f(x)=sin²x+√3tanθcosx+﹙√3/8﹚tanθ-3/2其中x∈[0,π/2] θ∈[0,π/3]①当θ=π/3时,求f(x)的最大值及相应的x的值②是否存在实数θ,使得函数f(x)最大值是-1/8?若存在,求出对应的θ值,若不存在,试说明
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 16:39:02
![f(x)=sin²x+√3tanθcosx+﹙√3/8﹚tanθ-3/2其中x∈[0,π/2] θ∈[0,π/3]①当θ=π/3时,求f(x)的最大值及相应的x的值②是否存在实数θ,使得函数f(x)最大值是-1/8?若存在,求出对应的θ值,若不存在,试说明](/uploads/image/z/2738920-40-0.jpg?t=f%28x%29%3Dsin%26%23178%3Bx%2B%E2%88%9A3tan%CE%B8cosx%2B%EF%B9%99%E2%88%9A3%2F8%EF%B9%9Atan%CE%B8-3%2F2%E5%85%B6%E4%B8%ADx%E2%88%88%5B0%2C%CF%80%2F2%5D+%CE%B8%E2%88%88%5B0%2C%CF%80%2F3%5D%E2%91%A0%E5%BD%93%CE%B8%3D%CF%80%2F3%E6%97%B6%2C%E6%B1%82f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%8F%8A%E7%9B%B8%E5%BA%94%E7%9A%84x%E7%9A%84%E5%80%BC%E2%91%A1%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E5%AE%9E%E6%95%B0%CE%B8%2C%E4%BD%BF%E5%BE%97%E5%87%BD%E6%95%B0f%28x%29%E6%9C%80%E5%A4%A7%E5%80%BC%E6%98%AF-1%2F8%3F%E8%8B%A5%E5%AD%98%E5%9C%A8%2C%E6%B1%82%E5%87%BA%E5%AF%B9%E5%BA%94%E7%9A%84%CE%B8%E5%80%BC%2C%E8%8B%A5%E4%B8%8D%E5%AD%98%E5%9C%A8%2C%E8%AF%95%E8%AF%B4%E6%98%8E)
f(x)=sin²x+√3tanθcosx+﹙√3/8﹚tanθ-3/2其中x∈[0,π/2] θ∈[0,π/3]①当θ=π/3时,求f(x)的最大值及相应的x的值②是否存在实数θ,使得函数f(x)最大值是-1/8?若存在,求出对应的θ值,若不存在,试说明
f(x)=sin²x+√3tanθcosx+﹙√3/8﹚tanθ-3/2其中x∈[0,π/2] θ∈[0,π/3]
①当θ=π/3时,求f(x)的最大值及相应的x的值②是否存在实数θ,使得函数f(x)最大值是-1/8?若存在,求出对应的θ值,若不存在,试说明理由.
f(x)=sin²x+√3tanθcosx+﹙√3/8﹚tanθ-3/2其中x∈[0,π/2] θ∈[0,π/3]①当θ=π/3时,求f(x)的最大值及相应的x的值②是否存在实数θ,使得函数f(x)最大值是-1/8?若存在,求出对应的θ值,若不存在,试说明
①θ=π/3 tanθ=tan(π/3)=√3
f(x)=sin²x+3cosx+3/8-3/2=1-cos²x+3cosx-9/8=-(cosx-3/2)²+9/4-1/8
f(x)最大值为-(1-3/2)²+9/4-1/8=-1/4+9/4-1/8=15/8
f(x)取最大值时需cosx=1 即 x=0
②f(x)=sin²x+√3tanθcosx+﹙√3/8﹚tanθ-3/2
=-cos²x+√3tanθcosx+﹙√3/8﹚tanθ-1/2
=-(cosx-√3tanθ/2)²+3tan²θ/4+﹙√3/8﹚tanθ-1/2
若θ∈[0,arctan(2/√3)] √3tanθ/2∈[0,1]
则f(x)的最大值为 3tan²θ/4+﹙√3/8﹚tanθ-1/2
若3tan²θ/4+﹙√3/8﹚tanθ-1/2=-1/8 解得tanθ=-√3/2 或tanθ=√3/3
因tanθ∈[0,(2/√3)] 所以 θ=π/6
若θ∈[arctan(2/√3),π/3] √3tanθ/2∈[1,√3]
则f(x)的最大值为 -(1-√3tanθ/2)²+3tan²θ/4+﹙√3/8﹚tanθ-1/2=9√3/8 tanθ-3/2
若9√3/8 tanθ-3/2=-1/8 得tanθ= 11√3/27 ,但 θ∈[arctan(2/√3),π/3]时 tanθ∈[2/√3,√3]
而 11√3/27