高中數(shù)學(xué)一輪復(fù)習(xí)課時(shí)作業(yè)梯級(jí)練六函數(shù)的奇偶性對(duì)稱性與周期性課時(shí)作業(yè)理含解析新人教A版

一輪復(fù)習(xí)精品資料(高中)PAGE1-課時(shí)作業(yè)梯級(jí)練六函數(shù)的奇偶性、對(duì)稱性與周期性一、選擇題(每小題5分,共25分)1.下列函數(shù)為奇函數(shù)的是 ()A.y=QUOTE B.y=|sinx|C.y=cosx D.y=ex-e-x〖解析〗選D.因?yàn)楹瘮?shù)y=QUOTE的定義域?yàn)椤?,+∞),不關(guān)于原點(diǎn)對(duì)稱,所以函數(shù)y=QUOTE為非奇非偶函數(shù),排除A;因?yàn)閥=|sinx|為偶函數(shù),所以排除B;因?yàn)閥=cosx為偶函數(shù),所以排除C;因?yàn)閥=f(x)=ex-e-x,f(-x)=e-x-ex=-(ex-e-x)=-f(x),所以函數(shù)y=ex-e-x為奇函數(shù).2.(2021·六盤(pán)水模擬)若函數(shù)f(x)=QUOTE為奇函數(shù),則實(shí)數(shù)a= ()A.-1 B.0 C.1 D.2〖解析〗選A.因?yàn)楹瘮?shù)f(x)為奇函數(shù),所以f(-x)+f(x)=QUOTE+QUOTE=0,化為(a+1)x=0,所以a+1=0,解得a=-1.3.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,則g(-1)+g(-2)= ()A.-13 B.-11 C.-9 D.-7〖解析〗選B.因?yàn)閤>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,所以x>0時(shí),f(x)=2x,所以x>0時(shí),g(x)=2x+x2,又g(x)是奇函數(shù),所以g(-1)+g(-2)=-〖g(1)+g(2)〗=-(2+1+4+4)=-11.〖加練備選·拔高〗x為實(shí)數(shù),〖x〗表示不超過(guò)x的最大整數(shù),則函數(shù)f(x)=x-〖x〗在R上為 ()A.奇函數(shù) B.偶函數(shù)C.增函數(shù) D.周期函數(shù)〖解析〗選D.函數(shù)f(x)=x-〖x〗在R上的圖象如圖:所以f(x)在R上是周期為1的函數(shù).4.(2021·德州模擬)已知定義在R上的奇函數(shù)f(x)滿足f(x+2)=-f(x),當(dāng)0≤x≤1時(shí),f(x)=x2,則f(2023)= ()A.20192 B.1 C.0 D.-1〖解析〗選D.根據(jù)題意,函數(shù)f(x)滿足f(x+2)=-f(x),則有f(x+4)=-f(x+2)=f(x),即函數(shù)是周期為4的周期函數(shù),則f(2023)=f(-1+2024)=f(-1),又函數(shù)y=f(x)為奇函數(shù),且x∈〖0,1〗時(shí),f(x)=x2,則f(-1)=-f(1)=-1,故f(2023)=-1.5.(2021·眉山模擬)定義在R上的偶函數(shù)f(x)滿足f(x)=f(x+2),當(dāng)x∈〖3,4〗時(shí),f(x)=2x,則下列不等式中正確的是 ()A.fQUOTE<fQUOTE B.fQUOTE>fQUOTEC.f(sin1)<f(cos1) D.fQUOTE<fQUOTE〖解析〗選C.因?yàn)閤∈〖3,4〗時(shí),f(x)=2x,且f(x)=f(x+2),故偶函數(shù)f(x)在〖3,4〗上是增函數(shù),T=2,所以偶函數(shù)f(x)在(-1,0)上是增函數(shù),所以f(x)在(0,1)上是減函數(shù),對(duì)于A,sinQUOTE<cosQUOTE,所以fQUOTE>fQUOTE,對(duì)于B,sinQUOTE>cosQUOTE,所以fQUOTE<fQUOTE,對(duì)于C,sin1>cos1,所以f(sin1)<f(cos1),對(duì)于D,cosQUOTE<sinQUOTE,所以fQUOTE>fQUOTE.二、填空題(每小題5分,共15分)6.(2020·江蘇高考)已知y=f(x)是奇函數(shù),當(dāng)x≥0時(shí),f(x)=QUOTE,則f(-8)的值是________.

