Report to the GMO of the teacher of mathematics MBOU SOSH №9
Molchanova Elena Vladimirovna
"Preparation for the exam in mathematics: problems with parameters."
Since there is no definition of a parameter in school textbooks, I propose to take its following simplest version as a basis.
Definition ... A parameter is an independent variable, the value of which in the problem is considered to be given, fixed or arbitrary. real number, or a number belonging to a predetermined set.
What does "solve a problem with a parameter" mean?
Naturally, it depends on the question in the problem. If, for example, it is required to solve an equation, inequality, their system or combination, then this means presenting a reasonable answer either for any parameter value, or for a parameter value belonging to a predetermined set.
If it is required to find the values of the parameter for which the set of solutions to an equation, inequality, etc. satisfies the declared condition, then, obviously, the solution to the problem consists in finding the indicated values of the parameter.
A more transparent understanding of what it means to solve a problem with a parameter will form in the reader after reading the examples of problem solving on the following pages.
What are the main types of tasks with parameters?
Type 1. Equations, inequalities, their systems and sets that need to be solved either for any value of the parameter (s), or for the values of the parameter belonging to a predetermined set.
This type of problem is basic when mastering the topic "Tasks with parameters", since the invested work predetermines success in solving problems of all other basic types.
Type 2. Equations, inequalities, their systems and sets, for which it is required to determine the number of solutions depending on the value of the parameter (s).
I draw your attention to the fact that when solving problems of this type, there is no need either to solve given equations, inequalities, their systems and totality, etc., nor provide these solutions; In most cases, such unnecessary work is a tactical mistake that leads to unjustified waste of time. However, it is not necessary to absolutize what has been said, since sometimes a direct solution in accordance with type 1 is the only reasonable way to get an answer when solving a type 2 problem.
Type 3. Equations, inequalities, their systems and sets, for which it is required to find all those values of the parameter for which the indicated equations, inequalities, their systems and sets have a given number of solutions (in particular, do not have or have an infinite set of solutions).
It is easy to see that problems of type 3 are in some sense inverse to problems of type 2.
Type 4. Equations, inequalities, their systems and sets, for which, for the sought-for values of the parameter, the set of solutions satisfies the specified conditions in the domain of definition.
For example, find the parameter values for which:
1) the equation is fulfilled for any value of the variable from a given interval;
2) the set of solutions to the first equation is a subset of the set of solutions to the second equation, etc.
A comment. The variety of problems with a parameter covers the entire course of school mathematics (both algebra and geometry), but the overwhelming majority of them on the final and entrance exams belong to one of the four listed types, which for this reason are called basic.
The most widespread class of problems with a parameter is problems with one unknown and one parameter. The next paragraph indicates the main ways to solve problems of this particular class.
What are the main ways (methods) of solving problems with a parameter?
Method I (analytical). This is a method of the so-called direct solution, repeating the standard procedures for finding an answer in problems without a parameter. It is sometimes said that this is a way of forcefulness, in good sense"Impudent" decision.
A comment. Analytical way solving problems with a parameter is the most the hard way, requiring high literacy and the greatest efforts to master it.
Method II (graphic). Depending on the task (with variable x and parametera ) plots are considered either in the coordinate plane (x; y), or in the coordinate plane (x;a ).
A comment. The exceptional clarity and beauty of the graphical method for solving problems with a parameter captivates students of the topic "Problems with a parameter" so much that they begin to ignore other methods of solving, forgetting the well-known fact: for any class of problems, their authors can formulate one that is brilliantly solved by this method and with colossal difficulties in other ways. Therefore on initial stage It is dangerous to start studying with graphical techniques for solving problems with a parameter.
Method III (decision regarding parameter). When solving in this way, the variables x and a are assumed to be equal, and the variable with respect to which the analytical solution is recognized as simpler is selected. After natural simplifications, we return to the original meaning of the variables x and a and complete the solution.
I will now turn to the demonstration of these methods for solving problems with a parameter, since this is my favorite method for solving problems of this type.
Having analyzed all the tasks with the parameters solved by the graphical method, I begin my acquaintance with the parameters with the tasks of the Unified State Examination B7 of 2002:
At what is the integer value of k equation 45x - 3x 2 - NS 3 + 3k = 0 has exactly two roots?
