The Unified State Examination in mathematics (profile) is optional. This exam is necessary for those who plan to further study this discipline, enter the Faculty of Economics, Mathematics, or continue their studies at technical universities. The profile level, unlike the basic level, requires in-depth knowledge. The exam focuses on skills practical application skills acquired over the years of study, but knowledge of theory for the Unified State Examination in mathematics is no less important.

What do you need to know?

As with passing the Unified State Exam basic level knowledge gained from school courses algebra and geometry, the ability to work with various inequalities and equations, be fluent in terminology and know algorithms for solving various problems. To successfully complete tasks of increased complexity, knowledge in the following areas is required:

• planimetry;
• inequalities;
• interest;
• progression;
• stereometry;
• equations;
• parametric systems, equations, inequalities;
• financial mathematics.

You cannot do without theory in the preparation process: without knowing the rules, axioms and theorems, it is impossible to solve the problems presented in exam papers tasks. At the same time, it would be a mistake to study theory at the expense of practice. Simply memorizing the rules will not help in the exam - it is important to develop and improve the ability to apply acquired knowledge when solving problems.

How to prepare for the exam?

It is better to start preparing for the exam at the beginning school year. In this case, you can calmly, without haste, go through all the sections, and then repeat them, refreshing your knowledge immediately before testing.

The classic method of preparation - simply reading a textbook in a row, memorizing the rules - is ineffective. To remember information, you need to understand it. You can, for example, try, after reading the rule, retell it in your own words or explain it to yourself. This approach allows you to remember what you read for a long time.

Individual formulas and axioms will have to be memorized. To make the memorization process easier, you should make sure that the necessary data is visible at all times - on the wall near the bed, in the bathroom, on the refrigerator, above the desk. If tables with formulas are always in front of your eyes, they will gradually be remembered without much effort.

Those who are preparing for the Unified State Exam not alone, but in the company of other graduates, can be advised to explain the theory to each other. This method disciplines and helps to better understand the material.

When performing practical tasks, it is necessary to analyze the most common errors. If they are associated not with inattention, but with ignorance of certain rules, it is important to carefully study such topics. The whole theory is structured, and the search the necessary rules will take a minimum of time.

Theory is important, but practice is indispensable. During the exam, the ability to apply the acquired knowledge is tested. It is necessary to practice, practicing the same algorithms over and over again, repeating the same topics, until completing tasks ceases to cause difficulties. Without practical application, knowledge is useless and easily forgotten.

We wish you success in studying the theory and applying the acquired knowledge in the exam!

, is a mandatory exam for 11th grade graduates. Statistically, it is the most difficult.

We suggest that you familiarize yourself with general information about the exam and start preparing immediately. The 2019 exam is no different from last year - this applies to both the basic and specialized options.

Basic level of the Unified State Examination

This option is suitable for graduates in two cases if:

1. you will not need mathematics to enter a university;
2. you do not intend to continue your studies after graduation.

If your chosen specialty has a field with the subject “mathematics,” then the basic level is not your option.

Basic Exam Scoring

The formula for converting primary scores into test scores is updated every year and becomes known after the early period of the Unified State Exam. A decree from Rosobrnadzor has already been issued, which officially established the correspondence of primary and test scores in all subjects for 2019.

According to the order to hand over basic Unified State Exam in mathematics with at least a C, you need to score 12 primary points. This is equivalent correct execution any 12 tasks. Maximum primary score – 20.

Basic Exam Structure

The 2019 Basic Level Mathematics Test consists of 20 short answer questions, which is an integer, or finite number. decimal, or a sequence of numbers. The answer must either be calculated or choose one of the proposed options.

Profile level of the Unified State Examination

This Unified State Examination in 2019 is no different from Unified State Exam of the past of the year.

It is the profile level that graduates must pass for admission to universities, because in the vast majority of specialties mathematics is indicated as the main subject for admission.

