Systems of equations with a parameter. Systems of equations with a parameter How to find the value of a parameter

1. Systems linear equations with parameter

Systems of linear equations with a parameter are solved by the same basic methods as conventional systems of equations: substitution method, equation addition method, and graphical method. Knowledge of the graphic interpretation of linear systems makes it easy to answer the question of the number of roots and their existence.

Example 1.

Find all values \u200b\u200bfor the parameter a for which the system of equations has no solutions.

(x + (a 2 - 3) y \u003d a,
(x + y \u003d 2.

Decision.

Let's consider several ways to solve this task.

1 way. We use the property: the system has no solutions if the ratio of coefficients in front of x is equal to the ratio of coefficients in front of y, but not equal to the ratio of free terms (a / a 1 \u003d b / b 1 ≠ c / c 1) Then we have:

1/1 \u003d (a 2 - 3) / 1 ≠ a / 2 or system

(a 2 - 3 \u003d 1,
(a ≠ 2.

From the first equation a 2 \u003d 4, therefore, taking into account the condition that a ≠ 2, we get the answer.

Answer: a \u003d -2.

Method 2.We solve by substitution method.

(2 - y + (a 2 - 3) y \u003d a,
(x \u003d 2 - y,

((a 2 - 3) y - y \u003d a - 2,
(x \u003d 2 - y.

After placing the common factor y in the first equation outside the brackets, we get:

((a 2 - 4) y \u003d a - 2,
(x \u003d 2 - y.

The system has no solutions if the first equation has no solutions, that is

(a 2 - 4 \u003d 0,
(a - 2 ≠ 0.

Obviously, a \u003d ± 2, but taking into account the second condition, the answer is only an answer with a minus.

Answer: a \u003d -2.

Example 2.

Find all values \u200b\u200bfor the parameter a for which the system of equations has an infinite set of solutions.

(8x + ay \u003d 2,
(ax + 2y \u003d 1.

Decision.

By property, if the ratio of the coefficients at x and y is the same and is equal to the ratio of free members of the system, then it has an infinite set of solutions (i.e., a / a 1 \u003d b / b 1 \u003d c / c 1). Therefore 8 / a \u003d a / 2 \u003d 2/1. Solving each of the obtained equations, we find that a \u003d 4 - the answer in this example.

Answer: a \u003d 4.

2. Systems of rational equations with a parameter

Example 3.

(3 | x | + y \u003d 2,
(| x | + 2y \u003d a.

Decision.

Let's multiply the first equation of the system by 2:

(6 | x | + 2y \u003d 4,
(| x | + 2y \u003d a.

We subtract the second equation from the first, we get 5 | x | \u003d 4 - a. This equation will have a unique solution for a \u003d 4. In other cases, this equation will have two solutions (for a< 4) или ни одного (при а > 4).

Answer: a \u003d 4.

Example 4.

Find all values \u200b\u200bof the parameter a for which the system of equations has a unique solution.

(x + y \u003d a,
(y - x 2 \u003d 1.

Decision.

We will solve this system using the graphical method. So, the graph of the second equation of the system is a parabola lifted along the Oy axis up by one unit segment. The first equation defines a set of straight lines parallel to the straight line y \u003d -x (picture 1)... The figure clearly shows that the system has a solution if the line y \u003d -x + a is tangent to the parabola at the point with coordinates (-0.5; 1.25). Substituting these coordinates into the equation with a straight line instead of x and y, we find the value of the parameter a:

1.25 \u003d 0.5 + a;

Answer: a \u003d 0.75.

Example 5.

Using the substitution method, find out at what value of the parameter a, the system has a unique solution.

(ax - y \u003d a + 1,
(ax + (a + 2) y \u003d 2.

Decision.

From the first equation, we express y and substitute it into the second:

(y \u003d ax - a - 1,
(ax + (a + 2) (ax - a - 1) \u003d 2.

Let us bring the second equation to the form kx \u003d b, which will have a unique solution for k ≠ 0. We have:

ax + a 2 x - a 2 - a + 2ax - 2a - 2 \u003d 2;

a 2 x + 3ax \u003d 2 + a 2 + 3a + 2.

