2021高考數(shù)學(xué)(江蘇專(zhuān)用)一輪復(fù)習(xí)學(xué)案：第二章2.3函數(shù)的奇偶性與周期性(含解析)

1、§2.3函數(shù)的奇偶性與周期性1 .函數(shù)的奇偶性奇偶性定義圖象特點(diǎn)偶函數(shù)一般地，如果對(duì)于函數(shù)f (x)的定義域內(nèi)任個(gè)X,者B有f (x)= f (x),那么函數(shù)f (x)就叫做偶函數(shù)關(guān)于y軸對(duì)稱(chēng)奇函數(shù)一般地，如果對(duì)于函數(shù)f (x)的定義域內(nèi)任個(gè) x,者B 有f (x)= f (x),那么函數(shù)f (x)就叫做奇函數(shù)關(guān)于原點(diǎn)對(duì)稱(chēng)2 .周期性周期函數(shù)：對(duì)于函數(shù) y=f (x),如果存在一個(gè)非零常數(shù) T,使得當(dāng)x取定義域內(nèi)的任何值時(shí), 都有f (x+T)=f (x)，那么就稱(chēng)函數(shù)y=f (x)為周期函數(shù)，稱(chēng) T為這個(gè)函數(shù)的周期.(2)最小正周期：如果在周期函數(shù)f (x)的所有周期中存在一個(gè)最小

2、的正數(shù),那么這個(gè)最小正數(shù)就叫做f (x)的最小正周期.概 念 方 法 微 思 考1 .如果函數(shù)f (x)是奇函數(shù)或偶函數(shù)，則f(x)的定義域關(guān)于原點(diǎn)對(duì)稱(chēng).2 .已知函數(shù)f (x)滿足下列條件，你能否得到函數(shù)f (x)的周期？(1)f (x+a) = f (x)(aw0).1(2)f (x+a) = -(a0). x(3)f (x+a) = f (x+ b)(aw b).提示(1)T=2|a|; (2)T=2|a|; (3)T=|a-b|.3 .若f (x)對(duì)于定義域中任意x,均有f (x) = f (2a x),或f (a+x)=f (a-x),則函數(shù)f (x)關(guān)于直線x=a對(duì)稱(chēng).題組一思考辨

3、析1.判斷下列結(jié)論是否正確(請(qǐng)?jiān)诶ㄌ?hào)中打或“X”)(1)函數(shù) y=x2, xC(0, +8)是偶函數(shù).(x )(2)如果函數(shù)f (x), g(x)為定義域相同的偶函數(shù)，則 F(x) = f (x)+g(x)是偶函數(shù).(,)(3)若函數(shù)y= f (x+a)是偶函數(shù)，則函數(shù) y=f(x)關(guān)于直線x= a對(duì)稱(chēng).(V )若T是函數(shù)的一個(gè)周期，則nT(nC Z, nw0)也是函數(shù)的周期.(V )題組二教材改編2.下列函數(shù)中為奇函數(shù)的是.(填序號(hào)) f (x) = 2x4+3x2; f (x) = x3 2x; f (x) =x2 + 1 f (x) = x3+ 1.答案 3.已知函數(shù)f (x)是定義在R

4、上的奇函數(shù)，且當(dāng) x>0時(shí)，f (x)=x(1 + x),則f(1)=. 答案 2解析 f (1)=1X2=2,又f (x)為奇函數(shù)，f (-1) = -f (1) = -2.4 .設(shè)奇函數(shù)f (x)的定義域?yàn)?5,5,若當(dāng)xC 0,5時(shí)，f (x)的圖象如圖所示，則不等式 f (x)<0 的解集為.答案 (一2,0) U (2,5解析 由圖象可知，當(dāng)0<x<2時(shí)，f (x)>0;當(dāng)2<xW5時(shí)，f (x)<0,又f (x)是奇函數(shù)，.當(dāng)2<x<0 時(shí)，f (x)<0,當(dāng)一5< x<2 時(shí),f (x)>0.綜上，f

