Task 16. Place punctuation marks: indicate all the numbers that should be replaced by commas in the sentence.
1. From a distance (1) he saw a house (2) unlike the others (3) built (4) by some Italian architect.
2. Above the endless sea (3), which had not yet calmed down (1) after the recent storm (2), rose the sky (4) covered with (5) brightly twinkling stars.
3. A large pond (1) densely overgrown with water lilies (2) was located (3) in a part of the old park remote from the house (4).
4. Vladimir (1) waved his scythe without ceasing (2) cut the grass (3) without showing (4) the slightest effort.
5. Cloud (1) hanging (2) over high peaks the poplars (3) were already falling (4) with drizzling rain.
6. Among the eccentrics (1) who lived in Moscow in Griboyedov’s times (2) was a man (3) described in the comedy “Woe from Wit” under the name (4) Maxim Petrovich.
7. Having made our way (1) through wet ferns and some (2) creeping vegetation (3), we get out onto a barely noticeable path.
8. Anya (2) with her (1) head down, sat motionless in a downy scarf (3) that carefully covered (4) her shoulders.
9. Ippolit Matveevich (1) languishing in shame (2) stood under the acacia tree and (3) without looking at the walkers (4) repeated three memorized phrases.
10. Flipping through the pages (1) of the book (3) brought from the office (2), the father stopped at the slightly open door (4) listening to the conversation in the kitchen.
11. Already in our time, researchers of the work of E. A. Poe (1), having received at their disposal (2) previously hidden (3) materials (4), were able to establish a connection between the life and work of the American writer.
12. Words (1) formed from geographical names(2) Quite often they pose questions to the speaker and writer (3) (4) related to normative word usage.
13. The sparrow (1) unexpectedly took off (2) disappeared into the light greenery of the garden (3) transparently visible (4) in the early evening sky.
14. In bad weather, pine trees groan, and their branches (1) are bent by gusts of an angry wind (2) crack (3) sometimes scratching (4) needles on the tree bark.
15. Under the sun (1) competing with it (2), unusually tall, juicy and large-colored swimsuits (3) similar to yellow roses shone brightly.
16. Masha sat in the corner until lunch (1) carefully looking at older sister and (2) listening to (3) the words she speaks (4).
17. Immediately beyond the river (1) rising up (2) one could see rocky mountains (3) outlined below (4) by a broken line of blackened low bushes.
18. High grass(1) leaning towards the ground (2) softly wrapped around (3) rain-soaked (4) tree trunks.
19. Tree branches (1) intertwining with hard, frozen ends (2) ring sadly (3) experiencing (4) the winter cold.
20. Pushkin (1) brought up on the “History of the Russian State” by N.M. Karamzin (2) said his own about Russian history (3) own word(4) in many ways superior to Karamzin’s.
I passed the math test and ran carefree into the barbecue sunset. Not about you? That's right, because at the end of the tunnel you have a happy future flickering, and in the tunnel itself - physics, history, English and other attributes for entering a university. Today in Russia we took social studies and chemistry. And the impression was created that, according to at least, chemistry was taken by Zen Buddhists. But most of the feedback received about social studies was “a very difficult test, I’m sitting there crying”, “who will jump from the ninth floor with me?”, “I’m afraid I didn’t pass the threshold.” This is what they write to us.
Society
“The Unified State Exam was easier than Russian, and much easier than mathematics. In general, I think the kimas were well composed, but there was also a moment of humor. So, for example, I had one question about anti-inflation, and what was my surprise that the method of fighting inflation was to increase unemployment. Are we treating one and crippling the other? Or no money means no inflation?
(Elmira, Ryazan)
“The exam passed without incident. In general, the tasks were not difficult, on par with the samples and tests from “Solving the Unified State Exam” (surprisingly). I thought it would be more difficult. So far, of all the exams I have passed (at this moment I only passed Russian and mathematics), society seemed the simplest. Part C was also doable. The text was rightfully received, and the rest of the “tseshki” were on fairly popular topics in the course of society (factors of production, types of society, law, morality, etc.). I’m passing the exam because I plan to enroll in international relations.”
