The lesson deals with solution 12 USE assignments in computer science, including assignments for 2017

Topic 12 - "Network addresses" - is characterized as tasks of a basic level of complexity, the execution time is about 2 minutes, the maximum score is 1

Internet Addressing

The address of a document on the Internet (from English - URL - Uniform Resource Locator) consists of the following parts:

• data transfer protocol; may be:
• http(for Web pages) or
• ftp(for file transfer)
• there is also a secure protocol https;
• separator characters :// , separating the name of the protocol from the rest of the address;
• website domain name (or IP address);
• may also be present: the directory on the server where the file is located;
• File name.

Directories on the server are separated by a forward slash " / »

1. network service protocol name - defines the server type HTTP(Hypertext Transfer Protocol);
2. separator in the form of a colon character and two characters Slash;
3. fully qualified domain name of the server;
4. search path for a web document on a computer;
5. web server name;
6. top level domain "org";
7. national domain name "ru";
8. catalog main on the computer;
9. catalog news in the catalog main;
10. search target - file main_news.html.

Network addresses

Physical adress or MAC address- a unique address "sewn in" at the factory - a 48-bit network card code (in hexadecimal system):

00-17-E1-41-AD-73

IP address– computer address (32-bit number), consisting of: network number + computer number on the network (host address):

15.30.47.48

Subnet mask:

• needed to determine which computers are on the same subnet;

in the 10th view in the 16th view

255.255.255.0 -> FF.FF.FF.0

the mask in binary code always has a structure: first all ones, then all zeros:

1…10…0

when superimposed on an IP address (logical conjunction And) gives the network number:

The part of the IP address that corresponds to the mask bits equal to one refers to the network address, and the part corresponding to the mask bits zero is the numeric address of the computer

Thus, it is possible to determine what the last number of the mask:

if two nodes belong to the same network, then they have the same network address.

Calculating a network number from an IP address and a netmask

In the subnet mask high bits allocated in the computer's IP address for network number, have a value of 1 (255); low bits allocated in the computer's IP address for computer addresses on the subnet, matter 0 .

* Image taken from presentation by K. Polyakov

Number of computers on the network

The number of network computers is determined by the mask: the least significant bits of the mask - zeros - are reserved in the computer's IP address for the address of the computer on the subnet.

If mask:

The number of computers on the network:

2 7 = 128 addresses

Of these, 2 are special: network address and broadcast address

128 - 2 = 126 addresses

Solving tasks 12 USE in Informatics

USE in Informatics 2017 task 12 FIPI option 1 (Krylov S.S., Churkina T.E.):

In TCP/IP networking terminology, a netmask is a binary number that determines which part of a host's IP address refers to the network address and which part refers to the address of the host itself on that network. Usually, the mask is written according to the same rules as the IP address - in the form of four bytes, with each byte written as a decimal number. In this case, in the mask, first (in the highest digits) there are ones, and then from a certain digit - zeros. The network address is obtained by applying a bitwise conjunction to the given host IP address and mask.

For example, if the host IP address is 211.132.255.41 and the mask is 255.255.201.0, then the network address is 211.132.201.0

For host with IP address 200.15.70.23 network address is 200.15.64.0 . What is equal to least possible meaning of the third byte of the mask from the left? Write your answer as a decimal number.

✍ Solution:

• The third byte from the left corresponds to the number 70 in the IP address and 64 — in the network address.
• The network address is the result of the bitwise conjunction of the mask and the IP address in binary:

? ? ? ? ? ? ? ? -> third mask byte AND (&) 0 1 0 0 0 1 1 0 2 -> 70 10 = 0 1 0 0 0 0 0 0 2 -> 64 10

least possible outcome masks can be:

1 1 0 0 0 0 0 0 - the third byte of the mask AND (&) 0 1 0 0 0 1 1 0 2 -> 70 10 = 0 1 0 0 0 0 0 0 2 -> 64 10

Here, the most significant bit is taken as one, although zero could have been taken for the result of the conjunction (0 & 0 = 0). However, since there is a guaranteed unit next, it means that we also put in the most significant bit 1 . As you know, in the mask, first there are ones, and then zeros (it cannot be like this: 0100… , or maybe just like this: 1100… ).

Let's translate 11000000 2 into the 10th number system and get 192 .

Result: 192

A step-by-step solution to this 12 task of the exam in computer science is available in the video lesson:

12 task. Demo version of the exam 2018 informatics:

In TCP/IP networking terminology, a netmask is a binary number that determines which part of a host's IP address refers to the network address and which part refers to the address of the host itself on that network. Usually, the mask is written according to the same rules as the IP address - in the form of four bytes, with each byte written as a decimal number. At the same time, in the mask, first (in the highest digits) there are ones, and then from a certain digit - zeros.
The network address is obtained by applying a bitwise conjunction to the given host IP address and mask.

For example, if the host IP address is 231.32.255.131 and the mask is 255.255.240.0, then the network address is 231.32.240.0.

For host with IP address 57.179.208.27 network address is 57.179.192.0 . What is greatest possible number units in the ranks of the mask?

