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 teacherThe 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.
#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 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 |
|
S = 18 + |
6 | |
| 2 |
Read also: Unified State Exam in Physics: solving problems about oscillations
Task No. 4- the objective of the course “Probability Theory and Statistics”. The ability to calculate the probability of an event in the simplest situation is tested.
Example 4. There are 5 red and 1 blue dots marked on the circle. Determine which polygons are larger: those with all the vertices red, or those with one of the vertices blue. In your answer, indicate how many there are more of some than others.
Solution: 1) Let's use the formula for the number of combinations of n elements by k:
whose vertices are all red.
3) One pentagon with all vertices red.
4) 10 + 5 + 1 = 16 polygons with all red vertices.
which have red tops or with one blue top.
which have red tops or with one blue top.
8) One hexagon with red vertices and one blue vertex.
9) 20 + 15 + 6 + 1 = 42 polygons with all red vertices or one blue vertex.
10) 42 – 16 = 26 polygons using the blue dot.
11) 26 – 16 = 10 polygons – how many more polygons in which one of the vertices is a blue dot are there than polygons in which all the vertices are only red.
Answer: 10.
Task No. 5- the basic level of the first part tests the ability to solve simple equations (irrational, exponential, trigonometric, logarithmic).
Example 5. Solve equation 2 3 + x= 0.4 5 3 + x .
Solution. Divide both sides of this equation by 5 3 + X≠ 0, we get
| 2 3 + x | = 0.4 or | 2 | 3 + X | = | 2 | , | ||
| 5 3 + X | 5 | 5 |
whence it follows that 3 + x = 1, x = –2.
Answer: –2.
Task No. 6 in planimetry to find geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. Study of constructed models using geometric concepts and theorems. The source of difficulties is, as a rule, ignorance or incorrect application of the necessary theorems of planimetry.
Area of a triangle ABC equals 129. DE– midline parallel to the side AB. Find the area of the trapezoid ABED.
Solution. Triangle CDE similar to a triangle CAB at two angles, since the angle at the vertex C general, angle СDE equal to angle CAB as the corresponding angles at DE || AB secant A.C.. Because DE is the middle line of a triangle by condition, then by the property of the middle line | DE = (1/2)AB. This means that the similarity coefficient is 0.5. The areas of similar figures are related as the square of the similarity coefficient, therefore
Hence, S ABED = S Δ ABC – S Δ CDE = 129 – 32,25 = 96,75.
Task No. 7- checks the application of the derivative to the study of a function. Successful implementation requires meaningful, non-formal knowledge of the concept of derivative.
Example 7. To the graph of the function y = f(x) at the abscissa point x 0 a tangent is drawn that is perpendicular to the line passing through the points (4; 3) and (3; –1) of this graph. Find f′( x 0).
Solution. 1) Let’s use the equation of a line passing through two given points and find the equation of a line passing through points (4; 3) and (3; –1).
(y – y 1)(x 2 – x 1) = (x – x 1)(y 2 – y 1)
(y – 3)(3 – 4) = (x – 4)(–1 – 3)
(y – 3)(–1) = (x – 4)(–4)
–y + 3 = –4x+ 16| · (-1)
y – 3 = 4x – 16
y = 4x– 13, where k 1 = 4.
2) Find the slope of the tangent k 2, which is perpendicular to the line y = 4x– 13, where k 1 = 4, according to the formula:
3) The tangent angle is the derivative of the function at the point of tangency. Means, f′( x 0) = k 2 = –0,25.
Answer: –0,25.
Task No. 8- tests the exam participants’ knowledge of elementary stereometry, the ability to apply formulas for finding surface areas and volumes of figures, dihedral angles, compare the volumes of similar figures, be able to perform actions with geometric figures, coordinates and vectors, etc.
The volume of a cube circumscribed 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:
- converting numeric/letter trigonometric expressions.
transformation of numerical rational expressions;
converting algebraic expressions and fractions;
conversion of numeric/letter irrational expressions;
actions with degrees;
converting logarithmic 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 algebraTask 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 |
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.
