2023-2024學(xué)年高中數(shù)學(xué)人教A版2019選擇性課后習(xí)題第五章　一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用培優(yōu)課恒成立能成立問(wèn)題

培優(yōu)課——恒成立、能成立問(wèn)題必備知識(shí)基礎(chǔ)練1.若函數(shù)f(x)=x2+4x+blnx在(0,+∞)上是減函數(shù),則b的取值范圍是()A.(∞,2] B.(∞,2)C.(2,+∞) D.[2,+∞)2.已知函數(shù)f(x)=lnxax存在最大值0,則a的值為()A.1 B.2 C.e D.13.已知函數(shù)f(x)=exx+a,若f(x)>0恒成立,則實(shí)數(shù)a的取值范圍是()A.(1,+∞) B.(∞,1)C.[1,+∞) D.(∞,1]4.已知a≤4x3+4x2+1對(duì)任意x∈[2,1]都成立,則實(shí)數(shù)a的取值范圍是.

5.已知函數(shù)f(x)=x+aex,其中a∈R,e(1)當(dāng)a=1時(shí),求函數(shù)f(x)在區(qū)間[0,+∞)上的零點(diǎn)個(gè)數(shù);(2)若f(x)>2對(duì)任意的實(shí)數(shù)x恒成立,求a的取值范圍.6.已知函數(shù)f(x)=ex2ax(a∈R).(1)若a=12,求函數(shù)f(x)的單調(diào)區(qū)間(2)當(dāng)x∈[2,3]時(shí),f(x)≥0恒成立,求實(shí)數(shù)a的取值范圍.關(guān)鍵能力提升練7.(多選題)定義在(0,+∞)上的函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且(x+1)f'(x)f(x)<x2+2x對(duì)x∈(0,+∞)恒成立.下列結(jié)論正確的是()A.2f(2)3f(1)>5B.若f(1)=2,x>1,則f(x)>x2+12x+C.f(3)2f(1)<7D.若f(1)=2,0<x<1,則f(x)>x2+12x+8.若存在x∈1e,e,使得不等式2xlnx+x2mx+3≥0成立,則實(shí)數(shù)m的最大值為()A.1e+3e2 B.3e+eC.4 D.e219.已知函數(shù)f(x)=sin2x+π6x22mx在0,π6上是減函數(shù),則實(shí)數(shù)m的最小值是()A.3 B.32 C.32 D10.已知函數(shù)f(x)=lnx,x>0,kx,x≤0.若?x0∈R且x0≠0,使得f(x0)=f(xA.(∞,1] B.∞,1e C.[1,+∞) D.1e,+∞11.已知y=f(x)是奇函數(shù),當(dāng)x∈(0,2)時(shí),f(x)=lnxaxa>12,當(dāng)x∈(2,0)時(shí),f(x)的最小值為1,則a的值為.

12.設(shè)函數(shù)f(x)=ax33x+1(a>1),若對(duì)于任意的x∈[1,1],都有f(x)≥0成立,則實(shí)數(shù)a的值為.

