定義在(-∞,+∞)上的偶函數(shù)f(x)滿足:f(x+1)=-f(x),且在[-1,0]上是增函數(shù),下面關(guān)于f(x)的判斷:①f(2)=f(0);②f(x)的圖象關(guān)于直線x=1對(duì)稱;③f(x)在[0,1]是增函數(shù);④f(x)在[1,2]上是減函數(shù);
其中正確的判斷是________(把你認(rèn)為正確的判斷的序號(hào)都填上).
解:∵f(x+1)=-f(x)∴f(2)=-f(1)=-[-f(0)]=f(0)∴①正確;
∵f(x)是偶函數(shù),又f(x+1)=-f(x)=f(x-1)=f(1-x)∴f(x)的圖象關(guān)于直線x=1對(duì)稱;∴②正確;
∵f(x)是偶函數(shù),f(x)在[-1,0]上是增函數(shù)∴f(x)在[0,1]上是減函數(shù)∴③錯(cuò)誤;
∵f(x+1)=-f(x)=f(x-1)∴周期T=2,f(x)在[-1,0]上是增函數(shù)∴f(x)在[1,2]上也是增函數(shù).∴④錯(cuò)誤
故答案為:①②
分析:可以根據(jù)所給出的條件,逐個(gè)進(jìn)行分析與判斷,得出結(jié)論.
點(diǎn)評(píng):本題考查的知識(shí)點(diǎn)是,判斷命題真假,比較綜合的考查了抽象函數(shù)的性質(zhì).