高考數(shù)學(xué)培優(yōu)微專題《函數(shù)的周期性》解析版

么就稱函數(shù)y=f(x)為周期函數(shù)，稱T為這個(gè)函數(shù)的周期.如果在周期函數(shù)f(x)的所有周期中存在一個(gè)最小的正數(shù)，那么稱這個(gè)最小整數(shù)叫做f(x)的最小正周期.(1)通過雙f等式出周期：①若函數(shù)y=f(x)有兩條對(duì)稱軸x=a，x=b(a<b)②若函數(shù)y=f(x)的圖象有兩個(gè)對(duì)稱中心(a,c),(b,c)(a<b)，則函數(shù)y=f(x)是周期函數(shù)，且T=2(b-a)；③若函數(shù)y=f(x)有一條對(duì)稱軸x=a和一個(gè)對(duì)稱中心(b,0)(a<b)，則函數(shù)y=f(x)是周期函數(shù)，且T=4(b-a).EQ\*jc3\*hps21\o\al(\s\up5(1),2)2.定義在R上的奇函數(shù)f(x)滿足f(x+4)=f(x),當(dāng)x∈(0,2)時(shí),f(x)=3x-1,則f(2022)+f(2023)=__【解析】答案-13.(2018·江蘇·高考真題)函數(shù)f(x)滿足f(x+4)=f(x)(x∈R)，且在區(qū)間(-2,2]上，f(x)=函數(shù)解析式求結(jié)果.詳解：由f(x+4)=f(x)得函數(shù)f(x(的周期為4，所以f=|-1+4.設(shè)f(x)為定義在R上的奇函數(shù)，f(x+2)=-f(x)，當(dāng)0≤x≤1時(shí)，f(x)=x，則f(7.5)=【解析】思路：由f(x+2)=-f(x)可得：f(x(的周期T=4，∴考慮將f(7.5)用0≤x≤1中的函數(shù)值進(jìn)行表示：f(7.5)=f(3.5(=f(-0.5(，此時(shí)周期性已經(jīng)無法再進(jìn)行調(diào)整，考慮利用奇偶性進(jìn)行微調(diào)：f(-0.5(=-f(0.5(=-所以5.已知定義在R上的函數(shù)滿足f=-1,則ff(3)=7.定義在R上的函數(shù)f(x(對(duì)任意x∈R，都有f(x+2(=,f(2(=，則f(2024(等于8.(2024·金華調(diào)研)定義在R上的函數(shù)f(x)滿足f(x+6)=f(x)，當(dāng)-3≤x<-1時(shí)，f(x)=-(x+2)2，當(dāng)-1≤x<3時(shí)，f(x)=x，則f(1)+f(2)+f(3)+?+f(2025)=.解析：因?yàn)閒(x+6)=f(x)，所以f(x)的周期T=6，于是f(1)=1，f(2)=2，f(3)=f(-3)=-(-3+2)2=-1，f(4)=f(-2)=-(-2+2)2=0，f(5)=f(-1)=-1，f(6)=f(0)=0，所以f(1)+f(2)+f(3)+f(4)+f(5)+f(6)=1，而2025=6×337+3，所以f(1)+f(2)+f(3)+?+f(29.已知f(x(是定義在R上的函數(shù)，滿足f(x(+f(-x(=0,f(x-1(=f(x+1(，當(dāng)x∈(0,1(時(shí)，f(x(=-x2【解析】思路：由f(x-1(=f(x+1(可得f(x(是周期為2的周期函數(shù)，所以只需要求出一個(gè)周期內(nèi)的最-(x-所以f(x(max=f(而由于f(x(為奇函數(shù)，所以在x∈(-1,0(時(shí)，f(x(min=f(-(=-f(，所以f(-(即為f(x(在(-1,1(的最小值，從而也是f(x(在R上的最小值10.