（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 專題突破練10 專題二 函數(shù)與導(dǎo)數(shù)過關(guān)檢測(cè) 理

專題突破練10專題二函數(shù)與導(dǎo)數(shù)過關(guān)檢測(cè)一、選擇題1.已知函數(shù)f(x)=11-x的定義域?yàn)镸,g(x)=ln(1+x)的定義域?yàn)镹,則MN=()A.x|x>-1B.x|x<1C.x|-1<x<1D.2.(2019全國卷1,理3)已知a=log20.2,b=20.2,c=0.20.3,則()A.a<b<cB.a<c<bC.c<a<bD.b<c<a3.(2019全國卷1,理5)函數(shù)f(x)=sinx+xcosx+x2在-,的圖象大致為()4.已知f(x)是R上的奇函數(shù),當(dāng)x0時(shí),f(x)=x3+ln(1+x),則當(dāng)x<0時(shí),f(x)=()A.-x3-ln(1-x)B.x3+ln(1-x)C.x3-ln(1-x)D.-x3+ln(1-x)5.(2019全國卷3,文5)函數(shù)f(x)=2sin x-sin 2x在0,2的零點(diǎn)個(gè)數(shù)為()A.2B.3C.4D.56.(2019全國卷2,理6)若a>b,則()A.ln(a-b)>0B.3a<3bC.a3-b3>0D.|a|>|b|7.(2019全國卷3,理6)已知曲線y=aex+xln x在點(diǎn)(1,ae)處的切線方程為y=2x+b,則()A.a=e,b=-1B.a=e,b=1C.a=e-1,b=1D.a=e-1,b=-18.定義在R上的函數(shù)f(x)滿足f(-x)=-f(x),f(x)=f(x+4),且x(-1,0)時(shí),f(x)=2x+15,則f(log220)=()A.1B.45C.-1D.-459.設(shè)函數(shù)f(x)=xex,則()A.x=1為f(x)的極大值點(diǎn)B.x=1為f(x)的極小值點(diǎn)C.x=-1為f(x)的極大值點(diǎn)D.x=-1為f(x)的極小值點(diǎn)10.“a-1”是“函數(shù)f(x)=ln x+ax+1x在1,+)上為單調(diào)函數(shù)”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11.已知定義域?yàn)镽的奇函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),當(dāng)x>0時(shí),f'(x)+f(x)x>0,若a=12f12,b=-2f(-2),c=ln12fln12,則a,b,c的大小關(guān)系正確的是()A.a<c<bB.b<c<aC.a<b<cD.c<a<b12.(2019全國卷2,理12)設(shè)函數(shù)f(x)的定義域?yàn)镽,滿足f(x+1)=2f(x),且當(dāng)x(0,1時(shí),f(x)=x(x-1).若對(duì)任意x(-,m,都有f(x)-89,則m的取值范圍是()A.-,94B.-,73C.-,52D.-,83二、填空題13.(2019全國卷1,理13)曲線y=3(x2+x)ex在點(diǎn)(0,0)處的切線方程為. 14.已知曲線y=x24-3ln x的一條切線的斜率為-12,則切點(diǎn)的橫坐標(biāo)為. 15.(2019全國卷2,理14)已知f(x)是奇函數(shù),且當(dāng)x<0時(shí),f(x)=-eax.若f(ln 2)=8,則a=. 16.設(shè)邊長(zhǎng)為1 m的正三角形薄鐵皮,沿一條平行于某邊的直線剪成兩塊,其中一塊是梯形,記S=(梯形的周長(zhǎng))2梯形的面積,則S的最小值是. 三、解答題17.(2019山西太原二模,理21)已知x1,x2(x1<x2)是函數(shù)f(x)=ex+ln(x+1)-ax(aR)的兩個(gè)極值點(diǎn).(1)求a的取值范圍;(2)證明:f(x2)-f(x1)<2ln a.18.(2019湖南六校聯(lián)考,理21)已知f(x-1)=2ln(x-1)-kx+k(x>1).(1)判斷當(dāng)-1k0時(shí)f(x)的單調(diào)性;(2)若x1,x2(x1x2)為f(x)兩個(gè)極值點(diǎn),求證:xf(x1)+f(x2)(x+1)f(x)+2-2x.19.已知函數(shù)f(x)=(2+x+ax2)ln(1+x)-2x.(1)若a=0,證明:當(dāng)-1<x<0時(shí),f(x)<0;當(dāng)x>0時(shí),f(x)>0;(2)若x=0是f(x)的極大值點(diǎn),求a.20.(2019山東青島二模,理21)已知函數(shù)f(x)=(x2+a)ekx,e=2.718為自然對(duì)數(shù)的底數(shù).