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, let's transform the original inequality:

5x - ax > a + 3, we take x out of brackets on the left side of the inequality:

(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 the inequality.

Example 2

Solve the inequality x(a - 2) / (a - 1) - 2a / 3 ≤ 2x - a for a ≠ 1.

Solution.

Let's transform the original inequality:

x(a - 2) / (a - 1) - 2x ≤ 2a/3 - a;

Ah/(a – 1) ≤ -a/3. Multiply by (-1) both parts of the inequality, we get:

ax/(a – 1) ≥ a/3. Let's explore the possible cases for the parameter a:

1 case. Let a/(a – 1) > 0 or a € (-∞; 0)ᴗ(1; +∞). Then x ≥ (а – 1)/3.

2nd case. Let a/(а – 1) = 0, i.e. a = 0. Then x is any real number.

3rd case. Let a/(а – 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 side of the inequality ax must be non-negative, i.e. ax ≥ 0. By the rule of expansion of the module from the inequality |1 + x| ≤ ax we have a double inequality

Ax ≤ 1 + x ≤ ax. We rewrite the result in the form of a system:

(ax ≥ 1 + x;
(-ax ≤ 1 + x.

Let's transform to the form:

((а – 1)x ≥ 1;
((a + 1)x ≥ -1.

We 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)].

When a \u003d 0 x \u003d -1.

At 0< а ≤ 1 решений нет.

Graphical method for solving inequalities

Plotting greatly simplifies the solution of equations containing a parameter. The use of the graphical method in solving inequalities with a parameter is even clearer and more expedient.

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, it is always necessary 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 of the inequality will be those values of the variable x for 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 intersection point of the graphs of these functions is the solution of the equation x + 5 = bx, whence x = 5/(b - 1). The graph y \u003d bx is higher for x from the interval (5 / (b - 1); +∞), which means that this set is the solution to the inequality.

2) Similarly, we find that at -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;
there are 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, a 2 = -1.

2) Let's solve this inequality on each subset of real numbers: (-∞; -1); (-1); (-10); (0); (0; +∞).

a) a< -1, из данного неравенства следует, что х >(a + 4)/a;

b) a \u003d -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, this inequality implies that x > (a + 4)/a.

Example 3

Solve the inequality |2 – |x||< a – x.

Solution.

We plot the function y = |2 – |x|| (Fig. 3) and consider all possible cases of the location of the line y \u003d -x + a.

Answer: the inequality has no solutions for a ≤ -2;
x € (-∞; (a - 2)/2) with 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 open up, 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.

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Course work

Artist: Bugrov S.K.

The study of many physical processes and geometric patterns often leads to the solution of problems with parameters. Some Universities also include equations, inequalities and their systems in 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 optional classes.

Preparing this work, I set the goal of a deeper study of this topic, identifying the most rational solution that quickly leads to an answer. In my opinion, the graphical method is a convenient and fast way to solve equations and inequalities with parameters.

In my essay, frequently encountered types of equations, inequalities and their systems are considered, and I hope that the knowledge I have gained in the process of work will help me when passing school exams and entering 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,

under which both the left and right parts 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 be the set of all admissible values of b, etc., X be the set of all admissible values of x, i.e. aОА, bОB, …, xОX. If each of the sets A, B, C, …, K choose and fix, respectively, one value a, b, c, …, k and substitute them into equation (1), then we obtain an equation for x, i.e. equation with one unknown.

Variables a, b, c, ..., k, which are considered constant when solving the equation, are called parameters, and the equation itself is called an equation containing parameters.

The parameters are denoted by the first letters of the Latin alphabet: a, b, c, d, …, k, l, m, n, and the unknown ones by the letters x, y, z.

To solve an equation with parameters means to indicate at what values of the parameters 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 values of the parameters;

b) every solution of the first equation is a solution of the second and vice versa.

§ 2. Solution algorithm.

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 definition of this equation.

We find the points of intersection of the line a=c, where cÎ(-¥;+¥) with the graph of the function a=¦(x). If the 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) with respect to x.

We write down the answer.

I. Solve the equation

(1)

Since x \u003d 0 is not the root of the equation, then we can solve the equation for a:

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 О (-¥;-1]И(1;+¥)И

, then the line y=a intersects the graph of equation (1) at one point. We find the abscissa of this point when solving the equation for x.

