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, let’s take x out of brackets on the left side of the inequality:

(5 – a)x > a + 3. Now consider possible cases for 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;

Ах/(а – 1) ≤ -а/3. Multiplying both sides of the inequality by (-1), we get:

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

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

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

Case 3. 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 relative to x.

Solution.

It follows from the condition that the right-hand side of the inequality ax must be non-negative, i.e. ax ≥ 0. By the rule of revealing the module from the inequality |1 + x| ≤ ax we have a double inequality

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

(ax ≥ 1 + x;
(-ax ≤ 1 + x.

Let's transform it to:

((a – 1)x ≥ 1;
((a + 1)x ≥ -1.

We study 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 = 0 x = -1.

At 0< а ≤ 1 решений нет.

Graphical method for solving inequalities

Plotting graphs greatly simplifies solving equations containing a parameter. Using the graphical method when solving inequalities with a parameter is even clearer and more expedient.

Graphically solving 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 to 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 picture shows:

1) For b > 1 the lines intersect. The abscissa of the intersection point 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); +∞), which means 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 parameter a: a 1 = 0, and 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 = -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, from this inequality it follows that x > (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 discovered, which can then be successfully applied in any other branches of mathematics.

Problems with parameters play an 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? Don't know how to solve inequalities?
To get help from a tutor -.
The first lesson is free!

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State budget educational institution

Samara region secondary general education

School No. 2 named after. V. Maskina railway Art. Klyavlino

Klyavlinsky municipal district

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 external examination (the scientific and methodological expert council of the Ministry of Education and Science of the Samara region dated December 19, 2008 recommended for use in educational institutions Samara region)

Authors

Romadanova Irina Vladimirovna

mathematics teacher at Klyavlinskaya Secondary Educational Institution

School No. 2 named after. 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

Unified State Exam objectives………………………………………………………...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 specific numerical values, but are designated 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 need to:

Select 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 valid values– these are the values of the parameter at which the equation or inequality makes sense.

Solving an equation or inequality with parameters means:

1) determine at what parameter values solutions exist;

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

You can solve an equation with a parameter using the following methods: analytical or graphical.

Analytical method involves the task of studying an equation by considering several cases, none of which can be missed.

Solving equations and inequalities with parameters of each type using an analytical method involves detailed analysis situations and consistent research during which the need arises "careful handling" with parameter.

Graphical method involves constructing a graph of the equation, from which one can determine how a change in the parameter affects the solution of the equation, respectively. The graph sometimes allows you to analytically formulate the necessary and sufficient conditions for solving the problem. The graphical solution method is especially effective when you need to establish how many roots an equation has depending on a parameter and has the undoubted advantage of seeing this clearly.

§ 1. Linear equations and inequalities.

Linear equation A x = b , recorded in general view, 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 of the unknown becomes zero.

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. Set of solutions to 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 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 = 2a – 3x.

Solution: This is a linear equation, ax – 6 = 2a – 3x (1)

ax + 3x = 2a +6

Rewriting the equation as (a+3)x = 2(a+3), 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 parameter values A among the roots of the equation

2ah – 4kh – a 2 + 4a – 4 = 0 there are more roots 1 ?

Solution: Let's solve the equation 2ah – 4kh – a 2 + 4a – 4 = 0– linear equation

2(a - 2) x = a 2 – 4a +4

2(a - 2) x = (a – 2) 2

At a = 2 solving the equation 0x = 0 will be any number, including one greater than 1.

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

Answer: At A (2) U (4;∞).

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

Solution. ax = 8– linear equation.

y = a– family of horizontal lines;

y = - The graph is a hyperbola. Let's build 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| = ah – 1.

y =| x | ,

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

Let's build graphs of these functions.

Answer: When |a|>1- one root

at | a|≤1 – the equation has no roots.

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

Solution : ax + 4 > 2x + a 2
(a – 2) x >A 2 – 4. Let's 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 standard solution methods using the following formulas:

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

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

(X 1,2 =
)

Quadratic inequalities are called

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) can be found similarly.

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

If a quadratic 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 negative for all x (-; x 1 )
(X 2; +).

Example 1. Solve the equation ax² - 2 (a – 1)x – 4 = 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 = (a-1)² + 4a = (a+1)²

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

Let's find the root by substituting a = -1.

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

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

X 1 =2, x 2 = -.

Answer: At a=0 and a= -1 the 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 this equation x²-2x-8-a=0 depending on parameter values A.

Solution. Let's rewrite given equation as x²-2x-8=a

y = 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. A quadratic 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 parameter,

reducible to linear

The process of solving fractional equations is carried out according to the usual scheme: the fraction is replaced by an integer by multiplying both sides of the equation by the common denominator of its left and right sides. After which 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 complex. Here, in order to “eliminate” extraneous roots, it is necessary 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 = 0, x = a.