These tasks allow, firstly, to remember how to build graphs using the derivative, and secondly, to explain the meaning of the straight line y = k.
In subsequent lessons, I use a selection of easy and medium-level competitive problems with parameters for preparing for the exam, equations with a module. These tasks can be recommended to teachers in mathematics as a starting set of exercises for teaching how to work with the parameter enclosed under the module sign. Most of the numbers are solved graphically and provide the teacher ready plan lesson (or two lessons) with a strong student. Initial preparation for the exam in mathematics on exercises close in complexity to the real numbers C5. Many of the proposed tasks were taken from the materials for preparing for the 2009 USE, and some from the Internet from the experience of colleagues.
1) Specify all parameter valuesp
for which the equation has 4 roots?
Answer:
2) At what values of the parametera
the equation has no solutions?
Answer:
3) Find all values of a, for each of which the equation has exactly 3 roots?
Answer: a = 2
4) At what values of the parameterb the equation It has only decision? Answer:
5) Find all valuesm
for which the equation has no solutions.
Answer:
6) Find all values of a for which the equation has exactly 3 distinct roots. (If there is more than one value of a, then write down their sum in the answer.)
Answer: 3
7) At what valuesb
the equation has exactly 2 solutions?
Answer:
8) Specify these parametersk
for which the equation has at least two solutions.
Answer:
9) At what values of the parameterp
the equation has only one solution?
Answer:
10) Find all values of a, for each of which the equation (x + 1)has exactly 2 roots? If there are several values of a, then write down their sum in response.
Answer: - 3
11) Find all values of a for which the equation has exactly 3 roots? (If there are more than one values of a, then write down their sum in response).
Answer: 4
12) At what is the smallest natural value of the parameter a, the equation – = 11 has only positive roots?
Answer: 19
13) Find all values of a, for each of which the equation = 1 has exactly 3 roots? (If there is more than one value of a, then write down their sum in the answer).
Answer: - 3
14) Specify such parameter valuest for which the equation has 4 different solutions... Answer:
15) Find these parametersm for which the equation has two different solutions. Answer:
16) At what values of the parameterp the equation has exactly 3 extrema? Answer:
17) Specify all possible parameters n for which the function has exactly one minimum point. Answer:
The published kit is regularly used by me to work with a capable, but not the strongest student, who nevertheless claims a high USE score by solving the number C5. The teacher prepares such a student in several stages, highlighting individual lessons for training individual skills necessary for finding and implementing long-term solutions. This selection is suitable for the stage of forming representations of floating pictures, depending on the parameter. Numbers 16 and 17 are modeled after a real equation with a parameter on the 2011 Unified State Exam. The tasks are arranged in order of increasing difficulty.
Task C5 in Mathematics USE 2012
Here we have a traditional parameter problem requiring moderate knowledge of the material and the application of several properties and theorems. This task is one of the most difficult tasks Unified State Examination in Mathematics. It is designed primarily for those who are going to continue their education in universities with increased requirements for the mathematical training of applicants. To successfully solve the problem, it is important to freely operate with the studied definitions, properties, theorems, and apply them in different situations, analyze the condition and find possible ways solutions.
On the website of preparation for the exam by Alexander Larin since 11.05.2012, training options No. 1 - 22 with tasks of level "C" were proposed, C5 of some of them were similar to those tasks that were on the real exam. For example, find all values of the parameter a, for each of which the graphs of the functionsf(x) = andg(x) = a (x + 5) + 2 have no common points?
Let's analyze the solution to task C5 from the 2012 exam.
Task C5 from the exam-2012
For what values of the parameter a the equation has at least two roots.
Let's solve this problem graphically. Let's plot the left side of the equation: and the graph on the right side:and formulate the question of the problem as follows: for what values of the parameter a are the graphs of the functions andhave two or more points in common.
There is no parameter on the left side of the original equation, so we can plot the function.
We will build this graph using function:
1. Let's move the graph of the function3 units down along the OY axis, we get the graph of the function:
2. Let's plot the function ... For this, part of the graph of the function , located below the OX axis, will be displayed symmetrically about this axis:
So, the graph of the functionlooks like:
Function graph
1. Problem.
At what values of the parameter a the equation ( a - 1)x 2 + 2x + a- 1 = 0 has exactly one root?
