2025高考數(shù)學(xué)一輪復(fù)習(xí)-2.7-對數(shù)函數(shù)-專項訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-2.7-對數(shù)函數(shù)-專項訓(xùn)練【A級基礎(chǔ)鞏固】1.已知函數(shù)y=lg[x2+(k-3)x+94]的值域為R,那么實數(shù)k的取值范圍是(A.(0,6) B.[0,6)C.(-∞,0]∪[6,+∞) D.(-∞,0)∪(6,+∞)2.已知函數(shù)y=loga(x+b)(a,b為常數(shù),其中a>0且a≠1)的圖象如圖所示,則下列結(jié)論正確的是()A.a=0.5,b=2 B.a=2,b=2C.a=0.5,b=0.5 D.a=2,b=0.53.已知函數(shù)f(x)=lg(x2-4x-5)在(a,+∞)上單調(diào)遞增,則a的取值范圍是()A.(2,+∞) B.[2,+∞)C.(5,+∞) D.[5,+∞)4.設(shè)正實數(shù)a,b,c分別滿足1a=2a,1b=log2b,1c=log3A.a>b>c B.b>a>cC.c>b>a D.a>c>b5.已知函數(shù)f(x)=log2(2x-1),函數(shù)f(x)的單調(diào)遞增區(qū)間是;若x∈[1,92],則函數(shù)f(x)的值域是6.函數(shù)f(x)=ax-4+loga(x-3)-7(a>0,a≠1)的圖象必經(jīng)過定點.

7.不等式log12(x+1)-log12(x-1)<-8.已知定義在{x∈R|x≠0}上的函數(shù)f(x)具備下列性質(zhì),①f(x)是偶函數(shù),②f(x)在(0,+∞)上單調(diào)遞增,③對任意非零實數(shù)x,y都有f(xy)=f(x)+f(y),寫出符合條件的函數(shù)f(x)的一個解析式(寫一個即可).

9.已知函數(shù)f(x)=log2(ax2+4x+5).(1)若f(1)=3,求函數(shù)f(x)的單調(diào)區(qū)間;(2)是否存在實數(shù)a,使函數(shù)f(x)的最小值為0?若存在,求出a的值;若不存在,請說明理由.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級能力提升】10.設(shè)a,b,c均為正數(shù),且2a=log12a,12b=log12b,A.a<b<c B.c<b<aC.c<a<b D.b<a<c11.已知f(x)=loga(ax-2)(其中a>0且a≠1).(1)若a=2,f(x)<2,求實數(shù)x的取值范圍;(2)若x∈[4,6],f(x)的最大值大于1,求a的取值范圍.12.已知函數(shù)f(x)=3-2log2x,g(x)=log2x.(1)當(dāng)x∈[1,4]時,求函數(shù)h(x)=[f(x)+1]·g(x)的值域;(2)如果對任意的x∈[1,4],不等式f(x2)·f(x)>k·g(x)恒成立,求實數(shù)k的取值范圍.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級應(yīng)用創(chuàng)新練】13.(多選題)已知實數(shù)a,b,c滿足lna=2b=c-12A.c>b>a B.a>c>bC.c>a>b D.a>b>c14.設(shè)函數(shù)f(x)=x,(1)若a=3,則f[f(9)]=;

(2)若函數(shù)y=f(x)-2有兩個零點,則a的取值范圍是.

