（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 課時15 3.4 導(dǎo)數(shù)的綜合應(yīng)用夯基提能作業(yè).docx

3.4導(dǎo)數(shù)的綜合應(yīng)用A組基礎(chǔ)題組1.“函數(shù)f(x)=a+ln x(xe)存在零點”是“a<-1”的() A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件答案B由題意可知,函數(shù)f(x)單調(diào)遞增,且f(x)min=f(e)=1+a,若f(x)在e,+)上存在零點,則1+a0,即a-1,所以函數(shù)f(x)=a+ln x(xe)存在零點的充要條件為a-1,故選B.2.已知y=f(x)為R上的連續(xù)可導(dǎo)函數(shù),且xf (x)+f(x)>0,則函數(shù)g(x)=xf(x)+1(x>0)的零點個數(shù)為()A.0B.1C.0或1D.無數(shù)個答案A因為g(x)=xf(x)+1(x>0),所以g(x)=xf (x)+f(x)(x>0),由題意可知g(x)>0,所以g(x)在(0,+)上單調(diào)遞增,因為g(0)=1,y=f(x)為R上的連續(xù)可導(dǎo)函數(shù),所以g(x)為(0,+)上的連續(xù)可導(dǎo)函數(shù),g(x)>g(0)=1,所以g(x)在(0,+)上無零點.3.(2018麗水模擬)設(shè)函數(shù)f(x)=ax3-3x+1(xR),若對于任意x-1,1,都有f(x)0成立,則實數(shù)a的值為.答案4解析當(dāng)x=0時,無論a取何值, f(x)0顯然成立;當(dāng)x(0,1時, f(x)=ax3-3x+10可化為a3x2-1x3.設(shè)g(x)=3x2-1x3,則g(x)=3(1-2x)x4,令g(x)=0,得x=12,所以g(x)在區(qū)間0,12上單調(diào)遞增,在區(qū)間12,1上單調(diào)遞減,因此g(x)max=g12=4,從而a4.當(dāng)x-1,0)時,同理,a3x2-1x3,g(x)在區(qū)間-1,0)上單調(diào)遞增,所以g(x)min=g(-1)=4,從而a4.綜上可知,a=4.4.(2019紹興一中月考)已知函數(shù)f(x)=ex-3x+3a(e為自然對數(shù)的底數(shù),aR).(1)求f(x)的單調(diào)區(qū)間與極值;(2)求證:當(dāng)a>ln3e,且x>0時,exx>32x+1x-3a.解析(1)由f(x)=ex-3x+3a知, f (x)=ex-3.令f (x)=0,得x=ln 3,于是當(dāng)x變化時, f (x)和f(x)的變化情況如下表:x(-,ln 3)ln 3(ln 3,+)f (x)-0+f(x)單調(diào)遞減極小值單調(diào)遞增故f(x)的單調(diào)遞減區(qū)間是(-,ln 3),單調(diào)遞增區(qū)間是(ln 3,+),f(x)在x=ln 3處取得極小值,極小值為f(ln 3)=eln3-3ln 3+3a=3(1-ln 3+a).(2)證明:待證不等式等價于ex-32x2+3ax-1>0,設(shè)g(x)=ex-32x2+3ax-1,x>0,則g(x)=ex-3x+3a,x>0.由(1)及a>ln3e=ln 3-1知,g(x)的最小值為g(ln 3)=3(1-ln 3+a)>0.g(x)在(0,+)上為增函數(shù),g(0)=0,當(dāng)x>0時,g(x)>0,即ex-32x2+3ax-1>0,即exx>32x+1x-3a.5.已知函數(shù)f(x)=12ax2-ln x(x>0,aR).(1)若a=2,求點(1, f(1)處的切線方程;(2)若不等式f(x)a2對任意x>0恒成立,求實數(shù)a的值.