定義在R上的函數(shù)f(x(滿足f(x(=(x-2(,，則f(2024(的值為【解析】思路：所給f(x(的特點(diǎn)為x<0才有解析式能夠求值，而x>0只能通過f(x(=f(x-1(-f(x-2(減少自變量的取值，由所求f(2009(可聯(lián)想到判斷f(x(是否具有周期性，x>0時(shí)，f(x(=f(x-1(-f(x-2(，則有f(x-1(=f(x-2(-f(x-3(，兩式相加可得：f(x(=-f(x-3(，則f(x(=-f(x-3(=f(x-6(，即f(x(在x>0時(shí)周期是6，故f(2(=f(1(-f(0(=f(0(-f(-1(-f(0(=f(-1(=111.定義在R上的函數(shù)f(x)滿足f(x+1)=f(x)-2,則下列是周期函數(shù)的是()A.y=f(x)-xB.y=f(x)+xC.y=f(x)-2xD.y=f(x)+2x思路：由f(x+2(=2f(x(?f(x(=f(x+2(，可類比函數(shù)的周期性，f(-已知可得f(4(=所以f(2016(=解析：由f(x+π(=f(x(+sinx可知f((=f((+sin(+sin(=f((+sin所以可得：f(x(=x2-2x，所以f(x+4(=(x+4(2-2(x+4(=x2+6x+8,因?yàn)楹瘮?shù)f(x(滿足f(x+2(=3f(x(，所以f(x+4(=3f(x+2(=9f(x(，所以f(x(=f(x+4(=(x2+6x+8(，x∈[-4,-2[，解不等式可得t≥3或-1≤t<0.因?yàn)閒(x+y(+f(x-y(=f(x(f(y(，令x=1,y=0可得，2f(1(=f(1(f(0(，所以f(0(=2，令x=0可得，f(y(+f(-y(=2f(y(，即f(y(=f(-y(，所以函數(shù)f(x(為偶函數(shù)，令y=1得，f(x+1(+f(x-1(=f(x(f(1(=f(x(，即有f(x+2(+f(x(=f(x+1(，從而可知f(x+2(=-f(x-1(，f(x-1(=-f(x-4(，故f(x+2(=f(x-4(，即f(x(=f(x+6(，所以函數(shù)f(x(的一個(gè)周期為6．因?yàn)閒(2(=f(1(-f(0(=1-2=-1，f(3(=f(2(-f(1(=-1-1=-2，f(4(=f(-2(=f(2(=-1，f(5(=f(-1(=f(1(=1，f(6(=f(0(=2，所以一個(gè)周期內(nèi)的f(1(+f(2(+?+f(6(=0．由于22除以6余4，所以f(k(=f(1(+f(2(+f(3(+f(4(=1-1-2-1=-3．故選：A.由f(x+y(+f(x-y(=f(x(f(y(，聯(lián)想到余弦函數(shù)和差化積公式cos(x+y(+cos(x-y(=2cosxcosy，可設(shè)f(x(=acosωx，則由方法一中所以f(x(=2cos則f(x+y(+f(x-y(=2cos(y(+2cos(y(=4cosxcosy=f(x(f(y(，所以-2,f(4(=-1,f(5(=1,f(6(=2，所以f(1)+f(2)+f(3)+f(4)+f(5)+f(6)=0，所以f(k(=f(1(+f(2(+f(3(+f(4(=1-1-2-1=-3．故選：A.16.(2021全國(guó)甲文)設(shè)f(x)是定義域?yàn)镽的奇則f(-x)=-f(x)，則f(x)=-f(-x)=-f(6+x)=f(x+12)，則f(x)的最小正周期是12，故f(-16)=f(-4)=-f(4)=-f(2)=-(-2)=2.19.定義在R上的奇函數(shù)f(x)滿足f(2-x)=f(x)，且當(dāng)-1≤x<0時(shí)，f(x)=2x-1，則f(log220)=【解析】解析：B依題意，知f(2+x)=f(-x)=-f(x)，則f(4+x)=f(x)，所以f(x)是周期函數(shù)，且周期為4.