2 3　函數(shù)的奇偶性 周期性和對稱性-2026版53高考數(shù)學總復習A版精煉VIP

2.3函數(shù)的奇偶性、周期性和對稱性五年高考考點1函數(shù)的奇偶性1.(2024天津,4,5分,易)下列函數(shù)是偶函數(shù)的為()A.y=eC.y=e答案B2.(2023全國乙理,4,5分,易)已知f(x)=xexeax?1是偶函數(shù),則aA.-2B.-1C.1D.2答案D3.(2023新課標Ⅱ,4,5分,易)若f(x)=(x+a)·ln2x?12x+1為偶函數(shù),則aA.-1B.0C.12答案B4.(2021全國乙理,4,5分,易)設函數(shù)f(x)=1?x1+x,則下列函數(shù)中為奇函數(shù)的是(A.f(x-1)-1B.f(x-1)+1C.f(x+1)-1D.f(x+1)+1答案B5.(2021新高考Ⅱ,8,5分,中)設函數(shù)f(x)的定義域為R,且f(x+2)為偶函數(shù),f(2x+1)為奇函數(shù),則()A.f?12=0B.f(-1C.f(2)=0D.f(4)=0答案B6.(多選)(2023新課標Ⅰ,11,5分,難)已知函數(shù)f(x)的定義域為R,f(xy)=y2f(x)+x2f(y),則()A.f(0)=0B.f(1)=0C.f(x)是偶函數(shù)D.x=0為f(x)的極小值點答案ABC7.(2023全國甲,文14,理13,5分,易)若f(x)=(x-1)2+ax+sinx+π2為偶函數(shù),則a答案28.(2021新高考Ⅰ,13,5分,易)已知函數(shù)f(x)=x3(a·2x-2-x)是偶函數(shù),則a=.

答案1考點2函數(shù)的周期性1.(2021全國甲文,12,5分,中)設f(x)是定義域為R的奇函數(shù),且f(1+x)=f(-x).若f?13=13,則f5A.-5答案C2.(2021全國甲理,12,5分,難)設函數(shù)f(x)的定義域為R,f(x+1)為奇函數(shù),f(x+2)為偶函數(shù),當x∈[1,2]時,f(x)=ax2+b.若f(0)+f(3)=6,則f92=()A.-9答案D3.(2022新高考Ⅱ,8,5分,難)已知函數(shù)f(x)的定義域為R,且f(x+y)+f(x-y)=f(x)f(y),f(1)=1,則k=122f(k)=(A.-3B.-2C.0D.1答案A考點3函數(shù)的對稱性(多選)(2022新高考Ⅰ,12,5分,難)已知函數(shù)f(x)及其導函數(shù)f'(x)的定義域均為R,記g(x)=f'(x).若f32?2x,g(2+x)均為偶函數(shù),則A.f(0)=0B.g?1C.f(-1)=f(4)D.g(-1)=g(2)答案BC三年模擬基礎強化練1.(2025屆上海大學附中開學考,13)在下列函數(shù)中,值域為R的偶函數(shù)是()A.y=x13B.y=lg|C.y=ex+e-xD.y=x3cosx答案B2.(2025屆廣東八校聯(lián)考,4)已知函數(shù)f(x)=sinx(x2+2x)(x+A.-2B.-1C.0D.2答案A3.(2025屆重慶八中月考,5)下列函數(shù)的圖象不存在對稱中心的是()A.y=x3+1B.y=xC.y=e答案D4.(2025屆云南昆明摸底測試,6)函數(shù)f(x)=ln(x2+1+kx)是奇函數(shù)且在R上單調遞增,則k的取值集合為(A.{-1}B.{0}C.{1}D.{-1,1}答案C5.(2025屆大灣區(qū)第一次調研,6)已知函數(shù)f(x)的定義域為R,且f(1)=f(5),函數(shù)f(ax-1)的圖象關于直線x=2對稱,則a=()A.1B.2C.3D.4答案B6.(2025屆江蘇連云港第一次檢測,7)已知函數(shù)f(2x+1)為奇函數(shù),f(x+2)為偶函數(shù),且當x∈(0,1]時,f(x)=log2x,則f322=(A.2B.-2C.1D.-1答案A7.(多選)(2025屆江西吉安月考,9)下列函數(shù)為奇函數(shù)的是()A.f(x)=eB.h(x)=lgxC.g(x)=-12D.m(x)=ln(x2+1+答案ABD8.