Solving a system of equations depending on a parameter. Solving systems of linear equations
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 compatible, 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 can be presented - 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 nonzero. Such a system has a unique solution obtained by ;
b) if r< n, то число независимых уравнений меньше числа неизвестных.
Let's move the extra unknowns x r+1 , x r+2 ,..., x n , which are usually called free, to the right sides; our system 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 with respect to x 1, x 2,..., x r, since the determinant of this system (rth order) is nonzero. Giving the free unknowns arbitrary numerical values, we obtain, using 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 has the form:
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 the matrix. This, however, is immediately visible - system (5.5) certainly has a zero, or trivial, solution x 1 = x 2 =... = x n = 0. Let matrix A of system (5.5) have rank r. If r = n, then the zero solution will be the only solution to 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 eigenvector of linear transformation (square matrix A ), if there is a number λ such that the equality
The number λ is called eigenvalue of linear transformation (matrix A ), corresponding to the vector X. The matrix A is of order n. In mathematical economics, the so-called productive matrices. It has been proven that matrix A is productive if and only if all eigenvalues of matrix A are less than one in absolute value. To find the eigenvalues of 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 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 non-zero solutions if and only if the determinant of this system is equal to zero, i.e.
We obtained an equation of the nth degree relative to the unknown λ, which is called characteristic equation of the matrix A, the polynomial is called characteristic polynomial of the matrix A, and its roots are characteristic numbers, or eigenvalues, of the matrix A. To find the eigenmatrix A into the vector equation (A - λE)X = 0 or into 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. Explore 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 matrices A and A|b using the method of elementary transformations, simultaneously bringing the system to a stepwise form:
Obviously, r(A) = r( A|b) = 2. The original system is equivalent to the following, reduced to stepwise form: x 1 + x 2 - 2x 3 - x 4 + x 5 = 1, - 4x 2 + 7x 3 + 7x 4 = 1. Since the determinant for unknowns x 1 And x 2 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, Where x 2 = 7/4 x 3 + 7/4 x 4 -1/4, x 1 = 1/4 x 3 -3/4 x 4 - x 5 + 5/4 is the general solution of a system that has an infinite number of solutions . Giving free unknowns x 3 , x 4 , x 5 specific numerical values, we will obtain particular solutions. For example, when 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 a system of equations and find a 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 this: x 1 + 2x 2 - x 3 + 4x 4 = 2, 5x 2 - 3x 3 + 7x 4 = a-2,
. We have A ~ 
This shows that the system is compatible only for a=5. The general solution in this case is:
x 2 = 3/5 + 3/5x 3 - 7/5x 4, x 1 = 4/5 - 1/5x 3 - 6/5x 4.
Example 2.18. Find out whether 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, at least one of which is non-zero(see paragraph 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 have obtained a system of linear homogeneous equations. We solve it by eliminating the unknowns:
The system is reduced to a stepwise form, equal to 3, which means that the homogeneous system of equations has solutions different from zero (r< n). Определитель при неизвестных x 1, x 2, x 4 is different from zero, so they can be selected as the main ones and the system can be rewritten as:
x 1 + x 2 - 3x 4 = x 5, -2x 2 + 2x 4 = -2x 3 - x 5, - 3x 4 = - x 5.
We have: x 4 = 1/3 x 5, x 2 = 5/6x 5 +x 3, x 1 = 7/6 x 5 -x 3. The system has countless solutions; if free unknown x 3 And x 5 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
Theorem. A system of linear equations is consistent only if the rank of the extended matrix is equal to the rank of the system matrix itself.
Systems of linear equations
Consistent r(A)=r() and incompatible r(A)≠r().
Thus, systems of linear equations have either an infinite number of solutions, one solution, or no solutions at all.
End of work -
This topic belongs to the section:
Elementary matrix transformations. Cramer's method. Vector definition
Two elements of a permutation form an inversion if in the notation of the permutation the larger element precedes the smaller one.. there are n different permutations of the nth degree of n numbers. Let’s prove this.. a 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
Let's consider a system of linear equations with n unknowns: Let's create a matrix and an extended 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. a system of the form is called homogeneous
Property of solutions to a homogeneous SLE
A linear combination of solutions to a homogeneous system of equations is itself a solution to this system.
