一道高一数学函数题目.函数y=Asin(ωx+φ)(A>0,ω>0,|φ|小于π\2)在同一个周期内,当x=π\4时,y

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  • 函数y=Asin(ωx+φ)(A>0,ω>0,|φ|小于π2)在同一个周期内,当x=π4时,y取最大值2,当x=7π12时,y取最小值-2.

    最大值=2,最小值=-2 所以A=2

    则半个周期T/2=7π/12-π/4=π/3

    T=2π/3 所以 w=3

    f(x)=2sin(3x+φ)

    代入点 (π/4,2)

    则 sin(3π/4+φ)=1 3π/4+φ=π/2 所以φ=-π/4

    所以

    (1)f(x)=2sin(3x-π/4)

    (2)若x∈[0,2π],

    且f(x)=√3,

    sin(3x-π/4)=√3/2

    3x-π/4=2kπ+π/3或3x-π/4=2kπ+2π/3

    x=2kπ/3+7π/36或x=2kπ/3+11π/36 k∈Z

    2kπ/3+7π/36∈[0,2π] 或x=2kπ/3+11π/36∈[0,2π]

    24kπ/36+7π/36x∈[0,2π] 24kπ/36+11π/36∈[0,2π]

    k=0 x0=7π/36 k=0 x3=11π/36

    k=1 x1=31π/36 k=1 x4=35π/36

    (3)若函数f(x)满足方程f(x)=a(0<a<2,)求在[0,2π]内的所有实数根之和

    画图可知 在[0,2π]内方程f(x)=a(0<a<2)有6个根

    x1+x2=2*π/4=π/2

    x3+x4=2*(π/4+2π/3)=π/2+4π/3

    x5+x6=2*(π/4+4π/3)=π/2+8π/3

    所以所有实数根之和=3π/2+4π=11π/4