(1)若k=-1,aR,判斷函數(shù)f(x)在(0,+)上的單調(diào)性;(2)令a=0,k=1,若0<m2e,求證:關(guān)于x的方程f(x)-m(x+1)ln x=0無實(shí)根.21.(2019山東濟(jì)寧二模,理21)已知函數(shù)f(x)=x-a(ln x)2,aR.(1)當(dāng)a=1,x>1時(shí),試比較f(x)與1的大小,并說明理由;(2)若f(x)有極大值,求實(shí)數(shù)a的取值范圍;(3)若f(x)在x=x0處有極大值,證明1<f(x0)<e2.參考答案專題突破練10專題二函數(shù)與導(dǎo)數(shù)過關(guān)檢測(cè)1.C解析函數(shù)的定義域是指使函數(shù)式有意義的自變量x的取值范圍,由1-x>0,得M=x|x<1.由1+x>0,得N=x|x>-1,MN=x|-1<x<1.2.B解析因?yàn)閍=log20.2<0,b=20.2>20=1,又0<0.20.3<0.20=1,即c(0,1),所以a<c<b.故選B.3.D解析由f(-x)=-f(x),得f(x)是奇函數(shù),其圖象關(guān)于原點(diǎn)對(duì)稱,排除A.又f2=1+222=4+22>1,f()=-1+2>0,排除B,C.故選D.4.C解析當(dāng)x<0時(shí),-x>0,f(-x)=(-x)3+ln(1-x),f(x)是R上的奇函數(shù),當(dāng)x<0時(shí),f(x)=-f(-x)=-(-x)3+ln(1-x),f(x)=x3-ln(1-x).5.B解析由f(x)=2sinx-sin2x=2sinx-2sinxcosx=2sinx(1-cosx)=0,得sinx=0或cosx=1.x0,2,x=0或x=或x=2.故f(x)在區(qū)間0,2上的零點(diǎn)個(gè)數(shù)是3.故選B.6.C解析取a=2,b=1,滿足a>b,但ln(a-b)=0,排除A;取a=2,b=1,3a=9,3b=3,3a>3b,排除B;y=x3是增函數(shù),a>b,a3>b3,故C正確;取a=1,b=-2,滿足a>b,但|a|<|b|,排除D.故選C.7.D解析y'=aex+lnx+1,k=y'|x=1=ae+1=2,ae=1,a=e-1.將點(diǎn)(1,1)代入y=2x+b,得2+b=1,b=-1.8.C解析定義在R上的函數(shù)f(x)滿足f(-x)=-f(x),函數(shù)f(x)為奇函數(shù).f(x)=f(x+4),函數(shù)f(x)為周期為4的周期函數(shù).又log232>log220>log216,4<log220<5,f(log220)=f(log220-4)=flog254=-f-log254=-flog245.又x(-1,0)時(shí),f(x)=2x+15,flog245=1,故f(log220)=-1.9.D解析f'(x)=ex+xex=(1+x)ex,當(dāng)x>-1時(shí),f'(x)>0,函數(shù)f(x)遞增;當(dāng)x<-1時(shí),f'(x)<0,函數(shù)f(x)遞減,所以當(dāng)x=-1時(shí),f(x)有極小值.10.A解析f'(x)=1x+a-1x2=ax2+x-1x2,ax2+x-10對(duì)x1,+)恒成立或ax2+x-10對(duì)x1,+)恒成立,則a1-xx2max或a1-xx2min.記g(x)=1-xx2(x1,+),則g'(x)=-x2-2x(1-x)x4=x(x-2)x4,函數(shù)g(x)的單調(diào)遞減區(qū)間為1,2,單調(diào)遞增區(qū)間為2,+).當(dāng)x>1時(shí),g(x)<0,1-xx2max=g(1)=0,1-xx2min=g(2)=-14,a0或a-14.故選A.11.A解析設(shè)h(x)=xf(x),h'(x)=f(x)+x·f'(x).y=f(x)是定義在實(shí)數(shù)集R上的奇函數(shù),h(x)是定義在實(shí)數(shù)集R上的偶函數(shù).又當(dāng)x>0時(shí),f'(x)+f(x)x>0,當(dāng)x>0時(shí),h'(x)=f(x)+x·f'(x)>0,函數(shù)h(x)在區(qū)間(0,+)內(nèi)單調(diào)遞增.a=12f12=h12,b=-2f(-2)=2f(2)=h(2),c=ln12fln12=hln12=h(-ln2)=h(ln2),且2>ln2>12,b>c>a.12.B解析f(x+1)=2f(x),f(x)=2f(x-1).當(dāng)x(0,1時(shí),f(x)=x(x-1),f(x)的圖象如圖所示.當(dāng)2<x3時(shí),f(x)=4f(x-2)=4(x-2)(x-3),令4(x-2)(x-3)=-89,整理得9x2-45x+56=0,即(3x-7)(3x-8)=0,解得x1=73,x2=83.