20XX东北三校联考(一、二、三次汇总)

哈尔滨师大附中 东北师大附中 辽宁省实验中学2023年高三第二次联合模拟考试数学注意事项：1.答卷前，考生务必将自己的姓名、准考证号填写在答题卡上.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60分）一、选择题（本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项符合题目要求的）1.已知集合{}1,2,3A =，{}20B x x x m =-+=，若{}2AB =，则B =（ ）A.{}2,1B.{}2,4C.{}2,3D.{}2,1-2.已知复数z 满足24i z z +=+，则z =（ ） A.34i +B.34i -C.34i -+D.34i --3.已知向量()1,0a =，1,22b ⎛=-⎝⎭，则a b -=（ ） A.3C.14.有7名运动员（5男2女）参加A 、B 、C 三个集训营集训，其中A 集训营安排5人，B 集训营与C 集训营各安排1人，且两名女运动员不在同一个集训营，则不同的安排方案种数为（ ） A.18B.22C.30D.365.两条直线()0y kx k =>和2y kx =-分别与抛物线24y x =交于异于原点的A 、B 两点，且直线AB 过点()1,0，则k =（）A.12B.1D.26.如图，直角梯形ABCD 中，3AB CD =，30ABC ∠=︒，4BC =，梯形ABCD 绕AD 所在直线旋转一周，所得几何体的外接球的表面积为（ ）A.1123πB.48πC.128πD.208π7.定义在R 上的奇函数()f x 满足()()11f x f x +=-，且在[]0,1上单调递减，若方程()10f x +=在[)0,1有实数根，则方程()1f x =在区间[)1,11-上所有实数根之和是（ ） A.6B.12C.30D.568.已知三个互异的正数a ，b ，c 满足2ln cc aa=+，()21ab =+，则关于a ，b ，c 下列判断正确的是（ ） A.a b c <<B.a b c >>C.2a c b -<-D.2a c b ->-二、选择题（本大题共4小题，每小题5分，共20分，在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分）9.函数()sin cos f x x x =+，则下列说法正确的是（ ） A.()f x 为偶函数B.()f x 的最小正周期是πC.()f x 在0,2π⎛⎫⎪⎝⎭单调递增 D.()f x 的最小值为1-10.金枪鱼因为肉质柔嫩鲜美、营养丰富深受现代人喜爱，常被制作成罐头食用.但当这种鱼罐头中的汞含量超过1.0mg/kg 时，食用它就会对人体产生危害.某工厂现有甲、乙两条金枪鱼罐头生产线，现从甲、乙两条生产线中各随机选出10盒罐头并检验其汞含量（单位为mg/kg ），其中甲生产线数据统计如下：0.07，0.24，0.39，0.54，0.61，0.66，0.73，0.82，0.95，0.99，其方差为210.08s =.乙生产线统计数据的均值为20.4x =，方差为220.11s =，下列说法正确的是（ ）A.甲生产线的金枪鱼罐头汞含量数值样本的上四分位数是0.82B.甲生产线的金枪鱼罐头汞含量数值样本的上四分位数是0.775C.由样本估计总体，甲生产线生产的金枪鱼罐头汞含量平均值高于两条生产线生产的金枪鱼罐头汞含量平均值D.由样本估计总体，甲生产线生产的金枪鱼罐头汞含量数值较两条生产线生产的金枪鱼罐头汞含量数值更稳定11.已知正方体1111ABCD A B C D -E ，F 是棱1DD ，1CC 的中点，点M 是侧面11CDD C 内运动（包含边界），且AM 与面11CDD C 所成角的正切值为2，下列说法正确的是（ ）A.1MC 2B.存在点M ，使得AM CE ⊥C.存在点M ，使得AM ∥平面BDFD.所有满足条件的动线段AM 形成的曲面面积为612.已知函数()()1,*mn f x x m n N x=+∈，下列结论正确的是（ ） A.对任意m ，*n N ∈，函数()f x 有且只有两个极值点 B.存在m ，*n N ∈，曲线()y f x =有经过原点的切线 C.对于任意10x >，20x >且12x x ≠，均满足()()121222f x f x x x f ++⎛⎫<⎪⎝⎭D.当0x >时，()()f x f x -≤恒成立第Ⅱ卷（非选择题共90分）三、填空题（本大题共4小题，每小题5分，共20分）13.大气压强p =压力受力面积，它的单位是“帕斯卡”（Pa ，21Pa 1N/m =），已知大气压强()Pa p 随高度()m h 的变化规律是0khp p e -=，其中0p 是海平面大气压强，10.000126m k -=.当地高山上一处大气压强是海平面处大气压强的13，则高山上该处的海拔为______米.（答案保留整数，参考数据ln3 1.1≈） 14.曲线22x y x y +=+围成的图形的面积是______.15.已知双曲线()2222:10,0x y C a b a b-=>>的右焦点为(),0F c ，过点F 且斜率为2的直线与双曲线C 的两条渐近线分别交于M 、N 两点，若P 是线段MN 的中点，且PF =，则双曲线的离心率为______. 16.A 、B 、C 、D 、E 五个队进行单循环赛（单循环赛制是指所有参赛队在竞赛中均能相遇一次），胜一场得3分，负一场得0分，平局各得1分.若A 队2胜2负，B 队得8分，C 队得9分，E 队胜了D 队，则D 队得分为______.四、解答题（本大题共6小题，共70分.解答应写出文字说明、证明过程和演算步骤）17.（本小题满分10分）记ABC △的内角A 、B 、C 的对边分别为a 、b 、c ，已知()21cos 4bc A a +=.（1）证明：3b c a +=； （2）若2a =，7cos 9A =，角B 的内角平分线与边AC 交于点D ，求BD 的长. 18.（本小题满分12分）调查问卷中常常涉及到个人隐私或本人不愿正面回答的问题，被访人可能拒绝回答，即使回答，也不能期望答案是真实的.某小区要调查业主对物业工作是否满意的真实情况，现利用“随机化选答抽样”方法制作了具体调查方案，其操作流程如下：在一个箱子里放3个红球和2个白球，被调查者在摸到球后记住颜色并立即将球放回，如果抽到的是红球，则回答“你的性别是否为男性？”如果抽到的是白球，则回答“你对物业工作现状是否满意？”两个问题均用“是”或“否”回答.（1）共收取调查问卷100份，其中答案为“是”的问卷为60份，求一个业主对物业工作表示满意的概率，已知该小区共有业主500人，估计该小区业主对物业工作满意的人数；（2）现为了提高对物业工作满意的业主比例，对小区业主进行随机访谈，请表示不满意的业主在访谈中提出两个有待改进的问题.（ⅰ）若物业对每一个待改进的问题均提出一个相应的解决方案，该方案需要由5名业主委员会代表投票决定是否可行.每位代表投赞同票的概率均为13，方案需至少3人投赞成票，方能予以通过，并最终解决该问题，求某个问题能够被解决的概率0p ；（ⅱ）假设业主所提问题各不相同，每一个问题能够被解决的概率都为0p ，并且都相互独立.物业每解决一个问题，业主满意的比例将提高一个百分点.为了让业主满意的比例提高到80%，试估计至少要访谈多少位业主？ 19.（本小题满分12分）如图，已知斜四棱柱1111ABCD A B C D -，底面ABCD 为等腰梯形，AB CD ∥，点1A 在底面ABCD 的射影为O ，且11AD BC CD AA ====，2AB =，112AO =，1AA BC ⊥.（1）求证：平面ABCD ⊥平面11ACC A ；（2）若M 为线段11B D 上一点，且平面MBC 与平面ABCD 夹角的余弦值为7，求直线1A M 与平面MBC所成角的正弦值. 20.（本小题满分12分） 已知数列{}n a ，设()12*nn a a a m n N n+++=∈，若{}n a 满足性质Ω：存在常数c ，使得对于任意两两不等的正整数i 、j 、k ，都有()()()k i j i j m j k m k i m c -+-+-=，则称数列{}n a 为“梦想数列”. （1）若()2*nn b n N =∈，判断数列{}n b 是否为“梦想数列”，并说明理由； （2）若()21*n c n n N =-∈，判断数列{}n c 是否为“梦想数列”，并说明理由； （3）判断“梦想数列”{}n a 是否为等差数列，并说明理由. 21.（本小题满分12分）已知椭圆()22122:10x y C a b a b +=>>的离心率为3，x 轴被抛物线22:4x C y b =-截得的线段长与1C 长轴长的比为2:3.（1）求1C 、2C 的方程；（2）设2C 与y 轴的交点为M ，过坐标原点O 的直线l 与2C 相交于点A 、B ，直线MA 、MB 分别与1C 相交与D 、E .（ⅰ）设直线MD 、ME 的斜率分别为1k 、2k ，求12k k 的值； （ⅱ）记MAB △、MDE △的面积分别是1S 、2S ，求12S S 的最小值. 22.（本小题满分12分）已知函数()()ln 10f x x ax a =-->.