How to determine an even or odd function online. Function properties

How to insert mathematical formulas into a website?

If you ever need to add one or two mathematical formulas to a web page, then the easiest way to do this is as described in the article: mathematical formulas are easily inserted into the site in the form of pictures that Wolfram Alpha automatically generates. In addition to simplicity, this versatile method will help improve your site's visibility in search engines. It has been working for a long time (and, I think, it will work forever), but it is morally outdated.

If you regularly use math formulas on your site, then I recommend that you use MathJax, a special JavaScript library that displays math notation in web browsers using MathML, LaTeX, or ASCIIMathML markup.

There are two ways to get started using MathJax: (1) with simple code, you can quickly connect a MathJax script to your site, which will be in the right moment automatically download from a remote server (server list); (2) upload the MathJax script from a remote server to your server and connect it to all pages of your site. The second method, which is more complicated and time-consuming, will speed up the loading of your site's pages, and if the parent MathJax server for some reason becomes temporarily unavailable, this will not affect your own site in any way. Despite these advantages, I chose the first method as it is simpler, faster and does not require technical skills. Follow my example, and in 5 minutes you will be able to use all the features of MathJax on your site.

You can connect the script of the MathJax library from a remote server using two versions of the code taken from the main MathJax site or from the documentation page:

One of these code variants must be copied and pasted into the code of your web page, preferably between the tags and or right after the tag ... According to the first option, MathJax loads faster and slows down the page less. But the second option automatically tracks and loads the latest versions of MathJax. If you insert the first code, then it will need to be updated periodically. If you insert the second code, the pages will load more slowly, but you will not need to constantly monitor MathJax updates.

The easiest way to connect MathJax is in Blogger or WordPress: in your site's dashboard, add a widget designed to insert third-party JavaScript code, copy the first or second version of the loading code presented above into it, and place the widget closer to the beginning of the template (by the way, this is not necessary at all because the MathJax script is loaded asynchronously). That's all. Now, learn the MathML, LaTeX, and ASCIIMathML markup syntax, and you're ready to embed math formulas into your website's web pages.

Any fractal is built according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.

The iterative algorithm for constructing the Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 adjacent cubes are removed from it. The result is a set consisting of the remaining 20 smaller cubes. Doing the same with each of these cubes, we get a set, already consisting of 400 smaller cubes. Continuing this process endlessly, we get a Menger sponge.

To do this, use graph paper or a graphing calculator. Select any multiple of the numeric explanatory variable values x (\ displaystyle x) and plug them into the function to calculate the values ​​of the dependent variable y (\ displaystyle y)... Draw the found coordinates of the points on the coordinate plane, and then connect these points to build a graph of the function.

• Substitute positive numeric values ​​into the function x (\ displaystyle x) and corresponding negative numeric values. For example, given a function. Plug in the following values x (\ displaystyle x):
  • f (1) = 2 (1) 2 + 1 = 2 + 1 = 3 (\ displaystyle f (1) = 2 (1) ^ (2) + 1 = 2 + 1 = 3) (1, 3) (\ displaystyle (1,3)).
  • f (2) = 2 (2) 2 + 1 = 2 (4) + 1 = 8 + 1 = 9 (\ displaystyle f (2) = 2 (2) ^ (2) + 1 = 2 (4) +1 = 8 + 1 = 9)... Got a point with coordinates (2, 9) (\ displaystyle (2.9)).
  • f (- 1) = 2 (- 1) 2 + 1 = 2 + 1 = 3 (\ displaystyle f (-1) = 2 (-1) ^ (2) + 1 = 2 + 1 = 3)... Got a point with coordinates (- 1, 3) (\ displaystyle (-1,3)).
  • f (- 2) = 2 (- 2) 2 + 1 = 2 (4) + 1 = 8 + 1 = 9 (\ displaystyle f (-2) = 2 (-2) ^ (2) + 1 = 2 ( 4) + 1 = 8 + 1 = 9)... Got a point with coordinates (- 2, 9) (\ displaystyle (-2.9)).

