第4課時導(dǎo)數(shù)及其應(yīng)用課后訓(xùn)練案鞏固提升一、A組1.(2016廣東實(shí)驗(yàn)中學(xué)月考)已知f(x)=x3-92x2+6x-a,若對任意實(shí)數(shù)x,f'(x)m恒成立,則m的最大值為()A.3B.2C.1D.-34解析:f'(x)=3x2-9x+6,因?yàn)閷θ我鈱?shí)數(shù)x,f'(x)m恒成立,即3x2-9x+(6-m)0恒成立,所以=81-12(6-m)0,解得m-34,即m的最大值為-34,故選D.答案:D2.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)解析:f(x)與-f(-x)的圖象關(guān)于原點(diǎn)對稱,故x0(x00)是f(x)的極大值點(diǎn)時,-x0是-f(-x)的極小值點(diǎn),故選D.答案:D3.(2016海南?？诟叨z測)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,+)解析:由f'(x)=k-1x,又f(x)在(1,+)上單調(diào)遞增,則f'(x)0在x(1,+)上恒成立,即k1x在x(1,+)上恒成立.又當(dāng)x(1,+)時,0<1x<1,故k1.故選D.答案:D4.(2017山西晉中高二檢測)函數(shù)f(x)=1+x+x22+x33(xR)的零點(diǎn)的個數(shù)是()A.0B.1C.2D.3解析:對于f'(x)=1+x+x2=x+122+34>0,因此函數(shù)f(x)在R上單調(diào)遞增,且f(-2)=-53<0,f(2)=233>0,因此函數(shù)f(x)零點(diǎn)的個數(shù)為1,故選B.答案:B5.(2016山東泰安高二檢測)若0<x1<x2<1,則()A.ex2-ex1>ln x2-ln x1B.ex2-ex1<ln x2-ln x1C.x2ex1>x1ex2D.x2ex1<x1ex2解析:令f(x)=exx,則f'(x)=xex-exx2=ex(x-1)x2.當(dāng)0<x<1時,f'(x)<0,即f(x)在(0,1)上單調(diào)遞減,0<x1<x2<1,f(x2)<f(x1),即ex2x2<ex1x1.x2ex1>x1ex2,故選C.答案:C6.(2015陜西高考)函數(shù)y=xex在其極值點(diǎn)處的切線方程為. 解析:令y'=(x+1)ex=0,得x=-1,則切點(diǎn)為-1,-1e.函數(shù)在極值點(diǎn)處的導(dǎo)數(shù)為0,即切線斜率為0,則切線方程為y=-1e.答案:y=-1e7.(2015天津高考)已知函數(shù)f(x)=axln x,x(0,+),其中a為實(shí)數(shù),f'(x)為f(x)的導(dǎo)函數(shù),若f'(1)=3,則a的值為. 解析:因?yàn)閒(x)=axlnx,所以f'(x)=alnx+ax·1x=a(lnx+1).由f'(1)=3得a(ln1+1)=3,所以a=3.答案:38.已知函數(shù)f(x)=ex(ax2-2x+2),其中a>0.(1)若曲線y=f(x)在x=2處的切線與直線x+e2y-1=0垂直,求實(shí)數(shù)a的值;(2)討論f(x)的單調(diào)性.解:f'(x)=exax2+(2a-2)x(a>0).(1)由題意得f'(2)·-1e2=-1,解得a=58.(2)令f'(x)=0,得x1=0,x2=2-2aa.當(dāng)0<a<1時,f(x)的增區(qū)間為(-,0),2-2aa,+,減區(qū)間為0,2-2aa;當(dāng)a=1時,f(x)在(-,+)內(nèi)單調(diào)遞增;當(dāng)a>1時,f(x)的增區(qū)間為-,2-2aa,(0,+),減區(qū)間為2-2aa,0.9.已知函數(shù)f(x)=(4x2+4ax+a2)x,其中a<0.