課時(shí)規(guī)范練10對(duì)數(shù)與對(duì)數(shù)函數(shù)基礎(chǔ)鞏固組1.函數(shù)y=log23(2x-1)的定義域是()A.1,2B.1,2)C.12,1D.12,12.已知函數(shù)f(x)=log2x,x>0,3-x+1,x0,則f(f(1)+flog312的值是()A.2B.3C.4D.53.(2017廣西名校聯(lián)考)已知x=ln ,y=log1332,z=-12,則()A.x<y<zB.z<x<yC.z<y<xD.y<z<x4.(2017安徽淮南一模,文9)已知e是自然對(duì)數(shù)的底數(shù),a>0,且a1,b>0,且b1,則“l(fā)oga2>logbe”是“0<a<b<1”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件5.(2017福建龍巖模擬)已知y=loga(2-ax)(a>0,且a1)在區(qū)間0,1上是減函數(shù),則a的取值范圍是()A.(0,1)B.(0,2)C.(1,2)D.2,+)6.若函數(shù)f(x)=loga(ax-3)在1,3上單調(diào)遞增,則a的取值范圍是()A.(1,+)B.(0,1)C.0,13D.(3,+)7.已知函數(shù)f(x)=ax+logax(a>0,a1)在1,2上的最大值與最小值之和為loga2+6,則a的值為()A.12B.14C.2D.48.若函數(shù)y=f(x)是函數(shù)y=ax(a>0,且a1)的反函數(shù),且f(2)=1,則f(x)=()A.log2xB.12xC.log12xD.2x-29.已知定義在R上的奇函數(shù)f(x)滿足f(x+2)=-1f(x),且在區(qū)間(0,1)內(nèi)f(x)=3x,則f(log354)=()A.32B.23C.-32D.-23導(dǎo)學(xué)號(hào)2419087010.(2017湖北荊州模擬)若函數(shù)f(x)=logax,x>2,-x2+2x-2,x2(a>0,且a1)的值域是(-,-1,則實(shí)數(shù)a的取值范圍是.11.函數(shù)f(x)=log2xlog2(2x)的最小值為.12.已知函數(shù)f(x)=loga(ax2-x+3)在1,3上是增函數(shù),則a的取值范圍是.導(dǎo)學(xué)號(hào)24190871綜合提升組13.(2017全國)若x,y,z為正數(shù),且2x=3y=5z,則()A.2x<3y<5zB.5z<2x<3yC.3y<5z<2xD.3y<2x<5z14.已知定義在R上的函數(shù)f(x)滿足f(-x)=-f(x),f(x-2)=f(x+2),且x(-1,0)時(shí),f(x)=2x+15,則f(log220)等于()A.1B.45C.-1D.-4515.若a>b>0,0<c<1,則()A.logac<logbcB.logca<logcbC.ac<bcD.ca>cb導(dǎo)學(xué)號(hào)2419087216.已知定義在R上的奇函數(shù)f(x),當(dāng)x(0,+)時(shí),f(x)=log2x,則不等式f(x)<-1的解集是.創(chuàng)新應(yīng)用組17.(2017北京,文8)根據(jù)有關(guān)資料,圍棋狀態(tài)空間復(fù)雜度的上限M約為3361,而可觀測(cè)宇宙中普通物質(zhì)的原子總數(shù)N約為1080,則下列各數(shù)中與MN最接近的是()(參考數(shù)據(jù):lg 30.48)A.1033B.1053C.1073D.1093導(dǎo)學(xué)號(hào)2419087318.(2017安徽馬鞍山一模,文10)已知函數(shù)f(x)=x-aln x,當(dāng)x>1時(shí),f(x)>0恒成立,則實(shí)數(shù)a的取值范圍是()A.(1,+)B.(-,1)C.(e,+)D.(-,e)導(dǎo)學(xué)號(hào)24190874答案：1.D由log23(2x-1)00<2x-1112<x1.2.Dlog312<0,由題意得f(f(1)+flog312=f(log21)+3-log312+1=f(0)+3log32+1=30+1+2+1=5.3.Dx=ln >1,y=log1322<log1333=12,z=-12=112,1.x>z>y.故選D.4.B解 當(dāng)a>1,0<b<1時(shí),loga2>0,logbe<0,推不出0<a<b<1,不是充分條件;當(dāng)0<a<b<1時(shí),loga2>logb2>logbe,是必要條件,故選B.5.C因?yàn)閥=loga(2-ax)(a>0,且a1)在0,1上單調(diào)遞減,u=2-ax在0,1上是減函數(shù),所以y=logau是增函數(shù),所以a>1.