Profile test assessment

There is nothing specific here: as usual, you collect initial points, which are then converted into test scores. And already using a 100-point system you can determine the mark for the exam.

In order for the exam to be accepted, it is enough to score 6 primary points. To do this, you need to solve at least 6 tasks of part 1. The maximum initial score is 32.

Structure of the profile test

In 2019, the Unified State Exam test in mathematics at the profile level consists of two parts, including 19 tasks.

• Part 1: 8 tasks (1–8) of basic difficulty level with a short answer.
• Part 2: 4 tasks (9–12) higher level difficulty with a short answer and 7 tasks (13–19) of increased and high levels of difficulty with a detailed answer.

Preparation for the Unified State Exam

• Pass Unified State Exam tests online for free without registration and SMS. The tests presented are identical in complexity and structure to the actual exams conducted in the corresponding years.

• Download demo versions of the Unified State Examination in mathematics, which will allow you to better prepare for the exam and pass it easier. All proposed tests are developed and approved to prepare for Unified State Examination Federal Institute of Pedagogical Measurements (FIPI). In the same FIPI all official Unified State Exam options.
• Check out with basic formulas to prepare for the exam, they will help refresh your memory before proceeding with the demo and test options.

The tasks that you will see most likely will not appear on the exam, but there will be tasks similar to the demo ones, on the same topic or simply with different numbers.

General Unified State Examination figures

Year	Minimum Unified State Exam score	Average score	Number of participants	Failed, %	Qty< 100 points	Duration- Exam length, min.
2009	21
2010	21	43,35	864 708	6,1	160	240
2011	24	47,49	738 746	4,9	205	240
2012	24	44,6	831 068	7,5	56	240
2013	24	48,7	803 741	6,2	538	240
2014	20	46,4				240
2015	27	45,4				235
2016	27					235
2017	27					235

There are no changes at the Unified State Exam in mathematics at the profile level in 2019 - the exam program, as in previous years, is composed of materials from the main mathematical disciplines. The tickets will contain mathematical, geometric, and algebraic problems.

There are no changes in the KIM Unified State Exam 2019 in mathematics at the profile level.

Features of Unified State Examination tasks in mathematics 2019

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

Exam structure

Unified State Exam assignments specialized mathematics divided into two blocks.

1. Part - short answers, includes 8 problems that test basic mathematical preparation and the ability to apply mathematics knowledge in everyday life.
2. Part - short and detailed answers. It consists of 11 tasks, 4 of which require a short answer, and 7 - a detailed one with arguments for the actions performed.

• Advanced difficulty- tasks 9-17 of the second part of KIM.
• High level of difficulty- problems 18-19 –. This part of the exam tasks tests not only the level of mathematical knowledge, but also the presence or absence of a creative approach to solving dry “numerical” tasks, as well as the effectiveness of the ability to use knowledge and skills as a professional tool.

Important! Therefore, when preparing for the Unified State Exam, always support your theory in mathematics by solving practical problems.

How will points be distributed?

The tasks in the first part of the KIM in mathematics are close to the basic level Unified State Exam tests, so it is impossible to score a high score on them.

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

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

Duration of the exam and rules of conduct for the Unified State Exam

To complete the exam paper -2019 the student is assigned 3 hours 55 minutes(235 minutes).

During this time, the student should not:

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

For such actions, the examinee may be expelled from the classroom.

For the state exam in mathematics allowed to bring Bring only a ruler with you; the rest of the materials will be given to you immediately before the Unified State Exam. are issued on the spot.

Effective preparation is the solution to online tests in mathematics 2019. Choose and get the maximum score!

Secondary general education

Line UMK G.K. Muravin. Algebra and principles of mathematical analysis (10-11) (in-depth)

UMK Merzlyak line. Algebra and beginnings of analysis (10-11) (U)

Mathematics

Preparation for the Unified State Exam in mathematics (profile level): assignments, solutions and explanations

We analyze tasks and solve examples with the teacher

Examination paper profile level 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 the tasks:

• part 1 contains 8 tasks (tasks 1-8) with a short answer in the form of a whole number 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 (a complete record of the solution with justification for the actions taken).

