45分钟作业与单元评估数学必修一答案

45分钟作业与单元评估数学必修一答案篇一

必修一第一单元45分钟课时作业

必修二第一单元45分钟课时作业

蛟河市第二高级中学 周晶 张季青 韩桂云 韩秀杰 李杨 I.单词拼写

1. He ___________ （不理睬）the doctor’s advice and goes on smoking.

2. Two _________（系列）of English textbooks will be tried in our province next year.

3. My mother is still _____________（心烦的）about me because I forgot her birthday.

4. You can’t do it best if you don’t ___________ （完全地）put your heart into your study.

5. The house has to stay ___________ （确切地）as it was.

6. But now I am a ___________ （青少年）and I have more things to think about.

7. I should be ___________ （感激的）if you would send me your advice as soon as possible.

8. She seemed to ___________ （不同意）with this decision.

9. She came down from the mountain last year and ___________ （定居）in the village.

10. He soon ___________ （恢复）from the snake’s bite.

II.单项选择

11. ---Would you lend Lucy some money?

---________! She is always borrowing money and never thinking of paying back.

A. No problem B. Come on C. No way D. Go ahead

12.You’d better __________your score and see if you have passed the exam.

A. pack up B. ring up C. add up D. put up

13.The meeting was concerned ________ the development of the company and everyone was concerned ________ their futures.

A. for; about B. with; about C. with; with D. about; with

14.As a solider, he would ________ fire and water for his country.

A. go after B. go through C. go with D. go against

15.---Have you ever been here before?

---No, this is the first time that I________ here.

A. was B. am coming C. came D. have come

16. ________ get a better score, Li Lei has been studying hard these days.

A. So as to B. In order to C. So that D. In order that

17. ---Excuse me, is this Mr Green’s house?

---I’m sorry, but Mr Green ________ lives here.

A. no more B. no longer C. not any longer D. not any more

18. He had ________ a lot during the hard times, but now, it’s time for him to enjoy the life.

A. suffered B. suffered from C. been suffered D. been suffered from

19. Though Tom is often tired ________ his job, he is never tired________ it. In fact, he enjoys it.

A. with; form B. from; with C. of ;from D. with; of

20.Readers can ________ quite well with the article without knowing the exact meaning of each word.

A. get in B. get over C. get along D. get through

21.You’d better ________ your idea on paper or you’ll forget.

A. calm down B. settle down C. set down D. go down

22. He ________ the golf club and had lots of fun there.

A. joined B. joined in C. left D. disliked

23.The teacher said that light ________ faster than sound.

A. travels B. travelled C. have travelled D. had travelled

24.He asked me __________

A. had I ever been to Guilin? B. have I ever been to Guilin?

C. whether I had ever been to Guilin? D. if I have ever been to Guilin.

25.He asked ________ for the computer.

A. did I pay how much B. I paid how much

C. how much did I pay D. how much I paid

III.阅读理解

Nick Petrels is a doctor in Montreal. He works 60 hours a week. He takes care of 159 patients a week in the hospital and at his office. He’s been a doctor for ten years.

Dr. Petrels gives his patients good medical advice. But he doesn’t just tell his patients what to do. He also sings to them on television! Dr. Petrels has his own TV show. The show is in Italian, English and French. The doctor starts the show with a song and then gives medical advice. He explains a medical problem or disease in simple language. After that, he sings another song.

Dr. Petrels produces and performs in his own show every week. The program is very popular with his patients and with people who enjoy his singing. His dream is to perform(表演) in Las Vegas. His favorite songs are love songs, and he has a compact disk of love songs that he wrote. Dr. Petrels says, “I always loved to sing. All my problems are gone when I sing.” But when Dr. Petrels was young, his father didn’t want him to be a singer, so he went to medical school.

Some people tell Dr. Petrels he can help people more as a doctor. But Dr. Petrels says he helps people when he sings, too. “I like to make people smile. Sometimes it’s difficult to make a sick person smile. Medicine and entertainment(娱乐) both try to do the same thing. They try to make people feel good.”

