How to solve tasks with the ege parameter. The math I like

Report on the GMO teacher of mathematics MBOU secondary school №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 the following simplest variant as its basis.

Definition . A parameter is an independent variable whose value in the problem is considered a given fixed or arbitrary real number, or a number belonging to a predetermined set.

What does it mean to "solve a problem with a parameter"?

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 reasoned answer either for any parameter value or for a parameter value belonging to a predetermined set.

If it is necessary to find the parameter values \u200b\u200bat which the set of solutions of the equation, inequality, etc., satisfies the declared condition, then, obviously, the solution to the problem is to search for the specified parameter values.

A more transparent understanding of what it means to solve a problem with a parameter, the reader will form after reading the examples of solving problems on the following pages.

What are the main types of tasks with parameters?

Type 1.   Equations, inequalities, their systems and sets, which must be solved either for any parameter value (s), or for parameter values \u200b\u200bbelonging to a predetermined set.

This type of task is basic when mastering the topic “Tasks with parameters”, since the invested work determines success in solving problems of all other basic types.

Type 2.   Equations, inequalities, their systems and sets, for which it is necessary to determine the number of solutions depending on the value of the parameter (s).

I draw attention to the fact that when solving problems of this type there is no need to solve the given equations, inequalities, their systems and aggregates, etc., nor to bring these solutions; such superfluous work in most cases is a tactical error, leading to unjustified waste of time. However, do not 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 necessary to find all those parameter values \u200b\u200bfor which the indicated equations, inequalities, their systems and sets have a given number of solutions (in particular, they do not have or have an infinite number of solutions).

It is easy to see that problems of type 3 are in a sense inverse to problems of type 2.

Type 4.   Equations, inequalities, their systems and sets, for which, for the desired parameter values, the set of solutions satisfies the given conditions in the domain of definition.

For example, find the parameter values \u200b\u200bat which:

1) the equation is satisfied for any value of a variable from a given interval;
2) the set of solutions of the first equation is a subset of the set of solutions of the second equation, etc.

A comment.   A variety of problems with a parameter covers the entire course of school mathematics (both algebra and geometry), but the vast majority of them at the final and entrance exams belong to one of the four types listed, which for this reason are called basic.

The most widespread class of tasks with a parameter is tasks with one unknown and one parameter. The next paragraph indicates the main methods for solving problems of this particular class.

What are the main ways (methods) of solving problems with a parameter?

Method I   (analytical). This is the method of the so-called direct solution, repeating the standard procedures for finding the answer in tasks without a parameter. It is sometimes said that this is a way of a forceful, in a good sense, “impudent” decision.

A comment.   An analytical way to solve problems with a parameter is the most difficult way, requiring high literacy and the greatest effort to master it.

Method II   (graphic). Depending on the task (with variable x and parametera ) graphs are considered either in the coordinate plane (x; y), or in the coordinate plane (x;a ).

A comment. The exceptional visibility and beauty of the graphical method of solving problems with a parameter so captivates those studying the topic "Problems with a parameter" that they begin to ignore other methods of solution, forgetting the well-known fact: for any class of problems, their authors can formulate one that is brilliantly solved in this way and with colossal difficulties in other ways. Therefore, at the initial stage of study, it is dangerous to start with graphical techniques for solving problems with a parameter.

Method III   (decision regarding the parameter). When solving in this way, the variables x and a are accepted as equals and that variable is selected with respect to which the analytical solution is recognized as simpler. After natural simplifications, we return to the original meaning of the variables x and a and complete the solution.

I will now proceed to demonstrate the indicated methods for solving problems with a parameter, since this is my favorite method for solving problems of this type.

After analyzing all the tasks with the parameters solved by the graphical method, I begin acquaintance with the parameters with the tasks of the Unified State Exam V7 of 2002:

At what is the whole value of k equation 45x - 3x 2   - x 3   + 3k \u003d 0 has exactly two roots?