Evenness and oddness of a function are one of its main properties, and evenness occupies an impressive part of the school mathematics course. It largely determines the nature of the behavior of the function and greatly facilitates the construction of the corresponding graph.

Let us define the parity of the function. Generally speaking, the function under study is considered even if for opposite values ​​of the independent variable (x) located in its domain of definition, the corresponding values ​​of y (function) turn out to be equal.

Let us give a more rigorous definition. Consider some function f (x), which is given in the domain D. It will be even if for any point x located in the domain of definition:

• -x (opposite point) is also in this scope,

• f (-x) = f (x).

The above definition implies a condition necessary for the domain of definition of such a function, namely, symmetry with respect to the point O, which is the origin, since if some point b is contained in the domain of an even function, then the corresponding point b also lies in this domain. Thus, the conclusion follows from the above: even function has a form symmetrical with respect to the ordinate axis (Oy).

How to determine the parity of a function in practice?

Let it be given using the formula h (x) = 11 ^ x + 11 ^ (- x). Following the algorithm that follows directly from the definition, we first investigate its domain of definition. Obviously, it is defined for all values ​​of the argument, that is, the first condition is satisfied.

The next step is to substitute its opposite value (-x) for argument (x).
We get:
h (-x) = 11 ^ (- x) + 11 ^ x.
Since addition satisfies the commutative (transposable) law, it is obvious that h (-x) = h (x) and the given functional dependence is even.

Let us check the evenness of the function h (x) = 11 ^ x-11 ^ (- x). Following the same algorithm, we get that h (-x) = 11 ^ (- x) -11 ^ x. Taking out the minus, in the end, we have
h (-x) = - (11 ^ x-11 ^ (- x)) = - h (x). Therefore, h (x) is odd.

By the way, it should be recalled that there are functions that cannot be classified according to these criteria, they are called neither even nor odd.

Even functions have a number of interesting properties:

• as a result of the addition of such functions, an even one is obtained;
• as a result of the subtraction of such functions, an even one is obtained;
• even, also even;
• as a result of multiplication of two such functions, an even one is obtained;
• as a result of multiplying the odd and even functions, an odd one is obtained;
• as a result of dividing the odd and even functions, an odd one is obtained;
• the derivative of such a function is odd;
• if we square an odd function, we get an even one.

The parity function can be used when solving equations.

To solve an equation of the type g (x) = 0, where the left-hand side of the equation is an even function, it will be enough to find its solution for non-negative values ​​of the variable. The resulting roots of the equation must be combined with opposite numbers. One of them is subject to verification.

This is also successfully used to solve non-standard problems with a parameter.

For example, is there any value for the parameter a for which the equation 2x ^ 6-x ^ 4-ax ^ 2 = 1 will have three roots?

If we take into account that the variable enters the equation in even powers, then it is clear that replacing x with - x given equation will not change. It follows that if some number is its root, then the opposite number is also the same. The conclusion is obvious: the nonzero roots of the equation are included in the set of its solutions in “pairs”.

It is clear that the number 0 itself is not, that is, the number of roots of such an equation can only be even and, naturally, at no value of the parameter it cannot have three roots.

But the number of roots of the equation 2 ^ x + 2 ^ (- x) = ax ^ 4 + 2x ^ 2 + 2 can be odd, and for any value of the parameter. Indeed, it is easy to check that the set of roots this equation contains solutions in pairs. Let's check if 0 is a root. When substituting it into the equation, we get 2 = 2. Thus, besides the "paired" ones, 0 is also a root, which proves their odd number.

even if for all \ (x \) from its domain of definition it is true: \ (f (-x) = f (x) \).

The graph of an even function is symmetric about the \ (y \) axis:

Example: the function \ (f (x) = x ^ 2 + \ cos x \) is even, because \ (f (-x) = (- x) ^ 2 + \ cos ((- x)) = x ^ 2 + \ cos x = f (x) \).