5、(x)<0 的解集為(一2,0) U (2,5.題組三易錯(cuò)自糾5 .函數(shù)f僅)=?;，：是函數(shù).(填“奇” “偶” “非奇非偶”)|x十 3|一 3答案奇1-x2>0,解析由|x+3| 3W0,得1<x<0 或 0vx<1 ,即f(x)的定義域?yàn)?一1,0) U (0,1), f (x)=lg 1x2f (-x) =lg 1 x2=-f (x),.f (x)為奇函數(shù).3116.已知定義在 R上的奇函數(shù)f (x)滿足f (x+ 3)=f(x),且當(dāng)xC 0,-時(shí)，f (x)=-x3,則f - 答案18解析 由f (x+ 3) = f (x)知函數(shù)f(x)的周期為3,

6、又函數(shù)f (x)為奇函數(shù)，所以 f f =f -1 =- f 1 = 1 3=122228.x7,若函數(shù)f (x)=為奇函數(shù)，則實(shí)數(shù) a的值為,且當(dāng)x>4時(shí)，f (x)的最大x+ 2 xa值為.答案2 13解析由f (x)為奇函數(shù)易知a= 2,1當(dāng)x>4時(shí)，f (x) =j在4, +8)上單倜遞減,4x一 x當(dāng) x= 4 時(shí)，f (x)max= . 3函數(shù)的奇偶性命題點(diǎn)1判斷函數(shù)的奇偶性例1 (2020日照模擬)判斷下列函數(shù)的奇偶性：(1)f (x)=x3+x, xC 1,4;(2)f (x)=ln2 x2 + x'1, 1 l(3)f (x)=aX1 + 2(a>。

7、，且 a1);(4)f(x) =x2+x, x<0,x2+x, x>0.解(1)f (x)=x3+x, xC 1,4的定義域不關(guān)于原點(diǎn)對(duì)稱(chēng)，f(x)既不是奇函數(shù)也不是偶函數(shù).(2)f (x)的定義域?yàn)?一2,2),f ( x) = ln2 + x-=2-xIn2x2+ x-f (x),，函數(shù)f (x)為奇函數(shù).(3).f (x)的定義域?yàn)閤xCR,且xw。, 其定義域關(guān)于原點(diǎn)對(duì)稱(chēng)，并且有1, 11, 1f (-x)=a-7 +2=匚;+2 axax 11 - ax -1 1k2=-3+2 =1 +即 f (x) = f (x), ，f (x)為奇函數(shù).顯然函數(shù)f (x)的定義域?yàn)?

8、8, 0)U(0, +8),關(guān)于原點(diǎn)對(duì)稱(chēng).當(dāng) x<0 時(shí)，一x>0,則 f ( x) = ( x)2 x= x2 x= f (x);當(dāng) x>0 時(shí)，一x<0,則 f (x) = ( x)2_x= x2 x= f (x)；綜上可知，對(duì)于定義域內(nèi)的任意 x,總有f (-x) = - f (x),，函數(shù)f (x)為奇函數(shù).命題點(diǎn)2函數(shù)奇偶性的應(yīng)用例 2 (1)(2018 全國(guó)出)已知函數(shù) f (x)=ln(>j1+x2-x)+ 1, f (a) = 4,則 f (-a) =答案 2解析 f (x)+ f ( x)= ln(，1 + x2 x) + 1 + ln(-1 +

9、 x2 + x) + 1 = ln(1 + x2 x2) + 2 = 2,.f (a)+f(a)=2, .f (-a) = -2.(2)已知函數(shù) f (x) = asin x+ b/x + 4,若 f (Ig 3) = 3,則 f Ig ；3答案 5解析 由 f (lg 3) =asin(lg 3) +b§jg3 + 4 = 3 得 asin(lg 3) +b§Jg3 = 1,而 f lg 1 =f (lg 3) 3=asin(lg 3) b3/jg3 + 4=- asin(lg 3) + b/ig3 + 4= 1 + 4=5.命題點(diǎn)3函數(shù)的對(duì)稱(chēng)性例3 已知函數(shù)f (x)的

