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The first lecture of high school mathematics Olympic tutorial materials

A total of ten lectures have been uploaded for the Senior High School Math Orsay

Lecture 1 The concept of sets and operations on sets

Knowledge, methods, skills

The concept of set is given in the textbook of first grade math (high) (experimental) in high school; generally, objects that meet certain conditions (or have certain properties) are grouped together to become a set.

On this basis, the certainty, mutual dissimilarity, and disorder of the elements of the set are introduced. The finite set, the infinite set, the enumeration method of the set, the description method and the subset, the true subset, the empty set, the non- More than a dozen new nouns or concepts, such as empty sets, complete sets, complements, unions, and more than twenty new symbols. This formed the problem of operations on sets, formed the title of the set as the background, and used the set to represent space. Line and surface and their relationships, surface plane trajectories and their relationships, express the necessary and sufficient conditions, describe permutations and combinations, and use combination of properties to perform combination counting and other comprehensive topics.

Explaining the problem

Ⅰ. Determination of pending elements in a collection

Making full use of the nature of the elements in the set and the basic relationship between the sets, often can solve some high school mathematics competition problems with the set as the background. See the following examples. Example 1: Find the point set {(x, y) | lg (x

3

131

y) lgx lgy}. 39

3

[Analysis of the idea] The logarithm should first be transformed into an algebraic equation to solve it. [Slight solution] Known from the set x 0, y 0, and x

3

131

y xy, 39

From the mean inequality, there is x

13111

y 3 (x3) (y3) () xy, 3939

If and only if x

3

13111

y, that is, x, y (the imaginary root is truncated), the equal sign holds. 3993

So the given point set has only one element.

[Comment] In this equation, the necessary and sufficient conditions of inequality and equal sign are applied. It is an important method to solve the problem, and we should pay attention to it.

Example 2: Given A {y | yx 4x 3, x R}, B {y | yx 2x 2, x R}. Find A B. [Thinking Analysis] First determine the sets A and B.

[Simplified solution] y (x 2) 1 1, and y (x 1) 3 3. ∴A = {y | y 1}, B {y | y 3}, so AB {y | 1 y 3}. [Comment ] This question should avoid the following wrong solutions:

2

2

2

2

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