\ (\ blacktriangleright \) The \ (f (x) \) function is called odd if for all \ (x \) from its domain it is true: \ (f (-x) = - f (x) \).

The graph of an odd function is symmetric about the origin:

Example: the function \ (f (x) = x ^ 3 + x \) is odd because \ (f (-x) = (- x) ^ 3 + (- x) = - x ^ 3-x = - (x ^ 3 + x) = - f (x) \).

\ (\ blacktriangleright \) Functions that are neither even nor odd are called functions general view... Such a function can always be uniquely represented as a sum of an even and an odd function.

For example, the function \ (f (x) = x ^ 2-x \) is the sum of an even function \ (f_1 = x ^ 2 \) and an odd \ (f_2 = -x \).

\ (\ blacktriangleright \) Some properties:

1) The product and quotient of two functions of the same parity is an even function.

2) The product and quotient of two functions of different parity is an odd function.

3) The sum and difference of even functions is an even function.

4) The sum and difference of odd functions is an odd function.

5) If \ (f (x) \) is an even function, then the equation \ (f (x) = c \ (c \ in \ mathbb (R) \)) has a unique root if and only if, when \ (x = 0 \).

6) If \ (f (x) \) is an even or odd function, and the equation \ (f (x) = 0 \) has a root \ (x = b \), then this equation will necessarily have a second root \ (x = -b \).

\ (\ blacktriangleright \) A function \ (f (x) \) is called periodic on \ (X \) if \ (f (x) = f (x + T) \), where \ (x, x + T \ in X \). The smallest \ (T \) for which this equality holds is called the main (main) period of the function.

A periodic function has any number of the form \ (nT \), where \ (n \ in \ mathbb (Z) \) will also be a period.

Example: any trigonometric function is periodic;
for the functions \ (f (x) = \ sin x \) and \ (f (x) = \ cos x \), the principal period is \ (2 \ pi \), for the functions \ (f (x) = \ mathrm ( tg) \, x \) and \ (f (x) = \ mathrm (ctg) \, x \) the principal period is \ (\ pi \).

In order to plot a graph of a periodic function, you can plot its graph on any segment of length \ (T \) (main period); then the graph of the entire function is completed by shifting the constructed part by an integer number of periods to the right and left:

\ (\ blacktriangleright \) The domain \ (D (f) \) of a function \ (f (x) \) is a set consisting of all values ​​of the \ (x \) argument for which the function is meaningful (defined).

Example: the function \ (f (x) = \ sqrt x + 1 \) has scope: \ (x \ in

Task 1 # 6364

Task level: Equal to the exam

For what values ​​of the parameter \ (a \) the equation

It has only decision?

Note that since \ (x ^ 2 \) and \ (\ cos x \) are even functions, then if the equation has a root \ (x_0 \), it will also have a root \ (- x_0 \).
Indeed, let \ (x_0 \) be a root, that is, the equality \ (2x_0 ^ 2 + a \ mathrm (tg) \, (\ cos x_0) + a ^ 2 = 0 \) right. Substitute \ (- x_0 \): \ (2 (-x_0) ^ 2 + a \ mathrm (tg) \, (\ cos (-x_0)) + a ^ 2 = 2x_0 ^ 2 + a \ mathrm (tg) \, (\ cos x_0) + a ^ 2 = 0 \).

Thus, if \ (x_0 \ ne 0 \), then the equation will already have at least two roots. Therefore, \ (x_0 = 0 \). Then:

We got two values ​​for the \ (a \) parameter. Note that we have used the fact that \ (x = 0 \) is exactly the root of the original equation. But we have never used the fact that he is the only one. Therefore, it is necessary to substitute the resulting values ​​of the parameter \ (a \) into the original equation and check for which \ (a \) the root \ (x = 0 \) will really be unique.

