給出下列命題:
①如果函數(shù)f(x)對(duì)任意的x∈R,都有f(1+x)=f(1-x),那么函數(shù)f(x)必是偶函數(shù);
②要得到函數(shù)y=sin(1-x)的圖象,只要將函數(shù)y=sin(-x)的圖象向右平移1個(gè)單位即可;
③如果函數(shù)f(x)對(duì)任意的x1、x2∈R,且x1≠x2,都有(x1-x2)[f(x1)-f(x2)]>0,那么函數(shù)f(x)在R上是增函數(shù);
④函數(shù)y=f(x)和函數(shù)y=f(x-2)+1的圖象一定不能重合.其中真命題的序號(hào)是 .
【答案】
分析:①如果函數(shù)f(x)對(duì)任意的x∈R,都有f(1+x)=f(1-x),那么函數(shù)f(x)必是偶函數(shù),由對(duì)稱性判斷;
②要得到函數(shù)y=sin(1-x)的圖象,只要將函數(shù)y=sin(-x)的圖象向右平移1個(gè)單位即可,由平移規(guī)則進(jìn)行判斷;
③如果函數(shù)f(x)對(duì)任意的x
1、x
2∈R,且x
1≠x
2,都有(x
1-x
2)[f(x
1)-f(x
2)]>0,那么函數(shù)f(x)在R上是增函數(shù),由函數(shù)的單調(diào)性的定義判斷;
④函數(shù)y=f(x)和函數(shù)y=f(x-2)+1的圖象一定不能重合,有平移規(guī)則判斷.
解答:解:①如果函數(shù)f(x)對(duì)任意的x∈R,都有f(1+x)=f(1-x),那么函數(shù)f(x)必是偶函數(shù),此命題中的條件說(shuō)明函數(shù)關(guān)于直線x=1對(duì)稱,不能得出偶函數(shù)的結(jié)論,故錯(cuò)誤;
②要得到函數(shù)y=sin(1-x)的圖象,只要將函數(shù)y=sin(-x)的圖象向右平移1個(gè)單位即可,由平移規(guī)則知,此命題是正確命題;
③如果函數(shù)f(x)對(duì)任意的x
1、x
2∈R,且x
1≠x
2,都有(x
1-x
2)[f(x
1)-f(x
2)]>0,那么函數(shù)f(x)在R上是增函數(shù),由函數(shù)單調(diào)性的定義知,此命題是正確命題;
④函數(shù)y=f(x)和函數(shù)y=f(x-2)+1的圖象一定不能重合,存在一個(gè)函數(shù)右移兩個(gè)單位再產(chǎn)移一個(gè)單位可以重合,如y=
x,故此命題不正確.
綜上,②③是正解命題
故答案為②③
點(diǎn)評(píng):本題考查函數(shù)y=Asin(ωx+φ)的圖象變換,求解本題的關(guān)鍵是掌握住平移的規(guī)則以及函數(shù)的對(duì)稱性的,函數(shù)的單調(diào)性的判斷方法,圖象的變化等知識(shí),本題是基礎(chǔ)知識(shí)基本概念題,涉及到的知識(shí)點(diǎn)較多,判斷時(shí)思維轉(zhuǎn)換快,易出錯(cuò).