適用于新高考新教材廣西專版2025屆高考數(shù)學(xué)一輪總復(fù)習(xí)第四章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用課時規(guī)范練17利用導(dǎo)數(shù)研究函數(shù)的極值與最值

課時規(guī)范練17利用導(dǎo)數(shù)探討函數(shù)的極值與最值基礎(chǔ)鞏固組1.函數(shù)f(x)=3x2+lnx-2x的極值點的個數(shù)是()A.0 B.1 C.2 D.多數(shù)2.函數(shù)f(x)=(x-2)·ex的最小值為()A.-2 B.-e C.-1 D.03.(2024廣西南寧二模)已知函數(shù)f(x)=alnx-bx的極值點為1,且f'(2)=1,則f(x)的微小值為(A.-1 B.-a C.b D.44.(多選)初等函數(shù)是由基本初等函數(shù)與常數(shù)經(jīng)過有限次的有理運算及有限次函數(shù)復(fù)合所產(chǎn)生,并且能用一個解析式表示的函數(shù),如函數(shù)f(x)=xx(x>0),我們可以作變形:f(x)=xx=elnxx=exlnx=et,其中t=xlnx,所以f(x)可看作是由函數(shù)g(t)=et和t=xlnx復(fù)合而成的,即f(x)=xx(x>0)為初等函數(shù).依據(jù)以上材料,關(guān)于初等函數(shù)h(x)=x1xA.無微小值 B.有微小值1C.無極大值 D.有極大值e5.若方程x3-3x+m=0在區(qū)間[0,2]上有解,則實數(shù)m的取值范圍是()A.[-2,2] B.[0,2]C.[-2,0] D.(-∞,-2)∪(2,+∞)6.設(shè)直線x=t與函數(shù)f(x)=2x2,g(x)=lnx的圖象分別交于點M,N,則|MN|的最小值為()A.12+ln2 B.3ln2-C.e2-1 D.7.(2024全國乙,文8)若函數(shù)f(x)=x3+ax+2存在3個零點,則a的取值范圍是()A.(-∞,-2) B.(-∞,-3)C.(-4,-1) D.(-3,0)8.(2024全國乙,理21)已知函數(shù)f(x)=1x+aln(1(1)當(dāng)a=-1時,求曲線y=f(x)在點(1,f(1))處的切線方程.(2)是否存在a,b,使得曲線y=f1x關(guān)于直線x=b對稱?若存在,求a,b的值;若不存在,說明理由(3)若f(x)在(0,+∞)存在極值點,求a的取值范圍.綜合提升組9.已知函數(shù)f(x)=x3+ax2-x+a有兩個極值點x1,x2,且|x1-x2|=233,則f(x)的極大值為(A.39 B.2C.33 D.10.當(dāng)x=1時,函數(shù)f(x)=alnx+bx取得最大值-2,則f'(2)=(A.-1 B.-1C.12 D.11.設(shè)函數(shù)f(x)=x+a,x≤0,lnx,x>0,已知x1<x2且f(x1)=f(A.-1 B.1e-C.-1或1e-D.212.已知函數(shù)f(x)=x2-a2lnx-x2(a∈R)在區(qū)間116,1上不存在極值點,則實數(shù)a的取值范圍是.13.(2024新高考Ⅱ,22)(1)證明:當(dāng)0<x<1時,x-x2<sinx<x;(2)已知函數(shù)f(x)=cosax-ln(1-x2),若x=0是f(x)的極大值點,求a的取值范圍.創(chuàng)新應(yīng)用組14.(2024山東淄博一模)已知函數(shù)f(x)=|x+2|+1,x≤0,lnx,x>0,若存在實數(shù)a<b<c,滿意f(a)=f(b)=f(c

