河南省新郑第一中学高一上学期第四次月考试题

河南省新郑市高三上学期期末考试（第四次月考）语文试题姓名:________ 班级:________ 成绩:________一、现代文阅读 (共1题；共10分)1. （10分） (2019高三上·安顺期末) 阅读下面的文字，完成下列小题。

材料一：针对学生及其家长不太重视传统文化教育的问题，有建议指出，要在中高考中增加传统文化的考试内容。

这其实还是“考什么就教什么教什么才学什么”的应试思路，表面上看，这可以提高学校，家长、学生对传统文化教育的重视程度，但“应对”考试的传统文化教育会偏离初衷。

学生对传统文化的体验、感知，是很难通过考试考出来的。

我国中高考改革都提到要重视学生的综合素质评价。

学生的传统文化素养是综合素质非常重要的一部分。

要让学生重视传统文化，需要综合素质评价在大学。

高中招生中切实发挥作用。

而在推进中高考改革过程中，需要学校教育从传统的讲授、灌输，转变为更多的探究、交互，从过去重视结果评价，转变为关注对学生的过程性评价。

学校教育也不能圆于教材、课本，而且该采取多种方式，借助丰富多彩的课外活动，社团活动等，引领学生走向社会，了解社会，多感悟优秀的传统文化，进而帮助其做一个承继传统的人，成为有根的中国人。

（《中国教育报》2017年03月07日第2版）材料二：目前，中办国办印发《国家“十三五”时期文化发展改革规划纲要》，其中提到“厘清中华优秀传统文化的内涵，改造陈旧的表现形式，赋予新的时代内涵和现代表达形式”“普及中华诗词、音乐舞蹈，书法绘画等，举办经典诵读、国学讲堂、文化讲坛、专题展览等活动，鼓励媒体开办主题专栏，节目。

