四川省成都外国语学校2019-2020学年高一化学下学期期中试题[含答案]

合集下载

2024学年四川省成都外国语学校高一上学期期中考数学试题及答案

2024学年四川省成都外国语学校高一上学期期中考数学试题及答案

成都外国语学校2023-2024学年度上期半期考试高一数学试卷注意事项:1.本试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.2.本堂考试120分钟,满分150分;3.答题前,考生务必先将自己的姓名、学号填写在答题卡上,并使用2B 铅笔填涂.4.考试结束后,将答题卡交回.第Ⅰ卷 选择题部分,共60分一、单选题:本题共8个小题,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. 设集合{}1,2,3,4,5,6U =,{}1,3,6A =,{}2,3,4B =,则A B = ( )A. 3 B. {}1,3 C. {}3 D. {}2,32. 命题“3x ∃≥,2230x x -+<”的否定是( )A. 3x ∀≥,2230x x -+< B. 3x ∀≥,2230x x -+≥C. 3x ∀<,2230x x -+≥ D. 3x ∃<,2230x x -+≥3. 函数()f x =)A [)1,+∞ B. ()1,+∞C. [)1,2 D. [)()1,22,⋃+∞4. “1k >-”是“函数3y kx =+在R 上为增函数”的( )A. 充分不必要条件B. 必要不充分条件C. 充要条件D. 既不充分又不必要条件5. 若,,R,0a b c c ∈>且0a b >>,下列不等式一定成立的是( )A ac bc < B.11a b < C. a c b c -<- D. 11b b a a +>+6. 函数()2605y x x x =-+≤≤的值域是( )A. []0,5B. []0,9C. []5,9D. [)0,∞+..7. 函数()21x f x x -=的大致图象为( )A. B.C. D.8. 若函数()f x 是定义在R 上的偶函数,在区间(],0-∞上是减函数,且()10f =,则不等式()10f x x+≥的解集为( )A. [)2,-+∞ B. [)()2,00,-⋃+∞ C. [)0,∞+ D. [)(]2,00,2-U 二、多选题:本题共4个小题,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. 下列数学符号使用正确的是( )A. 1N-Ï B. {}1Z ⊆C. 0∈∅ D. ∅ {}010. 下列各选项给出两个函数中,表示相同函数的有( )A. ()1f x x =+与()0g x x x=+B. ()f x x =与()g x =C. ()f x x =与()2x g x x=D. ()f t t =与()g x x =11. 设正实数m n 、满足2m n +=,则( )A.12m n +的最小值为B.最小值为2C.的最大值为1的的D. 22m n +的最小值为212. 已知定义在R 的函数()f x 满足以下条件:(1)对任意实数,x y 恒有()()()()()f x y f x f y f x f y +=++;(2)当0x >时,()f x 的值域是()0,∞+(3)()11f =则下列说法正确的是( )A. ()f x 值域为[)1,-+∞B. ()f x 单调递增C. ()8255f =D. ()()()31f x f f x f x -⎡⎤≥⎣⎦+的解集为[)1,+∞第Ⅱ卷 非选择题部分,共90分三、填空题:本题共4小题,每小题5分,共20分.13. 已知集合{}{}21,,A B a a ==,且A B A = ,则a 的值为_________.14. 设函数()4,0,2,0,3x x x f x x x x ⎧-<⎪⎪=⎨⎪≥⎪+⎩则()()1f f -=__________.15. 一元二次不等式23280x x -++≤的解集为________.16. 设函数()f x 的定义域为D ,若存在实数()0T T >,使得对于任意x D ∈,都有()()f x f x T <+,则称()f x 为“T -严格增函数”,对于“T -严格增函数”,有以下四个结论:①“T -严格增函数”()f x 一定在D 上严格增;②“T -严格增函数”()f x 一定是“nT -严格增函数”(其中*N n ∈,且2n ≥)③函数()[]f x x =是“T -严格增函数”(其中[]x 表示不大于x 的最大整数)④函数()[]f x x x =-不是“T -严格增函数”(其中[]x 表示不大于x 的最大整数)其中,所有正确的结论序号是______.四、解答题:本题共6个小题,共70分.解答应写出文字说明,证明过程或演算步骤.17. 已知集合U =R ,集合{}23A x x =-≤≤,{1B x x =<-或}4x >(1)求A B ⋃;(2)求()U A B ∩ð18. 已知函数()b f x x x =+过点(1,2).(1)判断()f x 在区间(1,)+∞上的单调性,并用定义证明;(2)求函数()f x 在[]2,7上的最大值和最小值.19. (1)已知函数()212f x x =+,则()f x 的值域;(2)已知1)f x +=+,求()f x 的解析式;(3)已知函数()f x 对于任意的x 都有()2()32f x f x x +-=-,求()f x 的解析式.20. 已知关于x 的不等式230x bx c ++-<的解集为()1,2-.(1)当[]0,3x ∈时,求2x bx c x++的最小值;(2)当x ∈R 时,函数2y x bx c =++的图象恒在直线2y x m =+的上方,求实数m 的取值范围.21. 已知函数()21ax b f x x-=+是定义在[]1,1-上的奇函数,且()11f =-.(1)求函数()f x 的解析式;(2)判断()f x 在[]1,1-上单调性,并用单调性定义证明;(3)解不等式()()()210f t f t f -+>.22. 若函数()f x 在[],x a b ∈时,函数值y 的取值区间恰为11,b a⎡⎤⎢⎥⎣⎦,就称区间[],a b 为()f x 的一个“倒域区间”.已知定义在[]22-,上的奇函数()g x ,当[]0,2x ∈时,()22g x x x =-+.(1)求()g x 的解析式;(2)求函数()g x 在[]1,2内的“倒域区间”;(3)求函数()g x 在定义域内的所有“倒域区间”.的成都外国语学校2023-2024学年度上期半期考试高一数学试卷注意事项:1.本试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.2.本堂考试120分钟,满分150分;3.答题前,考生务必先将自己的姓名、学号填写在答题卡上,并使用2B 铅笔填涂.4.考试结束后,将答题卡交回.第Ⅰ卷 选择题部分,共60分一、单选题:本题共8个小题,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. 设集合{}1,2,3,4,5,6U =,{}1,3,6A =,{}2,3,4B =,则A B = ( )A. 3B. {}1,3C. {}3D. {}2,3【答案】C【解析】【分析】利用交集的运算求解即可.【详解】由题知,{}3A B ⋂=.故选:C2. 命题“3x ∃≥,2230x x -+<”的否定是( )A. 3x ∀≥,2230x x -+< B. 3x ∀≥,2230x x -+≥C. 3x ∀<,2230x x -+≥ D. 3x ∃<,2230x x -+≥【答案】B【解析】【分析】利用含有一个量词的命题的否定规律“改量词,否结论”分析判断即可得解.【详解】解:因为命题“3x ∃≥,2230x x -+<”为存在量词命题,所以其否定为“3x ∀≥,2230x x -+≥”.故选:B .3. 函数()f x = )A. [)1,+∞ B. ()1,+∞C. [)1,2 D. [)()1,22,⋃+∞【答案】D【解析】【分析】根据开偶数次发根号里的数大于等于零,分母不等于零计算即可.【详解】由()f x =得1020x x -≥⎧⎨-≠⎩,解得1x >且2x ≠,所以函数()f x =[)()1,22,⋃+∞.故选:D.4. “1k >-”是“函数3y kx =+在R 上为增函数”的( )A 充分不必要条件 B. 必要不充分条件C. 充要条件D. 既不充分又不必要条件【答案】B【解析】【分析】根据一次函数的性质与必要不充分条件的判定即可得到答案.【详解】当12k =-时,满足1k >-,但是函数3y kx =+在R 上为减函数,则正推无法推出;反之,若函数3y kx =+在R 上为增函数,则01k >>-,则反向可以推出,则“1k >-”是“函数3y kx =+在R 上为增函数”的必要不充分条件,故选:B .5. 若,,R,0a b c c ∈>且0a b >>,下列不等式一定成立的是( )A. ac bc< B. 11a b < C. a c b c -<- D. 11b b a a +>+【答案】B【解析】【分析】ACD 举反例确定错误,B 作差法可判断..【详解】A ,2,1a c b ===时,2212⋅>⋅,A 错误;B ,11110,0,b a a b a b ab a b->>∴-=<∴< ,B 正确;C ,2,1a c b ===时,2212->-,C 错误;D ,2,1a c b ===时,111221+<+,D 错误.故选:B6. 函数()2605y x x x =-+≤≤的值域是( )A. []0,5 B. []0,9 C. []5,9 D. [)0,∞+【答案】B【解析】【分析】根据二次函数的性质即可求解.【详解】函数26y x x =-+的图象是一条开口向下的抛物线,对称轴为3x =,所以该函数在(0,3)上单调递增,在(3,5)上单调递减,所以max 39x y y ===,又050,5x x y y ====,所以min 0y =,即函数的值域为[0,9].故选:B.7. 函数()21x f x x -=的大致图象为( )A. B.C. D.【答案】D【解析】【详解】根据函数的奇偶性以及函数的解析式判断出正确答案.【分析】()21x f x x -=的定义域为{}|0x x ≠,()()()2211x x f x f x x x ----==-=--,所以()f x 是奇函数,图象关于原点对称,所以A 选项错误.当0x >时,()210x f x x -=≥,所以C 选项错误.