The square trinomial a 2 + 3a + 2 is represented as a product of brackets

(a + 2) (a + 1), and on the left we take out x outside the brackets:

(a 2 + 3a) x \u003d 2 + (a + 2) (a + 1).

Obviously, a 2 + 3a should not be equal to zero, therefore,

a 2 + 3a ≠ 0, a (a + 3) ≠ 0, and therefore a ≠ 0 and ≠ -3.

Answer:a ≠ 0; ≠ -3.

Example 6.

Using the graphical solution method, determine at what value of the parameter a, the system has a unique solution.

(x 2 + y 2 \u003d 9,
(y - | x | \u003d a.

Decision.

Based on the condition, we build a circle with a center at the origin and a radius of 3 unit segments, it is it that is set by the first equation of the system

x 2 + y 2 \u003d 9. The second equation of the system (y \u003d | x | + a) is a broken line. Through figure 2 consider all possible cases of its location relative to the circle. It is easy to see that a \u003d 3.

Answer: a \u003d 3.

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1. Task.
At what values \u200b\u200bof the parameter a the equation ( a - 1)x 2 + 2x + a - 1 \u003d 0 has exactly one root?

1. Solution.
When a \u003d 1 the equation has the form 2 x \u003d 0 and obviously has a unique root x \u003d 0. If a № 1, then this equation is square and has a single root for those values \u200b\u200bof the parameter for which the discriminant of the square trinomial is zero. Equating the discriminant to zero, we obtain an equation for the parameter a 4a 2 - 8a \u003d 0, whence a \u003d 0 or a = 2.

1. Answer:the equation has a single root at a O (0; 1; 2).

2. The task.
Find all parameter values afor which the equation has two different roots x 2 +4ax+8a+3 = 0.
2. Solution.
The equation x 2 +4ax+8a+3 \u003d 0 has two distinct roots if and only if D = 16a 2 -4(8a+3)\u003e 0. We get (after reduction by a common factor of 4) 4 a 2 -8a-3\u003e 0, whence

2. Answer:

a O (-Ґ; 1 -

C 7 2

) AND (1 +

C 7 2

; Ґ ).

3. Task.
It is known that
f 2 (x) = 6x-x 2 -6.
a) Plot the function f 1 (x) at a = 1.
b) At what value a function graphs f 1 (x) and f 2 (x) have a single point in common?

3. Solution.
3.a. We transform f 1 (x) in the following way
The graph of this function at a \u003d 1 is shown in the figure on the right.
3.b. Note right away that the graphs of the functions y = kx+b and y = ax 2 +bx+c (a No. 0) intersect at a single point if and only if quadratic equation kx+b = ax 2 +bx+c has a single root. Using the view f 1 of 3.a, equate the discriminant of the equation a = 6x-x 2 -6 to zero. From equation 36-24-4 a \u003d 0 we get a \u003d 3. Doing the same with equation 2 x-a = 6x-x 2 -6 find a \u003d 2. It is easy to verify that these values \u200b\u200bof the parameter satisfy the conditions of the problem. Answer: a \u003d 2 or a = 3.

4. The challenge.
Find all values afor which the set of solutions to the inequality x 2 -2ax-3a і 0 contains a segment.

4. Solution.
The first coordinate of the vertex of the parabola f(x) = x 2 -2ax-3a equals x 0 = a... From the properties of a quadratic function, the condition f(x) і 0 on an interval is equivalent to a set of three systems
has exactly two solutions?

5. Solution.
We rewrite this equation as x 2 + (2a-2)x - 3a+7 \u003d 0. This is a quadratic equation, it has exactly two solutions if its discriminant is strictly greater than zero. Calculating the discriminant, we find that the condition for the presence of exactly two roots is the fulfillment of the inequality a 2 +a-6\u003e 0. Solving the inequality, we find a < -3 или a \u003e 2. The first of the inequalities, obviously, solutions in natural numbers does not have, and the least natural solution of the second is the number 3.

5. Answer: 3.

6. Problem (10 grades)
Find all values aat which the graph of the function or, after obvious transformations, a-2 = | 2-a| ... The last equation is equivalent to the inequality a i 2.

6. Answer: a Where \\ - variables, \\ - parameter;

\\ [y \u003d kx + b, \\] where \\ - variables, \\ - parameter;

\\ [ax ^ 2 + bx + c \u003d 0, \\] where \\ is a variable, \\ [a, b, c \\] is a parameter.

