3.2導(dǎo)數(shù)的應(yīng)用考綱解讀考點(diǎn)內(nèi)容解讀要求高考示例常考題型預(yù)測(cè)熱度1.利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性1.了解函數(shù)單調(diào)性和導(dǎo)數(shù)的關(guān)系2.能利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,會(huì)求函數(shù)的單調(diào)區(qū)間(其中多項(xiàng)式函數(shù)不超過三次)2017課標(biāo)全國(guó),21;2017課標(biāo)全國(guó),21;2017課標(biāo)全國(guó),21;2016課標(biāo)全國(guó),21選擇題解答題2.利用導(dǎo)數(shù)研究函數(shù)的極值與最值1.了解函數(shù)在某點(diǎn)取得極值的必要條件和充分條件2.會(huì)用導(dǎo)數(shù)求函數(shù)的極大值、極小值(其中多項(xiàng)式函數(shù)不超過三次);會(huì)求閉區(qū)間上函數(shù)的最大值、最小值(其中多項(xiàng)式函數(shù)不超過三次)2017北京,20;2017江蘇,20;2016山東,203.導(dǎo)數(shù)的綜合應(yīng)用會(huì)利用導(dǎo)數(shù)解決實(shí)際問題2017天津,19;2016課標(biāo)全國(guó),21;2015課標(biāo),21分析解讀函數(shù)的單調(diào)性是函數(shù)的一條重要性質(zhì),也是高中階段研究的重點(diǎn).一是直接用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、求函數(shù)的最值與極值,以及實(shí)際問題中的優(yōu)化問題等,這是新課標(biāo)的一個(gè)新要求.二是把導(dǎo)數(shù)與函數(shù)、方程、不等式、數(shù)列等知識(shí)相聯(lián)系,綜合考查函數(shù)的最值與參數(shù)的取值,常以解答題的形式出現(xiàn).本節(jié)內(nèi)容在高考中分值為17分左右,屬難度較大題.1)函數(shù)f(x)的定義域?yàn)?-,+), f (x)=2e2x-aex-a2=(2ex+a)(ex-a).若a=0,則f(x)=e2x,在(-,+)上單調(diào)遞增.若a>0,則由f (x)=0得x=ln a.當(dāng)x(-,ln a)時(shí), f (x)<0;當(dāng)x(ln a,+)時(shí), f (x)>0.故f(x)在(-,ln a)上單調(diào)遞減,在(ln a,+)上單調(diào)遞增.若a<0,則由f (x)=0得x=ln-a2.當(dāng)x-,ln-a2時(shí), f (x)<0;當(dāng)xln-a2,+時(shí), f (x)>0.故f(x)在-,ln-a2上單調(diào)遞減,在ln-a2,+上單調(diào)遞增.(2)若a=0,則f(x)=e2x,所以f(x)0.若a>0,則由(1)得,當(dāng)x=ln a時(shí), f(x)取得最小值,最小值為f(ln a)=-a2ln a,從而當(dāng)且僅當(dāng)-a2ln a0,即a1時(shí),f(x)0.若a<0,則由(1)得,當(dāng)x=ln-a2時(shí), f(x)取得最小值,最小值為fln-a2=a234-ln-a2.從而當(dāng)且僅當(dāng)a234-ln-a20,即a-2e34時(shí), f(x)0.綜上,a的取值范圍是-2e34,1.五年高考考點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性1.(2017山東,10,5分)若函數(shù)exf(x)(e=2.718 28是自然對(duì)數(shù)的底數(shù))在f(x)的定義域上單調(diào)遞增,則稱函數(shù)f(x)具有M性質(zhì).下列函數(shù)中具有M性質(zhì)的是()A.f(x)=2-xB.f(x)=x2C.f(x)=3-xD.f(x)=cos x答案A2.(2016課標(biāo)全國(guó),12,5分)若函數(shù)f(x)=x-13sin 2x+asin x在(-,+)單調(diào)遞增,則a的取值范圍是()A.-1,1B.-1,13 C.-13,13 D.-1,-13答案C3.(2015課標(biāo),12,5分)設(shè)函數(shù)f(x)=ln(1+|x|)-11+x2,則使得f(x)>f(2x-1)成立的x的取值范圍是()A.13,1B.-13(1,+)C.-13,13D.-,-1313+答案A4.(2014課標(biāo),11,5分)若函數(shù)f(x)=kx-ln x在區(qū)間(1,+)單調(diào)遞增,則k的取值范圍是()A.(-,-2B.(-,-1C.2,+)D.1,+)答案D5.(2017江蘇,11,5分)已知函數(shù)f(x)=x3-2x+ex-1ex,其中e是自然對(duì)數(shù)的底數(shù).若f(a-1)+f(2a2)0,則實(shí)數(shù)a的取值范圍是.答案-1,126.(2017課標(biāo)全國(guó),21,12分)設(shè)函數(shù)f(x)=(1-x2)ex.(1)討論f(x)的單調(diào)性;(2)當(dāng)x0時(shí), f(x)ax+1,求a的取值范圍.解析(1)f (x)=(1-2x-x2)ex.令f (x)=0,得x=-1-2或x=-1+2.當(dāng)x(-,-1-2)時(shí), f (x)<0;當(dāng)x(-1-2,-1+2)時(shí), f (x)>0;當(dāng)x(-1+2,+)時(shí), f (x)<0.所以f(x)在(-,-1-2),(-1+2,+)上單調(diào)遞減,在(-1-2,-1+2)上單調(diào)遞增.(2)f(x)=(1+x)(1-x)ex.當(dāng)a1時(shí),設(shè)函數(shù)h(x)=(1-x)ex,h(x)=-xex<0(x>0),因此h(x)在0,+)上單調(diào)遞減,而h(0)=1,故h(x)1,所以f(x)=(x+1)h(x)x+1ax+1.當(dāng)0<a<1時(shí),設(shè)函數(shù)g(x)=ex-x-1,g(x)=ex-1>0(x>0),所以g(x)在0,+)上單調(diào)遞增,而g(0)=0,故exx+1.