✍ Solution:

• Since the network address is obtained by applying a bitwise conjunction to the given host IP address and mask, we get:

255.255.?.? -> mask & 57.179.208.27 -> IP address = 57.179.192.0 -> network address

Since the first two bytes on the left in the host IP address and the network address are the same, it means that in the mask to obtain such a result with a bitwise conjunction in the binary system, there must be all ones. Those.:

11111111 2 = 255 10

In order to find the remaining two bytes of the mask, it is necessary to translate the corresponding bytes in the IP address and network address into the 2nd number system. Let's do it:

208 10 = 11010000 2 192 10 = 11000000 2

Now let's see what the mask for this byte can be. Let's number the mask bits from right to left:

7 6 5 4 3 2 1 0 1 1 1 0 0 0 0 0 -> mask & 1 1 0 1 0 0 0 0 = 1 1 0 0 0 0 0 0

For the 5th bit we get: ? & 0 = 0 -> the mask can contain both one and 0 . But since the assignment asks us greatest possible number of ones, then it is necessary to say that in the mask this bit is equal to 1 .

For the 4th bit we get: ? & 1 = 0 -> in the mask can only be 0 .

Since the mask first goes ones, and then all zeros, then after this zero in the 4th bit, all the rest will be zeros. And the 4th byte from the left of the mask will be 0 10 .

Let's get the mask: 11111111.11111111.11100000.00000000 .

Let's count the number of units in the mask:

8 + 8 + 3 = 19

Result: 19

For a detailed solution to task 12 of the USE demo version of 2018, see the video:

The solution of task 12 (Polyakov K., option 25):

In TCP/IP networking terminology, a netmask is a binary number that indicates how much of a host's IP address is a network address and how much is a host address on that network. The network address is obtained by applying a bitwise conjunction to the given host address and its mask.

By the specified host IP address and mask determine the network address:

IP address: 145.92.137.88 Mask: 255.255.240.0

When writing the answer, select the four elements of the IP address from the numbers given in the table and write down the corresponding letters without dots in the correct order.

A	B	C	D	E	F	G	H
0	145	255	137	128	240	88	92

✍ Solution:

• To solve the task, you need to remember that the network IP address as well as the network mask are stored in 4 bytes written with a dot. That is, each of the individual IP address and netmask numbers is stored in 8-bit binary form. To get the network address, you need to perform a bitwise conjunction of these numbers.
• Since the number 255 in binary representation is 8 units, then with a bitwise conjunction with any number, the result will be the same number. Thus, there is no need to take into account those bytes of the IP address that correspond to the number 255 in the netmask. Therefore, the first two numbers of the IP address will remain the same ( 145.92 ).
• It remains to consider the numbers 137 and 88 IP Dares and 240 masks. Number 0 in the mask matches eight zeros in binary representation, that is, a bitwise conjunction with any number will turn this number into 0 .
• Let's convert both numbers of the ip-address and netmask to the binary system and write the IP-address and the mask under each other to perform a bitwise conjunction:

137:10001001 88:1011000 - IP address 240:11110000 0:00000000 - netmask 10000000 00000000 - result of bitwise conjunction

Let's translate the result:

10000000 2 = 128 10

In total, for the network address we get bytes:

145.92.128.0

Match the letters in the table and get BHEA.

Result: BHEA

We offer you to watch a detailed video analysis:

The solution of task 12 (Polyakov K., option 33):

If the subnet mask 255.255.255.128 and the IP address of the computer on the network 122.191.12.189 , then the number of the computer on the network is _____.

✍ Solution:

• Single bits of the mask (equal to one) determine the subnet address, since the subnet address is the result of the bitwise conjunction (logical multiplication) of the mask bits with the IP address.
• The rest of the mask (starting from the first zero) defines the computer number.
• Since in binary representation the number 255 is eight units 11111111 ), then the bitwise conjunction with any number returns the same number (1 ∧ 0 = 0; 1 ∧ 1 = 1). Thus, those bytes in the mask that are equal to numbers 255 , we will not consider, because they define the subnet address.
• Let's start with a byte equal to 128 . It corresponds to a byte 189 IP addresses. Let's translate these numbers into the binary number system:

128 = 10000000 2 189 = 10111101 2

Those bits of the IP address that correspond to the zero bits of the mask are used to determine the computer number. Let's translate the resulting binary number into the decimal number system:

0111101 2 = 61 10

Result: 61

For a detailed solution of this task, see the video:

The solution of task 12 (Polyakov K., option 41):

In the terminology of TCP / IP networks, a subnet mask is a 32-bit binary number that determines which bits of the computer's IP address are common to the entire subnet - these bits of the mask contain 1. Masks are usually written as four decimal numbers- according to the same rules as IP addresses.

A mask is used for some subnet 255.255.255.192 . How many different computer addresses theoretically allows this mask if two addresses (network address and broadcast) do not use?

✍ Solution:

• Single bits of the mask (equal to one) determine the subnet address, the rest of the mask (starting from the first zero) determines the computer number. That is, there are as many options for the computer address as can be obtained from zero bits in the mask.
• In our case, we will not consider the first three bytes of the mask on the left, because number 255 in binary representation, these are eight units ( 11111111 ).
• Consider the last byte of the mask, which is 192 . Let's convert the number to the binary number system:

192 10 = 11000000 2

Total received 6 zeros in the netmask. This means that 6 bits are allocated for addressing computers, or, in other words, 2 6 computer addresses. But since two addresses are already reserved (by condition), we get:

2 6 - 2 = 64 - 2 = 62

Result: 62

Watch the video of the task below:

The solution of task 12 (Boundary work, Far East, 2018):

For host with IP address 93.138.161.94 network address is 93.138.160.0 .For how many different mask values is it possible?

