培優(yōu)增分拓展提升課三對(duì)稱(chēng)性與周期性的二級(jí)結(jié)論【結(jié)論總結(jié)】一、函數(shù)的對(duì)稱(chēng)性相關(guān)結(jié)論1.同一個(gè)函數(shù)的自身對(duì)稱(chēng)結(jié)論?:若函數(shù)y=f(x)滿(mǎn)足f(a+x)=f(b-x)或f(x)=f(a+b-x),則函數(shù)y=f(x)的圖象關(guān)于直線(xiàn)x=a+b[說(shuō)明]軸對(duì)稱(chēng)問(wèn)題:若函數(shù)f(x)關(guān)于x=a對(duì)稱(chēng),可得到如下結(jié)論中任意一個(gè):f結(jié)論?:若函數(shù)y=f(x)滿(mǎn)足f(x)=2b-f(2a-x),則函數(shù)y=f(x)的圖象關(guān)于點(diǎn)A(a,b)對(duì)稱(chēng).反之亦成立.特別地,當(dāng)a=b=0時(shí),變?yōu)?若函數(shù)y=f(x)滿(mǎn)足f(x)=-f(-x),則函數(shù)y=f(x)的圖象關(guān)于點(diǎn)(0,0)對(duì)稱(chēng),即為奇函數(shù)的性質(zhì).[說(shuō)明]點(diǎn)對(duì)稱(chēng)問(wèn)題:若函數(shù)f(x)關(guān)于(a,0)對(duì)稱(chēng),可得到如下結(jié)論中任意一個(gè):f2.同一“f”下,兩個(gè)不同函數(shù)的對(duì)稱(chēng)結(jié)論?:函數(shù)y=f(a+x)與y=f(b-x)的圖象關(guān)于直線(xiàn)x=b+a結(jié)論?:函數(shù)y=f(x)與y=-f(2a-x)+2b的圖象關(guān)于點(diǎn)A(a,b)成中心對(duì)稱(chēng).二、函數(shù)周期性的結(jié)論1.函數(shù)周期性的常用結(jié)論結(jié)論?:若f(x+a)=f(x-a),則f(x)的一個(gè)周期為2a;結(jié)論?:若f(x+a)=-f(x),則f(x)的一個(gè)周期為2a;結(jié)論?:若f(x+a)+f(x)=c(a≠0),則f(x)的一個(gè)周期為2a;結(jié)論?:若f(x)=f(x+a)+f(x-a)(a≠0),則f(x)的一個(gè)周期為6a;結(jié)論?:若f(x+a)=1f(x),則f(結(jié)論:若f(x+a)=-1f(x),則f(2.由對(duì)稱(chēng)性推得周期結(jié)論:若函數(shù)y=f(x)的圖象同時(shí)關(guān)于直線(xiàn)x=a和直線(xiàn)x=b成軸對(duì)稱(chēng)(a≠b),則y=f(x)是周期函數(shù),且2|a-b|是其一個(gè)周期.結(jié)論:若函數(shù)y=f(x)的圖象同時(shí)關(guān)于點(diǎn)A(a,c)和點(diǎn)B(b,c)成中心對(duì)稱(chēng)(a≠b),則y=f(x)是周期函數(shù),且2|a-b|是其一個(gè)周期.結(jié)論:若函數(shù)y=f(x)的圖象關(guān)于點(diǎn)A(a,c)成中心對(duì)稱(chēng),又關(guān)于直線(xiàn)x=b成軸對(duì)稱(chēng)(a≠b),則y=f(x)是周期函數(shù),且4|a-b|是其一個(gè)周期.[記法]兩次對(duì)稱(chēng)成周期,兩軸兩心二倍差,一軸一心四倍差.【結(jié)論應(yīng)用】[典例1]函數(shù)f(x)的定義域?yàn)镽,若f(x+1)與f(x-1)都是奇函數(shù),則下列正確的是(D)A.f(x)是偶函數(shù) B.f(x)是奇函數(shù)C.f(x+1)=f(x+2) D.f(x+3)是奇函數(shù)解析：函數(shù)f(x+1)的圖象是由函數(shù)f(x)的圖象向左平移1個(gè)單位得到的,由f(x+1)是奇函數(shù),可知f(x+1)的對(duì)稱(chēng)中心是(0,0),從而f(x)的對(duì)稱(chēng)中心是(1,0),又因?yàn)閒(x-1)是奇函數(shù),從而f(x)的對(duì)稱(chēng)中心是(-1,0),由結(jié)論12可知,f(x)是周期函數(shù),其周期是T=2[1-(-1)]=4,于是f(x+3)=f(x-1),由已知f(x-1)是奇函數(shù),故f(x+3)是奇函數(shù).[典例2]設(shè)函數(shù)f(x)在R上滿(mǎn)足f(2-x)=f(2+x),f(7-x)=f(7+x),且在區(qū)間0,7上只有f(1)=f(3)=0,則方程f(x)=0在閉區(qū)間-2023,2A.806 B.807C.810 D.811解析：因?yàn)閒(2-x)=f(2+x),由結(jié)論1可知,f(x)的圖象關(guān)于直線(xiàn)x=2對(duì)稱(chēng),又f(7-x)=f(7+x),同理可得f(x)的圖象關(guān)于直線(xiàn)x=7對(duì)稱(chēng).由結(jié)論11得,f(x)是周期函數(shù),其周期為T(mén)=2×(7-2)=10,因?yàn)閒(1)=f(3)=0,所以f(11)=f(13)=0,f(-9)=f(-7)=0.故f(x)=0在[0,10]和[-10,0]上均有兩個(gè)解,從而可知在[0,2023]上有406個(gè)解,在[-2023,0]上有404個(gè)解.所以方程f(x)=0在閉區(qū)間-2[典例3]已知函數(shù)f(x)的定義域?yàn)镽,滿(mǎn)足f(1+x)=f(1-x),f(x)=-f(4-x),若f(1)=2,則∑k=12024f(k)= A.-1 B.0 C.1 D.2解析：因?yàn)閒(1+x)=f(1-x),由結(jié)論1可知,函數(shù)f(x)的圖象關(guān)于x=1對(duì)稱(chēng),又f(x)=-f(4-x),由結(jié)論2可知函數(shù)f(x)的圖象關(guān)于點(diǎn)(2,0)對(duì)稱(chēng).由結(jié)論13可知,函數(shù)的周期T=4×|2-1|=4.由f(1)=2,可得f(3)=-f(1)=-2.令x=2,得f(2)=-f(2),解得f(2)=0,f(4)=-f(0)=0.所以一個(gè)周期內(nèi)f(1)+f(2)+f(3)+f(4)=2+0+(-2)+0=0,因?yàn)?024=506×4,所以∑k=12024f(k)=506[f(1)+f(2)+[典例4](2021·全國(guó)高考)已知函數(shù)f(x)的定義域?yàn)镽,f(x+2)為偶函數(shù),f(2x+1)為奇函數(shù),則 (B)A.f(-12)=0 B.fC.f(2)=0 D.f(4)=0解析：f(x+2)為偶函數(shù),所以其圖象關(guān)于x=0對(duì)稱(chēng),則函數(shù)f(x)的圖象關(guān)于直線(xiàn)x=2對(duì)稱(chēng),又f(2x+1)為奇函數(shù),所以其關(guān)于(0,0)對(duì)稱(chēng),通過(guò)函數(shù)圖象變換,可以得到f(2x)關(guān)于(12,0)對(duì)稱(chēng),進(jìn)而f(x)關(guān)于(1,0)對(duì)稱(chēng)由結(jié)論