Solution of the system of equations depending on the parameter. The number of solutions to a system of linear equations depending on the number of unknowns and the ranks of the matrix and the extended matrix of the system. Elementary matrix transformations. Cramer's method. Vector definition
Suppose you want to find all pairs of values of the variables x and y that satisfy the equation
xy - 6 = 0 and the equation y - x - 1 = 0, that is, it is necessary to find the intersection of the sets of solutions of these equations. In such cases, they say that it is necessary to solve the system of equations xy - 6 \u003d 0 and y - x - 1 \u003d 0.
It is customary to write a system of equations using curly brackets. For example, the system of equations under consideration can be written as follows:
(xy - 6 = 0,
(y - x - 1 = 0.
A pair of values of variables that turns each equation of the system into a true equality is called a solution of a system of equations with two variables.
Solving a system of equations means finding the set of its solutions.
Consider systems of two linear equations with two variables in which at least one of the coefficients in each equation is different from zero.
The graphical solution of systems of this type is reduced to finding the coordinates of the common points of two straight lines.
As you know, two straight lines in a plane can be intersecting or parallel. In the case of parallelism, the lines either have no common points or coincide.
Let's consider each of these cases.
Example 1
Let's solve the system of equations:
(2x + y = -11,
(x - 2y = 8.
Solution.
(y \u003d -3x - 11,
(y \u003d 0.5x - 4.
The slope coefficients of the lines - graphs of the equations of the system are different (-3 and 0.5), which means that the lines intersect.
The coordinates of the point of their intersection are the solution of this system, the only solution.
Example 2
Let's solve the system of equations:
(3x - 2y = 12,
(6x - 4y = 11.
Solution.
Expressing from each equation y in terms of x, we get the system:
(y \u003d 1.5x - 6,
(y \u003d 1.5x - 2.75.
The lines y \u003d 1.5x - 6 and y \u003d 1.5x - 2.75 have equal slopes, which means that these lines are parallel, and the line y \u003d 1.5x - 6 intersects the y axis at the point (0; -6), and line y \u003d 1.5x - 2.75 - at the point (0; -2.75), therefore, the lines do not have common points. Therefore, the system of equations has no solutions.
The fact that this system has no solutions can be verified by arguing as follows. Multiplying all terms of the first equation by 2, we get the equation 6x - 4y = 24.
Comparing this equation with the second equation of the system, we see that the left parts of the equations are the same, therefore, for the same values of x and y, they cannot take different values (24 and 11). Therefore, the system
(6x - 4y \u003d 24,
(6x - 4y = 11.
has no solutions, which means that the system has no solutions
(3x - 2y = 12,
(6x - 4y = 11.
Example 3
Let's solve the system of equations:
(5x - 7y = 16,
(20x - 28y = 64.
Solution.
Dividing each term of the second equation by 4, we get the system:
(5x - 7y = 16,
(5x - 7y = 16,
consisting of two identical equations. The graphs of these equations coincide, so the coordinates of any point on the graph will satisfy each of the equations of the system, that is, they will be the solution of the system. This means that this system has an infinite number of solutions.
If in each equation of a system of two linear equations with two variables at least one of the coefficients of the variable is not equal to zero, then the system either has a unique solution or has infinitely many solutions.
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If the system
a 11 x 1 + a 12 x 2 +... + a 1n x n = b 1 ,
a 21 x 1 + a 22 x 2 +... + a 2n x n = b 2 ,
a m1 x 1 + a m1 x 2 +... + a mn x n = b m . (5.1)
turned out to be consistent, i.e. the matrices of the system A and the matrix of the extended system (with a column of free terms) A|b have the same rank, then two possibilities may appear - a) r = n; b) r< n:
a) if r = n, then we have n independent equations with n unknowns, and the determinant D of this system is different from zero. Such a system has a unique solution obtained from ;
b) if r< n, то число независимых уравнений less than number unknown.
We move the extra unknowns x r+1 , x r+2 ,..., x n , which are commonly called free, to the right-hand side; our system of linear equations will take the form:
a 11 x 1 + a 12 x 2 +... + a 1r x r = b 1 - a 1 , r+1 x r+1 -... - a 1n x n,
a 21 x 1 + a 22 x 2 +... + a 2r x r = b 2 - a 2 , r+1 x r+1 -... - a 2n x n,
... ... ... ... ... ... ... ... ... ...
a r1 x 1 + a r2 x 2 +... + a rr x r = b r - a r , r+1 x r+1 -... - a rn x n.
It can be solved for x 1 , x 2 ,..., x r , since the determinant of this system (rth order) is nonzero. Giving arbitrary numerical values to the free unknowns, we obtain, by Cramer's formulas, the corresponding numerical values for x 1 , x 2 ,..., x r. Thus, for r< n имеем бесчисленное множество решений.
System (5.1) is called homogeneous, if all b i = 0, i.e. it looks like:
a 11 x 1 + a 12 x 2 +... + a 1n x n = 0, a 21 x 1 + a 22 x 2 +... + a 2n x n = 0, (5.5) ... ... . .. ... ... ... a m1 x 1 + a m1 x 2 +... + a mn x n = 0.
It follows from the Kronecker-Capelli theorem that it is always consistent, since adding a column of zeros cannot increase the rank of a matrix. This, however, can also be seen directly - system (5.5) certainly has a zero, or trivial, solution x 1 = x 2 =... = x n = 0. Let the matrix A of system (5.5) have rank r. If r = n, then the zero solution will be the only solution of system (5.5); at r< n система обладает решениями, отличными от нулевого, и для их разыскания применяют тот же прием, как и в случае произвольной системы уравнений. Всякий ненулевой вектор - столбец X= (x 1 , x 2 ,..., x n) T называется own vector linear transformation(square matrix A ), if there is a number λ such that the equality
The number λ is called eigenvalue of the linear transformation (matrices A ), corresponding to the vector X. The matrix A has order n. In mathematical economics, the so-called productive matrices. It is proved that the matrix A is productive if and only if all eigenvalues of the matrix A are less than one in absolute value. To find the eigenvalues of the matrix A, we rewrite the equality AX = λX in the form (A - λE)X = 0, where E is the identity matrix of the nth order or in the coordinate form:
(a 11 -λ)x 1 + a 12 x 2 +... + a 1n x n =0,
a 21 x 1 + (a 22 -λ)x 2 +... + a 2n x n = 0, (5.6)
... ... ... ... ... ... ... ... ... a n1 x 1 + a n2 x 2 +... + (a nn -λ)x n = 0 .We have obtained a system of linear homogeneous equations that has nonzero solutions if and only if the determinant of this system is equal to zero, i.e.