Thus, on this interval, equation (1) has the solution

. , then the 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 obtain and . , then the line y=a does not intersect the graph of equation (1), hence there are no solutions.

If a О (-¥;-1]И(1;+¥)И

, That ; , That , ; , then there are no solutions.

II. Find all values of the parameter a for which the equation

has three different roots.

Rewriting the equation as

and having considered a pair of functions , you can see that the desired values of the parameter a and only they will correspond to those positions of the graph of the function at which it has exactly three points of intersection with the graph of the function .

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

- this 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 if this 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

at Therefore, this equation defines a family of “semi-parabolas” - the right branches of the parabola “slide” with their vertices along the abscissa axis.

Select the full squares on the left side of the second equation and factorize it

The set of points of the plane

satisfying the second equation are two straight lines and

Let us find out for what values of the parameter a a curve from the “semi-parabolas” family has at least one common point with one of the straight lines obtained.

State budgetary educational institution

Samara region secondary general education

school number 2 im. V. Maskina railway Art. Klyavlino

municipal district Klyavlinsky

Samara region

« Equations

And

inequalities

with parameters"

tutorial

Klyavlino

Tutorial

"Equations and inequalities with parameters" for students in grades 10-11

this manual is an appendix to the program of the elective course "Equations and inequalities with parameters", which has passed an external examination (scientific and methodological expert council of the Ministry of Education and Science of the Samara Region dated December 19, 2008 was recommended for use in educational institutions of the Samara Region)

Authors

Romadanova Irina Vladimirovna

Mathematics teacher, Klyavlinskaya Secondary General Education

school number 2 them. V. Maskina, Klyavlinsky district, Samara region

Serbaeva Irina Alekseevna

Introduction……………………………………………………………3-4

Linear equations and inequalities with parameters……………..4-7

Quadratic equations and inequalities with parameters……………7-9

Fractional rational equations with parameters……………..10-11

Irrational equations and inequalities with parameters……11-13

Trigonometric equations and inequalities with parameters.14-15

Exponential Equations and Inequalities with Parameters………16-17

Logarithmic equations and inequalities with parameters ...... 16-18

Tasks of the Unified State Examination………………………………………………………...18-20

Tasks for independent work…………………………...21-28

Introduction.

Equations and inequalities with parameters.

If in an equation or inequality some coefficients are not given by specific numerical values, but are indicated by letters, then they are called parameters, and the equation or inequality itself parametric.

In order to solve an equation or inequality with parameters, you must:

Highlight special meaning- this is the value of the parameter in which or when passing through which the solution of the equation or inequality changes.

Define allowed values are the parameter values at which the equation or inequality makes sense.

Solving an equation or inequality with parameters means:

1) determine at what values of the parameters solutions exist;

2) for each admissible system of parameter values, find the corresponding set of solutions.

An equation with a parameter can be solved by the following methods: analytical or graphical.

Analytical method assumes the task of investigating the equation by considering several cases, none of which can be missed.

The solution of the equation and inequality with parameters of each type by the analytical method involves a detailed analysis of the situation and a consistent study, during which the need arises "gentle handling" with a parameter.

Graphic method involves the construction of a graph of the equation, by which it is possible to determine how, respectively, the change in the parameter affects the solution of the equation. The graph sometimes allows one to analytically formulate the necessary and sufficient conditions for solving the set tasks. The graphical solution method is especially effective when it is necessary to establish how many roots the equation has depending on the parameter and has the undoubted advantage of seeing this visually.

§ 1. Linear equations and inequalities.

Linear Equation A x = b , written in general form, can be considered as an equation with parameters, where x – unknown , a , b - options. For this equation, the special or control value of the parameter is the one at which the coefficient vanishes in the unknown.

When solving a linear equation with a parameter, cases are considered when the parameter is equal to its special value and different from it.

Special parameter value a is the value A = 0.

b = 0 is a special parameter value b .

At b ¹ 0 the equation has no solutions.

At b = 0 the equation will take the form: 0x = 0. The solution to this equation is any real number.

Inequalities of the form ah > b And ax < b (a ≠ 0) are called linear inequalities. The set of solutions to the inequality ah >b– interval

(; +), If a > 0 , And (-;) , If A< 0 . Similarly for the inequality

Oh< b set of solutions - interval(-;), If a > 0, And (; +), If A< 0.

Example 1 solve the equation ax = 5

Solution: This is a linear equation.

