新教材高中數(shù)學(xué)第6章冪函數(shù)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)1-3綜合拔高練含解析蘇教版必修第一冊(cè)

PAGEPAGE11綜合拔高練五年高考練考點(diǎn)1比較大小1.(2020天津,6,5分,)設(shè)a=30.7,b=13-0.8,c=log0.70.8,則a,b,cA.a<b<cB.b<a<cC.b<c<aD.c<a<b2.(2019課標(biāo)全國Ⅲ,11,5分,)設(shè)f(x)是定義域?yàn)镽的偶函數(shù),且在(0,+∞)單調(diào)遞減,則 ()A.flog314>f(2-B.flog314>f(2-C.f(2-32)>f(2D.f(2-23)>f(2考點(diǎn)2冪函數(shù)、指數(shù)函數(shù)、對(duì)數(shù)函數(shù)的圖象3.(2020北京,6,4分,)已知函數(shù)f(x)=2x-x-1,則不等式f(x)>0的解集是 ()A.(-1,1)B.(-∞,-1)∪(1,+∞)C.(0,1)D.(-∞,0)∪(1,+∞)4.(2019課標(biāo)全國Ⅲ,7,5分,)函數(shù)y=2x32x+2-5.(2018課標(biāo)全國Ⅲ,7,5分,)下列函數(shù)中,其圖象與函數(shù)y=lnx的圖象關(guān)于直線x=1對(duì)稱的是 ()A.y=ln(1-x)B.y=ln(2-x)C.y=ln(1+x)D.y=ln(2+x)6.(2018課標(biāo)全國Ⅱ,3,5分,)函數(shù)f(x)=ex-e-xABCD考點(diǎn)3冪函數(shù)、指數(shù)函數(shù)、對(duì)數(shù)函數(shù)的性質(zhì)7.(2020全國Ⅰ,12,5分,)若2a+log2a=4b+2log4b,則()A.a>2bB.a<2bC.a>b2D.a<b28.(2020全國Ⅱ文,10,5分,)設(shè)函數(shù)f(x)=x3-1x3,則f(x) (A.是奇函數(shù),且在(0,+∞)單調(diào)遞增B.是奇函數(shù),且在(0,+∞)單調(diào)遞減C.是偶函數(shù),且在(0,+∞)單調(diào)遞增D.是偶函數(shù),且在(0,+∞)單調(diào)遞減9.(2020全國Ⅱ理,9,5分,)設(shè)函數(shù)f(x)=ln|2x+1|-ln|2x-1|,則f(x) ()A.是偶函數(shù),且在12,B.是奇函數(shù),且在-1C.是偶函數(shù),且在-∞,-12D.是奇函數(shù),且在-∞,-110.(2020北京,11,5分,)函數(shù)f(x)=1x+1+lnx的定義域是11.(2020江蘇,7,5分,)已知y=f(x)是奇函數(shù),當(dāng)x≥0時(shí),f(x)=x23,則f(-8)的值是考點(diǎn)4反函數(shù)12.(2020上海,4,4分,)已知函數(shù)f(x)=x3,f-1(x)是f(x)的反函數(shù),則f-1(x)=.

13.(2018上海,4,4分,)設(shè)常數(shù)a∈R,函數(shù)f(x)=log2(x+a).若f(x)的反函數(shù)的圖象經(jīng)過點(diǎn)(3,1),則a=.

三年模擬練1.(2021江蘇泰州高一期末,)設(shè)函數(shù)f(x)=4x-2,x≤1,log2(x+3),xA.-C.12.(2021河南駐馬店高一期末,)設(shè)函數(shù)y=f(x)和y=f(-x),若兩函數(shù)在區(qū)間[m,n]上的單調(diào)性相同,則把區(qū)間[m,n]叫作y=f(x)的“穩(wěn)定區(qū)間”.已知區(qū)間[1,2021]為函數(shù)g(x)=13x+a的“穩(wěn)定區(qū)間”,則實(shí)數(shù)a的可能取值是A.-1033.(2021江蘇南通如東高級(jí)中學(xué)高一期末,)已知f(x)=2-3x-2x2,x≤0,|lgx|,x>0.若關(guān)于x的方程f(x)=m(m∈R)有四個(gè)不相等的實(shí)根x1,x2,xA.04.(多選)(2021浙江溫州高一期末,)在同一平面直角坐標(biāo)系中,函數(shù)f(x)=loga(x-b),g(x)=bx-a的圖象可能是 ()ABCD5.(多選)(2020江蘇淮安淮陰中學(xué)高二期末,)已知函數(shù)f(x)=2x+1,x≤0,|log2x|-1,x>0,則方程[A.2B.6C.5D.46.(2019吉林白山高一期末聯(lián)考,)定義新運(yùn)算:當(dāng)m≥n時(shí),mn=m;當(dāng)m<n時(shí),mn=n.若函數(shù)f(x)=[(2x2)-(1log2x)]·2x,則f(x)在(0,2)上的值域?yàn)?

7.(2020江蘇南通高一期末,)已知函數(shù)f(x)=(3-a)x+a-1,x<1,lo8.(2021河北唐山高一期末,)已知定義域?yàn)镽的函數(shù)f(x)=n-3x(1)求y=f(x)的解析式;(2)若flog4x·log28x+f9.(2020江蘇揚(yáng)州大學(xué)附屬中學(xué)高一上期中,)已知函數(shù)y=f(x),若對(duì)于給定的正整數(shù)k,f(x)在其定義域內(nèi)存在實(shí)數(shù)x0,使得f(x0+k)=f(x0)+f(k),則稱此函數(shù)f(x)為“保k值函數(shù)”.(1)若函數(shù)f(x)=2x為“保1值函數(shù)”,求x0的值;(2)①試判斷函數(shù)f(x)=x+1x是不是“保k值函數(shù)”,若是,求出k的值;若不是,請(qǐng)說明理由②試判斷函數(shù)f(x)=lnaex+1是不是“保2值函數(shù)”,若是,求出實(shí)數(shù)a的取值范圍;若不是答案全解全析6.1~6.3綜合拔高練五年高考練1.D由函數(shù)y=3x單調(diào)遞增,函數(shù)y=log0.7x(x>0)單調(diào)遞減,可知a=30.7>30=1,b=13-0.8=30.8>30.7=a,c=log0.70.8<log0.70.7=1,即c<1<a2.C∵f(x)是定義域?yàn)镽的偶函數(shù),∴f(-x)=f(x).∴flog314=f(-log34)=f∵log34>log33=1,且1>2-∴l(xiāng)og34>2-2∵f(x)在(0,+∞)上單調(diào)遞減,∴f(2-32)>f(2-23)>f(log34)=3.D不等式f(x)>0等價(jià)于不等式2x>x+1,作出函數(shù)y=2x和函數(shù)y=x+1的圖象,如圖所示,易知兩個(gè)函數(shù)圖象的交點(diǎn)坐標(biāo)為(1,2)和(0,1),觀察函數(shù)圖象可知,當(dāng)x>1或x<0時(shí),函數(shù)y=2x的圖象在函數(shù)y=x+1圖象的上方,此時(shí)2x>x+1,故不等式f(x)>0的解集為(-∞,0)∪(1,+∞),故選D.4.B設(shè)f(x)=2x32x+2-x(x∈[-6,6]),則f(-x)=2(-x)32-x+2x=-f(x),∴f(x)為奇函數(shù),排除選項(xiàng)C;當(dāng)x=-1時(shí),f(-1)=-45<0,排除選項(xiàng)5.B設(shè)所求函數(shù)圖象上一點(diǎn)的坐標(biāo)為(a,b),則點(diǎn)(a,b)關(guān)于直線x=1的對(duì)稱點(diǎn)(2-a,b)在函數(shù)y=lnx的圖象上,∴b=ln(2-a),故所求的函數(shù)為y=ln(2-x).6.B因?yàn)閒(x)的定義域關(guān)于原點(diǎn)對(duì)稱且f(-x)=-f(x),所以f(x)為奇函數(shù),排除A選項(xiàng);f(2)=e2-1e24>1,排除C、7.B2a+log2a=22b+log2b<22b+log2(2b),令f(x)=2x+log2x,則f(a)<f(2b),又易知f(x)在(0,+∞)上單調(diào)遞增,所以a<2b,故選B.8.A由函數(shù)y=x3和y=-1x3都是奇函數(shù),知函數(shù)f(x)=x3-1x3是奇函數(shù).由函數(shù)y=x3和y=-1x3都在區(qū)間(0,+∞)上單調(diào)遞增,知函數(shù)f(x)=x3-1x3在區(qū)間(0,+∞)上單調(diào)遞增9.D|2x+1|>0,|2x-1|>0?x∈xx≠±12,x∈R,∴函數(shù)f(x)的定義域關(guān)于原點(diǎn)對(duì)稱,又∵f(-x)=ln|-2x+1|-ln|-2x-1|=ln|2x-1|-ln|2x+1|=-f(x),∴f(x)是奇函數(shù),排除A、C;當(dāng)x∈-12,∴f(x)在-12,12上單調(diào)遞增,排除B;當(dāng)x∈-∞,-12時(shí),f(x)=ln(-2x-1)-ln(1-2x)=ln2x+12x-1=ln1+22x-1,∵10.答案(0,+∞)解析要使函數(shù)f(x)有意義,則x+1≠0,x因此函數(shù)f(x)的定義域?yàn)?0,+∞).11.答案-4解析由函數(shù)f(x)是奇函數(shù)得f(-8)=-f(8)=-823=-(23)12.答案3解析由f(x)=x3,得x=3y,所以f(x)的反函數(shù)是f-1(x)=313.答案7解析∵f(x)的反函數(shù)的圖象經(jīng)過點(diǎn)(3,1),∴函數(shù)f(x)=log2(x+a)的圖象經(jīng)過點(diǎn)(1,3),∴l(xiāng)og2(1+a)=3,解得a=7.三年模擬練1.D當(dāng)x-14>1,即x>54時(shí),當(dāng)x≤1時(shí),f(x)+fx-14=4x-2+4×x-14-2>2,解得當(dāng)x>1且x-14≤1,即1<x≤54時(shí),f(x)+fx-14=log2(x+3)+4x-14-2=log2(x+3)+4x-3,易知函數(shù)y=f(x)+fx-14為單調(diào)增函數(shù),所以f(綜上,不等式的解集為78,+2.Bg(x)=13x+a,則g(-x)=|3x①當(dāng)兩個(gè)函數(shù)都是增函數(shù)時(shí),13x+a≤0,3x+a≥0在區(qū)間[1,2021]上恒成立,即(-3x)max≤a②當(dāng)兩個(gè)函數(shù)都是減函數(shù)時(shí),13x+a≥0,3x+a≤0在區(qū)間[1,2021]上恒成立,即綜上,-3≤a≤-13.故選3.D函數(shù)f(x)的圖象如圖所示.由圖可知方程f(x)=m有四個(gè)不相等的實(shí)根時(shí)2≤m<258設(shè)y=m與y=2-3x-2x2=-2x+342+258(x≤0)圖象的交點(diǎn)的橫坐標(biāo)從左到右為x1,x2,則x1<0,x2≤0,且x1+x2=-32∴x1x2=(-x1)(-x2)<(-x設(shè)y=m與y=|lgx|(x>0)圖象的交點(diǎn)的橫坐標(biāo)從左到右為x3,x4,則x3>0,x4>0.由|lgx3|=|lgx4|得lgx3=-lgx4,∴x3x4=1.∴x1x2x3x4∈0,9164.AC選項(xiàng)A中,根據(jù)對(duì)數(shù)函數(shù)的圖象知f(x)在定義域上單調(diào)遞增,所以a>1,又f(x)的圖象過點(diǎn)(2,0),所以b=1,所以g(x)=1,故A符合;選項(xiàng)B中,由g(x)的圖象可知a>1,0<b<1,所以對(duì)數(shù)函數(shù)f(x)的圖象應(yīng)由y=logax的圖象向右平移b個(gè)單位長(zhǎng)度,故B不符合;選項(xiàng)C中,由f(x)的圖象知0<a<1且0<b<1,由g(x)的圖象知0<a<1且0<b<1,故C符合;選項(xiàng)D中,由f(x)的圖象知0<a<1,由g(x)的圖象知a=0,故D不符合.故選AC.5.ACD畫出f(x)的圖象如圖所示.令t=f(x),則t2-2t+a2-1=0,Δ=4(2-a2).當(dāng)Δ=0,即a2=2時(shí),t=1,此時(shí)f(x)=1,由圖可知,直線y=1與y=f(x)的圖象有2個(gè)交點(diǎn),所以方程[f(x)]2-2f(x)+a2-1=0的根的個(gè)數(shù)為2,A正確.當(dāng)Δ>0,即a2<2時(shí),t=1±2-a2,因?yàn)?<2-a2≤2當(dāng)t=1-2-a2時(shí),f(x)∈[1-2,1),由圖可知x有當(dāng)t=1+2-a2時(shí),若t∈(1,2],則x有3個(gè)解;若t∈(2,1+2],則x有故方程[f(x)]2-2f(x)+a2-1=0的根的個(gè)數(shù)為5或4,C、D正確.故選ACD.6.答案(1,12)解析當(dāng)2x≥2,即x≥1時(shí),2x2=2x;當(dāng)2x<2,即x<1時(shí),2x2=2.當(dāng)1≥log2x,即0<x≤2時(shí),1log2x=1;當(dāng)1<log2x,即x>2時(shí),1log2x=log2x.∴f(x)=2當(dāng)0<x<1時(shí),f(x)=2x是增函數(shù),∴1<f(x)<2;當(dāng)1≤x<2時(shí),f(x)=22x-2x=2x∵1≤x<2,∴2≤2x<4,∴2-122-14≤f(x)<4-1綜上,f(x)在(0,2)上的值域?yàn)?1,12).7.答案1解析因?yàn)楹瘮?shù)f(x)在R上是增函數(shù),所以y=(3-a)x+a-1在區(qū)間(-∞,1)上是增函數(shù)且y=logax2-ax+由函數(shù)y=(3-a)x+a-1在(-∞,1)上是增函數(shù),得3-a>0?a<3.①對(duì)于函數(shù)y=logax2-ax+114,x∈[1,+∞),令u=x當(dāng)0<a<1時(shí),a2<1,所以u(píng)=x2-ax+114在[1,+∞)又y=logau為定義域內(nèi)的減函數(shù),所以根據(jù)復(fù)合函數(shù)“同增異減”可得0<a<1時(shí),函數(shù)y=logax2-ax+114在區(qū)間[1,+∞)當(dāng)a>1時(shí),要使函數(shù)y=logax2-ax+114為定義域內(nèi)的增函數(shù),需函數(shù)u=x2-ax+11所以a2≤1,12又a>1,所以1<a≤2.②由①②得1<a≤2.因?yàn)閒(x)在R上是增函數(shù),所以f(x)的圖象在銜接點(diǎn)處的函數(shù)值應(yīng)滿足3-a+a-1≤loga12-a+114,即a2解得-52所以實(shí)數(shù)a的取值范圍是1,8.解析(1)因?yàn)楹瘮?shù)f(x)=n-3所以f(-x)=-f(x),即n-所以n·所以n·3x-1=-n+3x,即(n-1)(3x+1)=0,解得n=1.故函數(shù)f(x)=1-(2)由(1)知f(x)=1-所以f(x)在(-∞,+∞)上單調(diào)遞減.由flog4x·log28x+f(4-2a)>0,因?yàn)楹瘮?shù)f(x)是奇函數(shù),所以flog4x·lo所以log4x·(3-log2x)<2a-4,整理得12log2x·(3-log2x)<2a-4設(shè)t=log2x,t∈R,則12(3t-t2)<2a-4恒成立,令y=12(3t-t當(dāng)t=32時(shí),y=12(3t-t2)有最大值,且最大值為98,所以2a-4>98,所以9.解析(1)因?yàn)楹瘮?shù)f(x)=2x為“保1值函數(shù)”,所以存在x0,使得f(x0+1)=f(x0)+f(1),即2x0+1=2x0+2,故(2)①函數(shù)f(x)=