当0x >时,令()210x f x x -==,解得1x =,所以B 选项错误.所以正确的是D.故选:D 8. 若函数()f x 是定义在R 上的偶函数,在区间(],0-∞上是减函数,且()10f =,则不等式()10f x x+≥的解集为( )A. [)2,-+∞ B. [)()2,00,-⋃+∞ C. [)0,∞+ D. [)(]2,00,2-U 【答案】B【解析】【分析】确定函数的单调性,考虑0x >和0x <两种情况,将问题转化为(1)0f x +≥或(1)0f x +≤,再根据函数值结合函数单调性得到答案.【详解】函数()f x 是定义在实数集R 上的偶函数,()f x 在区间(],0-∞上是严格减函数,故函数()f x 在()0,∞+上单调递增,且(1)(1)0f f -==,当0x >时,由(1)0f x x+≥,即(1)0f x +≥,得到11x +≥或11x +≤-(舍弃),所以0x >,当0x <时,由(1)0f x x +≥,即(1)0f x +≤,得到111x -≤+≤,所以20x -≤<,综上所述,20x -≤<或0x >,故选:B.二、多选题:本题共4个小题,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. 下列数学符号使用正确的是( )A. 1N-Ï B. {}1Z ⊆C. 0∈∅D. ∅ {}0【答案】ABD【解析】【分析】根据集合与元素之间关系符号和集合与集合之间的关系符号来判断即可.【详解】对于A ,N 表示自然数集,1-不是自然数,故1N -Ï成立,则A 选项正确;对于B ,Z 表示整数集,1Z ∈,故{}1Z ⊆成立,则B 选项正确;对于C ,∅表示空集,没有任何一个元素,即0∉∅,故C 选项不正确;对于D ,空集是任何一个非空集合的真子集,故∅ {}0成立,则D 选项正确.故选:ABD.10. 下列各选项给出的两个函数中,表示相同函数的有( )A. ()1f x x =+与()0g x x x=+B. ()f x()g x =C. ()f x x =与()2x g x x=D. ()f t t =与()g x x =【答案】BD【解析】【分析】根据函数的“三要素”一一判断每个选项中的函数,看定义域和对应关系是否相同,即可得答案.【详解】对于A ,函数()1f x x =+的定义域为R ,()0g x x x =+的定义域为{|0}x x ≠,故二者不是相同函数,A 错误;对于B ,()f x x =的定义为域为R ,()||g x x ==的定义域为R ,二者对应关系也相同,值域都为[0,)+∞,故二者表示相同函数,B 正确;对于C ,()f x x =的定义域为R ,()2x g x x=的定义域为{|0}x x ≠,故二者不是相同函数,C 错误;对于D ,()f t t =与()g x x =的的定义域均为(,0]-∞,对应关系相同,的值域均为(,0]-∞,故二者表示相同函数,D 正确;故选:BD11. 设正实数m n 、满足2m n +=,则( )A.12m n +的最小值为B.的最小值为2C. 的最大值为1D. 22m n +的最小值为2【答案】CD【解析】【分析】由已知条件结合基本不等式及其相关变形,分别检验各个选项即可判断正误.【详解】对于选项A ,322121222m n n m m n m n m n ⎛⎫+=++⎛⎫=+ ⎪⎪⎭⎭+⎝⎝32≥+= ,当且仅当=m n 且2m n +=时,即2m =-,4n =-时取等号,则A 错误;对于选项B , 22m n =++=+24m n ≤++=,当且仅当1m n ==2+≤+的最大值为2,则B 错误;对于选项C ,m n +≥212m n mn +⎛⎫≤= ⎪⎝⎭,当且仅当1m n ==时,等号成立,则C 正确;对于选项D , ()222242m n m n mn mn +=+-=-24222m n +⎛⎫≥-= ⎪⎝⎭,当且仅当1m n ==时,等号成立,则D 正确,故选: CD .12. 已知定义在R 的函数()f x 满足以下条件:(1)对任意实数,x y 恒有()()()()()f x y f x f y f x f y +=++;(2)当0x >时,()f x 的值域是()0,∞+(3)()11f =则下列说法正确的是( )A. ()f x 值域为[)1,-+∞B. ()f x 单调递增C. ()8255f =D. ()()()31f x f f x f x -⎡⎤≥⎣⎦+的解集为[)1,+∞【答案】BCD 【解析】【分析】计算()00f =得到()()1111f x f x =-+>--+,A 错误,根据单调性的定义得到B 正确,计算()23f =,()415f =,()8255f =得到C 正确,题目转化为()()2f x f x f ⎡⎤+≥⎣⎦得到()2x f x +≥,根据函数的单调性得到D 正确,得到答案.【详解】对选项A :令1,0x y ==可得()()()()()11001f f f f f =++,故()00f =,令y x =-可得()()()()()0f f x f x f x f x =-++-,()1f x -≠-,()()()()1111f x f x f x f x --==-+-+-+,当0x <时,()0f x ->,则()()1111f x f x =-+>--+,综上所述:()()1,f x ∈-+∞,错误;对选项B :任取12,R x x ∈且12x x >,()120f x x ->,()21f x >-,则()()()()()()()12122212212f x f x f x x x f x f x x f x f x x -=-+-=-+-()()12210f x x f x ⎡⎤=-+>⎣⎦,所以函数()y f x =在R 上单调递增,正确;对选项C :取1x y ==得到()()()()()211113f f f f f =++=;取2x y ==得到()()()()()4222215f f f f f =++=;取4x y ==得到()()()()()84444255f f f f f =++=,正确;对选项D :()()()31f x f f x f x -⎡⎤≥⎣⎦+,()()()13f f x f x f x ⎡⎤⎡⎤+≥-⎣⎦⎣⎦,即()()()()()()2f f x f x f x f f x f x f x f ⎡⎤⎡⎤⎡⎤++=+≥⎣⎦⎣⎦⎣⎦,即()2x f x +≥,函数()()g x x f x =+单调递增,且()1112g =+=,故1x ≥,正确;故选:BCD【点睛】关键点睛:本题考查了抽象函数问题,意在考查学生的计算能力,转化能力和综合应用能力,其中根据题目信息转化得到()()2f x f x f ⎡⎤+≥⎣⎦,再利用函数的单调性解不等式是解题的关键.第Ⅱ卷 非选择题部分,共90分三、填空题:本题共4小题,每小题5分,共20分.13. 已知集合{}{}21,,A B a a ==,且A B A = ,则a 的值为_________.【答案】1-【解析】【分析】由A B A = 得A B ⊆,列式求解,然后检验元素的互异性.【详解】∵A B A = ,∴A B ⊆,又{}{}21,,A B a a ==,∴1a =或21a =,解得1a =或1a =-,当1a =不满足元素的互异性,舍去,所以1a =-.故答案为:1-.14. 设函数()4,0,2,0,3x x xf x x x x ⎧-<⎪⎪=⎨⎪≥⎪+⎩则()()1f f -=__________.【答案】1【解析】【分析】分段函数求值,根据自变量的取值范围代入相应的对应关系.【详解】当=1x -时,()f -=--=-41131,则()()231(3)133ff f ⋅-===+.故答案为:115. 一元二次不等式23280x x -++≤的解集为________.【答案】(][),47,-∞-+∞【解析】【分析】由一元二次不等式的解法进行求解即可.【详解】()()22328032804707x x x x x x x -++≤⇒--≥⇒+-≥⇒≥,或4x ≤-所以一元二次不等式23280x x -++≤的解集为(][),47,-∞-+∞ ,故答案为:(][),47,-∞-+∞ 16. 设函数()f x 的定义域为D ,若存在实数()0T T >,使得对于任意x D ∈,都有()()f x f x T <+,则称()f x 为“T -严格增函数”,对于“T -严格增函数”,有以下四个结论:①“T -严格增函数”()f x 一定在D 上严格增;②“T -严格增函数”()f x 一定是“nT -严格增函数”(其中*N n ∈,且2n ≥)③函数()[]f x x =是“T -严格增函数”(其中[]x 表示不大于x 的最大整数)④函数()[]f x x x =-不是“T -严格增函数”(其中[]x 表示不大于x 的最大整数)其中,所有正确的结论序号是______.【答案】②③④【解析】【分析】根据“T -严格增函数”的定义对四个结论逐一分析,从而确定正确答案.【详解】①,函数(),01,0x x f x x x <⎧=⎨-≥⎩,定义域为R ,存在2T =,对于任意x ∈R ,都有()()2f x f x <+,但()f x 在R 上不单调递增,所以①错误.②,()f x 是“T -严格增函数”,则存在0T >,使得对任意x D ∈,都有()()f x f x T <+,因为2,0n T ≥>,所以()()f x T f x nT +<+,故()()f x f x nT <+,即存在实数0nT >,使得对任意x D ∈,都有()()f x f x nT <+,所以()f x 是“nT -严格增函数”, ②正确.③,()[]f x x =,定义域为R ,当1T =时,对任意的x ∈R ,都有[][]1x x <+,即()()1f x f x <+,所以函数()[]f x x =是“T -严格增函数”.④,对于函数()[]f x x x =-,()[][][]()11111f x x x x x x x f x +=+-+=+--=-=,所以()f x 是周期为1的周期函数,11112222f ⎛⎫⎡⎤=-= ⎪⎢⎥⎝⎭⎣⎦,若1T =,则133********f f ⎛⎫⎡⎤⎛⎫+=-== ⎪ ⎪⎢⎥⎝⎭⎣⎦⎝⎭,不符合题意.当0T >且1T ≠时,若()()f x f x T <+,则[][]x x x T x T -<+-+,即[][]T x T x >+-(*),其中,若01T <<,则总存在,2n n ∈≥*N ,使得1nT >,当1T >时,若T 是正整数,则[][]x T x T +-=,(*)不成立,若T 不是正整数,[][]T x T x >+-不恒成立,所以函数()[]f x x x =-不是“T -严格增函数”.