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河北省定州中学2020-2021学年高一上学期开学考试数学试题 答案和解析

河北省定州中学2020-2021学年高一上学期开学考试数学试题 答案和解析
河北省定州中学2020-2021学年高一上学期开学考试数学试题 答案和解析

河北省定州中学【最新】高一上学期开学考试数学试题

学校:___________姓名:___________班级:___________考号:___________

一、单选题

1.已知集合21,01,2A =--{,,},{}|(1)(2)0B x x x =-+<,则A

B =( )

A .{}1,0-

B .{}0,1

C .{}1,0,1-

D .{}0,1,2

2.集合{}

2

|4?A x x =≤, {}2|log 2,?B x x x N =<∈,则A

B =( )

A .[]1,2

B .()1,2

C .(]

0,2 D .{}1,2

3.集合{}{}

2

0,1,2,|60,A B x x x x z ==+-<∈,则A

B =( )

A .{}2,1,0,1,2--

B .

1,0,1,2

C .{}0,1

D .{}0,1,2

4.下列函数()f x 中,满足“对任意的()12,0,x x ∈+∞时,均有

()()()12120x x f x f x -->????”的是(

)

A .()12

x

f x =()

B .()2

44f x x x =-+

C .()2x

f x =

D .()12

log f x x =

5.设集合M ={m ∈Z|?3

6.设集合{}

42M x x =-<<,{

N x y ==,则M

N =( )

A .()0,4

B .[)0,4

C .()0,2

D .[)0,2

7.设函数()ln(1)ln(1)f x x x =+--,则()f x 是( ) A .奇函数,且在(0,1)上是增函数 B .奇函数,且在(0,1)上是减函数 C .偶函数,且在(0,1)上是增函数 D .偶函数,且在(0,1)上是减函数

8.函数y =1

2x 2?lnx 的单调减区间是( ) A .(0,1)

B .(0,1)∪(?∞,?1)

C .(?∞,1)

D .(?∞,+∞)

9.定义区间12[,]x x 长度为21x x -,21()x x >,已知函数

22

()1()(,0)a a x f x a R a a x

+-=∈≠的定义域与值域都是[,]m n ,则区间[,]m n 取最大长度时a 的取值构成的集合为( )

A .

B .{|13}a a a ><-或

C . {|1}a a >

D .{3}

10.已知集合2

{|320}A x x x =-+≥,{|2,}B x x x z =≤∈,则()R C A B =∩( ) A .? B .{1} C .{2} D .{1,2}

11.已知全集{}1,2,3,4,5,6,7U =,{}3,4,5M =,{}1,3,6N =,则集合{}2,7等于( )

A .M N ?

B .()()U U

C M C N ? C .()()U U C M C N ?

D .M N ?

12.设集合{}1,0,1A =-,{}0B x R x =∈>,则A B ?=( ) A .{}1,0- B .{}1- C .{}0,1 D .{}1

二、填空题

13.集合A ={1,2,3,4},B ={x |x =n 2,n ∈A },则A ∩B __________. 14.集合{}{}1,0,1,2,|21,A B x x n n A =-==+∈,则A B __________.

15.已知集合1

{|

0}ax A x x a

-=<-,且2A ∈,3A ?,则a 的取值范围是_______. 16.在映射:f A B →中,如果()f a b =,那么称b 为a 的像.设

{(,)|3,,}A x y x y x y N =+<∈,:f A B →使2(,)x y x y →+,则A 中所有元素的

像构成的集合是______.(用列举法表示)

三、解答题

17.已知全集U={1,2,3,4},集合{}

{}21,2,1,4A x B ==与是它的子集, (1)求U C B ;(2)若A B ?=B,求x 的值;(3)若A B ?= U ,求x .

参考答案

1.A 【详解】

由已知得{}|21B x x =-<<,

因为21,01,2A =--{,,},

所以{}1,0A B ?=-,故选A . 2.D 【解析】

{}

2|4{|22}A x x x x =≤=-≤≤, {}{}2|log 2,1,2,3B x x x N =<∈=.

{}1,2A B ?=.

故选D. 3.A 【解析】

集合{}{}

{}{}2

0,1,2,|60,|32,2,1,0,1A B x x x x Z x x x Z ==+-<∈=-<<∈=--.

{}2,1,0,1,2A B ?=--.

故选A. 4.C 【解析】

对任意的()12,0,x x ∈+∞时,均有()()()12120x x f x f x ??-->??, 即函数()f x 是定义在()0,+∞上的增函数.

