河南省豫西名校2020-2021学年高二10月联考数学试题

【市级联考】河南省豫西名校2020-2021学年高二上学期第二次联考数学（文）试卷学校:___________姓名：___________班级：___________考号：___________一、单选题1．已知集合{}2=|20A x x x -≤，{}1,0,1,2B =-，则AB 等于（ ）A ．[]0,2B ．{}0,1,2C ．()1,2-D ．{}1,0,1-2．命题“x 1∀>，x 11()22<”的否定是( )A ．x 1∀>，x 11()22≥B ．x 1∀≤，x 11()22≥C ．0x 1∃>，0x 11()22≥D ．0x 1∃≤，0x 11()22≥3．已知等差数列{a n }的前n 项和为S n ，且S 10=5，a 7=1，则a 1=（ ） A ．-1 B ．−12 C ．14 D ．124．已知1F ，2F 为椭圆C ：22x y 195+=的左、右焦点，点P 是椭圆上任意一点(非左右顶点)，则12PFF 的周长为( ) A ．12B ．10C ．8D ．65．王昌龄《从军行》中两句诗为“黄沙百战穿金甲，不破楼兰终不还”，其中后一句中“攻破楼兰”是“返回家乡”的（ ） A ．充分不必要条件 B ．必要不充分条件 C ．充分必要条件D ．既不充分也不必要条件6．若实数x ，y 满足条件x y 10x 30y 20--≤⎧⎪+≥⎨⎪-≤⎩，则z 2x y =-的最大值为( )A ．8-B ．6-C ．2-D ．47．已知命题p ：“∀x ∈[0,1]，a ≥e x ”，命题q ：“∀x ∈R ，x 2+4x +a ≠0”，若命题p ∧￢q 是真命题，则实数a 的取值范围是( ) A ．[1,4]B ．[e,4]C ．(4,+∞)D ．(−∞,1]8．已知椭圆C ：22221x y a b+=（0a b >>）的右焦点为F ，过点F 的直线交椭圆交于A ，B 两点，若AB 的中点11,2P ⎛⎫-⎪⎝⎭，且直线AB 的倾斜角为4π，则此椭圆的方程为（ ）A ．2224199x y +=B ．22194x y +=C ．22195x y +=D ．222199x y +=9．已知直线2x y 10k -+=与椭圆22x y 19m +=恒有公共点，则实数m 的取值范围为()A ．(]1,9B ．[)1,∞+C ．[)()1,99,∞⋃+ D ．()9,∞+10．若ABC ∆的三个内角A ，B ，C 成等差数列，且BC 边上的中线AD =，又2AB =，则ABC S ∆=（ ）A ．6B ．C ．D ．311．已知在ABC 中，内角A 、B 、C 所对的边分别为a 、b 、c ，若ABC 的面积为S ，且222()S a b c =+-，则tan C =（ ） A ．43-B ．34-C ．34D ．4312．斜率为1的直线l 与椭圆22x y 14+=相交于A 、B 两点，则AB 的最大值为()A ．2B ．5C ．5D ．5二、填空题13．已知ABC 的三个内角A 、B 、C 的对边分别是a 、b 、c ，且acosB bcosA 3a +=，则ca=______． 14．若命题“∃x∈R，x 2﹣2x+m≤0”是假命题，则m 的取值范围是__．15．已知点1F ，2F 是椭圆2222:1(0)x y C a b a b+=>>的两个焦点，P 为椭圆C 上一点，且122F PF π∠=．若△12PF F 的面积为9，则b =_______16．椭圆2222x y 1(a b 0)a b+=>>的中心在原点，1F ，2F 分别为左、右焦点，A ，B 分别是椭圆的上顶点和右顶点，P 是椭圆上一点，且1PF x ⊥轴，2PF //AB ，则此椭圆的离心率为______．三、解答题17．设命题p ：a 0>；命题q ：关于x 的不等式a x 0-≥对一切[]x 2,1∈--均成立．(Ⅰ)若命题q 为真命题，求实数a 的取值范围(用集合表示)；(Ⅱ)若命题p q ∨为真命题，且命题p q ∧为假命题，求实数a 的取值范围．18．在ABC 中，角A ，B ，C 的对边分别为a ，b ，c.已知asinB =．()1求角A 的大小； ()2若a b 2==，求ABC 的面积．19．已知0m >，:p ()()260x x +-≤，:q 22m m -≤+. （1）已知p 是q 成立的必要不充分条件，求实数m 的取值范围； （2）若q ⌝是p ⌝成立的充分不必要条件，求实数m 的取值范围.20．已知m R ∈，命题p ：对[]x 0,8∀∈，不等式()213log x 1m 3m +≥-恒成立；命题q ：对()x ,1∞∀∈--，不等式22x x 2mx +>+恒成立．()1若命题p 为真命题，求实数m 的取值范围； ()2若p q ∧为假，p q ∨为真，求实数m 的取值范围．21．设n S 为数列{}n a 的前n 项和，已知1a 2=，对任意*n N ∈，都有()n n 2S n 1a =+．(Ⅰ)求数列{}n a 的通项公式；(Ⅱ)若数列()n n 4a a 2⎧⎫⎪⎪⎨⎬+⎪⎪⎩⎭的前n 项和为n T ，求证：n 1T 12≤<．22．已知点A(0,1)与B(√3,12)都在椭圆C ：x 2a 2+y 2b 2=1(a >b >0)上，直线AB 交x 轴于点M ．(1)求椭圆C 的方程，并求点M 的坐标；(2)设O 为原点，点D 与点B 关于x 轴对称，直线AD 交x 轴于点N ，问：y 轴上是否存在点E ，使得∠OEM =∠ONE ？若存在，求点E 的坐标；若不存在，说明理由．23．已知椭圆C：2222x y1(a b0)a b+=>>的左、右顶点分别为A，B其离心率1e2=，点M为椭圆上的一个动点，MAB面积的最大值是()1求椭圆C的方程；()2若过椭圆C右顶点B的直线l与椭圆的另一个交点为D，线段BD的垂直平分线与y轴交于点P，当PB PD0⋅=时，求点P的坐标．参考答案1．B 【分析】220x x -≤，02x ∴≤≤，{}0,1,2A B ⋂=，选B2．C 【解析】因为“1x ∀>,1122x⎛⎫< ⎪⎝⎭”是全称命题，所以依据含一个量词的命题的否定可知：其否定是存在性命题，即“01x ∃>,01122x⎛⎫≥ ⎪⎝⎭”，应选答案C ．