高考數(shù)學(xué)一輪復(fù)習(xí)第二篇函數(shù)及其應(yīng)用（）第6節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)習(xí)題理（含解析）.docx

第6節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)【選題明細(xì)表】知識(shí)點(diǎn)、方法題號(hào)對(duì)數(shù)的運(yùn)算2,6,9,13對(duì)數(shù)函數(shù)的圖象3,12對(duì)數(shù)函數(shù)的性質(zhì)1,5,8,10綜合應(yīng)用4,7,11,14基礎(chǔ)鞏固(時(shí)間:30分鐘)1.(2018鄭州質(zhì)檢)函數(shù)f(x)=log2(x2+2x-3)的定義域是(D)(A)-3,1(B)(-3,1)(C)(-,-31,+)(D)(-,-3)(1,+)解析:使函數(shù)f(x)有意義需滿足x2+2x-30,解得x1或x0,且a1)的圖象大致為(A)解析:若0a2,選項(xiàng)C,D不滿足.當(dāng)a1時(shí),由2-ax=0,得x=0,a1)的定義域和值域都是0,1,則loga+loga等于(C)(A)1(B)2(C)3(D)4解析:由題意可得a-ax0,axa,定義域?yàn)?,1,所以a1,y=在定義域0,1上單調(diào)遞減,由值域0,1,所以f(0)=1,f(1)=0,所以a=2,所以loga+loga=log2+log2=log28=3,故選C.5.已知a,b0且a1,b1,若logab1,則(D)(A)(a-1)(b-1)0(C)(b-1)(b-a)0解析:因?yàn)閍0,b0且a1,b1,由logab1得loga0,所以a1,且1或0a1且0a1或0ba0.6.lg +2lg 2-()-1=.解析:lg +2lg 2-()-1=lg +lg 22-2=lg(4)-2=1-2=-1.答案:-17.(2018昆明診斷)設(shè)f(x)=lg(+a)是奇函數(shù),則使f(x)0的x的取值范圍是.解析:由f(x)是奇函數(shù)可得a=-1,所以f(x)=lg ,定義域?yàn)?-1,1).由f(x)0,可得01,所以-1x0在區(qū)間(-,-2上恒成立且函數(shù)y=x2-ax-3a在(-,-2上遞減,則-2且(-2)2-(-2)a-3a0,解得實(shí)數(shù)a的取值范圍是-4,4).答案:-4,4)能力提升(時(shí)間:15分鐘)9.設(shè)2a=5b=m,且+=2,則m等于(A)(A) (B)10 (C)20 (D)100解析:由已知,得a=log2m,b=log5m,則+=+=logm2+logm5=logm10=2,解得m=.10.(2018衡水中學(xué)模擬)設(shè)a=log54-log52,b=ln+ln 3,c=1,則a,b,c的大小關(guān)系為(A)(A)abc(B)bca(C)cab(D)balog2 e1,所以01,即0ab1,故ab0且a1)是R上的奇函數(shù),則不等式f(x)aln a的解集是(C)(A)(a,+)(B)(-,a)(C)當(dāng)a1時(shí),解集是(a,+),當(dāng)0a1時(shí),解集是(-,a),當(dāng)0aaln axln aaln a.當(dāng)a1時(shí),xa;當(dāng)0a1時(shí),x0)的兩個(gè)根x1,x2(x13(B)x1x22(C)x1x2=1 (D)1x1+x22解析:在同一坐標(biāo)系中作出y=|log2x|與y=a的大致圖象,由圖可知0x11,所以log2x1=-a,log2x2=a,即log2x1+log2x2=0,log2(x1x2)=0,故x1x2=1.故選C.13.已知函數(shù)f(x)=若f(2-a)=1,則f(a)=.解析:由題設(shè)若2-a0時(shí),f(2-a)=-log2(1+a)=1,解得a=-,不合題意;當(dāng)2-a2,即a0時(shí),f(2-a)=2-a-1=1,即2-a=2a=-1,符合題意.所以f(a)=f(-1)=-log24=-2.答案:-214.(2018武邑中學(xué)模擬)已知函數(shù)f(x)=lg(mx2+2mx+1),若f(x)的值域?yàn)镽,則實(shí)數(shù)m的取值范圍是.解析:令g(x)=mx2+2mx+1值域?yàn)锳,因?yàn)楹瘮?shù)f(x