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)4.3指數(shù)函數(shù)與對(duì)數(shù)函數(shù)的關(guān)系課時(shí)素養(yǎng)評(píng)價(jià)含解析新人教B版必修第二冊(cè)

PAGE指數(shù)函數(shù)與對(duì)數(shù)函數(shù)的關(guān)系(15分鐘30分)1.(2024·成都高一檢測(cè))已知函數(shù)f(x)=log2x,若函數(shù)g(x)是f(x)的反函數(shù),則fQUOTE= ()A.1 B.2 C.3 D.4【解析】選B.由函數(shù)y=f(x)=log2x,把x與y互換,得x=log2y(x∈R),可得y=2x,即g(x)=2x,所以g(2)=22=4,則fQUOTE=f(4)=log24=2.【補(bǔ)償訓(xùn)練】(2024·日照高一檢測(cè))已知函數(shù)f(x)=1+2lgx,則f(1)+f-1(1)= ()A.0 B.1 C.2 D.3【解析】選C.依據(jù)題意:fQUOTE=1+2lg1=1,若f(x)=1+2lgx=1,解得x=1,則f-1QUOTE=1,故fQUOTE+f-1QUOTE=1+1=2.2.函數(shù)y=QUOTE的反函數(shù)是 ()A.y=QUOTE(x∈R且x≠-4)B.y=QUOTE(x∈R且x≠3)C.y=QUOTE(x∈R且x≠Q(mào)UOTE)D.y=QUOTE(x∈R且x≠-QUOTE)【解析】選B.由y=QUOTE,得x=QUOTE.故所求反函數(shù)為y=QUOTE(x∈R且x≠3).3.設(shè)y=QUOTE+m和y=nx-9互為反函數(shù),那么m,n的值分別是 ()A.-6,3 B.2,1 C.2,3 D.3,3【解析】選D.求出y=QUOTE+m的反函數(shù)y=3x-3m,再與y=nx-9對(duì)比系數(shù),得m=3,n=3.4.函數(shù)f(x)是增函數(shù),它的反函數(shù)是f-1(x),若a=f(2)+f-1(2),b=f(3)+f-1(3),則下面結(jié)論中正確的是 ()A.a<b B.a=bC.a>b D.無法確定【解析】選A.因?yàn)閒(x)是增函數(shù),故其反函數(shù)f-1(x)也是增函數(shù),所以f(3)>f(2),f-1(3)>f-1(2),即b>a.5.函數(shù)f(x)=QUOTE的反函數(shù)是________.