（1）当1a =时，求过原点且与()f x 相切的直线方程；（2）若()()()0axg x x e f x a =+⋅>有两个不同的零点1x 、()2120x x x <<，不等式212mx x e ⋅>恒成立，求实数m 的取值范围.三省三校第二次模拟答案一、单选题二、多选题三、填空题：13、873014、2π+15 16、18.2ln 2ln c c a a -=-考虑：()()2ln 0f x x x x =->，则()221x f x x x-'=-= ()f x 在()0,2递减；()f x 在()2,+∞递增()()()min 221ln 20f x f ==->（1）当02a <<，2c >时，21a+=设()x xg x =+，是减函数，且()21g =()()2121aaag a g b a =+>=⇒=+>⇒> 2212152a b =+<+=⇒<所以，22c b a a c b >>>⇒->-（2）当02c <<，2a >时，同理可得：22a b c a c b >>>⇒->- 综上可得：2a c b ->-成立. 12.如图：（1）在第一象限+都是凹函数（二阶导数大于零） （2）图二、图三有过原点的切线 （3）极值点的个数是一个或两个（4）当m ，n 同奇数或同偶数时，()()f x f x =-；当m ，n 是一奇，一偶数时，()()f x f x >-； 15.设()11,M x y ，()22,N x y ，()00,P x y2211222222222200MN OP x y b a b k k a x y a b ⎧-=⎪⎪⇒⋅=⎨⎪-=⎪⎩，则OP 的方程为222b y x a =，MN 的方程为：()2y x c =- ()222224242P b y xa c x c OP e a ab y xc ⎧=⎪⇒==+⇒=⎨-⎪=-⎩16.A 队：2胜2负（无平局） C 队：3胜1负（无平局）B 队：2胜2平，则B 队和D 、E 是平局；B 队胜了A 、C这样找到了C 队负的一场，输给B 队 这样B 、C 结束；A 队赢D 、E 最后，E 胜D ，则D 的1分.四、解答题17.（本题满分10分）（1）证明：()222221cos 4142b c a bc A a bc a bc ⎛⎫+-+=⇒+= ⎪⎝⎭()229b c a +=，则3b c a +=……5'（2）由余弦定理得：2222cos a b c b A =+-，则9bc =，又3b c a +=，则3b c ==由角分线可得，95AD =所以，在ABD △中，由余弦定理得：2222cos BD AD c AD c A =+-⋅，BD =10'18.（本题满分12分）（1）记：事件A =“业主对物业工作表示满意”，则()()2316035521004P A P A ⋅+⋅=⇒= 所以，35003754⨯=（人）……4' 答：该小区业主对物业工作表示满意的人数约为375人.（2）（ⅰ）3245345055512121173333381P C C C ⎛⎫⎛⎫⎛⎫⎛⎫=⋅+⋅+= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭……8' （ⅱ）设至少要访谈n 位业主31738101280%10047.6481417n n ⎛⎫⎛⎫⋅-⋅⋅≥-⨯⇒≥≈ ⎪ ⎪⎝⎭⎝⎭答：至少要访谈48位业主.……12' 19.（本题满分12分）（1）证明：等腰梯形ABCD 中，2AB =，1BC CD AD ===则，60ABC ∠=︒……2'1BC ACBC BC AA ⊥⎧⇒⊥⎨⊥⎩平面11A ACC ，BC ⊂平面ABCD ，则平面ABCD ⊥平面11A ACC ，……4' （2）建立如图所示空间直角坐标系C xyz -，则)A，()0,1,0B，2O ⎛⎫ ⎪ ⎪⎝⎭，1122A ⎛⎫ ⎪ ⎪⎝⎭，131,0222CD BA ⎛⎫==-⎪ ⎪⎝⎭ 1133,022B DBD ⎛⎫==- ⎪ ⎪⎝⎭，1112DD AA ⎛⎫== ⎪ ⎪⎝⎭，1110,,22D⎛⎫- ⎪⎝⎭ 设111,0D M D B λ⎛⎫== ⎪ ⎪⎝⎭，131,,222M λ⎛⎫-+ ⎪ ⎪⎝⎭ (6)'设平面MBC 的法向量为(),,n x y z =131022220n CM y z n CB y λλ⎧⎛⎫⎧⋅-++=⎪⎪ ⎪⇒⎨⎨⎝⎭⋅⎪⎪⎩=⎩，取1x =，则()1,0,n =-……8' 取平面ABCD 的法向量()0,0,1m =221cos ,417m n m n m nλ⋅==⇒=，则12λ= 即：11,04A M ⎛⎫= ⎪⎪⎝⎭，1,0,n ⎛= ⎝⎭……10' 设直线1A M 与平面MBC 所成的角为θ，则1113sin cos ,7A M n A M n A M nθ⋅===⋅所以，直线1A M 与平面MBC……12' 20.（本题满分12分）（1）()()()k i j i j m j k m k i m c -+-+-=()()()k j i j i m i k m k j m c -+-+-=所以，0c =当2nn b =时，12m =，23m =，3143m =()()()142612232313033-+-⋅+-⋅=≠所以，{}n b 不是“梦想数列”……4' （2）21i a i =-，21j a j =-，21k a k =-()()()2220k i j i j j k k i k i j-+-+-=所以，{}n c 不是“梦想数列”……6'（3）①令1i =，2j =，3k = ()()()1231121223310312a a a a a a +++-+-+-= 所以，1322a a a +=，即：1a 、2a 、3a 成等差数列……8' ②令1i =，2j =，()3k n n =≥ ()()()21122102n S S n a n n -+-+-= ()()2122310n S n n a n n a +---= ()()21122210n S n n a n n a ++---+= 所以，11121122220n n a na a na a a nd +++--=⇒=+ 所以，()()114n a a n d n =+-≥，当1,2,3n =时也成立. 综上可得，“梦想数列”{}n a 是等差数列. ……12' 21.（本题满分12分）（1）椭圆方程：()222210x y a b a b+=>>13323c b a a ⎧=⎪=⎧⎪⇒⎨=⎩=，所以，221:19x C y +=，221:14C y x =-……4' （2）设直线l 的方程为y kx =，()11,A x y ，()22,B x y22440114y kxx kx y x =⎧⎪⇒--=⎨=-⎪⎩，则121244x x k x x +=⎧⎨⋅=-⎩……6' 又111114y x k x +==，12121164x x k k ==- 联立122114014y k x x k x x y =-⎧⎪⇒-=⎨=-⎪⎩，则114x k =，同理：224x k = 联立()1221122191180990y k x k x k x x y =-⎧⇒+-=⎨+-=⎩ 13211891k x k =+，同理：24221891k x k =+……8' ()()2211221sin 429191181sin 2MA MB AMBS k k S MD ME DME ∠==++∠……10' 2121481916919811616324k k ⎛⎫=+++≥ ⎪⎝⎭，当且仅当112k =±时，取等号 所以，12S S 的最小值为169324. ……12' 22.（本题满分12分）（1）()f x 的定义域为()0,+∞ ()111f x a x x'=-=- 设切点坐标()000,ln 1x x x -+，则切线方程为：()()00001ln 11y x x x x x ⎛⎫--+=--⎪⎝⎭把点()0,0带入切线得：20x e =所以，()f x 的切线方程为：221e y x e-=……4' （2）()()ln 1axg x x ex ax =+--有两个不同零点，则()()()ln ln 10ln 1ln 10ax x ax ax xx e x ax x ax e x ax e-+--=⇒+--=+--=……6' 构造函数()1xu x e x =+-，()1xu x e '=+()u x 为(),-∞+∞增函数，且()00u =即：ln 0x ax -=有两个不等实根1122ln ln ax x ax x =⎧⎨=⎩令1122ln ln x x t x x ==，()01t <<，则12ln ln x t x =，12ln ln ln x x t =+ 122ln 2ln ln 1t x x t t ++=-……8' 设()()2ln 011x v x x x x +=<<-，()()22123ln 1x x v x x x x ⎡⎤+-'=-+⎢⎥-⎣⎦ 设()23ln 1x x x xφ=-+-+，()()()212x x x x φ--'= ()x φ在()0,1递增，()10φ=，则()v x 在()0,1递减，且()10v =所以，()v x 的最小值()1v ，……10' ()()()112ln lim 2ln 31x x x x x x x =→+'=+=-所以，()v x 的最小值为3，即：m 的取值范围为(],3-∞. ……12'。

