Write the following sentences in four tenses: the present tense, the present progressive tense, the future tense and the past tense 1.I clean the blackboard. 2.He eats bananas. 3.Amy sends anemail. In fact, this is equivalent to English sentence making. I'll do you a favor,

Write the following sentences in four tenses: the present tense, the present progressive tense, the future tense and the past tense 1.I clean the blackboard. 2.He eats bananas. 3.Amy sends anemail. In fact, this is equivalent to English sentence making. I'll do you a favor,

1.I clean the blackboard.I am cleaning the blackboard.I cleaned the blackboard.I will clean the blackboard.2.He eats bananas.He is eating the bananas.He will eat the banbanas.He ate the bananas.3.Amy ...
How to distinguish English sentences is the general present tense, the general past tense, the general future tense and the present continuous tense
There are three simple changes in the present tense, the past tense and the future tense
Given the function f (x) = (ax2-2x + 1) · E-X (a ∈ R, e is the base of natural logarithm); (I) when a = 1, find the extreme value of F (x); (II) if f (x) monotonically decreases on [- 1,1], find the range of A
(1) When a = 1, f (x) = (x2-2x + 1) · E-X, f '(x) = (2x-2) · E-X - (x2-2x + 1) · E-X = - (x-1) (x-3) · E-X (2) when x changes, f (x), f '(x) changes as follows: X (- ∞, 1) 1 (1,3) 3 (3, + ∞) f' (x) - 0 + 0 - f (x) decreases, the minimum value increases, and the maximum value decreases. Therefore, when a = 1, the minimum value of function f (x) is f (1) = 0, and the maximum value is f (3) = 4e-3 (5 points) (II) f '(x) = (2ax-2) · E-X - (ax2-2x + 1) · E-X = - E-X [ax2-2ax-2x + 3] Let G (x) = ax2-2 (a + 1) x + 3. If a = 0, then G (x) = - 2x + 3, in (- 1,1), G (x) > 0, that is, f' (x) < 0, the function f (x) monotonically decreases in the interval [- 1,1] (7 points) ② if a > 0, then G (x) = ax2-2 (a + 1) x + 3, its image is a parabola with the opening upward, and the axis of symmetry is x = a + 1A > 1. If and only if G (1) ≥ 0, that is, 0 < a ≤ 1, in (- 1,1), G (x) > 0, f '(x) < 0, the function f (x) is monotonically decreasing in the interval [- 1,1] (9 points) ③ if a < 0, then G (x) = ax2-2 (a + 1) x + 3, its image is a parabola with opening downward, if and only if G (− 1) ≥ 0g (1) ≥ 0, that is, − 53 ≤ a < 0, in (- 1,1), G (x) > 0, f '(x) < 0, the function f (x) monotonically decreases in the interval [- 1,1] (11 points) to sum up, when the function f (x) monotonically decreases in the interval [- 1,1], the value range of a is − 53 ≤ a ≤ 1 (12 points)
Given the mapping f: a → B, where a = b = R, the corresponding rule F: X → y = x2-2x + 2, if there is no original image in the set a for the real number k ∈ B, then the value range of K is
I've seen a lot of people ask this question,
But still did not understand, there is an answer should be k > 1
The corresponding rule is f: X → y = x & # 178; - 2x + 2
To understand "for the real number k ∈ B, there is no prime image in the set a", which means that those numbers in the set B can not find the prime image? We can take a look first, for example, if we take a number 2 in B, does this 2 have the prime image? Yes, how to find the prime image of 2? It's just to solve the equation: X & # 178; - 2x + 2 = 2, and get x = 0 or x = 2. In this way, it's easy to find the prime image
So, which one can't find the primitive image? That is, the function y = x & # 178; - 2x + 2 outside the range. Y = (x-1) &# 178; + 1 ≥ 1, so far, you will find that you can find the primitive image of 0? Why? Because the equation x & # 178; - 2x + 2 = 0 has no solution. That is, you can't find the primitive image of the element smaller than 1, so K
If the function f (x) satisfies the relation f (x) + 2F (1 △ x) = 3x, what is the value of F (2)?
