（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 第三章 導(dǎo)數(shù)及其應(yīng)用 考點(diǎn)規(guī)范練13 導(dǎo)數(shù)與函數(shù)的單調(diào)性.docx

考點(diǎn)規(guī)范練13導(dǎo)數(shù)與函數(shù)的單調(diào)性基礎(chǔ)鞏固組1.函數(shù)f(x)=(x-3)ex的單調(diào)遞增區(qū)間是()A.(-,2)B.(0,3)C.(1,4)D.(2,+)答案D解析因?yàn)閒(x)=(x-3)ex,則f(x)=ex(x-2),令f(x)>0,得x>2,所以f(x)的單調(diào)遞增區(qū)間為(2,+).2.(2017浙江嘉興調(diào)研)已知函數(shù)f(x)=12x3+ax+4,則“a>0”是“f(x)在R上單調(diào)遞增”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件答案A解析f(x)=32x2+a,當(dāng)a0時(shí),f(x)0恒成立,故“a>0”是“f(x)在R上單調(diào)遞增”的充分不必要條件.3.設(shè)f(x)是函數(shù)f(x)的導(dǎo)函數(shù),y=f(x)的圖象如圖所示,則y=f(x)的圖象最有可能是()答案C解析由y=f(x)的圖象易知當(dāng)x<0或x>2時(shí),f(x)>0,故函數(shù)y=f(x)在區(qū)間(-,0)和(2,+)上單調(diào)遞增;當(dāng)0<x<2時(shí),f(x)<0,故函數(shù)y=f(x)在區(qū)間(0,2)上單調(diào)遞減.4.設(shè)函數(shù)f(x)=12x2-9ln x在區(qū)間a-1,a+1上單調(diào)遞減,則實(shí)數(shù)a的取值范圍是()A.1<a2B.a4C.a2D.0<a3答案A解析f(x)=12x2-9lnx,f(x)=x-9x(x>0),當(dāng)x-9x0,即0<x3時(shí),函數(shù)f(x)是減函數(shù),a-1>0,且a+13,解得1<a2.5.若函數(shù)f(x)=x2+ax+1x在12,+上是增函數(shù),則a的取值范圍是()A.-1,0B.-1,+)C.0,3D.3,+)答案D解析由f(x)=x2+ax+1x,得f(x)=2x+a-1x2=2x3+ax2-1x2,令g(x)=2x3+ax2-1,要使函數(shù)f(x)=x2+ax+1x在12,+上是增函數(shù),則g(x)=2x3+ax2-1在x12,+上大于等于0恒成立,g(x)=6x2+2ax=2x(3x+a),當(dāng)a=0時(shí),g(x)0,g(x)在R上為增函數(shù),則有g(shù)120,解得14+a4-10,a3(舍);當(dāng)a>0時(shí),g(x)在(0,+)上為增函數(shù),則g120,解得14+a4-10,a3;當(dāng)a<0時(shí),同理分析可知,滿足函數(shù)f(x)=x2+ax+1x在12,+是增函數(shù)的a的取值范圍是a3(舍).故選D.6.函數(shù)f(x)=lnxx的單調(diào)遞增區(qū)間是.答案(0,e)解析由f(x)=lnxx=1-lnxx2>0(x>0),可得1-lnx>0,x>0,解得x(0,e).7.(2017浙江麗水模擬)已知函數(shù)f(x)=ln x+2x,若f(x2+2)<f(3x),則實(shí)數(shù)x的取值范圍是.答案(1,2)解析由題可得函數(shù)定義域?yàn)?0,+),f(x)=1x+2xln2,所以在定義域內(nèi)f(x)>0,函數(shù)單調(diào)遞增,所以由f(x2+2)<f(3x)得x2+2<3x,所以1<x<2.8.已知函數(shù)f(x)=x3+3x對(duì)任意的m-2,2,f(mx-2)+f(x)<0恒成立,則x.答案-2,23解析由題意得,函數(shù)的定義域是R,且f(-x)=(-x)3+3(-x)=-(x3+3x)=-f(x),所以f(x)是奇函數(shù),又f(x)=3x2+3>0,所以f(x)在R上單調(diào)遞增,所以f(mx-2)+f(x)<0可化為f(mx-2)<-f(x)=f(-x),由f(x)在R上單調(diào)遞增可知mx-2<-x,即mx+x-2<0,則對(duì)任意的m-2,2,f(mx-2)+f(x)<0恒成立,等價(jià)于對(duì)任意的m-2,2,mx+x-2<0恒成立,所以-2x+x-2<0,2x+x-2<0,解得-2<x<23,即x的取值范圍是-2,23,故答案為-2,23.