Question

f(x)=1-2a-2acosx-2sin^2x的最小值为g(a). (1)求g(a).

(2)求能使g(a)=1/2的a值，并求当a取此值时f(x)的最大值

Answer 1

f(x) = 1 - 2a - 2a cosx - 2sin²x
= 1 - 2a - 2a cosx - 2(1 - cos²x)
= 1 - 2a - 2a cosx - 2 + 2cos²x
= 2cos²x - 2a cosx - 2a - 1
= 2(cos²x - acosx) - 2a - 1
= 2[cos²x - 2(cosx)(a/2) + (a/2)² - (a/2)²] - 2a - 1
= 2(cosx - a/2)² - 2(a/2)² - 2a - 1
= 2(cosx - a/2)² - a²/2 - 2a - 1
抛物线开口向上，有最小值g(a) = -a²/2 - 2a - 1
-----------------------------------------------------------------------------------------------------------------------
g(a) = 1/2
-a²/2 - 2a - 1 = 1/2
a² + 4a + 3 = 0
a = -1 或 a = -3
-----------------------------------------------------------------------------------------------------------------------
当a = -1时
f(x) = 2(cosx + 1/2)² + 1/2
最大值 = 2(1 + 1/2)² + 1/2 = 5
当a = -3时
f(x) = 2(cosx + 3/2)² + 1/2
最大值 = 2(1 + 3/2)² + 1/2 = 13