當(dāng)0<x<1時(shí), f(x)>(1-x)(1+x)2,(1-x)(1+x)2-ax-1=x(1-a-x-x2),取x0=5-4a-12,則x0(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0)>ax0+1.當(dāng)a0時(shí),取x0=5-12,則x0(0,1), f(x0)>(1-x0)(1+x0)2=1ax0+1.綜上,a的取值范圍是1,+).7.(2017課標(biāo)全國(guó),21,12分)已知函數(shù)f(x)=ln x+ax2+(2a+1)x.(1)討論f(x)的單調(diào)性;(2)當(dāng)a<0時(shí),證明f(x)-34a-2.解析(1)f(x)的定義域?yàn)?0,+), f (x)=1x+2ax+2a+1=(x+1)(2ax+1)x.若a0,則當(dāng)x(0,+)時(shí), f (x)>0,故f(x)在(0,+)上單調(diào)遞增.若a<0,則當(dāng)x0,-12a時(shí), f (x)>0;當(dāng)x-12a,+時(shí), f (x)<0,故f(x)在0,-12a上單調(diào)遞增,在-12a,+上單調(diào)遞減.(2)由(1)知,當(dāng)a<0時(shí), f(x)在x=-12a處取得最大值,最大值為f-12a=ln-12a-1-14a.所以f(x)-34a-2等價(jià)于ln-12a-1-14a-34a-2,即ln-12a+12a+10.設(shè)g(x)=ln x-x+1,則g(x)=1x-1.當(dāng)x(0,1)時(shí),g(x)>0;當(dāng)x(1,+)時(shí),g(x)<0.所以g(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減.故當(dāng)x=1時(shí),g(x)取得最大值,最大值為g(1)=0.所以當(dāng)x>0時(shí),g(x)0.從而當(dāng)a<0時(shí),ln-12a+12a+10,即f(x)-34a-2.8.(2016課標(biāo)全國(guó),21,12分)設(shè)函數(shù)f(x)=ln x-x+1.(1)討論f(x)的單調(diào)性;(2)證明當(dāng)x(1,+)時(shí),1<x-1lnx<x;(3)設(shè)c>1,證明當(dāng)x(0,1)時(shí),1+(c-1)x>cx.解析(1)由題設(shè)知, f(x)的定義域?yàn)?0,+), f (x)=1x-1,令f (x)=0,解得x=1.當(dāng)0<x<1時(shí), f (x)>0, f(x)單調(diào)遞增;當(dāng)x>1時(shí), f (x)<0, f(x)單調(diào)遞減.(4分)(2)證明:由(1)知f(x)在x=1處取得最大值,最大值為f(1)=0.所以當(dāng)x1時(shí),ln x<x-1.故當(dāng)x(1,+)時(shí),ln x<x-1,ln1x<1x-1,即1<x-1lnx<x.(7分)(3)證明:由題設(shè)c>1,設(shè)g(x)=1+(c-1)x-cx,則g(x)=c-1-cxln c,令g(x)=0,解得x0=lnc-1lnclnc.當(dāng)x<x0時(shí),g(x)>0,g(x)單調(diào)遞增;當(dāng)x>x0時(shí),g(x)<0,g(x)單調(diào)遞減.(9分)由(2)知1<c-1lnc<c,故0<x0<1.又g(0)=g(1)=0,故當(dāng)0<x<1時(shí),g(x)>0.所以當(dāng)x(0,1)時(shí),1+(c-1)x>cx.(12分)教師用書專用(924)9.(2013浙江,8,5分)已知函數(shù)y=f(x)的圖象是下列四個(gè)圖象之一,且其導(dǎo)函數(shù)y=f (x)的圖象如圖所示,則該函數(shù)的圖象是()答案B10.(2015四川,21,14分)已知函數(shù)f(x)=-2xln x+x2-2ax+a2,其中a>0.(1)設(shè)g(x)是f(x)的導(dǎo)函數(shù),討論g(x)的單調(diào)性;(2)證明:存在a(0,1),使得f(x)0恒成立,且f(x)=0在區(qū)間(1,+)內(nèi)有唯一解.解析(1)由已知,得函數(shù)f(x)的定義域?yàn)?0,+),g(x)=f (x)=2(x-1-ln x-a),所以g(x)=2-2x=2(x-1)x.當(dāng)x(0,1)時(shí),g(x)<0,g(x)單調(diào)遞減;當(dāng)x(1,+)時(shí),g(x)>0,g(x)單調(diào)遞增.(2)證明:由f (x)=2(x-1-ln x-a)=0,解得a=x-1-ln x.令(x)=-2xln x+x2-2x(x-1-ln x)+(x-1-ln x)2=(1+ln x)2-2xln x,則(1)=1>0,(e)=2(2-e)<0.于是,存在x0(1,e),使得(x0)=0.令a0=x0-1-ln x0=u(x0),其中u(x)=x-1-ln x(x1).由u(x)=1-1x0知,函數(shù)u(x)在區(qū)間(1,+)上單調(diào)遞增.故0=u(1)<a0=u(x0)<u(e)=e-2<1.即a0(0,1).當(dāng)a=a0時(shí),有f (x0)=0, f(x0)=(x0)=0.再由(1)知, f (x)在區(qū)間(1,+)上單調(diào)遞增,當(dāng)x(1,x0)時(shí), f (x)<0,從而f(x)>f(x0)=0;當(dāng)x(x0,+)時(shí), f (x)>0,從而f(x)>f(x0)=0;又當(dāng)x(0,1時(shí), f(x)=(x-a0)2-2xln x>0.故x(0,+)時(shí), f(x)0.綜上所述,存在a(0,1),使得f(x)0恒成立,且f(x)=0在區(qū)間(1,+)內(nèi)有唯一解.11.(2015天津,20,14分)已知函數(shù)f(x)=4x-x4,xR.(1)求f(x)的單調(diào)區(qū)間;(2)設(shè)曲線y=f(x)與x軸正半軸的交點(diǎn)為P,曲線在點(diǎn)P處的切線方程為y=g(x),求證:對(duì)于任意的實(shí)數(shù)x,都有f(x)g(x);(3)若方程f(x)=a(a為實(shí)數(shù))有兩個(gè)實(shí)數(shù)根x1,x2,且x1<x2,求證:x2-x1-a3+413.解析(1)由f(x)=4x-x4,可得f (x)=4-4x3.當(dāng)f (x)>0,即x<1時(shí),函數(shù)f(x)單調(diào)遞增;當(dāng)f (x)<0,即x>1時(shí),函數(shù)f(x)單調(diào)遞減.