Replace x with 1 / X to get
f(1/x)+2f(x)=3/x
Multiply by 2, that is, 4f (x) + 2F (1 / x) = 6 / X
This formula - the original formula, 3f (x) = 6 / x-3x
That is, f (x) = 2 / X-X
So f (2) = 1-2 = - 1
solution
Because f (x) + 2F (1 △ x) = 3x
therefore
Let x = 2, then f (2) + 2F (1 / 2) = 6
Let x = 1 / 2, then f (1 / 2) + 2F (2) = 3 / 2
The solution is f (2) = - 1
F (x) is replaced by F (1 / x)
f(1/X)+2F(X)=3/X （2）
Multiply 2 by 2 - (1) to get 3f (x) = 6 / x-3x
F (x) = 2 / x-x 2 = - 1
f(2) + 2f(1/2) = 6;
f(1/2)+2f(2) = 3/2;
Simultaneous solution
f(2) = -1
The known function f (x) = LG (AX2 + 2x + 1)
Given function f (x) = LG (AX ^ 2 + 2x + 1)
① If the domain of function f (x) is r, the value range of real number a is obtained;
② If the value range of function f (x) is r, find the value range of real number a
If the definition field is r, then ax ^ 2 + 2x + 1 > 0 holds
If a = 0, then true number = 2x + 1 > 0 is not constant
If a is not equal to 0, ax ^ 2 + 2x + 1 > 0 is constant, then it has no intersection with X axis, that is, the discriminant is less than 0
So 4-4a1
If the value range is r, then the true number will get all positive numbers
If a = 0, then true number = 2x + 1 can get all positive numbers
A is not equal to 0, ax ^ 2 + 2x + 1 is a quadratic function, to take all positive numbers
Then the opening is upward, and the minimum value cannot be greater than 0, otherwise the positive number between 0 and the minimum value cannot be obtained
So the function and the right common point of X axis, the discriminant is greater than or equal to 0
So 4-4a > = 0
A
1, if the domain of F (x) is r
Ax ^ 2 + 2x + 1 is always greater than 0, that is, a is greater than 0, and the discriminant is less than 0
4-4a1
2, the range of function f (x) is r
Ax ^ 2 + 2x + 1 > 0, so △ 0, △ = 4-4a ≥ 0
∴0
Given the mapping f: a → B, where a = b = R, the corresponding rule F: X → y = | x | 12, if for the real number k ∈ B, there is no element X in the set a such that F: X → K, then the value range of K is ()
A. k≤0B. k＞0C. k≥0D. k＜0
From the meaning of the problem, we can get that k = | x | ≥ 0, ∵ for the real number k ∈ B, there is no original image in the set a, ∵ K < 0, so we choose D
If the function f (x) satisfies 2F (x) + F (1 / x) = 3x, then the value of F (2) is () and the solution is given
Answer: 2F (x) + F (1 / x) = 3x (1) Let t = 1 / x, x = 1 / t be substituted into the above formula (1) to get: 2F (1 / T) + F (T) = 3 / T. the function has nothing to do with the sign, and the above formula is changed to: 2F (1 / x) + f (x) = 3 / X (2) From the solution of (1) and (2): 3f (x) = 6x-3 / XF (x) = 2x-1 / x, so: F (2) = 4-1 / 2 = 7 /
2f(x)+f(1/x)=3x
Let x = 2, and we get
2f(2)+f(1/2)=6 (1)
Taking x = 1 / 2, we get
2f(1/2)+f(2)=3/2 (2)
2 × (1) - (2)
3f(2)=21／2
therefore
f(2)=7／2
Known function f (x) = ex, X ≥ 0, known function f (x) = {e ^ x, x > = 0; - 2x, x = 0; - 2x, X
This is a piecewise function. In fact, f (x) in F (x) is y, and then take y as the independent variable to draw the image corresponding to f (x) function
Given that a and B are real numbers, the set M = {Ba, 1}, n = {a, 0}, F: X → x means that if the element X in M is mapped to the set N and is still x, then a + B=______ .
∵ a, B are real numbers, set M = {Ba, 1}, n = {a, 0}, F: X → x means that the element X in M is still X in set n, through mapping, we can get 1 ∈ n, the solution is a = 1, BA → Ba ∈ n, we can get Ba = 0, the solution is b = 0, ∵ a + B = 1, so the answer is 1;