3．B 【解析】试题分析：S 10=5,a 7=1⇒10a 1+45d =5,a 1+6d =1⇒a 1=−1，选B. 考点：等差数列基本量运算 4．B 【分析】根据椭圆的标准方程求得,,a b c 的值，所求三角形周长为22a c +，由此求得正确选项. 【详解】由22195x y +=知，3a =，b =2c =，∴12AF F ∆周长为226410a c +=+=.故选B.【点睛】本小题主要考查椭圆的标准方程和几何性质，考查焦点三角形的周长，属于基础题. 5．B 【分析】由诗句可分析,“返回家乡”之前一定是“攻破楼兰”的,但“攻破楼兰”后还是否有其他任务需要完成诗句中并未提及,进而得到结论. 【详解】由题,“不破楼兰终不还”意味着如果“返回家乡”,则一定“攻破楼兰”；但“攻破楼兰”后,是否还有其他任务,诗句中并未提及,无法判断此时可否“返回家乡”； 故选:B 【点睛】本题考查充分条件和必要条件的判定,属于基础题. 6．D 【解析】作出可行域，如图ABC ∆内部（含边界），作直线:20l x y -=，当直线l 向下平移时，2z x y =-增大，因此当l 过(3,2)A 时，22324z x y =-=⨯-=为最大值，故选D ．7．B 【解析】试题分析：若p 是真命题则a ≥e .若q 是真命题则16−4a <0.∴a >4.所以¬q:a ≤4.所以p ∧¬q:e ≤a ≤4.故选B.本小题考查命题的相关知识.含特称和全称的命题的运算.涉及对数函数函数和二次函数的知识.考点：1.特称命题和全称命题.2.命题的否定.3.命题的交集的运算. 8．A 【解析】 【分析】利用直线AB 的斜率和倾斜角的对应关系列方程，求得c 的值.利用点差法求得22,a b 的关系式，结合222a b c =+求得,a b 的值，进而求得椭圆方程. 【详解】∵1211c =-，∴32c =，令()11,A x y ，()22,B x y ，则22221x y a b +=，∴()()()()12121212220x x x x y y y y a b +⋅-+⋅-+=，22210a b -+=，∴292a =，294b =.故选A. 【点睛】本小题主要考查直线和椭圆的位置关系，考查椭圆标准方程的求法，以及有关点差法的运用.题目给出直线和椭圆相交所得所得弦的中点坐标，还有直线的倾斜角，这里可以根据焦点的坐标列方程求得c 的值.点差法主要用在有关直线和圆锥曲线相交，所得弦的中点有关的题目.属于中档题. 9．C 【分析】先求得直线过的定点，根据这个定点在椭圆内或者椭圆上列不等式，解不等式求得m 的取值范围. 【详解】直线2kx y 10-+=恒过定点()P 0,1，直线2kx y 10-+=与椭圆22x y 19m +=恒有公共点，即点()P 0,1在椭圆内或椭圆上，0119m∴+≤，即m 1≥，又m 9≠， 1m 9∴≤<或m 9>．故选C ． 【点睛】本小题主要考查含有参数的直线过定点，考查直线和椭圆的位置关系，属于基础题. 10．B 【分析】三角形内角成等差数列，可求得60B =，利用余弦定理列方程可求得BD 的长，由此得到BC 的长，利用三角形的面积公式可求得三角形面积.【详解】因为ABC ∆的三个内角A ，B ，C 成等差数列，则60B =︒，在ABC ∆中，由余弦定理得：2222cos AD AB BD AB BD B =+-⋅⋅，即2742BD BD =+-，所以3BD =或-1（舍去）， 可得6BC =，所以11sin 2622ABC S AB BC B ∆=⋅⋅=⨯⨯=.故选B. 【点睛】本小题主要考查等差中项的性质，考查利用余弦定理解三角形，考查三角形的面积公式，属于基础题. 11．A 【分析】由三角形面积公式和余弦定理可得C 的等式，利用二倍角公式求得tan 2C，从而求得tan C ． 【详解】∵222222()2S a b c a b ab c =+-=++-，即22212sin 22ab C a b ab c ⨯⋅=++-， ∴222sin 2ab C ab a b c ⋅-=+-，又222sin 2sin cos 1222a b c ab C ab CC ab ab +-⋅-===-，∴sin cos 12C C +=， 即22cos sin cos 222C C C =，则tan 22C =，∴222tan2242tan 1231tan 2CC C ⨯===---， 故选：A ． 【点睛】本题考查三角形面积公式，余弦定理，考查二倍角公式，同角间的三角函数关系，掌握相应的公式即可求解．属于中档题，考查了学生的运算求解能力． 12．C 【分析】设出直线的方程，代入椭圆方程中消去y ，根据判别式大于0求得t 的范围，进而利用弦长公式求得|AB |的表达式，利用t 的范围求得|AB |的最大值． 【详解】解：设直线l 的方程为y ＝x +t ，代入24x +y 2＝1，消去y 得54x 2+2tx +t 2﹣1＝0，由题意得△＝（2t ）2﹣5（t 2﹣1）＞0，即t 2＜5．弦长|AB |＝≤． 故选：C ． 【点睛】本题主要考查了椭圆的应用，直线与椭圆的关系．常需要把直线与椭圆方程联立，利用韦达定理，判别式找到解决问题的突破口． 13．3 【分析】利用正弦定理将题目所给已知条件转化为角的形式，化简后再次利用正弦定理将角的形式转化为边的形式，由此求得ca的值. 【详解】法一：由已知及正弦定理得sin cos sin cos 3sin A B B A A +=，∴()sin 3sin A B A +=， ∴sin 3sin C A =，∴3ca=. 法二：cos cos 3ac B bc A c a +==，∴3ca=. 【点睛】本小题主要考查利用正弦定理进行边角互化，求得边的比值.属于基础题. 14．m ＞1 【分析】结合命题的否定与原命题真假对立，将原命题转化为命题的否定，结合二次函数的性质，即可计算m 的范围． 【详解】若命题“∃x∈R，x 2﹣2x+m≤0”是假命题， 则命题“∀x∈R，x 2﹣2x+m ＞0”是真命题， 即判别式△=4﹣4m ＜0， 解得m ＞1，故答案为m ＞1 【点睛】本道题考查了命题的否定与原命题的关系，可以通过命题的否定找出解题切入点，即可． 