所以, f(x)的單調(diào)遞增區(qū)間為(-,1),單調(diào)遞減區(qū)間為(1,+).(2)證明:設(shè)點(diǎn)P的坐標(biāo)為(x0,0),則x0=413, f (x0)=-12.曲線y=f(x)在點(diǎn)P處的切線方程為y=f (x0)(x-x0),即g(x)=f (x0)(x-x0).令函數(shù)F(x)=f(x)-g(x),即F(x)=f(x)-f (x0)(x-x0),則F(x)=f (x)-f (x0).由于f (x)=-4x3+4在(-,+)上單調(diào)遞減,故F(x)在(-,+)上單調(diào)遞減.又因?yàn)镕(x0)=0,所以當(dāng)x(-,x0)時(shí),F(x)>0,當(dāng)x(x0,+)時(shí),F(x)<0,所以F(x)在(-,x0)上單調(diào)遞增,在(x0,+)上單調(diào)遞減,所以對(duì)于任意的實(shí)數(shù)x,F(x)F(x0)=0,即對(duì)于任意的實(shí)數(shù)x,都有f(x)g(x).(3)證明:由(2)知g(x)=-12(x-413).設(shè)方程g(x)=a的根為x2,可得x2=-a12+413.因?yàn)間(x)在(-,+)上單調(diào)遞減,又由(2)知g(x2)f(x2)=a=g(x2),因此x2x2.類似地,設(shè)曲線y=f(x)在原點(diǎn)處的切線方程為y=h(x),可得h(x)=4x.對(duì)于任意的x(-,+),有f(x)-h(x)=-x40,即f(x)h(x).設(shè)方程h(x)=a的根為x1,可得x1=a4.因?yàn)閔(x)=4x在(-,+)上單調(diào)遞增,且h(x1)=a=f(x1)h(x1),因此x1x1.由此可得x2-x1x2-x1=-a3+413.12.(2015福建,22,14分)已知函數(shù)f(x)=ln x-(x-1)22.(1)求函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)證明:當(dāng)x>1時(shí), f(x)<x-1;(3)確定實(shí)數(shù)k的所有可能取值,使得存在x0>1,當(dāng)x(1,x0)時(shí),恒有f(x)>k(x-1).解析(1)f (x)=1x-x+1=-x2+x+1x,x(0,+).由f (x)>0得x>0,-x2+x+1>0.解得0<x<1+52.故f(x)的單調(diào)遞增區(qū)間是0,1+52.(2)證明:令F(x)=f(x)-(x-1),x(0,+).則有F(x)=1-x2x.當(dāng)x(1,+)時(shí),F(x)<0,所以F(x)在1,+)上單調(diào)遞減,故當(dāng)x>1時(shí),F(x)<F(1)=0,即當(dāng)x>1時(shí), f(x)<x-1.(3)由(2)知,當(dāng)k=1時(shí),不存在x0>1滿足題意.當(dāng)k>1時(shí),對(duì)于x>1,有f(x)<x-1<k(x-1),則f(x)<k(x-1),從而不存在x0>1滿足題意.當(dāng)k<1時(shí),令G(x)=f(x)-k(x-1),x(0,+),則有G(x)=1x-x+1-k=-x2+(1-k)x+1x.由G(x)=0得,-x2+(1-k)x+1=0.解得x1=1-k-(1-k)2+42<0,x2=1-k+(1-k)2+42>1.當(dāng)x(1,x2)時(shí),G(x)>0,故G(x)在1,x2)內(nèi)單調(diào)遞增.從而當(dāng)x(1,x2)時(shí),G(x)>G(1)=0,即f(x)>k(x-1),綜上,k的取值范圍是(-,1).13.(2015重慶,19,12分)已知函數(shù)f(x)=ax3+x2(aR)在x=-43處取得極值.(1)確定a的值;(2)若g(x)=f(x)ex,討論g(x)的單調(diào)性.解析(1)對(duì)f(x)求導(dǎo)得f (x)=3ax2+2x,因?yàn)閒(x)在x=-43處取得極值,所以f -43=0,即3a169+2-43=16a3-83=0,解得a=12.(2)由(1)得g(x)=12x3+x2ex,故g(x)=32x2+2xex+12x3+x2ex=12x3+52x2+2xex=12x(x+1)(x+4)ex.令g(x)=0,解得x=0,x=-1或x=-4.當(dāng)x<-4時(shí),g(x)<0,故g(x)為減函數(shù);當(dāng)-4<x<-1時(shí),g(x)>0,故g(x)為增函數(shù);當(dāng)-1<x<0時(shí),g(x)<0,故g(x)為減函數(shù);當(dāng)x>0時(shí),g(x)>0,故g(x)為增函數(shù).綜上,知g(x)在(-,-4)和(-1,0)內(nèi)為減函數(shù),在(-4,-1)和(0,+)內(nèi)為增函數(shù).14.(2014安徽,20,13分)設(shè)函數(shù)f(x)=1+(1+a)x-x2-x3,其中a>0.(1)討論f(x)在其定義域上的單調(diào)性;(2)當(dāng)x0,1時(shí),求f(x)取得最大值和最小值時(shí)的x的值.解析(1)f(x)的定義域?yàn)?-,+), f (x)=1+a-2x-3x2.令f (x)=0,得x1=-1-4+3a3,x2=-1+4+3a3,x1<x2,所以f (x)=-3(x-x1)(x-x2).當(dāng)x<x1或x>x2時(shí), f (x)<0;當(dāng)x1<x<x2時(shí), f (x)>0.故f(x)在(-,x1)和(x2,+)內(nèi)單調(diào)遞減,在x1,x2內(nèi)單調(diào)遞增.(2)因?yàn)閍>0,所以x1<0,x2>0.(i)當(dāng)a4時(shí),x21,由(1)知, f(x)在0,1上單調(diào)遞增,所以f(x)在x=0和x=1處分別取得最小值和最大值.(ii)當(dāng)0<a<4時(shí),x2<1.由(1)知, f(x)在0,x2上單調(diào)遞增,在x2,1上單調(diào)遞減,因此f(x)在x=x2=-1+4+3a3處取得最大值.又f(0)=1, f(1)=a,所以當(dāng)0<a<1時(shí), f(x)在x=1處取得最小值;當(dāng)a=1時(shí), f(x)在x=0和x=1處同時(shí)取得最小值;當(dāng)1<a<4時(shí), f(x)在x=0處取得最小值.15.