Given the function f (x) = Sin & sup2; X + acosx + 5 / 8a-3 / 2, a ∈ R, the maximum value of function f (x) is obtained when a = 1 For any X in the interval [0, π / 2], f (x) ≤ 1 holds. Find the value range of real number a

Given the function f (x) = Sin & sup2; X + acosx + 5 / 8a-3 / 2, a ∈ R, the maximum value of function f (x) is obtained when a = 1 For any X in the interval [0, π / 2], f (x) ≤ 1 holds. Find the value range of real number a

(1) Let's: F ′ (x) = 2 SiNx cosx-sinx-sinx-sinx = SiNx (2cosx-1) Let f ′ (x) = 0, let: SiNx = 0 or cosx = 1 / 2 analyze one of its periodic x ∈ [0,2 π] when x ∈ (0, π / 3) is x ∈ (0,0,2 π] when x ∈ (0, π / 3) is (0, π / 3), f ′ (x) ＞ 0, f (x) is monotone increasing. When x ∈ (π / 3, π / 3, π (π / 3, π / 3, π) is the f ′ (x) \ (x) \\\\\\\\\\\\\\monotonicallyincreasing when x ∈ (5 π / 3, Let's compare two extreme values f (π / 3) and f (5 π / 3) get: F (5 π / 3): F (5 π / 3) is: F (5 π / 3) = f (5 π / 3) = f (π / 3) = f (π / 3) = 3 (3 π / 3) = 3 / 8, so when a = 1, the maximum value of F (x) is 3 / 8 (2) let t = cosx, then 1-T \\\\\\\\\35\\\\\\\\\\\\\\\\\\\\\letg (T) = - T & # 178; + at + (5 / 8) A-1 / 2, when G (T) reaches the maximum, When a ≤ 0, for t ∈ [0,1], G ′ (T) has g ′ (T) ≤ 0, and G (T) monotonically decreases on t ∈ [0,1], so when t = 0, G (T) has the maximum g (0) = (5 / 8) A-3 / 2 ＜ 0, which is consistent with the problem that when a > 2, G ′ (T) is positive on t ∈ [0,1], and G (T) monotonically increases on t ∈ [0,1], Therefore, when t = 1, G (T) has the maximum value, G (1) = (13 / 8) A-3 / 2, G (1) = (13 / 8) A-3 / 2 (13 / 8) A-3 / 2 ≤ 1, we can get: a ≤ 20 / 13 < 2, does not conform to 0 < a ≤ 2, when 0 < a ≤ 2, G ′ (T) is positive at t ∈ [0, a / 2 (A / 2,1] and negative at t ∈ (A / 2,1,1,1] when t ∈ [0, a / 2,1), G (t) monotone; when t ∈ (A / 2,2,1] is (A / 2,1,1,1,1], G (T) (g (T) has the maximum value, G (g (g (g) (g (A / 2) has the maximum, G (g (g (g) (g (A / 2) g (A / 2) g (a) (g) (g) (g) is the A & # / 4 + (5 / 8) A-3 / 2 ≤ 0, That is to say, 2A & # 178; + 5a-12 ≤ 0, (2a-3) (a + 4) ≤ 0, the solution is - 4 ≤ a ≤ 3 / 2, then 0 < a ≤ 3 / 2, the range of a is a ≤ 3 / 2

Given the function f (x) = Sin & # 178; X + acosx-1 / 2a-3 / 2, X belongs to R
(1) When a = 1, find the minimum value of function f (x)
(2) If the maximum value of F (x) is 1, find the value of real number a
(3) For any inequality f (x) ≥ 1 / 2-A / 2 belonging to [0, π / 3], the range of real number a is obtained

A:
f(x)=sin²x+acosx-a/2-3/2
=1-cos²x+acosx-a/2-3/2
=-(cosx-a/2)²+a²/4-a/2-1/2
1）
a=1,f(x)=-(cosx-1/2)²-3/4
When cosx = - 1, the minimum value of F (x) is f (x) min = - 3
2) The maximum value of F (x) is 1
2.1) when the axis of symmetry cosx = A / 2 = 5 / 2

It is known that one of the zeros of the function f (x) = SiNx acosx is π / 4
Let g (x) = f (x) f (- x) + (2 radical sign 3) sinxcos + 1, find the center of symmetry of G (x)

sin¼π－acos¼π＝0,
½·√2·(1-a)＝0,∴a=1.∴f(x)=sinx-cosx,f(-x)=-sinx-cosx,
Let g (x) = f (x) f (- x) + (2 radical 3) sinxcosx + 1, then G (x) = (COS & # 178; x-sin & # 178; x) + √ 3.2 · sinxcosx + 1,
∴g(x)＝√3sin2x＋cos2x＋1＝2sin(2x+30º)＋1.
Let 2x + 30 & # 186; = 180 & # 186; K, K be integers, then we can find the value of X. we can do it by ourselves. Note: the coordinate of the center of symmetry is (a, 1)