2024高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)規(guī)范練7函數(shù)的奇偶性與周期性文含解析新人教A版

PAGE課時(shí)規(guī)范練7函數(shù)的奇偶性與周期性基礎(chǔ)鞏固組1.函數(shù)f(x)=1x-x的圖象關(guān)于(A.y軸對(duì)稱(chēng) B.直線y=-x對(duì)稱(chēng)C.坐標(biāo)原點(diǎn)對(duì)稱(chēng) D.直線y=x對(duì)稱(chēng)2.(2024廣東湛江模擬)已知函數(shù)g(x)=f(2x)-x2為奇函數(shù),且f(2)=1,則f(-2)=()A.-2 B.-1 C.1 D.23.若函數(shù)y=f(2x-1)是偶函數(shù),則函數(shù)y=f(2x+1)的圖象的對(duì)稱(chēng)軸是()A.x=-1 B.x=0C.x=12 D.x=-4.已知定義域?yàn)镽的奇函數(shù)f(x)的圖象關(guān)于直線x=1對(duì)稱(chēng),且當(dāng)0≤x≤1時(shí),f(x)=x3,則f52=(A.-278 B.-18 C.185.已知函數(shù)y=f(x)滿意y=f(-x)和y=f(x+2)是偶函數(shù),且f(1)=π3,設(shè)F(x)=f(x)+f(-x),則F(3)=(A.π3 B.2C.π D.46.已知函數(shù)f(x)=ln(x2+1-x)+sinx-2,則f(2020)+f(-2020)=(A.2 B.0 C.-2 D.-47.已知定義在R上的函數(shù)f(x)滿意fx+32=f12-x,且當(dāng)x<1時(shí),f'(x)<0,若a=f(-log132),b=f(log32),c=f(21.5),則(A.a>c>b B.c>b>aC.a>b>c D.c>a>b8.已知函數(shù)f(x)是定義在R上的奇函數(shù),其最小正周期為4,且當(dāng)x∈-32,0時(shí),f(x)=log2(-3x+1),則f(2021)A.4 B.2 C.-2 D.log279.(2024全國(guó)2,理14)已知f(x)是奇函數(shù),且當(dāng)x<0時(shí),f(x)=-eax.若f(ln2)=8,則a=.

10.(2024山東濰坊臨朐模擬一,14)已知定義在R上的奇函數(shù)f(x)滿意f(x+4)=f(x),且當(dāng)x∈(0,2)時(shí),f(x)=x2+1,則f(7)的值為.

綜合提升組11.(2024河北衡水中學(xué)質(zhì)檢)定義在R上的奇函數(shù)f(x)滿意f(x+2)=f(-x),且當(dāng)x∈[0,1]時(shí),f(x)=2x-cosx,則下列結(jié)論正確的是()A.f20203<f20192<f(2018)B.f(2018)<f20203<f20192C.f(2018)<f20192<f20203D.f20192<f20203<f(2018)12.已知函數(shù)g(x)=f(x)+x2是奇函數(shù),當(dāng)x>0時(shí),函數(shù)f(x)的圖象與函數(shù)y=log2x的圖象關(guān)于y=x對(duì)稱(chēng),則g(-1)+g(-2)=()A.-7 B.-9 C.-11 D.-1313.已知函數(shù)f(x)=ex-1-e-x+1,則下列說(shuō)法正確的是()A.函數(shù)f(x)的最小正周期是1B.函數(shù)f(x)是單調(diào)遞減函數(shù)C.函數(shù)f(x)關(guān)于直線x=1軸對(duì)稱(chēng)D.函數(shù)f(x)關(guān)于(1,0)中心對(duì)稱(chēng)14.若定義在R上的偶函數(shù)f(x)滿意f(x)>0,f(x+2)=1f(x)對(duì)隨意x∈R恒成立,則f(2023)=15.函數(shù)y=f(x)對(duì)隨意x∈R都有f(x+2)=f(-x)成立,且函數(shù)y=f(x-1)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱(chēng),f(1)=4,則f(2020)+f(2021)+f(2022)的值為.

