云南省2015届高中毕业生第一次复习统一检测语文试题

秘密★考试结束前丽江市2025届高中毕业生复习统一检测历史试卷（全卷两个大题，共20个小题，共8页；满分100分，考试用时75分钟）注意事项：1．本卷为试题卷。

考生必须在答题卡上解题作答。

答案应书写在答题卡的相应位置上，在试题卷、草稿纸上作答无效。

2．考试结束后，请将答题卡交回。

一、选择题：本大题共16小题，每小题3分，共48分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1．如图是陕西临潼姜寨聚落遗址建筑分布图，该聚落遗址是由五个大家族组成的一个氏族公社，考古发现该聚落各个墓葬的随葬品不多且差别不大。

这反映了（）A．母系氏族公社的生产力水平状况B．中国大地上出现国家的初始形态C．中华文明即将迈入阶级社会门槛D．长江流域新石器时代的文化面貌2．下表是两汉各区域市场面积、人口、商业都市一览表，该表反映出当时（）平均每商业都市区域市场面积（平方公里）人口商业都市面积人口关陇市场9693156661849长安、姑臧4846573330924巴蜀市场8969484248040成都8969484248040关东市场89464931584084陶、洛阳、邯郸、蓟、淮阳、涿、温、荥阳、阳翟、平阳、轵、扬、临淄688192429545楚越市场168240714461202宛、江陵、寿春、吴、陈、合肥、番禺2403442065886A ．丝绸贸易沿线商业都市最先崛起B ．区域经济基础影响商业发展水平C ．关东地区的抑商政策有所松弛D ．南北方商业市场发展趋向平衡3．魏晋时期，皇帝对高门子弟多有倚重，部分望族相继掌权把持朝政；而到了南北朝，各国君主不论贤能与否，都做到了威福自专，“不肯假权与人”。

这一变化（）A ．满足南方经济开发的需要B ．体现了士族衰落庶族崛起C ．保障了政治的稳定D ．源于民族交融加强4．宋代文献中佃客“舍其耒耜而游于四方，择其所利而居之”等记载甚多。

