The purpose of this work is to study in various ways solving problems with parameters. The ability and ability to solve problems with parameters demonstrates mastery of methods for solving equations and inequalities, a meaningful understanding of theoretical information, level logical thinking, stimulate cognitive activity. To develop these skills, longer efforts are required, which is why in specialized grades 10-11 with in-depth study of the exact sciences, the course “Mathematical Practicum” has been introduced, part of which is the solution of equations and inequalities with parameters. The course is one of the disciplines included in the school's curriculum component.
Successful study of methods for solving problems with parameters can be helped by elective or elective courses, or a component behind the grid on the topic: “Problems with parameters.”
Let's consider four large classes of problems with parameters:
- Equations, inequalities and their systems that must be solved for any parameter value, or for parameter values belonging to a certain set.
- Equations, inequalities and their systems for which it is necessary 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 parameter values for which the above equations(systems, inequalities) have a given number of solutions.
- Equations, inequalities and their systems for which, for the required parameter values, the set of solutions satisfies the given conditions in the domain of definition.
Methods for solving problems with parameters.
1. Analytical method.
This is the way direct solution, repeating standard procedures for finding the answer in problems without a parameter.
Example 1: Find all values of a parameter 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 is sorted separately.
If a= 1/2, then the equation takes the form 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 for the discriminant to 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. Graphic method.
Depending on the task (with variable x and parameter a) graphs 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 = a.
Graph of a function 
shown in Fig. 1.
y = a is a horizontal line. Using the graph, it is easy to determine 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 solving with respect to a parameter.
When solving this way, the variables X And A are accepted as 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 finish the solution.
Example 3: Find all values of a parameter A, for each of which the equation = - ax +3a+2 has the only solution.
We will solve this equation by changing variables. Let = t , t≥ 0, then x = t 2 + 8 and the equation becomes at 2 +t + 5a– 2 = 0. Now the challenge is to find everything A, for which the equation at 2 +t + 5a– 2 = 0 has a unique non-negative solution. This occurs in the following cases.
1) If A= 0, then the equation has a unique solution t = 2.
Solving some types of equations and inequalities with parameters.
Problems with parameters help in the formation of logical thinking and in acquiring research skills.
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.Problem No. 1. At what values of the parameter b does the equation have no roots?
Task No. 2. Find all parameter values a, for which the set of solutions to the inequality is:
contains the number 6, and also contains two segments of length 6 that have no common points.
Let's 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 met: 
Fig.4
At a> 6 set of solutions to the inequality: 
.
The interval (0;5) cannot contain any segment of length 6. This means that two disjoint segments of length 6 must be contained in the interval (5; a).
. Exponential equations, inequalities and systems.Problem No. 3. In the area of defining a function 
take all the positive integers and add them up. Find all values for which this sum is greater than 5 but less than 10.
1) Graph of a linear fractional function 
is a hyperbole. By condition x> 0. With unlimited increase X the fraction monotonically decreases and approaches zero, and the function values z increase and approach 5. Moreover, z(0) = 1.
2) By definition of degree, domain of definition D(y) consists of solutions to the inequality. At a= 1 we obtain an inequality that has no solutions. Therefore the function at not defined anywhere.
3) At 0< a< 1 показательная функция с основанием A decreases and inequality is equivalent to inequality. Because x> 0, then z(x) > z(0) = 1 . So every 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 inequality is equivalent to 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 amount 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) .
Solving irrational equations and inequalities, as well as equations, inequalities and systems containing modules are discussed in Appendix 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 contain parameters whose numerical values are not specifically indicated, but are considered known and specified on a certain numerical set. In this case, the parameter values significantly influence the logical and technical course of solving the problem and the form of the answer.
According to statistics, many graduates do not start solving problems with parameters on the Unified State Exam. According to FIPI, only 10% of graduates begin to solve such problems, and their percentage the right decision low: 2–3%, so the acquisition of skills for solving difficult, non-standard tasks, including problems with parameters, by school students still remains relevant.
Task 1 #6329
Task level: Equal to the Unified State Exam
Find all values of the parameter \(a\) , for each of which the system \[\begin(cases) (x-2a-2)^2+(y-a)^2=1\\ y^2=x^2\end(cases)\]
has exactly four solutions.
(USE 2018, main wave)
The second equation of the system can be rewritten as \(y=\pm x\) . Therefore, we consider two cases: when \(y=x\) and when \(y=-x\) . Then the number of solutions of the system will be equal to the sum of the number of solutions in the first and second cases.
