The known function f (x) = 2Sin (AX - π / 6) sin (AX + π / 3)

The known function f (x) = 2Sin (AX - π / 6) sin (AX + π / 3)

The minimum positive period of the function f (x) = 2Sin (AX - π / 6) sin (AX + π / 3) (where a is a normal number, X ∈ R) is π (1). Find the value of a (2) in △ ABC, if a < B, and f (a) = f (b) = 1 / 2, find BC / Ab1) f (x) = 2Sin (AX - π / 6) sin (AX + π / 3) = - [cos (2aX + π

Given that the function f (x) is differentiable on R, and f (x) = x ^ 2 + 2x · f '(2), then the analytic expression of the function is?

Note that f '(2) is a number, let it be c
Therefore, Let f (x) = x & # 178; + 2cx
f'(x)=2x+2c
Substituting x = 2 has
c=f'(2) =4+2c
The solution is C = - 4
So f (x) = x & # 178; - 8x

Let f (x) = 2Sin (π 2x + π 5). If f (x1) ≤ f (x) ≤ f (x2) holds for any x ∈ R, then the minimum value of | x1-x2 | is______ ．

The period T of function f (x) = 2Sin (π 2x + π 5) is 2 π π 2 = 4. For any x ∈ R, f (x1) ≤ f (x) ≤ f (x2) holds. It shows that f (x1) has the minimum value, f (x2) has the maximum value, | x1-x2 | min = T2 = 2