10、定義域?yàn)镽,當(dāng)xC 2,2時(shí)，f (x)單調(diào)遞減，且函數(shù)y=f (x+2)為偶函數(shù)，則下列結(jié)論正確的是()A. f(7)<f (3)<f (V2)B. f(7)<f h/2)<f (3)C. f (>/2)<f (3)<f (力D. f (V2)<f(Ti)<f (3)答案 C解析 y= f (x+2)為偶函數(shù)，.f (-x+2)=f (x + 2), f c3)=f (i), f (m=f (4一步,0<4-兀 <1影,當(dāng)xC2,2時(shí)，f (x)單調(diào)遞減， f (4-1>f (1)>f h/2), .f (2)&l

11、t;f (3)<f (t),故選 C.思維升華(1)定義域關(guān)于原點(diǎn)對(duì)稱(chēng)是函數(shù)具有奇偶性的必要條件.(2)利用函數(shù)的奇偶性可畫(huà)出函數(shù)在另一對(duì)稱(chēng)區(qū)間上的圖象，確定函數(shù)在另一區(qū)間上的解析式，解決某些求值或參數(shù)問(wèn)題.由函數(shù)奇偶性延伸可得到一些對(duì)稱(chēng)性結(jié)論，如函數(shù)f (x+ a)為偶函數(shù)(奇函數(shù))，則y=f (x)的圖象關(guān)于直線x= a對(duì)稱(chēng)(關(guān)于點(diǎn)(a,0)對(duì)稱(chēng)).跟蹤訓(xùn)練1 (1)(2019黃岡模擬)下列函數(shù)中，既不是奇函數(shù)也不是偶函數(shù)的是()A . f (x) = x+ sin 2xB. f (x)=x2cos x71C. f (x)=3x-TxD . f(x)=x2+tan x3答案 D解析

12、對(duì)于選項(xiàng) A ,函數(shù)的定義域?yàn)?R, f(x)=x+sin 2(x)=(x+ sin 2x) = f (x),所 以f(x) = x+ sin 2x為奇函數(shù)；對(duì)于選項(xiàng) B,函數(shù)的定義域?yàn)?R , f (-x)= (-x)2- cos(-x)=x2 -cos x= f (x),所以f (x) =x2- cos x為偶函數(shù)；對(duì)于選項(xiàng) C,函數(shù)的定義域?yàn)?R , f (- x)=3x-1 = -f (x),所以f (x)=3x /為奇函數(shù)；只有f (x) = x2+tan x既不是奇函數(shù)也不是偶函數(shù).故選 D.(2)設(shè)f (x)=ex+ e x, g(x)=ex-e x, f (x), g(x)的定

13、義域均為 R,下列結(jié)論錯(cuò)誤的是()A . g(x)|是偶函數(shù)B. f (x)g(x)是奇函數(shù)C. f (x)|g(x)|是偶函數(shù)答案 DD . f (x)+g(x)是奇函數(shù)f (x)為偶函數(shù).解析 f ( x)= e x+ ex= f (x) g( - x) = e x- ex= g(x), g(x)為奇函數(shù).|g(x)|= |一g(x)|= g(x)|, |g(x)|為偶函數(shù)，A 正確；f (-x)g(-x) = f (x)-g(x) = -f (x)g(x),所以f (x)g(x)為奇函數(shù)，B正確；f (-x)|g(-x)|=f (x)|g(x)|,所以f (x)|g(x)|是偶函數(shù)，C正

14、確；f (x) + g(x) = 2ex,f (x) + g( x) = 2e xw f (x) + g(x),所以f (x) + g(x)不是奇函數(shù)，D錯(cuò)誤，故選D.(3)設(shè)函數(shù)f (x)在1,+0°)上為增函數(shù)，f (3)= 0,且g(x)= f (x+1)為偶函數(shù)，則不等式g(2-2x)<0 的解集為.答案 (0,2)解析 由已知g(x)在0, +8)上為增函數(shù)，g(2)=0,又g(x)為偶函數(shù)，.g(2-2x)<0 可化為 g(2-2x)<g(2),|2-2x|<2,2<2x 2<2,解得 0<x<2.函數(shù)的周期性4x2 + 2