1) \(y=x\) . Substitute into the first equation and get: \
(note that in the case of \(y=-x\) we will do the same and also obtain a quadratic equation)
For the original system to have 4 different solutions, it is necessary that in each of the two cases 2 solutions are obtained.
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 a 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 \(x_0\) be the general solution of equations (1) and (2), then \
From here we get that either \(x_0=0\) or \(a=0\) .
If \(a=0\) , then equations (1) and (2) are the same, therefore, they have the same roots. This case does not suit us.
If \(x_0=0\) is their common root, then \(2x_0^2-2(3a+2)x_0+(2a+2)^2+a^2-1=0\), from which \((2a+2)^2+a^2-1=0\) , from which \(a=-1\) or \(a=-0.6\) . Then the entire original system will have 3 different solutions, which does not suit us.
Considering all this, the answer will be:
Answer:
\(a\in\left(\frac(-2-\sqrt2)3; -1\right)\cup\left(-1; -0.6\right)\cup\left(-0.6; - 2+\sqrt2\right)\)
Task 2 #4032
Task level: Equal to the Unified State Exam
Find all values of \(a\) , 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 a unique solution.
Let's rewrite the system in the form: \[\begin(cases) ax^2+2ax+a\leqslant x^2-4\\ ax^2+2ax+a\geqslant -2x-1 \end(cases)\] Let's 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 that is specified by the conditions: “above” the graph \(h\) but “below” the graph \(g\):
(we will call the “left” region region I, the “right” region region II)
Note that for each fixed \(a\ne 0\) the graph of \(y\) is a parabola, the vertex of which is at the point \((-1;0)\), and the branches are directed either up or down. If \(a=0\) , then the equation looks like \(y=0\) and the graph is a straight line coinciding with the x-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 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 \(y\) face upward. In order for the original system to have a unique solution, it is necessary that the parabola \(y\) touch the boundary of region I or the boundary of region II, that is, touch the parabola \(g\), and the abscissa of the tangency point must be \(\leqslant -3\) or \(\geqslant 2\) (that is, the parabola \(y\) must touch the boundary of one of the regions that is located above the abscissa axis, since the parabola \(y\) lies above the abscissa axis).
\(y"=2a(x+1)\) , \(g"=2x\) . Conditions for the graphs \(y\) and \(g\) to touch at the point with the 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 \(x_0=-4\) , \(a=\frac43\) .
We got the first value of the parameter \(a\) .
2) \(a=0\) . Then \(y=0\) and it is clear that the straight line has an infinite number of common points with region II. Therefore, this parameter value does not suit 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's 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 with this value of the parameter the second point of intersection of the parabola \(y=-\frac34(x+1)^2\) with the straight line \(h=-2x-1\) is the point with coordinates \(\left(-\frac13; -\frac13\right)\).
Thus, we received another parameter value.
Since we have considered all possible cases for \(a\) , the final answer is: \
Answer:
\(\left\(-\frac34; \frac43\right\)\)
Task 3 #4013
Task level: Equal to the Unified State 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 equal to \(D=9y^2\) , therefore, \
Then the equation can be rewritten as \[(x-2y)\cdot (2x-y)=0\] Therefore, the entire system can be rewritten as \[\begin(cases) \left[\begin(gathered)\begin(aligned) &y=2x\\ &y=0.5x\end(aligned)\end(gathered)\right.\\ (x-a)^2 +(y-a)^2=5a^4\end(cases)\] The set defines two straight lines, the second equation of the system defines a circle with center at \((a;a)\) 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 \(y=kx\) has a tangent of the angle of inclination of this line to the positive direction of the axis \(Ox\) is equal to \(k\), then the tangent of the angle of inclination of the straight line \(y=0.5x\) is equal to \ (0.5\) (let's call it \(\mathrm(tg)\,\alpha\)), the straight line \(y=2x\) is equal to \(2\) (let's call it \(\mathrm(tg)\ ,\beta\) ). Note 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 the 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 symmetrical to each other with respect to \(y=x\), therefore, if the circle touches one of them , then it necessarily touches the second line.
Note that if \(a=0\) , then the circle degenerates into the point \((0;0)\) and has only one point of intersection with both lines. That is, this case does not suit us.
Thus, in order for a circle to have 2 points of intersection with lines, it must touch these lines: 
We see that the case when the circle is located in the third quarter is symmetrical (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. 
Note 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 hand, \[\mathrm(tg)\,\angle QOK=\mathrm(tg)\,(45^\circ-\alpha)=\dfrac(\mathrm(tg)\, 45^\circ-\mathrm(tg) \,\alpha)(1+\mathrm(tg)\,45^\circ\cdot \mathrm(tg)\,\alpha)\] 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 \(a\) . Therefore, the answer is:\
Answer:
\(\{-0,2;0,2\}\)
Task 4 #3278
Task level: Equal to the Unified State Exam
Find all values of \(a\) , for each of which the equation \
has a unique solution.