能力提升組9.函數(shù)y=f(x)在其定義域-32,3內(nèi)可導(dǎo),其圖象如圖所示,記y=f(x)的導(dǎo)函數(shù)為y=f(x),則不等式f(x)0的解集為.答案-13,12,3)解析由圖象可知,f(x)在區(qū)間-13,1和2,3)上單調(diào)遞減,所以f(x)0的解集為-13,12,3).10.若f(x)=-12x2+bln(x+2)在-1,+)上是減函數(shù),則實(shí)數(shù)b的取值范圍是()A.-1,+)B.1,+)C.(-,-1D.(-,1答案C解析由已知得f(x)=-x+bx+20在-1,+)上恒成立,b(x+1)2-1在-1,+)上恒成立,b-1.11.定義在(0,+)上的函數(shù)f(x)滿足x2f(x)+1>0,f(2)=92,則不等式f(lg x)<1lgx+4的解集為()A.(10,100)B.(0,100)C.(100,+)D.(1,100)答案D解析令g(x)=f(x)-1x,則g(x)=f(x)+1x2>0,g(x)在(0,+)遞增,而g(2)=f(2)-12=4,故由f(lgx)<1lgx+4,得g(lnx)<g(2),故0<lnx<2,解得1<x<100,故選D.12.已知函數(shù)f(x)(xR)滿足f(1)=1,且f(x)的導(dǎo)函數(shù)f(x)<12,則f(x)<x2+12的解集為()A.x|-1<x<1B.x|x<-1C.x|x<-1,或x>1D.x|x>1答案D解析設(shè)F(x)=f(x)-x2+12,則F(1)=f(1)-12+12=1-1=0,又F(x)=f(x)-12,對(duì)任意xR,有F(x)=f(x)-12<0,即函數(shù)F(x)在R上單調(diào)遞減,則F(x)<0的解集為(1,+),即f(x)<x2+12的解集為(1,+),故選D.13.已知方程ln|x|-ax2+32=0有4個(gè)不同的實(shí)數(shù)根,則實(shí)數(shù)a的取值范圍是()A.0,e22B.0,e22C.0,e23D.0,e23答案A解析由ln|x|-ax2+32=0得ax2=ln|x|+32,x0,方程等價(jià)于a=ln|x|+32x2,設(shè)f(x)=ln|x|+32x2,則函數(shù)f(x)是偶函數(shù),當(dāng)x>0時(shí),f(x)=lnx+32x2,則f(x)=1xx2-lnx+322xx4=x-2xlnx-3xx4=-2x(1+lnx)x4,由f(x)>0得-2x(1+lnx)>0,得1+lnx<0,即lnx<-1,得0<x<1e,此時(shí)函數(shù)單調(diào)遞增,由f(x)<0得-2x(1+lnx)<0,得1+lnx>0,即lnx>-1,得x>1e,此時(shí)函數(shù)單調(diào)遞減,即當(dāng)x>0時(shí),x=1e時(shí),函數(shù)f(x)取得極大值f1e=ln1e+32(1e)2=-1+32e2=12e2,作出函數(shù)f(x)的圖象如圖.要使a=ln|x|+32x2有4個(gè)不同的交點(diǎn),則滿足0<a<12e2,故選A.14.已知函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x<0時(shí),f(x)=xex,給出下列命題:當(dāng)x>0時(shí),f(x)=-xe-x;函數(shù)f(x)的單調(diào)遞減區(qū)間是(-,-1)和(1,+);對(duì)x1,x2R,都有|f(x1)-f(x2)|2e.其中正確的序號(hào)是.答案解析當(dāng)x>0時(shí),f(x)=-f(-x)=-(-x)e-x=xe-x,x>0,故錯(cuò)誤;當(dāng)x<0時(shí),f(x)=ex(x+1),則在(-,-1)單調(diào)遞減,由奇函數(shù)對(duì)稱性可知,在(1,+)也單調(diào)遞減,故正確;由導(dǎo)函數(shù)分析可知,f(x)min=f(-1)=-1ef(x)max=f(1)=1e,所以|f(x1)-f(x2)|f(x)max-f(x)min|=2e,故正確.所以正確的命題是.15.