解析(1)當(dāng)a=2時, f(x)=x2-ln x, f (x)=2x2-1x,f(1)=1, f (1)=1,所求的切線方程為y=x.(2)易得f (x)=ax2-1x.當(dāng)a0時, f (x)<0,當(dāng)x>1時, f(x)<a2,故此時不合題意;當(dāng)a>0時, f(x)在0,1a上單調(diào)遞減,在1a,+上單調(diào)遞增,f(x)min=f1a=12-ln1a,12-ln1aa2,即1+ln a-a0.設(shè)g(x)=1+ln x-x,則g (x)=1x-1=1-xx,g(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減,g(x)g(1)=0,即1+ln x-x0,故1+ln a-a=0,a=1.6.(2018浙江金華十校第二學(xué)期調(diào)研)設(shè)函數(shù)f(x)=ex-x,h(x)=-kx3+kx2-x+1.(1)求f(x)的最小值;(2)設(shè)h(x)f(x)對任意x0,1恒成立時k的最大值為,求證:4<<6.解析(1)因為f(x)=ex-x,所以f (x)=ex-1,當(dāng)x(-,0)時, f (x)<0, f(x)單調(diào)遞減,當(dāng)x(0,+)時, f (x)>0, f(x)單調(diào)遞增,所以f(x)min=f(0)=1.(2)證明:由h(x)f(x),化簡可得k(x2-x3)ex-1,當(dāng)x=0,1時,kR,當(dāng)x(0,1)時,kex-1x2-x3,要證4<<6,則需證以下兩個問題:ex-1x2-x3>4對任意x(0,1)恒成立;存在x0(0,1),使得ex0-1x02-x03<6成立.先證ex-1x2-x3>4,即證ex-1>4(x2-x3),由(1)可知,ex-x1恒成立,所以ex-1x,又x0,所以ex-1>x,即證x4(x2-x3)14(x-x2)(2x-1)20,(2x-1)20顯然成立,所以ex-1x2-x3>4對任意x(0,1)恒成立;再證存在x0(0,1),使得ex0-1x02-x03<6成立,取x0=12,e-114-18=8(e-1),因為e<74,所以8(e-1)<834=6,所以存在x0(0,1),使得ex0-1x02-x03<6成立,由可知,4<<6.7.(2019臺州中學(xué)月考)設(shè)f(x)=x-a-1x-aln x(aR).(1)當(dāng)a=1時,求曲線y=f(x)在點12, f12處的切線方程;(2)當(dāng)0<a<1時,在1e,e內(nèi)是否存在一實數(shù)x0,使f(x0)>e-1成立?請說明理由.解析(1)當(dāng)a=1時, f(x)=x-ln x, f (x)=1-1x.易知f12=12+ln 2,所以曲線y=f(x)在點12,12+ln2處的切線的斜率為f 12=1-112=-1.故所求的切線方程為y-12+ln2=-x-12,即x+y-ln 2-1=0.(2)存在.理由如下:假設(shè)當(dāng)0<a<1時,在1e,e內(nèi)存在一實數(shù)x0,使f(x0)>e-1成立,此時只需證明當(dāng)x1e,e時, f(x)max>e-1即可.f (x)=1+a-1x2-ax=x2-ax+(a-1)x2=(x-1)x-(a-1)x2(x>0),令f (x)=0得,x1=1,x2=a-1,當(dāng)0<a<1時,a-1<0,故當(dāng)x1e,1時, f (x)<0;當(dāng)x(1,e)時, f (x)>0.故函數(shù)f(x)在1e,1上遞減,在(1,e)上遞增,所以f(x)max=maxf1e, f(e).于是,只需證明f(e)>e-1或f1e>e-1即可.因為f(e)-(e-1)=e-a-1e-a-(e-1)=(e+1)(1-a)e>0,所以f(e)>e-1,所以假設(shè)成立,故當(dāng)0<a<1時,在x1e,e上至少存在一實數(shù)x0,使f(x0)>e-1成立.B組提升題組1.