These tasks allow, firstly, to recall how to build graphs using the derivative, and secondly, to explain the meaning of the straight line y \u003d k.

In subsequent classes, I use a selection of light and medium-level competitive tasks with parameters for preparing for the exam, equations with a module. These tasks can be recommended to mathematics teachers as a starting set of exercises for learning how to work with a parameter enclosed by a module sign. Most of the numbers are solved graphically and provide the teacher with a ready-made lesson plan (or two lessons) with a strong student. Initial preparation for the exam in mathematics in exercises that are close in complexity to real numbers C5. Many of the proposed tasks are taken from materials for preparing for the Unified State Exam 2009, and some from the Internet from the experience of colleagues.

1) Indicate all parameter valuesp for which the equation   has 4 roots?
Answer:

2) At what parameter valuesa   the equation doesn't have solutions?
Answer:

3) Find all the values \u200b\u200bof a, for each of which the equation   has exactly 3 roots?
Answer: a \u003d 2

4) At what parameter valuesb   the equation has a single solution? Answer:

5) Find all the valuesm for which the equation has no decisions.
Answer:

6) Find all the values \u200b\u200bof a for which the equation   has exactly 3 different roots. (If the values \u200b\u200bof a are more than one, then write down their sum in the answer.)

Answer: 3

7) At what valuesb   the equation has exactly 2 solutions?
Answer:

8) Indicate such parametersk for which the equation has at least two solutions.
Answer:

9) At what parameter valuesp   the equation   only has one solution?
Answer:

10) Find all the values \u200b\u200bof a, for each of which the equation (x + 1)   has exactly 2 roots? If there are several values \u200b\u200bof a, then write down their sum in response.

Answer: - 3

11) Find all the values \u200b\u200bof a for which the equation has exactly 3 roots? (If the values \u200b\u200bof a are more than one, then write down their sum in response).

Answer: 4

12) For which the smallest natural value of the parameter a equation – \u003d 11 has only positive roots?

Answer: 19

13) Find all the values \u200b\u200bof a, for each of which the equation \u003d 1 has exactly 3 roots? (If the values \u200b\u200bof a are more than one, 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 such parametersm for which the equation   has two different solutions. Answer:

16) At what parameter valuesp   the equation   has exactly 3 extrema? Answer:

17) Indicate 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 most powerful student, who nevertheless claims a high exam score by solving the number C5. The teacher prepares such a student in several stages, highlighting for the training of individual skills necessary for the search and implementation of long solutions, individual lessons. This selection is suitable for the stage of formation of ideas about floating patterns depending on the parameter. Numbers 16 and 17 are modeled after a real equation with a parameter at the USE 2011. Tasks are arranged in order of increasing complexity.

Task C5 in mathematics USE 2012

Here we have a traditional problem with a parameter that requires moderate ownership of the material and the application of several properties and theorems. This task is one of the most difficult tasks of the Unified State Exam in Mathematics. It is intended, first of all, for those who are going to continue their education in universities with increased requirements for the mathematical preparation of applicants. For a successful solution of the problem, it is important to freely operate on the studied definitions, properties, theorems, apply them in various situations, analyze the condition and find possible solutions.

Since 11/11/2012, training options No. 1 - 22 with tasks of level “C” were offered on the site for preparing for the Unified State Exam of Alexander Larin, C5 of some of them were similar to those that were on the real exam. For example, find all values \u200b\u200bof the parameter a, for each of which function graphsf(x) \u003d   andg(x) \u003d a (x + 5) + 2 do not have common points?

We will analyze the solution of task C5 from the 2012 exam.

Task C5 from the exam-2012

For what values \u200b\u200bof the parameter a has at least two roots.

We solve this problem graphically. We plot the left side of the equation:   and the graph of the right side:   and formulate the problem question as follows: for what values \u200b\u200bof the parameter a of the function graph   and   have two or more common points.

There is no parameter on the left side of the original equation, so we can plot the function.

