The permanence and existence problems in senior three mathematics review involve the properties and images of functions such as linear function and quadratic function, and permeate the thinking methods such as substitution, conversion, combination of numbers and shapes, function and equation, which is conducive to examining students' comprehensive problem-solving ability and has a positive effect on cultivating students' flexibility and creativity. Therefore, it has become a hot spot in college entrance examination over the years. There are many ways to deal with permanent and existing problems. Let's talk about it through examples.

First, the issue of continuous compilation.

1.？ Where x∈D, f (x) > g (x)

For appearance? To solve the problem of ∈ x∈D, f (x) > g (x), we need to set the function y = f (x)-g (x) first, and then convert it into? X∈D，y > 0。

Example 1: Given the function f(x) = x | x-a |+2x, find all real numbers A, so that the image of function f(x) is always under the image of function g (x) = 2x+ 1 for any x ∈ [65438].

Solution: From the meaning of the question, it is found that any real number x ∈ [1 2] and f (x) < g (x) are always true, that is, x | x- a | < 1 on [1 2] is always true, that is, x- ∴ when a < 0, f ′ (x) > 0 holds, the function f(x) is increasing function at (0, 1), and the function y = at (0, 1).

Let 0 < x ≤ x ≤ 1, then | f (x)-f (x) | = f (x)-f (x),-=-,so |f (x)-f(x)|≤4- is equivalent to f(x)-f (.

Let h(x) = f (x)+= x- 1-alnx+. Then |f(x)-f(x)|≤4- is equivalent to the function h (x) being a decreasing function in the interval (0, 1). Because h'(x). 1] is a constant, that is, a is not less than y=x- the maximum value in the interval (0, 1). The function y=x- is increasing function in the interval (0, 1), so the maximum value of y=x- is -3, so a≥-3.

[opinion]? Because of the lack of theoretical support, we can't directly use | x, x∈D, |f(x)-f(x)|≤a|x-x| to solve problems, which is equivalent to k=≤a, and further equivalent to the method of f ′ (x) ≤ a. In addition, the question (2) of this question cannot be | f.

When dealing with the frequent staffing problems, we should first distinguish the types of problems. If it is about the establishment of univariate constants, parameter separation should be considered first. If the parameters can't be separated or the function formed after the separation of parameters can't be processed, then you can choose classified discussion to deal with it; If it is about the identity of two independent variables, we only need to deal with it according to the processing type mentioned in the above inquiry point.

Second, the existing problems.

1.？ What's the matter with you? Pot x∈D, f(x)=g(x) type

For what? What's the matter with you? In the study of x∈D and f(x)=g(x), if the range of function f(x) is c and the range of function g(x) is c, then the problem is equivalent to c? C.

Example 1: Let the function f(x)=-x-x+x-4.

Let a≥ 1 and the function g(x)=x-3ax-2a. If for any x ∈ [0, 1], there is always x ∈ [0, 1], so f(x)=g(x) holds.

Solution: f ′ (x) =-x-x+,making f ′ (x) > 0, we can see that when x ∈ [0, 1] and f(x) increase monotonically,

When x ∈ [0, 1], f (x) ∈ [f (0), f (1)], it is f (x) ∈ [-4, -3].

And g'(x)=3x-3a, and a≥ 1, ∴ when x ∈ [0, 1], g'(x)≤0, and g(x) monotonically decreases.

∴ when x ∈ [0, 1], g (x) ∈ [g (1), g (0)], that is, g (x) ∈ [-3a-2a+ 1.