高中數(shù)學(xué)北師大版練習(xí)第四章3-2對(duì)數(shù)函數(shù)y=log2x的圖象和性質(zhì)

3.2對(duì)數(shù)函數(shù)y=log2x的圖象和性質(zhì)水平11.函數(shù)y=logxQUOTE是對(duì)數(shù)函數(shù). ()2.函數(shù)y=log2(x+1)的定義域是(0,+∞). ()3.y=log2x2在[0,+∞)上為增函數(shù). ()4.y=loQUOTEx2在(0,+∞)上為增函數(shù). ()【解析】1.提示:×.對(duì)數(shù)函數(shù)中自變量x在真數(shù)的位置上,且x>0,所以錯(cuò)誤.2.提示:×.由對(duì)數(shù)式log2(x+1)的真數(shù)x+1>0,可得x>1,所以函數(shù)的定義域?yàn)?1,+∞),所以錯(cuò)誤.3.提示:×.函數(shù)y=log2x2的定義域?yàn)閧x|x≠0}.4.提示:×.函數(shù)y=loQUOTEx2在(0,+∞)上為減函數(shù).·題組一對(duì)數(shù)函數(shù)的概念與圖象1.函數(shù)f(x)=|log2x|的圖象是 ()【解析】選A.易知函數(shù)值恒大于等于零,同時(shí)在(0,1)上單調(diào)遞減,且此時(shí)的圖象是對(duì)數(shù)函數(shù)y=log2x的圖象關(guān)于x軸的對(duì)稱圖形,在(1,+∞)上單調(diào)遞增.2.函數(shù)f(x)=log2(1x)的圖象為 ()【解析】選A.由題中函數(shù)知,當(dāng)x=0時(shí),y=0,圖象過原點(diǎn),又依據(jù)對(duì)數(shù)函數(shù)的性質(zhì)知,此函數(shù)是減函數(shù),根據(jù)此兩點(diǎn)可得答案.觀察四個(gè)圖的不同發(fā)現(xiàn),A,C圖中的圖象過原點(diǎn),故排除B,D;剩下A和C.又由函數(shù)的單調(diào)性知,原函數(shù)是減函數(shù),排除C.3.已知函數(shù)f(x)=QUOTE則函數(shù)f(1x)的圖象是()【解析】選D.由題意得f(1x)=QUOTE=QUOTE所以函數(shù)f(1x)在(0,+∞)上是減函數(shù),在(∞,0]上是增函數(shù),由圖易知D正確.4.若對(duì)數(shù)函數(shù)f(x)的圖象過點(diǎn)(4,2),則f(8)=________.

【解析】由題意設(shè)f(x)=logax(a>0且a≠1),則f(4)=loga4=2,所以a2=4,故a=QUOTE,f(x)=loQUOTEx,所以f(8)=loQUOTE8=3.答案:3·題組二對(duì)數(shù)型函數(shù)的定義域與值域1.不等式log2x<QUOTE的解集是 ()A.QUOTE B.QUOTEC.{x|x>QUOTE} D.QUOTE【解析】選B.依題意log2x<log2QUOTE,由于y=log2x是定義域上的遞增函數(shù),故0<x<QUOTE.2.已知集合A={x|log2x<1},B={x|2x2x>0},則A∩B= ()A.QUOTE B.(0,2)C.QUOTE D.(∞,0)∪QUOTE【解析】選C.因?yàn)锳={x|log2x<1},B={x|2x2x>0},所以A={x|0<x<2},B=QUOTE,所以A∩B=QUOTE=QUOTE.3.函數(shù)f(x)=QUOTE的定義域?yàn)開_______.

【解析】由題意有QUOTE解得x>QUOTE且x≠0,則函數(shù)的定義域?yàn)镼UOTE∪(0,+∞).答案:QUOTE∪(0,+∞)4.要使函數(shù)f(x)=log(x+1)(164x)有意義,則x的取值范圍是________.

