高考數(shù)學(xué)（理）一輪規(guī)范練【15】導(dǎo)數(shù)在函數(shù)單調(diào)性、極值中的應(yīng)用（含答案）

1、課時(shí)規(guī)范練15導(dǎo)數(shù)在函數(shù)單調(diào)性、極值中的應(yīng)用課時(shí)規(guī)范練第29頁(yè) 一、選擇題1.函數(shù)f(x)=的單調(diào)遞減區(qū)間是() A.e,+)B.1,+)C.0,eD.0,1答案:A解析:令f'(x)=0,則xe.又因?yàn)楫?dāng)xe時(shí),f'(x)不恒等于0,所以f(x)的單調(diào)遞減區(qū)間是e,+).2.函數(shù)f(x)=x3-3x2+3x的極值點(diǎn)的個(gè)數(shù)是()A.0B.1C.2D.3答案:A解析:由題知f(x)的導(dǎo)函數(shù)值恒大于或等于零,所以函數(shù)f(x)在定義域上單調(diào)遞增.3.(2013福建高考)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)

2、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:D解析:函數(shù)f(x)的極大值f(x0)不一定是最大值,故A錯(cuò);f(x)與-f(-x)關(guān)于原點(diǎn)對(duì)稱(chēng),故x0(x00)是f(x)的極大值點(diǎn)時(shí),-x0是-f(-x)的極小值點(diǎn),故選D.4.在R上可導(dǎo)的函數(shù)f(x)的圖象如圖所示,則關(guān)于x的不等式x·f'(x)<0的解集為()A.(-,-1)(0,1)B.(-1,0)(1,+)C.(-2,-1)(1,2)D.(-,-2)(2,+)答案:A解析:在(-,-1)和(1,+)上f(x)遞增,所以f'(x)>0

3、,使xf'(x)<0的范圍為(-,-1);在(-1,1)上f(x)遞減,所以f'(x)<0,使xf'(x)<0的范圍為(0,1).5.若a>0,b>0,且函數(shù)f(x)=4x3-ax2-2bx+2在x=1處有極值,則ab的最大值等于()A.2B.3C.6D.9答案:D解析:函數(shù)的導(dǎo)數(shù)為f'(x)=12x2-2ax-2b.由函數(shù)f(x)在x=1處有極值,可知函數(shù)f(x)在x=1處的導(dǎo)數(shù)值為零,即12-2a-2b=0,所以a+b=6,由題意知a,b都是正實(shí)數(shù),所以ab=9.當(dāng)且僅當(dāng)a=b=3時(shí)取到等號(hào),故選D.6.(2013浙江高考)已知

4、e為自然對(duì)數(shù)的底數(shù),設(shè)函數(shù)f(x)=(ex-1)·(x-1)k(k=1,2),則()A.當(dāng)k=1時(shí),f(x)在x=1處取到極小值B.當(dāng)k=1時(shí),f(x)在x=1處取到極大值C.當(dāng)k=2時(shí),f(x)在x=1處取到極小值D.當(dāng)k=2時(shí),f(x)在x=1處取到極大值答案:C解析:當(dāng)k=1時(shí),f(x)=(ex-1)(x-1),f'(x)=xex-1,f'(1)=e-10,2 / 5f(x)在x=1處不能取到極值;當(dāng)k=2時(shí),f(x)=(ex-1)(x-1)2,f'(x)=(x-1)(xex+ex-2),令H(x)=xex+ex-2,則H'(x)=xex+2ex

5、>0,x(0,+).說(shuō)明H(x)在(0,+)上為增函數(shù),且H(1)=2e-2>0,H(0)=-1<0,因此當(dāng)x0<x<1(x0為H(x)的零點(diǎn))時(shí),f'(x)<0,f(x)在(x0,1)上為減函數(shù).當(dāng)x>1時(shí),f'(x)>0,f(x)在(1,+)上是增函數(shù).x=1是f(x)的極小值點(diǎn),故選C.二、填空題7.已知函數(shù)f(x)=(m-2)x2+(m2-4)x+m是偶函數(shù),函數(shù)g(x)=-x3+2x2+mx+5在(-,+)內(nèi)單調(diào)遞減,則實(shí)數(shù)m的值為. 答案:-2解析:f(x)=(m-2)x2+(m2-4)x+m是偶函數(shù),m2-

6、4=0,m=±2.又函數(shù)g(x)=-x3+2x2+mx+5在(-,+)內(nèi)單調(diào)遞減,g'(x)=-3x2+4x+m,g'(x)恒小于等于0.=16+12m0.m-,m=-2.8.已知函數(shù)f(x)=x3+ax2+bx+a2在x=1處取得極值10,則f(2)=. 答案:18解析:f'(x)=3x2+2ax+b,由題意得即但當(dāng)a=-3,b=3時(shí),f'(x)=3x2-6x+30,故不存在極值.因此,a=4,b=-11,f(2)=18.9.已知某質(zhì)點(diǎn)的運(yùn)動(dòng)方程為s(t)=t3+bt2+ct+d,如圖所示是其運(yùn)動(dòng)軌跡的一部分,若t時(shí),s(t)<3d2

7、恒成立,則d的取值范圍為. 答案:d>或d<-1解析:因?yàn)橘|(zhì)點(diǎn)的運(yùn)動(dòng)方程為s(t)=t3+bt2+ct+d,所以s'(t)=3t2+2bt+c.由題圖可知,s(t)在t=1和t=3處取得極值.則s'(1)=0,s'(3)=0,即所以則s'(t)=3t2-12t+9=3(t-1)(t-3).當(dāng)t時(shí),s'(t)>0;當(dāng)t(1,3)時(shí),s'(t)<0;當(dāng)t(3,4)時(shí),s'(t)>0,故當(dāng)t=1時(shí),s(t)取得極大值4+d.又因?yàn)閟(4)=4+d,所以當(dāng)t時(shí),s(t)的最大值為4+d.因?yàn)楫?dāng)t時(shí),s(t)

8、<3d2恒成立,故4+d<3d2,即d>或d<-1.三、解答題10.設(shè)f(x)=kx-2ln x,若f(x)在其定義域內(nèi)為單調(diào)增函數(shù),求k的取值范圍.解:由題意可得f'(x)=k+.令h(x)=kx2-2x+k,要使f(x)在其定義域(0,+)上單調(diào)遞增,只需h(x)在(0,+)內(nèi)滿足h(x)0恒成立.由h(x)0得kx2-2x+k0,即k在x(0,+)上恒成立,x>0,x+2.1.k1.11.已知函數(shù)f(x)=aln x+x(a為實(shí)常數(shù)).(1)若a=-1,求函數(shù)f(x)的單調(diào)遞減區(qū)間;(2)若直線y=2x-1是曲線y=f(x)的切線,求a的值.解:(1

9、)當(dāng)a=-1時(shí),f(x)=x-ln x,故f'(x)=1-,x>0.令f'(x)0,得0<x1.故f(x)的單調(diào)遞減區(qū)間為(0,1.(2)設(shè)切點(diǎn)(x0,2x0-1),可知f'(x0)=1+,即1+=2x0=a.又2x0-1=aln x0+x0,2a-1=aln a+a,即aln a-a+1=0.令h(x)=xln x-x+1,則h'(x)=ln x.因此,x>1時(shí),h'(x)>0,h(x)=xln x-x+1單調(diào)遞增.0<x<1時(shí),h'(x)<0,h(x)=xln x-x+1單調(diào)遞減.故h(x)=xln

10、x-x+1有唯一零點(diǎn)x=1,即a=1.12.已知函數(shù)f(x)=x2e-x.(1)求f(x)的極小值和極大值;(2)當(dāng)曲線y=f(x)的切線l的斜率為負(fù)數(shù)時(shí),求l在x軸上截距的取值范圍.解:(1)f(x)的定義域?yàn)?-,+),f'(x)=-e-xx(x-2).當(dāng)x(-,0)或x(2,+)時(shí),f'(x)<0;當(dāng)x(0,2)時(shí),f'(x)>0.所以f(x)在(-,0),(2,+)單調(diào)遞減,在(0,2)單調(diào)遞增.故當(dāng)x=0時(shí),f(x)取得極小值,極小值為f(0)=0;當(dāng)x=2時(shí),f(x)取得極大值,極大值為f(2)=4e-2.(2)設(shè)切點(diǎn)為(t,f(t),則l的方程為y=f'(t)(x-t)+f(t).所以l在x軸上的截距為m(t)=t-=t+=t-2+3.由已知和得t<0