We will build this graph using   the functions:

1. Move the function graph   3 units down along the OY axis, we get a graph of the function:

2. We plot the function . To do this, part of the function graph located below the axis OX, display symmetrically with respect to this axis:

So, the function graph   has the form:

Function graph

For some reason, recently, tasks with parameters have caused almost sacred horror in schoolchildren, sometimes quiet, and sometimes not very. The problem, apparently, again, is that they are taught that way. In general, poor children ... To memorize a bunch of problems with one, two or more parameters, solve them countless times, it is not clear why, and on the same notorious exam, get a condition for such a problem with a parameter that you have never seen before and fall into a stupor from the impossibility to even begin to solve it, to understand which way to move. Well, how can you not regret graduates!

Since I really like to describe my school years, my studies, (which, however, you probably already noticed))), then I will write how it was with us. Attention, you will not believe: no one in our life has ever taught us how to solve problems with parameters! Here, I wrote another sedition))) We were just taught to solve problems, and that’s all. There was no separate class / type / group of tasks that would be called tasks with parameters. And at the same time, such tasks did not surprise anyone and did not make them tremble. All of them simply solved, like any other tasks. Like this.

And there were no various textbooks in which it was written what to do when looking at the parameters, in which direction to transfer and where to substitute ... Just for each task it was necessary to understand how to come to its solution, what, why and why, in what sequence to do, to get an answer. And it was precisely understanding why and why that was the main thing. There is nothing tricky in these tasks, believe me, please! There are no special special methods for solving them either. Yes, you can show some methods that, with a complete misunderstanding of what is happening (why and why) will help to cope with ten, fifteen, one hundred identical tasks, but there is one hundred and one that cannot be solved by this method!

What follows from here? That's what. If for some reason you are afraid of tasks with parameters, if your knees start to tremble when you mention them, you need to take tasks without parameters on the same topic that you think you can solve, and try to understand what's what, character by character understanding what , why, why and how to do it. If you deal with this in detail and thoroughly, clearly begin to imagine what is happening, you will not need any special teaching aids that provide such “useful” methods of solution, or tutors, many of whom are taught using the same manuals . And as a bonus, you can, without fear and without trembling, begin to solve any problem that has such seemingly scary parameters, but in fact - just letters that only ordinary numbers can stand behind, and nothing more !

Unfortunately, I cannot promise that everything will be easy. Moreover, if you yourself have never tried to ask these insidious questions: why? what for? where did it come from? and what follows from this? However, if you want to learn how to solve problems, you want to understand, you should do it. Yes, it’s hard to think, but you cannot do without it! Try it and you will see how much more interesting life has become!

Attention:   small saturated graphics can be increased by clicking on them with the mouse.

The study and solution of equations with parameters is considered not the easiest section of school mathematics. However, a parameter, as a concept, is often perceived by schoolchildren to be much more complex than it actually is. Here, the first paragraph presents very simple introductory examples of the use of parameters in equations. Those for whom this concept is not very difficult can immediately proceed to the solution of the tasks presented below.

What is an equation with a parameter?

Consider an example.

Let's say we need to solve the equation 2x + 5 = 2 − x.
  Decision: 2x + x = 2 − 5; 3x = −3; x = −3/3 = −1.

Now you need to solve the equation 2x + 5 = 3 − x.
  Decision: 2x + x = 3 − 5; 3x = −2; x = −2/3 ~ −0,67.

Then you need to solve the equation 2x + 5 = 0,5 − x.
  Decision: 2x + x = 0,5 − 5; 3x = −4,5; x = −4,5/3 = −1,5.

And then you may need to solve the equation 2x + 5 = 10,7 − x   or equation 2x + 5 = −0,19 − x.
It is clear that the equations are similar, and therefore their solution will be accompanied by the same actions as above. A natural question arises - how much can one do the same thing?

Reduce our labor costs. Note that all these equations differ in only one number on the right side. Denote this number by the symbol a .
  We get the equation 2x + 5 = a − x,
  Where a - a variable, instead of which you can substitute the desired numerical value and obtain the desired equation.   This variable is called a parameter.

