高三数学答案-1.pdf

该【高三数学答案-1 】是由【和合】上传分享，文档一共【6】页，该文档可以免费在线阅读，需要了解更多关于【高三数学答案-1 】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。:..公众号:高中试卷站、助力高中绝密★考试结束前2023-2024学年第二学期天域全国名校协作体联考高三年级数学学科参考答案一、单项选择题:本题共8小题,每小题5分,,、多项选择题:本题共3小题,每小题6分,,,部分选对的得部分分,、填空题:本题共3小题,每小题5分,、解答题:、?ax2?x?a15.(1)?f(x)?lnx?ax?(x?0),?f?(x)??a?2?2,xxxx若f?x?在定义域内是减函数,则f?(x)?0对?x?(0,??)恒成立,即?ax2?x?a?0恒成立,21所以a?0,??1?4a?0,解得a?..................................6分2(2)当a?0时,f￠(x)>0,?f(x)在(0,??)上单调递增,?f(x)无极值点;...................8分1?1?1?4a21?1?4a2?当0?a?时,??1?4a2?0,令f￠x>0,解得x??,?,()??2?2a2a??1?1?4a2??1?1?4a2?令f??x??0,解得x??0,???,???,?2a??2a??????1?1?4a2??1?1?4a21?1?4a2?则f(x)在?0,?上是单调递减,在?,?上是单调递增,在?2a??2a2a??????1?1?4a2?22?,????f(x)1?1?4a1?1?4a??上是单调递减,的极小值点为,极大值点为.?2a?2a2a.........................................12分高三数学参考答案第1页(共6页):..公众号:高中试卷站、助力高中11?1?4a2综上,当a?0时,f(x)无极值点;当0?a?时,f(x)的极小值点为,极大值点为22a1?1?4a2..................................13分2a88?(xi?x)(yi?y)?xiyi?8xy?i?1i?182743?8?86?11216.(1)由题意b?8?8?2?........................3分22262680?8?86?(xi?x)?xi?8xi?1i?1a??y?b?x??26.......................6分故y关于x的线性回归方程为y???26..........................................7分(2)设零假设为H0:是否报废与是否保养无关由题意,报废推进器中保养过的共20×30%=6台,未保养的推进器共20-6=×2列联表如下:保养未保养合计报废61420未报废542680合计6040100..........................11分100?(6?26?14?54)2则K2???..................................14分20?40?60?80根于小概率值α=,我们推断H0不成立,即认为是否报废与是否保养有关,...........................................15分17(.1)作BH?PE交PE于H,因为平面PAE?平面PBE,且平面PAE?平面PBE?PE,所以BH?平面PAE,又因为AE?平面PAE,所以BH?AE,因为PB?平面ABC,且AE?平面ABC,所以PB?AE,因为BH?AE,PB?AE,PB、BH?平面PBE,PB?BH?B,所以AE?平面PBE,又因为BE?平面PBE,所以AE?BE..........................................3分分别以直线BA,BC,BP为x轴,y轴,z轴建立空间直角坐标系,如图,高三数学参考答案第2页(共6页):..公众号:高中试卷站、助力高中则B?0,0,0?,P?0,0,2?,C?0,2,0?,A?2,0,0?????????????????设E?x,y,0?,因为AE?BE,所以AE?BE?0,AE?(x?2,y,0),BE?(x,y,0),22所以(x?2)?x?y?y?0,即(x?1)?y?1..........................................5分设AB中点为N,则N(1,0),如图:?ππ??π2π?又?ABE??,?,所以?ANE??,?,?83??43?2ππ5π因此,E的轨迹为圆弧QE,其长度为(?)?1?...........................................7分3412????????(2)由(1)知,可设E(x,y,0),PA??2,0,?2?,AE?(x?2,y,0),?设平面PAE的一个法向量为n??a,b,c?,?????????n?PA?0,?2a?2c?0?则?????即?,令a?y则b?2?x,c?y,n??y,2?x,y?..................?a?x?2??by?010分????n?AE?0,?BC??0,2,0?为平面PAB的一个法向量,令二面角E?PA?B为角?n?BCx?2x?23cos???,因为22所以cos???2(x?1)?y?12nBC?x?2??2y24?x3高三数学参考答案第3页(共6页):..公众号:高中试卷站、助力高中解得x?2,y?0(舍去)或x?1,y??1则E(1,1,0),或E?1,?1,0?..........................................13分从而可得三棱锥E?PCB的体积1112VE?PCB?S?PCB?h???2?2?1?..........................................15分332322221e1cb?c118.