仁宗时文彦博说：“不数年间，租佃之户或退或逃……”。

大理州2025届高中毕业生第一次复习统一检测数学（全卷四个大题，共19个小题，共8页；满分150分，考试用时120分钟）考生注意：1.答卷前，考生务必用黑色碳素笔将自己的姓名､准考证号､考场号､座位号填写在答题卡上，并认真核准条形码上的准考证号､姓名､考场号､座位号及科目，在规定的位置贴好条形码.2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑.如需改动，用橡皮擦干净后，再选涂其他答案标号.回答非选择题时，将答案写在答题卡上.写在本试卷上无效.3.考试结束后，将本试卷和答题卡一并交回.第I 卷（选择题共58分）一､单项选择题（本大题共8小题，每小题5分，共40分.在每小题所给的四个选项中，只有一项是符合题目要求的）1.已知集合，则（ ）A. B.C. D.2.已知复数满足，则的共轭复数在复平面中的对应点位于（ ）A.第一象限B.第二象限C.第三象限D.第四象限3.在中，是上一点，满足是的中点，若，则（ ）A.B.1C.D.4.下图是我国2018-2023年纯电动汽车销量统计情况，则下列说法错误的是（）A.我国纯电动汽车销量呈现逐年增长趋势(){30},{013}A xx x B x x =-<=<-<∣∣A B ⋂={34}xx -<<∣{10}xx -<<∣{13}xx <<∣{43}xx -<<∣z ()()i 1i 3i z --=+z z ABC D BC 3,BD DC M = AD BM BA BC λμ=+λμ+=547858B.这六年销量的第60百分位数为536.5万辆C.2022年销量低于这六年销量的平均值D.这六年增长率最大的为2019年至2020年5.已知等比数列中，，则（ ）A.26B.32C.512D.10246.已知，则（ ）A. B. C. D.7.抛物线的焦点为，过点的直线交抛物线于两点，则的最小值为（ ）A.5B.9C.8D.108.一个三角形纸板的三个顶点为，以边上的高所在直线为旋转轴，将三角形纸板旋转，则纸板扫过的空间所形成的几何体的体积为（ ）A.B. C. D.二､多项选择题（本大题共3小题，每小题6分，共18分.在每小题所给的四个选项中，有多项符合题目要求，全部选对得6分，部分选对得部分分，有选错的得0分）9.已知函数在区间上的最大值为4，则下列说法正确的是（ ）A.的最小正周期为B.C.在区间上单调递减D.点是图象的一个对称中心10.已知函数，则下列说法正确的是（ ）A.有最大值B.当时，的图象在点处的切线方程是C.在区间上单调递减{}n a 13462,16a a a a +=+=1012a a +=2sin 3αα+=πcos 23α⎛⎫-= ⎪⎝⎭6365-79-24254524y x =F F ,A B 4AF BF +3,AB BC AC ===AB 180 5π6π5π32π()π2sin 216f x x m ⎛⎫=+++ ⎪⎝⎭π0,2⎡⎤⎢⎥⎣⎦()f x π1m =()f x ππ,36⎡⎤-⎢⎥⎣⎦π,212⎛⎫-⎪⎝⎭()f x ()e xf x x a =-()f x 1ea --1a =()f x ()()0,0f 1y x =-()f x []2,0-D.关于的方程有两个不等实根，则的取值范围是11.法国数学家加斯帕尔蒙日被称为“画法几何创始人”“微分几何之父”.他发现与椭圆相切的两条互相垂直的切线的交点的轨迹是以该椭圆中心为圆心的圆，这个圆称为该椭圆的蒙日圆.若椭圆的蒙日圆为圆，过上的动点作的两条互相垂直的切线，分别与交于两点，直线交于两点，则（）A.椭圆的蒙日圆方程为B.面积的最大值为7C.的最小值为D.若动点在上，将直线的斜率分别记为，则第II 卷（非选择题共92分）三､填空题（本大题共3小题，每小题5分，共15分）12.在的展开式中，含的项的系数是__________.13.已知数列满足，则__________.14.设函数是的导函数，函数是的导函数，经过研究发现，任意一个三次函数的图象都有对称中心，其中满足.已知函数，当函数图象的对称中心为时，__________，当函数图象的对称中心为时，__________.四､解答题（本大题共5小题，共77分.解答应写出文字说明，证明过程或演算步骤）15.（本题13分）在中，角的对边分别为.已知，且.（1）求角和；x ()0f x =a 1,0e⎛⎫- ⎪⎝⎭22Γ:143x y +=C C M ΓC ,P Q PQ Γ,A B Γ227x y +=MPQ AB D Γ,DA DB 12,k k 1234k k =812⎛ ⎝3x {}n a 12221,2,n na a a a +===-2025a =()y f x ='()y f x =()y f x =''()y f x ='()()320f x ax bx cx d a =+++≠()()00,x f x 0x ()00f x ''=()3223f x x ax b =-+()f x 1,02⎛⎫⎪⎝⎭b =()f x 1,12⎛⎫ ⎪⎝⎭122023202420242024f f f ⎛⎫⎛⎫⎛⎫+++= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ABC ,,A B C ,,a b c sin sin sin sin a A b B c C b A +=-73c b =C sin B（2）若，求.16.（本题15分）如图，在四棱锥中，底面为平行四边形，侧棱底面，.点是棱的中点，点为棱上的一点，且.（1）求证：平面平面；（2）求直线与平面所成角的正弦值.17.（本题15分）已知函数.（1）当时，证明：；（2）若函数有极小值，且的极小值小于，求的取值范围.18.（本题17分）已知椭圆的两个焦点为，且椭圆.（1）求椭圆的标准方程；（2）已知为坐标原点，斜率为的直线与椭圆有两个不同的交点，且弦的中点为，直线的斜率为，求；（3）直线与椭圆有两个不同的交点，椭圆在点处的切线分别为与交于点，点在直线上.请你判断直线是否经过定点，并说明理由.19.（本题17分）今年立秋以后，我国西南地区持续性高温登上热搜，引发关注讨论､根据专家推测，主要是由于大陆高压和西太平洋副热带高压呈现非常强大，在高压的控制下，西南地区上空晴朗少云，在太阳辐射增温和气流下沉增温的共同作用下，两个地区的气温出现了直接攀升的状态.西南地区某城市一室内游泳馆，为给顾客更好的体验，推出了和两个套餐服务，顾客可自由选择和两个套餐之一；该游泳馆在App 平台上推出了优惠券活动，下表是App 平台统计某周内周一至周五销售优惠券情况.ABC a P ABCD -ABCD PD ⊥ABCD 2,45PD CD BD BDC ∠====E PCF PB 23BF BP =PBC ⊥PCD DC DEF ()ln 1a f x x x=+-1a =()0f x ≥()f x ()f x 2a a -a C ())12,F F C C O ()110k k ≠C ,A B AB E OE 2k 12k k ⋅L C ,P Q C ,P Q 121,,L L L 2L T T 4x =L A B A B星期12345销售量（张）218224230232236经计算可得：.