已知函數(shù)f(x)=x2ex,若f(x)在t,t+1上不單調(diào),則實(shí)數(shù)t的取值范圍是.答案(-3,-2)(-1,0)解析由題意,得f(x)=ex(x2+2x),f(x)在(-,-2),(0,+)上單調(diào)遞增,(-2,0)上單調(diào)遞減,又f(x)在t,t+1上不單調(diào),t<-2,t+1>-2或t<0,t+1>0,即實(shí)數(shù)t的取值范圍是(-3,-2)(-1,0),故填:(-3,-2)(-1,0).16.(2017江蘇高考)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-1,12解析因?yàn)閒(-x)=-x3+2x+1ex-ex=-f(x),所以函數(shù)f(x)是奇函數(shù),因?yàn)閒(x)=3x2-2+ex+e-x3x2-2+2exe-x0,所以f(x)在R上單調(diào)遞增,又f(a-1)+f(2a2)0,即f(2a2)f(1-a),所以2a21-a,即2a2+a-10,解得-1a12,故實(shí)數(shù)a的取值范圍為-1,12.17.設(shè)f(x)=ex1+ax2,其中a為正實(shí)數(shù).(1)當(dāng)a=43時(shí),求f(x)單調(diào)區(qū)間;(2)若f(x)為R上的單調(diào)函數(shù),求實(shí)數(shù)a的取值范圍.解對(duì)f(x)求導(dǎo)得f(x)=ex1+ax2-2ax(1+ax2)2.(1)當(dāng)a=43時(shí),若f(x)=0,則4x2-8x+3=0,解得x1=32,x2=12.結(jié)合,可知x-,121212,323232,+f(x)+0-0+f(x)極大值極小值所以增區(qū)間為-,12,32,+;減區(qū)間為12,32.(2)若f(x)為R上的單調(diào)函數(shù),則f(x)在R上不變號(hào),結(jié)合與條件a>0,知ax2-2ax+10在R上恒成立,即=4a2-4a=4a(a-1)0,又a>0,得0<a1.所以實(shí)數(shù)a的取值范圍為a|0<a1.18.設(shè)函數(shù)f(x)=aln x+x-1x+1,其中a為常數(shù).(1)若a=0,求曲線y=f(x)在點(diǎn)(1,f(1)處的切線方程;(2)討論函數(shù)f(x)的單調(diào)性.解(1)由題意知a=0時(shí),f(x)=x-1x+1,x(0,+).此時(shí)f(x)=2(x+1)2.可得f(1)=12,又f(1)=0,所以曲線y=f(x)在(1,f(1)處的切線方程為x-2y-1=0.(2)函數(shù)f(x)的定義域?yàn)?0,+).f(x)=ax+2(x+1)2=ax2+(2a+2)x+ax(x+1)2.當(dāng)a0時(shí),f(x)>0,函數(shù)f(x)在(0,+)上單調(diào)遞增.當(dāng)a<0時(shí),令g(x)=ax2+(2a+2)x+a,由于=(2a+2)2-4a2=4(2a+1).當(dāng)a=-12時(shí),=0,f(x)=-12(x-1)2x(x+1)20,函數(shù)f(x)在(0,+)上單調(diào)遞減.當(dāng)a<-12時(shí),<0,g(x)<0,f(x)<0,函數(shù)f(x)在(0,+)上單調(diào)遞減.當(dāng)-12<a<0時(shí),>0.設(shè)x1,x2(x1<x2)是函數(shù)g(x)的兩個(gè)零點(diǎn),則x1=-(a+1)+2a+1a,x2=-(a+1)-2a+1a.由x1=a+1-2a+1-a=a2+2a+1-2a+1-a>0,所以x(0,x1)時(shí),g(x)<0,f(x)<0,函數(shù)f(x)單調(diào)遞減;x(x1,x2)時(shí),g(x)>0,f(x)>0,函數(shù)f(x)單調(diào)遞增;x(x2,+)時(shí),g(x)<0,f(x)<0,函數(shù)f(x)單調(diào)遞減.綜上可得:當(dāng)a0時(shí),函數(shù)f(x)在(0,+)上單調(diào)遞增;當(dāng)a-12時(shí),函數(shù)f(x)在(0,+)上單調(diào)遞減;當(dāng)-12<a<0時(shí),f(x)在0,-(a+1)+2a+1a,-(a+1)-2a+1a,+上單調(diào)遞減,在-(a+1)+2a+1a,-(a+1)-2a+1a上單調(diào)遞增.