Odz complex functions. Range of permissible values ​​(APV), theory, examples, solutions. Find the domain of the function yourself and then look at the solution

In mathematics there is a fairly small number of elementary functions, the scope of which is limited. All other "complex" functions are just combinations and combinations of them.

1. Fractional function - restriction on the denominator.

2. Root of even degree - restriction on radical expression.

3. Logarithms - restrictions on the base of the logarithm and sublogarithmic expression.

3. Trigonometric tg(x) and ctg(x) - restriction on the argument.

For tangent:

4. Inverse trigonometric functions.

arcsine	arc cosine	Arctangent, Arctangent

Next, the following examples are solved on the topic “Domain of definition of functions”.

An example of finding the domain of definition of function No. 1

Finding the domain of definition of any linear function, i.e. functions of the first degree:

y = 2x + 3 - the equation defines a straight line on a plane.

Let's look carefully at the function and think about what numerical values ​​we can substitute into the equation instead of the variable x?

Let's try to substitute the value x=0

Since y = 2 0 + 3 = 3 - received a numerical value, therefore the function exists for the given value of the variable x=0.

Let's try to substitute the value x=10

since y = 2·10 + 3 = 23 - the function exists for the given value of the variable x=10.

Let's try to substitute the value x=-10

since y = 2·(-10) + 3 = -17 - the function exists for the given value of the variable x = -10.

The equation defines a straight line on a plane, and a straight line has neither beginning nor end, therefore it exists for any values ​​of x.

Note that no matter what numerical values ​​we substitute into a given function instead of x, we will always get the numerical value of the variable y.

Therefore, the function exists for any value x ∈ R, or we write it like this: D(f) = R

Forms of writing the answer: D(f)=R or D(f)=(-∞:+∞)or x∈R or x∈(-∞:+∞)

Let's conclude:

For any function of the form y = ax + b, the domain of definition is the set of real numbers.

An example of finding the domain of definition of function No. 2

A function of the form:

y = 10/(x + 5) - hyperbola equation

When dealing with a fractional function, remember that you cannot divide by zero. Therefore the function will exist for all values ​​of x that are not

set the denominator to zero. Let's try to substitute some arbitrary values ​​of x.

At x = 0 we have y = 10/(0 + 5) = 2 - the function exists.

For x = 10 we have y = 10/(10 + 5) = 10/15 = 2/3- the function exists.

At x = -5 we have y = 10/(-5 + 5) = 10/0 - the function does not exist at this point.

Those. if the given function is fractional, then it is necessary to equate the denominator to zero and find a point at which the function does not exist.

In our case:

x + 5 = 0 → x = -5 - at this point the given function does not exist.

x + 5 ≠ 0 → x ≠ -5

For clarity, let's depict it graphically:

On the graph we also see that the hyperbola comes as close as possible to the straight line x = -5, but does not reach the value -5 itself.

We see that the given function exists at all points of the real axis, except for the point x = -5

Response recording forms: D(f)=R\(-5) or D(f)=(-∞;-5) ∪ (-5;+∞) or x ∈ R\(-5) or x ∈ (-∞;-5) ∪ (-5;+∞)

If the given function is fractional, then the presence of a denominator imposes the condition that the denominator is not equal to zero.

An example of finding the domain of definition of function No. 3

Let's consider an example of finding the domain of definition of a function with a root of even degree:

Since we can only extract the square root from a non-negative number, therefore, the function under the root is non-negative.

2x - 8 ≥ 0

Let's solve a simple inequality:

2x - 8 ≥ 0 → 2x ≥ 8 → x ≥ 4

The specified function exists only for the found values ​​of x ≥ 4 or D(f)= . It remains to find the intersection of sets of values ​​x such that x∈D(f 2) and f 2 (x)∈D(f 1) :

To arcsinx>0, remember the properties of the arcsine function. The arcsine increases throughout the entire domain of definition [−1, 1] and goes to zero at x=0, therefore, arcsinx>0 for any x from the interval (0, 1] .

Let's return to the system:

Thus, the required domain of definition of the function is the half-interval (0, 1].

Answer:

(0, 1] .

Now let's move on to complex functions of the general form y=f 1 (f 2 (...f n (x)))). The domain of definition of the function f in this case is found as .

Example.

Find the domain of a function .

Solution.

A given complex function can be written as y=f 1 (f 2 (f 3 (x))), where f 1 – sin, f 2 – fourth-degree root function, f 3 – log.

We know that D(f 1)=(−∞, +∞) , D(f 2)=∪ ∪ [ 3 , + ∞) .

When preparing the Unified State Exam and the Unified State Exam, you can come across many combined tasks for functions where it is necessary to first of all pay attention to DL. Only after it has been determined can further decisions be made.