Panova Svetlana Anatolevna, mathematics teacher of the highest category of school, work experience 20 years:

“In order to receive a school certificate, a graduate must pass two mandatory exams in the form of the Unified State Examination, one of which is mathematics. In accordance with the Concept for the Development of Mathematical Education in the Russian Federation, the Unified State Examination in mathematics is divided into two levels: basic and specialized. Today we will look at profile-level options.”

Task No. 1- tests the Unified State Exam participants’ ability to apply the skills acquired in the 5th to 9th grade course in elementary mathematics in practical activities. The participant must have computational skills, be able to work with rational numbers, be able to round decimals, and be able to convert one unit of measurement to another.

Example 1. In the apartment where Peter lives, a cold water flow meter (meter) was installed. On May 1, the meter showed a 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 in May, if the price is 1 cubic meter? 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 (cubic m)

2) Let’s find how much money they will pay for wasted water:

34.17 5 = 170.85 (rub)

Answer: 170,85.

Task No. 2- is one of the simplest exam tasks. The majority of graduates successfully cope with it, which indicates knowledge of the definition of the concept of function. Type of task No. 2 according to the requirements codifier is a task on the use of 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 No. 2 tests the ability to extract information presented in tables, diagrams, and graphs. Graduates need to be able to determine the value of a function from the value of the argument in various ways of specifying the function and describe the behavior and properties of the function based on its graph. You also need to be able to find the greatest or smallest value and build graphs of the studied functions. Errors made are random in reading the conditions of the problem, reading the diagram.

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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 shares he purchased, and on April 13, he sold all the remaining shares. How much did the businessman lose as a result of these operations?

Solution:

2) 1000 · 3/4 = 750 (shares) - constitute 3/4 of all shares purchased.

6) 247500 + 77500 = 325000 (rub) - the businessman received 1000 shares after selling.

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

Answer: 15000.

Task No. 3- is a task at the basic level of the first part, tests the ability to perform actions with geometric shapes on the content of the course “Planimetry”. Task 3 tests the ability to calculate the area of ​​a figure on checkered paper, the ability to calculate degree measures of angles, calculate perimeters, etc.

Example 3. Find the area of ​​a rectangle drawn on checkered paper with a cell size of 1 cm by 1 cm (see figure). Give your answer in square centimeters.

Solution: To calculate the area of ​​a given figure, you can use the Peak formula:

To calculate the area of ​​a given rectangle, we use Peak’s formula:

S= B +	G
	2

where B = 10, G = 6, therefore

S = 18 +	6
	2

Answer: 20.

Read also: Unified State Exam in Physics: solving problems about oscillations

Task No. 4- the objective of the course “Probability Theory and Statistics”. The ability to calculate the probability of an event in the simplest situation is tested.

Example 4. There are 5 red and 1 blue dots marked on the circle. Determine which polygons are larger: those with all the vertices red, or those with one of the vertices blue. In your answer, indicate how many there are more of some than others.

Solution: 1) Let's use the formula for the number of combinations of n elements by k:

whose vertices are all red.

3) One pentagon with all vertices red.

4) 10 + 5 + 1 = 16 polygons with all red vertices.

which have red tops or with one blue top.

which have red tops or with one blue top.

8) One hexagon with red vertices and one blue vertex.

9) 20 + 15 + 6 + 1 = 42 polygons with all red vertices or one blue vertex.

10) 42 – 16 = 26 polygons using the blue dot.

11) 26 – 16 = 10 polygons – how many more polygons in which one of the vertices is a blue dot are there than polygons in which all the vertices are only red.

Answer: 10.

Task No. 5- the basic level of the first part tests the ability to solve simple equations (irrational, exponential, trigonometric, logarithmic).

Example 5. Solve equation 2 3 + x= 0.4 5 3 + x .