1) If \ (a = 0 \), then the equation takes the form \ (2x ^ 2 = 0 \). Obviously, this equation has only one root \ (x = 0 \). Therefore, the value \ (a = 0 \) suits us.

2) If \ (a = - \ mathrm (tg) \, 1 \), then the equation takes the form \ We rewrite the equation as \ Because \ (- 1 \ leqslant \ cos x \ leqslant 1 \), then \ (- \ mathrm (tg) \, 1 \ leqslant \ mathrm (tg) \, (\ cos x) \ leqslant \ mathrm (tg) \, 1 \)... Consequently, the values ​​of the right-hand side of equation (*) belong to the segment \ ([- \ mathrm (tg) ^ 2 \, 1; \ mathrm (tg) ^ 2 \, 1] \).

Since \ (x ^ 2 \ geqslant 0 \), the left side of the equation (*) is greater than or equal to \ (0+ \ mathrm (tg) ^ 2 \, 1 \).

Thus, equality (*) can only hold when both sides of the equation are \ (\ mathrm (tg) ^ 2 \, 1 \). This means that \ [\ begin (cases) 2x ^ 2 + \ mathrm (tg) ^ 2 \, 1 = \ mathrm (tg) ^ 2 \, 1 \\ \ mathrm (tg) \, 1 \ cdot \ mathrm (tg) \ , (\ cos x) = \ mathrm (tg) ^ 2 \, 1 \ end (cases) \ quad \ Leftrightarrow \ quad \ begin (cases) x = 0 \\ \ mathrm (tg) \, (\ cos x) = \ mathrm (tg) \, 1 \ end (cases) \ quad \ Leftrightarrow \ quad x = 0 \] Therefore, the value \ (a = - \ mathrm (tg) \, 1 \) suits us.

Answer:

\ (a \ in \ (- \ mathrm (tg) \, 1; 0 \) \)

Quest 2 # 3923

Task level: Equal to the exam

Find all values ​​of the parameter \ (a \), for each of which the graph of the function \

symmetrical about the origin.

If the graph of a function is symmetric about the origin, then such a function is odd, that is, \ (f (-x) = - f (x) \) holds for any \ (x \) from the domain of the function. Thus, it is required to find those values ​​of the parameter for which \ (f (-x) = - f (x). \)

\ [\ begin (aligned) & 3 \ mathrm (tg) \, \ left (- \ dfrac (ax) 5 \ right) +2 \ sin \ dfrac (8 \ pi a + 3x) 4 = - \ left (3 \ mathrm (tg) \, \ left (\ dfrac (ax) 5 \ right) +2 \ sin \ dfrac (8 \ pi a-3x) 4 \ right) \ quad \ Rightarrow \ quad -3 \ mathrm (tg) \ , \ dfrac (ax) 5 + 2 \ sin \ dfrac (8 \ pi a + 3x) 4 = - \ left (3 \ mathrm (tg) \, \ left (\ dfrac (ax) 5 \ right) +2 \ sin \ dfrac (8 \ pi a-3x) 4 \ right) \ quad \ Rightarrow \\ \ Rightarrow \ quad & \ sin \ dfrac (8 \ pi a + 3x) 4+ \ sin \ dfrac (8 \ pi a- 3x) ​​4 = 0 \ quad \ Rightarrow \ quad2 \ sin \ dfrac12 \ left (\ dfrac (8 \ pi a + 3x) 4+ \ dfrac (8 \ pi a-3x) 4 \ right) \ cdot \ cos \ dfrac12 \ left (\ dfrac (8 \ pi a + 3x) 4- \ dfrac (8 \ pi a-3x) 4 \ right) = 0 \ quad \ Rightarrow \ quad \ sin (2 \ pi a) \ cdot \ cos \ frac34 x = 0 \ end (aligned) \]

The last equation must be satisfied for all \ (x \) from the domain \ (f (x) \), therefore, \ (\ sin (2 \ pi a) = 0 \ Rightarrow a = \ dfrac n2, n \ in \ mathbb (Z) \).