又2-a>0,所以1<a<2.6.Da>0,且a1,u=ax-3為增函數(shù),若函數(shù)f(x)為增函數(shù),則f(x)=logau必為增函數(shù),因此a>1.又y=ax-3在1,3上恒為正,a-3>0,即a>3,故選D.7.C顯然函數(shù)y=ax與y=logax在1,2上的單調(diào)性相同,因此函數(shù)f(x)=ax+logax在1,2上的最大值與最小值之和為f(1)+f(2)=(a+loga1)+(a2+loga2)=a+a2+loga2=loga2+6,故a+a2=6,解得a=2或a=-3(舍去).故選C.8.A由題意知f(x)=logax.f(2)=1,loga2=1.a=2.f(x)=log2x.9.C由奇函數(shù)f(x)滿足f(x+2)=-1f(x),得f(x+4)=-1f(x+2)=f(x),所以f(x)的周期為4,f(log354)=f(3+log32)=f(-1+log32)=-f(1-log32)=-31-log32=-312=-32.10.12,1當(dāng)x2時(shí),f(x)=-x2+2x-2=-(x-1)2-1,f(x)在(-,1)內(nèi)遞增,在(1,2上遞減,f(x)在(-,2上的最大值是-1.又f(x)的值域是(-,-1,當(dāng)x>2時(shí),logax-1,故0<a<1,且loga2-1,12a<1.11.-14顯然x>0,f(x)=log2xlog2(2x)=12log2xlog2(4x2)=12log2x(log24+2log2x)=log2x+(log2x)2=log2x+122-14-14.當(dāng)且僅當(dāng)x=22時(shí),有f(x)min=-14.12.0,16(1,+)令t=ax2-x+3,則原函數(shù)可化為y=f(t)=logat.當(dāng)a>1時(shí),y=logat在定義域內(nèi)單調(diào)遞增,故t=ax2-x+3在1,3上也是單調(diào)遞增,所以12a1,a-1+3>0,a>1,可得a>1;當(dāng)0<a<1時(shí),y=logat在定義域內(nèi)單調(diào)遞減,故t=ax2-x+3在1,3上也是單調(diào)遞減,所以12a3,9a-3+3>0,0<a<1,可得0<a16.故a>1或0<a16.13.D由2x=3y=5z,同時(shí)取自然對(duì)數(shù),得xln 2=yln 3=zln 5.由2x3y=2ln33ln2=ln9ln8>1,可得2x>3y;再由2x5z=2ln55ln2=ln25ln32<1,可得2x<5z;所以3y<2x<5z,故選D.14.C由f(x-2)=f(x+2),得f(x)=f(x+4).因?yàn)?<log220<5,所以f(log220)=f(log220-4)=-f(4-log220)=-flog245=-2log245+15=-1.15.B對(duì)于A,logac=1logca,logbc=1logcb.0<c<1,對(duì)數(shù)函數(shù)y=logcx在(0,+)內(nèi)為減函數(shù),若0<b<a<1,則0<logca<logcb,1logca>1logcb,即logac>logbc;若0<b<1<a,則logca<0,logcb>0,1logca<1logcb,即logac<logbc;若1<b<a,則logca<logcb<0,1logca>1logcb,即logac>logbc.故A不正確;由以上解析可知,B正確;對(duì)于C,0<c<1,冪函數(shù)y=xc在(0,+)內(nèi)為增函數(shù).a>b>0,ac>bc,故C不正確;對(duì)于D,0<c<1,指數(shù)函數(shù)y=cx在R上為減函數(shù).a>b>0,ca<cb,故D不正確.16.(-,-2)0,12由已知條件可知,當(dāng)x(-,0)時(shí),f(x)=-log2(-x).當(dāng)x(0,+)時(shí),f(x)<-1,即為log2x<-1,解得0<x<12;當(dāng)x(-,0)時(shí),f(x)<-1,即為-log2(-x)<-1,解得x<-2.所以f(x)<-1的解集為(-,-2)0,12.17.D設(shè)MN=x=33611080,兩邊取對(duì)數(shù),得lg x=lg33611080=lg 3361-lg 1080=361lg 3-8093.28,所以x1093.28,即與MN最接近的是1093.故選D.18.Df(x)=1-ax=x-ax,當(dāng)a1時(shí),f(x)0在(1,+)內(nèi)恒成立,則f(x)是單調(diào)遞增的,則f(x)>f(1)=1恒成立,a1.當(dāng)a>1時(shí),令f(x)>0,解得x>a;令f(x)<0,解得1<x<a,故f(x)在(1,a)內(nèi)單調(diào)遞減,在(a,+)內(nèi)單調(diào)遞增.所以只需f(x)min=f(a)=a-aln a>0,解得1<x<e.綜上,a<e,故選D.