✍ Solution:

Result: 5

Video analysis of the task:

In the twelfth task of the OGE in mathematics of the Algebra module, we test our knowledge of transformations - the rules for opening brackets, taking variables out of brackets, bringing fractions to a common denominator and knowledge of abbreviated multiplication formulas.

The essence of the task is to simplify the expression given in the condition: do not immediately substitute values ​​in the original expression. You must first simplify it, and then substitute the value - all tasks are designed in such a way that after simplification, you only need to perform one or two simple actions.

It is necessary to take into account the allowable values ​​of variables included in algebraic expressions, use the properties of the degree with an integer exponent, the rules for extracting roots and the formulas for reduced multiplication.

The answer in the task is an integer or a final decimal fraction.

Theory for task number 12

First of all, let us remember what a degree is and

In addition, we will need abbreviated multiplication formulas:

sum square

(a + b) 2 = a 2 + 2ab + b 2

The square of the difference

(a - b) 2 = a 2 - 2ab + b 2

Difference of squares

a 2 - b 2 \u003d (a + b) (a - b)

sum cube

(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3

difference cube

(a - b) 3 = a 3 - 3a 2 b + 3ab 2 - b 3

Sum of cubes

a 3 + b 3 = (a + b)(a 2 - ab + b 2)

Difference of cubes

a 3 - b 3 \u003d (a - b) (a 2 + ab + b 2)

Rules operations with fractions :

Analysis of typical options for task No. 12 OGE in mathematics

The first version of the assignment

Find the value of the expression: (x + 5) 2 - x (x- 10) with x = - 1/20

Solution:

In this case, as in almost all tasks No. 7, you must first simplify the expression, for this we will open the brackets:

(x + 5) 2 - x (x - 10) = x 2 + 2 5 x + 25 - x 2 + 10x

Then we give like terms:

x2 + 25 x + 25 - x 2 + 10x = 20x + 25

20 x + 25 = 20 (-1/20) + 25 = - 1 + 25 = 24

The second version of the task

Find the value of the expression:

at a = 13, b = 6.8

Solution:

In this case, unlike the first, we will simplify the expression by taking it out of the brackets, and not expanding them.

You can immediately notice that b is present in the first fraction in the numerator, and in the second - in the denominator, so we can reduce them. Seven and fourteen are also reduced by seven:

We shorten (a-b):

And we get:

Substitute the value a = 13:

The third version of the task

Find the value of the expression:

at x = √45, y = 0.5

Solution:

So, in this task, when subtracting fractions, we need to bring them to a common denominator.

The common denominator is 15 x y To do this, multiply the first fraction by 5 y- and numerator and denominator, of course:

Let's calculate the numerator:

5y - (3x + 5y) = 5 y- 3 x - 5 y= - 3 x

Then the fraction will take the form:

Performing simple reductions of the numerator and denominator by 3 and by x, we get:

Substitute the value y = 0.5:

1 / (5 0,5) = - 1 / 2,5 = - 0,4

Answer: - 0.4

Demo version of the OGE 2019

Find the value of an expression

where a = 9, b = 36

Solution:

First of all, in tasks of this type, it is necessary to simplify the expression, and then substitute the numbers.

We bring the expression to a common denominator - this is b, for this we multiply the first term by b, after which we get in the numerator:

9b² + 5a - 9b²

We give similar terms - these are 9b² and - 9b², 5a remains in the numerator.

Let's write the final fraction:

Let's calculate its value by substituting the numbers from the condition:

Answer: 1.25

The fourth option

Find the value of the expression:

at x = 12.

Solution:

Let's perform identical transformations of the expression to simplify it.

1st step - the transition from dividing fractions to multiplying them:

now we reduce the expression (in the numerator of the first fraction and in the denominator of the second) and come to the finally simplified form:

We substitute the numerical value for x into the resulting expression and find the result:

In task No. 12 of the USE in mathematics at the profile level, we need to find the largest or smallest value functions. To do this, it is necessary to use, obviously, the derivative. Let's look at a typical example.

Analysis of typical options for assignments No. 12 USE in mathematics at the profile level

The first version of the task (demo version 2018)

Find the maximum point of the function y = ln(x+4) 2 +2x+7.

Solution algorithm:

1. We find the derivative.
2. We write down the answer.

Solution:

1. We are looking for x values ​​for which the logarithm makes sense. To do this, we solve the inequality:

Since the square of any number is non-negative. The only solution to the inequality is the value of x for which x + 4≠ 0, i.e. at x≠-4.

2. Find the derivative:

y'=(ln(x+4) 2 + 2x + 7)'

By the property of the logarithm, we get:

y'=(ln(x+4) 2)'+(2x)'+(7)'.