(USE 2017, official trial 04/21/2017)
Let's make the change \(t=5^x, t>0\) and move all the terms into one part: \
We obtained a quadratic equation, the roots of which, according to Vieta’s theorem, are \(t_1=a+6\) and \(t_2=5+3|a|\) . In order for the original equation to have one root, it is sufficient that the resulting equation with \(t\) also has one (positive!) root.
Let us immediately note that \(t_2\) for all \(a\) will be positive. Thus, we get two cases:
1) \(t_1=t_2\) : \ &a=-\dfrac14 \end(aligned) \end(gathered) \right.\]
2) Since \(t_2\) is always positive, then \(t_1\) must be \(\leqslant 0\) : \
Answer:
\((-\infty;-6]\cup\left\(-\frac14;\frac12\right\)\)
Task 5 #3252
Task level: Equal to the Unified State 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 must satisfy 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 when they coincide: \
So it suits us \(a\in \left[-\frac13; 0\right)\) and \(a=1\) .
Answer:
\(a\in \left[-\frac13;0\right)\cup\(1\)\)
Task 6 #3238
Task level: Equal to the Unified State Exam
Find all values of the parameter \(a\) , for each of which the equation \
has a single root on the segment \(.\)
(USE 2017, reserve day)
The equation is equivalent: \ ODZ equations: \[\begin(cases) x\geqslant 0\\ x-a\geqslant 0\\3a(1-x) \geqslant 0\end(cases)\] On the 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: \ Does not fit \(a<0\) . Следовательно, эти значения \(a\) не подходят.
2) Let \(a=0\) . Then the ODZ equation: \(x\geqslant 0\) . The equation will be rewritten as: \ The resulting root fits the ODZ and is included in the segment \(\) . Therefore, \(a=0\) is suitable.
3) Let \(a>0\) . Then ODZ: \(x\geqslant a\) and \(x\leqslant 1\) . Therefore, if \(a>1\) , then the ODZ is an empty set. Thus, \(0
The derivative is equal to \(y"=3x^2-2ax+3a\). Let's determine what sign the derivative can have. To do this, 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, \(y\) increases. Thus, by the property of an increasing function, the equation \(y(x)=0\) can have no more than 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 \[\begin(cases) y(1)\geqslant 0\\ y(a)\leqslant 0 \end(cases)\quad\Rightarrow\quad a\in \] Considering that initially in the case under consideration \(a\in (0;1]\) , then the answer is \(a\in (0;1]\). 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 \(\) .
Let's 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 coincides with \(x_1\) , or satisfies \((1)\) , but 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 both be less than zero and satisfy \((1)\) (that is, be greater than \(\frac35\) ). Given this remark, the cases are recorded in the following 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> 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 that has two roots on \(\) . This value of \(a\) 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:
\(\left(-\frac(13)5;-\frac(12)5\right] \cup\left[\frac(12)5;\frac(13)5\right)\)
Report on the GMO of a mathematics teacher at MBOU Secondary School No. 9
Molchanova Elena Vladimirovna
“Preparation for the Unified State Exam in mathematics: problems with parameters.”
Since there is no definition of the parameter in school textbooks, I propose to take the following simplest version as a basis.
Definition . A parameter is an independent variable, the value of which in the problem is considered to be 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, this depends on the question in the problem. If, for example, it is necessary to solve an equation, an inequality, a system or a set of them, then this means presenting a reasoned answer either for any value of a parameter or for a value of a parameter belonging to a predetermined set.
If you need to find parameter values for which the set of solutions to an equation, inequality, etc. satisfies the declared condition, then, obviously, the solution to the problem consists of finding the specified parameter values.
The reader will develop a more transparent understanding of what it means to solve a problem with a parameter after reading the examples of problem solving on the following pages.
What are the main types of problems with parameters?
Type 1. Equations, inequalities, their systems and sets that must be solved either for any value of the parameter (parameters) or for parameter values belonging to a predetermined set.
This type of problem is basic when mastering the topic “Problems 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 necessary to determine the number of solutions depending on the value of the parameter (parameters).
I draw your attention to the fact that when solving problems of this type there is no need to solve given equations, inequalities, their systems and aggregates, etc., nor provide these solutions; In most cases, such unnecessary work is a tactical mistake that leads to unnecessary waste of time. However, one should not make this absolute, since sometimes a direct solution in accordance with type 1 is the only reasonable way to obtain an answer when solving a problem of type 2.