(2018浙江,22,15分)已知函數(shù)f(x)=x-ln x.(1)若f(x)在x=x1,x2(x1x2)處導(dǎo)數(shù)相等,證明: f(x1)+f(x2)>8-8ln 2;(2)若a3-4ln 2,證明:對于任意k>0,直線y=kx+a與曲線y=f(x)有唯一公共點.解析本題主要考查函數(shù)的單調(diào)性,導(dǎo)數(shù)的運算及其應(yīng)用,同時考查邏輯思維能力和綜合應(yīng)用能力.(1)函數(shù)f(x)的導(dǎo)函數(shù)f (x)=12x-1x,由f (x1)=f (x2)得12x1-1x1=12x2-1x2,因為x1x2,所以1x1+1x2=12.由基本不等式得12x1x2=x1+x224x1x2,因為x1x2,所以x1x2>256.由題意得f(x1)+f(x2)=x1-ln x1+x2-ln x2=12x1x2-ln(x1x2).設(shè)g(x)=12x-ln x,則g(x)=14x(x-4),所以x(0,16)16(16,+)g(x)-0+g(x)2-4ln 2所以g(x)在256,+)上單調(diào)遞增,故g(x1x2)>g(256)=8-8ln 2,即f(x1)+f(x2)>8-8ln 2.(2)令m=e-(|a|+k),n=|a|+1k2+1,則f(m)-km-a>|a|+k-k-a0,f(n)-kn-a<n1n-an-kn|a|+1n-k<0,所以,存在x0(m,n)使f(x0)=kx0+a,所以,對于任意的aR及k(0,+),直線y=kx+a與曲線y=f(x)有公共點.由f(x)=kx+a得k=x-lnx-ax.設(shè)h(x)=x-lnx-ax,則h(x)=lnx-x2-1+ax2=-g(x)-1+ax2,其中g(shù)(x)=x2-ln x.由(1)可知g(x)g(16),又a3-4ln 2,故-g(x)-1+a-g(16)-1+a=-3+4ln 2+a0,所以h(x)0,即函數(shù)h(x)在(0,+)上單調(diào)遞減,因此方程f(x)-kx-a=0至多有1個實根.綜上,當(dāng)a3-4ln 2時,對于任意k>0,直線y=kx+a與曲線y=f(x)有唯一公共點.2.設(shè)函數(shù)f(x)=a2x2-(ax+1)ln x+ax.(1)若a0,且函數(shù)y=f(x)有且僅有一個零點,求a的值;(2)是否存在實數(shù)a,使得不等式f(x)0對定義域內(nèi)的任意x恒成立?若存在,求出實數(shù)a的取值范圍;若不存在,請說明理由.解析函數(shù)f(x)的定義域為(0,+),f(x)=(ax+1)(ax-ln x),(1)當(dāng)a=0時, f(x)=-ln x,滿足題意.當(dāng)a>0時,ax+1>0,所以f(x)=(ax+1)(ax-ln x)=0ax-ln x=0.因為函數(shù)y=f(x)有且僅有一個零點,所以當(dāng)直線y=ax與y=ln x相切時,a的值即為所求.令(ln x)=1x=a,即x=1a,故切點坐標(biāo)為1a,-lna,將其代入y=ax,得a=1e.綜上可得a=0或1e.(2)存在.假設(shè)存在實數(shù)a,使得不等式f(x)0對定義域內(nèi)的任意x恒成立,當(dāng)a=0時,原不等式化為ln x0,不滿足題意.當(dāng)a>0時,ax+1>0,由ax-ln x0,得alnxx.令F(x)=lnxx,則F (x)=1-lnxx2,所以函數(shù)F(x)在(0,e)上單調(diào)遞增,在(e,+)上單調(diào)遞減,所以函數(shù)F(x)在x=e處取得極大值,也為最大值,最大值為1e,由此可得a1e.當(dāng)a<0時,在0,-1a上,ax+1>0,在-1a,+上,ax+1<0,令g(x)=ax-ln x,則g(x)=a-1x<0,所以g(x)=ax-ln x在(0,+)上為減函數(shù),所以只要當(dāng)x=-1a時,ax-ln x=0成立即可,此時解得a=-e.