Task 4 USE in Russian

Placement of stress in words. Let's go over a little theory first.

Stresses of commonly used words that you should always remember:
Paragraph, agent, alibi, analogue, watermelon, arrest, athlete, bows, dishes, gas pipeline, blessing, fear, hyphen, contract, document, assistant professor, leisure, drowsiness, confessor, gospel, blinds, vent, clog, malice, sign Iconography, invention, sophistication, tool, spark, confession, rubber, quarter, self-interest, vine, ache, medicines, youth, ordeal, intention, illness, dumbness, provision, Adolescence, plateau, portfolio, sheet, percentage, pullOver, purple, revolver, belt, beets, silo, convocation, funds, customs, dancer, strengthening, chain, gypsy, porcelain, expert.

Nouns:
1) If words with the root -log- are offered, then know that it is stressed: dialogue, catalog, epilogue, obituary.
Exceptions are “analog” and words that name professions and occupations: philologist, biologist, archaeologist.
2) If the word ends in -mia, then [o] under stress: astronomy, economy, except for the words-terms (anemia, metonymy).
3) If the word has the second part -mania or -aria, then [a] is under stress: drug addiction, English; seminaryAria, culinaryAria, veterinaryAria.

Adjectives:
1) If the adjective is in the feminine form, then the ending is stressed: bad, fast, young, expensive.
2) Forms of the middle gender and plural require an emphasis on the stem: bad, fast, young, expensive; bad, fast, young, expensive.
3) The ending is always stressed in adjectives-exceptions: funny, heavy, hot, light, equal, dark, warm, smart, black, good. (Funny, funny, funny; heavy, heavy, heavy, etc.)

Verbs:
1) Remember that the prefix - you is always percussion (jump out, lay out), and the root - ringing - is always unstressed (phone, call, call).
2) In the verb-infinitive, the stress most often falls on the suffix: bestow, splash, seal.
3) As with nouns, in the feminine form, the stressed ending (waited, removed, accepted), and in the neuter gender and plural, the stem is stressed (waited, waited, understood, understood).
Exceptions: put, sent, stole, sent.
4) Prefixes for-, for-, pro-, co-tighten the stress (took, took, took).
Exceptions are verbs in which the stress falls on the root: called, called, called; tore, tore, tore.

Participles:
1) For full participles, the suffixes -ann- and -yann- are unstressed (broken, scattered).
2) The suffix -enn- is unstressed at the participle, if in the form of the future tense the stress is on the basis (wake up - awakened),
3) and the suffix -yonn- happens only if in the form of the future tense the emphasis is on the ending (bring in - brought in).
4) If in the full form of the short participle the suffix is ​​-yonn-, then in the short form -yon- (brought - brought),
another option is also possible (Given - Given, Given, Given, BUT given).
5) Prefixes pull the accents: Named - named, named, named, named. Collected - collected, collected, collected, collected.
6) In the feminine and neuter, as well as in the plural, the stress is on the ending (brought, brought, brought).

There are no uniform rules for pronunciation of adverbs ...
Now you can try to apply the acquired knowledge to solve several options for task 4 from the Unified State Examination in the Russian language.

Test options for task 4 from the Unified State Examination in Russian:

Try to solve them yourself and compare with the answers at the end of the page

Example 1:

briefcase
Clala
plum
called
chain

Example 2:

In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

delivered
will present
plucked
catalog
utterly

Example 3:

In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

accepted
started
called
cakes
arrived

Example 4:

In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

plucked
get through
(squirrel) agility
endowIt
include

Example 5:

In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

mosaic
call
removed
beard
facilities

Example 6:

In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

let's call them
rvalA
citizenship
old
selected

Example 7:

In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

disabled
adolescence
chauffeur
Wholesale
news

Example 8:

In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

arrived
bottom
accepted
click
started

Example 9:

In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

dry
aches
bows
beard
adolescence

Answers:

2. get through

   call you

   citizenship

3. Secondary general education

   Line UMK G.K. Muravina. Algebra and the beginnings of 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, teacher of mathematics of the highest category of the school, work experience of 20 years:

   “In order to get 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 consider options for the profile level.

