適用于老高考舊教材2023屆高考數(shù)學(xué)二輪總復(fù)習(xí)理專題檢測(cè)六函數(shù)與導(dǎo)數(shù)含解析

Page12專題檢測(cè)六函數(shù)與導(dǎo)數(shù)一、選擇題:本題共12小題,每小題5分,共60分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的.1.(2022·陜西咸陽(yáng)一模)下列函數(shù)中,既是偶函數(shù)又在(0,+∞)上單調(diào)遞增的是()A.y=x13 B.y=x2C.y=x D.y=2-|x|2.(2022·河北滄州三模)設(shè)f(x)是定義域?yàn)镽的偶函數(shù),且在(0,+∞)上單調(diào)遞減,則()A.flog314>f(2-B.flog314>f(2-C.f(2-32)>f(2D.f(2-23)>f(23.(2022·寧夏六盤山高級(jí)中學(xué)二模)定義在R上的偶函數(shù)f(x)在[0,+∞)上單調(diào)遞減,若f(x)>f(4),則實(shí)數(shù)x的取值范圍是()A.(-∞,4)B.(4,+∞)C.(-∞,-4)∪(4,+∞)D.(-4,4)4.(2022·河南開(kāi)封三模)函數(shù)f(x)=x-1xln|x|5.(2022·甘肅平?jīng)龆?函數(shù)f(x)在(-∞,+∞)上單調(diào)遞增,且為奇函數(shù),若f(2)=1,則滿足-1≤f(x-1)≤1的x的取值范圍是()A.[-2,2] B.[-1,3]C.[0,2] D.[1,3]6.(2022·北京·7)在北京冬奧會(huì)上,國(guó)家速滑館“冰絲帶”使用高效環(huán)保的二氧化碳跨臨界直冷制冰技術(shù),為實(shí)現(xiàn)綠色冬奧作出了貢獻(xiàn).如圖描述了一定條件下二氧化碳所處的狀態(tài)與T和lgP的關(guān)系,其中T表示溫度,單位是K;P表示壓強(qiáng),單位是bar.下列結(jié)論中正確的是()A.當(dāng)T=220,P=1026時(shí),二氧化碳處于液態(tài)B.當(dāng)T=270,P=128時(shí),二氧化碳處于氣態(tài)C.當(dāng)T=300,P=9987時(shí),二氧化碳處于超臨界狀態(tài)D.當(dāng)T=360,P=729時(shí),二氧化碳處于超臨界狀態(tài)7.已知函數(shù)f(x)=ln(x2+1+x)-1x3+2且f(3a)+f(-2a-3)>4,則正實(shí)數(shù)A.(1,+∞) B.(3,+∞)C.32,+∞ D.8.(2022·安徽合肥二模)函數(shù)f(x)=ex+4-e-x(e是自然對(duì)數(shù)的底數(shù))的圖象關(guān)于()A.直線x=-e對(duì)稱 B.點(diǎn)(-e,0)對(duì)稱C.直線x=-2對(duì)稱 D.點(diǎn)(-2,0)對(duì)稱9.設(shè)x=ln2,y=lg2,則()A.x-y>xy>tan(x+y)B.x-y>tan(x+y)>xyC.tan(x+y)>xy>x-yD.tan(x+y)>x-y>xy10.(2022·全國(guó)乙·文11)函數(shù)f(x)=cosx+(x+1)sinx+1在區(qū)間[0,2π]的最小值、最大值分別為()A.-π2,π2C.-π2,π2+2 D.11.(2022·北京西城二模)若函數(shù)f(x)=2x+3,x≤0A.(0,1] B.(0,1)C.(1,4) D.(2,4)12.(2022·陜西咸陽(yáng)二模)已知a∈(e,+∞),則函數(shù)f(x)=alnx+ax-xex的零點(diǎn)個(gè)數(shù)為()A.0 B.1 C.2 D.3二、填空題:本題共4小題,每小題5分,共20分.13.(2022·陜西西安四區(qū)縣聯(lián)考)若曲線f(x)=mxex-n在點(diǎn)(1,f(1))處的切線為y=ex,則mn=.

14.(2021·新高考Ⅰ·13)已知函數(shù)f(x)=x3(a·2x-2-x)是偶函數(shù),則a=.