【解析】要使函數(shù)有意義,需滿足QUOTE解得1<x<0或0<x<4.故x的取值范圍為(1,0)∪(0,4).答案:(1,0)∪(0,4)·題組三對(duì)數(shù)函數(shù)的性質(zhì)問題1.已知log2a>log2b,則下列不等式一定成立的是 ()A.QUOTE>QUOTE B.log2(ab)>0C.QUOTE>1 D.QUOTE<QUOTE【解析】選D.由于y=log2x在(0,+∞)上為增函數(shù),所以a>b>0.所以QUOTE<QUOTE,A選項(xiàng)錯(cuò)誤.ab>0,故log2(ab)符號(hào)無法判斷,B選項(xiàng)錯(cuò)誤.ab>0,y=QUOTE在R上遞減,故QUOTE<QUOTE=1,C選項(xiàng)錯(cuò)誤.由于a>b>0,當(dāng)α>0時(shí),y=xα在(0,+∞)上遞增,所以QUOTE<QUOTE<QUOTE,即QUOTE<QUOTE,故D選項(xiàng)正確.2.已知f(x)是奇函數(shù),當(dāng)x>0時(shí),f(x)=log2x,則fQUOTE= ()A.2 B.1C.1 D.2【解析】選B.由函數(shù)是奇函數(shù)可得fQUOTE=fQUOTE=log2QUOTE=1.3.已知函數(shù)f(x)是定義在R上的偶函數(shù),且在[0,+∞)上單調(diào)遞減,f(2)=0,則不等式f(log2x)>0的解集為 ()A.QUOTE B.(2,2)C.QUOTE D.(4,+∞)【解析】選A.根據(jù)題意,函數(shù)f(x)是定義在R上的偶函數(shù),且f(2)=0,則f(log2x)>0?f(|log2x|)>f(2),又由f(x)在[0,+∞)上單調(diào)遞減,得f(|log2x|)>f(2)?|log2x|<2,變形可得2<log2x<2,解得QUOTE<x<4,所以不等式的解集為QUOTE.易錯(cuò)點(diǎn)一求對(duì)數(shù)不等式忽視定義域不等式loQUOTE(2x+3)<loQUOTE(5x6)的解集為 ()A.(∞,3) B.QUOTEC.QUOTE D.QUOTE【解析】選D.由題意可得QUOTE解得QUOTE<x<3.【易錯(cuò)誤區(qū)】忽略函數(shù)本身的定義域2x+3>0且5x6>0致錯(cuò).易錯(cuò)點(diǎn)二求復(fù)合函數(shù)的單調(diào)性忽視定義域函數(shù)f(x)=log2(x2x2)的單調(diào)遞減區(qū)間是________.

