Question

已知函數(shù)f(x)=sinωx·sin(-φ)-sin(+ωx)sin(π+φ)是R上的偶函數(shù).其中ω>0,0≤φ≤π,其圖象關(guān)于點(diǎn)M(,0)對(duì)稱,且在區(qū)間[0,]上是單調(diào)函數(shù),求φ和ω的值.

Answer 1

ω=或ω=2

由已知得f(x)=sinωxcosφ+cosωxsinφ
=sin(ωx+φ),
∵f(x)是偶函數(shù),∴φ=kπ+,k∈Z.
又∵0≤φ≤π,∴φ=.
∴f(x)=sin(ωx+)=cosωx.
又f(x)關(guān)于(,0)對(duì)稱,
故ω=kπ+,k∈Z.
即ω=+,k∈Z.
又ω>0,故k=0,1,2,…
當(dāng)k=0時(shí),ω=,f(x)=cosx在[0,]上是減函數(shù).
當(dāng)k=1時(shí),ω=2,f(x)=cos2x在[0,]上是減函數(shù).
當(dāng)k=2時(shí),ω=,f(x)=cosx在[0,]上不是單調(diào)函數(shù),
當(dāng)k>2時(shí),同理可得f(x)在[0,]上不是單調(diào)函數(shù),
綜上,ω=或ω=2.