利用互联网，推动中华优秀传统文化网络传播”，这为传统文化的开发挖掘指明了道路：既保有传统的魅力，又带着时代的创意。

谈及传统文化，言必及深邃与厚重。

同时，任何一种文化，其生命力也来源于开放性与多义性，来源于多维度阐释的可能，古语云：“爱如一炬之大，万火引之，其火如故。

”传统文化也是如此，任何人取一星火光，都可收获智慧的启迪。

2023-2024学年河南省高一上册第一次月考数学试题一、单选题1．已知集合{}220A x x x =-≤，{}1,0,3B =-，则()R A B ⋂=ð（）A ．∅B ．{}0,1C ．{}1,0,3-D ．{}1,3-【正确答案】D【分析】先由一元二次不等式的解法求得集合A ，再由集合的补集和交集运算可求得答案.【详解】因为{}{}22002A x x x x x =-≤=≤≤，所以{R |0A x x =<ð或}2x >，又{}1,0,3B =-，所以(){}1,3R A B ⋂=-ð，故选：D ．2．已知函数()f x =()()3y f x f x =+-的定义域是（）A ．[-5，4]B ．[-2，7]C ．[-2，1]D ．[1，4]【正确答案】D【分析】由函数解析式可得2820x x +-≥，解不等式可得24x -≤≤，再由24234x x -≤≤⎧⎨-≤-≤⎩即可求解.【详解】由()f x =2820x x +-≥，解得24x -≤≤，所以函数()()3y f x f x =+-的定义域满足24234x x -≤≤⎧⎨-≤-≤⎩，解得14x ≤≤，所以函数的定义域为[1，4].故选：D 3．不等式3112x x-≥-的解集是（）A ．3{|2}4x x ≤≤B ．3{|2}4x x ≤<C ．{>2x x 或3}4x ≤D ．3{|}4x x ≥【正确答案】B【分析】把原不等式的右边移项到左边，通分计算后，然后转化为()()432020x x x ⎧--⎨-≠⎩，求出不等式组的解集即为原不等式的解集．【详解】解：不等式3112x x --可转化为31102x x ---，即4302x x --，即4302x x --，所以不等式等价于()()432020x x x ⎧--⎨-≠⎩，解得：324x <，所以原不等式的解集是3{|2}4x x <．故选：B ．4．命题“∀x ∈R ，∃n ∈N+，使n ≥2x+1”的否定形式是（）A ．∀x ∈R ，∃n ∈N+，有n<2x+1B ．∀x ∈R ，∀n ∈N+，有n<2x+1C ．∃x ∈R ，∃n ∈N+，使n<2x+1D ．∃x ∈R ，∀n ∈N+，使n<2x+1【正确答案】D【分析】根据全称命题、特称命题的否定表述：条件中的∀→∃、∃→∀，然后把结论否定，即可确定答案【详解】条件中的∀→∃、∃→∀，把结论否定∴“∀x ∈R ，∃n ∈N+，使n ≥2x+1”的否定形式为“∃x ∈R ，∀n ∈N+，使n<2x+1”故选：D本题考查了全称命题、特称命题的否定形式，其原则是将原命题条件中的∀→∃、∃→∀且否定原结论5．已知12a b ≤-≤，24a b ≤+≤，则32a b -的取值范围是（）A ．3,92⎡⎤⎢⎥⎣⎦B ．5,82⎡⎤⎢⎥⎣⎦C ．5,92⎡⎤⎢⎥⎣⎦D ．7,72⎡⎤⎢⎥⎣⎦【正确答案】D【分析】令32()()a b m a b n a b -=-++求,m n ，再利用不等式的性质求32a b -的取值范围.【详解】令32()()()()a b m a b n a b m n a n m b -=-++=++-，∴32m n n m +=⎧⎨-=-⎩，即51,22m n ==，∴55()5,121()222a b a b ≤-≤≤+≤，故73272a b ≤-≤.故选：D6．如图，ABC 中，90ACB ∠=︒，30A ∠=︒，16AB =，点P 是斜边AB 上任意一点，过点P 作PQ AB ⊥，垂足为P ，交边AC （或边CB ）于点Q ，设AP x =，APQ △的面积为y ，则y 与x 之间的函数图象大致是（）A ．B ．C ．D ．【正确答案】D【分析】首先过点C 作CD AB ⊥于点D ，由ABC 中，90ACB ∠= ，30A ∠= ，可求得B ∠的度数与AD 的长度，再分别从当012AD ≤≤与当1216x <≤时，去分析求解即可求得y 与x 之间的函数关系式，进一步选出图象.【详解】过点C 作CD AB ⊥于点D ，因为90ACB ∠= ，30A ∠= ，16AB =，所以60B ∠= ，142BD BC ==，12AD AB BD =-=.如图1，当012AD ≤≤时，AP x =，tan 30PQ AP x =⋅ ，所以21236y x x x ==，如图2：当1216x <≤时，16BP AB AP x =-=-，所以)tan 6016PQ BP x =⋅=-，所以)211622y x x x =-=-+，故选：D此题考查了动点问题，注意掌握含30 直角三角形的性质与二次函数的性质；注意掌握分类讨论的思想.属于中档题.7．已知函数221111x xf x x --⎛⎫= ⎪++⎝⎭，则()f x 的解析式为（）A ．()()2211x f x x x =≠-+B ．()()2211xf x x x =-≠-+C ．()()211xf x x x =≠-+D ．()()211xf x x x =-≠-+【正确答案】A 【分析】令11x t x -=+，则11tx t-=+，代入已知解析式可得()f t 的表达式，再将t 换成x 即可求解.【详解】令11x t x -=+，则11tx t-=+，所以()()222112111111t t t f t t t t t -⎛⎫- ⎪+⎝⎭==≠-+-⎛⎫+ ⎪+⎝⎭，所以()()2211xf x x x=≠-+，故选：A.8．已知0x >，0y >，且2121x y+=+，若2231x y m m +>--恒成立，则实数m 的取值范围是（）A ．1m ≤-或4m ≥B ．4m ≤-或m 1≥C ．14-<<mD ．41m -<<【正确答案】C 由2121x y +=+得121y x=+，利用基本不等式求出2x y +的最小值，再将不等式恒成立转化为最值，解不等式可得结果.【详解】由2121x y +=+得212(1)y x x y ++=+，所以12x xy +=，所以121y x=+，所以121x y x x +=++13≥=，当且仅当1,1x y ==时，等号成立，所以()min 23x y +=，所以2231x y m m +>--恒成立，可化为2331m m >--，即2340m m --<，解得14-<<m .故选：C结论点睛：本题考查不等式的恒成立与有解问题，可按如下规则转化：①若()k f x ≥在[,]a b 上恒成立，则max ()k f x ≥；②若()k f x ≤在[,]a b 上恒成立，则min ()k f x ≤；③若()k f x ≥在[,]a b 上有解，则min ()k f x ≥；④若()k f x ≤在[,]a b 上有解，则max ()k f x ≤；二、多选题9．有以下判断，其中是正确判断的有（）.A ．()xf x x =与()1,01,0x g x x ≥⎧=⎨-<⎩表示同一函数B ．函数()22122x f x x =+++的最小值为2C ．函数()y f x =的图象与直线1x =的交点最多有1个D ．若()1f x x x =--，则112f f ⎛⎫⎛⎫= ⎪⎪⎝⎭⎝⎭【正确答案】CD【分析】根据函数的定义域可判断A 的正误，根据基本不等式可判断B 的正误，根据函数的定义可判断C 的正误，根据函数解析式计算对应的函数值可判断D 的正误.【详解】对于A ，()xf x x=的定义域为()(),00,∞-+∞U ，而()1,01,0x g x x ≥⎧=⎨-<⎩的定义域为R ，两个函数的定义域不同，故两者不是同一函数.对于B ，由基本不等式可得()221222f x x x =++≥+，但221x +=无解，故前者等号不成立，故()2f x >，故B 错误.对于C ，由函数定义可得函数()y f x =的图象与直线1x =的交点最多有1个，故C 正确.对于D ，()1012f f f ⎛⎫⎛⎫== ⎪⎪⎝⎭⎝⎭，故D 正确.故选：CD.10．下面命题正确的是（）A ．“3x >”是“5x >"的必要不充分条件B ．“0ac <”是“一元二次方程20ax bx c ++=有一正一负两个实根”的充要条件C ．“1x ≠”是“2430x x -+≠”的必要不充分条件D ．设,R x y ∈，则“4x y +≥”是“2x ≥且2y ≥”的充分不必要条件【正确答案】ABC【分析】利用充分条件，必要条件的定义逐项判断作答.【详解】对于A ，3x >不能推出5x >，而5x >，必有3x >，“3x >”是“5x >"的必要不充分条件，A 正确；对于B ，若0ac <，一元二次方程20ax bx c ++=判别式240b ac ∆=->，方程有二根12,x x ，120cx x a=<，即12,x x 一正一负，反之，一元二次方程20ax bx c ++=有一正一负两个实根12,x x ，则120cx x a=<，有0ac <，所以“0ac <”是“一元二次方程20ax bx c ++=有一正一负两个实根”的充要条件，B 正确；对于C ，当1x ≠时，若3x =，有2430x x -+=，当2430x x -+≠时，1x ≠且3x ≠，因此“1x ≠”是“2430x x -+≠”的必要不充分条件，C 正确；对于D ，,R x y ∈，若4x y +≥，取1,4x y ==，显然“2x ≥且2y ≥”不成立，而2x ≥且2y ≥，必有4x y +≥，设,R x y ∈，则“4x y +≥”是“2x ≥且2y ≥”的必要不充分条件，D 不正确.故选：ABC11．函数()1,Q0,Qx D x x ∈⎧=⎨∉⎩被称为狄利克雷函数，则下列结论成立的是（）A ．函数()D x 的值域为[]0,1B ．若()01D x =，则()011D x +=C ．若()()120D x D x -=，则12x x -∈Q D ．x ∃∈R ，(1D x =【正确答案】BD【分析】求得函数()D x 的值域判断选项A ；推理证明判断选项B ；举反例否定选项C ；举例证明x ∃∈R ，(1D x =.判断选项D.【详解】选项A ：函数()D x 的值域为{}0,1.判断错误；选项B ：若()01D x =，则0Q x ∈，01Q x +∈，则()011D x +=.判断正确；选项C ：()()2ππ000D D -=-=，但2ππ=πQ -∉.判断错误；选项D ：当x =时，((()01D x D D ===.则x ∃∈R ，(1D x =.判断正确.故选：BD12．已知集合{}20,0x x ax b a ++=>有且仅有两个子集，则下面正确的是（）A ．224a b -≤B ．214a b+≥C ．若不等式20x ax b +-<的解集为()12,x x ，则120x x >D ．若不等式2x ax b c ++<的解集为()12,x x ，且124x x -=，则4c =【正确答案】ABD【分析】根据集合{}20,0x x ax b a ++=>子集的个数列方程，求得,a b 的关系式，对A ，利用二次函数性质可判断；对B ，利用基本不等式可判断；对CD ，利用不等式的解集及韦达定理可判断.【详解】由于集合{}20,0x x ax b a ++=>有且仅有两个子集，所以2240,4a b a b ∆=-==，由于0a >，所以0b >.A ，()22224244a b b b b -=-=--+≤，当2,b a ==时等号成立，故A 正确.B ，21144a b b b +=+≥=，当且仅当114,,2b b a b ===时等号成立，故B 正确.C ，不等式20x ax b +-<的解集为()12,x x ，120x x b =-<，故C 错误.D ，不等式2x ax b c ++<的解集为()12,x x ，即不等式20x ax b c ++-<的解集为()12,x x ，且124x x -=，则1212,x x a x x b c +=-=-，则()()22212121244416x x x x x x a b c c -=+-=--==，4c ∴=，故D 正确，故选：ABD三、填空题13．已知21,0()2,0x x f x x x ⎧+≥=⎨-<⎩,求()1f f -=⎡⎤⎣⎦________.【正确答案】5【分析】先求()1f -，再根据()1f -值代入对应解析式得()1.f f ⎡⎤-⎣⎦【详解】因为()()1212,f -=-⨯-=所以()[]1241 5.f f f ⎡⎤-==+=⎣⎦求分段函数的函数值，要先确定要求值的自变量属于哪一段区间，然后代入该段的解析式求值，当出现(())f f a 的形式时，应从内到外依次求值.14．已知正实数a 、b 满足131a b+=，则()()12a b ++的最小值是___________.【正确答案】13+13+【分析】由已知可得出3ba b =-且3b >，化简代数式()()12a b ++，利用基本不等式可求得结果.【详解】因为正实数a 、b 满足131a b +=，则03b a b =>-，由0b >可得3b >，所以，()()()()()()32312122222333b b a b b b b b b b +⎛⎫⎛⎫++=++=++=++⎪ ⎪---⎝⎭⎝⎭()()()33515222313131333b b b b b -+=++=-++≥+=+--当且仅当62b =时，等号成立.因此，()()12a b ++的最小值是13+.故答案为.13+15．对于[]1,1a ∈-，()2210x a x a +-+->恒成立的x 取值________.【正确答案】()(),02,-∞+∞ 【分析】设()()()2221121f a x a x a x a x x =+-+-=-+-+关于a 的一次函数，只需()()1010f f ⎧>⎪⎨->⎪⎩即可求解.【详解】令()()()2221121f a x a x a x a x x =+-+-=-+-+，因为对于[]11a ∈-，，不等式()2210x a x a +-+->恒成立，所以()()1010f f ⎧>⎪⎨->⎪⎩即220320x x x x ⎧->⎨-+>⎩解得：0x <或2x >.故答案为.()()02-∞⋃+∞，，方法点睛：求不等式恒成立问题的方法（1）分离参数法若不等式(),0f x λ≥()x D ∈（λ是实参数）恒成立，将(),0f x λ≥转化为()g x λ≥或()()g x x D λ≤∈恒成立，进而转化为()max g x λ≥或()()min g x x D λ≤∈，求()g x 的最值即可.（2）数形结合法结合函数图象将问题转化为函数图象的对称轴、区间端点的函数值或函数图象的位置关系（相对于x 轴）求解.此外，若涉及的不等式转化为一元二次不等式，可结合相应一元二次方程根的分布解决问题.（3）主参换位法把变元与参数变换位置，构造以参数为变量的函数，根据原变量的取值范围列式求解，一般情况下条件给出谁的范围，就看成关于谁的函数，利用函数的单调性求解.16．若函数2()2f x x x =+，()2(0)g x ax a =+>，对于1x ∀∈[]1,2-，[]21,2x ∃∈-，使12()()g x f x =，则a 的取值范围是_____________.【正确答案】(]0,3【分析】由题意可知函数()g x 在区间[]1,2-的值域是函数()f x 在区间[]1,2-的值域的子集，转化为子集问题求a 的取值范围.【详解】()()20g x ax a =+>在定义域上是单调递增函数，所以函数在区间[]1,2-的值域是[]2,22a a -+函数()22f x x x =+在区间[]1,2-是单调递增函数，所以函数()f x 的值域是[]1,8-，由题意可知[][]2,221,8a a -+⊆-，所以21228a a -≥-⎧⎨+≤⎩，解得.3a ≤故答案为.(]0,3本题考查双变量等式中任意，存在问题求参数的取值范围，重点考查函数的值域，转化与化归的思想，属于中档题型.四、解答题17．已知{|13}A x x =-<≤，{|13}B x m x m =≤<+（1）若1m =时，求A B ⋃；（2）若R B A ⊆ð，求实数m 的取值范围．【正确答案】（1）(1,4)A B =-U ；（2）()1,3,2m ⎛⎤∈-∞-+∞ ⎥⎝⎦ ．（1）利用集合的并集定义代入计算即可;（2）求出集合R A ð，利用集合包含关系，分类讨论B =∅和B ≠∅两种情况，列出关于m 的不等式，求解可得答案.【详解】（1）当1m =时，{|14}B x x =≤<，则{|14}A B x x ⋃=-<<即(1,4)A B =-U ．（2）{|1R A x x =≤-ð或}(]()3,13,x >=-∞-⋃+∞，由R B A ⊆ð，可分以下两种情况：①当B =∅时，13m m ≥+，解得：12m ≤-②当B ≠∅时，利用数轴表示集合，如图由图可知13131m m m <+⎧⎨+≤-⎩或133m m m <+⎧⎨>⎩，解得3m >；综上所述，实数m 的取值范围是：12m ≤-或3m >，即()1,3,2m ⎛⎤∈-∞-+∞ ⎥⎝⎦ 易错点睛：本题考查利用集合子集关系确定参数问题，易错点是要注意：∅是任何集合的子集，所以要分集合B =∅和集合B ≠∅两种情况讨论，考查学生的逻辑推理能力，属于中档题.18．（1）已知a b c <<，且0a b c ++=，证明：a a a c b c<--．（2213a a a a ---(3)a ≥【正确答案】（1）证明见解析；（2）证明见解析【分析】(1)利用不等式的性质证明即可；(2)a 3a -<1a -2a -，对不等式两边同时平方后只需证明()3a a -<()()12a a --.【详解】证明：（1）由a b c <<，且0a b c ++=，所以0a <，且0,a cbc -<-<所以()()0a c b c -->，所以()()a c a c b c -<--()()b c a c b c ---，即1b c -<1a c -；所以a b c ->a a c -，即a a c -<a b c-．（2213a a a a ---，(3)a ≥a 3a -<1-a 2a -，即证(3)(3)(1)(2)2(1)(2)a a a a a a a a +-+--+-+--()3a a -<()()12a a --即证(3)(1)(2)a a a a -<--；即证02<，显然成立；213a a a a ---19．已知二次函数y ＝ax 2+bx ﹣a +2．(1)若关于x 的不等式ax 2+bx ﹣a +2＞0的解集是{x |﹣1＜x ＜3}，求实数a ，b 的值；(2)若b ＝2，a ＞0，解关于x 的不等式ax 2+bx ﹣a +2＞0．【正确答案】(1)a ＝﹣1，b ＝2(2)见解析【分析】（1）根据一元二次不等式的解集性质进行求解即可；（2）根据一元二次不等式的解法进行求解即可.【详解】（1）由题意知，﹣1和3是方程ax 2+bx ﹣a +2＝0的两根，所以132(1)3b a a a ⎧-+=-⎪⎪⎨-+⎪-⨯=⎪⎩，解得a ＝﹣1，b ＝2；（2）当b ＝2时，不等式ax 2+bx ﹣a +2＞0为ax 2+2x ﹣a +2＞0，即（ax ﹣a +2）（x +1）＞0，所以()210a x x a -⎛⎫-+> ⎪⎝⎭，当21a a-=-即1a =时，解集为{}1x x ≠-；当21a a -<-即01a <<时，解集为2a x x a -⎧<⎨⎩或}1x >-；当21a a ->-即1a >时，解集为2a x x a -⎧>⎨⎩或}1x <-.20．（1）求函数()3f x x 在区间[]2,4上的值域.（2）已知二次函数2()1(R)f x x mx m m =-+-∈.函数在区间[]1,1-上的最小值记为()g m ，求()g m 的值域；【正确答案】（1）12,4⎤-⎦；（2）(]0-∞，【分析】(1)t =，可得函数()22()36318g t t tt t =--=+-，讨论其值域即可求解；(2)分类讨论二次函数的对称轴与给定区间[]1,1-的关系，分别表示出函数的最小值，表示为分段函数形式，作出图象即可求解.【详解】(1)函数()3f x x =，t =，则26x t =-∵[]2,4x ∈2t ≤≤那么函数()f x 转化为()22()36318g t t t t t =--=+-其对称轴16t =-，2t ≤≤时()g t 单调递增，∴()(2)g g t g ≤≤，12()4g t -≤≤-，故得()f x的值域为12,4⎤--⎦.（2）2()1f x x mx m =-+-，二次函数对称轴为2m x =，开口向上①若12m <-，即2m <-，此时函数()f x 在区间[]1,1-上单调递增，所以最小值()(1)2g m f m =-=.②若112m -≤≤，即22m -≤≤，此时当2m x =时，函数()f x 最小，最小值2()124m m g m f m ⎛⎫==-+- ⎪⎝⎭.③若12m >，即m>2，此时函数()f x 在区间[]1,1-上单调递减，所以最小值()(1)0g m f ==.综上22,2()1,2240,2m m m g m m m m <-⎧⎪⎪=-+--≤≤⎨⎪>⎪⎩，作出分段函数的图像如下，所以当2m <-时，()(,4);g m ∈-∞-当22m -≤≤时，[]4,0;g(m)∈-当m>2时，()0g m =，综上知()g m 的值域为(]0.，-∞21．今年，我国某企业为了进一步增加市场竞争力，计划在2023年利用新技术生产某款新手机.通过市场分析，生产此款手机全年需投入固定成本250万，每生产x （千部）手机，需另投入成本()R x 万元，且()2101001000,040100007018450,40x x x R x x x x ⎧++<<⎪=⎨+-≥⎪⎩，由市场调研知，每部手机售价0.7万元，且全年内生产的手机当年能全部销售完.(1)求2023年的利润()W x （万元）关于年产量x （千部）的函数关系式；(2)2023年产量为多少（千部）时，企业所获利润最大？最大利润是多少？【正确答案】(1)()2106001250,040100008200,40x x x W x x x x ⎧-+-<<⎪=⎨⎛⎫-++≥ ⎪⎪⎝⎭⎩(2)2023年产量为100（千部）时，企业所或利润最大，最大利润是8000万元【分析】（1）根据已知条件求得分段函数()W x 的解析式.（2）结合二次函数的性质、基本不等式求得()W x 的最大值以及此时的产量.【详解】（1）当040x <<时，()()22700101001000250106001250W x x x x x x =-++-=-+-；当40x ≥时，()100001000070070184502508200W x x x x x x ⎛⎫⎛⎫=-+--=-++ ⎪ ⎪⎝⎭⎝⎭；∴()2106001250,040100008200,40x x x W x x x x ⎧-+-<<⎪=⎨⎛⎫-++≥ ⎪⎪⎝⎭⎩；（2）若040x <<，()()210307750W x x =--+，当30x =时，()max 7750W x =万元；若40x ≥，()10000820082008000W x x x ⎛⎫=-++≤-= ⎪⎝⎭，当且仅当10000x x=即100x =时，()max 8000W x =万元.答：2023年产量为100（千部）时，企业所或利润最大，最大利润是8000万元.22．已知()11282,0,11f x f x x x x x ⎛⎫+=+-≠≠ ⎪-⎝⎭，(1)求()f x 的解析式；(2)已知()()()22,22g x mx mx g x x f x m =--<-+在()1,3上有解，求m 的取值范围．【正确答案】(1)1()2f x x=+，0,1x x ≠≠；(2)3m <.【分析】（1）根据给定条件，用11,1x x x--依次替换x ，再消元求解作答.（2）由（1）结合已知，变形不等式，分离参数构造函数，求出函数在()1,3的最大值作答.【详解】（1）0,1x x ≠≠，11()2()821f x f x x x +=+--，用11x-替换x 得：11()2912()1x f f x x x x -+=-+--，则有1114()4()8222(9)1011x f x f x x x x x x x --=+---+=-+---，用1x x-替换x 得：1112()2()82(1)711x f f x x x x x x x -+=+--=++--，于是得99()18f x x =+，则1()2f x x=+，所以()f x 的解析式为1()2f x x=+，0,1x x ≠≠.（2）(1,3)x ∈，2221()()22(2)22g x x f x m mx mx x m x-<-+⇔--+<-+，即22(2)22m x x x x -+<++，于是得22222x x m x x ++<-+，令2222(),132x x h x x x x ++=<<-+，依题意，(1,3)x ∈，()m h x <有解，当(1,3)x ∈时，222223()22323()22222222[()][()]23333x x x x h x x x x x x x -++-==+=+-+-+-+--++322316219(2333x x =+≤+-++-，当且仅当1629233x x -=-，即2x =时取等号，因此当2x =时，max ()(2)3h x h ==，则3m <，所以m 的取值范围是3m <．。