（1）已知关于的经验回归方程为，求关于的经验回归方程；（2）若购买优惠券的顾客选择套餐的概率为，选择套餐的概率为，并且套餐包含两张优惠券，套餐包含一张优惠券，记App 平台累计销售优惠券为张的概率为.（i ）求及；（ii ）求及的最值.参考公式：.大理州2025届高中毕业生第一次复习统一检测数学参考答案及评分标准一､单选题（本大题共8小题，每小题5分，共40分）y 55521111228,3464,555i i i i i i i y y t y t =======∑∑∑y ˆˆˆy bt a =+y A 13B 23A B n n P 12P P ､3P n P n P ()()()1122211ˆˆ,n niii ii i nniii i x x y y x y nx ybayx x x xnx ====---⋅===--∑∑∑∑题号12345678答案CDCCDBBA1.C【详解】由于，故.故选：C.2.D【详解】，故，其对应的点为，该点在第四象限.故选：D.3.C【详解】由题可知，，，所以有，所以，得.故选：C 4.C【详解】对于A ，从条形图中看出，纯电动汽车销量逐年递增，故A 正确；对于B ，因为，将所有汽车销量数据从小到大排序，所以销量的第60百分位数为第4个数据，即536.5，故B 正确；对于C ，这六年销量的平均数为，故C 错误；对于D ，因为2019年至2020年的增长率为，超过其他年份的增长率，故D 正确.故选：C.5.D【详解】设等比数列的公比为，因为，所以，由，则，得，解得，所以.(){30}{03},{013}{14}A xx x x x B x x x x =-<=<<=<-<=<<∣∣∣∣{13}A B xx ⋂=<<∣()()3i 1i 3ii i 13i 1i 2z +++=+=+=+-13i z =-()1,3-11122222AM AD BM BA BD BA BM BA BD =⇒-=-⇒=+()3334BD DC BC BD BD BC ==-⇒= 11132228BM BA BD BA BC =+=+ 13,28λμ==78λμ+=0.66 3.6⨯=410.35668.5<291.6111.51.6111.5-≈{}n a q 13462,16a a a a +=+=23511112,16a a q a q a q +=+=()231116a a q q+=38q =2q =125a =()91191110121122210245a a a q a q +=+=+=故选：D.6.B【详解】，故选：B.7.B【详解】由抛物线焦点弦性质可得，则，所以，令，所以，当且仅当，即时等号成立.所以的最小值为9.故选：B.8.A 【详解】，而A 为三角形内角，故故，故，故，故几何体的体积为故选：A.2π2sin ,sin 2sin 333ααααα⎛⎫+=∴+=+= ⎪⎝⎭ 2π12ππ7π7sin ,cos 212sin ,cos 23333939αααα⎛⎫⎛⎫⎛⎫⎛⎫∴+=∴+=-+=∴-=- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭1121AF BF p +==1BFAF BF =-441BF AF BF BF BF +=+-,1m BF m =>()114441411m m AF BF m m m m -++=+=+-+--()1541591m m =+-+≥+=-()1411m m -=-32m =4AF BF +222cos 2AC AB BC A AC AB +-===⨯sin A =sin 1CD AC A ===2AD ==1DB =()22115π1π2π1236⨯⨯⨯+⨯=二､多选题（本大题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项是符合题目要求的.全部选对的得6分，部分选对的得部分分，有选错的得0分）题号91011答案ABDBDABC9.ABD【详解】因为，选项A ，的最小正周期，故A 正确；选项B ，由，知，所以，所以的最大值为，而得，故B 正确；选项C ，由得，所以在上单调递增，故C 错误；选项D ，令，则，所以图象的对称中心为，所以点是图象的一个对称中心，故D 正确.10.BD【详解】因为，选项A ，当时，，当时，.所以在区间上单调递减，在区间上单调递增，所以有最小值，无最大值，故A 错误；选项B ，当时，，所以的图象在点处的切线方程是，故B 正确；选项C ，因为在区间上单调递减，在区间上单调递增，故C 错误；()π2sin 216f x x m ⎛⎫=+++ ⎪⎝⎭()f x 2ππ2T ==π0,2x ⎡⎤∈⎢⎥⎣⎦ππ7π2,666x ⎡⎤+∈⎢⎥⎣⎦π1sin 2,162x ⎛⎫⎡⎤+∈- ⎪⎢⎥⎝⎭⎣⎦()f x 3m +34m +=1m =ππ,36x ⎡⎤∈-⎢⎥⎣⎦πππ2,622x ⎡⎤+∈-⎢⎥⎣⎦()f x ππ,36⎡⎤-⎢⎥⎣⎦π2π,6x k k +=∈Z ππ,212k x k =-∈Z ()f x ππ,2,212k k ⎛⎫-∈⎪⎝⎭Z π,212⎛⎫- ⎪⎝⎭()f x ()()e 1xf x x =+'1x <-()0f x '<1x >-()0f x '>(),1∞--()f x ()1,∞-+()f x ()f x ()11ef a -=--1a =()()01,01f f ==-'()f x ()()0,0f 1y x =-(),1∞--()f x ()1,∞-+()f x选项D ，方程，即，令，而，当时，，当时，.所以在区间上单调递减，在区间上单调递增，当时，且，如图得的范围是，故D 正确.11.ABC【详解】选项A ，依题意，可设圆C 方程为，过椭圆的上顶点作轴的垂线，过椭圆的右顶点作轴的垂线（图略），则这两条垂线的交点在圆C 上，所以，即，所以椭圆的蒙日圆方程为，故A 正确；选项，因为点都在圆上，且，所以为圆的直径，所以面积的最大值为，故B 正确；选项，由于为圆的直径，过坐标原点，即过坐标原点所以C 正确；选项D ，由直线经过坐标原点，易得点关于原点对称，设，则，又，所以，所以，故D 错误.