In this video lesson, I analyzed in detail a rather serious problem 15 from the Unified State Examination in mathematics, which contains both logarithmic and fractional-rational inequalities. Particular attention is paid to Bezout's theorem (for finding the roots of a polynomial), as well as the method of dividing polynomials with a corner (for factoring).
In this lesson we will analyze a system of two inequalities from the Unified State Examination in mathematics:
⎧⎩⎨⎪⎪ log7−2x(x+6) ≤0x− x−3x+6−x2 +27x+90x2 +8x+12≤−1 \left\( \begin(align)& ((\log )_(7-2x))\left(x+6 \right)\le 0 \\& x-\frac(x-3)(x+6 )-\frac(((x)^(2))+27x+90)(((x)^(2))+8x+12)\le -1 \\\end(align) \right.
Solving the system of inequalities
As you can see, the system consists of a logarithmic inequality, as well as a classical fractional-rational inequality, but in the process of solving we will discover that this inequality is not as simple as it might seem at first glance. Let's start with logarithmic. To do this, we will write it out separately:
log7−2x(x+6) ≤ 0
((\log )_(7-2x))\left(x+6 \right)\le \text( )0
Like any logarithmic inequality, this construction is reduced to canonical form, i.e. on the left we leave everything unchanged, but on the right we write it as follows:
log7−2x(x+6) ≤ log7−2x 1
((\log )_(7-2x))\left(x+6 \right)\le ((\log )_(7-2x))1
How to use the rationalization method
Now let's use the rationalization method. Let me remind you that if we have an inequality of the form
logk (x) f(x)⋃ logk (x) g(x) ,
((\log )_(k\left(x \right)))f\left(x \right)\bigcup ((\log )_(k\left(x \right)))g\left(x \ right),
then we can move on to this construction:
(f (x) −g(x) )(k (x)−1)⋃0
\left(f\left(x \right)-g\left(x \right) \right)\left(k\left(x \right)-1 \right)\bigcup 0
Of course, this inequality does not take into account the domain of definition of the logarithm:
f (x) >0
f\left(x\right)>0
g (x) >0
g\left(x\right)>0
1≠k (x) >0
1\ne k\left(x\right)>0
So, in the role f (x) f\left(x\right) acts linear function x+6 x+6, and in the role g (x) g\left(x\right) is simply 1. Therefore, we rewrite our logarithmic inequality of the system as follows:
(x+6−1) (7−2x−1)
\left(x+6-1 \right)\left(7-2x-1 \right)
The last 1 is the one x−1 x-1, which is in the second bracket. All of these are less than or equal to 0. The inequality sign when executed of this transformation is saved. Here are similar ones in each bracket:
(x+5) (6−2x) ≤0
\left(x+5 \right)\left(6-2x \right)\le 0
Application of the interval method
Obviously, we have a simple inequality that can be easily solved by the interval method. Let's equate each bracket to 0:
(+5) =0→= −5
\left(+5 \right)=0\to =-5
6−2=0→2=6
x=3
Let's mark all these points (there are two such points) on the coordinate line. Note that they are shaded:
Let's note the signs. To do this, take any number greater than 3. The first one will be “minus”. Then the signs alternate everywhere, because there are no roots of even multiplicity. We are interested in the less than or equal sign, i.e. the minus sign. Paint over the required areas. Let me remind you that when solving inequalities using the interval method, we substitute 1 billion into the last expression that we obtained before moving on to the equations.
So we have found sets. But, as you understand, this is not yet a solution to inequality. Now we are required to find the domain of definition of the logarithm. To do this, we write the following functions:
Erroneous nesting of equation structures
\left[ \begin(align)& x+6>0 \\& 7-2x>0 \\& 7-2x\ne 1 \\\end(align) \right.=>\left[ \begin(align )& x>-6 \\& 7>2x \\& 6\ne 2x \\\end(align) \right.=>\left[ \begin(align)& \\& x<\text{ }3,5 \\& x\ne \text{ }3 \\\end{align} \right.