We got an equation of the nth degree with respect to the unknown λ, which is called matrix characteristic equation A, the polynomial is called characteristic polynomial of the matrix A, and its roots are characteristic numbers, or eigenvalues, matrices A. To find the eigenmatrix A in the vector equation (A - λE)X = 0 or in the corresponding system of homogeneous equations (5.6), you need to substitute the found values of λ and solve in the usual way. Example 2.16. Investigate a system of equations and solve it if it is consistent.
x 1 + x 2 - 2x 3 - x 4 + x 5 =1, 3x 1 - x 2 + x 3 + 4x 4 + 3x 5 =4, x 1 + 5x 2 - 9x 3 - 8x 4 + x 5 =0 .
Solution. We will find the ranks of the matrices A and A|b by the method of elementary transformations, simultaneously reducing the system to a stepwise form:
Obviously, r(A) = r( A|b) = 2. The original system is equivalent to the following reduced to a stepped form: x 1 + x 2 - 2x 3 - x 4 + x 5 = 1, - 4x 2 + 7x 3 + 7x 4 = 1. Since the determinant for unknown x 1 and x2 is different from zero, then they can be taken as the main ones and the system can be rewritten in the form: x 1 + x 2 = 2x 3 + x 4 - x 5 + 1, - 4x 2 = - 7x 3 - 7x 4 + 1, Whence x 2 \u003d 7/4 x 3 + 7/4 x 4 -1/4, x 1 \u003d 1/4 x 3 -3/4 x 4 - x 5 + 5/4 - the general solution of a system that has an infinite number of solutions . Giving free to the unknown x 3 , x 4 , x 5 specific numerical values, we will obtain particular solutions. For example, at x 3 = x 4 = x 5 = 0 x 1 = 5/4, x 2 = - 1/4. The vector C(5/4, - 1/4, 0, 0, 0) is a particular solution of this system. Example 2.17. Explore the system of equations and find the general solution depending on the value of the parameter a.
2x 1 - x 2 + x 3 + x 4 = 1, x 1 + 2x 2 - x 3 + 4x 4 = 2, x 1 + 7x 2 - 4x 3 + 11x 4 = a. Solution. This system corresponds to the matrix therefore, the original system is equivalent to: x 1 + 2x 2 - x 3 + 4x 4 = 2, 5x2 - 3x3 + 7x4 = a-2,
. We have A~ 
This shows that the system is consistent only for a=5. Common decision in this case looks like:
x 2 \u003d 3/5 + 3/5x 3 - 7/5x 4, x 1 \u003d 4/5 - 1/5x 3 - 6/5x 4.
Example 2.18. Find out if the system of vectors will be linearly dependent:a 1 =(1, 1, 4, 2),
a 2 = (1, -1, -2, 4),
a 3 = (0, 2, 6, -2),
a 4 =(-3, -1, 3, 4),
a 5 =(-1, 0, - 4, -7),
Solution. A system of vectors is linearly dependent if there are such numbers x 1 , x 2 , x 3 , x 4 , x 5 , of which at least one is different from zero(see item 1, section I) that the vector equality holds:
x 1 a 1 + x2 a 2+x3 a 3 + x4 a 4+x5 a 5 = 0.
In coordinate notation, it is equivalent to the system of equations:
x 1 + x 2 - 3x 4 - x 5 = 0, x 1 - x 2 + 2x 3 - x 4 = 0, 4x 1 - 2x 2 + 6x 3 +3x 4 - 4x 5 = 0, 2x 1 + 4x 2 - 2x 3 + 4x 4 - 7x 5 = 0.
So, we got a system of linear homogeneous equations. We solve it by eliminating the unknowns:
The system is reduced to a stepped form, equal to 3, which means that the homogeneous system of equations has solutions that are different from zero (r< n). Определитель при неизвестных x 1 , x 2 , x 4 is different from zero, so they can be chosen as the main ones and the system can be rewritten in the form:
x 1 + x 2 - 3x 4 = x 5 , -2x 2 + 2x 4 = -2x 3 - x 5 , - 3x 4 = - x 5 .
We have: x 4 \u003d 1/3 x 5, x 2 \u003d 5/6x 5 + x 3, x 1 \u003d 7/6 x 5 -x 3. The system has an infinite number of solutions; if free unknowns x 3 and x5 are not equal to zero at the same time, then the main unknowns are also different from zero. Therefore, the vector equation
x 1 a 1 + x2 a 2+x3 a 3 + x4 a 4+x5 a 5 = 0
To tasks with parameter include, for example, the search for a solution to linear and quadratic equations in a general form, the study of the equation for the number of roots available, depending on the value of the parameter.
Without citing detailed definitions, consider the following equations as examples:
y = kx, where x, y are variables, k is a parameter;
y = kx + b, where x, y are variables, k and b are parameters;
ax 2 + bx + c = 0, where x are variables, a, b and c are parameters.
To solve an equation (inequality, system) with a parameter means, as a rule, to solve an infinite set of equations (inequalities, systems).
Tasks with a parameter can be conditionally divided into two types:
a) the condition says: solve the equation (inequality, system) - this means, for all values of the parameter, find all solutions. If at least one case remains unexplored, such a solution cannot be considered satisfactory.
b) it is required to indicate the possible values of the parameter for which the equation (inequality, system) has certain properties. For example, it has one solution, has no solutions, has solutions that belong to the interval, etc. In such tasks, it is necessary to clearly indicate at what value of the parameter the required condition is satisfied.
The parameter, being an unknown fixed number, has, as it were, a special duality. First of all, it must be taken into account that the alleged fame suggests that the parameter must be perceived as a number. Secondly, the freedom to handle a parameter is limited by its unknown. So, for example, the operations of dividing by an expression in which there is a parameter or extracting a root of an even degree from a similar expression require preliminary research. Therefore, care must be taken in handling the parameter.
For example, to compare two numbers -6a and 3a, three cases need to be considered:
1) -6a will be greater than 3a if a is a negative number;
2) -6a = 3a in the case when a = 0;
3) -6a will be less than 3a if a is a positive number 0.
The decision will be the answer.
Let the equation kx = b be given. This equation is shorthand for an infinite set of equations in one variable.
When solving such equations, there may be cases:
1. Let k be any non-zero real number and b any number from R, then x = b/k.
2. Let k = 0 and b ≠ 0, the original equation will take the form 0 · x = b. Obviously, this equation has no solutions.
3. Let k and b be numbers equal to zero, then we have the equality 0 · x = 0. Its solution is any real number.
The algorithm for solving this type of equations:
1. Determine the "control" values of the parameter.
2. Solve the original equation for x with the values of the parameter that were determined in the first paragraph.
3. Solve the original equation for x with parameter values that differ from those selected in the first paragraph.
4. You can write down the answer in the following form:
1) when ... (parameter value), the equation has roots ...;
2) when ... (parameter value), there are no roots in the equation.
