已知函数f(x）=asinwx+bcoswx的最小正周期为2,当x=3分之1时,f（x）取得最大值为2.
1个回答
f(x) = a sin(wx) + b cos(wx)
T = 2π/w = 2 => w = π
f(x) = √(a² + b²) sin(wx + arctan(b/a))
f(1/3) = 2
π/3 + arctan(b/a) = π/2
arctan(b/a) = π/6
b/a = 1/√3
√3 b = a
√(a² + b²) = 2
a² + b² = 4
(√3 b)² + b² = 4
3b² + b² = 4
b = 1
a = √3
f(x) = √3 sin(πx) + cos(πx) = 2sin(πx + π/6)
-----------------------------------------------------------------------------------------------------------------------
对称轴方程是πx + π/6 = π/2 + kπ
即x = 1/3 + k,k∈Z
21/4 ≤ 1/3 + k ≤ 23/4
4又11/12 ≤ k ≤ 5又5/12
所以有整数解k = 5
对称轴方程为x = 16/3
相关问题