If a = 0, then the equation 0 × x = 5 has no solution.

If A¹ 0, x =- solution of the equation.

Answer: at A¹ 0, x=

for a = 0 there is no solution.

Example 2 solve the equation ax - 6 \u003d 2a - 3x.

Solution: This is a linear equation ax - 6 \u003d 2a - 3x (1)

ax + 3x = 2a +6

Rewriting the equation as (a+3)x = 2(a+3) Let's consider two cases:

a= -3 And A¹ -3.

If a= -3, then any real number X is the root of equation (1). If A¹ -3 , equation (1) has a single root x = 2.

Answer: At a = -3, x R ; at A ¹ -3, x = 2.

Example 3 At what values of the parameter A among the roots of the equation

2x - 4x - a 2 + 4а – 4 = 0 there are more roots 1 ?

Solution: Solve the equation 2x - 4x - a 2 + 4а – 4 = 0– linear equation

2 (a - 2) x \u003d a 2 - 4a +4

2(a - 2) x \u003d (a - 2) 2

At a = 2 solution of the equation 0x = 0 be any number, even greater than 1.

At A¹ 2 x =
. By condition x > 1, that is
>1, a > 4.

Answer: At A (2) U(4;∞).

Example 4 . For each parameter value A find the number of roots of the equation ax=8.

Solution. ax = 8 is a linear equation.

y = a– a family of horizontal lines;

y = - the graph is a hyperbola. We construct graphs of these functions.

Answer: If a = 0, then the equation has no solutions. If a ≠ 0, then the equation has one solution.

Example 5 . Using graphs, find out how many roots the equation has:

|x| = ax - 1.

y=| x | ,

y = ax - 1- the graph is a straight line passing through a point (0;-1).

We construct graphs of these functions.

Answer: When |a|>1- one root

at | a|≤1 The equation has no roots.

Example 6 . Solve the inequality ax + 4 > 2x + a 2

Solution : ax + 4 > 2x + a 2
(а – 2) х >A 2 – 4. Consider three cases.

Answer. x > a + 2 at a > 2; X<а + 2, at A< 2; at a=2 there are no solutions.

§ 2. Quadratic equations and inequalities

Quadratic equation is an equation of the form Oh ² + b x + c = 0 , Where a≠ 0,

A, b , With - options.

To solve quadratic equations with a parameter, you can use the standard methods of solving using the following formulas:

1 ) discriminant of the quadratic equation: D = b ² - 4 ac , (
²- ac)

2) formulas of the roots of the quadratic equation:X 1 =
, X 2 =
,

(X 1,2 =
)

Square inequalities are called inequalities of the form

a X 2 + b x + c > 0,a X 2 + b x + c< 0, (1), (2)

a X 2 + b x + c ≥ 0,a X 2 + b x + c ≤ 0,(3), (4)

The set of solutions to inequality (3) is obtained by combining the sets of solutions to inequality (1) and the equation , a X 2 + b x + c=0. The set of solutions to inequality (4) is found similarly.

If the discriminant of a square trinomial a X 2 + b x + c less than zero, then for a > 0 the trinomial is positive for all x R.

If the square trinomial has roots (x 1 < х 2 ), then for a > 0 it is positive on the set(-; x 2 )
(X 2; +) and negative on the interval

(x 1; x 2 ). If a< 0, то трехчлен положителен на интервале (х 1 ; x 2 ) and is negative for all x (-; x 1 )
(X 2; +).

Example 1 solve the equation ax² - 2 (a - 1) x - 4 \u003d 0.

This is a quadratic equation

Solution: Special meaning a = 0.

At a = 0 we get a linear equation 2x - 4 = 0. It has a single root x = 2.

At a ≠ 0. Let's find the discriminant.

D \u003d (a-1)² + 4a \u003d (a + 1)²

If a = -1, That D = 0 - one root.

Find the root by substituting a = -1.

-x² + 4x - 4 \u003d 0, that is x² -4x + 4 = 0, we find that x=2.

If a ≠ - 1, That D >0 . According to the root formula, we get:x=
;

X 1 =2, x 2 = -.

Answer: At a=0 and a=-1 equation has one root x = 2; at a ≠ 0 and

A ≠ - 1 equation has two rootsX 1 =2, x 2 =-.

Example 2 Find the number of roots of the given equation x²-2x-8-a=0 depending on parameter values A.

