设函数f(x)=cos(根号3倍的x+a)(0

设函数f(x)=cos(根号3倍的x+a)(0
设函数f(x)=cos(根号3倍的x+a)(0

设函数f(x)=cos(根号3倍的x+a)(0
f(x)=acos^2(ωx)-sqrt(3)asin(ωx)cos(ωx)+b
=a*(1+cos(2ωx))/2-sqrt(3)*a*sin(2ωx)/2+b
=a*[ cos(2ωx))*1/2-sin(2ωx)*sqrt(3)/2 ] +a/2+b
=a*[ cos(2ωx))*cos(π/3)-sin(2ωx)*sin(π/3) ] +a/2+b
=a*cos(2ωx+π/3)+a/2+b
(1)由于f(x)的最小正周期为π且ω>0,得ω=1.
(2)f(x)的定义域为[-π/3,π/6],意味着-π/3<=x<=π/6,此时
-π/3< =2x+π/3 <= 2π/3,
所以cos(2x+π/3)<=1=cos(0),
且cos(2x+π/3)>=cos(2π/3)=-1/2,
即 -1/2<= cos(2x+π/3) <=1,
所以f(x)=a*cos(2x+π/3)+a/2+b的取值范围为
当a>0时
f(x) >= a*(-1/2)+a/2+b = b (1)
f(x) <= a*1+a/2+b = 3a/2+b (2)
当a>0时
f(x) >= a*1+a/2+b = 3a/2+b (3)
f(x) <= a*(-1/2)+a/2+b = b (4)
由题设f(x)的值域为[-1,5],故
当a>0时,由(1)(2)两式得
b=-1,
3a/2+b=5,
解出a=4,b=-1；
当a<0时,由(3)(4)两式得
3a/2+b=-1,
b=5,
解出a=-4,b=5；
由f(x)=a*cos(2x+π/3)+a/2+b的表达式可知f(x)的单调区间是
-π/3 <= x <= -π/6（当a<0时f(x)是单调递减,当a>0时f(x)是单调递增）
和-π/6 <= x <= π/6（当a<0时f(x)是单调递增,当a>0时f(x)是单调递减）.
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f(x)=acos^2(ωx)-sqrt(3)asin(ωx)cos(ωx)+b
=a*(1+cos(2ωx))/2-sqrt(3)*a*sin(2ωx)/2+b
=a*[ cos(2ωx))*1/2-sin(2ωx)*sqrt(3)/2 ] +a/2+b
=a*[ cos(2ωx))*cos(π/3)-sin(2ωx)*sin(π/3) ] +a/2+b
=a*c...

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f(x)=acos^2(ωx)-sqrt(3)asin(ωx)cos(ωx)+b
=a*(1+cos(2ωx))/2-sqrt(3)*a*sin(2ωx)/2+b
=a*[ cos(2ωx))*1/2-sin(2ωx)*sqrt(3)/2 ] +a/2+b
=a*[ cos(2ωx))*cos(π/3)-sin(2ωx)*sin(π/3) ] +a/2+b
=a*cos(2ωx+π/3)+a/2+b
由于f(x)的最小正周期为π且ω>0，得ω=1
f(x)的定义域为[-π/3,π/6],意味着-π/3<=x<=π/6，此时
-π/3< =2x+π/3 <= 2π/3
所以cos(2x+π/3)<=1=cos(0)
且cos(2x+π/3)>=cos(2π/3)=-1/2，
即 -1/2<= cos(2x+π/3) <=1，
所以f(x)=a*cos(2x+π/3)+a/2+b的取值范围为
当a>0时
f(x) >= a*(-1/2)+a/2+b = b (1)
f(x) <= a*1+a/2+b = 3a/2+b (2)
当a>0时
f(x) >= a*1+a/2+b = 3a/2+b (3)
f(x) <= a*(-1/2)+a/2+b = b (4)
由题设f(x)的值域为[-1,5]，故
当a>0时，由(1)(2)两式得
b=-1
3a/2+b=5
解出a=4, b=-1
当a<0时，由(3)(4)两式得
3a/2+b=-1，
b=5，
解出a=-4, b=5；
由f(x)=a*cos(2x+π/3)+a/2+b的表达式可知f(x)的单调区间是
-π/3 <= x <= -π/6（当a<0时f(x)是单调递减，当a>0时f(x)是单调递增）
和-π/6 <= x <= π/6（当a<0时f(x)是单调递增，当a>0时f(x)是单调递减）
我可以帮助你，你先设置我最佳答案后，我百度Hii教你。

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