2022东北三校联考(一二三次汇总)篇一：2022东北三省三校联合考试二模参考答案(语文)东北三省三校联考二模答案（语文）1．D（以偏概全，原文中“对人的处境有真切的关心，对人生在这个世界上的命运有深刻周彻的肯认，对人内心经验有感同身受的体谅，并精骛八极，鞭辟入里”是许多人文社科类著作，包括一部分自然科学著作成为经典的原因，依据原文不能作为“衡量著作是不是经典”的标准。

）2．C（曲解原文，“它们都可以诠释和演绎人生”曲解文意。

）3．C（强加因果，卡夫卡将巴尔扎克“我摧毁了每一个障碍”置换成“每一个障碍都摧毁了我”，是“人生到处无往而非问题”的例子，二者并非因果关系。

）4.B5.B江东，指长江以东地区，古人以东为左，故又称江左。

6.C“他建议朝廷让有关官吏专门批阅奏折”错。

原文是“应当命令有关部门详细审察奏章”。

7.(1)君主遭忧是臣子的耻辱，按道义应当如此（或“这样”），反而凭借这个接受赏赐吗？（“主忧臣辱”译为判断句；“义”译为“按道义、道义上”；“尔”译为“如此”、“这样”；“顾邪”译为“反而（难道）吗？”以上四处，每译对一处给1分，译出大意给1分。

）(2)高宗不断告知辅弼大臣宽慰爱惜民力，大概是惩戒于秦桧的苛刻政治，希望这样使百姓安定。

（“宽恤”译为“宽慰体恤、宽慰爱惜”；“盖”译为“大概”；“期”译为“希望”；“安”译为“使安定”；以上四处，每译对一处给1分，译出大意给1分。

）【参考译文】8．首联运用用典、对比手法。

（2分）首先，诗人借司马相如的典故来写老友的才华横溢；整联又将高才与不遇形成鲜明的对比，一是“凌云”，一是“陆沉”。

（2分）从而表达了说明：手法2分，分析手法2分，思想感情1分。

手法还可答反问、借代。

“何意陆沉黄绶间？”以反诘之语表现对朋友的同情。

“黄绶”，根据注解，是借代修辞。

9．①才华横溢，位沉下僚：首句说老友有像司马相如那样的才华却只是身处下僚。

（2分）②迫于生计，委身官场：“头白眼花”却仍奔走仕途，“儿婚女嫁”之后，才可挂冠归去。

哈尔滨师大附中 东北师大附中 辽宁省实验中学2024年高三第二次联合模拟考试化学试卷性气味气体的产生，设计了由灯座、灯盏、烟管三部分组成的结构。

下列说法错误本试卷共19题，共100分。

考试用时75分钟。

注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

2．答选择题时，选出每小题答案后，用铅笔把答题卡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。

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。

本卷可能用到的相对原子质量：H1Li7C12O16F19Mg24S32K39一、选择题（本题共15小题，每小题3分，共45分。

在每小题给出的四个选项中，只有一个选项符合要求。

）1．东汉错银铜牛灯采用铜、银二种材质制作，常以动物油脂或植物油为燃料，为减少燃烧过程烟尘和刺激的是（ ）A ．烟管的作用是将燃烧产生的烟气导入铜牛灯座腹腔中B ．古人常用草木灰浸泡液代替牛腹中的水，吸收烟气的效果更佳C ．银、铜的导热性能好，可以使燃料充分燃烧D ．灯具的设计包含了装置、试剂、环保等实验要素 2．下列化学用语或表述错误的是（ ）A ．乙烯的球棍模型：B ．基态Al 原子最高能级的电子云轮廓图：C ．在()346Ni NH SO 中，阴离子的VSEPR 模型名称：正四面体形D ．次氯酸钠中含有的化学键类型：极性键、离子键 3．下列有关物质的工业制备反应错误的是（ ）A ．侯氏制碱：23234NaCl H O NH CO NaHCO NH Cl +++↓+B ．工业合成氨：223N 3H 2NH →+← 高温、高压催化剂C ．氯碱工业：2222NaCl 2H O2NaOH H Cl ++↑+↑电解D ．冶炼金属铝：322AlCl 2Al 3Cl +↑电解4．穴醚是一类可以与碱金属离子发生配位的双环或多环多齿配体。