(1)當(dāng)a=-4時,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:(1)當(dāng)a=-4時,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)因?yàn)閒'(x)=(10x+a)(2x+a)2x,a<0,由f'(x)=0得x=-a10或x=-a2.當(dāng)x0,-a10時,f(x)單調(diào)遞增;當(dāng)x-a10,-a2時,f(x)單調(diào)遞減;當(dāng)x-a2,+時,f(x)單調(diào)遞增.易知f(x)=(2x+a)2x0,且f-a2=0.當(dāng)-a21時,即-2a<0時,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時,即-8a<-2時,f(x)在1,4上的最小值為f-a2=0,不符合題意.當(dāng)-a2>4時,即a<-8時,f(x)在1,4上的最小值可能在x=1或x=4上取得,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,f(x)在(1,4)單調(diào)遞減,f(x)在1,4上的最小值為f(4)=8,符合題意.綜上有,a=-10.10.已知函數(shù)f(x)=ax2+x-xln x.(1)若a=0,求函數(shù)f(x)的單調(diào)區(qū)間;(2)若f(1)=2,且在定義域內(nèi)f(x)bx2+2x恒成立,求實(shí)數(shù)b的取值范圍.解:(1)當(dāng)a=0時,f(x)=x-xlnx,函數(shù)定義域?yàn)?0,+).f'(x)=-lnx,由-lnx=0,得x=1.當(dāng)x(0,1)時,f'(x)>0,f(x)在(0,1)上是增函數(shù);當(dāng)x(1,+)時,f'(x)<0,f(x)在(1,+)上是減函數(shù).(2)由f(1)=2,得a+1=2,所以a=1,因此f(x)=x2+x-xlnx.由f(x)bx2+2x,得(1-b)x-1lnx.因?yàn)閤>0,所以b1-1x-lnxx恒成立.令g(x)=1-1x-lnxx,可得g'(x)=lnxx2,因此g(x)在(0,1)上單調(diào)遞減,在(1,+)上單調(diào)遞增,所以g(x)min=g(1)=0,故b的取值范圍是(-,0.二、B組1.(2017山東日照高二檢測)已知函數(shù)f(x)=x(ln x-ax)有兩個極值點(diǎn),則實(shí)數(shù)a的取值范圍是()A.(-,0)B.0,12C.(0,1)D.(0,+)解析:由題意知,x>0,f'(x)=lnx+1-2ax,由于函數(shù)f(x)有兩個極值點(diǎn),則f'(x)=0有兩個不相等的正根,顯然a0時不合題意,必有a>0.令g(x)=lnx+1-2ax,則g'(x)=1x-2a,令g'(x)=0,得x=12a,故g(x)在0,12a上單調(diào)遞增,在12a,+上單調(diào)遞減,所以g(x)在x=12a處取得最大值,即f'12a=ln12a>0,所以0<a<12.答案:B2.(原創(chuàng)題)若函數(shù)f(x)定義域?yàn)镽,且xf'(x)<0,則下列結(jié)論成立的是()A.f(-1)+f(1)>2f(0)B.f(-1)+f(1)<2f(0)C.f(-1)+f(1)=2f(0)D.f(-1)+f(1)與2f(0)的大小不確定解析:由于xf'(x)<0,所以當(dāng)x>0時f'(x)<0,當(dāng)x<0時f'(x)>0,即函數(shù)f(x)在(-,0)上遞增,在(0,+)上遞減,因此f(-1)<f(0),f(1)<f(0),故f(-1)+f(1)<2f(0).答案:B3.(2016淮南二中檢測)已知函數(shù)y=f(x)的圖象在點(diǎn)(1,f(1)處的切線方程是x-2y+1=0,若g(x)=xf(x),則g'(1)=. 