1. Solution.
At a= 1 the equation has the form 2 x= 0 and obviously has a unique root x= 0. If a No. 1, then this equation is square and has a single root for those values of the parameter for which the discriminant square trinomial is 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. The task.
Find all parameter values a for which the equation x 2 +4ax+8a+3 = 0.
2. Solution.
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. The challenge.
It is known that
f 2 (x) = 6x-x 2 -6.
a) Plot 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 point in common?
3. Solution.
3.a. We transform f 1 (x) in the following way
The graph of this function at a= 1 is shown in the figure on the right.
3.b. We note right away that the graphs of the functions y =
kx+b and y = ax 2 +bx+c
(a No. 0) intersect at a single point if and only if quadratic equation kx+b =
ax 2 +bx+c has a single root. Using the 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 values of the parameter satisfy the conditions of the problem. Answer: a= 2 or a = 3.
4. The challenge.
Find all values a for which the set of solutions to the inequality x 2 -2ax-3aі 0 contains a 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) і 0 on an interval is equivalent to a set of three systems
has exactly two solutions?
5. Solution.
We rewrite this equation as 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 find that the condition for the presence of 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, obviously, solutions in natural numbers does not have, and the smallest natural solution to the second is the number 3.
5. Answer: 3.
6. Problem (10 grades)
Find all values a at 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
has exactly four solutions.
(USE 2018, main wave)
The second equation of the system can be rewritten as \ (y = \ pm x \). Therefore, consider two cases: when \ (y = x \) and when \ (y = -x \). Then the number of solutions to the system will be equal to the sum of the number of solutions in the first and second cases.
1) \ (y = x \). Substitute in the first equation and get: \
(note that in the case \ (y = -x \) we will do the same and also get a quadratic equation)
In order for the original system to have 4 different solutions, it is necessary that in each of the two cases there are 2 solutions.
A quadratic equation has two roots when its \ (D> 0 \). Let us find the discriminant of equation (1):
\ (D = -4 (a ^ 2 + 4a + 2) \).
Discriminant greater than zero: \ (a ^ 2 + 4a + 2<0\)
, откуда \ (a \ in (-2- \ sqrt2; -2+ \ sqrt2) \).
2) \ (y = -x \). We get the quadratic equation: \
The discriminant is greater than zero: \ (D = -4 (9a ^ 2 + 12a + 2)> 0 \), whence \ (a \ in \ left (\ frac (-2- \ sqrt2) 3; \ frac (-2+ \ sqrt2) 3 \ right) \).
It is necessary to check whether the solutions in the first case coincide with the solutions in the second case.
Let \ (x_0 \) - common decision equations (1) and (2), then \
Hence we obtain that either \ (x_0 = 0 \) or \ (a = 0 \).
If \ (a = 0 \), then equations (1) and (2) are the same, therefore, they have identical roots... This case does not suit us.
If \ (x_0 = 0 \) is their common root, then \ (2x_0 ^ 2-2 (3a + 2) x_0 + (2a + 2) ^ 2 + a ^ 2-1 = 0 \), whence \ ((2a + 2) ^ 2 + a ^ 2-1 = 0 \), whence \ (a = -1 \) or \ (a = -0.6 \). Then the whole original system will have 3 different solutions, which does not suit us.
Considering all this, the answer will be:
Answer:
\ (a \ in \ left (\ frac (-2- \ sqrt2) 3; -1 \ right) \ cup \ left (-1; -0.6 \ right) \ cup \ left (-0.6; - 2+ \ sqrt2 \ right) \)
Quest 2 # 4032
Task level: Equal to the exam
Find all the values \ (a \), for each of which the system \ [\ begin (cases) (a-1) x ^ 2 + 2ax + a + 4 \ leqslant 0 \\ ax ^ 2 + 2 (a + 1) x + a + 1 \ geqslant 0 \ end (cases) \ ]
has only one solution.
