Denis Valentinovich Manturov Minister of Industry and Trade of the Russian ...
The average general education
UMK line by G. K. Muravin. Algebra and the beginnings of mathematical analysis (10-11) (in-depth)
UMK Merzlyak line. 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 a teacherExamination paper the profile level 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 assignments:
- 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 (a complete record of the solution with the justification of the actions performed).
Panova Svetlana Anatolievna, teacher of mathematics of the highest category of the school, work experience 20 years:
“In order to receive a school certificate, a graduate must pass two compulsory exams in USE form one of which is math. In accordance with the Concept for the Development of Mathematical Education in Russian Federation The exam in mathematics is divided into two levels: basic and specialized. Today we will consider options for the profile level. "
Task number 1- tests the USE participants' ability 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 decimals, be able to convert one unit of measurement to another.
Example 1. An expense 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 June 1 - 177 cubic meters. m. How much should Peter pay for cold water in May, if the price of 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 meters)
2) Let's find how much money will be paid for the water spent:
34.17 5 = 170.85 (rub)
Answer: 170,85.
Task number 2-is one of the simplest exam tasks. Most graduates successfully cope with it, which testifies to the possession of the definition of the concept of function. Type of task number 2 according to the requirements codifier is a task on the use of acquired knowledge and skills in practical activities and Everyday life... Task number 2 consists of describing with the help of functions various real dependencies between quantities and the interpretation of 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 in various ways of defining a function and describe the behavior and properties of a function according to its graph. It is also necessary to be able to find the largest or smallest value on the graph of the function and plot the graphs of the studied functions. The mistakes made are random in reading the problem statement, reading the diagram.
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Example 2. The figure shows the change in the market value of one share of a mining company in the first half of April 2017. On April 7, the businessman acquired 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 rest. 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 1000 shares after the sale.
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 basic level the first part, checks the ability to perform actions with geometric shapes on the content of the course "Planimetry". In task 3, the ability to calculate the area of a figure on checkered paper, the ability to calculate the degree measures of angles, calculate the perimeters, etc. is tested.
Example 3. Find the area of a rectangle depicted 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 Pick formula:
To calculate the area of this rectangle, we will use the Peak formula:
|
S= B + |
G | |
| 2 |
|
S = 18 + |
6 | |
| 2 |
See also: Unified State Exam in Physics: Solving Oscillation 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 points marked on the circle. Determine which polygons are more: those with all the vertices are red, or those with one of the vertices blue. In the answer, indicate how many of some are more than others.
Solution: 1) We use the formula for the number of combinations from n elements by k:
in which all the vertices are red.
3) One pentagon with all vertices red.
4) 10 + 5 + 1 = 16 polygons with all vertices red.
whose vertices are red or with one blue vertex.
whose vertices are red or with one blue vertex.
8) One hexagon, with red peaks with one blue peak.
9) 20 + 15 + 6 + 1 = 42 polygons in which all vertices are red or with one blue vertex.
10) 42 - 16 = 26 polygons using the blue point.
11) 26 - 16 = 10 polygons - how many polygons with one of the vertices - a blue point, more than polygons with all vertices 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 the equation 2 3 + x= 0.4 5 3 + x .
Solution. Divide both sides of this equation by 5 3 + NS≠ 0, we get
| 2 3 + x | = 0.4 or | 2 | 3 + NS | = | 2 | , | ||
| 5 3 + NS | 5 | 5 |
whence it follows that 3 + x = 1, x = –2.
Answer: –2.
Task number 6 on planimetry for finding geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. 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 planimetry theorems.
Area of a triangle ABC is equal to 129. DE- the middle line parallel to the side AB... Find the area of a trapezoid ABED.
Solution. Triangle CDE like a triangle CAB in two corners, since the apex angle C general, angle CDE equal to the angle CAB as the corresponding angles at DE || AB secant AC... Because DE- the middle line of the triangle by the condition, then by the property of the middle line | DE = (1/2)AB... This means that the coefficient of similarity is 0.5. The areas of such figures are related as the square of the coefficient of similarity, therefore
Hence, S ABED = S Δ ABC – S Δ CDE = 129 – 32,25 = 96,75.
Task number 7- checks the application of the derivative to the study of the function. For successful implementation, a meaningful, non-formal knowledge of the concept of a derivative is required.
Example 7. Go to function graph 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 | · (-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 straight line y = 4x- 13, where k 1 = 4, according to the formula:
3) Slope tangent - the derivative of the function at the point of tangency. Means, f′( x 0) = k 2 = –0,25.