15．3 【分析】利用椭圆的标准方程定义及其三角形面积计算公式、勾股定理即可得出． 【详解】 解：122F PF π∠=,12PF F ∆的面积为9，设1||PF m =，2||PF n =．则22221924m n a mn m n c +=⎧⎪⎪=⎨⎪+=⎪⎩可得：224364c a +=， 即2229a c b -==， 解得3b =． 故答案为：3． 【点睛】本题考查了椭圆的标准方程定义及其性质、三角形面积计算公式、勾股定理，考查了推理能力与计算能力，属于中档题． 16【分析】先求得P 点的坐标，根据两直线平行，斜率相等列出方程，化简这个方程后可求得离心率. 【详解】如图所示，把x c =-代入椭圆方程22221x ya b +=（0a b >>）可得2,b P c a ⎛⎫- ⎪⎝⎭，又()0,A b ，(),0B a ，()2,0F c ，∴2AB bk ac=-，∵2PF AB ，∴22b b a ac-=-，化简得2b c =.∴22224c b a c ==-，即225a c =，∴5e ==.【点睛】本小题考查椭圆的标准方程和几何性质.通过椭圆上常见点的坐标和两直线平行这个条件，列方程后，将方程转化为ca的形式，由此求得离心率.属于基础题. 17．（Ⅰ）[)1,-+∞；（Ⅱ） []1,0- 【解析】 试题分析：(Ⅰ)由题意可知0a x -≥对一切[]2,1x ∈--均成立，结合一次函数的性质可得实数a 的取值范围是[)1,-+∞；(Ⅱ)由题意可得命题p q 、一真一假，据此分类讨论可得实数a 的取值范围是[]1,0-. 试题解析：（Ⅰ）当命题q 为真命题时，不等式0a x -≥对一切[]2,1x ∈--均成立，∴1a ≥ ∴实数a 的取值范围是[)1,-+∞；（Ⅱ）由命题p q ∨为真，且p q ∧为假，得命题p q 、一真一假 当p 真q 假时，则01a a >⎧⎨<-⎩，a ∈∅；当p 假q 真时，则01a a ≤⎧⎨≥-⎩，得10a -≤≤，∴实数a 的取值范围是[]1,0-18．（1）3π；（2. 【解析】试题分析：（1）因为正弦定理，所以sin 3cos 0a B b A -=化为sin 3cos 0sinA B sinB A -=，因为三角形内角有，所以即tan 3A =，所以3A π=；（2）由余弦定理，得2222cos a b c bc A =+-，而7,2a b ==，3A π=，得，即2230c c --=，因为三角形的边0c >，所以3c =，则133sin 22bc A =．试题解析：（1）因为sin cos 0a B A =由正弦定理，得sin cos 0sinA B A =，又sin 0B ≠，从而tan A =0A π<<所以3A π=（2）解法一：由余弦定理，得2222cos a b c bc A =+-，而2a b ==，3A π=，得，即2230c c --=因为0c >，所以3c =，故ABC ∆面积为1sin 2bc A =2sin sin3B =从而sin 7B =又由a b >知A B >，所以cos 7B =故sin sin()sin()3C A B B π=+=+sin coscos sin3314B B ππ=+=所以ABC ∆面积为1sin 2ab C =． 考点：1．正弦定理与余弦定理；2．三角形的面积公式． 19．(1)(0,4)；(2) (4，＋∞).【解析】 【分析】（1）求出p 的等价条件，结合充分条件和必要条件的定义转化为不等式关系进行求解即可； （2）根据￢q 是￢p 成立的充分不必要条件，转化为p 是q 的充分不必要条件进行求解即可. 【详解】（1）由（x+2）（x ﹣6）≤0得﹣2≤x≤6，即p ：﹣2≤x≤6∵p 是q 成立的必要不充分条件，则[2﹣m ，2+m]是[﹣2，6]的真子集，有222226m m m m -<+⎧⎪--⎨⎪+⎩，解得0＜m≤4， 又当m ＝4时，[2﹣m ，2+m]＝[﹣2，6]，不合题意， ∴m 的取值范围是（0，4）．（2）∵￢q 是￢p 的充分不必要条件，∴p 是q 的充分不必要条件，则[﹣2，6]是[2﹣m ，2+m]的真子集，则02226m m m >⎧⎪-<-⎨⎪+⎩，解得m≥4，又当m ＝4时，不合题意． ∴m 的取值范围为（4，+∞）． 【点睛】本题主要考查充分条件和必要条件的应用，结合不等式的关系是解决本题的关键,属于中档题．20．（1）[]1,2（2）()2,+∞ 【分析】（1）利用单调性求得()13log 1x +的最小值，利用23m m -小于或等于这个最小值求得m 的取值范围.（2）利用分离常数法，将命题q 所给不等式分离常数后，求得m 的取值范围.根据题目所给已知条件“p q ∧为假，p q ∨为真，”可知,p q 一真一假，分成p 真q 假，和p 假q 真两类，列不等式组求得m 的取值范围. 【详解】（1）令()()13log 1f x x =+，则()f x 在()1,-+∞上为减函数，因为[]0,8x ∈，所以当8x =时，()()min 82f x f ==-，不等式()213log 13x m m +≥-恒成立，等价于223m m -≥-，解得12m ≤≤，故命题p 为真，实数m 的取值范围为[]1,2. （2）若命题q 为真，则221m x x>-+，对(),1x ∀∈-∞-上恒成立， 令()21g x x x =-+，因为()g x 在(),1x ∈-∞-上为单调增函数， 则()()11g x g <-=，故1m ≥，即命题q 为真，1m ≥ 若p q ∧为假，p q ∨为真，则命题p ，q 中一真一假； ①若p 为真，q 为假，那么121m m <<⎧⎨<⎩，则无解；②若p 为假，q 为真，那么121m m m 或⎧⎨≥⎩，则2m >.综上m 的取值范围为()2,+∞. 【点睛】本小题主要考查不等式恒成立问题的主要解题策略，考查已知含有逻辑连接词命题真假性来求参数的取值范围.属于中档题. 21．(1) 2n a n =；(2）证明见解析. 【分析】（1）运用数列的递推式，化简整理即可得到所求通项公式； （2）b n ()()()44111222211n n a a n n n n n n ====-++++，由裂项相消求和即可得到所求和． 