新教材2023-2024學(xué)年高中數(shù)學(xué)第4章指數(shù)函數(shù)對(duì)數(shù)函數(shù)與冪函數(shù)4.2對(duì)數(shù)與對(duì)數(shù)函數(shù)4.2.3對(duì)數(shù)函數(shù)的性質(zhì)與圖象分層作業(yè)新人教B版必修第二冊(cè)

第四章4.2.3對(duì)數(shù)函數(shù)的性質(zhì)與圖象A級(jí)必備知識(shí)基礎(chǔ)練1.[探究點(diǎn)二]函數(shù)f(x)=1ln(x-A.(1,3] B.(1,2)∪(2,3]C.(1,3)∪(3,+∞) D.(-∞,3)2.[探究點(diǎn)二·2023安徽合肥階段練習(xí)]若a=ln0.4,b=0.23,c=log23,則a,b,c的大小關(guān)系正確的是()A.b<a<c B.a<c<bC.b<c<a D.a<b<c3.[探究點(diǎn)三]函數(shù)y=x+a與函數(shù)y=logax的圖象可能是()4.[探究點(diǎn)三]函數(shù)y=loga(x+2)+1(a>0且a≠1)的圖象過(guò)定點(diǎn)()A.(1,2) B.(2,1)C.(-2,1) D.(-1,1)5.[探究點(diǎn)二·2023四川成都高一]若函數(shù)f(x)=ln(e2x-2ex-a)對(duì)x∈R恒有意義,則實(shí)數(shù)a的取值范圍是()A.(-∞,+∞) B.(1,+∞)C.(-1,1) D.(-∞,-1)6.[探究點(diǎn)三](多選題)函數(shù)f(x)=loga(x+2)(0<a<1)的圖象過(guò)()A.第一象限 B.第二象限C.第三象限 D.第四象限7.[探究點(diǎn)一]若對(duì)數(shù)函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)P(8,3),則f12=8.[探究點(diǎn)二]函數(shù)f(x)=log2(3x+1)的值域?yàn)?

9.[探究點(diǎn)一、二]已知對(duì)數(shù)函數(shù)y=f(x)的圖象經(jīng)過(guò)點(diǎn)P(9,2).(1)求y=f(x)的解析式;(2)若x∈(0,1),求f(x)的取值范圍;(3)若函數(shù)y=g(x)的圖象與函數(shù)y=f(x)的圖象關(guān)于x軸對(duì)稱(chēng),求y=g(x)的解析式.B級(jí)關(guān)鍵能力提升練10.(多選題)已知函數(shù)f(x)=logax(a>0且a≠1)的圖象經(jīng)過(guò)點(diǎn)(4,2),則下列說(shuō)法正確的有()A.函數(shù)為增函數(shù)B.函數(shù)為偶函數(shù)C.若x>1,則f(x)>0D.若0<x1<x2,則f(x1)11.(多選題)已知函數(shù)f(x)=|lnx|,0<a<b,且f(a)=f(b),下列結(jié)論正確的是()A.1aB.a-2b>22C.2a+b>D.(a+1)2+(b+1)2>812.已知a=log32,b=e0.1,c=lne33,則a,b,c的大小關(guān)系是(A.a<b<c B.a<c<bC.c<b<a D.c<a<b13.函數(shù)f(x)的定義域是[-1,1],則函數(shù)f(log12x)的定義域?yàn)?4.設(shè)a=log3π,b=log23,c=log32,則a,b,c的大小關(guān)系是.

15.已知函數(shù)f(x)=log2x,x>0,3x,x≤016.作出函數(shù)y=|log2x|+2的圖象,并根據(jù)圖象寫(xiě)出函數(shù)的單調(diào)區(qū)間及值域.17.已知函數(shù)f(x)=log2(x-1)的定義域?yàn)锳,函數(shù)g(x)=1(1)求A∩B;(2)若C={y|y≤a-1},且B?C,求a的取值范圍.C級(jí)學(xué)科素養(yǎng)創(chuàng)新練18.已知函數(shù)f(x)=loga(x+1),g(x)=loga(1-x)(a>0且a≠1).(1)求函數(shù)F(x)=f(x)+g(x)的定義域;(2)試討論關(guān)于x的不等式f(x)≥g(x)的解集.

