九师联盟2024届高三教学质量监测10月联考(全国卷)文科数学试题及参考精品2309.pdf

该【九师联盟2024届高三教学质量监测10月联考(全国卷)文科数学试题及参考精品2309 】是由【小屁孩】上传分享，文档一共【11】页，该文档可以免费在线阅读，需要了解更多关于【九师联盟2024届高三教学质量监测10月联考(全国卷)文科数学试题及参考精品2309 】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。:..九师联盟2024届高三教学质量监测10月联考(全国卷)文科数学试题及参考答案一、选择题:本题12小题,每小题5分,,:????0,??,sin??0?,则为()??????????A.??0,,sin?0B.??0,,sin?0??????????C.??0,,sin?0D.??0,,sin?0?????????xy?lnx?3,B?xx??1,则CA?B?()R?????????1?x??1?x??1?x??1?x?3P?1,m??tan???2sin??,,则()255525A.?B.???????,b和实数?,则“a??b”是“a与b共线”的(),,,后来慢慢演变为纳凉、娱乐、,,是用竹木做扇骨,(如图2),若图2中?AOB??,C,D分别在OA,OB上,AC?BD?m,弧AB的长为l,则该折扇的扇面ABDC的面积为()m?l???m?l??m?m?2l???m?2l??m?:..3????????,b?log,c???,则()2143?????c??a??a??c?b???1????????,则sin?2????()?6?3?6??x???x2?ax?1?1,2??1,2?,则的取值范围是()????????,??,??,,41f?x??3sinxcosx?sin2x?????0?,若将其图象向左平移个单位长度2得到的图象关于原点对称,则?的最小值为()????,已知两个单位向量OA,OB和向量OC,OC??,且cos??,OB与OC的夹角为10?OC?xOA?yOB?x,y?R?x?y?,若,则()411A.?1B.??ABC中,D为BC上一点,?BAD??CAD,若AC?AD?AB?2,则BC?2()?x?R?x?Rf?4?x??f??x??0f?x?1?,若,,且为偶函f?1???1f?2023??数,,则().?.?22:..二、填空题:本题共4小题,每小题5分,?x??log?x?1??2x(a?0,且a?1)??????????????,b满足a?5,b?4,a与b的夹角为120°,若ka?2b?a?b,则k?.x?1f?x??y?f?x??0,f?0??,???sinxcosx三、解答题:本题共6小题,、证明过程或演算步骤.???1???????17.(10分)已知a?2cos2x,sinx,b??,3cosx?,fx?a?b.?2?f?x?(1)求函数的最小正周期和单调递减区间;7?ABCA?B??f?A??1BC?23AC(2)在中,,,,????18.(12分)已知函数fx?logmx?1?x(m?0,且m?1)(1)求m的值;1????f?x??x?x?a(2)若关于x的不等式?3?33?3??0在R上有解,求实数a的最2????????.(12分)已知sin?是方程5x2?7x?6?0的根.?3??3?sin?????cos????cos?2?????tan?????????22????(1)求的值;??????cos?????cos????22???????5??????(2)若?是第四象限角,sin??????0????,求sin??????的值.?6?13?2??6?3:..f?x??alnx?x?a?R?20.(12分)?x?(1)讨论的单调性;?1?f?x?,e2a(2)若在??上有2个零点,求的取值范围.?e?21.(12分)南京玄武湖称“金陵明珠”,,每到夏季,荷花飘香,,小胡和朋友游玄武湖,发现观赏荷花只能在岸边,无法深入其中,影响观赏荷花的乐趣,于是他便有了一个愿景:若在玄武湖一个盛开荷花的一角(该处岸边近似半圆形,如图所示)设计一些栈道和一个观景台,观景台P在半圆形的中轴线OC上(图中OC与直径AB垂直,P与O,C不重合),通过栈道把PA,PB,PC,AB连接起来,使人行在其中,犹如置身花海AB?200m?PAB??f???,,???(1)求;(2)若栈道的造价为每米5万元,试确定观景台P的位置,使实现该愿景的建造费用最小(观景台的建造费用忽略不计),.(12分)在锐角?ABC中,角A,B,C的对边分别为a,b,c,S为?ABC的面积,且a2?2S??b?c?2.(1)求tanA的值;(2)若a?8,证明:16?b?c?:..参考答案一、选择题?p????0,??,sin??:由含有量词的命题的否定的特点知为.????:由题意得A?xx?3,A?B?xx??1或x?3,则????CA?B?x?1?x??:由三角函数的定义知tan??m??2,∴sin????.55????????:若a??b,由共线向量定理知a与b共线,知“a??b”是“a与b共线”????????a与ba??1,2?,b??0,0?a??ba??b的充分条件;若共线,如,则不成立,故“”??????