The lesson discusses the solution to task 12 of the Unified State Exam in computer science, including tasks from 2017
Topic 12 - “Network addresses” - is characterized as tasks of a basic level of complexity, completion time - approximately 2 minutes, maximum score - 1
Internet addressing
The address of a document on the Internet (from English - URL - Uniform Resource Locator) consists of the following parts:
- data transfer protocol; May be:
- http(for Web pages) or
- ftp(for file transfer)
- there is also a secure protocol https;
- delimiter characters :// , separating the protocol name from the rest of the address;
- website domain name (or IP address);
- may also be present: the directory on the server where the file is located;
- file name.
Directories on the server are separated by forward slash " / »
- network service protocol name – defines the server type HTTP(Hypertext Transfer Protocol);
- delimiter in the form of a colon and two characters Slash;
- fully qualified domain name of the server;
- search path for a web document on a computer;
- web server name;
- top level domain "org";
- country code name "ru";
- catalog main on the computer;
- catalog news in the catalog main;
- the final goal of the search is a file main_news.html.
Network addresses
Physical adress or MAC address– a unique address, “hardwired” at production – 48-bit code of the network card (in hexadecimal):
00-17-E1-41-AD-73
IP address– computer address (32-bit number), consisting of: network number + computer number in the network (node address):
15.30.47.48
Subnet mask:
- necessary to determine which computers are on the same subnet;
255.255.255.0 -> FF.FF.FF.0
1…10…0
The part of the IP address that corresponds to the mask bits equal to one refers to the network address, and the part corresponding to the mask bits equal to zero is the numerical address of the computer
Calculation of network number by IP address and network mask
In the subnet mask most significant bits, allocated in the computer's IP address for network number, have a value of 1 (255); least significant bits, allocated in the computer's IP address for computer addresses in the subnet, matter 0
.
* Image taken from the presentation by K. Polyakov
Number of computers on the network
The number of computers on the network is determined by the mask: the low-order bits of the mask - zeros - are reserved in the computer's IP address for the address of the computer in the subnet.
If the mask:
The number of computers on the network:
2 7 = 128 addresses
Of these, 2 are special: network address and broadcast address
128 - 2 = 126 addresses
Solving tasks 12 Unified State Exam in computer science
Unified State Examination in Informatics 2017 task 12 FIPI option 1 (Krylov S.S., Churkina T.E.):
In the terminology of TCP/IP networks, a network mask is a binary number that determines which part of the IP address of a network host refers to the network address, and which part refers to the address of the host itself on this network. Typically, the mask is written according to the same rules as the IP address - as four bytes, with each byte written as a decimal number. In this case, the mask first contains ones (in the highest digits), and then from a certain digit there are zeros. The network address is obtained by applying a bitwise conjunction to the given host IP address and mask.
For example, if the host IP address is 211.132.255.41 and the mask is 255.255.201.0, then the network address is 211.132.201.0
For a node with an IP address 200.15.70.23
network address is 200.15.64.0
. What is equal to least possible value of the third byte from the left of the mask? Write your answer as a decimal number.
✍ Solution:
- The third byte from the left corresponds to the number 70 in the IP address and 64 - in the network address.
- The network address is the result of the bitwise conjunction of the mask and the IP address in binary:
Result: 192
A step-by-step solution to this 12th task of the Unified State Exam in computer science is available in the video tutorial:
Task 12. Demo version of the Unified State Exam 2018 computer science:
In the terminology of TCP/IP networks, a network mask is a binary number that determines which part of the IP address of a network host refers to the network address, and which part refers to the address of the host itself on this network. Typically, the mask is written according to the same rules as the IP address - in the form of four bytes, with each byte written as a decimal number. In this case, the mask first contains ones (in the highest digits), and then from a certain digit there are zeros.
The network address is obtained by applying a bitwise conjunction to the given host IP address and mask.
For example, if the host IP address is 231.32.255.131 and the mask is 255.255.240.0, then the network address is 231.32.240.0.
For a node with an IP address 57.179.208.27 network address is 57.179.192.0 . What's it like greatest possible quantity units in the ranks of the mask?
✍ Solution:
- Since the network address is obtained as a result of applying a bitwise conjunction to a given host IP address and mask, we get:
Result: 19
For a detailed solution to task 12 of the demo version of the Unified State Exam 2018, watch the video:
Solution to task 12 (Polyakov K., option 25):
In TCP/IP network terminology, a network mask is a binary number that indicates which part of the IP address of a network host relates to the network address, and which part to the host address on this network. The network address is obtained by applying a bitwise conjunction to a given node address and its mask.
