What is the standard format of a math set The correct answer is D Other why is it not right

a(f) The reason is that 2 is the element, and on the right is the set element, and the set is useful in general to belong to and not belong to the symbolic connection;

The question of b is coming, b is not wrong, for the time being, c(f) should be equal, and the title shows that it is unequal, so it is a false proposition;

d(t)d is absolutely correct, looking back and evaluating b, b is holding the question, thinking that the reason for the true is that it represents an empty set, thinking that the false such x cannot be found, and since the multiple-choice question can only choose one correct answer, so [d] is chosen

Why is it correct?

The 3k+1 class is the same as 3k-2, which simply means that the remainder of a division by 3 is 1, and a division by 3 and two more is enough to remove, which is the same meaning.

If it is a little more difficult from the perspective of the equality condition of the set, if you want to know, you can ask;

a should be used to represent the relationship between elements and sets, and the inclusion in the question is to represent the relationship between sets and sets.

b should be an empty set.

cBoth sets represent odd numbers.

If you have any questions, you can ask them.

Representation of sets: (commonly used are enumeration and description).

1.Enumeration method: It is often used to represent a finite set, listing all the elements in the set one by one and writing them in curly brackets, this method of representing the set is called the enumeration method.

2.Descriptive method: It is often used to represent an infinite set, and the common attributes of the elements in the set are described in words, symbols or formulas, etc., and written in curly brackets.

x is the general form of the elements of the set, and p is the common property of the elements of the set) e.g., the set of positive real numbers less than is expressed as: {x|02 or x<1.

c: The left and right sides are sets, and the relationship cannot be represented by an unequal sign, but can only be represented by an equal sign, contained, or contained.

d: The elements in the two sets are exactly the same, when k is taken as an integer, 3k+1 is the same as the number represented by 3k-2, because k in the two sets does not have to take the same value at the same time, the two ks have nothing to do with each other, so the two sets are actually the same, which can be represented by =.

The derivative can push monotonicity, and vice versa is not true, and needs to be found by definition. Counterexample y=x 3, derivative greater than or equal to 0

From the meaning of the title, it can be seen that f(x) is an increasing function.

The function -f(-x)=f(x) is the same as the odd function.

Therefore, the error of choosing da is that f'(x) can be equal to 0

i.e. f'(x) >=0 when f(x) is the increment function.

It can only be stated that it is greater than or equal to 0

I really don't know the meaning of doing this question.

According to the title, when x1 > x2, both f(x1) > f(x2), i.e., f(x) is increasing monotonically in (-

If f(x) is monotonically increasing on (a,b), then f(x) is the derivative f'(x) > = 0 is constant on (a, b).

a) but (a) for any x, f'(x) >0, so wrong.

A counter-example: f(x)=x3, is a single increase, but at x=0, f'(x)=0。

b) According to the title, when x1 > x2, there are f(x1) > f(x2) for any x'(-x) 0,x (-then -x (-.)

Obviously for arbitrary x, f'(-x) 0 is not true.

c) When x1>x2, f(x1) > f(x2), then when -x1<-x2, f(-x1)x2, f(x1>) f(x2) f(x) single increment.

f'(x)≥0

x1≤-x2

f(-x1)≤f(-x2)

i.e. -f(x1) -f(x2).

f(-x) single increment.

Obviously, D should be chosen.

If you don't understand, please ask.

It is implicated that f(x) is an increasing function in the defined domain.

To arbitrary x, f'(-x)≥0

Answer: F(x) is derivable on r, then f'(x) is always meaningful on r.

When x1 > x2 there is always f(x1) > f(x2), which means that f(x) is a monotonic increasing function.

So: f'(x)>=0

x1>x2, -x1<-x2, so f(-x1)-f(-x2) so f(-x) is a monotonic subtraction, and -f(-x) is a monotonic addition so: f'(-x)<=0

So: Answers A and C are incorrect, and answers B and D are correct.

The answer is D, and I can only suspect that there is a typographical error in option B, and there is no negative sign in parentheses: F'(x) <=0 is incorrect.

B Answer: A monotonically increasing function whose derivative is greater than or equal to 0 at any position. So B is false, D answers: f(x) monotonically increasing -f(x) should be symmetrically monotonically decreasing with f(x)x-axis The greater the value of x, the smaller -x is, so it also increases monotonically, but its derivatives are all small< = 0.

Actually, -f(-x) should be the origin (0, 0) transformation of f(x), so it should be said.

The stem can only show that f(x) is monotonically increasing and the x-axis is continuous.

If the image is only in the quadrant, option b doesn't work.

The derivative at any point is "0", not "0, so b is wrong.

Because f(x) is a monotonically increasing function, f'(x) is greater than or equal to 0, and b just treats the whole -x as x, which should be like this.

You can't use an equal sign, an equal sign, when there is x1 is not equal to x2, there is always f(x1)=f(x2).

f(-x) is a monotonically decreasing function, so how can the derivative be zero, only the constant function derivative is zero, b should be f'(-x)<0

b is not monotonically decreasing in the strict sense of the word, there are extremums, whereas f(-x) is a monotonic function.

The derivative of this function will not be equal to 0, silly boy.

Forehead? Is it wrong to determine B? It seems that B is right, the title is wrong, right?

This question is by definition.

This question is answered using definitions. g(x) is a primitive function of f(x), which is solved using this property I wrote. f(x) is integrable in the interval, so g(x) is continuous in the interval, so there is no discontinuity.

The reason is that your integration result is wrong, you give a primitive function, ignoring the +c part of the indefinite integral, because of the existence of the +c part, the piecewise functional number ends up continuous, in fact, g(1)=f(x) on (0,1) integral, is 2 3, and on the g+ side, he is equal to f(x) on (0,1+), and the +c part must adjust the function so that g+=2 3

So in the second function g(x)=x 2 6-x 3 +5 6, you are missing this 5 6

1 This is all a matter of understanding the definition, which is actually two sets, one is m straight line and the other is n = circle. They are different types, and when they intersect, they are naturally 0. Empty set!

Is the set the area of the circle, the piece, or the circumference? Are elements points? If the circle is infinite, because the intersecting segments are line segments, and there are infinitely many points on the line segments, if it is a circumference, then the intersection points are two.

Do empty sets count as intersections?

What is A?

m=2 + 3

then 2+1<2 + 3<2+2

That is: choose B out of 3, the answer is wrong.

Representing typical figures, it is easy to verify that BC is wrong.

Take -1 2 and -1 3 to prove that option b is wrong.

Take -1 10 and -1 100 to prove that option C is wrong.