Let's rewrite the system as: \ [\ begin (cases) ax ^ 2 + 2ax + a \ leqslant x ^ 2-4 \\ ax ^ 2 + 2ax + a \ geqslant -2x-1 \ end (cases) \] Consider three functions: \ (y = ax ^ 2 + 2ax + a = a (x + 1) ^ 2 \), \ (g = x ^ 2-4 \), \ (h = -2x-1 \). It follows from the system that \ (y \ leqslant g \), but \ (y \ geqslant h \). Therefore, for the system to have solutions, the graph \ (y \) must be in the area specified by the conditions: “above” the graph \ (h \), but “below” the graph \ (g \):
(we will call the “left” area area I, the “right” area - area II)
Note that for each fixed \ (a \ ne 0 \) the graph \ (y \) is a parabola, the vertex of which is at the point \ ((- 1; 0) \), and the branches are turned either up or down. If \ (a = 0 \), then the equation looks like \ (y = 0 \) and the graph is a straight line coinciding with the abscissa axis.
Note that in order for the original system to have a unique solution, the graph \ (y \) must have exactly one common point with region I or with region II (this means that the graph \ (y \) must have a single common point with the border of one of these areas).
Let's consider several cases separately.
1) \ (a> 0 \). Then the branches of the parabola \ (y \) are directed upwards. For the original system to have a unique solution, it is necessary that the parabola \ (y \) touches the boundary of region I or the boundary of region II, that is, touches the parabola \ (g \), and the abscissa of the tangency point must be \ (\ leqslant -3 \) or \ (\ geqslant 2 \) (that is, the \ (y \) parabola must touch the border of one of the regions that is above the abscissa axis, since the \ (y \) parabola lies above the abscissa axis).
\ (y "= 2a (x + 1) \), \ (g" = 2x \). Conditions for touching the graphs \ (y \) and \ (g \) at a point with abscissa \ (x_0 \ leqslant -3 \) or \ (x_0 \ geqslant 2 \): \ [\ begin (cases) 2a (x_0 + 1) = 2x_0 \\ a (x_0 + 1) ^ 2 = x_0 ^ 2-4 \\ \ left [\ begin (gathered) \ begin (aligned) & x_0 \ leqslant - 3 \\ & x_0 \ geqslant 2 \ end (aligned) \ end (gathered) \ right. \ end (cases) \ quad \ Leftrightarrow \ quad \ begin (cases) \ left [\ begin (gathered) \ begin (aligned) & x_0 \ leqslant -3 \\ & x_0 \ geqslant 2 \ end (aligned) \ end (gathered) \ right. \\ a = \ dfrac (x_0) (x_0 + 1) \\ x_0 ^ 2 + 5x_0 + 4 = 0 \ end (cases) \] From this system \ (x_0 = -4 \), \ (a = \ frac43 \).
Received the first value of the \ (a \) parameter.
2) \ (a = 0 \). Then \ (y = 0 \) and it is clear that the line has infinitely many common points with region II. Therefore, this parameter value is not suitable for us.
3) \ (a<0\) . Тогда ветви параболы \(y\) обращены вниз. Чтобы у исходной системы было единственное решение, нужно, чтобы парабола \(y\) имела одну общую точку с границей области II, лежащей ниже оси абсцисс. Следовательно, она должна проходить через точку \(B\) , причем, если парабола \(y\) будет иметь еще одну общую точку с прямой \(h\) , то эта общая точка должна быть “выше” точки \(B\) (то есть абсцисса второй точки должна быть \(<1\) ).
Let us find \ (a \) for which the parabola \ (y \) passes through the point \ (B \): \ [- 3 = a (1 + 1) ^ 2 \ quad \ Rightarrow \ quad a = - \ dfrac34 \] We make sure that for this value of the parameter, the second point of intersection of the parabola \ (y = - \ frac34 (x + 1) ^ 2 \) with the line \ (h = -2x-1 \) is a point with coordinates \ (\ left (- \ frac13; - \ frac13 \ right) \).
Thus, we got one more parameter value.
Since we have considered all possible cases for \ (a \), the final answer is: \
Answer:
\ (\ left \ (- \ frac34; \ frac43 \ right \) \)
Quest 3 # 4013
Task level: Equal to the exam
Find all values of the parameter \ (a \), for each of which the system of equations \ [\ begin (cases) 2x ^ 2 + 2y ^ 2 = 5xy \\ (x-a) ^ 2 + (y-a) ^ 2 = 5a ^ 4 \ end (cases) \]
has exactly two solutions.