So, we have received three simultaneous requirements, i.e. all these inequalities must be satisfied simultaneously. Let's draw a line parallel to our candidate answer:
We received the final answer for the first element of the system:
(−6;−5] ⋃(3;3,5)
\left(-6,-5 \right]\bigcup \left(3,3,5 \right). At this point, many students have a question. Look, 3 - on the one hand it is gouged out, but on the other hand, this the point is colored in. So how to mark it as a result? In order to correctly and once and for all deal with this issue, remember one simple rule.
What does intersection of sets mean? This is a set that is simultaneously included in both the first set and the second. In other words, when filling out the picture drawn below, we are looking for points that simultaneously belong to both the first and second lines. Consequently, if any point does not belong to at least one of these lines, then no matter how it looks on the second line, it does not suit us. And, in particular, with 3, exactly the following story happens: on the one hand, in the candidates for the answer, point 3 suits us, because it is shaded, but on the other hand, 3 is removed due to the domain of definition of the logarithm, and, therefore, in the final set this point must be gouged out. That's it, the answer to the first logarithmic inequality of the system is completely justified. To be safe, I'll duplicate it again:
(−6;−5] ⋃(3;3,5)
\left(-6,-5 \right]\bigcup \left(3,3.5 \right)
Solving a fractional rational inequality
x− x−3x+6−x2 +27x+90x2 +8x+12≤−1 x-\frac(x-3)(x+6)-\frac(((x)^(2))+27x+90)(((x)^(2))+8x+12)\le - 1
Now move -1 to the left:
x+1− x−3x+6−x2 +27x+90(x+6) (x+2)≤0 x+1-\frac(x-3)(x+6)-\frac(((x)^(2))+27x+90)(\left(x+6 \right)\left(x+2 \right))\le 0
x+1 1 −x−3x+6−x2 +27x+90(x+6) (x+2)≤0 \frac(x+1)(1)-\frac(x-3)(x+6)-\frac(((x)^(2))+27x+90)(\left(x+6 \right )\left(x+2 \right))\le 0
We bring the entire structure to a common denominator:
(x+1) (x+6) (x+2) −(x−3) (x+2) − (x2 +27x+90)(x+6) (x+2)≤0 \frac(\left(x+1 \right)\left(x+6 \right)\left(x+2 \right)-\left(x-3 \right)\left(x+2 \right)- \left(((x)^(2))+27x+90 \right))(\left(x+6 \right)\left(x+2 \right))\le 0
Let's expand the brackets:
(x+2) ( (x+1) (x+6) −(x−3) )−x2 −27x−90(x+6) (x+2)≤0 \frac(\left(x+2 \right)\left(\left(x+1 \right)\left(x+6 \right)-\left(x-3 \right) \right)-((x )^(2))-27x-90)(\left(x+6 \right)\left(x+2 \right))\le 0
x3 +6x2 +9x+2 x2 +12x+18− x2 −27x−90(x+6) (x+2)≤0 \frac(((x)^(3))+6((x)^(2))+9x+2((x)^(2))+12x+18-((x)^(2)) -27x-90)(\left(x+6 \right)\left(x+2 \right))\le 0
x3 +7x2 −6x−72(x+6) (x+2)≤0 \frac(((x)^(3))+7((x)^(2))-6x-72)(\left(x+6 \right)\left(x+2 \right))\le 0
What can you say about the resulting inequality? Firstly, it is fractional-rational, with the denominator already factorized. Therefore, the best option would be to solve this inequality using the interval method. However, in order to solve it using the interval method, it is necessary to factorize the numerator. This is the main difficulty, because the numerator is a polynomial of the third degree. Who remembers the formula for third-degree roots? Personally, I don't remember. But we don't need this.