河南省新郑市重点高中2020—2021学年高一数学上学期12月月考试题（无答案）考试时间:120分钟； 一、单选题（本大题共12小题，每小题5分，共60分.在每小题给出的四个选项中,只有一项是符合题目要求）1．已知集合()(){}|310M x x x =+-≤,{}2|log1N x x =≤,则M N ⋃=（）A ．[]3,2-B ．[)-3,2C ．[]1,2D ．(0,2] 2．已知12132111,log ,log 332a b c ⎛⎫=== ⎪⎝⎭，则（ ）A ．c b a >>B ．b c a >>C ．a b c>>D ．b a c >>3．下列命题中,错误的是 ( ）A ．平行于同一个平面的两个平面平行B ．平行于同一条直线的两个平面平行C ．一个平面与两个平行平面相交，交线平行D ．一条直线与两个平行平面中的一个相交，则必与另一个相交4．函数()e e ln --=xxf x x的图象大致为（ ）A ．B ．C ．D ．5．如图所示，一个水平放置的平面图形的斜二测直观图是等腰梯形OA B C '''，且直观图OA B C '''的面积为2，则该平面图形的面积为（ )A ．2B ．42C ．4D ．226．在三棱锥P ABC -中,2PA PB PC ===，且,,PA PB PC 两两互相垂直,则三棱锥P ABC -的外接球的体积为( ）A ．43πB ．83πC ．163πD ．23π7．已知函数(),142,12x a x f x a x x ⎧>⎪=⎨⎛⎫-+≤ ⎪⎪⎝⎭⎩是R 上的单调递增函数，则实数a 的取值范围是（ ) A ．()1,+∞ B ．[)4,8C ．()4,8D ．()1,88．函数f （x ）=｜x 2﹣6x+8｜的单调递增区间为（ ) A ．［3，+∞） B ．（﹣∞,2）,(4,+∞)C ．（2,3),（4，+∞）D ．(﹣∞，2]，［3，4］ 9．某三棱锥的三视图如图所示，如果网格纸上小正方形的边长为1， 则该三棱锥的体积为( )A ．4B ．8C ．12D ．2410．已知定义在R 上的函数()f x ，都有()()1f x f x =-,且函数()1f x +是奇函数，若1142f ⎛⎫-=- ⎪⎝⎭,则20194f ⎛⎫⎪⎝⎭的值为（ ） A ．1-B ．1C ．12-D ．1211．正方体1111ABCD A B C D -棱长为2点M ,N 分别是1,BC CC 的中点,动点P 在正方形11BCC B 内运动,且1//PA AMN则1PA 的长度范围为( ）A ．51,2⎡⎤⎢⎥⎣⎦B ．32,52⎡⎤⎢⎥⎣⎦ C ．32,32⎡⎤⎢⎥⎣⎦D ．31,2⎡⎤⎢⎥⎣⎦12．已知函数()f x 是定义在R 上的奇函数，当0x ≥时， ()()11232f x x x =-+--，若x R ∀∈， ()()f x a f x -≤，则a 的取值范围是（ ）A ．3a ≥B ．33a -≤≤C ．6a ≥D ．66a -≤≤二、填空题(本大题共4小题，每小题5分，共20分）13．已知)12(+x f 定义域为)5,3(,则)1-4(x f 定义域为 。