Example 1
Solve the equation with the parameter |6 – x| = a.
Solution.
It is easy to see that here a ≥ 0.
By the rule of modulo 6 – x = ±a, we express x:
Answer: x = 6 ± a, where a ≥ 0.
Example 2
Solve the equation a(x - 1) + 2(x - 1) = 0 with respect to the variable x.
Solution.
Let's open the brackets: ax - a + 2x - 2 \u003d 0
Let's write the equation in standard form: x(a + 2) = a + 2.
If the expression a + 2 is not zero, i.e. if a ≠ -2, we have the solution x = (a + 2) / (a + 2), i.e. x = 1.
If a + 2 is equal to zero, i.e. a \u003d -2, then we have the correct equality 0 x \u003d 0, therefore x is any real number.
Answer: x \u003d 1 for a ≠ -2 and x € R for a \u003d -2.
Example 3
Solve the equation x/a + 1 = a + x with respect to the variable x.
Solution.
If a \u003d 0, then we transform the equation to the form a + x \u003d a 2 + ax or (a - 1) x \u003d -a (a - 1). The last equation for a = 1 has the form 0 · x = 0, therefore, x is any number.
If a ≠ 1, then the last equation will take the form x = -a.
This solution can be illustrated on the coordinate line (Fig. 1)
Answer: there are no solutions for a = 0; x - any number at a = 1; x \u003d -a with a ≠ 0 and a ≠ 1.
Graphic method
Consider another way to solve equations with a parameter - graphical. This method is used quite often.
Example 4
How many roots, depending on the parameter a, does the equation ||x| – 2| = a?
Solution.
To solve by a graphical method, we construct graphs of functions y = ||x| – 2| and y = a (Fig. 2).
The drawing clearly shows the possible cases of the location of the line y = a and the number of roots in each of them.
Answer: the equation will have no roots if a< 0; два корня будет в случае, если a >2 and a = 0; the equation will have three roots in the case a = 2; four roots - at 0< a < 2.
Example 5
For which a the equation 2|x| + |x – 1| = a has a single root?
Solution.
Let's draw graphs of functions y = 2|x| + |x – 1| and y = a. For y = 2|x| + |x - 1|, expanding the modules by the gap method, we get:
(-3x + 1, at x< 0,
y = (x + 1, for 0 ≤ x ≤ 1,
(3x – 1, for x > 1.
On the Figure 3 it is clearly seen that the equation will have a unique root only when a = 1.
Answer: a = 1.
Example 6
Determine the number of solutions of the equation |x + 1| + |x + 2| = a depending on the parameter a?
Solution.
Graph of the function y = |x + 1| + |x + 2| will be a broken line. Its vertices will be located at the points (-2; 1) and (-1; 1) (picture 4).
Answer: if the parameter a is less than one, then the equation will have no roots; if a = 1, then the solution of the equation is an infinite set of numbers from the interval [-2; -one]; if the values of the parameter a are greater than one, then the equation will have two roots.
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1 1 Number of solutions to the system of equations Graphical dynamic method To find the number of solutions to a system of equations containing a parameter, the following trick is useful. We build graphs of each of the equations for a certain fixed value of the parameter and find the number of common points of the constructed graphs. Each common point is one of the solutions to the system. Then we mentally change parameter and imagine how the graph of the equation with the parameter is transformed, how the common points of the graphs appear and disappear Such a study requires a developed imagination To train the imagination, consider a number of typical tasks touch each other or the corner point of one of the graphs falls on another graph As a rule, when passing through a singular point, the number of solutions changes by two, and at such a point itself it differs by one from the number of solutions for little change parameter Consider problems in which it is required to find the number of solutions to a system of equations, one of which depends on the parameter a, and the other does not depend Variables in systems x and y We consider the numbers xi, yi, r to be given constants In the course of each solution, we build graphs of both equations We investigate , how the graph of the equation with a parameter changes when the value of the parameter changes Then we draw a conclusion about the number of solutions (common points of the constructed graphs) In the interactive figure, the graph of the equation without a parameter is shown in blue, and the dynamic graph of the equation with a parameter is shown in red To study the topic (tasks 1 7 ) use file InMA 11, 5 Number of system solutions with parameter For research (task 8) use GInMA file Number of system solutions with parameter (x x0) + (y y0) = r ; 1 Find the number of solutions to the system (x x1) + y = a (x x0) + (y y 0) = r ; Find the number of solutions to the system y = kx + a (x x0) + (y y0) = r ; 3 Find the number of solutions to the system y = ax + y1 (x x0) + (y y0) = r ; 4 Find the number of solutions to the system (x x1) + y = a (x x0) + y y0 = r ; 5 Find the number of solutions to the system (x x0) + (y y0) = a (x x0) + (y y0) = r ; 6 Find the number of solutions to the system y = x a + y1 x x0 + y y0 = r; 7 Find the number of solutions to the system (x x0) + (y y0) = a f (x, y) = 0; g (x, y, a) = 0 8 Find the number of solutions of the system VV Shelomovsky Thematic sets, cmdru/
2 1 Graphs of equations smooth curves (x x0) + (y y0) = r ; 1 Task Find the number of solutions to the system (x x1) + y \u003d a Solution: The graph of the first equation is a circle of radius r centered at point O (x0; y0) The graph of the second equation is a circle of radius a centered on the x-axis at point A (x1 ; 0) The center of the circle is fixed, the radius determines the parameter When the modulus of the parameter increases, the circle "swells" Special meanings parameter, those values at which the number of roots changes, that is, the parameter values at which the circle of the second graph touches the circle of the first the value of the variables and the parameter, find the number of solutions to the system. It is desirable to start the study with the simplest cases y0 = 0, when the common axis of the circles is horizontal, and x0 = x1, when the common axis of the circles is vertical In the general case, use Pythagorean triangles For example, x0 x1 = 3, y0 = ±4 Typically, both for small modulus and large modulo values of the parameter, there are no solutions. intermediate values of the parameter to two Creative task Find the value of the parameter at which three different points (x 1) + (y y0) = 9; are solutions of the system of equations (x x1) + y = a (x x0) + (y y0) = r ; Task Find the number of solutions to the system y \u003d kx + a Solution: The graph of the first equation is a circle of radius r centered at the point O (x0; y0) The graph of the second equation is a family of parallel lines passing through the points A (0; a) and having a constant slope The tangent of the inclination angle of the straight lines is equal to k As the parameter increases, the straight lines move upward Special parameter values are those values at which the number of roots changes, that is, the parameter values at which the straight lines touch the circle The tangency condition is found by equating the tangents of the inclination angle of the circle and the straight line cmdru/
