Solving systems of simple inequalities with a parameter. §2
Solving inequalities with a parameter.
Inequalities that have the form ax> b, ax< b, ax ≥ b, ax ≤ b, где a и b – действительные числа или выражения, зависящие от параметров, а x – неизвестная величина, называются linear inequalities.
The principles for solving linear inequalities with a parameter are very similar to the principles for solving linear equations with a parameter.
Example 1.
Solve the inequality 5x - a> ax + 3.
Solution.
First, we transform the original inequality:
5x - ax> a + 3, we take out the x in the left-hand side of the inequality outside the brackets:
(5 - a) x> a + 3. Now consider the possible cases for the parameter a:
If a> 5, then x< (а + 3) / (5 – а).
If a = 5, then there are no solutions.
If a< 5, то x >(a + 3) / (5 - a).
This solution will be the answer to inequality.
Example 2.
Solve the inequality x (a - 2) / (a - 1) - 2a / 3 ≤ 2x - a for a ≠ 1.
Solution.
We transform the original inequality:
x (a - 2) / (a - 1) - 2x ≤ 2a / 3 - a;
Ax / (a - 1) ≤ -a / 3. Multiplying both sides of the inequality by (-1), we get:
ax / (a - 1) ≥ a / 3. Let us investigate the possible cases for the parameter a:
1 case. Let a / (a - 1)> 0 or a € (-∞; 0) ᴗ (1; + ∞). Then x ≥ (a - 1) / 3.
2 case. Let a / (a - 1) = 0, that is, a = 0. Then x is any real number.
3 case. Let a / (a - 1)< 0 или а € (0; 1). Тогда x ≤ (а – 1)/3.
Answer: x € [(a - 1) / 3; + ∞) for a € (-∞; 0) ᴗ (1; + ∞);
x € [-∞; (a - 1) / 3] for a € (0; 1);
x € R for a = 0.
Example 3.
Solve the inequality | 1 + x | ≤ ax with respect to x.
Solution.
It follows from the condition that right part inequality ax must not be negative, i.e. ax ≥ 0. According to the modulus expansion rule from the inequality | 1 + x | ≤ ax we have the double inequality
Ax ≤ 1 + x ≤ ax. Let's rewrite the result as a system:
(ax ≥ 1 + x;
(-ax ≤ 1 + x.
We transform to the form:
((a - 1) x ≥ 1;
((a + 1) x ≥ -1.
Let us investigate the resulting system on intervals and at points (fig. 1):
For a ≤ -1 x € (-∞; 1 / (a - 1)].
At -1< а < 0 x € [-1/(а – 1); 1/(а – 1)].
For a = 0 x = -1.
At 0< а ≤ 1 решений нет.
Graphical method for solving inequalities
Plotting makes it much easier to solve equations that contain a parameter. The use of the graphical method in solving inequalities with a parameter is even clearer and more expedient.
The graphical solution of inequalities of the form f (x) ≥ g (x) means finding the values of the variable x for which the graph of the function f (x) lies above the graph of the function g (x). To do this, you always need to find the intersection points of the graphs (if they exist).
Example 1.
Solve the inequality | x + 5 |< bx.
Solution.
We build graphs of functions y = | x + 5 | and y = bx (fig. 2)... The solution to the inequality will be those values of the variable x at which the graph of the function y = | x + 5 | will be below the graph of the function y = bx. 
The figure shows:
1) For b> 1, the lines intersect. The abscissa of the point of intersection of the graphs of these functions is the solution to the equation x + 5 = bx, whence x = 5 / (b - 1). The graph y = bx is located above at x from the interval (5 / (b - 1); + ∞), so this set is the solution to the inequality.
2) Similarly, we find that for -1< b < 0 решением является х из интервала (-5/(b + 1); 5/(b – 1)).
3) For b ≤ -1 x € (-∞; 5 / (b - 1)).
4) For 0 ≤ b ≤ 1, the graphs do not intersect, which means that the inequality has no solutions.
Answer: x € (-∞; 5 / (b - 1)) for b ≤ -1;
x € (-5 / (b + 1); 5 / (b - 1)) at -1< b < 0;
no solutions for 0 ≤ b ≤ 1; x € (5 / (b - 1); + ∞) for b> 1.
Example 2.
Solve the inequality a (a + 1) x> (a + 1) (a + 4).
Solution.
1) Let's find the "control" values for the parameter a: a 1 = 0, and 2 = -1.