又2<log25<3，則-1<2-log25<0，所以f(log220)=f(2+log25)=f(log25-2)=+f(1)+f(2)+?+f(2022)的值為21.已知f(x)是定義域?yàn)?-∞,+∞)的奇函數(shù)，且滿足f(1-x)=f(1+x).若f(1)=2，則f(1)+f(2)+f(3)+?+f(50)=【解析】(1)法一∵f(x)在R上是奇函數(shù)，且f(1-x)=f(1+x).∴f(x+1)=-f(x-1)，即f(x+2)=-f(x).因此f(x+4)=f(x)，則函數(shù)f(x)是周期為4的函數(shù)，由于f(1-x)=f(1+x)，f(1)=2，故令x=1，得f(0)=f(2)=0，令x=2，得f(3)=f(-1)=-f(1)=-2，令x=3，得f(4)=f(-2)=-f(2)=0，故f(1)+f(2)+f(3)+f(4)=2+0-2+0=0，∴f(1)+f(2)+f(3)+?+f(50)=12×0+f(1)+f(2)=2.期為4，∴f(1)+f(2)+f(3)+?+f(50)=12[f(1)+f(2)+f(3)+f(4)]+f(49)+f(50)=12×0+f(1)+f(2)=2.22.已知定義在R上的函數(shù)f(x)滿足①f(x+2)=f(x),②f(x-2)為奇函數(shù),③當(dāng)x∈[0,1)時(shí),恒成立,則f(-的大小關(guān)系正確的是答案f(-A．B．f(-1(=0C．f(2(=0D．f(4(=0【詳解】因?yàn)楹瘮?shù)f(x+2(為偶函數(shù)，則f(2+x(=f(2-x(，可得f(x+3(=f(1-x(，因?yàn)楹瘮?shù)f(2x+1(為奇函數(shù)，則f(1-2x(=-f(2x+1(，所以，f(1-x(=-f(x+1(，所以，f(x+3(=-f(x+1(=f(x-1(，即f(x(=f(x+4(，因?yàn)楹瘮?shù)F(x(=f(2x+1(為奇函數(shù)，則F(0(=f(1(=0，故f(-1(=-f(1(=0，其它三個(gè)選項(xiàng)未知.24.(2021·全國(guó)甲卷·高考真題)設(shè)函數(shù)f(x(的定義域?yàn)镽，ff=ax2+b．若f(0(+f(3(=6，則f(【分析】通過f(x+1(是奇函數(shù)和f(x+2(是偶函數(shù)條件，可以確定出函數(shù)解析式f(x(=-2x2+因?yàn)閒(x+1(是奇函數(shù)，所以f(-x+1(=-f(x+1(①;因?yàn)閒(x+2(是偶函數(shù)，所以f(x+2(=f(-x+2(②.因?yàn)閒(0(+f(3(=6，所以-(4a+b(+a+b=6?a=-2，f(-=f(-+1(=-f+1(=-f所以(=-f(因?yàn)閒(x+1(是奇函數(shù)，所以f(-x+1(=-f(x+1(①;因?yàn)閒(x+2(是偶函數(shù)，所以f(x+2(=f(-x+2(②.因?yàn)閒(0(+f(3(=6，所以-(4a+b(+a+b=6?a=-2，所以(=f((=-f(A．B．f(x+7)為奇函數(shù)A.曲線y=f(x)關(guān)于直線x=1軸對(duì)稱B.f(x)是以4為周期的周期函數(shù)C.f(1)+f(2)+?+f(2023)=0D.曲線y=f(x)關(guān)于點(diǎn)(3,0)對(duì)稱27.