課時規(guī)范練17利用導(dǎo)數(shù)探討函數(shù)的極值與最值1.A解析函數(shù)f(x)的定義域為(0,+∞),導(dǎo)數(shù)f'(x)=6x+1x-2=6x2-2x+1x.令g(x)=6x2-2x+1,則Δ=-20<0,所以g(x)>0恒成立,又x>0,所以f'(x2.B解析f'(x)=ex+(x-2)ex=(x-1)ex,令f'(x)=0,解得x=1,易得f(x)在(-∞,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,故f(x)的最小值為f(1)=-e,故選B.3.D解析f'(x)=ax+bx2=ax+bx2,f'(1)=0,f'(2)=1,所以a+b=0,2a+b4=1,解得a=4,b=-4,則f(x)=4lnx+4x,f'(x)=4x-4.AD解析依據(jù)材料知,h(x)=x1x=elnx1x令h'(x)=0,得x=e.當(dāng)0<x<e時,h'(x)>0,函數(shù)h(x)單調(diào)遞增;當(dāng)x>e時,h'(x)<0,函數(shù)h(x)單調(diào)遞減.所以當(dāng)x=e時,h(x)取得極大值,且極大值為h(e)=e1e5.A解析由題意得-m=x3-3x,x∈[0,2].令y=x3-3x,x∈[0,2],則y'=3x2-3.令y'=0,解得x=1,易得函數(shù)在[0,1)上單調(diào)遞減,在(1,2]上單調(diào)遞增.又因為當(dāng)x=0時,y=0;當(dāng)x=1時,y=-2;當(dāng)x=2時,y=2,所以函數(shù)y=x3-3x,x∈[0,2]的值域是[-2,2],因此-m∈[-2,2],即m∈[-2,2].故選A.6.A解析由題意M(t,2t2),N(t,lnt),所以|MN|=|2t2-lnt|,令h(t)=2t2-lnt,t>0,則h'(t)=4t-1t令h'(t)=0,得t=12當(dāng)0<t<12時,h'(t)<0,函數(shù)h(t當(dāng)t>12時,h'(t)>0,函數(shù)h(t所以h(t)min=h12=12+ln2,即|MN|的最小值為12+ln27.B解析令f(x)=0,得-ax=x3+2,易知x≠0,所以-a=x3+2x.設(shè)g(x)則函數(shù)f(x)存在3個零點等價于函數(shù)g(x)=x3+2x的圖象與直線g'(x)=2(x3-1)x2.當(dāng)x>1時,g'(x)>0,函數(shù)g(x)在(1,+∞)內(nèi)單調(diào)遞增,當(dāng)x<1且x≠0時,g'(x)<0,函數(shù)g(x)在(-∞,0),(0,1)內(nèi)單調(diào)遞減,且g(1)=3,當(dāng)x從左側(cè)趨近于0時,g(x)→-∞,當(dāng)x從右側(cè)趨近于0時,g(x)→+∞,當(dāng)x→+∞時,g(x)→由圖知,當(dāng)-a>3時,函數(shù)g(x)=x3+2x的圖象與直線y=-a有三個交點,即函數(shù)f(x)有3個零點,所以a<-38.解(1)當(dāng)a=-1時,f(x)=1x-1ln(x+1)(x>-1且x≠0),f'(x)=-ln(x∵f(1)=11-1×ln(1+1)=0,f'(1)=-ln(1+1)∴y=f(x)在點(1,f(1))處的切線方程為y-f(1)=f'(1)(x-1),即(ln2)x+y-ln2=0.(2)∵f1x=(x+a)ln1x+1,要使函數(shù)有意義,則1x+1>0,解得x<-1或x>0,∴函數(shù)f1x的定義域關(guān)于x=-12對稱,∴若函數(shù)f1x的圖象關(guān)于直線x=b對稱,則b=-12.設(shè)g(x)=f1x,由對稱的性質(zhì)可知g(x)=g(-1-x).