The domain of definition of the sum, difference and product of functions

Before starting the solution, you need to learn how to correctly determine the domain of determination of the sum of functions. To do this, the following statement must hold:

When the function f f is considered the sum of n functions f 1, f 2, ..., f n , in other words, this function is given using the formula y = f 1 (x) + f 2 (x) + ... + f n (x) , then its domain of definition is considered to be the intersection of the domains of definition of the functions f 1, f 2, ..., f n . This statement can be written as:

D (f) = D (f 1) D (f 2) . . . D(fn)

Example 1

Find the domain of definition of a function of the form y = x 7 + x + 5 + t g x .

Solution

The given function is represented as the sum of four: power with exponent 7, power with exponent 1, constant, tangent function.

From the definition table we see that D (f 1) = (− ∞ , + ∞), D (f 2) = (− ∞ , + ∞), D (f 3) = (− ∞ , + ∞) , moreover, the domain of definition of the tangent includes all real numbers except π 2 + π · k, k ∈ Z.

The domain of definition of a given function f is the intersection of the domains of f 1, f 2, f 3 and f4. That is, for a function there is a number of real numbers that does not include π 2 + π · k, k ∈ Z.

Answer: all real numbers except π 2 + π · k, k ∈ Z.

To find the domain of definition of a product of functions, it is necessary to apply the following rule:

Definition 2

When a function f is considered to be the product of n functions f 1 , f 2 , f 3 and f n , then there is a function f that can be specified using the formula y = f 1 (x) f 2 (x) ... f n (x) , then its domain of definition is considered to be the domain of definition for all functions.

It will be written D (f) = D (f 1) D (f 2). . . D(fn)

Example 2

Find the domain of definition of the function y = 3 · a r c t g x · ln x .

Solution

The right side of the formula is considered as f 1 (x) f 2 (x) f 3 (x) , where for f 1 is a constant function, f 2 is the arctangent f 3 – logarithmic function with base e. By condition we have that D (f 1) = (− ∞ , + ∞), D (f 2) = (− ∞ , + ∞) and D (f 3) = (0 , + ∞) . We get that

D (f) = D (f 1) D (f 2) D (f n) = (- ∞ , + ∞) (- ∞ , + ∞) D (0 , + ∞) = (0 , + ∞)

Answer: domain of definition y = 3 · a r c t g x · ln x – the set of all real numbers.

It is necessary to focus on finding the domain of definition y = C f (x) , where C is a real number. From this it is clear that its domain of definition and the domain of f coincide.

Function y = C f (x) is the product of a constant function and f. The domain of definition is all the real numbers of the domain of definition. D(f). From this we see that the domain of definition of the function y = C f (x) is - ∞ , + ∞ D (f) = D (f) .

We found that the domain of definition y = f(x) And y = C f (x) , where C is some real number, coincide. This can be seen in the example of the definition of the root y = x is considered [ 0 , + ∞) , because the domain of definition of the function y = - 5 · x - [ 0 , + ∞) .

Domains y = f(x) and y = − f (x) coincide, which indicates that its domain of definition of the difference of the function is the same as the domain of definition of their sum.

Example 3

Find the domain of the function y = log 3 x − 3 · 2 x .

Solution

It is necessary to consider as the difference of two functions f 1 And f 2.

f 1 (x) = log 3 x And f 2 (x) = 3 2 x. Then we get that D (f) = D (f 1) D (f 2).

The domain of definition is written as D (f 1) = (0, + ∞) . Let's proceed to the domain of definition of f 2. in this case it coincides with the domain of definition of the exponential one, then we obtain that D (f 2) = (− ∞ , + ∞) .

To find the domain of definition of the function y = log 3 x − 3 2 x we get that

D (f) = D (f 1) D (f 2) = (0, + ∞) - ∞, + ∞

Answer: (0 , + ∞) .

It is necessary to voice the statement that the domain of definition is y = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 is the set of real numbers.

Consider y = a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 , where on the right side there is a polynomial with one variable of standard form in the form of degree n with real coefficients. It can be considered as the sum of the (n + 1)th function. The domain of definition for each of these functions includes a set of real numbers called R.

Example 4

Find the domain of definition f 1 (x) = x 5 + 7 x 3 - 2 x 2 + 1 2.

Solution

Let's take the notation f to be the difference of two functions, then we get that f 1 (x) = x 5 + 7 x 3 - 2 x 2 + 1 2 and f 2 (x) = 3 x - ln 5. It was shown above that D (f 1) = R. The domain of definition for f 2 coincides with the power domain for the exponent – ​​ln 5, in other words, that D (f 2) = (0, + ∞) .

We get that D (f) = D (f 1) D (f 2) = - ∞, + ∞ (0, + ∞) = (0, + ∞).

Answer: (0 , + ∞) .

Domain of a complex function

To solve this issue, it is necessary to consider a complex function of the form y = f 1 (f 2 (x)) . It is known that D(f) is the set of all x from the definition of the function f 2, where the domain of f 2 (x) belongs to the domain of definition f 1 .