26. Dr. Petrels works 60 hours a week, because he _____________.

A. gives his patients medical advice B. takes care of 159 patients a week

C. sings on television D. has his own TV show

27. Dr. Petrels ___________, so he is called a singing doctor.

A. has been a doctor for ten years B. always loved to sing

C. is popular with his patients D. also sings to his patients on TV

28. In his TV show, Dr. Petrel ____________.

A. sings and gives medical advice B. sings about different diseases

C. starts to explain diseases with a song D. sings love songs he wrote

29. Dr. Petrels’ show is popular ____________.

A. in Las Vegas B. at medical school

C. with people who like his singing D. with patients in Montreal

30. Dr. Petrels says he likes to ____________.

A. help people sing B. make people feel better

C. do the same thing D. make difficult people smile

IV．短文改错

多一个词：把多余的词用斜线（\）划掉；缺一个词：在缺词处加一个漏字符号（∧）；错一个词：在错的词下划一横线。

Last year, LiHua buys several golden fish. He fed them every day. He didn’t know that

the fish couldn' t eat much. But he gave them a lot food. One day . when he came back

to school, he found some fish floating on the water. he thought they were hungry

again, but he threw more food into the water, and then went to do her homework happily.

One and a half hours later, he went to see the fish. To his surprises, all the fish were dead.

He felt so sadly that he kept silent during the supper. He asked his mother that why

that happened. his mother said, " too many love kills a child "

必修一第一单元练习题答案

I.单词拼写

1.ignores 2.series 3.upset 4.entirely 5.exactly

6.teenager 7.grateful 8.disagree 9.settled 10.recovered

II.单项选择

11. C. 考查情景交际。题意：“你借给露西点钱好吗？”“没门！她总是借钱，却从不考虑还钱。”No way “没门”，表示拒绝，符合题意。 No problem“没问题，表示答应请求； Come on “来吧；加油”，表示鼓励；Go ahead“干吧；用吧”，表示许可。

12. C.考查动词短语辨析。 题意：你最好把分数加起来，看看是否通过了考试。add up “合计；把…加起来”， 符合题意。pack up “把… 打包”；ring up “给…打电话” ；put up “搭起；举起；张贴”。

13. B. 考查动词短语与介词的搭配。题意：这个会议与公司的发展有关，每个人都关心自己的命运。be concerned with“与…有关；涉及” ；be concerned about“关心；挂念”。

14.B. 考查动词短语的辨析。题意：作为一名战士， 他甘愿为祖国赴汤蹈火。go through“经历；经受”，符合题意。go after “追赶；追求”；go with “伴随；陪...一起去” go against “违背；违反”。

15. D. 考查定语从句的句型。题意：“你以前来过这里吗？”“没有， 这是我第一次来这里。” This/That/It is the first time that…结构中的从句部分要用现在完成时态。

16. B. 考查不定式作目的状语。题意：为了取得更好的成绩，李雷这些天一直努力学习。in order to “为了… ”，可位于句首和句中，符合题意。so as to “为了… ” ，不能位于句首； so that 和in order that都有“为了… ”的意思，但其后应接从句，且so that 一般不位于句首。

17. B. 考查固定短语的用法。题意：“打扰了，请问这是格林先生的家吗？” “不好意思， 格林先生已经不在这住了， 他三个月前搬到伦敦去了。” 此处表示过去的情况不再延续，故排除no more ；BCD三项都可表示过去情况不再延续，若选C项 和D项，则原句应为. “Mr Green does not live here any longer/any more”，故B项正确。

18. A. 考查动词的用法。题意：在困难时期他受了不少苦， 但现在他该好好地享受生活了。 suffer 作及物动词， 表示“遭受（苦痛、损失等）；忍受（侮辱等）”， 此处 a lot 作宾语，符合题意。 suffer from “表示遭受战争、自然灾害等带来的苦难或患病之苦”；suffer 与suffer from均不能用于被动语态。

19. D. 考查短语辨析。题意：尽管汤姆经常因为他的工作而感到疲惫， 但他从不对工作感到厌倦。事实上他很喜欢他的工作。be tired with/from sth. “因某事感到疲劳”；be tired of sth. “对某事感到厌烦”, 故D项正确。

20. C. 考查动词短语辨析。题意：尽管读者不知道每个单词的确切意思，但他们能够狠顺利

抵读这篇文章。get along “进展情况”，符合题意。get in “进入；到达”；get over“克服；恢复”； get through

“通过；接通电话”。

21. C. 考查动词短语辨析。题意：你最好把你的想法卸载纸上，不然就会忘记。set down “记

下；登记”，符合题意。calm down “平静下来；镇定下来”； settle down “定居下来；安下心来 ；安静下来”；go down“下降；下去”。

22.A. 考查动词辨析。题意：他加入了高尔夫俱乐部并在那里玩得很开心。join “指加入某团体并成为其中的成员”，符合题意。join in “参加（活动）”；leave“离开”； leave for “动身去”。

