Maths
How to solve the 18th task

How to solve the 18th task

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

M.: 2017 .-- 128 s.

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

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CONTENT
Introduction 4
§1. Linear equations and systems of linear equations 5
Tasks for an independent solution 11
§2. The study of a square trinomial using discriminant 12
Tasks for an independent solution 19
§3. Vieta Theorem 20
Tasks for an independent solution 26
§4. The location of the roots of the square trinomial 28
Tasks for an independent solution 43
§5. The use of graphic illustrations
to study the quadratic trinomial 45
Tasks for the independent solution 55
§6. Limited function. Finding Value Range 56
Tasks for the independent solution 67
§7. Other function properties 69
Tasks for the independent solution 80
§8. Logic tasks with parameter 82
Tasks for the independent solution 93
Illustrations on the coordinate plane 95
Tasks for the independent solution 108
The Okha Method 110
Tasks for the independent solution 119
Answers 120

This book is devoted to problems similar to problem 18 of the USE in mathematics (task with a parameter). Along with task 19 (a problem in which the properties of integers are used), problem 18 is the most difficult in version. Nevertheless, an attempt is made in the book to systematize problems of this type by various methods for solving them.
A few paragraphs seem to be devoted to such a popular topic as the study of a 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 of 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 regarding such properties of functions as boundedness, parity, continuity; After using examples, the application of these properties to problem solving is demonstrated.

USE in mathematics profile level

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

The lead time is 3 hours 55 minutes.

Examples of tasks of the exam

The solution of the exam tests 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, it showed 1,282 kilowatt hours.
How much do I need to pay for electricity for November?
Give the answer in rubles.

Problem with solution:

In the regular triangular pyramid ABCS with the base ABC, edges 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 ribs AS and BC.

Decision:

1. Since SABC is a regular pyramid, ABC is an equilateral triangle, and the remaining 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 be the midpoint of BC, E the midpoint of AS, SH the height lowered from point S to the base of the pyramid, EP the 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 correct, the point H is the intersection point of the heights / medians / bisectors of the triangle ABC, which means it divides AD in the ratio 2: 1 (AH \u003d 2 AD).

5. Find SH from the right 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. The triangles AEP and ASH are both rectangular and have a common angle A, therefore, similar. By hypothesis, AE \u003d AS / 2, which means that AP \u003d AH / 2 and EP \u003d SH / 2.

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

Angle tangent EDP \u003d EP / DP \u003d 6/5,
EDP \u200b\u200bAngle \u003d arctg (6/5)

Answer:

At the exchange point 1 hryvnia costs 3 rubles 70 kopecks.
Vacationers exchanged rubles for hryvnias and bought 3 kg of tomatoes at a price of 4 hryvnias per 1 kg.
How much rubles did this purchase cost them? Round the answer to an integer.

Masha sent SMS messages with New Year's greetings to her 16 friends.
The cost of one SMS-message 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 on a camping trip with 20 people?

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

Do you know that?

Among all the figures, with the same perimeter, the circle will have the largest area. And vice versa, among all the figures with the same area, the circle will have the smallest perimeter.

Leonardo da Vinci deduced a rule according to which the square of the diameter of a tree trunk is the sum of the squares of the diameters of the branches taken at a common fixed height. Later studies confirmed it with only one difference - the degree in the formula does not necessarily equal 2, but lies in the range from 1.8 to 2.3. It was traditionally 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 Eloy 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 can choose the best route. After localization of the flowers placed in different places, the bee flies and comes back in such a way that the final path is the shortest. Thus, these insects effectively cope with the classic “salesman problem” from computer science, which modern computers, depending on the number of points, can spend more than one day to solve.

If you multiply your age by 7, then multiply by 1443, the result will be 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 III century, but were used only for exceptional cases, as they were considered, in general, meaningless. A little later, negative numbers began to be used in India to denote debts, but to the west they did not take root - the famous Diophantus of Alexandria argued that the equation 4x + 20 \u003d 0 is absurd.

The American mathematician George Danzig, being a graduate student at the university, was once late for a lesson and accepted the equations written on the board for his homework. It seemed to him more complicated 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 fought.

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 used by us arose due to the fact that a person has 10 fingers on his hands. The ability to abstract counting appeared in people not immediately, but it was the most convenient to use fingers precisely for counting. Mayan civilization and independently of them the Chukchi historically used the twenty-decimal number system, using fingers not only of hands, but also of feet. The basis of the duodecimal and hexadecimal systems common in ancient Sumer and Babylon was also the use of hands: the phalanges of the other fingers of the palm, the number of which is 12, were counted with the thumb.

One lady friend asked Einstein to call her, but warned 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 largest theoretical physicists and a popularizer of science. In a story about himself, Hawking mentioned that he had become a professor of mathematics, having not received 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 Schwartzman rules (rules for writing Roman numerals) is 3999 (MMMCMXCIX) - you cannot write more than three digits in a row.

Many parables are known about how one person offers another to pay him for a certain service in the following way: he puts one rice grain on the first square of the chessboard, two on the second and so on: twice as much on the next cell as on the previous one. As a result, one who pays in this way will certainly go broke. This is not surprising: it is estimated that the total weight of rice will be more than 460 billion tons.

In many sources, there is a statement that Einstein overwhelmed mathematics at school or, moreover, generally studied very poorly in all subjects. In fact, everything was not so: Albert at an early age began to show talent in mathematics and knew it far beyond the school curriculum.

Unified State Exam 2019 in mathematics task 18 with a solution

Demo USE 2019 in mathematics

USE in mathematics 2019 in pdf format Basic level | Profile level

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

Unified State Exam 2019 in mathematics task 18

Unified State Exam 2019 in mathematics profile level task 18 with a solution

USE in mathematics

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

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

The function g (x) is continuous, strictly increasing throughout 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 unity, 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
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. The function h (x) decreases to the left of this value, and increases to the right.

Therefore, it either does not have 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 a value a at 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).

Substitute again in the second equation:
(e (1 / x)) x (1 / x) \u003d 1
e 1 \u003d x
x \u003d e.

And this is substituted into the first equation:
a e \u003d e
a \u003d e (1 / e)

Answer:

(0; 1] (e (1 / e))

USE in mathematics

Find all values \u200b\u200bof the parameter a for which the function
f (x) \u003d x 2 - | x-a 2 | - 9x
has at least one maximum point.

Decision:

We will reveal the module:

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

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

Both the left and right parts can have only a minimum. Therefore, the only maximum for the function f (x) can be if and only if at the point x \u003d a 2 the left-hand side increases (i.e. 2x-8\u003e 0), and the right-hand side decreases (i.e. 2x-10< 0).

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

Where from
4 < a 2 < 5

a ~ (-sqrt (5); -2) ~ (2; sqrt (5))

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

In task 18 - the penultimate task of the profile level of the 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 parameter a, and it is necessary to find such values \u200b\u200bat which the system will behave in a predetermined way - to have two or one or have no solutions at all.

Analysis of typical options for tasks №18 USE in mathematics of a profile level

The first version of the assignment (demo version 2018)

Find all the 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:

We consider the second equation, establish what is its graph.
We determine the condition for the uniqueness of a solution.
We find the distance between the centers, determine the parameter values.
We write down the answer.

Decision:

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

2. The second equation is one circle of radius "a" with the coordinates of the center: C (-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, i.e., the coordinates of the tangency of the external and internal image.

But it is worth noting that we will only be interested in the roots that determine the external touch of the left circle and the internal touch of the right circle. Since two other equations contradict the condition and will have more than one solution. Just look at the attached figure: