2024-2025學(xué)年新教材高中數(shù)學(xué)第四章指數(shù)函數(shù)對(duì)數(shù)函數(shù)與冪函數(shù)4.2.3.2對(duì)數(shù)函數(shù)的性質(zhì)與圖像的應(yīng)用課時(shí)素養(yǎng)評(píng)價(jià)含解析新人教B版必修第二冊(cè)

PAGE對(duì)數(shù)函數(shù)的性質(zhì)與圖像的應(yīng)用(15分鐘30分)1.已知函數(shù)f(x)=loga(x-m)的圖像過點(diǎn)(4,0)和(7,1),則f(x)在定義域上是()A.增函數(shù) B.減函數(shù)C.奇函數(shù) D.偶函數(shù)【解析】選A.由題意,QUOTE解得QUOTE所以f(x)=log4(x-3),所以f(x)是增函數(shù),因?yàn)閒(x)的定義域是(3,+∞),不關(guān)于原點(diǎn)對(duì)稱.所以f(x)為非奇非偶函數(shù).【補(bǔ)償訓(xùn)練】已知函數(shù)f(x)=loga(x-2),若圖像過點(diǎn)(11,2),則f(5)的值為 ()A.-1 B.1 C.-2 D.2【解析】選B.由函數(shù)圖像過點(diǎn)(11,2),則loga(11-2)=2,解得a=3.故f(5)=log3(5-2)=1.2.已知a=21.1,b=log23,c=QUOTE,則a,b,c的大小關(guān)系為 ()A.a>b>c B.a>c>bC.b>a>c D.b>c>a【解析】選A.21.1>2,QUOTE=QUOTE.又2>log23>log2QUOTE=log2QUOTE=QUOTE,所以a>b>c.3.函數(shù)f(x)=2+log6(6x+1),x∈R的值域?yàn)?()A.(0,1] B.(0,+∞)C.[1,+∞) D.(2,+∞)【解析】選D.因?yàn)?x+1>1,所以log6(6x+1)>0,故f(x)=2+log6(6x+1)>2.4.(2024·南昌高一檢測(cè))已知函數(shù)f(x)=log2(-x2+6x+7)的值域記為集合A,函數(shù)g(x)=QUOTE的值域?yàn)锽,則有()A.B?RA B.A?RBC.A?B D.B?A【解析】選D.令t=-x2+6x+7,t>0,當(dāng)x=3時(shí),tmax=-32+6×3+7=16,此時(shí)f(x)max=log216=4,所以函數(shù)f(x)=log2QUOTE的值域?yàn)?A=QUOTE.在函數(shù)g(x)=QUOTE中,可得:0≤16-x2≤16,所以函數(shù)g(x)=QUOTE的值域?yàn)?B=QUOTE,所以B?A.5.已知f(x)=lgQUOTE,x∈(-1,1),則函數(shù)f(x)是________函數(shù)(填奇或偶或非奇非偶).若f(a)=2,則f(-a)=________.

【解析】因?yàn)閘gQUOTE=lgQUOTE,所以x∈(-1,1),且f(-x)=lgQUOTE=lgQUOTE=-lgQUOTE=-f(x),所以f(x)為奇函數(shù),所以f(-a)=-f(a)=-2.答案:奇-26.函數(shù)f(x)=loga(3-ax)(a>0且a≠1)在區(qū)間(a-2,a)上單調(diào)遞減,則a的取值范圍為________.

【解析】因?yàn)楹瘮?shù)在區(qū)間(a-2,a)上單調(diào)遞減,所以QUOTE解得1<a≤QUOTE.答案:{a|1<a≤QUOTE}(30分鐘60分)一、單選題(每小題5分,共20分)1.已知a<b,函數(shù)f(x)=(x-a)·(x-b)的圖像如圖所示,則函數(shù)g(x)=logb(x+a)的圖像可能是 ()【解析】選B.由題圖可知0<a<1<b,故函數(shù)g(x)單調(diào)遞增,解除A、D,結(jié)合a的范圍可知選B.2.已知函數(shù)y=|logQUOTEx|的定義域?yàn)镼UOTE,值域?yàn)閇0,1],則m的取值范圍為()A.QUOTE B.QUOTEC.[1,2] D.[1,+∞)【解析】選C.作出y=|logQUOTEx|的圖像(如圖),可知fQUOTE=f(2)=1,由題意結(jié)合圖像知:1≤m≤2.3.(2024·牡丹江高一檢測(cè))已知函數(shù)f(x)=log2(x2-2x+a)的最小值為2,則a=()A.4 B.5 C.6 D.7【解析】選B.內(nèi)層函數(shù)為u=x2-2x+a,外層函數(shù)為y=log2u,由于內(nèi)層函數(shù)u=x2-2x+a的減區(qū)間為QUOTE,增區(qū)間為QUOTE,且外層函數(shù)為增函數(shù),所以,函數(shù)f(x)=log2QUOTE的單調(diào)遞減區(qū)間為QUOTE,單調(diào)遞增區(qū)間為QUOTE,所以,函數(shù)y=f(x)在x=1處取得最小值,即f(x)min=fQUOTE=log2QUOTE=2,解得a=5.【補(bǔ)償訓(xùn)練】(2024·辛集高一檢測(cè))若-3≤QUOTEx≤-QUOTE,求f(x)=QUOTE·QUOTE的最大值和最小值.【解析】由題意,依據(jù)對(duì)數(shù)的運(yùn)算性質(zhì),可得函數(shù)f(x)=QUOTE=QUOTE-3log2x+2=QUOTE-QUOTE,又-3≤QUOTEx≤-QUOTE,所以QUOTE≤log2x≤3.所以當(dāng)log2x=3,即x=8時(shí),f(x)max=fQUOTE=2;當(dāng)log2x=QUOTE,即x=2QUOTE時(shí),f(x)min=fQUOTE=-QUOTE.4.(2024·嘉興高一檢測(cè))函數(shù)y=QUOTE(x2-3x+2)的單調(diào)遞減區(qū)間為 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選A.因?yàn)閥=QUOTE(x2-3x+2),所以x2-3x+2>0,解得x<1或x>2,令t=x2-3x+2,因?yàn)閠=x2-3x+2的圖像開口向上,對(duì)稱軸方程為x=QUOTE,所以內(nèi)函數(shù)t=x2-3x+2在QUOTE上單調(diào)遞增,外函數(shù)y=QUOTEt單調(diào)遞減,所以由復(fù)合函數(shù)單調(diào)性的性質(zhì)可知函數(shù)y=QUOTE(x2-3x+2)的單調(diào)遞減區(qū)間為QUOTE.二、多選題(每小題5分,共10分,全部選對(duì)得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)5.已知f(x)=lg(10+x)+lg(10-x),則 ()A.f(x)是奇函數(shù)B.f(x)是偶函數(shù)C.f(x)在(0,10)上單調(diào)遞增D.f(x)在(0,10)上單調(diào)遞減【解析】選BD.由QUOTE得x∈(-10,10),故函數(shù)f(x)的定義域?yàn)?-10,10),關(guān)于原點(diǎn)對(duì)稱,又由f(-x)=lg(10-x)+lg(10+x)=f(x),故函數(shù)f(x)為偶函數(shù),而f(x)=lg(10+x)+lg(10-x)=lg(100-x2),y=100-x2在(0,10)上遞減,y=lgx在(0,10)上遞增,故函數(shù)f(x)在(0,10)上遞減.6.關(guān)于函數(shù)f(x)=lgQUOTE(x≠0),有下列結(jié)論,其中正確的是 ()A.其圖像關(guān)于y軸對(duì)稱B.f(x)的最小值是lg2C.當(dāng)x>0時(shí),f(x)是增函數(shù);當(dāng)x<0時(shí),f(x)是減函數(shù)D.f(x)的增區(qū)間是(-1,0),(1,+∞)【解析】選ABD.f(-x)=lgQUOTE=f(x),f(x)是偶函數(shù),選項(xiàng)A正確;令t=QUOTE=|x|+QUOTE≥2,當(dāng)且僅當(dāng)|x|=1時(shí)取等號(hào),y=lgt在(0,+∞)上單調(diào)遞增,y=lgt≥lg2,所以f(x)的最小值為lg2,選項(xiàng)B正確;當(dāng)x>0時(shí),t=QUOTE=x+QUOTE,依據(jù)對(duì)勾函數(shù)可得,t=x+QUOTE的單調(diào)遞減區(qū)間是(0,1),單調(diào)遞增區(qū)間是(1,+∞),y=lgt在(0,+∞)上單調(diào)遞增,所以f(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,選項(xiàng)C錯(cuò)誤;依據(jù)偶函數(shù)的對(duì)稱性,f(x)在(-∞,-1)上單調(diào)遞減,在(-1,0)上單調(diào)遞增,f(x)的增區(qū)間是(-1,0),(1,+∞),選項(xiàng)D正確.【補(bǔ)償訓(xùn)練】(2024·濟(jì)南高一檢測(cè))已知函數(shù)f(x)=QUOTE,則下面結(jié)論正確的有 ()A.f(x)的圖像關(guān)于原點(diǎn)對(duì)稱B.f(x)的圖像關(guān)于y軸對(duì)稱C.f(x)的值域?yàn)?-1,1)D.?x1,x2∈R,且x1≠x2,QUOTE<0恒成立【解析】選ACD.A.f(x)=QUOTE,則f(-x)=QUOTE=QUOTE=-f(x),則f(x)的圖像關(guān)于原點(diǎn)對(duì)稱;B.計(jì)算fQUOTE=-QUOTE,fQUOTE=QUOTE≠fQUOTE,故f(x)的圖像不關(guān)于y軸對(duì)稱;C.f(x)=QUOTE=-1+QUOTE,令1+2x=t,t∈QUOTE,y=-1+QUOTE,易知:-1+QUOTE∈(-1,1),故f(x)的值域?yàn)?-1,1);D.f(x)=QUOTE=-1+QUOTE,在定義域上單調(diào)遞減,故?x1,x2∈R,且x1≠x2,QUOTE<0恒成立.三、填空題(每小題5分,共10分)7.(2024·運(yùn)城高一檢測(cè))已知函數(shù)f(x)=log2QUOTE·log4QUOTE,x∈QUOTE,則f(x)的最小值為________.

【解析】將函數(shù)化簡為:f(x)=QUOTE,設(shè)log2x=t,則y=QUOTE,因?yàn)閤∈QUOTE,所以t∈QUOTE.依據(jù)二次函數(shù)的性質(zhì)得到:當(dāng)t=-1時(shí),y取得最小值0,故f(x)的最小值為0.答案:08.已知定義域?yàn)镽的偶函數(shù)f(x)在[0,+∞)上單調(diào)遞增,且fQUOTE=0,則不等式f(log4x)<0的解集是________.

【解析】由題意可知,由f(log4x)<0,得-QUOTE<log4x<QUOTE,即log4QUOTE<log4x<log4QUOTE,得QUOTE<x<2.答案:QUOTE四、解答題(每小題10分,共20分)9.已知f(x)=log2(x+1),當(dāng)點(diǎn)(x,y)在函數(shù)y=f(x)的圖像上時(shí),點(diǎn)QUOTE在函數(shù)y=g(x)的圖像上.(1)寫出y=g(x)的解析式.(2)求方程f(x)-g(x)=0的根.【解析】(1)依題意,得QUOTE則gQUOTE=QUOTElog2(x+1),故g(x)=QUOTElog2(3x+1).(2)由f(x)-g(x)=0,得log2(x+1)=QUOTElog2(3x+1),所以QUOTE解得x=0或x=1.10.設(shè)f(x)=loga(3+x)+loga(3-x)(a>0,a≠1),且f(0)=2.(1)求實(shí)數(shù)a的值及函數(shù)f(x)的定義域.(2)求函數(shù)f(x)在區(qū)間[0,QUOTE]上的最小值.【解析】(1)由題意,f(0)=loga3+loga3=2loga3=2,所以a=3,所以f(x)=log3(3+x)+log3(3-x),所以QUOTE解得-3<x<3,所以f(x)的定義域是(-3,3).(2)因?yàn)閒(x)=log3(3+x)+log3(3-x)=log3[(3+x)(3-x)]=log3(9-x2)且x∈(-3,3),所以當(dāng)x=QUOTE時(shí),f(x)在區(qū)間[0,QUOTE]上取得最小值,f(x)min=log33=1.1.若ax>1的解集為{x|x<0}且函數(shù)y=logaQUOTE的最大值為-1,則實(shí)數(shù)a的值為()A.2 B.QUOTE C.3 D.QUOTE【解析】選B.因?yàn)閍x>1=a0的解集為{x|x<0},所以0<a<1,因?yàn)閥=logaQUOTE的最大值為-1,由對(duì)勾函數(shù)的性質(zhì)可知,x>0時(shí),x+QUOTE≥2,所以loga2=-1,解得a=QUOTE.2.已知函數(shù)f(x)=loga(-x2+ax-9)(a>0,a≠1).(1)當(dāng)a=10時(shí),求f(x)的值域和單調(diào)減區(qū)間;(2)若f(x)存在單調(diào)遞增區(qū)間,求a的取值范圍.【解析】(1)當(dāng)a=10時(shí),f(x)=log10QUOTE=lgQUOTE,設(shè)t=-x2+10x-9=-QUOTE+16,由-x2+10x-9>0,得x2-10x+9<0,得1<x<9,即函數(shù)的定義域?yàn)镼UOTE,此時(shí)t=-QUOTE+16∈QUOTE,則y=lgt≤lg16,即函數(shù)的值域?yàn)?/p>