參考答案4.2.3對(duì)數(shù)函數(shù)的性質(zhì)與圖象1.B由題意知x-1>0,x-1≠1,3-x≥0,2.Da=ln0.4<ln1=0,b=0.23=0.008,c=log23>log22=1,故a<b<c.故選D.3.C因?yàn)閍為對(duì)數(shù)函數(shù)y=logax的底數(shù),所以a>0且a≠1.同時(shí)a為直線y=x+a在y軸上的截距,故排除A,D.當(dāng)a>1時(shí),y=logax為增函數(shù),y=x+a在y軸上的截距大于1,故排除B.故選C.4.D令x+2=1,得x=-1,此時(shí)y=1.5.D由題意得,e2x-2ex-a>0對(duì)x∈R恒成立,即a<e2x-2ex=ex(ex-2)恒成立,令y=ex(ex-2),當(dāng)且僅當(dāng)ex=1,即x=0時(shí),有最小值-1,故a<-1,即實(shí)數(shù)a的取值范圍為(-∞,-1).故選D.6.BCD因?yàn)?<a<1,所以函數(shù)y=logax單調(diào)遞減,在y軸右側(cè),過(guò)定點(diǎn)(1,0).函數(shù)f(x)=loga(x+2)的圖象是把y=logax的圖象向左平移2個(gè)單位長(zhǎng)度,所以圖象過(guò)第二、三、四象限.7.-1設(shè)f(x)=logax(a>0且a≠1),則loga8=3,∴a3=8,∴a=2.∴f(x)=log2x,故f12=log212=-8.(0,+∞)∵3x>0,∴3x+1>1,∴l(xiāng)og2(3x+1)>0,∴f(x)的值域?yàn)?0,+∞).9.解(1)設(shè)f(x)=logax(a>0且a≠1).由題意,f(9)=loga9=2,故a2=9,解得a=3或a=-3.又因?yàn)閍>0,所以a=3.故f(x)=log3x.(2)因?yàn)?>1,所以當(dāng)x∈(0,1)時(shí),f(x)<0,即f(x)的取值范圍為(-∞,0).(3)因?yàn)楹瘮?shù)y=g(x)的圖象與函數(shù)y=log3x的圖象關(guān)于x軸對(duì)稱(chēng),所以g(x)=log110.ACD由題意知2=loga4,解得a=2,故f(x)=log2x.對(duì)A,函數(shù)為增函數(shù)正確;對(duì)B,f(x)=log2x不為偶函數(shù);對(duì)C,當(dāng)x>1時(shí),f(x)=log2x>log21=0成立;對(duì)D,∵f(x1)+f(x2)2=log2x1x2,fx1∴f(x111.CD對(duì)于A,由題意得a<1<b,且-lna=lnb,則ab=1,故1a=b,故A錯(cuò)誤對(duì)于B,a-2b=a-2a,而0<a<1,故a-2a<-1,故B對(duì)于C,2a+b=3b>3,故C正確對(duì)于D,(a+1)2+(b+1)2=a2+b2+2(a+b)+2>2ab+4ab+2=8,故D正確.故選CD.12.Ba=log32<log33=12,b=e0.1>e0=1,c=lne33=13.12,2由題得-1≤log12x≤1,所以log122≤log12x≤log1212,所以12≤x≤2,所以函數(shù)f14.a>b>c由題意,a=log3π>log33=1,12=log33>log32=c,12=log22<b=log23<log22∴c<12<b<1<a.∴15.(0,1]函數(shù)f(x)的圖象如圖所示,要使直線y=a與f(x)的圖象有兩個(gè)不同的交點(diǎn),則0<a≤1.16.解先作出函數(shù)y=log2x的圖象,如圖1.再將y=log2x在x軸下方的圖象關(guān)于x軸對(duì)稱(chēng)翻折到x軸上方(原來(lái)在x軸上方的圖象不變),得函數(shù)y=|log2x|的圖象,如圖2;然后將y=|log2x|的圖象向上平移2個(gè)單位長(zhǎng)度,得函數(shù)y=|log2x|+2的圖象,如圖3.由圖3得函數(shù)y=|log2x|+2的單調(diào)遞增區(qū)間是[1,+∞),單調(diào)遞減區(qū)間是(0,1],值域是[2,+∞).17.解(1)由題意知,x-1>0,∴A={x|x≥2}.易知B={y|1≤y≤2},∴A∩B={2}.(2)由(1)知B={y|1≤y≤2},若要使B?C,則有a-1≥2.所以a≥3.即a的取值范圍為[3,+∞).18.解(1)由題意可得x+1>0,1-x故函數(shù)