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.](/uploads/image/z/5630509-37-9.jpg?t=%E9%AB%98%E4%B8%80%E6%95%B0%E5%AD%A6%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E6%9C%80%E5%80%BC%E9%97%AE%E9%A2%98%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D2sin%26sup2%3B%EF%BC%88%CF%80%2F4%2Bx%EF%BC%89-%E2%88%9A3+cos2x.x%E2%88%88%5B%CF%80%2F4%2C%CF%80%2F2%5D1.%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC2.%E8%8B%A5%E4%B8%8D%E7%AD%89%E5%BC%8F%7Cf%EF%BC%88x%EF%BC%89-m%7C%EF%BC%9C2%E5%9C%A8x%E2%88%88%5B%CF%80%2F4%2C%CF%80%2F2%5D%E4%B8%8A%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
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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