3 3 Solving the resulting equation, we find the coordinates of two touch points: kr x = x0 ± ; x0 x 1 + k = k k (y y0) + (y y0) = r r y y0 y = y0 1+ k : By changing the value of the variables and the parameter, find the number of solutions of the system. It is desirable to start the study with the simplest case k = 0, when the lines are parallel to the x-axis. Then consider the cases when the root is extracted (for example, k = 3), pay attention to the popular case k = 1. For small and at large values there is no solution parameter Since a straight line and a circle can have no more than two common points, the number of solutions is no more than two For parameter values corresponding to tangency, the number of solutions is equal to one, for intermediate values of the parameter, two Creative task It is known that this system of equations has no more than one solution Find the value of the parameter for which the system of equations has a solution: (x) + (y 3) = r; y = x + a (x x0) + (y y0) = r ; 3 Find the number of solutions to the system y \u003d ax + y1 Solution: The graph of the first equation is a circle of radius r centered at the point O (x0; y0) The graph of the second equation is a family of lines passing through the point A (0; y1) The tangent of the slope of the lines ( a) determines the value of the parameter As the parameter increases, the angle between the graph and the positive direction of the abscissa increases. Special values of the parameter are those values at which the number of roots changes, that is, the parameter values at which the lines touch the circle If the point A (0; y1) is inside the circle , then any possible straight line intersects the circle at two points. The tangency condition is found by equating the tangents of the inclination of the circle and the straight line. Solving the resulting equation, we find the coordinates of the two tangent points: VV Shelomovsky
4 4 ar x = x0 ± ; x0 x 1 + a = a a (y y0) + (y y0) = r r y y0 y = y0 1+ a singular values of the parameter a = ± r If y0 = y1, x0 r, then singular values of the parameter a = ± (y1 y 0) r r x0 If x0 = ± r, then the circle touches the vertical line passing through the point r (y1 y 0) A(0; y1) and parameter value a = In other cases x0 (y1 y 0) a= x0 (y 0 y1) ± r (x0 + (y 0 y1) r) r x0 Research: Changing the value of variables and parameter, find the number of solutions of the system It is desirable to start the study with the simplest case y0 = y1, x0< r, когда точка А(0; у1) внутри окружности и число решений всегда равно двум Рассмотрите случай х0 = r, когда число решений легко найти (х0 = r =, y0 = 3, y1 =) Затем рассмотрите случаи, когда корень хорошо извлекается (например, х0 = 3, y0 = 4, r =, y1 =) Поскольку прямая и окружность могут иметь не более двух общих точек, число решений не более двух При значениях параметра, соответствующих касанию, число решений равно единице, при остальных значениях параметра нулю или двум (x + 3) + (y 5) = r ; при всех y = ax + 1 Творческое задание Известно, что система уравнений значениях параметра, кроме одного, имеет два решения Найдите то значение параметра, при котором система уравнений имеет единственное решение (x x0) + (y y0) = r ; 4 Задание Найдите число решений системы (x x1) + y = a Решение: В ходе решения строим графики каждого из уравнений и исследуем число общих точек построенных графиков График первого уравнения это пара окружностей одинакового радиуса r Центры окружностей O и Q имеют одинаковую ординату y0 и ВВ Шеломовский Тематические комплекты, cmdru/
5 5 abscissas of the same modulus but different in sign ±x0 Graphs are shown in blue and purple The graph of the second equation is a circle of radius a centered on the abscissa axis at point A(x1; 0) Special values of the parameter are those values at which the number of roots changes , that is, the values of the parameter at which the circle of the second graph touches the circles of the first. Conditions for touching the sum or difference of the radii of the circles is equal to the center-to-center distance: a ± r = AO, a ± r = AQ Investigation: By changing the value of the variables and the parameter, find the number of solutions to the system values for one center-to-center distance (for example, x0 = 6, y0 = 3, r = 3, x1 =) Typically, for small modulus and large values of the parameter, there are no solutions. At the points of contact, the number of roots is odd, at other points the number of roots is even ( x 6) + (y y 0) = r; Creative task It is known that the system of equations for (x x1) + y = a has exactly two solutions for a certain value of the parameter. At this value of the parameter, the graphs touch Find this value of the parameter (x x0) + y y0 = r; 5 Find the number of solutions to the system (x x0) + (y y0) = a Solution: The graph of the first equation consists of a pair of parabolas that meet at y = y0 Equations of parabolas y = y0 ± (r (x x0)) They have a horizontal axis of symmetry y \u003d y0, the vertical axis of symmetry x \u003d x0 Center of symmetry point (x0, y0) The second graph is a circle with radius a, the center of which is located at the center of symmetry of the parabolas At the point of contact: x = x0, y = y0 ± r = y = y0 ± а, therefore, а = ± r from a system of equations to an equation with one variable: (y y 0) = a (x x0) = (r (x x0)) This is a quadratic equation for (x x 0) It has one root if the discriminant is zero: VV Shelomovsky Thematic sets, cmdru/