【解析】由x2x2>0,解得x<1或x>2.令t=x2x2,則y=log2t為增函數(shù),所以當(dāng)x<1時(shí),函數(shù)f(x)=log2(x2x2)單調(diào)遞減.答案:(∞,1)【易錯(cuò)誤區(qū)】忽略原函數(shù)的定義域驗(yàn)證條件x2x2>0,導(dǎo)致錯(cuò)把函數(shù)t=x2x2的遞減區(qū)間QUOTE當(dāng)作答案.水平1、2限時(shí)30分鐘分值60分戰(zhàn)報(bào)得分____一、選擇題(每小題5分,共30分)1.下列函數(shù)是對(duì)數(shù)函數(shù)的是 ()A.y=loga(2x) B.y=log22xC.y=log2x+1 D.y=log2x【解析】選D.選項(xiàng)A,B,C中的函數(shù)都不是“y=logax(a>0且a≠1)”的形式,只有D選項(xiàng)符合.【變式備選】函數(shù)y=1+loQUOTE(x1)的圖象恒過定點(diǎn) ()A.(1,1) B.(1,0) C.(2,1) D.(2,0)【解析】選C.令x1=1,得x=2,此時(shí)y=1,故函數(shù)y=1+loQUOTE(x1)的圖象一定經(jīng)過點(diǎn)(2,1).2.函數(shù)y=QUOTE的定義域?yàn)?()A.(∞,2) B.(2,+∞)C.(2,3)∪(3,+∞) D.(2,4)∪(4,+∞)【解析】選C.要使原函數(shù)有意義,則QUOTE解得2<x<3或x>3,所以原函數(shù)的定義域?yàn)?2,3)∪(3,+∞).3.(金榜原創(chuàng)題)下列各式中錯(cuò)誤的是 ()A.30.8>30.7 B.log0.50.4>log0.50.60.1<0.750.1 D.log2QUOTE>log2QUOTE【解析】選C.A.因?yàn)閥=3x在R上為增函數(shù),又因?yàn)?.8>0.7,所以30.8>30.7,故A正確;B.因?yàn)閥=log0.5x在(0,+∞)上為減函數(shù),又因?yàn)?.4<0.6,所以log0.50.4>log0.50.6,故B正確;C.因?yàn)閥=0.75x在R上為減函數(shù),又因?yàn)?.1<0.1,所以0.750.1>0.750.1,故C錯(cuò)誤;D.因?yàn)閥=log2x在(0,+∞)上為增函數(shù),又因?yàn)镼UOTE>QUOTE,所以log2QUOTE>log2QUOTE,故D正確.4.(金榜原創(chuàng)題)函數(shù)f(x)=QUOTE,則不等式f(2)<f(log2x)的解集是 ()A.(0,4) B.QUOTEC.(4,+∞) D.QUOTE∪(4,+∞)【解析】選C.因?yàn)閒(x)=QUOTE,可知f(x)的定義域?yàn)镽,所以f(x)=QUOTE=QUOTE=f(x),則f(x)為奇函數(shù),當(dāng)x>0時(shí),f(x)=QUOTE=1QUOTE單調(diào)遞增,根據(jù)奇函數(shù)的性質(zhì),可知f(x)在R上單調(diào)遞增,由f(2)<f(log2x),可得2<log2x,解得:x>4,即不等式f(2)<f(log2x)的解集為(4,+∞).【變式備選】函數(shù)y=log2(|x|+1)的圖象大致是 ()【解析】選B.y=log2(|x|+1)是偶函數(shù),當(dāng)x≥0時(shí),y=log2(x+1)是對(duì)數(shù)型函數(shù)且為增函數(shù),且過點(diǎn)(0,0),(1,1),只有選項(xiàng)B滿足.5.(練拓展)下列函數(shù)中,既是偶函數(shù),又在(∞,0)上單調(diào)遞增的是 ()A.f(x)=(x1)2 B.f(x)=log2QUOTEC.f(x)=2|x| D.f(x)=log2|x1|【解析】選B.對(duì)于選項(xiàng)A,函數(shù)f(x)不是偶函數(shù),所以不滿足題意;對(duì)于選項(xiàng)B,f(x)=log2QUOTE=log2QUOTE=f(x),所以函數(shù)是偶函數(shù),根據(jù)復(fù)合函數(shù)的單調(diào)性得函數(shù)在(∞,0)上單調(diào)遞增,所以滿足題意;對(duì)于選項(xiàng)C,函數(shù)是偶函數(shù),根據(jù)復(fù)合函數(shù)單調(diào)性得它在(∞,0)上單調(diào)遞減,所以不滿足題意;對(duì)于選項(xiàng)D,f(x)=log2|x1|=log2|x+1|≠f(x),所以D不滿足題意.6.(多選題)下列不等號(hào)連接正確的是 ()A.log0.52.2>log0.52.3 B.log34<log65C.log34>log56 D.logπe>logeπ【解析】選AC.對(duì)A,根據(jù)y=log0.5x為減函數(shù)可知正確.對(duì)B,由log34>log33=1=log55>log65可知錯(cuò)誤.對(duì)C,由log34=1+log3QUOTE>1+log3QUOTE>1+log5QUOTE=log56可知正確.對(duì)D,由π>e>1得,logeπ>1>logπe可知錯(cuò)誤.【變式備選】已知函數(shù)f(x)=QUOTE,下列結(jié)論正確的是 ()A.若f(a)=1,則a=3B.fQUOTE=2020C.若f(a)≥2,則a≤1或a≥5D.函數(shù)f(x)的值域?yàn)镽【解析】選BCD.對(duì)于A:由f(a)=1,得QUOTE或QUOTE解得a=3或a=0,故A錯(cuò)誤;對(duì)于B:fQUOTE=log2QUOTE=log2QUOTE=loQUOTE2020,因?yàn)閘oQUOTE2020<0,所以fQUOTE=f(loQUOTE2020)=QUOTE=2020,故B正確;對(duì)于C:由f(a)≥2,得QUOTE或QUOTE,解得a≥5或a≤1,故C正確;對(duì)于D:作出f(x)的圖象,如圖所示:可知D正確.二、填空題(每小題5分,共20分)7.函數(shù)y=log2QUOTE的定義域?yàn)開_______.