故答案为:②③④【点睛】本题主要考查新定义函数的理解,对于新定义函数的题,解题方法是通过转化法,将“新”转化为“旧”来解题,选择题中,可利用特殊值进行举反例来排除.四、解答题:本题共6个小题,共70分.解答应写出文字说明,证明过程或演算步骤.17. 已知集合U =R ,集合{}23A x x =-≤≤,{1B x x =<-或}4x >(1)求A B ⋃;(2)求()U A B∩ð【答案】(1){3x x ≤或}4x > (2){}13x x -≤≤【解析】【分析】(1)根据并集概念进行计算;(2)先求出{}14U B x x =-≤≤ð,进而利用交集概念进行计算.【小问1详解】{}{|231A B x x x x ⋃=-≤≤⋃<-或}4x >{3x x =≤或}4x >;【小问2详解】{}14U B x x =-≤≤ð,(){}{}{}|231413U A B x x x x x x ⋂=-≤≤⋂-≤≤=-≤≤ð18. 已知函数()bf x x x=+过点(1,2).(1)判断()f x 在区间(1,)+∞上的单调性,并用定义证明;(2)求函数()f x 在[]2,7上的最大值和最小值.【答案】(1)()f x 在区间(1,)+∞上单调递增,证明见解析 (2)最大值507,最小值为52【解析】【分析】(1)求出函数的表达式,利用单调性定义即可判断函数的单调性;(2)根据单调性即可得出函数()f x 在[]2,7上最大值和最小值.【小问1详解】单调递增,由题意证明如下,由函数()b f x x x=+过点(1,2),有121b+=,解得1b =,所以()f x 的解析式为:1()f x x x=+.设12,(1,)x x ∀∈+∞,且12x x <,有()()()()121212121212111x x x x f x f x x x x x x x --⎛⎫⎛⎫-=+-+= ⎪ ⎪⎝⎭⎝⎭.由1212,(1,),x x x x ∈+∞<,得121210,0x x x x ->-<.则()()12121210x x x x x x --<,即()()12f x f x <.∴()f x 在区间(1,)+∞上单调递增.【小问2详解】由()f x 在(1,)+∞上是增函数,所以()f x 在区间[2,7]上的最小值为5(2)2f =,最大值为50(7)7f =.19. (1)已知函数()212f x x =+,则()f x 的值域;为的(2)已知1)f x +=+,求()f x 的解析式;(3)已知函数()f x 对于任意的x 都有()2()32f x f x x +-=-,求()f x 的解析式.【答案】(1)1|02y y ⎧⎫<≤⎨⎬⎩⎭;(2)2()1f x x =-,其中1x ≥;(3)2()33f x x =--【解析】【分析】(1)根据函数的性质即可得函数的值域;(2)配凑法或换元法求函数的解析式(3)列方程组法求函数的解析式【详解】(1)由于220,22x x ≥+≥,故211022x <≤+,故函数的值域为1|02y y ⎧⎫<≤⎨⎬⎩⎭(2))221)1111f =++-=+-,,故所求函数的解析式为2()1f x x =-,其中1x ≥.(3)∵对于任意的x 都有()2()32f x f x x +-=-,∴将x 替换为-x ,得()2()32f x f x x -+=--,联立方程组:()2()32()2()32f x f x x f x f x x +-=-⎧⎨-+=--⎩消去()f x -,可得2()33f x x =--.20. 已知关于x 的不等式230x bx c ++-<的解集为()1,2-.(1)当[]0,3x ∈时,求2x bx cx++的最小值;(2)当x ∈R 时,函数2y x bx c =++的图象恒在直线2y x m =+的上方,求实数m 的取值范围.【答案】(1)1 (2)5,4⎛⎫-∞-⎪⎝⎭【解析】【分析】(1)依题意可得,1-和2是方程230x bx c ++-=的两根,从而可求得b ,c 的值,再利用基本不等式即可求解;(2)依题意可得,已知条件等价于212x x x m -+>+在(),-∞+∞上恒成立,分离参数转化为最值问题即可求解.【小问1详解】因为关于x 的不等式230x bx c ++-<的解集为()1,2-,所以1-和2是方程230x bx c ++-=的两根,所以12123b c -+=-⎧⎨-⨯=-⎩,解得11b c =-⎧⎨=⎩,由2x bx c x++可知,0x ≠,所以当(]0,3x ∈时,2211111x bx c x x x x x x ++-+==+-≥-=,当且仅当1x =时,等号成立,所以2x bx c x ++的最小值为1.【小问2详解】结合(1)可得221y x bx c x x =++=-+,对于R x ∀∈,函数2y x bx c =++的图象恒在函数2y x m =+的图象的上方,等价于212x x x m -+>+在(),x ∈-∞+∞上恒成立,即231m x x <-+在(),x ∈-∞+∞上恒成立,则()2min31m x x <-+即可,因为2235531()244x x x -+=--≥-,所以54m <-,所以实数m 的取值范围为5,4⎛⎫-∞- ⎪⎝⎭.21. 已知函数()21ax bf x x -=+是定义在[]1,1-上的奇函数,且()11f =-.(1)求函数()f x 的解析式;(2)判断()f x 在[]1,1-上的单调性,并用单调性定义证明;(3)解不等式()()()210f t f t f -+>.【答案】21. ()221xf x x -=+,[]1,1x ∈- 22. 减函数;证明见解析;23. ⎡⎢⎣【解析】【分析】(1)根据奇函数的性质和()11f =求解即可.(2)利用函数单调性定义证明即可.(3)首先将题意转化为解不等式()()21f t f t >-,再结合()f x 的单调性求解即可.【小问1详解】函数()21ax bf x x-=+是定义在[]1,1-上的奇函数,()()f x f x -=-;2211ax b ax bx x---=-++,解得0b =,∴()21axf x x =+,而()11f =-,解得2a =-,∴()221xf x x-=+,[]1,1x ∈-.小问2详解】函数()221xf x x-=+在[]1,1-上为减函数;证明如下:任意[]12,1,1x x ∈-且12x x <,则()()()()()()121212122222121221221111x x x x x x f x f x x x x x ------=-=++++因为12x x <,所以120x x -<,又因为[]12,1,1x x ∈-,所以1210x x ->,所以()()120f x f x ->,即()()12f x f x >,所以函数()()12f x f x >在[]1,1-上为减函数.【小问3详解】由题意,()()()210f t f t f -+>,又()00f =,所以()()210f t f t -+>,即解不等式()()21f tf t >--,所以()()21f t f t >-,所以22111111t t t t⎧-≤≤⎪-≤-≤⎨⎪<-⎩,解得0t ≤<,所以该不等式的解集为⎡⎢⎣.22. 若函数()f x 在[],x a b ∈时,函数值y 的取值区间恰为11,b a⎡⎤⎢⎥⎣⎦,就称区间[],a b 为()f x 的一个“倒【域区间”.已知定义在[]22-,上的奇函数()g x ,当[]0,2x ∈时,()22g x x x =-+.(1)求()g x 的解析式;(2)求函数()g x 在[]1,2内的“倒域区间”;(3)求函数()g x 在定义域内的所有“倒域区间”.【答案】(1)()222,022,20x x x g x x x x ⎧-+≤≤=⎨+-≤<⎩(2)⎡⎢⎣(3)⎡⎢⎣和1⎤-⎥⎦【解析】【分析】(1)设[)2,0x ∈-,利用奇函数的定义可求得函数()g x 在[)2,0-上的解析式,由此可得出函数()g x 在[]22-,上的解析式;(2)设12a b ≤<≤,分析函数()g x 在[]1,2上的单调性,可出关于a 、b 的方程组,解之即可;(3)分析可知0a bab <⎧⎨>⎩,只需讨论02a b <<≤或20a b -≤<<,分析二次函数()g x 的单调性,根据题中定义可得出关于实数a 、b 的等式组,求出a 、b 的值,即可得出结果.【小问1详解】解:当[)2,0x ∈-时,则(]0,2x -∈,由奇函数的定义可得()()()()2222x g x g x x x x ⎡⎤=--=---=⎣⎦++-,所以,()222,022,20x x x g x x x x ⎧-+≤≤=⎨+-≤<⎩.【小问2详解】解:设12a b ≤<≤,因为函数()g x 在[]1,2上递减,且()g x 在[],a b 上的值域为11,b a⎡⎤⎢⎥⎣⎦,所以,()()22121212g b b b bg a a a a a b ⎧=-+=⎪⎪⎪=-+=⎨⎪≤<≤⎪⎪⎩,解得1a b =⎧⎪⎨=⎪⎩,所以,函数()g x 在[]1,2内的“倒域区间”为⎡⎢⎣.【小问3详解】解:()g x 在[],a b 时,函数值()g x 的取值区间恰为11,b a⎡⎤⎢⎥⎣⎦,其中a b ¹且0a ≠,0b ≠,所以,11a bb a<⎧⎪⎨<⎪⎩,则0a b ab <⎧⎨>⎩,只考虑02a b <<≤或20a b -≤<<,①当02a b <<≤时,因为函数()g x 在[]0,1上单调递增,在[]1,2上单调递减,故当[]0,2x ∈时,()()max 11g x g ==,则11a≤,所以,12a ≤<,所以,12a b ≤<≤,由(2)知()g x 在[]1,2内的“倒域区间”为⎡⎢⎣;②当20a b -≤<<时,()g x 在[]2,1--上单调递减,在[]1,0-上单调递增,故当[]2,0x ∈-时,()()min 11g x g =-=-,所以,11b≥-,所以,21b -<≤-.21a b ∴-≤<≤-,因为()g x 在[]2,1--上单调递减,则()()22121221g a a a ag b b b b a b ⎧=+=⎪⎪⎪=+=⎨⎪-≤<≤-⎪⎪⎩,解得1a b ⎧=⎪⎨⎪=-⎩,所以,()g x 在[]2,1--内的“倒域区间”为1⎤-⎥⎦.综上所述,函数()g x 在定义域内的“倒域区间”为⎡⎢⎣和1⎤-⎥⎦.【点睛】关键点点睛:本题考查函数的新定义,解题的关键在于分析函数的单调性,结合题意得出关于参数的方程,进行求解即可.。