某种穴醚的键线式如图。

2024年高三第一次联合模拟考试数学参考答案一.单项选择题1-4 CABD 5-8 CBBB 二.多项选择题9.ACD 10.ABD 11.ABD 三.填空题12. 3274四.解答题15.解：（1）()2cos 22sin f x x x '=− 2' (0)2,(0)2f f '== 4'∴()f x 在0x =处的切线方程为22(0)y x −=−，即22y x =+ 6'（2）22()2cos 22sin 2(1sin )2sin 2(2sin sin 1)f x x x x x x x '=−=−−=−+− 8'()0f x '<则22(2sin sin 1)0x x −+−< 10'即2(2sin 1)(sin 1)0x x −−+<即1sin 2x >解得5(2,2),66x k k k Z ππππ∈++∈ 12' 故()f x 的单调递减区间为5(2,2),66k k k Z ππππ++∈ 13' 16.解：（1）底面ABCD 为平行四边形，120ADC ∠=,60DAB ∴∠=. 4,8DA AB ==由余弦定理可得：2222cos 6048DB AB AD AB AD =+−⨯=DB ∴=则222DA DB AB +=，DA DB ∴⊥ 2' 侧棱1DD ABCD ⊥平面，DB ABCD ⊂平面1DD DB ∴⊥4'111111,,DA ADD A DD ADD A DA DD D ⊂⊂=又平面平面且11DB ADD A ∴⊥平面6' 111AA ADD A ⊂又平面1DB AA ∴⊥7'（2）四棱台中1111ABCD A B C D −的体积为2833111111111()3ABCD A B C D ABCD A B C D V S S S S ∴=++1111111112831()33DD AD DB A D D B AD DB A D D B ∴=++ 1283128333DD ∴=,解得：11DD = 9'如图，以点D 为原点，1,,DA DB DD 所在直线为x 轴，y 轴，z 轴， 建立如图的空间直角坐标系，则1(4,0,0),(0,43,0),(4,43,0),(0,23,1)A B C B −1(4,0,0),(0,23,1)BC BB ∴=−=−11'设平面11BCC B 的法向量为(,,)n x y z =，则有140230n BC x n BB y z ⎧=−=⎪⎨=−+=⎪⎩所以(0,1,23)n =13'平面11ADD A 的法向量为(0,1,0)m =，设平面11ADD A 与平面11BCC B 所成锐二面角为θ 则113cos |cos ,|1313m n m n m nθ⋅=<>=== 15'17.解：（1）由图估计甲班平均分较高3'（2）由图可知，甲班中有12的学生分数低于128分； 乙班中有34的学生分数低于128分 设从两班中随机抽取一人， “该同学来自甲班为事件A ”，“该同学分数低于128分为事件B ”，则1113(),(),(),(),2224P A P A P B A P B A ==== 5' ()()()()()()()P B P AB P AB P B A P A P B A P A ∴=+=⋅+⋅1131522428=⨯+⨯=7'11()()()222()5()()58P A P B A P AB P A B P B P B ⨯==== 8'13()()()324()5()()58P A P B A P AB P A B P B P B ⨯====9'所以，该同学来自甲乙两班的概率分别为23,55(3)依题X 的所有可能取值为0,1,2,310'30643101(0)6C C P X C === 11'21643101(1)2C C P X C === 12'12643103(2)10C C P X C ===13'03643101(4)30C C P X C ===14'所以X 的分布列为:15'18.解：（1）设1122(,),(,)M x y N x y ，则12122,6x x y y +=+=,M N 两点在双曲线C 上22112222222211x y a b x y a b ⎧−=⎪⎪∴⎨⎪−=⎪⎩①②，由−①②得22221212220x x y y a b −−−= 即2221222212y y b x x a −=−， ()()()()2121221212y y y y b x x x x a+−∴=+− 2'22OQ MNb k k a∴⋅=，即222213,3b b a a ∴⋅=∴=又21,3a b =∴=，∴双曲线C 的方程为：2213y x −=4'（2）由已知可得，直线MN 的方程为：31(1)y x −=⋅−，即2y x =+联立22222470,1656720330y x x x x y =+⎧⇒−−=∆=+=>⎨−−=⎩ 6' 则121272,2x x x x +==− 8'11221212(1,)(1,)(1)(1)EM EN x y x y x x y y ⋅=−⋅−=−−+12121212(1)(1)(2)(2)2()5x x x x x x x x =−−+++=+++72()2502=⨯−++=EM EN ∴⊥，EMN ∴∆为直角三角形 10'（3）由题意可知，若直线AB 有斜率则斜率不为0，故设直线AB 方程为：x my n =+ 设334455(,),(,),(,)P x y A x y B x y34345353,(,)(,)AP PB x x y y x x y y λλ=∴−−=−−45334533453453()1()1x x x x x x x y y y y y y y λλλλλλ+⎧=⎪−=−⎧⎪+∴⇒⎨⎨−=−+⎩⎪=⎪+⎩点P 在双曲线C 上， 22454511113x x y y λλλλ++⎛⎫⎛⎫ ⎪ ⎪++⎝⎭⎝⎭∴−= 22245453()()3(1)x x y y λλλ∴+−+=+22222244554545(3)(3)2(3)3(1)x y x y x x y y λλλ∴−+−+−=+③又2222445530,30x y x y −=−=，245452(3)3(1)x x y y λλ∴−=+，245453(1)32x x y y λλ+∴−=④ 联立2222230(31)630x y m y mny n x my n ⎧−=⇒−++=⎨=+⎩2222231033612(31)0m m m n n m ⎧−≠⇒≠±⎨∆=−−>⎩245452263,3131mn n y y y y m m −+==−−⑤⑥14',A B 分别在第一象限和第四象限，2450,310y y m ∴<∴−<由④式得：245453(1)3()()2my n my n y y λλ+++−=22245453(1)(31)3()32m y y mn y y n λλ+∴−+++=⑦将⑤⑥代入⑦得：222222363(1)(31)3331312n mn m mn n m m λλ−+∴−++=−− 22263(1)312n m λλ−+∴=−121sin 2AOB S OA OB AOB y y ∆∴=⋅⋅∠=221223(1)12312n y m λλλλ+⎫=====++⎪−⎭15'令11(),[,2]3h λλλλ=+∈ 221(1)(1)1()1,[,2]3h λλλλλλ+−'=−=∈ 1,1,()03h λλ⎡⎫'∴∈<⎪⎢⎣⎭，()h λ单调递减(]1,2,()0h λλ'∈>，()h λ单调递增10()[2,]3h λ∴∈， 16'3AOB S ∆∴∈⎦17'19.（1）证明：32310183222121k k k n a a a +++=⋅+⋅++⋅+⋅+01(83)11()2k S n a a a S n ∴+=+++++=+ 3'21210143222121k k k n a a a +++=⋅+⋅++⋅+⋅+01(43)11()2k S n a a a S n ∴+=+++++=+6' (83)(43)S n S n ∴+=+7'（2）（Ⅰ）解：260321684(111100)=+++=(60)2I ∴= 10'（Ⅱ）解： 21(1)=，2511(111111111)=，故从1n =到511n =中 I(n)=0有9个，I(n)=1有C 11+C 21+⋯C 81=C 92个， I(n)=2有C 22+C 32+⋯C 82=C 93个，……，I(n)=9有C 88=C 99=1个， ∑2I(n)511n=1=9×20+C 92×21+C 93×22+⋯C 99×2813'=C91×21+C92×22+C93×23+⋯C99×292=C90×20+C91×21+C92×22+C93×23+⋯C99×29−1216'=(1+2)9−12=984117'。