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

   Example 1 In the apartment where Petr lives, a cold water meter (meter) was installed. 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.

   Decision:

   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 the 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?

   
   

   Decision:

   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.

   Answer: 15000.

   Task number 3- is a task of the basic level of the first part, it checks the ability to perform actions with geometric shapes according to 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.

   

   Decision: To calculate the area of ​​this figure, you can use the Peak formula:

   To calculate the area of ​​this rectangle, we use the Peak formula:
   

   S= B +	G
   	2

   where V = 10, G = 6, therefore
   

   S = 18 +	6
   	2

   Answer: 20.
   
   See also: Unified State Examination in Physics: solving vibration problems
   
   Task number 4- the task 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 on the circle. Determine which polygons are larger: those with all red vertices, or those with one of the blue vertices. In your answer, indicate how many more of one than the other.

   Decision: 1) We use the formula for the number of combinations from n elements by k:

   all of whose vertices are red.

   3) One pentagon with all red vertices.

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

   whose vertices are red or with one blue vertex.

   whose vertices are red or with one blue vertex.

   8) One hexagon whose vertices are red with one blue vertex.

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

   10) 42 - 16 = 26 polygons that use the blue dot.

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

   Answer: 10.

   Task number 5- the basic level of the first part tests the ability to solve the simplest equations (irrational, exponential, trigonometric, logarithmic).

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

   Decision. 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 number 6 in planimetry for finding geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. The study of the 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- median line parallel to side AB. Find the area of ​​the trapezoid ABED.

   
   

   Decision. Triangle CDE similar to a triangle CAB at two corners, since the corner at the vertex C general, angle CDE equal to the angle CAB as the corresponding angles at DE || AB secant AC. As DE is the middle line of the triangle by the condition, then by the property of the middle line | DE = (1/2)AB. So the similarity coefficient is 0.5. The areas of similar figures are related as the square of the similarity coefficient, so

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

   Task number 7- checks the application of the derivative to the study of the function. For successful implementation, a meaningful, non-formal possession of the concept of a derivative is necessary.

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

   Decision. 1) Let's use the equation of a straight line passing through two given points and find the equation of a straight 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| · (-one)

   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 slope of the tangent is the derivative of the function at the point of contact. Means, f′( x 0) = k 2 = –0,25.

   Answer: –0,25.

   Task number 8- checks the knowledge of elementary stereometry among the participants of the exam, 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.

   
   

   Decision. 1) V cube = a 3 (where a is the length of the edge of the cube), so

   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 number 9- requires the graduate to transform and simplify algebraic expressions. Task No. 9 of an increased level of complexity with a short answer. Tasks from the section "Calculations and transformations" in the USE are divided into several types:

   transformations of numerical rational expressions;

   transformations of algebraic expressions and fractions;

   transformations of numerical/letter irrational expressions;

   actions with degrees;

   transformation of logarithmic expressions;

   1. conversion of numeric/letter trigonometric expressions.

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

   3π	< α < π.
   4

   Decision. 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

   Hence, tan 2 α = ± 0.5.

   3) By condition

   3π	< α < π,
   4

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

   Answer: –0,5.

   #ADVERTISING_INSERT# Task number 10- checks the ability of students to use the 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 tasks are reduced 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 in the form of a whole number or a final 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?
   Decision. 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 sin2α ≥ 50

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

   We represent the solution of the inequality graphically:

   
   

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

   Task number 11- is typical, but it turns out to be difficult for students. The main source of difficulties is the construction of a mathematical model (drawing up an equation). Task number 11 tests the ability to solve word problems.

   Example 11. During spring break, 11-grader Vasya had to solve 560 training problems to prepare for the 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 on the last day of vacation.

   Decision: Denote a 1 = 5 - the number of tasks that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 - the number of days from March 18 to April 2 inclusive, S 16 = 560 - the total number of tasks, a 16 - the number of tasks that Vasya solved on April 2. Knowing that every day Vasya solved the same number of tasks more than the previous day, then you can use the 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 number 12- check students' ability to perform actions with functions, be able to apply the derivative to the study of the function.

