某數(shù)學(xué)興趣小組對(duì)偶函數(shù)f(x)的性質(zhì)進(jìn)行研究,發(fā)現(xiàn)函數(shù)f(x)在定義域R上滿足f(x+2)=f(x)+f(1),且在區(qū)間[0,1]上為增函數(shù),在此基礎(chǔ)上,本組同學(xué)得出如下結(jié)論:
①函數(shù)y=f(x)的圖象關(guān)于直線x=1對(duì)稱;
②函數(shù)y=f(x)的周期為2;
③當(dāng)x∈[-3,-2]時(shí),f′(x)≥0;
④函數(shù)y=f(x)的圖象上橫坐標(biāo)為偶數(shù)的點(diǎn)都是函數(shù)的極小值點(diǎn).其中正確結(jié)論的序號(hào)是________.
①②④
分析:①先確定f(1)=f(-1)=0,從而f(x+2)=f(x),利用f(x)為偶函數(shù),可得f(1+x)=f(1-x);
②根據(jù)f(x+2)=f(x),可知f(x) 的周期為2;
③由f(x)在[0,1]上單調(diào)遞增,f(x)為偶函數(shù)可推知f(x)在[-1,0]上單調(diào)遞減;又因?yàn)閒(x)是周期為2的函數(shù),所以f(x)在[-1+2k,2k]k∈Z上單調(diào)遞減,從而f(x)在[-3,-2]上單調(diào)遞減,故f′(x)≤0;
④根據(jù)R上的偶函數(shù)在區(qū)間[0,1]上為增函數(shù),可知0是函數(shù)的極小值點(diǎn),根據(jù)f(x) 的周期為2,可知函數(shù)y=f(x)的圖象上橫坐標(biāo)為偶數(shù)的點(diǎn)都是函數(shù)的極小值點(diǎn),
故可得結(jié)論.
解答:①,f(-1+2)=f(-1)+f(1),∴f(-1)=0,又知f(x)為偶函數(shù),
∴f(1)=f(-1)=0,∴f(x+2)=f(x)
∵f(x)為偶函數(shù),∴f(x+2)=f(-x),∴f(1+x)=f(1-x),
∴函數(shù)y=f(x)的圖象關(guān)于直線x=1對(duì)稱,故①正確;
②,根據(jù)f(x+2)=f(x),可知f(x) 的周期為2,故②正確;
③,由f(x)在[0,1]上單調(diào)遞增,f(x)為偶函數(shù)可推知f(x)在[-1,0]上單調(diào)遞減;
又因?yàn)閒(x)是周期為2的函數(shù),所以f(x)在[-1+2k,2k]k∈Z上單調(diào)遞減,從而f(x)在[-3,-2]上單調(diào)遞減,故f′(x)≤0,所以③不正確;
④,根據(jù)R上的偶函數(shù)在區(qū)間[0,1]上為增函數(shù),可知0是函數(shù)的極小值點(diǎn),根據(jù)f(x) 的周期為2,可知函數(shù)y=f(x)的圖象上橫坐標(biāo)為偶數(shù)的點(diǎn)都是函數(shù)的極小值點(diǎn),所以④正確
故正確結(jié)論的序號(hào)是①②④
故答案為:①②④
點(diǎn)評(píng):本題綜合考查偶函數(shù)的性質(zhì),考查函數(shù)的周期性,函數(shù)的對(duì)稱性,合理運(yùn)用條件進(jìn)行轉(zhuǎn)化是解題的關(guān)鍵.
科目:高中數(shù)學(xué)
來(lái)源:2011-2012學(xué)年福建省莆田四中高三(上)第一次月考數(shù)學(xué)試卷(解析版)
題型:填空題
某數(shù)學(xué)興趣小組對(duì)偶函數(shù)f(x)的性質(zhì)進(jìn)行研究,發(fā)現(xiàn)函數(shù)f(x)在定義域R上滿足f(x+2)=f(x)+f(1),且在區(qū)間[0,1]上為增函數(shù),在此基礎(chǔ)上,本組同學(xué)得出如下結(jié)論:
①函數(shù)y=f(x)的圖象關(guān)于直線x=1對(duì)稱;
②函數(shù)y=f(x)的周期為2;
③當(dāng)x∈[-3,-2]時(shí),f′(x)≥0;
④函數(shù)y=f(x)的圖象上橫坐標(biāo)為偶數(shù)的點(diǎn)都是函數(shù)的極小值點(diǎn).其中正確結(jié)論的序號(hào)是 .
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