According to the formula for the derivative of a complex function:

(lnf)'=(1/f)∙f'. We have f=(x+4) 2

y, = (ln(x+4) 2)'+ 2 + 0 = (1/(x+4) 2)∙((x+4) 2)' + 2=(1/(x+4) 2 2) ∙ (x 2 + 8x + 16) ' + 2 \u003d 2 (x + 4) / ((x + 4) 2) + 2

y'= 2/(x + 4) + 2

3. Equate the derivative to zero:

y, = 0 → (2+2∙(x + 4))/(x + 4)=0,

2 + 2x +8 = 0, 2x + 10 = 0,

The second version of the task (from Yaschenko, No. 1)

Find the minimum point of the function y = x - ln(x+6) + 3.

Solution algorithm:

1. We define the scope of the function.
2. We find the derivative.
3. We determine at what points the derivative is equal to 0.
4. We exclude points that do not belong to the domain of definition.
5. Among the remaining points, we are looking for x values ​​at which the function has a minimum.
6. We write down the answer.

Solution:

2. Find the derivative of the function:

3. Equate the resulting expression to zero:

4. We got one point x=-5, which belongs to the domain of the function.

5. At this point, the function has an extremum. Let's see if this is the minimum. At x=-4

At x = -5.5, the derivative of the function is negative, since

Hence, the point x=-5 is the minimum point.

The third version of the task (from Yaschenko, No. 12)

Find highest value functions on the segment [-3; one].

Solution algorithm:.

1. We find the derivative.
2. We determine at what points the derivative is equal to 0.
3. We exclude points that do not belong to a given segment.
4. Among the remaining points, we are looking for the x values ​​at which the function has a maximum.
5. We find the values ​​of the function at the ends of the segment.
6. We are looking for the largest among the obtained values.
7. We write down the answer.

Solution:

1. We calculate the derivative of the function, we get

Single State exam in mathematics of the basic level consists of 20 tasks. In task 12, the skills of choosing the best option from the proposed ones are tested. The student should be able to evaluate the possible options and choose the most optimal of them. Here you can learn how to solve task 12 of the Unified State Examination in mathematics at a basic level, as well as study examples and solutions based on detailed tasks.

All tasks USE database all tasks (263) USE database task 1 (5) USE database task 2 (6) USE database task 3 (45) USE database task 4 (33) USE database task 5 (2) USE database task 6 (44) ) USE base task 7 (1) USE base task 8 (12) USE base task 10 (22) USE base task 12 (5) USE base task 13 (20) USE base task 15 (13) USE base task 19 (23) USE base task 20 (32)

On average, citizen A. consumes electricity per month during the daytime

On average, citizen A. consumes K kWh of electricity per month during the daytime, and L kWh of electricity per month at night. Previously, A. had a one-rate meter installed in his apartment, and he paid for all electricity at the rate of M rubles. per kWh. A year ago, A. installed a two-tariff meter, while daily electricity consumption is paid at the rate of N rubles. per kWh, and night consumption is paid at the rate P rub. per kWh. During R months, the mode of consumption and tariffs for payment for electricity did not change. How much more would A. have paid during this period if the meter had not changed? Give your answer in rubles.

When building a rural house, you can use one of two types of foundation

When building a rural house, one of two types of foundation can be used: stone or concrete. A stone foundation requires A tons of natural stone and B bags of cement. For a concrete foundation, C tons of crushed stone and D bags of cement are needed. A ton of stone costs E rubles, rubble costs F rubles per ton, and a sack of cement costs G rubles. How many rubles will the material for the foundation cost if you choose the cheapest option?

The task is part of the USE in mathematics of the basic level for grade 11 at number 12.

How many rubles will you have to pay for the cheapest trip for three

A family of three plans to travel from St. Petersburg to Vologda. You can travel by train, or you can drive your own car. A train ticket for one person costs N rubles. The car consumes K liters of gasoline for L kilometers, the distance along the highway is M km, and the price of gasoline is P rubles per liter. How many rubles will you have to pay for the cheapest trip for three?

The task is part of the USE in mathematics of the basic level for grade 11 at number 12.

When building a house, the company uses one of the types of foundation

When building a house, the company uses one of the types of foundation: concrete or foam block. For a foundation of foam blocks, you need K cubic meters of foam blocks and L bags of cement. A concrete foundation requires M tons of crushed stone and N bags of cement. A cubic meter of foam blocks costs A rubles, crushed stone costs B rubles per ton, and a bag of cement costs C rubles. How much will the material cost if you choose the cheapest option?

Average general education

UMK line G. K. Muravina. Algebra and the beginnings of mathematical analysis (10-11) (deep)

Line UMK Merzlyak. Algebra and the Beginnings of Analysis (10-11) (U)

Maths

Preparation for the exam in mathematics ( profile level): tasks, solutions and explanations

We analyze tasks and solve examples with the teacher

The profile-level examination paper lasts 3 hours 55 minutes (235 minutes).

Minimum Threshold- 27 points.

The examination paper consists of two parts, which differ in content, complexity and number of tasks.

The defining feature of each part of the work is the form of tasks:

• part 1 contains 8 tasks (tasks 1-8) with a short answer in the form of an integer or a final decimal fraction;
• part 2 contains 4 tasks (tasks 9-12) with a short answer in the form of an integer or a final decimal fraction and 7 tasks (tasks 13-19) with a detailed answer (full record of the decision with the rationale for the actions performed).

Panova Svetlana Anatolievna, teacher of mathematics of the highest category of the school, work experience of 20 years:

“In order to obtain a school certificate, a graduate must pass two mandatory exams in USE form, one of which is mathematics. In accordance with the Concept for the Development of Mathematical Education in Russian Federation The USE in mathematics is divided into two levels: basic and specialized. Today we will consider options for the profile level.