【详解】（1）因为()21n n S n a =+，当2n ≥时，112n n S na --= 两式相减得：()121n n n a n a na -=+- 即()11n n n a na --=， 所以当2n ≥时，11n n a a n n -=-.所以121n a a n ==，即2n a n =. （2）因为2n a n =，()42n n n b a a =+，*N n ∈，所以()()411122211n b n n n n n n ===-+++.所以12112n n T b b b ⎛⎫=+++=-+ ⎪⎝⎭ 11111123111n n n n n ⎛⎫⎛⎫-++-=-= ⎪ ⎪-++⎝⎭⎝⎭， 因为101n >+，所以1111n -<+. 又因为()11f n n =+在*N 上是单调递减函数，所以111n -+在*N 上是单调递增函数.所以当1n =时，n T 取最小值12，所以112n T ≤<.【点睛】裂项相消法是最难把握的求和方法之一，其原因是有时很难找到裂项的方向，突破这一难点的方法是根据式子的结构特点，常见的裂项技巧：(1)()1111n n k k n n k ⎛⎫=- ⎪++⎝⎭；（2） 1k=； （3）()()1111212122121n n n n ⎛⎫=- ⎪-+-+⎝⎭；（4）()()11122n n n =++ ()()()11112n n n n ⎡⎤-⎢⎥+++⎢⎥⎣⎦；此外，需注意裂项之后相消的过程中容易出现丢项或多项的问题，导致计算结果错误.22．（Ⅰ）x 24+y 2=1，M(2√3,0)（Ⅱ）在y 轴上存在点E ，使得∠OEM =∠ONE ，且点E 的坐标为(0,2)或(0,−2)． 【解析】试题分析：（Ⅰ）将两点坐标代入椭圆方程，解方程组得{a 2=4,b 2=1.（Ⅱ）求定点问题，一般以算代定. 解几中角的问题，一般转化成坐标问题：∠OEM =∠ONE⇒|OM||OE|=|OE||ON|⇒y E 2=|x M ||x N |，从而确定y E =±2试题解析：（Ⅰ）由题意得∴{a 2=4,b 2=1.故椭圆C 的方程为x 24+y 2=1．直线AB 方程为y =2√3+1，与x 轴交点M(2√3,0)．（Ⅱ）因为点D 与点B 关于x 轴对称，所以D(√3,−12)， 直线AD 的方程为y =−√32x +1，与x 轴交于点N(2√33,0)． “存在点E(0,y E )使得∠OEM =∠ONE ”等价于“存在点E(0,y E )使得|OM||OE|=|OE||ON|”，即y E 满足y E 2=|x M ||x N |，∴y E 2=2√3×2√33=4，∴y E =±2，故在y 轴上存在点E ，使得∠OEM =∠ONE ，且点E 的坐标为(0,2)或(0,−2)． 考点：椭圆方程，定点问题【思路点睛】定点、定值问题通常是通过设参数或取特殊值来确定“定点”是什么、“定值”是多少，或者将该问题涉及的几何式转化为代数式或三角问题，证明该式是恒定的. 定点、定值问题同证明问题类似，在求定点、定值之前已知该值的结果，因此求解时应设参数，运用推理，到最后必定参数统消，定点、定值显现.23．（1）22143x y +=（2）当34k =时，20,7P ⎛⎫ ⎪⎝⎭，当34k =-时，20,7P ⎛⎫- ⎪⎝⎭【分析】（1）由题意可知2221,2122,c e a ab a b c ⎧==⎪⎪⎪⨯=⎨⎪=+⎪⎪⎩解方程即可得解；（2）设直线BD 的方程为()2y k x =-，()11,D x y ，由直线与椭圆联立得()2222241616120k xk x k +-+-=，由根与系数的关系可得1x ，从而得BD 中点的坐标，进而得BD 的垂直平分线方程，令x=0可得P ，再由0PB PD ⋅=，用坐标表示即可解k . 【详解】（1）由题意可知2221,2122,c e a ab a b c ⎧==⎪⎪⎪⨯=⎨⎪=+⎪⎪⎩解得2a =，b =所以椭圆方程为22143x y +=.（2）由（1）知()2,0B ，设直线BD 的方程为()2y k x =-，()11,D x y ，把()2y k x =-代入椭圆方程22143x y +=，整理得()2222241616120kxk x k +-+-=，所以221122168623434k k x x k k -+=⇒=++，则2228612,3434k k D k k ⎛⎫-- ⎪++⎝⎭， 所以BD 中点的坐标为22286,3434k k k k ⎛⎫- ⎪++⎝⎭， 则直线BD 的垂直平分线方程为2226183434k k y x k k k ⎛⎫--=-- ⎪++⎝⎭，得220,34k P k ⎛⎫ ⎪+⎝⎭又0PB PD ⋅=，即2222286142,,0343434k k k k k k ⎛⎫--⎛⎫-⋅= ⎪ ⎪+++⎝⎭⎝⎭， 化简得()424226428360642836034k k k k k+-=⇒+-=+， 解得34k =± 故当34k =时，20,7P ⎛⎫⎪⎝⎭，当34k =-时，20,7P ⎛⎫- ⎪⎝⎭.【点睛】本题主要考查了直线与椭圆的位置关系，用到了向量问题坐标化，坐标通过设而不求的方程灵活处理，考查了学生的运算能力，属于中档题.。

豫西名校2020－2021学年上期第一次联考高二英语试题(考试时间：100分钟试卷满分：120分)第一部分听力(略)第二部分阅读理解(共两节，满分40分)第一节(共15小题；每小题2分，满分30分)阅读下列短文，从每题所给的四个选项(A、B、C、和D)中，选出最佳选项。

