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)
Maths
Preparation for the exam in mathematics (profile level): tasks, solutions and explanations
We analyze tasks and solve examples with the teacherThe 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 Exam 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.
Solution:
1) Find the amount of water spent per month:
177 - 172 = 5 (cu m)
2) Find how much money will be paid for the spent water:
34.17 5 = 170.85 (rub)
Answer: 170,85.
Task number 2- is one of the simplest tasks of the exam. The majority of graduates successfully cope with it, which indicates the possession of the definition of the concept of function. Task type No. 2 according to the requirements codifier is a task for using 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?
Solution:
2) 1000 3/4 = 750 (shares) - make up 3/4 of all purchased shares.
6) 247500 + 77500 = 325000 (rubles) - the businessman received after the sale of 1000 shares.
7) 340,000 - 325,000 = 15,000 (rubles) - the businessman lost as a result of all operations.
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.
Solution: 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 |
|
S = 18 + |
6 | |
| 2 |
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.
Solution: 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 .
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 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.
Solution. 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. Because 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
Consequently, 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).
Solution. 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.
Solution. 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:
- conversion of numeric/letter trigonometric expressions.
transformations of numerical rational expressions;
transformations of algebraic expressions and fractions;
transformations of numerical/letter irrational expressions;
actions with degrees;
transformation of logarithmic expressions;
Example 9 Calculate tgα 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 |
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?
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 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.
Solution: 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.
Solution: 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 manualsTask 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.
Solution: 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 |
The circumference diameter 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.
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 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 .
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; +∞). 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.
Solution: 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.
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
(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?
Solution: 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 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.
Solution: 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.
Grade 11
Task Conditions
- The price of an electric kettle was increased by 14% and amounted to 1,596 rubles. How much was the kettle worth before the price increase?
- The graph shows the dependence of the engine torque on the number of revolutions per minute. The number of revolutions per minute is plotted on the abscissa axis, and the torque in N∙m is plotted on the ordinate axis. Vehicle speed (in km/h) is approximated by the formula
where n is the number of engine revolutions per minute. What is the minimum speed the car must be moving in order for the torque to be 120 N∙m? Give your answer in kilometers per hour.
- A triangle ABC is depicted on checkered paper with cell size x. Find the length of its height dropped to side BC.
- The scientific conference is held in 5 days. A total of 75 reports are planned - the first three days, 17 reports each, the rest are distributed equally between the fourth and fifth days. At the conference, a report by Professor M. is planned. The order of reports is determined by a drawing of lots. What is the probability that Professor M.'s report will be scheduled for the last day of the conference?
- Find the root of the equation
- Quadrilateral ABCD is inscribed in a circle. Angle ABC is equal to 105 o , angle CAD is equal to 35 o . Find angle ABD. Give your answer in degrees.
- The figure shows a graph of the derivative of a function defined on the interval . Find the number of maximum points of the function that belong to the segment .
- The ball is inscribed in a cylinder. The surface area of the sphere is 111. Find the total surface area of the cylinder.
- Find the value of an expression
- To obtain an enlarged image of a light bulb on the screen, a converging lens with a main focal length cm is used in the laboratory. The distance from the lens to the light bulb can vary from 30 to 50 cm, and the distance from the lens to the screen - from 150 to 180 cm the screen will be clear if the ratio is met. Indicate the smallest distance from the lens that a light bulb can be placed so that its image on the screen is clear. Express your answer in centimeters.
- The distance between piers A and B is 120 km. From A to B, a raft set off down the river, and an hour later a yacht set off after it, which, having arrived at point B, immediately turned back and returned to A. By this time, the raft had covered 24 km. Find the speed of the yacht in still water if the speed of the river is 2 km/h. Give your answer in km/h.
- Find the maximum point of the function.
- a) Solve the equation
; b) Indicate the roots of this equation that belong to the segment. - Points M and N are marked on the edges AB and BC of the triangular pyramid ABCD, respectively, with AM:MB = CN:NB = 3:1. The points P and Q are the midpoints of the edges DA and DC, respectively.
a) Prove that the points P,Q,M and N lie in the same plane;
b) Find in what ratio this plane divides the volume of the pyramid. - Solve the inequality
- Point E is the midpoint of the lateral side CD of the trapezoid ABCD. On its side, AB took a point K so that the lines SC and AE are parallel. Segments SK and BE intersect at point O.
a) Prove that CO=CO.
b) Find the ratio of the bases of the trapezoid BC: AD, if the area of the triangle BCK is 9/64 of the area of the entire trapezoid ABCD. - In July, it is planned to take a loan from a bank for a certain amount. The conditions for its return are as follows:
- each January the debt increases by r% compared to the end of the previous year;
- From February to June of each year, part of the debt must be repaid.
Find r if it is known that if you pay 777,600 rubles each, then the loan will be repaid in 4 years, and if you pay 1,317,600 rubles each year, then the loan will be fully repaid in 2 years? - Find all values of the parameter for each of which the equation has exactly one root on the interval .