Based on the specified host IP address and mask determine the network address:
IP address: 145.92.137.88 Mask: 255.255.240.0
When writing down the answer, select the four elements of the IP address from the numbers given in the table and write down the letters corresponding to them without dots in the required order.
| A | B | C | D | E | F | G | H |
| 0 | 145 | 255 | 137 | 128 | 240 | 88 | 92 |
✍ Solution:
- To solve the problem, you need to remember that the IP address of the network, as well as the network mask, are stored in 4 bytes written with a dot. That is, each of the individual IP address and netmask numbers are stored in 8-bit binary form. To obtain the network address, it is necessary to perform a bitwise conjunction of these numbers.
- Since the number 255 in binary representation this is 8 units, then with a bitwise conjunction with any number, the result will be the same number. Thus, there is no need to take into account those bytes of the IP address that correspond to the number 255 in the network mask. Therefore, the first two numbers of the IP address will remain the same ( 145.92 ).
- It remains to consider the numbers 137 And 88 IP addresses and 240 masks. Number 0 in the mask matches eight zeros in binary representation, that is, a bitwise conjunction with any number will turn this number into 0 .
- Let's convert both numbers of the IP address and network mask into the binary system and write the IP address and mask below each other to carry out the bitwise conjunction:
Result: BHEA
We invite you to watch a detailed video analysis:
Solution to task 12 (Polyakov K., option 33):
If the subnet mask 255.255.255.128 and IP address of the computer on the network 122.191.12.189 , then the computer number on the network is _____.
✍ Solution:
- The single bits of the mask (equal to one) determine the subnet address, because The subnet address is the result of the bitwise conjunction (logical multiplication) of the mask bits with the IP address.
- The rest of the mask (starting with the first zero) specifies the computer number.
- Since in binary representation the number 255 - this is eight units ( 11111111 ), then with a bitwise conjunction with any number, the same number is returned (1 ∧ 0 = 0; 1 ∧ 1 = 1). Thus, those bytes in the mask that are equal to numbers 255 , we will not consider, because they define the subnet address.
- Let's start with a byte equal to 128 . It corresponds to a byte 189 IP addresses. Let's convert these numbers to the binary number system:
Result: 61
For a detailed solution to this task, watch the video:
Solution to task 12 (Polyakov K., option 41):
In the terminology of TCP/IP networks, a subnet mask is a 32-bit binary number that determines which bits of the computer's IP address are common to the entire subnet - these bits of the mask contain 1. Typically, masks are written as a quadruple of decimal numbers - according to the same rules , the same as IP addresses.
A mask is used for some subnet 255.255.255.192 . How many different computer addresses theoretically allows this mask if two addresses (network and broadcast address) are not used?
✍ Solution:
- The single bits of the mask (equal to one) determine the subnet address, the rest of the mask (starting with the first zero) determines the computer number. That is, there are as many options for the computer address as can be obtained from the zero bits in the mask.
- In our case, we will not consider the first three bytes of the mask on the left, because number 255 in binary representation it is eight units ( 11111111 ).
- Consider the last byte of the mask, equal to 192 . Let's convert the number to the binary number system:
Result: 62
Watch the video description of the task below:
Solution to task 12 (Regional work, Far East, 2018):
For a node with an IP address 93.138.161.94 network address is 93.138.160.0 .For how many different mask values is this possible?
✍ Solution:
Result: 5
Video analysis of the task:
In task No. 12 of the Unified State Exam in mathematics at the profile level, we need to find the largest or smallest value of the function. To do this, it is necessary to use, obviously, a derivative. Let's look at a typical example.
Analysis of typical options for tasks No. 12 of the Unified State Exam in mathematics at the profile level
First version of the task (demo version 2018)
Find the maximum point of the function y = ln(x+4) 2 +2x+7.
Solution algorithm:
- Finding the derivative.
- We write down the answer.
Solution:
1. We are looking for values of x for which the logarithm makes sense. To do this, we solve the inequality:
Because the square of any number is non-negative. The solution to the inequality will be only the value of x at which x+4≠ 0, i.e. at x≠-4.
