AT last years at the entrance exams, at the final testing in the form of the USE, tasks with parameters are offered. These tasks allow diagnosing the level of mathematical and, most importantly, logical thinking applicants, the ability to carry out research activities, as well as simply knowledge of the main sections of the school mathematics course.
The view of the parameter as an equal variable is reflected in graphical methods. Indeed, since the parameter is “equal in rights” with the variable, then, of course, it can “allocate” its own coordinate axis. Thus, there is a coordinate plane. The rejection of the traditional choice of letters and for the designation of axes, defines one of the most effective methods for solving problems with parameters - "domain method". Along with other methods used in solving problems with parameters, I introduce my students to graphical techniques, paying attention to how to recognize “such” problems and what the process of solving a problem looks like.
The most common signs that will help you recognize tasks that are suitable for the method in question are:
Task 1. “For what values of the parameter does the inequality hold for all ?”
Decision. 1).
Let's expand the modules taking into account the sign of the submodule expression: 
2). We write down all systems of the resulting inequalities:
a) 
b) 
in) 
G) 
3). Let us show the set of points that satisfy each system of inequalities (Fig. 1a).
4). Combining all the areas shown in the figure by hatching, you can see that the inequality does not satisfy the points lying inside the parabolas.
The figure shows that for any value of the parameter, you can find the area where the points lie, the coordinates of which satisfy the original inequality. The inequality holds for all if . Answer: at .
The considered example is an “open problem” - you can consider the solution of a whole class of problems without changing the expression considered in the example 
, in which the technical difficulties of plotting have already been overcome.
Task. For what values of the parameter does the equation have no solutions? Answer: at .
Task. For what values of the parameter does the equation have two solutions? Write down both solutions you find.
Answer: then 
, 
;
Then 
; , then 
, .
Task. At what values of the parameter does the equation have one root? Find this root. Answer: at at .
Task. Solve the inequality.
(“Working” points lying inside the parabolas).
, ; , there are no solutions;
Task 2. Find all parameter values a, for each of which the system of inequalities 
forms a segment of length 1 on the number line.
Decision. We rewrite the original system in this form 
All solutions of this system (pairs of the form ) form a certain area bounded by parabolas 
and 
(Figure 1).
Obviously, the solution to the system of inequalities will be a segment of length 1 for and for . Answer: ; .
Task 3. Find all values of the parameter for which the set of solutions to the inequality 
contains the number , and also contains two segments of length , which do not have common points.
Decision. According to the meaning of inequality ; rewrite the inequality, multiplying both its parts by (), we obtain the inequality:
, ,
(1)
Inequality (1) is equivalent to the combination of two systems:
(Fig. 2).
Obviously, an interval cannot contain a segment of length . This means that two non-intersecting length segments are contained in the interval. This is possible for , i.e. at . Answer: .
Task 4. Find all values of the parameter , for each of which the set of solutions to the inequality 
contains a segment of length 4 and is also contained in some segment of length 7.
Decision. Let us carry out equivalent transformations, taking into account that and .
, ,
; the last inequality is equivalent to the combination of two systems:
Let's show the areas that correspond to these systems (Fig. 3).
1) For a set of solutions is an interval of length less than 4. For a set of solutions is the union of two intervals . Only an interval can contain a segment of length 4 . But then , and the union is no longer contained in any segment of length 7. Hence, such ones do not satisfy the condition.
2) the set of solutions is the interval . It contains a segment of length 4 only if its length is greater than 4, i.e. at . It is contained in a segment of length 7 only if its length is not greater than 7, i.e. at , then . Answer: .
Task 5. Find all values of the parameter for which the set of solutions to the inequality 
contains the number 4, and also contains two non-intersecting segments of length 4 each.
Decision. By the terms. We multiply both parts of the inequality by (). We get an equivalent inequality in which we group all the terms on the left side and transform it into a product:
, ,
, 
.
From the last inequality follows:
1) 
2)
Let's show the areas that correspond to these systems (Fig. 4).
a) For , we obtain an interval that does not contain the number 4. For , we obtain an interval that also does not contain the number 4.
b) For , we obtain the union of two intervals. Non-intersecting segments of length 4 can only be located in the interval . This is possible only if the length of the interval is greater than 8, i.e. if . For such, another condition is also fulfilled: . Answer: .
Problem 6. Find all values of the parameter for which the set of solutions to the inequality 
contains some segment of length 2, but does not contain
no segment of length 3.
Decision. According to the meaning of the task, we multiply both parts of the inequality by , group all the terms on the left side of the inequality and transform it into a product:
, 
. From the last inequality follows:
1) 
2) 
Let's show the area that corresponds to the first system (Fig. 5).
Obviously, the condition of the problem is satisfied if 
. Answer: .
Problem 7. Find all values of the parameter for which the set of solutions to the inequality 1+ 
is contained in some segment of length 1 and at the same time contains some segment of length 0.5.
Decision. one). Specify the ODZ of the variable and parameter: 
2). Let us rewrite the inequality in the form
, 
,
(one). Inequality (1) is equivalent to the combination of two systems:
1) 
2) 
Taking into account the ODZ, the solutions of the systems look like this:
a) 
b) 
(Fig. 6).
a) 
b)
Let us show the area corresponding to the system a) (Fig. 7). Answer: .
Problem 8. Six numbers form an increasing arithmetic progression. The first, second and fourth terms of this progression are solutions to the inequality 
, and the rest
are not solutions to this inequality. Find the set of all possible values of the first term of such progressions.
Decision. I. Find all solutions of the inequality
a). ODZ: 
, i.e.
(we took into account in the solution that the function increases by ).
b). On the ODZ inequality 
is equivalent to the inequality 
, i.e. 
, what gives:
1). 
2). 
Obviously, the solution of the inequality serves as a set of values 
.
II. Let us illustrate the second part of the problem about the terms of an increasing arithmetic progression with a figure ( rice. eight , where is the first term, is the second, etc.). Notice, that:
Or we have a system of linear inequalities:
Let's solve it graphically. We construct lines and , as well as lines
Then, .. The first, second and sixth terms of this progression are solutions to the inequality 
, and the rest are not solutions of this inequality. Find the set of all possible values of the difference of this progression.
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 of the square trinomial is equal to 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 common factor 4) 4a 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 the properties of a quadratic function, the 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: aО where \ - variables, \ - parameter;
\[y = kx + b,\] where \ - variables, \ - parameter;
\[ax^2 + bx + c = 0,\] where \ is a variable, \[a, b, c\] is a parameter.
Solving an equation with a parameter means, as a rule, solving an infinite set of equations.
However, adhering to a certain algorithm, one can easily solve the following equations:
1. Determine the "control" values of the parameter.
2. Solve the original equation for [\x\] with the parameter values specified in the first paragraph.
3. Solve the original equation for [\x\] with parameter values that differ from those selected in the first paragraph.
Let's say the following equation is given:
\[\mid 6 - x \mid = a.\]
After analyzing the initial data, it is clear that a \[\ge 0.\]
By the modulus rule \ we express \
Answer: \ where \
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To tasks with parameter include, for example, the search for solutions to linear and quadratic equations in general view, the study of the equation for the number of roots available depending on the value of the parameter.
Without giving 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 real number non-zero and b is 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.
Decision.
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.
Decision.
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.
Decision.
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?
Decision.
To solve by a graphical method, we construct graphs of functions y = ||x| – 2| and y = a (Fig. 2).
The drawing clearly shows possible cases 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?
Decision.
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?
Decision.
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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