A,()1

2

x

f x =

()为减函数,不成立; B,()2

44f x x x =-+在()2,+∞单调递增,不成立; C ,()2x

f x =在()0,+∞上为增函数,成立;

D,()12

log f x x =在()0,+∞上为减函数,不成立.

故选C.

点睛:本题属于对函数单调性应用的考察,若函数()f x 在区间上单调递增,则

1212,,()()x x D f x f x 且∈>时,若()f x 在区间上单调递减,则当

1212,,()()x x D f x f x 且∈>时.对于()()()12120x x f x f x ??-->??则等价于函数单增.

5.B 【解析】

由M ={m ∈z|?3

{N x y ===[0,)+∞ ,所以[)0,2M N ?=,选D.

7.A 【解析】

试题分析:由题意得,函数的定义域为10

{10

x x +>->,解得11x -<<, 又()ln(1)ln(1)[ln(1)ln(1)]()f x x x x x f x -=--+=-+--=-,所以函数()f x 的奇函数, 由1()ln(1)ln(1)ln

1x f x x x x +=+--=-,令()11x

g x x

+=-,又由1201x x <<<,则 ()()2121212121112()

011(1)(1)

x x x x g x g x x x x x ++--=

-=>----,即,所以函数

()11x

g x x

+=

-为单调递增函数,根据复合函数的单调性可知函数()ln(1)ln(1)f x x x =+--在(0,1)上增函数,故选A.

考点:函数的单调性与奇偶性的应用.

【方法点晴】本题主要考查了函数的单调性与奇偶性的应用,其中解答中涉及到函数的奇偶性的判定、函数的单调性的判定与应用、复合函数的单调性的判定等知识点的综合考查,着重考查了学生分析问题和解答问题的能力,以及推理与运算能力,本题的解答中确定函数的定义域是解答的一个易错点,属于基础题. 8.A

【详解】

试题分析:令f′(x )=x ?1

x <0?0

试题分析:设[],m n 是已知函数定义域的子集,[]()[]()0,,,0,0,x m n m n ≠?-∞?+∞或,故函数()211

a f x a a x

+=

-在[],m n 上单调递增,则()()

f m m

f n n =???

=??,故,m n 是方程211

a x a a x

+-=的同号的相异实数根,即()22210a x a a x -++=的同号的相异实数根,21

,,mn m n

a

=∴同号,只需

()()2310,13,a a a a a n m ?=+->∴><--==或,

n m ∴-取最大值为

33

a =此时,故选D. 考点:1.函数的性质;2.函数与方程.

【方法点晴】本题考查学生的是函数与方程思想,属于中档题目.根据题目中的已知定义域和值域区间相同,先判出给定函数的单调性为单调递增,因此在左端点取到最小值,右端点取最大值,即转化为方程()f x x = 至少有两个不等的实数根,因此问题成为二次类型的函数的方程根问题,写出韦达定理讨论? 求出参数的取值. 10.A 【解析】

试题分析:集合{}(){}

()12,1,2,2,,R R A x x x C A B x x x Z C A B φ=≤≥==≤∈∴=或,

故选A.

考点:集合的交并补运算. 11.B 【解析】

试题分析:由题意可得}7,5,4,2{},7,6,2,1{==N C M C U U ,故{2,7}U U C M

C N =,应选B.

考点:集合的补集和交集运算. 12.D 【解析】

试题分析:{}1A B ?=,故选D. 考点:集合的运算. 13.{1,4} 【解析】

{}1,4,9,16B =,∴{}1,4A B ?= .

14.{}1,1- 【解析】

B ={}|21,x x n n A =+∈={1,1,3,5}- ,∴{}1,1,A B ?=-,故答案为{}1,1- .

15.

【解析】

试题分析:∵关于的不等式

的解集为

,若2A ∈,3A ?,故有

,化简可得,解得,故实数的

取值范围是,故答案为.

考点:分式不等式的解法. 【方法点晴】根据集合

,且

,

,知道.满足不等式

,

不满足该不等式,即,解此不等式组即可求得实数的取值范围为

.一方面时的等价不等式要注意,千万不要写成,

这样就不等价了,这是一个易错点.另一方面不等式组的交并运算最好结合数轴. 16.{}0,1,2,4 【解析】 试题分析:集合

.

考点:集合的列举法表示,映射.

17.①U C B ={2,3};②2x =±;③x =【解析】试题分析:(1)欲求U C B 只需在U 中去掉B 中的元素即可得到,(2)

,故有24x =;

A B ?= U ,由

可知.

试题解析:(1) U C B ={2,3} 4分 (2)若A B ?=B,则24x =6分 ∴2x =±8分

(3)若A B ?= U ,则23x =10分

∴x =分 考点:集合运算.