Answer:

\ (\ dfrac n2, n \ in \ mathbb (Z) \)

Quest 3 # 3069

Task level: Equal to the exam

Find all values ​​of the parameter \ (a \), for each of which the equation \ has 4 solutions, where \ (f \) is an even periodic function with period \ (T = \ dfrac (16) 3 \) defined on the whole number line , and \ (f (x) = ax ^ 2 \) for \ (0 \ leqslant x \ leqslant \ dfrac83. \)

(Challenge from subscribers)

Since \ (f (x) \) is an even function, its graph is symmetric about the ordinate axis, therefore, for \ (- \ dfrac83 \ leqslant x \ leqslant 0 \)\ (f (x) = ax ^ 2 \). Thus, for \ (- \ dfrac83 \ leqslant x \ leqslant \ dfrac83 \), and this is a segment of length \ (\ dfrac (16) 3 \), function \ (f (x) = ax ^ 2 \).

1) Let \ (a> 0 \). Then the graph of the function \ (f (x) \) will look like this:


Then, in order for the equation to have 4 solutions, it is necessary that the graph \ (g (x) = | a + 2 | \ cdot \ sqrtx \) passes through the point \ (A \):


Hence, \ [\ dfrac (64) 9a = | a + 2 | \ cdot \ sqrt8 \ quad \ Leftrightarrow \ quad \ left [\ begin (gathered) \ begin (aligned) & 9 (a + 2) = 32a \\ & 9 (a +2) = - 32a \ end (aligned) \ end (gathered) \ right. \ quad \ Leftrightarrow \ quad \ left [\ begin (gathered) \ begin (aligned) & a = \ dfrac (18) (23) \\ & a = - \ dfrac (18) (41) \ end (aligned) \ end ( gathered) \ right. \] Since \ (a> 0 \), then \ (a = \ dfrac (18) (23) \) is suitable.

2) Let \ (a<0\) . Тогда картинка окажется симметричной относительно начала координат:


It is necessary that the graph \ (g (x) \) goes through the point \ (B \): \ [\ dfrac (64) 9a = | a + 2 | \ cdot \ sqrt (-8) \ quad \ Leftrightarrow \ quad \ left [\ begin (gathered) \ begin (aligned) & a = \ dfrac (18) (23 ) \\ & a = - \ dfrac (18) (41) \ end (aligned) \ end (gathered) \ right. \] Since \ (a<0\) , то подходит \(a=-\dfrac{18}{41}\) .

3) The case when \ (a = 0 \) does not fit, since then \ (f (x) = 0 \) for all \ (x \), \ (g (x) = 2 \ sqrtx \) and the equation will only have 1 root.

Answer:

\ (a \ in \ left \ (- \ dfrac (18) (41); \ dfrac (18) (23) \ right \) \)

Quest 4 # 3072

Task level: Equal to the exam

Find all values ​​\ (a \), for each of which the equation \

has at least one root.

(Challenge from subscribers)

We rewrite the equation as \ and consider two functions: \ (g (x) = 7 \ sqrt (2x ^ 2 + 49) \) and \ (f (x) = 3 | x-7a | -6 | x | -a ^ 2 + 7a \ ).
The function \ (g (x) \) is even, has a minimum point \ (x = 0 \) (moreover, \ (g (0) = 49 \)).
The function \ (f (x) \) for \ (x> 0 \) is decreasing, and for \ (x<0\) – возрастающей, следовательно, \(x=0\) – точка максимума.
Indeed, for \ (x> 0 \) the second module expands positively (\ (| x | = x \)), therefore, regardless of how the first module expands, \ (f (x) \) will be equal to \ ( kx + A \), where \ (A \) is an expression from \ (a \), and \ (k \) is either \ (- 9 \) or \ (- 3 \). For \ (x<0\) наоборот: второй модуль раскроется отрицательно и \(f(x)=kx+A\) , где \(k\) равно либо \(3\) , либо \(9\) .
Find the value \ (f \) at the maximum point: \

In order for the equation to have at least one solution, the graphs of the functions \ (f \) and \ (g \) must have at least one intersection point. Therefore, you need: \ \\]

Answer:

\ (a \ in \ (- 7 \) \ cup \)

Task 5 # 3912

Task level: Equal to the exam

Find all values ​​of the parameter \ (a \), for each of which the equation \

has six different solutions.