哈尔滨师大附中 东北师大附中 辽宁省实验中学2024年高三第一次联合模拟考试数学注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上．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，定在．本试卷上无效．3．考试结束后，将本试卷和答题卡一并交回．一、单项选择题：本大题共8小题，每小题5分，共40分．在每小题给出的四选项中，只有一项是符合题目要求的．1．已知集合{}1,2M =，(){}2log 212N x x =∈−≤R ，则M N = （ ） A ．{}1B ．{}2C ．{}1,2D ．∅2．已知复数z 的共轭复数是z ，若i 1i z ⋅=−，则z =（ ） A ．1i −+B ．1i −−C ．1i −D ．1i +3．已知函数()y f x =是定义在R 上的奇函数，且当0x <时，()2af x x x=+，若()38f =−，则a =（ ） A ．3−B ．3C ．13D ．13−4．已知平面直角坐标系xOy 中，椭圆C ：22221x y a b+=（0a b >>）的左顶点和上顶点分别为A ，B ，过左焦点F 且平行于直线AB 的直线交y 轴于点D ，若2OD DB =，则椭圆C 的离心率为（ ）A ．12B C ．13D ．235．()521x x y y −−的展开式中32x y 的系数为（ ） A ．55B ．70−C ．30D ．25−6．已知正四棱锥P ABCD −各顶点都在同一球面上，且正四棱锥底面边长为4，体积为643，则该球表面积为（ ） A ．9πB ．36πC ．4πD ．4π37．已知函数()22e e xx f x ax −=−−，若0x ≥时，恒有()0f x ≥，则a 的取值范围是（ ）A ．(],2−∞B ．(],4−∞C ．[)2,+∞D ．[)4,+∞8．设1033e a =，11ln 10b =，ln 2.210c =，则（ ） A ．a b c <<B ．c b a <<C ．b c a <<D ．a c b <<二、多项选择题：本大题共3小题，每小题6分，共18分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得6分，部分选对的得部分分，有选错的得0分．9．等差数列{}n a 中，10a >，则下列命题正确的是（ ） A ．若374a a +=，则918S =B ．若150S >，160S <，则2289a a > C ．若211a a +=，349a a +=，则7825a a += D ．若810a S =，则90S >，100S <10．在平面直角坐标系xOy 中，抛物线C ：24y x =的焦点为F ，点P 在抛物线C 上，点Q 在抛物线C 的准线上，则以下命题正确的是（ ） A ．PQ PF +的最小值是2 B ．PQ PF ≥C ．当点P 的纵坐标为4时，存在点Q ，使得3QF FP =D ．若PQF △是等边三角形，则点P 的橫坐标是311．在一个只有一条环形道路的小镇上，有2家酒馆A ，一个酒鬼家住在D ，其相对位置关系如图所示．小镇的环形道路可以视为8段小路，每段小路需要步行3分钟时间．某天晚上酒鬼从酒馆喝完酒后离开，因为醉酒，所以酒鬼在每段小路的起点都等可能的选择顺时针或者逆时针的走完这段小路。

2024届东北三省三校第一次联考数学试题
+答案
2024年2月东北三省三校高三一模数学试题
2024年2月东北三省三校高三一模数学试题及参考答案
东北三省三校联考的意义
东北地区最具影响力的高考联合模拟考试，一个是三省四市联合模拟考试，另一个是三省三校联合模拟考试，三省四市模考由东北F4市教研院组织，参加学校基本为四市市重点高中。

三省三校联考由东北排名第一的高中东北师大附中，哈尔滨市排名第二的高中哈师大附中以及辽宁省实验中学联合举办，参加学校多为省重点高中，两类考试在东北享有很高的声誉，为莘莘学子备考提供重要参考!
高三模拟考试和高考哪个更难
这个回答没有绝对的答案，因为每年的模拟卷内容不通、高考的考卷难度也不同。