1) Consider the first equation of the system as quadratic with respect to \ (x \): \ The discriminant is \ (D = 9y ^ 2 \), therefore, \
Then the equation can be rewritten as \ [(x-2y) \ cdot (2x-y) = 0 \] Therefore, the whole system can be rewritten as \ [\ begin (cases) \ left [\ begin (gathered) \ begin (aligned) & y = 2x \\ & y = 0.5x \ end (aligned) \ end (gathered) \ right. \\ (xa) ^ 2 + (ya) ^ 2 = 5a ^ 4 \ end (cases) \] The set defines two lines, the second equation of the system defines a circle with center \ ((a; a) \) and radius \ (R = \ sqrt5a ^ 2 \). For the original equation to have two solutions, the circle must intersect the population graph at exactly two points. Here is a drawing when, for example, \ (a = 1 \): 
Note that since the coordinates of the center of the circle are equal, the center of the circle “runs” along the straight line \ (y = x \).
2) Since the straight line \ (y = kx \) the tangent of the angle of inclination of this straight line to the positive direction of the axis \ (Ox \) is \ (k \), the tangent of the angle of inclination of the straight line \ (y = 0.5x \) is \ (0,5 \) (let's call it \ (\ mathrm (tg) \, \ alpha \)), the straight line \ (y = 2x \) - equals \ (2 \) (let's call it \ (\ mathrm (tg) \ , \ beta \)). notice, that \ (\ mathrm (tg) \, \ alpha \ cdot \ mathrm (tg) \, \ beta = 1 \), hence, \ (\ mathrm (tg) \, \ alpha = \ mathrm (ctg) \, \ beta = \ mathrm (tg) \, (90 ^ \ circ- \ beta) \)... Therefore, \ (\ alpha = 90 ^ \ circ- \ beta \), whence \ (\ alpha + \ beta = 90 ^ \ circ \). This means that the angle between \ (y = 2x \) and the positive direction \ (Oy \) is equal to the angle between \ (y = 0.5x \) and the positive direction \ (Ox \): 
And since the straight line \ (y = x \) is the bisector of the I coordinate angle (that is, the angles between it and the positive directions \ (Ox \) and \ (Oy \) are equal in \ (45 ^ \ circ \)), then the angles between \ (y = x \) and straight lines \ (y = 2x \) and \ (y = 0,5x \) are equal.
We needed all this in order to say that the lines \ (y = 2x \) and \ (y = 0.5x \) are symmetric to each other with respect to \ (y = x \), therefore, if the circle touches one of them , then it necessarily touches the second straight line.
Note that if \ (a = 0 \), then the circle degenerates into the point \ ((0; 0) \) and has only one intersection point with both lines. That is, this case does not suit us.
Thus, in order for a circle to have 2 points of intersection with straight lines, it must be tangent to these straight lines: 
We see that the case when the circle is located in the third quarter is symmetric (relative to the origin) to the case when it is located in the first quarter. That is, in the first quarter \ (a> 0 \), and in the third \ (a<0\)
(но такие же по модулю).
Therefore, we will consider only the first quarter. 
notice, that \ (OQ = \ sqrt ((a-0) ^ 2 + (a-0) ^ 2) = \ sqrt2a \), \ (QK = R = \ sqrt5a ^ 2 \). Then \ Then \ [\ mathrm (tg) \, \ angle QOK = \ dfrac (\ sqrt5a ^ 2) (\ sqrt (2a ^ 2-5a ^ 4)) \] But on the other side, \ [\ mathrm (tg) \, \ angle QOK = \ mathrm (tg) \, (45 ^ \ circ- \ alpha) = \ dfrac (\ mathrm (tg) \, 45 ^ \ circ- \ mathrm (tg) \, \ alpha) (1+ \ mathrm (tg) \, 45 ^ \ circ \ cdot \ mathrm (tg) \, \ alpha) \] hence, \ [\ dfrac (1-0.5) (1 + 1 \ cdot 0.5) = \ dfrac (\ sqrt5a ^ 2) (\ sqrt (2a ^ 2-5a ^ 4)) \ quad \ Leftrightarrow \ quad a = \ pm \ dfrac15 \] Thus, we have already immediately obtained both positive and negative values for \ (a \). Hence the answer is: \
Answer:
\(\{-0,2;0,2\}\)
Quest 4 # 3278
Task level: Equal to the exam
Find all the values \ (a \), for each of which the equation \
has only one solution.