Solving an equation with a parameter means, as a rule, solving an infinite number of equations.

However, adhering to a certain algorithm, you can easily solve the following equations:

1. Determine the "control" values \u200b\u200bof the parameter.

2. Solve the original equation for [\\ x \\] at the parameter values \u200b\u200bdefined in the first paragraph.

3. Solve the original equation for [\\ x \\] with parameter values \u200b\u200bdifferent from those selected in the first paragraph.

Let's say the following equation is given:

\\ [\\ mid 6 - x \\ mid \u003d a. \\]

After analyzing the initial data, it can be seen that a \\ [\\ ge 0. \\]

By the modulus rule \\ express \\

Answer: \\ where \\

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For what values \u200b\u200bof the parameter $ a $ does the inequality $ () - x ^ 2 + (a + 2) x - 8a - 1\u003e 0 $ have at least one solution?

Decision

Let us reduce this inequality to a positive coefficient for $ x ^ 2 $:

$ () - x ^ 2 + (a + 2) x - 8a - 1\u003e 0 \\ quad \\ Leftrightarrow \\ quad x ^ 2 - (a + 2) x + 8a + 1< 0 .$

Let's calculate the discriminant: $ D \u003d (a + 2) ^ 2 - 4 (8a + 1) \u003d a ^ 2 + 4a + 4 - 32a - 4 \u003d a ^ 2 - 28a $. For this inequality to have a solution, it is necessary that at least one point of the parabola lies below the $ x $ axis. Since the branches of the parabola are directed upwards, for this it is necessary that the square trinomial on the left side of the inequality has two roots, that is, its discriminant is positive. We come to the need to solve the square inequality $ a ^ 2 - 28a\u003e 0 $. The square trinomial $ a ^ 2 - 28a $ has two roots: $ a_1 \u003d 0 $, $ a_2 \u003d 28 $. Therefore, the inequality $ a ^ 2 - 28a\u003e 0 $ is satisfied by the intervals $ a \\ in (- \\ infty; 0) \\ cup (28; + \\ infty) $.

Answer. $ a \\ in (- \\ infty; 0) \\ cup (28; + \\ infty) $.

For what values \u200b\u200bof the parameter $ a $ does the equation $ (a-2) x ^ 2-2ax + a + 3 \u003d 0 $ have at least one root, and all the roots are positive?

Decision

Let $ a \u003d 2 $. Then the equation takes the form $ () - 4x +5 \u003d 0 $, whence we obtain that $ x \u003d \\ dfrac (5) (4) $ is a positive root.

Now let $ a \\ ne 2 $. It turns out a quadratic equation. Let us first determine for what values \u200b\u200bof the parameter $ a $ this equation has roots. Its discriminant must be non-negative. I.e:

$ D \u003d 4a ^ 2 - 4 (a-2) (a + 3) \u003d () -4a + 24 \\ geqslant 0 \\ Leftrightarrow a \\ leqslant 6. $

The roots by the condition must be positive, therefore, from the Vieta theorem we obtain the system:

$ \\ begin (cases) x_1 + x_2 \u003d \\ dfrac (2a) (a - 2)\u003e 0, \\\\ x_1x_2 \u003d \\ dfrac (a + 3) (a - 2)\u003e 0, \\\\ a \\ leqslant 6 \\ end (cases) \\ quad \\ Leftrightarrow \\ quad \\ begin (cases) a \\ in (- \\ infty; 0) \\ cup (2; + \\ infty), \\\\ a \\ in (- \\ infty; -3) \\ cup ( 2; + \\ infty), \\\\ a \\ in (- \\ infty; 6] \\ end (cases) \\ quad \\ Leftrightarrow \\ quad a \\ in (- \\ infty; -3) \\ cup (2; 6]. $

Combining the answers, we get the required set: $ a \\ in (- \\ infty; -3) \\ cup $.

Answer. $ a \\ in (- \\ infty; -3) \\ cup $.

For what values \u200b\u200bof the parameter $ a $ does the inequality $ ax ^ 2 + 4ax + 5 \\ leqslant 0 $ have no solutions?