〖解析〗y(tǒng)=f(x)是奇函數(shù),當(dāng)x≥0時(shí),f(x)=QUOTE,則f(-8)=-f(8)=-QUOTE=-4.〖答案〗:-47.設(shè)f(x)是定義在R上且周期為2的函數(shù),在區(qū)間〖-1,1〗上f(x)=QUOTE(a,b∈R),若fQUOTE=fQUOTE,則a+3b=________.

〖解析〗由題知,f(x)的周期為2,所以fQUOTE=fQUOTE=1-QUOTEa,fQUOTE=QUOTE,又因?yàn)閒QUOTE=fQUOTE,所以1-QUOTEa=QUOTE①,又f(-1)=f(1),所以2a+b=0②,由①②解得a=2,b=-4,所以a+3b=-10.〖答案〗:-10〖加練備選·拔高〗已知函數(shù)f(x)=QUOTE的圖象關(guān)于原點(diǎn)對(duì)稱,g(x)=ln(ex+1)-bx是偶函數(shù),則logab=______.

〖解析〗由題意得f(0)=0,所以a=2.因?yàn)間(x)為偶函數(shù),所以g(1)=g(-1),即ln(e+1)-b=lnQUOTE+b,所以b=QUOTE,所以log2QUOTE=-1.〖答案〗:-18.若定義在R上的函數(shù)fQUOTE滿足fQUOTE=-fQUOTE,fQUOTE是奇函數(shù),現(xiàn)給出下列4個(gè)論斷:①fQUOTE是周期為4的周期函數(shù);②fQUOTE的圖象關(guān)于點(diǎn)QUOTE對(duì)稱;③fQUOTE是偶函數(shù);④fQUOTE的圖象經(jīng)過(guò)點(diǎn)QUOTE;其中正確論斷的個(gè)數(shù)是________.

〖解析〗①:由fQUOTE=-fQUOTE得:fQUOTE=-fQUOTE=fQUOTE,所以函數(shù)fQUOTE的周期為4,故①正確;②:由fQUOTE是奇函數(shù),知fQUOTE的圖象關(guān)于原點(diǎn)對(duì)稱,所以函數(shù)fQUOTE的圖象關(guān)于點(diǎn)QUOTE對(duì)稱,故②正確;③:由fQUOTE是奇函數(shù),得fQUOTE=-fQUOTE,又fQUOTE=-fQUOTE,所以fQUOTE=-fQUOTE=-fQUOTE=fQUOTE=fQUOTE,所以函數(shù)fQUOTE是偶函數(shù),故③正確;④:fQUOTE=-fQUOTE=-fQUOTE,無(wú)法判斷其值,故④錯(cuò)誤.綜上,正確論斷的序號(hào)是:①②③.故正確的有3個(gè).〖答案〗:3三、解答題(每小題10分,共20分)9.已知函數(shù)f(x)=QUOTE是奇函數(shù).(1)求實(shí)數(shù)m的值.(2)若函數(shù)f(x)在區(qū)間〖-1,a-2〗上單調(diào)遞增,求實(shí)數(shù)a的取值范圍.〖解析〗(1)設(shè)x<0,則-x>0,所以f(-x)=-(-x)2+2(-x)=-x2-2x.又f(x)為奇函數(shù),所以f(-x)=-f(x).于是x<0時(shí),f(x)=x2+2x=x2+mx,所以m=2.(2)要使f(x)在〖-1,a-2〗上單調(diào)遞增,結(jié)合f(x)的圖象知QUOTE所以1<a≤3,故實(shí)數(shù)a的取值范圍是(1,3〗.10.設(shè)函數(shù)f(x)是定義在R上的奇函數(shù),對(duì)任意實(shí)數(shù)x都有fQUOTE=-fQUOTE成立.(1)證明y=f(x)是周期函數(shù),并指出其周期.(2)若f(1)=2,求f(2)+f(3)的值.(3)若g(x)=x2+ax+3,且y=|f(x)|·g(x)是偶函數(shù),求實(shí)數(shù)a的值.〖解析〗(1)由fQUOTE=-fQUOTE,且f(-x)=-f(x),知f(3+x)=fQUOTE=-fQUOTE=-f(-x)=f(x),所以y=f(x)是周期函數(shù),且T=3是其一個(gè)周期.