(USE 2017, official trial 04/21/2017)
We make the change \ (t = 5 ^ x, t> 0 \) and transfer all terms to one part: \
We got a quadratic equation, the roots of which, according to Vieta's theorem, are \ (t_1 = a + 6 \) and \ (t_2 = 5 + 3 | a | \). For the original equation to have one root, it is sufficient that the resulting equation with \ (t \) also has one (positive!) Root.
Note right away that \ (t_2 \) will be positive for all \ (a \). Thus, we get two cases:
1) \ (t_1 = t_2 \): \ & a = - \ dfrac14 \ end (aligned) \ end (gathered) \ right. \]
2) Since \ (t_2 \) is always positive, \ (t_1 \) must be \ (\ leqslant 0 \): \
Answer:
\ ((- \ infty; -6] \ cup \ left \ (- \ frac14; \ frac12 \ right \) \)
Task 5 # 3252
Task level: Equal to the exam
\ [\ sqrt (x ^ 2-a ^ 2) = \ sqrt (3x ^ 2- (3a + 1) x + a) \]
has exactly one root on the segment \ (\).
(USE 2017, reserve day)
The equation can be rewritten as: \ [\ sqrt ((x-a) (x + a)) = \ sqrt ((3x-1) (x-a)) \] Thus, we note that \ (x = a \) is the root of the equation for any \ (a \), since the equation takes the form \ (0 = 0 \). In order for this root to belong to the segment \ (\), it is necessary that \ (0 \ leqslant a \ leqslant 1 \).
The second root of the equation is found from \ (x + a = 3x-1 \), that is, \ (x = \ frac (a + 1) 2 \). In order for this number to be the root of the equation, it is necessary that it satisfies the ODZ of the equation, that is: \ [\ left (\ dfrac (a + 1) 2-a \ right) \ cdot \ left (\ dfrac (a + 1) 2 + a \ right) \ geqslant 0 \ quad \ Rightarrow \ quad - \ dfrac13 \ leqslant a \ leqslant 1 \] In order for this root to belong to the segment \ (\), it is necessary that \
Thus, for the root \ (x = \ frac (a + 1) 2 \) to exist and belong to the segment \ (\), it is necessary that \ (- \ frac13 \ leqslant a \ leqslant 1 \).
Note that then for \ (0 \ leqslant a \ leqslant 1 \) both roots \ (x = a \) and \ (x = \ frac (a + 1) 2 \) belong to the segment \ (\) (that is, the equation has two roots on this segment), except for the case when they coincide: \
Thus, we are suited \ (a \ in \ left [- \ frac13; 0 \ right) \) and \ (a = 1 \).
Answer:
\ (a \ in \ left [- \ frac13; 0 \ right) \ cup \ (1 \) \)
Task 6 # 3238
Task level: Equal to the exam
Find all values of the parameter \ (a \), for each of which the equation \
has a unique root on the segment \ (. \)
(USE 2017, reserve day)
The equation is equivalent to: \ ODZ equations: \ [\ begin (cases) x \ geqslant 0 \\ x-a \ geqslant 0 \\ 3a (1-x) \ geqslant 0 \ end (cases) \] On ODZ, the equation will be rewritten as: \
1) Let \ (a<0\) . Тогда ОДЗ уравнения: \(x\geqslant 1\) . Следовательно, для того, чтобы уравнение имело единственный корень на отрезке \(\) , этот корень должен быть равен \(1\) . Проверим: \ Doesn't match \ (a<0\) . Следовательно, эти значения \(a\) не подходят.
2) Let \ (a = 0 \). Then the ODZ of the equation: \ (x \ geqslant 0 \). The equation will be rewritten as: \ The resulting root fits the ODZ and is included in the segment \ (\). Therefore, \ (a = 0 \) - fits.