15.(2022·陜西寶雞三模)已知函數(shù)f(x)是定義域?yàn)镽的奇函數(shù),當(dāng)x<0時(shí),f(x)=2x,則f(log27)=.

16.(2022·山東臨沂一模)已知函數(shù)f(x)=ex-1-e1-x+x,則不等式f(2-x)+f(4-3x)≤2的解集是.

三、解答題:本題共6小題,共70分.解答應(yīng)寫出文字說(shuō)明、證明過(guò)程或演算步驟.17.(10分)已知函數(shù)f(x)=x2-lnx.(1)求函數(shù)f(x)在x=1處的切線方程;(2)若f(x)+ex-1-ax≥0,求實(shí)數(shù)a的取值范圍.18.(12分)(2022·北京·20)已知函數(shù)f(x)=exln(1+x).(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)設(shè)g(x)=f'(x),討論函數(shù)g(x)在[0,+∞)上的單調(diào)性;(3)證明:對(duì)任意的s,t∈(0,+∞),有f(s+t)>f(s)+f(t).19.(12分)(2022·河南洛陽(yáng)一模)已知函數(shù)f(x)=xex+a(lnx-x)(a∈R(1)當(dāng)a=1時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)討論函數(shù)f(x)的極值點(diǎn)的個(gè)數(shù).20.(12分)(2022·陜西商洛一模)已知函數(shù)f(x)=exa+(b-1)x+(1)當(dāng)a=14,b=-1時(shí),求曲線y=f(x)在點(diǎn)(0,f(2)當(dāng)0<a≤e2,且x>2時(shí),f(x)>bln[a(x-1)]恒成立,求實(shí)數(shù)b的取值范圍.21.(12分)已知函數(shù)f(x)=1+lnx(1)求函數(shù)y=f(x)的最大值;(2)若關(guān)于x的方程lnx=xex-ex2+kx-1有實(shí)數(shù)根,求實(shí)數(shù)k的取值范圍;(3)證明:ln222+ln332+…+lnnn222.(12分)(2022·新高考Ⅰ·22)已知函數(shù)f(x)=ex-ax和g(x)=ax-lnx有相同的最小值.(1)求a;(2)證明:存在直線y=b,其與兩條曲線y=f(x)和y=g(x)共有三個(gè)不同的交點(diǎn),并且從左到右的三個(gè)交點(diǎn)的橫坐標(biāo)成等差數(shù)列.

專題檢測(cè)六函數(shù)與導(dǎo)數(shù)1.B解析:對(duì)于A,y=x13=3x是冪函數(shù),是奇函數(shù),不是偶函數(shù),不符合題意;對(duì)于B,y=x2+1是二次函數(shù),既是偶函數(shù)又在(0,+∞)上單調(diào)遞增,符合題意;對(duì)于C,y=x是冪函數(shù),既不是奇函數(shù)也不是偶函數(shù),不符合題意;對(duì)于D,y=2-|x|=2-x,2.C解析:∵f(x)是R上的偶函數(shù),∴flog314=f∵log34>1,1=20>2-23>2-32>0,且f(x)在(0,+∞)上單調(diào)遞減,∴f(2-33.D解析:∵偶函數(shù)滿足f(x)=f(-x)=f(|x|),∴f(x)=f(|x|)>f(4).又f(x)在[0,+∞)上單調(diào)遞減,∴|x|<4,即-4<x<4,∴x∈(-4,4).故選D.4.A解析:因?yàn)閒(x)的定義域?yàn)閧x|x≠0},f(-x)=-x-1-xln|-x|=-f(x),所以f(x)為奇函數(shù),圖象關(guān)于原點(diǎn)對(duì)稱,故C,D不正確;當(dāng)0<x<1時(shí),f(x)5.B解析:f(x)是奇函數(shù),故f(-2)=-f(2)=-1.又f(x)是增函數(shù),-1≤f(x-1)≤1,∴f(-2)≤f(x-1)≤f(2),則-2≤x-1≤2,解得-1≤x≤3.