## It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (4a-3) > 0, find the value range of the real number a

Because f (x) is an odd function,

Therefore, f (2a + 1) + F (4a-3) > 0 can be transformed into f (2a + 1) > - f (4a-3) = f (3-4A),

And f (x) is the subtraction function on the definition field [- 2,2],

So there are

2a+1＜3−4a

−2≤2a+1≤2

−2≤4a−3≤2 ， Solution 1

4≤a＜1

3，

So the value range of real number a is 1

4≤a＜1

3．

### It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (- 4a-3) > 0, find the value of the real number a It is known that the odd function f (x) is a subtractive function on the definition field [- 2,2]. If f (2a + 1) + F (- 4a-3) > 0, find the value range of the real number a

Because it's an odd function

So f (- 4a-3) = - f (4a + 3)

So f (2a + 1) > F (4a + 3) is a subtractive function

So - 2

### The function f (x) = x2 + ax + 3. When x ∈ [- 2,2], f (x) ≥ A is constant. Find the value range of real number a

∵ function f (x) = x2 + ax + 3. When x ∈ [- 2,2], f (x) ≥ A is constant,

‡ (x-1) a ≥ - x2-3 is constant when x ∈ [- 2,2],

① When x ∈ (1,2],

∴a≥−x2− three

X − 1 is constant in X ∈ (1,2]

Let g (x) = − x2 − 3

X − 1, X ∈ (1, 2], i.e. a ≥ g (x) max

∵g′（x）=−(x−3)(x+1)

(x − 1) 2, ‡ (1, 2] is the increasing range, G (2) is the largest and - 7

∴a≥-7；

② When x ∈ [- 2, 1),

∴a≤−x2− three

X − 1 is constant in X ∈ [- 2, 1)

Let g (x) = − x2 − 3

x−1，x∈[-2，1），

I.e. a ≤ g (x) min

And G (x) = − x2 − 3

The minimum value of X − 1 on ∈ [- 2, 1) is g (- 1) = 2,

∴a≤2；

To sum up, the value range of real number A: [- 7, 2]

### Given the function f (x) = (x ^ 2-ax + 3) / (2 ^ x + 1), when x ∈ [2,3], f (x) > = 0 is always true, find the value range of real number a

Because the denominator is always greater than 0, only molecules can be considered

y=x^2-ax+3=(x-a/2)^2+3-a^2/4

When 4 = A6 or a = 0, --- > A = 0, --- > A

### Known inequality 1 x+1 y+m If x + y ≥ 0 is constant for any positive real number x and y, the minimum value of real number m is __

∵ inequality 1

x+1

y+m

X + y ≥ 0 is constant for any positive real number x and y,

‡ inequality (x + y) (1)

x+1

y) ≥ - M is constant for any positive real number x and y

And (x + y) (1)

x+1

y）=2+y

x+x

y≥4

‡ - M ≤ 4, i.e. m ≥ - 4

So the answer is: - 4

### A function and equation problem Find the number of zeros of function y = LNX + 2x-6

From the question, we can get LNX + 2x-6 = 0, that is, LNX = 6-2x

So

Draw the images of y = LNX and y = 6-2x in the same coordinate system

It can be seen from the figure that the two images have only one intersection

So there's only one zero