∵g(x)=(x+a)ln1x+1,g(-1-x)=(-1-x+a)ln1-1-x+1則(x+a)ln1x+1=(-1-x+a)ln1-1-x+1,解得a=1∴存在a=12,b=-12,使函數(shù)f1x圖象關(guān)于直線x=-12(3)由題意,f(x)=1x+aln(x+1),x>0,則f'(x)=-ln(x+1)x2+ax+1x設(shè)H(x)=ln(x+1)-x(ax+1則H'(x)=1x當(dāng)a≤0時,H'(x)>0,H(x)在(0,+∞)上單調(diào)遞增,故H(x)>ln1-0=0,即f'(x)<0,f(x)在(0,+∞)上單調(diào)遞減,f(x)在(0,+∞)上無極值;當(dāng)a≥12時,H'(x)<0,H(x)在(0,+∞)上單調(diào)遞減,故H(x)<ln1-0=0,即f'(x)>0,f(x)在(0,+∞)上單調(diào)遞增,f(x當(dāng)0<a<12時,令H'(x)=0,解得x=1a-2,則當(dāng)x∈0,1a-2時,H'(x)>0,當(dāng)x∈1a-2,+∞時,H'(x)<0,∴H(x)在0,1a-2內(nèi)單調(diào)遞增,在1a-2,+∞上單調(diào)遞減,H1a-2=ln1a-1-(1a-2)(2-2a)1a-1=ln(1設(shè)φ(a)=ln(1-a)-lna-2+4a,0<a<12,則φ'(a)=1a-1-1a+4=4a2-4∴φ(a)>ln1-12-ln12-2+4×12=0,即H1a-2>0.又H(x)=ln(x+1)-x(ax+1)x+1=ln(x+1)-ax-(設(shè)h(x)=ln(x+1)-ax,x>0,則h(e1a2-1)=lne1a2-a(e1a2-1)=1設(shè)t=1a∈(2,+∞),G(t)=t3-et2+1,則G'(t)=3t2-2tet2<3t2-2t(t2+1)<0,故G(t)在(2,+∞)上單調(diào)遞減,G(t)<8-e4+1<0,故h(e1a2-1)<0,故存在x0∈1a-2,e1a2-1,使得H(x0)=0,即存在f'(x0)=0,且f'(x)在x0兩側(cè)異號,故f(x)在(0,+∞綜上,滿意條件的a的取值范圍為0,12.9.B解析因為f'(x)=3x2+2ax-1,Δ=4a2+12>0,所以f'(x)=0有兩個不同的實數(shù)解x1,x2,且由根與系數(shù)的關(guān)系得x1+x2=-2a3,x1x2=-由題意可得|x1-x2|=(x1+此時f(x)=x3-x,f'(x)=3x2-1.當(dāng)x∈-∞,-33,x∈33,+∞時,f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈-33,33時,f'(x)<0,f(故當(dāng)x=-33時,f(x)取得極大值210.B解析函數(shù)f(x)的定義域為(0,+∞).因為f(1)=aln1+b=-2,所以b=-2.因為f'(x)=ax所以f'(1)=a-b=0,所以a=b=-2.所以f(x)=-2lnx-2x,f'(x)=-2閱歷證,符合題意.所以f'(2)=-1+24=-1故選B.11.A解析令f(x1)=f(x2)=t,作出函數(shù)f(x)的大致圖象,如圖.由圖象可知t∈(-∞,a],因為x1<x2,所以x1+a=t,lnx2=t,得x1=t-a,x2=et,所以x2-x1=et-t+a.令g(t)=et-t+a(t≤a),則g'(t)=et-1.當(dāng)a≤0時,g'(t)≤0,g(t)在(-∞,a]上單調(diào)遞減,所以g(t)min=g(a)=ea-a+a=ea,由g(t)min=ea=1e,解得a=-當(dāng)a>0時,令g'(t)=0,得t=0,易得g(t)在區(qū)間(-∞,0]上單調(diào)遞減,在區(qū)間(0,a]上單調(diào)遞增,所以g(t)min=g(0)=e0-0+a=1+a,由g(t)min=1+a=1e,解得a=1e-1<綜上可得a=-1,故選A.12.