2. Find the derivative:
y’=(ln(x+4) 2 + 2x + 7)’
By the property of the logarithm we get:
y’=(ln(x+4) 2)’+(2x)’+(7)’.
According to the formula for the derivative of a complex function:
(lnf)’=(1/f)∙f’. We have f=(x+4) 2
y, = (ln(x+4) 2)'+ 2 + 0 = (1/(x+4) 2)∙((x+4) 2)' + 2=(1/(x+4) 2 2)∙(x 2 + 8x + 16)' +2=2(x + 4) /((x + 4) 2) + 2
y’= 2/(x + 4) + 2
3. We equate the derivative to zero:
y, = 0 → (2+2∙(x + 4))/(x + 4)=0,
2 +2x +8 =0, 2x + 10 = 0,
Second version of the task (from Yashchenko, No. 1)
Find the minimum point of the function y = x – ln(x+6) + 3.
Solution algorithm:
- We determine the domain of definition of the function.
- Finding the derivative.
- We determine at which points the derivative is equal to 0.
- We exclude points that do not belong to the domain of definition.
- Among the remaining points, we look for x values at which the function has a minimum.
- We write down the answer.
Solution:
2. Find the derivative of the function:
3. We equate the resulting expression to zero:
4. We received one point x=-5, belonging to the domain of definition of the function.
5. At this point the function has an extremum. Let's check if this is the minimum. At x=-4
At x=-5.5, the derivative of the function is negative, since
This means that point x=-5 is the minimum point.
Third version of the task (from Yashchenko, No. 12)
Find the largest value of the function 
on the segment [-3; 1].
Solution algorithm:
- Finding the derivative.
- We determine at which points the derivative is equal to 0.
- We exclude points that do not belong to a given segment.
- Among the remaining points, we look for the x values at which the function has a maximum.
- We find the values of the function at the ends of the segment.
- We are looking for the largest among the obtained values.
- We write down the answer.
Solution:
1. We calculate the derivative of the function, we get
The Unified State Exam in basic level mathematics consists of 20 tasks. Task 12 tests the skills of choosing the optimal option from those proposed. The student must be able to evaluate possible options and choose the most optimal one. Here you can find out how to solve task 12 of the Unified State Exam in basic level mathematics, as well as study examples and solutions based on detailed tasks.
All USE base tasks all tasks (263) USE base task 1 (5) USE base task 2 (6) USE base task 3 (45) USE base task 4 (33) USE base task 5 (2) USE base task 6 (44 ) Unified State Examination base assignment 7 (1) Unified State Examination base assignment 8 (12) Unified State Examination base assignment 10 (22) Unified State Examination base assignment 12 (5) Unified State Examination base assignment 13 (20) Unified State Examination base assignment 15 (13) Unified State Examination base assignment 19 (23) Unified State Exam base task 20 (32)
On average, a citizen A. consumes electricity per month during the daytime
On average, a citizen of A consumes K kWh of electricity per month during the day, and L kWh of electricity at night. Previously, A. had a single-tariff meter installed in his apartment, and he paid for all electricity at the rate of M rubles. per kWh. A year ago, A. installed a two-tariff meter, while daily electricity consumption is paid at the rate of N rubles. per kWh, and night consumption is paid at the rate P rub. per kWh. During R months, the consumption mode and electricity payment tariffs did not change. How much more would A. have paid for this period if the meter had not changed? Give your answer in rubles.
When building a rural house, you can use one of two types of foundation
When building a rural house, you can use one of two types of foundation: stone or concrete. For a stone foundation you need A tons of natural stone and B bags of cement. For a concrete foundation you need C tons of crushed stone and D bags of cement. A ton of stone costs E rubles, crushed stone costs F rubles per ton, and a bag of cement costs G rubles. How many rubles will the foundation material cost if you choose the cheapest option?
The problem is part of the Unified State Examination in basic level mathematics for grade 11 under number 12.
How many rubles will you have to pay for the cheapest trip for three
A family of three is planning to travel from St. Petersburg to Vologda. You can go by train, or you can go by car. A train ticket for one person costs N rubles. A car consumes K liters of gasoline per L kilometers, the distance along the highway is M km, and the price of gasoline is P rubles per liter. How many rubles will you have to pay for the cheapest trip for three?
The problem is part of the Unified State Examination in basic level mathematics for grade 11 under number 12.