Type 3. Equations, inequalities, their systems and collections, for which it is required to find all those parameter values for which the specified equations, inequalities, their systems and collections 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 some sense the inverse of problems of type 2.
Type 4. Equations, inequalities, their systems and sets, for which, for the required values of the parameter, the set of solutions satisfies the specified conditions in the domain of definition.
For example, find parameter values at which:
1) the equation is satisfied 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.
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 in 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 are 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 standard procedures for finding the answer in problems without a parameter. Sometimes they say that this is a method of forceful, in a good sense, “arrogant” solution.
Comment. The analytical method of solving problems with a parameter is the most difficult method, 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 ).
Comment. Exceptional visibility and beauty graphic method solving problems with a parameter captivates students of the topic “Problems with a parameter” so much 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 can be solved brilliantly in this way and with enormous 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 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 considered 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 move on to demonstrating 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 parameters solved graphically, I begin my acquaintance with the parameters with the tasks of the Unified State Exam B7 2002:
At what is the integer value for the equation 45x – 3x 2 - X 3 + 3k = 0 has exactly two roots?
These tasks allow, firstly, to remember how to construct graphs using the derivative, and secondly, to explain the meaning of the straight line y = k.
In subsequent classes, I use a selection of easy and medium-level competitive problems with parameters for preparing for the Unified State Exam, equations with a module. These tasks can be recommended to mathematics teachers as a starting set of exercises for learning to work with the parameter enclosed under the module sign. Most of the numbers are solved graphically and provided to the teacher ready plan lesson (or two lessons) with a strong student. Initial training for the Unified State Examination in mathematics using exercises close in complexity to real C5 numbers. Many of the proposed tasks are taken from materials for preparing for the Unified State Exam 2009, and some are 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
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 parameter valuesb equation has a single solution? 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 different roots. (If there is more than one value of a, then write down their sum in your answer.)
Answer: 3
7) At what valuesb
equation has exactly 2 solutions?
Answer:
8) Specify these parametersk
, for which the equation has at least two solutions.
Answer:
9) At what parameter valuesp
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 is more than one value of a, then write down their sum in response).
Answer: 4
12) At what smallest natural value of parameter a is 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 your answer).
Answer:- 3
14) Specify the following 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 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 strongest student, who nevertheless aspires to high score Unified State Exam by solving number C5. The teacher prepares such a student in several stages, allocating separate lessons for training individual skills necessary for finding and implementing long-term solutions. This selection is suitable for the stage of forming ideas about floating patterns, depending on the parameter. Numbers 16 and 17 are based on the model of a real equation with a parameter on the Unified State Exam 2011. The tasks are arranged in order of increasing difficulty.
Assignment C5 in mathematics Unified State Exam 2012
Here we have a traditional parameter problem that requires a moderate mastery 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 designed primarily for those who intend to continue their education at universities with increased requirements for the mathematical preparation 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 solutions.
On the site of preparation for the Unified State Exam of Alexander Larin from May 11, 2012, training options No. 1 – 22 with tasks of level “C”, 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 look at the solution to task C5 from the 2012 exam.
Task C5 from the Unified State Exam 2012
For what values of the parameter a does 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 problem question as follows: at 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 functions:
1. Shift the graph of the function3 units down along the OY axis, we get the graph of the function:
2. Let's plot the function . To do this, part of the graph of the function , located below the OX axis, will be displayed symmetrically relative to this axis:
So, the graph of the functionhas the form:
Graph of a function
MKOU "Lodeynopolskaya secondary secondary school No. 68"
_________________________________________________________________________________________________________________________________
Speech at a meeting of the Moscow Region
Problem solving methods
with parameters
Prokusheva Natalya Gennadievna
Lodeynoye Pole
2013-2014
Problems with parameters
Problems with parameters are among the most difficult of the problems offered both at the Unified State Exam and at additional competitive examinations at universities.
They play an important role in the formation of logical thinking and mathematical culture. The difficulties that arise when solving them are due to the fact that each problem with parameters represents a whole class of ordinary problems, for each of which a solution must be obtained.
If in an equation (inequality) some coefficients are not given by specific numerical values, but are designated by letters, then they are called parameters, and the equation (inequality) is parametric.
As a rule, unknowns are denoted by the last letters of the Latin alphabet: x, y, z, ..., and parameters by the first: a, b, c, ...
To solve an equation (inequality) with parameters means to indicate at what values of the parameters solutions exist and what they are. Two equations (inequalities) containing the same parameters are called equivalent if:
a) they make sense for the same parameter values;
b) every solution to the first equation (inequality) is a solution to the second and vice versa.
