已知函数f(x)-cos(2x-φ),(0已知函数f(x)=1+2*sin(2x-π/3)(1)求f(x)的单调减区间(2)若不等式m>f(x)-2 m<f(x)+2 对x∈[π/4,π/2]恒成立,求实数m的取值范围
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![已知函数f(x)-cos(2x-φ),(0已知函数f(x)=1+2*sin(2x-π/3)(1)求f(x)的单调减区间(2)若不等式m>f(x)-2 m<f(x)+2 对x∈[π/4,π/2]恒成立,求实数m的取值范围](/uploads/image/z/8633227-67-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29-cos%282x-%CF%86%29%2C%280%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D1%2B2%2Asin%282x-%CF%80%2F3%29%281%29%E6%B1%82f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%87%8F%E5%8C%BA%E9%97%B4%282%29%E8%8B%A5%E4%B8%8D%E7%AD%89%E5%BC%8Fm%EF%BC%9Ef%28x%29-2+m%EF%BC%9Cf%28x%29%2B2+%E5%AF%B9x%E2%88%88%5B%CF%80%2F4%EF%BC%8C%CF%80%2F2%5D%E6%81%92%E6%88%90%E7%AB%8B%EF%BC%8C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知函数f(x)-cos(2x-φ),(0已知函数f(x)=1+2*sin(2x-π/3)(1)求f(x)的单调减区间(2)若不等式m>f(x)-2 m<f(x)+2 对x∈[π/4,π/2]恒成立,求实数m的取值范围
已知函数f(x)-cos(2x-φ),(0
已知函数f(x)=1+2*sin(2x-π/3)
(1)求f(x)的单调减区间
(2)若不等式m>f(x)-2 m<f(x)+2 对x∈[π/4,π/2]恒成立,求实数m的取值范围
已知函数f(x)-cos(2x-φ),(0已知函数f(x)=1+2*sin(2x-π/3)(1)求f(x)的单调减区间(2)若不等式m>f(x)-2 m<f(x)+2 对x∈[π/4,π/2]恒成立,求实数m的取值范围
(1)将(π/6,根号3/2)代入f(x),有根号3/2=cos(π/3-φ).
因为根号3/2=cos(π/6)或根号3/2=cos(-π/6)
所以π/3-φ=π/6 或 π/3-φ=-π/6
因为0
(1)f(x)=cos(2x-φ)其图像过点(π/6,根号3/2)
则 f(π/6)=cos(π/3-φ)=根号3/2, 所以π/3-φ=±π/6
又因为 0<φ<π/2,所以 -π/6< π/3-φ<π/3
所以 π/3-φ= π/6
φ= π/6
(2)原函数为 f(x)=cos(2x-π/6)
则 g(x)=cos(4x-π/6)
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(1)f(x)=cos(2x-φ)其图像过点(π/6,根号3/2)
则 f(π/6)=cos(π/3-φ)=根号3/2, 所以π/3-φ=±π/6
又因为 0<φ<π/2,所以 -π/6< π/3-φ<π/3
所以 π/3-φ= π/6
φ= π/6
(2)原函数为 f(x)=cos(2x-π/6)
则 g(x)=cos(4x-π/6)
区间[0,π/4]
x=π/24时, g(x)max=1
x=π/4时, g(x)min= - √3/2
已知函数f(x)=1+2*sin(2x-π/3)
(1)求f(x)的单调减区间
2kπ+π/2<2x-π/3<2kπ+3π/2时,f(x)单调递减
f(x)的单调减区间为【k+5π/12,k+11π/12】
x∈[π/4,π/2]时,f(x)max=3(x=5π/12时),f(x)min=1+√3(x=π/2时)
m>f(x)-2恒成立,则m>3-2,故m>1
m<f(x)+2恒成立,则m<1+√3+2,故m<3+√3
所以:1
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