文档介绍：该【河南省南阳市2023-2024学年高三上学期期中数学试题及答案 】是由【小屁孩】上传分享，文档一共【9】页，该文档可以免费在线阅读，需要了解更多关于【河南省南阳市2023-2024学年高三上学期期中数学试题及答案 】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。:..2023年秋期高中三年级期中质量评估数学试题注意事项:(选择题)和第II卷(非选择题),在本试卷上答题无效。,考生务必先将自己的姓名、,(签字)笔或碳素笔书写,字体工整,(黑色线框)内作答,,不折叠、不破损。第I卷选择题(共60分)一、选择题(本题共8小题,每小题5分,,只有一项是符合题目要求的),表示空集的是??????2,且x?2?????Nx2?1??“?x?R,x2?x?1?0”的否定为000A.?x?R,x2?x?1?0B.?x?R,x2?x?1?0C.?x?R,x2?x?1?0D.?x?R,x2?x?1?0z?1?z?i?2z?z?,则A.?.??a?aa?aa?16aaaa?,若,?x??4x??a?1?2x?a2?,则实数的取值范围为?7????7??1?55?A.?1,B.?1,,D.?,????????3??3?23??????0,x?sin??sin?y?cos??sin?z?sin??cos????,?,?,?,则?4??y??z??x??x?,b,c分别为△ABC的三个内角A,B,C的对边,若点P在△ABC的内部,且满足:..?PAB??PBC??PCA??,则称P为△ABC的布洛卡(Brocard)点,?:PAPBPCcot??cotA?cotB?cotC(注:tanxcotx?1).则???????1f?x??aex?x2?ax?0,???,则实数的取值范围为2???,?1????,?1??0,????0,???、选择题(本题共4小题,每小题5分,,,部分选对的得2分,有选错的得0分.)f?x??sin??x???f?x??,则函数????????2x?????3??3?????5????2x?????6??6?S?a?n3S?a?,,则nnnn?a?a?a???,y?R,若4x2?y2?xy?1,则x?y的值可能为A.?2B.??0x?af?x??a?x?a?2?x?b?,若为函数的极小值点,?0且a??0且a??0且a??0且a?b第II卷非选择题(共90分)三、填空题(本题共4小题,每小题5分,共20分),整数部分和自身成等差数列,则这个正实数是______.????????,AD?2,CD?3,BD是四边形ABCD的外接圆的直径,则AC?BD?______.:..f?x?f?2?x??f?1?x?f??1??2023f?2023??,,?x?ax2b?,,满足是,的等比中项,则函数???、解答题(本题共6小题,、证明过程或演算步骤.)17.(本小题满分10分)????设数列?a?为等差数列,其前n项和为Sn?N*,?b?1,nnn11a?3b,S??a??b?(1)求数列和的通项公式;nn(2)求数列?a?b?.(本小题满分12分)1f?x??3sin?xcos?x?sin2?x???0xxf?x??f?x??2已知函数,其中,若实数,满足时,21212?x??f?x?(1)求的值及的单调递减区间;2???????(2)若不等式?f?x???2acos2x??2a?2?0对任意x??,时恒成立,求实数a的取值范围.???????6??126?19.(本小题满分12分)2SS?a????.nnnn?a?是等差数列;(1)证明:n?1?(2)若a,a,a成等比数列,求数列的前2024项的和.??137aa?nn1??20.(本小题满分12分)在△ABC中,角A,B,C的对边分别为a,b,c,且满足_____.(从以下两个条件中任选一个补充在上面横线上作为已知,将其序号写在答题卡的横线上并作答.)条件①:?b?c??sinB?sinC??asinA?3bsinC???5条件②:cos2?A?cosA????2?4(1)求角A;(2)若△ABC为锐角三角形,c?1,求△ABC面积的取值范围.:..21.(本小题满分12分)????????已知函数f?x??x3?x,gx?x2?a,a?R,曲线y?fx在点x,fx处的切线也是曲线11y?g?x?的切线.(1)若x?1,求a;1(2).(本小题满分12分)f?x??xlnxg?x??f?1?x??f?1?x?(1)已知函数,判断函数的单调性并证明;11?n1?1??n1?1?n2(2)设n为大于1的整数,证明:?n?n?.:1-::.?515.?:?a?d?b?:(1)设等差数列的公差为,等比数列的公比为,nnS?4S4a?6d?4?2a?d?由可得,4211即6d?4?4?d?2?,解得d?2,a?a??n?1?d?1?2?n?1??2n?1,所以,n13b?3q?a?9,∴q?325则b?bqn?1?3n?1;n1(2)ab??2n?1??3n?1,nn则T?1?30?3?31?5?32??????2n?1??3n?1①,n:..可得3T?1?31?3?32??????2n?3??3n?1??2n?1??3n②,n6?13n?1??2T12?31323n?1??2n1?3n1?2n1?3n①?②得:?????????????????n1?3??2?2n??3n?2,因此,T??n?1??3n?1n1f?x??3sin?xcos?x?sin2?x?:(1)231?cos2?x1?sin2?x??22231?sin2?x?cos2?x22????sin2?x????6??????因为实数x,x满足fx?fx?2时,x??f?x?T?????1所以的最小正周期,解得,2??????3?所以f?x??sin2x?,由2k???2x??2k???k?Z?,得???6?262??2??f?x?k??,k???k?Z?