已知定義域?yàn)镽的函數(shù)y=f(x(在[0,7[上只有1和3兩個(gè)零點(diǎn)，且y=f(x+2(與y=f(x+7(都是偶f(x+2(,f(x+7(為偶函數(shù)可得f(x(關(guān)于x=2,x=7軸對(duì)稱，從而判斷出f(x(是周期函數(shù)，且f(x+7(為偶函數(shù)∴f(x+2(=f(-x+2(,f(x+7(=f(-x+7(∴f(x(關(guān)于x=2,x=7軸對(duì)稱∵f(x(關(guān)于x=2,x=7軸對(duì)稱∴f(x(=f(4-x(,f(x(=f(14-x(∵f(1(=f(6(=0f(8(=f(14-8(=f(6(=0f(3(=f(4-3(=f(1(=0f(2011(=f(1(=0,f(2013(=f(3(=0共兩個(gè)答案ABC【解析】由題設(shè)f(-x-1)=-f(x-1)，則f(x)關(guān)于(-1,0)對(duì)稱，即f(x)=-f(-x-2)，f(x+1)=f(-x+1)，則f(x)關(guān)于x=1對(duì)稱，即f(x)=f(2-x)，所以f(2-x)=-f(-x-2)，則f(2+x)=-f(x-2)，故f(x)=-f(x-4)，所以f(x-4)=-f(x-8)，即f(x)=f(x-8)，故f(x)=f(x+8)，f(x)+lgx=0的根等價(jià)于f(x)與y=-lgx交點(diǎn)橫坐標(biāo)，28.(多選)已知函數(shù)f(x)對(duì)任意x∈R都有f(x)=f(x+4)+f(2),若函數(shù)y=f(x+3)的圖象關(guān)于直線x=2-x1)(f(x2)-f(x1))>0,則下列結(jié)論正確的是()A.f(2)=0B.f(x)是奇函數(shù)C.f(x)是周期為4的周期函數(shù)D.f(3)<f(-4)29.(多選題)(2024·山東煙臺(tái)·統(tǒng)考二模)定義在R上的函數(shù)f(x)滿足f(x)+f(4+x)=0，f(2+2x)是偶函A.f(x(是奇函數(shù)B.f(2023(=-1C.f(x(的圖象關(guān)于直線x=1對(duì)稱D.=-100∴函數(shù)f(x)關(guān)于直線x=2對(duì)稱，∴f(-x(=f(4+x(，∵f(x)+f(4+x)=0，∴f(-x(=-f(x(，∴f(x(是奇函數(shù)，則A正確；對(duì)于選項(xiàng)B，∵f(4+x)=-f(x)，∴f(8+x)=-f(4+x)，∴f(8+x)=f(x),:f(x(的周期為8，:f(2023(=f(253×8-1(=f(-1(=-f(1(=-1，則B正確；對(duì)于選項(xiàng)C，若f(x(的圖象關(guān)于直線x=1對(duì)稱，則f(3(=f(-1(，但是f(-1(=-f(1(=-1，f(3(=f(1(=1，即f(3(≠f(-1(對(duì)于選項(xiàng)D，將x=代入f，得f(3(=f(1(=1，將x=1，代入f(x)+f(4+x)=0，得f(5(=-f(1(=-1，同理可知f(7(=-f(3(=-1，又“f(x(的周期為8，:f(x(正奇數(shù)項(xiàng)的周期為4，30.(多選題)(2024·山東濱州·統(tǒng)考二模)函數(shù)y=f(x(在區(qū)間(-∞,+∞(上的圖象是一條連續(xù)不斷的曲線，且滿足f(3+x(-f(3-x(+6x=0，函數(shù)f(1-2x(的圖象關(guān)于點(diǎn)(0,1(對(duì)稱，則()A．f(x(的圖象關(guān)于點(diǎn)(1,1(對(duì)稱B．8是f(x(的一個(gè)周期C．f(x(一定存在零點(diǎn)D．f(101(=-299【解析】對(duì)于A,由于f