已知函数f(x)=sin(wx+π/6)+sin(wx-π/6)+2cos^2(wx/2),w使f(x)能在π/3处取得最大值的最小正整数.设△ABC的三边a,b,c满足b^2=ac,且边b所对的角O的取值集合为P,当x∈P是 求f(x)值域
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![已知函数f(x)=sin(wx+π/6)+sin(wx-π/6)+2cos^2(wx/2),w使f(x)能在π/3处取得最大值的最小正整数.设△ABC的三边a,b,c满足b^2=ac,且边b所对的角O的取值集合为P,当x∈P是 求f(x)值域](/uploads/image/z/8575733-29-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28wx%2B%CF%80%2F6%29%2Bsin%28wx-%CF%80%2F6%29%2B2cos%5E2%28wx%2F2%29%2Cw%E4%BD%BFf%28x%29%E8%83%BD%E5%9C%A8%CF%80%2F3%E5%A4%84%E5%8F%96%E5%BE%97%E6%9C%80%E5%A4%A7%E5%80%BC%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E6%95%B4%E6%95%B0.%E8%AE%BE%E2%96%B3ABC%E7%9A%84%E4%B8%89%E8%BE%B9a%2Cb%2Cc%E6%BB%A1%E8%B6%B3b%5E2%3Dac%2C%E4%B8%94%E8%BE%B9b%E6%89%80%E5%AF%B9%E7%9A%84%E8%A7%92O%E7%9A%84%E5%8F%96%E5%80%BC%E9%9B%86%E5%90%88%E4%B8%BAP%2C%E5%BD%93x%E2%88%88P%E6%98%AF+%E6%B1%82f%28x%29%E5%80%BC%E5%9F%9F)
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已知函数f(x)=sin(wx+π/6)+sin(wx-π/6)+2cos^2(wx/2),w使f(x)能在π/3处取得最大值的最小正整数.设△ABC的三边a,b,c满足b^2=ac,且边b所对的角O的取值集合为P,当x∈P是 求f(x)值域
已知函数f(x)=sin(wx+π/6)+sin(wx-π/6)+2cos^2(wx/2),w使f(x)能在π/3处取得最大值的最小正整数.
设△ABC的三边a,b,c满足b^2=ac,且边b所对的角O的取值集合为P,当x∈P是 求f(x)值域
已知函数f(x)=sin(wx+π/6)+sin(wx-π/6)+2cos^2(wx/2),w使f(x)能在π/3处取得最大值的最小正整数.设△ABC的三边a,b,c满足b^2=ac,且边b所对的角O的取值集合为P,当x∈P是 求f(x)值域
f(x)=2sin(2ωx/2)cosπ/3 +1+cosωx
=√3sinωx+cosωx+1
=2sin(ωx+π/6)+1 因为π/3处取得最大值,把ω提出f(x)=2sinω(x+π/(6ω))+1,可以看出它是由f(x)=sinx图象横坐标缩小ω倍,再向左平移π/(6ω)得到的,所以得(π/2)÷ω-(π/6)÷ω=π/3,得ω=1,所以f(x)=2sin(x+π/6)+1
由于a+c>b,所以(a+c)²>b²,(a+c)²/b²>1,又b²=ac,由余弦定理得cosB=(a²+c²-b²)/2ac
=1/2 (a+c)²/b²-3/2>-1,所B∈(0,π),f(x)∈(-1,3)
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