【解析】函數(shù)的值域?yàn)閇0,+∞),令y=QUOTE,將其中的x,y對(duì)調(diào)得x=QUOTE,解得y=4-x2,所以反函數(shù)f-1(x)=4-x2(x≥0).答案:f-1(x)=4-x2(x≥0)6.若函數(shù)y=f(x)的反函數(shù)是y=-QUOTE(-1≤x≤0),求原函數(shù)的定義域和f(-1)的值.【解析】因?yàn)樵瘮?shù)的定義域?yàn)榉春瘮?shù)的值域,又-1≤x≤0,所以1≤2-x2≤2,即y∈[-QUOTE,-1].令-QUOTE=-1,解得x=±1,因?yàn)樵瘮?shù)的定義域?yàn)閇-QUOTE,-1],所以x=-1.所以原函數(shù)y=f(x)的定義域?yàn)閇-QUOTE,-1],f(-1)=-1.(30分鐘60分)一、單選題(每小題5分,共20分)1.若f(x)為y=2-x的反函數(shù),則f(x-1)的圖像大致是 ()【解析】選C.由題意,f(x)的圖像與y=2-x的圖像關(guān)于y=x對(duì)稱,即f(x)=loQUOTEx,所以f(x-1)的圖像就是將f(x)=loQUOTEx右移一個(gè)單位得到.2.函數(shù)f(x)與g(x)=QUOTE互為反函數(shù),則函數(shù)f(4-x2)的單調(diào)增區(qū)間是()A.(-∞,0] B.[0,+∞)C.(-2,0] D.[0,2)【解析】選D.因?yàn)閒(x)與g(x)=QUOTE互為反函數(shù),所以f(x)=loQUOTEx,所以f(4-x2)=loQUOTE(4-x2),x>0,由4-x2>0,所以-2<x<2,所以0<x<2,因?yàn)閥=4-x2在[0,2)上是減函數(shù),所以f(x)在[0,2)上是增函數(shù).【補(bǔ)償訓(xùn)練】(2024·辛集高一檢測(cè))若y=f(x)是函數(shù)y=2x的反函數(shù),則函數(shù)y=f(-x2+2x+3)的單調(diào)遞增區(qū)間是 ()A.(-∞,1) B.(-3,-1)C.(-1,1) D.(1,+∞)【解析】選C.由y=f(x)是函數(shù)y=2x的反函數(shù),得y=f(x)=log2x,則y=f(-x2+2x+3)=log2(-x2+2x+3),由-x2+2x+3>0,解得-1<x<3,所以函數(shù)y=f(-x2+2x+3)的定義域?yàn)?-1,3),因?yàn)閥=log2u是增函數(shù),u=-x2+2x+3在(-∞,1)上是增函數(shù),所以y=log2(-x2+2x+3)的遞增區(qū)間為(-1,1).3.(2024·濰坊高一檢測(cè))已知函數(shù)g(x)=f(x)+x2是奇函數(shù),當(dāng)x>0時(shí),函數(shù)f(x)的圖像與函數(shù)y=log2x的圖像關(guān)于y=x對(duì)稱,則gQUOTE+gQUOTE= ()A.-7 B.-9 C.-11 D.-13【解析】選C.因?yàn)閤>0時(shí),f(x)的圖像與函數(shù)y=log2x的圖像關(guān)于y=x對(duì)稱,所以x>0時(shí),f(x)=2x,所以x>0時(shí),g(x)=2x+x2,又g(x)是奇函數(shù),所以g(-1)+g(-2)=-[g(1)+g(2)]=-(2+1+4+4)=-11.4.若函數(shù)y=QUOTE的圖像關(guān)于直線y=x對(duì)稱,則a的值為 ()A.1 B.-1 C.2 D.-2【解析】選B.由y=QUOTE,將x,y的位置互換得x=QUOTE,整理得y=QUOTE,則原函數(shù)的反函數(shù)是y=QUOTE,由題意知原函數(shù)與反函數(shù)相同,所以QUOTE=QUOTE,解得a=-1.二、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.函數(shù)y=2|x|在下面的區(qū)間上,不存在反函數(shù)的是 ()A.[-1,1] B.(-∞,0]C.[-2,4] D.[2,4]【解析】選AC.函數(shù)若在區(qū)間上單調(diào),則存在反函數(shù),易知函數(shù)y=2|x|在[-1,1],[-2,4]上不單調(diào).6.函數(shù)y=f(x)是y=ax(a>0,且a≠1)的反函數(shù),則下列結(jié)論正確的是 ()A.f(x2)=2f(|x|)B.f(2x)=f(x)+f(2)C.fQUOTE=f(x)-f(2)D.f(2x)=2f(x)【解析】選ABC.因?yàn)楹瘮?shù)y=f(x)是y=ax(a>0,且a≠1)的反函數(shù),所以f(x)=logax,所以fQUOTE=loga2x=loga2+logax=f(x)+f(2)≠2f(x),B對(duì)D錯(cuò);fQUOTE=logax2=2loga|x|=2f(|x|),A對(duì);fQUOTE=logaQUOTE=logax-loga2=f(x)-f(2),C對(duì).【補(bǔ)償訓(xùn)練】已知函數(shù)f(x)=ex+x-2,g(x)=lnx+x-2,且f(a)=g(b)=0,則下列結(jié)論錯(cuò)誤的是 ()A.a>b B.g(a)<0<f(b)C.g(a)>0>f(b) D.a+b=2【解析】選AC.因?yàn)楹瘮?shù)y=ex,y=lnx,y=x-2都是增函數(shù),所以f(x)=ex+x-2,g(x)=lnx+x-2都是增函數(shù),又f(0)=e0+0-2=-1<0,f(1)=e1+1-2=e-1>0,所以0<a<1,g(1)=ln1+1-2=-1<0,g(2)=ln2+2-2=ln2>0,所以1<b<2,所以0<a<1<b<2,故A錯(cuò)誤;因?yàn)閍<b,所以g(a)<g(b)=0,f(a)<f(b),即f(b)>0,所以g(a)<0<f(b),故B正確,C錯(cuò)誤;令f(x)=ex+x-2=0,g(x)=lnx+x-2=0,則ex=2-x,lnx=2-x,由于函數(shù)y=ex,y=lnx都和y=2-x相交,且y=ex和y=lnx關(guān)于直線y=x對(duì)稱,又y=2-x和y=x的交點(diǎn)為(1,1),所以a+b=2,所以D正確.三、填空題(每小題5分,共10分)7.設(shè)a>0且a≠1,若函數(shù)f(x)=QUOTE+2的反函數(shù)的圖像經(jīng)過定點(diǎn)P,則點(diǎn)P的坐標(biāo)是________.

【解析】因?yàn)楹瘮?shù)f(x)=QUOTE+2經(jīng)過定點(diǎn)(1,3),所以函數(shù)f(x)的反函數(shù)的圖像經(jīng)過定點(diǎn)P(3,1).答案:(3,1)【補(bǔ)償訓(xùn)練】若函數(shù)f(x)=log2(x+1)+a的反函數(shù)的圖像經(jīng)過點(diǎn)(4,1),則實(shí)數(shù)a=________.