23.A.考查间接引语中的时态。 题意：老师说光比声音传播得快。“光比声音传播得快”是

客观真理；直接引语如果是客观真理，变为间接引语时，时态不变。

24.C. 考查含有一般疑问句的直接引语变为间接引语的连接词。题意：他问我是否去过桂林。

直接引语时一般疑问句，间接引语要变成有whether/if引导的宾语从句；引述动词用了过去时态，直接引语中的现在完成时在变为间接引语时要用过去完成时。

25.D.考查含有特殊疑问句的直接引语变为间接引语。 题意：他问我买电脑花了多少钱。直

接引语时是特殊疑问句，间接引语要变成由原特殊疑问词引导的宾语从句；引述动词用了过去时态，直接引语中的一般现在时在变为间接引语时要用一般过去时。 III．阅读理解

这篇文章描绘在Montreal一个有趣的医生，他不但给病人治病，还在自己的节目中为病人唱歌，给他们带来欢笑。

26. B。第1段告诉我们He takes care of 159 patients a week, 如此多的病人，难怪He works 60 hours a week。

27. D。由第2段的He also sings to them on television，我们可得知他被称为a singing doctor的原因。

28. A。根据第2段的The doctor starts the show with a song and then gives medical advice，我们便可知道Dr. Petrel在他的节目中干些什么。

29. C。根据第3段的第2句The program is very popular with his patients and with people who enjoy his singing 可推知答案为C。Dr. Petrel受欢迎的不只在Las Vegas这些地方和Patient这些人。

30. B。最后一句的They try to make people feel good告诉了我们Dr. Petrels想干什么。 IV．短文改错 ’t know that bought

the fish couldn' t eat much. But he gave them a lot∧ food. One day . when he came back of

from

So his

A surprise

sad

much

第一句 ：buys 改为bought 。由时间状语last year可知，句子应用一般过去时。

45分钟作业与单元评估数学必修一答案篇二

高中北师版数学A版必修1(45分钟课时作业与单元测试卷)：阶段性检测 Word版含解析{45分钟作业与单元评估数学必修一答案}.

45分钟作业与单元评估数学必修一答案篇三

高中北师版数学A版必修1(45分钟课时作业与单元测试卷)：单元测试三 Word版含解析

45分钟作业与单元评估数学必修一答案篇四

高中北师版数学A版必修1(45分钟课时作业与单元测试卷)：1.1集合的含义与表示 Word版含解析

以下情况：

①{4}，{0,8}，{1,7}，{2,6}，{3,5}，共5个；

②{4,0,8}，{4,1,7}，{4,2,6}，{4,3,5}，{0,8,1,7}，{0,8,2,6}，{0,8,3,5}，{1,7,2,6}，{1,7,3,5}，{2,6,3,5}，共10个；

③{4,0,8,1,7}，{4,0,8,2,6}，{4,0,8,3,5}，{4,1,7,2,6}，{4,1,7,3,5}，{4,2,6,3,5}，{0,8,1,7,2,6}，{0,8,1,7,3,5}，{1,7,2,6,3,5}，{0,8,2,6,3,5}，共10个；

④{4,0,8,1,7,2,6}，{4,0,8,1,7,3,5}，{4,0,8,2,6,3,5}，{4,1,7,2,6,3,5}，{0,8,1,7,2,6,3,5}，共5个；

⑤{4,0,8,1,7,2,6,3,5}，共1个．

于是满足题设条件的集合M共有5＋10＋10＋5＋1＝31个．

45分钟作业与单元评估数学必修一答案篇五

高一数学必修一 单元质量评估(一)

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单元质量评估(一)

第一章

(120分钟 150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.(2012·山东高考)已知全集U={0,1,2,3,4},集合A={1,2,3},B={2,4},则(ðUA)∪B为( )

A.{1,2,4} B.{2,3,4}

C.{0,2,4} D.{0,2,3,4}

2.如图可作为函数y=f(x)的图象的是(

)

3.已知集合P={x|x2=1},集合Q={x|ax=1},若Q⊆P,那么a的值是( )

A.1 B.-1

C.1或-1 D.0,1或-1

4.方程x2-px+6=0的解集为M,方程x2+6x-q=0的解集为N,且M∩N={2},那么 p+q=( ){45分钟作业与单元评估数学必修一答案}.

A.21 B.8 C.6 D.7

5.(2012·安徽高考)下列函数中,不满足f(2x)=2f(x)的是( )

A.f(x)=|x| B.f(x)=x-|x|

C.f(x)=x+1 D.f(x)=-x

6.(2013·衡水高一检测)下列各组中的两个函数是同一函数的为( ) (1)y=

(2)y=(3)y=x,y=

(4)y=x,y=

(5)y=(. . )2,y=2x-5. ,y=x-5. ,y=. A.(1),(2) B.(2),(3)

C.(3),(5) D.(4)

7.下面4个结论:

①偶函数的图象一定与y轴相交;②奇函数的图象一定通过原点;③偶函数的图象关于y轴对称;④既是奇函数,又是偶函数的函数一定是f(x)=0(x∈R),上述正确说法的个数是( ){45分钟作业与单元评估数学必修一答案}.

