Question

已知函数f(x)=cosx(asinx-cosx)+sin²x？

已知函数f(x)=cosx(asinx-cosx)+sin²x，满足f(-π/3)=f(0)
(Ⅰ)求函数f(x)的最小正周期
(Ⅱ)求函数f(x)在［π/4,11π/24］上的最大值和最小值

Answer 1

f(x)=cosx(asinx-cosx)+sin²x=acosxsinx-cos²x+sin²x=(a/2)sin2x-cos2x
f(-π/3)=f(0)
(a/2)sin(-2π/3)-cos(-2π/3)=(a/2)sin0-cos0
(a/2)sin*(-√3/2)+1/2=0-1
a=2√3
f(x)=√3sin2x-cos2x=2(sin2xcosπ/6-cos2xsinπ/6)=2sin(2x-π/6)
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第一问：
f(x)=2sin(2x-π/6)
最小正周期 T=2π/2 = π
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第二问：
f(x)=2sin(2x-π/6)
x∈［π/4,11π/24］
2x∈ [π/2,11π/12］
2x-π/6∈ [π/3,3π/4］
当2x-π/6=π/2时，有最大值=2sinπ/2=2
当2x-π/6=3π/4时，有最小值=2sin3π/4=√2

Answer 2

(Ⅰ)
f(x)=cosx.(asinx-cosx)+(sinx)^2
f(-π/3)=f(0)
cosx.(asinx-cosx)+(sinx)^2=-1
(1/2)( -a√3/2 -1/2 ) + 3/4 = -1
(1/2)( -a√3/2 -1/2 ) = -7/4
a√3 +1 =7
a√3 =2√3
f(x)
=cosx.(2√3sinx-cosx)+(sinx)^2
=2√3.sinx.cosx - (cosx)^2 +(sinx)^2
=√3.sin2x -(1/2)(1+cos2x) +(1/2)(1-cos2x)
=√3.sin2x - cos2x
=2[(√3/2).sin2x -(1/2)cos2x]
=2sin(2x-π/6)
最小正周期=π
(II)

f(x)=2sin(2x-π/6)
在［π/4, 11π/24］
最大值 f(x) =f(π/3) = 2
最小值 f(x)
= f(11π/24)
=2sin(11π/12 -π/6)
=2sin(3π/4)
=√2

Answer 3