.4.2對(duì)數(shù)函數(shù)的圖象和性質(zhì)基礎(chǔ)過(guò)關(guān)練題組一對(duì)數(shù)(型)函數(shù)的圖象1.函數(shù)y=f(x)的圖象如圖所示,則f(x)的解析式可能是f(x)=()A.2-2x,x>0B.12-C.lnxD.-x2+8x-7,x>02.(多選題)若logab<0(a>0且a≠1,b>0),則函數(shù)f(x)=ax+b的大致圖象是()3.如圖所示,①②③④中不對(duì)應(yīng)函數(shù)y=log12x,y=log13x,y=log24.函數(shù)y=loga(x-1)+1(a>0,且a≠1)的圖象恒過(guò)點(diǎn)A,則點(diǎn)A的坐標(biāo)為;若點(diǎn)A在函數(shù)y=mx+n-1(m,n>0)的圖象上,則mn的最大值為.

5.(1)函數(shù)y=log2(x-1)的圖象是由函數(shù)y=log2x的圖象如何變化得到的?(2)在直角坐標(biāo)系中作出y=|log2(x-1)|的圖象;(3)設(shè)函數(shù)y=12x與y=|log2(x-1)|的圖象的兩個(gè)交點(diǎn)的橫坐標(biāo)分別為x1,x2,M=(x1-2)·(x2-2),請(qǐng)判斷M題組二對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用6.已知A={x|y=lgx},B=x1x≥12,則A∩B=()A.{x|0≤x≤2}B.{x|0<x≤2}C.{x|1≤x≤2}D.?7.設(shè)函數(shù)f(x)=|lgx|,則下列說(shuō)法正確的是()A.f(x)在(0,+∞)上單調(diào)遞增B.f(x)在(0,+∞)上單調(diào)遞減C.f(x)在[1,+∞)上單調(diào)遞增D.f(x)在(0,1]上單調(diào)遞增8.已知a=20.3,b=log32,c=log52,則()A.a>b>cB.a>c>bC.c>b>aD.b>c>a9.(易錯(cuò)題)函數(shù)f(x)=log2(x2-4x+3)的單調(diào)遞增區(qū)間是.

10.若對(duì)數(shù)函數(shù)f(x)=log(3a-1)x和函數(shù)g(x)=a-1x在區(qū)間(0,+∞)上均單調(diào)遞增,則實(shí)數(shù)a的取值范圍是11.已知a>2,函數(shù)f(x)=log4(x-2)-log4(a-x).(1)求f(x)的定義域;(2)當(dāng)a=4時(shí),求不等式f(2x-5)≤f(3)的解集.題組三對(duì)數(shù)函數(shù)的最大(小)值與值域問(wèn)題12.函數(shù)f(x)=log2(x2-2x+3)的值域?yàn)?)A.[0,+∞)B.[1,+∞)C.RD.[2,+∞)13.已知函數(shù)f(x)=(1-a)x+3,x<1,lnA.(-∞,-4]B.(-4,1)C.[-4,1)D.(0,1)14.已知函數(shù)f(x)=logax(0<a<1)在[2,4]上的最大值比最小值大2,則a的值為.

15.已知x滿足12≤x≤(1)求log2x的取值范圍;(2)求函數(shù)f(x)=log2(2x)·log2x4的最小值16.已知函數(shù)f(x)=log4(ax2+2x+3).(1)若f(1)=1,求函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)若函數(shù)f(x)的最小值是0,求實(shí)數(shù)a的值;(3)若函數(shù)f(x)的值域?yàn)镽,求實(shí)數(shù)a的取值范圍.題組四反函數(shù)17.若對(duì)數(shù)函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)(4,2),則它的反函數(shù)g(x)的解析式為()A.g(x)=2xB.g(x)=12C.g(x)=4xD.g(x)=x218.若函數(shù)f(x)與g(x)的圖象關(guān)于直線y=x對(duì)稱,且f(x+2)=xa+3,則g(x)的圖象必過(guò)定點(diǎn)()A.(4,0)B.(4,1)C.(4,2)D.(4,3)19.(多選題)已知函數(shù)f(x)=ax(a>0,且a≠1)的反函數(shù)為g(x),則()A.g(x)=logax(a>0,且a≠1),且定義域是(0,+∞)B.函數(shù)f(x)與g(x)的圖象關(guān)于直線y=x對(duì)稱C.若f(2)=14,則g22D.當(dāng)a>1時(shí),函數(shù)f(x)與g(x)的圖象的交點(diǎn)個(gè)數(shù)可能是0,1,2能力提升練題組一對(duì)數(shù)函數(shù)的圖象1.函數(shù)f(x)=x3·ln|x|的圖象大致是()2.已知實(shí)數(shù)a,b,c滿足:12a=log2a,12b=log3b,13c=logA.a<b<cB.c<a<bC.c<b<aD.b<c<a3.(多選題)已知f(x)=|lgx|,若a>b>c,且f(c)>f(a)>f(b),則()A.a>1B.b>1C.0<c<1D.0<ac<14.如圖,曲線C1,C2,C3依次為y=2log2x,y=log2x,y=klog2x的圖象,其中k為常數(shù),0<k<1,點(diǎn)A是曲線C1上位于第一象限的點(diǎn),過(guò)A分別作x軸、y軸的平行線交曲線C2于點(diǎn)B、D,過(guò)點(diǎn)B作y軸的平行線交曲線C3于點(diǎn)C,若四邊形ABCD為矩形,則k的值是.