6 6 D = (r 0.5) (r a) = 0, a = ± r 1 4 The number of roots changes at such a value of the parameter at which the circle and the parabola intersect at the break points of the first graph, that is, at y = y0 Research : By changing the value of the variables and the parameter, find the number of solutions of the system Use the values r = 1, 4 and 9 Note that the parameters x0 and y0 do not affect the answer of the problem For small and large values of the parameter, there are no solutions x x0 + y y0 = r; 6 Find the number of solutions of the system (x x0) + (y y0) = a Solution: The graph of the first equation is a square inclined at an angle of 45 to the coordinate axes, the length of half of the diagonal of which is r The second graph is a circle of radius a, the center of which is located in the center symmetry of the square The number of roots changes at the value of the parameter at which the circle passes through the vertices of the square In this case, y = y0, a = ±r The number of roots changes at the value of the parameter at which the circle touches the sides of the square internally To find this value, we pass from a system of equations to an equation with one variable: (y y 0) = a (x x0) = (r x x0) This is a quadratic equation for x x 0 It has one root if the discriminant is zero In this case a = ± r The radius of the circle in this case refers to the radius in the previous case, as sin 45: 1 VV Shelomovsky Thematic sets, cmdru/
7 7 (x x0) + (y y0) = r ; 7 Find the number of solutions to the system y \u003d x a + y1 The graph of the first equation is a circle with center O (x0; y0) The graph of the second equation consists of two rays with a common beginning - “bird, wings up”, the top of the graph is located at point A (a; y1) The number of roots changes at the value of the parameter at which the “wing” of the second graph touches the circle or the vertex of the graph lies on this circle. this wing touches the circle at points (xk; yk) such that r yk = y0 Tangency condition yk = xk a + y1 a = xk yka + y1= x0 y0 + y1 ± r Since the "wing" is a ray going up , the condition is added that the vertex ordinate should not be greater than the tangent point ordinate, that is, y1 yk y0 y1 ± r Similarly, we write down the conditions for tangency with the “left wing” If the vertex of the graph lies on a circle, then its coordinates satisfy the circle equation: (a x0) + (y1 y0) = r lo solutions of the system, that is, the number of common points of the graphs At singular points, the number of roots is odd, at other points the number of roots is even (x) + (y y 0) = r, Creative task It is known that the system of equations for y = x a + y1, some value parameter has three solutions Find this value of the parameter if it is known that the ordinates of the two solutions coincide f (x, y) = 0; g (x, y, a) = 0 8 Find the number of solutions of the system Set the functions yourself according to the model and explore the number of solutions VV Shelomovsky Thematic sets, cmdru/
8 8 VV Shelomovsky Thematic sets, cmdru/
9 9 Assignments С5 (Semyonov Yashchenko) Option 1 Find all values of a, for each of which the set of solutions of the inequality 4 x 1 x+ 3 a 3 is the segment 3 a 4 x Thinking Let's perform transformations x b 1, 1 x b 1, 4 x 1 x+ 3 a x b 3=, b=3 a 3 a 4 x x (x) 0, (x +1) b 1 0 The boundary lines of the x 3a plane are: x = 0, x =, x= 3a, x=± 3 a a= (x+ 1) 1 4 If 0 x, then b< 4x, b (x +1) 1 Так как 4x >(x +1) 1, then b (x +1) 1 If 0 > x then b > 4x, (x +1) 1 b There is a solution for 1 b For example, x = 1 If x > then b > 4x, (x +1) 1 b Since 4x< (x +1) 1, то (x +1) 1 b Значит, решения таковы Если 3а >8, then x [ 3 a + 1 1.0] [, 3 a + 1 1] If 0< 3а < 8, то Если 3а = 0, то х [,0) (0, ] Если 1< 3а < 0, то х [ 3 a +1 1, 3 a+1 1] [ 0, ] Если 1 = 3а, то х 1 } Если 1 >3a, then x Solution Let 1 3a Then x = 1 satisfies the inequality, 4 x 1 x+ 3 a 16+3 a 3 a 3 = 3 =, a contradiction, this number is outside the segment 3 a 4 x 3 a+ 4 3 a +4 Let 1 > 3а Then x b 1, 4 x 1 x+3 a x b 3=, b=3 a< 1 3 a 4 x 1 x b 1, x (x) 0, (x +1) b 1 0 Числа из промежутка 0 х удовлетворяют обоим неравенствам Если x >, then the first inequality is not satisfied VV Shelomovsky Thematic sets, cmdru/
10 10 If 0 > x, then b (x +1) 1, the second inequality is not satisfied Answer: 1 > 3a Option 3 Find all values of a, for each of which the equation a +7 x x + x +5 has at least one root = a+ 3 x 4 a +1 Thinking Let f (a, x)=a +7 x x + x +5 a 3 x 4 a+1 Singular point of the function x + 1 = 0 If x = 1, then the equation is a +10 a 1 a =0 It is easy to find its four solutions It is necessary to prove that the original function is always greater than this one Solution Let f (a, x)=a + 7 x x + x +5 a 3 x 4 a+1 Equation f (a, x)=0 Then f (a, 1)=a +10 a 1 a =0 Difference f (a, x) f (a, 1)=7 x +1 +5(x + x +5)+ 3 4 a 3 x 4 a+1 3(x a 4 a x 1) 0 Therefore, the equation f (a, x)=0 has roots only if f (a, 1) 0 The equation f (a, 1)=0 has four roots a 1= , a = , a 3= , a 4 = Function f (a, 1) 0 (not positive) for a For example, if a = 10, that is, the root x) f (a, 1)>0 No roots Answer: [ 5 15, 5+ 15] Option 5 Find all values of a, each of which has at least one root ur equation a +11 x+ +3 x + 4 x +13=5 a+ x a + Use the function f (a,)=a +9 5 a 4 a =0 and the inequality f (a, x) f (a,) (x+ + a x a+) 0 Answer: [ , ] Variant 9 Find the number of roots of the equation x + 4x 5 3a = x + a the derivative of one is greater on the interval than the other Let the difference of the values of the functions on the left end have one sign, on the right end the other Then the equation f(x) = g(x) has exactly one root on the interval Solution Denote f(x, a) = 3а + x + a, g(x) = x + 4x Equation f(x, a) = g(x) VV Shelomovsky Thematic sets, cmdru/
11 11 Singular points of the function g(x) are minima at x = 1 and x = 5 and maximum at x = Values g(1) = g(5) = 1, g() = 10 The function has an axis of symmetry x = 3 At For values of x greater in modulus, the quadratic function g(x) is greater than the linear function f(x, a) The slope of the function outside the interval [5,1] is determined by the derivative (x + 4x 5)" = x for x > 1 The function g(x) for x > 1 monotonically increases with a factor greater than 6 Due to symmetry, the function g(x) monotonically decreases with a factor greater than 6 at x< 5 Наклон g(x) равен 1 только на промежутке (5, 1) При этом производная (x 4x + 5)" = x 4 = 1 Значит, в точке x = 5 наклон равен 1 Функция f(x, a) = 3а + x + a монотонно убывает с коэффициентом 1 при x + а < 0 и монотонно возрастает с коэффициентом 1 при x + а >0 Values at a number of points f(a, a) = 3a, f(5, a) = 3a + 5 a, f(, a) = 3a + a, f(1, a) = 3a + 1+ a Plots f (x, a) and g(x) touch if their slopes are equal Touching is possible at x = 5 In this case, g(x) = 39/4 f(x, a) = 4a + x = 39/4, 4a = 49 /4, a = 49/16 We analyze the roots of the equation f(x, a) = g(x) If a<, f(5, a) = а +5 < 1, f(1, a) = а 1 < 5 f(x, a) < g(x), так как в промежутке 5 < x < 1 f(x, a) < 1 < g(x) Если x >1, g(x) grows faster than f(x, a), that is, everywhere f(x, a)< g(x) Если x < 5, g(x) убывает быстрее, чем f(x, a), то есть всюду f(x, a) < g(x) Других корней нет Если a =, f(5, a) = 1, f(1, a) = 5 f(5,) = g(5) Один корень х = 5 Во всех других точках f(x, a) < g(x), как и в предыдущем случае Если < a < 0, f(5, a) = а +5 >1, f(1, a) = 4a + 1< 1f(, a) = а + < 10 При x >f(x, a)< g(x), корней нет При x < f(1,a) >1 At x< 5 быстро убывающая g(x) пересекает медленно убывающую левую ветвь f(x,а), один корень При 5 < x < возрастающая g(x) пересекает убывающую f(x,а), один корень, всего корней два, один при x < 5, второй при 5 < x < Если a = 0, f(5, a) = 5, f(1, a) = 1 f(1, a) = g(1), один корень х = 1 Как и раньше, один корень при x < 5, один корень при 5 < x < Всего корней три Если 0 < a < 3, корней 4, два на левой ветке f(х, a) при x <, два на правой при x >If a = 3, f(3, 3) = 8 = g(3), f(, 3) = 10 = g(), roots 4, one two on the left branch of f(x, a) at x< 5, один в вершине f(х, 3) при x = 3, один в вершине g(x) при x =, один при x >1 If 3< a < 49/16, корней 4, один на левой ветке f(х, a) при x < 5, два на правой ветви g(x) при 3 < x <, один при x >1 If a = 49/16, then the number of roots is 3, one on the left branch of f(x, a) at x< 5, один в точке касания при x = 5, один при x >1 If a > 49/16, then the number of roots, one on the left branch of f(x, a) at x< 5, один на правой при x >1 Answer: no roots for a< ; один корень при a =, два корня при < a < 0 или 49/16 < a, три корня при a = 0 или а = 49/16, четыре корня при 0 < a < 49/16 ВВ Шеломовский Тематические комплекты, cmdru/
12 1 Option 10 Find all values of the parameter a, for each of which the equation 4x 3x x + a = 9 x 3 has two roots Solution Denote f(x, a) = 4x 3x x + a, g(x) = 9 x 3 The singular point of the function g(x) is x = 3 The function decreases monotonically by a factor of 9 as x< 3 и монотонно возрастает с коэффициентом 9 при x >3 The function f(x, a) is piecewise linear with coefficients 8, 6, or 0 Therefore, it does not decrease in x, its growth rate is less than that of the right branch of the function 9 x 3 f(3, a) = a Graph of this the expression is a polyline with vertices (1, 1), (3, 3), (6, 1) The values of the function are positive for a (4, 18) It follows from what was found If f(3, a)< 0, уравнение не может иметь корней, так как g(x) >f(x, a) If f(3, a) = 0, the equation has exactly one root x = 3 For other x's g(x) > f(x, a) If f(3, a) > 0, the equation has exactly two roots, one for x< 3, когда пересекаются убывающая ветвь g(x) и монотонно не убывающая f(x, a) Другой при x >3, when the rapidly increasing branch g(x) intersects the slowly increasing branch f(x, a) Answer: a (4, 18) Option 11 Find all values of the parameter a, for each of which, for any value of the parameter b, has at least one solution system of equations (1+ 3 x)a +(b 4 b+5) y =, x y +(b) x y+ a + a=3 Thinking The system looks like (1+ 3 x)a +(1+(b) ) y =, Conveniently x y +(b) x y=4 (a+ 1) a (1+3 x) =1, The solution x = y = 0 and x y =4 (a +1) is seen corresponding parameter values a = 1 and a = 3 analyze singular point b = Then (1+ 3 x)a +(1+(b)) y =, x y +(b) x y=4 (a+ 1) Solution We write the system as Solution x = y = 0 always exists for a = 1 or a = 3 If b =, then the system has the form (1+ 3 x)a +1 y =, or x y =4 (a +1) (1+3 x)a=1, x y =4 (a +1) If a > 1 or a< 3 система не имеет решений, так как их не имеет второе уравнение Если 1 < a < 3, из второго уравнения получим, что x >0, from the first we find a = 0 Let a = 0 Then for b = 4 from the first equation we get that y = 0 In this case, the second equation has no solution Answer: 1 or 3 VV Shelomovsky Thematic sets, cmdru/
13 13 Option 14 Find all values of the parameter, for each of which the modulus of the difference of the roots of the equation x 6x a 4a = 0 takes highest value Solution Let's write the equation in the form (x 3) = 1 (a) Its solution = 0 due to the periodicity of the sine and cosine functions, the problem can be solved for the segment x=3± 1 (a) The largest difference of the roots is when a = Answer: Option 15 Find all values of the parameter, for each of which the equation (4 4 k) sin t =1 has at least one solution on the interval [ 3 π ; 5 π ] cos t 4 sin t Solution Due to the periodicity of the sine and cosine functions, the problem can be solved for the interval t [ π ; 15 π ], then subtract 4π from each solution obtained. Transform the equation to the form + 4 k sin t cos t \u003d 0 cos t 4 sin t On the segment t [ π ; 15 π] the sine monotonically decreases from zero to minus one, the cosine monotonically increases from minus one to zero The denominator vanishes at 4tgt = 1, that is, at sin t = 1 4, cos t = t = 15π is equal to 4k If k 0, the numerator is positive and the equation has no roots If k > 0, both variable terms of the numerator decrease, that is, the numerator changes monotonically So, the numerator takes a zero value exactly once, if k 05 and is positive for smaller values k The equation has a root if the numerator is zero and the denominator is not zero, that is, in the case of 4k =+ 4 k sin t cos t + k Answer: k [ 05,+)\1+ ) Option 18 of which the system of equations (x a 5) + (y 3 a +5) \u003d 16, (x a) + (y a + 1) \u003d 81 has a unique solution We think Each equation describes a circle The solution is unique in the case of touching circles Solution The first equation defines a circle centered at (a + 5, 3a 5) and radius 4 The second equation is circular centered at the point (a +, a 1) with a radius of 9 VV Shelomovsky Thematic sets, cmdru/
14 14 The system has a unique solution if the circles are tangent In this case, the distance between the centers is = 13 or 0 4 = 5 The square of the center distance: ((a + 5) (a +)) + ((3a 5) (a 1)) = a a + 5 If the distance is 5, then a = 0 or a = 1 If the distance is 13, then a = 8 or a = 9 Answer: 8, 0, 1, 9 Option 1 Find all values of the parameter, each of which has exactly two non-negative solutions equation 10 0.1 x a 5 x + a =004 x Solution Perform transformations 5 x a 5 x + a =5 x Denote t = 5x 1 exponential function 5x, each root t 1 generates exactly one root x 0 The equation becomes t a t+ a t =0 If a t, then t + 3t + a = 0 there are no roots greater than 1 If t > a t/, then t t + 3a = 0 For t > 1, the function monotonically increases, there is only one root If 1/ > t/ > a, then t 3t a = 0 For t > 1, the function t 3t monotonically decreases from at t = 1 to 5 at t = 15 and then monotonically increases This means that for 5 > a there are two roots, for smaller a there are no roots, for large a there is exactly one root Answer: 5 > a Variant Find, depending on the parameter, the number of solutions of the system x (a+1) x+ a 3= y, y (a +1) y + a 3= x We think The system has the form f(x)= y, f(y)= x, or f(f(x)) = x One of the solutions f(x)= x We find the second solution, subtracting the equations Solution Subtract the second equation from the first equation We get (x + y a)(x y) = 0 Let x = y Substitute into the first equation, transform We get (x a 1) = 4 + a Let x + y = a Substitute into the first equation, transform : (x a) = 3 + a If a<, корней нет Если a =, то x = y = a + 1, единственное решение Если 15 >a >, that is, a pair of solutions x= y =a+ 1± 4+ a If a = 15, then two solutions: x = y = a, x = y = a + If 15< a то решения x= y =a+ 1± 4+ a, x=a± 3+ a, y= a x Ответ: a < нет решений, а = одно, 15 a >, two solutions, a > 15 four solutions VV Shelomovsky Thematic sets, cmdru/