Let's make the replacement \ ((\ sqrt2) ^ (x ^ 3-3x ^ 2 + 4) = t \), \ (t> 0 \). Then the equation takes the form \ We will gradually write down the conditions under which the original equation will have six solutions.
Note that the quadratic equation \ ((*) \) can have at most two solutions. Any cubic equation \ (Ax ^ 3 + Bx ^ 2 + Cx + D = 0 \) can have at most three solutions. Therefore, if the equation \ ((*) \) has two different solutions (positive !, since \ (t \) must be greater than zero) \ (t_1 \) and \ (t_2 \), then, having made the reverse substitution, we we get: \ [\ left [\ begin (gathered) \ begin (aligned) & (\ sqrt2) ^ (x ^ 3-3x ^ 2 + 4) = t_1 \\ & (\ sqrt2) ^ (x ^ 3-3x ^ 2 +4) = t_2 \ end (aligned) \ end (gathered) \ right. \] Since any positive number can be represented as \ (\ sqrt2 \) to some extent, for example, \ (t_1 = (\ sqrt2) ^ (\ log _ (\ sqrt2) t_1) \), then the first equation of the set will be rewritten as \ As we already said, any cubic equation has at most three solutions, therefore, each equation from the set will have at most three solutions. This means that the entire set will have no more than six solutions.
This means that for the original equation to have six solutions, the quadratic equation \ ((*) \) must have two different solutions, and each obtained cubic equation (from the set) must have three different solutions (moreover, no solution of one equation must coincide with which one - or by the decision of the second!)
Obviously, if the quadratic equation \ ((*) \) has one solution, then we will not get six solutions of the original equation.

Thus, the solution plan becomes clear. Let's write down the conditions that must be met, point by point.

1) For the equation \ ((*) \) to have two different solutions, its discriminant must be positive: \

2) You also need both roots to be positive (since \ (t> 0 \)). If the product of two roots is positive and their sum is positive, then the roots themselves will be positive. Therefore, you need: \ [\ begin (cases) 12-a> 0 \\ - (a-10)> 0 \ end (cases) \ quad \ Leftrightarrow \ quad a<10\]

Thus, we have already provided ourselves with two different positive roots \ (t_1 \) and \ (t_2 \).

3) Let's take a look at such an equation \ For which \ (t \) will it have three different solutions?
Consider the function \ (f (x) = x ^ 3-3x ^ 2 + 4 \).
Can be factorized: \ Therefore, its zeros are \ (x = -1; 2 \).
If we find the derivative \ (f "(x) = 3x ^ 2-6x \), then we get two extremum points \ (x_ (max) = 0, x_ (min) = 2 \).
Hence, the graph looks like this:


We see that any horizontal line \ (y = k \), where \ (0 \ (x ^ 3-3x ^ 2 + 4 = \ log _ (\ sqrt2) t \) had three different solutions, it is necessary that \ (0<\log_ {\sqrt2}t<4\) .
Thus, you need: \ [\ begin (cases) 0<\log_{\sqrt2}t_1<4\\ 0<\log_{\sqrt2}t_2<4\end{cases}\qquad (**)\] Let's also immediately notice that if the numbers \ (t_1 \) and \ (t_2 \) are different, then the numbers \ (\ log _ (\ sqrt2) t_1 \) and \ (\ log _ (\ sqrt2) t_2 \) will be different, hence, the equations \ (x ^ 3-3x ^ 2 + 4 = \ log _ (\ sqrt2) t_1 \) and \ (x ^ 3-3x ^ 2 + 4 = \ log _ (\ sqrt2) t_2 \) will have mismatched roots.
The \ ((**) \) system can be rewritten as follows: \ [\ begin (cases) 1