(2014重慶,19,12分)已知函數(shù)f(x)=x4+ax-ln x-32,其中aR,且曲線y=f(x)在點(diǎn)(1, f(1)處的切線垂直于直線y=12x.(1)求a的值;(2)求函數(shù)f(x)的單調(diào)區(qū)間與極值.解析(1)對(duì)f(x)求導(dǎo)得f (x)=14-ax2-1x,由f(x)在點(diǎn)(1, f(1)處的切線垂直于直線y=12x知f (1)=-34-a=-2,解得a=54.(2)由(1)知f(x)=x4+54x-ln x-32,則f (x)=x2-4x-54x2,令f (x)=0,解得x=-1或x=5.因x=-1不在f(x)的定義域(0,+)內(nèi),故舍去.當(dāng)x(0,5)時(shí), f (x)<0,故f(x)在(0,5)內(nèi)為減函數(shù);當(dāng)x(5,+)時(shí), f (x)>0,故f(x)在(5,+)內(nèi)為增函數(shù).由此知函數(shù)f(x)在x=5時(shí)取得極小值f(5)=-ln 5.16.(2014湖北,21,14分)為圓周率,e=2.718 28為自然對(duì)數(shù)的底數(shù).(1)求函數(shù)f(x)=lnxx的單調(diào)區(qū)間;(2)求e3,3e,e,e,3,3這6個(gè)數(shù)中的最大數(shù)與最小數(shù).解析(1)函數(shù)f(x)的定義域?yàn)?0,+).因?yàn)閒(x)=lnxx,所以f (x)=1-lnxx2.當(dāng)f (x)>0,即0<x<e時(shí),函數(shù)f(x)單調(diào)遞增;當(dāng)f (x)<0,即x>e時(shí),函數(shù)f(x)單調(diào)遞減.故函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,e),單調(diào)遞減區(qū)間為(e,+).(2)因?yàn)閑<3<,所以eln 3<eln ,ln e<ln 3,即ln 3e<ln e,ln e<ln 3.于是根據(jù)函數(shù)y=ln x,y=ex,y=x在定義域上單調(diào)遞增,可得3e<e<3,e3<e<3.故這6個(gè)數(shù)的最大數(shù)在3與3之中,最小數(shù)在3e與e3之中.由e<3<及(1)的結(jié)論,得f()<f(3)<f(e),即ln <ln33<ln ee.由ln <ln33,得ln 3<ln 3,所以3>3;由ln33<ln ee,得ln 3e<ln e3,所以3e<e3.綜上,6個(gè)數(shù)中的最大數(shù)是3,最小數(shù)是3e.17.(2014湖南,21,13分)已知函數(shù)f(x)=xcos x-sin x+1(x>0).(1)求f(x)的單調(diào)區(qū)間;(2)記xi為f(x)的從小到大的第i(iN*)個(gè)零點(diǎn),證明:對(duì)一切nN*,有1x12+1x22+1xn2<23.解析(1)f (x)=cos x-xsin x-cos x=-xsin x.令f (x)=0,得x=k(kN*).當(dāng)x(2k,(2k+1)(kN)時(shí),sin x>0,此時(shí)f (x)<0;當(dāng)x(2k+1),(2k+2)(kN)時(shí),sin x<0,此時(shí)f (x)>0,故f(x)的單調(diào)遞減區(qū)間為(2k,(2k+1)(kN),單調(diào)遞增區(qū)間為(2k+1),(2k+2)(kN).(2) 由(1)知, f(x)在區(qū)間(0,)上單調(diào)遞減,又f2=0,故x1=2,當(dāng)nN*時(shí),因?yàn)閒(n)f(n+1)=(-1)nn+1(-1)n+1(n+1)n+1<0,且函數(shù)f(x)的圖象是連續(xù)不斷的,所以f(x)在區(qū)間(n,(n+1)內(nèi)至少存在一個(gè)零點(diǎn).又f(x)在區(qū)間(n,(n+1)上是單調(diào)的,故n<xn+1<(n+1).因此當(dāng)n=1時(shí),1x12=42<23;當(dāng)n=2時(shí),1x12+1x22<12(4+1)<23;當(dāng)n3時(shí),1x12+1x22+1xn2<124+1+122+1(n-1)2<125+112+1(n-2)(n-1)=125+1-12+12-13+1n-2-1n-1=126-1n-1<62<23.綜上所述,對(duì)一切nN*,1x12+1x22+1xn2<23.18.(2014江西,18,12分)已知函數(shù)f(x)=(4x2+4ax+a2)x,其中a<0.(1)當(dāng)a=-4時(shí),求f(x)的單調(diào)遞增區(qū)間;(2)若f(x)在區(qū)間1,4上的最小值為8,求a的值.解析(1)f(x)的定義域?yàn)?,+).當(dāng)a=-4時(shí),由f (x)=2(5x-2)(x-2)x=0得x=25或x=2,由f (x)>0得x0,25或x(2,+),故函數(shù)f(x)的單調(diào)遞增區(qū)間為0,25和(2,+).(2)f (x)=(10x+a)(2x+a)2x,a<0,由f (x)=0得x=-a10或x=-a2.當(dāng)x0,-a10時(shí),f(x)單調(diào)遞增;當(dāng)x-a10,-a2時(shí),f(x)單調(diào)遞減;當(dāng)x-a2,+時(shí),f(x)單調(diào)遞增.易知 f(x)=(2x+a)2x0,且f-a2=0.當(dāng)-a21,即-2a<0時(shí),f(x)在1,4上的最小值為f(1),由f(1)=4+4a+a2=8,得a=22-2,均不符合題意.當(dāng)1<-a24,即-8a<-2時(shí), f(x)在1,4上的最小值為f-a2=0,不符合題意.當(dāng)-a2>4,即a<-8時(shí),f(x)在1,4上的最小值可能在x=1或x=4處取得,而f(1)8,由f(4)=2(64+16a+a2)=8得a=-10或a=-6(舍去),當(dāng)a=-10時(shí),f(x)在(1,4)上單調(diào)遞減, f(x)在1,4上的最小值為f(4)=8,符合題意.綜上,a=-10.19.(2013課標(biāo)全國(guó),20,12分)已知函數(shù)f(x)=ex(ax+b)-x2-4x,曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程為y=4x+4.(1)求a,b的值;(2)討論f(x)的單調(diào)性,并求f(x)的極大值.解析(1)f (x)=ex(ax+a+b)-2x-4.