【全程復(fù)習(xí)方略】2020年人教A版數(shù)學(xué)理(廣東用)課時(shí)作業(yè)：第二章-第五節(jié)對(duì)-數(shù)-函-數(shù)

溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)整合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。課時(shí)提升作業(yè)(八)一、選擇題1.(2021·鄭州模擬)函數(shù)f(x)=QUOTE的定義域?yàn)?)(A)(0,+∞) (B)(1,+∞)(C)(0,1) (D)(0,1)∪(1,+∞)2.下列函數(shù)中,既是偶函數(shù),又在(0,1)上單調(diào)遞增的函數(shù)是()(A)y=|log3x| (B)y=x3(C)y=e|x| (D)y=cos|x|3.(2021·茂名模擬)0<x<y<1,則()(A)3y<3x (B)logx3<logy3(C)log4x<log4y (D)(QUOTE)x<(QUOTE)y4.若點(diǎn)(a,b)在y=lgx的圖象上,a≠1,則下列點(diǎn)也在此圖象上的是()(A)(QUOTE,b) (B)(10a,1-b)(C)(QUOTE,b+1) (D)(a2,2b)5.(2021·陽(yáng)江模擬)已知函數(shù)f(x)=QUOTE若f(2-x2)>f(x),則實(shí)數(shù)x的取值范圍是()(A)(-∞,-1)∪(2,+∞)(B)(-∞,-2)∪(1,+∞)(C)(-1,2)(D)(-2,1)6.已知偶函數(shù)f(x)在[0,2]上遞減,則a=f(1),b=f(loQUOTE),c=f(log2QUOTE)的大小關(guān)系是()(A)a>b>c (B)a>c>b(C)b>a>c (D)c>a>b7.若loga(a2+1)<loga2a<0,則a的取值范圍是()(A)(0,1) (B)(0,QUOTE)(C)(QUOTE,1) (D)(0,1)∪(1,+∞)8.已知函數(shù)f(x)=|log2x|,正實(shí)數(shù)m,n滿(mǎn)足m<n,且f(m)=f(n),若f(x)在區(qū)間[m2,n]上的最大值為2,則m,n的值分別為()(A)QUOTE,2 (B)QUOTE,4 (C)QUOTE,QUOTE (D)QUOTE,49.(2021·濟(jì)南模擬)設(shè)函數(shù)f(x)定義在實(shí)數(shù)集上,f(2-x)=f(x),且當(dāng)x≥1時(shí),f(x)=lnx,則有()(A)f(QUOTE)<f(2)<f(QUOTE) (B)f(QUOTE)<f(2)<f(QUOTE)(C)f(QUOTE)<f(QUOTE)<f(2) (D)f(2)<f(QUOTE)<f(QUOTE)10.(力氣挑戰(zhàn)題)設(shè)函數(shù)f(x)=QUOTE若f(a)>f(-a),則實(shí)數(shù)a的取值范圍是()(A)(-1,0)∪(0,1) (B)(-∞,-1)∪(1,+∞)(C)(-1,0)∪(1,+∞) (D)(-∞,-1)∪(0,1)二、填空題11.計(jì)算:lgQUOTE-lgQUOTE+lg7QUOTE=.12.函數(shù)y=loga(x-1)+2(a>0,且a≠1)的圖象恒過(guò)定點(diǎn).13.已知f(3x)=4xlog23+233,則f(2)+f(4)+f(8)+…+f(28)的值是.14.(力氣挑戰(zhàn)題)已知f(x)=QUOTE則f(2022)=.三、解答題15.(2021·長(zhǎng)春模擬)設(shè)f(x)=loga(1+x)+loga(3-x)(a>0,a≠1),且f(1)=2.(1)求a的值及f(x)的定義域.(2)求f(x)在區(qū)間[0,QUOTE]上的最大值.16.(力氣挑戰(zhàn)題)若f(x)=x2-x+b,且f(log2a)=b,log2f(a)=2(a(1)求f(log2x)的最小值及對(duì)應(yīng)的x值.(2)x取何值時(shí),f(log2x)>f(1),且log2f答案解析1.【解析】選D.由QUOTE得QUOTE∴0<x<1或x>1,故選D.2.【解析】選C.函數(shù)y=e|x|與y=cos|x|是偶函數(shù),函數(shù)y=e|x|在(0,1)上單調(diào)遞增,故選C.3.【解析】選C.對(duì)于A,由于y=3x在(-∞,+∞)上遞增,又x<y,故3x<3y,因此A錯(cuò)誤;對(duì)于D,由y=(QUOTE)x在(-∞,+∞)上遞減,又x<y,故(QUOTE)x>(QUOTE)y,因此D錯(cuò)誤;對(duì)于B,由于log3x<log3y<0,所以QUOTE>QUOTE,即logx3>logy3,故B錯(cuò)誤;對(duì)于C,由于y=log4x在(0,+∞)上遞增,又0<x<y,故log4x<log4y,因此C正確.4.【解析】選D.∵點(diǎn)(a,b)在函數(shù)y=lgx的圖象上,∴b=lga,則2b=2lga=lga2,故點(diǎn)(a2,2b)也在函數(shù)y=lgx的圖象上.5.【解析】選D.畫(huà)出函數(shù)f(x)的圖象如圖,由圖象可知y=f(x)在(-∞,+∞)上是增函數(shù),又f(2-x2)>f(x),故2-x2>x,解得-2<x<1.6.