而且因为考生的成绩不同，对于考卷的难易程度判断也不同。

所以这个问题没有确切的答案。

如果按照总体的水平来评估，高考试卷的难度不会高于模拟卷，高考中基础部分和中级难度的题目占比在80%左右，只有20%是拔高题。

所以如果基础知识打的好，那么对于考生来说，高考题目不难。

高考的题目的难度是在问法、提问方式上，而并不是运用了超纲的知识点，与大家传统意义上的难度不通。

相比高考的难度，模拟考要更难一点。

因为模拟考的目的是希望通过考试来判断学生对知识点的掌握情况，如果过于简单就起不到探底的目的。

2020年东北三省三校高三第二次联合模拟英语试题注意事项：1．答题前填写好自己的姓名、班级、考号等信息；2．请将答案正确填写在答题卡上。

第I卷（选择题)一、阅读理解Moringa Farm Internship ProgramDATE：5 May〜30 November 2019LOCATION: Jaipur IndiaEVENT TYPE：TwiningEVENT INDUSTRY/TOPIC: Clean Energy, Climate & Environment, Food & Agriculture and Quality/Process ManagemMoringa (辣木), a kind of miracle tree, has come a long way today to become a symbol of the fight against growing nutrition deficiency (营养不足) across the globe. It is also a crop suited to the climatic conditions of the region, which is unusually rich in its nutrition content literally from top to bottom.Supporting learning：The Moringa Farm Internship Program is sort of like a live, interactive slideshow of agriculture making of Moringa from “Soil to Super food” that touches all the participant’s senses.The Program’s Objectives：The Moringa Farm Internship Program is devoted to enriching the knowledge and lives of young people by providing them with a unique educational experience on an organic Moringa Farm in rural Jaipur, India. Program participants can learn about conservation, maintenance (维护), care, and management of Moringa Farming by means of hands-on work experience with local Moringa Farm staff.Participants will be able to interact with Moringa specialists and educators, on all issuesthat may affect their operations. Experts will let you know management strategies and equipment operation up close on hand to answer your specific questions.The Moringa Farm Internship will provide an excellent opportunity to learn about sustainable tools, techniques, and approaches that can be used in Moringa Production and value-added production systems, righting (突出) pest and disease management, leaf/seed production, and tools and equipment for commercial production.If you’re trying to learn about Moringa, there’s nothing like going to a farm and getting your hands dirty. So click here to register now!1．Why is Moringa specifically described in the text?A．To stress the importance of the crop.B．To encourage people to plant Moringa.C．To show the significance of the program.D．To draw attention to nutrition deficiency.2．What can we know about the Moringa Farm Internship Program?A．It is intended for farmers. B．It lasts more than half a year.C．It provides vocational training. D．It is available on the Internet.3．How do the program’s participants learn about Moringa?A．By attending various courses about it.B．By involving themselves in farming it.C．By observing specialists and educators.D．By watching slideshows about the crop.We had two dogs, Lucky and Lily Bit. Lucky loved to find ways to leave the confines (束缚) of her indoor living routine, which made her a masterful escape artist. Lily Bit, on the other hand, did exactly as she was told, never straying, always staying close to home... except when she came under the influence of Lucky.Whenever the two got out together, they could be seen out of sights often staying away for hours. More often than not, they’d arrive home near suppertime, covered in mud and leaves and smelling to high heaven from their afternoon outing in the woods.I remember one such occasion w hen they’d once again mysteriously escaped, except this time, they returned clean and no worse for wear. It wasn’t until a few days later that we learned why. My wife ran into our neighbor Carl who lived a few houses down from us.“I saw your two dogs the other day,” Carl said with a smile,“Oh, yes? I’m so sorry. They’d been so good lately, so it was such a surprise when they got out. I sure hope they didn’t cause any trouble.” my wife replied,“Oh, no, no trouble at all. Did you have any idea where they went?”“No,” my wife answered.“They attended my dog’s funeral (葬礼)” Carl said sadly.“What?” Ann exclaimed, “Oh, no. I didn’t know Toby had died.“Yep, it died a couple days ago. I was burying him in the back lot when your two dogs came running up. They sa t down not far from me while. I finished burying him, then they left.”“Wow that’s amazing! ” my wife replied. “Toby was Lily Bit’s father, you know. ”“No, I didn’t know that,” Carl replied, “but I guess that explains why she felt like she needed to be th ere. She had to say her final goodbye.”4．What words can best explain the meaning of the word “straying” underlined in Para. 1? A．Wandering away. B．Staying stillC．Shouting loudly. D．Disobeying impolitely.5．What difference between Lucky and Lily Bit is suggested in the first paragraph? A．Their abilities were different. B．Their hobbies were different. C．Their personalities were different D．Their physical features were different. 6．What would generally happen to the two dogs after they got home from outside? A．They got more united. B．They looked in a mess.C．They smelled nice as usual. D．They got willing to do as told to. 7．What is the author’s purpose of writing the text?A．To show his respect for dogs. B．To be in memory of his dogs.C．To suggest learning from dogs, D．To think highly of his two dogs.Two years ago, Kursat Ceylan was in New York to give a talk about disability rights at the U. N.. Blind since birth, the Turkish man was struggling to find his hotel, holding a cane in one hand and pulling his luggage with the other.“Not surprisingly, all of a sudden I bumped into a pole,” he says. “It was a little bit bloody.”The problem with a cane is that, while it can tell you what's on the grounds it doesn't help with objects at the body or head level. It wasn't the first time Ceylan had run into something,injuring himself.“I have no problem with my scars, they make me more handsome I guess,” Ceylan says, laughing. “But I don’t need new ones.”With WeWalk, a new smart cane, Ceylan hopes to help other blind people navigate their environments more easily. The cane uses an ultrasonic(超声波)sensor which detects objects at body or head level and gives a warning vibration. WeWalk users pair the cane with their smartphones and then use the cane's touchpad to access features like voice assistant or navigation. Before leaving home, they can plug their destination into Google Maps and get spoken directions as they walk. In the future, Ceylan hopes to connect WeWalk with public transportation and ridesharing services.The GPS- and voice technology-enabled cane can tell users what’s around them—they’re passing a McDonald’s, they’re at Exit D of the subway station. An ultrasonic sensor detects objects at body or head level and gives a warning vibration.Assistive technology is often expensive for blind people, says Eelke Folmer, a computer science professor at the University of Nevada! Reno. But developers fail to realize their devices are out of reach for many blind people. To Folmer, the price point—$500——sets the WeWalk cane apart from other technologies.Ceylan sees WeWalk as part of an attempt to help blind people achieve greater freedom of movement, which he believes will give them greater access to education and jobs. The canes are already having an impact on users, Ceylan says. He recently received an email from a teacher in Ireland who had become blind as an adult. He'd been depressed and housebound. But since getting a WeWalk cane, “your device forced me to go out. It became my anti- depressant. ” he wrote.8．Why is a story about Kursat Ceylan given at the beginning of the text?A．To stress the difficulty caused by blindness.B．To show his reason for developing his cane.C．To indicate the problems with present canes.D．To show his positive and humorous character.9．What can WeWalk do at present according to the text?A．Provide fast Internet access. B．