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

   Decision: 1) Find the domain 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; ∞). We define 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 work program in mathematics to the line of UMK G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free algebra manuals
   
   Task number 13- an increased level of complexity with a detailed answer, which tests 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.

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

   	log3(2cos x) =	2	⇔		2cos x = 9	⇔		cos x =	4,5	⇔ because |cos x| ≤ 1,
   	log3(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 .

   
   

   It can be seen from the figure that the given segment has roots

   11π	and	13π	.
   6		6

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

   Task number 14- advanced level refers to the tasks of the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes. The task contains two items. In the first paragraph, the task must be proved, and in the second paragraph, it must be 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 bases 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 the same side of this plane.

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

   Decision: 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 on a plane parallel to the bases of the cylinders is either 8 + 6 = 14, or 8 − 6 = 2.

   

   Then the distance between 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 axis of the cylinder. 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's 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 the perpendicular bisector to this chord (it has a length of 8, as already noted) and from the center of the other base to another chord. They lie in the same plane β perpendicular to these chords. Let's call the midpoint of the smaller chord B, greater than 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 line of intersection of the base with the given plane.

   So the required angle is

   ∠ABH = arctan	AH	= arctg	28	= arctg14.
   	BH		8 – 6

   Task number 15- an increased level of complexity with a detailed answer, checks the ability to solve inequalities, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

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

   Decision: 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; +∞). In this case, this inequality can be rewritten in the form ( 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 a positive expression 3 x – x 2 , we get log 2 ( x + 1) ≤ 1, x + 1 ≤ 2, x≤ 1. Taking into account the area, we have x ∈ (0; 1].

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

   Answer: (–1; –0.5] ∪ ∪ {3}.

   Task number 16- advanced level refers to the tasks of 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 items. In the first paragraph, the task must be proved, and in the second paragraph, it must be calculated.

   In an isosceles triangle ABC with an angle of 120° at the 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 ​​the rectangle DEFH if AB = 4.

   Decision: a)

   
   

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

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

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

   BD is the bisector of ∠B, so ∠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 number 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 task with economic content.
   

   Example 17. The deposit in the amount 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 depositor annually replenishes the deposit by X million rubles, where X - whole number. Find the highest value X, at which the bank will add less than 17 million rubles to the deposit in four years.

   Decision: 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

   (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 number 18- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is not a task for applying one solution method, but for a combination of different methods. For the successful completion of task 18, in addition to solid mathematical knowledge, a high level of mathematical culture is also required.

   At what a system of inequalities

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

   has exactly two solutions?

   Decision: This system can be rewritten as

   	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 centered at the point (0, a). The set of solutions of the second inequality is the part of the plane that lies 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 of this system is the intersection of the solution sets of each of the inequalities.

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

   
   

   The points of contact between the circle and 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 the triangle PQR- rectangular isosceles. Dot Q has coordinates (0, a), and the point R– coordinates (0, – a). In addition, cuts PR and PQ are equal to the circle radius equal to 1. Hence,

   QR= 2a = √2, a =	√2	.
   	2

   Answer: a =	√2	.
   	2

   
   
   Task number 19- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is not a task for applying one solution method, but for a combination of different methods. For the successful completion of task 19, it is necessary to be able to search for a solution, choosing various approaches from among the known ones, modifying the studied methods.
   

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

   a) Give the formula P th member of this progression.

   b) Find the smallest modulo sum S n.

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

   Decision: a) Obviously, a n = S n – S n- one . 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) because S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Her graph can be seen in the figure.

   
   

   It is obvious that the smallest value is reached 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) It follows from the previous paragraph that sn positive since 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, with 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 full square is not achieved.

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

   ________________

   *Since May 2017, the DROFA-VENTANA joint publishing group has been part of the Russian Textbook Corporation. The corporation also included 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, LECTA digital educational platform) has been appointed General Director. Prior to joining the DROFA publishing house, he held the position of Vice President for Strategic Development and Investments of the EKSMO-AST publishing holding. Today, the Russian Textbook Publishing Corporation has the largest portfolio of textbooks included in the Federal List - 485 titles (approximately 40%, excluding textbooks for correctional schools). The corporation's publishing houses own the sets of textbooks in physics, drawing, biology, chemistry, technology, geography, astronomy, most in demand by Russian schools - areas of knowledge that are needed to develop the country's production potential. The corporation's portfolio includes textbooks and teaching aids for elementary schools awarded the President's Prize in Education. These are textbooks and manuals on subject areas that are necessary for the development of the scientific, technical and industrial potential of Russia.

