The purpose of this work is to study various ways solving problems with parameters. The ability and ability to solve problems with parameters demonstrate mastery of methods for solving equations and inequalities, a meaningful understanding of theoretical information, the level logical thinking, stimulate cognitive activity. Longer efforts are needed to develop these skills, which is why in profile grades 10-11 with in-depth study exact sciences a course was introduced: “Mathematical Practice”, part of which is the solution of equations and inequalities with parameters. The course is one of the disciplines included in the curriculum component of the school.
The successful study of methods for solving problems with parameters can be helped by elective or optional courses, or a component behind a grid on the topic: “Problems with parameters”.
Consider four large classes of problems with parameters:
- Equations, inequalities and their systems that need to be solved for any parameter value, or for parameter values that belong to a certain set.
- Equations, inequalities and their systems, for which it is required to determine the number of solutions depending on the value of the parameter.
- Equations, inequalities and their systems, for which it is required to find all those values of the parameter for which the indicated equations (systems, inequalities) have a given number of solutions.
- Equations, inequalities and their systems, for which, for the desired values of the parameter, the set of solutions satisfies the given conditions in the domain of definition.
Methods for solving problems with parameters.
1. Analytical method.
This is a direct solution method that repeats the standard procedures for finding an answer in problems without a parameter.
Example 1: Find all parameter values a, for which the equation:
(2a – 1)x 2 + ax + (2a – 3) =0 has at most one root.
At 2 a– 1 = 0 this equation is not quadratic, so the case a=1/2 are analyzed separately.
If a= 1/2, then the equation becomes 1/2 x– 2 = 0, it has one root.
If a≠ 1/2, then the equation is quadratic; for it to have at most one root it is necessary and sufficient that the discriminant be nonpositive:
D= a 2 – 4(2a – 1)(2a – 3) = -15a 2 + 32a – 12;
In order to write down the final answer, it is necessary to understand
2. Graphical method.
Depending on the task (with a variable x and parameter a) graphs are considered in the coordinate plane ( x;y) or in the plane ( x;a).
Example 2. For each parameter value a quantify the solutions of the equation 
.
Note that the number of solutions to the equation equal to the number of intersection points of function graphs 
and y = a.
Function Graph 
shown in Fig.1.
y=a is a horizontal line. According to the graph, it is easy to establish the number of intersection points depending on a(for example, when a= 11 – two points of intersection; at a= 2 – eight points of intersection).
Answer: when a < 0 – решений нет; при a= 0 and a= 25/4 – four solutions; at 0< a < 6 – восемь решений; при a= 6 – seven solutions; at
6 < a < 25/4 – шесть решений; при a> 25/4 - two solutions.
3. Method of decision regarding the parameter.
When solving this way, the variables X and a are taken equal, and the variable with respect to which the analytical solution becomes simpler is selected. After simplifications, you need to return to the original meaning of the variables X and a and complete the solution.
Example 3: Find all parameter values a, for each of which the equation = - ax +3a+2 has a unique solution.
We will solve this equation by change of variables. Let = t , t≥ 0 then x = t 2 + 8 and the equation becomes at 2 +t + 5a– 2 = 0 . Now the task is to find all a, for which the equation at 2 +t + 5a– 2 = 0 has a unique non-negative solution. This takes place in the following cases.
1) If a= 0, then the equation has a unique solution t = 2.
Solution of some types of equations and inequalities with parameters.
Tasks with parameters help in the formation of logical thinking, in acquiring research skills.
The solution of each problem is unique and requires an individual, non-standard approach, since there is no single way to solve such problems.
. Linear equations.Task number 1. For what values of the parameter b equation has no roots?
Task number 2. Find all parameter values a, for which the set of solutions to the inequality:
contains the number 6, and also contains two segments of length 6 that do not have common points.
Let us transform both sides of the inequality.
In order for the set of solutions to the inequality to contain the number 6, it is necessary and sufficient that the following condition be satisfied: 
Fig.4
At a> 6 set of solutions to the inequality: 
.
