Trigonometric equations task 13. Examples of Unified State Examination tasks

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 of a high level of difficulty.

Running time - 3 hours 55 minutes.

Examples of Unified State Examination tasks

Solving the Unified State Examination task in mathematics.

Problem with solution:

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

Solution:

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 equal to 5 sqrt(3), and all side edges are equal to 13.

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

3. Find AD from the right triangle CAD using the Pythagorean theorem. It turns out 15/2 = 7.5.

4. Since the pyramid is regular, point H is the point of intersection of the altitudes/medians/bisectors of triangle ABC, and therefore divides AD in the ratio 2:1 (AH = 2 AD).

5. Find SH from right triangle ASH. AH = AD 2/3 = 5, AS = 13, according to the Pythagorean theorem SH = sqrt(13 2 -5 2) = 12.

6. Triangles AEP and ASH are both right angles and have a common angle A, hence similar. By condition, AE = AS/2, which means AP = AH/2 and EP = SH/2.

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

Angle tangent EDP = EP/DP = 6/5,
Angle EDP = arctan(6/5)

Answer:

Do you know what?

Among all the figures 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 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 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.

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

We think of negative numbers as something natural, but this was not always the case. Negative numbers were first legalized in China in the 3rd 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 in the west they did not take root - the famous Diophantus of Alexandria argued that the equation 4x+20=0 was absurd.

American mathematician George Dantzig, while a graduate student at the university, was late for class one day and mistook the equations written on the blackboard for homework. It seemed more difficult to him than usual, but after a few days he was able to complete it. It turned out that he solved two “unsolvable” problems in statistics that many scientists had struggled with.

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 because humans have 10 fingers. People did not immediately develop the ability to count abstractly, and it turned out to be most convenient to use their fingers for counting. The Mayan civilization and, independently of them, the Chukchi historically used the twenty-digit number system, using fingers not only on the hands, but also on the toes. The duodecimal and sexagesimal systems common in ancient Sumer and Babylon were also based on 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 leading theoretical physicists and popularizer of science. In his story about himself, Hawking mentioned that he became a mathematics professor without receiving any mathematics 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 in the following way: on the first square of the chessboard he will put one grain of rice, on the second - two, and so on: on each subsequent square 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, often with the purpose of encouraging poorly performing students, 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.

The lesson discusses the solution to task 13 of the Unified State Exam in computer science.

Topic 13 - “Amount of information” - is characterized as tasks of an increased level of complexity, completion time - approximately 3 minutes, maximum score - 1

when working with text

By using K bit can be encoded Q = 2 K various symbols:
Q- power of the alphabet
K Q character options
2 — binary number system (data is stored in binary form)

N=2i

I, you need to multiply the number of characters N by the number of bits to store one character K:
I
N— message length (number of characters),
K— the number of bits to store one character.
These two formulas use the same variable:

Q = 2 K I = N * K

Let's look at an example using two formulas at the same time:

Example:
Message volume – 7.5 KB 7680 characters. What is the power of the alphabet?

✍ Solution:

Let's use the formula:

I = N*K;
I— message size = 7.5 KB;
N— number of characters = 7680;
K- number of bits per character

Let's find the number of bits required to store 1 character (first convert the value into bits):

\[ K= \frac (7.5 * 2^(13))(7680) = \frac (7.5 * 2^(13))(15 * 2^9) = \frac (7.5 * 16 )(15) = 8\]

those. K = 8 bits per character

Next we use the formula:

Q = 2 K
K— the number of bits to store one character from Q character options (= 8)
Q— the power of the alphabet, i.e. number of character options

8 bits per character allows you to encode:

2 8 = 256 different characters
256 characters - that's power

Answer: 256

Measuring the amount of information
when working with various systems

By using K bit can be encoded Q = 2 K various (numbers) of objects of some system:
Q- the total number of objects in a certain system, data about which is stored in a computer or transmitted in a message,
K— the number of bits for storing one object out of the total number Q,
2 — binary number system (data is stored in binary form).