ALake Tekapo lies in the center of the South Island in an area known as the Mackenzie Basin in New Zealand.Tekapo Tourism provides one of the most comprehensive accommodation(住宿)listings in the region.All accommodation providers in Tekapo are listed as follows.●The Chalet Boutique MotelIt's only a short distance from the Church of the Good Shepherd.You'll find the Chalet Boutique Motel very close to the shores of Lake Tekapo.At the Chalet Boutique Motel you have the choice of six self-contained(独立的)apartments varying in size from2-bedroom apartments to studio apartments(单间公寓).Some offer excellent views of the lake and mountains.Address:Pioneer Drive,PO Box2,Lake Tekapo●Tekapo B2ApartmentTekapo B2apartment is a self-contained,one-bedroom,upstairs apartment with views overlooking Lake Tekapo.It's only a short walk from the town center.The bedroom is suitable for two people.A double sofa bed is available for a third person or child.Address:2Murray Place,Lake Tekapo●Mt John Historic HomesteadBuilt in1911,the Mount John Homestead has recently been renovated(整修).An original collection of art by a local artist and thousands of local photos are displayed throughout the rooms.The yard is also available for special occasions such as weddings,family reunions or small parties.A minimum(最少的)booking of2nights is required.Address:State Highway8,PO Box91,Lake Tekapo21.If Lily wants to take a walk along the shores of Lake Tekapo,she'd better stay in.A.the Church of the Good ShepherdB.the Chalet Boutique MotelC.Tekapo B2ApartmentD.Mt John Historic Homestead22.What do the Chalet Boutique Motel and Tekapo B2Apartment have in common?A.Both have a long history.B.Both have recently been renovated.C.Both have local photos in the rooms.D.Both have rooms with a lake view.23.What is special about Mt John Historic Homestead?A.It has studio apartments.B.It is near the town center.C.It can hold a wedding.D.It is the cheapest of the three.BJaye Gardiner loves comic books and she also loves science.Sensing an opportunity,she decided to combine the two.In2015,she and her two friends,Khoa Tran and Kelly Montgomery,founded an online publishing company called JKX Comics.They decided to use a cartoon format and light humor to explain science to students.Their first comic book came out in2016.In the book,a virus goes out with friends to a human cell.The colorful visuals help illustrate the science,explains Train.Through comics,his JKX coworkers hope to inspire the next generation to pursue STEM.STEM is short for science,technology,engineering and math.A lot of research has suggested that comic books can introduce a variety of people to science ics can also make information more accessible.Text,pictures and stories can help people understand scientific ideas more easily.Their latest comic is called"Gilbert's Glitch Switch."It is about a biochemist named Gilbert who gets sucked into a video