A.1 B.2 C.3 D.4

8.已知A={0,1},B={-1,0,1},f是从A到B映射的对应关系,则满足f(0)>f(1)的映射有( )

A.3个 B.4个 C.5个 D.6个

9.若函数y=f(x)的定义域是[-2,4],则函数g(x)=f(x)+f(-x)的定义域是( )

A.[-4,4] B.[-2,2]

C.[-4,-2] D.[2,4]

10.若f(x)= 则f(x)的最大值,最小值分别为( )

A.10,6 B.10,8 C.8,6 D.8,8

11.函数f(x)是定义在R上的奇函数,下列说法:

①f(0)=0;

②若f(x)在[0,+∞)上有最小值为-1,则f(x)在(-∞,0]上有最大值为1; ③若f(x)在[1,+∞)上为增函数,则f(x)在(-∞,-1]上为减函数;

④若x>0时,f(x)=x2-2x,则x<0时,f(x)=-x2-2x.其中正确说法的个数是( )

A.1个 B.2个 C.3个 D.4个

12.f(x)满足对任意的实数a,b都有f(a+b)=f(a)·f(b)且f(1)=2,

则+

++„+=( ){45分钟作业与单元评估数学必修一答案}.

A.1 006 B.2 014 C.2 012 D.1 007

二、填空题(本大题共4小题,每小题5分,共20分.请把正确答案填在题中的横线上)

13.(2012·广东高考)函数y=

14.若函数f(x)=的定义域为 . 则f(-3)= .

15.已知二次函数f(x)=ax2+2ax+1在区间[-3,2]上的最大值为4,则a的值为 .

16.若函数f(x)同时满足①对于定义域上的任意x,恒有f(x)+f(-x)=0;②对于定义域上的任意x1,x2,当x1≠x2时,恒有<0,则称函数f(x)为“理想函数”.

给出下列三个函数中:(1)f(x)=.

(2)f(x)=x2.(3)f(x)=

号).

三、解答题(本大题共6小题,共70分,解答时写出必要的文字说明、证明过程或演算步骤)

17.(10分)已知A={x|x2-ax+a2-19=0},B={x|x2-5x+6=0},C={x|x2+2x-8=0},且能被称为“理想函数”的有 (填相应的序(A∩B),A∩C=,求a的值.

18.(12分)已知函数f(x-1)=x2-4x,求函数f(x),f(2x+1)的解析式.

19.(12分)某省两相近重要城市之间人员交流频繁,为了缓解交通压力,特修一条专用铁路,用一列火车作为交通车,已知该车每次拖4节车厢,一天能来回16次,如果每次拖7节车厢,则每天能来回10次.

(1)若每天来回的次数是车头每次拖挂车厢节数的一次函数,求此一次函数解析式.

(2)在(1)的条件下,每节车厢能载乘客110人.问这列火车每天来回多少次才能使运营人数最多?并求出每天最多运营人数.

20.(12分)已知函数f(x)=,

(1)判断函数在区间[1,+∞)上的单调性,并用定义证明你的结论.

(2)求该函数在区间[1,4]上的最大值与最小值.

21.(12分)(能力挑战题)设f(x)是定义在R上的函数,对任意x,y∈R,恒有f(x+y)=f(x)+f(y).

(1)求f(0)的值.

(2)求证:f(x)为奇函数.

(3)若函数f(x)是R上的增函数,已知f(1)=1,且f(2a)>f(a-1)+2,求a的取值范围.

22.(12分)(能力挑战题)已知二次函数f(x)的最小值为1,且f(0)=f(2)=3.

(1)求f(x)的解析式.

(2)若f(x)在区间[2a,a+1]上不单调,求实数a的取值范围.

(3)在区间[-1,1]上,y=f(x)的图象恒在y=2x+2m+1的图象上方,试确定实数m的取值范围.

答案解析

1. 【解题指南】先求集合A关于全集U的补集,再求它与集合B的并集即可.

【解析】选C.(ðUA)∪B={0,4}∪{2,4}={0,2,4}.

2.【解析】选D.只有选项D中对定义域内任意x都有唯一的y值与之对应.

