数学题f(x)=2sin^(π/4 x)-√3cos2x-1已知f(x)=2sin^(π/4 x)-√3cos2x-1 x属于R,1.求函数f(x)的最小正周期2.若对任意X属于{π/4,π/2},不等式f(x)>m^2-3m+3恒成立,求实数m的取值范围!题目是f(x)=2sin^2(π/4 +x)-√

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数学题f(x)=2sin^(π/4 x)-√3cos2x-1已知f(x)=2sin^(π/4 x)-√3cos2x-1 x属于R,1.求函数f(x)的最小正周期2.若对任意X属于{π/4,π/2},不等式f(x)>m^2-3m+3恒成立,求实数m的取值范围!题目是f(x)=2sin^2(π/4 +x)-√
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数学题f(x)=2sin^(π/4 x)-√3cos2x-1已知f(x)=2sin^(π/4 x)-√3cos2x-1 x属于R,1.求函数f(x)的最小正周期2.若对任意X属于{π/4,π/2},不等式f(x)>m^2-3m+3恒成立,求实数m的取值范围!题目是f(x)=2sin^2(π/4 +x)-√
数学题f(x)=2sin^(π/4 x)-√3cos2x-1
已知f(x)=2sin^(π/4 x)-√3cos2x-1 x属于R,
1.求函数f(x)的最小正周期
2.若对任意X属于{π/4,π/2},不等式f(x)>m^2-3m+3恒成立,求实数m的取值范围!
题目是f(x)=2sin^2(π/4 +x)-√3cos2x-1

数学题f(x)=2sin^(π/4 x)-√3cos2x-1已知f(x)=2sin^(π/4 x)-√3cos2x-1 x属于R,1.求函数f(x)的最小正周期2.若对任意X属于{π/4,π/2},不等式f(x)>m^2-3m+3恒成立,求实数m的取值范围!题目是f(x)=2sin^2(π/4 +x)-√
1、f(x)=2sin^2 (x+π/4)-√3cos2x-1
=-cos(2x+π/2)-√3cos2x
=sin2x-√3cos2x
=2sin(2x-π/3)
当x属于R时,函数f(x)的最小正周期T=2π/2=π
2、X属于{π/4,π/2},
所以:π/6≤2x-π/3≤2π/3
f(x)=2sin(2x-π/3)的最小值为:1,即:当π/4的时候.
所以:m^2-3m+3≤1
解得:1≤m≤2

1f(x)=2sin^2 (x+π/4)-√3cos2x-1
=-cos(2x+π/2)-√3cos2x
=sin2x-√3cos2x
=2sin(2x-π/3)
当x属于R时,函数f(x)的最小正周期T=2π/2=π

f(x)=2sin²(π/4+x)-√3cos2x-1=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2(0.5sin2x-0.5√3cos2x)=2sin(2x-π/3)
f(x)最小正周期为T=π
∵x∈(π/4,π/2)
∴2x-π/3∈(π/6,2π/3)
∴2sin(2x-π/3)∈(1,2)
∵f(x)>m...

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f(x)=2sin²(π/4+x)-√3cos2x-1=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2(0.5sin2x-0.5√3cos2x)=2sin(2x-π/3)
f(x)最小正周期为T=π
∵x∈(π/4,π/2)
∴2x-π/3∈(π/6,2π/3)
∴2sin(2x-π/3)∈(1,2)
∵f(x)>m²-3m+3恒成立
∴m²-3m+3≤1
∴实数m的取值范围为{m|1≤m≤2}

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