game.Gilbert uses amino acids to get proteins to communicate with each other.Proteins are an essential part of all living organisms.They form the basis of living cells, muscle and tissues;they also do the work inside of cells.By understanding how proteins communicate,scientists can learn how to modify them to interact better.Certain diseases develop when proteins communicate incorrectly.Alzheimer's,a disorder that causes brain cells to die,is one example.Translating a complex science topic into a comic can be challenging.The creators have to not only be accurate but also tell a story that interests readers.The aat1.ors hope to spark readers' curiosity.There's a lot of misinformation circulating around the internet.Some of it"scares people away from science,"Montgomery says.The goal of the new comics is to"limit some of those misunderstandings,"she adds.24.What caused Jaye Gardiner and her friends to create JKX Comics?A.Their love of comics and the difficulty of explaining science to students.B.The opportunity to earn a great deal of money while being creating.C.Their love of comics and the lack of good visual science resources.D.The opportunity to obtain better jobs after graduating from college.25.What is the paragraph4mainly about?A.Why science is hard.B.The advantages of comics.C.The purpose of JKX Comics.D.The people who began JKX Comics.26.Why did the author include the information about"Gilbert's Glitch Switch"?A.To describe JKX Comic's most popular comic to date.B.To illustrate that science can help stop illness and diseases.C.To show how important amino acids and proteins are in biochemistry.D.To provide an example of how comics can convey scientific information.27.What can be the best title for this passage?ics are Designed to Teach Children about ScienceB.Errors in Protein Communication Cause Certain DiseasesC.The Science Topics are being Explored by JKX ComicsD.Jaye Gardiner and Her Friends who Began JKX ComicsCMany people who live on Norway's far-northern island of Sommaroey are campaigning to go "time-free".They note that between May and July,the sun does not set there for69days.To enjoy the bright summer months,they would like to ignore traditional business hours.For example,stores would be open when people are there to work at them.Schools would operate when teachers and students are ready.Kjell Ove Hveding is one of the local leaders of the campaign.He said the350people who live on the island already go mostly time-free.He said that in the summer at 2in the morning,you might see"children playing soccer,people painting their houses or mowing their lawns,and teens going for a swim".To