We solve this equation in the same way as all the previous ones.
  Decision: 2x + 5 = a − x; 2x + x = a − 5; 3x = a − 5; x = (a − 5)/3.

Now, in order to find the answers for the last two examples, we can not completely repeat the entire solution of each equation, but simply substitute in the resulting formula for x   numerical value of the parameter a:
x = (10,7 − 5)/3 = 5,7/3 = 1,9;
x = (−0,19 − 5)/3 = −5,19/3 = −1,73.

Thus, under the term "equation with parameter", in fact, lies a whole family of "almost identical equations" which differ from each other by only one number (one term or one coefficient) and are solved equally. A parameter is a number that changes from equation to equation.
  We can program the resulting formula for the root of the equation on a computer. It will be enough to enter the parameter value ato get a solution to any such equation.

Consider another example.

It is necessary to solve several equations:
2x + 5 = 2 − x;
3x + 5 = 2 − x;
−4x + 5 = 2 − x;
17x + 5 = 2 − x;
0,5x + 5 = 2 − x.

We notice that they are similar to each other and differ only in the first coefficient. Denote it, for example, by the symbol k.
  Solve the equation kx + 5 = 2 − x   with parameter k.

Decision:
kx + 5 = 2 − x;
kx + x = 2 − 5;
(k + 1)x = −3;
x = −3/(k + 1).

Using this formula, we calculate all the answers for the above equations.
x = −3/(2 + 1) = −1
x = −3/(3 + 1) = −0,75
x = −3/(−4 + 1) = 1
x = −3/(17 + 1) = −1/6 ~ −0,167
x = −3/(0,5 + 1) = −2

Can we now program this formula and say that it can be used to solve any similar equation?
  We can program. The computer will cope with both very large coefficient values \u200b\u200band very small ones.
  For example, if we introduce k   \u003d 945739721, then for an equation of a given form a root approximately equal to −0.0000000031721201195353831188 will be obtained if k   \u003d 0.0000004, then we get the root ≈ −2,9999988000004799998080000768.
  But, if we introduce into the program a seemingly simpler meaning k   \u003d −1, then the computer will freeze.
  Why?

Let's take a closer look at the formula x = −3/(−1 + 1) = −3/0.   Division by zero ?!!
  Let's look at the corresponding equation −1 x + 5 = 2 − x.
  Transform it −x + x = 2 − 5.
  It turns out that it is equivalent to the equation 0 \u003d −3 ( ?!! ) and cannot have roots.
Thus, there may be exceptions to the general approach to solving “almost identical equations” that must be taken care of separately. Those. conduct a preliminary study of the entire family of equations. This is what they learn in mathematics with the help of so-called problems with parameters.

Graphic methods for solving equations

First, recall what constitutes graphic way to solve the usual equation (without parameter).
  Let an equation of the form f (x) \u003d g (x) be given. We plot the graphs of the functions y \u003d f (x) and y \u003d g (x) and find the intersection points of these graphs. The abscissas of the intersection points are the roots of the equation.

For quick sketching of graphs, repeat again which are studied in the school course of mathematics, and

Let's look at some examples.

1.   Solve the equation
2x + 5 = 2 − x

Answer: x = −1 .

2.   Solve the equation
2x 2 + 4x − 1 = 2x + 3

Answer: x 1 = -2; x 2 = 1 .

3.   Solve the equation
log 2 x = −0,5x + 4

Answer: x = 2 .

You can solve the first two of these equations analytically, since these are ordinary linear and quadratic equations. The second equation contains functions of different classes - power-law (linear here) and transcendental (here logarithmic). For such cases, the choice of solutions for students is very limited. In fact, the only available way is the graphical solution.