不是“a与b共线”,“a??b”是“a与b共线”:由弧长公式可知,l???OA,∴OA?,则OC??m,∴该折扇的扇面??1l1l2m?2l??m???的面积为:l??????m??.2?2???2303?????????:1????????????,即1?a?c,?2??2??3??3?1又log?log4?1,∴b?a????????????????:sin?2????sin2??????cos2?????1?2sin2?????.6?62?669??????????aa??f??x??a?2xf??x??0x?,由题意得?1,2a??2,4?:,令,得,∴.22131???f?x??3sinxcosx?sin2x??sin2x?cos2x?sin2x?,:∵??2226?????????0?y?sin2x?2??将其图象向左平移个单位长度得到函数??的图象,?6?????y?sin2x?2??2???k??k?Z?,∵??的图象关于原点对称,∴66??5:..?k??????k?Z???0k?0?即,由于,当时,?2?72cos??,???0,??,得sin??1????:由,10?10?10?????3?cos??????OB?OC?1?2cos?1,∴,由题意得?4?5413OA?OC?1?2cos??,OA?OB??,55在OC?xOA?yOB两边分别点乘OA,OB,得313OA?OC?x?y?,OB?OC??x?y?1,555?5x?????41两式联立并解得?,∴x?y??.72?y?????:设?BAC??,由S?S?S,得?ABC?BADCAD11?1???AB?ACsin??AB?ADsin?AC?ADsin,即2sin??2sin?sin,2222222?????34sincos?3sin???0,??sin?0cos?∴,∵,∴,∴.222224?1cos??2cos2?1?∴,81∴BC2?AB2?AC2?2AB?ACcos??16?4?2?4?2??18,∴BC??x?1?f??x?1??f?x?1?f?x??f?2?x?∵为偶函数,即,∴,f?4?x??f??x??0f?x?2???f?2?x???f?x?又由,∴,f?x?4??f?x?f?x?∴,故为周期函数且4是一个周期,f?2023??f?3???f?1??1∴.二、填空题?0,1?x?0?0,1???1,???f?x?:当时,a在上无论取何值,的值总为1,故函f?x??0,1?:..4?????1?a?b?abcos120??5?4??????:?2?????????由ka?2b?a?b,得????????????ka?2b?a?b?ka2?2b2??k?2?a?b?25k?2?16?10?k?2??0,4解得k?.5ex??x?1?ex2?x2x?y?1?0f??x???f??0??2f?0???:,∴,又,??2exexy???1??2?x?0?2x?y?1?0故所求切线方程为,即.?2222?1?t2??16.?,解析:令t?sinx?cosx,则sinxcosx??2?t?2,??552??2t??∴y??2?t?2,当t?0时,y?0,3?t223当?2?t?2,且t?0时,y?,令u?t?,3tt?t?22??22?已知u的值域为???,????,???,?55?????2?22??22?y??,0???0,∴的取值范围为??.35??5t?????t?2222?综上所述,所求函数的值域为??,?.55??三、:(1)由题意得113???1f?x??cos2x?3sinxcosx??cos2x?sin2x?sin2x??,??222?6?22?f?x?T???,∴的最小正周期2??3??2??2k??2x???2k??k?Z??k??x??k??k?Z?令,解得,262637:..??2??f?x??k?,?k??k?Z?∴的单调递减区间为??.?63????1f?A??sin2A????1,(2)由(1)知??6?2???1?????13??sin2A???,又A?0,?,∴2A??,?∴??,?6?26?66??5??7?∴2A??,∴A?.∵A?B??,∴B?,663124223?ACBCBCsinB2由正弦定理得?,∴AC????x?f?x??f??x?x?:(1)∵为偶函数,∴对任意的恒成立,????即logm?x?1?x?logmx?1?x对任意的x?R恒成立,33mx?1??????又logm?x?1?log?logmx?1?logmx?logmx?1?xlogm,33mx3333????∴logmx?1?xlogm?x?logmx?1?x对任意的x?R恒成立,333即x?logm?2??0对任意的x?R恒成立,3必须logm?2?0,即m?9,故m?????1f?x??log9x?1?x3f?x??3log9x?1?x?3x?(2)由(1)知,,????11tx?x?t?t23x3x22设?3?3?2,则???2,即??t?,3x3x12??∴圆原问题等价于关于t的不等式t?3t?a?1?0在2,??上有解,2?1?1111231yt23t1?t3?2t??,∴a??t?t??,又???????,?2,???2?222max1111∴当t?3时,y?,∴a?,?5x2?7x?6?0sin???或sin??2(舍),:(1)由是方程的根,得58:..?3??3??sin????cos????cos????tan??????22????原式?sin????sin???sin??cos?cos?????cos???sin??cos???tan??cos???cos?????.?sin2?sin?3sin???,∴?是第三象限或第四象限角,由544?cos???,此时?cos??;若是第三象限角,则554444?cos??,此时?cos???.故所求式子的值为?若是第四象限角,?sin???,cos??,(2)由(1)知,当是第四象限角时,55???5??????12由sin??????0????,得cos?????,?6?13?2??6?13??????????????∴sin???????sin???????sin?cos?????cos?sin????6?6?66??????????3124556???????.51351365aa?xf?x??0,???f??x???1??x?0?.:(1)函数的的定义域为,且xxa?0f??x??0?0,???f?x??0,???当时,在上恒成立,故在上单调递减;a?0f??x??0?x?0?0?x?af??x??0x?a当时,令,得,令,得,f?x??0,a??a,???∴在上单调递增,?0f?x??0,???综上所述,当时,在上单调递减;a?0f?x??0,a??a,???当时,在上单调递增,在上单调递减.?1?a?0f?x???x,e2(2)若,,在??上无零点,不合题意;?e?1lnxa?0f?x??0?若,由,得,axlnx1?1?g?x??y?g?x?,e2令,则直线与函数在??上的图象有两个交点,xa?e?9:..1?lnx1????2???g?x?,当?x?e时,g?x?0,当e?x?e时,gx?0,x2e?1???????1g?x?,ee,e2上单调递减.∴gx?ge?∴在??上单调递增,在,emaxe???1???2g????e,ge2?又,?e?e21?1?211y?g?x?,e2??∴要使直线与函数在??上的图象有两个交点,则,a?e?e2aee2?e2?∴e?a?,即实数a的取值范围为?e,.??22???:(1)由题意知?PAB??,0???,OC?AB,OA?OB?100,4100则PA?PB?,PO?100tan?,∴PC?100?100tan?,cos?200f????PA?PB?PC?AB??100?100tan??200∴cos??2?sin??????100??3??0????.?cos???4??2?sin??2sin??1F????5f????500?3F?????500?(2)建造栈道的费用??,,?cos??cos2?1??F?????0sin??,又0?????令,得,∴.246???0???F?????0???F?????0当时,,当时,,664???????F???0,,∴在??上单调递减,在??上单调递增,?6??64???????1003F????F?5003?3PC?100?100tan?100?,∴??,此时min663???1003???故观景台位于离岸边半圆弧中点距离?100??米时,建造费用5003?3万元.?3???1?ABCS?bcsinA2S?a2??b?c?2,:(1)∵的面积为,2∴bcsinA?a2?b2?c2?2bc,由余弦定理得a2?b2?c2??osA,∴bcsinA?2bc?osA,∵bc?0,∴sinA?2cosA?2,10:..???A??0,?sin2A?cos2A?1sinA?21?sin2A?2又,,∴,?2?4化简得5sin2A?4sinA?0,解得sinA?或sinA?0(不合题意,舍去).5???3sinA4A??0,?cosA?1?sin2A?tanA??.∵,∴,?2?5cosA3bca8(2)证明:由正弦定理,得????10,sinBsinCsinA45b?10sinB,c?10sinC?10sin?A?B?∴,b?c?10sinB?10sin?A?B??16sinB?8cosB?85sin?B???∴,525?sin??,cos??.其中为锐角,且55????????∵A??0,?,???0,?,∴??A???,?2??2?22???sinA?sin?A??0?A???0??A???又,∴,∴,∴,222??????0?C?0???A?B??A?B???A????2????2????2又?,即?,∴?,????0?B??0?B??0?B?????2????2????2????∴?A?B?.∴?A???B?????,2222??????∵函数y?sinx在?0,?上单调递增,在?,??上单调递减,?2??2????2535425sin?A???cos???A??cos?cosA?sin?sinA且????????2?55555???25sin?????cos??,?2?5285??85sin?B????85,即16?b?c?85.∴511