題組二對(duì)數(shù)函數(shù)單調(diào)性的應(yīng)用5.已知a=log32,b=ln3ln4,c=23,則a,b,cA.a<b<cB.a<c<bC.c<a<bD.b<a<c6.函數(shù)f(x)=lg(x2-ax-1)在(1,+∞)上單調(diào)遞增的一個(gè)充分不必要條件是()A.a≤0B.a<2C.-1≤a<2D.-1≤a≤07.已知正數(shù)a,b滿足2a-4b=log2ba,則A.a≥2bB.a≤2bC.a>2bD.a<2b8.若函數(shù)f(x)=loga(6-ax)在[0,2]上單調(diào)遞減,則a的取值范圍是.

9.已知函數(shù)f(x)=log21?bxx+1(b≠-1)是定義在(1)求實(shí)數(shù)b的值;(2)寫出函數(shù)f(x)的單調(diào)區(qū)間;(3)若f(a)-f(1-a)≤2a-1,求實(shí)數(shù)a的取值范圍.題組三對(duì)數(shù)函數(shù)的最大(小)值與值域問(wèn)題10.函數(shù)f(x)=loga(x+1)+loga(1-x)a>0,且a≠1,x∈0,22,若f(x)max-f(x)min=1,則a的值為()A.4B.4或14C.2或111.已知f(x)=loga(x2-ax+a),a>0且a≠1,若?x0∈R,使得f(x)≥f(x0)恒成立,則實(shí)數(shù)a的取值范圍是()A.1<a<4B.0<a<4,a≠1C.0<a<1D.a≥412.已知函數(shù)f(x)=(lgx)2+algx+54(a∈R(1)當(dāng)a=1時(shí),求函數(shù)f(x)在區(qū)間110,100(2)若存在x0∈(1,+∞),使得f(x0)=a成立,求a的取值范圍.13.已知函數(shù)f(x)=ln(ax2+2ax+1)的定義域?yàn)镽.(1)求a的取值范圍;(2)若a≠0,函數(shù)f(x)在[-2,1]上的最大值與最小值的和為0,求實(shí)數(shù)a的值.題組四對(duì)數(shù)函數(shù)的綜合運(yùn)用14.設(shè)函數(shù)f(x)=ln|2x+1|-ln|2x-1|,則f(x)()A.是偶函數(shù),且在12B.是奇函數(shù),且在-1C.是偶函數(shù),且在-∞,-1D.是奇函數(shù),且在-∞,-115.(多選題)關(guān)于函數(shù)f(x)=ln(e2x+1)-x,下列說(shuō)法正確的有()A.f(x)在R上是增函數(shù)B.f(x)為偶函數(shù)C.f(x)的最小值為ln2,無(wú)最大值D.?x1,x2∈(0,+∞),都有fx1+16.已知函數(shù)f(x)=|log5x|,若f(x)<f(2-x),則x的取值范圍是.

17.已知函數(shù)f(x)=log2(x2+a-x)是定義在R上的奇函數(shù)(1)求a的值;(2)求使不等式f(x)≥1成立的x的取值范圍.