Solution. Divide both sides of this equation by 5 3 + X≠ 0, we get

2 3 + x	= 0.4 or		2		3 + X	=	2	,
5 3 + X			5				5

whence it follows that 3 + x = 1, x = –2.

Answer: –2.

Task No. 6 in planimetry to find geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. Study of constructed models using geometric concepts and theorems. The source of difficulties is, as a rule, ignorance or incorrect application of the necessary theorems of planimetry.

Area of ​​a triangle ABC equals 129. DE– midline parallel to the side AB. Find the area of ​​the trapezoid ABED.

Solution. Triangle CDE similar to a triangle CAB at two angles, since the angle at the vertex C general, angle СDE equal to angle CAB as the corresponding angles at DE || AB secant A.C.. Because DE is the middle line of a triangle by condition, then by the property of the middle line | DE = (1/2)AB. This means that the similarity coefficient is 0.5. The areas of similar figures are related as the square of the similarity coefficient, therefore

Hence, S ABED = S Δ ABC – S Δ CDE = 129 – 32,25 = 96,75.

Task No. 7- checks the application of the derivative to the study of a function. Successful implementation requires meaningful, non-formal knowledge of the concept of derivative.

Example 7. To the graph of the function y = f(x) at the abscissa point x 0 a tangent is drawn that is perpendicular to the line passing through the points (4; 3) and (3; –1) of this graph. Find f′( x 0).

Solution. 1) Let’s use the equation of a line passing through two given points and find the equation of a line passing through points (4; 3) and (3; –1).

(y – y 1)(x 2 – x 1) = (x – x 1)(y 2 – y 1)

(y – 3)(3 – 4) = (x – 4)(–1 – 3)

(y – 3)(–1) = (x – 4)(–4)

–y + 3 = –4x+ 16| · (-1)

y – 3 = 4x – 16

y = 4x– 13, where k 1 = 4.

2) Find the slope of the tangent k 2, which is perpendicular to the line y = 4x– 13, where k 1 = 4, according to the formula:

3) The tangent angle is the derivative of the function at the point of tangency. Means, f′( x 0) = k 2 = –0,25.

Answer: –0,25.

Task No. 8- tests the exam participants’ knowledge of elementary stereometry, the ability to apply formulas for finding surface areas and volumes of figures, dihedral angles, compare the volumes of similar figures, be able to perform actions with geometric figures, coordinates and vectors, etc.

The volume of a cube circumscribed about a sphere is 216. Find the radius of the sphere.

Solution. 1) V cube = a 3 (where A– length of the edge of the cube), therefore

A 3 = 216

A = 3 √216

2) Since the sphere is inscribed in a cube, it means that the length of the diameter of the sphere is equal to the length of the edge of the cube, therefore d = a, d = 6, d = 2R, R = 6: 2 = 3.

Task No. 9- requires the graduate to have the skills to transform and simplify algebraic expressions. Task No. 9 of an increased level of difficulty with a short answer. The tasks from the “Calculations and Transformations” section in the Unified State Exam are divided into several types:

transformation of numerical rational expressions;

converting algebraic expressions and fractions;

conversion of numeric/letter irrational expressions;

actions with degrees;

converting logarithmic expressions;

1. converting numeric/letter trigonometric expressions.

Example 9. Calculate tanα if it is known that cos2α = 0.6 and

3π	< α < π.
4

Solution. 1) Let’s use the double argument formula: cos2α = 2 cos 2 α – 1 and find

tan 2 α =	1	– 1 =	1	– 1 =	10	– 1 =	5	– 1 = 1	1	– 1 =	1	= 0,25.
	cos 2 α		0,8		8		4		4		4

This means tan 2 α = ± 0.5.

3) By condition

3π	< α < π,
4

this means α is the angle of the second quarter and tgα< 0, поэтому tgα = –0,5.

Answer: –0,5.