的单调递减区间为???63?2????????f?x???2acos2x??2a?2?0x??,(2)不等式????对任意??时恒成立,?6??126?2????f?x???2acos2x??2a?2?????6????????sin22x??2acos2x??2a?2?????6??6?????????cos22x??2acos2x??2a?1,?????6??6???????????t?cos2x?2x??0,cos2x???0,1?令??,??,∴???6?6?2??6?:..?t2?2at?2a?1?0t??0,1?,t2?12a?t?1??t2?1,2a?恒成立t?1t2?1m2?2m?22令m?t?1???1,0?,??m??2??1t?1mm1∴2a??1,解得:a??,2?1?故实数a的取值范围是?,?????2?:(1)因为n?n?2a?1,即2S?n2?2na?n①,nnnn2S?n1?22?n1?a?n1?当n?2时,??????②,n?1n?12Sn22S?n1?22nan2?n1?a?n1?①?②得,??????????,nn?1nn?12a?2n?1?2na?2?n?1?a?1即,nnn?12?n?1?a?2?n?1?a?2?n?1?a?a?1n?2n?N*,即,所以,且nn?1nn?1所以?a?(2)由(1)可得a?a?2,a?a?63171aaa?a?2?2?a??a?6?a?2又,,成等比数列,所以,解得,1371111所以a?n?1n1111∴???.aa?n?1??n?2?n?1n?2nn?1?1?∴数列??的前2024项和为:aa?nn1???11??11??11??11?11506???????????????????????23??34??45??20252026?:解析:(1)选择条件①:?b?c?2?a2?3bc,由题意及正弦定理知:..b2?c2?a21∴a2?b2?c2?bc,∴cosA??2bc2?∵0?A??,∴A?.3???55选择条件②:因为cos2?A?cosA?,所以sin2A?cosA?,???2?44511?cos2A?cosA?,解得cosA?,又0?A??,即42?所以A?3bc(2)由?可得sinBsinC???sin?C??sinB3b????sinCsinC31cosC?sinC22131????sinC22tanC?2△ABCA?A?B?C??B?C??,因为是锐角三角形,由(1)知,得到33??0?C??2??1?故,解得?C?,所以?b?2.?2??622?0??C?????3213?33?S?bcsinA?b,S?,??△ABC24△ABC?82???f?1??0f??x??3x2?1f??1??3?1?:(1)由题意知,,,,则y?f?x?在点?1,0?处的切线方程为y?2?x?1?,y?2x?2???x,g?x??g??x??2xg??x??2x?2x?1设该切线与gx切于点,,则,解得,22222g?1??1?a?2?2?0a??1则,解得;????????????f??x??3x2?1,则y?fx在点x,fx处的切线方程为y?x3?x?3x2?1x?x,(2)因为111111y??3x2?1?x?2x3整理得,11:..???x,g?x??g??x??2xg??x??2x设该切线与gx切于点,,则,2222y??x2?a??2x?x?x?y?2xx?x2?a则切线方程为,整理得,222223x212x2????3x21?931则12,整理得a?x2?2x3?1??2x3?x4?2x3?x2?,????2x3??x2?a21221411214?12??931h?x??x4?2x3?x2?h??x??9x3?6x2?3x?3x?3x?1??x?1?令,则,4241令h??x??0,解得??x?0或x?1,31令h??x??0,解得x??或0?x?1,3则x变化时,h??x?,h?x?的变化情况如下表:?1?1?1?x,,0?0,1??1,????????????01?3?3?3?h??x?-0+0-0+51h?x?????1?274h?x???1,???a??1,???则的值域为,故的取值范围为f?x??0,???g?x???1,1?:(1)函数的定义域为,函数的定义域为g?x???1?x?ln?1?x???1?x?ln?1?x???1,0?函数在上单调递减,在?0,1?上单调递增证明:g??x???1?x?ln?1?x???1?x?ln?1?x?,∴g??x??g?x?所以g?x?为??1,1??x?2?g??x??ln?1?x??ln?1?x??ln?ln?1??0对?x??0,1?恒成立.??1?x?x?1?所以函数g?x?在??1,0?上单调递减,在?0,1?上单调递增11?n1?1??n1?1?11?n??n?n1?1??n1?1?n21(2)(证法一)要证明?n?n?,需证明?111?1?nn?nn:..?11??n1?1??n1?1?1111?n?n????????ln??01ln11ln10即证明??,即??????????????,?11?nnnn1?1??????????nnnn????1?由(1)可知即证g?0.???n?1?1???0,1?g?x??0,1?g?g?0??0∵且在单调递增,∴??n?n?11?n1?1??n1?1?n2n?N*n?1所以?n?n?对,?1??1??n1?1??n1?1?n21ln?n1?1ln?n1?2lnn(证法二)要证明?n?n?即证明??????????,?n??n?即证?n?1?ln?n?1???n?1?ln?n?1??2nlnn,即证?n?1?ln?n?1??nlnn?nlnn??n?1?ln?n?1?g?x???x?1?ln?x?1??xlnx设函数g??x??ln?x?1??lnx?0g?x??0,???,故函数在上单调递增又n?n?1,∴g?n??g?n?1?,故原不等式成立.