The interval (0;5) cannot contain any segment of length 6. Hence, two non-intersecting segments of length 6 must be contained in the interval (5; a).
. exponential equations, inequalities and systems.Task number 3. In the domain of the function definition 
Take all positive integers and add them up. Find all values for which such a sum is greater than 5 but less than 10.
1) Graph of a linear-fractional function 
is a hyperbole. By condition x> 0. With an unlimited increase X fraction decreases monotonically and approaches zero, and the values of the function z increase and approach 5. In addition, z(0) = 1.
2) By definition of the degree, the domain of definition D(y) consists of solutions to the inequality . At a= 1 we get an inequality that has no solutions. Therefore, the function at nowhere defined.
3) At 0< a< 1 exponential function with base a decreases and the inequality is equivalent to the inequality . Because x> 0 , then z(x) > z(0) = 1 . So each positive value X is a solution to the inequality . Therefore, for such a the amount specified in the condition cannot be found.
4) When a> 1 exponential function with base a increases and the inequality is equivalent to the inequality . If a≥ 5, then any positive number is its solution, and the sum specified in the condition cannot be found. If 1< a < 5, то множество положительных решений – это интервал (0;x 0) , where a = z(x 0) .
5) Integers are located in this interval in a row, starting from 1. Let's calculate the sums of consecutive natural numbers, starting from 1: 1; 1+2 = 3; 1+2+3 = 6; 1+2+3+4 = 10;… Therefore, the indicated sum will be greater than 5 and less than 10 only if the number 3 lies in the interval (0; x 0), and the number 4 does not lie in this interval. So 3< x 0 ≤ 4 . Since it increases by , then z(3) < z(x 0) ≤ z(4) .
The solution of irrational equations and inequalities, as well as equations, inequalities and systems containing modules are considered in Annex 1.
Problems with parameters are complex because there is no single algorithm for solving them. The specificity of such problems is that, along with unknown quantities, they include parameters whose numerical values are not specified specifically, but are considered known and given on a certain numerical set. At the same time, the values of the parameters significantly affect the logical and technical course of solving the problem and the form of the answer.
According to statistics, many of the graduates do not start solving problems with parameters for the USE. According to FIPI, only 10% of graduates start solving such problems, and the percentage of their correct solution is low: 2–3%, so the acquisition of skills in solving difficult, non-standard tasks, including tasks with parameters, by school students is still relevant.
Type equation f(x; a) = 0 is called variable equation X and parameter a.
Solve an equation with a parameter a This means that for every value a find values X satisfying this equation.
Example 1 Oh= 0
Example 2 Oh = a
Example 3
x + 2 = ax
x - ax \u003d -2
x (1 - a) \u003d -2
If 1 - a= 0, i.e. a= 1, then X 0 = -2 no roots
If 1 - a 0, i.e. a 1, then X =
Example 4
(a 2 – 1) X = 2a 2 + a – 3
(a – 1)(a + 1)X = 2(a – 1)(a – 1,5)
(a – 1)(a + 1)X = (1a – 3)(a – 1)
If a= 1, then 0 X = 0
X- any real number
If a= -1, then 0 X = -2
no roots
If a 1, a-1 then X= (the only solution).
This means that for every valid value a matches a single value X.
For instance:
if a= 5, then X = = ;
if a= 0, then X= 3 etc.
Didactic material
1. Oh = X + 3
2. 4 + Oh = 3X – 1
3. a = +
at a= 1 there are no roots.
at a= 3 no roots.
at a = 1 X any real number except X = 1
at a = -1, a= 0 there are no solutions.
at a = 0, a= 2 no solutions.
at a = -3, a = 0, 5, a= -2 no solutions
at a = -With, With= 0 there are no solutions.
Quadratic equations with a parameter
Example 1 solve the equation
(a – 1)X 2 = 2(2a + 1)X + 4a + 3 = 0
At a = 1 6X + 7 = 0
When a 1 select those values of the parameter for which D goes to zero.
