How to introduce the solution of inequality with absolute value into the topic?

B)∈M, and for other elements (c, d) in m, there is always c≥a, then a = _ _ _.

Analysis: Understanding and revealing the mathematical essence of the problem will be the breakthrough to solve the problem. How to understand "for other elements (C, D) in M, there is always c≥a"? What are the characteristics of the elements in M?

Solution: According to the problem, this problem is equivalent to finding the function x=f(y)=(y+3)? 6? 1|y- 1|+(y+3)

② When 1≤y≤3,

So when y= 1, = 4.

Comments: The form of set appears in the conditions of topic setting, so it is necessary to recognize the essential attributes of set elements and then reveal them in combination with conditions.

Its mathematical essence is to find the elements in the set m that satisfy the relationship.

Example 2. If a non-negative real number is known, and the maximum value is ().

A.B. C. D。

Solution: draw an image, which can be obtained from linear programming knowledge, and choose D.

Example 3. This order is determined by the following conditions:

(1) Proof: For,

(2) Proof: Yes.

Proof: (1)

(2) When,

= 。

Example 4. Solve inequalities about:

Analysis: This example mainly reviews the idea of solving inequality with absolute value and classified discussion. The key to this problem is not to discuss the parameters, but to discuss the unknowns when the absolute values are removed, and two groups of inequalities are obtained. Finally, the solution sets of the two groups of inequalities are combined to get the solution set of the original inequality.

Solution: When

Example 5. If the image of quadratic function y=f(x) crosses the origin, and 1≤f(- 1)≤2, 3≤f( 1)≤4, find the range of f(-2).

Analysis: If you want the range of f(-2), you only need to find the inequality (group) containing f(-2). Because y=f(x) is a quadratic function, you should write the expression of f(x) first, then you can get the expression of f(-2), and then list the inequalities containing f(-2) according to the conditions.

Solution: Because the image of y=f(x) passes through the origin, it can be set as y = f (x) = AX2+Bx. therefore

Solution 1 (using the properties of basic inequalities)

The inequality set (Ⅰ) is deformed.

(Ⅰ)

So the range of f(-2) is [6, 10].

Solution 2 (combination of numbers and shapes)

Establish a rectangular coordinate system aob so that the area is represented by the inequality group (I), as shown in the shaded part of Figure 6. Because f(-2)=4a-2b, 4a-2b-f(-2)=0 represents a linear system with a slope of 2. As shown in fig. 6, when the straight line 4a-2b-f(-2)=0 passes through the point A(2,

Solution 3 (Using the Idea of Equation)

F (-2) = 4a-2b = 3f (-1)+f (1), and

1≤f(- 1)≤2，3≤f( 1)≤4，①

So 3 ≤ 3f (- 1) ≤ 6。 ②.

①+② 4 ≤ 3f (-1)+f (1) ≤10, that is, 6 ≤ f (-2) ≤ 10.

Note: (1) When solving inequalities, the deformation of the same solution is needed. One of the following error solutions should be avoided:

2b, 8≤4a≤ 12, -3≤-2b≤- 1, so 5 ≤ f (-2) ≤1.

(2) The key step to solve this kind of problem is to find the mathematical structure of f(-2), then reveal its algebraic and geometric essence according to its mathematical structure characteristics, and solve the same problem from different angles by using the basic properties of inequality, the combination of numbers and shapes, equations and other mathematical thinking methods. If you think like this for a long time, your math literacy will definitely improve quickly.

Example 6. Let f(x)=ax2+bx+c and two straight lines y=x, y= x disjoint image. Try to prove that it is true for everything.

Analysis: Because x∈R, if the minimum value of |f(x)| exists, the minimum value is determined by the vertex, so let f (x) = a (x-x0) 2+f (x0).

Proof: A ≠ 0. Let f(x)=a(x-x0)2+f(x0), then

The quadratic equation AX2+BX+C = X has no real root, so

δ 1 =(b+ 1)2-4ac < 0，δ2 =(b- 1)2-4ac < 0。

Therefore, (b+1) 2+(b-1) 2-8ac < 0, that is, 2b2+2-8ac < 0, that is, B2-4ac 1.

Comments: As can be seen from the above examples, when proving the inequality problem related to quadratic function, if different forms of quadratic function are reasonably adopted according to the conditions of the problem, then an effective proof method can be found.

Example 7. Let the quadratic function f (x) = ax2+bx+c (a > 0), the two roots of the equation f (x)-x = 0, x 1 and x2 satisfy 0 < x 1 < x2.