Task number 1- checks the ability of USE participants to apply the skills acquired in the course of 5-9 grades in elementary mathematics in practical activities. The participant must have computational skills, be able to work with rational numbers, be able to round decimals be able to convert one unit of measurement to another.

Example 1 In the apartment where Petr lives, a cold water meter (meter) was installed. On the first of May, the meter showed an consumption of 172 cubic meters. m of water, and on the first of June - 177 cubic meters. m. What amount should Peter pay for cold water for May, if the price of 1 cu. m of cold water is 34 rubles 17 kopecks? Give your answer in rubles.

Solution:

1) Find the amount of water spent per month:

177 - 172 = 5 (cu m)

2) Find how much money will be paid for the spent water:

34.17 5 = 170.85 (rub)

Answer: 170,85.

Task number 2- is one of the simplest tasks of the exam. The majority of graduates successfully cope with it, which indicates the possession of the definition of the concept of function. Task type No. 2 according to the requirements codifier is a task for using the acquired knowledge and skills in practical activities and everyday life. Task No. 2 consists of describing, using functions, various real relationships between quantities and interpreting their graphs. Task number 2 tests the ability to extract information presented in tables, diagrams, graphs. Graduates need to be able to determine the value of a function by the value of the argument with various ways of specifying the function and describe the behavior and properties of the function according to its graph. It is also necessary to be able to find the largest or smallest value from the function graph and build graphs of the studied functions. The mistakes made are of a random nature in reading the conditions of the problem, reading the diagram.

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Example 2 The figure shows the change in the exchange value of one share of a mining company in the first half of April 2017. On April 7, the businessman purchased 1,000 shares of this company. On April 10, he sold three-quarters of the purchased shares, and on April 13 he sold all the remaining ones. How much did the businessman lose as a result of these operations?

Solution:

2) 1000 3/4 = 750 (shares) - make up 3/4 of all purchased shares.

6) 247500 + 77500 = 325000 (rubles) - the businessman received after the sale of 1000 shares.

7) 340,000 - 325,000 = 15,000 (rubles) - the businessman lost as a result of all operations.

Answer: 15000.

Task number 3- is a task of the basic level of the first part, checks the ability to perform actions with geometric shapes on the content of the course "Planimetry". Task 3 tests the ability to calculate the area of ​​a figure on checkered paper, the ability to calculate degree measures of angles, calculate perimeters, etc.

Example 3 Find the area of ​​a rectangle drawn on checkered paper with a cell size of 1 cm by 1 cm (see figure). Give your answer in square centimeters.

Solution: To calculate the area of ​​this figure, you can use the Peak formula:

To calculate the area of ​​this rectangle, we use the Peak formula:

S= B +	G
	2

where V = 10, G = 6, therefore

S = 18 +	6
	2

Answer: 20.

See also: Unified State Examination in Physics: solving vibration problems

Task number 4- the task of the course "Probability Theory and Statistics". The ability to calculate the probability of an event in the simplest situation is tested.

Example 4 There are 5 red and 1 blue dots on the circle. Determine which polygons are larger: those with all red vertices, or those with one of the blue vertices. In your answer, indicate how many more of one than the other.

Solution: 1) We use the formula for the number of combinations from n elements by k:

all of whose vertices are red.

3) One pentagon with all red vertices.

4) 10 + 5 + 1 = 16 polygons with all red vertices.

whose vertices are red or with one blue vertex.

whose vertices are red or with one blue vertex.

8) One hexagon whose vertices are red with one blue vertex.

9) 20 + 15 + 6 + 1 = 42 polygons that have all red vertices or one blue vertex.

10) 42 - 16 = 26 polygons that use the blue dot.

11) 26 - 16 = 10 polygons - how many polygons, in which one of the vertices is a blue dot, are more than polygons, in which all vertices are only red.

Answer: 10.

Task number 5- the basic level of the first part tests the ability to solve the simplest equations (irrational, exponential, trigonometric, logarithmic).

Example 5 Solve Equation 2 3 + x= 0.4 5 3 + x .

Solution. Divide both sides of this equation by 5 3 + X≠ 0, we get

2 3 + x	= 0.4 or		2		3 + X	=	2	,
5 3 + X			5				5

whence it follows that 3 + x = 1, x = –2.

Answer: –2.

Task number 6 in planimetry for finding geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. The study of the constructed models using geometric concepts and theorems. The source of difficulties is, as a rule, ignorance or incorrect application of the necessary theorems of planimetry.

Area of ​​a triangle ABC equals 129. DE- median line parallel to side AB. Find the area of ​​the trapezoid ABED.

Solution. Triangle CDE similar to a triangle CAB at two corners, since the corner at the vertex C general, angle CDE equal to the angle CAB as the corresponding angles at DE || AB secant AC. Because DE is the midline of the triangle by condition, then by property middle line | DE = (1/2)AB. So the similarity coefficient is 0.5. The areas of similar figures are related as the square of the similarity coefficient, so

Consequently, S ABED = S Δ ABC – S Δ CDE = 129 – 32,25 = 96,75.