15、, 1W x<0,1.設(shè)f (x)是定義在R上的周期為2的函數(shù)，當(dāng)xC 1,1)時(shí)，f (x) =x, 0< x<1,則 f 3 =.答案1311-斛析 f 2 =f 2= - 4x 2 2+2=1.12.已知定義在 R上的函數(shù)f (x)滿足f (2) = 2J3,且對(duì)任意的x都有f(x+ 2) = TV，則 xf (2 020) =.答案 2 一 十解析 由f (x+ 2) = ,得f (x+ 4)= fVo=f (x),所以函數(shù)f (x)的周期為4,所以 f xfx+ 2f (2 020) = f (4).因?yàn)?f (2 + 2)=f",所以 f «)

16、 = 一六=2j3=2 V3.故 f (2 020) =2- - 3.3- (2019石家莊模擬)已知f (x)是定義在R上的奇函數(shù)，且滿足 f (x) = f (2-x),當(dāng)xC 0,1時(shí),5f (x) = 4x-1,則 f 2 =.答案 151解析 因?yàn)?f (x) = f (2-x),所以 f 2 =f 2 ,又f (x)是定義在R上的奇函數(shù)，所以 f5=f_1=_f1222 -因?yàn)楫?dāng) xC 0,1時(shí),f (x)=4x1,i所以 f 1 =42-1 = 1,則 f 5 = 1.4.設(shè)定義在R上的函數(shù)f(x)同時(shí)滿足以下條件： f (x) + f(x)=0;f (x)=f (x+2);當(dāng)一

17、.135o<x<1 時(shí)，f (x) = 2x1,則 f 2+f (1) + f 2+f (2) + f 2=.答案V2-1解析依題意知：函數(shù)f (x)為奇函數(shù)且周期為2,則 f (1) + f (1)=0, f (-1)=f (1),即 f (1)=0.1 35f2+f (1)+f2+f (2) + f211 .1=f2+o+f2+f (0) + f21 11=f 2 -f 2 +f (0) + f 21=f 1 +f (0)= 22 - 1 + 20- 1 = a/2-1.思維升華 利用函數(shù)的周期性，可將其他區(qū)間上的求值、求零點(diǎn)個(gè)數(shù)、求解析式等問(wèn)題，轉(zhuǎn)化到已知區(qū)間上，進(jìn)而解決問(wèn)題

18、.函數(shù)性質(zhì)的綜合應(yīng)用命題點(diǎn)1函數(shù)的奇偶性與單調(diào)性相結(jié)合例4 (2017全國(guó)I改編)函數(shù)f (x)在( 8, +oo )上單調(diào)遞減，且為奇函數(shù).若 f(1) = 1,則 滿足一1 W f (x 2) W 1的x的取值范圍是.答案1,3解析 因?yàn)楹瘮?shù)f (x)在(一00, + 8)上單調(diào)遞減，且f(1)= 1,所以f ( 1)= f (1)= 1 ,由 一1 w f (x 2)w 1)得一 1Wx2W 1 , 所以 1 w xW3.命題點(diǎn)2函數(shù)的奇偶性與周期性相結(jié)合例5 設(shè)f (x)是定義在R上周期為 4的奇函數(shù)，若在區(qū)間2,0) U (0,2上，f (x) =則 f (2 019) =ax+ b

19、, 2 < x<0, ax1, 0<x< 2,答案12解析 設(shè)0<xw 2,則一2Wx<0,f (x)= ax+b.因?yàn)閒 (x)是定義在R上周期為4的奇函數(shù),所以 f (x)= f (x) = - ax+ 1 = ax+ b,所以 b= 1.而 f ( 2)= f (2),所以一2a+ 1 = 2a- 1,111斛得 a = 2，所以 f (2 019) = f (1) = 1 * 2+1 =2.命題點(diǎn)3函數(shù)的奇偶性與對(duì)稱(chēng)性相結(jié)合例6已知定義在R上的函數(shù)f (x),對(duì)任意實(shí)數(shù)x有f (x+4) = f (x),若函數(shù)f(x1)的圖象 關(guān)于直線 x= 1 對(duì)

