已知函数f(x)=2sin²(π/4+x)-(根号3*cos2x),x∈[π/4,π/2] (1).求f(x)的最大值和最小值(2).若不等式-2<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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![已知函数f(x)=2sin²(π/4+x)-(根号3*cos2x),x∈[π/4,π/2] (1).求f(x)的最大值和最小值(2).若不等式-2<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/1042142-14-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2sin%26%23178%3B%28%CF%80%2F4%2Bx%29-%EF%BC%88%E6%A0%B9%E5%8F%B73%2Acos2x%29%2Cx%E2%88%88%5B%CF%80%2F4%2C%CF%80%2F2%5D+%EF%BC%881%29.%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%BC%282%29.%E8%8B%A5%E4%B8%8D%E7%AD%89%E5%BC%8F-2%EF%BC%9Cf%EF%BC%88x%EF%BC%89-m%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)
已知函数f(x)=2sin²(π/4+x)-(根号3*cos2x),x∈[π/4,π/2] (1).求f(x)的最大值和最小值(2).若不等式-2<f(x)-m<2在x∈[π/4,π/2]上恒成立,求实数m的取值范围
已知函数f(x)=2sin²(π/4+x)-(根号3*cos2x),x∈[π/4,π/2] (1).求f(x)的最大值和最小值
(2).若不等式-2<f(x)-m<2在x∈[π/4,π/2]上恒成立,求实数m的取值范围
已知函数f(x)=2sin²(π/4+x)-(根号3*cos2x),x∈[π/4,π/2] (1).求f(x)的最大值和最小值(2).若不等式-2<f(x)-m<2在x∈[π/4,π/2]上恒成立,求实数m的取值范围
sin方转换成两倍角,然后和化积.
f(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,
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f(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.
|f(x)-m|<2
m <= 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。
综合,知,
0 < m < 5,满足题意。
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