It can be seen that the domain of definition of a complex function of the form y = f 1 (f 2 (x)) is at the intersection of two sets such where x ∈ D (f 2) and f 2 (x) ∈ D (f 1) . In standard notation this will take the form

x ∈ D (f 2) f 2 (x) ∈ D (f 1)

Let's look at solving several examples.

Example 5

Find the domain of definition y = ln x 2.

Solution

We represent this function in the form y = f 1 (f 2 (x)), where we have that f 1 is a logarithm with base e, and f 2 is a power function with exponent 2.

To solve it is necessary to use known domains of definition D (f 1) = (0 , + ∞) And D (f 2) = (− ∞ , + ∞) .

Then we obtain a system of inequalities of the form

x ∈ D (f 2) f 2 (x) ∈ D (f 1) ⇔ x ∈ - ∞ , + ∞ x 2 ∈ (0 , + ∞) ⇔ ⇔ x ∈ (- ∞ , + ∞) x 2 > 0 ⇔ x ∈ (- ∞ , + ∞) x ∈ (- ∞ , 0) ∪ (0 , + ∞) ⇔ ⇔ x ∈ (- ∞ , 0) ∪ (0 , + ∞)

The required domain of definition has been found. The entire real number axis except zero is the domain of definition.

Answer: (− ∞ , 0) ∪ (0 , + ∞) .

Example 6

Find the domain of definition of the function y = (a r c sin x) - 1 2.

Solution

Since this is a complex function, it is necessary to consider it as y = f 1 (f 2 (x)) , where f 1 is a power function with exponent - 1 2, and f 2 is an arcsine function, now it is necessary to look for its domain of definition. Need to consider D (f 1) = (0 , + ∞) And D (f 2) = [ − 1 , 1 ] . Now let’s find all sets of values ​​x, where x ∈ D (f 2) and f 2 (x) ∈ D (f 1) . We obtain a system of inequalities of the form

x ∈ D (f 2) f 2 (x) ∈ D (f 1) ⇔ x ∈ - 1, 1 a r c sin x ∈ (0, + ∞) ⇔ ⇔ x ∈ - 1, 1 a r c sin x > 0

To solve a r c sin x > 0, it is necessary to resort to the properties of the arcsine function. Its increase occurs in the domain of definition [−1, 1], and it goes to zero at x = 0, which means that arc sin x > 0 from the definition of x belongs to the interval (0, 1].

Let's transform the system of the form

x ∈ - 1 , 1 a r c sin x > 0 ⇔ x ∈ - 1 , 1 x ∈ (0 , 1 ] ⇔ x ∈ (0 , 1 ]

The domain of definition of the desired function has an interval equal to (0, 1].

Answer: (0 , 1 ] .

We gradually came to the point that we would work with complex functions of the general form y = f 1 (f 2 (... f n (x)))) . The domain of definition of such a function is sought from x ∈ D (f n) f n (x) ∈ D (f n - 1) f n - 1 (f n (x)) ∈ D (f n - 2) . . . f 2 (f 3 (... (f n (x))) ∈ D (f 1) .

Example 7

Find the domain of definition y = sin (l g x 4) .

Solution

The given function can be written as y = f 1 (f 2 (f 3 (x))) , where we have f 1 – sine function, f 2 – function with a 4th root, f 3 – logarithmic function.

We have that by condition D (f 1) = (− ∞ , + ∞) , D (f 2) = [ 0 , + ∞) , D (f 3) = (0 , + ∞) . Then the domain of definition of the function is the intersection of sets of such values, where x ∈ D (f 3), f 3 (x) ∈ D (f 2), f 2 (f 3 (x)) ∈ D (f 1) . We get that

x ∈ D (f 3) f 3 (x) ∈ D (f 2) f 2 (f 3 (x)) ∈ D (f 1) ⇔ x ∈ (0 , + ∞) log x ∈ [ 0 , + ∞ ) log x 4 ∈ - ∞ , + ∞

The condition lg x 4 ∈ - ∞, + ∞ is similar to the condition lg x ∈ [ 0, + ∞), which means

x ∈ (0 , + ∞) log x ∈ [ 0 , + ∞) log x 4 ∈ - ∞ , + ∞ ⇔ x ∈ (0 , + ∞) log x ∈ [ 0 , + ∞) log x ∈ [ 0 , + ∞) ⇔ ⇔ x ∈ (0 , + ∞) log x ∈ [ 0 , + ∞) ⇔ x ∈ (0 , + ∞) log x ≥ 0 ⇔ ⇔ x ∈ (0 , + ∞) log x ≥ log 1 ⇔ x ∈ (0 , + ∞) x ≥ 1 ⇔ ⇔ x ∈ [ 1 , + ∞)

Answer: [ 1 , + ∞) .

When solving examples, functions were taken that were composed using elementary functions in order to consider the domain of definition in detail.