(1) When x ∈ [0, x 1, it is proved that x < f (x) < x1;

(2) Let the image of the function f(x) be symmetrical about the straight line x=x0, and prove x0.

Solution: (1) Let f (x) = f (x)-x, because x 1 and x2 are the roots of the equation f (x)-x = 0, so f (x) = a (x-x 1) (x-x2).

And a > 0, f (x) = a (x-x 1) (x-x2) > 0, that is, x < f (x).

x 1-F(x)= x 1-〔x+F(x)〕= x 1-x+a(x 1-x)(x-x2)=(x 1-x)〔 1+a(x-x2)〕

∫0 < x < x 1 < x2 < ,∴x 1-x>0, 1+a(x-x2)= 1+ax-ax2> 1-ax2>0

∴ x 1-f (x) > 0, so f (x) < x 1

(2)X0 =- because x 1 and x2 are the two roots of equation F (x)-X = 0, that is, x 1 and x2 are the roots of equation AX2+(B- 1) X+C = 0.

∴x 1+x2=-

X0 =-,because ax2 < 1,

∴x0 0, if there is-1 ≤ x ≤ 1, the maximum value of g(x) is 2, and then find f(x).

Proposition intention: This topic mainly examines the nature of quadratic function, the nature of absolute inequality, and the ability to comprehensively use mathematical knowledge to analyze and solve problems. Belong to the category of ★★★★★.

Knowledge: the related properties of quadratic function and monotonicity of function are drug introduction, and the flexible use of the properties of absolute inequality is the soul of this question.

Error analysis: This problem is comprehensive, and the key to solving it is a deep understanding of the monotonicity of function f(x) and the application of the condition "-1≤ x≤1| f (x) |≤1"; Improper use of the nature of absolute inequality will make the problem-solving process empty and imprecise, thus bringing the problem to a deadlock.

Skills and methods: There are three proofs for this question (2). The first proof makes use of the monotonicity of g(x); Prove that 2 is an absolute inequality: | | A |-| B | |≤| A B |≤| A |+| B |；; Prove 3 deals with the relationship between g(x) and f(x) as a whole.

(1) Prove that if the condition = 1≤x≤ 1, |f(x)|≤ 1, take x=0 to get |c|=|f(0)|≤ 1, that is.

(2) Proof 1: According to the topic |f(0)|≤ 1 and f(0)=c, so |c|≤ 1. When a > 0, g(x)=ax+b is [- 1, 65438+].

g(- 1)≤g(x)≤g( 1)，(- 1≤x≤ 1)。

∫| f(x)|≤ 1，(- 1≤x≤ 1)，|c|≤ 1，

∴g( 1)=a+b=f( 1)-c≤|f( 1)|+|c|=2，

g(- 1)=-a+b =-f(- 1)+c≥-(| f(-2)|+| c |)≥-2，

Therefore | g (x) |≤ 2 (-1≤ x ≤1);

When a < 0, g(x)=ax+b is the decreasing function in [- 1, 1], so g (- 1) ≥ g (x) ≥ g (1), (-/kloc-)

∵| f(x)|≤ 1(- 1≤x≤ 1)，|c|≤ 1

∴|g(x)|=|f( 1)-c|≤|f( 1)|+|c|≤2.

Based on the above results, when-1 ≤ x ≤ 1, there is |g(x)|≤2.

Method 2: ∫| f (x) |≤1(-1≤ x≤1)

∴|f(- 1)|≤ 1,|f( 1)|≤ 1,|f(0)|≤ 1，

∵f(x)=ax2+bx+c,∴|a-b+c|≤ 1,|a+b+c|≤ 1,|c|≤ 1，

Therefore, according to the nature of absolute inequality:

| a-b | = |(a-b+c)-c |≤| a-b+c |+| c |≤2，

| a+b | = |(a+b+c)-c |≤| a+b+c |+| c |≤2，

∵g(x)=ax+b,∴|g( 1)| = | a+b | = | a b |≤2，

The image of the function g(x)=ax+b is a straight line, so the maximum value of |g(x)| on [- 1, 1] can only be obtained at the endpoint of the interval x =- 1 or x= 1, so it is determined by | g (.

When-1 ≤ x ≤ 1, there are 0 ≤1,-1 ≤ 0,

∵|f(x)|≤ 1,(- 1≤x≤ 1),∴|f |≤ 1，| f()|≤ 1；

Therefore, when-1 ≤ x ≤ 1, |g(x)|≤|f |+|f( )|≤2.