(Ivan, Tyumen)
“I chose social studies because it is needed for admission to economics. The exam was extremely easy, even for me, a person who last time the lessons on society were six years ago, and which were absolutely not prepared. I already have the highest unfinished, but I am not going to finish it. I prepared for exactly one day, read jurisprudence, and passed it on general knowledge, fortunately, social studies allows you to do this. Does it help? life experience? I think that a university education, even if incomplete, can be considered an advantage. At the university they taught me how to really use my head.”
(Christina, bashkortostan)
“When you are preparing for one thing, and you come across something else, it’s very disappointing. I, like everyone else, probably, everything is shaking, I’m afraid that I haven’t crossed the threshold, I’m very worried! I was preparing for “Solving the Unified State Exam”, very little coincided, very little. The school solved tests for 2012-14, which was of little use. Yes, and in Russian a lot of things coincided, but there is no math at all, our girls were crying after the exam.”
(Pavel, Satka)
“It’s really possible to prepare for society. I prepared productively for a year (4 hours a week), and felt comfortable during the exam. I can say that there is enough time if you know what to write. Again, the tasks are the same as on Internet portals. Nothing new.
The organization is excellent. The test technicians are very attentive, do not interfere or distract. Overall, the exam went well, I hope I’m not the only one.”
(Angelina, Rostov-on-Don)
“The test is not so much difficult as it is stupid, there are a lot of dubious and ambiguous tasks. But somehow I wasn’t very lucky with part C (where has it been seen that already at 28 there were 3 sub-questions?!) In general, I’m dissatisfied with the result, I’ll wait for the numbers without much hope.”
(Daria, St. Petersburg)
Chemistry
“The exam in chemistry went well, without any incidents. In general, everything was quite calm and quiet: in my town, chemistry and society were taken in the same school, because... There are few graduates, only 10 chemists. The tasks were not particularly difficult; I would even say that they were easy. There was no failure like in mathematics. True, I myself, through carelessness, made a mistake in C5, since both supposed substances had the same mass fraction chemical elements(there was acetic acid, but I wrote formaldehyde). In general, everything is not bad, there was no such jitters as in Russian and mathematics.”
(Rome, Agidel, Bashkiria)
“The exam was relatively simple (purely subjective), there were no difficulties. I chose chemistry because I want to enroll in the Faculty of Chemistry (I am passionate about chemistry). There were 4 people from my physics and biology class and 24 people from my chemistry and biology class. Everyone approaches chemistry differently. I like both chemistry and literature. Here we can talk about “techies” and “humanities”, but why? And the critical judgment that almost everyone considers chemistry a hateful subject somehow does not really fit with my life experience.”
(Denis, Arkhangelsk)
“The Unified State Exam in Chemistry passed quickly and completely unnoticed, but I had been waiting for this moment for a long time, because after the exams you feel freer. There was no excitement at all, because I was completely immersed in work. But waiting for results is a more serious thing than the exam itself. There was more than enough time, the observers were friendly, the audiences were cozy. In general, compared to mathematics and Russian, it is more difficult than expected. Problem C4 caused difficulties: I had to think and reason. Organics, my love, made me happy. But there is already an error with acetylene. It's a shame, stupid mistake! If we compare preparation for Russian and mathematics, then in chemistry the tasks are at a higher level. In Russian, the assignments were at the preparation level. But perhaps this is all because chemistry is not so easy for me. For me, it’s much more interesting when you have to “comprehend” a subject. It’s probably strange, but even when things don’t work out, I still have an interest in chemistry.”
(Marina, Magnitogorsk)
“I’ll be happy to tell you about my exam. I chose chemistry because I want to connect my life with medicine. The exam was not so difficult, of course, for those who had been preparing for it all this time. academic year. Yes. There were difficulties in task 39 and 40, but I think I coped well enough with the exam. At 39, for example, I thought about the reaction itself and therefore I could not solve the problem further. I analyzed similar tests for task 40 very carefully, but for some reason exam task I found it difficult because, in my opinion, there were several answers.”
(Arthur, Omsk)
Average 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 an integer or finite number decimal;
- 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 decisions with justification for the actions taken).