D = (2(2 a + 1)) 2 – 4(a – 1)(4a + 30 = 16a 2 + 16a + 4 – 4(4a 2 + 3a – 4a – 3) = 16a 2 + 16a + 4 – 16a 2 + 4a + 12 = 20a + 16
20a + 16 = 0
20a = -16
If a < -4/5, то D < 0, уравнение имеет действительный корень.
If a> -4/5 and a 1, then D > 0,
X = 
If a= 4/5, then D = 0,
Example 2 At what values of the parameter a the equation
x 2 + 2( a + 1)X + 9a– 5 = 0 has 2 different negative roots?
D = 4( a + 1) 2 – 4(9a – 5) = 4a 2 – 28a + 24 = 4(a – 1)(a – 6)
4(a – 1)(a – 6) > 0
according to t. Vieta: X 1 + X 2 = -2(a + 1)
X 1 X 2 = 9a – 5
By condition X 1 < 0, X 2 < 0 то –2(a + 1) < 0 и 9a – 5 > 0
| Eventually | 4(a – 1)(a – 6) > 0 - 2(a + 1) < 0 9a – 5 > 0 |
a < 1: а > 6 a > - 1 a > 5/9 |
 (Rice. one) < a < 1, либо a > 6 |
Example 3 Find values a for which this equation has a solution.
x 2 - 2( a – 1)X + 2a + 1 = 0
D = 4( a – 1) 2 – 4(2a + 10 = 4a 2 – 8a + 4 – 8a – 4 = 4a 2 – 16a
4a 2 – 16 0
4a(a – 4) 0
a( a – 4)) 0
a( a – 4) = 0
a = 0 or a – 4 = 0
a = 4
(Rice. 2)
Answer: a 0 and a 4
Didactic material
1. At what value a the equation Oh 2 – (a + 1) X + 2a– 1 = 0 has one root?
2. At what value a the equation ( a + 2) X 2 + 2(a + 2)X+ 2 = 0 has one root?
3. For what values of a is the equation ( a 2 – 6a + 8) X 2 + (a 2 – 4) X + (10 – 3a – a 2) = 0 has more than two roots?
4. For what values of a equation 2 X 2 + X – a= 0 has at least one common root with equation 2 X 2 – 7X + 6 = 0?
5. For what values of a do the equations X 2 +Oh+ 1 = 0 and X 2 + X + a= 0 have at least one common root?
1. When a = - 1/7, a = 0, a = 1
2. When a = 0
3. When a = 2
4. When a = 10
5. When a = - 2
Exponential Equations with a Parameter
Example 1.Find all values a, for which the equation
9 x - ( a+ 2) * 3 x-1 / x +2 a*3 -2/x = 0 (1) has exactly two roots.
Solution. Multiplying both sides of equation (1) by 3 2/x, we obtain an equivalent equation
3 2(x+1/x) – ( a+ 2) * 3 x + 1 / x + 2 a = 0 (2)
Let 3 x+1/x = at, then equation (2) takes the form at 2 – (a + 2)at + 2a= 0, or
(at – 2)(at – a) = 0, whence at 1 =2, at 2 = a.
If at= 2, i.e. 3 x + 1/x = 2 then X + 1/X= log 3 2 , or X 2 – X log 3 2 + 1 = 0.
This equation has no real roots because it D= log 2 3 2 – 4< 0.
If at = a, i.e. 3 x+1/x = a then X + 1/X= log 3 a, or X 2 –X log 3 a + 1 = 0. (3)
Equation (3) has exactly two roots if and only if
D = log 2 3 2 – 4 > 0, or |log 3 a| > 2.
If log 3 a > 2, then a> 9, and if log 3 a< -2, то 0 < a < 1/9.
Answer: 0< a < 1/9, a > 9.
Example 2. At what values of a equation 2 2x - ( a - 3) 2 x - 3 a= 0 has solutions?
In order to given equation has solutions, it is necessary and sufficient that the equation t 2 – (a - 3) t – 3a= 0 has at least one positive root. Let's find the roots using Vieta's theorem: X 1 = -3, X 2 = a = >
a is a positive number.