Attention:   For roots found graphically, a check is required! Are you sure that in the third figure the intersection is exactly at the point x   \u003d 4, but not at 3.9 or 4.1? And if in a real exam you do not have the opportunity to build a schedule accurately enough? In the freehand drawing, the spread may be even greater. Therefore, the algorithm of actions should be as follows:

1. Preliminary conclusion: x ≈ 4.
2. Verification: log 2 4   \u003d −0.5 · 4 + 4; 2 \u003d −2 + 4; 2 ≡ 2.
3. Final conclusion x = 4.

To graphically solve equations with parameters, it is necessary to build not separate graphs, but their families.

Solving equations with parameters using graphs.

Task 1

q   for which the equation |x + 1| − |x − 3| − x = q 2 − 8q + 13   has exactly 2 roots.

For each parameter value q   can calculate the value of the expression q 2 − 8q + 13 . Denote the result by the variable a.
  Those. will accept q 2 − 8q + 13 = a   and solve the equation with the parameter |x + 1| − |x − 3| − x = a

Build a function graph y = |x + 1| − |x − 3| − x located on the left side of the equation.
  To do this, we divide the numerical axis into segments by points at which each of the encountered modules assumes a zero value.

|x + 1| = 0; x = −1;
|x − 3| = 0; x = 3.

  For each of these sections, we will open the modules taking into account the signs.
Recall:   by definition |x| = x ,   if x ≥ 0 , and |x| = −x ,   if x < 0 . To check the signs of the modules on the site, just substitute any intermediate value x   from this segment, for example, −2, 0 and 4.

So on the plot Iwhere −∞ < x ≤ −1,   we have −(x + 1) + (x − 3) − x = − x − 4.
Therefore, you must plot the function y = − x − 4 .
  This is a linear function. Her graph is a straight line that can be built on two points, for example, x = 0, y = −4   and at = 0, x = −4.   We draw the entire line with a pale line, and then select the part of the graph that applies only to the area under consideration.

Similarly, we deal with the remaining two sections.

Location on II, where −1 < x ≤   3, we have (x + 1) + (x − 3) − x = x − 2
y = x − 2 .

Location on IIIwhere 3 < x ≤ ∞ , we have (x + 1) − (x − 3) − x = − x + 4
  and should build the corresponding part of the function graph y = − x + 4 .

The sequential construction of the final graph is shown below. (To enlarge a drawing, you need to click on it with the left mouse button.)

Comment:   if you have mastered the topic, then you can cope with this part of the task faster than shown in the example.

So, we have completed the construction of the graph of the function located on the left side of the equation. Let's see what is on the right side.

Function graph y = a    represents a straight line parallel to the abscissa ( Ox), and the intersecting ordinate ( Oy) at the point a . Because a   is a parameter that can take different values, then you need to build a whole family of such parallel lines intersecting the ordinate axis at different heights. Obviously, we cannot build all the graphs of the family, since there are an infinite number of them. For example, let’s draw a few pieces in the area of \u200b\u200ba function graph that has already been built. Below are direct families y = a    shown in red.

  The figure shows that the number of intersection points of each of the red lines with the previously constructed (green) graph depends on the height at which this line is located, i.e. from parameter a . Direct below y   \u003d −3, they cross the graph at one point, which means that these equations have only one solution. Lines passing at level −3< y < 1 имеют по три точки пересечения, значит соответствующие уравнения будут иметь по три решения. Прямые, расположенные выше точки y   \u003d 1, again have only one intersection point.
  Exactly two intersection points with a green graph will have only straight lines y   \u003d 1 and y   \u003d −3. The corresponding equations will have exactly two roots, which was required to be determined in the task.

However, we found the values \u200b\u200bof the parameter introduced by us ain which the given equation has 2 roots, and the question of the problem was to find all the values \u200b\u200bof the parameter q . To do this, you have to solve the following set of equations:

  These are ordinary quadratic equations that are solved through discriminant or by the Vieta theorem.


  Thus the final answer: (2; 4; 6).

Task 2

Find all parameter values a for which the equation (2 − x)x(x − 4) = a   has exactly 3 roots.

