Literature      06/10/2020

Solution 7 tasks. USE in Mathematics (profile). Numbers and their properties

The exam program, as in previous years, is made up of materials from the main mathematical disciplines. The tickets will include mathematical, geometric, and algebraic problems.

Changes in KIM USE 2020 in mathematics profile level No.

Features of USE assignments in mathematics-2020

  • When preparing for the exam in mathematics (profile), pay attention to the basic requirements of the examination program. It is designed to test the knowledge of an in-depth program: vector and mathematical models, functions and logarithms, algebraic equations and inequalities.
  • Separately, practice solving tasks for.
  • It is important to show non-standard thinking.

Exam Structure

Tasks USE profile mathematics divided into two blocks.

  1. Part - short answers, includes 8 tasks that test basic mathematical training and the ability to apply knowledge of mathematics in everyday life.
  2. Part - brief and detailed answers. It consists of 11 tasks, 4 of which require a short answer, and 7 - a detailed one with an argumentation of the actions performed.
  • Increased complexity- tasks 9-17 of the second part of KIM.
  • High level difficulties- tasks 18-19 –. This part examination tasks checks not only the level of mathematical knowledge, but also the presence or absence of a creative approach to solving dry "digital" tasks, as well as the effectiveness of the ability to use knowledge and skills as a professional tool.

Important! Therefore, in preparation for USE theory in mathematics, always support the solution of practical problems.

How will points be distributed?

The tasks of the first part of KIMs in mathematics are close to USE tests basic level, That's why high score it's impossible to get them.

The points for each task in mathematics at the profile level were distributed as follows:

  • for correct answers to tasks No. 1-12 - 1 point each;
  • No. 13-15 - 2 each;
  • No. 16-17 - 3 each;
  • No. 18-19 - 4 each.

The duration of the exam and the rules of conduct for the exam

To complete the exam -2020 the student is assigned 3 hours 55 minutes(235 minutes).

During this time, the student should not:

  • be noisy;
  • use gadgets and other technical means;
  • write off;
  • try to help others, or ask for help for yourself.

For such actions, the examiner can be expelled from the audience.

On State exam mathematics allowed to bring only a ruler with you, the rest of the materials will be given to you immediately before the exam. issued on the spot.

Effective preparation is the solution online tests Math 2020. Choose and get the highest score!

In task No. 7 of the profile USE level in mathematics, it is necessary to demonstrate knowledge of the function of the derivative and the antiderivative. In most cases, simply defining the concepts and understanding the meanings of the derivative is sufficient.

Analysis of typical options for tasks No. 7 USE in mathematics of a profile level

The first version of the task (demo version 2018)

The figure shows a graph of a differentiable function y = f(x). Nine points are marked on the x-axis: x 1 , x 2 , …, x 9 . Among these points, find all points where the derivative of the function y = f(x) is negative. In your answer, indicate the number of points found.

Solution algorithm:
  1. Let's look at the graph of the function.
  2. We are looking for points at which the function decreases.
  3. We count their number.
  4. We write down the answer.
Solution:

1. On the graph, the function periodically increases, periodically decreases.

2. In those intervals where the function decreases, the derivative has negative values.

3. These intervals contain points x 3 , x 4 , x 5 , x 9 . There are 4 such points.

The second version of the task (from Yaschenko, No. 4)

Solution algorithm:
  1. Let's look at the graph of the function.
  2. We consider the behavior of the function at each of the points and the sign of the derivative at them.
  3. Finding points in highest value derivative.
  4. We write down the answer.
Solution:

1. The function has several intervals of decreasing and increasing.

2. Where the function decreases. The derivative has a minus sign. Such points are among those indicated. But there are points on the graph where the function increases. Their derivative is positive. These are the points with abscissas -2 and 2.

3. Consider a graph at points with x=-2 and x=2. At the point x = 2, the function goes up steeper, which means that the tangent at this point has a larger slope. Therefore, at the point with the abscissa 2. The derivative has the greatest value.

The third version of the task (from Yaschenko, No. 21)

Solution algorithm:
  1. We equate the equations of the tangent and the function.
  2. We simplify the obtained equality.
  3. We find the discriminant.
  4. Define the parameter A, for which the solution is unique.
  5. We write down the answer.
Solution:

1. The coordinates of the tangent point satisfy both equations: the tangent and the function. So we can equate the equations. We will receive.

Average general education

UMK line G. K. Muravina. Algebra and beginnings mathematical analysis(10-11) (deep)

Line UMK Merzlyak. Algebra and the Beginnings of Analysis (10-11) (U)

Mathematics

Preparation for the exam in mathematics (profile level): tasks, solutions and explanations

We analyze tasks and solve examples with the teacher

The profile-level examination paper lasts 3 hours 55 minutes (235 minutes).