   The fourth task of the Unified State Examination in the Russian language tests the ability of graduates to correctly place stress in various words. For its correct implementation, you can get one primary point; To do this, you need to choose a word with the wrong accent. Stressing often causes difficulties even for adults and educated people - the orthoepic norm does not always coincide with the usual pronunciation for us.

   In order to correctly complete this task, you need to make some efforts in preparation. The rules below will help.

   Theory for task No. 4 USE in the Russian language

   In verbs that end in "-it", the stress falls on the endings -ish, -it, -im, -ite, -at (-yat):

   • turn on - turn on, turn on, turn on, turn on, turn on;
   • call - call, call, call, call, call;
   • ease - make it easier, make it easier, make it easier, make it easier, make it easier;
   • strengthen - strengthen, strengthen, strengthen, strengthen, strengthen;
   • lend - lend, lend, borrow, borrow, lend;
   • hand over - hand over, hand over, hand over, hand over, hand over;
   • tame - tame, tame, tame, tame, tame;
   • pinch - pinch;
   • roll over - roll over.

   Exceptions in which the accent does not fall on the ending: vulgarize, inquire .

   In feminine verbs in the past tense, the stress falls on the ending "a":

   • took (took), removed (removed), understood, tore off, overtook, started, lied, left

   Exceptions: past tense verbs with the prefix "you" - the stress in them goes to the prefix, as well as the following words: put, stole, sent, sent, sent .

   In short passive feminine participles in the past tense, the stress also falls on the ending:

   • occupied, created, withdrawn, populated

   In verbs formed from adjectives, the stress falls on "-it":

   • easy - lighten
   • deep - deepen
   • complex - complicate

   Exception: Evil - embitter.

   In real past participles that have the suffix "-vsh-", the stress falls on the vowel before this suffix; The same rule applies to adverbs:

   • started, understood, completed, bored
   • starting, understanding, completing, giving, arriving

   Exception: exhausted.

   In the following words, the stress falls on the prefix:

   • bent, bent, bent

   The stress does not fall on the root "-bal-", therefore:

   • spoiled, spoiled, spoiled, spoiled, spoiled

   AT In past participles formed with the suffix "-yonn-", the stress falls on this suffix in the short form of the masculine gender, and in the short form of the feminine and neuter gender it passes to the ending:

   • disabled - disabled - disabled A - disabled
   • repeated - repeated - repeated A - repeated
   • tamed - tamed - tamedA - tamedO
   • populated - populated - populatedA - populated
   • enabled - enabled - enabledA - enabledO

   In nouns of foreign (mainly French) origin, the stress falls on the last syllable:

   • blinds, parter, bureau, jury, heretic, dispensary, quarter, obituary

   In verbal nouns, the stress usually coincides with the stress in the original verb:

   • provide - ensure
   • BUT CARRIES GAS – GAS PIPELINE

   However: lighten - relieve .

   In the following words, the stress is fixed and in all cases remains on the root:

   • airport - airports
   • scarf - scarves
   • cake - cakes
   • crane - cranes
   • bow - bows

   The stress falls on the prefix "for-" in words such as:

   • ahead of time, after dark, before dawn

   It is important to remember that this rule does not apply to the word enviably.

   The stress falls on the prefix "do-" in words such as:

   top, bottom, dry.

   It is important to remember that this rule does not apply to words red-hot, white-hot, utterly .

   You also need to remember the stress of the following words:

   • prettier, prettier, plum, kitchen

   Task execution algorithm

   • Read the assignment carefully.
   • We mentally pronounce the words proposed in the answer options, putting stress on different syllables.
   • Words in which the stress is correct are not taken into account.
   • When in doubt, we recall the rules for placing stresses in the words of the Russian language and the exceptions to these rules.
   • Write down the correct answer.