答案與分層梯度式解析4.4.2對(duì)數(shù)函數(shù)的圖象和性質(zhì)基礎(chǔ)過(guò)關(guān)練1.C2.BC6.C7.C8.A12.B13.C17.A18.D19.ABD1.C對(duì)于A,當(dāng)x∈(0,+∞)時(shí),y=2-2x<2,A錯(cuò)誤;對(duì)于B,當(dāng)x∈(0,+∞)時(shí),y=12-12x<12,B錯(cuò)誤;對(duì)于C,y=lnx的圖象經(jīng)過(guò)點(diǎn)(1,0),且在(0,+∞)上單調(diào)遞增,符合題意,C正確;對(duì)于D,y=-x2+8x-7的圖象開(kāi)口向下,顯然不符合題意,D2.BC由logab<0得,當(dāng)0<a<1時(shí),b>1,此時(shí)f(x)=ax+b>1,且f(x)單調(diào)遞減,B滿足;當(dāng)a>1時(shí),0<b<1,此時(shí)f(x)=ax+b>b,且f(x)單調(diào)遞增,C滿足.故選BC.3.答案③解析函數(shù)y=log12x,y=log13x在(0,+∞)上均單調(diào)遞減,只有函數(shù)y=log2x在(0,+∞)上單調(diào)遞增,因此排除①②;易知函數(shù)y=log12x與y=log當(dāng)x=12時(shí),y=log12x=1,當(dāng)x=13時(shí),y=log13x=1,因此函數(shù)y=log2x的圖象為④,從而③無(wú)對(duì)應(yīng)函數(shù).故答案為③.解題模板函數(shù)圖象的辨識(shí)可從以下幾個(gè)方面入手:根據(jù)函數(shù)的定義域判斷圖象的左右位置,根據(jù)函數(shù)的值域判斷圖象的上下位置;根據(jù)函數(shù)的單調(diào)性,判斷圖象的變化趨勢(shì);根據(jù)函數(shù)的奇偶性,判斷圖象的對(duì)稱性;根據(jù)函數(shù)的特征點(diǎn),排除不符合要求的圖象.4.答案(2,1);1解析函數(shù)y=loga(x-1)+1(a>0,且a≠1)中,令x-1=1,得x=2,則y=1,即點(diǎn)A(2,1).依題意得2m+n=2,又m>0,n>0,所以2=2m+n≥22mn,當(dāng)且僅當(dāng)2m=n=1時(shí)取“=”,即mn≤12,所以mn的最大值為解題模板解決函數(shù)圖象過(guò)定點(diǎn)問(wèn)題,應(yīng)從定值入手,如a0=1(a≠0),logb1=0(b>0且b≠1),由此確定定點(diǎn)坐標(biāo).5.解析(1)函數(shù)y=log2(x-1)的圖象是由函數(shù)y=log2x的圖象向右平移1個(gè)單位長(zhǎng)度得到的.(2)y=|log2(x-1)|的圖象如圖①所示.圖①圖②(3)不妨設(shè)x1<x2,在同一坐標(biāo)系中作出y=12x與y=|log2(x-1)|的圖象,如圖②所示.由圖②知1<x1<2<x2<3,所以M=(x1-2)(x2-2)<0,故M6.C∵A={x|y=lgx}={x|lgx≥0}={x|x≥1},B=x1x≥12={x|0<x≤2},∴A∩B={x|1≤x≤2}.故選C.7.Cf(x)=|lgx|=-lgx,0<x≤1,lgx,x>1,因?yàn)閥=lgx在(0,+∞)上單調(diào)遞增,所以f(x)在(0,1]8.Aa=20.3>20=1,b=log32>log33=12,且b=log32<log33=1,c=log52<log55=12,∴a>b>c,故選9.答案(3,+∞)解析對(duì)于f(x)=log2(x2-4x+3),令x2-4x+3>0,得x<1或x>3.設(shè)t=x2-4x+3,則t=(x-2)2-1,易知其在(-∞,1)上單調(diào)遞減,在(3,+∞)上單調(diào)遞增,又y=log2t在(0,+∞)上單調(diào)遞增,所以函數(shù)f(x)=log2(x2-4x+3)的單調(diào)遞增區(qū)間是(3,+∞).易錯(cuò)警示求解由對(duì)數(shù)函數(shù)復(fù)合而成的函數(shù)的單調(diào)區(qū)間時(shí),首先要考慮函數(shù)的定義域,由單調(diào)性與定義域結(jié)合求解單調(diào)區(qū)間.10.