How to solve task 15 in the exam. Examples of Unified State Examination tasks

The article is devoted to the analysis of tasks 15 of profile Unified State Examination in mathematics for 2017. In this task, schoolchildren are asked to solve inequalities, most often logarithmic ones. Although there may be indicative ones. 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 the open bank of Unified State Examination tasks in mathematics (profile), so such inequalities are likely to come across in the exam as task 15. Ideal for those who want to learn how to solve task 15 from the second part of the profile Unified State Exam in a short period of time in mathematics to get more marks in the exam.

Analysis of tasks 15 from the profile Unified State Examination in mathematics

Example 1. Solve the inequality:

In tasks 15 of the Unified State Exam in mathematics (profile), logarithmic inequalities are often encountered. Solving logarithmic inequalities begins with determining the range of acceptable values. IN in this case There is no variable at the base of both logarithms, there is only the number 11, which greatly simplifies the problem. So the only limitation we have here is that both expressions under the logarithm sign are positive:

Title="Rendered by QuickLaTeX.com">!}

The first inequality in the system is the quadratic inequality. To solve it, we would really like to factorize the left-hand side. I think you know that any quadratic trinomial of the form is 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 equal to 1, and the coefficient is the dummy term, it is equal to -20. The roots of a trinomial are most easily determined using Vieta's theorem. The equation we have given means 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's easy to guess that the roots will be -5 and 4.

Now the left side of the inequality can be factorized: title="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. This means that the required solution to the inequality is the interval . For those who do not understand what is written here, you can watch 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 resolved. Moreover, the answer is exactly the same as for the first inequality of the system. That is, the set written above is the region of permissible values of the inequality.

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

Using the formula, we add 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, 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 , whose solution is the interval . All that remains is to intersect it with the region of acceptable values of the inequality, and this will be the answer to the entire task.

So, the required answer to the task looks like:

We have dealt with this task, now we move on to the next example of task 15 of the Unified State Exam in mathematics (profile).

Example 2. Solve the inequality:

We begin the solution by determining the range of acceptable values of this inequality. At the base of each logarithm there must be a positive number that is not equal to 1. All expressions under the sign of the logarithm must be positive. The denominator of the fraction must not contain zero. The last condition is equivalent to the fact that , since only otherwise both logarithms in the denominator vanish. All these conditions determine the range of permissible values of this inequality, given by the following system of inequalities:

Title="Rendered by QuickLaTeX.com">!}

In the range of acceptable values, we can use logarithm conversion formulas to simplify the left side of the inequality. Using formula we get rid of the denominator:

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

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

Let's use one more replacement: . As a result, we arrive at the following result:

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

Unified State Examination in mathematics profile level

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

Running time - 3 hours 55 minutes.

Examples of Unified State Examination tasks

Solving Unified State Examination tasks in mathematics.

To solve it yourself:

1 kilowatt-hour of electricity costs 1 ruble 80 kopecks.
The electricity meter showed 12,625 kilowatt-hours on November 1, and 12,802 kilowatt-hours on December 1.
How much should I pay for electricity for November?
Give your answer in rubles.

At 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 a whole number.

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 left after sending all the messages?

The school has three-person camping tents.
What is the smallest number of tents you need to take on a camping trip involving 20 people?

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

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 came up with a rule according to which the square of the diameter of a tree trunk equal to the sum 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 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 its branches with nutrients. However, in 2010 American physicist Christophe Alloy 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 optimal route. After localizing the flowers placed in different places, the bee makes a flight and returns back in such a way that the final path turns out to be the shortest. Thus, these insects effectively cope with the classic “traveling salesman problem” from computer science, which modern computers, depending on the number of points, can spend more than one day solving.

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 leading 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 the textbook two weeks ahead of his own students.

The maximum number that can be written in Roman numerals without violating Shvartsman's rules (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 some service as follows: on the first square chessboard he will put one grain of rice, on the second - two, and so on: on each next cell twice as much as on the previous one. As a result, the one who pays in this way will certainly go bankrupt. 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 failed mathematics at school or, moreover, generally studied very poorly in all subjects. In fact, everything was not like that: Albert began to show talent in mathematics at an early age and knew it far beyond the school curriculum.

Unified State Exam 2020 in mathematics task 15 with solution

Demo Unified State Exam option 2020 in mathematics

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

Assignments for preparing for the Unified State Exam in mathematics: basic and specialized level with answers and solutions.

| | home

Unified State Exam 2020 in mathematics task 15

Unified State Exam 2020 in mathematics profile level task 15 with solution

Unified State Exam 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:
Let's deal with 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,< = 1, следовательно,
7 (-x 2)< = -2 < 0

7 (-x 2) - 3
This means that for the first condition on the ODZ to be satisfied, it is necessary that< 0
7 (-(x 2)+16) - 1< 1 = 7 0
7 (-(x 2)+16)< 0
-(x 2)+16
x 2 > 16

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 brackets.

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 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 sqrt(7-log_7(x)) is approximately equal to.
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 redundant, since in paragraph (1) we have already excluded the interval including 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 ODZ:

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

This means that the inequality holds for any x belonging to the ODZ.