   Analysis of typical options for task No. 4 USE in the Russian language

   The fourth task of the 2018 demo

   1. amassed
   2. adolescence
   3. living
   4. correct
   5. took

   Execution algorithm:

   • Acquired - the stress is set correctly, in the real participles of the past tense with the suffix -vsh- the stress falls on the vowel before this suffix; Adolescence - right, you need to remember; lived - right, in the verbs of the past tense, the emphasis is on the ending; true - the stress is set correctly, since in short adjectives the stress is placed on the ending.
   • The last word raises doubts: did you take it or did you take it? We recall the rule: in verbs of the 3rd person of the feminine, the stress falls on the ending. So the accent is wrong.

   Answer: got it.

   The first version of the assignment

   In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

   1. overtaken
   2. busy
   3. self-interest
   4. will make it easier
   5. joined

   Execution algorithm:

   • You need to find a word in which a mistake was made in the formulation of stress.
   • According to the rule about feminine verbs of the past tense, the stress in the first two words is set correctly; the same applies to option number 5. The emphasis in the word "self-interest" is also correct, you just need to remember it.
   • The above words are stressed correctly.
   • Option 4 is wrong; this is also confirmed by the rule about verbs with the infinitive ending in "-it-" - the correct stress in this word is on the letter "and". So, the answer is - make it easier.

   Answer: make it easier.

   The second version of the task

   In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

   1. ailment
   2. encouraged
   3. fruit
   4. beet
   5. poured

   Execution algorithm:

   • You need to find a word in which a mistake was made in the formulation of stress.
   • The stress in word number 2 is correct, according to the rule about short participles of the past tense, formed from words with the suffix "yonn": encouraged - encouraged. In word number 5, everything is also correct: it is a passive past participle of the feminine gender, the stress in which falls on the ending. In word number 4, there is no error in stress: in words with the letter Y, the stress often falls on it. Then, in the word "bearing" the emphasis falls on "and", you just need to remember
   • The above words are stressed correctly.
   • The word ailment raises doubts. It must be remembered that the stress in it falls on the letter U. Therefore, this will be the correct answer.

   Answer: illness.

   The third version of the task

   In one of the words below, a mistake was made in setting the stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

   1. cakes
   2. put
   3. document
   4. tamed
   5. get to know

   Execution algorithm:

   • You need to find a word in which a mistake was made in the formulation of stress.
   • Put - the stress is true, in verbs on -it the stress falls on And, the document - the word must be remembered, tamed - in the past participles with the suffix -yonn- the stress falls on this suffix, you will learn - the word must be remembered.
   • The above words are stressed correctly.
   • Doubt causes cakes. In fact, the correct emphasis in it is cakes. This must be remembered.

   Answer: cakes.

   It turns out that for confident knowledge of the rules for setting stresses in Russian, you need to periodically look into the spelling dictionary; we provide such a dictionary, which contains the words used to compile the variants of the exam.

   Task Formulation:

   4. In one of the words below, a mistake was made in the formulation of stress: the letter denoting the stressed vowel is highlighted INCORRECTLY. Write out this word.

   adopted

   kitchen

   dispensary

   Answer: drill.

   What do students need to know in order to complete the task correctly?

   ORPHEPIC NORM OF THE RUSSIAN LANGUAGE.

   Distinctive features of Russian stress are its heterogeneity and mobility. The diversity lies in the fact that the stress in Russian can be on any syllable of the word (book, signature - on the first syllable; lantern, underground - on the second; hurricane, orthoepy - on the third, etc.). In some words, the stress is fixed on a certain syllable and does not move during the formation of grammatical forms, in others it changes from place (compare: ton - tons and wall - wall - walls and walls).

   Stress in adjectives.

   In full forms of adjectives, only a fixed stress is possible on the basis or on the ending. Little-used and bookish words often have an accent on the basis, and frequent, stylistically neutral or lowered ones - on the ending.

   The degree of mastery of the word is manifested in the variants of the place of stress: circle and circle, spare and spare, near-earth and near-earth, minus and minus, clearing and clearing. Such words are not included in the USE assignments, since both options are considered correct.!!!