故選B.6.D解析:對(duì)于選項(xiàng)A,lgP=lg1026>3,T=220,根據(jù)圖象可知二氧化碳處于固態(tài);對(duì)于選項(xiàng)B,lgP=lg128>2,T=270,根據(jù)圖象可知二氧化碳處于液態(tài);對(duì)于選項(xiàng)C,lgP=lg9987≈3.999,T=300,根據(jù)圖象可知二氧化碳處于固態(tài);對(duì)于選項(xiàng)D,lgP=lg729>2,T=360,根據(jù)圖象可知二氧化碳處于超臨界狀態(tài).故選D.7.B解析:函數(shù)f(x)的定義域?yàn)?-∞,0)∪(0,+∞).令g(x)=f(x)-2,由g(-x)=ln((-x)2+1-x)-1(-x)3=-ln(x2+1+x)所以f(3a)+f(-2a-3)>4等價(jià)于g(3a)>g(2a+3),而a>0,由y=ln(x2+1+x),y=-1x3在(0,+∞)上單調(diào)遞增,故g(x所以3a>2a+3,可得a>3.故選B.8.D解析:由f(-2e-x)=e-x-2e+4-e-(-2e-x)=e-x-2e+4-e2e+x,它與f(x)之間沒(méi)有恒等關(guān)系,f(-2e-x)與f(x)相加也不為0,故A,B錯(cuò);而f(-4-x)=e-4-x+4-e-(-4-x)=e-x-e4+x=-f(x),所以f(x)的圖象關(guān)于點(diǎn)(-2,0)對(duì)稱.故選D.9.D解析:x-y=ln2-lg2=lg2-lg2lgelge=lg2(1-因?yàn)閘g10e>lg2,所以x-y>xy因?yàn)閤=ln2∈(0,1),y=lg2∈(0,1),所以x+y>x-y.由x+y∈(0,2),得tan(x+y)>x+y>x-y.故選D.10.D解析:函數(shù)f(x)的導(dǎo)數(shù)f'(x)=-sinx+sinx+(x+1)cosx=(x+1)cosx,x∈[0,2π].令f'(x)=0,得x=π2或x=3π2.當(dāng)x∈0,π2時(shí),f'(x)>0,函數(shù)f(x)單調(diào)遞增;當(dāng)x∈π2,3π2時(shí),f'(x)<0,函數(shù)f(x)單調(diào)遞減;當(dāng)x∈3π2,2π時(shí),f'(x)>0,函數(shù)f(x)單調(diào)遞增.故當(dāng)x=π2時(shí),函數(shù)f(x)有極大值fπ2=π2+2;當(dāng)x=3π2時(shí),函數(shù)f(x又因?yàn)閒(0)=2,f(2π)=2,所以函數(shù)f(x)在區(qū)間[0,2π]上的最小值為-3π2,最大值為π211.B解析:由題意f(x)的定義域?yàn)?-∞,a],a>0,當(dāng)x≤0時(shí),f(x)=2x+3,則f(x)在(-∞,0]上單調(diào)遞增,所以f(x)∈(3,4].要使定義域和值域的交集為空集,顯然0<a≤3.當(dāng)0<x≤a時(shí),f(x)=(x-2)2,若a≥2,則f(2)=0,此時(shí)顯然不滿足定義域和值域的交集為空集;若0<a<2時(shí),f(x)在(0,a]上單調(diào)遞減,此時(shí)f(x)∈[(a-2)2,4),則f(x)∈[(a-2)2,4)∪(3,4],所以a<(a-2)2,012.C解析:函數(shù)f(x)=alnx+ax-xex的定義域?yàn)?0,+∞),f'(x)=(x+1)ax令g(x)=ax-ex,x>0,顯然g(x)在(0,+∞)上單調(diào)遞減,由a>e,g(a)=1-ea<0,g(1)=a-e>則存在x0∈(1,a),使得g(x0)=0,即ax0=ex0,當(dāng)0<x<x0時(shí),g(x)>0,f'當(dāng)x>x0時(shí),g(x)<0,f'(x)>0,因此f(x)在(0,x0)上單調(diào)遞增,在(x0,+∞)上單調(diào)遞減,f(x)max=f(x0)=alnx0+ax0-x0ex0=a(lnx0+x0-1)而f1a=aln1a+1-1ae1a=-alna+1-1ae1a<a+1-1ae1a即f(x)在(0,x0)上存在唯一零點(diǎn).