()0f x =e 0e x x x a x a -=⇔=()e xg x x =()()e e e1xxxg x x x ='+=+1x <-()0g x '<1x >-()0g x '>(),1∞--()g x ()1,∞-+()g x 0x <()0g x <()00g =a 1,0e ⎛⎫-⎪⎝⎭222x y r +=Γ(y Γ()2,0x (2222r +=27r =Γ227x y +=B ,,M P Q C 90PMQ ∠= PQ C MPQ 112722PQ ⨯=⨯=C PQ C AB min ||2AB b ==PQ ,A B ()()1122,,,A x y D x y ()121211121212,,,y y y y B x y k k x x x x -+--==-+22112222143143x y x y ⎧+=⎪⎪⎨⎪+=⎪⎩22221212043x x y y --+=221212221234y y k k x x -==--三､填空题（本大题共3小题，每小题5分，共15分）题号121314答案71（2分）2023（3分）12.7【详解】在展开式的通项为，当时，.所以含的项的系数是7.13.1【详解】方法一：由题意知，，，，因此数列是周期为4的周期数列，所以.方法二：把看作，则因此数列是周期为4的周期数列，所以.14.，【详解】因为，且图象的对称中心为，所以，解得，12812⎛ ⎝88218811C (C (1)22kkk k kk kk T x --+⎛⎫⎛⎫==- ⎪⎪⎝⎭⎝⎭6k =263381C 72x x ⎛⎫= ⎪⎝⎭3x 121,2a a ==3451232222,1,1a a a a a a =-=-=-=-=-=6422,a a =-= {}n a 20255064111a a a ⨯+===22n na a +=-()()22f n f n +=-()()()242f n f n f n +=-=+{}n a 20255064111a a a ⨯+===122023()()266,126f x x ax f x x a =''=--'()f x 1,02⎛⎫⎪⎝⎭11126022f a ⎛⎫='⨯-⎝⎭'=⎪1a =而，解得；因为函数图象的对称中心为，即，所以，同理设①②由①+②得，所以.四､解答题（本大题共5小题，共77分.解答应写出文字说明，证明过程或演算步骤）15.【详解】（1）由正弦定理得，，整理可得，.由余弦定理得，.又，则.由正弦定理得，，即，.（2）方法一：由可知，.32111230222f b ⎛⎫⎛⎫⎛⎫=⨯-⨯+= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭12b =()f x 1,12⎛⎫ ⎪⎝⎭()()12f x f x +-=1112023122024202420242024f f f f ⎛⎫⎛⎫⎛⎫⎛⎫+-=+= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭22022320212,2,2024202420242024f f f f ⎛⎫⎛⎫⎛⎫⎛⎫+=+=⋯⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭122023202420242024S f f f ⎛⎫⎛⎫⎛⎫=+++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭202320221202420242024S f f f ⎛⎫⎛⎫⎛⎫=+++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭222023S =⨯2023S =2sin sin sin a b c R A B C ===2222222a b c ab R R R R+=-222a b c ab +-=-2221cos 22a b c C ab +-==-()0,πC ∈2π3C =2sin sin sin a b c R A B C ===72sin 2sin 3R C R B =⨯3sin sin 7C B ==2πsin 3B C ==13cos 14B ==()2π2πsin sin sin coscos sin 33A B C B B =+=+=21sin sin 2sin B C S a A ==.方法二：由可知，.由得，16.【详解】（1）在中，，即.又，则有，即.平面平面平面（2）由（1）可知，.以D 为坐标原点，所在直线分别为轴，轴，轴，建立如图空间直角坐标系.22142a a ⇒=⇒=2a ∴=2πsin 3B C ==13cos14B ==()2π2πsin sin sin coscos sin 33A B C B B =+=+=2sin sin sin a b c R A B C===,,a b c ===211sin 24S ab C R ==⇒=⇒=2.a R ∴===BCD 2222cos 4BC BD CD BD CD BDC ∠=+-⋅⋅=2BC =2BD CD BC ===222BC CD BD +=BC CD ⊥PD ABCD PD BC BC ABCD ⊥⎫⇒⊥⎬⊂⎭平面平面,,PD BC CD BC CD PD ABCD BC CD PD D ⊥⊥⎫⎪⊂⇒⊥⎬⎪⋂=⎭平面PCDBC PCD BC PBC ⊥⎫⇒⎬⊂⎭平面平面PBC ⊥PCD ,,DA DC DA DP DC DP ⊥⊥⊥,,DA DC DP x y z依题意得，.设点的坐标为，.由三点共线，则有，.又，，可得，..设平面的法向量为.，取一个法向量.设直线与平面所成角为，则所以，直线与平面17.【详解】（1）要证，只需证.当时，令.()()()()()2,2,0,0,2,0,0,0,0,0,0,2,0,1,1B C D P E F (),,F a b c ,,P F B 23BF BP = ()()2,2,,2,2,2BF a b c BP =--=-- ()()22,2,2,2,23a b c ∴--=--224,,333a b c ===224,,333F ⎛⎫∴ ⎪⎝⎭DEF ()()224,,,,,,0,1,1333n x y z DF DE ⎛⎫=== ⎪⎝⎭ 222240033300n DF x z x y z y z n DE y z ⎧⎧⋅==-++=⎧⎪⎪⇒⇒⎨⎨⎨=-⋅=⎩⎪⎪+=⎩⎩()1,1,1n =- DC DEF θ()()0,2,0,1,1,1DC n ==- 2,2,DC n DC n ⋅=== sin cos ,n DC n DC n DCθ⋅=<>== DC DEF ()0f x ≥min ()0f x ≥1a =()1ln 1,0f x x x x=+->，令得，.的变化情况如下表所示：1-0+递减极小值0递增，即.，当时，等号成立.（2），令得，.（1）当时，又，则，此时，在单调递增，无极小值，不符题意.（2）当时，的变化情况如下表所示：-0+递减极小值递增，即.令，且.不等式即为，.，当且仅当，即.在单调递增，又，则.