2022-2023学年四川省成都外国语学校高一下学期期中考试英语试题

2022-2023学年四川省成都外国语学校高一下学期期中考试英语试题

2022-2023学年四川省成都外国语学校高一下学期期中考试英语试题1. Artificial intelligence (Al) is practically everywhere today. There are so many products out there which use Al. Some are being developed, some are already in use, and some failed and are being improved, so it’s very difficult to name a few of them and regard them as the best.ViIt is an Al personal trainer which is mainly concerned with fitness and coaching. It, however, requires the use of bio-sensing earphones and other fitness tracking equipment! It can play your favourite music while you work out and all you have to worry about is the exercise you're doing.Deep TextDo you ever wonder how an ad appears suddenly just when you are looking for something similar? This is because of Deep Text. It uses real-time consumer information to produce data which in turn is used to target consumers. Thus, if you search online for flight tickets from Bangalore to Delhi, it is very likely that an ad relating to hotels in Delhi will soon follow.Hello EggIf you live alone and miss your mother because you always miss your breakfast or don’t know what to eat for dinner, then Hello Egg is exactly what you are looking for. A very healthy choice of the 2-minute noodles and oats, Hello Egg provides you with a detailed weekly meal plan about the needs of your body. It is truly a modern AI-powered home cooking tool for the young.WordsmithYou can put Mr. Smith into your Microsoft Excel using their free API, and let it write up detailed analysis of the stories behind your numbers. It can produce detailed reports on thousands of pages of spreadsheets in seconds.1. What can we learn about Vi from the text?A.It is an AI music player.B.It is a bio-sensing earphone.C.It doesn't work without bio-sensing earphones.D.It can make you more energetic while you work out.2. Which can help you improve cooking skill?A.Deep Text. B.Vi.C.Wordsmith. D.Hello Egg.3. What can Wordsmith do for us?A.Produce a detailed report. B.Provide us with a detailed meal plan.C.Book a ticket ahead of time. D.Offer us information on hotels fortraveling.2. I began to grow up that winter night when my parents and I were returning from my aunt’s house, and my mother said that we might soon be leaving for America. We were on the bus then. I was crying, and some people on the bus were turning around to look at me. I remember that I could not bear the thought of never hearing again the radio program for school children to which I listened every morning.I do not remember myself crying for this reason again. In fact, I think I cried very little when I was saying goodbye to my friends a nd relatives. When we were leaving I thought about all the places I was going to see—the strange and magical places I had known only from books and pictures. The country I was leaving and never to come back was hardly in my head then.The four years that followed taught me the importance of optimism(乐观), but the idea did not come to me at once. For the first two years in New York I was really lost—having to study in three schools as a result of family moves. I did not quite know what I was or what I should be. Mother remarried, and things became even more complex for me. Some time passed before my stepfather and I got used to each other. I was often sad, and saw no end to “the hard times.”My responsibilities in the family increased a lot since I knew English better than everyone else at home. I wrote letters, filled out forms, translated at interviews with Immigration officers(移民局官员), took my grandparents to the doctor and translated there, and even discussed telephone bills with company representatives.From my experiences I have learned one important rule: Almost all common troubles go away at last! Something good is certain to happen in the end when you do not give up, and just wait a little! I believe that my life will turn out all right, even though it will not be that easy.1. How did the author get to know America?A.From radio programs B.From books and picturesC.From her mother D.From her relatives2. Upon leaving for America the author felt __________.A.excited B.confusedC.worried D.amazed3. For the first two years in New York, the author __________.A.often lost her wayB.did not think about her futureC.studied in three different schoolsD.got on well with her stepfather4. What can we learn about the author from Paragraph 4?A.She worked as a translator.B.She attended a lot of job interviews.C.She paid telephone bills for her family.D.She helped her family with her English.3. High school often serves to prepare students for a college education, or it might be the last stop on the educational journey for some people. No matter which path one may choose, high school life offers students the opportunities to achieve a great deal of learning, get to know themselves, or simply have fun.In America, high school is a special experience with a culture of its own. Public high schools exist in all communities(社区). Every community in the United States has a public school system, teaching children at primary and secondary levels. These schools are free and supported mainly by taxes, with additional aid from states and governments. Private high schools are a choice for parents who want to give their child a better education. They are often very expensive. Private schools are usually famous for their academic reputation (学术声望). This academic reputation is related to college preparation.In high school, students study various academic subjects such as English, literature, foreign languages, maths, science, history, geography and so on. Some large high schools have classes that teach job-related subjects like auto mechanics, cooking and others, in order to provide students with skills needed for certa in jobs. High school might also offer “honours classes” for particularly smart and sharp students.Some states give basic skills exams during the last year of the high school process to make sure that certain academic standards are met, but, again, high school curricula (课程) vary greatly among communities.High school sports, particularly basketball and football, are often organised. Inter-city and cross-town competitions are often held. These competitions are popular, especially in communities that do not have their own professional sports teams. Nearly all high school teams have nicknames: the Knights and the Tigers being some famous examples.1. Which of the following is not the supporting source of public school?A.Taxes. B.States.C.Community. D.Governments.2. Which of the following statements is true about private schools in the United States?A.There is a private school system in every community.B.Private schools can usually provide better college preparation.C.All private high schools in the United States are very expensive.D.Private schools are a better choice for parents than public schools.3. What does the author say about American high schools?A.Sports competitions are only in the city or town.B.Some states set academic standards in the first year.C.Some schools have special classes for smart students.D.All communities have their own professional sports teams.4. What might be the best title for the text?A.Life in the USAB.Education in the USAC.High Schools in the USAD.Public Schools and Private Schools in the USA4. Hundreds of friends on Facebook can’t replace a handful of close friends in real life, a study has found. In a recent study, researchers discovered that people with only a few friends were at least as happy as those with far more if many of theirs were online.Social media, the researchers said, has encouraged younger people to have larger but more impersonal networks of “friends”. But instead of trying to amass friends, they added, a better cure for lo neliness might be spending time with those you’re closest to.Scientists from the University of Leeds did their study using data from two online surveys on 1,496 people by a non-profit research organization. People taking part in the study showed their ages, the make-up of their social networks, how often they had different types of social interactions, and their own feelings of well-being. They included details of how often and how they interacted (交流) with families or neighbors, and whether they included people who provided services to them in their networks.The number of close friends someone had appeared to be the only thing which influenced how satisfied they were with their social life.“Loneliness has less to do with the number of friends you have, a nd more to do with how you feel about your friends,” said Dr Wändi Bruine de Bruin. Actually, it's often the younger adults who admit to having a bad opinion of their friends.If you feel lonely, it may be more helpful to make a positive connection with a friend than to try and seek out new people to meet.1. Which of the following can replace the underlined word “amass” in Paragraph 2?A.Play the role of. B.Think highly of.C.Take advantage of. D.Increase the number of.2. What can be learned from the scientists’ study?A.Scientists did one online survey on 1,496 people.B.People surveyed mentioned new forms of social networks.C.People surveyed showed their ways to interact with families.D.The number of friends was the only cause of people’s happin ess.3. What makes people happy in social life?A.Living alone.B.Having close friends.C.Having good appearance.D.Making as many friends as possible.4. What’s the author’s purpose in writing the text?A.To advise us to make friends online.B.To show us the importance of friends.C.To present us with the findings of a study.D.To tell us the problems causes by loneliness.5. Travelling is a very enjoyable experience as it provides an opportunity to see new things. 1 The following article discusses the advantages of traveling. Traveling gives you the opportunity to disconnect from your regular life. People all have crazy schedules, work and a family to take care of, and traveling alone or with some friends can give you distance and perhaps even make you realize how important these people are to you. 2Another great benefit is the relaxation you get. 3 When you come back you feel energetic and you are happy to be back to your daily routine. 4 They will create a bond that nothing can erase no matter what happens to the friendship/relationship. You can create photo albums and when you feel nostalgic (对往事怀恋的) you can experience the trips again by looking at your pictures.It’s never been this cheap to travel. With increasing oil prices the cheap travel era might be coming to an end. 5With the Internet and all the new technology, you can plan your trip exactly the way you want it. You can choose your budget, the duration of the trip and what you want to do.6. A video about a homeless man went viral. Johal, a video-maker known for his social ________, decided to give a homeless man some money and ________ his reaction.In the video, approaching a ________ homeless man, Johal carefully places some money on his bag. Minutes later, when the man wakes up, he’s surprised to see the cash and looks excited and grateful. The man walks to a store, steps out with a bag of ________, probably to face the cold nights, and returns to his bench.Johal sits beside the man, talking on the phone with a friend, asking if he can borrow some money for his daughter. The homeless man ________ listens to Johal. After he hangs up, the man asks if he’s okay. Johal explains he’s getting his daughter some ________ while having a hard time financially.In a(n) ________ turn of events, the man asks Johal to ________ his bag and leaves with the shopping bag. He returns ________ after returning the items and offers the money to Johal, who is a total ________, comforting he can go by without a blanket but the girl needs the treatment. Touched with the homeless man’s kindness, Johal ________ the experiment, offering $500 to the man, who instantly starts crying in ________.________ but true, the result proves sometimes, those with less are the ones ________ most. Not all heroes wear capes. In fact, heroes are the ones with a ________.1.A.service B.skill C.activities D.experiments2.A.filmed B.predicted C.developed D.commented3.A.hungry B.sleeping C.freezing D.depressed4.A.blankets B.food C.clothes D.candles5.A.casually B.respectfully C.patiently D.attentively6.A.gifts B.toys C.medicines D.meals7.A.lucky B.unexpected C.rough D.awkward8.A.watch for B.pack up C.tie up D.look for9.A.hard-headed B.heavy-footed C.empty-handed D.low-spirited 10.A.actor B.parent C.failure D.stranger11.A.continues B.explains C.conducts D.analyzes12.A.disbelief B.relief C.sorrow D.anger13.A.Heartbroken B.Imaginative C.Incredible D.Artificial14.A.losing B.asking C.owning D.giving15.A.will B.heart C.thought D.cause7. The writer was so a________ in his writing that his cigar burned his fingers. (根据首字母单词拼写)8. The o________ sculpture was destroyed in the war, so yours must be a fake. (根据首字母单词拼写)9. When books are a________ to movies, certain changes are often necessary in order to make it more visually appealing. (根据首字母单词拼写)10. It was the c________ and selflessness of the brave firefighters that touched us deeply. (根据首字母单词拼写)11. Lauran Bath, whose p________ job was a chef, uses her photography to support environmental protection as a professional photo blogger. (根据首字母单词拼写)12. When visiting foreign countries, you should be a________ of cultural differences and treat them with respect. (根据首字母单词拼写)13. In order to reduce carbon emissions, the government introduced electric v________ as public transport. (根据首字母单词拼写)14. The artist combined western and eastern singing t________ in the same song which was creative and incredible. (根据首字母单词拼写)15. The l________ lining the route is changing dramatically and passengers are drawn to the beautiful scenery outside the train window. (根据首字母单词拼写)16. If you’re on a tight b________, you could consider borrowing one from library when you needan expensive textbook. (根据首字母单词拼写)17. The well is ________ deep that kids are warned against playing beside it. (用适当的词填空)18. I’ve checked the comments ________ this restaurant and it’s highly rated. (用适当的词填空)19. I’m used to ________ (plan) every trip in details. I don’t just leave on a whim. (所给词的适当形式填空)20. Chinese Spring Gala is an entertainment that appeals ________ people of all ages. (用适当的词填空)21. The new student is quite ________ (determine) to catch up with others as soon as possible. (所给词的适当形式填空)22. Please inform the ________(contest) to gather in the auditorium at 9:00 for the competition.(所给词的适当形式填空)23. The ________(run) water in the river sounded soothing to my ears. (所给词的适当形式填空)24. Summer is around the corner, and I can’t wait ________ (get) a suntan and start swimmin g.(所给词的适当形式填空)25. ________ (see) all the films and read the books many times, I am quite familiar with the story of Harry Potter and its characters. (所给词的适当形式填空)26. The temperature dropped ________(dramatic) when the storm hit, causing the streets to floodand trees to swing violently in the wind. (所给词的适当形式填空)27. 用方框中所给短语的适当形式填空(每个短语只能用一次,其中有两个短语是多余的)get across put on date back to result into some extent live up to in advance set offin particular be aware of pick up make an impact on1. With luggage and our vehicle ________ at the airport, we started our road trip.2. The pub manager has persuaded the band into ________ an extra show for the eager fans.3. Peking Opera ________ the 18th century, and has been a beloved art form in China for centuries.4. If you combine a clear presentation with a simple PPT, you can easily ________ your ideas.5. The film has incredible special effects but the plot is weak, ________ mixed reviews from the viewers.6. The foreign languages festival received much positive feedback. The English dramas, ________, attracted the most attention.7. ________ his parents’ expectations, he worked hard in his study.8. ChatGPT has strong functions of writing essays and translating, but ________, its capabilities are limited by its training data.9. It’s always a good idea to make arrangements ________ when planning a trip, such as booking flights or hotels early to avoid price hikes.10. We got out of the bathroom, only to find that our travel group ________ for the next town already without us.28. 阅读下面短文,在空白处填入 1 个适当的单词或括号内单词的正确形式。

