課時(shí)規(guī)范練8函數(shù)的概念及其表示基礎(chǔ)鞏固練1.(2024·湖南郴州模擬)函數(shù)f(x)=1-2xxA.(-∞,12] B.(-∞,0)∪(0,1C.(-∞,0) D.[12,1)∪(1,+∞2.(2024·山東煙臺(tái)模擬)已知函數(shù)f(x-1)=x2-2x,且f(a)=3,則實(shí)數(shù)a的值等于()A.2 B.±2 C.2 D.±23.(2024·河南襄城模擬)已知函數(shù)f(x)=log2x,0<x<1,4A.-4 B.-2 C.2 D.44.(2024·福建寧德模擬)已知函數(shù)f(x)=lg1-x1+x,則函數(shù)g(x)=f(x-1)+2A.(-∞,0)∪(2,+∞) B.[12,C.(2,+∞) D.[12,+∞5.(多選題)(2024·江蘇徐州模擬)記無理數(shù)e=2.718281828459045…小數(shù)點(diǎn)后第n位上的數(shù)字為m,則m是關(guān)于n的函數(shù),記作m=f(n),其定義域?yàn)锳,值域?yàn)锽,則()A.f(5)=8B.函數(shù)f(n)的圖象是一群孤立的點(diǎn)C.n是關(guān)于m的函數(shù)D.B?A6.(多選題)(2024·福建福州模擬)已知f(x)是定義在R上的不恒為零的函數(shù),對(duì)于任意a,b∈R都滿足f(ab)=af(b)+bf(a),則下列說法正確的是()A.f(0)=0B.f(1)=0C.f(x)是偶函數(shù)D.若f(2)=2,則f(-12)=7.(2024·陜西渭南模擬)函數(shù)f(x)=3-3-x+ln|x|8.(2024·河南鄭州模擬)已知f(1-sinx)=cos2x,則f(x)的解析式為.

9.(2024·江蘇淮安模擬)已知函數(shù)f(x)=-x2+2,x≤1,x+1綜合提升練10.(2024·湖南岳陽模擬)已知f(x)=2x,x<0,a+3x,x≥0,若f(fA.-178 B.-4或-C.-4 D.不存在11.(2024·山東濰坊模擬)存在函數(shù)f(x)滿足:對(duì)任意x∈R都有()A.f(|x|)=x3 B.f(sinx)=x2C.f(x2+2x)=|x| D.f(|x|)=x2+112.(2024·成都七中模擬)設(shè)f(x)定義在R上且f(x)=log2(2-x)(x<13.(2024·廣東茂名模擬)已知函數(shù)f(x)=log2(x+4),-4<x<0,4x-1創(chuàng)新應(yīng)用練14.(2024·華大新高考聯(lián)盟質(zhì)檢)已知函數(shù)f(x)滿足f(ex-1)=2x-1,f(a)+f(b)=0,則下列說法正確的是()A.a+b=1 B.a+b=1C.ab=1 D.ab=115.(2024·山東濰坊模擬)已知函數(shù)f(x)=2x-1-2,x≤1,log2(x+1),x>1,試舉出一個(gè)a的值

課時(shí)規(guī)范練8函數(shù)的概念及其表示1.B解析要使函數(shù)有意義,應(yīng)滿足1-2x≥0,x2+x≠0,解得x≤12.D解析令x-1=t,則x=t+1,則f(t)=(t+1)2-2(t+1)=t2-1,則f(a)=a2-1=3,解得a=±2,故選D.3.B解析f(1)=41-2=14,∴f(f(1))=f(14)=log214=-2,4.B解析要使f(x)=lg1-x1+x有意義,則1-x1+x>0,即(1-x)(1+x)>0,解得-1<x<1,所以函數(shù)f(x)的定義域?yàn)?-1,1),要使g(x)=f(x-1)+2x-1有意義,則-1<x-15.AB解析根據(jù)函數(shù)的定義可知,定義域A=N*,對(duì)應(yīng)關(guān)系:數(shù)位n對(duì)應(yīng)數(shù)字7,1,8,2,8,1,8,2,8,…,f(5)=8,函數(shù)f(n)的圖象是一群孤立的點(diǎn),故A,B正確;對(duì)于C,n不是關(guān)于m的函數(shù),如m=8時(shí),n可能為3,5,7,9,不符合函數(shù)的定義,故C錯(cuò)誤;因?yàn)?