已知函数f(x)=√3cos²ωx+sinωxcosωx+a(ω>0,a∈R)图像的两相邻对称轴间距离为π/2(1)求ω值(2)求函数y=f(x)的单调递减区间(3)已知f(x)在区间[0,π/2]上的最小值为1,求a的值
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![已知函数f(x)=√3cos²ωx+sinωxcosωx+a(ω>0,a∈R)图像的两相邻对称轴间距离为π/2(1)求ω值(2)求函数y=f(x)的单调递减区间(3)已知f(x)在区间[0,π/2]上的最小值为1,求a的值](/uploads/image/z/3996783-63-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%E2%88%9A3cos%26%23178%3B%CF%89x%2Bsin%CF%89xcos%CF%89x%2Ba%EF%BC%88%CF%89%EF%BC%9E0%2Ca%E2%88%88R%EF%BC%89%E5%9B%BE%E5%83%8F%E7%9A%84%E4%B8%A4%E7%9B%B8%E9%82%BB%E5%AF%B9%E7%A7%B0%E8%BD%B4%E9%97%B4%E8%B7%9D%E7%A6%BB%E4%B8%BA%CF%80%2F2%EF%BC%881%EF%BC%89%E6%B1%82%CF%89%E5%80%BC%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0y%3Df%EF%BC%88x%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%E5%8C%BA%E9%97%B4%EF%BC%883%EF%BC%89%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2C%CF%80%2F2%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA1%2C%E6%B1%82a%E7%9A%84%E5%80%BC)
已知函数f(x)=√3cos²ωx+sinωxcosωx+a(ω>0,a∈R)图像的两相邻对称轴间距离为π/2(1)求ω值(2)求函数y=f(x)的单调递减区间(3)已知f(x)在区间[0,π/2]上的最小值为1,求a的值
已知函数f(x)=√3cos²ωx+sinωxcosωx+a(ω>0,a∈R)图像的两相邻对称轴间距离为π/2
(1)求ω值
(2)求函数y=f(x)的单调递减区间
(3)已知f(x)在区间[0,π/2]上的最小值为1,求a的值
已知函数f(x)=√3cos²ωx+sinωxcosωx+a(ω>0,a∈R)图像的两相邻对称轴间距离为π/2(1)求ω值(2)求函数y=f(x)的单调递减区间(3)已知f(x)在区间[0,π/2]上的最小值为1,求a的值
f(x)=(√3/2)(1+cos2ωx)+(1/2)sin2ωx+a
=(√3/2)cos2ωx+(1/2)sin2ωx+a+√3/2
=sin(2ωx+π/3)+a+√3/2
(1)因为两相邻对称轴间距离为π/2,即周期的一半T/2=π/2,T=π
所以 2π/2ω=π,ω=1
(2)令 π/2+2kπ≤2x+π/3≤3π/2+2kπ,
得 π/12+kπ≤x≤7π/12+kπ,
即单调递减区间为[π/12+kπ,7π/12+kπ],k∈Z
(3)易得,f(x)在区间[0,π/12]为增,在区间[π/12,π/2]为减,
最小值为f(π/2)=sin(π+π/3)+a+√3/2=1,
-sinπ/3+a+√3/2=1,得 a=1