show their feelings about time,many islanders take off their watches.They leave them on a bridge that connects the island to the mainland.Earlier this month,Hveding met with a Norwegian lawmaker.Hveding presented a formal request signed by people who live on the island.The request asked the government to declare the area a"time-free zone".Hveding noted that he understands going completely time-free would be too complex in a world that operates on schedules.And,as an article in the Smithsonian magazine says,people's health depends on keeping a usual rhythm of being asleep and being awake.But Hveding argues that letting go of time limits can also be good for people.He said the time-free proposal would permit the islanders to get used to the daylight and feel calmer."The idea is also to chill out,"he said,"I have seen people suffering from stress because they were pressed by time."28.Paragraph2is mainly about people's ways to.A.work during summer.B.spend business hours.C.enjoy time-free months.D.entertain during summer.29.What do we know about the formal request presented by Hveding?A.It was put forward by a lawmaker.B.It was approved by the islanders.C.It was rejected by the government.D.It was about a"trade-free zone".30.Why does the author mention Hveding's opinion in the last paragraph?A.To prove going time-free would be beneficial to the islanders.B.To show the article's view in the magazine is wrong.C.To introduce the lifestyle of the islanders.D.To present people's suffering from time pressure.31.Where is this text most likely from?A.A news report.B.A geography book.C.A health magazine.D.A travel brochure.DTo someone stuck in a traffic jam,the ability to soar(高飞)right over other cars seems like a dream.Inventors and businessmen tried to make flying cars,but have never been able to.That may be changing.Many companies are competing to be the first to develop a flying car.One of the ideas being developed is a car with wings that unfold for flight.But most are morelike helicopters instead of cars and they are operated by a battery-powered motor.Still,some companies are moving forward.They see a need for"air taxis",especially as cities grow more congested.These taxis would send passengers to the rooftops of office buildings during rush hours.Charles Eastlake is a professor at Embry-Riddle Aeronautical University.He says advances in technology have opened the door to a whole new world of flying machines.But he also expresses some hesitancy,"It's possible that people actually use self-flying air taxis in the next10or15years. But the challenges are big."The key will be making longer-lasting