2022-2023学年四川省成都外国语学校高一下学期期末考试化学试题

2022-2023学年四川省成都外国语学校高一下学期期末考试化学试题

2022-2023学年四川省成都外国语学校高一下学期期末考试化学试题1.下列所涉及的材料不属于无机非金属材料的是A.我国华为公司的5G核心专利在全球排名第一,5G通讯技术离不开光导纤维,制造光导纤维使用的材料二氧化硅B.中国第一艘深海载人潜水器“蛟龙号”使用的氮化硅陶瓷发动机C.中国自主研发的飞机的风挡材料有机玻璃D.“嫦娥五号”成功着陆月球时使用碳化硅作钻杆材料2.下列说法正确的是A.淀粉和纤维素都属于天然高分子,且两者互为同分异构体B.植物油中含有不饱和脂肪酸酯,能使溴的四氯化碳溶液褪色C.煤的干馏、石油的裂化裂解均属于化学变化;煤的气化、液化属于物理变化D.绿色化学的核心是治理产品在生产和应用中涉及的有害化学物质3. N A为阿伏伽德罗常数的值。

下列说法正确的是A.常温常压下,28g乙烯中所含碳氢键数为4N AB.标准状况下,11.2LSO 3中含有的原子数为2N AC.18g由与D 2 O组成的混合物中所含中子数为10N AD.一定条件下,2molSO 2与1molO 2充分反应后生成SO 3分子总数为2N A4.下列装置能达到实验目的的是吸收氨气石油的蒸馏制取乙酸乙酯实验室制取少量氨5.下列化学用语表达正确的是A.羟基的电子式为B.葡萄糖的结构简式为CH 2 OH(CHOH) 4 CHOC.CH 4的空间填充模型:D.含174个中子的117号元素Ts的核素:6.经研究发现,新药瑞德西韦对抗冠状病毒有效,其结构如下图,下列关于该有机物说法正确的是A.不能使酸性高锰酸钾溶液褪色B.与碱溶液反应,与酸溶液不反应C.分子中所有碳原子一定在同一平面上D.一定条件下,可以发生取代、加成和氧化反应7.化学电源在日常生活和高科技领域中都有广泛应用。

下列说法正确的是A.图甲:锂电池放电时,电解质溶液中向锂电极迁移B.图乙:锌筒作正极C.图丙:铅蓄电池放电时电子流入铅所在电极D.图丁:该电池工作时,标况下每消耗22.4LO 2转移4mol电子8.下列反应过程不正确的是A.向NaAlO 2溶液中通入过量CO 2的离子反应:B.碳酸氢铵中加入过量氢氧化钠溶液的离子反应:C.以水杨酸为原料制备阿司匹林: ++D.工业生产环氧乙烷:2CH 2 =CH 2 +O 2 29.根据下列实验操作、实验现象和得出的结论均正确的是稀硝酸将Fe氧化为淀粉未发生水解原溶液中一定含有浓硫酸有脱水性和10.氮、硫元素的“价—类”二维图如图所示。

四川省成都外国语学校2019-2020学年高二下学期5月月考生物试题 Word版含答案

四川省成都外国语学校2019-2020学年高二下学期5月月考生物试题 Word版含答案

成都外国语学校2019—2020学年度下学期5月月考高二生物试卷(理科)注意事项1 本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分2本堂考试时间: 100分钟,满分100分3 答题前,考生务必将自己的姓名,学号填写在试卷上,并用2B铅笔填涂4考试结束后,将答题卷交回Ⅰ卷一、选择题(1-40题每题1分,共40分。