Fraction scope

Consider a function of the form f 1 (x) f 2 (x). It is worth paying attention to the fact that this fraction is determined from the set of both functions, and f 2 (x) should not go to zero. Then we find that the domain of definition of f for all x is written in the form x ∈ D (f 1) x ∈ D (f 2) f 2 (x) ≠ 0.

Let's write the function y = f 1 (x) f 2 (x) in the form y = f 1 (x) · (f 2 (x)) - 1. Then we obtain a product of functions of the form y = f 1 (x) with y = (f 2 (x)) - 1. Domain of the function y = f 1 (x) is the set D (f 1) , and for a complex y = (f 2 (x)) - 1 we define from a system of the form x ∈ D (f 2) f 2 (x) ∈ (- ∞ , 0) ∪ (0 , + ∞) ⇔ x ∈ D (f 2) f 2 (x) ≠ 0.

This means x ∈ D (f 1) x ∈ D (f 2) f 2 (x) ∈ (- ∞ , 0) ∪ (0 , + ∞) ⇔ x ∈ D (f 1) x ∈ D (f 2) f 2 (x) ≠ 0.

Example 8

Find the domain of definition y = t g (2 x + 1) x 2 - x - 6.

Solution

The given function is fractional, so f 1 is a complex function, where y = t g (2 x + 1) and f 2 is an entire rational function, where y = x 2 − x − 6, and the domain of definition is considered to be the set of all numbers. We can write this in the form

x ∈ D (f 1) x ∈ D (f 2) f 2 (x) ≠ 0

Complex Function Representation y = f 3 (f 4 (x)), Where f 3 is a tangent function, where the domain of definition includes all numbers except π 2 + π k, k ∈ Z, and f 4 is an entire rational function y = 2 x + 1 with domain D (f 4) = (− ∞ , + ∞) . Then we proceed to find the domain of definition of f 1:

x ∈ D (f 4) 2 x + 1 ∈ D (f 3) ⇔ x ∈ (- ∞ , + ∞) 2 x + 1 ≠ π 2 + π k, k ∈ Z ⇔ x ≠ π 4 - 1 2 + π 2 k, k ∈ Z

It is also necessary to consider the lower domain of definition y = t g (2 x + 1) x 2 - x - 6. Then we get that

x ∈ D (f 1) x ∈ D (f 2) f 2 (x) ≠ 0 ⇔ x ≠ π 4 - 1 2 + π 2 k, k ∈ Z x ∈ - ∞, + ∞ x 2 - x - 6 ≠ 0 ⇔ ⇔ x ≠ π 4 - 1 2 + π 2 k, k ∈ Z x ≠ - 2 x ≠ 3

Answer: set of real numbers, except - 2, 3 and π 4 - 1 2 + π 2 · k, k ∈ Z.

Domain of a logarithm with a variable in the base

The definition of logarithm exists for positive bases not equal to 1. This shows that the function y = log f 2 (x) f 1 (x) has a domain that looks like this:

x ∈ D (f 1) f 1 (x) > 0 x ∈ D (f 2) f 2 (x) > 0 f 2 (x) ≠ 1

A similar conclusion can be reached when the function can be depicted in this form:

y = log a f 1 (x) log a f 2 (x), a > 0, a ≠ 1. After which you can proceed to the domain of definition of a fractional function.

The domain of definition of a logarithmic function is the set of real positive numbers, then the domain of definition of complex functions of the type y = log a f 1 (x) and y = log a f 2 (x) can be determined from the resulting system of the form x ∈ D (f 1) f 1 ( x) > 0 and x ∈ D (f 2) f 2 (x) > 0 . Otherwise, this area can be written in the form y = log a f 1 (x) log a f 2 (x), a > 0, a ≠ 1, which means finding y = log f 2 (x) f 1 (x) from the system itself of the form

x ∈ D (f 1) f 1 (x) > 0 x ∈ D (f 2) f 2 (x) > 0 log a f 2 (x) ≠ 0 = x ∈ D (f 1) f 1 (x) > 0 x ∈ D (f 2) f 2 (x) > 0 f 2 (x) ≠ 1

Example 9

Designate the domain of definition of the function y = log 2 x (x 2 - 6 x + 5) .

Solution

The notation should be adopted f 1 (x) = x 2 − 6 x + 5 And f 2 (x) = 2 x, from here D (f 1) = (− ∞ , + ∞) And D (f 2) = (− ∞ , + ∞). It is necessary to start searching for a set x where the condition x ∈ D (f 1), f 1 (x) > 0, x ∈ D (f 2), f 2 (x) > 0, f 2 (x) ≠ 1 is satisfied . Then we get a system of the form

x ∈ (- ∞ , + ∞) x 2 - 6 x + 5 > 0 x ∈ (- ∞ , + ∞) 2 x > 0 2 x ≠ 1 ⇔ x ∈ (- ∞ , + ∞) x ∈ ( - ∞ , 1) ∪ (5 , + ∞) x ∈ (- ∞ , + ∞) x > 0 x ≠ 1 2 ⇔ ⇔ x ∈ 0 , 1 2 ∪ 1 2 , 1 ∪ (5 , + ∞)

From here we see that the desired domain of the function y = log 2 x (x 2 - 6 x + 5) is considered to be a set that satisfies the condition 0, 1 2 ∪ 1 2, 1 ∪ (5, + ∞).