Answer: –0,25.
Task number 8- tests the participants' knowledge of elementary stereometry, the ability to apply formulas for finding the areas of surfaces and volumes of figures, dihedral angles, to compare the volumes of similar figures, to be able to perform actions with geometric figures, coordinates and vectors, etc.
The volume of the cube described around the 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), 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 number 9- requires the graduate to convert and simplify algebraic expressions. Task number 9 increased level difficulty with a short answer. Tasks from the section "Calculations and Transformations" in the exam are divided into several types:
- converting numeric / alphabetic trigonometric expressions.
converting numerical rational expressions;
transformations of algebraic expressions and fractions;
converting numeric / alphabetic 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 formula of the double argument: cos2α = 2 cos 2 α - 1 and find
| tg 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, tg 2 α = ± 0.5.
3) By condition
| 3π | < α < π, |
| 4 |
hence, α is the angle of the II quarter and tgα< 0, поэтому tgα = –0,5.
Answer: –0,5.
# ADVERTISING_INSERT # Task number 10- tests students' ability to use early acquired knowledge and skills in practice 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 linear or square inequality. Therefore, it is necessary to be able to solve such equations and inequalities, and determine the answer. The answer should be either an integer or a final decimal fraction.
Two bodies weighing 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. What is the smallest angle 2α (in degrees) should 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 of the inequality graphically:
Since, by hypothesis, α ∈ (0 °; 90 °), it means 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 difficulty is the construction of a mathematical model (drawing up an equation). Task number 11 tests the ability to solve word problems.
Example 11. During the spring break, 11-grader Vasya had to solve 560 training 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 tasks more than the previous day. Determine how many problems Vasya solved on April 2 on the last day of the vacation.
Solution: We denote a 1 = 5 - the number of tasks that Vasya solved on March 18, d- the 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 problems that Vasya solved on April 2. Knowing that every day Vasya solved the same number of problems 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- test students' ability to perform actions with functions, be able to apply a derivative to the study of a 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; ∞). Let us determine the signs of the derivative of the function and depict the behavior of the function in the figure:
Seeking maximum point x = –8.
Download free of charge a work program in mathematics for the line of teaching methods of G.K. Muravina, K.S. Muravina, O. V. Muravina 10-11 Download free teaching aids on algebraTask number 13-increased level of difficulty 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,
|
|
log 3 (2cos x) = | 2 | ⇔ |
|
2cos x = 9 | ⇔ |
|
cos x = | 4,5 | ⇔ since | 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.
It can be seen from the figure that the roots
| 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 circumference 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 with a length of 12 is at a distance = 8 from the center of the base circle, and a chord with a length of 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.
By 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 was required to be proved.
b) Let's designate the centers of the bases for O 1 and O 2. Let us draw from the center of the base with a chord of length 12 a middle perpendicular to this chord (it has a length of 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. We call the midpoint of the lesser 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.
Hence, the required angle is
| ∠ABH = arctg | AH | = arctg | 28 | = arctg14. |
| BH | 8 – 6 |
Task number 15- an increased level of difficulty with a detailed answer, tests the ability to solve inequalities, the 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 this inequality is the interval (–1; + ∞). Consider three cases separately:
1) Let x 2 – 3x= 0, i.e. NS= 0 or NS= 3. In this case, this inequality becomes true, therefore, these values are included in the solution.
2) Now let 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 positive 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 as (3 x – x 2) log 2 ( x + 1) ≤ 3x – x 2. After division by positive expression 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 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 apex 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 by the property of the leg, which lies 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, it means that ∠B = ∠C = 30˚.
BD is the bisector of ∠B, so ∠ABD = ∠DBC = 15˚.
4) Consider ΔDBH - rectangular, since 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 practice and everyday life, the ability to build and explore mathematical models. This assignment is a text problem 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 its 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 NS million rubles, where NS - whole number. Find the largest value NS, in which the bank will charge the deposit less than 17 million rubles 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 + NS), and at the end - (24,2 + NS) + (24,2 + NS) 0.1 = (26.62 + 1.1 NS). At the beginning of the fourth year, the contribution will be (26.62 + 2.1 NS), and at the end - (26.62 + 2.1 NS) + (26,62 + 2,1NS) 0.1 = (29.282 + 2.31 NS). By hypothesis, 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 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 training of applicants. A task of a high level of complexity is not a task for using 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.
Under 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 of 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 to 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 function graph
y = |
x|
shifted down by a... The solution to this system is the intersection of the solution sets for each of the inequalities.