2) We solve this inequality on each subset of real numbers: (-∞; -1); (-1); (-ten); (0); (0; + ∞).
a) a< -1, из данного неравенства следует, что х >(a + 4) / a;
b) a = -1, then this inequality will take the form 0 x> 0 - there are no solutions;
c) -1< a < 0, из данного неравенства следует, что х < (a + 4)/a;
d) a = 0, then this inequality has the form 0 x> 4 - there are no solutions;
e) a> 0, it follows from this inequality that х> (a + 4) / a.
Example 3.
Solve the inequality | 2 - | x ||< a – x.
Solution.
We build a graph of the function y = | 2 - | x || (fig. 3) and consider all possible cases of the location of the straight line y = -x + a. 
Answer: the inequality has no solutions for a ≤ -2;
x € (-∞; (a - 2) / 2) for a € (-2; 2];
x € (-∞; (a + 2) / 2) for a> 2.
When deciding different tasks, equations and inequalities with parameters opens up a significant number of heuristic techniques that can then be successfully applied in any other areas of mathematics.
Tasks with parameters play important role in the formation logical thinking and mathematical culture. That is why, having mastered the methods of solving problems with parameters, you will successfully cope with other problems.
Still have questions? Not sure how to deal with inequalities?
To get help from a tutor - register.
The first lesson is free!
site, with full or partial copying of the material, a link to the source is required.
The study of many physical processes and geometric laws often leads to the solution of problems with parameters. Some universities also include equations, inequalities and their systems on exam tickets, which are often very complex and require a non-standard approach to solving. At school, this one of the most difficult sections of the school mathematics course is considered only in a few elective classes.
Cooking this work, I set the goal of a deeper study of this topic, identifying the most rational solution quickly leading to an answer. In my opinion, the graphical method is convenient and fast way solutions of equations and inequalities with parameters.
In my essay, common types of equations, inequalities and their systems are considered, and I hope that the knowledge I gained in the process of work will help me when passing school exams and when applying for a university.
§ 1. Basic definitions
Consider the equation
¦ (a, b, c,…, k, x) = j (a, b, c,…, k, x), (1)
where a, b, c,…, k, x are variables.
Any system of variable values
a = a0, b = b0, c = c0, ..., k = k0, x = x0,
at which both the left and right sides of this equation take real values, is called the system of admissible values of the variables a, b, c,…, k, x. Let A be the set of all admissible values of a, B the set of all admissible values of b, etc., X be the set of all admissible values of x, i.e. aÎA, bÎB,…, xÎX. If for each of the sets A, B, C,…, K we choose and fix, respectively, one value of a, b, c,…, k and substitute them into equation (1), then we get an equation for x, i.e. equation with one unknown.
The variables a, b, c,…, k, which are considered constant when solving the equation, are called parameters, and the equation itself is called the equation containing the parameters.
Parameters are designated by the first letters of the Latin alphabet: a, b, c, d,…, k, l, m, n and unknowns - by letters x, y, z.
To solve an equation with parameters means to indicate at what values of the parameters the solutions exist and what they are.
Two equations containing the same parameters are said to be equivalent if:
a) they make sense for the same parameter values;
b) each solution to the first equation is a solution to the second and vice versa.
§ 2. Algorithm for solving.
Find the domain of the equation.
We express a as a function of x.
In the xOa coordinate system, we build a graph of the function a = ¦ (x) for those values of x that are included in the domain of this equation.
We find the intersection points of the straight line a = c, where cÎ (- ¥; + ¥) with the graph of the function a = ¦ (x). If the straight line a = c intersects the graph a = ¦ (x), then we determine the abscissas of the intersection points. To do this, it is enough to solve the equation a = ¦ (x) for x.
We write down the answer.
I. Solve the equation
(1)Since x = 0 is not a root of the equation, it is possible to solve the equation for a:
or 
The function graph is two “glued” hyperbolas. The number of solutions to the original equation is determined by the number of intersection points of the constructed line and the straight line y = a.
If a Î (- ¥; -1] È (1; + ¥) È
, then the straight line y = a intersects the graph of equation (1) at one point. We find the abscissa of this point by solving the equation for x.Thus, on this interval, Eq. (1) has a solution
... , then the straight line y = a intersects the graph of equation (1) at two points. The abscissas of these points can be found from the equations and 
, we get and 
... , then the straight line y = a does not intersect the graph of equation (1), therefore there are no solutions. If a Î (- ¥; -1] È (1; + ¥) È
, then ; , then ,; , then there are no solutions. II. Find all values of the parameter a for which the equation
has three distinct roots.Rewriting the equation in the form
and having considered a couple of functions, one can notice that the desired values of the parameter a and only they will correspond to those positions of the function graph, at which it has exactly three points of intersection with the function graph.In the xOy coordinate system, we plot the function
). To do this, we can represent it in the form and, having considered four arising cases, we write this function in the form
Since the graph of the function
Is a straight line that has an angle of inclination to the Ox axis equal to, and intersects the Oy axis at a point with coordinates (0, a), we conclude that the three indicated intersection points can be obtained only when this straight line touches the graph of the function. Therefore, we find the derivative.III. Find all values of the parameter a, for each of which the system of equations
has solutions.