每题只有一个答案正确。

请将答案填涂于机读卡上,否则不得分。

)1.组成人体细胞的蛋白质和脱氧核糖核酸( )A.都由C、H、O、N、P等元素组成B.都由基本组成单位聚合而成C.都可作为细胞的能源物质D.合成过程中可互相作为模板2.下列细胞亚显微结构示意图,正确的是( )A.AB.BC.CD.D3.在不同的放大倍数下,所呈现的视野分别为甲和乙(如图所示),下列相关叙述正确的是A.若使用相同的光圈,则甲比乙亮B.在甲中所观察到的细胞,在乙中均可被观察到C.若玻片右移,则甲的物像会右移而乙的物像左移D.若在甲中看到的物像模糊,则改换成乙就可以看到清晰的物像4.生命科学研究中常用图示表示微观物质的结构,图中1~3分别表示植物细胞中常见的三种有机物,则图中1~3可分别表示()A.多肽、RNA、淀粉B.DNA、RNA、纤维素C.DNA、蛋白质、糖原D.蛋白质、核酸、糖原5.如图所示是生物体内某种有机大分子的基本组成单位的模式图,相关叙述正确的是( )A.a、b、c结合而成的基本单位,在人体中共有4 种B.若b为脱氧核糖,则c有4 种,分别是A、U、C、GC.在HIV病毒中可以检测到的c有5 种D.若c为T,则该有机小分子的名称是胸腺嘧啶脱氧核苷酸6.蛋白质在生物体内具有重要作用。

下列叙述正确的是( )A.蛋白质化学结构的差异只是R基团的不同B.某些化学物质可使蛋白质的空间结构发生改变C.蛋白质控制和决定着细胞及整个生物体的遗传特性D."检测生物组织中的蛋白质"需同时加入双缩脲试剂A和B7.下列关于脂质的叙述中正确的是( )A.磷脂是构成细胞膜的重要成分,在动物的脑、卵细胞、肝脏及大豆种子中含量丰富B.脂肪组成元素只有C、H、O,且与糖类物质相比,C、H比例低,O比例较高C.胆固醇也是构成细胞膜的重要成分,能够有效促进人和动物肠道对钙磷的吸收D.性激素可由性腺产生并分泌,可以促进生殖器官的发育,也是重要的储能物质8.下列有关细胞膜制备及观察的叙述,正确的是( )A.家鸡的红细胞是最佳的实验材料B.若选用洋葱鳞片叶表皮细胞应先用镊子去除细胞壁C.制备细胞膜应先利用吸水涨破法,再利用差速离心法获取D.可以用高倍镜直接观察9.玉米叶肉细胞和蝗虫的体细胞中都具有的结构是()①细胞壁②细胞膜③线粒体④叶绿体⑤中心体⑥核糖体⑦内质网A.①②⑥⑦B.②④⑤⑦C.②③⑥⑦D.①③⑤⑥10.下列与真核生物细胞核有关的叙述,错误的是()A.细胞中的染色质存在于细胞核中B.细胞核是遗传信息转录和翻译的场所C.细胞核是细胞代谢和遗传的控制中心 D.细胞核内遗传物质的合成需要能量11.古生物学家推测: 被原始真核生物吞噬的蓝藻有些未被消化,反而能依靠原始真核生物的“生活废物”制造营养物质,逐渐进化为叶绿体。

四川省成都外国语学校2020-2021学年高一上学期期中考试数学试题 Word版含答案

四川省成都外国语学校2020-2021学年高一上学期期中考试数学试题 Word版含答案

成都外国语学校2020一2021学年度上期期中考试高一数学试卷注意事项:1.本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.2.本堂考试时间120分钟,满分150分.3.答题前,考生务必先将自己的姓名、学号填写在答题卡上,并使用2B 铅笔填涂.4.考试结束后,将答题卡交回.第Ⅰ卷 (选择题,共60分)一、选择题: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的.1.设集合{1,2,6},{2,4},{1,2,3,4}A B C ===,则()A B C ⋃⋂=( )A .{2}B .{1,2,4}C .{1,2,4,6}D .{1,2,3,4,6}2.下列各组函数中,表示同一个函数的是( )A.112--=x x y 与1+=x y B.x y =与)1,0(log ≠>=a a a y xaC.12-=x y 与1-=x y D.x y lg =与2lg 21x y =3.设全集U R =,集合1284x A x⎧⎫=<<⎨⎬⎩⎭,{}05B x x =<<,则韦恩图中阴影部分表示的集合是( )A.{}25x x -<<B.{}20x x -<≤C.{}35x x -<<D.{}35x x ≤<4.已知函数=⎩⎨⎧≥-<-=)2(,0)5(0)(log )(3f x x f x x x f 则( ) A .-1 B .1 C .0 D .2 5.已知函数)(54)12(R x x x f ∈+=-,若13)(=a f ,则实数a 的值为( )A.5B.4C.3D.2 6.已知5log 3=a ,23log 2b =,2.05-=c ,则c b a ,,的大小关系为( )A.a c b >>B. a b c >>C.c b a >>D.c a b >>7.函数2121xy =+-的部分图象大致为( ) A .B .C .D .8.已知函数0(1,1,3)(>⎩⎨⎧≥<+-=a x a x a x x f x且)1≠a 在R 上是减函数,则a 的范围为( ) A.)1,0( B.]21,0( C.)1,21[ D.),21[+∞9.已知)(x f y =是定义在R 上的函数,且)()4(x f x f -=+,如果当(]4,0∈x 时,x x f 3)(-=,则=)985(f ( ) A .9B .-9C .3D .-310.若函数)2(log 2+-=ax x y a 在区间(]1,∞-上为减函数,则a 的取值范围是( ) A .()1,0 B .[)3,2 C .[)+∞,2 D .)3,1(11.高斯是德国著名的数学家,近代数学奠基者之一,享有“数学王子”的称号,用其名字命名的“高斯函数”为:设x ∈R ,用[]x 表示不超过x 的最大整数,则[]y x =称为高斯函数,例如:[]2.13-=-, []3.13=,已知函数()121123x x f x +=-+,则函数[()]y f x =的值域是( )A .{}0,1B .{}1,1-C .{}1,0-D .{}1,0,1-12.已知函数⎩⎨⎧>+-≤<=3,430,log )(3x x x x x f ,若函数m y x f y ==与)(有三个不同的交点,其横坐标依次为,21,x x 3x ,且321x x x <<,则()3211x x x m-+的取值范围是( )A .)1,3(--B .)2,0(C .)3,1(-D .)0,3(第Ⅱ卷 (非选择题,共90分)二、填空题:本大题共4小题,每小题5分,共20分.把答案填在答题卡上. 13.已知集合},1{2a A =,}1,{-=a B ,若}1,,1{a B A -= ,则=a .14.函数)10(2)(2020≠>+=-a a a x f x 且的图象必经过定点 .15.已知函数)(x f 是奇函数,当0>x 时,)6()(x x x f +=,则当0<x 时=)(x f .16.定义域为R 的函数()f x 满足()()22f x f x +=,当[)0,2x ∈时,()[)[)2 1.5,0,10.5,1,2x x x x f x x -⎧-∈⎪=⎨-∈⎪⎩,若[)4,2x ∈--时,()142t f x t≥-恒成立,则实数t 的取值范围是 . 三、解答题:本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.17.(本小题满分10分)已知函数A x x x f 的定义域为集合)2lg()(2++-=,}2|{+≤≤=m x m x B . (Ⅰ)当2-=m 时,求B A ;(Ⅱ)若B B A = ,求实数m 的取值范围.18.(本小题满分12分)化简求值:(Ⅰ)2log 432302155327log 25.0)32()1613(+-+---π; (Ⅱ)已知52121=+-xx ,求54122-+++--x x x x 的值.19.(本小题满分12分)已知函数)10(≠>=a a a y x 且在[1,2]上的最大值与最小值之和为20,记2)(+=x xa a x f .(Ⅰ)求a 的值;(Ⅱ)求)20212020(......)20213()20212()20211(f f f f +++的值. 20.(本小题满分12分)某手机品牌公司生产某款手机的年固定成本为40万美元,每生产1万部还需另投入16万美元.设该公司一年内共生产该款手机x 万部并全部销售完,每万部的销售收入为)(x R 万美元,且⎪⎩⎪⎨⎧>-≤<-=404000074004006400)(2x x xx x x R(Ⅰ)写出年利润W (万美元)关于年产量x (万部)的函数解析式;(Ⅱ)当年产量为多少万部时,公司在该款手机的生产中所获得的利润最大?并求出最大利润.21.(本小题满分12分)已知()21,f x log a a R x ⎛⎫⎪⎝⎭=+∈. (Ⅰ)当1a =时,解不等式()1f x >;(Ⅱ)设0a >,若对任意1,12t ⎡⎤∈⎢⎥⎣⎦,函数()f x 在区间[],1t t +上的最大值与最小值的差不超过1,求a 的取值范围.22.(本小题满分12分)已知函数()()9log 91xf x kx =++,()k R ∈是偶函数.(Ⅰ)求k 的值;(Ⅱ)若()102b x x f ⎛⎫-+>⎪⎝⎭对于任意x 恒成立,求b 的取值范围; (Ⅲ)若函数[]8log ,0,1329)(921)(∈+⋅+=+x m x h x xx f ,是否存在实数m 使得)(x h 的最小值为0?若存在,求出m 的值,若不存在,请说明理由.成都外国语学校2020-2021学年度上期期中考试高一数学参考答案一、选择题二、填空题13.0 14.(2020,3) 15.)6(x x - 16.(](]1,02, -∞- 三、解答题17.(Ⅰ)}{21<<-=x x A ,}{22<≤-=x x B A 5分 (Ⅱ)01<<-m 10分 18.(Ⅰ)11(Ⅱ)211-各6分 19.(Ⅰ)4 5分 (Ⅱ)1010 12分 20.(1)利用利润等于收入减去成本,可得当040x <时,2()(1640)638440W xR x x x x =-+=-+-;当40x >时,40000()(1640)167360W xR x x x x=-+=--+ ∴W ={−6x 2+384x −40,0<x ⩽40−40000x−16x +7360,x >40; 6分(2)当040x <时,226384406(32)6104W x x x =-+-=--+,32x ∴=时,(32)6104max W W ==;当40x >时,400004000016736027360W x x x=--+-, 当且仅当4000016x x=,即50x =时,(50)5760max W W == 61045760>32x ∴=时,W 的最大值为6104万美元. 12分21.(Ⅰ)当1a =时,()22111f x log a log x x ⎛⎫⎛⎫⎪ ⎪⎝⎭⎝=+=⎭+ ()211112101111log x x x x f x ⎛⎫>++>∴>∴∴ ⎪⎝>⎭<∴<不等式解集为(0,1). 5分(Ⅱ)因为()f x 在(0,)+∞上单调递减,所以函数()f x 在区间[],1t t +上的最大值与最小值的差为()(1)t f f t -+,因此2211()(1)log log 11f t f t a a t t ⎛⎫⎛⎫-+=+-+≤ ⎪⎪+⎝⎭⎝⎭即2(1)10at a t ++-≥对任意1,12t ⎡⎤∈⎢⎥⎣⎦恒成立,因为0a >,所以2(1)1y at a t =++-在1,12t ⎡⎤∈⎢⎥⎣⎦上单调递增,所以21131(1)1(1)1=4242y at a t a a a =++-≥⨯++⨯-- 因此3120423a a -≥∴≥ 12分 22.(Ⅰ)函数()()9log 91xf x kx =++,()k R ∈是偶函数则满足()()f x f x =-所以()()99log 91log 91x xkx kx -++=-++即()()99919912log log log 991991x x x xx x kx x --++====-++ 所以21k =- 解得12k =-3分 (Ⅱ)由(1)可知,()()91log 912x f x x =-++,()102b x x f ⎛⎫-+> ⎪⎝⎭对于任意x 恒成立 代入可得()9log 910x x b +-->所以()9log 91xb x <+-对于任意x 恒成立令()()()999log 91log 91log 9xxxg x x =+-=+-99911log log 199x x x +⎛⎫==+ ⎪⎝⎭因为1119x +>所以由对数的图像与性质可得91log 109x ⎛⎫+> ⎪⎝⎭所以0b ≤ 7分 (Ⅲ)()()129231f x xxh x m +=+⋅+,[]90,log 8x ∈,且()()91log 912xf x x =-++代入化简可得()9232xxh x m =+⋅+令3x t =,因为[]90,log 8x ∈,所以t ⎡∈⎣则()()222222,p t t mt t m m t ⎡=++=++-∈⎣①当1m -≤,即1m ≥-时,()p t在⎡⎣上为增函数,所以()()min 1230p t p m ==+=,解得32m =-,不合题意,舍去 ②当1m <-<即1m -<-时,()p t 在[]1,m -上为减函数,()p t在m ⎡-⎣上为增函数,所以()()2min 20p t p m m =-=-=,解得m =,所以m =③当m ≤-,即m ≤-, ()p t 在⎡⎣上为减函数,所以()(min 100p t p ==+=解得m =不合题意,舍去,综上可知,m = 12分。

