If f (x) = sin (2x-a) + 1 satisfies f ((π / 3) + x) = f ((π / 3) - x, what is a equal to?

If f (x) = sin (2x-a) + 1 satisfies f ((π / 3) + x) = f ((π / 3) - x, what is a equal to?

Let x = π / 3 be replaced by F ((π / 3) + x) = f ((π / 3) - x) f (0) = Sina + 1 = f (2 π / 3) = sin ((4 π / 3) - a) + 1sina = sin ((4 π / 3) - a) = sin (4 π / 3) cosa cos (4 π / 3) Sina / 2 ((1-sina) ^ 0.5 = - 3 ^ 0.5 / 2 ((1 - (Sina) ^ 2) ^ 0.5) + 0.5sina, then 0.5sina = - 3 ^ 0.5

The minimum value of function f (x) = x Λ 3-x (- 2x + X Λ 2-1)

A:
f(x)=x^3-x(-2x+x^2-1)
=x^3+2x^2-x^3+x
=2x^2+x
=2(x+1/4)^2-1/8
If and only if x = - 1 / 4, the minimum value of F (x) is - 1 / 8

If f (x) is a differentiable function on R and for any x with XF '(x) + F (x) > 0, then f (- 2) + F (2) is greater than or equal to or less than 0?

Let f (x) = XF (x)
Then f '(x) = XF' (x) + F (x) > 0
So f (x) is an increasing function
So f (2) > F (- 2)
That is 2F (2) > - 2F (- 2)
2(f(2)+f(-2))>0,
So f (2) + F (- 2) > 0

F (x) is a nonnegative differentiable function defined on (0, positive infinity) and satisfies that XF '(x) + F (x) is less than or equal to 0
F (x) is a nonnegative differentiable function defined on (0, positive infinity), and it is constant if XF '(x) - f (x) > 0. If a > b > 0, then there must be
A af(a)
It's XF '(x) - f (x) > 0. Can you give me a new answer?

Which pair is XF '(x) + F (x) less than or equal to 0 or XF' (x) - f (x) > 0?
It should be XF '(x) + F (x) less than or equal to 0
[xf(x)]'=x'*f(x)+x*f'(x)=f(x)+x*f'(x)≤0
So XF (x) is a decreasing function
a>b
So AF (a) 0
So f (x) + X * f '(x) > 0
That is [XF (x)] '> 0
So XF (x) is an increasing function
a>b
So AF (a) > BF (b)
Choose D

It is known that f (x) is a differentiable function defined on (0, + ∞) and satisfies XF ′ (x) - f (x) ≥ 0. For any positive number a, B, if a > b, there must be ()
A. af（a）≤bf（b）B. bf（b）≤af（a）C. af（b）≤bf（a）D. bf（a）≤af（b）

F (x) = f (x) x, we can get f '(x) = 1x2 [XF' (x) - f (x)], and from XF '(x) - f (x) ≥ 0, we discuss it in two cases: ① XF' (x) - F (x) > 0, so F '(x) > 0, that is, f (x) is an increasing function, that is, when a > b > 0, f (a) > F (b), (f (b) B < f (a) a, thus AF (b) < BF (a); ② XF' (x) - f (x) = 0, so f (x) is a constant function, where f (b) B = f (a) a, that is, AF (b) = BF (a); in general, AF (b) ≤ BF (a); therefore, C is selected;

For any differentiable function f (x) on R, if (x-1) f ′ (x) ≥ 0, then ()
A. f（0）+f（2）＜2f（1）B. f（0）+f（2）≤2f（1）C. f（0）+f（2）≥2f（1）D. f（0）+f（2）＞2f（1）

According to the meaning of the problem, when x ≥ 1, f '(x) ≥ 0, the function f (x) is an increasing function on (1, + ∞); when x < 1, f' (x) ≤ 0, f (x) is a decreasing function on (- ∞, 1), so when x = 1, the minimum value of F (x) is also the minimum value, that is, f (0) ≥ f (1), f (2) ≥ f (1), f (0) + F (2) ≥ 2F (1)

For any differentiable function f (x) on R, if (x-1) f '(x) is greater than or equal to 0, then f (0) + F (2) must be greater than or equal to 0,
I can't understand the answer of Souna

The key to this problem lies in the transformation condition (x-1) f ′ (x) greater than or equal to 0 = > x > = 1, f ′ (x) > = 0 or XF (1) - f (0), that is, f (0) + F (2) greater than 2F (1)

He passes through points (2, - 3) and has a common focus with the ellipse 9x2 + 4y2 = 36. What is the standard equation of his ellipse

c^2=5
x^2/a^2+y^2/b^2=1
b^2=5+a^2
4/a^2+9/(5+a^2)=1
a^2=10,b^2=15
x^2/10+y^2/15=1

The standard equation of an ellipse passing through points (2, - 3) and having a common focus with the ellipse 9x ^ + 2Y ^ = 36?

x²/4+y²/18=0
c²=18-4=14
Let X & # 178; / (A & # 178; - 14) + Y & # 178; / A & # 178; = 1
4/(a²-14)+9/a²=1
4a²+9a²-126=a^4-14a²
a^4-27a+126=0
(a²-21)(a²-6)=0
a²>c²=14
So x & # 178 / 7 + Y & # 178 / 21 = 1

The standard equation of ellipse passing through point (3. - 2) and having the same focus as ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1
It is helpful for the responder to give an accurate answer

C = √ (a Λ 2-B Λ 2) = √ 5. Let the standard equation of an ellipse passing through a point (3. - 2) and having the same focus as the ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1 be x Λ 2 / (M + 5) + y Λ 2 / M = 1. (M > 0). Because the ellipse passes through a point (3. - 2), so 9 / (M + 5) + 4 / M = 1. The solution is m = 10. So the equation of the ellipse is x Λ 2 / 15 + y Λ 2 / 10 = 1