①已知函数y=1/2sin(2x+π/6),x∈R用五点法作出它在一个周期内的简图:(下面是列表, x2x+π/6y=1/22sin(2x+π/6) ②函数y=Asin(ωx+ψ),(A>0,ω>0,|ψ|<π/2)的图像上两相邻最高点和最低点坐标分别
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![①已知函数y=1/2sin(2x+π/6),x∈R用五点法作出它在一个周期内的简图:(下面是列表, x2x+π/6y=1/22sin(2x+π/6) ②函数y=Asin(ωx+ψ),(A>0,ω>0,|ψ|<π/2)的图像上两相邻最高点和最低点坐标分别](/uploads/image/z/8740098-18-8.jpg?t=%E2%91%A0%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0y%3D1%2F2sin%282x%2B%CF%80%2F6%EF%BC%89%2Cx%E2%88%88R%E7%94%A8%E4%BA%94%E7%82%B9%E6%B3%95%E4%BD%9C%E5%87%BA%E5%AE%83%E5%9C%A8%E4%B8%80%E4%B8%AA%E5%91%A8%E6%9C%9F%E5%86%85%E7%9A%84%E7%AE%80%E5%9B%BE%EF%BC%9A%EF%BC%88%E4%B8%8B%E9%9D%A2%E6%98%AF%E5%88%97%E8%A1%A8%2C+x2x%2B%CF%80%2F6y%3D1%2F22sin%282x%2B%CF%80%2F6%29+%E2%91%A1%E5%87%BD%E6%95%B0y%3DAsin%28%CF%89x%2B%CF%88%29%2C%28A%EF%BC%9E0%2C%CF%89%EF%BC%9E0%2C%EF%BD%9C%CF%88%EF%BD%9C%EF%BC%9C%CF%80%2F2%EF%BC%89%E7%9A%84%E5%9B%BE%E5%83%8F%E4%B8%8A%E4%B8%A4%E7%9B%B8%E9%82%BB%E6%9C%80%E9%AB%98%E7%82%B9%E5%92%8C%E6%9C%80%E4%BD%8E%E7%82%B9%E5%9D%90%E6%A0%87%E5%88%86%E5%88%AB)
①已知函数y=1/2sin(2x+π/6),x∈R用五点法作出它在一个周期内的简图:(下面是列表, x2x+π/6y=1/22sin(2x+π/6) ②函数y=Asin(ωx+ψ),(A>0,ω>0,|ψ|<π/2)的图像上两相邻最高点和最低点坐标分别
①已知函数y=1/2sin(2x+π/6),x∈R
用五点法作出它在一个周期内的简图:(下面是列表,
x
2x+π/6
y=1/22sin(2x+π/6)
②函数y=Asin(ωx+ψ),(A>0,ω>0,|ψ|<π/2)的图像上两相邻最高点和最低点坐标分别是:(π/9,1/2)(4π/9,-1/2),则该函数的解析式是
③设函数f(x)=3sin(2x-π/3)的图像为C,如下结论正确的是
图像C关于直线x=11π/2对称
图像C关于点(2π/3,0)对称
函数f(x)在区间(-π/12,5π/12)内是增函数
由函数y=3sin2x的图像向右平移π/3个单位长度可以得到图像C
①已知函数y=1/2sin(2x+π/6),x∈R用五点法作出它在一个周期内的简图:(下面是列表, x2x+π/6y=1/22sin(2x+π/6) ②函数y=Asin(ωx+ψ),(A>0,ω>0,|ψ|<π/2)的图像上两相邻最高点和最低点坐标分别
(1) x -π/12 π/6 5π/12 2π/3 11π/12
2x+π/6 0 π/2 π 3π/2 2π
y=1/22sin(2x+π/6) 0 1/2 0 -1/2 0
(2) 由题意,A=1/2
设最小正周期为T,则T/2=4π/9-π/9=π/3,T=2π/3 ω=2π/T=3
所以,y=1/2sin(3x+ψ),代入点(π/9,1/2)解得 ψ=π/6
所以,y=1/2sin(3x+π/6),
(3) ① 将x=11π/2 代入 f(x)=3sin(2x-π/3)得不到f(x)的最值,错
② 将x=2π/3 代入 f(x)=3sin(2x-π/3),得 f(x)=0 对
③ 解 -π/2<2x-π/3<π/2 得 -π/12<x<5π/12,对
④ f(x)=3sin(2x-π/3)=3sin[2(x-π/6)],即
函数y=3sin2x的图像向右平移π/6个单位长度可以得到图像C 错