Let f (x) be a function on R and satisfy f (0) = 1, and for any real number x, y has f (X-Y) = f (x) - Y (2x-y + 1), and find the expression of F (x)

Let f (x) be a function on R and satisfy f (0) = 1, and for any real number x, y has f (X-Y) = f (x) - Y (2x-y + 1), and find the expression of F (x)

For abstract functions, the most important method is assignment. This is the problem
Obviously, it wants you to make use of the condition that f (0) = 1, so you can think that x = y, so that the following formula appears 0. After substitution, f (0) = f (x) - x (2x-x + 1), that is, 1 = f (x) - x ^ 2-x
So the final result is f (x) = x ^ 2 + X + 1

Given that the function f (x) has f (x + y) = f (x) + F (y) + 2Y (x + y) + 1 for any real number x, y, and f (1) = 1, if x ∈ n *, find the expression of F (x) RT (2) If x ∈ n * and X ≥ 2, the inequality f (x) ≥ (a + 7) x - (a + 10) always holds, and the value range of real number a is obtained.

Y=1
f(x+1)=f(x)+2x+4
therefore
f(2)=f(1)+2*1+4
f(3)=f(2)+2*2+4
f(4)=f(3)+2*3+4
.
.
f(x)=f(x-1)+2(x-1)+4
Add left = add right
So f (x) = x ^ 2 + 3x-3 (x ∈ n *)
When x ∈ n * and X ≥ 2
x^2+3x-3≥(a+7)x-(a+10）
That is, a ≤ (x ^ 2-4x + 7) / (x-1)
Let y = (x ^ 2-4x + 7) / (x-1)
Then x ^ 2 - (4 + y) x + 7 + y = 0
∆=(4+y)^2-4*(7+y)≥0
(y+2)^2≥16
Y ≥ 2 or Y ≤ - 6 (omitted)
So a ≤ 2

Let the function f (x) on R have f (x + y) = f (x) + 2Y (x + y) for any real number x and y, and satisfy f (1) = 1, find the expression of F (x) The answer is f (x) = 2x 2 - 1 But I calculate it like this: if x + y = 1, then y = 1-x, ∵ f (1) = 1, ᙽ f (x + y) = f (x) + 2 (1-x) × 1 = 1 ∴f(x)=2x-1 I think it makes sense, but why is it wrong?

There is a problem with the subject itself
It should be said that no function can satisfy the question condition

Let f (x) = asinx bcosx image be x = π 4, then the inclination angle of the line ax by + C = 0 is () A. π Four B. 3π Four C. π Three D. 2π Three

When x is the axis of symmetry, the value of function is maximum or minimum
That is, a − B
2＝
a2+b2，
The solution is: a + B = 0
Slope k = a
b＝−1，
Ψ the inclination angle of the line ax by + C = 0 is α = 3 π
4．
Therefore, B

Given that the function g (x) = asinx + bcosx + C, when a = 1, C = 0, the function g (x) is symmetric about 5 π / 3. Find the symmetry axis of function y = bsinx + acosx

F (x) = SiNx + bcos is about x = 5 π / 3 symmetry. F (x) = g (- x) will be x = - 5 π / 3 symmetry. H (x) = bsinx + cosx = bcos (π / 2-x) + sin (π / 2-x) = f (π / 2-x) = f (π / 2-x) = f (x-π / 2) is from F (x) to move parallel from F (x) to the right parallel shift π / 2, so h (x) about x = - 5 π / 3 + 3 + π / 2 = - 7 π / 6 or about x = 2 π - 7 π / 6 or about x = 2 π - 7 π - 7 π / 6 or about x = 2 π - 7 π - 7 π - 6 6 or about x= 5 π

Given that a, B satisfy A2 + b2-4a + 3 = 0, the maximum value of function f (x) = asinx + bcosx + 1 is denoted as φ (a, b), then the minimum value of φ (a, b) is () A. 1 B. 2 C. 3+1 D. 3

∵ the real numbers a and B satisfy A2 + b2-4a + 3 = 0,

The maximum value of the function f (x) = asinx + bcosx + 1 is denoted as t (a, b), Then the minimum value of T (a, b) is?

It can be concluded that: - 1 ≤ A-2 ≤ 1; 1 ≤ a ≤ 3
a²+b²=4a-3∈[1,9]
f(x)=asinx+bcosx+1=√(a²+b²)sin(x+w)+1;
So t (a, b) = √ (a 2 + B 2) + 1 ∈ [2,4]
So the minimum value of T (a, b) is 2

Let f (x) = - sin2x - asinx + B + 1 have the maximum value of 0 and the minimum value of - 4. If the real number a > 0, find the value of a and B A = 2 b = - 2 don't ignore the discussion of interval

F '(x) = - 2cos2x acosx = - 2 (2 (cosx) ^ 2-1) - acosx = - 4 (cosx) ^ 2-acosx + 2 = 0cosx = (a + - sqrt (a ^ 2 + 32)) / - 8sinx = sqrt (1 - (cosx) ^ 2) is substituted by F (x) = - sin2x-asinx + B + 1, making it = 0 and = - 4, and then two values a and B are solved

Given the function f (x) = asinx + bcosx and f (60 °) = 1, then for any real numbers a and B, the maximum value range of function f (x) is 2. When y = 2cosx-3sinx reaches the maximum value, what is the value of TaNx?

1. F (60) = 1, we can get the root sign 3 * a + B = 2. Since the maximum value is equal to the root sign (a ^ 2 + B ^ 2) = root sign (a ^ 2 + (2-radical 3 * a) ^ 2), we can simplify 2 * root sign ((a-radical 3 / 2) ^ 2 + 1 / 4) > = 12. There are many methods, y = 2cosx-3sinx = radical (13) * sin (x + Z), where Tanz = - 2 / 3

Known function f (x)= 3x+2，x＜1 If f (f (0)) = 4A, then the real number a=______ .

∵f（0）=2，
∴f（f（0））=f（2）=4+2a=4a，
So a = 2
So the answer is: 2