Consider the function y = (2 − x)x(x − 4) . It can be seen that if you open the brackets, then the senior term will be −x 3 . Those. the graph of the function should be a cubic parabola, and at xtending to + ∞, y   → −∞, and as xtending to −∞, y → +∞.
  Since the equation (2 − x)x(x − 4) = 0   has three roots 2, 0, and 4, then the function graph will cross the abscissa axis three times.
  It is clear that under the mentioned conditions, the graph of a continuous function should have a section with a “wave”. Build a sketch graphic by hand.

Right side of the equation y = a    same as in the previous task. Therefore, further constructions do not require comments. See the pictures. To zoom, use the mouse click.

It can be seen from the figures that the lines separating the lines with three intersection points from other cases pass through the extrema of the cubic function. Therefore, we determine the values y   max and y   min through the derivative. (It is not necessary to fully investigate the function, since we see the approximate position of the points of the extremum on the sketch of the graph.) Note that the exact values \u200b\u200bare used to calculate the values \u200b\u200bof the function x   and abbreviated multiplication formulas. Approximate values \u200b\u200bare not used in intermediate calculations.

Answer:

The challenge for an independent solution

Task 3.

At what is the largest negative value of the parameter a   the equation   has one root?

Show solution.

To enlarge a drawing, left-click on it.
  Move 2 x   to the right side of the equation, as a result we get two elementary functions whose graphs were studied at school.
  From the figure, we see that the line of the graph meets the condition of the task. Therefore, for further calculations, we use the conditions:
  1) the tangent of the angle of inclination of the tangent is equal to the derivative of the function at the point of tangency;
  2) the desired parametric line and the graph have a common point.
  In the calculations, we ignore the module, since we spend them for the right section of the curve ( x > 0 ).

Answer: -1,625

The task of the real exam ZNO-2013 (http://www.osvita.ua/).

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The purpose of this work is to study various ways of 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 of logical thinking, and stimulate cognitive activity. The development of these skills requires longer efforts, which is why in specialized 10-11 classes with in-depth study of the exact sciences a course was introduced: “Mathematical Workshop”, part of which is the solution of equations and inequalities with parameters. The course is among the disciplines included in the school curriculum component.

Elective or optional courses, or a component behind the grid on the topic: “Tasks with parameters,” can help a successful study of methods for solving problems with parameters.

Consider four large classes of tasks with parameters:

1. Equations, inequalities and their systems, which must be solved for any parameter value, or for parameter values \u200b\u200bbelonging to a certain set.
2. Equations, inequalities and their systems, for which it is necessary to determine the number of solutions depending on the parameter value.
3. Equations, inequalities and their systems, for which it is required to find all those parameter values \u200b\u200bfor which the indicated equations (systems, inequalities) have a given number of solutions.
4. Equations, inequalities and their systems, for which, for the desired parameter values, the set of solutions satisfies the given conditions in the domain of definition.

Methods for solving problems with parameters.

1. The analytical method.

This is a direct solution method, repeating standard procedures for finding an answer in tasks without a parameter.

Example 1. Find all parameter values ain which the equation:

(2a - 1) x 2 + ax + (2a - 3) \u003d 0 has at most one root.

At 2 a   - 1 \u003d 0 this equation is not quadratic, therefore the case   a   \u003d 1/2 disassemble separately.

If a   \u003d 1/2, then the equation takes the form 1/2 x   - 2 \u003d 0, it has one root.

If   a   ≠ 1/2, then the equation is quadratic; so that it has no more than one root, it is necessary and sufficient that the discriminant be non-positive:

D= a 2 – 4(2a – 1)(2a – 3) = -15a 2 + 32a – 12;

To write down the final answer, you need to understand

2. The graphical method.

Depending on the task (with 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   determine the number of solutions to the equation .

Note that the number of solutions to the equation   equal to the number of intersection points of the function graphs   and y \u003d a.

