IDZ 6.4 – Option 13. Solutions Ryabushko A.P.

1. Solution tasks 1.13:

   Given: the cost of rail transportation of cargo per 1 km (AB) - k1 p., by road (PC) - k2 p. (k1 2); |AB| = a, |BC| = b.

   We need to find a place P where we need to start building a highway so that delivery of cargo from point A to C can be done cheaper.

   Answer:

   Let the distance from point P to point B be x.

   Then the distance from point P to point A will be equal to a - x, and the distance from point P to point C will be equal to b + x.

   Therefore, the cost of delivering cargo from point A to point C via point P will be:

   k1 (a - x) + k2 (b + x) = (k1 a + k2 b) + (k2 - k1) x.

   This function is a parabola with its vertex at point x = (k2 b - k1 a) / (k1 + k2).

   If x is in the range [0, a], then delivery will be cheaper if x = 0, that is, point P is at point A, if x = a, that is, point P is at point B.

   If x is in the range [-b, 0], then shipping will be cheaper if x = 0, that is, point P is at point C.

   If x is in the range [a, b], then shipping will be cheaper if x = (k2 b - k1 a) / (k1 + k2).

2. Solution tasks 2.13:

   Know: y = (x2 - x - 1) / (x2 - 2x).

   We need to find a general solution (general integral) of the differential equation.

   Answer:

   Let's divide the numerator by the denominator and get:

   y = 1 + (x - 1) / (x2 - 2x).

   Let's decompose the fraction into simpler fractions:

   (x - 1) / (x2 - 2x) = A / (x - 2) + B / x.

   Solving the system of equations, we obtain A = 1 and B = 1:

   y = 1 + 1 / (x - 2) + 1 / x.

   The general solution to the differential equation will be:

   y = C + ln|x - 2| +ln|x| = C + ln|x (x - 2)|, where C is an arbitrary constant.

3. Solution tasks 3.13:

   Given: y = (x + 2)e1-x.

   It is necessary to conduct a complete study of the function and construct its graph.

   Answer:

   Let's find the derivatives of the function:

   y' = (1 - x) e1-x, y'' = (x - 2) e1-x.

   OX axis intersection points:

   (x + 2) e1-x = 0, x = -2.

   OY axis intersection points:

   y(0) =2, since y = (0 + 2) e1-0 = 2.

   Behavior of the function in the vicinity of the point x = -2:

   For x < -2, the y function increases; for x > -2, the y function decreases. The point x = -2 is the local maximum point of the function.

   Behavior of the function in the vicinity of the point x = 0:

   For x < 0, the function y decreases; for x > 0, the function y increases. The point x = 0 is the global minimum point of the function.

   Asymptotes:

   Horizontal asymptote: y = 0, since lim x→+∞ (x + 2) e1-x = 0.

   Slant asymptote: y = x + 2, since lim x→-∞ (y - (x + 2)) = lim x→-∞ (x + 2) e1-x = -∞, and lim x→+∞ ( y - (x + 2)) / x = lim x→+∞ e1-x = 0.

   Function graph:

4. Solution tasks 4.13:

   Hopefully: y = (x - 1) e-x, [0; 3].

   We need to find the smallest and largest values ​​of the function y=f(x) on the segment [a; b].

   Answer:

   Let's calculate the derivatives of the function:

   y' = -x e-x + e-x, y'' = x e-x - 2 e-x.

   OX axis intersection points:

   (x - 1) e-x = 0, x = 1.

   OY axis intersection points:

   y(0) = 1 - e0 = 0, since e0 = 1.

   Behavior of the function in the vicinity of the point x = 1:

   For x < 1, the function y decreases; for x > 1, the function y increases. The point x = 1 is the local minimum point of the function.

   On the segment [0; 3] the smallest value of the function is achieved at point x = 3, and the largest value at point x = 1.

   Minimum function value:

   y(3) = -2 e-3 ≈ 0.0498.

   Maximum function value:

   y(1) = 0.

This digital product is a collection of solutions to problems in mathematics called “IDZ 6.4 – Option 13. Solutions by A.P. Ryabushko.” It contains detailed and clear solutions to problems on various topics in mathematics, which can be useful for students, schoolchildren and anyone interested in mathematics.

The design of this digital product is made in a beautiful and convenient HTML format, which makes it easy to find the information you need and convenient to read the text. This product can be useful both for independent work and for preparing for exams, tests and olympiads. Additionally, it can be used as supplementary material for educators and teachers to help their students better understand math concepts and improve their academic performance.

Digital product "IDZ 6.4 – Option 13. Solutions by Ryabushko A.P." is a collection of solutions to problems in mathematics, which contains detailed and understandable solutions on various topics in mathematics. It contains solutions to problems:

1.13 - the problem of finding a place P where it is necessary to begin construction of a highway is being solved so that delivery of cargo from point A to C is possible cheaper. The solution uses a formula to find the local extremum point of the function.

2.13 - the problem of finding a general solution to a differential equation is solved. The solution uses the method of decomposition into simple fractions.

3.13 - the problem of fully studying the function and constructing its graph is solved. The solution contains the derivatives of the function, determines the intersection points of the axes, determines the behavior of the function in the vicinity of the local extremum points, finds the asymptotes and plots the graph of the function.

4.13 - the problem of finding the smallest and largest values ​​of a function on a given segment is solved. The solution uses the method of finding local extrema points of a function and determining the values ​​of the function at the ends of the segment.

The collection is designed in a beautiful and convenient HTML format, which makes it easy to find the information you need and convenient to read the text. Solutions to problems are prepared in Microsoft Word 2003 using the formula editor. This product can be useful both for independent work and for preparing for exams, tests and olympiads. Additionally, it can be used as supplementary material for educators and teachers to help their students better understand math concepts and improve their academic performance.

***

IDZ 6.4 – Option 13. Solutions Ryabushko A.P. is a set of solutions to problems from various fields of mathematics, made by the author Ryabushko A.P. The product description indicates that this set of problems contains problems on mathematical analysis and probability theory. The set includes tasks on determining the cost of rail and road transportation of goods, finding a general solution to a differential equation, conducting a complete study of the specified functions and finding the smallest and largest values ​​of the function on a given segment. All solutions were made in Microsoft Word 2003 using the formula editor.

***

1. A great solution for preparing for a math exam!
2. Decisions Ryabushko A.P. helped me understand complex IPD topics.
3. Thanks to the author for clear and concise solutions to problems.
4. Good structure of tasks, convenient to work with the material.
5. IPD solutions are great for self-study.
6. A large number of tasks of varying complexity is excellent training for improving problem-solving skills.
7. High-quality digital product, I recommend it to anyone preparing for the exam!

Peculiarities:

Solutions of IDZ 6.4 - Option 13 from Ryabushko A.P. helped me to better understand the material and prepare for the exam.

It is very convenient that IDZ 6.4 - Option 13 is available in electronic form, you can easily open it on a tablet or computer.

Solutions Ryabushko A.P. in IPD 6.4 - Option 13, each task is analyzed in detail, which helps to quickly understand the topic.

IDZ 6.4 - Option 13 from Ryabushko A.P. contains many useful tips and tricks to help you cope with the tasks.

Solutions in IDZ 6.4 - Option 13 from Ryabushko A.P. very clear and easy to read, which makes the learning process more enjoyable.

With the help of IDZ 6.4 - Option 13 from Ryabushko A.P. I was able to strengthen my knowledge in mathematics and prepare for the exam perfectly well.

I am very grateful to Ryabushko A.P. for IDZ 6.4 - Option 13, which helped me cope with difficult tasks and better assimilate the material.