Start conversations with users.C．Tell users what is around them. D．Connect with ridesharing services. 10．What does Folmer think is the advantage of WeWalk over other assistive technologies?A．It is easily affordable, B．It is easily controllable.C．It works better for users, D．It looks more attractive.11．What is the text mainly about?A．The increasing demand for smart canes.B．A blind man’ s devotion to smart canes.C．A smart cane’s effects on blind people.D．An assistive technology for the blind.When you think of a national park, you generally picture fresh air and wild animals, right? Well, now you're going to have to add tea shops and something called "the Tube" to your definition, because London, England has signed up to be the first "National Park City."London was established by the Romans around 2,000 years ago and has been continually inhabited (居住于) since then. In all that time, however, nobody had the idea to replace all the parks with big box stores or high buildings, which means London already has a much lower urban density (密度)than most of the worlds cities. Nowadays about a third of the city is green space,In July 2019, London announced its willingness to become the world’s first National Park City. Now the city is moving toward the goal of achieving 50% green space by the year 2050 by connecting and expanding public parks, greening up unused parking lots and the private yards of existing and new houses, fixing some green roofs on existing buildings and even cutting holes in fences for wildlife to pass through."Inspired by the aims and values of our precious rural national parks, the London National Park City is basically about making life better in the capital through both small everyday things and long-term strategic thinking," Daniel Raven-Ellison, who began the campaign to make London a National Park City six years ago said in a press release. "We've been doing that in London for centuries, which is why London is so green and diverse. "London will have a much easier job achieving this type of green transformation than more densely-urbanized cities like Paris and New York, which have 10% and 27% greenspace, respectively. But that doesn't mean it's not possible——the National Park City Foundation hopes to employ 25 more cities in addition to London by the year 2025. Glasgow, Scotland and Newcastle upon Tyne in northeast England are both currently considering becoming National Park Cities.12．What makes London more likely to become a national park city than other cities? A．Its smaller population. B．The government’s efforts.C．Its less dense urbanization, D．Its citizens’ great support.13．Where can we find the data on London’s measures to achieve its goal?A．In Paragraph 1. B．In Paragraph 2.C．In Paragraph 3. D．In Paragraph 4.14．What is Daniel Raven-Ellison trying to talk about in the fourth paragraph? A．London’s long-term strategic thinking.B．The significance of London’s campaign.C．The effects of national parks on London.D．The resources of London’s green space.15．What can we infer from the last paragraph?A．Trying to be a national park city is turning new trend.B．National park cities are springing up around the world.C．It is so easy for London to become a national park city.D．National park cities are making improvements to our life.二、完形填空When I was growing up, there was a pizza shop right down the street from me, The pizza shop was 16 by a Chinese woman who opened the place around 11:00 a. m. and closed it around midnight every day. The food was good, and its 17 were a right size. As far as I knew, the menu never 18 . The woman would take a（n）19 and then return to the kitchen and make it while the customer stood 20 at the counter.One day the woman 21 the pizza shop to a man. The man 22 on running the pizza shop with his wife. I was around fifteen and one day I went inside the pizza shop and began talking with the new owners. Within minutes of my starting the 23 , the man and his wife offered me a(n) 24 They’d printed many paper menus and they wanted me to hang them on the doors in the area.I enjoyed the work as it was fun. As I did this with them one Sunday afternoon, they told me about how 25 the Chinese woman had been. They told me she used too expensive cheese. They were planning to stop 26 chicken because chicken was expensive andpizza was really 27 to make. They were 28 the business was going to be a cash cow for them.As I listened to all this, I decided these people must be very smart. 29 , within several months, they were out of 30 completely, and the pizza place sat there vacant for over a year 31 it was turned into flooring store. The new owners thought only about how the business could 32 them, and how they could reduce 33 to fatten their own wallets. This is why they ended up 34 . After all, everyone loves people and businesses that are more 35 with serving others than themselves. 16．A．purchased B．expanded C．operated D．frequented 17．A．populations B．portions C．measures D．surroundings 18．A．improved B．extended C．worsened D．altered 19．A．order B．chance C．walk D．dish 20．A．resting B．admiring C．waiting D．observing 21．A．transferred B．sold C．referred D．donated 22．A．focused B．went C．insisted D．planned 23．A．trade B．argument C．discussion D．deal 24．A．recommendation B．job C．assignment D．reward 25．A．stupid B．intelligent C．awkward D．hardworking 26．A．preparing B．buying C．serving D．consuming 27．A．swift B．cheap C．interesting D．convenient 28．A．worried B．lucky C．hopeful D．sure 29．A．Instead B．Additionally C．However D．Consequently 30．A．business B．sight C．place D．control 31．A．Unfortunately B．Instantly C．Suddenly D．Eventually 32．A．influence B．benefit C．promote D．involve 33．A．costs B．bills C．prices D．ingredients 34．A．declining B．quitting C．failing D．compromising 35．A．satisfied B．concerned C．patient D．familiar第II卷（非选择题）三、七选五Continuous learning benefits us in many ways. First of all, it allows the increase in knowledge and ability in our career. For example, watching someone work can make us a better worker. 36．For example, we can learn about the general workforce and how the application process works to better prepare us for job searching. This can help if for some reason you lose your job and need to find other work.What’s more, continuous learning can open o ur mind. Having an open mind and willingness to take on new ideas can do wonders. First of all it builds your attitude towards change. 37．Second, when you take continuous learning into account, you can begin to understand how others feel about a particular issue.38．They are always looking for new experiences and do different things. Moreover,they not only have knowledge on various topics that aren’t always related to present roles but also know about the latest trends and technologies in the industry.To develop continuous learning, you have to begin with setting a clear and specific goal. 39．Once your goal is set, build a system to help support your strategy. You want to be looking for different sources of information, but also to be picky about it, In other words, try to learn within your specific field but ensure the information is coming from a trustworthy source- At the same time, do use tools to help improve your learning system. 40．Seminars (研讨会) workshops, and live classes are the tools that modern learners need as they make learning effective.A．Learning can improve other areas in our lives.B．Besides, learning can prepare us for the unexpected.C．If always learning, you are always improving.D．Excellent continuous learners behave in a specific manner.E．Being excited about change can affect others around you positively.F．There’re various tools to help you present information and learning.G．Knowing what you want to achieve can encourage you to keep learnings.四、语法填空阅读下列材料，在空白处填入适当的内容（1个单词）或括号内单词的正确形式。