創(chuàng)新應(yīng)用組16.(2024全國(guó)百?gòu)?qiáng)名校聯(lián)考,理11)已知對(duì)隨意實(shí)數(shù)x,滿意f(1+x)=f(1-x),當(dāng)x∈(1,+∞)時(shí),函數(shù)f(x)=sin2x-x,設(shè)a=f-12,b=f(3),c=f(0),則a,b,c的大小關(guān)系為()A.a<b<c B.c<a<bC.b<c<a D.b<a<c17.(2024湖南常德一模,文10)已知定義在R上的函數(shù)y=f(x),對(duì)隨意x∈R,都有f(x+2)=1f(x),且當(dāng)x∈(0,4]時(shí),f'(x)>f(x)x,則6f(2017),3A.6f(2017)<3f(2018)<2f(2019) B.3f(2018)<6f(2017)<2f(2019)C.2f(2019)<3f(2018)<6f(2017) D.2f(2019)<6f(2017)<3f(2018)參考答案課時(shí)規(guī)范練7函數(shù)的奇偶性與周期性1.C∵f(-x)=-1x+x=-1x-x=-f(x),且定義域?yàn)?-∞,0)∴f(x)為奇函數(shù).∴f(x)的圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱(chēng).2.C因?yàn)間(x)為奇函數(shù),且f(2)=1,所以g(-1)=-g(1),即f(-2)-1=-f(2)+1=-1+1=0,所以f(-2)=1.3.A因?yàn)楹瘮?shù)y=f(2x-1)是偶函數(shù),所以函數(shù)y=f(2x-1)的圖象關(guān)于y軸對(duì)稱(chēng),因?yàn)楹瘮?shù)y=f(2x+1)的圖象是由函數(shù)y=f(2x-1)的圖象向左平移一個(gè)單位長(zhǎng)度得到的,故y=f(2x+1)的圖象關(guān)于x=-1對(duì)稱(chēng).4.B∵f(x)是奇函數(shù),且圖象關(guān)于x=1對(duì)稱(chēng),∴f(2-x)=f(x).又0≤x≤1時(shí),f(x)=x3,∴f52=f2-52=f-12=-f15.B由y=f(-x)和y=f(x+2)是偶函數(shù)知f(-x)=f(x),且f(x+2)=f(-x+2),則f(x+2)=f(x-2).∴f(x+4)=f(x),則y=f(x)的周期為4.所以F(3)=f(3)+f(-3)=2f(3)=2f(-1)=2f(1)=2π6.D由題意f(x)=ln(x2+1-x)+sinx-2,則f(-x)=ln(x2+1+x)-sinx-2,所以f(x)+f(-x)=ln(x2+1-x)+ln(x2+1+x)-4=ln1-4=-4,所以f(2024)+f(7.D由fx+32=f12-x,得f(x+1)=f(1-x),故直線x=1為函數(shù)f(x)圖象的一條對(duì)稱(chēng)軸.易知函數(shù)f(x)在(-∞,1)上單調(diào)遞減,故在(1,+∞)上單調(diào)遞增,a=f(-log132)=f(log32)=f(2-log32),b=f(log32)=f(log34)因?yàn)?-log32-log34=2-log38>0,所以21.5>2>2-log32>log34>1,故c>a>b.8.C因?yàn)楹瘮?shù)f(x)是定義在R上的奇函數(shù),其最小正周期為4,所以f(2024)=f(4×505+1)=f(1)=-f(-1).因?yàn)?1∈-32,0,且當(dāng)xf(x)=log2(-3x+1),所以f(-1)=log2[-3×(-1)+1]=2,所以f(2024)=-f(-1)=-2.9.-3∵ln2∈(0,1),f(ln2)=8,f(x)是奇函數(shù),∴f(-ln2)=-8.∵當(dāng)x<0時(shí),f(x)=-eax,∴f(-ln2)=-e-aln2=-8,∴e-aln2=8,∴-aln2=ln8,∴-a=3,∴a=-3.10.-2因?yàn)閒(x+4)=f(x),所以f(x)的周期為4.又因?yàn)閒(x)是奇函數(shù),所以f(7)=f(8-1)=f(-1)=-f(1),由題意f(1)=12+1=2,所以f(7)=-2,故答案為-2.11.C因?yàn)閒(x)是R上的奇函數(shù),所以f(x+2)=f(-x)=-f(x),故f(x+4)=f(x),f(x)的周期為4.因此f(2024)=f(2)=f(0),f20192=f32=f12,f20203=f43=f23.又因?yàn)楫?dāng)x∈[0,1]時(shí),f(x)=2x-cosx單調(diào)遞增,所以f(0)<f12<f23,故f(2024)<f20192<f20203.12.C∵x>0時(shí),f(x)的圖象與函數(shù)y=log2x的圖象關(guān)于y=x對(duì)稱(chēng),∴x>0時(shí),f(x)=2x,∴x>0時(shí),g(x)=2x+x2.又g(x)是奇函數(shù),∴g(-1)+g(-2)=-[g(1)+g(2)]=-(2+1+4+4)=-11.故選C.13.D函數(shù)f(x)=ex-1-e-x+1,即f(x)=ex-1-1ex-1,可令t=ex-1,由y=t-1t在(0,+∞)上遞增,t=ex-1在R上遞增,可得函數(shù)f(x)在R上為增函數(shù),則A,B均錯(cuò)由函數(shù)f(x)的圖象向左平移1個(gè)單位長(zhǎng)度,得函數(shù)的解析式為y=ex-e-x,明顯此函數(shù)為奇函數(shù),圖象關(guān)于原點(diǎn)對(duì)稱(chēng),所以函數(shù)f(x)的圖象關(guān)于(1,0)中心對(duì)稱(chēng),則C錯(cuò)誤,D正確.故選D.14.1因?yàn)閒(x)>0,f(x+2)=1f(x),所以f(x+4)=f[(x+2)+2]=1則函數(shù)f(x)的周期為4,所以f(2024)=f(506×4-1)=f(-1).因?yàn)楹瘮?shù)f(x)為偶函數(shù),所以f(2024)=f(-1)=f(1).當(dāng)x=-1時(shí),f(-1+2)=1f(-1),得f由f(x)>0,得f(1)=1,所以f(2024)=f(1)=1.15.4因?yàn)楹瘮?shù)y=f(x-1)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱(chēng),所以函數(shù)y=f(x)的圖象關(guān)于原點(diǎn)對(duì)稱(chēng),所以f(x)是R上的奇函數(shù).因?yàn)閒(x+2)=-f(x),所以f(x+4)=-f(x+2)=f(x),故f(x)的周期為4.所以f(2024)=f(505×4+1)=f(1)=4,所以f(2024)+f(2024)=f(2024)-f(2024)=0.所以f(2024)+f(2024)+f(2024)=4.16.D由f(1+x)=f(1-x)可得f(x)的圖象關(guān)于x=1對(duì)稱(chēng).當(dāng)x∈(1,+∞)時(shí),可得f'(x)=2sinxcosx-1=sin2x-1≤0,所以f(x)在(1,+∞)上單調(diào)遞減,結(jié)合對(duì)稱(chēng)性可得距離對(duì)稱(chēng)軸x=1越近,函數(shù)值越大,所以f(3)<f-12<f(0).故選D.17.A由f(x+2)=1f(x),