Naturally, such small class problems does not allow many to grasp the main thing: the parameter, being a fixed but unknown number, has a kind of dual nature. Firstly, the supposed fame allows you to “communicate” with the parameter as a number, and secondly, the degree of freedom of communication is limited by its obscurity. Thus, dividing by an expression containing a parameter and extracting the root of an even degree from such expressions require preliminary research. Typically, the results of these studies influence both the decision and the answer.
How to start solving such problems? Don't be afraid of problems with parameters. First of all, you need to do what is done when solving any equation or inequality - bring the given equation (inequality) to a more simple view, if possible: factor a rational expression, factor a trigonometric polynomial, get rid of modules, logarithms, etc.. then you need to carefully read the task again and again.
When solving problems containing a parameter, there are problems that can be divided into two large classes. The first class includes problems in which it is necessary to solve an inequality or equation for all possible values parameter. The second class includes tasks in which not everything needs to be found. possible solutions, but only those that satisfy some additional conditions.
The most understandable way for schoolchildren to solve such problems is to first find all the solutions and then select those that satisfy additional conditions. But this is not always possible. There are a large number of problems in which it is impossible to find all the many solutions, and we are not asked to do so. Therefore, we have to look for a way to solve the problem without having at our disposal the entire set of solutions to a given equation or inequality, for example, to look for the properties of the functions included in the equation that will allow us to judge the existence of a certain set of solutions.
Main types of tasks with parameters
Type 1. Equations, inequalities, their systems and sets that must be solved either for any value of the parameter (parameters) or for parameter values belonging to a predetermined set.
This type of problem is basic when mastering the topic “Problems 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 necessary to determine the number of solutions depending on the value of the parameter (parameters).
We draw attention to the fact that when solving problems of this type, there is no need either to solve given equations, inequalities, their systems and combinations, etc., or to provide these solutions; In most cases, such unnecessary work is a tactical mistake that leads to unnecessary waste of time. However, one should not make this absolute, since sometimes a direct solution in accordance with type 1 is the only reasonable way to obtain an answer when solving a problem of type 2.
Type 3. Equations, inequalities, their systems and collections, for which it is required to find all those parameter values for which the specified equations, inequalities, their systems and collections 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 some sense the inverse of problems of type 2.
Type 4. Equations, inequalities, their systems and sets, for which, for the required values of the parameter, the set of solutions satisfies the specified conditions in the domain of definition.
For example, find parameter values at which:
1) the equation is satisfied 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.
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 in 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 are problems with one unknown and one parameter. The next paragraph indicates the main ways to solve problems of this particular class.
Basic methods for solving problems with a parameter
Method I(analytical). This is a method of the so-called direct solution, repeating standard procedures for finding the answer in problems without a parameter. Sometimes they say that this is a method of forceful, in a good sense, “arrogant” solution.
Method II(graphic). Depending on the task (with variable x and parameter a) graphs are considered or in the coordinate plane ( x; y), or in the coordinate plane ( x; a).
Comment. The exceptional clarity and beauty of the graphical method of solving problems with a parameter captivates students of the topic “Problems with a parameter” so much 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 parameter). When solving this way, the variables x And a are accepted as equal and the variable with respect to which the analytical solution is considered simpler is selected. After natural simplifications, we return to the original meaning of the variables x And a and finish the solution.
Let us now move on to demonstrating these methods for solving problems with a parameter.
1. Linear equations and inequalities with parameters
Linear function: – equation of a straight line with slope coefficient . The angular coefficient is equal to the tangent of the angle of inclination of the straight line to the positive direction of the axis .
Linear equations with parameters of the form
If , the equation has the only thing solution.
If , that equation has no solutions, When , and the equation has infinitely many solutions, When .
Example 1. Solve the equation | x | = a .
Solution:
a > 0, => x 1.2 = ± a
a = 0, => x = 0
a < 0, =>there are no solutions.
Answer: x 1.2 = ± a at a > 0; x= 0 at a= 0; there are no solutions for a < 0.
Example 2. Solve equation |3 – x | = a .
Solution:
a > 0, => 3 – x = ± a , => x= 3 ± a
a = 0, => 3 – x = 0. => x = 3
a < 0, =>there are no solutions.
Answer: x 1.2 = 3 ± a at a > 0; x= 3 at a= 0; there are no solutions for a < 0.
Example 3. Solve the equation m ² x – m = x + 1.
