The function f (x) is an odd function defined on R, and f (2) = 0. F (x) increases monotonically on [0,1] and decreases monotonically on (1, + ∞). The solution set of inequality f (x) ≥ 0 is Step by step

The function f (x) is an odd function defined on R, and f (2) = 0. F (x) increases monotonically on [0,1] and decreases monotonically on (1, + ∞). The solution set of inequality f (x) ≥ 0 is Step by step

[make an image for easy understanding]
Because the function f (x) is an odd function defined on R
So f (0) = 0, f (- 2) = 0, f (2) = 0
When x ≥ 0, it increases monotonically on [0,1] and decreases monotonically on (1, + ∞)
Because f (2) = 0, f (0) = 0
So on [0,2] ≥ 0
At 0
Because it is an odd function, it is ≥ 0 on (- ∞, - 2]
(0,2) upper

If an odd function f (x) has f (x) = f (2-x) for any X in the domain of definition, then f (x) is a periodic function. What is the minimum positive period of F (x)?

Let me tell you two formulas: if f (x + a) = f (x + b), then the minimum positive period is t = | a-b|
If f (x + a) = - f (x + b), then the minimum positive period is t = 2|a-b|
If the function is odd, then f (2-x) = - f (X-2), so: F (x) = - f (X-2)
According to the above formula, the minimum positive period is 2 | 0 - (- 2) | = 4

If the domain of definition of odd function f (x) = 3sinx + C is [a, b], then a + B + C equals ()
A. 3b. - 3C. 0d. Cannot calculate

Because the definition field of function f (x) = 3sin & nbsp; X + C is [a, b], and it is an odd function, so f (0) = 0, and a + B = 0, that is, 3sin0 + C = 0, we get C = 0, the image of odd function is symmetric about the origin, so a + B = 0, so a + B + C = 0; so we choose C