新版高中數(shù)學(xué)北師大版必修1習(xí)題第三章指數(shù)函數(shù)和對(duì)數(shù)函數(shù)3.5.1

§5對(duì)數(shù)函數(shù)5.1對(duì)數(shù)函數(shù)的概念課時(shí)過關(guān)·能力提升1已知全集U=R,集合A為函數(shù)f(x)=ln(x1)的定義域,則?UA=()A.(∞,1] B.[1,+∞)C.(∞,2] D.[2,+∞)解析:函數(shù)f(x)=ln(x1)的定義域?yàn)閧x|x>1},所以?UA={x|x≤1}.答案:A2若函數(shù)y=ex的圖像與函數(shù)y=f(x)的圖像關(guān)于直線y=x對(duì)稱,則()A.f(x)=lgx B.f(x)=log2xC.f(x)=lnx D.f(x)=xe解析:易知y=f(x)是y=ex的反函數(shù),則f(x)=lnx.故選C.答案:C3已知f(x)=x3,x≤0,log2x,x>0,A.1或2 B.1 C.2 D.1或2解析:當(dāng)a≤0時(shí),f(a)=a3=1,解得a=1,1>0,故a=1舍去;當(dāng)a>0時(shí),f(a)=log2a=1,解得a=2,2>0,故a=2.答案:C4若函數(shù)y=f(x)是函數(shù)y=ax(a>0且a≠1)的反函數(shù),且f(x)的圖像經(jīng)過點(diǎn)(a,a),則f(x)=()A.log2x B.log12x C.12x解析:因?yàn)楹瘮?shù)y=f(x)是函數(shù)y=ax(a>0且a≠1)的反函數(shù),所以f(x)=logax.又因?yàn)閒(x)=logax的圖像經(jīng)過點(diǎn)(a,a),所以logaa=a?a=12,即f(x)=log1答案:B5設(shè)f(x)=logax(a>0,a≠1),對(duì)于任意的正實(shí)數(shù)x,y都有()A.f(xy)=f(x)f(y) B.f(x+y)=f(x)f(y)C.f(x+y)=f(x)+f(y) D.f(xy)=f(x)+f(y)解析:因?yàn)閒(x)=logax(a>0,a≠1),所以f(xy)=loga(xy).又f(x)+f(y)=logax+logay=loga(xy),所以f(xy)=f(x)+f(y).答案:D6對(duì)數(shù)函數(shù)f(x)滿足f(9)=2,則f(3)=()A.0 B.1 C.3 D.4解析:設(shè)對(duì)數(shù)函數(shù)為f(x)=logax,所以2=loga9.所以a=3.所以解析式為y=log3x.所以f(3)=log33=1.答案:B7已知f(x)=log3x,則f827+f38=.解析:f827+f38=log3827+=log3827×38=log3答案:28若f(x)=12x,x∈(-∞,1],log81x解析:因?yàn)楫?dāng)x≤1時(shí),f(x)=12x=14,當(dāng)x∈(1,+∞)時(shí),log81x=14,解得x=481=答案:39函數(shù)f(x)=lg(4-x)解析:由4-x>0,x-所以定義域?yàn)閧x|x<4且x≠3}.答案:{x|x<4且x≠3}★10若函數(shù)y=log2(a-1)x2+2x解由題意得,(a1)x2+2x+14>0在R上恒成立當(dāng)a=1時(shí),顯然不成立,所以a≠1,所以a-1>0,所以a的取值范圍為a>5.★11已知函數(shù)f(x)=log2(x-1)的定義域?yàn)锳,函數(shù)g(x)=12x((1)求A∩B;(2)若C={y|y≤a1},且B?C,求a的取值范圍.解(1)由題意知,x解得x≥2,120≤12x所以A={x|