答案2解析由題意可得3a-1>1,a-1<0,解得23<a<1,11.解析(1)由題意得x-2>0,a故函數(shù)f(x)的定義域?yàn)?2,a).(2)∵a=4,∴f(x)=log4(x-2)-log4(4-x),∴f(2x-5)=log4(2x-7)-log4(9-2x),f(3)=log41-log41=0,則f(2x-5)≤f(3)即log4(2x-7)-log4(9-2x)≤0,即log4(2x-7)≤log4(9-2x),因此2x-7>0,9?2x>0,2x-7≤9-212.B∵x2-2x+3=(x-1)2+2≥2,∴f(x)=log2(x2-2x+3)≥log22=1,因此,函數(shù)f(x)的值域是[1,+∞),故選B.13.C當(dāng)x≥1時(shí),f(x)=lnx-2a≥-2a,∵f(x)的值域?yàn)镽,∴當(dāng)x<1時(shí),f(x)=(1-a)x+3的取值范圍需包含(-∞,-2a),∴1?a>0,1?a+3≥?2a,14.答案2解析∵0<a<1,∴函數(shù)f(x)在[2,4]上單調(diào)遞減,∴f(x)max=f(2)=loga2,f(x)min=f(4)=loga4,則loga2-loga4=2,即loga12=2,∴a=215.解析(1)由12≤x≤8,得-1≤log2x≤3,因此log2x的取值范圍是(2)f(x)=log2(2x)·log2x4=(log2x+1)(log2x-2)=log2x-122-94∴當(dāng)log2x=12時(shí),f(x)取得最小值,為-916.解析(1)∵f(1)=1,∴l(xiāng)og4(a+5)=1,解得a=-1,∴f(x)=log4(-x2+2x+3).由-x2+2x+3>0,解得-1<x<3,∴f(x)的定義域?yàn)?-1,3).令t=-x2+2x+3,易知函數(shù)t=-x2+2x+3在(-1,1)上單調(diào)遞增,在(1,3)上單調(diào)遞減,而y=log4t是定義域上的增函數(shù),∴函數(shù)f(x)的單調(diào)遞增區(qū)間為(-1,1).(2)∵函數(shù)f(x)=log4(ax2+2x+3)的最小值為0,∴函數(shù)y=ax2+2x+3有最小值1,∴a>0,12a-4(3)∵函數(shù)f(x)=log4(ax2+2x+3)的值域?yàn)镽,∴函數(shù)y=ax2+2x+3能夠取到(0,+∞)上的所有實(shí)數(shù),則a=0或a>0,22-12a≥0,17.A設(shè)f(x)=logax(a>0且a≠1),∵函數(shù)f(x)的圖象過(guò)點(diǎn)(4,2),∴f(4)=loga4=2,∴a=2,∴f(x)=log2x,故f(x)的反函數(shù)g(x)的解析式為g(x)=2x.故選A.18.D由f(x+2)=xa+3可得f(x)=(x-2)a+3,當(dāng)x=3時(shí),無(wú)論a為何值,都有f(3)=4,即f(x)的圖象恒過(guò)點(diǎn)(3,4),因?yàn)楹瘮?shù)f(x)與g(x)的圖象關(guān)于直線y=x對(duì)稱,所以f(x)與g(x)互為反函數(shù),所以g(x)的圖象恒過(guò)點(diǎn)(4,3),故選D.19.ABD對(duì)于A,根據(jù)同底數(shù)的指數(shù)函數(shù)與對(duì)數(shù)函數(shù)互為反函數(shù)可得g(x)=logax(a>0,且a≠1),且定義域是(0,+∞),故A正確.對(duì)于B,根據(jù)反函數(shù)的特點(diǎn)知函數(shù)f(x)與g(x)的圖象關(guān)于直線y=x對(duì)稱,故B正確.對(duì)于C,若f(2)=14,則a2=14,解得a=12或a=-12(舍去),則g(x)=log12x,則g22=log1222=12,故C錯(cuò)誤.