Task number 7- checks the application of the derivative to the study of the function. For successful implementation, a meaningful, non-formal possession of the concept of a derivative is necessary.

Example 7 To the graph of the function y = f(x) at the point with the abscissa x 0 a tangent is drawn, which is perpendicular to the straight line passing through the points (4; 3) and (3; -1) of this graph. Find f′( x 0).

Solution. 1) We use the equation of a straight line passing through two given points and find the equation of a straight line passing through the points (4; 3) and (3; -1).

(y – y 1)(x 2 – x 1) = (x – x 1)(y 2 – y 1)

(y – 3)(3 – 4) = (x – 4)(–1 – 3)

(y – 3)(–1) = (x – 4)(–4)

–y + 3 = –4x+ 16| · (-one)

y – 3 = 4x – 16

y = 4x– 13, where k 1 = 4.

2) Find the slope of the tangent k 2 which is perpendicular to the line y = 4x– 13, where k 1 = 4, according to the formula:

3) The slope of the tangent is the derivative of the function at the point of contact. Means, f′( x 0) = k 2 = –0,25.

Answer: –0,25.

Task number 8- checks the knowledge of elementary stereometry among the participants of the exam, the ability to apply formulas for finding surface areas and volumes of figures, dihedral angles, compare the volumes of similar figures, be able to perform actions with geometric shapes, coordinates and vectors, etc.

The volume of a cube circumscribed around a sphere is 216. Find the radius of the sphere.

Solution. 1) V cube = a 3 (where a is the length of the edge of the cube), so

a 3 = 216

a = 3 √216

2) Since the sphere is inscribed in a cube, it means that the length of the diameter of the sphere is equal to the length of the edge of the cube, therefore d = a, d = 6, d = 2R, R = 6: 2 = 3.

Task number 9- requires the graduate to transform and simplify algebraic expressions. Task number 9 advanced level Difficulty with short answers. Tasks from the section "Calculations and transformations" in the USE are divided into several types:

transformations of numerical rational expressions;

transformations of algebraic expressions and fractions;

transformations of numerical/letter irrational expressions;

actions with degrees;

transformation logarithmic expressions;

1. conversion of numeric/letter trigonometric expressions.

Example 9 Calculate tgα if it is known that cos2α = 0.6 and

3π	< α < π.
4

Solution. 1) Let's use the formula double argument: cos2α = 2 cos 2 α – 1 and find

tan 2 α =	1	– 1 =	1	– 1 =	10	– 1 =	5	– 1 = 1	1	– 1 =	1	= 0,25.
	cos 2 α		0,8		8		4		4		4

Hence, tan 2 α = ± 0.5.

3) By condition

3π	< α < π,
4

hence α is the angle of the second quarter and tgα< 0, поэтому tgα = –0,5.

Answer: –0,5.

#ADVERTISING_INSERT# Task number 10- checks the ability of students to use the acquired early knowledge and skills in practical activities and everyday life. We can say that these are problems in physics, and not in mathematics, but all the necessary formulas and quantities are given in the condition. The problems are reduced to solving a linear or quadratic equation, or a linear or quadratic inequality. Therefore, it is necessary to be able to solve such equations and inequalities, and determine the answer. The answer must be in the form of a whole number or a final decimal fraction.

Two bodies of mass m= 2 kg each, moving at the same speed v= 10 m/s at an angle of 2α to each other. The energy (in joules) released during their absolutely inelastic collision is determined by the expression Q = mv 2 sin 2 α. At what smallest angle 2α (in degrees) must the bodies move so that at least 50 joules are released as a result of the collision?
Solution. To solve the problem, we need to solve the inequality Q ≥ 50, on the interval 2α ∈ (0°; 180°).

mv 2 sin 2 α ≥ 50

2 10 2 sin 2 α ≥ 50

200 sin2α ≥ 50

Since α ∈ (0°; 90°), we will only solve

We represent the solution of the inequality graphically:

Since by assumption α ∈ (0°; 90°), it means that 30° ≤ α< 90°. Получили, что наименьший угол α равен 30°, тогда наименьший угол 2α = 60°.

Task number 11- is typical, but it turns out to be difficult for students. The main source of difficulties is the construction of a mathematical model (drawing up an equation). Task number 11 tests the ability to solve word problems.

Example 11. On the spring break 11-grader Vasya had to solve 560 training problems to prepare for the exam. On March 18, on the last day of school, Vasya solved 5 problems. Then every day he solved the same number of problems more than the previous day. Determine how many problems Vasya solved on April 2 on the last day of vacation.

Solution: Denote a 1 = 5 - the number of tasks that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 - the number of days from March 18 to April 2 inclusive, S 16 = 560 – total tasks, a 16 - the number of tasks that Vasya solved on April 2. Knowing that every day Vasya solved the same number of tasks more than the previous day, then you can use the formulas for finding the sum arithmetic progression:

560 = (5 + a 16) 8,

5 + a 16 = 560: 8,

5 + a 16 = 70,

a 16 = 70 – 5

a 16 = 65.

Answer: 65.

Task number 12- check students' ability to perform actions with functions, be able to apply the derivative to the study of the function.

Find the maximum point of a function y= 10ln( x + 9) – 10x + 1.

Solution: 1) Find the domain of the function: x + 9 > 0, x> –9, that is, x ∈ (–9; ∞).

