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1、抽象函數(shù)的奇偶性、對(duì)稱(chēng)性和周期性【自我診斷】1.(1)在函數(shù)f(x)對(duì)于定義域內(nèi)的任取x1,都有x2滿(mǎn)足x1+x2=0,且f(x1)=f(x2)恒成立,則該函數(shù)是____函數(shù);(2)在函數(shù)f(x)對(duì)于定義域內(nèi)的任取x1,都有x2滿(mǎn)足x1+x2=0,且f(x1)=-f(x2)恒成立,則該函數(shù)是____函數(shù).【答案】(1)偶;(2)奇.2.已知f(x+y)+f(x-y)=2f(x)f(y)對(duì)一切實(shí)數(shù)x,y恒成立,且f(0)≠0,判斷f(x)的奇偶性.【答案】偶函數(shù).【解析】令x=0,y=0,代入式子f(
2、x+y)+f(x-y)=2f(x)f(y)可得2f(0)=2f(0)·f(0),且f(0)≠0,所以f(0)=1.令x=0,代入式子f(x+y)+f(x-y)=2f(x)f(y)可得f(y)+f(-y)=2f(0)f(y),f(y)+f(-y)=2f(y),化簡(jiǎn)得f(-y)=f(y).所以f(x)是偶函數(shù).【反思】背景函數(shù)f(x)=cosx.3.函數(shù)f(x)的值域不是單元素集合,定義域?yàn)镽,且對(duì)一切實(shí)數(shù)x,y恒有f(2x)+f(2y)=2f(x+y)f(x-y),且f(1)=0,則f(2016)的值
3、是_________.【解析】令x=+1,y=,代入已知式得f(u+2)+f(u)=2f(u+1)f(1)=0,所以f(u+2)=-f(u).所以f(u+4)=f[(u+2)+2]=-f(u+2)=f(u).所以f(x)是周期為4的周期函數(shù).所以f(2016)=f(4×504+0)=f(0).在f(2x)+f(2y)=2f(x+y)f(x-y)中,令x=y(tǒng),則2f(2x)=2f(2x)f(0)恒成立,所以f(0)=1.所以f(2016)=1.【反思】背景函數(shù)f(x)=cosx.4.已知函數(shù)f(x)(
4、x∈R)滿(mǎn)足f(x)=f(2-x),若函數(shù)y=
5、x-2x-3
6、與y=f(x)圖象的交點(diǎn)為(x1,y1),(x2,y2),…,(xm,ym),則x1+x2+…xm的值為().(A)0(B)m(C)2m(D)4m【答案】B.【解析】因?yàn)閥=f(x),y=
7、x-2x-3
8、都關(guān)于直線x=1對(duì)稱(chēng),所以它們的交點(diǎn)也關(guān)于直線x=1對(duì)稱(chēng).當(dāng)m為偶數(shù)時(shí),所求的和為2×=m;當(dāng)m為奇數(shù)時(shí),所求的和為2×+1=m.因此選B.5.已知函數(shù)f(x)(x∈R)滿(mǎn)足f(-x)=2-f(x),若函數(shù)y=與y=f(x)圖象的交點(diǎn)為(
9、x1,y1),(x2,y2),…,(xm,ym),則x1+y1+x2+y2+…xm+ym的值為().(A)0(B)m(C)2m(D)4m【答案】B.【解析】由f(-x)=2-f(x),得f(x)關(guān)于點(diǎn)(0,1)對(duì)稱(chēng),函數(shù)y==1+也關(guān)于點(diǎn)(0,1)對(duì)稱(chēng),所以?xún)珊瘮?shù)圖象的交點(diǎn)也關(guān)于點(diǎn)(0,1)對(duì)稱(chēng),將除點(diǎn)(0,1)外的其余交點(diǎn)兩兩配對(duì),對(duì)于每一組對(duì)稱(chēng)點(diǎn)有xi+xj=0,yi+yj=0,所以x1+y1+x2+y2+…xm+ym=(x1+x2+…+xm)+(y1+y2+…+ym)=0+2×=m.6.設(shè)定義
10、在R上的函數(shù)f(x)滿(mǎn)足f(x)=f(x+1)-f(x+2),證明f(x)是周期函數(shù).【證明】由已知f(x)=f(x+1)-f(x+2),得f(x+1)=f(x+2)-f(x+3),上面兩式相加,得f(x)=-f(x+3),由此式可得f(x+3)=-f(x+6),從而,得f(x)=f(x+6).所以f(x)是周期為6的函數(shù).7.設(shè)f(x)是定義在R上的偶函數(shù),其圖象關(guān)于直線x=1對(duì)稱(chēng),證明f(x)是周期函數(shù).【證明】函數(shù)y=f(x)關(guān)于直線x=1對(duì)稱(chēng),所以f(x)=f(2-x),x∈R.又由f(x)
11、是偶函數(shù)得f(x)=f(-x),所以f(-x)=f(2-x),x∈R.將上式中的-x用x代換,得f(x)=f(2+x),x∈R.這表明f(x)是以2為周期的周期函數(shù).【跟蹤訓(xùn)練】1.若f(x+y)=f(x)+f(y)對(duì)一切實(shí)數(shù)x,y恒成立,且不恒等于0,試判斷函數(shù)f(x)的奇偶性.【答案】奇函數(shù).2.函數(shù)f(x)的定義域?yàn)槿w實(shí)數(shù),對(duì)任意實(shí)數(shù)a,b都有f(a+b)+f(a-b)=2f(a)f(b),且存在c>0,使得f()=0,求證:f(x)是周期函數(shù).【答案】f(x)是以2c為周期的周期函數(shù).