Solution. Let us rewrite this equation in the form x²-2x-8=a

y \u003d x²-2x-8- the graph is a parabola;

y =a- a family of horizontal lines.

Let's build graphs of functions.

Answer: When A<-9 , the equation has no solutions; when a=-9, the equation has one solution; at a>-9, the equation has two solutions.

Example 3 At what A inequality (a - 3) x 2 – 2ax + 3a – 6 >0 holds for all values of x?

Solution. The square trinomial is positive for all values of x if

a-3 > 0 and D<0, т.е. при а, удовлетворяющих системе неравенств

, whence it follows thata > 6 .

Answer.a > 6

§ 3. Fractional-rational equations with a parameter,

reduced to linear

The process of solving fractional equations is carried out according to the usual scheme: the fractional is replaced by an integer by multiplying both parts of the equation by the common denominator of its left and right parts. After that, the whole equation is solved, excluding extraneous roots, that is, numbers that turn the denominator to zero.

In the case of equations with a parameter, this problem is more complicated. Here, in order to “eliminate” extraneous roots, it is required to find the value of the parameter that turns the common denominator to zero, that is, to solve the corresponding equations for the parameter.

Example 1 solve the equation
= 0

Solution: D.Z: x +2 ≠ 0, x ≠ -2

x - a \u003d 0, x \u003d a.

Answer: At a ≠ - 2, x=a

At a = -2 there are no roots.

Example 2 . solve the equation
-
=
(1)

This is a fractional rational equation

Solution: Meaning a = 0 is special. At a = 0 the equation loses its meaning and, therefore, has no roots. If a ≠ 0, then after transformations the equation will take the form: x² + 2 (1-a) x + a² - 2a - 3 = 0 (2)- quadratic equation.

Let's find the discriminant \u003d (1 - a)² - (a² - 2a - 3) \u003d 4, find the roots of the equationX 1 = a + 1, x 2 = a - 3.

When passing from equation (1) to equation (2), the domain of definition of equation (1) expanded, which could lead to the appearance of extraneous roots. Therefore, verification is necessary.

Examination. Exclude from found values X those in which

x 1 +1=0, x 1 +2=0, x 2 +1=0, x 2 +2=0.

If X 1 +1=0, that is (a+1) + 1= 0, That a = -2. Thus,

at a= -2 , X 1 -

If X 1 +2=0, that is (a+1)+2=0, That a = - 3. Thus, at a \u003d - 3, x 1 - extraneous root of the equation. (1).

If X 2 +1=0, that is (a - 3) + 1= 0, That a = 2. Thus, at a = 2 x 2 - extraneous root of equation (1).

If X 2 +2=0, that is ( a – 3) + 2 = 0, That a=1. Thus, at a = 1,

X 2 - extraneous root of equation (1).

In accordance with this, a = - 3 we get x \u003d - 3 - 3 \u003d -6;

at a \u003d - 2 x \u003d -2 – 3= - 5;

at a \u003d 1 x \u003d 1 + 1 \u003d 2;

at a \u003d 2 x \u003d 2 + 1 \u003d 3.

You can write down the answer.

Answer: 1) if a= -3, That x= -6; 2) if a= -2, That x= -5; 3) if a=0, then there are no roots; 4) if a=1, That x=2; 5) if a=2, That x=3; 6) if a ≠ -3, a ≠ -2, a ≠ 0, a ≠ 1, a ≠ 2, then x 1 = a + 1, x 2 = a-3.

§4. Irrational equations and inequalities

Equations and inequalities in which the variable is contained under the root sign are called irrational.

The solution of irrational equations is reduced to the transition from an irrational to a rational equation by raising both sides of the equation to a power or by changing the variable. When both sides of the equation are raised to an even power, extraneous roots may appear. Therefore, when using this method, all found roots should be checked by substitution into the original equation, taking into account changes in the parameter values.

Type equation
=g (x ) is equivalent to the system

The inequality f (x) ≥ 0 follows from the equation f (x) = g 2 (x).

When solving irrational inequalities, we will use the following equivalent transformations:

≤ g(x)

≥g(x)

Example 1 Solve the Equation
= x + 1 (3)

This is an irrational equation

Solution: By definition of the arithmetic root, equation (3) is equivalent to the system
.

At a = 2 the first equation of the system has the form 0 x = 5, that is, it has no solutions.