#ADVERTISING_INSERT# Task No. 10- tests students’ ability to use acquired early knowledge and skills in practical activities and everyday life. We can say that these are problems in physics, and not in mathematics, but all the necessary formulas and quantities are given in the condition. The problems boil down to solving a linear or quadratic equation, or a linear or quadratic inequality. Therefore, it is necessary to be able to solve such equations and inequalities and determine the answer. The answer must be given as a whole number or a finite decimal fraction.

Two bodies of mass m= 2 kg each, moving at the same speed v= 10 m/s at an angle of 2α to each other. The energy (in joules) released during their absolutely inelastic collision is determined by the expression Q = mv 2 sin 2 α. At what smallest angle 2α (in degrees) must the bodies move so that at least 50 joules are released as a result of the collision?
Solution. To solve the problem, we need to solve the inequality Q ≥ 50, on the interval 2α ∈ (0°; 180°).

mv 2 sin 2 α ≥ 50

2 10 2 sin 2 α ≥ 50

200 sin 2 α ≥ 50

Since α ∈ (0°; 90°), we will only solve

Let us represent the solution to the inequality graphically:

Since by condition α ∈ (0°; 90°), it means 30° ≤ α< 90°. Получили, что наименьший угол α равен 30°, тогда наименьший угол 2α = 60°.

Task No. 11- is typical, but turns out to be difficult for students. The main source of difficulty is the construction of a mathematical model (drawing up an equation). Task No. 11 tests the ability to solve word problems.

Example 11. During spring break, 11th-grader Vasya had to solve 560 practice problems to prepare for the Unified State Exam. On March 18, on the last day of school, Vasya solved 5 problems. Then every day he solved the same number of problems more than the previous day. Determine how many problems Vasya solved on April 2, the last day of the holidays.

Solution: Let's denote a 1 = 5 – the number of problems that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 – number of days from March 18 to April 2 inclusive, S 16 = 560 – total number of tasks, a 16 – the number of problems that Vasya solved on April 2. Knowing that every day Vasya solved the same number of problems more compared to the previous day, we can use formulas for finding the sum arithmetic progression:

560 = (5 + a 16) 8,

5 + a 16 = 560: 8,

5 + a 16 = 70,

a 16 = 70 – 5

a 16 = 65.

Answer: 65.

Task No. 12- they test students’ ability to perform operations with functions, and to be able to apply the derivative to the study of a function.

Find the maximum point of the function y= 10ln( x + 9) – 10x + 1.

Solution: 1) Find the domain of definition of the function: x + 9 > 0, x> –9, that is, x ∈ (–9; ∞).

2) Find the derivative of the function:

4) The found point belongs to the interval (–9; ∞). Let's determine the signs of the derivative of the function and depict the behavior of the function in the figure:

The desired maximum point x = –8.

Download for free the working program in mathematics for the line of teaching materials G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free teaching aids on algebra

Task No. 13-increased level of complexity with a detailed answer, testing the ability to solve equations, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

a) Solve the equation 2log 3 2 (2cos x) – 5log 3 (2cos x) + 2 = 0

b) Find all the roots of this equation, belonging to the segment.

Solution: a) Let log 3 (2cos x) = t, then 2 t 2 – 5t + 2 = 0,

	log 3(2cos x) =	2	⇔		2cos x = 9	⇔		cos x =	4,5	⇔ because |cos x| ≤ 1,
	log 3(2cos x) =	1			2cos x = √3			cos x =	√3
		2							2

then cos x =	√3
	2

	x =	π	+ 2π k
		6
	x = –	π	+ 2π k, k ∈ Z
		6

b) Find the roots lying on the segment .

The figure shows that the roots of the given segment belong to

11π	And	13π	.
6		6

Answer: A)	π	+ 2π k; –	π	+ 2π k, k ∈ Z; b)	11π	;	13π	.
	6		6		6		6

Task No. 14-advanced level refers to tasks in the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes. The task contains two points. In the first point, the task must be proven, and in the second point, calculated.