3) Let \ (a> 0 \). Then ODZ: \ (x \ geqslant a \) and \ (x \ leqslant 1 \). Consequently, if \ (a> 1 \), then the GCD is an empty set. Thus, \ (0
The derivative is \ (y "= 3x ^ 2-2ax + 3a \). Determine what sign the derivative can be. To do this, we find the discriminant of the equation \ (3x ^ 2-2ax + 3a = 0 \): \ (D = 4a ( a-9) \). Therefore, for \ (a \ in (0; 1] \) the discriminant \ (D<0\)
. Значит, выражение \(3x^2-2ax+3a\)
положительно при всех \(x\)
. Следовательно, при \(a\in (0;1]\)
производная \(y">0 \). Therefore, \ (y \) is increasing. Thus, by the property of an increasing function, the equation \ (y (x) = 0 \) can have at most one root.
Therefore, in order for the root of the equation (the point of intersection of the graph \ (y \) with the abscissa axis) to be on the segment \ (\), it is necessary that \ [\ begin (cases) y (1) \ geqslant 0 \\ y (a) \ leqslant 0 \ end (cases) \ quad \ Rightarrow \ quad a \ in \] Taking into account that initially in the considered case \ (a \ in (0; 1] \), then the answer \ (a \ in (0; 1] \). Note that the root \ (x_1 \) satisfies \ ((1) \), the roots \ (x_2 \) and \ (x_3 \) satisfy \ ((2) \). Also note that the root \ (x_1 \) belongs to the segment \ (\).
Consider three cases:
1) \ (a> 0 \). Then \ (x_2> 3 \), \ (x_3<3\)
, следовательно, \(x_2\notin .\)
Тогда уравнение будет иметь один корень на \(\)
в одном из двух случаях:
- \ (x_1 \) satisfies \ ((2) \), \ (x_3 \) does not satisfy \ ((1) \), or matches \ (x_1 \), or satisfies \ ((1) \), but is not included in the segment \ (\) (that is, less than \ (0 \));
- \ (x_1 \) does not satisfy \ ((2) \), \ (x_3 \) satisfies \ ((1) \) and is not equal to \ (x_1 \).
Note that \ (x_3 \) cannot simultaneously be less than zero and satisfy \ ((1) \) (that is, it cannot be greater than \ (\ frac35 \)). With this observation in mind, cases are recorded in the following aggregate: \ [\ left [\ begin (gathered) \ begin (aligned) & \ begin (cases) \ dfrac9 (25) -6 \ cdot \ dfrac35 + 10-a ^ 2> 0 \\ 3-a \ leqslant \ dfrac35 \ end (cases) \\ & \ begin (cases) \ dfrac9 (25) -6 \ cdot \ dfrac35 + 10-a ^ 2 \ leqslant 0 \\ 3-a> Solving this set and taking into account that \ (a> 0 \), we get: \
2) \ (a = 0 \). Then \ (x_2 = x_3 = 3 \ in. \) Note that in this case \ (x_1 \) satisfies \ ((2) \) and \ (x_2 = 3 \) satisfies \ ((1) \), then there is an equation with two roots on \ (\). This value \ (a \) does not suit us.
3) \ (a<0\) . Тогда \(x_2<3\) , \(x_3>3 \) and \ (x_3 \ notin \). Reasoning similarly to point 1), you need to solve the set: \ [\ left [\ begin (gathered) \ begin (aligned) & \ begin (cases) \ dfrac9 (25) -6 \ cdot \ dfrac35 + 10-a ^ 2> 0 \\ 3 + a \ leqslant \ dfrac35 \ end (cases) \\ & \ begin (cases) \ dfrac9 (25) -6 \ cdot \ dfrac35 + 10-a ^ 2 \ leqslant 0 \\ 3 + a> \ dfrac35 \ end (cases) \ end (aligned) \ end (gathered) \ right. \] Solving this set and taking into account that \ (a<0\) , получим: \\]
Answer:
\ (\ left (- \ frac (13) 5; - \ frac (12) 5 \ right] \ cup \ left [\ frac (12) 5; \ frac (13) 5 \ right) \)
1. Systems of linear equations with parameter
Systems of linear equations with a parameter are solved by the same basic methods as ordinary systems of equations: the substitution method, the method of adding equations, and graphical method... Knowledge of the graphic interpretation of linear systems makes it easy to answer the question of the number of roots and their existence.
Example 1.
Find all values for the parameter a for which the system of equations has no solutions.
(x + (a 2 - 3) y = a,
(x + y = 2.