由已知得f(0)=4, f (0)=4.故b=4,a+b=8.從而a=4,b=4.(2)由(1)知f(x)=4ex(x+1)-x2-4x,f (x)=4ex(x+2)-2x-4=4(x+2)ex-12.令f (x)=0,得x=-ln 2或x=-2.從而當(dāng)x(-,-2)(-ln 2,+)時(shí), f (x)>0;當(dāng)x(-2,-ln 2)時(shí), f (x)<0.故f(x)在(-,-2),(-ln 2,+)上單調(diào)遞增,在(-2,-ln 2)上單調(diào)遞減.當(dāng)x=-2時(shí),函數(shù)f(x)取得極大值,極大值為f(-2)=4(1-e-2).20.(2013大綱全國(guó),21,12分)已知函數(shù)f(x)=x3+3ax2+3x+1.(1)當(dāng)a=-2時(shí),討論f(x)的單調(diào)性;(2)若x2,+)時(shí), f(x)0,求a的取值范圍.解析(1)當(dāng)a=-2時(shí), f(x)=x3-32x2+3x+1,f (x)=3x2-62x+3.令f (x)=0,得x1=2-1,x2=2+1.(3分)當(dāng)x(-,2-1)時(shí), f (x)>0, f(x)在(-,2-1)上是增函數(shù);當(dāng)x(2-1,2+1)時(shí), f (x)<0, f(x)在(2-1,2+1)上是減函數(shù);當(dāng)x(2+1,+)時(shí), f (x)>0, f(x)在(2+1,+)上是增函數(shù).(6分)(2)由f(2)0得a-54.(8分)當(dāng)a-54,x(2,+)時(shí),f (x)=3(x2+2ax+1)3x2-52x+1=3x-12(x-2)>0,所以f(x)在(2,+)上是增函數(shù),于是當(dāng)x2,+)時(shí),f(x)f(2)0.綜上,a的取值范圍是-54,+.(12分)21.(2013山東,21,12分)已知函數(shù)f(x)=ax2+bx-ln x(a,bR).(1)設(shè)a0,求f(x)的單調(diào)區(qū)間;(2)設(shè)a>0,且對(duì)任意x>0, f(x)f(1).試比較ln a與-2b的大小.解析(1)由f(x)=ax2+bx-ln x,x(0,+),得f (x)=2ax2+bx-1x.當(dāng)a=0時(shí), f (x)=bx-1x.(i)若b0,當(dāng)x>0時(shí), f (x)<0恒成立,所以函數(shù)f(x)的單調(diào)遞減區(qū)間是(0,+).(ii)若b>0,當(dāng)0<x<1b時(shí), f (x)<0,函數(shù)f(x)單調(diào)遞減,當(dāng)x>1b時(shí), f (x)>0,函數(shù)f(x)單調(diào)遞增.所以函數(shù)f(x)的單調(diào)遞減區(qū)間是0,1b,單調(diào)遞增區(qū)間是1b,+.當(dāng)a>0時(shí),令f (x)=0,得2ax2+bx-1=0.由=b2+8a>0得x1=-b-b2+8a4a,x2=-b+b2+8a4a.顯然,x1<0,x2>0.當(dāng)0<x<x2時(shí), f (x)<0,函數(shù)f(x)單調(diào)遞減;當(dāng)x>x2時(shí), f (x)>0,函數(shù)f(x)單調(diào)遞增.所以函數(shù)f(x)的單調(diào)遞減區(qū)間是0,-b+b2+8a4a,單調(diào)遞增區(qū)間是-b+b2+8a4a,+.綜上所述,當(dāng)a=0,b0時(shí),函數(shù)f(x)的單調(diào)遞減區(qū)間是(0,+);當(dāng)a=0,b>0時(shí),函數(shù)f(x)的單調(diào)遞減區(qū)間是0,1b,單調(diào)遞增區(qū)間是1b,+;當(dāng)a>0時(shí),函數(shù)f(x)的單調(diào)遞減區(qū)間是0,-b+b2+8a4a,單調(diào)遞增區(qū)間是-b+b2+8a4a,+.(2)由題意,函數(shù)f(x)在x=1處取得最小值,由(1)知-b+b2+8a4a是f(x)的唯一極小值點(diǎn),故-b+b2+8a4a=1,整理得2a+b=1,即b=1-2a.令g(x)=2-4x+ln x.則g(x)=1-4xx.令g(x)=0,得x=14.當(dāng)0<x<14時(shí),g(x)>0,g(x)單調(diào)遞增;當(dāng)x>14時(shí),g(x)<0,g(x)單調(diào)遞減.因此g(x)g14=1+ln14=1-ln 4<0.故g(a)<0,即2-4a+ln a=2b+ln a<0,即ln a<-2b.22.(2013天津,20,14分)設(shè)a-2,0,已知函數(shù)f(x)=x3-(a+5)x,x0,x3-a+32x2+ax,x>0.(1)證明f(x)在區(qū)間(-1,1)內(nèi)單調(diào)遞減,在區(qū)間(1,+)內(nèi)單調(diào)遞增;(2)設(shè)曲線y=f(x)在點(diǎn)Pi(xi, f(xi)(i=1,2,3)處的切線相互平行,且x1x2x30.證明x1+x2+x3>-13.證明(1)設(shè)函數(shù)f1(x)=x3-(a+5)x(x0),f2(x)=x3-a+32x2+ax(x0), f 1(x)=3x2-(a+5),由a-2,0,從而當(dāng)-1<x<0時(shí),f 1(x)=3x2-(a+5)<3-a-50,所以函數(shù)f1(x)在區(qū)間(-1,0內(nèi)單調(diào)遞減. f 2(x)=3x2-(a+3)x+a=(3x-a)(x-1),由于a-2,0,所以當(dāng)0<x<1時(shí), f 2(x)<0;當(dāng)x>1時(shí), f 2(x)>0.即函數(shù)f2(x)在區(qū)間0,1)內(nèi)單調(diào)遞減,在區(qū)間(1,+)內(nèi)單調(diào)遞增.綜合,及f1(0)=f2(0),可知函數(shù)f(x)在區(qū)間(-1,1)內(nèi)單調(diào)遞減,在區(qū)間(1,+)內(nèi)單調(diào)遞增.(2)由(1)知f (x)在區(qū)間(-,0)內(nèi)單調(diào)遞減,在區(qū)間0,a+36內(nèi)單調(diào)遞減,在區(qū)間a+36,+內(nèi)單調(diào)遞增.因?yàn)榍€y=f(x)在點(diǎn)Pi(xi, f(xi)(i=1,2,3)處的切線相互平行,從而x1,x2,x3互不相等,且f (x1)=f (x2)=f (x3).