Question

已知函数f(x)=1-2a-2acosx-2sin^2x的最小值为g(a),a∈R （1）求g(a)，（2）若g(a)=1/2,求此时f(x)的最大值

Answer 1

f(x)=1-2a-2acosx-2sin^2x
解：
f(x)=1-2a-2acosx-2sin^2x
=1-2a-2acosx-2(1-cos^2x)
=1-2a-2acosx-2+2cos^2x
=2cos^2x-2acosx-2a-1
=2(cos^2x-acosx)-2a-1
=2(cos^2x-acosx+a^2/4)-a^2/2-2a-1
=2(cosx-a/2)^2-a^2/2-2a-1

当cosx-a/2=0时，f(x)的值最小（-2<=a<=2）
g(a)=-a^2/2-2a-1
-a^2/2-2a-1=1/2
a^2/2+2a+3/2=0
a^2+4a+3=0
(a+3)(a+1)=0
a=-1或a=-3(舍去）

f(x)=2(cosx-a/2)^2-a^2/2-2a-1
=2(cosx+1/2)^2-1/2+2-1
=2(cosx+1/2)^2+1/2
当cosx=1时，f(x)的值最大
f(x)=2(cosx+1/2)^2+1/2
=2(1+1/2)^2+1/2
=2*9/4+1/2
=9/2+1/2
=5

Answer 2

f(x)=1-2a-2acosx-2sin^2x
解：
f(x)=1-2a-2acosx-2sin^2x
=1-2a-2acosx-2(1-cos^2x)
=1-2a-2acosx-2+2cos^2x
=2cos^2x-2acosx-2a-1
=2(cos^2x-acosx)-2a-1
=2(cos^2x-acosx+a^2/4)-a^2/2-2a-1
=2(cosx-a/2)^2-a^2/2-2a-1

当cosx-a/2=0时，f(x)的值最小（-2<=a<=2）
g(a)=-a^2/2-2a-1
-a^2/2-2a-1=1/2
a^2/2+2a+3/2=0
a^2+4a+3=0
(a+3)(a+1)=0
a=-1或a=-3(舍去）

f(x)=2(cosx-a/2)^2-a^2/2-2a-1
=2(cosx+1/2)^2-1/2+2-1
=2(cosx+1/2)^2+1/2
当cosx=1时，f(x)的值最大
f(x)=2(cosx+1/2)^2+1/2
=2(1+1/2)^2+1/2
=2*9/4+1/2
=9/2+1/2
=5