At a≠ 2 x=
. Let us find out for what valuesA found valueX satisfies the inequalityx ≥ -1:
≥ - 1,
≥ 0,

where a ≤ or a > 2.

Answer: At a≤, a > 2 x=
, at < а ≤ 2 the equation has no solutions.

Example 2 solve the equation
= a (Annex 4)

Solution. y =

y = a is a family of horizontal lines.

Let's build graphs of functions.

Answer: at A<0 - there are no solutions

at A≥ 0 - one solution.

Example 3 . Let's solve the inequality(a+1)
<1.

Solution. O.D.Z. x ≤ 2. If a+1 ≤0, then the inequality holds for all admissible values X. If a+1>0, That

(a+1)
<1.

<

where X (2-
2

Answer. X (- ;2at a (-;-1, X (2-
2

at A (-1;+).

§ 5. Trigonometric equations and inequalities.

Here are the formulas for solving the simplest trigonometric equations:

Sinx = a
x= (-1) n arcsin a+πn, n Z, ≤1, (1)

Cos x = a
x = ± arccos a + 2 πn, n Z, ≤1. (2)

If >1, then equations (1) and (2) have no solutions.

tan x = a
x= arctg a + πn, n Z, a R

ctg x = a
x = arcctg a + πn, n Z, a R

For each standard inequality, we indicate the set of solutions:

1. sin x > a
arcsin a + 2 n Z,

at a <-1, x R ; at a ≥ 1, there are no solutions.

2. . sin x< a
π - arcsin a + 2 πnZ,

for a≤-1, there are no solutions; when a>1,x R

3. cos x > a
- arccos a + 2 pn < x < arccos a + 2 pn , n Z ,

at A<-1, x R ; at a ≥ 1 , there are no solutions.

4. cos x arccos a+ 2 nZ,

at a≤-1 , there are no solutions; ata > 1, x R

5. tg x > a, arctg a + πnZ

6.tg x< a, -π/2 + πn Z

Example1. Find A, for which this equation has a solution:

Cos 2 x + 2(a-2) cosx + a 2 - 4a - 5 \u003d 0.

Solution. We write the equation in the form

Withos 2 x + (2 a -4) cosx +(a – 5)(а+1) =0, solving it as a square, we get cosx = 5-A And cosx = -a-1.

The equation cosx = 5- A has solutions provided -1≤ 5-A ≤1
4≤ A≤ 6, and the equation cosx = - a-1 provided -1≤ -1-A ≤ 1
-2 ≤ A ≤0.

Answer. A -2; 0
4; 6

Example 2 At what bthere exists a such that the inequality
+ b> 0 is satisfied for all x ≠pn , n Z .

Solution. Let's put A= 0. The inequality holds for b >0. Let us now show that no b ≤0 satisfies the conditions of the problem. Indeed, it suffices to put x = π /2, If A <0, и х = - π /2 at A ≥0.

Answer.b > 0

§ 6. Exponential equations and inequalities

1. Equation h(x) f ( x ) = h(x) g ( x) at h(x) > 0 is equivalent to the combination of two systems
And

2. In a particular case (h (x)= a ) the equation A f(x) = A g(x) at A> 0, is equivalent to the combination of two systems

And

3. Equation A f(x) = b , Where A > 0, a ≠1, b>0, is equivalent to the equation

f(x)= log a b . Happening A=1 are considered separately.

The solution of the simplest exponential inequalities is based on the degree property. Inequality of the formf(a x ) > 0 by changing the variablet= a x reduces to solving the system of inequalities
and then to the solution of the corresponding simplest exponential inequalities.

When solving a non-strict inequality, it is necessary to add the roots of the corresponding equation to the set of solutions of a strict inequality. As with solving equations in all examples containing the expression A f (x ) , we assume A> 0. Case A= 1 are considered separately.

Example 1 . At what A equation 8 x =
has only positive roots?

Solution. By the property of an exponential function with a base greater than one, we have x>0
8 X >1

>1

>0, whencea (1,5;4).

Answer. a (1,5;4).

Example 2 Solve the inequality a 2 ∙2 x > a

Solution. Consider three cases:

1. A< 0 . Since the left side of the inequality is positive and the right side is negative, the inequality holds for any x R.

2. a=0. There are no solutions.

3. A > 0 . a 2 ∙2 x > a
2 x >
x > - log 2 a

Answer. X R at A > 0; no solutions for a =0; X (- log 2 a; +) ata > 0 .