(2)因?yàn)閒(x)為定義在R上的奇函數(shù),所以f(0)=0,且f(-1)=-f(1)=-2,又T=3是y=f(x)的一個(gè)周期,所以f(2)+f(3)=f(-1)+f(0)=-2+0=-2.(3)因?yàn)閥=|f(x)|·g(x)是偶函數(shù),且|f(-x)|=|-f(x)|=|f(x)|,所以|f(x)|為偶函數(shù).故g(x)=x2+ax+3為偶函數(shù),即g(-x)=g(x)恒成立,于是(-x)2+a(-x)+3=x2+ax+3恒成立.于是2ax=0恒成立,所以a=0.1.(5分)已知定義在R上的奇函數(shù)f(x)滿足f(x-4)=-f(x),且在區(qū)間〖0,2〗上是增函數(shù),則 ()A.f(-25)<f(11)<f(80)B.f(80)<f(11)<f(-25)C.f(11)<f(80)<f(-25)D.f(-25)<f(80)<f(11)〖解析〗選D.因?yàn)閒(x)滿足f(x-4)=-f(x),所以f(x-8)=f(x),所以函數(shù)f(x)是以8為周期的周期函數(shù),則f(-25)=f(-1),f(80)=f(0),f(11)=f(3).由f(x)是R上的奇函數(shù),且滿足f(x-4)=-f(x),得f(11)=f(3)=-f(-1)=f(1).因?yàn)閒(x)在區(qū)間〖0,2〗上是增函數(shù),f(x)在R上是奇函數(shù),所以f(x)在區(qū)間〖-2,2〗上是增函數(shù),所以f(-1)<f(0)<f(1),即f(-25)<f(80)<f(11).2.(5分)(2020·濰坊模擬)關(guān)于函數(shù)f(x)=QUOTE,下列結(jié)論不正確的是()A.圖象關(guān)于y軸對(duì)稱B.圖象關(guān)于原點(diǎn)對(duì)稱C.在QUOTE上單調(diào)遞增D.fQUOTE恒大于0〖解析〗選B.函數(shù)f(x)=QUOTE定義域?yàn)?-∞,0)∪(0,+∞),因?yàn)閒(x)=QUOTE=QUOTE·QUOTE,f(-x)=QUOTE·QUOTE=-QUOTE·QUOTE=QUOTE·QUOTE=f(x),故函數(shù)fQUOTE為偶函數(shù),所以A正確,B不正確;當(dāng)x>0時(shí),y=QUOTE>0,且y=QUOTE在QUOTE單調(diào)遞減,當(dāng)x>0時(shí),y=1+QUOTE>0,且y=1+QUOTE在QUOTE單調(diào)遞減,而f(x)=QUOTE,故fQUOTE在QUOTE單調(diào)遞減,又由fQUOTE為偶函數(shù),故fQUOTE在QUOTE上單調(diào)遞增,所以C正確;x>0時(shí),f(x)=QUOTE>0,f(x)為偶函數(shù),所以當(dāng)x<0時(shí),f(x)>0,故fQUOTE>0,D正確.3.(5分)(2021·昭通模擬)已知函數(shù)f(x)為定義在R上的奇函數(shù),且滿足f(x+4)=f(-x),又當(dāng)x∈(0,2〗時(shí),f(x)=2x,則f(0)+f(1)+…+f(2020)=()A.0 B.2 C.6 D.8〖解析〗選D.因?yàn)閒(x+4)=f(-x),且f(x)為奇函數(shù),所以f(x+4)=-f(x),所以周期T=8,所以f(0)=0,f(1)=2,f(2)=4,f(3)=f(-1+4)=f(1)=2,f(4)=f(0)=0,f(5)=f(1+4)=f(-1)=-f(1)=-2,f(6)=f(2+4)=f(-2)=-f(2)=-4,f(7)=f(3+4)=f(-3)=-f(3)=-2,所以f(0)+f(1)+…+f(7)=0+2+4+2+0-2-4-2=0,f(0)+f(1)+…+f(2020)=252〖f(0)+f(1)+…+f(7)〗+f(0)+f(1)+f(2)+f(3)+f(4)=252×0+0+2+4+2+0=8.〖加練備選·拔高〗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(x-1)的圖象關(guān)于直線x=1對(duì)稱,f(-2)=2,則f(2018)=________.