()22111x f x x x x-=-='()0f x '=1x =()(),f x f x 'x ()0,1()1,∞+()f x '()f x ()min ()10f x f ∴==()0f x ≥()0f x ∴≥1x =()221a x a f x x x x-=-='()0f x '=x a =0a ≤0x >()20x a f x x -=>'()f x ()0,∞+()f x 0a >()(),f x f x 'x ()0,a a (),a ∞+()f x '()f x ln a 2ln a a a ∴<-+2ln 0a a a +-<()2ln ,0g a a a a a =+->()10g =2ln 0a a a +-<()()1g a g <()121110g a a a '=+-≥-=>12a a =a =min ()1g a ='()g a ∴()0,∞+()()1g a g <01a <<的取值范围是：.18.【详解】（1）设椭圆的标准方程为：椭圆的标准方程为：.（2）方法一：点差法设，则，①又在椭圆上，则，两式相减得：，即：②由①②得，.而，.方法二：椭圆方程代换设，直线则，①a ∴()0,1C ()222210x y a b a b+=>>22221c e a a c c a b c b ⎧==⎪=⎧⎪⎪⎪=⇒=⎨⎨⎪⎪=+=⎩⎪⎪⎩∴C 2214x y +=()()()112200,,,,,A x y B x y E x y 12012022x x x y y y +=⎧⎨+=⎩,A B C 221122221414x y x y ⎧+=⎪⎪⎨⎪+=⎪⎩2222212104x x y y -+-=()()()()2121212104x x x x y y y y +-++-=()()021*******x x x y y y -+-=021********,,4y y y k k k k x x x -==∴⋅=--()()()112200,,,,,A x y B x y E x y 1:l y k x t =+12012022x x x y y y +⎧=⎪⎪⎨+⎪=⎪⎩②又，即③由①②③得，方法三：联立方程设，直线则，①联立方程得，②由①②得，，则，.又，.（3）设，先求椭圆在点处的切线的方程.方法一：根据判别式求解椭圆在点处的切线，设21220212121122221210212122y y y y y y y y y k k x x x x x x x x x +---=⋅=⋅=+---221122221414x y x y ⎧+=⎪⎪⎨⎪+=⎪⎩221122221414x y x y ⎧=-⎪⎪⎨⎪=-⎪⎩()22121222211144x x k k x x -==--()()()112200,,,,,A x y B x y E x y 1:l y k x t =+12012022x x x y y y +⎧=⎪⎪⎨+⎪=⎪⎩22114x y y k x t ⎧+=⎪⎨⎪=+⎩()()22211418440k x k tx t +++-=11221841k t x x k ∴+=-+1021441k t x k =-+21010221144141k t t y k x t t k k -=+=+=++020114y k x k ==-1214k k ∴⋅=-()()3344,,,P x y Q x y C ,P Q 12,L L Δ0=C ()33,P x y 1L 1:L y rx s=+联立方程得，.，即.同理可得，.，可得T 点的横坐标，即又，可得，由题意可知直线的斜率不为0，设.则，整理得，，即.又，则.，即直线恒过定点.方法二：导数的几何意义.当点在或2214x y y rx s ⎧+=⎪⎨⎪=+⎩()()222418440r x rsx s +++-=()2222Δ6441041r s s r =-+=⇒=+()3228842241rs rs r x s s r ---===+22233441r s r y rx s s s s s--=+=+==33331,44sx x s r y y ∴==-=-31331:4x x L y y y ∴=-+313:14x x L y y +=424:14x x L y y +=()333444334414414x x y y y y x x x x y x y y y ⎧+=⎪-⎪⇒=⎨-⎪+=⎪⎩()3443344T y y x x y x y -=-4T x =4334430x y x y y y -+-=L :L x my n =+()()4334430my n y my n y y y +-++-=()()34110n y n y ---=()()3410n y y --=34y y ≠1n =:1L x my ∴=+L ()1,0221224x y y x +=⇔=-≤≤()33,P x y y =223314x y +=3y =，则即.当点在时，同理可得.，同理可得，.后续同方法一.19.【详解】（1）由题意，，则.所以关于的经验回归方程为.（2）（i ）由题意，可知，，（求解另一种方法：）（ii ）当时，，即，又，所以当时，数列为各项都为1的常数列，即，所以，又，所以数列为首项为公比为的等比数列，y ='3334x x x k y y ='-===切线()22223333313333333:144444x x x x x x L y y x x y y y x y y y y ∴-=--⇒-=-+⇒+=+=313:14x x L y y +=()33,P x y y =313:14x x L y y +=313:14x x L y y ∴+=424:14x x L y y +=1234535t ++++==122221400469053228 4.49165ˆ3n i i i n i i t y nt yb t nt =-=∑-⋅-⨯-⨯⨯===--⨯∑-‖8ˆˆ228 4.43214.ay bt =-=-⨯=y ˆ 4.4214.8y t =+123P =222173339P =⨯+=3222122120333333327P =⨯⨯+⨯+⨯=3P 31212214203392727P P P =+=+=3n ≥122133n n n P P P --=+1121133n n n n P P P P ---+=+21171213933P P +=+⨯=2n ≥113n n P P -⎧⎫+⎨⎬⎩⎭()11123n n P P n -+=≥1313,2434n n P P n -⎛⎫-=--≥ ⎪⎝⎭1323143412P -=-=-34n P ⎧⎫-⎨⎬⎩⎭112-13-所以，即.当为偶数时，，且随的增大而减小，因此的最大值为；当为奇数时，，且随的增大而增大，因此的最小值为，综上所述，的最大值为，最小值为.13114123n n P -⎛⎫-=-⨯- ⎪⎝⎭311443n n P ⎛⎫=+⨯- ⎪⎝⎭n 31134434n n P ⎛⎫=+⨯> ⎪⎝⎭n P n n P 279P =n 31134434n n P ⎛⎫=-⨯< ⎪⎝⎭n P n n P 123P =n P 7923。