The diameter of the circle of the base of the cylinder is 20, the generatrix of the cylinder is 28. The plane intersects its base along chords of length 12 and 16. The distance between the chords is 2√197.

a) Prove that the centers of the bases of the cylinder lie on one side of this plane.

b) Find the angle between this plane and the plane of the base of the cylinder.

Solution: a) A chord of length 12 is at a distance = 8 from the center of the base circle, and a chord of length 16, similarly, is at a distance of 6. Therefore, the distance between their projections onto a plane parallel to the bases of the cylinders is either 8 + 6 = 14, or 8 − 6 = 2.

Then the distance between the chords is either

= = √980 = = 2√245

= = √788 = = 2√197.

According to the condition, the second case was realized, in which the projections of the chords lie on one side of the cylinder axis. This means that the axis does not intersect this plane within the cylinder, that is, the bases lie on one side of it. What needed to be proven.

b) Let us denote the centers of the bases as O 1 and O 2. Let us draw from the center of the base with a chord of length 12 a perpendicular bisector to this chord (it has length 8, as already noted) and from the center of the other base to the other chord. They lie in the same plane β, perpendicular to these chords. Let's call the midpoint of the smaller chord B, the larger chord A and the projection of A onto the second base - H (H ∈ β). Then AB,AH ∈ β and therefore AB,AH are perpendicular to the chord, that is, the straight line of intersection of the base with the given plane.

This means that the required angle is equal to

∠ABH = arctan	A.H.	= arctan	28	= arctg14.
	B.H.		8 – 6

Task No. 15- increased level of complexity with a detailed answer, tests the ability to solve inequalities, which is most successfully solved among tasks with a detailed answer of an increased level of complexity.

Example 15. Solve inequality | x 2 – 3x| log 2 ( x + 1) ≤ 3x – x 2 .

Solution: The domain of definition of this inequality is the interval (–1; +∞). Consider three cases separately:

1) Let x 2 – 3x= 0, i.e. X= 0 or X= 3. In this case, this inequality becomes true, therefore, these values ​​are included in the solution.

2) Let now x 2 – 3x> 0, i.e. x∈ (–1; 0) ∪ (3; +∞). Moreover, this inequality can be rewritten as ( x 2 – 3x) log 2 ( x + 1) ≤ 3x – x 2 and divide by a positive expression x 2 – 3x. We get log 2 ( x + 1) ≤ –1, x + 1 ≤ 2 –1 , x≤ 0.5 –1 or x≤ –0.5. Taking into account the domain of definition, we have x ∈ (–1; –0,5].

3) Finally, let's consider x 2 – 3x < 0, при этом x∈ (0; 3). In this case, the original inequality will be rewritten in the form (3 x – x 2) log 2 ( x + 1) ≤ 3x – x 2. After dividing by positive 3 x – x 2 , we get log 2 ( x + 1) ≤ 1, x + 1 ≤ 2, x≤ 1. Taking into account the region, we have x ∈ (0; 1].

Combining the solutions obtained, we obtain x ∈ (–1; –0.5] ∪ ∪ {3}.

Answer: (–1; –0.5] ∪ ∪ {3}.

Task No. 16- advanced level refers to tasks in the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes, coordinates and vectors. The task contains two points. In the first point, the task must be proven, and in the second point, calculated.

IN isosceles triangle ABC with an angle of 120° at vertex A, a bisector BD is drawn. Rectangle DEFH is inscribed in triangle ABC so that side FH lies on segment BC, and vertex E lies on segment AB. a) Prove that FH = 2DH. b) Find the area of ​​rectangle DEFH if AB = 4.

Solution: A)

1) ΔBEF – rectangular, EF⊥BC, ∠B = (180° – 120°): 2 = 30°, then EF = BE by the property of the leg lying opposite the angle of 30°.

2) Let EF = DH = x, then BE = 2 x, BF = x√3 according to the Pythagorean theorem.

3) Since ΔABC is isosceles, it means ∠B = ∠C = 30˚.