(3) solution: since a > 0, g(x) is the increasing function on [- 1, 1], when x= 1, the maximum value of 2 is obtained, that is.

g( 1)= a+b = f( 1)-f(0)= 2。 ①

∵- 1≤f(0)=f( 1)-2≤ 1-2=- 1,∴c=f(0)=- 1.

Because when-1 ≤ x ≤ 1, f (x) ≥- 1, that is, f(x)≥f(0),

According to the property of quadratic function, the straight line x=0 is the symmetry axis of the image of f(x).

This gives -< 0, that is, b=0.

A=2 from ①, so f (x) = 2x2- 1.

Example 9. At the end of 200 1, the number of cars in a city was 300,000. It is estimated that 6% cars will be scrapped at the end of last year every year, and so will the number of new cars every year. In order to protect the urban environment, it is required that the number of cars in the city should not exceed 600 thousand, so how many cars should be added every year?

Solution: Let 200 1 car ownership at the end of the year be, and then the car ownership at the end of each year be, and increase 1 10,000 cars every year. Judging from the meaning of the question.

First, multiple choice questions

1.(★★★★★★) odd function f(x) defined on r is increasing function, and the image of even function g(x) in the interval [0, +∞) coincides with that of f(x). Let a > b > 0, and give the following inequality, where the sequence number of the correct inequality is ().

①f(b)-f(-a)> g(a)-g(-b)②f(b)-f(-a)< g(a)-g(-b)

③f(a)-f(-b)> g(b)-g(-a)④f(a)-f(-b)< g(b)-g(-a)

A.①③ B.②④ C.①④ D.②③

Second, fill in the blanks

2. (★★★★★★) Among the following four propositions: ① A+B ≥ 22Sin2X+≥ 4③ Let X and Y be positive numbers, if = 1, and the minimum value of x+y is124 IF | X-2 | < ε, | Y-2 | < ε.

3. (★★★★★★★) A company rents land to build a warehouse, and the monthly land occupation fee y 1 is inversely proportional to the distance from the garage to the station, while the monthly freight y2 of the inventory goods is directly proportional to the distance to the station. If the warehouse is built at a distance of 0/0 km from the station/kloc-0, then the two costs y 1 and y2 are 20,000 yuan and 80,000 yuan respectively, so we should make them into one.

Third, answer questions.

4. (★★★★★★) Given the quadratic function f(x)=ax2+bx+ 1(a, b∈R, A > 0), let the two real roots of the equation f(x)=x be x 1, x2.

(1) If x 1 < 2 < x2 < 4, let the symmetry axis of function f(x) be x=x0, and verify x0 >-1;

(2) If | x 1 | < 2, | x2-x 1 | = 2, find the range of B. 。

5. (★★★★★★) The original price of a commodity is P yuan, and N pieces will be sold every month. If the price increases by X% (where X% means 0 < X ≤ 10), the monthly sales volume decreases by Y% and the sales amount increases by z times.

(1) let y=ax, where a is a constant that satisfies ≤ A < 1, and a is used to represent the value of x when the sales volume is maximum;

(2) If y= x, find the value range of x for increasing sales.

6. (★★★★★★) Let the function f(x) be defined on r, and there is always f(m+n)=f(m) for any m and n? 6? 1f(n), and when x > 0, 0 < f (x) < 1.

(1) verification: f(0)= 1, when x < 0, f (x) >1;

(2) Verification: f(x) monotonically decreases on r;

(3) Let the set A={ (x, y)|f(x2)? 6? 1f (y2) > f (1)}, let B={(x, y) | f (ax-g+2) = 1, a∈R}, if A∩B=, find a.

7. (★★★★★★) The range of the known function f (x) = (b < 0) is [1, 3],

(1) Find the values of b and c;

(2) Judge the monotonicity of the function F(x)=lgf(x) when x ∈ [- 1, 1], and prove your conclusion;

(3) If t∈R, it is proved that LG ≤ f (| t-| | t+|) ≤ LG.

Inequality relations in mathematics

Mathematics is a science that studies the relationship between spatial form and quantity. Engels pointed out in Dialectics of Nature that mathematics is an auxiliary tool and manifestation of dialectics, which contains extremely rich dialectical materialism factors. The relationship between equality and inequality is a vivid manifestation of this. They are unity of opposites, they are interrelated and influence each other. Equality and inequality are the most basic relations in middle school mathematics.

