解:(1)函數(shù)f(x)=
sin2x+2cos
2x+m=
sin2x+cos2x+1+m=2sin(2x+
)+1+m,函數(shù)f(x)=
sin2x+2cos
2x+m在區(qū)間[0,
]上的最大值為6,所以m=3,函數(shù)的表達式為f(x)=2sin(2x+
)+4;它的對稱中心為(
,0),k∈Z.
(2)函數(shù)f(x)關(guān)于y軸的對稱圖象得函數(shù)f
1(x)=2sin(-2x+
)+4的圖象,函數(shù)f
1(x)的圖象向右平移
個單位得函數(shù)f
2(x)=2sin(-2x+
+
)+4=2cos(2x-
)+4的圖象;
函數(shù)f
2(x)的單調(diào)遞減區(qū)間為:2kπ≤2x-
≤2kπ+π,kπ
≤x≤kπ+
k∈Z.
分析:(1)利用二倍角公式,兩角和的正弦函數(shù)化簡函數(shù)的表達式,通過函數(shù)f(x)=
sin2x+2cos
2x+m在區(qū)間[0,
]上的最大值為6,求出m的值.即可求出函數(shù)的對稱中心.
(2)求出函數(shù)f
1(x)的表達式,再把函數(shù)f
1(x)的圖象向右平移
個單位得函數(shù)f
2(x)的圖象對應(yīng)的表達式,利用余弦函數(shù)的單調(diào)減區(qū)間求函數(shù)f
2(x)的單調(diào)遞減區(qū)間.
點評:本題是中檔題,考查函數(shù)的最值,對稱中心的求法,函數(shù)圖象的變換,考查計算能力.