(多選)(2024江蘇南京二模,10)已知函數(shù)f(x)滿足f(x)f(y)=f(xy)+|x|+|y|,則()A.f(0)=1B.f(1)=-1C.f(x)是偶函數(shù)D.f(x)是奇函數(shù)答案AC9.(2025屆四川瀘州月考,13)已知函數(shù)f(x)=2022x3+2x2+3x+6x2+3,且f(a)答案-1010.(2025屆重慶南開中學月考,13)已知函數(shù)f(x)=sinx2cosx?3+1(-2025≤x≤2025)的值域為[m,M],則M+答案211.(2025屆黑龍江龍東十校開學考,15)已知函數(shù)f(2x)=4x(1)求f(x)的解析式;(2)判斷函數(shù)g(x)=|f(x)|的奇偶性,并說明理由;(3)求f(1)+f(log32)+f(log30.5)的值.解析(1)因為f(2x)=(2x所以f(x)=x2(2)g(x)是偶函數(shù).理由如下:函數(shù)g(x)=|f(x)|的定義域為(-∞,0)∪(0,+∞),關于原點對稱,且f(-x)=(?x)2+1?因此g(-x)=|f(-x)|=|-f(x)|=g(x),即g(x)為偶函數(shù).(3)log30.5=-log32,由(2)可得f(x)是奇函數(shù),因此f(log32)+f(log30.5)=f(log32)+f(-log32)=0,所以f(1)+f(log32)+f(log30.5)=f(1)=73能力拔高練1.(2024廣東深圳羅湖開學?？?4)已知函數(shù)f(x)=ln(eax+1)x?32為奇函數(shù),A.12答案D2.(2024重慶七校開學考,3)定義在R上的奇函數(shù)f(x)滿足f(x+4)=f(x),當x∈(0,2)時,f(x)=3x-1,則f(2022)+f(2023)=()A.-2023B.-1C.1D.32022答案B3.(2025屆重慶南開中學質檢,5)在同一直角坐標系內,存在一條直線l,使得函數(shù)y=f(x)與函數(shù)y=g(x)的圖象關于直線l對稱,就稱函數(shù)y=g(x)是函數(shù)y=f(x)的“軸對稱函數(shù)”.已知函數(shù)f(x)=ex(e是自然對數(shù)的底數(shù)),則下列函數(shù)不是函數(shù)y=f(x)的“軸對稱函數(shù)”的是()A.y=2-exB.y=e2-xC.y=-e-xD.y=lnx答案C4.(2025屆江蘇如東開學考,6)已知f(x)是奇函數(shù),g(x)是偶函數(shù),且f(x)+g(x)=ex,則2f(x)+4g(x)的最小值是()A.2B.23C.4D.25答案B5.(2025屆湖南長沙聯(lián)考,7)已知函數(shù)f(x)的定義域為(-∞,+∞),滿足f(x+2)是奇函數(shù),且f(1-x)=f(1+x),若f(1)=3,則f(0)+f(1)+f(2)+f(3)+…+f(2025)=()A.-100B.-3C.3D.2025答案C6.(2024湖南師大附中摸底考,8)已知函數(shù)f(x)=x2+2x+2-x,若不等式f(1-ax)<f(2+x2)對任意x∈R恒成立,則實數(shù)a的取值范圍是()A.(-23,2)B.(-2,23)C.(-23,23)D.(-2,2)答案D7.(多選)(2025屆河北部分地區(qū)開學考,7)已知函數(shù)f(x)的定義域為R,當x>0時,f(x+2)·f(x)=1,且當x≤2時,f(x)=x2-3x+4,則下列結論中一定正確的是()A.f(10)=1B.f(11)=1C.f(12)<10D.f(13)>20答案BC8.(多選)(2024山東部分學校開學摸底聯(lián)考,10)已知函數(shù)f(x)對任意x∈R都有f(x)=f(x+4)+f(2),若函數(shù)y=f(x+3)的圖象關于直線x=-3對稱,且?x1,x2∈[0,2],當x1≠x2時,都有(x2-x1)(f(x2)-f(x1))>0,則下列結論正確的是()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)答案AC創(chuàng)新風向練(概念深度理解)(多選)(2024江蘇揚州模擬,11)定義:若存在非零常數(shù)k,T,使得函數(shù)f(x)滿足f(x+T)=f(x)+k對定義域內的任意實數(shù)x恒成立,則稱函數(shù)f(x)為“k距周期函數(shù)”