【解析】選D.由題意知,b=f(loQUOTE)=f(2),c=f(log2QUOTE)=f(-QUOTE)=f(QUOTE),又函數(shù)f(x)在[0,2]上是減函數(shù),因此f(2)<f(1)<f(QUOTE),∴c>a>b.7.【解析】選C.∵loga(a2+1)<0=loga1,a2+1>1,∴0<a<1,∴a2+1>2a,又loga2a<0,即2a>1,∴QUOTE解得QUOTE<a<1.【誤區(qū)警示】本題易忽視loga2a<0這一條件,而誤選A.8.【解析】選A.f(x)=|log2x|=QUOTE則函數(shù)f(x)在(0,1)上是減函數(shù),在(1,+∞)上是增函數(shù),又m<n且f(m)=f(n),則0<m<1,n>1,∴0<m2<m<1,∴f(m2)>f(m)=f(n),即函數(shù)f(x)在區(qū)間[m2,n]上的最大值為f(m2).由題意知f(m2)=2,即-log2m2=2,∴m=QUOTE,由f(m)=f(n)得-log2QUOTE=log2n,∴n=2.9.【解析】選C.由f(2-x)=f(x)知f(QUOTE)=f(2-QUOTE)=f(QUOTE),f(QUOTE)=f(2-QUOTE)=f(QUOTE),又函數(shù)f(x)=lnx在[1,+∞)上是增函數(shù),∴f(QUOTE)<f(QUOTE)<f(2),即f(QUOTE)<f(QUOTE)<f(2),故選C.10.【思路點(diǎn)撥】a的范圍不確定,故應(yīng)分a>0和a<0兩種狀況求解.【解析】選C.①當(dāng)a>0時(shí),-a<0,由f(a)>f(-a)得log2a>loQUOTEa,∴2log2a>0,∴a>1.②當(dāng)a<0時(shí),-a>0,由f(a)>f(-a)得loQUOTE(-a)>log2(-a),∴2log2(-a)<0,∴0<-a<1,即-1<a<0,由①②可知-1<a<0或a>1.11.【解析】原式=lg4+QUOTElg2-lg7-QUOTElg8+lg7+QUOTElg5=2lg2+QUOTE(lg2+lg5)-2lg2=QUOTE.答案:QUOTE12.【解析】∵loga1=0,∴x-1=1,即x=2,此時(shí)y=2,因此函數(shù)恒過(guò)定點(diǎn)(2,2).答案:(2,2)13.【解析】令3x=t,則x=log3t,∴f(t)=4log23·log3t+233=4log2t+233,∴f(2)+f(4)+f(8)+…+f(28)=4(log22+log24+log28+…+log228)+8×233=4·log2(2·22·23·…·28)+8×233=4·log2236+1864=4×36+1864=2008.答案:200814.【思路點(diǎn)撥】由當(dāng)x≥0時(shí),f(x)=f(x-7)知f(x)是周期為7的函數(shù),由此可對(duì)f(2022)進(jìn)行化簡(jiǎn).【解析】當(dāng)x≥0時(shí),f(x)=f(x-7),即f(x+7)=f(x),從而f(2022)=f(3)=f(-4)=log44=1.答案:115.【解析】(1)∵f(1)=2,∴l(xiāng)oga4=2(a>0,a≠1),∴a=2.由QUOTE得x∈(-1,3),∴函數(shù)f(x)的定義域?yàn)?-1,3).(2)f(x)=log2(1+x)+log2(3-x)=log2(1+x)(3-x)=log2[-(x-1)2+4],∴當(dāng)x∈(-1,1]時(shí),f(x)是增函數(shù);當(dāng)x∈(1,3)時(shí),f(x)是減函數(shù),函數(shù)f(x)在[0,QUOTE]上的最大值是f(1)=log24=2.【變式備選】已知函數(shù)f(x)=loga(3-ax).(1)當(dāng)x∈[0,2]時(shí),函數(shù)f(x)恒有意義,求實(shí)數(shù)a的取值范圍.(2)是否存在這樣的實(shí)數(shù)a,使得函數(shù)f(x)在區(qū)間[1,2]上為減函數(shù),并且最大值為1?假如存在,試求出a的值;假如不存在,請(qǐng)說(shuō)明理由.【解析】(1)由題設(shè),3-ax>0對(duì)一切x∈[0,2]恒成立,設(shè)g(x)=3-ax,∵a>0,且a≠1,∴g(x)=3-ax在[0,2]上為減函數(shù).從而g(2)=3-2a>0,∴a<QUOTE.∴a的取值范圍為(0,1)∪(1,QUOTE).(2)假設(shè)存在這樣的實(shí)數(shù)a,由題設(shè)知f(1)=1,即loga(3-a)=1,∴a=QUOTE.此時(shí)f(x)=loQUOTE(3-QUOTEx),當(dāng)x=2時(shí),f(x)沒(méi)有意義,故這樣的實(shí)數(shù)a不存在.16.【思路點(diǎn)撥】(1)由題設(shè)中的兩個(gè)等式先求出a,b的值,確定f(x),進(jìn)而求最小值及對(duì)應(yīng)的x值.(2)依據(jù)題意列出不等式組求解.【解析】(1)∵f(x)=x2-x+b,∴f(log2a)=(log2a)2-log2a+b,由已知(log2a)2-log2a+b=b,∴l(xiāng)og2a(log2a-1)=0.∵a≠1,∴