Decision

1. If $ a \u003d 0 $, then this inequality degenerates into the inequality $ 5 \\ leqslant 0 $, which has no solutions. Therefore, the value $ a \u003d 0 $ satisfies the condition of the problem.
2. If $ a\u003e 0 $, then the graph of the square trinomial on the left side of the inequality is a parabola with branches pointing up. Calculate $ \\ dfrac (D) (4) \u003d 4a ^ 2 - 5a $. The inequality has no solutions if the parabola is located above the abscissa axis, that is, when the square trinomial has no roots ($ D< 0$). Решим неравенство $4a^2 - 5a < 0$. Корнями квадратного трёхчлена $4a^2 - 5a$ являются числа $a_1 = 0$ и $a_2 = \dfrac{5}{4}$, поэтому $D < 0$ при $0 < a < \dfrac{5}{4}$. Значит, из положительных значений $a$ подходят числа $a \in \left(0; \dfrac{5}{4}\right)$.
3. If $ a< 0$, то график квадратного трехчлена в левой части неравенства - парабола с ветвями, направленными вниз. Значит, обязательно найдутся значения $х$, для которых трёхчлен отрицателен. Следовательно, все значения $a < 0$ не подходят.

Answer. $ a \\ in \\ left $ lies between the roots, so there must be two roots (hence $ a \\ ne 0 $). If the branches of the parabola $ y \u003d ax ^ 2 + (a + 3) x - 3a $ are directed upwards, then $ y (-1)< 0$ и $y(1) < 0$; если же они направлены вниз, то $y(-1) > 0 $ and $ y (1)\u003e 0 $.

Case I. Let $ a\u003e 0 $. Then

$ \\ left \\ (\\ begin (array) (l) y (-1) \u003d a- (a + 3) -3a \u003d -3a-3<0 \\ y(1)=a+(a+3)-3a=-a+3<0 \\ a>0 \\ end (array) \\ right. \\ quad \\ Leftrightarrow \\ quad \\ left \\ (\\ begin (array) (l) a\u003e -1 \\\\ a\u003e 3 \\\\ a\u003e 0 \\ end (array) \\ right. \\ quad \\ Leftrightarrow \\ quad a\u003e 3. $

That is, in this case it turns out that all $ a\u003e 3 $ are suitable.

Case II. Let $ a< 0$. Тогда

$ \\ left \\ (\\ begin (array) (l) y (-1) \u003d a- (a + 3) -3a \u003d -3a-3\u003e 0 \\\\ y (1) \u003d a + (a + 3) -3a \u003d -a + 3\u003e 0 \\\\ a<0 \end{array} \right.\quad \Leftrightarrow \quad \left\{ \begin{array}{l} a<-1 \\ a<3 \\ a<0 \end{array} \right.\quad \Leftrightarrow \quad a<-1.$

That is, in this case, it turns out that all $ a< -1$.

Answer. $ a \\ in (- \\ infty; -1) \\ cup (3; + \\ infty) $

Find all values \u200b\u200bof the parameter $ a $, for each of which the system of equations

$ \\ begin (cases) x ^ 2 + y ^ 2 \u003d 2a, \\\\ 2xy \u003d 2a-1 \\ end (cases) $

has exactly two solutions.

Decision

Subtract the second from the first: $ (x-y) ^ 2 \u003d 1 $. Then

$ \\ left [\\ begin (array) (l) x-y \u003d 1, \\\\ x-y \u003d -1 \\ end (array) \\ right. \\ quad \\ Leftrightarrow \\ quad \\ left [\\ begin (array) (l) x \u003d y + 1, \\\\ x \u003d y-1. \\ end (array) \\ right. $

Substituting the obtained expressions into the second equation of the system, we obtain two quadratic equations: $ 2y ^ 2 + 2y - 2a + 1 \u003d 0 $ and $ 2y ^ 2 - 2y - 2a + 1 \u003d 0 $. The discriminant of each of them is $ D \u003d 16a-4 $.

Note that it cannot happen that the pair of roots of the first of the quadratic equations coincides with the pair of roots of the second quadratic equation, since the sum of the roots of the first is $ -1 $, and the second is 1.

This means that each of these equations must have one root, then the original system will have two solutions. That is, $ D \u003d 16a - 4 \u003d $ 0.