2) Find the derivative of the function:

4) The found point belongs to the interval (–9; ∞). We define the signs of the derivative of the function and depict the behavior of the function in the figure:

The desired maximum point x = –8.

Download for free the work program in mathematics to the line of UMK G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free algebra manuals

Task number 13- an increased level of complexity with a detailed answer, which tests the ability to solve equations, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

a) Solve the equation 2log 3 2 (2cos x) – 5log 3 (2cos x) + 2 = 0

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

Solution: a) Let log 3 (2cos x) = t, then 2 t 2 – 5t + 2 = 0,

	log3(2cos x) =	2	⇔		2cos x = 9	⇔		cos x =	4,5	⇔ because |cos x| ≤ 1,
	log3(2cos x) =	1			2cos x = √3			cos x =	√3
		2							2

then cos x =	√3
	2

	x =	π	+ 2π k
		6
	x = –	π	+ 2π k, k ∈ Z
		6

b) Find the roots lying on the segment .

It can be seen from the figure that the given segment has roots

11π	and	13π	.
6		6

Answer: a)	π	+ 2π k; –	π	+ 2π k, k ∈ Z; b)	11π	;	13π	.
	6		6		6		6

Task number 14- advanced level refers to the tasks of the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes. The task contains two items. In the first paragraph, the task must be proved, and in the second paragraph, it must be calculated.

The circumference diameter of the base of the cylinder is 20, the generatrix of the cylinder is 28. The plane intersects its bases along chords of length 12 and 16. The distance between the chords is 2√197.

a) Prove that the centers of the bases of the cylinder lie on the same side of this plane.

b) Find the angle between this plane and the plane of the base of the cylinder.

Solution: a) A chord of length 12 is at a distance = 8 from the center of the base circle, and a chord of length 16, similarly, is at a distance of 6. Therefore, the distance between their projections onto a plane, parallel to the bases cylinders is either 8 + 6 = 14 or 8 − 6 = 2.

Then the distance between chords is either

= = √980 = = 2√245

= = √788 = = 2√197.

According to the condition, the second case was realized, in which the projections of the chords lie on one side of the axis of the cylinder. So the axis does not intersect given plane within the cylinder, that is, the bases lie on one side of it. What needed to be proven.

b) Let's denote the centers of the bases as O 1 and O 2. Let us draw from the center of the base with a chord of length 12 the perpendicular bisector to this chord (it has a length of 8, as already noted) and from the center of the other base to another chord. They lie in the same plane β perpendicular to these chords. Let's call the midpoint of the smaller chord B, greater than A, and the projection of A onto the second base H (H ∈ β). Then AB,AH ∈ β and, therefore, AB,AH are perpendicular to the chord, that is, the line of intersection of the base with the given plane.

So the required angle is

∠ABH = arctan	AH	= arctg	28	= arctg14.
	BH		8 – 6

Task number 15- an increased level of complexity with a detailed answer, checks the ability to solve inequalities, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

Example 15 Solve the inequality | x 2 – 3x| log 2 ( x + 1) ≤ 3x – x 2 .

Solution: The domain of definition of this inequality is the interval (–1; +∞). Consider three cases separately:

1) Let x 2 – 3x= 0, i.e. X= 0 or X= 3. In this case, this inequality becomes true, therefore, these values ​​are included in the solution.

2) Let now x 2 – 3x> 0, i.e. x∈ (–1; 0) ∪ (3; +∞). In this case, this inequality can be rewritten in the form ( x 2 – 3x) log 2 ( x + 1) ≤ 3x – x 2 and divide by a positive expression x 2 – 3x. We get log 2 ( x + 1) ≤ –1, x + 1 ≤ 2 –1 , x≤ 0.5 -1 or x≤ -0.5. Taking into account the domain of definition, we have x ∈ (–1; –0,5].

3) Finally, consider x 2 – 3x < 0, при этом x∈ (0; 3). In this case, the original inequality will be rewritten in the form (3 x – x 2) log 2 ( x + 1) ≤ 3x – x 2. After dividing by a positive expression 3 x – x 2 , we get log 2 ( x + 1) ≤ 1, x + 1 ≤ 2, x≤ 1. Taking into account the area, we have x ∈ (0; 1].

Combining the obtained solutions, we obtain x ∈ (–1; –0.5] ∪ ∪ {3}.

Answer: (–1; –0.5] ∪ ∪ {3}.

Task number 16- advanced level refers to the tasks of the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes, coordinates and vectors. The task contains two items. In the first paragraph, the task must be proved, and in the second paragraph, it must be calculated.

In an isosceles triangle ABC with an angle of 120° at the vertex A, a bisector BD is drawn. Rectangle DEFH is inscribed in triangle ABC so that side FH lies on segment BC and vertex E lies on segment AB. a) Prove that FH = 2DH. b) Find the area of ​​the rectangle DEFH if AB = 4.

Solution: a)

1) ΔBEF - rectangular, EF⊥BC, ∠B = (180° - 120°) : 2 = 30°, then EF = BE due to the property of the leg opposite the angle of 30°.

2) Let EF = DH = x, then BE = 2 x, BF = x√3 by the Pythagorean theorem.

3) Since ΔABC is isosceles, then ∠B = ∠C = 30˚.