【解析】函數(shù)f(x)=log2(x+1)+a的反函數(shù)的圖像經(jīng)過點(diǎn)(4,1),即函數(shù)f(x)=log2(x+1)+a的圖像經(jīng)過點(diǎn)(1,4),所以4=log2(1+1)+a,所以4=1+a,a=3.答案:38.設(shè)點(diǎn)(9,3)在函數(shù)f(x)=loga(x-1)(a>0,a≠1)的圖像上,則f(x)的反函數(shù)f-1(x)=________.

【解析】因?yàn)辄c(diǎn)(9,3)在函數(shù)f(x)=loga(x-1)(a>0,a≠1)的圖像上,所以loga(9-1)=3,可得a=2,則令函數(shù)f(x)=y=log2(x-1),那么x=2y+1.把x與y互換可得y=2x+1,所以f(x)的反函數(shù)f-1(x)=2x+1.答案:2x+1四、解答題(每小題10分,共20分)9.已知函數(shù)f(x)=QUOTE的定義域?yàn)?-∞,2].(1)求函數(shù)f(x)的值域;(2)求函數(shù)f(x)的反函數(shù).【解析】(1)f(x)的定義域?yàn)閤≤2,0<2x≤4,所以4≤8-2x<8,所以QUOTE8<QUOTE≤QUOTE4,即-3<QUOTE≤-2,故函數(shù)的值域?yàn)?-3,-2].(2)由y=QUOTE,將x,y互換,x=loQUOTE(8-2y)(-3<x≤-2).整理得2y=8-QUOTE,y=log2QUOTE,所以f-1(x)=log2QUOTE(-3<x≤-2).10.已知函數(shù)f(x)=lg(x+1).(1)當(dāng)x∈[1,9]時(shí),求函數(shù)f(x)的反函數(shù).(2)若0<f(1-2x)-f(x)<1,求x的取值范圍.【解析】(1)令y=f(x)=lg(x+1),所以當(dāng)x∈[1,9]時(shí),y∈[lg2,1],且x+1=10y,即x=10y-1,互換x,y得,y=10x-1,所以,f-1(x)=10x-1,x∈[lg2,1].(2)不等式0<f(1-2x)-f(x)<1可化為:0<lgQUOTE<1,等價(jià)為1<QUOTE<10且x+1>0,解得x∈QUOTE,所以,原不等式中x的取值范圍為:QUOTE.1.已知函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x<0時(shí),f(x)=QUOTE,那么f-1(-9)的值為 ()A.2 B.-2 C.3 D.-3【解析】選A.設(shè)f-1(-9)=x,則f(x)=-9,設(shè)x>0,則-x<0.因?yàn)楫?dāng)x<0時(shí),f(x)=QUOTE,所以f(-x)=QUOTE=3x.因?yàn)楹瘮?shù)f(x)是定義在R上的奇函數(shù),所以f(x)=-f(-x)=-3x.所以-3x=-9,故x=2.2.已知函數(shù)f(x)=x2+bx+4滿意f(1+x)=f(1-x),且函數(shù)g(x)=ax(a>0,且a≠1)與函數(shù)y=log3x互為反函數(shù).(1)求函數(shù)f(x),g(x)的解析式;(2)函數(shù)y=f(g(x))-m在x∈[-1,2]上有零點(diǎn),求實(shí)數(shù)m的取值范圍.【解析】(1)由f(1+x)=f(1-x)知函數(shù)的對(duì)稱軸-QUOTE=1,故b=-2;所以f(x)=x2-2x+4.由函數(shù)g(x)=ax(a>0,且a≠1)與函數(shù)y=log3x互為反函數(shù),知a=3.故g(x)=3x.(2)令t=3x,因?yàn)閤∈[-1,2],所以t∈QUOTE,則y=f(g(x))-m=t2-2t+4-m在t∈QUOTE上有零點(diǎn),即函數(shù)y=m的圖像與y=t2-2t+4的圖像在t∈QUOTE上有交點(diǎn),而y=(t-1)2+3,t∈QUOTE,所以當(dāng)t=1時(shí),此時(shí)x=0,f(g(x))min=3,當(dāng)t=9時(shí),此時(shí)x=2,f(g(x))max=67,因此m的取值范圍是QUOTE.【補(bǔ)償訓(xùn)練】已知函數(shù)f(x)=2x的反函數(shù)為f-1(x).(1)若f-1(x)-f-1(1-x)=1,求實(shí)數(shù)x的值;(2)若關(guān)于x的方程f(x)+f(1