Answer. $ a \u003d \\ dfrac (1) (4) $

Find all values \u200b\u200bof the parameter $ a $, for each of which the equation $ 4x- | 3x- | x + a || \u003d 9 | x-3 | $ has two roots.

Decision

Let's rewrite the equation as:

$ 9 | x-3 | -4x + | 3x- | x + a || \u003d 0. $

Consider the function $ f (x) \u003d 9 | x-3 | -4x + | 3x- | x + a || $.

When $ x \\ geqslant 3 $, the first module is expanded with a plus sign, and the function takes the form: $ f (x) \u003d 5x-27 + | 3x- | x + a || $. Obviously, any expansion of the modules will eventually result in a linear function with the coefficient $ k \\ geqslant 5-3-1 \u003d 1\u003e 0 $, that is, this function increases indefinitely on this interval.

Consider now the interval $ x<3$. В этом случае первый модуль раскрывается с минусом, и функция принимает следующий вид: $f(x) = - 13x+27+|3x-|x+a||$. При любом раскрытии модулей в итоге будет получаться линейная функция с коэффициентом $k\leqslant - 13+3+1 = - 9<0$, то есть на этом промежутке функция убывает.

So, we got that $ x \u003d 3 $ is the minimum point of this function. This means that in order for the original equation to have two solutions, the value of the function at the minimum point must be less than zero. That is, the inequality holds: $ f (3)<0$.

$ 12- | 9- | 3 + a ||\u003e 0 \\ quad \\ Leftrightarrow \\ quad | 9- | 3 + a ||< 12 \quad \Leftrightarrow \quad -12 < 9-|3+a| < 12 \quad \Leftrightarrow \quad$

$ \\ Leftrightarrow \\ quad | 3 + a |< 21 \quad \Leftrightarrow \quad - 21 < 3+a < 21 \quad \Leftrightarrow \quad -24

Answer. $ a \\ in (-24; 18) $

For what values \u200b\u200bof the parameter $ a $ does the equation $ 5 ^ (2x) -3 \\ cdot 5 ^ x + a-1 \u003d 0 $ have a single root?

Decision

Let's make the change: $ t \u003d 5 ^ x\u003e 0 $. Then the original equation takes the form of a quadratic equation: $ t ^ 2-3t + a-1 \u003d 0 $. The original equation will have a single root in the event that this equation has one positive root or two roots, one of which is positive, the other negative.

The discriminant of the equation is: $ D \u003d 13-4a $. This equation will have one root if the resulting discriminant is equal to zero, that is, for $ a \u003d \\ dfrac (13) (4) $. Moreover, the root $ t \u003d \\ dfrac (3) (2)\u003e 0 $, so the given value $ a $ is suitable.

If there are two roots, one of which is positive, the other is non-positive, then $ D \u003d 13-4a\u003e 0 $, $ x_1 + x_2 \u003d 3\u003e 0 $ and $ x_1x_2 \u003d a - 1 \\ leqslant 0 $.

That is, $ a \\ in (- \\ infty; 1] $

Answer. $ a \\ in (- \\ infty; 1] \\ cup \\ left \\ (\\ dfrac (13) (4) \\ right \\) $

Find all values \u200b\u200bof the parameter $ a $ for which the system

$ \\ begin (cases) \\ log_a y \u003d (x ^ 2-2x) ^ 2, \\\\ x ^ 2 + y \u003d 2x \\ end (cases) $

has exactly two solutions.

Decision

We transform the system to the following form:

$ \\ begin (cases) \\ log_a y \u003d (2x-x ^ 2) ^ 2, \\\\ y \u003d 2x-x ^ 2. \\ end (cases) $

Since the $ a $ parameter is at the base of the logarithm, the following restrictions are imposed on it: $ a\u003e 0 $, $ a \\ ne 1 $. Since the variable $ y $ is the logarithm argument, then $ y\u003e 0 $.