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2020学年湖南省长沙市长郡中学高一下学期入学考试数学 试题 一、单选题 1.已知}3{1 A =,,5{}34 B =,,,则集合A B =I ( ) A .{}3 B .{4}5, C .15}2{4,,, D .{345}, , 【答案】A 【解析】由交集的定义直接求解即可. 【详解】 Q }3{1A =,,5{}34B =,,, ∴{}3A B ?=. 故选:A. 【点睛】 本题考查交集的求法,属于基础题. 2.已知函数31(0) ()2(0) x a x f x x x -?+≤=?+>?,若((1))18f f -=,那么实数a 的值是( ) A .4 B .1 C .2 D .3 【答案】C 【解析】先求出(1)4f -=,((1))18f f -=变成(4)18f =,可得到4218a +=,解方程即可得解. 【详解】 (1)4f -=,((1))18f f -=变成(4)18f =,即4218a +=,解之得:2a =. 故选:C. 【点睛】 本题考查已知函数值求参数的问题,考查分段函数的知识,考查计算能力,属于常考题. 3.已知()f x =22x x -+,若()3f a =,则()2f a 等于 A .5 B .7 C .9 D .11 【答案】B

【解析】因为()f x =22x x -+,所以()f a =223a a -+=,则 ()2f a =2222a a -+=2(22)2a a -+-=7. 选B. 4.设α是第三象限角,且cos cos 2 2 α α =-,则 2 α 所在象限是( ) A .第一象限 B .第二象限 C .第三象限 D .第四象限 【答案】B 【解析】先由α是第三象限角,得出2 α 可能在第二、四象限,进一步由cos cos 22αα=-再判断出 2 α 所在的象限. 【详解】 αQ 是第三象限角, ∴3222 k k π ππαπ+<<+ ,k Z ∈, 32 2 4 k k π α π ππ∴+< <+ ,k Z ∈, ∴ 2 α 在第二、四象限, 又cos cos 22 α α =-,∴cos 02 α <, ∴ 2 α 在第二象限. 故选:B. 【点睛】 本题考查由三角函数式的符号判断角所在象限的问题,考查逻辑思维能力和分析能力,属于常考题. 5 ) A .0 sin15cos15 B .2 2 cos sin 1212ππ - C .0 1tan151tan15+- D 【答案】B 【解析】A.0 011sin15cos15sin 3024 = =

2020届上海市市北中学高三英语上学期期中检测试卷

2019-2020学年市北中学高三上英语期中考试 Ⅱ. Grammar and Vocabulary Section A Directions: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank. A really smart students of mine who has been getting excellent grades in economics was considering it was his major. He wanted to know (21) ________ he could do with it after graduation. I have had that question regarding every subject you can name. The reality is (22) ________ it is a bad question, (23) ________ it assumes that the subject is what you will do. If your major is history, you will be a historian. If you major in philosophy, people will laugh and ask you what a philosopher’s job is. It does not work that way. The history course for a student (24) ________ (seek) to become a crime scene invest gator is a great way to learn forensic(法医)skills. Likewise, philosophy (25) ________ (often consider) a good gateway to law or other careers where logic is required. Therefore, I pointed out to my students that majors deliver a bundle of skill sets that can be used in the course of their careers. All academic ares require skills (26) ________ ________ reading, critical thinking research in the lab or the library, and the ability to analyze data as well as to report conclusion. The Association of American Medical Colleges has announced that the Medical College Admission Test will include a new behavioral science section. It means that it so far (27) ________ (recognize) the importance of the humanities to the future of medicine. I feel frustrated that universities do not share these facts with students. (28) ________ list of caree r being pursued by graduates seems to make it easy to answer the question “What can I do with this major?” However, it is far from satisfactory. It would be good (29) ________ (explain) to the students what skills they can get through courses or assignments at different stages. To sum up, it is the skills learned through the course but not the major (30) ________ that matters in your future career. Section B Directions: Complete the following passage by using the words in the box. Each word can only No metal plates or screws(螺丝钉)needed: a new 3D-printed ceramic(陶瓷)implant mends broken legs by holding the broken parts together, then turning into __31__ bones. The implant has the same strength as real bone, and is made by Hala Zreiqat at the University of Sydney in Australia and her colleagues. In __32__ studies, they showed that the material could heal broken leg bones in rabbits. Now, in work yet to be __33__ they have shown it can also repair large broken legs in sheep. The eight sheep in the study were able to walk on the implants immediately after surgery, with plaster casts(石膏模)helping to __34__ their legs for the first four weeks. The researchers saw

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