又f(a)=a(lna+a-ea),令h(x)=lnx+x-ex,x>e,則h'(x)=1x+1-ex<則h(x)在(e,+∞)上單調(diào)遞減,?x>e,h(x)<h(e)=1+e-ee<1+e-e2<0,于是得f(a)<0,則存在x2∈(x0,a)使得f(x2)=0,即f(x)在(x0,+∞)上存在唯一零點(diǎn).綜上,函數(shù)f(x)=alnx+ax-xex的零點(diǎn)個(gè)數(shù)為2.故選C.13.-e4解析:將x=1代入y=ex,得切點(diǎn)為(1,e),將切點(diǎn)坐標(biāo)代入曲線f(x)=mxex-n,得e=me-n又f'(x)=mex(x+1),f'(1)=2me=e,則m=12,所以e=12e-n,n=-e2,所以14.1解析:∵函數(shù)f(x)=x3(a·2x-2-x)是偶函數(shù),∴f(x)=f(-x),即x3(a·2x-2-x)=(-x)3[a·2-x-2-(-x)].整理得,a·2x-2-x=-(a·2-x-2x),即(a-1)·2x+(a-1)·2-x=0.(a-1)(2x+2-x)=0.∴a=1.15.-17解析:由f(x)是定義域?yàn)镽的奇函數(shù),且當(dāng)x<0時(shí),f(x)=2x,得f(log27)=-f(-log27)=-flog21716.[1,+∞)解析:構(gòu)造函數(shù)g(x)=f(x)-1=ex-1-1ex-1+(x-1),那么g(x)是增函數(shù),且其圖象向左移動(dòng)一個(gè)單位長(zhǎng)度后得到h(x)=g(x+1)=ex-h(x)的定義域?yàn)镽,且h(-x)=1ex-ex-x=-h(∴h(x)為奇函數(shù),圖象關(guān)于原點(diǎn)對(duì)稱,∴g(x)的圖象關(guān)于(1,0)對(duì)稱,即g(x)+g(2-x)=0.不等式f(2-x)+f(4-3x)≤2等價(jià)于f(2-x)-1+f(4-3x)-1≤0,等價(jià)于g(2-x)≤-g(4-3x)=g(2-4+3x)=g(3x-2),結(jié)合g(x)單調(diào)遞增可知2-x≤3x-2,解得x≥1.17.解(1)定義域?yàn)?0,+∞),f'(x)=2x-1x,則f'(1)=f(1)=1,故切線方程為y-f(1)=f'(1)(x-1),即x-y=0.(2)(方法一)記F(x)=x2-lnx+ex-1-ax,由F(1)≥0,得1-0+1-a≥0,即a≤2.當(dāng)a≤2時(shí),由x>0,F(x)≥x2-lnx+ex-1-2x,令G(x)=x2-lnx+ex-1-2x,則G'(x)=2x-1x+ex-1-2=2(x-1)+e當(dāng)x∈(0,1)時(shí),G'(x)<0;當(dāng)x∈(1,+∞)時(shí),G'(x)>0,所以G(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增.G(x)≥G(1)=0,即F(x)≥G(x)≥0.綜上可知,實(shí)數(shù)a的取值范圍為(-∞,2].(方法二)由條件,x2-lnx+ex-1-ax≥0,x>0,所以a≤x2-lnx+ex-則F'(x)=2x當(dāng)x∈(0,1)時(shí),F'(x)<0;當(dāng)x∈(1,+∞)時(shí),F'(x)>0,所以F(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,F(x)min=F(1)=2,則實(shí)數(shù)a的取值范圍為(-∞,2].18.(1)解由題得f'(x)=ex·ln(1+x)+ex·11+x=ex·ln(1+x)+11+x,故f'(0)=e0·ln(1+0)+11+0=1,f(0)=e0ln(1+0)=0,因此曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=x(2)解由(1)知g(x)=f'(x)=ex·ln(1+x)+11+x,x∈[0,+∞),則g'(x)=ex·ln(1+x)+21+x-1(1+x)2=ex·ln(1+x)+2x+1(1+x)2由于x∈[0,+∞),故ln(1+x)≥0,11+x>0,所以對(duì)任意x∈[0,+∞),g'(x)>0恒成立,故g(x)在[0,+∞)上單調(diào)遞增.