Solution.
Let's consider several ways to solve this task.
1 way. We use the property: the system has no solutions if the ratio of the coefficients in front of x is equal to the ratio of the coefficients in front of y, but not equal to the ratio of free terms (a / a 1 = b / b 1 ≠ c / c 1). Then we have:
1/1 = (a 2 - 3) / 1 ≠ a / 2 or system
(a 2 - 3 = 1,
(a ≠ 2.
From the first equation a 2 = 4, therefore, taking into account the condition that a ≠ 2, we get the answer.
Answer: a = -2.
Method 2. We solve by substitution method.
(2 - y + (a 2 - 3) y = a,
(x = 2 - y,
((a 2 - 3) y - y = a - 2,
(x = 2 - y.
After taking the common factor y out of the brackets in the first equation, we get:
((a 2 - 4) y = a - 2,
(x = 2 - y.
The system has no solutions if the first equation has no solutions, that is
(a 2 - 4 = 0,
(a - 2 ≠ 0.
Obviously, a = ± 2, but taking into account the second condition, the answer is only an answer with a minus.
Answer: a = -2.
Example 2.
Find all values for the parameter a for which the system of equations has an infinite set of solutions.
(8x + ay = 2,
(ax + 2y = 1.
Solution.
By property, if the ratio of the coefficients at x and y is the same and is equal to the ratio of free members of the system, then it has an infinite set of solutions (i.e., a / a 1 = b / b 1 = c / c 1). Therefore 8 / a = a / 2 = 2/1. Solving each of the obtained equations, we find that a = 4 - the answer in this example.
Answer: a = 4.
2. Systems of rational equations with a parameter
Example 3.
(3 | x | + y = 2,
(| x | + 2y = a.
Solution.
Let's multiply the first equation of the system by 2:
(6 | x | + 2y = 4,
(| x | + 2y = a.
We subtract the second equation from the first, we get 5 | x | = 4 - a. This equation will have a unique solution for a = 4. In other cases, this equation will have two solutions (for a< 4) или ни одного (при а > 4).
Answer: a = 4.
Example 4.
Find all values of the parameter a for which the system of equations has a unique solution.
(x + y = a,
(y - x 2 = 1.
Solution.
We will solve this system using the graphical method. So, the graph of the second equation of the system is a parabola lifted along the Oy axis up by one unit segment. The first equation defines a set of straight lines parallel to the straight line y = -x (picture 1)... It is clearly seen from the figure that the system has a solution if the straight line y = -x + a is tangent to the parabola at the point with coordinates (-0.5; 1.25). Substituting these coordinates into the equation with a straight line instead of x and y, we find the value of the parameter a: 
1.25 = 0.5 + a;
Answer: a = 0.75.
Example 5.
Using the substitution method, find out at what value of the parameter a, the system has a unique solution.
(ax - y = a + 1,
(ax + (a + 2) y = 2.
Solution.
From the first equation, we express y and substitute it into the second:
(y = ax - a - 1,
(ax + (a + 2) (ax - a - 1) = 2.
Let us bring the second equation to the form kx = b, which will have a unique solution for k ≠ 0. We have:
ax + a 2 x - a 2 - a + 2ax - 2a - 2 = 2;
a 2 x + 3ax = 2 + a 2 + 3a + 2.
The square trinomial a 2 + 3a + 2 can be represented as a product of brackets
(a + 2) (a + 1), and on the left we take out x outside the brackets:
(a 2 + 3a) x = 2 + (a + 2) (a + 1).
Obviously, a 2 + 3a should not be equal to zero, therefore,
a 2 + 3a ≠ 0, a (a + 3) ≠ 0, and therefore a ≠ 0 and ≠ -3.
Answer: a ≠ 0; ≠ -3.
Example 6.
Using the graphical solution method, determine at what value of the parameter a, the system has a unique solution.
(x 2 + y 2 = 9,
(y - | x | = a.
Solution.
Based on the condition, we build a circle with a center at the origin and a radius of 3 unit segments, it is it that is set by the first equation of the system
x 2 + y 2 = 9. The second equation of the system (y = | x | + a) is a broken line. By using Figure 2 we consider all possible cases of its location relative to the circle. It is easy to see that a = 3. 
Answer: a = 3.
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