x=and y=
Relationship between solutions of homogeneous and inhomogeneous systems of linear equations
Let's consider both systems: I and
Axiomatic approach to defining linear space
Previously, the concept of n-dimensional vector space was introduced as a collection of ordered systems of n-real numbers, for which the operations of addition and multiplication by a real number were introduced
Corollaries from the axioms
1. Uniqueness of the zero vector 2. Uniqueness of the opposite vector
Proof of consequences
1. Let's assume that. -zero
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
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1 1 Number of solutions to a system of equations Graphical dynamic method To find the number of solutions to a system of equations containing a parameter, the following technique is useful. We construct 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 of the system. Next, mentally change parameter and imagine how the graph of an equation with a parameter is transformed, how common points of the graphs appear and disappear. Such research requires a developed imagination. To train the imagination, let’s consider a number of typical problems. Let us call special values of the parameter those values at which the number of solutions changes. These values correspond to situations where the graphs of solutions 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 with a slight change in the parameter. Consider problems in which you need 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 the systems x and y. The numbers xi, yi, r are considered given constants. During each solution, we build graphs of both equations. We study how the graph of the equation with the parameter changes at changing the value of the parameter. 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 the InMA file 11, 5 Number solutions of a system with a parameter For research (task 8), use the GInMA file Number of solutions of a system with a 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 to the system VV Shelomovsky Thematic kits, 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 = a Solution: The graph of the first equation is a circle of radius r with center at point O(x0; y0) The graph of the second equation is a circle of radius a with center on the x-axis at point A(x1 ; 0) The center of the circle is stationary, the radius determines the parameter. When the modulus of the parameter increases, the circle “inflates” Special values of the parameter are 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. Condition for the circles to touch the modulus of the sum or difference. radii of the circles is equal to the intercenter distance: a ± r = AO a = ± AO ± r Research: By changing the value of the variables and the parameter, find the number of solutions to the system. It is advisable 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, for both small and large modulus values of the parameter, there are no solutions Since two divergent circles can have no more than two common points, the number of solutions in the general case is no more than two. At points of tangency, the number of solutions is equal to one, with intermediate values of the parameter two. Creative task Find the value of the parameter at which three different points (x 1) + (y y0) = 9; are solutions to the system of equations (x x1) + y = a (x x0) + (y y0) = r; Task Find the number of solutions to the system y = kx + a Solution: The graph of the first equation is a circle of radius r with the center at the point O(x0; y0) The graph of the second equation is a family of parallel lines passing through points A(0; a) and having a constant slope The tangent of the angle of inclination of the straight lines is equal to k. As the parameter increases, the straight lines move upward. 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 straight lines touch the circle. We find the tangency condition by equating the tangents of the inclination angle of the circle and the straight line. VV Shelomovsky Thematic sets, cmdru/
3 3 Solving the resulting equation, we find the coordinates of the two tangent points: kr x = x0 ± ; x0 x 1 + k = k k (y y0) + (y y0) = r r y y0 y = y0 1+ k Substituting the obtained expressions into the equation of the straight line, we find the value of the parameter at the singular points: a = y 0 kx0 ± r 1 + k Research : By changing the value of the variables and the parameter, find the number of solutions to the system. It is advisable to start the study with the simplest case k = 0, when the lines are parallel to the abscissa axis. Then consider cases when the root is extracted (for example, k = 3), pay attention to the popular case k = 1 For small and for large values of the parameter there are no solutions 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 values of the parameter 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 equations has no more than one solution Find the value of the parameter at 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 = ax + y1 Solution: The graph of the first equation is a circle of radius r with the center 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 angle of inclination 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 x-axis increases. Special values of the parameter are those values at which the number of roots changes, that is, parameter values at which the straight lines touch the circle. If point A(0; y1) is located inside the circle , then any possible straight line intersects the circle at two points. We find the tangency condition by equating the tangents of the angle of inclination of the circle and the straight line. Solving the resulting equation, we find the coordinates of two points of tangency: VV Shelomovsky Thematic kits, cmdru/