13.已知函數(shù)f(x)=aexx,x∈[1,3],且?x1,x2∈[1,3],x1≠x2,f(x1)-f(x14.已知函數(shù)f(x)=xlnx.(1)求f(x)的最小值;(2)若對(duì)任意x≥1都有f(x)≥ax1,求實(shí)數(shù)a的取值范圍.學(xué)科素養(yǎng)創(chuàng)新練15.已知函數(shù)f(x)=(x+a)ex,其中a為常數(shù).(1)若函數(shù)f(x)在區(qū)間[1,+∞)上是增函數(shù),求實(shí)數(shù)a的取值范圍;(2)若f(x)≥e3xex在x∈[0,1]時(shí)恒成立,求實(shí)數(shù)a的取值范圍.參考答案培優(yōu)課——恒成立、能成立問(wèn)題1.A∵f(x)=x2+4x+blnx在(0,+∞)上是減函數(shù),∴f'(x)≤0在(0,+∞)上恒成立,即f'(x)=2x+4+bx≤0,即b≤2x24x∵2x24x=2(x1)22≥2,∴b≤2.2.D∵f'(x)=1xa,x>0,∴當(dāng)a≤0時(shí),f'(x)>0恒成立,故函數(shù)f(x)單調(diào)遞增,不存在最大值;當(dāng)a>0時(shí),令f'(x)=0,得x=1a∴當(dāng)x∈0,1a時(shí),f'(x)>0,函數(shù)f(x)單調(diào)遞增,當(dāng)x∈1a,+∞時(shí),f'(x)<0,函數(shù)f(x)單調(diào)遞減,∴f(x)max=f1a=ln1a1=0,解得a=1e3.Af'(x)=ex1,令f'(x)>0,解得x>0,令f'(x)<0,解得x<0,故f(x)在(∞,0)上是減函數(shù),在(0,+∞)上是增函數(shù),故f(x)min=f(0)=1+a.若f(x)>0恒成立,則1+a>0,解得a>1,故選A.4.(∞,15]設(shè)f(x)=4x3+4x2+1,x∈[2,1],則f'(x)=12x2+8x,令f'(x)=0,得x=0或x=23.所以在區(qū)間2,23上,f'(x)>0,f(x)為增函數(shù),在區(qū)間23,0上,f'(x)<0,f(x)為減函數(shù),在區(qū)間(0,1)上,f'(x)>0,f(x)為增函數(shù),因此在閉區(qū)間[2,1]上,函數(shù)f(x)在x=23處取得極大值f23,在x=0時(shí)函數(shù)取得極小值f(0),且f(0)=1,f(1)=9,f(2)=15,所以f(2)=15是最小值,所以實(shí)數(shù)a≤15.5.解(1)當(dāng)a=1時(shí),f(x)=x1e則f'(x)=1+1ex∴f(x)在[0,+∞)上是增函數(shù),又f(0)=1<0,f(1)=11e>故?x0∈(0,1),使得f(x0)=0,∴函數(shù)f(x)在區(qū)間[0,+∞)上有1個(gè)零點(diǎn).(2)若f(x)>2對(duì)任意的實(shí)數(shù)x恒成立,即a>ex(2x)恒成立,令g(x)=ex(2x),則g'(x)=ex(1x),令g'(x)>0,得x<1;令g'(x)<0,得x>1,∴g(x)在(∞,1)上是增函數(shù),在(1,+∞)上是減函數(shù),∴g(x)max=g(1)=e,∴a的取值范圍為(e,+∞).6.解(1)當(dāng)a=12時(shí),f(x)=exx,f'(x)=ex令f'(x)=0,得x=0;令f'(x)>0,得x>0;令f'(x)<0,得x<0.所以函數(shù)f(x)=exx的單調(diào)遞增區(qū)間為(0,+∞),單調(diào)遞減區(qū)間為(∞,0).(2)當(dāng)x∈[2,3]時(shí),f(x)=ex2ax≥0恒成立,等價(jià)于2a≤exx對(duì)任意的x∈[2,3]即2a≤exxmin,設(shè)g(x)=exx,則g'(x)=ex(x-1)x2,顯然當(dāng)x∈[2,3]∴g(x)在[2,3]上是增函數(shù),∴g(x)min=g(2)=e2∴2a≤e22,即a≤故實(shí)數(shù)a的取值范圍為∞,e24.7.CD設(shè)函數(shù)g(x)=f(則g'(x)=[=(x因?yàn)?x+1)f'(x)f(x)<x2+2x,所以g'(x)<0,故g(x)在(0,+∞)上是減函數(shù),從而