對(duì)于D,在同一坐標(biāo)系中作出a>1時(shí)兩函數(shù)的圖象,如圖1,2,3,可知能力提升練1.D2.B3.ACD5.B6.D7.D10.C11.A14.D15.BCD1.D函數(shù)f(x)=x3·ln|x|的定義域?yàn)閧x|x≠0},關(guān)于原點(diǎn)對(duì)稱,又f(-x)=-x3·ln|x|=-f(x),∴f(x)為奇函數(shù),則f(x)的圖象關(guān)于原點(diǎn)對(duì)稱,故排除選項(xiàng)A、C;當(dāng)x>1時(shí),f(x)>0,故排除選項(xiàng)B.故選D.2.B在同一直角坐標(biāo)系中畫出y=12x,y=13x,y=log2x,y=log3x設(shè)y=12x與y=log2x的圖象相交于點(diǎn)A,則其橫坐標(biāo)為a,設(shè)y=12x與y=log3x的圖象相交于點(diǎn)B,則其橫坐標(biāo)為b,由圖可知a<b;設(shè)y=13x與y=log2x綜上,c<a<b.故選B.3.ACD由已知得f(x)=-lgx,0<x<1,lgx,x≥1,定義域?yàn)?0,+∞),且f(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,由a>b>c,且f(c)>f(a)>f(b),結(jié)合函數(shù)圖象可知,0<c<1,且a>1,b≥1或0<b<1,故f(c)=-lgc,f(a)=lga,由f(c)>f(a)得-lgc>lga,即lg4.答案1解析設(shè)A(t,2log2t),則D(t,log2t),其中t>1,設(shè)B(x,log2x),則2log2t=log2x,得x=t2,所以B(t2,2log2t),則點(diǎn)C的坐標(biāo)為(t2,log2t),將點(diǎn)C的坐標(biāo)代入函數(shù)y=klog2x的解析式,得log2t=klog2t2,∴2k=1,k=125.B∵c=23=log3323=log339>log3又c=23=log4423=log4316<log4327=log43=6.D設(shè)t=x2-ax-1,該函數(shù)的圖象開(kāi)口向上,且對(duì)稱軸方程為x=a2所以t=x2-ax-1在a2,+∞又y=lgt在定義域上單調(diào)遞增,所以要使f(x)在(1,+∞)上單調(diào)遞增,只需(1,+∞)?a2,+∞,所以a2≤又12-a-1≥0,所以a≤0.故a≤0是函數(shù)f(x)在(1,+∞)上單調(diào)遞增的充要條件,根據(jù)充分不必要條件的定義可得D滿足.故選D.7.D因?yàn)檎龜?shù)a,b滿足2a-4b=log2ba=log2b-log2所以2a+log2a=22b+log2b,(變量分離)因此2a+log2a=22b+log2(2b)-1<22b+log2(2b),(通過(guò)放縮將兩邊化為相同的結(jié)構(gòu))令f(x)=2x+log2x,則f(x)在(0,+∞)上單調(diào)遞增,(構(gòu)造函數(shù),利用單調(diào)性解決問(wèn)題)所以a<2b.故選D.8.答案(1,3)解析令u=6-ax,x∈[0,2],則y=logau,a>0且a≠1,因?yàn)楹瘮?shù)f(x)=loga(6-ax)在[0,2]上單調(diào)遞減且u=6-ax在[0,2]上單調(diào)遞減,所以y=logau單調(diào)遞增且u>0在x∈[0,2]時(shí)恒成立,所以a>1,6?2a>0,解得1<a<3,易錯(cuò)警示求含對(duì)數(shù)函數(shù)的復(fù)合函數(shù)的單調(diào)性時(shí),既要考慮內(nèi)、外兩層函數(shù)的單調(diào)性,還要考慮函數(shù)的定義域(單調(diào)區(qū)間是函數(shù)定義域的子集).9.解析(1)因?yàn)閒(x)是定義在(-1,1)上的奇函數(shù),所以f(-x)=-f(x),即log21+bx-x+1因此1+bx-x+1=x+11?bx,即1-b2x2=1-x2,故b2經(jīng)檢驗(yàn)b=1符合題意.(2)由(1)得f(x)=log21?x任取x1,x2∈(-1,1),且x1<x2,則f(x1)-f(x2)=log21?x1x=log21?x1x1由于(x2-x1+1-x1x2)-(x1-x2+1-x1x2)=2(x2-x1)>0,∴x2-x1+1-x1x2>x1-x2+1-x1x2,又x1-x2+1-x1x2=(x1+1)(1-x2)>0,∴x2-x1+1?x1x2x故函數(shù)f(x)的單調(diào)遞減區(qū)間為(-1,1),無(wú)單調(diào)遞增區(qū)間.