Minimum Threshold- 27 points.

The examination paper consists of two parts, which differ in content, complexity and number of tasks.

The defining feature of each part of the work is the form of tasks:

  • part 1 contains 8 tasks (tasks 1-8) with a short answer in the form of an integer or a final decimal fraction;
  • part 2 contains 4 tasks (tasks 9-12) with a short answer in the form of an integer or a final decimal fraction and 7 tasks (tasks 13-19) with a detailed answer (full record of the decision with the rationale for the actions performed).

Panova Svetlana Anatolievna, mathematic teacher the highest category schools, 20 years of work experience:

"In order to receive school certificate, the graduate must pass two mandatory exams in USE form, one of which is mathematics. In accordance with the Development Concept mathematics education V Russian Federation The USE in mathematics is divided into two levels: basic and specialized. Today we will consider options for the profile level.

Task number 1- checks with USE participants the ability to apply the skills acquired in the course of 5-9 classes in elementary mathematics in practical activities. The participant must have computer skills, be able to work with rational numbers, be able to round decimals be able to convert one unit of measurement to another.

Example 1 An expense meter was installed in the apartment where Petr lives cold water(counter). On the first of May, the meter showed an consumption of 172 cubic meters. m of water, and on the first of June - 177 cubic meters. m. What amount should Peter pay for cold water for May, if the price of 1 cu. m of cold water is 34 rubles 17 kopecks? Give your answer in rubles.

Solution:

1) Find the amount of water spent per month:

177 - 172 = 5 (cu m)

2) Find how much money will be paid for the spent water:

34.17 5 = 170.85 (rub)

Answer: 170,85.


Task number 2- is one of the simplest tasks of the exam. The majority of graduates successfully cope with it, which indicates the possession of the definition of the concept of function. Task type No. 2 according to the requirements codifier is a task for using acquired knowledge and skills in practical activities and Everyday life. Task No. 2 consists of describing, using functions, various real relationships between quantities and interpreting their graphs. Task number 2 tests the ability to extract information presented in tables, diagrams, graphs. Graduates need to be able to determine the value of a function by the value of the argument with various ways of specifying the function and describe the behavior and properties of the function according to its graph. It is also necessary to be able to find the largest or smallest value from the function graph and build graphs of the studied functions. The mistakes made are of a random nature in reading the conditions of the problem, reading the diagram.

#ADVERTISING_INSERT#

Example 2 The figure shows the change in the exchange value of one share of a mining company in the first half of April 2017. On April 7, the businessman purchased 1,000 shares of this company. On April 10, he sold three-quarters of the purchased shares, and on April 13 he sold all the remaining ones. How much did the businessman lose as a result of these operations?


Solution:

2) 1000 3/4 = 750 (shares) - make up 3/4 of all purchased shares.

6) 247500 + 77500 = 325000 (rubles) - the businessman received after the sale of 1000 shares.

7) 340,000 - 325,000 = 15,000 (rubles) - the businessman lost as a result of all operations.

I present the solution of task 7 of the OGE-2016 in informatics from the demo project. Compared to the 2015 demo, task 7 has not changed. This is a task for the ability to encode and decode information (Coding and decoding information). The answer to task 7 is a sequence of letters, which should be written in the answer field.

Screenshot of task 7.

Exercise:

The scout sent a radiogram to the headquarters
– – – – – – – –
This radiogram contains a sequence of letters in which only the letters A, D, G, L, T occur. Each letter is encoded using Morse code. There are no separators between letter codes. Write down the given sequence of letters in the answer.
The required fragment of Morse code is given below.

Answer: __

Such a task is best solved sequentially, closing each possible code.
1. (-) - - - - - - -, the first two positions can only be the letter A
2.
a) (-) (-) - - - - - -, the following three positions can be the letter D
b) (-) (-) - - - - - -, or one position letter L, but if we take the following combination (-) (-) (-) - - - - -, (letter T) then we will not choose more we can (there are simply no such combinations starting with two points), so we have reached a dead end and we conclude that this path is wrong
3. We return to option a)
(-) (- ) (- ) - - - - -, this is the letter Zh
4. (-) (-) (-) (-) - - - -, this is the letter L
5. (-) (-) (-) (-) (-) - - -, this is the letter D
6. ( -) (- ) ( - ) (-) (- ) (-) - -, and this is the letter L
7. (-) (- ) (- ) (-) (- ) (-) (-) -, letter A
8. (-) (- ) (- ) (-) (- ) (-) (-) (-), letter L
9. We collect all the letters that we got: AJLDLAL.

Answer: AJLDLAL