batteries.Batteries now could keep an air taxi in the sky for about15to30minutes.Depending on how fast it flies,that probably isn't enough time to transport panies will also have to get permission from the government.The government office in charge of flying says further study is needed on flying cars.It thinks the ideas being worked on are good ones,but they must be safe.Reducing noise is another challenge because air taxis will be taking off and landing in areas with a lot of people.Creating enough landing pads to deal with many flying cars at the same time is also a challenge.Though there are so many challenges for humans to deal with,inventors and businessmen think their dream of making flying cars will come true soon.32.What do we know about flying cars now?A.They can fly as fast as helicopters.B.They mainly depend on the sunlight.C.They resemble helicopters very much.D.They can fold their wings easily.33.What does the underlined word"congested"in paragraph2probably mean?A.Crowded.B.Developed.C.Modern.D.Convenient.34.What can we infer from Charles Eastlake's words?A.He will devote himself to designing flying machines.B.He supposes that making flying cars is beyond his power.,C.He still has confidence in making flying cars into reality.D.He advises to apply the advanced technology to flying cars.35.The challenges of making flying cars are mentioned to show.A.making flying cars is only a dreamB.safety is of importance to flying carsC.there is a long way to make real flying carsD.noise is the biggest problem第二节(共5小题；每小题2分，满分10分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

2024-2025学年河南省百师联盟高二上学期10月联考数学试题一、单选题：本题共8小题，每小题5分，共40分。

在每小题给出的选项中，只有一项是符合题目要求的。

1.设直线l ：2x +4y +3=0的倾斜角为α，则cos α的值为A.55B. −55C. 255D. −2552.已知直线l 1的方向向量为a 1=(k,1)，直线l 2的方向向量为a 2=(2−k,k)，若l 1// l 2，则k =A. −2B. 1C. −2或1D. 0或23.在四面体OABC 中，OA =a ，OB =b ，OC =c ，OM =34OA ，BN =λBC(λ>0)，若MN =−34a +12b +12c ，则λ=A. 12B. 13C. 23D. 144.若{e 1,e 2,e 3}是空间的一个基底，那么对任意一个空间向量a ，存在唯一的有序实数组(x,y,z)，使得a =x e 1+ye 2+ze 3，我们把有序实数组(x,y,z)叫做基底{e 1,e 2,e 3}下向量a 的斜坐标．设向量p 在基底{a ,b ,c }下的斜坐标为(−1,2,3)，则向量p 在基底{a−b +c ,a−b ,a +c }下的斜坐标为A. (2,−4,−1)B. (−2,−4,1)C. (−2,4,1)D. (2,−4,1)5.平行六面体ABCD−A 1B 1C 1D 1的底面ABCD 是矩形，其中AD =2，AB =4，且∠A 1AD =∠A 1AB =60°，AA 1=4，M 为A 1C 1，B 1D 1的交点，则线段BM 的长为A.15 B. 4 C.17 D. 326.已知从点(3,3)发出的一束光线，经过直线2x−y +2=0反射，反射光线恰好过点(4,0)，则反射光线所在的直线方程为A. 3x +y−12=0 B. 3x +7y−12=0C. x +y−4=0D. x =47.圆C ：(x−1)2+y 2=1与圆D ：x 2+y 2−2x +8y +8=0的公切线的条数为A. 0B. 1C. 2D. 38.已知圆C ：(x +2)2+y 2=4，直线l ：(m +1)x +4y−1+m =0(m ∈R)，则A. 直线l恒过定点(−1,1)B. 直线l与圆C有两个交点C. 当m=1时，圆C上恰有两个点到直线l的距离等于1D. 过直线l的平行线3x+4y−7m=0上一动点P作圆C的一条切线，切点为A，则|PA|min=4二、多选题：本题共3小题，共18分。