Analysis of exam tasks profile level. Preparation for the Unified State Exam in mathematics (profile level): assignments, solutions and explanations

Author Bagmenova T.A. mathematic teacherMBOU Secondary School No. 14, Novocherkassk, Rostov Region.

When solving tasks on the use of derivatives in preparation for the Unified State Exam, there is a wide variety of tasks, which prompts the need to divide the tasks into groups, accompanied by theoretical material on the topic “Derivative”.

Let's look at examples of tasks No. 7 on the topic “Derivative” of a profile level in mathematics, dividing them into groups.

1 . Let the function f(x) be continuous on the interval [ a ; b ] and is differentiable on the interval (a;b). Then if the derivative of a function is greater than zero for all x belonging to [ a ; b ], then the function increases by [ a ; b ], and if the derivative of a function is less than zero, then it decreases on this segment.

Examples:

1)

Solution.

At points and points the function decreases, therefore the derivative of the function at these points is negative.

Answer: 2.

2)

Solution.

On the intervals (-2;2), (6;10) the derivative of the function is negative; therefore, the function decreases on these intervals. The length of both intervals is 4.

Answer: 4.

3)

Solution.

On the segment, the derivative of the function is positive, therefore the function increases on this interval, therefore the function takes its smallest value at point 3.

Answer: 3.

4)

Solution.

On the interval [-2;3] the derivative of the function is negative, therefore the function decreases on this interval, therefore the function takes on its greatest value at point -2.

Answer: -2.

2 . If at a point the derivative of a function changes sign from “-” to “+”, then this is the minimum point of the function; if at a point the derivative of a function changes sign from “+” to “-”, then this is the point of maximum of the function.

Example:

Solution.

At point x=3; x=13 the derivative of the function changes sign from “-” to “+”, therefore these are the minimum points of the function.

Answer: 2.

3. Condition( x )=0 is a necessary condition for the extremum of the differentiable function f ( x ). Since at the points of intersection of the graph of the derivative of a function with the Ox axis, the derivative of the function is equal to zero, then these points are extremum points.

Example:

Solution.

There are 4 points of intersection of the graph of the derivative of a function with the Ox axis on a given segment, therefore there are 4 extremum points.

Answer: 4.

4 . The derivative of a function is equal to zero at the extremum points of the function. In this problem, these are the points where the function switches from increasing to decreasing or vice versa.

Example:

Solution.

At points the derivative is zero.

Answer: 4.

5. Find the value of the derivative of the function at a point, this means finding the tangent of the angle of inclination of the tangent to the Ox axis or to a straight line parallel to the Ox axis. If the angle of inclination of the tangent to the Ox axis is acute, then the tangent of the angle is positive; if the angle of inclination of the tangent to the Ox axis is obtuse, then the tangent of the angle is negative.

Example:

Solution.

Let's construct a right triangle in which the hypotenuse will lie on the tangent, and one of the legs will lie on the Ox axis or on a straight line parallel to the Ox axis, then we will count the lengths of the legs and calculate the tangent of the acute angle of the right triangle. The opposite side is equal to 2, the adjacent side is equal to 8, therefore the tangent of the acute angle of a right triangle is equal to 0.25. The angle of inclination of the tangent to the Ox axis is obtuse, therefore the tangent of the angle of inclination of the tangent is negative, therefore the value of the derivative of the function at the point is -0.25.

Answer: - 0.25.

6. 1) The angular coefficients of parallel lines are equal.

2) The value of the derivative of the function f ( x y = f ( x ) at point (; f ()).

Example.

Solution.

The slope of the straight line is 2. Sincevalue of the derivative of the functionf( x) at a point is equal to the slope of the tangent to the graph of the functiony= f( x) at point (;f()), then we find the points at which the derivative of the functionf( x) is equal to 2.There are 4 such points on this graph. Therefore, the number of points at which the tangent to the graph of the functionf( x) is parallel to a given line or coincides with it equals 4.

Answer: 4.

Used Books:

Kolyagin Yu. M., Tkacheva M. V., Fedorova N. E. et al. Algebra and the beginnings of mathematical analysis (basic and advanced level). 10 grades – Enlightenment. 2014

Unified State Exam: 4000 problems with answers in mathematics. All tasks are “Closed segment”. Basic and profile level. Edited by I. V. Yashchenko. - M.: Publishing House "Exam", - 2016. - 640 p.

To successfully solve the profile variants of the Unified State Examination in mathematics, it is worth abandoning such an algorithm. When preparing for an exam, you need to focus not on passing it as an end in itself, but on increasing the student’s level of knowledge. To do this, you need to study theory, practice skills, solving various options for the profile Unified State Examination in mathematics in non-standard ways with detailed answers, and monitor the dynamics of learning. And the Shkolkovo educational project will help you with all this.

Why should you choose our resource?

