Given the function f (x) = sin2x + acos2x (a ∈ R), and π / 4 is the zero point of the function y = f (x). (1) find the value of a, and find the function f (x) It is known that the function f (x) = sin2x + acos2x (a ∈ R), and π / 4 is the zero of the function y = f (x) (1) Find the value of a and the minimum positive period of function f (x); (2) If x ∈ [0, π / 2], find the range of function f (x)

Given the function f (x) = sin2x + acos2x (a ∈ R), and π / 4 is the zero point of the function y = f (x). (1) find the value of a, and find the function f (x) It is known that the function f (x) = sin2x + acos2x (a ∈ R), and π / 4 is the zero of the function y = f (x) (1) Find the value of a and the minimum positive period of function f (x); (2) If x ∈ [0, π / 2], find the range of function f (x)

That 2 is the square? Otherwise f (π / 4) = sin π / 2 + ACOS π / 2 = 1 cannot be 0
f(x)=sin^2 x+acos^2 x
1) From F (π / 4) = 1 / 2 + A / 2 = 0, a = - 1
f(x)=sin^2 x-cos^2 x=-cos2x
Period T = π
2) If x ∈ [0, π / 2], the range of F (x) [- 1,1]

If the function f (x) defined on [- 20102010] satisfies that for any x1, X2, f (x1 + x2) = f (x1) + F (x2) + 2009, Let f (x) be the maximum and the maximum
The values are m and N respectively. Find m + n

Let X1 = x2 = 0, f (0) = f (0) + F (0) + 2009
f(0)=-2009
f(0)=f(x-x)=f(x)+f(-x)+2009
f(x)=-f(-x)-4018
Since the maximum value of function f (x) is m, f (- x) ≤ M
f(x)=-f(-x)-4018≥-M-4018
The minimum value of F (x) is n = - m-4018
So, M + n = - 4018

If the function f (x) defined on [- 20102010] satisfies: for any x1, X2 ∈ [- 20102010], f (x1 + x2) = f (x1) + F (x2) - 2009, and if x > 0, f (x) > 2009, the maximum and minimum values of f (x) are m and N respectively, then M + n = ()
A. 2009B. 2010C. 4020D. 4018

Let g (x) = f (x) - 2009, then G (x1 + x2) = g (x1) + G (x2) can be obtained for any x1, X2 ∈ [- 20102010] with F (x1 + x2) = f (x1) + F (x2) - 2009, f (x1 + x2) - 2009 = [f (x1) - 2009] + [f (x2) - 2009], and G (x1 + x2) = g (x1) + G (x2) when x > 0, G (x) > 0, let X1 = x2 = 0, G (0) = 0 & nbsp; Let X1 = x, X2 = - x, then G (0) = g (- x) + G (x) = 0, then G (- x) = - G (x), so g (x) is an odd function. If the maximum value of G (x) is m, then the minimum value is - M. therefore, from F (x) = g (x) + 2009, the maximum value of F (x) is m + 2009, and the minimum value is - M + 2009, so m + n = m + 2009 + (- M) + 2009 = 4018, so D is selected

If the function f (x) defined on [- 20102010] satisfies that for any x1, X2, f (x1 + x2) = f (x1) + F (x2), let the maximum and minimum values of F (x) be m
, N, M + n