Having combined both equations of the system, we pass to the equation: $ \\ log_a y \u003d y ^ 2 $. Depending on what values \u200b\u200bthe $ a $ parameter takes, two cases are possible:

1. Let $ 0< a < 1$. В этом случае функция $f(y) = \log_a y$ убывает на области определения, а функция $g(y)=y^2$ возрастает в той же области $y > 0 $. From the behavior of the graphs it is obvious that the root of the equation is one, while it is less than 1. The second equation of the system and the entire system as a whole have, therefore, two solutions, since the discriminant of the equation $ x ^ 2-2x + y \u003d 0 $ at $ 0
2. Now let $ a\u003e 1 $. In this case, the function $ f (y) \u003d \\ log_a y \\ leqslant 0 $ for $ y< 1$, а функция $g(y) = y^2 > 0 $ for the same $ y $. This means that if there are solutions, then only for $ y\u003e 1 $, but the second equation of the system of solutions will not have, since the discriminant of the equation $ x ^ 2 - 2x + y \u003d 0 $ for $ y\u003e 1 $ is negative.

Answer. $ a \\ in (0; 1) $

Consider the case when $ a\u003e 1 $. Since for large modulo values \u200b\u200bof $ t $ the graph of the function $ f (t) \u003d a ^ t $ lies above the straight line $ g (t) \u003d t $, the only common point can only be a point of contact.

Let $ t_0 $ be the tangency point. At this point, the derivative to $ f (t) \u003d a ^ t $ is equal to one (tangent of the angle of inclination of the tangent), in addition, the values \u200b\u200bof both functions coincide, that is, the system takes place:

$ \\ begin (cases) a ^ (t_0) \\ ln a \u003d 1, \\\\ a ^ (t_0) \u003d t_0 \\ end (cases) \\ quad \\ Leftrightarrow \\ quad \\ begin (cases) a ^ (t_0) \u003d \\ dfrac (1) (\\ ln a), \\\\ a ^ (\\ tau) \u003d \\ tau \\ end (cases) $

From where $ t_0 \u003d \\ dfrac (1) (\\ ln a) $.

$ a ^ (\\ frac (1) (\\ ln a)) \\ ln a \u003d 1 \\ quad \\ Leftrightarrow \\ quad a ^ (\\ log_a e) \u003d \\ frac (1) (\\ ln a) \\ quad \\ Leftrightarrow \\ quad a \u003d e ^ (\\ frac (1) (e)). $

At the same time, the line and the exponential function obviously have no other common points.

Answer. $ a \\ in (0; 1] \\ cup \\ left \\ (e ^ (e ^ (- 1)) \\ right \\) $

In recent years, in the entrance exams, in the final testing in the form of the USE, problems with parameters have been proposed. These tasks make it possible to diagnose the level of mathematical and, most importantly, logical thinking of applicants, the ability to carry out research activities, as well as simply knowledge of the main sections of the school mathematics course.

Looking at the parameter as an equal variable is reflected in graphical methods. Indeed, since the parameter is “equal in rights” with the variable, then, naturally, it can “select” its own coordinate axis. Thus, a coordinate plane appears. The rejection of the traditional choice of letters and for the designation of axes determines one of the most effective methods for solving problems with parameters - "Area method". Along with other methods used in solving problems with parameters, I introduce my students to graphical techniques, paying attention to how to recognize “such” problems and how the process of solving a problem looks like.

The most common signs that will help you recognize tasks suitable for the method under consideration are:

Problem 1. “For what values \u200b\u200bof the parameter does the inequality hold for all?”

Decision. 1). Let's expand the modules taking into account the sign of the submodular expression:

2). We write down all the systems of the resulting inequalities:

and)

b) in)

d)

3). Let us show the set of points satisfying each system of inequalities (Fig. 1a).

4). Combining all the areas shown in the figure by hatching, one can notice that the points lying inside the parabolas do not satisfy the inequality.

The figure shows that for any value of the parameter, one can find the area where the points lie whose coordinates satisfy the original inequality. The inequality holds for all, if. Answer: at.

The considered example is an "open problem" - you can consider the solution of a whole class of problems without changing the expression considered in the example , in which the technical difficulties of plotting have already been overcome.

A task. For what values \u200b\u200bof the parameter does the equation have no solutions? Answer: at.

A task. For what values \u200b\u200bof the parameter does the equation have two solutions? Write down both solutions you find.

Answer: then , ;

Then ; then , .

A task. For what values \u200b\u200bof the parameter does the equation have one root? Find this root. Answer: when.

A task. Solve inequality.

(Points lying inside parabolas "work").