We do not offer you typical examples of profile problems of the Unified State Exam in mathematics, which wander on the Internet from one site to another. Our experts have independently developed a database of tasks, which consists of interesting and unique exercises and is updated daily. All USE problems in mathematics at the profile level contain answers and detailed solutions. They allow you to identify strengths and weaknesses in a student’s preparation and teach him to think freely and outside the box.

In order to complete tasks and view solutions to USE tasks in mathematics at a profile level, select an exercise in the “Catalogue”. This is quite easy to do because it has a clear structure that includes topics and subtopics. All tasks are arranged in ascending order from simple to more complex and contain answers to the profile Unified State Exam in mathematics with solutions.

In addition, the student is given the opportunity to independently create variants of problems. Using the “Constructor”, he can select USE tasks in mathematics at a profile level on any topic that interests him and view their solutions. This will allow you to practice skills in a specific section, for example, geometry or algebra.

Also, a student can analyze the tasks of a specialized Unified State Examination in mathematics in the “Personal Account of the Student.” In this section, the student will be able to monitor his own dynamics and communicate with the teacher.

All this will help you effectively prepare for the specialized Unified State Exam in mathematics and easily find solutions to even the most complex problems.

Practice shows that problems on finding the area of ​​a triangle appear in the Unified State Exam every year. That is why, if students want to get decent scores on the assessment test, they should definitely review this topic and understand the material again.

How to prepare for the exam?

The Shkolkovo educational project will help you learn to solve problems on finding the area of ​​a triangle, similar to those found in the Unified State Examination. Here you will find all the necessary material to prepare for passing the certification test.

To ensure that exercises on the topic “Area of ​​a triangle in Unified State Exam problems” do not cause difficulties for graduates, we recommend that you first refresh your memory of basic trigonometric concepts and rules. To do this, just go to the “Theoretical Information” section. It presents basic definitions and formulas that will help in finding the correct answer.

To consolidate the learned material and practice solving problems, we suggest doing exercises that were selected by specialists of the Shkolkovo educational project. Each task on the site has a correct answer and a detailed description of how to solve it. Students can practice with both simple and more complex problems.

Schoolchildren can “pump up” their skills in performing such exercises online both in Moscow and in any other city in Russia. If necessary, the completed task can be saved in the “Favorites” section in order to return to it later and discuss the progress of the solution with the teacher.

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

The 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.

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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:

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

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:

transformation of numerical rational expressions;

converting algebraic expressions and fractions;

conversion of numeric/letter irrational expressions;

actions with degrees;

converting logarithmic expressions;

1. converting numeric/letter trigonometric expressions.

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

Solution. 1) Let’s use the double argument formula: cos2α = 2 cos 2 α – 1 and find

This means tan 2 α = ± 0.5.

3) By condition

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.

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.

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,

b) Find the roots lying on the segment .

The figure shows that the roots of the given segment belong to

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

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.

√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.

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

The largest integer solution to this inequality is the number 24.

Answer: 24.

At what a system of inequalities

has exactly two solutions?

Solution: This system can be rewritten in the form

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

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.

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This article presents an analysis of tasks 9-12 of part 2 of the Unified State Exam in mathematics at a specialized level from a tutor in mathematics and physics. The tutor's video lesson with an analysis of the proposed tasks contains detailed and understandable comments on each of them. If you have just started preparing for the Unified State Exam in mathematics, this article may be very useful for you.

Using the properties of logarithms, which you can familiarize yourself with in detail in the video tutorial above, we transform the expression:

10. A spring pendulum oscillates with a period T= 16 s. Suspended weight m= 0.8 kg. The speed of movement of the load changes over time in accordance with the formula . At the same time m/s. The defining formula for kinetic energy (in joules) is: , where m taken in kilograms, - in meters per second. What is the kinetic energy of the load in joules 10 s after the start of the oscillatory motion?

The speed of movement of the load 10 s after the start of the oscillatory motion will be equal to:

Then the kinetic energy at this moment in time will be equal to:

J.

Let x- the price of one candy, and y- price of chocolate. Then 6 lollipops cost 6 x, and 2% of the cost of a chocolate bar is equal to 0.02 y. Since it is known that 6 lollipops cost 2% less than a chocolate bar, the first equation holds: 6 x + 0,02y = y, from which we obtain that x = 0,98/6 y = 98/600 y = 49/300 y. In turn, 9 lollipops cost 9 x, that is 9·49/300 y = 49/300 y = 1,47 y. The task comes down to determining by what percentage 1.47 y more than y. If y is 100%, then 1.47 y is 1.47·100% = 147%. That is 1.47 y more than y by 47%.

1) DL is given by the inequality: title="Rendered by QuickLaTeX.com" height="23" width="106" style="vertical-align: -5px;"> (так выражение, стоящее под знаком логарифма, должно быть больше нуля), откуда получаем, что .!}

2) We are looking for the derivative of the function. For a detailed description of how the derivative of this function is calculated, see the video above. The derivative of the function is equal to:

3) Looking for values x, for which the derivative is equal to 0 or does not exist. It does not exist for , since in this case the denominator goes to zero. The derivative is set to zero when.