The article is devoted to the analysis of tasks 15 of profile exam in mathematics for 2017. In this task, students are offered to solve inequalities, most often logarithmic ones. Although there may be indicative. This article provides an analysis of examples logarithmic inequalities, including those containing a variable at the base of the logarithm. All examples are taken from open bank tasks of the exam in mathematics (profile), so that such inequalities are likely to come across you on the exam as task 15. Ideal for those who, in a short period of time, want to learn how to solve task 15 from the second part of the profile exam in mathematics in order to get more points on the exam.

Analysis of 15 tasks from the profile exam in mathematics

In the tasks of the 15th exam in mathematics (profile), logarithmic inequalities are often encountered. The solution of logarithmic inequalities begins with the definition of the area acceptable values... In this case, there is no variable at the base of both logarithms, there is only the number 11, which greatly simplifies the task. Therefore, the only restriction we have here is that both expressions under the sign of the logarithm are positive:

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The first inequality in the system is the square inequality. To solve it, we would really not hurt to decompose left side by factors. I think you know that any square trinomial of the form factorized as follows:

where and are the roots of the equation. In this case, the coefficient is 1 (this is the numeric coefficient in front of). The coefficient is also 1, and the coefficient is an intercept, it is -20. The roots of a trinomial are most easily determined by Vieta's theorem. The equation we have given, then the sum of the roots will be equal to the coefficient with the opposite sign, that is, -1, and the product of these roots will be equal to the coefficient, that is, -20. It is easy to guess that the roots will be -5 and 4.

Now the left side of the inequality can be factorized: title = "(! LANG: Rendered by QuickLaTeX.com" height="20" width="163" style="vertical-align: -5px;"> Решаем это неравенство. График соответствующей функции — это парабола, ветви которой направлены вверх. Эта парабола пересекает ось !} X at points -5 and 4. Hence, the desired solution to the inequality is an interval. For those who do not understand what is written here, you can see the details in the video, starting from this moment. There you will also find a detailed explanation of how the second inequality of the system is solved. It is being solved. Moreover, the answer is exactly the same as for the first inequality of the system. That is, the set written above is the range of admissible values ​​of inequality.

So, taking into account the factorization, the original inequality takes the form:

Using the formula, we bring 11 to the power of the expression under the sign of the first logarithm, and move the second logarithm to the left side of the inequality, while changing its sign to the opposite:

After reduction we get:

The last inequality, due to the increase of the function, is equivalent to the inequality , the solution of which is the interval ... It remains to intersect it with the range of admissible values ​​of inequality, and this will be the answer to the entire task.

So, the desired answer to the task is:

We figured out this task, now we turn to the next example of the 15 USE task in mathematics (profile).

We begin the solution by determining the range of admissible values ​​of this inequality. At the base of each logarithm should be positive number, which is not equal to 1. All expressions under the sign of the logarithm must be positive. There should be no zero in the denominator of the fraction. The last condition is equivalent to that, since only otherwise both logarithms in the denominator vanish. All these conditions determine the range of admissible values ​​of this inequality, which is defined by the following system of inequalities:

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In the range of valid values, we can use the transformation formulas for the logarithms in order to simplify the left side of the inequality. Using the formula get rid of the denominator:

Now we have only base logarithms. This is already more convenient. Next, we use the formula, as well as the formula, in order to bring the expression worth glory to the following form:

In the calculations, we used what is in the range of acceptable values. Using the replacement, we arrive at the expression:

We use one more replacement:. As a result, we come to the following result:

So, we gradually return to the original variables. First to the variable:

Unified State Exam 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 high level difficulties.

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, it showed 12802 kilowatt-hours.
How much should I pay for electricity for November?
Give your answer in rubles.

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 figures with the same perimeter, the circle will have the most big square... Conversely, among all shapes 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 equal to the sum 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 does not necessarily equal 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 the 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 localization of flowers placed in different places, the bee flies around and returns 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.

One lady friend asked Einstein to call her, but warned her that her phone number was very difficult to remember: - 24-361. 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, having not received any mathematical education since high school... When Hawking began teaching mathematics at Oxford, he was reading 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 cell of the chessboard, two on the second, and so on: on each next cell 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 2020 in mathematics task 15 with solution

Demonstration version of the exam 2020 in mathematics

Unified State Exam in Mathematics 2020 in pdf format Basic level | Profile level

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USE 2020 in mathematics task 15

USE 2020 in mathematics profile level task 15 with solution

USE in mathematics task 15

Condition:

Solve inequality:
log 2 ((7 -x 2 - 3) (7 -x 2 +16 -1)) + log 2 ((7 -x 2 -3) / (7 -x 2 +16 - 1))> log 2 ( 7 7-x 2 - 2) 2

Solution:

We deal with the ODZ:
1. The expression under the first sign of the logarithm must be greater than zero:
(7 (- (x 2)) - 3) (7 (- (x 2) + 16) -1)> 0

X 2 is always less than or equal to zero, therefore,
7 (-x 2)< = 1, следовательно,
7 (-x 2) - 3< = -2 < 0

This means that for the first condition on the ODD to be fulfilled, it is necessary that
7 (- (x 2) +16) - 1< 0
7 (- (x 2) +16)< 1 = 7 0
- (x 2) +16< 0
x 2> 16
x belongs to (-infinity; -4) U (4, + infinity)

2. The expression under the second sign of the logarithm must be greater than zero. But there the result will be the same as in the first paragraph, since the same expressions are in parentheses.

3. The expression under the third sign of the logarithm must be greater than zero.
(7 (7-x 2) -2) 2> 0
This inequality is always true, except for the case when
7 (7-x 2) -2 = 0
7 (7-x 2) = 7 (log_7 (2))
7-x 2 = log_7 (2)
x 2 = 7 - log_7 (2)
x = (+ -) sqrt (7-log_7 (x))

Let's estimate what is roughly equal to sqrt (7-log_7 (x)).
1/3 = log_8 (2)< log_7(2) < log_4(2) = 1/2
2 = sqrt (4)< sqrt(7-1/2) < sqrt(7-log_7(2)) < sqrt(7-1/3) < sqrt(9) = 3

That is, the condition x is not equal to (+ -) sqrt (7-log_7 (x)) is already superfluous, since in item (1) we have already thrown out the interval that includes these points from the ODZ.

So, once again, ODZ:
x belongs to (- infinity; -4) U (4, + infinity)

4. Now, using the properties of the logarithm, the original inequality can be transformed like this:
log_2 ((7 (-x 2) - 3) 2)> log_2 ((7 (7 - x 2) - 2) 2)

Log_2 (x) is an increasing function, so we get rid of the logarithm without changing the sign:
(7 (-x 2) -3) 2> (7 (7-x 2) -2) 2

Let us estimate from above and below the expressions (7 (-x 2) -3) 2 and (7 (7-x 2) -2) 2 taking into account the DHS:

X 2< -16
0 < 7 (-x 2) < 1
-3 < 7 (-x 2) -3 < -2
4 < (7 (-x 2) -3) 2 < 9

X 2< -16
0 < 7 (7-x 2) < 1
-2 < 7 (-x 2) -2 < -1
1 < (7 (-x 2) -3) 2 < 4

Hence, the inequality holds for any x belonging to the GDZ.