(3)證明設(shè)函數(shù)F(x)=f(x+t)-f(x)(x>0),F'(x)=f'(x+t)-f'(x)=g(x+t)-g(x).因?yàn)閠>0,所以x+t>x>0.因?yàn)間(x)在[0,+∞)上單調(diào)遞增,所以g(x+t)>g(x),即g(x+t)-g(x)>0,F'(x)>0,所以F(x)在(0,+∞)上單調(diào)遞增.又因?yàn)閟>0,所以F(s)>F(0),即f(s+t)-f(s)>f(t)-f(0).又f(0)=0,所以f(s+t)>f(s)+f(t).19.解(1)當(dāng)a=1時(shí),f(x)=xex+lnx-x(f'(x)=1-xex+1x-1=(1-x)1又f(1)=1e-1,∴曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y=1e-(2)由題知f'(x)=1-xex+a1x-1=1-xxxex+a(x>0)令g'(x)>0,得0<x<1;令g'(x)<0,得x>1,∴g(x)的單調(diào)遞增區(qū)間為(0,1),單調(diào)遞減區(qū)間為(1,+∞),∴g(x)max=g(1)=1e+a,∴a<g(x)≤1e①當(dāng)a≥0時(shí),g(x)>0恒成立.令f'(x)>0,得0<x<1;令f'(x)<0,得x>1,∴f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,此時(shí)有且只有1個(gè)極值點(diǎn).②當(dāng)a≤-1e時(shí),g(x)≤0恒成立令f'(x)>0,得x>1;令f'(x)<0,得0<x<1,∴f(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增.此時(shí)f(x)有且只有1個(gè)極值點(diǎn).③當(dāng)-1e<a<0時(shí),方程g(x)=0有兩個(gè)不相等的實(shí)數(shù)根x1,x2,不妨設(shè)0<x1<1<x2,則當(dāng)0<x<x1時(shí),f'(x)<0;當(dāng)x1<x<1時(shí),f'(x)>0;當(dāng)1<x<x2時(shí),f'(x)<0;當(dāng)x2<x時(shí),f'(x)>0,∴f(x)在(0,x1)上單調(diào)遞減,在(x1,1)上單調(diào)遞增,在(1,x2)上單調(diào)遞減,在(x2,+∞此時(shí)f(x)有3個(gè)極值點(diǎn).綜上,當(dāng)a≥0或a≤-1e時(shí),f(x)有1個(gè)極值點(diǎn);當(dāng)-1e<a<0時(shí),f(x20.解(1)當(dāng)a=14,b=-1時(shí),f(x)=4ex-2x+則f'(x)=4ex-2,又因?yàn)閒(0)=5,f'(0)=2,所以曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=2x+5.(2)f(x)>bln[a(x-1)]恒成立,即exa+bx-x+1>bln(x-1)+bln等價(jià)于exa+b(x-lna)>x-1+bln(x-構(gòu)造函數(shù)h(x)=x+blnx(x>1),則exa+blnexa>x-1+bln(x-1)在x∈(2,+∞)上恒成立等價(jià)于hexa>h(x-1)在x∈(2,+∞)上恒成立.因?yàn)?<a≤e2令函數(shù)H(x)=ex-2-x+1(x>2),則H'(x)=ex-2-1,顯然H'(x)是增函數(shù),則H'(x)>H'(2)=0,H(x)在(2,+∞)上單調(diào)遞增,所以H(x)>H(2)=0,故exa-x+1≥ex-2-x+1>0,從而exa>x-又hexa>h(x-1)在x∈(2,+所以h(x)在(1,+∞)上單調(diào)遞增,所以當(dāng)x∈(1,+∞)時(shí),h'(x)=1+bx所以b≥(-x)max=-1,即實(shí)數(shù)b的取值范圍是[-1,+∞).21.