綜上可得,a1e,+-e.3.(2019紹興一中月考)設(shè)函數(shù)f(x)=ex-1x,求證:(1)當(dāng)x<0時, f(x)<1;(2)對任意a>0,當(dāng)0<|x|<ln(1+a)時,|f(x)-1|<a.證明(1)設(shè)(x)=ex-1-x,xR,則(x)=ex-1,令(x)>0,得x>0,令(x)<0,得x<0,故(x)在(-,0)內(nèi)遞減,在(0,+)內(nèi)遞增,所以對任意xR,都有(x)(0)=0,即ex-1-x0(當(dāng)且僅當(dāng)x=0時,等號成立).所以當(dāng)x<0時,ex-1>x,即f(x)<1.(2)要證當(dāng)0<|x|<ln(1+a)時,|f(x)-1|<a,即證當(dāng)0<x<ln(1+a)時,ex-1-(1+a)x<0;當(dāng)-ln(1+a)<x<0時,ex-1-(1-a)x<0.令函數(shù)g(x)=ex-1-(1+a)x,h(x)=ex-1-(1-a)x.注意到g(0)=h(0)=0,故要證與,只需證明g(x)在(0,ln(1+a)內(nèi)遞減,h(x)在(-ln(1+a),0)內(nèi)遞增.事實上,當(dāng)x(0,ln(1+a)時,g(x)=ex-(1+a)<eln(1+a)-(1+a)=0;當(dāng)x(-ln(1+a),0)時,h(x)=ex-(1-a)>e-ln(1+a)-(1-a)=11+a-(1-a)=a21+a>0.綜上,對任意a>0,當(dāng)0<|x|<ln(1+a)時,|f(x)-1|<a.4.已知函數(shù)f(x)=-12ax2+(1+a)x-ln x(aR).(1)當(dāng)a>0時,求函數(shù)f(x)的單調(diào)遞減區(qū)間;(2)當(dāng)a=0時,設(shè)函數(shù)g(x)=xf(x)-k(x+2)+2.若函數(shù)g(x)在區(qū)間12,+上有兩個零點,求實數(shù)k的取值范圍.解析(1)f(x)的定義域為(0,+),f (x)=-ax+1+a-1x=-(ax-1)(x-1)x(a>0),當(dāng)a(0,1)時,1a>1.由f (x)<0,得x>1a或x<1,所以f(x)的單調(diào)遞減區(qū)間為(0,1),1a,+;當(dāng)a=1時,恒有f (x)0,所以f(x)的單調(diào)遞減區(qū)間為(0,+);當(dāng)a(1,+)時,1a<1.由f (x)<0,得x>1或x<1a,所以f(x)的單調(diào)遞減區(qū)間為0,1a,(1,+).綜上,當(dāng)a(0,1)時,f(x)的單調(diào)遞減區(qū)間為(0,1),1a,+;當(dāng)a=1時, f(x)的單調(diào)遞減區(qū)間為(0,+);當(dāng)a(1,+)時, f(x)的單調(diào)遞減區(qū)間為0,1a,(1,+).(2)g(x)=x2-xln x-k(x+2)+2在x12,+上有兩個零點,即關(guān)于x的方程k=x2-xlnx+2x+2在x12,+上有兩個不相等的實數(shù)根.令函數(shù)h(x)=x2-xlnx+2x+2,x12,+,則h(x)=x2+3x-2lnx-4(x+2)2,令函數(shù)p(x)=x2+3x-2ln x-4,x12,+.則p(x)=(2x-1)(x+2)x在12,+上有p(x)0,故p(x)在12,+上單調(diào)遞增.因為p(1)=0,所以當(dāng)x12,1時,p(x)<0,即h(x)<0,故h(x)單調(diào)遞減;當(dāng)x(1,+)時,p(x)>0,即h(x)>0,故h(x)單調(diào)遞增.因為h12=910+ln25,h(1)=1,所以k的取值范圍是1,910+ln25.