g(1)>g(2)>g(3),整理得2f(2)3f(1)<5,f(3)2f(1)<7,故A錯(cuò)誤,C正確;當(dāng)0<x<1時(shí),若f(1)=2,因?yàn)間(x)在(0,+∞)上是減函數(shù),所以g(x)>g(1)=12,即f即f(x)>x2+12x+12,故D正確,B8.A∵2xlnx+x2mx+3≥0,∴m≤2lnx+x+3x設(shè)h(x)=2lnx+x+3x則h'(x)=2x+13當(dāng)1e≤x<1時(shí),h'(x)<0,h(x)單調(diào)遞減當(dāng)1<x≤e時(shí),h'(x)>0,h(x)單調(diào)遞增.∵存在x∈1e,e,m≤2lnx+x+3x成立,∴m≤h(x)max.∵h(yuǎn)1e=2+1e+3e,h(e)=2+e+3e∴h1e>h(e).∴m≤1e+3e∴m的最大值是1e+3e29.D由f(x)=sin2x+π6x22mx在0,π6上是減函數(shù),得f'(x)=2cos2x+π6xm≤0x∈0,π6,即2cos2x+π6x≤mx∈0,π6,令g(x)=2cos2x+π6xx∈0,π6,則g'(x)=4sin2x+π61x∈0,π6,當(dāng)x∈0,π6時(shí),π6≤2x+π6則2≤4sin2x+π6≤4,所以5≤4sin2x+π61≤3,即g'(x)<0,所以g(x)在x∈0,π6上是減函數(shù),g(x)max=g(0)=3,所以m≥3,m的最小值為3.10.D由題意可得,存在實(shí)數(shù)x0≠0,使得f(x0)=f(x0)成立,假設(shè)x0>0,則x0<0,所以有kx0=lnx0,則k=ln令h(x)=lnxx,則h'(x)=令h'(x)>0,即lnx>1,解得x>e,令h'(x)<0,即lnx<1,解得0<x<e,則h(x)在(0,e)上是減函數(shù),在(e,+∞)上是增函數(shù),所以h(x)≥h(x)min=h(e)=lnee所以k≥1e11.1由題意知,當(dāng)x∈(0,2)時(shí),f(x)的最大值為1.令f'(x)=1xa=0,得x=1∵a>12,∴0<1a<當(dāng)0<x<1a時(shí),f'(x)>當(dāng)1a<x<2時(shí),f'(x)<0∴f(x)max=f1a=lna1=1.解得a=1.12.4由題意得,f'(x)=3ax23,當(dāng)a>1時(shí),令f'(x)=3ax23=0,解得x=±aa,±aa∈(①當(dāng)1≤x<aa時(shí),f'(x)>0,f(x)單調(diào)遞增②當(dāng)aa<x<aa時(shí),f'(x)<0,f(x)③當(dāng)aa<x≤1時(shí),f'(x)>0,f(x)單調(diào)遞增所以只需faa≥0,且f(1)≥0即可,由faa≥0,得a·aa33·aa+1≥0,解得a≥4,由f(1)≥0,可得a≤4.綜上可得a=4.13.∞,9e3設(shè)x1>x2,f(x1)-f(x2)x1-x2<2可化為f(x可得函數(shù)g(x)=f(x)2x=aexx2x在x∈∴g'(x)=aex(x-1)x2即a≤2x2ex(x令h(x)=2x2ex則h'(x)=-2x[(x-1)2∴函數(shù)h(x)在x∈(1,3]內(nèi)單調(diào)遞減,∴a≤h(3)=9e則實(shí)數(shù)a的取值范圍是∞,9e3.14.解(1)f(x)的定義域?yàn)?0,+∞),f'(x)=1+lnx,令f'(x)>0,解得x>1e,令f'(x)<0,解得0<x<1所以當(dāng)x=1e時(shí),f(x)取得最小值,最小值為f1e=1e(2)依題意知,f(x)≥ax1在[1,+∞)上恒成立,即不等式a≤lnx+1x對(duì)于x∈[1,+∞)恒成立令g(x)=lnx+1x,則g'(x)=1當(dāng)x>1時(shí),g'(x)>0,故g(x)在(1,+∞)內(nèi)單調(diào)遞增,所以g(x)的最小值是g(1)=1.因此a≤g(x)min=g(1)=1,故a的取值范圍為(∞,1].15.解(1)由f(x)=(x+a)ex,得f'(x)=(x+a+1)ex,∵函數(shù)f(x)在區(qū)間[1,+∞)上是增函數(shù),∴f'(x)=(x+a+1)ex≥0在區(qū)間[1,+∞)上恒成立,即a≥x1在區(qū)間[1,