Preparation for the Unified State Exam in mathematics at basic and specialized levels. Exam in mathematics

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 pays attention to the skills of practical application of the skills acquired over the years of study, but knowledge of the theory for the Unified State Exam in mathematics is no less important.

What do you need to know?

As with passing the basic level Unified State Exam, you will need knowledge gained from school courses in 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 the exam papers. 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 of the 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 finding 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!

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

The profile level examination 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 largest or smallest value from a function graph 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 basic level task of the first part, tests the ability to perform actions with geometric figures according to the content of the Planimetry course. 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.

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 around 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;

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 of an 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 that belong 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, 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 an isosceles triangle ABC with an angle of 120°, the bisector BD is drawn at vertex A. 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 is a text 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 the greatest 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 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 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.

Consequently, this system will have two solutions 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. Alexander Brychkin, a graduate of the Financial Academy under the Government of the Russian Federation, Candidate of 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) was appointed General Director. 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 schools, which were awarded the Presidential Award 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.

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

The profile level examination 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:

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.

#ADVERTISING_INSERT#

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.

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.

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.

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.

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

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.

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.

The volume of a cube circumscribed around 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.

transformation of numerical rational expressions;

converting algebraic expressions and fractions;

conversion of numeric/letter irrational expressions;

actions with degrees;

converting logarithmic expressions;

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.

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°.

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.

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

b) Find all the roots of this equation that belong 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

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.

Then the distance between the chords is either

= = √980 = = 2√245

= = √788 = = 2√197.

This means that the required angle is equal to

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

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.

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

Answer: (–1; –0.5] ∪ ∪ {3}.

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.

(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.

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

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

Qr= 2a = √2, a =	√2	.
	2

Answer: a =	√2	.
	2

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.

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.

________________

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Unified State Examination 2017 in mathematics, basic level

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