〖解析〗由函數(shù)y=f(x-1)的圖象關(guān)于直線x=1對(duì)稱可知,函數(shù)f(x)的圖象關(guān)于y軸對(duì)稱,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(2018)=f(2+252×8)=f(2)=f(-2)=2.〖答案〗:24.(10分)設(shè)fQUOTE是定義在R上的函數(shù),且對(duì)任意實(shí)數(shù)x,恒有fQUOTE=fQUOTE,fQUOTE=-fQUOTE,當(dāng)x∈QUOTE時(shí),fQUOTE=2x-x2.(1)當(dāng)x∈QUOTE時(shí),求fQUOTE的〖解析〗式;(2)計(jì)算f(0)+f(1)+f(2)+…+fQUOTE.〖解析〗(1)將fQUOTE=-fQUOTE中的x用-x代換得fQUOTE=-fQUOTE,又fQUOTE=fQUOTE,得fQUOTE=-fQUOTE,將x用x-2替換得fQUOTE=-fQUOTE,所以周期為4,由fQUOTE=-fQUOTE得函數(shù)fQUOTE的對(duì)稱中心是QUOTE,此函數(shù)是奇函數(shù),在QUOTE的〖解析〗式為fQUOTE=-fQUOTE=x2+2x,向右移4個(gè)單位得fQUOTE=QUOTE+2(x-4)=x2-6x+8.(2)f(0)=0,f(1)=1,f(2)=0,f(3)=-1,由周期是4知f(0)+f(1)+f(2)+f(3)+f(4)+…+fQUOTE=f(1)+f(2)+f(3)=0.5.(10分)設(shè)函數(shù)f(x)對(duì)任意實(shí)數(shù)x滿足f(2+x)=f(2-x),f(7+x)=f(7-x)且f(0)=0,判斷函數(shù)f(x)圖象在區(qū)間QUOTE上與x軸至少有多少個(gè)交點(diǎn).〖解析〗由題設(shè)知函數(shù)f(x)圖象關(guān)于直線x=2和x=7對(duì)稱,又由函數(shù)的性質(zhì)得f(x)是以10為周期的函數(shù).在一個(gè)周期區(qū)間QUOTE上,f(0)=0,f(4)=f(2+2)=f(2-2)=f(0)=0且f(x)不能恒為零,故f(x)圖象與x軸至少有2個(gè)交點(diǎn).而區(qū)間QUOTE有6個(gè)周期,故在閉區(qū)間QUOTE上f(x)圖象與x軸至少有13個(gè)交點(diǎn).1.已知函數(shù)y=f(x)的定義域?yàn)镽,f(x+1)為偶函數(shù),且對(duì)?x1<x2≤1,滿足QUOTE<0.若f(3)=1,則不等式fQUOTE<1的解集為 ()A.QUOTE B.(1,8)C.QUOTE∪(8,+∞) D.(-∞,1)∪(8,+∞)〖解析〗選A.因?yàn)閷?duì)?x1<x2≤1,滿足QUOTE<0,所以y=f(x)當(dāng)x≤1時(shí),是單調(diào)遞減函數(shù),又因?yàn)閒(x+1)為偶函數(shù),所以y=f(x)關(guān)于直線x=1對(duì)稱,所以函數(shù)y=f(x)當(dāng)x>1時(shí),是單調(diào)遞增函數(shù),又因?yàn)閒(3)=1,所以有f(-1)=1,當(dāng)log2x≤1,即當(dāng)0<x≤2時(shí),fQUOTE<1?fQUOTE<f(-1)?log2x>-1?x>QUOTE,所以QUOTE<x≤2;當(dāng)log2x>1,即當(dāng)x>2時(shí),fQUOTE<1?fQUOTE<f(3)?log2x<3?x<8,所以2<x<8,綜上所述:不等式fQUOTE<1的解集為QUOTE.2.對(duì)于函數(shù)y=f(x),若存在x0,使f(x0)+f(-x0)=0,則稱點(diǎn)(x0,f(x0))是曲線f(x)的“優(yōu)美點(diǎn)”.已知f(x)=QUOTE若曲線f(x)存在“優(yōu)美點(diǎn)”,則實(shí)數(shù)k的取值范圍為_(kāi)_______.