All we need is Bezout’s theorem, or rather, not the theorem itself, but one of its most important corollaries, which states the following: if a polynomial with integer coefficients has a root x1 ((x)_(1)), and it is an integer, then the free coefficient (in our case 72) will necessarily be divided by x1 ((x)_(1)). In other words, if we want to find the roots of this cubic equation, then all we need to do is just dig into the factors that factor 72 into.
Let's factor the number 72 into prime factors:
72=8⋅9=2⋅2⋅2⋅3⋅3
72=8\cdot 9=2\cdot 2\cdot 2\cdot 3\cdot 3
So, we need to go through all combinations of twos and threes to get at least one root of our cubic expression. At first glance it may seem that this is a combinatorial problem, but in reality everything is not so scary. Let's start with the minimum number:
x=2
Let's check if 2 is the answer. To do this, let’s remember what a root is. This is a number that, when substituted into a polynomial, turns it into 0. Let's substitute:
(2) =8+28−12−72<0
\left(2 \right)=8+28-12-72<0
We get that x−2 x-2 is not suitable. Let's move on. Let's take 4:
(4) =64+112−24−72>0
\left(4 \right)=64+112-24-72>0
x=4 x=4 is also not a root of our construction.
Let's move on. Which one is next? x x we will disassemble? To answer this question, let's note an interesting fact: when x−2 x-2 our polynomial was negative, and at x=4 x=4 it turned out to be positive. This means that somewhere between points 2 and 4 our polynomial intersects the axis x x. In other words, somewhere on this segment ours turns to 0. This means that this point will be the desired number. Let's think about what integer lies between 4 and 2. Obviously, only 3, and 3 is present in the expansion, therefore, it really can be the root of our expression. Consider this option:
x=3
(3) =27+63−18−72=90−90=0
\left(3 \right)=27+63-18-72=90-90=0
Great, our hypothesis was confirmed. Really, x=3 x=3 is the root of our construction. But how does this help us factor this polynomial? Very simple. All from the same Bezout’s theorem it follows that if x1 ((x)_(1)) is the root of the polynomial p (x) p\left(x \right), this means that we can write the following:
x1 :p(x) =Q(x) (x− x1 )
((x)_(1)):p\left(x \right)=Q\left(x \right)\left(x-((x)_(1)) \right)
In other words, knowing x1 ((x)_(1)) we can assert that in the decomposition of our expression into factors there will necessarily be a factor x1 ((x)_(1)). In our case, we can write that our polynomial necessarily has a factor in its expansion (x−3)\left(x-3 \right) because 3 is its root.
x3 +7x2 −6x−72x−3=x2 +10x+24\frac(((x)^(3))+7((x)^(2))-6x-72)(x-3)=((x)^(2))+10x+24
In other words, we can rewrite our inequality from the system as follows:
(x+3) (x2 +10x+24)(x+6) (x+2)≤0 \frac(\left(x+3 \right)\left(((x)^(2))+10x+24 \right))(\left(x+6 \right)\left(x+2 \right ))\le 0
Note that in the second bracket of the numerator there is a square trinomial, which can also be factorized very simply, we get:
(x+3) (x+6) (x+4)(x+6) (x+2)≤0 \frac(\left(x+3 \right)\left(x+6 \right)\left(x+4 \right))(\left(x+6 \right)\left(x+2 \right) )\le 0
That's all, all that remains is to write out the roots:
x=3
≠−6(2k)
\ne -6\left(2k \right)
=−4
≠−2
Let's mark all these points that could be a solution to the system on the coordinate line x x:
In order to determine the signs, we take any number greater than 3, substitute it into each of these brackets and get five positive numbers, that is, to the right of 3 there is a plus sign. Then the signs change everywhere, but in -6 nothing changes, because -6 is the root of the second multiplicity. We are interested in those areas where the sign of the function is negative, so we shade the “minuses”.
