到大排列10,12,14,14,16,20,24,30,40,共9个数,?:e???,a?:由等差数列性质可得a?a?2a?6,∴a?3,3755S?8?a?a??8?a?a??8??3?17??:A:举反例,????p,若m∥p,l∥p,m∥?,l∥?,得m∥l,故A错;B:仅有一组平行直线得不出两个平面平行,故B错;D:垂直于同一条直线(一组平行直线)的两个平面平行,:分两种情况:乙丙站1,4号位或者站2,5号位,先排乙丙,再排甲(去掉一端2?A2?C1?A2?16仅有两个选择),?x?x?1????????0Px,yQx,y,QP?1,?3?x?x,y?y,:设,?,0000y?y?3?0x?2y?1?0x?1?2?y?3??1?0x?2y?6?0由,故,?3????????,???1?tan??0tan2???4tan????:∵,∴,∵,?4??4?2tan?tan??11??4tan???或tan???2(舍),∴,∴1?tan2?1?tan2?21?2sin2?sin2??cos2??2sin?cos?tan2??1?2tan?1∴???.2cos2??sin2?2cos2??2sin?cos?2?2tan?:FB?2FA?2k,则FB?FB?2a??FB?4a2?2a?4acos?AFB,cos?AFB?,(∵对角线互相平分,可知四22222边形AFBF为平行四边形,故有).215:..?4a?2??2a?2??2c?21cos?FBF???,解得c2?7a2,e??4a?2a2二、选择题3?3?3?3?f?x??sin2xcos?cos2xsin?cos2xcos?:44442222??sin2x?cos2x?cos2x?sin2x??2sin2x,2222??????A:f?x????2sin?2x???2cos2x为偶函数,故A对;?4??2??????2??f?k????2sin2k??0x?,2x?,?,故C对;B:,故B错;C:??时,???32??3?f?x?2D:的最小值为,:令z?a?bi,w?c?di,a,b,c,d?R,222z2?a2?b2?z2A:z?a?b?2abi,,故A错误;za?bi?a?bi??a?bi?z2B:???,故B正确;za?bi?a?bi??a?bi?z2z?w??a?c???b?d?iz?w??a?c???b?d?iz?a?biw?c?diC:,故,又,,z?w??a?bi???c?di???a?c???b?d?i?a?w故,故C正确;zabi?ac?bd???bc?ad?i?acbd?2?bcad?2????D:???wc?dic2?d2c2?d2????za2c2?b2d2?b2c2?a2d2a2?b2?c2?d2a2?b2????,c2?d2c2?d2c2d2w??1??1??1???x?0y?f?f?0?f?0,f?0,f0??1,:①令,得??????2?2??2??2?11?1??1??1?x?y??f?0??ff????1f??0②令,,得???,??,故A正确;22222??????1?1??1??1??1?fx??f??f?x??4??x??2xfx???2x③令y??,则??????,即??,22222????????6:..故C错误;?1?④令x?1,则f????2,故B正确;?2??1?x?x?1fx???2?x?1?⑤令,则??,故D正确.?2?三、:∵A?B?A,则A?B,故x?3??2?3?m,故m?:3;1解析:设圆锥底半径为r,则母线长l?2r,高h?3r圆锥体积13V??r2?h??r3,表面积S??r2??r?l?3?r2,1331343O2R?3rR?rV??R3??r2,球的表面积球的直径,即,球的体积2232S?4?R2?3?r2,故V:V?2:3S:S?1:1,.212121?a?1?m?n?:令b?a?m,c?b?n,1?c?p,m,n,p?0,则?,5b?1?n?p?①令b?2a,即1?n?p?2?2m?2n?2p,∴2m?n?p?1,m?max?b?a,c?b,1?c??max?m,n,p?令,?2M?2m?1∴M?n,即4M?2m?n?p?1,∴M?;?4?Mp??a?b?12?m?2n?2p?1m?2n?2p?1M?max?m,n,p?