(1)解f'(x)=-lnxx2.在區(qū)間(0,1)上,f'(x)>0,f(x)單調(diào)遞增;在區(qū)間(1,+∞)上,f'(x)<0,f(x)單調(diào)遞減,所以f(x)max=f(1)=1,即當(dāng)x=1時(shí),f(2)解(方法一)lnx=xex-ex2+kx-1有實(shí)數(shù)根等價(jià)于1+lnxx=ex-ex+k由于f(x)=1+lnxx,由(1)得,在區(qū)間(0,1)上,f(x)單調(diào)遞增;在區(qū)間(1,+∞)上,f(x)單調(diào)遞減,f(x)max=f(1)=令g(x)=ex-ex+k(x>0),則g'(x)=ex-e.在區(qū)間(0,1)上,g'(x)<0,g(x)單調(diào)遞減;在區(qū)間(1,+∞)上,g'(x)>0,g(x)單調(diào)遞增,所以g(x)min=g(1)=k,且x→+∞時(shí),f(x)→0,g(x)→+∞,于是當(dāng)滿足k≤1時(shí),方程lnx=xex-ex2+kx-1有實(shí)數(shù)根.所以實(shí)數(shù)k的取值范圍是(-∞,1].(方法二)lnx=xex-ex2+kx-1可變?yōu)閗=1+lnxx+(ex-ex).令g(x)=1+lnxx+(e則g'(x)=-lnxx2+(e-ex).在(0,1)上,g'(x)>0,g(x)單調(diào)遞增;在(1,+∞)上,g'(x)<0,gx→+∞時(shí),g(x)→-∞.又g(1)=1,可得g(x)∈(-∞,1].所以實(shí)數(shù)k的取值范圍是(-∞,1].(3)證明由(1)得f(x)≤1,當(dāng)且僅當(dāng)x=1時(shí)取等號(hào),即lnx<x-1(x>1),所以當(dāng)n≥2時(shí),lnn2<n2-1,lnnn2=12·lnn2n2<12·n2-1n2=121-1n2<121-1n(22.(1)解由題意,得f'(x)=ex-a,g'(x)=a-1x=ax-當(dāng)a≤0時(shí),f'(x)>0,f(x)為增函數(shù),f(x)無(wú)最小值,g'(x)<0,g(x)在區(qū)間(0,+∞)上單調(diào)遞減,g(x)無(wú)最小值.當(dāng)a>0時(shí),令f'(x)=0,即ex-a=0,解得x=lna.當(dāng)x變化時(shí),f'(x),f(x)的變化情況如下表.x(-∞,lna)lna(lna,+∞)f'(x)—0+f(x)↘極小值a-alna↗由表可知,當(dāng)x=lna時(shí),f(x)取得極小值也是最小值,為a-alna.由g'(x)=0得x=1a>0.當(dāng)x在區(qū)間(0,+∞)上變化時(shí),g'(x),g(x)的變化情況如下表x011g'(x)—0+g(x)↘極小值1+lna↗由上表可知,當(dāng)x=1a時(shí),g(x)取得極小值也是最小值,為1+lna.則a-alna=1+lna.令h(a)=a-alna-1-lna,則h'(a)=1-(lna+1)-1a=-lna-1a=令1a=t,則lnt-t<0∴h(a)在區(qū)間(0,+∞)上單調(diào)遞減,且h(1)=0.∴a=1.(2)證明由(1)知f(x)=ex-x,g(x)=x-lnx,且f(x)在區(qū)間(-∞,0)上單調(diào)遞減,在區(qū)間(0,+∞)上單調(diào)遞增,g(x)在區(qū)間(0,1)上單調(diào)遞減,在區(qū)間(1,+∞)上單調(diào)遞增,且f(x)min=g(x)min=1.①當(dāng)b<1時(shí),此時(shí)f(x)min=g(x)min=1>b.顯然直線y=b與兩條曲線y=f(x)和y=g(x)沒(méi)有交點(diǎn),不符合題意;②當(dāng)b=1時(shí),此時(shí)f(x)min=g(x)min=1=b,直線y=b與兩條曲線y=f(x)和y=g(x)共有2個(gè)交點(diǎn),交點(diǎn)的橫坐標(biāo)分別為0和1;③當(dāng)b>1時(shí),首先,證明y=b與曲線y=f(x)有2個(gè)交點(diǎn),