* other designations are also accepted: N=2i

To find the information volume of a message I, you need to multiply the number of objects in the message - N- by the number of bits K to store one object:
I- information volume of the message,
N— number of objects in the message
K— the number of bits for storing one system object.

Example:
In production, there is an automatic system for informing the warehouse about the need to deliver certain groups of consumables to the workshop. The system is designed in such a way that via the communication channel to the warehouse the conditional number of consumables is transmitted(this uses the same, but the minimum possible number of bits in the binary representation of this number). It is known that a delivery request has been sent 9 groups materials from 19 used in production. Determine the volume of the message sent (Give your answer in bits)

✍ Solution:

Let's use the formula:

K— number of bits for storing one material group number
Q— total number of numbers for various groups of consumables = 19

To store the number of one group, a bit is required:

2 5 < 19 =>5 bits

Degree 4 We are not satisfied, because 2 4 = 16 , and groups 19 .

Next we use the formula:

I = N*K;
I— message volume = ? bit;
N— number of transmitted group numbers (= 9);
K— number of bits per 1 number (= 5)

Let's find the information volume of the message:

I = 9 * 5 = 45 bits

Answer: 45

Solving tasks 13 Unified State Exam in computer science

Unified State Examination in Informatics 2017 task 13 FIPI option 1 (Krylov S.S., Churkina T.E.):

7 33 -character alphabet. The database allocates the same and smallest possible integer to store information about each user byte bit. In addition to your own password, additional information is stored in the system for each user, for which an integer number of bytes are allocated; this number is the same for all users.

To store information about 60 users required 900 byte.

How many bytes are allocated to store additional information about one user?
In response, write down only an integer - the number of bytes.

✍ Solution:

First, let's decide on a password. According to the formula Q = M N we get:

33 = 2 N -> N = 6 bits per character

The password consists of 7 characters:

-> 7*6 =42 bit just for the password

Since all user data is stored in bytes, let’s take the nearest number larger 42 and multiple 8 :

48/8 = 6 42 bits ~ 6 bytes

Now let’s find how many bytes are allocated to store information about one user:

900 bytes / 60 (users) = 15 bytes per user

Let's get the amount of memory to store additional information:

15 bytes (to store all information) - 6 bytes (to store the password) = 9 bytes for additional information

Result: 9

A step-by-step solution to this 13th task of the Unified State Exam in computer science is also available in the video tutorial:

Unified State Exam 2017 collection by D.M. Ushakova “10 training options...” option 1:

The cable network is voting on which of four films they would like to watch that night. They use the cable network 2000 Human. Participated in the voting 1200 Human.
What is the amount of information ( in bytes) recorded by an automated voting system?

✍ Solution:

Since the four movie numbers are stored in the computer system, we can find the number of bits needed to store the movie number:

Q = 2 k -> 4 = 2 k -> k = 2 bat

Since all 1200 people will vote for one of the films, the same amount of memory must be allocated for each vote (i.e. 2 bits).

Let's find the number of bits required to store all 1200 votes:

1200 * 2 = 2400 bits = 2400/8 bytes = 300 byte

Result: 300

Unified State Exam 2017 collection by D.M. Ushakova “10 training options...” option 6:

When registering in a computer system, each user is given a password consisting of 15 characters and containing only characters from 12 - character set A, B, C, D, E, F, G, H, I, K, L, M, N. The database allocates the same and smallest possible integer to store information about each user byte. In this case, character-by-character encoding of passwords is used, all characters are encoded with the same and the minimum possible number bit. In addition to the password itself, additional information is stored in the system for each user, for which 12 bytes per user.

Determine the amount of memory ( in bytes), necessary for storing information about 30 users.
In your answer, write down only an integer - the number of bytes.

