f(x)=Acos^2(ωx+φ)+1(A>0,ω>0)的最大值为3,与y轴的焦点坐标为(0,2),其相邻两条对称轴之间的距离为2则f(1)+f(2)+f(3)+.+f(2010)=?老师给的答案是4019和4021，

Question

f(x)=Acos^2(ωx+φ)+1(A>0,ω>0)的最大值为3,与y轴的焦点坐标为(0,2),其相邻两条对称轴之间的距离为2
则f(1)+f(2)+f(3)+.+f(2010)=?
老师给的答案是4019和4021，

Answer 1

f(x)= Acos^2(ωx+φ)+1 = (A/2) cos(2ωx+2φ) + A/2 + 1 的最大值为3
=> A+1 = 3 => A=2
x=0, y = cos(2φ) + 2 = 2 => cos(2φ) =0 2φ=Kπ/2,K不等于零的整数
其相邻两条对称轴之间的距离为 2 ,周期为4,2π/2ω=4,ω = π/4
可得,f(x)=cos(π/2x+Kπ/2)+2
f(1)+f(2)+f(3)+.+f(2010)=4020+cosπ/2(1+K)+cosπ/2(2+K)+cosπ/2(3+K)+cosπ/2(4+K)+————+cosπ/2(1+K)+cosπ/2(2+K)=4020+cosπ/2(1+K)+cosπ/2(2+K)=4020+0+1或4020-1+0
可得4019和4021
每相邻4项抵消为0.