Answer: when a > 0
Didactic material
1. Find all values of a for which the equation
25 x - (2 a+ 5) * 5 x-1 / x + 10 a* 5 -2/x = 0 has exactly 2 solutions.
2. For what values of a does the equation
2 (a-1) x? + 2 (a + 3) x + a \u003d 1/4 has a single root?
3. For what values of the parameter a the equation
4 x - (5 a-3) 2 x +4 a 2 – 3a= 0 has a unique solution?
Logarithmic Equations with a Parameter
Example 1 Find all values a, for which the equation
log 4x (1 + Oh) = 1/2 (1)
has a unique solution.
Solution. Equation (1) is equivalent to the equation
1 + Oh = 2X at X > 0, X 1/4 (3)
X = at
au 2 - at + 1 = 0 (4)
The (2) condition from (3) is not satisfied.
Let a 0, then au 2 – 2at+ 1 = 0 has real roots if and only if D = 4 – 4a 0, i.e. at a 1. To solve inequality (3), we construct graphs of functions Galitsky M.L., Moshkovich M.M., Shvartsburd S.I. In-depth study of the course of algebra and mathematical analysis. - M.: Enlightenment, 1990
1. Task.
At what values of the parameter a the equation ( a - 1)x 2 + 2x + a- 1 = 0 has exactly one root?
1. Decision.
At a= 1 equation has the form 2 x= 0 and obviously has a single root x= 0. If a No. 1, then this equation is quadratic and has a single root for those values of the parameter for which the discriminant square trinomial zero. Equating the discriminant to zero, we obtain an equation for the parameter a
4a 2 - 8a= 0, whence a= 0 or a = 2.
1. Answer: the equation has a single root at a O(0; 1; 2).
2. Task.
Find all parameter values a, for which the equation has two different roots x 2 +4ax+8a+3 = 0.
2. Decision.
The equation x 2 +4ax+8a+3 = 0 has two distinct roots if and only if D =
16a 2 -4(8a+3) > 0. We get (after reduction by a common factor of 4) 4 a 2 -8a-3 > 0, whence
2. Answer:
| a O (-Ґ ; 1 - | C 7 2 |
) AND (1 + | C 7 2 |
; Ґ ). |
3. Task.
It is known that
f 2 (x) = 6x-x 2 -6.
a) Graph the function f 1 (x) at a = 1.
b) At what value a function graphs f 1 (x) and f 2 (x) have a single common point?
3. Solution.
3.a. Let's transform f 1 (x) in the following way
The graph of this function a= 1 is shown in the figure on the right.
3.b. We immediately note that the function graphs y =
kx+b and y = ax 2 +bx+c
(a No. 0) intersect at a single point if and only if the quadratic equation kx+b =
ax 2 +bx+c has a single root. Using View f 1 of 3.a, we equate the discriminant of the equation a = 6x-x 2 -6 to zero. From Equation 36-24-4 a= 0 we get a= 3. Doing the same with equation 2 x-a = 6x-x 2 -6 find a= 2. It is easy to verify that these parameter values satisfy the conditions of the problem. Answer: a= 2 or a = 3.
4. Task.
Find all values a, under which the set of solutions of the inequality x 2 -2ax-3a i 0 contains the segment .
4. Solution.
The first coordinate of the vertex of the parabola f(x) =
x 2 -2ax-3a is equal to x 0 =
a. From properties quadratic function condition f(x) i 0 on the interval is equivalent to the totality of three systems
has exactly two solutions?
5. Decision.
Let's rewrite this equation in the form x 2 + (2a-2)x - 3a+7 = 0. This is a quadratic equation, it has exactly two solutions if its discriminant is strictly greater than zero. Calculating the discriminant, we get that the condition for having exactly two roots is the fulfillment of the inequality a 2 +a-6 > 0. Solving the inequality, we find a < -3 или a> 2. The first of the inequalities is obviously solutions in natural numbers does not have, and the smallest natural solution of the second is the number 3.
5. Answer: 3.
6. Task (10 cells)
Find all values a, for which the graph of the function or, after obvious transformations, a-2 = |
2-a| . The last equation is equivalent to the inequality a i 2.