【解析】由題意,函數(shù)y=log2QUOTE有意義,則滿足x1>0,解得x>1,即函數(shù)y=log2QUOTE的定義域?yàn)?1,+∞).答案:(1,+∞)8.函數(shù)f(x)=QUOTE的值域?yàn)開_______.

【解析】當(dāng)x≥1時(shí),f(x)=loQUOTEx≤0;當(dāng)x<1時(shí),f(x)=2x∈(0,2),故函數(shù)的值域?yàn)?∞,2).答案:(∞,2)9.(練情境)已知p:log2(x+2)≤3,q:x22x+1m2≤0(m>0),若p是q的充分不必要條件,則實(shí)數(shù)m的取值范圍是________.

【解析】不等式log2(x+2)≤3可化為log2(x+2)≤log28,由對(duì)數(shù)函數(shù)的單調(diào)性可得0<x+2≤8,解得2<x≤6,同理不等式x22x+1m2≤0化為[x(1+m)][x(1m)]≤0,因?yàn)閙>0,所以1+m>1m,所以1m≤x≤1+m,因?yàn)閜是q的充分不必要條件,所以{x|2<x≤6}?{x|1m≤x≤1+m},所以QUOTE解得m≥5.答案:m≥510.設(shè)函數(shù)f(x)=QUOTE,若函數(shù)y=f(x)k有且只有兩個(gè)零點(diǎn),則實(shí)數(shù)k的取值范圍是________.

【解析】由題意得:當(dāng)x<1時(shí),f(x)∈QUOTE,且單調(diào)遞減;當(dāng)x≥1時(shí),f(x)∈[0,+∞),且單調(diào)遞增;所以要使y=f(x)與y=k有且只有兩個(gè)交點(diǎn),需k>QUOTE.答案:QUOTE三、解答題11.(10分)已知函數(shù)f(x)=log2(3+x)+log2(3x).(1)求f(1)的值;(2)求函數(shù)f(x)的定義域;(3)判斷函數(shù)f(x)的奇偶性,并證明.【解析】(1)令x=1,則f(1)=log24+log22=2+1=3.(2)由題意:QUOTE解得3<x<3,故定義域?yàn)?3,3).(3)函數(shù)f(x)為偶函數(shù).證明:對(duì)任意x∈(3,3),f(x)=log2(3x)+log2(3+x)=f(x),由偶函數(shù)的定義可得函數(shù)f(x)為偶函數(shù).【變式備選】已知函數(shù)f(x)=loQUOTE(a為常數(shù)).(1)若常數(shù)a<2且a≠0,求f(x)的定義域;(2)若f(x)在區(qū)