- Each of the 32 students either wrote one of the two tests, or wrote both tests. For each work it was possible to get an integer number of points from 0 to 20 inclusive. For each of the two test papers separately, the average score was 14. Then each student named the highest of his scores (if the student wrote one paper, he named the score for it). The arithmetic mean of the named scores was equal to S.
a) Give an example when S<14
b) Could the value of S be equal to 17?
c) What is the smallest value S could take if both tests were written by 12 students?
Passing a unified state exam is not only a necessity at the end of general secondary education, but also part of the entrance examinations to universities. Schoolchildren who decide to enter specialties with a mathematical or technical bias pass not only the basic level of mathematics, but also the profile one. Consider its features, timing and verification, and some points related to the results.
The procedure for conducting the exam is established by Federal Law No. 273 "On Education in the Russian Federation".
When will the exam results be known?
The official timetable determined the surrender USE in mathematics 2018 profile direction on Friday, June 1. As reserve day date is highlighted in the main loop June 25, and July 2 remains a spare day for the delivery of all items.
Separation math exam on the levels happened last year. They differ on a number of grounds:
- Grading system. The basic level of knowledge of the subject is assessed on a five-point scale (3 points are set as a minimum). The assessment in the profile subject is evaluated on a scale of 100 points;
- The next difference is in the admission of basic and profile level exams for admission to educational institutions senior and middle professional level. So, the basic level is enough for colleges, schools, liberal arts universities. The presence of mathematics in the entrance exams for technical specialties requires the applicant to pass the profile level;
- Differ exam structures. The base consists of 20 problems with short answers. The profile exam is much more difficult and consists of 2 parts.
The USE system allows school graduates to take the basic and profile part of the subject without restrictions. This significantly increases the chances of getting into universities.
Processing the results of the exam has a certain time frame and order:
- Scanning and processing of forms in the regions - up to 4 days;
- Processing of results at the federal level - up to 7 days;
- Sending results to the regions - 1 day;
- Confirmation of the results by the state examination committee - no longer than 1 day;
- Announcement of results - 1 day.
Thus, the period for checking and publishing the results is no more than 2 weeks. The results of the USE 2018 in mathematics at the profile level will be known no later than June 17.
How to know your result?
Find out the results of the last exam can be done in several ways:
- Official portal of the Unified State Examination www.ege.edu.ru;
- At information stands in schools or other institutions where the exam was held;
- In regional departments or committees of education;
- A number of regions create specialized websites or hotlines.
Check your score available if available:
- Full name of the subject;
- Number of the passport or other document used during the identity exam;
- An identification code assigned to each participant in the exam.
Information about the results of the exam is free and is provided free of charge to USE participants and their parents.
Pre-term USE exam in mathematics
A number of schoolchildren have already passed the USE in mathematics in the so-called early period. Participation in it is allowed if the student cannot take part in the main stage. The reasons may be:
- Planned treatment;
- Rest in health-improving establishments;
- Participation in competitions, olympiads and other educational or creative events.
In 2017, early delivery of mathematics took place March 31 and April 14(reserve day). 4.8 thousand schoolchildren passed the basic level, and about 17 thousand specialized ones.
According to the plan, the results of the early USE in mathematics 2017 were supposed to be available on April 11, but were made public much earlier - on the 7th.
Where to see your work
You can view your work after passing the exam in electronic form. Her scan is available in your personal account on the USE portal. Access to it is issued when:
- The presence of the identification code of the participant of the unified state exam;
- Full name and passport number.
If, after the announcement of the results, the participant does not agree with the points given, then he has 2 days to file an appeal to the Examining Committee. The application is written in 2 copies and submitted to the commission for consideration. By June 5, the solutions to the problems will be reviewed again and a decision will be made to change the assessment or confirm it.
How is the exam graded? The USE system for evaluating results uses primary and test scores, as well as a special scale for translating them into each other. Solutions of KIMs (control and measuring materials) are evaluated in primary points and then transferred according to the table into test ones. The final result of the exam is the number of test points scored.
The development of a scale for converting primary scores into test scores is carried out every year and takes into account the general level of preparation of schoolchildren.
For successful passing profile mathematics in 2018 you need to type the minimum:
- 6 primary points;
- 27 test points.
Date of retaking the exam in mathematics in 2018
There are a number additional deadlines for passing the exam. They are available if, for good reason, the student was unable to pass the subject on the main day. For profile mathematics, this is:
- June 25– reserve day within the framework of the main stage;
- July 2- a reserve day of the main part of the exam, when you can pass any subject.
The opportunity to retake profile mathematics in September has a number of conditions:
- If a student has passed basic mathematics, then he will not be allowed to retake the profile level this year. The opportunity to retake the exam will arise only next year;
- If both exams in mathematics (basic and profile) are failed, the student can decide which one he will retake.
Math retake appointed in September September 7. September 15 is listed as a reserve day.
