2020版高考數(shù)學(xué)復(fù)習(xí) 第二單元 第9講 對(duì)數(shù)與對(duì)數(shù)函數(shù)練習(xí) 文（含解析）新人教A版

第9講對(duì)數(shù)與對(duì)數(shù)函數(shù)1.函數(shù)y=loga(3x-2)(a>0且a1)的圖像經(jīng)過定點(diǎn)A,則點(diǎn)A的坐標(biāo)是()A.0,23B.23,0C.(1,0)D.(0,1)2.設(shè)a=log123,b=130,3,c=ln,則a,b,c的大小關(guān)系為()A.a<b<cB.a<c<bC.c<a<bD.b<a<c3.2018·菏澤一模 若log12m<log12n,則下列不等式一定成立的是()A.14m<13nB.1m>1nC.ln(m-n)>0D.3m-n<14.2018·廣州三模 若函數(shù)f(x)=a+log2x在區(qū)間1,a上的最大值為6,則a=. 5.2018·江門一模 設(shè)x表示不超過x的最大整數(shù),如=3,-3.2=-4,則lg1+lg2+lg3+lg100=. 6.已知函數(shù)f(x)=log2x,x>0,2x,x0,則ff12的值是()A.2B.-2C.22D.-227.2018·福州模擬 已知函數(shù)f(x)=log2x+a,x>0,4x-2-1,x0,若f(a)=3,則f(a-2)=()A.-1516B.3C.-6364或3D.-1516或38.已知為銳角,且logasin>logbsin>0,則a和b的大小關(guān)系為()A.a>b>1B.b>a>1C.0<a<b<1D.0<b<a<19.已知函數(shù)y=log13x的定義域?yàn)閍,b,值域?yàn)?,1,則b-a的取值范圍為()A.(0,3B.13,3C.0,83D.23,8310.已知x1=log132,x2=2-12,13x3=log3x3,則()A.x1<x2<x3B.x1<x3<x2C.x2<x1<x3D.x3<x1<x211.2018·烏魯木齊模擬 已知f(x)=ln(x2+1),g(x)=12x-m,若對(duì)于任意x10,2,存在x20,2,使得f(x1)g(x2),則實(shí)數(shù)m的取值范圍是()A.14,+B.-,14C.12,+D.-,1212.已知x>0,y>0,x+y2=2,則log2x+2log2y的最大值為. 13.若對(duì)于任意的實(shí)數(shù)x0,12,都有2-2x-logax<0恒成立,則實(shí)數(shù)a的取值范圍是. 14.已知直線y=m(m>0)與函數(shù)y=|log2x|的圖像交于A(x1,y1),B(x2,y2)(x1<x2)兩點(diǎn),則下列說法中正確的是.(填序號(hào)) 0<x1<1<x2;x1x2=1;2x1+2x2<4;2x1+2x2>4.15.如果一個(gè)點(diǎn)是一個(gè)指數(shù)函數(shù)和一個(gè)對(duì)數(shù)函數(shù)的圖像的交點(diǎn),那么稱這個(gè)點(diǎn)為“好點(diǎn)”.下列四個(gè)點(diǎn)P1(1,1),P2(1,2),P312,12,P4(2,2)中,“好點(diǎn)”的個(gè)數(shù)為()A.1B.2C.3D.416.已知直線l與函數(shù)f(x)=ln(ex)-ln(1-x)的圖像交于P,Q兩點(diǎn).若點(diǎn)M12,m是線段PQ的中點(diǎn),則實(shí)數(shù)m的值為()A.2B.1C.12D.145課時(shí)作業(yè)(九)1.C解析 當(dāng)3x-2=1,即x=1時(shí),y=loga1=0,故定點(diǎn)A的坐標(biāo)是(1,0).2.A解析a=log123<log121=0,0<130.3<130=1,即0<b<1,c=ln>lne=1,a<b<c.3.A解析 因?yàn)閘og12m<log12n,所以m>n>0,由冪函數(shù)的性質(zhì)得14n<13n,由指數(shù)函數(shù)的性質(zhì)得14m<14n,因此14m<13n,故選A.4.4解析 因?yàn)閒(x)=a+log2x在區(qū)間1,a上單調(diào)遞增,所以f(x)max=f(a)=a+log2a=6,解得a=4.5.92解析lg1=0,lg10=1,lg100=2,故原式=0×9+1×90+2×1=92.6.C解析 由題意可得f12=log212=log22-12=-12,ff12=f-12=2-12=22.7.A解析 若a>0,則f(a)=log2a+a=3,解得a=2,f(a-2)=f(0)=4-2-1=-1516;若a0,則f(a)=4a-2-1=3,解得a=3,不合題意舍去.