高中物理学习材料（马鸣风萧萧**整理制作）叶县二高2015-2016学年上学期第四次月考试题高一物理答案1.AC2.D3.A4.C5.BC6.C7.B8.C9.D 10.A 11.BC 12.D13.ABDF （6分，对而不全得3分）14.答案：(1)3.79(2)0.19(3)498 （每空3分）解析：由题中上图可知：Δx1＝15.89 m－8.31 m＝2a1T2Δx2＝12.11 m－4.52 m＝2a2T2所以a＝＝3.79 m/s2。

(2)由题中下图可知：Δx′1＝10.17 m－9.79 m＝2a′1T2Δx′2＝9.98 m－9.61 m＝2a′2T2所以a′＝＝0.19 m/s2(3)由牛顿第二定律知：加速阶段：F－f＝ma减速阶段：f＝ma′其中m＝50 kg＋75 kg＝125 kg解得牵引力F＝498 N。

15.解：以物体为研究对象，进行受力分析：重力G、a绳的拉力T a和b绳的拉力T b，作出力图如图所示．以水平方向为x轴，竖直方向为y轴建立直角坐标系，如图所示．由共点力的平衡条件得：T b﹣T a sin37°=0…①……………3分T a cos37°°﹣G=0…②……………3分由②得：T a==N=50N…③……………2分将③代入①得：T b=T a sin37°=50×0.6N=30N……………2分答：绳子a和b对物体的拉力分别是50N和30N．16.解：木箱的加速度为，代入已知得：a=0．5 m/s2……………3分由v=at得4 s后木箱的速度v1=2 m/s……………3分由得，代入已知得：a1=-1 m/s2……………3分由，代入已知得：F1=40 N……………3分17.答案：(1)36m(2)6.5s解析：(1)已知足球的初速度为v1＝12m/s，加速度大小为a1＝2m/s2…………3分足球做匀减速运动的时间为：t1＝＝6s 运动位移为：x1＝t1＝36m…………3分(2)已知前锋队员的加速度为a2＝2m/s2，最大速度为v2＝8m/s，前锋队员做匀加速运动达到最大速度的时间和位移分别为：t2＝＝4s ; x2＝16m…………2分之后前锋队员做匀速直线运动，到足球停止运动时，其位移为：x3＝v2(t1－t2)＝16m…………2分由于x2＋x3<x1，故足球停止运动时，…………2分前锋队员没有追上足球，然后前锋队员继续以最大速度匀速运动追赶足球，利用公式x1－(x2＋x3)＝v2t3，得：t3＝0.5s…………2分前锋队员追上足球的时间t＝t1＋t3＝6.5s…………1分。