參考答案【A級基礎(chǔ)鞏固】1.解析:函數(shù)y=lg[x2+(k-3)x+94可得Δ=(k-3)2-9≥0,解得k∈(-∞,0]∪[6,+∞).故選C.2.解析:由圖象可得函數(shù)在定義域上單調(diào)遞增,所以a>1,排除A,C;又因為函數(shù)過點(0.5,0),所以b+0.5=1,解得b=0.5.故選D.3.解析:由x2-4x-5>0得x>5或x<-1,所以f(x)的定義域為(-∞,-1)∪(5,+∞),因為y=x2-4x-5在(5,+∞)上單調(diào)遞增,所以f(x)=lg(x2-4x-5)在(5,+∞)上單調(diào)遞增,所以a≥5.故選D.4.解析:在同一坐標(biāo)系中,作出y=2x,y=1x,y=log2x,y=log35.解析:因為函數(shù)f(x)=log2(2x-1)的定義域為(12,+∞),令t=2x-1,易知t=2x-1在(12,+∞)上單調(diào)遞增,而y=log2t在(0,+∞)上單調(diào)遞增,故函數(shù)f(x)的單調(diào)遞增區(qū)間是(因為函數(shù)f(x)=log2(2x-1)在[1,92所以f(1)≤f(x)≤f(92所以0≤f(x)≤3,故所求函數(shù)的值域為[0,3].答案:(126.解析:因為f(4)=a0+loga1-7=-6恒成立,所以f(x)的圖象必過定點(4,-6).答案:(4,-6)7.解析:因為log12(x+1)-log12(x-1)<-12可化為log即x+1x-得1<x<3+22,所以x∈(1,17+122),即原不等式的解集為(1,17+122).答案:(1,17+122)8.解析:函數(shù)f(x)=ln|x|的定義域為{x∈R|x≠0},對任意的x∈{x∈R|x≠0},f(-x)=ln|-x|=ln|x|=f(x),即函數(shù)f(x)=ln|x|為偶函數(shù),f(x)=ln|x|滿足①;當(dāng)x>0時,f(x)=lnx,則函數(shù)f(x)=ln|x|在(0,+∞)上為增函數(shù),f(x)=ln|x|滿足②;對任意的非零實數(shù)x,y,f(xy)=ln|xy|=ln|x|+ln|y|=f(x)+f(y),f(x)=ln|x|滿足③.故滿足條件的一個函數(shù)解析式為f(x)=ln|x|.答案:f(x)=ln|x|(答案不唯一)9.解:(1)因為f(1)=3,所以a+9=23,即a=-1,f(x)=log2(-x2+4x+5),由-x2+4x+5>0,x2-4x-5=(x-5)(x+1)<0,解得-1<x<5,所以函數(shù)f(x)的定義域為(-1,5),因為函數(shù)t=-x2+4x+5在(-1,2)上單調(diào)遞增,在[2,5)上單調(diào)遞減,又因為y=log2t在(0,+∞)上為增函數(shù),所以函數(shù)f(x)的單調(diào)遞增區(qū)間為(-1,2),單調(diào)遞減區(qū)間為[2,5).(2)設(shè)存在實數(shù)a,使函數(shù)f(x)的最小值為0,令h(x)=ax2+4x+5,因為函數(shù)f(x)的最小值為0,所以函數(shù)h(x)的最小值為1,所以a>0,①且20a聯(lián)立①②解得,a=1.所以存在實數(shù)a=1,使函數(shù)f(x)的最小值為0.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級能力提升】10.解析:在同一坐標(biāo)系中分別畫出y=2x,y=(12)x,y=log2x,y=logy=2x與y=log12x的交點的橫坐標(biāo)為a,y=12x與y=log111.解:(1)當(dāng)a=2時,f(x)=log2(2x-2)<2,即有l(wèi)og2(2x-2)<log24,所以2x故實數(shù)x的取值范圍是(1,3).(2)因為a>0,則x∈[4,6]時,4a-2≤ax-2≤6a-2.當(dāng)a>1時,函數(shù)f(x)的最大值f(x)max=loga(6a-2)>1,6a-2>a,解得a>1;當(dāng)0<a<1時,函數(shù)f(x)的最大值f(x)max=loga(4a-2)>1,0<4a-2<a,解得12<a<2綜上所述,a的取值范圍是(12,212.解:(1)h(x)=(4-2log2x)log2x=2-2(log2x-1)2.因為x∈[1,4],所以log2x∈[0,2],故函數(shù)h(x)的值域為[0,2].(2)由f(x2)·f(x)>k·g(x),得(3-4log2x)(3-log2x)>k·log2x,令t=log2x,因為x∈[1,4],所以t=log2x∈[0,2],所以(3-4t)(3-t)>k·t對任意t∈[0,2]恒成立,①當(dāng)t=0時,k∈R;②當(dāng)t∈(0,2]時,k<(3即k<4t+9t因為4t+9t≥12,當(dāng)且僅當(dāng)4t=9t,即t=32所以k<-3.綜上,實數(shù)k的取值范圍為(-∞,-3).INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級應(yīng)用創(chuàng)新練】13.解析:根據(jù)題意,設(shè)lna=2b=c-12=t,其中t>0,則a=et,b=log2在同一坐標(biāo)系中分別畫出函數(shù)y=ex,y=log2x,y=1x當(dāng)t=x1時,c>a>b;當(dāng)t=x2時,a>c>b;當(dāng)