In total, we can write down the solution to our original inequality - it will be as follows:
(−∞;−6) ⋃(−6;−4] ⋃(−2;3]
\left(-\infty ;-6 \right)\bigcup \left(-6;-4 \right]\bigcup \left(-2;3 \right]
Final Steps
We have solved the second inequality of our system, and now it remains to solve the system itself, that is, to intersect the sets we obtained. To do this, I propose to build another line parallel to our two old lines responsible for the logarithmic inequality from the system:
We can write down the final answer of the second element of the system of inequalities: (−6;−5] \left(-6;-5 \right]. Now we can return to our system and write down the final set:
x∈ (−6; −5]
x\in \left(-6;\text( )-5 \right]
Key Points
There are several key points in this task:
- You need to be able to solve logarithmic inequalities using the transition to the canonical form.
- You need to be able to work with fractional rational inequalities. This is generally 8th-9th grade material, so if you work with logarithms, then you will understand fractional rational inequalities.
- Bezout's theorem. The most important consequence of this theorem is the fact that the roots of a polynomial with integer coefficients are divisors of its free term.
Otherwise, this is a simple, although rather voluminous, task of solving a system of equations. Certain difficulties in solving the system can also arise in the intersection of all sets, especially those associated with point 3. Everything is very simple here: just remember that intersection means the requirement for the simultaneous fulfillment of all inequalities, i.e. the required point must be shaded on all three axes. If on at least one axis it is not painted over or punctured, then such a point cannot be part of the answer.
Unified State Exam 2018. Mathematics. Profile level. Solving equations and inequalities. Sadovnichy Yu.V.
M.: 2018. - 96 p.
This book is devoted to problems similar to problem 15 of the Unified State Exam in mathematics (solving equations and inequalities). Various methods for solving such problems, including original ones, are considered. The book will be useful for high school students, mathematics teachers, and tutors.
Format: pdf
Size: 860 KB
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TABLE OF CONTENTS
INTRODUCTION 4
CHAPTER 1. INTERVAL METHOD FOR SOLVING INEQUALITIES 6
Problems for independent solution 10
CHAPTER 2. DISCLOSURE OF MODULES IN EQUATIONS AND INEQUALITIES 13
Problems for independent solution 23
CHAPTER 3. IRRATIONAL EQUATIONS AND INEQUALITIES 25
Problems for independent solution 33
CHAPTER 4. EXPONENTARY AND LOGARITHMIC EQUATIONS AND INEQUALITIES 35
4.1. Basic formulas and solutions to simple equations and inequalities 35
4.2. Converting the sum and difference of logarithms 36
Problems for independent solution 41
4.3. Variable replacement method 42
Problems for independent solution 47
4.4. Splitting inequalities 49
Problems for independent solution 55
4.5. Transition to a new foundation 56
Problems for independent solution 60
CHAPTER 5. EQUATIONS AND INEQUALITIES OF MIXED TYPE 61
Problems for independent solution 68
CHAPTER 6. LOGARITHMIC INTERVAL METHOD 70
Problems for independent solution 75
CHAPTER 7. SYSTEMS OF ALGEBRAIC EQUATIONS AND INEQUALITIES 76
Problems for independent solution 84
ANSWERS TO PROBLEMS FOR INDEPENDENT SOLUTION 88
This book is devoted to problems similar to problem 15 of the profile Unified State Examination in mathematics (equations and inequalities). The book is divided into chapters by topic, the material in each chapter is presented “from simple to complex.”
It is no secret that problems 16-19 (planimetry, word problem, parameter problem, integer problem) are difficult for the vast majority of high school graduates. The same can be said about problem 14 (stereometry). Therefore, solved problem 15 (along with problem 13) is an opportunity to increase your score on the Unified State Exam to a good level.
The first three chapters are preparatory and cover solving inequalities by the interval method, equations and inequalities containing modulus, irrational equations and inequalities.
The fourth chapter is the main one in this book, since the problems in it are closest to the real problem of the 15th profile Unified State Exam in mathematics. This chapter is divided into several paragraphs, each of which explores a method for solving such a problem.