Answer: 0 , 1 2 ∪ 1 2 , 1 ∪ (5 , + ∞) .

Domain of definition of exponential function

The exponential function is given by a formula of the form y = (f 1 (x)) f 2 (x). Its domain of definition includes those values ​​of x that satisfy the system x ∈ D (f 1) x ∈ D (f 2) f 1 (x) > 0.

This area allows you to move from exponential to complex form y = a log a (f 1 (x)) f 2 (x) = a f 2 (x) log a f 1 (x) , where where a > 0, a ≠ 1.

Example 10

Find the domain of definition of the exponential function y = (x 2 - 1) x 3 - 9 · x.

Solution

Let us take the notation f 1 (x) = x 2 − 1 and f 2 (x) = x 3 - 9 · x.

Function f 1 is defined on the set of real numbers, then we obtain a domain of definition of the form D (f 1) = (− ∞ , + ∞) . Function f 2 is complex, so its representation will take the form y = f 3 (f 4 (x)), A f 3– square root with domain of definition D (f 3) = [ 0 , + ∞), and the function f 4– rational integer, D (f 4) = (− ∞ , + ∞) . We get a system of the form

x ∈ D (f 4) f 4 (x) ∈ D (f 3) ⇔ x ∈ (- ∞ , + ∞) x 3 - 9 x ≥ 0 ⇔ ⇔ x ∈ (- ∞ , + ∞) x ∈ - 3 , 0 ∪ [ 3 , + ∞) ⇔ x ∈ - 3 , 0 ∪ [ 3 , + ∞)

This means that the domain of definition for the function f 2 has the form D (f 2) = [ − 3 , 0 ] ∪ [ 3 , + ∞) . After which it is necessary to find the domain of definition of the exponential function according to the condition x ∈ D (f 1) x ∈ D (f 2) f 1 (x) > 0.

We obtain a system of the form x ∈ - ∞ , + ∞ x ∈ - 3 , 0 ∪ [ 3 , + ∞) x 2 - 1 > 0 ⇔ x ∈ - ∞ , + ∞ x ∈ - 3 , 0 ∪ [ 3 , + ∞) x ∈ (- ∞ , - 1) ∪ (1 , + ∞) ⇔ ⇔ x ∈ - 3 , - 1 ∪ [ 3 , + ∞)

Answer: [ − 3 , − 1) ∪ [ 3 , + ∞)

In general

To solve it, it is imperative to look for the domain of definition, which can be presented as the sum or difference of functions and their products. The areas of definition of complex and fractional functions are often challenging. Thanks to the above rules, you can correctly determine the DL and quickly solve the problem on the domain of definition.

Main results tables

For convenience, we will place all the studied material in a table for convenient arrangement and quick memorization.F

Let us arrange the functions and their domains of definition.

Note that transformations can be performed starting from the right side of the expression. From this it is clear that identical transformations are allowed, which do not affect the domain of definition. For example, y = x 2 - 4 x - 2 and y = x + 2 are different functions, since the first is defined on (− ∞, 2) ∪ (2, + ∞), and the second is from the set of real numbers. From the transformation y = x 2 - 4 x - 2 = x - 2 x + 2 x - 2 = x + 2 it is clear that the function makes sense for x ≠ 2.

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Job type: 13

Condition

A) Solve equation 2(\sin x-\cos x)=tgx-1.

b) \left[ \frac(3\pi )2;\,3\pi \right].

Solution

A) Opening the brackets and moving all the terms to the left side, we get the equation 1+2 \sin x-2 \cos x-tg x=0. Considering that \cos x \neq 0, the term 2 \sin x can be replaced by 2 tan x \cos x, we obtain the equation 1+2 tg x \cos x-2 \cos x-tg x=0, which by grouping can be reduced to the form (1-tg x)(1-2 \cos x)=0.

1) 1-tg x=0, tan x=1, x=\frac\pi 4+\pi n, n \in \mathbb Z;

2) 1-2 \cos x=0, \cos x=\frac12, x=\pm \frac\pi 3+2\pi n, n \in \mathbb Z.

b) Using the number circle, select the roots belonging to the interval \left[ \frac(3\pi )2;\, 3\pi \right].

x_1=\frac\pi 4+2\pi =\frac(9\pi )4,

x_2=\frac\pi 3+2\pi =\frac(7\pi )3,

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

Answer

A) \frac\pi 4+\pi n, \pm\frac\pi 3+2\pi n, n \in \mathbb Z;

b) \frac(5\pi )3, \frac(7\pi )3, \frac(9\pi )4.