   1. The choice of the place of stress causes difficulties most often in short forms of adjectives. The stressed syllable of the full form of a number of common adjectives remains stressed in the short form: beautiful - beautiful - beautiful - beautiful - beautiful; unthinkable - unthinkable - unthinkable - unthinkable - unthinkable, etc.

   2. Emphasis often falls on the stem in the form of masculine, neuter and many others. numbers and ending in the feminine form: right - right - right - right - right - right; gray - gray - gray - gray - gray; slender - slender - slender - slender - slender.

   3. It should also be said about the pronunciation of adjectives in a comparative degree. There is such a norm: if the stress in the short form of the feminine falls on the ending, then in a comparative degree it will be on the suffix -her: strong - stronger, sick - sicker, alive - livelier, slender - slenderer, right - right; if the stress in the feminine gender is on the basis, then to a comparative extent it is preserved on the basis: beautiful - more beautiful, sad - sadder, nasty - more nasty. The same applies to the superlative form.

   Stress in verbs.

   1. The stress in the past tense usually falls on the same syllable as in the infinitive: sit - sat, moan - moaned. hide - hide, start - start.

   2. The group of common verbs (about 300) obeys a different rule: the stress in the feminine form goes to the ending, and in other forms it remains on the stem. These are verbs to take. be, take, twist, lie, drive, give, wait, live, call, lie, pour, drink, tear, etc. It is recommended to say: live - lived - lived - lived - lived; wait - waited - waited - waited - waited; pour - lil - lilo - lili-lila. Derivative verbs are pronounced in the same way (live, pick up, drink up, spill, etc.).

   3. Verbs with the prefix you-, have an accent on the prefix: survive - survived, pour out - poured out, call out - called out.

   4. For verbs to put, steal, send, send, the stress in the form of the feminine gender of the past tense remains on the basis: krala, slala, sent, stlala.

   5. Quite often, in reflexive verbs (in comparison with irrevocable ones), the stress in the form of the past tense passes to the ending: begin - began I, began, began, began; accepted - accepted, accepted, accepted, accepted.

   6. About the pronunciation of the verb to call in conjugated form. Spelling dictionaries of recent times quite rightly continue to recommend stress on the ending: you call, call, call, call, call.

   Emphasis in some participles and participles.

   1. The most frequent fluctuations of stress are recorded when pronouncing short passive participles. If the stress in full form is on the suffix -yonn-, then it remains on it only in the masculine form, in other forms it goes to the ending: conducted - conducted, conducted, conducted; imported - imported, imported, imported, imported.

   2. A few remarks about the pronunciation of full participles with the suffix -t-. If the suffixes of an indefinite form -o-, -nu- are stressed, then in participles it will go one syllable forward: weed - weeded, pricked - pricked, bend - bent, wrap - wrapped.

   3. The participles often have an accent on the same syllable as in the indefinite form of the corresponding verb: putting, setting, baying, taking, drinking, exhausting (DO NOT: exhausted), starting, raising, living, watering, putting, understanding, preAv, undertaking, having arrived, having accepted, having sold, having cursed, having spilled, having penetrated, having drunk, having created.

   Stress in adverbs should mainly be studied by memorizing and referring to the orthoepic dictionary.

   I give a list of words that occur in task number 4 (you need to learn it).

   Nouns

   airports, fixed stress on the 4th syllable

   bows, fixed stress on 1st syllable

   beard, win.p., only in this form singular. stress on 1st syllable

   accountants, rod.p.mn.ch., motionless. stress on 2nd syllable

   religion, from faith to confess

   water pipes

   gas pipeline

   citizenship

   hyphen, from German, where the stress is on the 2nd syllable

   cheapness

   dispensary, the word came from English. lang. through French, where the blow. always on the last syllable

   agreement

   document

   blinds, from French lang., where is the blow. always on the last syllable

   significance, from adj. significant

   X, im.p. pl., motionless stress

   catalog, in the same row with the words dialogue, monologue, obituary, etc.

   quarter, from it. lang., where the stress is on the 2nd syllable

   kilometer, on a par with the words centimeter, decimeter, millimeter ...

   cones, cones, motionless stress on the 1st syllable in all cases in singular and plural.

   cranes, fixed stress on 1st syllable

   flint, flint, blow. in all forms on the last syllable, as in the word fire

   lecturers, lecturers, see the word bow(s)

   localities, genus p.pl., on a par with the word form of honors, jaws ... but news

   garbage chute, in the same row with the words gas pipeline, oil pipeline, water pipeline

   intention

   obituary, see catalog

   hatred

   pipeline

   news, news, but: see localities

   nail, nail, motionless. stress in all forms singular.