Thus, we have determined that both roots of the equation \ ((*) \) must lie in the interval \ ((1; 4) \). How do you write this condition?
We will not write out the roots explicitly.
Consider the function \ (g (t) = t ^ 2 + (a-10) t + 12-a \). Its graph is a parabola with upward branches, which has two points of intersection with the abscissa axis (we wrote this condition in point 1)). How should its graph look like so that the points of intersection with the abscissa axis are in the interval \ ((1; 4) \)? So:


First, the values ​​\ (g (1) \) and \ (g (4) \) of the function at the points \ (1 \) and \ (4 \) must be positive, and secondly, the vertex of the parabola \ (t_0 \ ) must also be in the range \ ((1; 4) \). Therefore, we can write the system: \ [\ begin (cases) 1 + a-10 + 12-a> 0 \\ 4 ^ 2 + (a-10) \ cdot 4 + 12-a> 0 \\ 1<\dfrac{-(a-10)}2<4\end{cases}\quad\Leftrightarrow\quad 4\ (a \) always has at least one root \ (x = 0 \). Hence, to fulfill the condition of the problem, it is necessary that the equation \

had four different nonzero roots representing, together with \ (x = 0 \), an arithmetic progression.

Note that the function \ (y = 25x ^ 4 + 25 (a-1) x ^ 2-4 (a-7) \) is even, so if \ (x_0 \) is the root of the equation \ ((*) \ ), then \ (- x_0 \) will also be its root. Then it is necessary that the roots of this equation are numbers ordered in ascending order: \ (- 2d, -d, d, 2d \) (then \ (d> 0 \)). It is then that these five numbers will form an arithmetic progression (with the difference \ (d \)).

For these roots to be the numbers \ (- 2d, -d, d, 2d \), it is necessary that the numbers \ (d ^ (\, 2), 4d ^ (\, 2) \) be the roots of the equation \ (25t ^ 2 +25 (a-1) t-4 (a-7) = 0 \). Then by Vieta's theorem:

We rewrite the equation as \ and consider two functions: \ (g (x) = 20a-a ^ 2-2 ^ (x ^ 2 + 2) \) and \ (f (x) = 13 | x | -2 | 5x + 12a | \) ...
The function \ (g (x) \) has a maximum point \ (x = 0 \) (moreover, \ (g _ (\ text (vert)) = g (0) = - a ^ 2 + 20a-4 \)):
\ (g "(x) = - 2 ^ (x ^ 2 + 2) \ cdot \ ln 2 \ cdot 2x \)... Derivative zero: \ (x = 0 \). For \ (x<0\) имеем: \(g">0 \), for \ (x> 0 \): \ (g "<0\) .
The function \ (f (x) \) for \ (x> 0 \) is increasing, and for \ (x<0\) – убывающей, следовательно, \(x=0\) – точка минимума.
Indeed, for \ (x> 0 \) the first module will open positively (\ (| x | = x \)), therefore, regardless of how the second module will open, \ (f (x) \) will be equal to \ ( kx + A \), where \ (A \) is an expression from \ (a \), and \ (k \) is equal to either \ (13-10 = 3 \) or \ (13 + 10 = 23 \). For \ (x<0\) наоборот: первый модуль раскроется отрицательно и \(f(x)=kx+A\) , где \(k\) равно либо \(-3\) , либо \(-23\) .
Find the value \ (f \) at the minimum point: \

In order for the equation to have at least one solution, the graphs of the functions \ (f \) and \ (g \) must have at least one intersection point. Therefore, you need: \ Solving this set of systems, we get the answer: \\]

Answer:

\ (a \ in \ (- 2 \) \ cup \)