§ 7. Logarithmic equations and inequalities

Let us present some equivalences used in solving logarithmic equations and inequalities.

1. The equation log f (x) g (x) \u003d log f (x) h (x) is equivalent to the system

In particular, if A >0, A≠1, then

log a g(x)= log a h(x)

2. The equation log a g(x)=b
g(x)=a b ( A >0, a ≠ 1, g(x) >0).

3. Inequality log f ( x ) g (x) ≤ log f ( x ) h(x) is equivalent to the combination of two systems:
And

If a, b are numbers, a >0, a ≠1, then

log a f(x) ≤ b

log a f(x) > b

Example 1 Solve the Equation

Solution. Let's find ODZ: x > 0, x ≠ A 4 , a > 0, A≠ 1. Transform the equation

log x - 2 = 4 - log a x
log x + log a x– 6 = 0, whence log a x = - 3

x = A-3 and log a x = 2
x = A 2. Condition x = A 4
A – 3 = A 4 or A 2 = A 4 not performed on the ODZ.

Answer: x = A-3, x = A 2 at A (0; 1)
(1; ).

Example 2 . Find the highest value A, for which the equation

2 log -
+ a = 0 has solutions.

Solution. Let's replace
= tand get the quadratic equation 2t 2 – t + a = 0. Solving, we findD = 1-8 a . Consider D≥0, 1-8 A ≥0
A ≤.

At A = quadratic equation has a roott= >0.

Answer. A =

Example 3 . Solve the inequalitylog(x 2 – 2 x + a ) > - 3

Solution. Let's solve the system of inequalities

Roots of square trinomials x 1,2 = 1 ±
their 3,4 = 1 ±
.

Critical parameter values: A= 1 and A= 9.

Let X 1 and X 2 be the solution sets of the first and second inequalities, then

X 1
X 2 = X is the solution to the original inequality.

At 0< a <1 Х 1 = (- ;1 -
)
(1 +
; +), atA> 1 x 1 = (-;+).

At 0< a < 9 Х 2 = (1 -
; 1 +
), atA≥9 Х 2 – no solutions.

Consider three cases:

1. 0< a ≤1 X = (1 -
;1 -
)
(1 +
;1 +
).

2. 1 < a < 9 Х = (1 -
;1 +
).

3. a≥ 9 Х – no solutions.

USE tasks

High level C1, C2

Example 1 Find all values R, for which the equation

R ∙ ctg 2x+2sinx+ p= 3 has at least one root.

Solution. Let's transform the equation

R ∙ (
-1)+2sinx+ p\u003d 3, sinx \u003d t, t
, t 0.

- p+ 2t + p = 3, + 2t = 3, 3 -2t = , 3t2 – 2t3 = p .

Let f(y) = 3 t 2 – 2 t 3 . Let's find the set of function valuesf(x) on

. at / = 6 t – 6 t 2 , 6 t - 6 t 2 = 0, t 1 =0, t 2 = 1. f(-1) = 5, f(1) = 1.

At t
, E(f) =
,

At t
, E(f) =
, that is, when t

, E(f) =
.

To Equation 3t 2 – 2 t 3 = p (hence the given) had at least one root necessary and sufficientp E(f), that is p
.

Answer.
.

Example 2

At what values of the parameterA the equation log
(4 x 2 – 4 a + a 2 +7) = 2 has exactly one root?

Solution. Let's transform the equation into an equivalent one:

4x 2 - 4 a + a 2 +7 \u003d (x 2 + 2) 2.

Note that if a certain number x is the root of the resulting equation, then the number - x is also the root of this equation. By condition, this is not feasible, so the only root is the number 0.

Let's find A.

4∙ 0 2 - 4a + a 2 +7 = (0 2 + 2) 2 ,

a 2 - 4a +7 = 4, a 2 - 4a +3 = 0, a 1 = 1, a 2 = 3.

Examination.

1) a 1 = 1. Then the equation has the form:log
(4 x 2 +4) =2. We solve it

4x 2 + 4 \u003d (x 2 + 2) 2, 4x 2 + 4 \u003d x 4 + 4x 2 + 4, x 4 \u003d 0, x \u003d 0 is the only root.

2) a 2 = 3. The equation looks like:log
(4 x 2 +4) =2
x = 0 is the only root.

