高考文數(shù)一輪夯基作業(yè)本3第三章導(dǎo)數(shù)及其應(yīng)用夯基提能作業(yè)本3

第三節(jié)導(dǎo)數(shù)與函數(shù)的極值與最值A(chǔ)組基礎(chǔ)題組1.設(shè)函數(shù)f(x)在定義域R上可導(dǎo),其導(dǎo)函數(shù)為f'(x),若函數(shù)y=(1x)f'(x)的圖象如圖所示,則下列結(jié)論中一定成立的是()A.函數(shù)f(x)有極大值f(2)和極小值f(1)B.函數(shù)f(x)有極大值f(2)和極小值f(1)C.函數(shù)f(x)有極大值f(2)和極小值f(2)D.函數(shù)f(x)有極大值f(2)和極小值f(2)2.設(shè)函數(shù)f(x)=2xA.x=12為f(x)的極大值點(diǎn) B.x=1C.x=2為f(x)的極大值點(diǎn) D.x=2為f(x)的極小值點(diǎn)3.函數(shù)f(x)=12x2A.12 B.1C.0 D.不存在4.已知f(x)=2x36x2+m(m為常數(shù))在[2,2]上有最大值3,那么此函數(shù)在[2,2]上的最小值為()A.37 B.73 C.10 D.375.已知函數(shù)f(x)=x3px2qx的圖象與x軸切于(1,0)點(diǎn),則f(x)的極大值、極小值分別為()A.427,0 B.0,C.427,0 D.0,6.若函數(shù)f(x)=2x2lnx在區(qū)間(k1,k+1)上有定義且不是單調(diào)函數(shù),則實(shí)數(shù)k的取值范圍是()A.[1,+∞) B.1C.[1,2) D.37.函數(shù)f(x)=xsinx+cosx在π6,π上的最大值為8.已知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9.(2017北京朝陽期中)已知函數(shù)f(x)=ax+1(1)若曲線y=f(x)在點(diǎn)(0,f(0))處的切線的斜率為2,求函數(shù)f(x)的最小值;(2)若函數(shù)f(x)在區(qū)間(0,1)上無極值,求a的取值范圍.B組提升題組10.已知函數(shù)f(x)=-(1)求f(x)在區(qū)間(∞,1)上的極大值點(diǎn)和極小值;(2)求f(x)在[1,e](e為自然對(duì)數(shù)的底數(shù))上的最大值.11.(2016北京海淀一模)已知函數(shù)f(x)=1-(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)求函數(shù)f(x)的零點(diǎn)和極值;(3)若對(duì)任意x1,x2∈[a,+∞),都有f(x1)f(x2)≥1e12.(2016北京豐臺(tái)二模)設(shè)函數(shù)f(x)=exa(x1).(1)求函數(shù)f(x)的單調(diào)區(qū)間和極值;(2)若函數(shù)f(x)在區(qū)間(0,2]上存在唯一零點(diǎn),求a的取值范圍.答案精解精析A組基礎(chǔ)題組1.D2.D因?yàn)閒(x)=2x+lnx,所以f'(x)=2x2+1此時(shí)f(x)為增函數(shù);當(dāng)0<x<2時(shí),f'(x)<0,此時(shí)f(x)為減函數(shù),據(jù)此知x=2為f(x)的極小值點(diǎn).3.Af'(x)=x1x=x令f'(x)>0,得x>1;令f'(x)<0,得0<x<1.∴f(x)在x=1處取得極小值,即最小值,且f(1)=12ln1=14.D由題意知,f'(x)=6x212x,令f'(x)=0,得x=0或x=2,當(dāng)x<0或x>2時(shí),f'(x)>0,當(dāng)0<x<2時(shí),f'(x)<0,∴f(x)在[2,0]上單調(diào)遞增,在(0,2]上單調(diào)遞減,由條件知f(0)=m=3,∴f(2)=5,f(2)=37,∴所求最小值為37.5.C由題意知,f'(x)=3x22pxq,由f'(1)=0,f(1)=0得3-2p-q=0f'(x)=3x24x+1=0,得x=13或x=1,易得當(dāng)x=13時(shí),f(x)取得極大值6.B由f'(x)=4x1x=(得x=12x=-12舍去.當(dāng)x∈0,12時(shí),f'(x)<0;當(dāng)x∈12,+∞7.答案π2解析因?yàn)閒'(x)=sinx+xcosxsinx=xcosx,所以f'(x)=0在x∈π6,π上的解為x=π2.又fπ6=π12+f(π)=1,所以函數(shù)f(x)=xsinx+cosx在π6,π8.