解析:函數(shù)y=f(x)的圖象在點(diǎn)(1,f(1)處的切線方程是x-2y+1=0,f(1)=1,f'(1)=12.g(x)=xf(x),g'(x)=f(x)-xf'(x)f(x)2.g'(1)=f(1)-f'(1)f(1)2=12.答案:124.在平面直角坐標(biāo)系xOy中,直線y=2x+b是曲線y=aln x的切線,則當(dāng)a>0時,實(shí)數(shù)b的最小值是. 解析:設(shè)切點(diǎn)為(x0,alnx0),則y=alnx上此點(diǎn)處的切線為y=ax0x+alnx0-a,故ax0=2,alnx0-a=b,b=alna2-a=alna-aln2-a(a>0),b'=lna2,b在(0,2)上單調(diào)遞減,在(2,+)上單調(diào)遞增.b的最小值為-2.答案:-25.函數(shù)f(x)=1x+2x2+1x3.(1)求y=f(x)在-4,-12上的最值;(2)若a0,求g(x)=1x+2x2+ax3的極值點(diǎn).解:(1)f'(x)=-(x+1)(x+3)x4,令f'(x)=0,得x=-1或x=-3.當(dāng)x變化時,f(x),f'(x)的變化情況如下:x-4(-4,-3)-3(-3,-1)f'(x)-0+f(x)-964單調(diào)遞減極小值-427單調(diào)遞增x-1-1,-12-12f'(x)0-f(x)極大值0單調(diào)遞減-2y=f(x)在-4,-12上的最大值為0,最小值為-2.(2)g'(x)=-x2+4x+3ax4,設(shè)u=x2+4x+3a,=16-12a,當(dāng)a43時,0,g'(x)0,y=g(x)沒有極值點(diǎn).當(dāng)0<a<43時,x1=-2-4-3a,x2=-2+4-3a<0,減區(qū)間為(-,x1),(x2,0),增區(qū)間為(x1,x2),有兩個極值點(diǎn)x1,x2.當(dāng)a=0時,g(x)=1x+2x2,g'(x)=-x+4x3,減區(qū)間為(-,-4),增區(qū)間為(-4,0).有一個極值點(diǎn)x=-4.綜上所述,a=0時,有一個極值點(diǎn)x=-4;0<a<43時,有兩個極值點(diǎn)x1=-2-4-3a,x2=-2+4-3a;a43時,沒有極值點(diǎn).6.導(dǎo)學(xué)號59254065(2016山東威海一模)已知函數(shù)f(x)=xlnx+ax,x>1.(1)若f(x)在(1,+)上單調(diào)遞減,求實(shí)數(shù)a的取值范圍;(2)若a=2,求函數(shù)f(x)的極小值;(3)若方程(2x-m)ln x+x=0在(1,e上有兩個不等實(shí)根,求實(shí)數(shù)m的取值范圍.解:(1)f(x)=xlnx+ax,x>1.f'(x)=lnx-1ln2x+a.由題意可得f'(x)0,即a1ln2x-1lnx=1lnx-122-14,對x(1,+)恒成立.x(1,+),lnx(0,+),1lnx-12=0時,函數(shù)t(x)=1lnx-122-14的最小值為-14,a-14.(2)當(dāng)a=2時,f(x)=xlnx+2x,f'(x)=lnx-1+2ln2xln2x=(2lnx-1)(lnx+1)ln2x,由f'(x)=0,x>1,得x=e12.f(x)與f'(x)在(1,+)上的情況如下表:x(1,e12)e12(e12,+)f'(x)-0+f(x)極小值4e函數(shù)f(x)的極小值為42.(3)x>1,(2x-m)lnx+x=02x-m+xlnx=0m=xlnx+2x,方程(2x-m)lnx+x=0在(1,e上有兩個不等實(shí)根,即函數(shù)f(x)與函數(shù)y=m在(1,e上有兩個不同的交點(diǎn).由(2)可知,f(x)在(1,e12)上單調(diào)遞減,在(e12,e上單調(diào)遞增且f(e12)=4e,f(e)=3e,當(dāng)x1時,xlnx+,4e<m3e,故實(shí)數(shù)m的取值范圍是(4e,3e.5