Solution:
m ² x – m = x + 1
m ² x – x = m + 1
(m² – 1)x = m + 1
Answer: 
at m
≠ ±
1; x
Є
R at m= –1; there are no solutions for m
= 1.
Example 4. A solve the equation: ( a 2 – 4) x = a + 2 .
Solution: Let us factorize the coefficient. .
If , the equation has the only thing solution: .
If , equation has no solutions.
If , then the equation has infinitely many solutions .
Example 6. For all parameter values a
solve the equation: 
.
Solution: ODZ: .
Under this condition, the equation is equivalent to the following: 
.
Let's check whether you belong to the ODZ: ,
If .
If ,
then the equation has no solutions.
Example 7. For all parameter values A solve the equation: | X + 3| – a | x – 1| = 4.
Solution: Let's divide the number line into 3 parts by points at which the expressions under the modulus sign vanish and solve 3 systems:
1) 
,
If .
Found will be the solution if .
2) 
,
If .
The one found satisfies the required inequality, therefore, is a solution for .
If ,
then the solution is any .
3) 
,
If .
Found Not satisfies the required inequality, therefore, Not is a solution when .
If ,
then the solution is any x > 1.
Answer: at ; at ;
n ri ; is also a solution for all .
Example 8. Find all A, for each of which at least one of the solutions to equation 15 x – 7a = 2 – 3ax + 6a less 2 .
Solution: Let us find solutions to the equation for each .
, If .
Let's solve the inequality: 
.
When the equation has no solutions.
Answer : AÎ (–5 , 4) .
Linear inequalities with parameters
For example: Solve inequality: kx < b .
If k> 0, then 
. If k
< 0, то 
. If k= 0, then when b> 0 solution is any x
Є R, and when 
there are no solutions.
Solve the remaining inequalities in the box in the same way.
Example 1. For all values of parameter a, solve the inequality 
.
Solution:
. If the parenthesis is before x is positive, i.e. at 
, That 
. If the parenthesis is before x negative, i.e. at 
, That 
. If a= 0 or a = , then there are no solutions.
Answer: at ; at ;
there are no solutions for a= 0 or a = .
Example 2. For all parameter values A solve inequality | X– a| – | x + a| < 2a .
Solution:
At a=0 we have an incorrect inequality 0< 0, т.е. решений нет. Пусть a >0, then at x< –a both modules are expanded with a minus and we get the incorrect inequality 2 a < 2a, i.e. there are no solutions. If x Є [– a ; a] , then the first module opens with a minus, and the second with a plus, and we get the inequality –2 x < 2a, i.e. x > –a, i.e., the solution is any x Є (– a ; a]. If x > a both modules open with a plus and we get the correct inequality –2 a < 2a, i.e. , the solution is any x Є ( a; +∞). Combining both answers, we get that when a > 0 x Є (– a ; +∞).
Let a < 0, тогда первое слагаемое больше, чем второе, поэтому разность в левой части неравенства положительна и, следовательно, не может быть меньше отрицательного числа 2a. Thus, with a < 0 решений нет.
Answer: x
Є (– a; +∞) at a> 0, there are no solutions for 
.
Comment. The solution to this problem is faster and simpler if you use the geometric interpretation of the modulus of the difference of two numbers as the distance between points. Then the expression on the left side can be interpreted as the difference in distances from the point X to points A And - A .
Example 3. Find all A, for each of which all solutions of the inequality 
satisfy inequality 2 x
– a² + 5< 0.
Solution:
The solution to the inequality |x | ≤ 2 is a set A=[–2; 2], and the solution to inequality 2 x
– a² + 5< 0 является множество B
= (–∞; 
) . To satisfy the conditions of the problem, it is necessary that set A be included in set B (). This condition will be satisfied if and only if .
Answer: a Є (–∞; –3)U (3; +∞).
Example 4. Find all values of a for which the inequality 
runs for everyone x from the segment.
Solution:
A fraction is less than zero between the roots, so you need to figure out which root is larger.
–3a
+ 2 < 2a
+ 4 
and –3 a
+ 2 > 2a
+ 4 
. Thus, with 
xЄ (–3 a
+ 2; 2a+ 4) and for the inequality to hold for all x from the segment , it is necessary that
At 
xЄ (2 a
+ 4; –3a+ 2) and so that the inequality holds for all x from the segment , it is necessary that
When a = – (when the roots coincide) there are no solutions, because in this case the inequality takes the form: .
Answer: 
.
Example 5. A the inequality is valid for all negative values X?
Solution:
The function increases monotonically if the coefficient at x non-negative, and it decreases monotonically if the coefficient at x negative.
