已知函数f(x)=sin(ωx+π/3),f(π/6)=f(π/4),且f(x)在区间(π/6,π/4)有最小值无最大值
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已知函数f(x)=sin(ωx+π/3),f(π/6)=f(π/4),且f(x)在区间(π/6,π/4)有最小值无最大值,则ω=
![已知函数f(x)=sin(ωx+π/3),f(π/6)=f(π/4),且f(x)在区间(π/6,π/4)有最小值无最大值](/uploads/image/z/14973495-15-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28%CF%89x%2B%CF%80%2F3%29%2Cf%28%CF%80%2F6%29%3Df%EF%BC%88%CF%80%2F4%EF%BC%89%2C%E4%B8%94f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%EF%BC%88%CF%80%2F6%2C%CF%80%2F4%EF%BC%89%E6%9C%89%E6%9C%80%E5%B0%8F%E5%80%BC%E6%97%A0%E6%9C%80%E5%A4%A7%E5%80%BC)
f(π/6)=sin(ωπ/6+π/3),f(π/4)=sin(ωπ/4+π/3),sin(ωπ/6+π/3)=sin(ωπ/4+π/3),ωπ/6+π/3=ωπ/4+π/3,ωπ/6+π/3=π-ωπ/4-π/3,解得:ω=0,ω=4/5,取ω=4/5,f(x)在区间(π/6,π/4)有最大值无最小值.
再问: 答案还有52/5、148/5
再答: 正弦函数在一、二和三、四象限函数值相等,ωπ/6+π/3<ωπ/4+π/3,当在一、二象限时,ωπ/6+π/3-2kπ=2kπ+π-(ωπ/4+π/3),ω=48k/5+4/5,ωπ/6+π/3-2kπ<π/2,ω<4k+2,k<1/4;当在三、四象限时,ωπ/6+π/3-(2kπ+π)=2kπ+2π-(ωπ/4+π/3),ω=48k/5+28/5,ωπ/6+π/3-(2kπ+π)<π/2,ω<12k+7,k>-7/12;综上k的取值范围:-7/12<k<1/4,因为k为整数,所以取k=0,ω=4/5或ω=28/5
再问: 答案还有52/5、148/5
再答: 正弦函数在一、二和三、四象限函数值相等,ωπ/6+π/3<ωπ/4+π/3,当在一、二象限时,ωπ/6+π/3-2kπ=2kπ+π-(ωπ/4+π/3),ω=48k/5+4/5,ωπ/6+π/3-2kπ<π/2,ω<4k+2,k<1/4;当在三、四象限时,ωπ/6+π/3-(2kπ+π)=2kπ+2π-(ωπ/4+π/3),ω=48k/5+28/5,ωπ/6+π/3-(2kπ+π)<π/2,ω<12k+7,k>-7/12;综上k的取值范围:-7/12<k<1/4,因为k为整数,所以取k=0,ω=4/5或ω=28/5
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