Answer. 1; 3

High level C4, C5

Example 3 Find all values R, under which the equation

x 2 - ( R+ 3)x + 1= 0 has integer roots and these roots are solutions to the inequality: x 3 - 7 R x 2 + 2x 2 - 14 R x - 3x +21 R ≤ 0.

Solution. Let x 1, X 2 are the integer roots of the equation x 2 – (R + 3)x + 1= 0. Then, by the Vieta formula, x 1 + x 2 = R + 3, x 1 ∙ x 2 = 1. The product of two integers x 1 , X 2 can be equal to one only in two cases: x 1 = x 2 = 1 or x 1 = x 2 = - 1. If x 1 = x 2 = 1, thenR + 3 = 1+1 = 2
R = - 1; if x 1 = x 2 = - 1, thenR + 3 = - 1 – 1 = - 2
R = - 5. Check whether the roots of the equation x 2 – (R + 3)x + 1= 0 in the cases described by the solutions of this inequality. For the caseR = - 1, x 1 = x 2 = 1 we have

1 3 - 7 ∙ (- 1) ∙ 1 2 +2 ∙ 1 2 - 14 ∙ (- 1) ∙ 1 - 3 ∙ 1 + 21 ∙ (- 1) = 0 ≤ 0 - true; for the case R\u003d - 5, x 1 \u003d x 2 \u003d - 1 we have (- 1) 3 - 7 ∙ (- 5) ∙ (-1) 2 + 2 ∙ (-1) 2 - 14 ∙ (-5) × (- 1) - 3 ∙ (- 1) + 21 ∙ (-5) \u003d - 136 ≤ 0 - right. So, the condition of the problem is satisfied only R= - 1 and R = - 5.

Answer.R 1 = - 1 and R 2 = - 5.

Example 4 Find all positive parameter values A, for which the number 1 belongs to the domain of the function

at = (A
- A
).

Job type: 18

Condition

For what values of the parameter a does 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 Solution

Solution

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 hold for all values of -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).

The 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 when -\frac(2)(5) \leq a \leq 0 .

Answer

\left [-\frac(2)(5); 0\right]

Source: "Mathematics. Preparation for the exam-2016. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 18
Topic: Inequalities with a parameter

Condition

Find all values of the parameter a , for each of which the inequality

x^2+3|x-a|-7x\leqslant -2a

has a unique solution.

Show Solution

Solution

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 \left[\!\!\begin(array)(l) \begin(cases) x \geqslant a, \\ x^2-4x-a\leqslant0; \end(cases) \\ \begin(cases)x \left[\!\!\begin(array)(l) \begin(cases) a \leqslant x, \\ a\geqslant x^2-4x; \end(cases) \\ \begin(cases)a>x, \\ a\leqslant -\frac(x^2)(5)+2x. \end(cases)\end(array)\right.

In the Oxa coordinate system, we construct graphs of functions a=x, a=x^2-4x, a=-\frac(x^2)(5)+2x.

The resulting set is satisfied by the points enclosed between the function graphs a=x^2-4x, a=-\frac(x^2)(5)+2x at 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 ordinate a equal to -4 and equal to 5.

In this lesson, we will study the algorithm for solving inequalities with parameters and learn how to apply it when solving this type of tasks.

Definition one.

To solve an inequality with a parameter means, for each value of the parameter, to find the set of all solutions of this inequality or to prove that there are no solutions.

Consider linear inequalities.

Definition two.

Inequalities of the form a x plus be greater than zero, greater than or equal to zero, less than zero, less than or equal to zero, where a and b are real numbers, X— a variable are called inequalities of the first degree (linear inequalities).

An algorithm for solving a linear inequality with a parameter, for example, the inequality x plus b is greater than zero, where a and b are real numbers, X- variable. Consider the following cases:

First case:a greater than zero, then x is greater than minus ba divided by a.

Consequently, the set of solutions to the inequality is an open numerical ray from minus be divided by a to plus infinity.

Second case:a less than zero, then x is less than minus ba divided by a

and, consequently, the set of solutions to the inequality is an open numerical ray from minus infinity to minus be divided by a.

Third case: a is equal to zero, then the inequality will take the form: zero multiplied by x plus be is greater than zero and for bae greater than zero, any real number is a solution to the inequality, and when bae less than or equal to zero, the inequality has no solutions.

The remaining inequalities are solved similarly.

Consider examples.

Exercise 1

Solve the inequality and x is less than or equal to one.