Equality relation embodies the beauty of symmetry and unity in mathematics, while inequality relation embodies the beauty of singularity in mathematics. Unequal relation originated from the properties of real numbers, which resulted in the size relation of real numbers, simple inequalities and basic properties of inequalities. If the real numbers in simple inequalities are abstracted into mathematical expressions that integrate various mathematical symbols, inequalities will develop into a prosperous family, from simple to complex, with different forms. If the variables in inequality are given specific values and specific relationships, obviously they are not, which leads to two extremely important problems: understanding inequality and proving inequality. Solving inequality is to find the range or conditions that variables should meet when inequality is established, and different types of inequalities have different solutions; The proof of inequality is a problem of reasoning or exploration. Reasoning refers to expounding the process of argument and revealing the internal laws under certain conditions. The basic methods are comparison, synthesis and analysis. Exploratory problems are mostly related to the proof of natural number n, and the idea of observation-induction-conjecture-proof is often adopted, and the proof is completed by mathematical induction In addition, the proof methods of inequality include method of substitution, method of substitution, reduction to absurdity, construction and so on.

Mathematical science is an inseparable organic whole, and its vitality lies in the relationship between its parts. Inequality knowledge permeates all branches of mathematics and is inextricably linked with each other, so inequality can be used as a tool to solve other problems in mathematics, such as set problem, discussion of solutions of equations (groups), study of monotonicity of functions, determination of function definition domain, triangle, sequence, complex number, solid geometry and so on. Inequalities can also solve mathematical problems reflected in the real world. The common basic ideas and methods in inequality are equivalent transformation, classified discussion, combination of numbers and shapes, functions and equations. In a word, the application of inequality embodies certain comprehensiveness, flexibility and diversity.

Equality and inequality are inseparable, and there is a conceptual kinship, which is the most extensive and universal relationship in middle school mathematics. The basic characteristics of mathematics are the universality of application, the abstraction of theory and the rigor of logic, and the inequality relationship is its profound and vivid embodiment. Although inequality is not as gentle, harmonious, perfect and seamless as equality, it is as tall and straight as a mountain, with beautiful peaks and vast expanses.

Reference answer

Hard magnetic field

Solution: (1) Let f (x) = f (x)-x, because x 1 and x2 are the roots of the equation f (x)-x = 0, so f (x) = a (x-x 1) (x-x2).

And a > 0, f (x) = a (x-x 1) (x-x2) > 0, that is, x < f (x).

x 1-F(x)= x 1-〔x+F(x)〕= x 1-x+a(x 1-x)(x-x2)=(x 1-x)〔 1+a(x-x2)〕

∫0 < x < x 1 < x2 < ,∴x 1-x>0, 1+a(x-x2)= 1+ax-ax2> 1-ax2>0

∴ x 1-f (x) > 0, so f (x) < x 1

(2)X0 =- because x 1 and x2 are the two roots of equation F (x)-X = 0, that is, x 1 and x2 are the roots of equation AX2+(B- 1) X+C = 0.

∴x 1+x2=-

X0 =-,because ax2 < 1,

∴x0 0，f (b) = g (b) > 0，f (a) > f (b)，g (a) > g (b)。

∴f(b)-f(-a)=f(b)+f(a)=g(a)+g(b)

And g (a)-g (-b) = g (a)-g (b) ∴ g (a)+g (b)-[g (a)-g (b)]

=2g(b)>0,∴f(b)-f(-a)>g(a)-g(-b)

The same can be proved: f (a)-f (-b) > g (b)-g (-a).

A: A.

2. Analysis: ① ② ③ The use conditions of the mean inequality "positive, definite and equal" are not met. ④ Formula: | x-y | =| (x-2)-(y-2) |≤| (x-2)-(y-2) |≤||

Answer: ④

3. Analysis: From the known y1=; Y2=0.8x(x is the distance from the warehouse to the station). Sum of expenses y=y 1+y2=0.8x+ ≥2 =8.

"=" is true if and only if 0.8x= x=5.

Answer: 5 kilometers away

3.4. Proof: (1) Let g (x) = f (x)-x = AX2+(b-1) x+1,and x > 0.

∫x 1 < 2 < x2 < 4, ∴ (x 1-2) (x2-2) < 0, namely x 1x2 < 2 (x 1+x2)-4,

(2) solution: from the equation g (x) = ax2+(b-1) x+1= 0, we can know that x 1? 6? 1x2 = > 0, so x 1 and x2 have the same number?

1 If 0 < x 1 < 2, then x2-x 1 = 2, ∴ x2 = x 1+2 > 2,

∴ g (2) < 0, namely 4a+2b- 1 < 0 ①.

(x2-x 1) 2 =

Substitute ∴2a+ 1 =(∫a > 0) into formula ①.

2 0，n=0，f(m)=f(m)？ 6? 1f(0)。 ∵f(m)≠0,∴f(0)= 1