郑州一中27届（高一）第一次模拟测试数学试题卷第I 卷（选择题）一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1. 设全集，，则如图中阴影部分表示的集合为（ ）A. B. C. D. 2. 命题“，”的否定是（ ）A. ， B. ，C. ， D. ，3. 已知函数的值为（ ）A. B. 0 C. 2 D. 44. 已知，若，，，且，，，则的值（ ）A. 大于0B. 等于0C. 小于0D. 不能确定5. 函数的部分图象大致为（ ）A.B.U R =(){}{}30,1M x x x N x x =+<=<-{|1}x x ≥-{|30}-<<x x {|3}x x ≤-{|10}x x -≤<x ∃∈R 310x x +>x ∃∈R 310x x +≥x ∃∈R 310x x+≤x ∀∈R 310x x+≤x ∀∈R 310x x +>()()2,1,2,1x x f x f x x -≤⎧=⎨>⎩2-3()2f x x x =+a b c ∈R 0a b +>0a c +>0b c +>()()()f a f b f c ++()22111x f x x +=-+C. D.6. 已知，则下列不等式一定成立的是（ ）A. B. C D. 7. 已知，关于的一元二次不等式的解集中有且仅有3个整数，则的值不可能是（ ）A 13 B. 14 C. 15 D. 168. 已知函数，若的值域为，则实数的取值范围是（ ）A. B. C. D. 二、选择题：本题共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. 下列函数中，既是奇函数，又在上单调递增的是（ ）A. B. C. D. 10. 命题“，”为真命题的一个充分不必要条件可以是（ ）A. B. C. D. 11. 设为实数，不超过的最大整数称为的整数部分，记作.例如，.称函数为取整函数，下列关于取整函数的结论中正确的是（ ）A. 在上是单调递增函数B. 对任意，都有C. 对任意，，都有..0a b >>22a b a b +>+2()4a b ab+≤2b a a b +<22b b a a +<+Z a ∈x 280x x a -+≤a 212,()23,3x c f x x x x c x ⎧-+<⎪=⎨⎪-+≤≤⎩()f x [2,6]c 11,4⎡⎤--⎢⎥⎣⎦1,04⎡⎫-⎪⎢⎣⎭[1,0)-11,2⎡⎤--⎢⎥⎣⎦(0,)+∞()f x =()||f x x x =2()1x x f x x -=-3()f x x =[1,2)x ∀∈20x a -≤4a ≥5a >6a ≥7a >x x x []x [1.2]1=[ 1.4]2-=-()[]f x x =()f x ()f x R x ∈R ()1f x x >-x ∈R k ∈Z ()()f x k f x k+=+D 对任意，，都有第II 卷（非选择题）三、填空题：本题共3小题，每小题5分，共15分.12. 用列举法表示______．13. 函数是上的偶函数， 且当时，函数的解析式为，则______；当时，函数的解析式为___________.14. 已知，为非负实数，且，则的最小值为______.四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或验算步骤.15. 已知全集，集合，.（1）求；（2）求.16. 设命题，使得不等式恒成立；命题，不等式成立．（1）若为真命题，求实数取值范围；（2）若命题、有且只有一个是真命题，求实数取值范围．17. 设函数为定义在上的奇函数.（1）求实数的值；（2）判断函数的单调性，并用定义法证明在(0,+∞)上的单调性.18. 已知某园林部门计划对公园内一块如图所示的空地进行绿化，用栅栏围4个面积相同的小矩形花池，一面可利用公园内原有绿化带，四个花池内种植不同颜色的花，呈现“爱我中华”字样．（1）若用48米长的栅栏围成小矩形花池（不考虑用料损耗），则每个小矩形花池的长、宽各为多少米时，才能使得每个小矩形花池的面积最大？.的的x y ∈R ()()()f xy f x f y =6N N 1a a ⎧⎫∈∈=⎨⎬-⎩⎭∣()f x R 0x >2()1f x x=-(1)f -=0x <a b 21a b +=22211a b a b+++R U ={}2|560A x x x =-+>{|230}B x x =->A B ⋂()()U U A B ðð[]:1,1p x ∀∈-2230x x m --+<[]:0,1q x ∃∈2223x m m -≥-p m p q m ()22a f x x a x+=-+(,0)(0,)-∞+∞ a ()f x ()f x（2）若每个小矩形的面积为平方米，则当每个小矩形花池的长、宽各为多少米时，才能使得围成4个小矩形花池所用栅栏总长度最小？19. 已知集合中含有三个元素，同时满足①；②；③为偶数，那么称集合具有性质.已知集合，对于集合的非空子集，若中存在三个互不相同的元素，使得均属于，则称集合是集合的“期待子集”.（1）试判断集合是否具有性质，并说明理由；（2）若集合具有性质，证明：集合是集合的“期待子集”；（3）证明：集合具有性质的充要条件是集合是集合的“期待子集”.983A ,,x y z x y z <<x y z +>x y z ++A P {}1,2,3,,2n S n = *(N ,4)n n ∈≥n SB n S ,,a b c ,,+++a b b c c a B B n S {}1,2,3,5,7,9A =P {}3,4,B a =P B 4S M P M n S郑州一中27届（高一）第一次模拟测试数学试题卷第I卷（选择题）一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.【1题答案】【答案】D【2题答案】【答案】C【3题答案】【答案】D【4题答案】【答案】A【5题答案】【答案】A【6题答案】【答案】D【7题答案】【答案】D【8题答案】【答案】A二、选择题：本题共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题答案】【答案】BD【10题答案】【答案】BCD【11题答案】【答案】BC第II卷（非选择题）三、填空题：本题共3小题，每小题5分，共15分.【12题答案】【答案】【13题答案】【答案】 ①. ②. 【14题答案】【答案】2四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或验算步骤.【15题答案】【答案】（1）或 （2）【16题答案】【答案】（1）（2）【17题答案】【答案】（1）（2）在上单调递减，在(0,+∞)上单调递减，证明见解析【18题答案】【答案】（1）长为6米、宽为4米（2）长为7米、宽为米【19题答案】【答案】（1）不具有，理由见解析（2）证明见解析 （3）证明见解析{}1,2,3,61()21f x x=--{3|22x x <<3}x >3|232x x x ⎧⎫≤≤≤⎨⎬⎩⎭或(,0)-∞(,3]-∞0a =(,0)-∞143。