四川省成都外国语学校2020-2021学年高二下学期期中考试化学试题

四川省成都外国语学校2020-2021学年高二下学期期中考试化学试题

成都外国语学校2020—2021学年度下期期中考试高二化学试卷注意事项:1、本试卷分Ⅰ卷(选择题)和Ⅰ卷(非选择题)两部分。

2、本堂考试100分钟,满分100分。

3、答题前,考生务必先将自己的姓名、学号填写在答卷上,并使用2B铅笔填涂。

4、考试结束后,将答题卡交回。

可能用到的相对原子质量: H-1 N-14 Al-27 O-16 C-12 Co-59Ⅰ卷(50分)选择题:本题共20小题,共50分。

1至10题,每小题2分。

11至20题,每小题3分。

每小题只有一个选项符合题目要求。

1.化学与生活、社会密切相关,下列说法正确的是()。

A.晶体硅常用做光导纤维的主要材料B.近年来已发布“空气质量日报”中,将CO2、SO2、NO2和可吸入颗粒物等列入了首要污染物C.煤炭经气化、液化和干馏等过程,可获得清洁能源和重要的化工原料D.SO2可以用来漂白纸浆、毛、丝、草帽辫、增白食品等2.下列有关化学用语的表示正确的是()A. 14C 的原子结构示意图:B.四氯化碳分子的比例模型:C. HClO 的结构式:H–Cl–OD.PH3的电子式:3.下列各组物质的分类正确的是()①同位素:1H、2H2、3H ②同素异形体:C80、金刚石、石墨③酸性氧化物:CO2、NO、SO3 ④混合物:水玻璃、水银、水煤气⑤电解质:明矾、冰醋酸、石膏⑥干冰、液氯、乙醇都是非电解质A.②⑤B.②⑤⑥C.②④⑤⑥D.①②③④⑤⑥4.下列有关共价键和键参数的说法不正确的是()A.一个丙烯(C3H6)分子中含有9个σ键和1个π键B.碳碳双键的键能不为碳碳单键键能的两倍C.C-H 键比S i-H 键键长更短,故C H4比S iH4 更稳定D.由于孤电子对的存在,H2O 分子的键角小于109°28′5.下列分子或离子中,中心原子含有孤电子对的是A.NH4+B.SiH4C.H3O+D.SO42-6.下列物质的转化在给定条件下能实现的是()①Al2O3NaAlO2(aq ) Al(OH)3②S SO3H2SO4③Fe2O3FeCl3 (aq) 无水FeCl3④饱和NaCl(aq) NaHCO3Na2CO3⑤MgCl2 (aq) Mg(OH)2MgOA.①③⑤B.①④⑤C.②④⑤D.②③④7.N A代表阿伏加德罗常数的值。

成都七中2019-2020学年度高一下期期末化学考试试卷

成都七中2019-2020学年度高一下期期末化学考试试卷

成都七中2019—2020学年度下期高一年级期末考试化学试卷命题人:向仲云审题人:闵丽根可能用到的相对原子质量:H—1 C—12 N—14 O—16 Fe—56第Ⅰ卷选择题(共40分)一、选择题(本题共20小题,每小题2分,每题只有一个选项符合题意)1.下列化学用语正确的是()A.四氯化碳的电子式:B.HCN的结构式:H—N=CC.丙烷的球棍模型:D.2.化学与生产、生活、社会密切相关,下列有关说法中不正确的是()A.地沟油经过加工处理后可用来制生物柴油和肥皂B.考古时利用14C测定一些文物的年代C.高铁“复兴号”车厢连接关键部位所使用的增强聚四氟乙烯属于合成材料D.“春蚕到死丝方尽,蜡炬成灰泪始干”中的蜡炬属于油脂3.用下列装置进行实验,能达到目的的是()选项 A B C D实验装置实验目的模拟石油分馏制备硝基苯制备乙酸乙酯制备纯净的四氯化碳4.设N A为阿伏加德罗常数的值,下列说法正确的是()A.H2与Cl2化合时断开1 mol H-H键,转移的电子数为N A B.100 g 9%葡萄糖溶液中所含的氧原子的数目为0.3 N AC.28 g聚乙烯所含的质子数为16 N AD.等物质的量的14NO和13CO所含中子数均为15 N A5.下列关于元素周期表和周期律的说法正确的是()A.现行元素周期表有七个周期,18个族B.熔点:Li>K,沸点:I2>Br2C.盐酸的酸性比氢硫酸酸性强,可推测非金属性:Cl>SD.At与I属于同主族元素,可推测AgAt为难溶于水的白色固体6.下列关于常见有机物的说法中正确的是()A.分子式为C3H6FCl的有机物共6种B.蔗糖与麦芽糖分子式不同,结构也不同C.乙醇和乙烯都能使酸性高锰酸钾溶液褪色,均被氧化生成CO2和H2OD.1,3—丁二烯所有原子可能共平面7.迄今为止,煤、石油、天然气仍是人类使用的主要能源,同时它们也是重要的化工原料。

以下说法正确的个数是()①石油分馏得到的产品是混合物,无固定的沸点②煤中含苯、甲苯、二甲苯等有机物,可通过干馏获得③天然气可与水在一定条件下转化为CO和H2,再在催化剂作用下合成CH3OH④石油裂解气可用于生产合成塑料与合成纤维,如聚乙烯和聚氯乙烯,二者被广泛用于食品包装。

2023-2024学年四川省成都外国语学校高一上学期期中考试英语试题

2023-2024学年四川省成都外国语学校高一上学期期中考试英语试题

2023-2024学年四川省成都外国语学校高一上学期期中考试英语试题1. What did the man do last night?A.He played a game. B.He watched a movie. C.He did some training.2. What does the woman want to do?A.Get some rest. B.Go for a walk. C.Read some magazines.3. What will the weather be like this weekend?A.Windy. B.Sunny. C.Rainy.4. How does the man feel now?A.Angry. B.Pleased. C.Sorry.5. Where does the conversation most probably take place?A.In a bookstore. B.In a music store. C.In a recording studio. 听下面一段较长对话,回答以下小题。

