2025高考數(shù)學(xué)一輪復(fù)習(xí)-2.6-對(duì)數(shù)與對(duì)數(shù)函數(shù)-專項(xiàng)訓(xùn)練【含答案】

2.6-對(duì)數(shù)與對(duì)數(shù)函數(shù)-專項(xiàng)訓(xùn)練1.函數(shù)f(x)=log12A.[0,1) B.(-∞,1)C.(1,+∞) D.[0,+∞)2.函數(shù)f(x)=2log4(1-x),則函數(shù)f(x)的大致圖象是()A BC D3.函數(shù)f(x)=log0.5(-x2+8x-15)的單調(diào)遞減區(qū)間為()A.(-∞,4) B.(4,+∞)C.(3,4) D.(4,5)4.(多選題)(2024蘇州調(diào)研)已知a>0,b>0,且滿足ba=9,a+log3b=3,則b的可能取值為()A.13 B.3 C.19 D5.已知函數(shù)f(x)=-lg(3-ax)(a≠1)在區(qū)間(0,4]上單調(diào)遞增,則實(shí)數(shù)a的取值范圍為()A.0,34 C.(0,1) D.(1,+∞)6.當(dāng)x∈(1,2)時(shí),不等式(x-1)2<logax恒成立,則實(shí)數(shù)a的取值范圍為()A.(2,3] B.[4,+∞)C.(1,2] D.[2,4)7.(多選題)已知函數(shù)f(x)=-log2x,則下列說法正確的有()A.函數(shù)f(|x|)為偶函數(shù)B.若f(a)=|f(b)|,其中a>0,b>0,a<1<b,則ab=1C.函數(shù)f(-x2+2x)在(1,3)上單調(diào)遞增D.若0<a<1,則|f(1+a)|<|f(1-a)|8.已知4x=5y=10,則12x+19.已知a=log372,b=1413,c=log1315,則a10.已知函數(shù)f(x)=ln(x+x2①f(x)的定義域?yàn)?-∞,+∞);②f(x)是奇函數(shù);③f(x)在(-∞,+∞)上單調(diào)遞增;④若實(shí)數(shù)a,b滿足f(a)+f(b-1)=0,則a+b=1.其中真命題的序號(hào)是.(寫出所有真命題的序號(hào))

11.已知函數(shù)f(x)=log2(x-4)-log2(x-2).(1)求f(x)的定義域;(2)求f(x)的值域.12.已知函數(shù)f(x)=log12(x2-ax+a)的值域?yàn)镽,且f(x)在(-3,-1)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是(A.2≤a≤0B.-12≤a≤0或a≥C.-2≤a≤0或a≥4D.0≤a≤413.已知函數(shù)f(x)=-x2+4x+a+16(a∈R),則關(guān)于x的不等式f(log2x)>f(1)的解集為()A.(2,+∞) B.(-∞,2)C.(2,6) D.(2,8)14.(多選題)已知函數(shù)f(x)=|logax|(a>0,且a≠1)的定義域?yàn)閇m,n](0<m<n),值域?yàn)閇0,1].若n-m的最小值為14,則實(shí)數(shù)a的值可以是(A.54 B.34 C.43 15.已知函數(shù)f(x)=loga(8-ax)(a>0,且a≠1),若f(x)>1在區(qū)間[1,2]上恒成立,則實(shí)數(shù)a的取值范圍為.

16.已知函數(shù)f(x)=(2log4x-2)log(1)當(dāng)x∈[1,16]時(shí),求該函數(shù)的值域;(2)求不等式f(x)>2的解集;(3)若f(x)<mlog4x在x∈[4,16]上恒成立,求實(shí)數(shù)m的取值范圍.17.若函數(shù)f(x)=loga(-3x2+4ax-1)有最小值,則實(shí)數(shù)a的取值范圍是()A.32,1 B.C.0,32 D.(3,參考答案1.A2.C3.C4.BD5.A6.C7.ABD8.19.b<a<c10.①②③④11.解(1)因?yàn)閒(x)=log2(x-4)-log2(x-2),所以x-4>0,x-2>0,解得x>4,所以f(x(2)因?yàn)閒(x)=log2(x-4)-log2(x-2)=log2x-4x-2=log2x-2-2x-2=log21-2x-2,由(1)知f(x)的定義域?yàn)?4,+∞),所以x-2>2,0<2x-2<1,0<1-2x-故f(x)的值域?yàn)?-∞,0).12.B13.D14.BC15.16.解(1)令t=log4x,x∈[1,16],則t∈[0,2],函數(shù)f(x)轉(zhuǎn)化為y=(2t-2)·t+12,t∈[0,2],則二次函數(shù)y=(2t-2)t+12在0,14上單調(diào)遞減,在14,2上單調(diào)遞增,所以當(dāng)t=14時(shí),y取到最小值-98,當(dāng)t=2時(shí)(2)由題得(2log4x-2)·log4x+12-2>0,令t=log4x,則(2t-2)t+12-2>0,即2t2-t-3>0,解得t>32或t<-1.當(dāng)t>32時(shí),log4x>32,解得x>故不等式f(x)>2的解集為x(3)由于(2log4x-2)log4x+12<mlog4x在x∈[4,16]上恒成立,令t=log4x,x∈[4,16],則t∈[1,2],即(2t-2)t+12<mt在t∈[1,2]上恒成立,所以m>2t-1t-1在t∈[1,