Let us find out the sign of the coefficient at 
a ≤ –3,
a ≥ 1; (a² + 2 a – 3) < 0 <=> –3 < a < 1.
a ≤ –3,
Let a≥ 1. Then the function f
(x
)
does not decrease monotonically, and the condition of the problem will be satisfied if f
(x
)
≤ 0 <=> 3a
² – a
– 14 ≤ 0 <=> 
.
a ≤ –3,
Along with the conditions a≥ 1; we get: 
Let -3< a < 1. Тогда функция f (x ) decreases monotonically, and the condition of the problem can never be satisfied.
Answer:
.
2. Quadratic equations and inequalities with parameters
.
In the set of real numbers, this equation is studied using the following scheme.
Example 1. At what values a equationx ² – ax + 1 = 0 has no real roots?
Solution:
x ² – ax + 1 = 0
D = a ² – 4 1 =a ² – 4
a ² – 4< 0 + – +
( a – 2)( a + 2) < 0 –2 2
Answer: ata Є (–2; 2)
Example 2.For what values of a does the equation A (X ² – X + 1) = 3 X + 5 has two different real roots?
Solution:
A (X ² – X + 1) = 3 X + 5, A ≠ 0
Oh ² – ah+ a – 3 X – 5 = 0
Oh ² – ( A + 3) X + A – 5 = 0
D = ( a +3)² – 4a ( a – 5) = a ² +6a + 9 – 4 a ² + 20a = –3 a ² + 26a + 9
–3 a ² + 26 a + 9 > 0
3 a ² – 26a – 9 < 0
D = 26² – 4 3 (–9) = 784
a
1
=
;
a
2
=
+ –
+
0 9
Answer:ataЄ (–1/3; 0)U (0; 9)
Example 3: Solve the equation 
.
Solution:
ODZ: x ≠1, x ≠ a
x
– 1 +
x
–
a
= 2, 2
x
= 3 +
a
, 
1)
; 3 +
a
≠ 2;
a
≠ –1
2) 
; 3 +
a
≠ 2
a
;
a
≠ 3
Answer:
ata
Є (–∞; –1)U
(–1; 3)
U
(3; +∞);
there are no solutions fora = –1; 3.
Example4 . Solve the equation | x ²–2 x –3 | = a .
Solution:
Let's look at the functions y = | x ²–2 x –3 | Andy = a .
At a
<
0
no solutions;
at a
=
0 and a> 4 two solutions;
at 0< a
< 4 – четыре решения;
at a= 4 – three solutions.
Answer:
at a
< 0 нет решений;
at a= 0 and a> 4 two solutions;
at 0< a
< 4 – четыре решения;
at a= 4 – three solutions.
Example 5.Find all values
a
, for each of which the equation
|
x
²–(
a
+2)
x
+2
a
| = |
3
x
–6
|
has exactly two roots. If such values
a
more than one, indicate their product in your answer.
Solution:
Let's decompose quadratic trinomial x
²–(
a
+2)
x
+2
a
by multipliers.
;
;
;
We get |
(
x
–2)(
x
–
a
)
| =
3
|
x
–2
|.
This equation is equivalent to the set
Therefore, this equation has exactly two roots if a+ 3 = 2 and a
– 3 = 2.
From here we find that the desired values a are a
1
= –1; a
2
= 5; a
1
·
a
2
= –5.
Answer: –5.
Example 6.Find all values a , for which the roots of the equation ax ² – 2( a + 1) x – a + 5 = 0 are positive.
Solution:
Checkpoint a= 0, because changes the essence of the equation.
1. a = 0 –2x + = 0;
Answer: a Є U .
Example 7.Atwhat parameter values a equation | x ² – 4 x + 3 | = ax has 3 roots.
Solution:
Let's build function graphs y = | x ² – 4 x + 3 | And y = ax .
The function is graphed on the segment 
.
This equation will have three roots if the graph of the function y
=
ax will be tangent to the graph y
= x
²+ 4
x
– 3
on
segment
The tangent equation has the form y
=
f
(x
0
) +
f
’(x
0
)(x
–
x
0
),
Because tangent equation y
=
a, we obtain a system of equations 
Because x 0 Є ,
Answer: at a
= 4 – 2
.
Quadratic inequalities with parameters
Example.Find all parameter values
a
, for each of which among the solutions to the inequalities 
there is not a single point on the line segment.
Solution:
First, let’s solve the inequality for all values of the parameter, and then find those for which there is not a single point of the segment among the solutions .
Let 
, ax
= t
²
t ≥ 0
With such a replacement of variables, the ODZ of inequality is performed automatically. x can be expressed through t, If a≠ 0. Therefore, the case when a
= 0, we will consider separately.