20、稱(chēng)，f (2)=2,則 f (2 018) =.答案 2解析 由函數(shù)y=f(x 1)的圖象關(guān)于直線x= 1對(duì)稱(chēng)可知，函數(shù)f (x)的圖象關(guān)于y軸對(duì)稱(chēng)，故f (x)為偶函數(shù).由 f (x+4) = f (x),得 f (x+4 + 4)=f (x+ 4) = f (x),所以 f (x)是周期 T= 8 的偶函數(shù)，所以 f (2 018) = f (2+252X 8) = f (2)=2.命題點(diǎn)4函數(shù)的周期性與對(duì)稱(chēng)性相結(jié)合例7 已知f (x)的定義域?yàn)镽,其函數(shù)圖象關(guān)于 x=- 1對(duì)稱(chēng)，且f (x+4) = f (x2).若當(dāng)xC -4, 1時(shí)，f (x)=6 x,則 f (919) =.答案

21、216解析 由 f (x+ 4) = f (x-2),得 f (x+ 6)=f (x).故f (x)是周期為6的函數(shù).所以 f (919)= f (6X 153+1)=f (1).因?yàn)閒 (x)的圖象關(guān)于x=- 1對(duì)稱(chēng)，所以f (1) = f (-3).又 xC 4, 1時(shí),f (x)=6 x,所以 f (3)= 6 ( 3)=216.從而 f (1) = 216,故 f (919)=216.思維升華函數(shù)的奇偶性、對(duì)稱(chēng)性、周期性和單調(diào)性是函數(shù)的四大性質(zhì)，在高考中常常將它們綜合在一起命題，解題時(shí)，往往需要借助函數(shù)的奇偶性、對(duì)稱(chēng)性和周期性來(lái)確定另一區(qū)間 上的單調(diào)性，即實(shí)現(xiàn)區(qū)間的轉(zhuǎn)換，再利用單調(diào)性解

22、決相關(guān)問(wèn)題.跟蹤訓(xùn)練2 (1)定義在R上的函數(shù)f (x)滿足f (x) = f (x),且f (x)=f (x+6),當(dāng)xC 0,3時(shí)，f (x) 單調(diào)遞增，則f (x)在下列哪個(gè)區(qū)間上單調(diào)遞減()A. 3,7 B. 4,5 C. 5,8 D. 6,10答案 B解析 依題意知，f (x)是偶函數(shù)，且是以6為周期的周期函數(shù).因?yàn)楫?dāng) xC 0,3時(shí)，f (x)單調(diào)遞 增，所以f (x)在3,0上單調(diào)遞減.根據(jù)函數(shù)周期性知，函數(shù) f (x)在3,6上單調(diào)遞減.又因?yàn)?4,5? 3,6,所以函數(shù)f (x)在4,5上單調(diào)遞減.(2)(2018全國(guó)n )已知f(x)是定義域?yàn)?8, +oo )的奇函數(shù)，滿

23、足 f (1-x) = f (1 + x).若f =2,則 f (1) + f (2)+f (3) + f (50)等于()A. 50 B. 0 C. 2 D. 50答案 C解析.f(x)是奇函數(shù)，.f (-x)=-f (x),.f (1-x)=- f (x-1). 1. f (1-x)=f (1+x),-f (x-1)=f (x+1), .-.f (x+ 2)=- f (x),.f (x+4)= f (x + 2) = f (x) = f (x),.函數(shù)f (x)是周期為4的周期函數(shù).由f (x)為奇函數(shù)且定義域?yàn)镽得f (0) = 0,又.f (1x) = f (1 + x),.f (x)

24、的圖象關(guān)于直線x=1對(duì)稱(chēng)，.f (2)=f (0)=0, .f (-2) = 0.又 f(1) = 2, .f (-1)=- 2,.f (1)+f (2) + f (3) + f (4) = f (1)+f (2)+f ( 1) + f (0) = 2+ 02+0=0,.f (1) +f (2) + f (3) + f (4) + + f (49) + f (50)=0X 12+f (49)+f (50) = f (1)+f (2) = 2+0=2.故選C.(3)(多選)已知函數(shù)y = f (x)是R上的奇函數(shù)，對(duì)任意 xCR,者B有f (2-x)=f (x) + f (2)成立，當(dāng)f Xi f x2_ 一一 一 一Xi , X2 0,1,且X1WX2時(shí)，都有XX >0,則下列結(jié)論正確的有 ()A . f (1) + f (2) + f (3)+ - + f (2 020) = 0B.直線x=5是函數(shù)y=f (x)圖象的一條對(duì)稱(chēng)軸C.函