6. What are the speakers talking about?A.The hotel in the city center.B.The woman’s trip to Kingston.C.The woman’s plan for a summer event.7. What does the woman think of her stay in the hotel?A.Unsatisfying. B.Comfortable. C.Wonderful.听下面一段较长对话,回答以下小题。

8. Why does the man look bad?C.He took three courses.A.He was sick. B.He stayed up latestudying.9. When do the speakers plan to meet?A.On Saturday evening. B.On Sunday morning. C.On Sunday noon. 10. What is the probable relationship between the speakers?A.Co-workers. B.Schoolmates. C.Teacher and student. 听下面一段较长对话,回答以下小题。

四川省成都外国语学校2023-2024学年高一下学期7月月考英语试题

四川省成都外国语学校2023-2024学年高一下学期7月月考英语试题

四川省成都外国语学校2023-2024学年高一下学期7月月考英语试题一、听力选择题1.What is the probable relationship between the speakers?A.Mum and son.B.Teacher and student.C.Doctor and patient. 2.When will the woman take her flight?A.On the 22nd.B.On the 23rd.C.On the 24th.3.What are the speakers mainly talking about?A.School activities.B.Favorite subjects.C.Test scores.4.What does the woman mean?A.The man has to leave.B.She offers another option.C.She can’t make it. 5.Where does the conversation take place?A.In a bookstore.B.At a restaurant C.In a library.听下面一段对话,回答以下小题。

6.Why does Monica choose Tai Chi instead of yoga?A.It enjoys popularity.B.It requires less tension.C.It reflects Chinese culture. 7.What’s the true significance of Tai Chi?A.Maintaining balance.B.Strengthening muscles.C.Preventing diseases.听下面一段对话,回答以下小题。

8.Why does George want to go to the book exhibition?A.To buy a book.B.To meet a penfriend.C.To do a book signing. 9.What will the woman do before going to the book exhibition?A.Visit her parents.B.Go to her son’s school.C.Pick up Sarah.听下面一段对话,回答以下小题。

四川省成都市实验外国语学校2022-2023学年高一下学期第一次周考化学考试

四川省成都市实验外国语学校2022-2023学年高一下学期第一次周考化学考试

成都市实验外国语学校高2022级高一下化学周考(一)班级姓名相对原子质量:H—1C—12N—14O—16S—32Zn—65Cu—64Ba —137一、单选题(每题3分,共45分)1.下列关于硫及其化合物说法错误的是A .工业上用98.3%的浓H 2SO 4吸收SO 3,目的是防止形成酸雾,以使SO 3吸收完全B .葡萄酒中添加适量SO 2可以起到抗氧化的作用C .硫酸钡可用作消化系统X 射线检查的内服药剂D .实验室常用98.3%的浓硫酸与Na 2SO 3固体制备SO 22.下列“类比”合理的是A .Cl 2能够从KI 溶液中置换出I 2,则F 2也能够从KI 溶液中置换出I 2B .3CaCO 溶解度小于()32Ca HCO ,则23Na CO 溶解度小于3NaHCO C .NaClO 溶液与2CO 反应生成3NaHCO 和HClO ,则NaClO 溶液与2SO 反应生成3NaHSO 和HClOD .2Cl 与Fe 反应生成3FeCl ,则2Cl 与Cu 反应生成2CuCl 3.1891年Engel 首次制得S 6分子,1912年Beckmann 获得S 8分子。

下列说法正确的是A .S 6和S 8互为同位素B .S 6和S 8分别与铜粉反应,所得产物均为CuSC .S 6和S 8分别与过量的氧气反应可以得到SO 3D .等质量的S 6和S 8分别与足量的热KOH 溶液反应,消耗KOH 的物质的量相同4.下列说法正确的是A .氨水显碱性,是因为氨水是一种弱碱,可以电离出OH -B .氨气易液化得液氨,液氨是纯净物常作制冷剂C .氨气遇浓硫酸能产生白烟D .NO 2能与水反应生成硝酸,所以NO 2是酸性氧化物5.大气中的氮循环(如图所示)可以减少环境的污染。

下列说法正确的是A .转化过程中-2NO 被+4NH 氧化为2NB .2N 转化成+4NH 发生了还原反应C .温度越高,2N 转变成+4NH 的速率越快D .图中含氮物质中氮元素的化合价共有三种6.将足量的2SO 通入2BaCl 溶液中,无明显现象,当加入(或通入)某试剂(或气体)X 后有沉淀生成。

成都外国语学校2020-2021学年高一下学期期中考试化学试题

成都外国语学校2020-2021学年高一下学期期中考试化学试题

成都外国语学校2020-2021学年度下期期中考试高一化学试卷注意事项:1.本试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分;2.本堂考试100分钟,满分100分;3.答题前,考生务必将自己的姓名、学号填写在答题卡上,并使用2B 铅笔填涂;4.考试结束后,将答题卡交回。

可能用到的相对原子质量:H -1C -12O -16第Ⅰ卷(选择题,共50分)一、单选题(1-25题,每题2分)1.下列表示不正确的是()A .丁烷的球棍模型:B .CO 2的电子式:C .乙烯的结构简式:CH 2=CH 2D .R 2+离子核外有a 个电子,b 个中子,R 原子可表示为:a+b+2a+2R2.下列叙述中正确的是()A .35Cl 2和37Cl 2都是氯气单质B .H 3O +和OH -中具有相同的质子数和电子数C .235U 和238U 互为同位素,物理性质几乎相同,化学性质不同D .质子数相同的微粒一定属于同一元素3.如图是某两种烃的结构,分析其结构,下列分析正确的是()A .这两种烃互为同系物B .这两种分子结构不同,分子式不同C .这两种分子中,碳原子的化合价都达到“饱和”D .这两种烃都属于烷烃,具有相似的化学性质和物理性质4.下列反应中前者属于取代反应,后者属于加成反应的是()A .光照甲烷与氯气的混合气体;乙烯使酸性高锰酸钾溶液褪色B .乙烯使溴的四氯化碳溶液褪色;苯与氢气在一定条件下反应生成环己烷C .苯滴入浓硝酸和浓硫酸的混合液中水浴加热;乙烯使溴水褪色D .在苯中滴入溴水,溴水褪色;乙烯与水蒸气在一定条件下反应生成乙醇5.设A N 为阿伏加德罗常数的数值﹐下列说法不正确的是()A .常温下,100g 46%的酒精中所含的氢原子数为12A NB .标准状况下,22.4L 22CH Cl 中含有的共价键数目为4AN C .20g 由182H O 与2D O 组成的混合物中所含的质子数为10AND .1mol Na 与O 2反应,生成Na 2O 和Na 2O 2的混合物,钠失去N A 个电子6.X 、Y 、Z 、W 四种短周期元素的原子半径和最高正化合价见下表。

2024届四川省成都外国语学校高三下学期高考模拟(二)理综化学试题

2024届四川省成都外国语学校高三下学期高考模拟(二)理综化学试题

2024届四川省成都外国语学校高三下学期高考模拟(二)理综化学试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.化学与生活、科技、社会发展息息相关。

下列有关说法不正确的是A.卡塔尔世界杯球馆屋顶采用了折叠式PTFE(聚四氟乙烯)板材,该板材属于合成高分子材料B.乙醇、过氧化氢、次氯酸钠等消毒剂均能将病毒氧化而达到消毒的目的C.砖瓦是用黏土烧制而成,黏土主要成分为含水的铝硅酸盐D.高铁酸钾(K2FeO4)在水处理过程中涉及的变化过程有:胶体聚沉、盐类水解、氧化还原反应等2.设N A为阿伏加德罗常数的值,下列说法正确的是A.46.0gC2H6O完全燃烧,有5N A个C-H断裂B.25℃时,1LpH=8的CH3COONa溶液中水电离出的OH-数目为10-8N AC.铅酸蓄电池中,当正极理论上增加3.2g时,电路中通过的电子数目为0.1N AD.32gCu将足量浓、稀硝酸分别还原为NO2和NO,浓、稀硝酸得到的电子数分别为N A、3N A3.为完成下列各组实验,所选玻璃仪器和试剂均准确、完整的是(不考虑存放试剂的容器) 实验目的玻璃仪器试剂A配制100mL一定物质的量浓度的NaCl溶液100mL容量瓶、胶头滴管、烧杯、量筒、玻璃棒蒸馏水、NaCl固体B 制备Fe(OH)3胶体烧杯、胶头滴管蒸馏水、饱和FeCl3溶液C 测定NaOH溶液浓度烧杯、锥形瓶、胶头滴管、酸式滴定管待测NaOH溶液、已知浓度的盐酸、甲基橙试剂D 制备乙酸乙酯试管、量筒、导管、酒精灯冰醋酸、无水乙醇、饱和Na2CO3溶液A .AB .BC .CD .D4.有机物X 可用于合成抗菌药物,X 的结构简式为。

下列有关X 的说法错误的是A .酸性条件下能发生水解反应得到羟基酸B .完全燃烧时生成的()()22CO :H O 18:7n n =C .分子中至少有8个碳原子共平面D .同一个苯环上的二溴代物有6种5.X 、Y 、Z 、M 、Q 为原子序数依次增大的短周期主族元素,其最外层电子数之和为19,Z 的原子序数是X 与Y 的原子序数之和,M 与X 同主族,X 与Q 可形成原子个数比为2:1的18e –分子。

  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
相关文档
最新文档