答案1解析因?yàn)閒(x)是奇函數(shù),所以f(x)在(0,2)上的最大值為1,當(dāng)x∈(0,2)時(shí),f'(x)=1x令f'(x)=0,得x=1a,因?yàn)閍>1所以0<1a令f'(x)>0,得x<1a所以f(x)在0,令f'(x)<0,得x>1a,所以f(x)在1a,2上單調(diào)遞減,所以當(dāng)x∈(0,2)時(shí),f(x)max=f1a=ln所以ln1a9.解析因?yàn)閒(x)=ax+1e所以f'(x)=-ax(1)依題意得f'(0)=a1=2,解得a=1.所以f(x)=-x+1e當(dāng)x>2時(shí),f'(x)>0,函數(shù)f(x)為增函數(shù);當(dāng)x<2時(shí),f'(x)<0,函數(shù)f(x)為減函數(shù).所以函數(shù)f(x)的最小值是f(2)=1e(2)①若a=0,則f'(x)=1e此時(shí)f(x)在(0,1)上單調(diào)遞減,滿足條件.②若a≠0,令f'(x)=0,得x=a-1a(i)若11a≤0,即0<a≤1,則f'(x)<0在(0,1)上恒成立此時(shí)f(x)在(0,1)上單調(diào)遞減,滿足條件.(ii)若0<11a由f'(x)>0得x<11a由f'(x)<0得x>11a此時(shí),f(x)在0,1-(iii)若11a≥1,即a則f'(x)<0在(0,1)上恒成立.此時(shí)f(x)在(0,1)上單調(diào)遞減,滿足條件.綜上,a的取值范圍為(∞,1].B組提升題組10.解析(1)當(dāng)x<1時(shí),f'(x)=3x2+2x=x(3x2),令f'(x)=0,解得x=0或x=23當(dāng)x變化時(shí),f'(x),f(x)的變化情況如下表:x(∞,0)0022f'(x)0+0f(x)↘極小值↗極大值↘故當(dāng)x=0時(shí),函數(shù)f(x)取得極小值,為f(0)=0,函數(shù)f(x)的極大值點(diǎn)為x=23(2)①當(dāng)1≤x<1時(shí),由(1)知,函數(shù)f(x)在[1,0]和23,1因?yàn)閒(1)=2,f23=4所以f(x)在[1,1)上的最大值為2.②當(dāng)1≤x≤e時(shí),f(x)=alnx,當(dāng)a≤0時(shí),f(x)≤0;當(dāng)a>0時(shí),f(x)在[1,e]上單調(diào)遞增,則f(x)在[1,e]上的最大值為f(e)=a.綜上所述,當(dāng)a≥2時(shí),f(x)在[1,e]上的最大值為a;當(dāng)a<2時(shí),f(x)在[1,e]上的最大值為2.11.解析(1)對(duì)f(x)求導(dǎo)得f'(x)=x-2e因?yàn)閒(0)=1,所以曲線f(x)在(0,f(0))處的切線方程為2x+y1=0.(2)令f(x)=1-由(1)知f'(x)=x-令f'(x)=0,解得x=2,當(dāng)x變化時(shí),f'(x)及f(x)的變化情況如下表:x(∞,2)2(2,+∞)f'(x)0+f(x)↘極小值1↗所以函數(shù)f(x)在x=2處取得極小值,極小值為1e(3)解法一:當(dāng)x>1時(shí),f(x)=1-當(dāng)x<1時(shí),f(x)=1-若a≤1,由(2)可知f(x)的最小值為f(2),f(x)的最大值為f(a),所以“對(duì)任意x1,x2∈[a,+∞),有f(x1)f(x2)≥1e2成立”等價(jià)于f(2)f(a)≥即1e21-若a>1,求出a的值顯然大于1.所以a的最小值為1.解法二:當(dāng)x>1時(shí),f(x)=1-當(dāng)x<1時(shí),f(x)=1-由(2)可知,f(x)的最小值為f(2)=1e若a<1,令x1=2,x2∈[a,1),則x1,x2∈[a,+∞).而f(x1)f(x2)<f(x1)0=f(x1)=f(2)=1e當(dāng)a=1時(shí),?x1,x2∈[1,+∞),f(x1)≤0,f(x2)≤0.所以f(x1)f(x2)≥f(x1)0≥f(2)=1e綜上,a的最小值為1.12.解析(1)由已知得f'(x)=exa,若a≤0,則在區(qū)間(∞,+∞)上,f'(x)>0,f(x)單調(diào)遞增.所以當(dāng)a≤0時(shí),f(x)的單調(diào)遞增區(qū)間為(∞,+∞),沒有極值點(diǎn).若a>0,令f'(x)=0,得ex=a,解得x=lna,所以在區(qū)間(∞,lna)上f'(x)<0,f(x)單調(diào)遞減,在區(qū)間(lna,+∞)上f'(x)>0,f(x)單調(diào)遞增.所以當(dāng)a>0時(shí),f(x)的單調(diào)遞減區(qū)間為(∞,lna),f(x)的單調(diào)遞增區(qū)間為(lna,+∞),當(dāng)x=lna時(shí),函數(shù)f(x)有極小值2aalna.(2)當(dāng)a≤0時(shí),由(1)可知,f(x)在(∞,+∞)上單調(diào)遞增,因?yàn)閒(0)=1+a,f(1)=