✍ Solution:

Result: 600

An example of solving this Unified State Exam task is available in the video tutorial:

Unified State Exam 2017 collection by D.M. Ushakova “10 training options...” option 10:

Taking a rehearsal exam at school 105 Human. Each of them is assigned a special number that identifies him in the automatic answer checking system. When registering a participant to record his number, the system uses the minimum possible number of bit, the same for each participant.

How much information is there? in bits, recorded by the device after registration 60 participants?

✍ Solution:

Result: 420

An example of solving this Unified State Exam task is available in the video tutorial:

Task 13. Demo version of the Unified State Exam 2018 computer science:

10 characters. Capital letters of the Latin alphabet are used as symbols, i.e. 26 various symbols. In the database, each password is stored in the same and smallest possible integer byte. In this case, character-by-character encoding of passwords is used, all characters are encoded with the same and the minimum possible number bit.

Determine the amount of memory ( in bytes), necessary for storing data about 50 users.
In your answer, write down only an integer - the number of bytes.

✍ Solution:

The basic formula for solving this problem is:

Where Q— the number of character variants that can be encoded using N bit.

To find the number of bits required to store one password, you first need to find the number of bits required to store 1 character in the password. Using the formula we get:

26 = 2 N -> N~5 bits

The password consists of 10 characters. This means you need to allocate a bit for the password:

10 * 5 = 50 bits total per password

Since password information is stored in bytes, we translate:

50 bits / 8 ~ 7 bytes (take the nearest number greater than 50 and a multiple of 8: 57/8 = 7)

Now let's find how many bytes are allocated to store information about 50 users:

7 bytes * 50 (users) = 350 byte

Result: 350

For a detailed solution to task 13 of the demo version of the Unified State Exam 2018, watch the video:

Solution 13 of the Unified State Examination task in computer science (diagnostic version of the examination paper, Unified State Examination simulator 2018, S.S. Krylov, D.M. Ushakov):

In some countries, the license plate consists of 7 characters. Each character can be one of 18 various letters or decimal number.

Each such number in a computer program is written in the minimum possible and identical integer quantity byte, in this case character-by-character encoding is used and each character is encoded with the same and minimum possible number bit.

Determine the amount of memory in bytes, allocated by this program for recording 50 numbers.

✍ Solution:

Since the number can use either one letter from 18 , or one digit from 10 , then just one character in the number can be used one of 28 characters:

18 + 10 = 28

Let's determine how many bits are needed to store one character in the number; for this we use the formula N=2i:

28 = 2 i => i = 5

Since the total number of characters in the number is 7 , then we get the required number of bits to store one number:

I = 7 * 5 = 35 bits

Since the same amount is allocated for storing the number byte, then convert it to bytes:

35 / 8 ~ 5 bytes

The problem asks how much memory is needed to store 50 numbers. We find:

I = 50 * 5 = 250 bytes for storing 50 numbers

Result: 250

Video analysis:

Solution 13 of the Unified State Examination task in computer science (control version No. 1 of the exam paper, Simulator 2018, S.S. Krylov, D.M. Ushakov):

Passing the rehearsal exam 9 flows by 100 a person in everyone. Each of them is assigned a special code consisting of a thread number and a number in the stream. When encoding these participant numbers, the verification system uses the minimum possible number of bit, the same for each participant, separately for the thread number and the number in the stream. In this case, the minimum possible and identically integer number is used to write the code bytes.
What is the amount of information in bytes recorded by the device after registration 80 participants?
Please indicate only the number in your answer.

✍ Solution:

The code consists of two components: 1. stream number (in bits) and 2. sequence number (in bits). Let's find the number of bits required to store them:

1. N = 2 i -> 9 = 2 i -> i = 4 bits (2 3 100 = 2 i -> i = 7 bits (2 6

Total we get 4 + 7 = 11 bits for one code. But according to the condition, an integer number of bytes are allocated to store the code. So let’s convert the resulting result into bytes:

11/ 8 ~ 2 bytes (one byte is not enough, 8

Since we need to obtain a volume of information after registration 80 participants, then we calculate:

2 * 80 = 160 byte

Result: 160

Video analysis of the task:

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 4):

Message volume – 7.5 KB. It is known that this message contains 7680 characters. What is the power of the alphabet?

