Question

已知函数f(x)=Asin(wx+φ)(其中A>0,w>0,-π/2<φ<0)的相邻对称轴之间的距离为π/2，且该函数图象的一个最

高点为（5/12π,4）（1）求函数f(x)的解析式和单调增区间
(2)若x属于[π/4,π/2],求函数f(x)的最大值和最小值

Answer 1

(1)
不妨取ωx₁+φ = -π/2, ωx₂+φ = π/2 (-π/2, π/2为相邻对称轴)
ωx₂+φ - (ωx₁+φ) = ω(x₂ - x₁) = ωπ/2 = π/2 - (-π/2) = π
ω = 2
最高点y = 4, A = 4
x = 5π/12, y取最大值, ω(5π/12) + φ = π/2, 5π/6 + φ = π/2, φ = -π/3
f(x) = 4sin(2x - π/3)

f(x)周期为π
x = 5π/12 + nπ时, y取最大值
x = -π/12 + nπ时, y取最小值
x∈(-π/12 + nπ, 5π/12 + nπ), 单调增

x∈(5π/12 + nπ, 11π/12 + nπ), 单调减

(2)
5π/12在[π/4,π/2]内, 函数f(x)的最大值=4
f(π/4) = 4sin(π/2 - π/3) = 4sin(π/6) = 2
f(π/2) = 4sin(π - π/3) = 2√3 > f(π/4)
最小值2