Job type: 13
Topic: Permissible value range (APV)

Condition

A) Solve the equation (2\sin ^24x-3\cos 4x)\cdot \sqrt (tgx)=0.

b) Indicate the roots of this equation that belong to the interval \left(0;\,\frac(3\pi )2\right] ;

Solution

A) ODZ: \begin(cases) tgx\geqslant 0\\x\neq \frac\pi 2+\pi k,k \in \mathbb Z. \end(cases)

The original equation on the ODZ is equivalent to a set of equations

\left[\!\!\begin(array)(l) 2 \sin ^2 4x-3 \cos 4x=0,\\tg x=0. \end(array)\right.

Let's solve the first equation. To do this we will make a replacement \cos 4x=t, t \in [-1; 1]. Then \sin^24x=1-t^2. We get:

2(1-t^2)-3t=0,

2t^2+3t-2=0,

t_1=\frac12, t_2=-2, t_2\notin [-1; 1].

\cos 4x=\frac12,

4x=\pm\frac\pi 3+2\pi n,

x=\pm \frac\pi (12)+\frac(\pi n)2, n \in \mathbb Z.

Let's solve the second equation.

tg x=0,\, x=\pi k, k \in \mathbb Z.

Using the unit circle, we find solutions that satisfy the ODZ.

The “+” sign marks the 1st and 3rd quarters, in which tg x>0.

We get: x=\pi k, k \in \mathbb Z; x=\frac\pi (12)+\pi n, n \in \mathbb Z; x=\frac(5\pi )(12)+\pi m, m \in \mathbb Z.

b) Let's find the roots belonging to the interval \left(0;\,\frac(3\pi )2\right].

x=\frac\pi (12), x=\frac(5\pi )(12); x=\pi ; x=\frac(13\pi )(12); x=\frac(17\pi )(12).

Answer

A) \pi k, k \in \mathbb Z; \frac\pi (12)+\pi n, n \in \mathbb Z; \frac(5\pi )(12)+\pi m, m \in \mathbb Z.

b) \pi; \frac\pi (12); \frac(5\pi )(12); \frac(13\pi )(12); \frac(17\pi )(12).

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 13
Topic: Permissible value range (APV)

Condition

A) Solve the equation: \cos ^2x+\cos ^2\frac\pi 6=\cos ^22x+\sin ^2\frac\pi 3;

b) List all roots belonging to the interval \left(\frac(7\pi )2;\,\frac(9\pi )2\right].

Solution

A) Because \sin \frac\pi 3=\cos \frac\pi 6, That \sin ^2\frac\pi 3=\cos ^2\frac\pi 6, This means that the given equation is equivalent to the equation \cos^2x=\cos ^22x, which, in turn, is equivalent to the equation \cos^2x-\cos ^2 2x=0.

But \cos ^2x-\cos ^22x= (\cos x-\cos 2x)\cdot (\cos x+\cos 2x) And

\cos 2x=2 \cos ^2 x-1, so the equation will take the form

(\cos x-(2 \cos ^2 x-1))\,\cdot(\cos x+(2 \cos ^2 x-1))=0,

(2 \cos ^2 x-\cos x-1)\,\cdot (2 \cos ^2 x+\cos x-1)=0.

Then either 2 \cos ^2 x-\cos x-1=0, or 2 \cos ^2 x+\cos x-1=0.

Solving the first equation as a quadratic equation for \cos x, we get:

(\cos x)_(1,2)=\frac(1\pm\sqrt 9)4=\frac(1\pm3)4. Therefore either \cos x=1 or \cos x=-\frac12. If \cos x=1, then x=2k\pi , k \in \mathbb Z. If \cos x=-\frac12, That x=\pm \frac(2\pi )3+2s\pi , s \in \mathbb Z.

Similarly, solving the second equation, we get either \cos x=-1 or \cos x=\frac12. If \cos x=-1, then the roots x=\pi +2m\pi , m \in \mathbb Z. If \cos x=\frac12, That x=\pm \frac\pi 3+2n\pi , n \in \mathbb Z.

Let's combine the solutions obtained:

x=m\pi , m \in \mathbb Z; x=\pm \frac\pi 3 +s\pi , s \in \mathbb Z.

b) Let's select the roots that fall within a given interval using a number circle.

We get: x_1 =\frac(11\pi )3, x_2=4\pi , x_3 =\frac(13\pi )3.

Answer

A) m\pi, m\in \mathbb Z; \pm \frac\pi 3 +s\pi , s \in \mathbb Z;

b) \frac(11\pi )3, 4\pi , \frac(13\pi )3.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 13
Topic: Permissible value range (APV)

Condition

A) Solve the equation 10\cos ^2\frac x2=\frac(11+5ctg\left(\dfrac(3\pi )2-x\right) )(1+tgx).

b) Indicate the roots of this equation that belong to the interval \left(-2\pi ; -\frac(3\pi )2\right).