4 4 ar x = x0 ± ; x0 x 1 + a = a a (y y0) + (y y0) = r r y y0 y = y0 1+ a Substituting the resulting expressions into the equation of the straight line, we find the value of the parameter at (y1 y 0) r singular points If x0 = 0, then special values of the parameter a = ± r If y0 = y1, x0 r, then special values of the parameter a = ± (y1 y 0) r r x0 If x0 = ± r, then the circle is tangent to the vertical line passing through the point r (y1 y 0) A(0; y1) and the value of the parameter a = In other cases x0 (y1 y 0) a= x0 (y 0 y1) ± r (x0 + (y 0 y1) r) r x0 Research: By changing the value of the variables and the parameter, find the number of solutions to the system. It is advisable 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 identical in magnitude, but different in abscissa sign ±x0 The graphs are shown in blue and purple The graph of the second equation is a circle of radius a with a center on the x-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 of tangency: sum or difference radii of circles is equal to the center-to-center distance: a ± r = AO, a ± r = AQ Research: By changing the value of the variables and the parameter, find the number of solutions to the system. Use integer values for one center-to-center distance (for example, x0 = 6, y0 = 3, r = 3 , x1 =) It is typical that for small absolute and large values of the parameter there are no solutions. At points of tangency, 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 certain value of the parameter has exactly two solutions. 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 = y0, vertical axis of symmetry x = x0 Center of symmetry point (x0, y0) The second graph is a circle of radius a, the center of which is located at the center of symmetry of the parabolas. The number of roots changes at a value of the parameter at which the circle of the second graph touches the vertices of the parabolas. At the point of contact: x = x0, y = y0 ± r = y = y0 ± a, therefore a = ± r The number of roots changes at the value of the parameter at which the circle of the second graph with parabolas touches internally. To find this value, go 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 a parameter value at which the circle and 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 to the system. Use the values r = 1, 4 and 9. Please note that the parameters x0 and y0 do not affect the answer to the problem. For small absolute and large values of the parameter, there are no solutions x x0 + y y0 = r; 6 Find the number of solutions to 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 equal to 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 internally touches the sides of the square. To find this value, go 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. Moreover, 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 kits, cmdru/
7 7 (x x0) + (y y0) = r ; 7 Find the number of solutions to the system y = 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 common beginning this is a “bird, wings up”, the top of the graph is located at point A(a; y1) The number of roots changes with the value of the parameter at which the “wing” of the second graph touches the circle or the top of the graph lies on this circle Tangent of the angle of inclination of the “right wing” to the abscissa axis is equal to 1, which means the straight line, r x = x ±, k 0 containing 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 upward, a condition is added that the ordinate of the vertex should not be greater than the ordinate of the point of contact, that is, y1 yk y0 y1 ± r We similarly write the conditions for contact with the “left wing” » If the vertex of the graph lies on a circle, then its coordinates satisfy the equation of the circle: (a x0) + (y1 y0) = r By changing the value of the parameter, examine the number of 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 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, a certain value of the 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 to the system Define the functions yourself according to the example and explore the number of solutions VV Shelomovsky Thematic kits, cmdru/
8 8 BB Shelomovsky Thematic kits, cmdru/
9 9 Tasks C5 (Semenov Yashchenko) Option 1 Find all values of a, for each of which the set of solutions to the inequality 4 x 1 x+ 3 a 3 is the segment 3 a 4 x Thinking Let’s carry out the 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 plane x 3a 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 3a = 8, then x [ 4.0] 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 interval 3 a 4 x 3 a+ 4 3 a +4 Let 1 > 3a 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 kits, 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 has the form 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 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 This means that 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, then there is a root For other values of a x = f (a, x) f (a, 1)>0 There are no roots Answer: [ 5 15, 5+ 15] Option 5 Find all values of a for each of which the equation a +11 x+ +3 x + 4 x + has at least one root 13=5 a+ x a + We 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: [ , ] Option 9 Find the number of roots of the equation x + 4x 5 3a = x + a 1 Let's think Let's consider the following (obvious) statement known Let the functions f(x) and g(x) be given on a certain interval Let the derivative of one be greater on the interval than the other Let the difference in the values of the functions on the left end has one sign, on the right another. Then the equation f(x) = g(x) has exactly one root on the interval. Solution Let us denote f(x, a) = 3a + x + a, g(x) = x + 4x Equation f(x, a) = g(x) VV Shelomovsky Thematic kits, 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 When For large values of x, 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 Function g(x) for x > 1 monotonically increases with a coefficient greater than 6 Due to symmetry, the function g(x) monotonically decreases with a coefficient 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(а, a) = 3а, f(5, a) = 3а + 5 a, f(, a) = 3а + a, f(1, a) = 3а + 1+ a Graphs f (x, a) and g(x) touch if their slopes are equal. Touching is possible at x = 5. Moreover, g(x) = 39/4 f(x, a) = 4a + x = 39/4, 4a = 49 /4, a = 49/16 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) increases faster than f(x, a), that is, f(x, a) everywhere< 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(), there are 4 roots, one or 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: there are 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 with 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. This means that 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 broken line 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 at 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, there is at least one solution system of equations (1+ 3 x)a +(b 4 b+5) y =, x y +(b) x y+ a + a=3 Think The system has the form (1+ 3 x)a +(1+(b) ) y =, Convenient 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 visible, the corresponding values of the parameter a = 1 and a = 3 analyze the singular point b = Then (1+ 3 x)a +(1+(b)) y =, x y +(b) x y=4 (a+ 1) Solution We write the system in the form 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 obtain that y = 0 At the same time, the second equation has no solution Answer: 1 or 3 VV Shelomovsky Thematic kits, cmdru/
13 13 Option 14 Find all values of the parameter, for each of which the modulus of the difference between the roots of the equation x 6x a 4a = 0 takes highest value Solution We 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 in roots is equal to 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 segment t [ π ; 15 π ], then subtract 4π from each solution obtained. Transform the equation to the form + 4 k sin t cos t =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 goes to zero at 4tgt = 1, that is, at sin t = 1 4, cos t = The numerator at t = π is equal to 1, at 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. This means that 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 Find all values of the parameter, for each of which the system of equations (x a 5) +(y 3 a +5) =16, (x a) +(y a+1)=81 has a unique solution. Let’s think Each equation describes a circle The solution is unique in the case of tangent circles Solution The first equation defines a circle with center at point (a + 5, 3a 5) and radius 4 The second equation is a circle with center at point (a +, a 1) with radius 9 BB Shelomovsky Thematic kits, cmdru/
14 14 The system has a unique solution if the circles touch. In this case, the distance between centers is equal to = 13 or 0 4 = 5 Square of the center-to-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 for each of which there are 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 Let us denote t = 5x 1 Due to the monotonicity of the exponential function 5x, each root t 1 generates exactly one root x 0 The equation will take the form 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 increases monotonically, 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 increases monotonically. 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 Option 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 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 by subtracting the equations Solution Subtract the second 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 kits, 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 Thinking 8a 4x = (4a x), 7x6 = (3x)3 This means that the equation includes the sum and the sum of 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 Let's 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. Isolating 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 of a for each of which the greatest value of the function x a x is not less than one Solution If x a, the function f(x,a) = x a x It is maximum at 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 the largest value of the function a + 0.5 1 at a 0.75 Answer: a 0.75 or 075 a Pair of functions Find the range positive values a, for each of which there is a b such that the system of equations: y = x4 + a, x = 8y + b has an even number of solutions Solution: From the first equation it follows that y > 0, the second equation can be transformed to the form: y =, x (b; +) Let's remove y: x b f (x) = x a = 0; f `(x) = 4 x 3 + x b (x b)3 Each root of the resulting equation generates exactly one solution to the original system. For b 0, the function f(x) is monotonically increasing and the equation has exactly one root. For negative b< 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 = x1, both roots coincide and the equation f(x) = 0 has only one root. The derivative f`(x) is positive at x b and at x + It is equal to zero under the condition f`(x) = 0 g (x) = x (x b) + 1 = 0 The last equation can have one or two roots, and only for negative x Let’s denote them x1 and x: g(x1) = g(x) = 0 Answer: a (0; 3) VV Shelomovsky Thematic kits, cmdru/
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QUADRATE EQUATIONS Contents QUADRATE EQUATIONS... 4. and study of quadratic equations... 4.. Quadratic equation with numerical coefficients... 4.. Solve and explore quadratic equations relatively
Equations, inequalities, systems with a parameter Answers to tasks are a word, phrase, number or sequence of words, numbers. Write your answer without spaces, commas or other additional characters.
SECTION PROBLEMS WITH PARAMETERS Comment Problems with parameters are traditionally complex tasks in the structure of the Unified State Examination, requiring the applicant not only to master all the methods and techniques for solving various
Mathematics. Collection of assignments (April 14, 01). Problems with parameter -. Problem 1. For what values of the parameter a is there a unique solution to the equation 4 + 1 = + a ax x x x a Problem. Find all valid
I. V. Yakovlev Materials on mathematics MathUs.ru Interval method The interval method is a method for solving so-called rational inequalities. General concept we will discuss rational inequality later, but now
Differential Calculus Introduction to mathematical analysis Limit of sequence and function. Uncovering uncertainties within limits. Derivative of a function. Rules of differentiation. Application of derivative
Part I (Option 609) A Enter a factor under the sign of the root 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 depict all the given numbers on the number line. 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 line there is a set of numbers that satisfy
6..N. Derivative 6..N. Derivative. Contents 6..0.N. Derivative Introduction.... 6..0.N. Derivative complex function.... 5 6..0.N. Derivatives of functions with modules.... 7 6..0.N. Ascending and descending
c) (xe+y"=1, d) (x"+y"=2a - 1,
(xy=a; (xy=a - 1?
9.198. Find the number of solutions to the system of equations ((x(+)y~=!,
depending on parameter a.
9.199. How many solutions does the system of equations have depending on a:
a) (x"+y"=9, b) (x"+y"+!Ox=0,
(~x~ =y - a; (y=~x - a~?
9.200. At what values of parameter a does the system of equations
has three solutions? Find these solutions.
9.201. At what values of the parameter p does the system of equations
(ру+х) (х - р УЗ)=О
has three solutions?
9.202. At what values of parameter b does the system of equations
a) 1 ~x~ +4)y~ = b, b) 1 x~ +2 ~y(= 1, c) (~y! +x =4
! ~y!+xr=1 ! ~y!+xr=b (x +Y =b
has four different solutions?
9.208. At what values of the parameter c does the system of equations
has eight different solutions?
9.204. Solve the system of equations
where a)O, and prove that if a is an integer, then for
For each solution (x; y) of a given system, the number 1+xy is the square of an integer.
9.205. At what values of parameter a does the system of equations
x"+ y"+ 2xy - bx - bu+ 10 - a = O,
x"+ y" - 2xy - 2x+ 2Y+ a = O
has at least one solution?
Solve the system for the found values of a.
9.206. Find all values of the parameter a for which the system
equations (x"+(y - 2)"=1, has at least one solution.
9.207. Find all values of the parameter a for which the circles x" + d" = 1 and (x - a) " + d" = 4 touch.
9.208. Find all values of the parameter a (a>O) for which the circles x"+d"=1 and (x - 3)"+(d - 4)"=a" touch.
Find the coordinates of the point of contact.
9.209. Find all values of a (a>0) for which the circle
x"+d"=a" touches the line 3x+4d=12. Find the coordinates of the point of contact.
D" - 2x+ 4d = 21. Find the coordinates of the intersection points
straight line and circle.
9.211. At what value of parameter a will the straight line ed=x+1 be
pass through the center of the circle (x - 1) + (d - a)"=8?
Find the coordinates of the intersection points of the line and the circle.
9 212. It is known that the straight line d = 12x - 9 and the parabola d = ax" have
only one common point. Find the coordinates of this point.
9.213. At what values of b and z (b>0, z>0) does the circle
(x - 1)"+(d - b)"=g" will touch the straight lines d=0 and d= - x?
Find the coordinates of the touch points.
9.214. Draw a set of points on the coordinate plane with
coordinates (a; b) such that the system of equations
has at least one solution.
9.215. At what values of parameter a does the system of equations
a (x"+ 1) = d - ~ x ~ + 1,
has a single solution?
9 1O. TEXT PROBLEMS
Word problems are usually solved according to the following scheme: unknowns are chosen; make up an equation or a system of equations, and in some problems - an inequality or a system of inequalities; solve the resulting system (sometimes it is enough to find some combination of unknowns from the system, and not solve it in the usual sense).
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 to these equations. In such cases, they say that it is necessary to solve the system of equations xy – 6 = 0 and y – x – 1 = 0.
It is customary to write a system of equations using curly braces. 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 makes each equation of the system true is called a solution to a system of equations with two variables.
Solving a system of equations means finding many of its solutions.
Let us consider systems of two linear equations with two variables, in which in each equation at least one of the coefficients is different from zero.
The graphical solution of systems of this type comes down to finding the coordinates of the common points of two straight lines.
As you know, two lines on a plane can be intersecting or parallel. In the case of parallelism, the lines either do not have 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 = -3x – 11,
(y = 0.5x – 4.
The angular coefficients of the straight lines - the graphs of the equations of the system - are different (-3 and 0.5), which means that the straight lines intersect.
The coordinates of their intersection point are the solution to this system, the only solution.
Example 2.
Let's solve the system of equations:
(3x – 2y = 12,
(6x – 4y = 11.
Solution.
Expressing y through x from each equation, we obtain the system:
(y = 1.5x – 6,
(y = 1.5x – 2.75.
The straight lines y = 1.5x – 6 and y = 1.5x – 2.75 have equal angular coefficients, which means these lines are parallel, and the straight line y = 1.5x – 6 intersects the y axis at point (0; -6), and straight line y = 1.5x – 2.75 – at point (0; -2.75), therefore, straight lines do not have common points. Therefore, the system of equations has no solutions.
The fact that this system has no solutions can be seen by reasoning 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 sides 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 = 24,
(6x – 4y = 11.
has no solutions, which means 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, therefore the coordinates of any point on the graph will satisfy each of the equations of the system, that is, they will be a solution to 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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