In task 18 - the penultimate task profile level Unified State Exam in Mathematics - it is necessary to demonstrate the ability to solve problems with parameters. In the overwhelming majority, this task is a system of two equations with the parameter a, and it is necessary to find such values \u200b\u200bfor which the system will behave in a given way - have two or one or no solutions at all.

Analysis of typical options for assignments No. 18 of the USE in mathematics of the profile level

The first variant of the task (demo version 2018)

Find all positive values \u200b\u200bof a, for each of which the system has a unique solution:

• (| x | –5) 2 + (y – 4) 2 \u003d 4
• (x – 2) 2 + y 2 \u003d a 2

Solution algorithm:

1. Consider the second equation, establish what is its graph.
2. We define the condition for the uniqueness of the solution.
3. Find the distance between the centers, determine the parameter values.
4. We write down the answer.

Decision:

1. The first equation is two circles with radii 3 and coordinates of the centers C 2 (5; 4) and C 2 (-5; 4). One circle is given by this equation at x≥0, and the second - at x<0. Они не пересекаются и не касаются.

2. The second equation is one circle of radius "a" with center coordinates: С (-2; 0).

3. The presence of a single solution means that one circle must touch one of the circles at one point. Therefore, two systems should be solved in pairs.

Naturally, in the first and second cases, a pair of roots is obtained, that is, the tangency coordinates are externally and internally.

But it is worth noting that we will only be interested in the roots defining the tangency of the outer left circle and the tangent of the inner right circle. Since the other two equations contradict the condition and will have more than one solution. Just look at the attached picture:

4. Let's use the attached drawing.

Let's draw the rays CC 1, and CC 2, marking the points of their intersection with the circles A1, B 1 and A 2, B 2.
Then

If a

5. We have: the original system has a unique solution for

Answer:

The second option (from Yashchenko, no. 1)

Find all values \u200b\u200bof a, for each of which the equation

has exactly one root.

Decision:

This equation is equivalent to the form:

Consider the case:

Provided

We receive.

With this value of x, the condition takes the form:

We have in this case: at.

Consider now the case:

,

wherein .

We solve the equation. We get:

From here .

Unified State Exam 2017. Mathematics. Task 18. Tasks with a parameter. Sadovnichy Yu.V.

M .: 2017 .-- 128 p.

This book is devoted to problems similar to the 18th exam in mathematics (problem with a parameter). Various methods of solving such problems are considered, and much attention is paid to graphic illustrations. The book will be useful for high school students, mathematics teachers, tutors.

Format: pdf

The size: 1.6 MB

Watch, download:drive.google

CONTENT
Introduction 4
§1. Linear equations and systems of linear equations 5
Tasks for self-solution 11
§2. Examination of the square trinomial using the discriminant 12
Solve tasks 19
§3. Vieta's theorem 20
Tasks for independent solution 26
§4. The location of the roots of a square trinomial 28
Solving Tasks 43
§five. Application of graphic illustrations
to the study of the square trinomial 45
Tasks for independent solution 55
§6. Limited function. Finding the Range 56
Tasks for independent solution 67
§7. Other properties of functions 69
Self-help problems 80
§8. Logic problems with parameter 82
Solving Tasks 93
Illustrations on the coordinate plane 95
Solving Tasks 108
Okha Method 110
Tasks for independent solution 119
120 responses

This book is devoted to problems similar to the 18th exam in mathematics (problem with a parameter). Along with Problem 19 (a problem that uses the properties of integers), Problem 18 is the most difficult in its version. Nevertheless, the book attempts to systematize problems of this type according to various methods of solving them.
Several paragraphs are devoted to a seemingly such a popular topic as the study of the square trinomial. However, sometimes such tasks require different, sometimes the most unexpected approaches to their solution. One such non-standard approach is demonstrated in example 7 in paragraph 2.
Often, when solving a problem with a parameter, it is necessary to investigate the function given in the condition. The book formulates some statements concerning such properties of functions as boundedness, parity, continuity; then the examples demonstrate the application of these properties to problem solving.

USE in mathematics profile level

The work consists of 19 tasks.
Part 1:
8 tasks with a short answer of a basic level of difficulty.
Part 2:
4 tasks with a short answer
7 tasks with a detailed answer of a high level of complexity.

Completion time - 3 hours 55 minutes.

Examples of exam assignments

Solving USE tasks in mathematics.

For an independent solution:

1 kilowatt-hour of electricity costs 1 ruble 80 kopecks.
The electricity meter on November 1 showed 12,625 kilowatt-hours, and on December 1, showed 12802 kilowatt-hours.
How much should I pay for electricity for November?
Give your answer in rubles.