From the first equation of the system, we obtain
for Consequently, this equation defines a family of “semi-parabolas” - the right branches of the parabola The vertices “slide” along the abscissa axis. Select the complete squares on the left side of the second equation and factor it into factors
Job type: 18
Condition
For what values of the parameter a the inequality
\ log_ (5) (4 + a + (1 + 5a ^ (2) - \ cos ^ (2) x) \ cdot\ sin x - a \ cos 2x) \ leq 1 holds for all values of x?
Show solutionSolution
This inequality is equivalent to the double inequality 0 < 4+a+(5a^{2}+\sin^{2}x) \sin x+ a (2 \ sin ^ (2) x-1) \ leq 5.
Let \ sin x = t, then we get the inequality:
4 < t^{3}+2at^{2}+5a^{2}t \leq 1 \: (*) which must be true for all values -1 \ leq t \ leq 1. If a = 0, then inequality (*) holds for any t \ in [-1; 1].
Let a \ neq 0. The function f (t) = t ^ (3) + 2at ^ (2) + 5a ^ (2) t increases on the interval [-1; 1], since the derivative f "(t) = 3t ^ (2) + 4at + 5a ^ (2)> 0 for all values of t \ in \ mathbb (R) and a \ neq 0 (discriminant D< 0 и старший коэффициент больше нуля).
Inequality (*) will hold for t \ in [-1; 1] under the conditions
\ begin (cases) f (-1)> -4, \\ f (1) \ leq 1, \\ a \ neq 0; \ end (cases) \: \ Leftrightarrow \ begin (cases) -1 + 2a-5a ^ (2)> -4, \\ 1 + 2a + 5a ^ (2) \ leq 1, \\ a \ neq 0; \ end (cases) \: \ Leftrightarrow \ begin (cases) 5a ^ (2) -2a-3< 0, \\ 5a^{2}+2a \leq 0, \\ a \neq 0; \end{cases}\: \Leftrightarrow - \ frac (2) (5) \ leq a< 0 .
So, the condition is satisfied for - \ frac (2) (5) \ leq a \ leq 0.
Answer
\ left [- \ frac (2) (5); 0 \ right]
Source: “Mathematics. Preparation for the exam-2016. Profile level ". Ed. FF Lysenko, S. Yu. Kulabukhova.
Job type: 18
Topic: Parameter Inequalities
Condition
Find all values of the parameter a, for each of which the inequality
x ^ 2 + 3 | x-a | -7x \ leqslant -2a
has only one solution.
Show solutionSolution
Inequality is equivalent to a set of systems of inequalities
\ left [\! \! \ begin (array) (l) \ begin (cases) x \ geqslant a, \\ x ^ 2 + 3x-3a-7x + 2a \ leqslant0; \ end (cases) \\ \ begin (cases) x
In the Oxa coordinate system, we plot the graphs of the functions a = x, a = x ^ 2-4x, a = - \ frac (x ^ 2) (5) + 2x.
The resulting set is satisfied by the points enclosed between the graphs of the functions a = x ^ 2-4x, a = - \ frac (x ^ 2) (5) + 2x at the interval x \ in (shaded area).
According to the graph, we determine: the original inequality has a unique solution for a = -4 and a = 5, since in the shaded area there will be a single point with the ordinate a equal to -4 and equal to 5.
Solving inequalities with a parameter.
Inequalities that have the form ax> b, ax< b, ax ≥ b, ax ≤ b, где a и b – действительные числа или выражения, зависящие от параметров, а x – неизвестная величина, называются linear inequalities.
The principles for solving linear inequalities with a parameter are very similar to the principles for solving linear equations with a parameter.
Example 1.
Solve the inequality 5x - a> ax + 3.
Solution.
First, we transform the original inequality:
5x - ax> a + 3, we take out the x in the left-hand side of the inequality outside the brackets:
(5 - a) x> a + 3. Now consider the possible cases for the parameter a:
If a> 5, then x< (а + 3) / (5 – а).
If a = 5, then there are no solutions.
If a< 5, то x >(a + 3) / (5 - a).
This solution will be the answer to inequality.
Example 2.
Solve the inequality x (a - 2) / (a - 1) - 2a / 3 ≤ 2x - a for a ≠ 1.
Solution.