∈B,0?A,所以D錯(cuò)誤,故選AB.6.ABD解析對(duì)于A,令a=b=0,則f(0)=0×f(0)+0×f(0)=0,故A正確;對(duì)于B,令a=b=1,則f(1)=1×f(1)+1×f(1)=2f(1),則f(1)=0,故B正確;對(duì)于C,令a=b=-1,則f(1)=-f(-1)-f(-1)=-2f(-1),因?yàn)閒(1)=0,所以f(-1)=0,又令a=-1,b=x,則f(-x)=-f(x)+xf(-1)=-f(x)+0=-f(x),所以f(x)是奇函數(shù),故C錯(cuò)誤;對(duì)于D,令a=2,b=-12,則f[2×(-12)]=f(-1)=2f(-12)-12f(2)=2f(-12)-1=0,所以f(-12)=12,7.[-1,0)∪(0,+∞)解析要使函數(shù)有意義,應(yīng)有3-3-x≥0,x≠0,8.f(x)=2x-x2(x∈[0,2])解析設(shè)1-sinx=t,則t∈[0,2],則sinx=1-t,因?yàn)閒(1-sinx)=cos2x=1-sin2x,所以f(t)=1-(1-t)2=2t-t2,t∈[0,2],即f(x)=2x-x2,x∈[0,2].9.[-1,+∞)解析當(dāng)x≤1時(shí),由f(x)≥1可得,-x2+2≥1,x2≤1,解得-1≤x≤1;當(dāng)x>1時(shí),由f(x)≥1可得,x+1x-1≥1,即x2-2x+1=(x-1)2≥0恒成立,所以x>1.綜上,不等式f(x)≥1的解集為[-1,+∞)10.B解析由題意,f(1)=a+3,f(-1)=12,即f(a+3)=12.當(dāng)a+3≥0,即a≥-3時(shí),f(a+3)=a+3(a+3)=4a+9=12,解得a=-178,滿足題意;當(dāng)a+3<0,即a<-3時(shí),f(a+3)=2a+3=12,解得a=-4,滿足題意,所以a=-1711.D解析對(duì)于A,當(dāng)x=1時(shí),f(|1|)=f(1)=1;當(dāng)x=-1時(shí),f(|-1|)=f(1)=-1,不符合函數(shù)定義,A錯(cuò)誤;對(duì)于B,令x=0,則f(sin0)=f(0)=0,令x=π,則f(sinπ)=f(0)=π2,不符合函數(shù)定義,B錯(cuò)誤;對(duì)于C,令x=0,則f(0)=0,令x=-2,則f((-2)2+2(-2))=f(0)=2,不符合函數(shù)定義,C錯(cuò)誤;對(duì)于D,f(|x|)=x2+1=|x|2+1,x∈R,則|x|≥0,則存在x≥0時(shí),f(x)=x2+1,符合函數(shù)定義,且對(duì)任意x∈R都有f(|x|)=x2+1,D正確,故選D.12.0解析依題意f(13)=f(12)-f(11)=f(11)-f(10)-f(11)=-f(10),f(10)=f(9)-f(8)=f(8)-f(7)-f(8)=-f(7),同理可得f(13)=f(7)=f(1)=log2(2-1)=0.13.(-2,0)∪(12,+∞)解析設(shè)u=f(a),于是f(f(a))>3即為f(u)>3,則log2(u+4)>3,-4<u<0或4u-14.D解析設(shè)t=ex-1,則x=lnt+1,∴f(t)=2lnt+1,t>0.由f(a)+f(b)=0,有2lna+1+2lnb+1=0,即ln(ab)=-1,∴ab=1e,故選D15.-1或7解析因?yàn)楹瘮?shù)f(x)=2x-1-2,x≤1,log2(x+1),x>1,可得當(dāng)x>1時(shí),f(x)=log2(x+1)