Consequently, this system will have two solutions only in the case shown in Fig. 1.
The points of tangency of the circle with straight lines will be 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. Point Q has coordinates (0, a), and the point R- coordinates (0, - a). In addition, the segments PR and PQ are equal to the radius of the circle 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 training of applicants. A task of a high level of complexity is not a task for using 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, modifying the studied methods.
Let be Sn sum NS members of the arithmetic progression ( a n). It is known that S n + 1 = 2n 2 – 21n – 23.
a) Specify the formula NS th member of this progression.
b) Find the least modulo sum S n.
c) Find the smallest NS 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 reached at the integer points that are closest to the zeros of the function. Obviously these are points NS= 1, NS= 12 and NS= 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 point it follows that Sn positively 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 NS= 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 = 2321, S 24 = 24 23.
It turns out that for smaller values NS full square is not achieved.
Answer: a) a n = 4n- 27; b) 12; c) 25.
________________
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In task number 12 of the USE in mathematics of the profile level, we need to find the largest or smallest value of the function. For this, it is necessary to use, obviously, a derivative. Let's look at a typical example.
Analysis of typical options for assignments No. 12 of the USE in mathematics of the profile level
The first variant of the task (demo version 2018)
Find the maximum point of the function y = ln (x + 4) 2 + 2x + 7.
Solution algorithm:
- Find the derivative.
- We write down the answer.
Solution:
1. We are looking for values of x at which the logarithm makes sense. To do this, we solve the inequality:
Since the square of any number is non-negative. The solution to the inequality will be only that value of x at which x + 4 ≠ 0, i.e. for 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)'.
By 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. Equate the derivative to zero:
y, = 0 → (2 + 2 ∙ (x + 4)) / (x + 4) = 0,
2 + 2x +8 = 0, 2x + 10 = 0,
The second variant of the task (from Yashchenko, no. 1)
Find the minimum point of the function y = x - ln (x + 6) + 3.
Solution algorithm:
- Determine the scope of the function.
- Find the derivative.
- Determine at what points the derivative is 0.
- We exclude points that do not belong to the definition area.
- Among the remaining points, we are looking for the values of x at which the function has a minimum.
- We write down the answer.
Solution:
2. Find the derivative of the function:
3. Equate the resulting expression to zero:
4. Received one point x = -5, belonging to the domain of the function.
5. At this point, the function has an extremum. Let's check if this is the minimum. When x = -4
For x = -5.5, the derivative of the function is negative, since
Hence, the point x = -5 is the minimum point.
The third variant of the task (from Yashchenko, no. 12)
Find the largest function value 
on the segment [-3; 1].
Solution algorithm:
- Find the derivative.
- Determine at what points the derivative is 0.
- We exclude points that do not belong to the specified segment.
- Among the remaining points, we are looking for the values of x at which the function has a maximum.
- 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. Calculate the derivative of the function, we get
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USE-2017 in mathematics, basic level
Dmitry Gushchin undertakes to help with such exams as the OGE and the Unified State Examination, using a very common method. It lies in the fact that all new knowledge is submitted and systematized by topic. The student can easily choose what he needs to repeat for the final consolidation of the material.
Assignments are available at basic and specialized levels. Mathematics is a prime example of such tasks. The main (basic) level covers the general school volume of knowledge. It requires the knowledge that each student receives in 11 years. The profile level is designed for graduates of specialized schools with a focus on a particular subject.
An interesting feature of the system is its similarity to a real exam. In the case of the final test assignments are submitted in the USE format. The student can also find out their final score after passing the test. This helps motivate a person to achieve new goals and to learn new material. Realizing your real chances on the exam helps you gather your thoughts and understand what exactly needs to be learned.
The most demanded subjects in the "Reshu Unified State Exam" are provided along with others. Dmitry Gushchin's Russian language includes the rules of grammar, punctuation and syntax, as well as vocabulary. Chemistry contains examples of solving specific problems, special formulas. Also, the chemistry section includes various compounds and concepts of chemicals. The section of biology covers the vital activity of all kingdoms of living organisms. It contains important theory that will ultimately help you pass the exam.
The next feature is that your progress is recorded and you can track your progress. This approach will help you motivate yourself even when you don't feel like learning anymore. Your own result always forces you to do more.
The system also has criteria for evaluating work. They will make your exam preparation planned and thoughtful. The prospective student will always be able to read them and understand what the examiner will pay attention to. This is important in order to pay attention to certain important aspects of the work. In general, the student is fully aware of the importance of his choice and remembers the assessment criteria.