Solution

A) 1. According to the reduction formula, ctg\left(\frac(3\pi )2-x\right) =tgx. The domain of definition of the equation will be such values ​​of x such that \cos x \neq 0 and tan x \neq -1. Let's transform the equation using the double angle cosine formula 2 \cos ^2 \frac x2=1+\cos x. We get the equation: 5(1+\cos x) =\frac(11+5tgx)(1+tgx).

Note that \frac(11+5tgx)(1+tgx)= \frac(5(1+tgx)+6)(1+tgx)= 5+\frac(6)(1+tgx), so the equation becomes: 5+5 \cos x=5 +\frac(6)(1+tgx). From here \cos x =\frac(\dfrac65)(1+tgx), \cos x+\sin x =\frac65.

2. Transform \sin x+\cos x using the reduction formula and the sum of cosines formula: \sin x=\cos \left(\frac\pi 2-x\right), \cos x+\sin x= \cos x+\cos \left(\frac\pi 2-x\right)= 2\cos \frac\pi 4\cos \left(x-\frac\pi 4\right)= \sqrt 2\cos \left(x-\frac\pi 4\right) = \frac65.

From here \cos \left(x-\frac\pi 4\right) =\frac(3\sqrt 2)5. Means, x-\frac\pi 4= arc\cos \frac(3\sqrt 2)5+2\pi k, k \in \mathbb Z,

or x-\frac\pi 4= -arc\cos \frac(3\sqrt 2)5+2\pi t, t \in \mathbb Z.

That's why x=\frac\pi 4+arc\cos \frac(3\sqrt 2)5+2\pi k,k \in \mathbb Z,

or x =\frac\pi 4-arc\cos \frac(3\sqrt 2)5+2\pi t,t \in \mathbb Z.

The found values ​​of x belong to the domain of definition.

b) Let us first find out where the roots of the equation fall at k=0 and t=0. These will be numbers accordingly a=\frac\pi 4+arccos \frac(3\sqrt 2)5 And b=\frac\pi 4-arccos \frac(3\sqrt 2)5.

1. Let us prove the auxiliary inequality:

\frac(\sqrt 2)(2)<\frac{3\sqrt 2}2<1.

Really, \frac(\sqrt 2)(2)=\frac(5\sqrt 2)(10)<\frac{6\sqrt2}{10}=\frac{3\sqrt2}{5}.

Note also that \left(\frac(3\sqrt 2)5\right) ^2=\frac(18)(25)<1^2=1, Means \frac(3\sqrt 2)5<1.

2. From inequalities (1) By the arc cosine property we get:

arccos 1

0

From here \frac\pi 4+0<\frac\pi 4+arc\cos \frac{3\sqrt 2}5<\frac\pi 4+\frac\pi 4,

0<\frac\pi 4+arccos \frac{3\sqrt 2}5<\frac\pi 2,

0

Likewise, -\frac\pi 4

0=\frac\pi 4-\frac\pi 4<\frac\pi 4-arccos \frac{3\sqrt 2}5< \frac\pi 4<\frac\pi 2,

0

For k=-1 and t=-1 we obtain the roots of the equation a-2\pi and b-2\pi.

\Bigg(a-2\pi =-\frac74\pi +arccos \frac(3\sqrt 2)5,\, b-2\pi =-\frac74\pi -arccos \frac(3\sqrt 2)5\Bigg). At the same time -2\pi

2\pi This means that these roots belong to the given interval \left(-2\pi , -\frac(3\pi )2\right).

For other values ​​of k and t, the roots of the equation do not belong to the given interval.

Indeed, if k\geqslant 1 and t\geqslant 1, then the roots are greater than 2\pi. If k\leqslant -2 and t\leqslant -2, then the roots are smaller -\frac(7\pi )2.

Answer

A) \frac\pi4\pm arccos\frac(3\sqrt2)5+2\pi k, k\in\mathbb Z;

b) -\frac(7\pi)4\pm arccos\frac(3\sqrt2)5.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 13
Topic: Permissible value range (APV)

Condition

A) Solve the equation \sin \left(\frac\pi 2+x\right) =\sin (-2x).

b) Find all the roots of this equation that belong to the interval ;

Solution

A) Let's transform the equation:

\cos x =-\sin 2x,

\cos x+2 \sin x \cos x=0,

\cos x(1+2 \sin x)=0,

\cos x=0,

x =\frac\pi 2+\pi n, n\in \mathbb Z;

1+2 \sin x=0,

\sin x=-\frac12,

x=(-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z.

b) We find the roots belonging to the segment using the unit circle.

The indicated interval contains a single number \frac\pi 2.

Answer

A) \frac\pi 2+\pi n, n \in \mathbb Z; (-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z;

b) \frac\pi 2.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 13
Topic: Permissible value range (APV)

Condition

A) Solve the equation \frac(\sin x-1)(1+\cos 2x)=\frac(\sin x-1)(1+\cos (\pi +x)).

b) Find all the roots of this equation that belong to the segment \left[ -\frac(3\pi )(2); -\frac(\pi )2 \right].