(1)由题意知2?2?2?22?22?2?1,abaababb解得b?1,所以a?2;x2所以椭圆W的方程为?y2?1;........................................4分4(2)①若直线AB的斜率存在,设AB的方程为y?kx?m1,A(x1,y1),B(x2,y2),因为ABCD,故可设CD方程设为y?kx?m2,?y?kx?m1?2222由?x2得(1?4k)x?8km1x?4m1?4?0,??y?1?4?8km1?x1?x2??222?1?4k则??8(2k?m1?1)?0,且?2,.......................................7分?4m1?4x1x2?2????1?4k2222244k?m1?1所以|AB|?1?k(x1?x2)?4x1x2?1?k,21?4k44k2?m2?1同理|CD|?1?k22,.......................................9分1?4k2因为|AB|?|CD|,所以m2?m2,因为m?m,所以m?m?0........................................11分121212|m1?m2||2m1|设两平行线AB,CD间的距离为d,则d?2,因为m1?m2?0,所以d??k1?k224k2?m2?1?m24k?m?1|2m|22211所以S?|AB|?d?41?k21?1(4k?m?1)?811?82?41?4k1?k221?4k1?4k4k2?1?2m2ABCDS所以当1时,的面积取得最大值为4........................................14分②若直线AB的斜率不存在,此时平行四边形ABCD为矩形,设A(x1,y1),易知S?4|x1y1|,x2x2又1?1?y2?21?y2?|xy|,所以S?4,当且仅当x?y时取等;...............................16分11111144综上所述:ABCD的面积S的最大值为4..........................:(1)由题设可知,正三角形R的对称变换如下:绕中心O作120°的旋转变换高三数学参考答案第4页(共6页):..公众号:高中试卷站、助力高中?123??123?m1???;绕中心O作240°的旋转变换m2???;绕中心O作360°的旋转变换?312??231??123??123?m3???;关于对称轴r1所在直线的反射变换l1???;关于对称轴r2所在直线的反?123??132??123??123?射变换l2???;关于对称轴r3所在直线的反射变换l3???.?321??213???123??123??123??123??123??123??综上,S????,??,??,??,??,???.(形式不唯??312??231??123??132??321??213??一)..............3分?a1a2a3??b1b2b3??a1a2a3??b1b2b3??a1a2a3?(2)①I.???,???S,?????????S;?b1b2b3??c1c2c3??b1b2b3??c1c2c3??c1c2c3??a1a2a3??b1b2b3??c1c2c3?II.???,??,???S,?b1b2b3??c1c2c3??d1d2d3???a1a2a3??b1b2b3???c1c2c3??a1a2a3??c1c2c3??a1a2a3??????????????????????b1b2b3??c1c2c3???d1d2d3??c1c2c3??d1d2d3??d1d2d3??a1a2a3???b1b2b3??c1c2c3???a1a2a3??b1b2b3??a1a2a3?????????????????????b1b2b3???c1c2c3??d1d2d3???b1b2b3??d1d2d3??d1d2d3?所以??a1a2a3??b1b2b3???c1c2c3??????????????b1b2b3??c1c2c3???d1d2d3??a1a2a3???b1b2b3??c1c2c3????????????;..............5分?b1b2b3???c1c2c3??d1d2d3???123??a1a2a3?III.????S,????S?123??b1b2b3??a1a2a3??a1a2a3??a1a2a3??a1a2a3??b1b2b3???????????????,?a1a2a3??b1b2b3??b1b2b3??b1b2b3??b1b2b3??a1a2a3??b1b2b3??123??123?而????????,所以e???;?a1a2a3??b1b2b3??123??123?高三数学参考答案第5页(共6页):..公众号:高中试卷站、助力高中IV.?a1a2a3??b1b2b3?????S,????S,?b1b2b3??a1a2a3??a1a2a3??b1b2b3??b1b2b3??a1a2a3?????????????e;......................................7分?b1b2b3??a1a2a3??a1a2a3??b1b2b3?综上可知,集合S对于给定的新运算*来说能作成一个群...................................8分②e?e?,a?1?a?,证明如下:先证明e?e?:由于H是G的子群,取a?H,则a?G,a?1?G,根据群的定义,有a?e?a,a?e??a,所以a?e?a?e?,所以a?1(a?e)?a?1(a?e?),即(a?1?a)?e?(a?1?a)?e?,即e?e?e?e?,所以e?e?....................................10分再证明a?1?a?:由于e?e?,e?a?1?a,e??a??a,所以?1,所以?1?1??1,a?a?a??aa?(a?a)?a?(a?a)所以?1?,所以?1?.....................................12分a?e?a?ea?a③S的所有子群如下:??123????123??123??H1?????,H2????,???,??123????123??132????123??123????123??123??H3????,???,H4????,???,..................................14分??123??321????123??213????123??123??123??H5????,??,???,??312??231??123????123??123??123??123??123??123??H6????,??,??,??,??,???.......................17分??312??231??123??132??321??213??高三数学参考答案第6页(共6页)