所以f(a-2)=-1516,故選A.8.D解析0<sin<1,logasin>logbsin>0,0<a<1,0<b<1,又logasin>logbsin,1logsina-1logsinb=logsinb-logsinalogsina·logsinb>0,可得logsinb>logsina,0<sin<1,a>b,故0<b<a<1,故選D.9.D解析 當(dāng)log13x=0時(shí),x=1;當(dāng)log13x=1時(shí),x=13或x=3.因?yàn)楹瘮?shù)y=log13x的定義域?yàn)閍,b,值域?yàn)?,1,所以當(dāng)a=13時(shí),b1,3;當(dāng)b=3時(shí),a13,1.所以b-a23,83,故選D.10.A解析 由題意可知x3是函數(shù)y1=13x與y2=log3x的圖像的交點(diǎn)的橫坐標(biāo),在同一平面直角坐標(biāo)系中作出函數(shù)y1=13x與y2=log3x的圖像,如圖所示,由圖可知x3>1,而x1=log132<0,0<x2=2-12<1,所以x3>x2>x1.故選A.11.A解析對(duì)于任意x10,2,存在x20,2,使得f(x1)g(x2),在0,2上,f(x)ming(x)min.f(x)=ln(x2+1)在0,2上單調(diào)遞增,f(x)min=f(0)=0.g(x)=12x-m在0,2上單調(diào)遞減,g(x)min=g(2)=122-m=14-m.由14-m0,得m14,實(shí)數(shù)m的取值范圍是14,+.故選A.12.0解析x>0,y>0,x+y2=2,xy2x+y222=1,當(dāng)且僅當(dāng)x=y2=1時(shí),等號(hào)成立,log2x+2log2y=log2x+log2y2=log2(xy2)log21=0.13.14<a<1解析 若對(duì)于任意的實(shí)數(shù)x0,12,都有2-2x-logax<0恒成立,即對(duì)于任意的實(shí)數(shù)x0,12,都有l(wèi)ogax>2-2x恒成立,即對(duì)任意的實(shí)數(shù)x0,12,y=logax的圖像恒在y=14x的圖像的上方,0<a<1,12<loga12,即a>12,a>14.綜上可得,14<a<1.14.解析 作出函數(shù)y=|log2x|的圖像,如圖所示,該函數(shù)先減后增,且在x=1處取得最小值0.在同一平面直角坐標(biāo)系中作出直線y=m,兩圖像交于A,B兩點(diǎn),由圖可知,0<x1<1<x2.y1=y2,-log2x1=log2x2,可得x1x2=1.x2-1>1-x1,x1+x2>2,2x1+2x2>22x1+x2>2×22=4.綜上可得,中說法正確,中說法正確,中說法錯(cuò)誤,中說法正確.15.B解析 設(shè)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)分別為y=ax(a>0,a1),y=logbx(b>0,b1).若P1為“好點(diǎn)”,則P1(1,1)在y=ax的圖像上,得a=1,與a>0且a1矛盾;P2(1,2)顯然不在y=logbx的圖像上;P312,12在y=ax,y=logbx的圖像上時(shí),a=14,b=14;易得P4(2,2)是“好點(diǎn)”.故選B.16.C解析f(x)=ln(ex)-ln(1-x)=12+lnx1-x,設(shè)P(x1,f(x1),Q(x2,f(x2),則f(x1)=12+lnx11-x1,f(x2)=12+lnx21-x2,依題意,得f(x1)+f(x2)=2m,x1+x2=1,所以2m=f(x1)+f(x2)=1+lnx1x2(1-x1)(1-x2)=1+lnx1x21-(x1+x2)+x1x2=1+ln1=1,可得m=12.故選C.