,; , there are no solutions;

Task 2: Find all parameter values and, for each of which the system of inequalities forms a segment of length 1 on the number line.

Decision. Let's rewrite the original system as follows

All solutions of this system (pairs of the form) form a certain region bounded by parabolas and (Figure 1).

Obviously, the solution to the system of inequalities will be a segment of length 1 for and for. Answer:; ...

Problem 3: Find all values \u200b\u200bof the parameter for which the set of solutions to the inequality contains a number, and also contains two line segments with no common points.

Decision. Within the meaning of inequality; we rewrite the inequality by multiplying both sides by (), we obtain the inequality:

, ,

(1)

Inequality (1) is equivalent to the combination of two systems:

(fig. 2).

Obviously, the interval cannot contain a length segment. This means that two disjoint segments of length are contained in the interval This is possible at, i.e. at. Answer:.

Problem 4: Find all values \u200b\u200bof the parameter, for each of which the set of solutions to the inequality contains a segment of length 4 and is contained in some segment of length 7.

Decision. Let's carry out equivalent transformations, taking into account that and.

, ,

; the last inequality is equivalent to the combination of two systems:

Let us show the areas that correspond to these systems (fig. 3).

1) For, the set of solutions is an interval of length less than 4. For, the set of solutions is the union of two intervals. Only an interval can contain a segment of length 4. But then, the union is no longer contained in any segment of length 7. Hence, such do not satisfy the condition.

2) the set of solutions is an interval. It contains a segment of length 4 only if its length is greater than 4, i.e. at. It is contained in a segment of length 7 only if its length is not more than 7, that is, for, then. Answer:.

Problem 5. Find all values \u200b\u200bof the parameter for which the set of solutions to the inequality contains the number 4, and also contains two disjoint segments of length 4 each.

Decision. By conditions. Let us multiply both sides of the inequality by (). We obtain an equivalent inequality in which we group all the terms on the left side and transform it into a product:

, ,

, .

The last inequality implies:

1) 2)

Let us show the areas that correspond to these systems (fig. 4).

a) For, we obtain an interval that does not contain the number 4. For, we obtain an interval that also does not contain the number 4.

b) For, we obtain the union of two intervals. Non-overlapping segments of length 4 can be located only in the interval. This is possible only if the interval length is greater than 8, that is, if. With these, another condition is also satisfied:. Answer:.

Problem 6. Find all values \u200b\u200bof the parameter for which the set of solutions to the inequality contains some segment of length 2, but does not contain no line length 3.

Decision. In the sense of the task, we multiply both sides of the inequality by, group all the terms on the left side of the inequality and transform it into a product:

, ... The last inequality implies:

1) 2)

Let us show the region corresponding to the first system (fig. 5).

Obviously, the condition of the problem is satisfied if . Answer:.

Problem 7. Find all values \u200b\u200bof the parameter for which the set of solutions to inequality 1+ is contained in some segment of length 1 and at the same time contains some segment of length 0.5.

Decision. 1). Let's indicate the ODZ of the variable and parameter:

2). We rewrite the inequality as

, ,

(1). Inequality (1) is equivalent to the combination of two systems:

1)

2)

Taking into account the ODZ, the solutions of the systems look like this:

and) b)

(fig. 6).

and) b)

Let us show the region corresponding to the system a) (fig. 7).Answer:.

Problem 8. Six numbers form an increasing arithmetic progression. The first, second and fourth terms of this progression are solutions to inequality and the rest

are not solutions to this inequality. Find the set of all possible values \u200b\u200bof the first term of such progressions.

Decision. I. Find all solutions to the inequality

and). ODZ:
, i.e.

(we took into account in the solution that the function increases by).

b). In the DHS, inequality tantamount to inequality , i.e. , what gives:

1).

2).

Obviously, by solving the inequality serves many meanings .

II. Let us illustrate the second part of the problem on the terms of an increasing arithmetic progression by drawing ( fig. 8 , where is the first term, is the second, etc.). Notice, that:

Or we have a system of linear inequalities:

let's solve it graphically. We build straight lines and as well as straight lines

That, .. The first, second and sixth terms of this progression are solutions to inequality , and the rest are not solutions to this inequality. Find the set of all possible values \u200b\u200bof the difference of this progression.