Problem with solution:

In a regular triangular pyramid ABCS with a base ABC, ribs are known: AB \u003d 5 roots of 3, SC \u003d 13.
Find the angle formed by the plane of the base and the straight line passing through the middle of the edges AS and BC.

Decision:

1. Since SABC is a regular pyramid, ABC is an equilateral triangle, and the rest of the faces are equal isosceles triangles.
That is, all sides of the base are 5 sqrt (3), and all side edges are 13.

2. Let D - midpoint of BC, E - midpoint of AS, SH - height lowered from point S to the base of the pyramid, EP - height lowered from point E to the base of the pyramid.

3. Find AD from the right-angled triangle CAD by the Pythagorean theorem. It turns out 15/2 \u003d 7.5.

4. Since the pyramid is regular, point H is the point of intersection of heights / medians / bisectors of triangle ABC, which means it divides AD in a ratio of 2: 1 (AH \u003d 2 AD).

5. Find SH from the right-angled triangle ASH. AH \u003d AD 2/3 \u003d 5, AS \u003d 13, by the Pythagorean theorem SH \u003d sqrt (13 2 -5 2) \u003d 12.

6. Triangles AEP and ASH are both rectangular and have a common angle A, therefore, similar. By hypothesis, AE \u003d AS / 2, so AP \u003d AH / 2 and EP \u003d SH / 2.

7. It remains to consider the right-angled triangle EDP (we are just interested in the angle EDP).
EP \u003d SH / 2 \u003d 6;
DP \u003d AD 2/3 \u003d 5;

Tangent of angle EDP \u003d EP / DP \u003d 6/5,
Angle EDP \u003d arctg (6/5)

Answer:

In the exchange office, 1 hryvnia costs 3 rubles 70 kopecks.
Vacationers exchanged rubles for hryvnia and bought 3 kg of tomatoes at a price of 4 hryvnia per 1 kg.
How many rubles did this purchase cost them? Round your answer to the nearest whole number.

Masha sent SMS-messages with New Year's greetings to her 16 friends.
The cost of one SMS is 1 ruble 30 kopecks. Before sending the message, Masha had 30 rubles in her account.
How many rubles will Masha have after sending all the messages?

The school has triple tourist tents.
What is the smallest number of tents to take on a hike with 20 people?

The Novosibirsk-Krasnoyarsk train leaves at 15:20 and arrives at 4:20 the next day (Moscow time).
How many hours does the train take?

Do you know what?

Among all the shapes with the same perimeter, the circle will have the largest area. Conversely, among all shapes with the same area, the circle will have the smallest perimeter.

Leonardo da Vinci derived a rule according to which the square of the diameter of a tree trunk is equal to the sum of the squares of the diameters of the branches taken at a fixed total height. Later studies confirmed it with only one difference - the degree in the formula is not necessarily equal to 2, but lies in the range from 1.8 to 2.3. Traditionally, it was believed that this pattern is explained by the fact that a tree with such a structure has an optimal mechanism for supplying branches with nutrients. However, in 2010, the American physicist Christoph Elloy found a simpler mechanical explanation for the phenomenon: if we consider a tree as a fractal, then Leonardo's law minimizes the likelihood of branches breaking under the influence of wind.

Laboratory studies have shown that bees are able to choose the best route. After localizing the flowers placed in different places, the bee flies around and comes back in such a way that the final path is the shortest. Thus, these insects effectively cope with the classic "traveling salesman problem" from computer science, on the solution of which modern computers, depending on the number of points, can spend more than one day.

If you multiply your age by 7, then multiply by 1443, the result is your age written three times in a row.

We consider negative numbers to be something natural, but this was not always the case. For the first time, negative numbers were legalized in China in the 3rd century, but were used only for exceptional cases, since they were considered, in general, meaningless. A little later, negative numbers began to be used in India to denote debts, but they did not take root to the west - the famous Diophantus of Alexandria argued that the equation 4x + 20 \u003d 0 is absurd.

American mathematician George Danzig, as a graduate student at the university, once arrived late for class and took the equations written on the blackboard for homework. It seemed to him more difficult than usual, but after a few days he was able to complete it. It turned out that he solved two "unsolvable" problems in statistics, over which many scientists were struggling.

In Russian mathematical literature, zero is not a natural number, but in Western literature, on the contrary, it belongs to the set of natural numbers.

