（福建專版）2019高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練10 對(duì)數(shù)與對(duì)數(shù)函數(shù) 文.docx

《（福建專版）2019高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練10 對(duì)數(shù)與對(duì)數(shù)函數(shù) 文.docx》由會(huì)員分享，可在線閱讀，更多相關(guān)《（福建專版）2019高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練10 對(duì)數(shù)與對(duì)數(shù)函數(shù) 文.docx（5頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

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,x0,3-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.xyzB.zxyC.zyxD.yz0,且a1,b0,且b1,則“l(fā)oga2logbe”是“0ab0,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(a0,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(a0,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,x2,-x2+2x-2,x2(a0,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(guó))若x,y,z為正數(shù),且2x=3y=5z,則()A.2x3y5zB.5z2x3yC.3y5z2xD.3y2xb0,0c1,則()A.logaclogbcB.logcalogcbC.accb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)002x-1112x1.2.Dlog3121,y=log1322zy.故選D.4.B解 當(dāng)a1,0b0,logbe0,推不出0ab1,不是充分條件;當(dāng)0ablogb2logbe,是必要條件,故選B.5.C因?yàn)閥=loga(2-ax)(a0,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a1.又2-a0,所以1a0,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a1.又y=ax-3在1,3上恒為正,a-30,即a3,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)x2時(shí),logax-1,故0a1,且loga2-1,12a0,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)a1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+30,a1,可得a1;當(dāng)0a0,0a1,可得01或01,可得2x3y;再由2x5z=2ln55ln2=ln25ln321,可得2x5z;所以3y2x5z,故選D.14.C由f(x-2)=f(x+2),得f(x)=f(x+4).因?yàn)?log2205,所以f(log220)=f(log220-4)=-f(4-log220)=-flog245=-2log245+15=-1.15.B對(duì)于A,logac=1logca,logbc=1logcb.0c1,對(duì)數(shù)函數(shù)y=logcx在(0,+)內(nèi)為減函數(shù),若0ba1,則0logca1logcb,即logaclogbc;若0b1a,則logca0,1logca1logcb,即logaclogbc;若1ba,則logcalogcb1logcb,即logaclogbc.故A不正確;由以上解析可知,B正確;對(duì)于C,0cb0,acbc,故C不正確;對(duì)于D,0cb0,cacb,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0x12;當(dāng)x(-,0)時(shí),f(x)-1,即為-log2(-x)-1,解得x-2.所以f(x)f(1)=1恒成立,a1.當(dāng)a1時(shí),令f(x)0,解得xa;令f(x)0,解得1x0,解得1xe.綜上,ae,故選D.