（適用于新高考新教材）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)研究函數(shù)的極值與最值基礎(chǔ)鞏固組1.函數(shù)f(x)=3x2+lnx2x的極值點的個數(shù)是()A.0 B.1 C.2 D.無數(shù)2.函數(shù)f(x)=(x2)·ex的最小值為()A.2 B.e C.1 D.03.已知函數(shù)f(x)=3xex,則f(xA.在(∞,+∞)上單調(diào)遞增B.在(∞,1)上單調(diào)遞減C.有極大值3eD.有極小值3e4.(多選)(2022廣東珠海模擬)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).根據(jù)以上材料,關(guān)于初等函數(shù)h(x)=x1xA.無極小值 B.有極小值1C.無極大值 D.有極大值e5.若方程x33x+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.(2022山東泰安模擬)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.C.e21 D.7.(2022湖南長郡中學(xué)高三檢測)函數(shù)f(x)=1x+ln|x|的極值點為.8.已知函數(shù)f(x)=x1+aex(a∈R(1)若曲線y=f(x)在點(1,f(1))處的切線平行于x軸,求實數(shù)a的值;(2)求函數(shù)f(x)的極值.綜合提升組9.(2022廣東深圳高三檢測)已知函數(shù)f(x)=x3+ax2x+a有兩個極值點x1,x2,且|x1x2|=233,則f(x)的極大值為(A.39 B.2C.33 D.10.(2022全國甲,文8)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.1eC.1或1eD.212.已知函數(shù)f(x)=x2a2lnxx2(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.設(shè)函數(shù)f(x)=ln(ax),已知x=0是函數(shù)y=xf(x)的極值點.(1)求a;(2)設(shè)函數(shù)g(x)=x+f(x)xf(x創(chuàng)新應(yīng)用組14.已知函數(shù)f(x)=ex+lnx,g(x)=4x+1x,且x滿足1≤x≤2,則g(x)f(x)的最大值為.

課時規(guī)范練17利用導(dǎo)數(shù)研究函數(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+1x2=6x2-2x+1x.令g(x)=6x22x+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+(x2)ex=(x1)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.C解析:由已知得f'(x)=3(1-x)ex.令f'(x)=0,得x=1.當(dāng)x<1時,f'(x)>0,f(x)單調(diào)遞增;當(dāng)x>1時,f'(x)<0,f(x)單調(diào)遞減,所以,當(dāng)x=1時,函數(shù)f(x)取得極大值,也是最大值,且f(1)=3e4.AD解析:根據(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=x33x,x∈[0,2].令y=x33x,x∈[0,2],則y'=3x23.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=x33x,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|=|2t2lnt|,令h(t)=2t2lnt,t>0,則h'(t)=4t1t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+ln27.1解析:當(dāng)x>0時,f(x)=1x+lnx,則f'(x)=1x2+1x=x-1x2.令f'(x)=0,得x=1,當(dāng)0<x<1時,f'(x)<0;當(dāng)x>1時,f'所以1為函數(shù)f(x)的極小值點.當(dāng)x<0時,f(x)=1x+ln(x),f'(x)=1x所以函數(shù)f(x)在區(qū)間(∞,0)內(nèi)單調(diào)遞減,所以f(x)無極值點.綜上所述,函數(shù)f(x)的極值點為1.8.解(1)由f(x)=x1+aex,得f'(x)=1因為曲線y=f(x)在點(1,f(1))處的切線平行于x軸,所以f'(1)=0,即1ae=0,解得a=e(2)由(1)得f'(x)=1ae當(dāng)a≤0時,f'(x)>0,所以f(x)在(∞,+∞)上單調(diào)遞增,因此f(x)無極值;當(dāng)a>0時,令f'(x)=0,得x=lna.當(dāng)x∈(∞,lna)時,f'(x)<0;當(dāng)x∈(lna,+∞)時,f'(x)>0.所以f(x)在區(qū)間(∞,lna)上單調(diào)遞減,在區(qū)間(lna,+∞)上單調(diào)遞增,故f(x)在x=lna處取得極小值,且極小值為f(lna)=lna,無極大值.綜上,當(dāng)a≤0時,f(x)無極值;當(dāng)a>0時,f(x)在x=lna處取得極小值lna,無極大值.9.B解析:因為f'(x)=3x2+2ax1,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|x1x2|=(x1+此時f(x)=x3x,f'(x)=3x21.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)=ab=0,所以a=b=2.所以f(x)=2lnx2xf'(x)=2x經(jīng)驗證,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=ta,x2=et,所以x2x1=ett+a.令g(t)=ett+a(t≤a),則g'(t)=et1.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)=eaa+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)=e00+a=1+a,由g(t)min=1+a=1e,解得a=1e1<綜上可得a=1,故選A.12.∞,116∪[3,+∞)解析:因為函數(shù)f(x)=x2a2lnxx2(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)=2xa2x-12=4x2-x-a2x,令g(x)=4x2xa,x∈116,1,則其圖象的對稱軸為直線x=18,所以g(x)min=4×18218a=116a,g(x)max=4×121a=3a.當(dāng)f'(x)≥0時,需滿足116綜上所述,a的取值范圍為∞,116∪[3,+∞).13.(1)解由題意,f(x)的定義域為(∞,a).令p(x)=xf(x),則p(x)=xln(ax),x∈(∞,a),p'(x)=ln(ax)+x·-1a-x=ln(ax因為x=0是函數(shù)y=xf(x)的極值點,則有p'(0)=0,即lna=0,所以a=1.當(dāng)a=1時,p'(x)=ln(1x)+-x1-x,則當(dāng)x<0時,p'(x)>0,函數(shù)p(x)單調(diào)遞增;當(dāng)0<x<1時,p'(x)<0,函數(shù)p(x)單調(diào)遞減,所以當(dāng)a=1時,x=0是函數(shù)y=xf(x)的極大值點.(2)證明由(1)可知,xf(x)=xln(1x),要證x+f(x)xf因為當(dāng)x∈(∞,0)時,xln(1x)<0,當(dāng)x∈(0,1)時,xln(1x)<0,所以需證明x+ln(1x)>xln(1x),即證x+(1x)ln(1x)>0.令h(x)=x+(1x)ln(1x),x<1,則h'(x)=1ln(1x)+(1x)·-11-x所以h'(0)=0,當(dāng)x∈(∞,0)時,h'(x)<0,函數(shù)h(x)單調(diào)遞減;當(dāng)x∈(0,1)時,h'(x)>0,函數(shù)h(x)單調(diào)遞增.所以x=0為h(x)的唯一極小值點,也是最小值點,所以當(dāng)x∈(∞,0)∪(0,1)時,h(x)>h(0)=0,即x+(1x)ln(1x)>0,所以x+ln(1-x)14.5e解析:令h(x)=g(x)f(x)=4x+1xexlnx,1≤x≤2,則h'(x)=41x2ex1x,令m(x)=41x2ex1x,1≤x≤2,則m