The decimal number system we use arose due to the fact that a person has 10 fingers on his hands. The ability for abstract counting did not appear in people immediately, and it turned out to be most convenient to use fingers for counting. The Maya civilization and, independently of them, the Chukchi historically used the twenty number system, using the fingers not only of the hands, but also of the feet. The duodecimal and hexadecimal systems that were widespread in ancient Sumeria and Babylon were also based on the use of hands: the phalanges of the other fingers of the palm were counted with the thumb, the number of which is 12.

One lady friend asked Einstein to call her, but warned her that her phone number was very difficult to remember: - 24-361. Do you remember? Repeat! Surprised Einstein replied: - Of course I remember! Two dozen and 19 squared.

Stephen Hawking is one of the greatest theoretical physicists and popularizer of science. In his story about himself, Hawking mentioned that he became a professor of mathematics without receiving any mathematical education since high school. When Hawking began teaching mathematics at Oxford, he read a textbook, two weeks ahead of his own students.

The maximum number that can be written in Roman numerals without violating the Schwarzman rules (the rules for writing Roman numerals) is 3999 (MMMCMXCIX) - you cannot write more than three digits in a row.

There are many parables about how one person invites another to pay him for a certain service as follows: he will put one grain of rice on the first square of the chessboard, two on the second, and so on: on each next square there is twice as much as on the previous one. As a result, those who pay in this way are bound to go broke. This is not surprising: it is estimated that the total weight of rice will be over 460 billion tons.

Many sources claim that Einstein flunked mathematics at school, or, moreover, generally studied very badly in all subjects. In fact, this was not the case: Albert, at an early age, began to show talent in mathematics and knew it far beyond the school curriculum.

USE 2019 in mathematics task 18 with solution

Demonstration version of the exam 2019 in mathematics

Unified State Exam in Mathematics 2019 in pdf format A basic level of | Profile level

Tasks for preparing for the exam in mathematics: basic and profile level with answers and a solution.

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USE 2019 in mathematics task 18

USE 2019 in mathematics profile level task 18 with solution

Unified State Exam in Mathematics

Find all positive values \u200b\u200bof the parameter a,
for each of which the equation a x \u003d x has only one solution.

Let f (x) \u003d a x, g (x) \u003d x.

The function g (x) is continuous, strictly increasing over the entire domain of definition and can take any value from minus infinity to plus infinity.

At 0< a < 1 функция f(x) - непрерывная, строго убывающая на всей области определения и может принимать значения в интервале (0;+бесконечность). Поэтому при любых таких a уравнение f(x) = g(x) имеет ровно одно решение.

For a \u003d 1, the function f (x) is identically equal to one, and the equation f (x) \u003d g (x) also has a unique solution x \u003d 1.

For a\u003e 1:
The derivative of the function h (x) \u003d (a x - x) is equal to
(a x - x) \u003d a x ln (a) - 1
Let's equate it to zero:
a x ln (a) \u003d 1
a x \u003d 1 / ln (a)
x \u003d -log_a (ln (a)).

The derivative has a single zero. To the left of this value, the function h (x) decreases, to the right, it increases.

Therefore, it either has no zeros at all, or has two zeros. And it has one root only if it coincides with the found extremum.

That is, we need to find such a value of a for which the function
h (x) \u003d a x - x reaches an extremum and vanishes at the same point. In other words, when the line y \u003d x is tangent to the graph of the function a x.

A x \u003d x
a x ln (a) \u003d 1

Substitute a x \u003d x into the second equation:
x ln (a) \u003d 1, whence ln (a) \u003d 1 / x, a \u003d e (1 / x).

Substituting it into the second equation again:
(e (1 / x)) x (1 / x) \u003d 1
e 1 \u003d x
x \u003d e.

And we substitute this into the first equation:
a e \u003d e
a \u003d e (1 / e)

Answer:

Unified State Exam in Mathematics

Decision:

Let's expand the module:

For x<= a 2: f(x) = x 2 - 8x - a 2 ,
for x\u003e a 2: f (x) \u003d x 2 - 10x + a 2.

Derivative of the left side: f "(x) \u003d 2x - 8
Derivative of the right side: f "(x) \u003d 2x - 10

Both the left and right sides can only have a minimum. This means that the only maximum for the function f (x) can be if and only if at the point x \u003d a 2 the left side increases (that is, 2x-8\u003e 0), and the right side decreases (that is, 2x-10< 0).

That is, we get the system:
2x-8\u003e 0
2x-10< 0
x \u003d a 2

From where
4 < a 2 < 5

Answer: (-sqrt (5); -2) ~ (2; sqrt (5))