We transform the original inequality:
x (a - 2) / (a - 1) - 2x ≤ 2a / 3 - a;
Ax / (a - 1) ≤ -a / 3. Multiplying both sides of the inequality by (-1), we get:
ax / (a - 1) ≥ a / 3. Let us investigate the possible cases for the parameter a:
1 case. Let a / (a - 1)> 0 or a € (-∞; 0) ᴗ (1; + ∞). Then x ≥ (a - 1) / 3.
2 case. Let a / (a - 1) = 0, that is, a = 0. Then x is any real number.
3 case. Let a / (a - 1)< 0 или а € (0; 1). Тогда x ≤ (а – 1)/3.
Answer: x € [(a - 1) / 3; + ∞) for a € (-∞; 0) ᴗ (1; + ∞);
x € [-∞; (a - 1) / 3] for a € (0; 1);
x € R for a = 0.
Example 3.
Solve the inequality | 1 + x | ≤ ax with respect to x.
Solution.
It follows from the condition that the right-hand side of the inequality ax must not be negative, i.e. ax ≥ 0. According to the modulus expansion rule from the inequality | 1 + x | ≤ ax we have the double inequality
Ax ≤ 1 + x ≤ ax. Let's rewrite the result as a system:
(ax ≥ 1 + x;
(-ax ≤ 1 + x.
We transform to the form:
((a - 1) x ≥ 1;
((a + 1) x ≥ -1.
Let us investigate the resulting system on intervals and at points (fig. 1):
For a ≤ -1 x € (-∞; 1 / (a - 1)].
At -1< а < 0 x € [-1/(а – 1); 1/(а – 1)].
For a = 0 x = -1.
At 0< а ≤ 1 решений нет.
Graphical method for solving inequalities
Plotting makes it much easier to solve equations that contain a parameter. The use of the graphical method in solving inequalities with a parameter is even clearer and more expedient.
The graphical solution of inequalities of the form f (x) ≥ g (x) means finding the values of the variable x for which the graph of the function f (x) lies above the graph of the function g (x). To do this, you always need to find the intersection points of the graphs (if they exist).
Example 1.
Solve the inequality | x + 5 |< bx.
Solution.
We build graphs of functions y = | x + 5 | and y = bx (fig. 2)... The solution to the inequality will be those values of the variable x at which the graph of the function y = | x + 5 | will be below the graph of the function y = bx.
The figure shows:
1) For b> 1, the lines intersect. The abscissa of the point of intersection of the graphs of these functions is the solution to the equation x + 5 = bx, whence x = 5 / (b - 1). The graph y = bx is located above at x from the interval (5 / (b - 1); + ∞), so this set is the solution to the inequality.
2) Similarly, we find that for -1< b < 0 решением является х из интервала (-5/(b + 1); 5/(b – 1)).
3) For b ≤ -1 x € (-∞; 5 / (b - 1)).
4) For 0 ≤ b ≤ 1, the graphs do not intersect, which means that the inequality has no solutions.
Answer: x € (-∞; 5 / (b - 1)) for b ≤ -1;
x € (-5 / (b + 1); 5 / (b - 1)) at -1< b < 0;
no solutions for 0 ≤ b ≤ 1; x € (5 / (b - 1); + ∞) for b> 1.
Example 2.
Solve the inequality a (a + 1) x> (a + 1) (a + 4).
Solution.
1) Let's find the "control" values for the parameter a: a 1 = 0, and 2 = -1.
2) We solve this inequality on each subset of real numbers: (-∞; -1); (-1); (-ten); (0); (0; + ∞).
a) a< -1, из данного неравенства следует, что х >(a + 4) / a;
b) a = -1, then this inequality will take the form 0 x> 0 - there are no solutions;
c) -1< a < 0, из данного неравенства следует, что х < (a + 4)/a;
d) a = 0, then this inequality has the form 0 x> 4 - there are no solutions;
e) a> 0, it follows from this inequality that х> (a + 4) / a.
Example 3.
Solve the inequality | 2 - | x ||< a – x.
Solution.
We build a graph of the function y = | 2 - | x || (fig. 3) and consider all possible cases of the location of the straight line y = -x + a.
Answer: the inequality has no solutions for a ≤ -2;
x € (-∞; (a - 2) / 2) for a € (-2; 2];
x € (-∞; (a + 2) / 2) for a> 2.
When solving various problems, equations and inequalities with parameters, a significant number of heuristic techniques are revealed, which can then be successfully applied in any other branches of mathematics.
Problems with parameters play an important role in the formation of logical thinking and mathematical culture. That is why, having mastered the methods of solving problems with parameters, you will successfully cope with other problems.
Still have questions? Not sure how to deal with inequalities?
To get help from a tutor -.
The first lesson is free!
blog. site, with full or partial copying of the material, a link to the source is required.
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