BD is the bisector of ∠B, so ∠ABD = ∠DBC = 15˚.

4) Consider ΔDBH - rectangular, because DH⊥BC.

2x	=	4 – 2x
2x(√3 + 1)		4

1	=	2 – x
√3 + 1		2

√3 – 1 = 2 – x

x = 3 – √3

EF = 3 - √3

2) S DEFH = ED EF = (3 - √3 ) 2(3 - √3 )

S DEFH = 24 - 12√3.

Answer: 24 – 12√3.

Task number 17- a task with a detailed answer, this task tests the application of knowledge and skills in practical activities and everyday life, the ability to build and explore mathematical models. This task is a text task with economic content.

Example 17. The deposit in the amount of 20 million rubles is planned to be opened for four years. At the end of each year, the bank increases the deposit by 10% compared to its size at the beginning of the year. In addition, at the beginning of the third and fourth years, the depositor annually replenishes the deposit by X million rubles, where X - whole number. Find the highest value X, at which the bank will add less than 17 million rubles to the deposit in four years.

Solution: At the end of the first year, the contribution will be 20 + 20 · 0.1 = 22 million rubles, and at the end of the second - 22 + 22 · 0.1 = 24.2 million rubles. At the beginning of the third year, the contribution (in million rubles) will be (24.2 + X), and at the end - (24.2 + X) + (24,2 + X) 0.1 = (26.62 + 1.1 X). At the beginning of the fourth year, the contribution will be (26.62 + 2.1 X), and at the end - (26.62 + 2.1 X) + (26,62 + 2,1X) 0.1 = (29.282 + 2.31 X). By condition, you need to find the largest integer x for which the inequality

(29,282 + 2,31x) – 20 – 2x < 17

29,282 + 2,31x – 20 – 2x < 17

0,31x < 17 + 20 – 29,282

0,31x < 7,718

x <	7718
	310

x <	3859
	155

x < 24	139
	155

The largest integer solution to this inequality is the number 24.

Answer: 24.

Task number 18- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. Exercise high level complexity is not a task for applying one solution method, but for a combination of different methods. For the successful completion of task 18, in addition to solid mathematical knowledge, a high level of mathematical culture is also required.

At what a system of inequalities

	x 2 + y 2 ≤ 2ay – a 2 + 1
	y + a ≤ |x| – a

has exactly two solutions?

Solution: This system can be rewritten as

	x 2 + (y– a) 2 ≤ 1
	y ≤ |x| – a

If we draw on the plane the set of solutions to the first inequality, we get the interior of a circle (with a boundary) of radius 1 centered at the point (0, a). The set of solutions of the second inequality is the part of the plane that lies under the graph of the function y = | x| – a, and the latter is the graph of the function
y = | x| , shifted down by a. The solution of this system is the intersection of the solution sets of each of the inequalities.

Consequently, this system will have two solutions only in the case shown in Fig. one.

The points of contact between the circle and the lines will be the two solutions of the system. Each of the straight lines is inclined to the axes at an angle of 45°. So the triangle PQR- rectangular isosceles. Dot Q has coordinates (0, a), and the point R– coordinates (0, – a). In addition, cuts PR and PQ are equal to the circle radius equal to 1. Hence,

QR= 2a = √2, a =	√2	.
	2

Answer: a =	√2	.
	2

Task number 19- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is not a task for applying one solution method, but for a combination of different methods. To successfully complete task 19, you must be able to search for a solution by choosing different approaches from among the known, modifying the studied methods.

Let sn sum P members of an arithmetic progression ( a p). It is known that S n + 1 = 2n 2 – 21n – 23.

a) Give the formula P th member of this progression.

b) Find the smallest modulo sum S n.

c) Find the smallest P, at which S n will be the square of an integer.

Solution: a) Obviously, a n = S n – S n- one . Using this formula, we get:

S n = S (n – 1) + 1 = 2(n – 1) 2 – 21(n – 1) – 23 = 2n 2 – 25n,

S n – 1 = S (n – 2) + 1 = 2(n – 1) 2 – 21(n – 2) – 23 = 2n 2 – 25n+ 27

means, a n = 2n 2 – 25n – (2n 2 – 29n + 27) = 4n – 27.

B) because S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Her graph can be seen in the figure.

It is obvious that the smallest value is reached at the integer points located closest to the zeros of the function. Obviously these are points. X= 1, X= 12 and X= 13. Since, S(1) = |S 1 | = |2 – 25| = 23, S(12) = |S 12 | = |2 144 – 25 12| = 12, S(13) = |S 13 | = |2 169 – 25 13| = 13, then the smallest value is 12.

c) It follows from the previous paragraph that sn positive since n= 13. Since S n = 2n 2 – 25n = n(2n– 25), then the obvious case when this expression is a perfect square is realized when n = 2n- 25, that is, with P= 25.

It remains to check the values ​​​​from 13 to 25:

S 13 = 13 1, S 14 = 14 3, S 15 = 15 5, S 16 = 16 7, S 17 = 17 9, S 18 = 18 11, S 19 = 19 13 S 20 = 20 13, S 21 = 21 17, S 22 = 22 19, S 23 = 23 21, S 24 = 24 23.

It turns out that for smaller values P full square is not achieved.

Answer: a) a n = 4n- 27; b) 12; c) 25.

________________

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