1.Let a
= 0, then X> 0, and the given segment is a solution.
2.Let a≠ 0, then 
and inequality will take the form 
, 
The solution to the inequality depends on the values a, so we have to consider two cases.
1) If a>0, then 
at 
, or in old variables,
The solution does not contain a single point of the given segment if and only if the conditions are met a ≤ 7,
16a≥ 96. Hence, a
Є .
2). If A< 0, то ; 
; tЄ (4 a
; a). Because t≥ 0, then there are no solutions.
Answer: .
Irrational equations with parameters
When solving irrational equations and inequalities with a parameter, firstly, the range of acceptable values should be taken into account. Secondly, if both sides of the inequality are non-negative expressions, then such an inequality can be squared while maintaining the sign of the inequality.
In many cases irrational equations and the inequalities after changing variables are reduced to quadratic ones.
Example 1. Solve the equation 
.
Solution:
ODZ: x + 1 ≥ 0, x ≥ –1, a ≥ 0.
x + 1 = a ².
If x = a² – 1, then the condition is satisfied.
Answer: x = a² – 1 at A≥ 0; there are no solutions for a < 0.
Example 2: Solve the equation 
.
Solution:
ODZ: x + 3 ≥ 0, x ≥ –3,
a–x ≥ 0; x ≤ a;
x + 3 = a–x,
2x = a – 3,
<=>
<=>
<=>
a
≥ –3.
Answer: at a≥ –3; there are no solutions for a
< –3.
Example 3. How many roots does the equation have? 
depending on parameter values A?
Solution:
Range of acceptable values of the equation: x Є [–2; 2]
Let's build graphs of functions. The graph of the first function is the upper half of the circle x² + y² = 4. The graph of the second function is the bisector of the first and second coordinate angles. From the graph of the first function, subtract the graph of the second and get the graph of the function 
. If you replace at on A, then the last graph of the function is the set of points (x; a) satisfying the original equation. 
According to the graph we see the answer.
Answer: at AЄ (–∞; –2) U (1; +∞), no roots;
at AЄ [–2; 2), two roots;
at A= 1, one root.
Example 4. At what parameter values A equation 
has a single solution?
Solution:
Method 1 (analytical):
Answer:
Method 2 (graphical):
Answer: for a ≥ –2 the equation has a unique solution
Example 5. For what values of the parameter a does the equation = 2 + x have a unique solution.
Solution:
Let's consider a graphical version of the solution to this equation, that is, we will construct two functions:
at 1 = 2 + X And at 2 =
The first function is linear and passes through the points (0; 2) and (–2; 0).
The graph of the second function contains a parameter. Let us first consider the graph of this function at A= 0 (Fig. 1). When changing the parameter value, the graph will move along the axis OH by the corresponding value to the left (for positive A) or to the right (for negative A) (Fig. 2)
From the figure it is clear that when A < –2 графики не пересекают друг друга, а следовательно не имеют general solutions. If the value of parameter a is greater than or equal to –2, then the graphs have one intersection point, and therefore one solution.
Answer: at a≥ –2 the equation has a unique solution.
Trigonometric equations with parameters.
Example 1.Solve the equation sin (– x + 2 x – 1) = b + 1.
Solution:
Given the oddness of the function 
, we reduce this equation to the equivalent 
.
1. b = –1
3. b =–2
4. | b + 1| > 1
There are no solutions.
5. bЄ(–1; 0)
6. bЄ(–2; –1)
Example 2.Find all values of the parameter p for which the equation
has no solutions.
Solution:
Let's express cos 2 x through sinx.
Let 
then the task was reduced to finding all the values p, for which the equation has no solutions on [–1; 1]. The equation cannot be solved algorithmically, so we will solve the problem using a graph. Let's write the equation in the form , and now a sketch of the graph of the left side 
easy to build.
The equation has no solutions if the straight line y
=
p+ 9 does not intersect the graph on the interval [–1; 1], i.e. 
Answer:p Є (–∞; –9) U (17; +∞).
Systems of equations with parameters
Systems of two linear equations with parameters
System of equations 
The solutions to a system of two linear equations are the points of intersection of two straight lines: and .
There are 3 possible cases:
1. Lines are not parallel . Then their normal vectors are not parallel, i.e. . In this case, the system has the only solution.
2. The lines are parallel and do not coincide. Then their normal vectors are parallel, but the shifts are different, i.e. .
In this case the system has no solutions .
3. The straight lines coincide. Then their normal vectors are parallel and the shifts coincide, i.e. . In this case, the system has infinitely many solutions - all points of a line .