〖解析〗由“優(yōu)美點(diǎn)”的定義,可知若點(diǎn)(x0,f(x0))是曲線y=f(x)的“優(yōu)美點(diǎn)”,則點(diǎn)(-x0,-f(x0))也在曲線y=f(x)上.如圖所示作出函數(shù)y=x2+2x(x<0)的圖象,然后作出其關(guān)于原點(diǎn)對(duì)稱的圖象,此圖象對(duì)應(yīng)的函數(shù)〖解析〗式為y=-x2+2x(x>0).設(shè)過(guò)定點(diǎn)(0,2)的直線y=k1x+2與曲線y=f(x)=-x2+2x(x>0)切于點(diǎn)A(x1,f(x1)),則k1=-2x1+2=QUOTE,解得x1=QUOTE或x1=-QUOTE(舍去),所以k1=-2QUOTE+2.由圖可知,若曲線y=f(x)存在“優(yōu)美點(diǎn)”,則k≤2-2QUOTE.〖答案〗:(-∞,2-2QUOTE〗〖加練備選·拔高〗設(shè)函數(shù)f(x)是定義在R上的偶函數(shù),且對(duì)任意的x∈R恒有f(x+1)=f(x-1),已知當(dāng)x∈〖0,1〗時(shí),f(x)=2x,則有①2是函數(shù)f(x)的周期;②函數(shù)f(x)在(1,2)上是減函數(shù),在(2,3)上是增函數(shù);③函數(shù)f(x)的最大值是1,最小值是0.其中所有正確命題的序號(hào)是________.

〖解析〗在f(x+1)=f(x-1)中,令x-1=t,則有f(t+2)=f(t),因此2是函數(shù)f(x)的周期,故①正確;當(dāng)x∈〖0,1〗時(shí),f(x)=2x是增函數(shù),根據(jù)函數(shù)的奇偶性知,f(x)在〖-1,0〗上是減函數(shù),根據(jù)函數(shù)的周期性知,函數(shù)f(x)在(1,2)上是減函數(shù),在(2,3)上是增函數(shù),故②正確;由②知,f(x)在〖0,2〗上的最大值f(x)max=f(1)=2,f(x)的最小值f(x)min=f(0)=20=1且f(x)是周期為2的周期函數(shù),所以f(x)的最大值是2,最小值是1,故③錯(cuò)誤.〖答案〗:①②課時(shí)作業(yè)梯級(jí)練六函數(shù)的奇偶性、對(duì)稱性與周期性一、選擇題(每小題5分,共25分)1.下列函數(shù)為奇函數(shù)的是 ()A.y=QUOTE B.y=|sinx|C.y=cosx D.y=ex-e-x〖解析〗選D.因?yàn)楹瘮?shù)y=QUOTE的定義域?yàn)椤?,+∞),不關(guān)于原點(diǎn)對(duì)稱,所以函數(shù)y=QUOTE為非奇非偶函數(shù),排除A;因?yàn)閥=|sinx|為偶函數(shù),所以排除B;因?yàn)閥=cosx為偶函數(shù),所以排除C;因?yàn)閥=f(x)=ex-e-x,f(-x)=e-x-ex=-(ex-e-x)=-f(x),所以函數(shù)y=ex-e-x為奇函數(shù).2.(2021·六盤(pán)水模擬)若函數(shù)f(x)=QUOTE為奇函數(shù),則實(shí)數(shù)a= ()A.-1 B.0 C.1 D.2〖解析〗選A.因?yàn)楹瘮?shù)f(x)為奇函數(shù),所以f(-x)+f(x)=QUOTE+QUOTE=0,化為(a+1)x=0,所以a+1=0,解得a=-1.3.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,則g(-1)+g(-2)= ()A.-13 B.-11 C.-9 D.-7〖解析〗選B.因?yàn)閤>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,所以x>0時(shí),f(x)=2x,所以x>0時(shí),g(x)=2x+x2,又g(x)是奇函數(shù),所以g(-1)+g(-2)=-〖g(1)+g(2)〗=-(2+1+4+4)=-11.