3.【解析】选D.P={-1,1},QP,所以(1)当Q=时,a=0.(2)当Q≠时,Q={},≨=1或=-1,解之得a=〒1.

【变式备选】(2012〃上海高考改编)若集合A={x|2x+1>0},B={x|-2<x-1<2},则A∩B= .

【解题指南】本题考查集合的交集运算知识,此类题的易错点是临界点的大小比较.

【解析】集合A={x|2x+1>0}={x|x>-},集合B={x|-2<x-1<2}={x|-1<x<3},所以A∩B={x|-<x<3}.

45分钟作业与单元评估数学必修一答案篇六

高一数学必修一作业本【答案】

高中新课程作业本 数学 必修1

答案与提示 仅供参考

第一章集合与函数概念

1．1集合

1 1 1集合的含义与表示

1.D.2.A.3.C.4.{1,-1}.5.{x|x=3n+1,n∈N}.6.{2,0,－2}.

7.A={(1,5),(2,4),(3,3),(4,2),(5,1)}.8.1.9.1,2,3,6.

10.列举法表示为{(-1,1),(2,4)},描述法的表示方法不唯一,如可表示为(x,y)|y=x+2,

y=x2.

11.-1,12,2.

1 1 2集合间的基本关系

1.D.2.A.3.D.4. ,{-1},{1},{-1,1}.5. .6.①③⑤.

7.A=B.8.15,13.9.a≥4.10.A={ ,{1},{2},{1,2}},B∈A.

11.a=b=1．

1 1 3集合的基本运算(一)

1.C.2.A.3.C.4.4.5.{x|-2≤x≤1}.6.4.7.{-3}.

8.A∪B={x|x＜3,或x≥5}.9.A∪B={-8,-7,-4,4,9}.10.1.

11.{a|a=3,或-22＜a＜22}．提示:∵A∪B=A，∴B A．而A={1，2}，对B进行讨论：①当B= 时，x2-ax+2=0无实数解，此时Δ=a2-8＜0，∴-22＜a＜22.②当B≠ 时，B={1,2}或B={1}或B={2};当B={1,2}时，a=3;当B={1}或B={2}时，

Δ=a2-8=0，a=±22，但当a=±22时，方程x2-ax+2=0的解为x=±2，不合题意． 1 1 3集合的基本运算(二)

1.A.2.C.3.B.4.{x|x≥2,或x≤1}.5.2或8.6.x|x=n+12,n∈Z.

7.{-2}.8.{x|x＞6,或x≤2}.9.A={2,3,5,7},B={2,4,6,8}．

10.A,B的可能情形

有:A={1,2,3},B={3,4};A={1,2,4},B={3,4};A={1,2,3,4},B={3,4}.

11.a=4,b=2.提示:∵A∩ 綂 UB={2}，∴2∈A，∴4+2a-12=0 a=4，

∴A={x|x2+4x-12=0}={2,-6},∵A∩ 綂 UB={2}，∴－6 綂 UB，∴－6∈B，将x=-6代入B,得b2-6b+8=0 b=2,或b=4.①当b=2

时,B={x|x2+2x-24=0}={-6,4},∴-6 綂 UB，而2∈ 綂 UB，满足条件A∩ 綂 UB={2}.②当b=4时,B={x|x2+4x-12=0}={-6,2},

∴2 綂 UB，与条件A∩ 綂 UB={2}矛盾．

1．2函数及其表示

1 2 1函数的概念（一）

1.C.2.C.3.D.4.22.5.-2,32∪32,+∞.6.［1,+∞).

7.(1)12,34.(2){x|x≠-1，且x≠-3}．8.-34.9.1.

10.(1)略.(2)72.11.-12,234.

1 2 1函数的概念（二）

1.C.2.A.3.D.4.{x∈R|x≠0,且x≠-1}.5.［0，+∞).6.0.

7.-15,-13,-12,13.8.(1)y|y≠25.(2)［-2,+∞).

9.(0,1］．10.A∩B=-2,12;A∪B=［-2,+∞).11.［-1,0).

1 2 2函数的表示法(一)

1.A.2.B.3.A.4.y=x100.5.y=x2-2x+2.6.1x.7.略.

8.

x1234y828589889.略.10.1.11.c=-3.

1 2 2函数的表示法(二)

1.C.2.D.3.B.4.1.5.3.6.6.7.略.

8.f(x)＝2x(-1≤x＜0),

-2x+2(0≤x≤1).

9.f(x)=x2-x+1.提示:设f(x)=ax2+bx+c，由f(0)=1,得c=1，又f(x+1)-f(x)=2x，即a(x+1)2+b(x+1)+c-(ax2+bx+c)=2x，展开得2ax+(a+b)=2x,所以2a=2, a+b=0,解得a=1，b=-1.

10.y=1.2(0＜x≤20),

2.4(20＜x≤40),

3.6(40＜x≤60),

4.8(60＜x≤80).11.略．

1．3函数的基本性质

1 3 1单调性与最大（小）值(一)

1.C.2.D.3.C.4.［-2,0),［0,1),［1,2］.5.-∞,32.6.k＜12．

7.略.8.单调递减区间为(-∞,1),单调递增区间为［1,+∞).9.略.10.a≥-1．

11.设－1＜x1＜x2＜1，则f(x1)－f(x2)＝x1x21-1－x2x22-1＝

(x1x2+1)(x2-x1)(x21-1)(x22-1)，∵x21－1＜0,x22－1＜0,x1x2＋1＜0,x2－x1＞0，∴(x1x2+1)(x2-x1)(x21-1)(x22-1)＞0，∴函数y＝f(x)在(－1，1)上为减函数．

1 3 1单调性与最大（小）值(二)

1.D.2.B.3.B.4.-5,5.5.25.

6.y=316(a+3x)(a-x)(0＜x＜

a),312a2,5364a2.7.12.8.8a2+15.9.(0,1］.10.2500m2.

11.日均利润最大，则总利润就最大．设定价为x元，日均利润为y元．要获利每桶定价必须在12元以上，即x＞12．且日均销售量应为440-(x-13)·40＞0，即x＜23,总利润y=(x-12)［440-(x-13)·40］-600(12＜x＜23),配方得

y=-40(x-18)2+840,所以当x=18∈(12,23)时，y取得最大值840元，即定价为18元时，日均利润最大.

1 3 2奇偶性

1.D.2.D.3.C.4.0.5.0.6.答案不唯一,如y=x2.

7.(1)奇函数.(2)偶函数.(3)既不是奇函数，又不是偶函数.(4)既是奇函数，又是偶函数.

8.f(x)=x(1+3x)(x≥0),

x(1-3x)(x＜0).9.略.

10.当a=0时，f(x)是偶函数；当a≠0时，既不是奇函数，又不是偶函数.

11.a=1，b=1，c=0.提示:由f(－x)=－f(x)，得c=0，

∴f(x)=ax2+1bx,∴f(1)=a+1b=2 a=2b-1.∴f(x)=(2b-1)x2+1bx.∵f(2)＜3，∴4(2b-1)+12b＜3 2b-32b＜0 0＜b＜32.∵a,b,c∈Z,∴b=1,∴a=1.

单元练习

1.C.2.D.3.D.4.D.5.D.6.B.7.B.8.C.9.A.

10.D.11.{0,1,2}.12.-32.13.a=-1,b=3.14.［1,3)∪(3,5］.

15.f12＜f(-1)＜f-72.16.f(x)=-x2-2x-3.

17.T(h)=19-6h(0≤h≤11),

-47(h＞11).18.{x|0≤x≤1}．

19.f(x)=x只有唯一的实数解，即xax+b=x(*)只有唯一实数解，当ax2+(b-1)x=0有相等的实数根x0，且ax0+b≠0时，解得f(x)=2xx+2，当ax2+(b-1)x=0有不相等的实数根，且其中之一为方程(*)的增根时，解得f(x)=1．

20.(1)x∈R,又f(-x)=(-x)2-2|-x|-3=x2-2|x|-3=f(x)，所以该函数是偶函数.(2)略.(3)单调递增区间是［-1,0］,［1,+∞)，单调递减区间是(-∞,-1］,［0,1］.

21.（1）f(4)=4×1

3=5.2,f(5.5)=5×1.3+0.5×3.9=8.45,f(6.5)=5×1.3+1×3.9+0.5×6 5=13.65.

（2）f(x)=1.3x(0≤x≤5),

3.9x-13(5＜x≤6),

6.5x-28.6(6＜x≤7).

22.（1）值域为［22,+∞).（2）若函数y=f(x)在定义域上是减函数，则任取x1,x2∈(0,1］且x1＜x2，都有f(x1)＞f(x2)成立，即(x1-x2)2+ax1x2＞0，只要a＜-2x1x2即可，由于x1,x2∈(0,1］，故-2x1x2∈(-2,0)，a＜-2，即a的取值范围是(-∞,-2)．

第二章基本初等函数(Ⅰ)

2．1指数函数

2 1 1指数与指数幂的运算(一)

1.B.2.A.3.B.4.y=2x(x∈N).5.(1)2.(2)5.6.8a7.

7.原式=|x-2|-|x-3|=-1(x＜2),

2x-5(2≤x≤3),

1(x＞3).8.0.9.2011.10.原式=2yx-y=2.

11.当n为偶数,且a≥0时,等式成立;当n为奇数时,对任意实数a,等式成立. 2 1 1指数与指数幂的运算(二)

1.B.2.B.3.A.4.94.5.164.6.55.

7.(1)-∞,32.(2)x∈R|x≠0,且x≠-52.8.原式=52-1+116+18+110=14380.

9.-9a.10.原式=(a-1+b-1)·a-1b-1a-1+b-1=1ab.

11.原式=1-2-181+2-181+2-141+2-121-2-18=12-827.

2 1 1指数与指数幂的运算(三)

1.D.2.C.3.C.4.36.55.5.1-2a.6.225.7.2.

8.由8a=23a=14=2-2,得a=-23,所以f(27)=27-23=19.9.4 7288,0 0885.

10.提示：先由已知求出x-y=-(x-y)2=-(x+y)2-4xy=-63,所以原式

=x-2xy+yx-y=-33.

11.23.{45分钟作业与单元评估数学必修一答案}.

2 1 2指数函数及其性质(一)

1.D.2.C.3.B.4.A B.5.(1,0).6.a＞0.7.125.

8.(1)图略.（2）图象关于y轴对称.

9.(1)a=3,b=-3.（2）当x=2时,y有最小值0;当x=4时,y有最大值6.10.a=1.

11.当a＞1时，x2-2x+1＞x2-3x+5，解得{x|x＞4}；当0＜a＜1时，x2-2x+1＜x2-3x+5，解得{x|x＜4}.

2 1 2指数函数及其性质(二)

1.A.2.A.3.D.4.(1)＜.(2)＜.(3)＞.(4)＞.

5.{x|x≠0},{y|y＞0，或y＜-1}.6.x＜0.7.56-0.12＞1=π0＞0.90.98.

8.(1)a=0.5.(2)-4＜x≤0.9.x2＞x4＞x3＞x1.

10.(1)f(x)=1(x≥0),

2x(x＜0).(2)略.11.am+a-m＞an+a-n.

2 1 2指数函数及其性质(三)

1.B.2.D.3.C.4.-1.5.向右平移12个单位.6.(-∞,0).

7.由已知得0.3(1-0.5)x≤0.08,由于0.51.91=0.2667,所以x≥1.91,所以2h后才可驾驶.

8.(1-a)a＞(1-a)b＞(1-b)b.9.815×(1+2%)3≈865(人).

10.指数函数y=ax满足f(x)·f(y)=f(x+y)；正比例函数y=kx(k≠0)满足f(x)+f(y)=f(x+y).

11.34,57.

2．2对数函数

2 2 1对数与对数运算(一)

1.C.2.D.3.C.4.0;0;0;0.5.(1)2.(2)-52.6.2.

7.(1)-3.(2)-6.(3)64.(4)-2.8.(1)343.(2)-12.(3)16.(4)2.

9.(1)x=z2y,所以x=(z2y)2=z4y(z＞0,且z≠1).(2)由x+3＞0,2-x＜0，且2-x≠1,得-3＜x＜2,且x≠1.

10.由条件得lga=0,lgb=-1，所以a=1,b=110,则a-b=910.

11.左边分子、分母同乘以ex，去分母解得e2x=3,则x=12ln3.

2 2 1对数与对数运算(二)

1.C.2.A.3.A.4.0 3980.5.2logay-logax-3logaz.6.4.

7.原式=log2748×12÷142=log212=-12.

8.由已知得(x-2y)2=xy，再由x＞0,y＞0,x＞2y,可求得xy=4.9.略.10.4.

11.由已知得(log2m)2-8log2m=0,解得m=1或16.

2 2 1对数与对数运算(三)

1.A.2.D.3.D.4.43.5.24.6.a+2b2a.

7.提示：注意到1-log63=log62以及log618=1+log63,可得答案为1.

8.由条件得3lg3lg3+2lg2=a,则去分母移项，可得(3-a)lg3=2alg2,所以lg2lg3=3-a2a.

9.2 5.10.a=log34+log37=log328∈(3，4).11.1.

2 2 2对数函数及其性质(一)

1.D.2.C.3.C.4.144分钟.5.①②③.6.-1.

7.-2≤x≤2.8.提示：注意对称关系.

9.对loga(x+a)<1进行讨论:①当a>1时,0<x+a<a,得-a<x<0;②当0<a<1时,x+a>a,得x>0.