✍ Solution:

Let's use the formula:

I - message volume N - number of characters K - number of bits per character

In our case N=7680 characters highlighted I = 7.5 KB of memory. Let's find the number of bits needed to store one character (first converting KB to bits):

I = 7.5 KB = 7.5 * 2 13 bits

\[ K = \frac (7.5 * 2^(13))(7680) = \frac (7.5 * 2^(13))(15 * 2^9) = \frac (7.5 * 16 )(15) = 8\]

8 bits per character allow you to encode:

2 8 = 256 various characters
(according to the formula Q = 2 N)

256 characters - that's power

Result: 256

A video analysis of the task is presented after the next task.

Message (text) encoding:

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 6):

The power of the alphabet is 256 . How many KB of memory will be required to save 160 pages of text, containing on average 192 characters on every page?

✍ Solution:

Let's find the total number of characters on all pages (for convenience, we will use powers of two):

160 * 192 = 15 * 2 11

According to the formula Q = 2n let's find the number of bits required to store one character (in our case Q=256):

256 = 2 n -> n = 8 bits per character

Let's use the formula I=N*K and find the required volume:

\[ I = (15 * 2^(11)) * 2^3 bits = \frac (15 * 2^(14))(2^(13)) KB = 30 KB \]

I = 30 KB

Result: 30

See a detailed analysis of text encoding tasks: from 1 to 2100), month number (day from 1 to 12) and the number of the day in the month (day from 1 to 31). Each field is written separately from other fields using the smallest possible number of bits.
Determine the minimum number of bits required to encode one record.

✍ Solution:

Formula needed Q = 2n.
Let's calculate the required number of bits to store each item of the entire record:

1. 2100 options: 2100 ~ 2 12 -> n = 12 bits 2. 12 options: 12 ~ 2 4 -> n = 4 bits 3. 31 options: 31 ~ 2 5 -> n = 5 bits

Let's find the total number of bits for the entire record:

12 + 4 + 5 = 21

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 33):

A license plate consists of several letters (the number of letters is the same in all license plates), followed by three digits. In this case, they are used 10 digits and only 5 letters: N, O, M, E And R. You must have at least 100 thousand different numbers.
What is the smallest number of letters that should be in a license plate number?

✍ Solution:

Formula needed Q = mn.

Q - number of options m - power of the alphabet n - length

Let's compose the right side of the formula based on the given conditions of the task (an unknown number of letters (from five options) and three numbers (from 10 options)):

5 ... 5 10 10 10 = 5 x * 10 3

This entire result, by condition, must be no less than 100000 . Let's substitute the rest of the data into the formula:

100000

From here we find the smallest suitable x:

x = 3 : 5 3 * 1000 = 125000 (125000 > 100000)

Result: 3

We invite you to watch a video analysis of the task:

Solution 13 of the Unified State Exam assignment in computer science (K. Polyakov, v. 58):

When registering in a computer system, each user is given a password consisting of 9 characters. The symbols are used uppercase and lowercase letters of the Latin alphabet (in it 26 characters), and also decimal digits. The database allocates the same and minimum possible integer number of bytes to store information about each user. In this case, character-by-character encoding of passwords is used, all characters are encoded with the same and minimum possible number of bits. In addition to the password itself, additional information is stored in the system for each user, for which purpose 18 bytes per user. In the computer system it is allocated 1 KB to store information about users.

What is the largest number of users that can be stored in the system? In your answer, write down only an integer - the number of users.