奉贤区高三语文第一次高中毕业生复习统一检测试卷姓名:________ 班级:________ 成绩:________一、选择题 (共1题；共6分)1. （6分） (2020高三下·柳州月考) 阅读下面的文字，完成下列小题。

常言“文如其人”，就是说一个人的性情、喜好等等会自然流露在文章的字里行间，文字和作者，非常相似。

如果作文之人风流倜傥，他的文章定如行云流水般一气呵成；如果作文之人不苟言笑，他的文章定然庄严肃穆，天然一般有威慑力。

（）。

每当阅读先贤的书籍，常常忍不住或者为先贤的，或者为先贤的深刻思想，恨不得立马和他当面交流所得、畅所欲言。

然而面对时间和空间的不可逆转，使所有的想法只能一人独对，先贤言谈的美好景象惟有在梦中出现，稍稍可以安慰的是，通过阅读先贤书籍，了解先贤的品性，知道先贤的喜好，体会到先贤的喜怒哀乐、嬉笑怒骂，自觉心意相通，感情更加贴近，未曾得缘一见，内心里却已把他当作最亲近的朋友。

（1）依次填入文中横线上的词语，全都恰当的一项是（）A . 休戚相关金玉良言拍手称快虽然B . 息息相关金玉良言拍案叫绝虽然C . 息息相关金科玉律拍案叫绝即使D . 休戚相关金科玉律拍手称快即使（2）下列在文中括号内补写的语句，最恰当的一项是（）A . 所以熟读一个人的文章，就大约能够揣摩出这人的性格；同样的，了解一个人，自然能够推断他文章的大体风貌。