2021届河南省新郑市第一中学高三英语第四次联考试题及参考答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项AWhile Jennifer was at home taking an online exam for her business law class, a monitor（监控器）a few hundred miles away was watching her every move.Using a web camera equipped in Jennifer'sLos Angelesapartment, the monitor inPhoenixtracked how frequently her eyes moved from the computer screen and listened for the secret sounds of a possible helper in the room. Her Internet access was locked remotely to prevent Internet searches, and her typing style was analyzed to make sure she was who she said she was: Did she enter her student number at the same speed as she had in the past？Or was she slowing down？In the battle against cheating, this is thecutting edgeand a key to encourage honesty in the booming field of online education. This technology gives trust to the entire system, to the institution and to online education in general. Only with solid measures against cheating, experts say, can Internet universities show that their exams and diplomas are valid — that students haven't searched the Internet to get the right answers.Although online classes have existed for more than a decade, the concern over cheating has become sharper in the last year with the growth of "open online courses". Private colleges, public universities and corporations are jumping into the online education field, spending millions of dollars to attract potential students, while also taking steps to help guarantee honesty at a distance.Aside from the web cameras, a number of other high-tech methods are becoming increasingly popular. Among them are programs that check students’ identities using personal information, such as the telephone numbers they once used.Other programs can produce unique exams by drawing on a large list of questions and can recognize possible cheaters by analyzing whether difficult test questions are answered at the same speed as easy ones. As in many university classes, term papers are scanned against some large Internet data banks for cheating.1. Why was Jennifer watched in an online exam?A. To correct her typing mistakes.B. To find her secrets in the room.C. To keep her from dishonest deeds.D. To prevent her from slowing down.2. What does the underlined expression "cutting edge" in Paragraph 3 probably mean?A. sharpening toolB. advanced techniqueC. effective ruleD. dividing line3. How can some programs find out possiblecheaters?A. By scanning the Internet test questions.B. By checking the question answering speed.C. By producing a large number of questions.D. By giving difficult test questions.BExperts are warning about the risks of extremely picky(挑剔的)eating after a teenager living on a diet of chips and crisps developed lasting sight loss. Eye doctors inBristolcared for the 17-year-old after his sight had gone to the point of blindness. Tests showed he had serious vitamin deficiency(缺乏). Dr. Denize Atan, who treated him at the hospital, said, “His diet was basically a portion of chips from the local fish and chip shop every day. He also used to snack on crisps and sometimes white bread and ham, and not really any fruit and vegetables.”The teenager saw his doctor at the age of 14 because he had been feeling tired and unwell. At that time he suffered from vitamin B12 deficiency, but he did not stick with the treatment or improve his poor diet. Three years later, he was taken to theBristolEyeHospitalbecause of progressive sight loss.He was not overweight or underweight, but he had lost minerals from his bones, which was really quite shocking for a boy of his age. In terms of his sight loss, he met the standards of being blind. “He had blind spots right in the middle of his sight,” said Dr Denize Atan, “That means he can’t drive and would find it reallyarduousto read, watch TV or recognize faces.”Dr Denize Atan said that parents should learn about the harm that can be caused by picky eating, and turn to experts for help. For those who are concerned , she advised, “It’s best not to be anxious about picky eating , and instead calmly introduce one or two new foods with every meal.” She said multivitamin tablets can supplement(补充) a diet, but cannot take the place of eating healthily. “It’s much better to take in vitamins through a varied and balanced diet,” she said, adding that too manycertain vitamins , including vitamin A, can be harmful ,“so you don’t want to overdo it.”4. What does Dr Denize Atan imply in paragraph 1?A. The diet of the boy is not balanced.B. Fruit and vegetables are rich in vitamins.C. Picky eating is common among teenagers.D. The cause of the boy’s disease is unknown.5. Why did the boy go to see his doctor at the age of 14?A. To improve his poor diet.B. To get some help to lose weight.C. To be treated for his discomfort.D. To slow down his progressive sight loss.6. What does the underlined word “arduous” in paragraph 3 probably mean?A. Important.B. Easy.C. Necessary.D. Difficult.7. What does the last paragraph mainly talk about?A. Reasons why the boy is seriously ill.B.Suggestions for the boy’s family to care for him.C. Advice for parents worried about picky eating.D. Waysof taking in enough vitamins and minerals.CWhat a day! I started at my new school this morning and had the best time. I made lots of new friends and really liked my teachers. I was nervous the night before, but I had no reason to be. Everyone was so friendly and polite. They made me feel at ease. It was like I'd been at the school for a hundred years!The day started very early at 7:00 am. I had my breakfast downstairs with my mom. She could tell that I was very nervous. Mom kept asking me what was wrong. She told me I had nothing to worry about and that everyone was going to love me. If they didn't love me, Mom said to send them her way for a good talking to. I couldn't stop laughing.My mom dropped me off at the school gates about five minutes before the bell. A little blonde girl got dropped off at the same time and started waving at me. She ran over and told me her name was Abigail. She was very nice and we became close straight away. We spent all morning together and began to talk to another girl called Stacey. The three of us sat together in class all day and we even made our way home together! It went so quickly. Our teacher told us that tomorrow we would really start learning and developing new skills.I cannot wait until tomorrow and feel as though I am really going to enjoy my time at my new school. I only hope that my new friends feel the same way too.8. How did the author feel the night before her new school?A. Tired.B. ConfidentC. Worried.D. homesick9. What did the author think of her mother’s advice?A. Clear.B. Funny.C. OptionalD. Respectable10. What happened on the author's first day of school?A. She met many nice people.B. She had a hurried breakfast.C. She learned tome new skills.D. She arrived at school very early.11. What can we infer about Abigail?A. She disliked Stacey.B. She was shy and quiet.C. She got on well with the author.D. She was an old friend of the author.DSophie became friends with the gray squirrels during her first week atPennState, after spotting them running around and wondering what they would look like with tiny hats on their heads. Today, everyone at the university knows her as the “Squirrel Girl”.Sophie tried bringing them food, and gradually they began to trust her. She managed to put a hat on a squirrel and take a picture. Thinking that her colleagues could do with something to lift their spirits, she started posting similar photos on Facebook. The response was greatly positive, and before long Sophie and her squirrels became an Internetsensation.Growing up in a neighborhood outside ofState College, Sophie was always fond of birds and animals around her home, but she didn't interact with people very much. She was later diagnosed (诊断) with Asperger's syndrome, but the squirrels changed that. “The squirrels help me break the ice, because I'll be sitting here patting a squirrel and other people will come over and well just start like feeding the squirrels together and chatting about them,” she said, “I am a lot more outgoing.”And in case you're wondering how Sophie is able to get the squirrels to do what she wants for her photos, it has a lot to do with food. For example, whenever she wants them to hold or play with something, she puts peanut butter on the prop (道具), and they'll grab it. In the beginning, she would throw peanuts up the trees on campus and invite the squirrels to come down and get them, but they hesitated to approach her. She had the patience to earn their trust, though.This year, Sophie is graduating with a degree in English and wildlife sciences. She wants to be a science writer and educate people on how to preserve the environment. As for her furry friends, Sophie plans to stay in the areaand visit them as often as she can.12. What does the underlined word “sensation” in paragraph 2 mean?A. Event.B. Hit.C. Service.D. Addiction.13. What can we learn about Sophie according to paragraph 3?A. She got lots of friends due to squirrels.B. She used to be a popular girl in her childhood.C. She lived in the far countryside when young.D. She was more outgoing than before.14. How did Mary manage to take photos of squirrels wearing hats?A. By attracting them with food.B. By putting them in cages.C. By playing music to them.D. By dressing like squirrels.15. What do you think of Sophie?A. Tolerant and capable.B. Sociable and aggressive.C. Patient and caring.D. Indifferent and appreciative.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

河南省郑州市新郑第一中学分校2020年高一物理月考试题含解析一、选择题：本题共5小题，每小题3分，共计15分．每小题只有一个选项符合题意1. （单选）神舟飞船”与“天宫一号”成功对接后，遨游太空。

下列说法正确的是（）A；“神舟飞船”相对于“天宫一号”是运动的B；“神舟飞船”和“天宫一号”相对地球是静止的C；“神舟飞船”和“天宫一号”相对于地球是运动的D；“神舟飞船”相对于地球是运动的，“天宫一号”相对于地球是静止的参考答案：CA、“天宫一号”和“神舟飞船”对接后，“神舟飞船”与“天宫一号'飞行器一起运动，相对位置没有发生变化，所以“神舟飞船”相对于“天宫一号”飞行器来说是静止的；BCD、“神舟飞船”和“天宫一号”相对于地球的位置在不断发生变化，所以“神舟飞船”和“天宫一号”相对于地球是运动的，故BD错误，C正确。

故选C。

2. （单选）一个物体沿粗糙斜面匀速滑下，则下列说法正确的是（）A．物体机械能不变，内能也不变B．物体机械能减小，内能不变C．物体机械能减小，内能也减小D．物体机械能减小，内能增大参考答案：D3. （单选）如图所示，总质量为460kg的热气球，从地面刚开始竖直上升时的加速度为0.5m/s2，当热气球上升到180m时，以5m/s的速度向上匀速运动。

若离开地面后热气球所受浮力保持不变，上升过程中热气球总质量不变，所受空气阻力与速度成正比，重力加速度g=10m/s2 。

关于热气球，下列说法正确的是（）A．所受浮力大小为4830NB．加速上升过程中热气球处于完全失重状态C．从地面开始上升10s后的速度大小为5m/sD．以5m/s匀速上升时所受空气阻力大小为200N 参考答案：A4. （多选）下列关于路程和位移的说法正确的是（）A．路程是标量，位移是矢量。