BD is the bisector of ∠B, which means ∠ABD = ∠DBC = 15˚.

4) Consider ΔDBH – rectangular, because DH⊥BC.

2x	=	4 – 2x
2x(√3 + 1)		4

1	=	2 – x
√3 + 1		2

√3 – 1 = 2 – x

x = 3 – √3

EF = 3 – √3

2) S DEFH = ED EF = (3 – √3 ) 2(3 – √3 )

S DEFH = 24 – 12√3.

Answer: 24 – 12√3.

Task No. 17- a task with a detailed answer, this task tests the application of knowledge and skills in practical activities and everyday life, the ability to build and explore mathematical models. This task - word problem with economic content.

Example 17. A deposit of 20 million rubles is planned to be opened for four years. At the end of each year, the bank increases the deposit by 10% compared to its size at the beginning of the year. In addition, at the beginning of the third and fourth years, the investor annually replenishes the deposit by X million rubles, where X - whole number. Find highest value X, in which the bank will accrue less than 17 million rubles to the deposit over four years.

Solution: At the end of the first year, the contribution will be 20 + 20 · 0.1 = 22 million rubles, and at the end of the second - 22 + 22 · 0.1 = 24.2 million rubles. At the beginning of the third year, the contribution (in million rubles) will be (24.2 + X), and at the end - (24.2 + X) + (24,2 + X)· 0.1 = (26.62 + 1.1 X). At the beginning of the fourth year the contribution will be (26.62 + 2.1 X), and at the end - (26.62 + 2.1 X) + (26,62 + 2,1X) 0.1 = (29.282 + 2.31 X). By condition, you need to find the largest integer x for which the inequality holds

(29,282 + 2,31x) – 20 – 2x < 17

29,282 + 2,31x – 20 – 2x < 17

0,31x < 17 + 20 – 29,282

0,31x < 7,718

x <	7718
	310

x <	3859
	155

x < 24	139
	155

The largest integer solution to this inequality is the number 24.

Answer: 24.

Task No. 18- a task of an increased level of complexity with a detailed answer. This assignment is for competitive selection to universities with increased requirements for the mathematical preparation of applicants. Exercise high level complexity - this task is not about using one solution method, but about a combination various methods. To successfully complete task 18, in addition to solid mathematical knowledge, you also need a high level of mathematical culture.

At what a system of inequalities

	x 2 + y 2 ≤ 2ay – a 2 + 1
	y + a ≤ |x| – a

has exactly two solutions?

Solution: This system can be rewritten in the form

	x 2 + (y– a) 2 ≤ 1
	y ≤ |x| – a

If we draw on the plane the set of solutions to the first inequality, we get the interior of a circle (with a boundary) of radius 1 with center at point (0, A). The set of solutions to the second inequality is the part of the plane lying under the graph of the function y = | x| – a, and the latter is the graph of the function
y = | x| , shifted down by A. The solution to this system is the intersection of the sets of solutions to each of the inequalities.

Therefore, two solutions this system will have only in the case shown in Fig. 1.

The points of contact of the circle with the lines will be the two solutions of the system. Each of the straight lines is inclined to the axes at an angle of 45°. So it's a triangle PQR– rectangular isosceles. Dot Q has coordinates (0, A), and the point R– coordinates (0, – A). In addition, the segments PR And PQ equal to the radius of the circle equal to 1. This means

Qr= 2a = √2, a =	√2	.
	2

Answer: a =	√2	.
	2

Task No. 19- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection into universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is a task not on the use of one solution method, but on a combination of various methods. To successfully complete task 19, you must be able to search for a solution, choosing different approaches from among the known ones, and modifying the studied methods.

Let Sn sum P terms of an arithmetic progression ( a p). It is known that S n + 1 = 2n 2 – 21n – 23.

a) Provide the formula P th term of this progression.

b) Find the smallest absolute sum S n.

c) Find the smallest P, at which S n will be the square of an integer.