Panova Svetlana Anatolevna, mathematic teacher highest category schools, work experience 20 years:
"In order to receive school certificate, the graduate must pass two mandatory exam V Unified State Examination form, one of which is mathematics. In accordance with the Development Concept mathematics education V 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- checks with Unified State Exam participants the ability to apply the skills acquired in the course of grades 5 - 9 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. A flow meter was installed in the apartment where Peter lives cold water(counter). 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 by the value of its argument when in various ways specifying a function and describing the behavior and properties of the function based on its graph. You also need to be able to find the greatest or smallest value and build graphs of the studied functions. Errors made are random in reading the conditions of the problem, reading the diagram.
#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 task at the basic level of the first part, tests the ability to perform actions with geometric shapes on the content of the course “Planimetry”. Task 3 tests the ability to calculate the area of a figure on checkered paper, the ability to calculate degree measures of angles, calculate perimeters, etc.
Example 3. Find the area of a rectangle drawn on checkered paper with a cell size of 1 cm by 1 cm (see figure). Give your answer in square centimeters.
Solution: To calculate the area of a given figure, you can use the Peak formula:
To calculate area given rectangle Let's 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 How 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. For successful implementation a meaningful, non-formal mastery of the concept of derivative is required.
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 straight line passing through two given points and find the equation of the straight line passing through the 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 higher level 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 are reduced to solving linear or quadratic equation, or linear or quadratic inequality. Therefore, it is necessary to be able to solve such equations and inequalities and determine the answer. The answer must be given as a whole number or a finite decimal fraction.
Two bodies of mass m= 2 kg each, moving at the same speed v= 10 m/s at an angle of 2α to each other. The energy (in joules) released during their absolutely inelastic collision is determined by the expression Q = mv 2 sin 2 α. At what smallest angle 2α (in degrees) must the bodies move so that at least 50 joules are released as a result of the collision?
Solution. To solve the problem, we need to solve the inequality Q ≥ 50, on the interval 2α ∈ (0°; 180°).
mv 2 sin 2 α ≥ 50
2 10 2 sin 2 α ≥ 50
200 sin 2 α ≥ 50
Since α ∈ (0°; 90°), we will only solve
Let us represent the solution to the inequality graphically:
Since by condition α ∈ (0°; 90°), it means 30° ≤ α< 90°. Получили, что наименьший угол α равен 30°, тогда наименьший угол 2α = 60°.
Task No. 11- is typical, but turns out to be difficult for students. The main source of difficulty is the construction of a mathematical model (drawing up an equation). Task No. 11 tests the ability to solve word problems.
Example 11. During spring break, 11th-grader Vasya had to solve 560 practice problems to prepare for the Unified State Exam. On March 18, on the last day of school, Vasya solved 5 problems. Then every day he solved the same number of problems more than the previous day. Determine how many problems Vasya solved on April 2, the last day of the holidays.
Solution: Let's denote a 1 = 5 – the number of problems that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 – number of days from March 18 to April 2 inclusive, S 16 = 560 – total number of tasks, a 16 – the number of problems that Vasya solved on April 2. Knowing that every day Vasya solved the same number of problems more compared to the previous day, we can use formulas for finding the sum arithmetic progression:560 = (5 + a 16) 8,
5 + a 16 = 560: 8,
5 + a 16 = 70,
a 16 = 70 – 5
a 16 = 65.
Answer: 65.
Task No. 12- they test students’ ability to perform operations with functions, and to be able to apply the derivative to the study of a function.
Find the maximum point of the function y= 10ln( x + 9) – 10x + 1.
Solution: 1) Find the domain of definition of the function: x + 9 > 0, x> –9, that is, x ∈ (–9; ∞).
2) Find the derivative of the function:
4) The found point belongs to the interval (–9; ∞). Let's determine the signs of the derivative of the function and depict the behavior of the function in the figure:
The desired maximum point x = –8.
Download for free the working program in mathematics for the line of teaching materials G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free teaching aids on 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 turns into 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 isosceles triangle ABC with an angle of 120° at vertex A, a bisector BD is drawn. Rectangle DEFH is inscribed in triangle ABC so that side FH lies on segment BC, and vertex E lies on segment AB. a) Prove that FH = 2DH. b) Find the area of rectangle DEFH if AB = 4.