15 15 Option 4 Find all values of a, for each of which the equation 7 x 6 +(4 a x)3 +6 x +8 a=4 x has no roots Thinking 8a 4x = (4a x), 7x6 = (3x)3 This means that the equation includes the sum and the sum of the cubes of the same expressions. This can be used Solution Let's transform the equation to the form (3 x)3 +(4 a x)3+ (3 x + 4 a x)=0 Expand the sum of cubes (3 x +4 a x) ( (3 x) 3 x (4 a x)+(4 a x) +)=0 The second factor is the incomplete square of the difference increased by It is positive. Selecting the square in the first factor, we get 1 1 3(x) + 4 a = This equation does not have roots, if 4 a > 0, a > 3 1 Answer: 1a > 1 Option 8 Find the values a, for each of which the largest value of the function x a x is not less than one Solution If x a, the function f (x, a) \u003d x a x It is maximum for x = 0.5, maximum is 0.5 a At a< 0,5 наибольшее значение функции 0,5 а 1 при 075 а Если x < a, функция f(x,a) = a x x Она максимальна при x = 0,5, максимум равен a + 05 При a >0.5 is the largest value of the function a + 0.5 1 with a 0.75 Answer: a 0.75 or 075 a a, x = 8y + b has even number solutions Solution: It follows from the first equation that y > 0, the second equation can be 8 transformed to the form: y=, x (b; +) Excluding y: x b f (x) = x a = 0; f `(x) = 4 x 3 + x b (x b)3 Each root of the obtained equation generates exactly one solution of the original system< 0 функция f(x) монотонно возрастает от минус бесконечности до f(х1), уменьшается до f(х) и вновь монотонно возрастает при положительных иксах до плюс бесконечности Уравнение может иметь чётное число корней два только если корень совпадает с минимумом или максимумом функции, то есть в точке корня производная равна нулю, то есть f(х1) = g(х1) = 0 Исключая корень из уравнений, найдём: а = (4х1 + х14) Полученная функция имеет максимум при х1 = 1 (а = 3; b = 1,5), поэтому для любого a (0; 3) существуют х1, х х1 и b, при которых число корней равно два Однако при а = 3 х ВВ Шеломовский Тематические комплекты, cmdru/
16 16 \u003d x1, both roots are the same and the equation f (x) \u003d 0 has only one root = x (x b) + 1 = 0 The last equation can have one or two roots, and only with negative x. Thematic kits, cmdru/
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System of tasks on the topic “Tangential Equation” Determine the sign of the slope of the tangent drawn to the graph of the function y f (), at points with abscissas a, b, c a) b) Indicate the points at which the derivative
EQUATIONS AND INEQUALITIES WITH MODULES Gushchin DD www.mathnet.spb.ru 1 0. The simplest equations. To the simplest (not necessarily simple) equations, we will refer to equations solved by one of the following
MODULE “Application of continuity and derivative. Application of the derivative to the study of functions. Application of continuity.. Method of intervals.. Tangent to the graph. Lagrange formula. 4. Application of the derivative
SOLUTION OF THE PROBLEM OF R E A L N O V A R I A N T A E G E - 2001 P O M A T E M A T I K E Part 1 A1. Find the value of the expression. 1. 15 2. 10 3. 5 4. Solution. Answer: 1. A2. Simplify the expression. one.
Methodology for the formation of the competence-based component of the mathematical culture of class students The system for studying educational modules in mathematics I. K. Sirotina, Senior Lecturer of the Department information technologies
Algebra 0 class Topic Trigonometric functions and transformations Basic concepts The letter Z denotes the set of integers: Z (0; ; ; ;) The arcsine of the number a belonging to the interval [- ; ], is called
111 Functions A basic level of Table of contents 11101 Coordinate systems 1110 Function concept 7 1110 Function domain 10 11104 Function range (set) 1 11105 Function increase and decrease
Chapter TESTS T-0 Investigation of a function according to the schedule T-0 Correspondence between the graph of a rational function and the formula T-0 Construction of a graph by properties T-04 Parallel transfer graphics T-05 Symmetrical
Single State exam Math Year 7 Demo Part A Find the value of the expression 6p p when p = Solution Use the power property: Substitute in the resulting expression Correct
Lesson 8 Basic trigonometric formulas (continued) Trigonometric functions Product transformation trigonometric functions to sum Formulas for converting the product of sine and cosine
FUNCTIONS. The concept of a function. Let's say the speed of a person is 5 km/h. If we take the travel time as x hours, and the distance traveled as y km, then the dependence of the distance traveled on the travel time can be
General information USE Profile level Task 0 Problems with parameters Quadratic equations and equations with a square trinomial Dikhtyar MB Equation f (a) x + g(a) x + ϕ (a) = 0, where f (a) 0, is
Around assignments 18 from the Unified State Examination 2017 A.V. Shevkin, [email protected] Annotation: The article analyzes various methods for solving a number of tasks with a parameter. Keywords: equation, inequality, parameter, function,
Curves of the second order Circle Ellipse Hyperbola Parabola Let a rectangular Cartesian coordinate system be given on the plane. A curve of the second order is a set of points whose coordinates satisfy
Various approaches for solving problems С С С5 Unified State Examination 9-years Preparation for the Unified State Examination (material for a lecture for teachers) Prokofiev AA [email protected] Tasks C Example (USE C) Solve the system of equations y si (si) (7 y)
1 Tickets 9 10. Solutions Ticket 9 1. A linear function f(x) is given. It is known that the distance between the intersection points of the graphs y = x and y = f(x) is equal to 10, and the distance between the intersection points of the graphs y =
Department of Mathematics and Informatics Mathematical Analysis Educational and methodological complex for HPE students studying with the use of distance technologies Module 4 Applications of the derivative Compiled by: Associate Professor
Lecture 5 on the plane. Definition. Any straight line on a plane can be given by a first-order equation, and the constants A, B are not equal to zero at the same time. This first order equation is called the general equation.