Solution: a) It is obvious that a n = S n – S n- 1 . Using this formula, we get:

S n = S (n – 1) + 1 = 2(n – 1) 2 – 21(n – 1) – 23 = 2n 2 – 25n,

S n – 1 = S (n – 2) + 1 = 2(n – 1) 2 – 21(n – 2) – 23 = 2n 2 – 25n+ 27

Means, a n = 2n 2 – 25n – (2n 2 – 29n + 27) = 4n – 27.

B) Since S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Its graph can be seen in the figure.

Obviously, the smallest value is achieved at the integer points located closest to the zeros of the function. Obviously these are points X= 1, X= 12 and X= 13. Since, S(1) = |S 1 | = |2 – 25| = 23, S(12) = |S 12 | = |2 · 144 – 25 · 12| = 12, S(13) = |S 13 | = |2 · 169 – 25 · 13| = 13, then the smallest value is 12.

c) From the previous paragraph it follows that Sn positive, starting from n= 13. Since S n = 2n 2 – 25n = n(2n– 25), then the obvious case, when this expression is a perfect square, is realized when n = 2n– 25, that is, at P= 25.

It remains to check the values ​​from 13 to 25:

S 13 = 13 1, S 14 = 14 3, S 15 = 15 5, S 16 = 16 7, S 17 = 17 9, S 18 = 18 11, S 19 = 19 13, S 20 = 20 13, S 21 = 21 17, S 22 = 22 19, S 23 = 23 21, S 24 = 24 23.

It turns out that for smaller values P a complete square is not achieved.

Answer: A) a n = 4n– 27; b) 12; c) 25.

________________

*Since May 2017, the united publishing group "DROFA-VENTANA" has been part of the Russian Textbook corporation. The corporation also includes the Astrel publishing house and the LECTA digital educational platform. General Director Alexander Brychkin, graduate of the Financial Academy under the Government of the Russian Federation, candidate economic sciences, head of innovative projects of the DROFA publishing house in the field of digital education (electronic forms of textbooks, Russian Electronic School, digital educational platform LECTA). Before joining the DROFA publishing house, he held the position of vice president for strategic development and investments of the publishing holding EKSMO-AST. Today, the publishing corporation "Russian Textbook" has the largest portfolio of textbooks included in the Federal List - 485 titles (approximately 40%, excluding textbooks for special schools). The corporation's publishing houses own the most popular sets of textbooks in Russian schools in physics, drawing, biology, chemistry, technology, geography, astronomy - areas of knowledge that are needed for the development of the country's productive potential. The corporation's portfolio includes textbooks and teaching aids For primary school, awarded the Presidential Prize in the field of education. These are textbooks and manuals in subject areas that are necessary for the development of the scientific, technical and production potential of Russia.

In this section we are preparing for the Unified State Examination in mathematics at a basic, profile level - we present analysis of problems, tests, description of the exam and useful recommendations. Using our resource, you will at least understand how to solve problems and be able to successfully pass the Unified State Exam in mathematics in 2019. Begin!

The Unified State Examination in mathematics is mandatory exam any student in 11th grade, so the information presented in this section is relevant for everyone. The mathematics exam is divided into two types - basic and specialized. In this section I provide an analysis of each type of task with detailed explanation for two options. The Unified State Exam tasks are strictly thematic, so for each issue you can give precise recommendations and provide the theory necessary specifically for solving this type of task. Below you will find links to assignments, by clicking on which you can study the theory and analyze examples. Examples are constantly replenished and updated.

Structure of the basic level of the Unified State Examination in mathematics

The examination paper in basic level mathematics consists of one piece , including 20 short-answer tasks. All tasks are aimed at testing the development of basic skills and practical skills in applying mathematical knowledge in everyday situations.

The answer to each of tasks 1–20 is integer, trailing decimal , or sequence of numbers .

A task with a short answer is considered completed if the correct answer is written down in answer form No. 1 in the form provided for in the instructions for completing the task.