Function graph   shown in fig. 1.

y \u003d a   Is a horizontal line. According to the schedule, it is easy to establish the number of intersection points depending on a   (e.g. when a   \u003d 11 - two points of intersection; at a   \u003d 2 - eight intersection points).

Answer: when a < 0 – решений нет; при a   \u003d 0 and a   \u003d 25/4 - four decisions; at 0< a < 6 – восемь решений; при a   \u003d 6 - seven decisions; at

6 < a < 25/4 – шесть решений; при a   \u003e 25/4 - two solutions.

3. The method of solving the parameter.

When solving this method, variables xand a   are accepted as equitable, and that variable is selected with respect to which the analytical solution becomes simpler. After simplifications, you need to return to the original meaning of the variables xand a   and complete the decision.

Example 3. Find all parameter values a   , for each of which the equation \u003d - ax +3a   +2 has a unique solution.

We will solve this equation by changing variables. Let \u003d   t , t   ≥ 0, then x = t   2 + 8 and the equation takes the form at 2 +   t + 5a   - 2 \u003d 0. Now the challenge is to find everything afor which the equation at 2 +   t + 5a   - 2 \u003d 0 has a unique non-negative solution. This occurs in the following cases.

1) If a   \u003d 0, then the equation has a unique solution t = 2.

The solution of certain types of equations and inequalities with parameters.

Tasks with parameters help in the formation of logical thinking, in acquiring the skills of research activities.

The solution to 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 parameter values b   the equation has no roots?

. Power equations, inequalities and their systems.

Task number 2. Find all parameter values aunder 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.

We transform both sides of the inequality.

In order for the set of solutions of the inequality to contain the number 6, the following condition is necessary and sufficient:

Fig. 4

At a   \u003e 6 many solutions to the inequality: .

The interval (0; 5) cannot contain any segment of length 6. Therefore, two disjoint segments of length 6 must be contained in the interval (5;   a).

. Exponential equations, inequalities and systems.

Task number 3. In the field of function definition took all positive integers and added them. Find all the values \u200b\u200bfor which such a sum will be more than 5, but less than 10.

1) A graph of a linear fractional function   is a hyperbole. By condition x   \u003e 0. With unlimited increase x   the fraction decreases monotonically and approaches zero, and the values \u200b\u200bof the function z increase and approach 5. In addition, z (0) \u003d 1.

2) By definition of degree, the scope D (y)   consists of solutions to inequality. At a\u003d 1 we obtain an inequality for which there are no solutions. Therefore, the function at   not defined anywhere.

3) At 0< a< 1 показательная функция с основанием a   Inequality also decreases, which is equivalent to inequality. Because x   \u003e 0, then z(x) > z(0) \u003d 1. So every positive value x   is a solution to inequality. Therefore, for such a   the amount indicated in the condition cannot be found.

4) When a   \u003e 1 exponential function with base a   and inequality is tantamount to inequality. If   a   ≥ 5, then any positive number is its solution, and the sum indicated 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. We calculate the sum of consecutive natural numbers starting from 1: 1; 1 + 2 \u003d 3; 1 + 2 + 3 \u003d 6; 1 + 2 + 3 + 4 \u003d 10; ... Therefore, the indicated amount will be more 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. Mean 3< x   0 ≤ 4. As 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 Appendix 1.

Tasks with parameters are difficult because there is no single algorithm for solving them. The specificity of such problems is that along with unknown quantities, parameters appear in them, the numerical values \u200b\u200bof which are not indicated specifically, but are considered known and given on a certain numerical set. Moreover, the parameter values \u200b\u200bsignificantly affect the logical and technical progress of the problem solution and the response form.

According to statistics, many of the graduates do not begin to solve problems with parameters on the exam. According to FIPI, only 10% of graduates begin to solve such problems, and the percentage of their correct solution is low: 2-3%, so the acquisition of skills to solve difficult, non-standard tasks, including tasks with parameters, school students still remains relevant.