不妨設(shè)x1<0<x2<x3,由3x12-(a+5)=3x22-(a+3)x2+a=3x32-(a+3)x3+a,可得3x22-3x32-(a+3)(x2-x3)=0,解得x2+x3=a+33,從而0<x2<a+36<x3. 設(shè)g(x)=3x2-(a+3)x+a,則ga+36<g(x2)<g(0)=a.由3x12-(a+5)=g(x2)<a,解得-2a+53<x1<0,所以x1+x2+x3>-2a+53+a+33,設(shè)t=2a+53,則a=3t2-52,因?yàn)閍-2,0,所以t33,153,故x1+x2+x3>-t+3t2+16=12(t-1)2-13-13,即x1+x2+x3>-13.23.(2013湖北,21,13分)設(shè)a>0,b>0,已知函數(shù)f(x)=ax+bx+1.(1)當(dāng)ab時(shí),討論函數(shù)f(x)的單調(diào)性;(2)當(dāng)x>0時(shí),稱f(x)為a、b關(guān)于x的加權(quán)平均數(shù).(i)判斷f(1),f ba,f ba是否成等比數(shù)列,并證明f baf ba;(ii)a、b的幾何平均數(shù)記為G.稱2aba+b為a、b的調(diào)和平均數(shù),記為H.若Hf(x)G,求x的取值范圍.解析(1)f(x)的定義域?yàn)?-,-1)(-1,+),f (x)=a(x+1)-(ax+b)(x+1)2=a-b(x+1)2.當(dāng)a>b時(shí), f (x)>0,函數(shù)f(x)在(-,-1),(-1,+)上單調(diào)遞增;當(dāng)a<b時(shí), f (x)<0,函數(shù)f(x)在(-,-1),(-1,+)上單調(diào)遞減.(2)(i)計(jì)算得f(1)=a+b2>0, fba=2aba+b>0,f ba=ab>0,故f(1)fba=a+b22aba+b=ab=fba2,即f(1)fba=fba2.所以f(1),f ba,f ba成等比數(shù)列.因?yàn)閍+b2ab,所以f(1)f ba.由得f baf ba.(ii)由(i)知f ba=H,f ba=G.故由Hf(x)G,得f baf(x)f ba.當(dāng)a=b時(shí),f ba=f(x)=f ba=a.這時(shí),x的取值范圍為(0,+);當(dāng)a>b時(shí),0<ba<1,從而ba<ba,由f(x)在(0,+)上單調(diào)遞增與式,得baxba,即x的取值范圍為ba,ba;當(dāng)a<b時(shí),ba>1,從而ba>ba,由f(x)在(0,+)上單調(diào)遞減與式,得baxba,即x的取值范圍為ba,ba.24.(2013江蘇,20,16分)設(shè)函數(shù)f(x)=ln x-ax,g(x)=ex-ax,其中a為實(shí)數(shù).(1)若f(x)在(1,+)上是單調(diào)減函數(shù),且g(x)在(1,+)上有最小值,求a的取值范圍;(2)若g(x)在(-1,+)上是單調(diào)增函數(shù),試求f(x)的零點(diǎn)個(gè)數(shù),并證明你的結(jié)論.解析(1)令f (x)=1x-a=1-axx<0,考慮到f(x)的定義域?yàn)?0,+),故a>0,進(jìn)而解得x>a-1,即f(x)在(a-1,+)上是單調(diào)減函數(shù).同理, f(x)在(0,a-1)上是單調(diào)增函數(shù).由于f(x)在(1,+)上是單調(diào)減函數(shù),故(1,+)(a-1,+),從而a-11,即a1.令g(x)=ex-a=0,得x=ln a.當(dāng)x<ln a時(shí),g(x)<0;當(dāng)x>ln a時(shí),g(x)>0.又g(x)在(1,+)上有最小值,所以ln a>1,即a>e.綜上,有a(e,+).(2)當(dāng)a0時(shí),g(x)必為單調(diào)增函數(shù);當(dāng)a>0時(shí),令g(x)=ex-a>0,解得a<ex,即x>ln a,因?yàn)間(x)在(-1,+)上是單調(diào)增函數(shù),類似(1)有l(wèi)n a-1,即0<ae-1.結(jié)合上述兩種情況,有ae-1.(i)當(dāng)a=0時(shí),由f(1)=0以及f (x)=1x>0,得f(x)存在唯一的零點(diǎn).(ii)當(dāng)a<0時(shí),由于f(ea)=a-aea=a(1-ea)<0, f(1)=-a>0,且函數(shù)f(x)在ea,1上的圖象不間斷,所以f(x)在(ea,1)上存在零點(diǎn).另外,當(dāng)x>0時(shí), f (x)=1x-a>0,故f(x)在(0,+)上是單調(diào)增函數(shù),所以f(x)只有一個(gè)零點(diǎn).(iii)當(dāng)0<ae-1時(shí),令f (x)=1x-a=0,解得x=a-1.當(dāng)0<x<a-1時(shí), f (x)>0,當(dāng)x>a-1時(shí), f (x)<0,所以,x=a-1是f(x)的最大值點(diǎn),且最大值為f(a-1)=-ln a-1.當(dāng)-ln a-1=0,即a=e-1時(shí), f(x)有一個(gè)零點(diǎn)x=e.當(dāng)-ln a-1>0,即0<a<e-1時(shí), f(x)有兩個(gè)零點(diǎn).實(shí)際上,對(duì)于0<a<e-1,由于f(e-1)=-1-ae-1<0, f(a-1)>0,且函數(shù)f(x)在e-1,a-1上的圖象不間斷,所以f(x)在(e-1,a-1)上存在零點(diǎn).另外,當(dāng)x(0,a-1)時(shí), f (x)=1x-a>0,故f(x)在(0,a-1)上是單調(diào)增函數(shù),所以f(x)在(0,a-1)上只有一個(gè)零點(diǎn).下面考慮f(x)在(a-1,+)上的情況.先證f(ea-1)=a(a-2-ea-1)<0.為此,我們要證明:當(dāng)x>e時(shí),ex>x2.設(shè)h(x)=ex-x2,則h(x)=ex-2x,再設(shè)l(x)=h(x)=ex-2x,則l(x)=ex-2.當(dāng)x>1時(shí),l(x)=ex-2>e-2>0,所以l(x)=h(x)在(1,+)上是單調(diào)增函數(shù).故當(dāng)x>2時(shí),h(x)=ex-2x>h(2)=e2-4>0,從而h(x)在(2,+)上是單調(diào)增函數(shù),進(jìn)而當(dāng)x>e時(shí),h(x)=ex-x2>h(e)=ee-e2>0.