东北三省三校一模2023届高三3月联考数学试题及答案东北三省三校一模2023届高三3月联考数学试题及答案东北三省三校一模于昨天(3月6日)进行，据悉，本次东北三省三校一模是同学们作为高考之前一次最大的练兵考试。

以下是关于东北三省三校一模2023高三3月联考数学试题及答案的相关内容，供大家参考!2023东北三省三校高三一模数学试题2023东北三省三校高三一模数学试题答案2023东北三省三校一模考试方向及内容三校联考试题命制依据教育部考试中心对2023年高考的基本定调，讨论近几年全国新课标卷、新高考卷的规律和方向，突出立德树人导向，体现学科核心素养。

把握“一核四层四翼”原则，参考2023年关于高考的最新消息准时调整。

试题强调“基础性、综合性、应用性”，并把握“以力量立意为主，贴近现实”的命题指导思想，体现出新课改、新高考精神，注意考查学科核心素养。

试题力保原创性，试题样式会依据已获知的2023年最新高考信息进行适当变动。

高三怎么数学复习1、立足基础学问高三复习数学的时候老师平常讲的大多数都是基础学问，很少讲特殊难的，由于只有高考考察的大部分内容还是基础，并且只有基础学问把握好了才能进一步学好难的。

再者平常考试结束以后，许多同学都会消失这种状况：明明是很简洁的题，但是不知道为什么当时考虑错了，这也是由于基础学问没有学好，考试的时候一紧急就会消失思维混乱，简洁的题就会做错。

2、做题注意审题削减错误审题是做题的第一步，只有读懂了题干，清晰了题目的要求才能连续分析解题，假如题干内容都不清晰就半猜想的做题，就很简单做错。

就像考试卷子发下来以后，发觉明明是会做的题却做错了，就是由于审题不清晰、不谨慎。

所以高三同学备考数学的时候不仅要注意学问的把握，还要改善自身的小毛病，那些可以避开的错误以后就不要再犯。

3、重总结归纳对做错的题、没有完全把握的内容、常常犯错的地方进行总结，该补的补改的改，不要把小毛病攒成大毛病，或者一个小的学问点攒成一个重大的弱点。