(3)f(a)-f(1-a)≤2a-1可化為f(a)-a≤f(1-a)-(1-a),令g(x)=f(x)-x,則g(x)=log21?xx+1則g(a)≤g(1-a),由(2)知,f(x)=log21?xx+1在因此g(x)=f(x)-x在(-1,1)上單調(diào)遞減,所以a≥1?a,-1<a<1,-1<1-a<1,10.C由題意可得f(x)=loga(1-x2),x∈0,2當(dāng)0<a<1時(shí),由復(fù)合函數(shù)的單調(diào)性可得f(x)=loga(1-x2)在區(qū)間0,22由f(x)max-f(x)min=1,可得f22即loga1?12-loga(1-0)=1,解得a=當(dāng)a>1時(shí),由復(fù)合函數(shù)的單調(diào)性可得f(x)=loga(1-x2)在區(qū)間0,22由f(x)max-f(x)min=1,可得f(0)-f22即loga(1-0)-loga1?12=1,綜上所述,a=2或a=12.故選C11.A由題知,函數(shù)f(x)存在最小值,所以u(píng)=x2-ax+a大于0恒成立,且函數(shù)y=logau在(0,+∞)上單調(diào)遞增,所以a>1,Δ故選A.12.解析(1)當(dāng)a=1時(shí),f(x)=(lgx)2+lgx+54設(shè)u=lgx,由x∈110,100,得u則f(x)可轉(zhuǎn)化為g(u)=u2+u+54=u+122+1,所以當(dāng)u=-12,即因此函數(shù)f(x)在區(qū)間110,100(2)設(shè)t=lgx0,由x0∈(1,+∞),得t∈(0,+∞),則f(x0)=a可化為t2+at+54即方程t2+at+54-a=0存在大于零的解所以Δ=a解得a≤-5或a>54故a的取值范圍為(-∞,-5]∪5413.解析(1)∵函數(shù)f(x)=ln(ax2+2ax+1)的定義域?yàn)镽,∴ax2+2ax+1>0對(duì)任意x∈R恒成立,當(dāng)a=0時(shí),可得1>0,恒成立,滿足題意;當(dāng)a≠0時(shí),要使ax2+2ax+1>0對(duì)任意x∈R恒成立,只需a>0,Δ綜上可得,a的取值范圍是[0,1).(2)由(1)及題意知0<a<1.令u=ax2+2ax+1,易知y=lnu是定義域內(nèi)的增函數(shù),函數(shù)u=ax2+2ax+1(0<a<1)在[-2,-1]上單調(diào)遞減,在(-1,1]上單調(diào)遞增,故f(x)在[-2,-1]上單調(diào)遞減,在(-1,1]上單調(diào)遞增,又f(-2)=ln(4a-4a+1)=0,f(1)=ln(a+2a+1)=ln(3a+1)>0,∴f(x)max=f(1)=ln(3a+1),f(x)min=f(-1)=ln(1-a),∵f(x)在[-2,1]上的最大值與最小值的和為0,∴l(xiāng)n(3a+1)+ln(1-a)=0,即ln[(3a+1)(1-a)]=0,可得(3a+1)(1-a)=1,解得a=0(舍去)或a=23故實(shí)數(shù)a的值為2314.D由2x+1≠0,2x-1≠0,得x≠±12,故f(x)的定義域?yàn)閤x∈R,且x≠±1又f(-x)=ln|-2x+1|-ln|-2x-1|=-(ln|2x+1|-ln|2x-1|)=-f(x),∴f(x)為奇函數(shù).f(x)=ln|2x+1|-ln|2x-1|=ln|2x+1||2x-1|=ln2則t=2x+12x-1=2畫出函數(shù)t=2x+12x由圖可知t=2x+12x-1在-∞,-12上單調(diào)遞減,在又對(duì)數(shù)函數(shù)y=lnx是定義域內(nèi)的增函數(shù),所以由復(fù)合函數(shù)的單調(diào)性可得f(x)在-∞,-12上單調(diào)遞減,在-12,12上單調(diào)遞增,15.BCDf(x)=ln(e2x+1)-x=lne2x+1ex=ln(ex+e-x),對(duì)于B選項(xiàng),易知f(x)的定義域?yàn)镽,又f(-x)=ln(e