-∞,-116∪[3,+∞)解析因為函數(shù)f(x)=x2-a2lnx-x2(a∈R)在區(qū)間116,1上不存在極值點,所以函數(shù)f(x)在區(qū)間116,1上單調(diào)遞增或單調(diào)遞減,所以f'(x)≥0或f'(x)≤0在區(qū)間116,1上恒成立.f'(x)=2x-a2x-12=4x2-x-a2x,令g(x)=4x2-x-a,x∈116,1,則其圖象的對稱軸為直線x=18,所以g(x)min=4×182-18-a=-116-a,g(x)max=4×12-1-a=3-a.當(dāng)f'(x)≥0時,需滿意-116綜上所述,a的取值范圍為-∞,-116∪[3,+∞).13.(1)證明設(shè)h(x)=sinx-x,x∈[0,1],則h'(x)=cosx-1≤0對?x∈[0,1]恒成立,且僅在x=0時有h'(0)=0,所以函數(shù)h(x)在[0,1]上單調(diào)遞減.所以對?x∈(0,1),有h(x)<h(0)恒成立.又因為h(0)=0,所以sinx-x<0恒成立.所以sinx<x,x∈(0,1).設(shè)g(x)=sinx-(x-x2),則g'(x)=cosx+2x-1.令G(x)=cosx+2x-1,則G'(x)=-sinx+2>0對?x∈[0,1]恒成立,所以g'(x)在x∈[0,1]上單調(diào)遞增,且因為g'(0)=1+0-1=0,所以對?x∈[0,1],g'(x)≥0恒成立,且僅在x=0時有g(shù)'(0)=0,所以函數(shù)y=g(x)在[0,1]上單調(diào)遞增.所以對?x∈(0,1),有g(shù)(x)>g(0)恒成立.又因為g(0)=0,所以sinx+x2-x>0對?x∈(0,1)恒成立.所以x-x2<sinx,x∈(0,1).綜上可知,x-x2<sinx<x,x∈(0,1)成立.(2)解若a≠0,當(dāng)0<x<min1,(ⅰ)若a>0,有a2x-a3x2<asinax<a2x;(ⅱ)若a<0,有a2x+a3x2<asinax<a2x.證明:(ⅰ)當(dāng)a>0時,由(1)知ax-a2x2<sinax<ax,0<x<min1,所以a2x-a3x2<asinax<a2x;(ⅱ)當(dāng)a<0時,由(1)知-ax-(-ax)2<sin(-ax)<-ax,0<x<min1,所以-a2x-a(-ax)2>asin(-ax)>-a2x,即a2x+a3x2<asinax<a2x.故上述說法成立.因為f(x)=cosax-ln(1-x2),x∈(-1,1),所以f'(x)=-asinax--2x1-x滿意f'(0)=-asin0+0=0.因為f'(-x)=asinax-2x1-x所以函數(shù)f'(x)為奇函數(shù).①若a=0,則f'(x)=2x當(dāng)x∈(-1,0)時,f'(x)<0;當(dāng)x∈(0,1)時,f'(x)>0.故x=0是函數(shù)f(x)的微小值點,不符合題意.②若a∈(2,+∞),則當(dāng)0≤x<min1,1a時,asinax>a2x-a3所以f'(x)<-a2x+a3x2+2x1-x2=x21-令F1(x)=21-x2+a3x-a2,0≤則F'1(x)=4x(1-故當(dāng)0≤x<min1,1|a|時,F又因為F1(0)=2-a2<0,所以存在t1>0,使得F1(x)在(0,t1)上恒小于0,此時f'(x)<xF1(x)<0.又因為函數(shù)f'(x)是奇函數(shù),所以在(-t1,0)內(nèi)有f'(x)>0.因此函數(shù)f(x)在(-t1,0)內(nèi)單調(diào)遞增,在(0,t1)內(nèi)單調(diào)遞減,所以x=0是函數(shù)f(x)的極大值點.③若a∈(-∞,-2),則當(dāng)0≤x<min1,1|a|時,a2x+a3x2<a所以