即當(dāng)x>e時(shí),ex>x2.當(dāng)0<a<e-1,即a-1>e時(shí), f(ea-1)=a-1-aea-1=a(a-2-ea-1)<0,又f(a-1)>0,且函數(shù)f(x)在a-1,ea-1上的圖象不間斷,所以f(x)在(a-1,ea-1)上存在零點(diǎn).又當(dāng)x>a-1時(shí), f (x)=1x-a<0,故f(x)在(a-1,+)上是單調(diào)減函數(shù),所以f(x)在(a-1,+)上只有一個(gè)零點(diǎn).綜合(i),(ii),(iii),當(dāng)a0或a=e-1時(shí), f(x)的零點(diǎn)個(gè)數(shù)為1,當(dāng)0<a<e-1時(shí),f(x)的零點(diǎn)個(gè)數(shù)為2.考點(diǎn)二利用導(dǎo)數(shù)研究函數(shù)的極值與最值1.(2016四川,6,5分)已知a為函數(shù)f(x)=x3-12x的極小值點(diǎn),則a=()A.-4B.-2C.4D.2答案D2.(2014遼寧,12,5分)當(dāng)x-2,1時(shí),不等式ax3-x2+4x+30恒成立,則實(shí)數(shù)a的取值范圍是()A.-5,-3B.-6,-98C.-6,-2D.-4,-3答案C3.(2015陜西,15,5分)函數(shù)y=xex在其極值點(diǎn)處的切線方程為.答案y=-1e4.(2017北京,20,13分)已知函數(shù)f(x)=excos x-x.(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)在區(qū)間0,2上的最大值和最小值.解析(1)因?yàn)閒(x)=excos x-x,所以f (x)=ex(cos x-sin x)-1, f (0)=0.又因?yàn)閒(0)=1,所以曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程為y=1.(2)設(shè)h(x)=ex(cos x-sin x)-1,則h(x)=ex(cos x-sin x-sin x-cos x)=-2exsin x.當(dāng)x0,2時(shí),h(x)<0,所以h(x)在區(qū)間0,2上單調(diào)遞減.所以對(duì)任意x0,2,有h(x)<h(0)=0,即f (x)<0.所以函數(shù)f(x)在區(qū)間0,2上單調(diào)遞減.因此f(x)在區(qū)間0,2上的最大值為f(0)=1,最小值為f2=-2.5.(2017江蘇,20,16分)已知函數(shù)f(x)=x3+ax2+bx+1(a>0,bR)有極值,且導(dǎo)函數(shù)f (x)的極值點(diǎn)是f(x)的零點(diǎn).(極值點(diǎn)是指函數(shù)取極值時(shí)對(duì)應(yīng)的自變量的值)(1)求b關(guān)于a的函數(shù)關(guān)系式,并寫出定義域;(2)證明:b2>3a;(3)若f(x), f (x)這兩個(gè)函數(shù)的所有極值之和不小于-72,求a的取值范圍.解析(1)由f(x)=x3+ax2+bx+1,得f (x)=3x2+2ax+b=3x+a32+b-a23.當(dāng)x=-a3時(shí), f (x)有極小值b-a23.因?yàn)閒 (x)的極值點(diǎn)是f(x)的零點(diǎn),所以f -a3=-a327+a39-ab3+1=0,又a>0,故b=2a29+3a.因?yàn)閒(x)有極值,故f (x)=0有實(shí)根,從而b-a23=19a(27-a3)0,即a3.當(dāng)a=3時(shí), f (x)>0(x-1),故f(x)在R上是增函數(shù), f(x)沒有極值;當(dāng)a>3時(shí), f (x)=0有兩個(gè)相異的實(shí)根x1=-a-a2-3b3,x2=-a+a2-3b3.列表如下:x(-,x1)x1(x1,x2)x2(x2,+)f (x)+0-0+f(x)極大值極小值故f(x)的極值點(diǎn)是x1,x2.從而a>3.因此b=2a29+3a,定義域?yàn)?3,+).(2)證明:由(1)知,ba=2aa9+3aa.設(shè)g(t)=2t9+3t,則g(t)=29-3t2=2t2-279t2.當(dāng)t362,+時(shí),g(t)>0,從而g(t)在362,+上單調(diào)遞增.因?yàn)閍>3,所以aa>33,故g(a a)>g(33)=3,即ba>3.因此b2>3a.(3)由(1)知, f(x)的極值點(diǎn)是x1,x2,且x1+x2=-23a,x12+x22=4a2-6b9.從而f(x1)+f(x2)=x13+ax12+bx1+1+x23+ax22+bx2+1=x13(3x12+2ax1+b)+x23(3x22+2ax2+b)+13a(x12+x22)+23b(x1+x2)+2=4a3-6ab27-4ab9+2=0.記f(x), f (x)所有極值之和為h(a),因?yàn)閒 (x)的極值為b-a23=-19a2+3a,所以h(a)=-19a2+3a,a>3.因?yàn)閔(a)=-29a-3a2<0,于是h(a)在(3,+)上單調(diào)遞減.因?yàn)閔(6)=-72,于是h(a)h(6),故a6.因此a的取值范圍為(3,6.6.(2015安徽,21,13分)已知函數(shù)f(x)=ax(x+r)2(a>0,r>0).(1)求f(x)的定義域,并討論f(x)的單調(diào)性;(2)若ar=400,求f(x)在(0,+)內(nèi)的極值.解析(1)由題意知x-r,所求的定義域?yàn)?-,-r)(-r,+).f(x)=ax(x+r)2=axx2+2rx+r2,f (x)=a(x2+2rx+r2)-ax(2x+2r)(x2+2rx+r2)2=a(r-x)(x+r)(x+r)4,所以當(dāng)x<-r或x>r時(shí),f (x)<0,當(dāng)-r<x<r時(shí),f (x)>0,因此,f(x)的單調(diào)遞減區(qū)間為(-,-r),(r,+);f(x)的單調(diào)遞增區(qū)間為(-r,r).(2)由(1)的解答可知f (r)=0,f(x)在(0,r)上單調(diào)遞增,在(r,+)上單調(diào)遞減.因此,x=r是f(x)的極大值點(diǎn),所以f(x)在(0,+)內(nèi)的極大值為f(r)=ar(2r)2=a4r=4004=100.教師用書專用(715)7.(2013福建,12,5分)設(shè)函數(shù)f(x)的定義域?yàn)镽,x0(x00)是f(x)的極大值點(diǎn),以下結(jié)論一定正確的是()A.xR, f(x)f(x0)B.