②若,即,∴,令,?M?m?1∴2M?2n,即5M?m?2n?2p?1,∴M?,?5?2M2p??1当m?2n?2p时可取等号,故M?.min5四、解答题f?x??lnx?x2?ax?2?2,f?2??:(1)设函数在点处的切线的斜率为,7:..119f??x???2x?ak?f??2???4?a??a,由题可得,则x22?2?3∵切线与直线2x?3y?0垂直,∴k??????1,解得k?,∴a????f?x??lnx?x2?3x?2x?0(2)由(1)可得,11f??x???2x?1??x?1??0,得x?令或1,x21???1???f??x??0x?10?x?,∴fx在0,?,1.??上单调递增;令得或?22??1?1?f??x??0?x?1f?x?,1令得,∴在??上单调递减,22??1?1?3????故fx在x?处有极大值f????ln2,在x?1处有极小值f1??2?4C3?234P?A??4?;:(1)法一:设事件A为取出的3个小球的数字两两不同,故C3784C124334法二:有两个相同的概率:P?6??,则两两不同的概率为1??.C3567778(2)X的取值可取1,2,3C1C2?C1369P?X?1??266??最小数是一个1或两个1,另外的数任选,则C356148C1C2?C1162P?X?2??244??最小数是一个2或两个2,另外的数从3或4中选,则C35678C1?C141P?X?3??22??.最小数是一个3或两个3,另外的数是4,则C356148则X的分布列为:X123921P1471492110XE?X????2??3?:(1)O中,由余弦定理易求得CO?2,118:..∴CO2?2,则CO?CO;111?CCB??CCD,CB?CD,则?BCD为等腰三角形,11111则CO?BD,CO?(2)以OB,OC,OC为x,y,z轴建立空间直角坐标系,1????????则B2,0,0,A0,?2,0,D?2,0,0,A0,?22,2,1????在平面BAA中,AA?0,?2,2,BA??2,?2,0,11???则可求得平面BAA的法向量m?1,?1,?1,1????在平面AAD中,AD??2,2,0,AA?0,?2,2,11???则可求得平面AAD的法向量n?1,1,1,1设二面角B?AA?D的平面角为?,11?1?1122cos????sin??.则,则3333?1F?1,0?l:x?ty?1,t?0l:x??y?:(1)由题意可得焦点,设,,ABDEt????????设Ax,y,Bx,y,Dx,y,Ex,?y2?4x?y?y?4t?y2?4ty?4?0?12??,x?ty?1yy??4??12?y2?4x?4?4?y?y??1?y2?y?4?0?34t??,x??y?1t??yy??4?t?3414x?x?t?y?y??2?4t2?2x?x???y?y??2??2,∵,121234t34t2?22??2?1∴M2t?1,2t,N??,??,则可求l的直线方程为:t2tMN??9:..2??2tt??1??y?2t?x?2t2?1?x?2t2?1,21?2t2?tt2t?1?l??整理得??t?y?x?3?0,从而直线过定点3,???GMNGx??1MN?3,0?(2)面积,由极点极线知在上,过定点,1?2?S??4??2t???8,当且仅当t?1时等号成立.?GMN2t??∴?,?就是210?1024210,??:(1),?p?b?mlog?p?c?m(2)设,,a1a2∴am1除以p的余数为b,am2除以p的余数为c,am??p?bam??p?cm,m??0,1,?,p?2??,??N1,2,,12?log?p??b?c??m?m证:,a12am?m??p?b???p?c???p2???c?b??p?bc12??,pppam?mpbcpb?c12除以的余数为除以的余数,即为,log?p??b?c??m?m?log?p?b?log?p?c.∴a12aamnbm?b?modn?(3)若除以的余数为,可记为,????由n?logpb,∴an?bmodp,an?p2?,??????n?p?2???∴y?y???y?ynp?2modp?x?bk?akmodp2121?xank?ank?p?2??modp??x?ank?p?1??modp??p?1?nk??p?1nk?????xamodp(∵a除以p的余数为1)?x?1modp??p2?,??n?p?∵y?y??,x?0,1,?,p?1,∴x?y?y?2212110