B . 所以了解一个人，就大约能够揣摩出这人的性格；同样的，熟读一个人的文章，自然能够推断他文章的大体风貌。

C . 所以了解一个人，自然能够推断他文章的大体风貌；同样的，熟读一个人的文章，就大约能够揣摩出这人的性格。

D . 所以能够推断他文章的大体风貌，自然可以了解这个人；同样的，能够揣摩出这人的性格，自然可读懂他的文章。

（3）下列对文中划线句子修改正确的一项是（）A . 然而时间和空间的不可逆转，所有的想法只能一人独对，先贤言谈的美好景象惟有在梦中出现。

2024届云南省第一次高中毕业生复习统一检测（一模）英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读理解2024 Travel Planner: The Best Things to See in the UK Icons of British Fashion, OxfordshireA fashion exhibition will open at Blenheim Palace in March. Icons of British Fashion is the largest event in the palace’s 300-year history and will showcase some of Britain’s most famous designers. Clothing, drawings, photographs and patterns will be on display. 23 March to 30 June. Palace entry tickets are valid for a year and cost £ 38 for adults, £ 22 for under-16s.Toulouse-Lautrec in BathA new exhibition, Toulouse-Lautrec and the Masters of Montmartre, opens at Victoria Art Gallery in Bath in April. More than 30 pieces from the artist’s colourful career will be shown alongside works by other artists of the day. This is the only chance to see Lautrec’s complete collection of posters in the UK before it moves to a permanent home at the Musée d’Ixelles in Belgium. 26 April to 29 September, adults £ 10, children £ 3.50.Back in Time at BeamishA truly recreated 1950s picture house at Beamish will offer visitors a chance to recall the golden age of movie-going. The cinema will be the latest addition to a “1950s Town” opened last summer at the living history museum. Adults £ 24.95, children £ 15.45, a family of four £ 63.50, valid for a year.National Gallery goes on tourThe National Gallery celebrates its 200th anniversary on 10 May. To mark the occasion，the gallery is lending 12 of its most iconic paintings to 12 venues across the UK. Paintings will travel to centres from Brighton to Edinburgh, and will all go on display on 10 May. Each of the 12 venues will run exhibitions and digital interventions to showcase its particular painting.1．How much should a couple with a 5-year-old pay for Blenheim’s exhibition?A．£ 23.50.B．£ 60.00.C．£ 63.50.D．£ 98.00. 2．What do Toulouse-Lautrec and National Gallery have in common?A．Recalling the artist’s career.B．Offering chances to make posters.C．Employing digital interventions.D．Enabling visitors to enjoy arts.3．Which of the following is a year-round event?A．Icons of British Fashion.B．Toulouse-Lautrec and the Masters of Montmartre.C．1950s Picture House Visit.D．National Gallery Painting Show.The memory of one particular summer evening is still burned in my brain as if it were yesterday. There was nothing but wide-open fields for miles and miles around our rural Minnesota home. We never saw strangers not ever and here on this hot evening was a real live one walking up our driveway.A young man, a slightly-built hitchhiker (搭便车的人) approached our door. He knew there was a storm coming, and he desperately needed shelter. Not wanting to intrude on our home and family, he asked my dad if he could sleep in our basement for the night for protection from the rain. Instead of saying yes, my dad loaded us all up in the 1959 Chevrolet: five kids, my mum, and the man.Our family consisted of three older children whose father had died young and three more children from the union of my mother and father. Our older brother Jerry was in the Navy, on a ship somewhere overseas. Our mum and dad worried about him.We drove him 10 miles to the next town, where Dad bought the man a room for the night along with a hot evening meal. In the car after we dropped off the stranger, I heard my dad say to my mum, “I just hope that if Jerry ever needs anything, this kindness will be returned to him.”Weeks later, Dad told my uncle about the young man. My uncle suggested that perhaps my dad shouldn’t have taken the risk of having a stranger in our car. My dad replied, “You are absolutely right. I should have invited him into our home.”4．What does the underlined phrase “intrude on” in paragraph 2 mean?A．Disturb.B．Scare.C．Embarrass.D．Upset. 5．How did the author’s father help the stranger?A．By providing a bed for him in the basement.B．By letting him stay for the night in the car.C．By taking him to a hotel in another town.D．By cooking a hot evening meal for him.6．What do we know about the author’s family?A．The family had five children in total.B．Jerry was serving in the army in the USA．C．They had no spare room for the hitchhiker.D．Dad wished his elder son to be treated kindly.7．What did the author’s uncle think of his father’s deed?A．Worthwhile.B．Inappropriate.C．Inspiring.D．Unbelievable.Almost all new parents struggle to get enough sleep while caring for their newborns. For some penguin (企鹅) parents, though, periods of sleep are especially short.Researchers recently found that Chinstrap penguins only sleep for about four seconds at a time in order to protect their eggs and newborn chicks. They do this thousands of times throughout the day. The short “microsleeps” total around 11 hours each day. The microsleeps appear to be enough to keep the parents going for many weeks.Niels Rattenborg, a sleep researcher at the Max Planck Institute for Biological Intelligence in Germany, said, “What’s surprising is that the penguins are able to function okay and successfully raise their young.”Chinstrap penguins usually lay their eggs in November in nests made up of small rocks. Mated (交配的) pairs share parenting responsibilities. One parent usually watches the eggs and chicks alone while the other goes off fishing for family meals. Adult penguins don’t face many natural enemies in the mating season. But large birds called brown skuas eat the penguin eggs and small chicks. Other adult penguins may also try to steal the small rocks from the nests, so penguin parents must always be on guard.For the first time, scientists followed the sleeping behavior of Chinstrap penguins. They did this by attaching devices that measure brain waves. They collected data on 14 adults over 11 days on King George Island off the coast of Antarctica.Won Young Lee, a biologist at the Korean Polar Research Institute, thought of the idea for the study when he saw mating penguins often blinking (眨) their eyes and possiblysleeping. But the team needed to record brain waves to confirm the animals were sleeping. “For these penguins, microsleeps have some restorative functions,” he said. He added that they would not survive without it.The researchers didn’t collect sleep data outside the mating season. But they suspect that the penguins may sleep for longer periods at other times of the year.8．What have researchers found about Chinstrap penguins?A．They take turns to take care of the young.B．They sleep for several seconds at a time.C．They rest for 11 hours in a row every day.D．They behave poorly due to lack of sleep.9．What is paragraph 4 mainly about?A．The habitats of the penguins.B．