-x0是f(-x)的極小值點(diǎn)C.-x0是-f(x)的極小值點(diǎn)D.-x0是-f(-x)的極小值點(diǎn)答案D8.(2016天津,20,14分)設(shè)函數(shù)f(x)=x3-ax-b,xR,其中a,bR.(1)求f(x)的單調(diào)區(qū)間;(2)若f(x)存在極值點(diǎn)x0,且f(x1)=f(x0),其中x1x0,求證:x1+2x0=0;(3)設(shè)a>0,函數(shù)g(x)=|f(x)|,求證:g(x)在區(qū)間-1,1上的最大值不小于.解析(1)由f(x)=x3-ax-b,可得f (x)=3x2-a.下面分兩種情況討論:當(dāng)a0時(shí),有f (x)=3x2-a0恒成立,所以f(x)的單調(diào)遞增區(qū)間為(-,+).當(dāng)a>0時(shí),令f (x)=0,解得x=3a3,或x=-3a3.當(dāng)x變化時(shí), f (x), f(x)的變化情況如下表:x-,-3a3-3a3-3a3,3a33a33a3,+f (x)+0-0+f(x)單調(diào)遞增極大值單調(diào)遞減極小值單調(diào)遞增所以f(x)的單調(diào)遞減區(qū)間為-3a3,3a3,單調(diào)遞增區(qū)間為-,-3a3,3a3,+.(2)證明:因?yàn)閒(x)存在極值點(diǎn),所以由(1)知a>0,且x00.由題意,得f (x0)=3x02-a=0,即x02=a3,進(jìn)而f(x0)=x03-ax0-b=-2a3x0-b.又f(-2x0)=-8x03+2ax0-b=-8a3x0+2ax0-b=-2a3x0-b=f(x0),且-2x0x0,由題意及(1)知,存在唯一實(shí)數(shù)x1滿足 f(x1)=f(x0),且x1x0,因此x1=-2x0.所以x1+2x0=0.(3)證明:設(shè)g(x)在區(qū)間-1,1上的最大值為M,maxx,y表示x,y兩數(shù)的最大值.下面分三種情況討論:當(dāng)a3時(shí),-3a3-1<13a3,由(1)知, f(x)在區(qū)間-1,1上單調(diào)遞減,所以f(x)在區(qū)間-1,1上的取值范圍為f(1), f(-1),因此M=max|f(1)|,|f(-1)|=max|1-a-b|,|-1+a-b|=max|a-1+b|,|a-1-b|=a-1+b,b0,a-1-b,b<0.所以M=a-1+|b|2.當(dāng)34a<3時(shí),-23a3-1<-3a3<3a3<123a3,由(1)和(2)知f(-1)f -23a3=f 3a3, f(1)f 23a3=f -3a3,所以f(x)在區(qū)間-1,1上的取值范圍為f 3a3, f -3a3,因此M=max,=max-2a93a-b,2a93a-b=max2a93a+b,2a93a-b=2a93a+|b|2934334=14.當(dāng)0<a<34時(shí),-1<-23a3<23a3<1,由(1)和(2)知f(-1)<f -23a3=f 3a3, f(1)>f 23a3=f -3a3,所以f(x)在區(qū)間-1,1上的取值范圍為f(-1), f(1),因此M=max|f(-1)|,|f(1)|=max|-1+a-b|,|1-a-b|=max|1-a+b|,|1-a-b|=1-a+|b|>14.綜上所述,當(dāng)a>0時(shí),g(x)在區(qū)間-1,1上的最大值不小于14.9.(2014天津,19,14分)已知函數(shù)f(x)=x2-23ax3(a>0),xR.(1)求f(x)的單調(diào)區(qū)間和極值;(2)若對(duì)于任意的x1(2,+),都存在x2(1,+),使得f(x1)f(x2)=1.求a的取值范圍.解析(1)由已知,有f (x)=2x-2ax2(a>0).令f (x)=0,解得x=0或x=1a.當(dāng)x變化時(shí), f (x), f(x)的變化情況如下表:x(-,0)00,1a1a1a,+f (x)-0+0-f(x)013a2所以, f(x)的單調(diào)遞增區(qū)間是0,1a;單調(diào)遞減區(qū)間是(-,0),1a,+.當(dāng)x=0時(shí), f(x)有極小值,且極小值f(0)=0;當(dāng)x=1a時(shí),f(x)有極大值,且極大值f1a=13a2.(2)由f(0)=f32a=0及(1)知,當(dāng)x0,32a時(shí), f(x)>0;當(dāng)x32a,+時(shí), f(x)<0.設(shè)集合A=f(x)|x(2,+),集合B=1f(x)|x(1,+), f(x)0.則“對(duì)于任意的x1(2,+),都存在x2(1,+),使得f(x1)f(x2)=1”等價(jià)于AB.顯然,0B.下面分三種情況討論:當(dāng)32a>2,即0<a<34時(shí),由f32a=0可知,0A,而0B,所以A不是B的子集.當(dāng)132a2,即34a32時(shí),有f(2)0,且此時(shí)f(x)在(2,+)上單調(diào)遞減,故A=(-, f(2),因而A(-,0);由f(1)0,有f(x)在(1,+)上的取值范圍包含(-,0),則(-,0)B.所以,AB.當(dāng)32a<1,即a>32時(shí),有f(1)<0,且此時(shí)f(x)在(1,+)上單調(diào)遞減,故B=1f(1),0,A=(-,f(2),所以A不是B的子集.綜上,a的取值范圍是34,32.10.(2014浙江,21,15分)已知函數(shù)f(x)=x3+3|x-a|(a>0).若f(x)在-1,1上的最小值記為g(a).(1)求g(a);(2)證明:當(dāng)x-1,1時(shí),恒有f(x)g(a)+4.解析(1)因?yàn)閍>0,-1x1,所以(i)當(dāng)0<a<1時(shí),若x-1,a,則f(x)=x3-3x+3a, f (x)=3x2-3<0,故f(x)在(-1,a)上是減函數(shù);若xa,1,則f(x)=x3+3x-3a, f (x)=3x2+3>0,故f(x)在(a,1)上是增函數(shù).所以g(a)=f(a)=a3.(ii)當(dāng)a1時(shí),有xa,則f(x)=x3-3x+3a, f (x)=3x2-3<0,故f(x)在(-1,1)上是減函數(shù),所以g(a)=f(1)=-2+3a.綜上,g(a)=a3,0<a<1,-2+3a,a1.(2)令h(x)=f(x)-g(a),(i)當(dāng)0<a<1時(shí),g(a)=a3,若xa,1,h(x)=x3+3x-3a-a3,得h(x)=3x2+3,則h(x)在(a