B．给定初末位置，路程有无数种可能，位移只有两种可能。

C．若物体作单一方向的直线运动，位移的大小就等于路程。

[来D．路程是物体运动径迹的长度，位移描述了物体位置移动的方向和距离。

2021年河南省新郑市第一中学高三生物第四次联考试卷及答案一、选择题：本题共15小题，每小题2分，共30分。

每小题只有一个选项符合题目要求。

1. 下列有关生物膜的结构特点的叙述，最恰当的一项是（）A. 构成生物膜的磷脂分子可以运动B. 构成生物膜的蛋白质分子可以运动C. 构成生物膜的磷脂分子和蛋白质分子是静止的D. 构成生物膜磷脂分子和大多数蛋白质分子可以运动2. 关于下列a、b、c、d四种生物的叙述，不正确的是()A. a和d不具有以核膜为界限的细胞核B. a和b都能进行光合作用C. a、b、c、d都能独立繁殖和代谢D. a属于原核生物，b、c属于真核生物，d属于病毒3. 兴奋在神经元之间的传递是A.先从树突传向细胞体再传向轴突B.通过电荷流动来实现的C.由突触小体来完成的D.单向的4. 用盐腌制的蔬菜可以保存较长时间而不腐败变质，这可能是因为（）A. 蔬菜细胞渗透失水过多而死亡，不能为微生物生长繁殖提供足够营养B. 蔬菜细胞吸收盐分过多而死亡，不能为微生物生长繁殖提供足够营养C. 微生物细胞渗透失水过多，难以生长繁殖D. 微生物细胞吸收盐分过多，难以生长繁殖5. 脂质是细胞和生物体的重要组成成分，下列叙述错误的是（）A.磷脂是构成细胞膜的重要成分B.脂肪仅由C、H、O三种元素组成C.维生素D在人体内参与血液中脂质的运输D.性激素能促进人和动物生殖器官的发育6. 对人群免疫接种是预防传染性疾病的重要措施。

下列叙述错误的是（）A.注射某种流感疫苗后可预防各种流感病毒的感染B.接种脊髓灰质炎疫苗可产生针对脊髓灰质炎病毒的抗体C.接种破伤风疫苗比注射抗破伤风血清可获得更长时间的免疫力D.感染过新型冠状病毒且已完全恢复者的血清可用于治疗新冠肺炎患者7. 在生物体内，某些重要化合物的元素组成和功能关系如图所示。

其中X、Y代表元素，A、B、C是生物大分子。

相关叙述不正确的是（）A.B在细胞中有三种，都能参与蛋白质的合成过程B. 人体中，单体a的种类有4种，其排列顺序决定了C中c的种类和排列C. 同一生物不同细胞中A、B、C均不同，A的多样性决定C的多样性D. 单体a、b、c在形成A、B、C化合物过程中都会消耗能量8. 下列物质中不属于内环境成分的是（）A.水通道蛋白B.HCO3−C.尿素D.激素9. 下列叙述不正确的（）A. 控制细胞器进行物质合成、能量转化等的指令，主要通过核孔从细胞核送到细胞质B. 染色体和染色质是同一物质在细胞不同时期的两种存在状态C. 细胞是生物体代谢和遗传的基本单位D. 在设计并制作细胞模型时，先考虑美观与否，然后考虑科学性10. 在棉花种植过程中，“打顶”的目的是（）A. 提高侧芽部位的生长素浓度，抑制侧芽发育成侧枝B. 降低侧芽部位的生长素浓度，抑制侧芽发育成侧枝C. 提高侧芽部位的生长素浓度，促进侧芽发育成侧枝D. 降低侧芽部位的生长素浓度，促进侧芽发育成侧枝11. 在“噬菌体侵染细菌”的实验中相关的方法与结果正确的是（）A. 分别用含35S和含32P的培养基培养细菌，再用此细菌培养出带相应标记的噬菌体B. 未标记的噬菌体在含32P标记的细菌体内复制三次，其子代噬菌体中含32P的个体占3/4C. 用含35S的培养基直接培养噬菌体，用以标记35SD. 若32P标记组的上清液有放射性，则可能原因是搅拌不充分12. 如图是细胞膜结构示意图，相关叙述中错误的是（）A.①是蛋白质，与糖类分子结合形成糖蛋白，与细胞表面的识别有密切关系B.①是磷脂分子的疏水端，水溶性分子或离子不能自由通过，因此具有屏障作用C.A面是细胞膜的外侧，B面是细胞膜的内侧D.细胞膜的流动性主要表现为磷脂分子可以运动而蛋白质不可以运动13. 下列有关生物学实验的叙述，正确的是（）A.用苏丹①染液可将花生子叶细胞染成橘红色B.观察洋葱根尖的成熟区细胞染色体清晰可见C.检测产生酒精时，向试管中缓慢滴加溶有重铬酸钾的浓硫酸溶液D.分离细胞中各种细胞器和叶绿体中的色素都用差速离心法14. 下图为某用15N标记的DNA分子片段，假设该DNA分子中有碱基5000对，A＋T占碱基总数的34%，若该DNA分子在14N的培养基中连续复制2次，下列叙述正确的是（）A.复制时作用于①处的酶为DNA聚合酶B.复制2次需游离的胞嘧啶脱氧核苷酸9900个C.①处指的是腺嘌呤核糖核苷酸D.子代中含15N的DNA分子占3/415. 下列关于无机盐在生物体中功能的叙述错误的是（）A. 镁是叶绿体中参与光合作用各种色素的组成元素B. 人体缺铁会影响正常的有氧呼吸功能C. 耕牛血液中浓度太低，会出现抽搐症状D. 细胞中的某些无机盐离子对维持细胞的酸碱平衡具有一定作用二、选择题:本题共5小题，每小题3分，共15分。

2021届河南省新郑市第一中学高三语文月考试卷及答案解析一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

将军和士兵聂鑫森胡家村村长胡大尊，做梦都没有想到，县长宋公义，会在这个冬天的早晨，出现在龙虎关！当他和村民走进望楼的休息室时，浓眉大眼的宋公义迎上来打招呼：“大尊，早啊。