The living habits of the penguins.C．The mating seasons of the penguins.D．The natural enemies of the penguins. 10．Which of the following might Won Young Lee agree with?A．Microsleeps of the penguins matter in their survival.B．It’s difficult to record brain waves of the penguins.C．The penguins may sleep longer than people believe.D．More funds are needed to further study the penguins.11．Where is the text most probably taken from?A．A book review.B．A travel brochure.C．A science journal.D．A chemistry textbook.When people think of word innovators throughout history, male writers likely come to mind. Shakespeare is credited with inventing more than 1,700 words, including “bedroom”, “courtship” and “swagger”. Charles Dickens is said to have first used the words “butterfingers” and “doormat”, and Dr. Seuss reportedly came up with “nerd”.But despite contributions from famous writers, historians say another group has an even greater impact on the development of language: teenage girls. Women lead up to 90 percent of linguistic (语言的) changes, as sociolinguist William Labov observed in the early 2000s. In fact, he wrote, women are often linguistically ahead of men “by a full generation”.Now women are leading the charge online. Though Oxford University Press’2023 word of the year, “rizz”, meaning charm, was coined by a man, several runners-up, including“situationship” and “swiftie” were inspired or first used by women. The term “goblin mode”, which refers to lazy behavior, was Oxford’s 2022 word of the year and appears to have been first used by a woman on Twitter in 2009.It’s often impossible to tell who first used new words. But whether or not young girls invent new phrases, they are more likely to be early adopters of the latest lingo (术语), says University of Toronto linguist Sali Tagliamonte, “They’re pushing changes forward.”There are a handful of possible reasons why girls lead lexical (词汇的) innovation. According to Gretchen McCulloch, author of Because Internet: Understanding the New Rules of Language, women tend to be more socially aware. They have larger social circles and may be exposed to more language diversity. And because women tend to be caregivers, boys usually learn language from their mothers, whereas women and girls learn words from other women.12．Why does the author mention some words coined by male writers?A．To compare male and female writers.B．To introduce some well-known writers.C．To lead in the topic to be talked about.D．To put forward a point to be proved.13．Which of the following words was invented by a male?A．Rizz.B．Situationship.C．Swiftie.D．Goblin mode. 14．Who thinks women’s social awareness contributes to lexical innovation?A．Dr. Seuss.B．William Labov.C．Sali Tagliamonte.D．Gretchen McCulloch.15．What can be a suitable title for the text?A．New Words Are Added to English DictionariesB．Female Teens Take a Lead in Linguistic ChangesC．Women Are More Sensitive to Language DiversityD．New Terms Make Communication More ColorfulHow to Deal with Awkward SilencesWe all know what it’s like when a conversation dies off and you’re left to sit in silence.16 Thankfully, filling these silences isn’t as difficult as it may seem. Read on to learn how to initiate small talk and deal with awkward silences.• Think of topics ahead of time.Before attending a social event, think of a few “go-to” topics to jumpstart a dead conversation. This will help you to fill the silences instead of struggling for words in the moment. Recent news, local events, popular books and television shows are all good fallback options. 17• Ask open-ended questions.Ask questions that have more than one possible answer. 18 They help avoid resulting in short response. For example, instead of asking, “Did you like your high school?”, you could ask, “What did you like about your high school?”• 19As with any good conversation, the biggest key is to listen. If they respond to your question with a short statement such as “Yes”or “No”, that might indicate that they aren’t comfortable talking about a particular subject. Instead, talk about something that you know they're interested in. For example, “I heard you won your football game the other night. I’d love to hear about it.”• Give a compliment (表扬).20 For example, you could say, “I really like your shirt. Where did you get it?” This could move the conversation in another direction while making the other person feel good. Stick to saying nice things if you’re trying to make small talk.A．Keep the conversation flowing.B．It also allows you to spend more time together.C．Listen carefully and respond to their reactions.D．There’s only one word to describe it: awkward.E．This is always a safe bet, as long as it’s appropriate.F．Such questions are likely to get the person talking more.G．However, avoid some subjects that may cause arguments.二、完形填空Growing up on a farm, Tori James spent much of her early years outdoors, building shelters and playing in muddy streams. She 21 herself as an ambitious and adventurous child and owes the Duke of Edinburgh Award to her love for 22 . Jamessays she could never have 23 that one day she might be the youngest British woman to 24 the peak of Mount Everest.Today, James is a(n) 25 adventurer. Alongside her successful climb of the world’s 26 peak in 2007, she was a member of the first all-female team to 27a tiring 650-kilometre ski race to the Magnetic North Pole in 2005, and in 2014 she set the28 for the longest non-stop open-sea kayak crossing in UK waters.Yet ahead of her first exploration at the age of 18, James had 29 her ability to last a month out in the cold without the comfort of a hot bath. “It’s only by 30 our comfort zone that we understand our true potential,” she explains, “That’s when the 31 happens.”James encourages would-be adventurers not to be 32 by the conventional image of the rugged (粗犷的), hardy explorer. “I am such a believer that, with the right support and skills, everyone can bring 33 to an exploration,” she adds, “Sometimes, it’s only out in the field that someone’s 34 talents emerge,35 in the process of exploration.”21．A．describes B．admits C．imagines D．praises 22．A．wildlife B．science C．exploration D．competition 23．A．explained B．understood C．declared D．predicted 24．A．measure B．approach C．admire D．reach 25．A．patient B．ordinary C．professional D．honest 26．A．coldest B．highest C．prettiest D．straightest 27．A．complete B．coach C．judge D．cancel 28．A．record B．date C．goal D．standard 29．A．lost B．doubted C．limited D．developed 30．A．entering B．restoring C．leaving D．expanding 31．A．emergency B．accident C．event D．magic 32．A．pushed ahead B．kept apart C．cheered up D．put off 33．A．something B．nothing C．anything D．everything 34．A．academic B．creative C．hidden D．artistic 35．A．eventually B．particularly C．generally D．obviously三、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。