Solution: A)
1) ΔBEF – rectangular, EF⊥BC, ∠B = (180° – 120°): 2 = 30°, then EF = BE by the property of the leg lying opposite the angle of 30°.
2) Let EF = DH = x, then BE = 2 x, BF = x√3 according to the Pythagorean theorem.
3) Since ΔABC is isosceles, it means ∠B = ∠C = 30˚.
BD is the bisector of ∠B, which means ∠ABD = ∠DBC = 15˚.
4) Consider ΔDBH – rectangular, because DH⊥BC.
| 2x | = | 4 – 2x |
| 2x(√3 + 1) | 4 |
| 1 | = | 2 – x |
| √3 + 1 | 2 |
√3 – 1 = 2 – x
x = 3 – √3
EF = 3 – √3
2) S DEFH = ED EF = (3 – √3 ) 2(3 – √3 )
S DEFH = 24 – 12√3.
Answer: 24 – 12√3.
Task No. 17- a task with a detailed answer, this task tests the application of knowledge and skills in practical activities and everyday life, the ability to build and explore mathematical models. This task 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 highest value X, in which the bank will accrue less than 17 million rubles to the deposit over four years.
Solution: At the end of the first year, the contribution will be 20 + 20 · 0.1 = 22 million rubles, and at the end of the second - 22 + 22 · 0.1 = 24.2 million rubles. At the beginning of the third year, the contribution (in million rubles) will be (24.2 + X), and at the end - (24.2 + X) + (24,2 + X)· 0.1 = (26.62 + 1.1 X). At the beginning of the fourth year the contribution will be (26.62 + 2.1 X), and at the end - (26.62 + 2.1 X) + (26,62 + 2,1X) · 0.1 = (29.282 + 2.31 X). By condition, you need to find the largest integer x for which the inequality holds
(29,282 + 2,31x) – 20 – 2x < 17
29,282 + 2,31x – 20 – 2x < 17
0,31x < 17 + 20 – 29,282
0,31x < 7,718
| x < | 7718 |
| 310 |
| x < | 3859 |
| 155 |
| x < 24 | 139 |
| 155 |
The largest integer solution to this inequality is the number 24.
Answer: 24.
Task No. 18- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection into universities with increased requirements for the mathematical preparation of applicants. Exercise high level complexity - this task is not about using one solution method, but about a combination various methods. To successfully complete task 18, in addition to solid mathematical knowledge, you also need a high level of mathematical culture.
At what a system of inequalities
| x 2 + y 2 ≤ 2ay – a 2 + 1 | |
| y + a ≤ |x| – a |
has exactly two solutions?
Solution: This system can be rewritten in the form
| x 2 + (y– a) 2 ≤ 1 | |
| y ≤ |x| – a |
If we draw on the plane the set of solutions to the first inequality, we get the interior of a circle (with a boundary) of radius 1 with center at point (0, A). The set of solutions to the second inequality is the part of the plane lying under the graph of the function y = |
x| –
a,
and the latter is the graph of the function
y = |
x|
, shifted down by A. The solution to this system is the intersection of the sets of solutions to each of the inequalities.
Therefore, two solutions this system will have only in the case shown in Fig. 1.
The points of contact of the circle with the lines will be the two solutions of the system. Each of the straight lines is inclined to the axes at an angle of 45°. So it's a triangle PQR– rectangular isosceles. Dot Q has coordinates (0, A), and the point R– coordinates (0, – A). In addition, the segments PR And PQ equal to the radius of the circle equal to 1. This means
| Qr= 2a = √2, a = | √2 | . |
| 2 |
| Answer: a = | √2 | . |
| 2 |
Task No. 19- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection into universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is a task not on the use of one solution method, but on a combination of various methods. To successfully complete task 19, you must be able to search for a solution by choosing different approaches from among the known ones, modifying the studied methods.