Solution

Depending on the sign a consider three cases.

First case: if a greater than zero, then x is less than or equal to one divided by a;

Second case: if a less than zero, then x is greater than or equal to one divided by a;

Third case: if a is equal to zero, then the inequality will take the form: zero multiplied by x is less than or equal to one and, therefore, any real number is a solution to the original inequality.

Thus, if A greater than zero, then x belongs to the ray from minus infinity to unity divided by a.

If a a equals zero,

That x

Answer: if A greater than zero, then x belongs to the ray from minus infinity to unity divided by a;

If a less than zero, then x belongs to the ray from unity divided by a to plus infinity, and if a equals zero,

That x x belongs to the set of real numbers.

Task 2

Solve the inequality mod x minus two is greater than the minus square of the difference between a and one.

Solution

Note that modulo x minus two is greater than or equal to zero for any real X and minus the square of the difference between a and unity is less than or equal to zero for any value of the parameter a. Therefore, if a is equal to one, then any X— a real number other than two is a solution to the inequality, and if a is not equal to one, then any real number is a solution to the inequality.

Answer: if a is equal to one, then x belongs to the union of two open numerical rays from minus infinity to two and from two to plus infinity,

and if a belongs to the union of two open numerical rays from minus infinity to one and from one to plus infinity, then X belongs to the set of real numbers.

Task 3

Solve the inequality three multiplied by the difference of four a and x less than two a x plus three.

Solution

After elementary transformations of this inequality, we get the inequality: x times the sum of two a's and three is greater than three times the difference of four a's and one.

First case: if two a plus three is greater than zero, that is a more than minus three second, then x is greater than a fraction, the numerator of which is three times the difference of four a and one, and the denominator is two a plus three.

Second case: if two a plus three is less than zero, that is a less than minus three second, then x is less than a fraction, the numerator of which is three times the difference of four a and one, and the denominator is two a plus three.

Third case: if two a plus three equals zero, that is a equals minus three second,

any real number is a solution to the original inequality.

Therefore, if a belongs to an open number ray from minus three second to plus infinity, then x

belongs to an open numerical ray from a fraction, the numerator of which is three times the difference of four a and one, and the denominator is two a plus three, up to plus infinity.

If a belongs to an open numerical ray from minus infinity to minus three second, then x belongs to an open numerical ray from minus infinity to a fraction whose numerator is three times the difference of four a and one, and the denominator is two a plus three;

If a equals minus three second, then X belongs to the set of real numbers.

Answer: if a belongs to an open number ray from minus three second to plus infinity, then x

belongs to an open numerical ray from a fraction, the numerator of which is three times the difference of four a and one, and the denominator is two a plus three to plus infinity;

if a belongs to an open numerical ray from minus infinity to minus three second, then x belongs to an open numerical ray from minus infinity to a fraction whose numerator is three times the difference of four a and one, and the denominator is two a plus three;

If a equals minus three second, then X belongs to the set of real numbers.

Task 4

For all valid parameter values A solve the inequality square root of x minus a plus square root of two a minus x plus square root of a minus one plus square root of three minus a greater than zero.

Solution

Find the domain of the parameter A. It is determined by a system of inequalities, solving which we find that a belongs to the segment from one to three.

This inequality is equivalent to a system of inequalities, solving which we find that x belongs to the segment from a to two a.

If a belongs to the segment from one to three, then the solution to the original inequality is the segment from a to two a.

Answer: if a belongs to the segment from one to three, then x belongs to the segment from a to two a.

Task 5

Find all A, for which the inequality

the square root of x squared minus x minus two plus the square root of a fraction whose numerator is two minus x and whose denominator is x plus four greater than or equal to and x plus two minus the square root of a fraction whose numerator is x plus one and whose denominator is five minus x has no solution.

Solution

First. Let us calculate the domain of definition of this inequality. It is determined by a system of inequalities, the solution of which is two numbers: x is equal to minus one and x is equal to two.

Second. Let us find all values of a for which this inequality has solutions. To do this, we will find everything A, for which x is equal to minus one and x is equal to two - this is the solution to this inequality. Consider and solve the set of two systems. The solution is to combine two numerical rays from minus infinity to minus one second, and from one to plus infinity.

Hence, this inequality has a solution if a belongs to the union of two numerical rays from minus

infinity to minus one second, and from one to plus infinity.