〖加練備選·拔高〗x為實(shí)數(shù),〖x〗表示不超過(guò)x的最大整數(shù),則函數(shù)f(x)=x-〖x〗在R上為 ()A.奇函數(shù) B.偶函數(shù)C.增函數(shù) D.周期函數(shù)〖解析〗選D.函數(shù)f(x)=x-〖x〗在R上的圖象如圖:所以f(x)在R上是周期為1的函數(shù).4.(2021·德州模擬)已知定義在R上的奇函數(shù)f(x)滿足f(x+2)=-f(x),當(dāng)0≤x≤1時(shí),f(x)=x2,則f(2023)= ()A.20192 B.1 C.0 D.-1〖解析〗選D.根據(jù)題意,函數(shù)f(x)滿足f(x+2)=-f(x),則有f(x+4)=-f(x+2)=f(x),即函數(shù)是周期為4的周期函數(shù),則f(2023)=f(-1+2024)=f(-1),又函數(shù)y=f(x)為奇函數(shù),且x∈〖0,1〗時(shí),f(x)=x2,則f(-1)=-f(1)=-1,故f(2023)=-1.5.(2021·眉山模擬)定義在R上的偶函數(shù)f(x)滿足f(x)=f(x+2),當(dāng)x∈〖3,4〗時(shí),f(x)=2x,則下列不等式中正確的是 ()A.fQUOTE<fQUOTE B.fQUOTE>fQUOTEC.f(sin1)<f(cos1) D.fQUOTE<fQUOTE〖解析〗選C.因?yàn)閤∈〖3,4〗時(shí),f(x)=2x,且f(x)=f(x+2),故偶函數(shù)f(x)在〖3,4〗上是增函數(shù),T=2,所以偶函數(shù)f(x)在(-1,0)上是增函數(shù),所以f(x)在(0,1)上是減函數(shù),對(duì)于A,sinQUOTE<cosQUOTE,所以fQUOTE>fQUOTE,對(duì)于B,sinQUOTE>cosQUOTE,所以fQUOTE<fQUOTE,對(duì)于C,sin1>cos1,所以f(sin1)<f(cos1),對(duì)于D,cosQUOTE<sinQUOTE,所以fQUOTE>fQUOTE.二、填空題(每小題5分,共15分)6.(2020·江蘇高考)已知y=f(x)是奇函數(shù),當(dāng)x≥0時(shí),f(x)=QUOTE,則f(-8)的值是________.

〖解析〗y(tǒng)=f(x)是奇函數(shù),當(dāng)x≥0時(shí),f(x)=QUOTE,則f(-8)=-f(8)=-QUOTE=-4.〖答案〗:-47.設(shè)f(x)是定義在R上且周期為2的函數(shù),在區(qū)間〖-1,1〗上f(x)=QUOTE(a,b∈R),若fQUOTE=fQUOTE,則a+3b=________.

〖解析〗由題知,f(x)的周期為2,所以fQUOTE=fQUOTE=1-QUOTEa,fQUOTE=QUOTE,又因?yàn)閒QUOTE=fQUOTE,所以1-QUOTEa=QUOTE①,又f(-1)=f(1),所以2a+b=0②,由①②解得a=2,b=-4,所以a+3b=-10.〖答案〗:-10〖加練備選·拔高〗已知函數(shù)f(x)=QUOTE的圖象關(guān)于原點(diǎn)對(duì)稱,g(x)=ln(ex+1)-bx是偶函數(shù),則logab=______.

〖解析〗由題意得f(0)=0,所以a=2.因?yàn)間(x)為偶函數(shù),所以g(1)=g(-1),即ln(e+1)-b=lnQUOTE+b,所以b=QUOTE,所以log2QUOTE=-1.〖答案〗:-18.若定義在R上的函數(shù)fQUOTE滿足fQUOTE=-fQUOTE,fQUOTE是奇函數(shù),現(xiàn)給出下列4個(gè)論斷:①fQUOTE是周期為4的周期函數(shù);②fQUOTE的圖象關(guān)于點(diǎn)QUOTE對(duì)稱;③fQUOTE是偶函數(shù);④fQUOTE的圖象經(jīng)過(guò)點(diǎn)QUOTE;其中正確論斷的個(gè)數(shù)是________.

