C
分析:由函數(shù)f(x)=Msin(ωx+∅)(ω>0)在區(qū)間[a,b]上是增函數(shù),且f(a)=-M,f(b)=M,可知函數(shù)f(x)為奇函數(shù)且M>0,從而可得區(qū)間[a,b]關(guān)于原點對稱,∅=0,代入g(x)中結(jié)合余弦函數(shù)的單調(diào)性判斷.
解答:∵函數(shù)f(x)在區(qū)間[a,b]上是增函數(shù),且f(a)=-M,f(b)=M
∴M>0且區(qū)間[a,b]關(guān)于原點對稱
從而函數(shù)函數(shù)f(x)為奇函數(shù)∅=2kπ
∴函數(shù)g(x)=Mcos(ωx+∅)=Mcoswx在區(qū)間[a,0]是增函數(shù),[0,b]減函數(shù)
∴函數(shù)g(x)=Mcos(ωx+∅)在區(qū)間[a,b]上取得最大值M,最小值為0
故選C.
點評:本題綜合考查了正弦函數(shù)與余弦函數(shù)的圖象及性質(zhì),利用整體思想進(jìn)行求值,在解題時要熟練運用相關(guān)結(jié)論:y=Asin(wx+∅)為奇(偶)函數(shù)?∅=kπ(
)(k∈Z)