Answer: At a ≠ - 2, x=a

At a = -2 there are no roots.

Example 2 . Solve the equation
-
=
(1)

This fractional-rational the equation

Solution: Meaning a = 0 is special. At a = 0 the equation makes no sense 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 = (1 – a)² - (a² - 2a – 3)= 4, find the roots of the equationX 1 = a + 1, x 2 = a - 3.

When moving 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. Let's exclude from the 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, when a = - 3, x 1 - extraneous root of the equation. (1).

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

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

X 2 - extraneous root of equation (1).

In accordance with this, when a = - 3 we get x = - 3 – 3 = -6;

at a = - 2 x = -2 – 3= - 5;

at a = 1 x =1 + 1= 2;

at a = 2 x = 2+1 = 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.

Solving irrational equations comes down to moving from an irrational to a rational equation by exponentiating both sides of the equation or replacing a variable. When both sides of the equation are raised to an even power, extraneous roots may appear. Therefore, when using this method, you should check all the roots found by substituting them into the original equation, taking into account changes in the parameter values.

Equation of the form
=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 an 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's find out at 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 (Appendix 4)

Solution. y =

y = a– 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= arctan 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 πn 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; for a > 1,x R

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

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

4. cos x arccos a+ 2 πnZ,

at a≤-1 , no solutions; ata > 1, x R

5. tan x > a, arctan a + πnZ

6.tg x< a, -π/2 + πn Z

Example 1. Find A, for which this equation has a solution:

Cos 2 x + 2(a-2)cosx + a 2 – 4a – 5 =0.

Solution. Let's write the equation in the form

Withos 2 x + (2 a -4) cosx +(a – 5)(a+1) =0, solving it as a quadratic, 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 Eq. cosx = - a-1 provided -1≤ -1-A ≤ 1
-2 ≤ A ≤0.

Answer. A -2; 0
4; 6

Example 2. At what bthere is such that the inequality
+ b> 0 holds for all x ≠πn , 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 is enough 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 a collection of two systems
And

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

And

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

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

The simplest solution exponential inequalities based on the property of degree. Inequality of the formf(a x ) > 0 using variable changet= a x reduces to solving the system of inequalities
and then to solving the corresponding simple exponential inequalities.

When solving a non-strict inequality, it is necessary to add the roots of the corresponding equation to the set of solutions to the strict inequality. As in 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, where froma (1,5;4).

Answer. a (1,5;4).

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

Solution. Let's 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; there are 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) = 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 a 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 the 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 is not performed on ODZ.

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

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

2 log -
+ a = 0 has solutions.

Solution. We'll make a replacement
= tand we get quadratic equation 2t 2 – t + a = 0. Solving, we findD = 1-8 a . Let's consider D≥0, 1-8 A ≥0
A ≤.

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

Answer. A =

Example 3 . Solve 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 sets of solutions to 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 X 2 – no solutions.

Let's consider three cases:

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

2. 1 < a < 9 Х = (1 -
;1 +
).

3. a≥ 9 X – no solutions.

Unified State Exam objectives

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= 3, sinx =t, t
, t 0.

- p+2t+ p = 3, + 2 t = 3, 3 -2t = , 3t 2 – 2t 3 = 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 parameter valuesA the equation log
(4 x 2 – 4 a + a 2 +7) = 2 has exactly one root?

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

4x 2 – 4 a + a 2 +7 = (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.

We'll 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 looks like:log
(4 x 2 +4) =2. Let's solve it

4x 2 + 4 = (x 2 + 2) 2, 4x 2 + 4 = x 4 + 4x 2 + 4, x 4 = 0, x = 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, for 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 – integer roots of the equation x 2 – (R + 3)x + 1= 0. Then, according to Vieta’s formula, the equalities x 1 + x 2 = R + 3, x 1 ∙ x 2 = 1. 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. Let's check whether the roots of the equation x 2 – (R + 3)x + 1= 0 in the described cases by solutions to this inequality. For the occasionR = - 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 occasion R= - 5, x 1 = x 2 = - 1 we have (- 1) 3 – 7 ∙ (- 5) ∙ (-1) 2 + 2 ∙ (-1) 2 – 14 ∙ (-5) × (- 1 ) – 3 ∙ (- 1) + 21 ∙ (-5) = - 136 ≤ 0 – correct. So, the conditions of the problem are satisfied only R= - 1 and R = - 5.

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

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

at = (A
- A
).

Application

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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 is satisfied 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 be executed 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).

Inequality (*) will be satisfied 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 Unified State 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 will construct graphs of functions a=x, a=x^2-4x, a=-\frac(x^2)(5)+2x.

The resulting set is satisfied by the points between the graphs of the functions a=x^2-4x, a=-\frac(x^2)(5)+2x on the interval x\in (shaded area).

From 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.