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)

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    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