Solution

A) Let's find the ODZ equation: \cos 2x \neq -1, \cos (\pi +x) \neq -1; From here the ODZ: x \neq \frac \pi 2+\pi k,

k \in \mathbb Z, x\neq 2\pi n, n \in \mathbb Z. Note that when \sin x=1, x=\frac \pi 2+2\pi k, k \in \mathbb Z.

The resulting set of x values ​​is not included in the ODZ.

Means, \sin x \neq 1.

Divide both sides of the equation by a factor (\sin x-1), different from zero. We get the equation \frac 1(1+\cos 2x)=\frac 1(1+\cos (\pi +x)), or equation 1+\cos 2x=1+\cos (\pi +x). Applying the reduction formula on the left side and the reduction formula on the right, we obtain the equation 2 \cos ^2 x=1-\cos x. This equation is by substitution \cos x=t, Where -1 \leqslant t \leqslant 1 reduce it to square: 2t^2+t-1=0, whose roots t_1=-1 And t_2=\frac12. Returning to the variable x, we get \cos x = \frac12 or \cos x=-1, where x=\frac \pi 3+2\pi m, m \in \mathbb Z, x=-\frac \pi 3+2\pi n, n \in \mathbb Z, x=\pi +2\pi k, k \in \mathbb Z.

b) Let's solve inequalities

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 ,

2) -\frac(3\pi )2 \leqslant -\frac \pi 3+2\pi n \leqslant -\frac \pi (2,)

3) -\frac(3\pi )2 \leqslant \pi+2\pi k \leqslant -\frac \pi 2 , m, n, k \in \mathbb Z.

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 , -\frac32\leqslant \frac13+2m \leqslant -\frac12 -\frac(11)6 \leqslant 2m\leqslant -\frac56 , -\frac(11)(12) \leqslant m \leqslant -\frac5(12).

\left [-\frac(11)(12);-\frac5(12)\right].

2) -\frac (3\pi) 2 \leqslant -\frac(\pi )3+2\pi n \leqslant -\frac(\pi )(2), -\frac32 \leqslant -\frac13 +2n \leqslant -\frac12 , -\frac76 \leqslant 2n \leqslant -\frac1(6), -\frac7(12) \leqslant n \leqslant -\frac1(12).

There are no integers in the range \left[ -\frac7(12) ; -\frac1(12)\right].

3) -\frac(3\pi )2 \leqslant \pi +2\pi k\leqslant -\frac(\pi )2, -\frac32 \leqslant 1+2k\leqslant -\frac12, -\frac52 \leqslant 2k \leqslant -\frac32, -\frac54 \leqslant k \leqslant -\frac34.

This inequality is satisfied by k=-1, then x=-\pi.

Answer

A) \frac \pi 3+2\pi m; -\frac \pi 3+2\pi n; \pi +2\pi k, m, n, k \in \mathbb Z;

b) -\pi .

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

(\sin x-\cos 2x)\cdot (\sin x+\cos 2x) and

\cos 2x=1-2 \sin ^2 x, so the equation will take the form

(\sin x-(1-2 \sin ^2 x))\,\cdot(\sin x+(1-2 \sin ^2 x))=0,

(2 \sin ^2 x+\sin x-1)\,\cdot (2 \sin ^2 x-\sin x-1)=0.

Then either 2 \sin ^2 x+\sin x-1=0, or 2 \sin ^2 x-\sin x-1=0.

Let's solve the first equation as a quadratic equation with respect to \sin x,

(\sin x)_(1,2)=\frac(-1 \pm \sqrt 9)4=\frac(-1 \pm 3)4. Therefore either \sin x=-1 or \sin x=\frac12. If \sin x=-1, then x=\frac(3\pi )2+ 2k\pi , k \in \mathbb Z. If \sin x=\frac12, either x=\frac\pi 6 +2s\pi , s \in \mathbb Z, or x=\frac(5\pi )6+2t\pi , t \in \mathbb Z.

Similarly, solving the second equation, we get either \sin x=1 or \sin x=-\frac12. Then x =\frac\pi 2+2m\pi , m\in \mathbb Z, or x=\frac(-\pi )6 +2n\pi , n \in \mathbb Z, or x=\frac(-5\pi )6+2p\pi , p \in \mathbb Z.

Let's combine the solutions obtained:

x=\frac\pi 2+m\pi,m\in\mathbb Z; x=\pm\frac\pi 6+s\pi,s \in \mathbb Z.

b) Let's select the roots that fall within a given interval using a number circle.

We get: x_1 =\frac(7\pi )2, x_2 =\frac(23\pi )6, x_3 =\frac(25\pi )6.

Answer

A) \frac\pi 2+ m\pi , m \in \mathbb Z; \pm \frac\pi 6 +s\pi , s \in \mathbb Z;

b) \frac(7\pi )2;\,\,\frac(23\pi )6;\,\,\frac(25\pi )6.