Let Sn sum P terms of an arithmetic progression ( a p). It is known that S n + 1 = 2n 2 – 21n – 23.
a) Provide the formula P th term of this progression.
b) Find the smallest absolute sum S n.
c) Find the smallest P, at which S n will be the square of an integer.
Solution: a) It is obvious that a n = S n – S n- 1 . Using this formula, we get:
S n = S (n – 1) + 1 = 2(n – 1) 2 – 21(n – 1) – 23 = 2n 2 – 25n,
S n – 1 = S (n – 2) + 1 = 2(n – 1) 2 – 21(n – 2) – 23 = 2n 2 – 25n+ 27
Means, a n = 2n 2 – 25n – (2n 2 – 29n + 27) = 4n – 27.
B) Since S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Its graph can be seen in the figure.
Obviously, the smallest value is achieved at the integer points located closest to the zeros of the function. Obviously these are points X= 1, X= 12 and X= 13. Since, S(1) = |S 1 | = |2 – 25| = 23, S(12) = |S 12 | = |2 · 144 – 25 · 12| = 12, S(13) = |S 13 | = |2 · 169 – 25 · 13| = 13, then the smallest value is 12.
c) From the previous paragraph it follows that Sn positive, starting from n= 13. Since S n = 2n 2 – 25n = n(2n– 25), then the obvious case, when this expression is a perfect square, is realized when n = 2n– 25, that is, at P= 25.
It remains to check the values from 13 to 25:
S 13 = 13 1, S 14 = 14 3, S 15 = 15 5, S 16 = 16 7, S 17 = 17 9, S 18 = 18 11, S 19 = 19 13, S 20 = 20 13, S 21 = 21 17, S 22 = 22 19, S 23 = 23 21, S 24 = 24 23.
It turns out that for smaller values P a complete square is not achieved.
Answer: A) a n = 4n– 27; b) 12; c) 25.
________________
*Since May 2017, the united publishing group "DROFA-VENTANA" has been part of the Russian Textbook corporation. The corporation also includes the Astrel publishing house and the LECTA digital educational platform. General Director appointed 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). 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 Russian Textbook Publishing Corporation has the largest portfolio of textbooks included in Federal list- 485 titles (approximately 40%, excluding textbooks for correctional school). The corporation's publishing houses own the most popular Russian schools sets of textbooks on physics, drawing, biology, chemistry, technology, geography, astronomy - areas of knowledge that are needed for the development of the country's production potential. The corporation's portfolio includes textbooks and teaching aids For primary school, awarded the Presidential Prize in the field of education. These are textbooks and manuals in subject areas that are necessary for the development of the scientific, technical and production potential of Russia.
The “Get an A” video course includes all the topics you need to successful completion Unified State Examination in mathematics for 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!
Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.
All the necessary theory. Quick ways solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.
The course contains 5 big topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.
Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Visual explanation complex concepts. Algebra. Roots, powers and logarithms, function and derivative. A basis for solving complex problems of Part 2 of the Unified State Exam.
Assessment
two parts, including 19 tasks. Part 1 Part 2
3 hours 55 minutes(235 minutes).
Answers
But you can make a compass Calculators on the exam not used.
passport), pass and capillary or! Allowed to take with myself water(in a transparent bottle) and I'm going
The examination paper consists of two parts, including 19 tasks. Part 1 contains 8 tasks of a basic difficulty level with a short answer. Part 2 contains 4 tasks of an increased level of complexity with a short answer and 7 tasks of a high level of complexity with a detailed answer.
For execution exam paper in mathematics is assigned 3 hours 55 minutes(235 minutes).
Answers for tasks 1–12 are written down as a whole number or finite decimal fraction. Write the numbers in the answer fields in the text of the work, and then transfer them to answer form No. 1, issued during the exam!
When performing work, you can use the ones issued along with the work. Only a ruler is allowed, but it's possible make a compass with your own hands. Do not use tools with markings on them. reference materials. Calculators on the exam not used.
You must have an identification document with you during the exam ( passport), pass and capillary or gel pen with black ink! Allowed to take with myself water(in a transparent bottle) and I'm going(fruit, chocolate, buns, sandwiches), but they may ask you to leave them in the corridor.