When building a house, the company uses one of the types of foundation
When building a house, the company uses one of the types of foundations: concrete or foam block. For a foundation made of foam blocks, you need K cubic meters of foam blocks and L bags of cement. For a concrete foundation you need M tons of crushed stone and N bags of cement. A cubic meter of foam blocks costs A rubles, crushed stone costs B rubles per ton, and a bag of cement costs C rubles. How many rubles will the material cost if you choose the cheapest option?
In the twelfth task of the OGE in mathematics of the Algebra module, our knowledge of transformations is tested - the rules for opening brackets, placing variables outside brackets, reducing fractions to a common denominator and knowledge of abbreviated multiplication formulas.
The essence of the task comes down to simplifying the expression specified in the condition: you should not immediately substitute values into the original expression. You must first simplify it and then substitute the value - all tasks are structured in such a way that after simplification you only need to perform one or two simple actions.
It is necessary to take into account the permissible values of variables included in algebraic expressions, use the properties of powers with an integer exponent, rules for extracting roots and abbreviated multiplication formulas.
The answer in the task is an integer or a finite decimal fraction.
Theory for task No. 12
First of all, let's remember what a degree is and
In addition, we will need abbreviated multiplication formulas:
Square of the sum
(a + b) 2 = a 2 + 2ab + b 2
Squared difference
(a - b) 2 = a 2 - 2ab + b 2
Difference of squares
a 2 – b 2 = (a + b)(a – b)
Cube of sum
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
Difference cube
(a - b) 3 = a 3 - 3a 2 b + 3ab 2 - b 3
Sum of cubes
a 3 + b 3 = (a + b)(a 2 - ab + b 2)
Difference of cubes
a 3 – b 3 = (a – b)(a 2 + ab + b 2)
Rules operations with fractions :
Analysis of typical options for task No. 12 OGE in mathematics
First version of the task
Find the value of the expression: (x + 5) 2 - x (x- 10) at x = - 1/20
Solution:
In this case, as in almost all tasks No. 7, you must first simplify the expression; to do this, open the brackets:
(x + 5) 2 - x (x - 10) = x 2 + 2 5 x + 25 - x 2 + 10x
Then we present similar terms:
x 2 + 2 5 x + 25 -x 2 + 10x = 20 x + 25
20 x + 25 = 20 (-1/20) + 25 = - 1 + 25 = 24
Second version of the task
Find the meaning of the expression:
at a = 13, b = 6.8
Solution:
In this case, unlike the first, we will simplify the expression by taking it out of brackets, rather than opening them.
You can immediately notice that b is present in the first fraction in the numerator, and in the second - in the denominator, so we can reduce them. Seven and fourteen are also reduced by seven:
Let's shorten (a-b):
And we get:
Substitute the value a = 13:
Third version of the task
Find the meaning of the expression:
at x = √45, y = 0.5
Solution:
So, in this task, when subtracting fractions, we need to bring them to a common denominator.
The common denominator is 15 x y, To do this, you need to multiply the first fraction by 5 y- both numerator and denominator, naturally:
Let's calculate the numerator:
5 y - (3 x + 5 y) = 5y- 3 x - 5 y= - 3 x
Then the fraction will take the form:
By performing simple reductions of the numerator and denominator by 3 and by x, we get:
Let's substitute the value y = 0.5:
1 / (5 0,5) = - 1 / 2,5 = - 0,4
Answer: - 0.4
Demo version of OGE 2019
Find the meaning of the expression
where a = 9, b = 36
Solution:
First of all, in tasks of this type, you need to simplify the expression and then substitute the numbers.
Let's reduce the expression to a common denominator - this is b, to do this we multiply the first term by b, after which we get in the numerator:
9b² + 5a - 9b²
Let us present similar terms - these are 9b² and - 9b², leaving 5a in the numerator.
Let's write the final fraction:
Let's calculate its value by substituting the numbers from the condition:
Answer: 1.25
Fourth version of the task
Find the meaning of the expression:
at x = 12.
Solution:
Let's perform identical transformations of the expression to simplify it.
Step 1 – transition from dividing fractions to multiplying them:
Now we reduce the expression (in the numerator of the first fraction and in the denominator of the second) and arrive at a finally simplified form:
We substitute the numerical value for x into the resulting expression and find the result:
