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PHYSICO-MATHEMATICAL SCIENCES
METHODS FOR SOLVING DEPENDING ON PARAMETERS Saipnazarov ShA.1, Musayeva Z.A.2 (Republic of Uzbekistan) Email: Saipnazarov572@scientifictext.ru
1Saipnazarov Shaylovbek Aktamovich - Candidate of Pedagogical Sciences, Docent; 2Musayeva Ziyoda Allayarovna - Senior Lecturer, DEPARTMENT HIGHER MATHEMATICS, TASHKENT UNIVERSITY OF ECONOMICS, TASHKENT, REPUBLIC OF UZBEKISTAN
Abstract: parameter problems are difficult because there is no single algorithm for solving them. The specificity of such problems is that, along with unknown values, they include parameters whose numerical values are not specified specifically, but are considered known and given on a certain numerical set. The article shown examples of decision problems with the parameters. It is clear that the roots of equation can not meet outside the areas of determining the equation. In all considered problems sufficient for the desired value parameter established as a result of checks. Keywords: parameter, the necessary conditions, sufficient conditions, root, equation.
МЕТОДЫ РЕШЕНИЯ В ЗАВИСИМОСТИ ОТ ПАРАМЕТРОВ Саипназаров Ш.А.1, Мусаева З.А.2 (Республика Узбекистан)
1Саипназаров Шайловбек Актамович - кандидат педагогических наук, доцент; 2Мусаева Зиёда Аллаяровна - старший преподаватель, кафедра высшей математики, Ташкентский экономический университет, г. Ташкент, Республика Узбекистан
Аннотация: задачи с параметрами являются сложными потому, что не существует единого алгоритма их решения. Спецификой подобных задач является то, что наряду с неизвестными величинами в них фигурируют параметры, численные значения которых не указаны конкретно, но считаются известными и заданными на некотором числовом множестве. В статье показаны примеры решения задач с параметрами. Ясно, что корни уравнения не могут встретиться за пределами областей определения уравнения. Во всех рассмотренных задачах достаточные условия для искомых значений параметров устанавливались в результате проверки.
Ключевые слова: параметр, необходимые условия, достаточные условия, корень, уравнение. 1. Extreme task
A large number of tasks reduces to finding the largest and smallest values of quadratic function. Since the expression ax2 + bx + c, a ^ 0 , can be represented in the form of [1]
2 A f b )2 b2
ax + bx + c = a\ x ч--I + c--
^ 2a) 4a
then it follows that:
a) at a > 0 function y = ax2 + bx + c has the smallest value equal to c _ !bL, and it is
4a
achieved at x = —— 2a
2 h 2 b) at a < 0 function y = ax + hx + c has the largest value equal to c _ _, and it
4a
achieved at
c
Task 1. For each value A(0 < A < 2) find the smallest value of expression
2
2 x2 + _ A(2 x + y) provided that [2]
tg
§ xy | = 0
v •
The solution. The smallest value of this expression M is achieved, obviously, with non-negative values of x and y - otherwise changing the sign of the negative value of the variable, we could reduce the value of M. Therefore, we will look for the smallest value of expression
N = 2M + 2A2 = (z — A)2 + (y — A)2 for non-negative values of z and y where z = 2x, provided that zy = n, where n - is an arbitrary positive integer or zero. [3] Expression N is the
square of the distance from point K(A, A) to the variable coordinate axes or on the positive
branch of one of the hyperbolas zy = 4n (n > 0). Tangent l to hyperbolas zy = 4 at point
p(2, 2) has equation z + y = 4 and everywhere, with the lies below the hyperbola: for z ^ 2
point (z, y), we have hyperbola z + y = z H— > 4 . On the other hand, a circle centered at
z
point K passing through the point P also touches the line l, but is located below the branches of all hyperbolas zy = 4n lying on opposite of the line l.
It follows that the distance from the point K to any point of the hyperbola zy = 4n is greater than or equal to the length of the segment KP. Therefore, the smallest value of N is equal to the numbers KP = (A — 2)2 + (A — 2)2 and the A2 — distance from point K to the coordinate axes. Since the 0 < A < 2 and
2(A — 2)2 < A2 « A2 — 8A + 8 < 0 «
4 — 2V2 < A < 4 + 2V2, then the smallest value of N 4 — 2^2 < A < 2 is 2(A — 2)2, and at 0 < A < 4 — 242 equals A . The corresponding value of expression M are equal
A2
4 — 4A and--.
2
2. Equations
Task 2. Determine A so that the sum of the squares of all solutions of equation log A\x — 2 A + log Ax = 2 is 4.
The solution. Given the equation is equivalent to the equation
x|x — 2 A| = A2, to address which should find positive roots equation
x2 (x — 2 A)2 = A4,
or
(x - 2Ax + A2)(x2 - 2Ax - A2)= 0.
Since the A > 0, this equation has a positive roots A and a(i + 42), where we obtain that
the sum of squares roots is equal 4 only at A = 42-V2 . Task 3. Find all values A, each of which equation
((x - A -1)2 - 2(x - A -1)2 = A2 -1 has a more positive roots than negative.
The solution. Transferring all members of the equation in the right part, be represented equation in the form of f (x) • g(x) = 0, where
f (x) = x 2 - 2(A + l)x + A(A +1),
g(x) = x2 - 2(A + l)x + A(A + 3).
Trinomial f (x) has roots in A > -1. At A > 0 its roots are positive, with -1 < A < 0
roots different characters ("special" value of A = 0 and A = 1 we consider separately). These facts schematically depicted in the figure (sec. fig. 1). Together with similar statements for three
trusted g(x). Then in figure we find that (excluding "special" values 1, 0, -1 and -3) the decisions
of tasks are values A > 0.
m
Fig. 1. With roots different characters
Of the "special" values, the condition of the problem is A > 0.
Task 4. For what integer values A and B, the product of two of the roots of equation
x4 + Ax3 + 6x2 + Ax +1 = 0
is 3?
The solution. Suppose that at some of entire values of A and B the product of the roots of p and q given equation is 3, and p ^ q.
Since 0 is not the root of the equation, the equation can be rewritten as
^x2 +-! 1 + A
c
2
V x
2
V
x +1 | + B = 0, x J
1 1 1
or y2 + Ay + B - 2 = 0 , where y = x + — . Number of p +--and q + — are the roots
x p q
of this equation, and since the
11 11 p - q
p + — = q + p - q =---^ p - q = —— ^ p = q,
pq q q p 3
then by theorem viete
j ! í i Y i Л
p + — + q + — = - A, p + — q + — = B - 2
p q \ p A q)
3A
Due to the fact that the pq = 3 , from the first equality get p + q = ——, and of the second
equality p + q = B -16 . So p2 + q2 = — - 6, p2 + q2 = 3B -16, i.e.
q p 3 16
9 A = 40B -160 . However, as much as A and B this equality impracticable, since the 160 is
not divided into 3. If p is equal to q, then both of then are equal to s , or . But s is the
root of the given equation when the equality
9 + 3aV3 + 3B + AV3
+1 = 0, location (as - number of irrational) that
A = 0, 3B +10 = 0, that is not possible for A whole B. Thus, the desired values of A and B does not exist.
References / Список литературы
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2. KushnirI.A. Geometric solutions of non-geometric problems // kvant, 1989. № 11. P. 61-63.
3. Boltyansky V.G. Coordinate direct as a means of clarity // Math at school, 1978. № 1. P. 13-18.
4. Genkin G.Z. Geometric solutions of algebraic problems // Math at school, 2001. № 7. P. 61-66.
5. Atanasyan L.S. Geometry. Part I. Textbook for students ped. institutes of Education, 1973. 480 pages.
6. Saipnazarov Sh.A., Gulamov A. // Analytic and graphical methods for the analysis of equations and their analysis. Physics, Mathematics and Informatics, 2016. № 3. P. 56-60.