✍ Solution:

Since both uppercase and lowercase letters are used, we get a total of character options for encoding:

26 + 26 + 10 = 62

From the formula Q = 2 n we obtain the number of bits required to encode 1 password character:

Q = 2 n -> 62 = 2 n -> n = 6

Since the password has 9 characters, we get the number of bits to store 1 password:

6 * 9 = 54

Let's convert it to bytes (since, by convention, passwords are stored in bytes):

54 / 8 = 7 bytes

18 bytes are allocated for storing additional information. Let's get the number of bytes to store all information for one user:

18 + 7 = 25 bytes

According to the condition, 1 KB is allocated to store information about all users. Let's convert this value into bytes:

1 KB = 1024 bytes

Let's get the possible number of users:

1024 / 25 = 40,96

Let's discard the fractional part: 40

Result: 40

Watch the video with the solution to the task:

In task 13 of the profile level of the Unified State Examination in mathematics, it is necessary to solve an equation, but of an increased level of complexity, since tasks of the former level C begin with task 13, and this task can be called C1. Let's move on to consider examples of typical tasks.

Analysis of typical options for tasks No. 13 of the Unified State Exam in mathematics at the profile level

First version of the task (demo version 2018)

a) Solve the equation cos2x = 1-cos(n/2-x)

b) Find all the roots of this equation belonging to the interval [-5n/2;-n].

Solution algorithm:

t
We make the reverse substitution and solve the simplest trigonometric equations.

We build a number axis.
We apply roots to it.
Mark the ends of the segment.
We select those values that lie inside the interval.
We write down the answer.

Solution:

1. Transform the right side of the equality using the reduction formula cos( π/ 2−x)=sin x. We have:

сos2x = 1 – sin x.

Let's transform the left side of the equation using the double argument cosine formula using sine:

cos(2x)=1−2sin 2 x

We get the following equation: 1−sin 2 x=1− sin x

Now there is only one trigonometric function sin in the equation x.

2. Enter the replacement: t= sin x. We solve the resulting quadratic equation:

1−2t 2 =1−t,

−2t 2 +t=0,

t(−2t+1)=0,

t = 0 or -2t + 1 = 0,

t 1 = 0 t 2 = 1/2.

3. Make a reverse replacement:

sin x= 0 or sin x = ½

Let's solve these equations:

sin x =0↔x=πn, nЄZ

sin( x)=1/2↔x= (-1) n ∙( π/6)+πn, nЄZ.

Consequently, we obtain two families of solutions.

1. In the previous paragraph, two families were obtained, each of which has infinitely many solutions. It is necessary to find out which of them are in a given interval. To do this, we build a number line.

2. We apply the roots of both families to it, marking them with green (the first) and blue (the second).

3. Mark the ends of the gap in red.

4. In the indicated interval there are three roots that are three roots: −2 π ;−11π/ 6 and −7 π/ 6.

A) πn, nЄZ;(-1) n ∙( π/6)+πn, nЄZ

b) −2 π ;−11π 6;−7π 6

Second version of the task (from Yashchenko, No. 1)

a) Solve the equation.

Solution algorithm:

We replace this function with a variable t and solve the resulting quadratic equation.
We make the reverse substitution and solve the simplest exponential, then trigonometric equations.

We construct a coordinate plane and a circle of unit radius on it.
We mark the points that are the ends of the segment.
We select those values that lie inside the segment.
We write down the answer.

Solution:

1. We introduce the replacement t = 4 cos x. then the equation will take the form:

We solve the quadratic equation using discriminant and root formulas:

D=b 2 – c = 81 – 4∙4∙2 =49,

t 1 = (9 – 7)/8= ¼, t 2 = (9+7)/8=2.

1. Construct a coordinate plane and a circle of unit radius on it.

2. Mark the points that are the ends of the segment.

3. Select those values that lie inside the segment..

These are the roots. There are two of them.

Third version of the task (from Yashchenko, No. 6)

a) Solve the equation .

b) Find all the roots of this equation that belong to the segment.

Solution algorithm:

Using trigonometric formulas, we reduce the equation to a form containing only one trigonometric function.
We replace this function with a variable t and solve the resulting quadratic equation.
We make the reverse substitution and solve the simplest exponential and then trigonometric equations.