最新版2017教师用书步步高大一轮复习讲义习题详细答案第3章第8讲

格式：pdf
大小：752.37 KB
文档页数：13

下载文档原格式

2017版新步步高高考数学大一轮复习讲义课件：第3章三角函数、解三角形 3.3

(k∈Z) 上递增；
(k∈Z)
＋2kπ，
3π
2
在 [2kπ，π＋2kπ]
(k∈Z)
上递减
＋2kπ]
π
2
x＝＋2kπ(k∈Z)
ymax＝1;
x＝－
π
2
＋2kπ(k∈Z)
ymin＝－1
上递增；
在(－
π
2
π
＋kπ，
2
(k∈Z)
＋kπ)
上递增
(k∈Z) 上递减
时,
时；
当x＝ 2kπ(k∈Z) 时，
ymax＝1；
2
π
5π
所以 2kπ＋ ≤x≤2kπ＋ (k∈Z).故选 B.
6
6
第十七页，编辑于星期六：三点十三分。
解析答案
(2)函数

π
π

f(x)＝3sin2x－在区间[0， ]上的值域为(
6
2

3 3

A.－，
2 2

3

B.－，3
2

3
3
3
3

C.－
＋
≥
，
2
4 2
所以
π 3π

ωπ＋4≤ 2 ，
1
5
解得2≤ω≤4.
第二十四页，编辑于星期六：三点十三分。
思维升华
解析答案

π

(1)函数 f(x)＝sin－2x＋的单调减区间为
3

π
5

，kπ＋12π，k∈Z
kπ－
12

2017版新步步高高考数学大一轮复习讲义课件：第3章导数及其应用 3.2 课时3

思第维十一升页，华编辑于星期六：三解点析十四答分。案
思维升华
(1)利用导数解不等式，一般可构造函数，利用已知条件确定函数单调性解不等式； (2)证明不等式f(x)<g(x)，可构造函数F(x)＝f(x)－g(x)，利用导数求F(x)的值域，得到F(x)<0即可； (3)利用导数研究不等式恒成立问题，首先要构造函数，利用导数研究函数的单调性，求出最值，进而得出相应的含参不等式，从而求出参数的取值范围；也可分离变量，构造函数，直接把问题转化为函数的最值问题.
第五十二页，编辑于星期六：解三点析十答四案分。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
8.若对于任意实数x≥0，函数f(x)＝ex＋ax恒大于零，则实数a的取值范围是________.
第五十四页，编辑于星期六：解三点析十答四案分。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
9.设a为实数，函数f(x)＝ex－2x＋2a，x∈R. (1)求f(x)的单调区间与极值；
解析答案第五十六页，编辑于星期六：三点十四分。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(2)求证：当a>ln 2－1且x>0时，ex>x2－2ax＋1. 证明设g(x)＝ex－x2＋2ax－1，x∈R，于是g′(x)＝ex－2x＋2a，x∈R. 由(1)知当a>ln 2－1时，g′(x)取最小值为g′(ln 2)＝2(1－ln 2＋a)>0. 于是对任意x∈R，都有g′(x)>0，所以g(x)在R内单调递增. 于是当a>ln 2－1时，对任意x∈(0，＋∞)，都有g(x)>g(0). 而g(0)＝0，从而对任意x∈(0，＋∞)，都有g(x)>0. 即ex－x2＋2ax－1>0，故当a>ln 2－1且x>0时，ex>x2－2ax＋1.

新步步高大一轮复习讲义英语人教版

新步步高大一轮复习讲义英语人教版新步步高大一轮复习讲义Unit 1 School lifeIn this unit, we will focus on school life and explore various topics related to education and learning. By the end of this unit, you should have a better understanding of different aspects of school life and be able to express your opinions and experiences in English. Let's get started!Part 1: School Subjects and Activities1.1 SubjectsIn this section, we will learn and review different school subjects. Vocabulary related to subjects such as mathematics, English, history, and biology will be introduced, along with common phrases and expressions used in the classroom.1.2 Extra-curricular ActivitiesBesides academic subjects, schools also offer a variety of extra-curricular activities. We will discuss popular activities such as sports, arts, and music, and talk about their benefits and the skills they develop.Part 2: School Rules and Regulations2.1 School RegulationsEvery school has its own set of rules and regulations. We will discuss common rules such as uniform requirements, punctuality, and behavior in the classroom, and learn how to talk about rules and regulations in English.2.2 Disciplinary ActionsWhen students break the rules, there are consequences. We will explore different disciplinary actions schools take, such as warnings, detentions, and suspensions, and practice discussing consequences in English.Part 3: School Life and Experiences3.1 Daily RoutineSchool life revolves around a daily routine. We will talk about typical activities in a school day, including attending classes, doing homework, and participating in various school events. You will also learn how to describe your own daily routine in English.3.2 School Events and CelebrationsThroughout the school year, there are special events and celebrations that bring the school community together. We will discuss events such as sports day, graduation ceremonies, and cultural festivals, and learn how to talk about these events in English.Part 4: School Facilities and Resources4.1 Classroom Tools and TechnologySchools are equipped with various tools and technology to facilitate learning. We will explore common classroom resources such as whiteboards, projectors, and computers, and practice using correct terminology in English.4.2 Library and Study SpacesA well-stocked library and comfortable study spaces are essential for a conducive learning environment. We will discuss the importance of libraries and study spaces, as well as how to ask for assistance and borrow books in English.Part 5: Exam Preparation5.1 Study SkillsPreparing for exams requires effective study skills. We will introduce different study techniques, such as note-taking, summarizing, and self-testing, and provide tips on how to improve your study habits.5.2 Exam StrategiesTo perform well in exams, it is important to develop exam strategies. We will discuss time management, question analysis, and exam tips to help you approach exams with confidence.ConclusionIn this unit, we have explored various aspects of school life, including subjects, activities, rules and regulations, daily routines, facilities, and exam preparation. By mastering the vocabulary and expressions introduced in this unit, you will be better equipped to communicate and navigate through your school life in English. Keep practicing and learning, and good luck with your studies!。

2017版新步步高高考数学大一轮复习讲义课件：第3章三角函数、解三角形 3.6

4－3 3 (2)已知 cosθ＋π4＝ 1100，θ∈0，π2，则 sin2θ－π3＝___1_0____. 解析由题意可得，cos2θ＋π4＝1＋cos22θ＋π2＝110，
cos2θ＋2π＝－sin 2θ＝－54，即 sin 2θ＝54.
因为 cosθ＋π4＝ 1100>0，θ∈0，π2，所以 0<θ<π4，2θ∈0，π2，根据同角三角函数基本关系式可得 cos 2θ＝35，由两角差的正弦公式可得
123 45
解析答案第九页，编辑于星期六：三点十三分。
4.若
f(x)＝2tan
2sin2 x－ x
2x－1x，则
sin 2cos 2
f1π2的值为___8___.
解析
1－2sin2 ∵f(x)＝2tan x＋ 1
x 2
2sin x
＝2tan
x＋2scionsxx＝sin
2 xcos
x＝sin42x，
得c2oassiθn21θ－－2siansiθn－θa＝＝01，，
由 θ∈－π2，π2知 cos θ≠0， a＝－1，
解得θ＝－π6.
思维升华第二十八页，编辑于星期六：解三析点十答三案分。
跟踪训练3
(1)(2014·课标全国Ⅱ)函数f(x)＝sin(x＋φ)－2sin φcos x的最大值为 ___1_____. 解析因为f(x)＝sin(x＋φ)－2sin φcos x ＝sin xcos φ－cos xsin φ＝sin(x－φ)，－1≤sin(x－φ)≤1，所以f(x)的最大值为1.
π
A.3
B. 3
C.6
D.4
解析由已知可得 tan A＋tan B＝ 3(tan A·tan B－1)，

2017版新步步高高考数学大一轮复习讲义课件：第8章平面解析几何 8.9 课时3

思维升华
已知椭圆 C 过点

M1，

6
，点 F(－ 2，0)是椭圆的左焦点，
2
跟踪训练1
点 P，Q 是椭圆 C 上的两个动点，且|PF|，|MF|，|QF|成等差数列.
(1)求椭圆C的标准方程；
x2 y2
解设椭圆 C 的方程为a2＋b2＝1 (a>b>0).

6

2

4
a
1

＝4，
解析答案
思维升华
1
1
如图，在平面直角坐标系 xOy 中，点 F(2，0)，直线 l：x＝－2，跟踪训练2
点 P 在直线 l 上移动，R 是线段 PF 与 y 轴的交点，RQ⊥FP，
PQ⊥l.
(1)求动点Q的轨迹C的方程；
解
依题意知，点R是线段FP的中点，且RQ⊥FP，
∴RQ是线段FP的垂直平分线.
题型二
例2
定值问题
x2 y2
1
已知椭圆 C：a2＋b2＝1 (a>b>0)的离心率是2，其左，右顶点分别为 A1，
A2，B 为短轴的一个端点，△A1BA2 的面积为 2 3.
(1)求椭圆C的方程；
解

c 1
e＝＝，
a 2

由已知，可得ab＝2 3，

a2＝b2＋c2，

解得 a＝2，b＝ 3.
第八章平面解析几何
§8.9
圆锥曲线的综合问题
第一页，编辑于星期六：三点十八分。
内容
索引
课时3 定点、定值、探索性问题
题型一定点问题
题型二
定值问题
题型三探索性问题

2017版新步步高高考数学大一轮复习讲义课件：第3章导数及其应用 3.2 课时1

题型一不含参数的函数的单调性
例1 求函数 f(x)＝lnxx的单调区间.
解函数f(x)的定义域为(0，＋∞).
因为
f(x)＝lnx
x，所以
1－ln f′(x)＝ x2
x .
当f′(x)>0，即0<x<e时，函数f(x)单调递增；
当f′(x)<0，即x>e时，函数f(x)单调递减.
故函数f(x)的单调递增区间为(0，e)，
4.定义在R上的函数f(x)满足：f′(x)>f(x)恒成立，若x1<x2，则ex1 f x2 与
ex2 f x1 的大小关系为_e_x_1 f__ _x2____e_x2_f__ x_1_.
解析设 g(x)＝fexx，
f′xex－fxex f′x－fx
则 g′(x)＝
ex2
＝
ex
，
由题意g′(x)>0，所以g(x)单调递增，
返回第三十九页，编辑于星期六：三点十四分。
练出高分
第四十页，编辑于星期六：三点十四分。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1.函数f(x)＝(x－3)ex的单调递增区间是____(2_，__＋__∞. ) 解析函数f(x)＝(x－3)ex的导数为f′(x)＝[(x－3)ex]′＝ex＋(x－3)ex＝(x －2)ex. 由函数导数与函数单调性的关系，得当f′(x)>0时，函数f(x)单调递增，此时由不等式f′(x)＝(x－2)ex>0，解得x>2.
解析答案第四十六页，编辑于星期六：三点十四分。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

最新版2017教师用书步步高大一轮复习讲义习题详细答案第3章第8讲

合集下载

最新版2017教师用书步步高大一轮复习讲义习题详细答案第3章第10讲