〖解析〗①:由fQUOTE=-fQUOTE得:fQUOTE=-fQUOTE=fQUOTE,所以函數(shù)fQUOTE的周期為4,故①正確;②:由fQUOTE是奇函數(shù),知fQUOTE的圖象關(guān)于原點(diǎn)對(duì)稱,所以函數(shù)fQUOTE的圖象關(guān)于點(diǎn)QUOTE對(duì)稱,故②正確;③:由fQUOTE是奇函數(shù),得fQUOTE=-fQUOTE,又fQUOTE=-fQUOTE,所以fQUOTE=-fQUOTE=-fQUOTE=fQUOTE=fQUOTE,所以函數(shù)fQUOTE是偶函數(shù),故③正確;④:fQUOTE=-fQUOTE=-fQUOTE,無(wú)法判斷其值,故④錯(cuò)誤.綜上,正確論斷的序號(hào)是:①②③.故正確的有3個(gè).〖答案〗:3三、解答題(每小題10分,共20分)9.已知函數(shù)f(x)=QUOTE是奇函數(shù).(1)求實(shí)數(shù)m的值.(2)若函數(shù)f(x)在區(qū)間〖-1,a-2〗上單調(diào)遞增,求實(shí)數(shù)a的取值范圍.〖解析〗(1)設(shè)x<0,則-x>0,所以f(-x)=-(-x)2+2(-x)=-x2-2x.又f(x)為奇函數(shù),所以f(-x)=-f(x).于是x<0時(shí),f(x)=x2+2x=x2+mx,所以m=2.(2)要使f(x)在〖-1,a-2〗上單調(diào)遞增,結(jié)合f(x)的圖象知QUOTE所以1<a≤3,故實(shí)數(shù)a的取值范圍是(1,3〗.10.設(shè)函數(shù)f(x)是定義在R上的奇函數(shù),對(duì)任意實(shí)數(shù)x都有fQUOTE=-fQUOTE成立.(1)證明y=f(x)是周期函數(shù),并指出其周期.(2)若f(1)=2,求f(2)+f(3)的值.(3)若g(x)=x2+ax+3,且y=|f(x)|·g(x)是偶函數(shù),求實(shí)數(shù)a的值.〖解析〗(1)由fQUOTE=-fQUOTE,且f(-x)=-f(x),知f(3+x)=fQUOTE=-fQUOTE=-f(-x)=f(x),所以y=f(x)是周期函數(shù),且T=3是其一個(gè)周期.(2)因?yàn)閒(x)為定義在R上的奇函數(shù),所以f(0)=0,且f(-1)=-f(1)=-2,又T=3是y=f(x)的一個(gè)周期,所以f(2)+f(3)=f(-1)+f(0)=-2+0=-2.(3)因?yàn)閥=|f(x)|·g(x)是偶函數(shù),且|f(-x)|=|-f(x)|=|f(x)|,所以|f(x)|為偶函數(shù).故g(x)=x2+ax+3為偶函數(shù),即g(-x)=g(x)恒成立,于是(-x)2+a(-x)+3=x2+ax+3恒成立.于是2ax=0恒成立,所以a=0.1.(5分)已知定義在R上的奇函數(shù)f(x)滿足f(x-4)=-f(x),且在區(qū)間〖0,2〗上是增函數(shù),則 ()A.f(-25)<f(11)<f(80)B.f(80)<f(11)<f(-25)C.f(11)<f(80)<f(-25)D.f(-25)<f(80)<f(11)〖解析〗選D.因?yàn)閒(x)滿足f(x-4)=-f(x),所以f(x-8)=f(x),所以函數(shù)f(x)是以8為周期的周期函數(shù),則f(-25)=f(-1),f(80)=f(0),f(11)=f(3).由f(x)是R上的奇函數(shù),且滿足f(x-4)=-f(x),得f(11)=f(3)=-f(-1)=f(1).因?yàn)閒(x)在區(qū)間〖0,2〗上是增函數(shù),f(x)在R上是奇函數(shù),所以f(x)在區(qū)間〖-2,2〗上是增函數(shù),所以f(-1)<f(0)<f(1),即f(-25)<f(80)<f(11).2.(5分)(2020·濰坊模擬)關(guān)于函數(shù)f(x)=QUOTE,下列結(jié)論不正確的是()A.圖象關(guān)于y軸對(duì)稱B.圖象關(guān)于原點(diǎn)對(duì)稱C.在QUOTE上單調(diào)遞增D.fQUOTE恒大于0〖解析〗選B.函數(shù)f(x)=QUOTE定義域?yàn)?