2022年秋高中數(shù)學(xué)第五章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用培優(yōu)課恒成立能成立問題課后習(xí)題新人教A版選擇性必修第二冊

歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！培優(yōu)課——恒成立、能成立問題必備知識(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)=lnx-ax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)=ex-x+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)=ex-2ax(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+x2-mx+3≥0成立,則實(shí)數(shù)m的最大值為()A.1e+3e-2 B.3e+eC.4 D.e2-19.已知函數(shù)f(x)=sin2x+π6-x22-mx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(A.(-∞,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)=lnx-axa>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)=ax3-3x+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)≥ax-1,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)≥e3-xex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≤2x2-4x∵2x2-4x=2(x-1)2-2≥-2,∴b≤-2.2.D∵f'(x)=1x-a,x>∴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=ln1a-1=0,解得a=1e3.Af'(x)=ex-1,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f-23,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)=x-1e則f'(x)=1+1ex∴f(x)在[0,+∞)上是增函數(shù),又f(0)=-1<0,f(1)=1-1e>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(2-x)恒成立,令g(x)=ex(2-x),則g'(x)=ex(1-x),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)=ex-x,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)=ex-x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)=ex-2ax≥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+x2-mx+3≥0,∴m≤2lnx+x+3x設(shè)h(x)=2lnx+x+3x則h'(x)=2x+1-3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+3e-29.D由f(x)=sin2x+π6-x22-mx在0,π6上是減函數(shù),得f'(x)=2cos2x+π6-x-m≤0x∈0,π6,即2cos2x+π6-x≤mx∈0,π6,令g(x)=2cos2x+π6-xx∈0,π6,則g'(x)=-4sin2x+π6-1x∈0,π6,當(dāng)x∈0,π6時(shí),π6≤2x+π6則2≤4sin2x+π6≤4,所以-5≤-4sin2x+π6-1≤-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=-lnx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=-1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)=1x-a=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=-lna-1=-1.解得a=1.12.4由題意得,f'(x)=3ax2-3,當(dāng)a>1時(shí),令f'(x)=3ax2-3=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·aa3-3·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(x1)可得函數(shù)g(x)=f(x)-2x=aexx-2x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)≥ax-1在[1,+∞)上恒成立,即不等式a≤