Grade 8 Decisions 017-018 Task Task 1 Find the sum of cubes of the roots of the equation (x x 7) (x x) 0. To solve the equation, we use the method of changing the variable. Denote y \u003d x + x 7, then x + x \u003d (x
APPLICATION OF THE DERIVATIVE FUNCTION Equation of the tangent Consider the following problem: it is required to write the equation of the tangent l drawn to the graph of the function at a point According to the geometric meaning of the derivative
RESEARCH OF FUNCTIONS Sufficient conditions for the increase and decrease of a function: If the derivative of a differentiable function is positive inside some interval X, then it increases on this interval If
Webinar 7 (6-7) Topic: USE parameters Profile Task 8 Find all parameter values, for each of which the set of function values 5 5 5 contains a segment Find all parameter values, for each
5.0. 014 Classwork. Equations and systems of equations with parameters. An experience entrance exams to universities shows that the solution of equations and inequalities containing parameters causes great difficulties
L.A. Strauss, I.V. Barinova Tasks with a parameter in the Unified State Examination Guidelines y=-x 0 -a- -a x -5 Ulyanovsk 05 Strauss L.A. Tasks with a parameter in the exam [Text]: guidelines/ L.A. Strauss, I.V.
Lecture 13 Topic: Curves of the second order Curves of the second order on the plane: ellipse, hyperbola, parabola. Derivation of equations of curves of the second order based on their geometric properties. Study of the shape of an ellipse,
Mathematics grade 8 2 PROGRAM CONTENT Section 1. Algebraic fractions (24 hours) The concept of algebraic fractions. The main property of an algebraic fraction. Reduction algebraic fractions. Addition and subtraction
Topic 10 "Graphics elementary functions". one. Linear function f(x) = kx + b. The graph is a straight line. 1) Domain of definition D(f) = R.) Domain of values E(f) = R. 3) Zeros of the function y = 0 for x = k/b. 4) Extremes
P0 Derivative Consider some function f () depending on the argument Let this function be defined at the point 0 and some of its neighborhood, continuous at this point and its neighborhood
Problems with parameters (grades 10 11) Parameters are the same numbers, just not known in advance 1 Linear equations and inequalities with parameters Linear function: - equation of a straight line with a slope
Option Find the domain of the function: y + The domain of the given function is determined by the inequality In addition, the denominator should not vanish Find the roots of the denominator: Combining the results
TICKET 15 Fiztekh 017. Tickets 15 16. Solution 1. It is known that for three consecutive natural values of the argument, the quadratic function f(x) takes the values 1, 1, and 5, respectively. Find the smallest
Construction of graphs of functions 1. Plan for the study of a function when plotting a graph 1. Find the domain of the function. It is often useful to consider multiple values of a function. Explore the special properties of a function:
geometric sense derivative Consider the graph of the function y=f(x) and the tangent at the point P 0 (x 0 ; f(x 0)). Find the slope of the tangent to the graph at this point. The angle of inclination of the tangent Р 0
The geometric meaning of the derivative, tangent 1. The figure shows the graph of the function y \u003d f (x) and the tangent to it at the point with the abscissa x 0. Find the value of the derivative of the function f (x) at the point x 0. Value
Ministry of Education and Science Russian Federation Moscow Institute of Physics and Technology ( State University) Correspondence physical and technical school MATHEMATICS Solving problems with parameters (01 015
QUADRATIC EQUATIONS quadratic equations relatively
Equations, inequalities, systems with a parameter Answers to tasks are a word, a phrase, a number or a sequence of words, numbers. Write your answer without spaces, commas, or other extra characters.
PROBLEM SECTION WITH PARAMETERS Comment Tasks with parameters are traditionally complex tasks in USE structure requiring from the applicant not only the possession of all methods and techniques for solving various
Maths. Collection of assignments (April 14, 01). Tasks with -. Problem 1. For what values of the parameter a does the equation have a unique solution 4 + 1 = + a ax x x x a Problem. Find all valid
IV Yakovlev Materials in mathematics MathUs.ru Interval method The interval method is a method for solving so-called rational inequalities. General concept rational inequality we will discuss later, but for now
Differential Calculus Introduction to mathematical analysis Sequence limit and function. Disclosure of uncertainties within. Function derivative. Differentiation rules. Application of the derivative
Part I (Option 609) A Factor under the root sign 8 q A) q 8) q 8) q 8) q 8 8 8 q q Correct answer) Find the value of the expression),5) Correct answer) 9 with a = a a)) 8 A log 8 Find the value
Solutions A Let's draw all these numbers on the number axis. The one that is located to the left of all and is the smallest This number is 4 Answer: 5 A Let's analyze the inequality On the number axis, the set of numbers satisfying
6..N. Derivative 6..H. Derivative. Table of contents 6..0.N. Derivative Introduction.... 6..0.N. Derivative complex function.... 5 6..0.N. Derivatives of functions with modules.... 7 6..0.Н. Ascending and descending
Theorem. A system of linear equations is consistent only if the rank of the extended matrix is equal to the rank of the matrix of the system itself.
Systems of linear equations
Joint r(A)=r() incompatible r(A)≠r().
Thus, systems of linear equations have either an infinite number of solutions, or one solution, or no solutions at all.
End of work -
This topic belongs to:
Elementary matrix transformations. Cramer's method. Vector definition
Two elements of a permutation form an inversion if the larger element precedes the smaller in the notation of the permutation.. there are n different permutations of the nth degree from n numbers. Let's prove this.. the permutation is called even if the total number of inversions is an even number and, accordingly, odd if..
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All topics in this section:
Kronecker-Capelli theorem
Consider a system of linear equations with n unknowns: Compose a matrix and an augmented matrix
The concept of a homogeneous system of linear equations
A system of linear equations in which all free terms are equal to 0, i.e. the species system is called homogeneous
Property of solutions of a homogeneous SLE
Linear combination of solutions homogeneous system equations itself is a solution to this system. x=and y=
Connection between solutions of homogeneous and inhomogeneous systems of linear equations
Consider both systems: I and
An axiomatic approach to the definition of a linear space
Previously, the concept of an n-dimensional vector space was introduced as a collection of ordered systems n-real numbers, for which the operations of addition and multiplication by real h were introduced
Consequences from the axioms
1. Uniqueness of the zero vector 2. Uniqueness of the opposite vector
Proof of Consequences
1. Suppose that. -null
Basis. Dimension. Coordinates
Definition 1. A basis of a linear space L is a system of elements belonging to L that satisfies two conditions: 1) the system
