作业帮高一数学三角函数最值问题已知函数f（x）=2sin²（π/4+x）-√3 cos2x.x∈[π/4,π/2]1.求f（x）的最大值和最小值2.若不等式|f（x）-m|＜2在x∈[π/4,π/2]上恒成立,求实数m的取值范围.
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高一数学三角函数最值问题已知函数f（x）=2sin²（π/4+x）-√3 cos2x.x∈[π/4,π/2]1.求f（x）的最大值和最小值2.若不等式|f（x）-m|＜2在x∈[π/4,π/2]上恒成立,求实数m的取值范围.
高一数学三角函数最值问题
已知函数f（x）=2sin²（π/4+x）-√3 cos2x.x∈[π/4,π/2]
1.求f（x）的最大值和最小值
2.若不等式|f（x）-m|＜2在x∈[π/4,π/2]上恒成立,求实数m的取值范围.
高一数学三角函数最值问题已知函数f（x）=2sin²（π/4+x）-√3 cos2x.x∈[π/4,π/2]1.求f（x）的最大值和最小值2.若不等式|f（x）-m|＜2在x∈[π/4,π/2]上恒成立,求实数m的取值范围.
（1）f(x)=2sin(2x-π/3)+1.π/6
（1）f（x）=1-cos(π/2+2x)-√3 cos2x=1+sin2x--√3 cos2x=1+2sin(2x-π/3)
2x-π/3∈[π/6,2π/3] 所以f（x）∈[2,3]
（2）-2
f(x)+2∈[4,5]
1
见图
⑴f(x) = [sin(x) + cos(x)]^2 - 3^(1/2)cos(2x)
= 1 + sin(2x) - 3^(1/2)cos(2x)
= 1 + 2[sin(2x)/2 - 3^(1/2)cos(2x)/2]
= 1 + 2sin(2x - π/3)
π/4 <= x <= π/2,
π/2 <= 2x <= π,
π/...
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⑴f(x) = [sin(x) + cos(x)]^2 - 3^(1/2)cos(2x)
= 1 + sin(2x) - 3^(1/2)cos(2x)
= 1 + 2[sin(2x)/2 - 3^(1/2)cos(2x)/2]
= 1 + 2sin(2x - π/3)
π/4 <= x <= π/2,
π/2 <= 2x <= π,
π/6 <= 2x - π/3 <= 2π/3
2 = 1 + 2sin(π/6) <= 1 + 2sin(2x - π/3) = f(x) <= 1 + 2sin(π/2) = 3,
当 x = π/4时，f(x)取得最小值2，
当 x = 5π/12时，f(x)取得最大值3.
⑵m <= 2时，2 > |f(x) - m| = f(x) - m >= 2 - m, m > 0.
0 < m <= 2满足要求。
m >= 3时，2 > |f(x) - m| = m - f(x) >= m - 3, m < 5.
3 <= m < 5满足要求。
2 < m < 3时，-2 > -m > -3
-1 = 2 - 3 < 2 - m <= f(x) - m <= 3 - m < 3 - 2 = 1,
|f(x) - m| < 1 < 2,满足要求。
综上，0 < m < 5
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