10.C1：a=32，C2:a=3，C3:a=110，C4:a=25.

11.由f(-1)=-2,得lgb=lga-1①,方程f(x)=2x即x2+lga·x+lgb=0有两个相等的实数根，可得lg2a-4lgb=0,将①式代入,得a=100,继而b=10.

2 2 2对数函数及其性质(二)

1.A.2.D.3.C.4.22,2.5.(-∞,1).6.log20 4＜log30.4＜log40.4.

7.logbab＜logba＜logab.8.(1)由2x-1＞0得x＞0.(2)x＞lg3lg2.

9.图略，y=log12(x+2)的图象可以由y=log12x的图象向左平移2个单位得到.

10.根据图象,可得0＜p＜q＜1.11.(1)定义域为{x|x≠1},值域为R.(2)a=2. 2 2 2对数函数及其性质(三)

1.C.2.D.3.B.4.0，12.5.11.6.1,53.

7.（1）f35=2,f-35=-2.(2)奇函数，理由略.8.{-1，0，1，2，3，4，5，6}.

9.(1)0.(2)如log2x.

10.可以用求反函数的方法得到,与函数y=loga(x+1)关于直线y=x对称的函数应该是y=ax-1,和y=logax+1关于直线y=x对称的函数应该是y=ax-1.

11.(1)f(-2)+f(1)=0.(2)f(-2)+f-32+f12+f(1)=0.猜想:f(-x)+f(-1+x)=0,证明略.

2 3幂函数

1.D.2.C.3.C.4.①④.5.6.2518＜0.5-12＜0.16-14.

6.(-∞,-1)∪23,32.7.p=1,f(x)=x2.

8.图象略,由图象可得f(x)≤1的解集x∈［-1,1］.9.图象略,关于y=x对称.

10.x∈0,3+52.11.定义域为(-∞,0)∪(0,∞),值域为（0，∞），是偶函数，图象略.

单元练习

1.D.2.D.3.C.4.B.5.C.6.D.7.D.8.A.9.D.

10.B.11.1.12.x＞1.13.④.14.25 8.提示：先求出h=10.

15.（1）-1.(2)1.

16.x∈R，y=12x=1+lga1-lga＞0,讨论分子、分母得-1＜lga＜1，所以a∈110,10.

17.（1）a=2.（2）设g(x)＝log12(10-2x)－12x,则g(x)在［3,4］上为增函数,g(x)＞m对x∈［3,4］恒成立，m＜g(3)=－178．

18.(1)函数y=x+ax(a＞0),在(0,a］上是减函数,［a,+∞）上是增函数,证明略.

(2)由(1)知函数y=x+cx(c＞0)在［1,2］上是减函数,所以当x=1时,y有最大值1+c;当x=2时,y有最小值2+c2.

19.y=(ax+1)2-2≤14，当a＞1时，函数在［-1，1］上为增函数，ymax=(a+1)2-2=14，此时a=3；当0＜a＜1时，函数［-1，1］上为减函数，ymax=(a-1+1)2-2=14，此时a=13.∴a=3,或a=13.

20.(1)F(x)=lg1-xx+1+1x+2,定义域为(-1,1).

(2)提示:假设在函数F(x)的图象上存在两个不同的点A,B，使直线AB恰好与y轴垂直,则设A(x1,y),B(x2,y)(x1≠x2),则f(x1)-f(x2)=0,而

f(x1)-f(x2)=lg1-x1x1+1+1x1+2-lg1-x2x2+1-1x2+2=lg(1-x1)(x2+1)(x1+1)(1-x2)+x2-x1(x1+2)(x2+2)=①+②,可证①，②同正或同负或同为零,因此只有当x1=x2时,f(x1)-f(x2)=0,这与假设矛盾,所以这样的两点不存在.(或用定义证明此函数在定义域内单调递减)第三章函数的应用

3 1函数与方程

3 1 1方程的根与函数的零点

1.A.2.A.3.C.4.如：f(a)f(b)≤0.5.4,254.6.3.

7.函数的零点为-1，1，2.提示：f(x)=x2(x-2)-(x-2)=(x-2)(x-1)(x+1).

8.(1)(-∞,-1)∪(-1,1).(2)m=12．

9.（1）设函数f(x)=2ax2-x-1,当Δ=0时，可得a=-18，代入不满足条件，则函数f(x)在（0，1）内恰有一个零点.∴f(0)·f(1)＝-1×(2a-1-1)＜0，解得a＞1.

（2）∵在［-2，0］上存在x0，使f(x0)=0,则f(-2)·f(0)≤0,∴（-6m-4）×(-4)≤0,