-∞,0)∪(0,+∞),因?yàn)閒(x)=QUOTE=QUOTE·QUOTE,f(-x)=QUOTE·QUOTE=-QUOTE·QUOTE=QUOTE·QUOTE=f(x),故函數(shù)fQUOTE為偶函數(shù),所以A正確,B不正確;當(dāng)x>0時(shí),y=QUOTE>0,且y=QUOTE在QUOTE單調(diào)遞減,當(dāng)x>0時(shí),y=1+QUOTE>0,且y=1+QUOTE在QUOTE單調(diào)遞減,而f(x)=QUOTE,故fQUOTE在QUOTE單調(diào)遞減,又由fQUOTE為偶函數(shù),故fQUOTE在QUOTE上單調(diào)遞增,所以C正確;x>0時(shí),f(x)=QUOTE>0,f(x)為偶函數(shù),所以當(dāng)x<0時(shí),f(x)>0,故fQUOTE>0,D正確.3.(5分)(2021·昭通模擬)已知函數(shù)f(x)為定義在R上的奇函數(shù),且滿足f(x+4)=f(-x),又當(dāng)x∈(0,2〗時(shí),f(x)=2x,則f(0)+f(1)+…+f(2020)=()A.0 B.2 C.6 D.8〖解析〗選D.因?yàn)閒(x+4)=f(-x),且f(x)為奇函數(shù),所以f(x+4)=-f(x),所以周期T=8,所以f(0)=0,f(1)=2,f(2)=4,f(3)=f(-1+4)=f(1)=2,f(4)=f(0)=0,f(5)=f(1+4)=f(-1)=-f(1)=-2,f(6)=f(2+4)=f(-2)=-f(2)=-4,f(7)=f(3+4)=f(-3)=-f(3)=-2,所以f(0)+f(1)+…+f(7)=0+2+4+2+0-2-4-2=0,f(0)+f(1)+…+f(2020)=252〖f(0)+f(1)+…+f(7)〗+f(0)+f(1)+f(2)+f(3)+f(4)=252×0+0+2+4+2+0=8.〖加練備選·拔高〗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(x-1)的圖象關(guān)于直線x=1對(duì)稱,f(-2)=2,則f(2018)=________.

〖解析〗由函數(shù)y=f(x-1)的圖象關(guān)于直線x=1對(duì)稱可知,函數(shù)f(x)的圖象關(guān)于y軸對(duì)稱,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(2018)=f(2+252×8)=f(2)=f(-2)=2.〖答案〗:24.(10分)設(shè)fQUOTE是定義在R上的函數(shù),且對(duì)任意實(shí)數(shù)x,恒有fQUOTE=fQUOTE,fQUOTE=-fQUOTE,當(dāng)x∈QUOTE時(shí),fQUOTE=2x-x2.(1)當(dāng)x∈QUOTE時(shí),求fQUOTE的〖解析〗式;(2)計(jì)算f(0)+f(1)+f(2)+…+fQUOTE.〖解析〗(1)將fQUOTE=-fQUOTE中的x用-x代換得fQUOTE=-fQUOTE,又fQUOTE=fQUOTE,得fQUOTE=-fQUOTE,將x用x-2替換得fQUOTE=-fQUOTE,所以周期為4,由fQUOTE=-fQUOTE得函數(shù)fQUOTE的對(duì)稱中心是QUOTE,此函數(shù)是奇函數(shù),在QUOTE的〖解析〗式為fQUOTE=-fQUOTE=x2+2x,向右移4個(gè)單位得fQUOTE=QUOTE+2(x-4)=x2-6x+8.(2)f(0)=0,f(1)=1,f(2)=0,f(3)=-1,由周期是4知f(0)+f(1)+f(2)+f(3)+f(4)+…+fQUOTE=f(1)+f(2)+f(3)=0.5.(10分)設(shè)函數(shù)f(x)對(duì)任意實(shí)數(shù)x滿足f(2+x)=f(2-x),f(7+x)=f(7-x)且f(0)=0,判斷函數(shù)f(x)圖象在區(qū)間QUOTE上與x軸至少有多少個(gè)交點(diǎn).〖解析〗由題設(shè)知函數(shù)f(x)圖象關(guān)于直線x=2和x=7對(duì)稱,又由函數(shù)的性質(zhì)得f(x)是以10為周期的函數(shù).在一個(gè)周期區(qū)間QUOTE