”“宋县长，你什么时候来的？”“我昨晚十点叫关，然后和胡四哥一起值班、聊天，累了就在木炭火边打了个盹。

”胡子巴叉、右脚有点跛的胡四，惊得张大了嘴巴，问：“你是宋县长？怠慢了，怠慢了。

你说你是路过这里，借个宿。

”“宋县长，吃过早饭了？”大尊问。

“胡四哥煮的粥，还有两个喷香的烤红薯，好吃。

”“第一批游客，要九点后才会到达。

先喝茶，歇歇憩，过下子请宋县长现场指导。

”宋公义淡淡一笑。

这个瑟缩在湘黔边界的胡家村，几个月来忽然热闹起来了。

就因为在这块地界上，老祖宗留下了一座古城堡，名叫龙虎关，县里拨下了专项扶贫款，把龙虎关修旧如旧，又修好公路，再经宣传，这里立即成了一个旅游热点。

胡家村的村民，祖祖辈辈靠种包谷为活，莽莽苍苍的大山，当然也产茶叶、野果、蔬菜，但交通不便，怎么往外运？换不来现钱啊。

于是，穷，且穷得很冷清。

龙虎关离胡家村不过三里地，左边是青龙山，右边是白虎山，两山之间是商旅的通道。

大概在清康熙年间便在这里设卡筑关，一是为防止边民作乱，二是为保证边贸的税收。

龙虎关的城墙都是粗犷的麻石砌成，城高且厚，城墙上有望楼、烽火台、行道、石级。

城垛与城垛，依次排列，像一个个的“凹”字。

村民万万没想到，这玩意城里人觉得新鲜，更没想到要花钱买票才能看；看了龙虎关，还要买他们地摊上摆着的茶叶、野果、蔬菜、腊肉、腊鱼，说这是百分之百的生态食品。

有古代的龙虎关，就不能没有守关的将军和兵卒。

县里的旅游局，为龙虎关免费捐赠了仿制的古代军装和兵器。

将军的装束最显眼，头盔、甲胄、护心镜、宝剑，威风凛凛。

兵卒军装的前胸后背，都印着一个粗黑的“兵”字，一手拿藤制的盾牌，一手握长矛或是大刀。

2021届河南省新郑市第一中学高三英语第四次联考试卷及答案解析第一部分阅读（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A、B、C、D四个选项中选出最佳选项AYou’re becoming an adult; your friends are changing; school is more challenging; and your life has more ups and downs than it used to. These books are just for you.Say Goodbye to Stressby Dr Jeff BrownKindle Edition $11.79Paperback $16.95Many have trouble getting their stress (压力) under control and want help. This new book will encourage stressed-out readers with its stories from people like them about how they resolved or rethought the stress in their lives, learned to let go of anxiety and worry, and improved their lives by dealing with stress.Find Your Inner Strengthby Amy NewmarkKindle Edition $7.99Paperback $12.75This powerful collection of stories will inspire (激励) you and help you find the inner strength to do with the challenges in your own life. We are stronger than we think.... when we have to be. These brave. courageous people are the role models that show us all what is possible.Random Acts of Kindnessby Amy NewmarkKindle Edition $12.99Paperback $17.77Make miracles happen for yourself and others. It’s easy. Just think outside the box and look around. There are so many ways that you can help—and it turns out the biggest beneficiary (受益人) may be you! Scientific studies have shown that “doing good” is not only good for others but also for the person doing it, making that person happier and healthier.Be the Best You Can Beby Amy NewmarkKindle Edition $10. 99Paperback $15.67This collection shows kids positive role models to follow in its stories about making good choices, havingconfidence, and doing the right things. Parents and grandparents will enjoy discussing the stories with children, making it a family event.1.How will you feel after reading Say Goodbye to Stress?A.Anxious.B.Awkward.C.Relaxed.D.Confused.2.What is unique about Find Your Inner Strength?A.It is written by a well-known author.B.It is the cheapest of the four books.C.It has role models for kids to follow.D.It shows one how to do good deeds.3.Which book is suitable for one who has no confidence?A.Say Goodbye to StressB.Find Your Inner StrengthC.Random Acts of KindnessD.Be the Best You Can BeBWhile the arts can' t stop the COVID-19 virus or the social unrest we see in the world today, they can give us insight into the choices we make when moving through crises and chaos. The arts invite everyone to think in new ways.We often experience works of art as something that's pleasing to our senses without a full understanding of the creative effort. Great art often shows us contradictions and crises, and we can learn a great deal from their resolutions(解决). Through our understanding of art, we can gain a deeper understanding of how we might overcome our own challenges. In understanding extremes of contrast, we can see the beauty in art with themes that are not simply pleasing for their magnificent features or qualities.Beethoven offers a wonderful example of moving artfully through crises and chaos. He composed his Symphony No. 9 as his hearing loss became more and more pronounced. The opening of the symphony seems to come out of nowhere, from near silence in the opening to a full expression of what many consider to be the joy of freedom and universal brotherhood with Schiller’s Ode to joy（欢乐颂). Beethoven appears to have created a work of art that not only freed him from his personal struggles, but one that also speaks to the joy of living together in peace and harmony.Have a dialogue between the two opposing parts and you will find that they always start out fighting each other until we come to an appreciation of difference—a oneness of the two opposingforces.The arts offer many lessons that can help us gain the knowledge we need to move more confidently in today’ s competitive anduncertain environment. An openness to arts-based solutions will give you more control over your future.4. What value does art have beyond pleasing people's senses?A. It brings people inner peace.B. It contributes to problem-solving.C. It reduces the possibility of crises.D. It deepens understanding of music.5. What can we learn about Beethoven's Symphony No. 9?A. It celebrates freedom and unity.B. It aims to show crises and chaos.C. It opens with Schiller's Ode to Joy.D. It is unfinished due to his hearing loss.6. What is the author's suggestion on dealing with conflicting forces?A. Leaving things as they are.B. Making a choice between them.C. Separating them from each other.D. Engaging them in a conversation.7. Which of the following can be the best title for the text?A. How COVID-19 changes artB. Essentials of Symphony No. 9C. Moving artfully through crisesD. Joy in the eyes of BeethovenCThose who are concerned that robots are taking over the world can rest easy—for now. Though the androids have proved useful at performing ordinary tasks, they are not ready for the greatest time. At least that appears to be the case atJapan’s Henn-na Hotel chain where over half of the robot staff are being replaced by humans.The first location of the unique hotel opened in July 2015 was atNagasaki’s Huis Ten Bosch Theme Park. The hotel’s owner, Hideo Sawada, promised the hotel to be managed primarily by robots. Guests were greeted and checked-in by a dinosaur robot, while a cute android called Churi, placed inside each room, provided information about attractions. Not surprisingly, the lodging, recognized in 2016 as the world’s first robot-staffed hotel by Guinness World Records, drew in curious visitors from all around the world.But as the years have passed, the hotel’s main draw is becoming less novel and more unsatisfactory. Also as the robots are “aging”, they are costing more to repair. Among the 283 androids being replaced are the chain’s two dinosaur receptionists. In addition to scaring young guests, they are also unable to photocopy guests’ passports, forcing human employees to step in each time. Also out are the cute Churi robots, which annoyed guests by interrupting their conversations. For example, one guest told The Wall Street Journal that Churi mistook his snoring for a command and kept asking him to repeat his request all night.Sawada told The Wall Street Journal, “When you actually use robots you realize there are places where they aren’t needed—or just annoy people.” While Sawada may be cutting back on his use of androids, the recently-opened Smart LYZ Hotel and the Fly Zoo Hotel inChina, are run entirely by robots, with not a human in sight. Whether the employees have more competence than those “hired” by the Henn-na Hotel chain remains to be seen.8. What makesJapan’s Henn-na Hotel unique?A. Its robot employees.B. Its advanced equipment.C. Its convenient location.D. Its successful management.9. What is the author’s purpose with the example in paragraph 3?A. To entertain readers.B. To prove Churi’s drawback.C. To introduce Churi’s functions.D. To persuade people not to book the hotel.10. What does the owner ofJapan’s Henn-na Hotel think of his robot staff now?A. Attractive.B. Costly.C. Pioneering.D. Disappointing.11. What is the best title for the text?A. Robots Are Taking Over the World.B. The Boom of Robots-staffed Hotel.C. Robot Staff Are Fired For No Competence.D. The First Robots-staffed Hotel Won Guinness World Record.DTOKYO—Japanese Prime Minister Yoshihide Suga told the media on Monday if any places hosting events of the Tokyo Olympics and Paralympics declare a state of emergency due to the COVID-19 epidemic during the games, the events will continue to beheld but without spectators (观众). With one month to go before the games are due to begin on July 23, Suga is again showing his administration’s determination to hold the Olympic Games as planned, despite so much pressure from various parties urging it tocancel the event.Although the Japanese government regards the Tokyo Olympics as an important opportunity to improve its soft power, the Japanese people’s enthusiasm for the Games has been continuously dented (挫伤) since they were postponed last year. The resurgence (再猖獗) of the novel coronavirus in some places is Japan in recent months has cast a shadow over people’s confidence that the Olympics will not give rise to new clusters (群) of infections, and there are fears that the Games will provide new channels for the virus’ global transmission.Some torchbearers from Japan have withdrawn from the Olympic torch relay in the country. And the latest survey indicates only 34 percent of Japanese people support holding the games as scheduled. Predictably, the Suga administration will do all it can to try to ensure the games go ahead. But it remains to be seen whether it can stand the tests of the uncertainties related to epidemic prevention and control that might happen during the Games.Since it has not yet got the virus under control at home, the people have reasons to question is ability to deal with the prevention and control work when large numbers of participants will be flocking to Japan from around the world in a short time. It is to be hoped that Japan can draw lessons from the organization of epidemic prevention and control work during the ongoing UEFA European Championship, carry out strict epidemic prevention and control measures, and be prepared for emergencies to guarantee the safety and success of the Olympics at this special time.It should be a common wish of the whole world that the Tokyo Olympics can become a stage showing unity and resolve of human beings in their fight against the virus. That willendowthe games with special meaning beyond sports.12. What is the second paragraph mainly about?A. The virus’ global transmission.B. People’ worry about the infections.C. The resurgence of the novel coronavirus.D. The benefit of holding the Tokyo Olympics.13. How do about one third of Japanese people like holding the games as planned?A. Uncertain.B. Negative.C. Approving.D. Indifferent.14. Which of the following words can replace the underlined word “endow” in the last paragraph?A. Compare.B. Equip.C. Provide.D. Charge.15. What can be the best title for the news report?A. Japan can ensure Olympics go aheadB. Olympics big test for Japanese governmentC. Japanese people’s enthusiasm for the GamesD. Japan to carry out strict epidemic prevention during the Games第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。