6. Answer: a O .
Let's sum up. The restriction on the parameter gives only the second condition from the DPV: a∈[−4; 4], and the requirement that the roots do not coincide is satisfied if we exclude from this interval a= ±3.
Answer: a ∈[−4;−3)∪(−3; 3)∪(3; 4]
As you can see, the coefficients here are chosen so that the algebraic operations are not difficult and do not take much time. But, if you forgot about the features square roots and lost sight of exactly condition 2) from the ODZ, then you will not get a solution at all.
I hope that many graduates nevertheless coped with this task, and I wish them further success in their elective exams.
Task 2
Find all values a , for each of which the equation
x − 2a _____ x + 2 + x − 1 ____ x − a = 1
Has a single root.
Solution.
We start, of course, with the ODZ: x≠ −2 and x ≠ a
.
Let's transform:
Brought fractions to common denominator and immediately discarded the denominator. The new equation will be equivalent to the given one only taking into account the limitations of the ODZ.
Why can this be done?
Because fractions with equal denominators are equal when their numerators are equal.
When can't you do this?
- When the inequality of the denominator to zero is not checked or the ODZ was forgotten to be written down first.
Who can and who can't do this?
- Accurate and thoughtful students can do it, inattentive students can't. The last one needs to transfer everything to the left side of the equation, simplify the expression in the form of a complete fraction, then proceed to the set of conditions: "a fraction is equal to zero if its numerator is equal to zero, and its denominator is not equal to zero."
After opening the brackets and bringing like terms, we get
x 2 − 2ax + 2a 2 − x − 2 = −2a.
Let us finally bring it to the form characteristic of quadratic equation:
x 2 − (2a+ 1) x + (2a 2 + 2a − 2) = 0.
The discriminant of this equation
D = (2a+ 1) 2 − 4 (2 a 2 + 2a − 2) = −4a 2 − 4a + 9.
The equation given in the condition of the problem can have a unique solution in two cases. Firstly, when the discriminant of the resulting quadratic equation is equal to zero, and its only root does not coincide with the restrictions of the ODZ. Otherwise, it will have to be discarded and there will be no solutions at all. Secondly, when the quadratic equation has two different roots (the discriminant is greater than zero), but one and only one of them does not satisfy the ODZ.
Case I.D = 0.
−4a 2 − 4a + 9 = 0 at a= (−1 ± √10 __ )/2.In this case, the root of the equation x = (2a + 1)/2 = a + 0,5
. Obviously, with the obtained values a it does not match with a, neither with −2.
Thus, two desired values of the parameter are obtained.
Case II.
Let's define those values a x = a.
a 2 − (2a+ 1) a + (2a 2 + 2a − 2) = 0.
a 2 + a − 2 = 0.
a= 1 and a = −2.
Let's define those values a, for which the root of the quadratic equation is x = −2.
(−2) 2 − (2a+ 1) (−2) + (2 a 2 + 2a − 2) = 0.
a 2 + 3a + 2 = 0.
a= −1 and a = −2.
With these parameter values a we can continue to study the discriminant and the second root of the quadratic equation. But it is easier to check them by substituting the conditions of the problem into the original equation.
a = 1x− 2 1 _______ x + 2 + x − 1 ____ x − 1 = 1; x − 2 _____ x + 2 + 1 = 1; x − 2 _____ x + 2 = 0; x = 2.
a = −1x− 2 (−1) _________ x + 2 + x − 1 _______ x − (−1) = 1; x + 2 ____ x + 2 + x − 1 ____ x + 1 = 1; 1 + x − 1 ____ x + 1 = 1; x − 1 ____ x + 1 = 0; x = 1.
a = −2x− 2 (−2) _________ x + 2 + x − 1 _______ x − (−2) = 1; x + 4 ____ x + 2 + x − 1 ____ x + 2 = 1; x + 4 + x − 1 = x + 2; x = −1.
Thus, all three values satisfy the condition of the problem.
Answer: a ∈{(−1 − √10__ )/2; −2; −1; 1; (−1 + √10__ )/2.}
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