   SUPPORT

   Adolescence, from Otrok - teenager

   parter, from French. lang., where is the blow. always on the last syllable

   briefcase

   dowry, noun

   call, in the same row with the words call, recall (ambassador), convocation, but: Review (for publication)

   orphans, im.p.pl., stress in all forms pl. only on the 2nd syllable

   funds, im.p.pl.

   carpenter, in the same poison with the words painter, doYar, shkolYar ...

   convocation, see call

   shorthand

   dancer

   cakes, cakes

   fluorography

   Christian

   scarves, see bows

   chauffeur, on a par with the words kioskёr, controller ...

   expert, from the French. lang., where the stress is always on the last syllable

   Adjectives

   correct, short adj. zh.r.

   pear

   old

   significant

   most beautiful, excellent

   kitchen

   agility, short adj. zh.r.

   salmon

   mosaic

   perspicacious, short adj. zh.r., on a par with the words cute, fussy, talkative ... but: gluttonous

   plum, derived from plum

   Verbs

   spoil, on a par with the words spoil, spoil, spoil ..., but: the minion of fate

   favor

   take-took

   take-take

   take-took

   take-took

   turn on, turn on

   turn on, turn on

   join-merged

   break in-break in

   perceive-perceived

   recreate-recreated

   hand-hand over

   drive-driven

   chasing-chasing

   get-dobrala

   get-got

   wait-wait

   get through-get through

   get through

   dose

   wait-waited

   live-lived

   cork up

   occupied-occupied, occupied, occupied, occupied

   lock up-locked up (with a key, with a lock, etc.)

   call-called

   call-call, call, call them

   exclude-exclude

   exhaust

   lay-lay

   sneak-stalked

   bleed

   lie-lie

   pour-lila

   pour-poured

   lie-lied

   endow-endowit

   overstrained-overstrained

   name-named

   bank-roll

   pour-poured

   narwhal-narwhala

   litter-litter

   start-started, started, started

   call-call-call

   facilitate-facilitate

   drenched-drenched

   hug-hugged

   overtake-overtaken

   rip-off

   encourage

   cheer up - cheer up

   exacerbate

   borrow-borrow

   embitter

   surround-surround

   seal, in the same row with the words form, normalize, sort ...

   vulgarize - vulgarize

   inquire - inquire

   depart-departed

   give-gave

   turn-off

   withdraw-revoked

   responded-responded

   call back-call back

   transfuse-transferred

   mold

   fruit

   repeat-repeat

   call-called

   call-call-call

   pour-watered

   put-put

   understand-understood

   send-sent

   arrive-arrived-arrived-arrived

   accept-accepted-accepted - accepted

   force

   tear-torn

   drill-drill-drill

   take off-taken off

   create-created

   pluck-plucked

   litter-litter

   remove-remove

   speed up

   deepen

   strengthen-strengthen

   move to hide

   pinch-pinch

   Communions

   pampered

   enabled-enabled, see relegated

   delivered

   folded

   busy-busy

   locked-locked

   inhabited-inhabited

   spoiled, see spoiled

   feeding

   bleeding

   amassed

   acquired-acquired

   poured - poured

   hired

   started

   relegated-reduced, see included…

   encouraged-encouraged-encouraged

   aggravated

   defined-defined

   disabled

   repeated

   divided

   understood

   adopted

   tamed

   lived

   removed-removed

   bent

   Participles

   clogged

   starting

   Adverbs

   utterly

   enviably, in the meaning of the predicate

   ahead of time, colloquial

   before dark

   prettier, adj. and adv. in comp.