當(dāng)x(-,m時(shí),f(x)-89恒成立,m73,故m-,73.13.y=3x解析由題意可知y'=3(2x+1)ex+3(x2+x)ex=3(x2+3x+1)ex,k=y'|x=0=3.曲線y=3(x2+x)ex在點(diǎn)(0,0)處的切線方程為y=3x.14.2設(shè)切點(diǎn)坐標(biāo)為(x0,y0),且x0>0,y'=12x-3x,k=12x0-3x0=-12,x0=2.15.-3解析ln2(0,1),f(ln2)=8,f(x)是奇函數(shù),f(-ln2)=-8.當(dāng)x<0時(shí),f(x)=-eax,f(-ln2)=-e-aln2=-8,e-aln2=8,-aln2=ln8,-a=3,a=-3.16.3233如圖所示,設(shè)AD=xm(0<x<1),則DE=AD=xm,梯形的周長(zhǎng)為x+2(1-x)+1=3-x(m),又SADE=34x2(m2),SABC=34×12=34(m2),梯形的面積為34-34x2(m2),S=433×x2-6x+91-x2(0<x<1),S'=-833×(3x-1)(x-3)(1-x2)2,令S'=0,得x=13或3(舍去),當(dāng)x0,13時(shí),S'<0,S遞減;當(dāng)x13,1時(shí),S'>0,S遞增.故當(dāng)x=13時(shí),S的最小值是3233.17.解(1)由題意得f'(x)=ex+1x+1-a,x>-1,令g(x)=ex+1x+1-a,x>-1,則g'(x)=ex-1(x+1)2,令h(x)=ex-1(x+1)2,x>-1,則h'(x)=ex+2(x+1)3>0,h(x)在(-1,+)上遞增,且h(0)=0,當(dāng)x(-1,0)時(shí),g'(x)=h(x)<0,g(x)遞減;當(dāng)x(0,+)時(shí),g'(x)=h(x)>0,g(x)遞增,g(x)g(0)=2-a.當(dāng)a2時(shí),f'(x)=g(x)g(0)=2-a0,f(x)在(-1,+)遞增,此時(shí)無極值;當(dāng)a>2時(shí),g1a-1=e1a-1>0,g(0)=2-a<0,x11a-1,0,g(x1)=0,當(dāng)x(-1,x1)時(shí),g(x)=f'(x)>0,f(x)遞增;當(dāng)x(x1,0)時(shí),g(x)=f'(x)<0,g(x)遞減;x=x1是f(x)的極大值點(diǎn);g(lna)=11+lna>0,g(0)=2-a<0,x2(0,lna),g(x2)=0.當(dāng)x(0,x2)時(shí),g(x)=f'(x)<0,f(x)遞減;當(dāng)x(x2,+)時(shí),g(x)=f'(x)>0,f(x)遞增,x=x2是f(x)的極小值點(diǎn);綜上所述,a(2,+).(2)證明由(1)得a(2,+),1a-1<x1<0<x2<lna,且g(x1)=g(x2)=0,x2-x1>0,1a<x1+1<1,1<x2+1<1+lna,ex2-ex1=x2-x1(x1+1)(x2+1),1(x1+1)(x2+1)-a<0,1<x2+1x1+1<a(1+lna)<a2,f(x2)-f(x1)=ex2-ex1+lnx2+1x1+1-a(x2-x1)=(x2-x1)1(x1+1)(x2+1)-a+lnx2+1x1+1<lna2=2lna.18.(1)解因?yàn)閒(x-1)=2ln(x-1)+k(x-1)x(x>1),所以f(x)=2lnx+kxx+1(x>0).f'(x)=2x+k(x+1)2=2x2+(4+k)x+2x(x+1)2.當(dāng)-1k0時(shí),=(4+k)2-16=k(k+8)0,2x2+(4+k)x+2>0恒成立.于是,f(x)在定義域上為單調(diào)增函數(shù).(2)證明f'(x)=2x+k(x+1)2=2x2+(4+k)x+2x(x+1)2,由題設(shè)知,f'(x)=0有兩個(gè)不相等的正實(shí)數(shù)根x1,x2,則x1+x2=-4+k2>0,x1x2=1>0,=(4+k)2-16>0,解得k<-8,而f(x1)+f(x2)=2lnx1+kx1x1+1+2lnx2+kx2x2+1=2ln(x1x2)+kx1x1+1+x2x2+1=2ln(x1x2)+k·2x1x2+x1+x2x1x2+x1+x2+1=k.又(x+1)f(x)-2lnxx=k,故欲證原不等式等價(jià)于證明不等式(x+1)f(x)-2lnxxx+1xf(x)-2(x-1),也就是要證明對(duì)任意x>0,有l(wèi)nxx-1.令g(x)=lnx-x+1(x>0),由于g(1)=0,并且g'(x)=1x-1,當(dāng)x>1時(shí),g'(x)<0,則g(x)在(1,+)上為減函數(shù);當(dāng)0<x<1時(shí),g'(x)>0,則g(x)在(0,1)上為增函數(shù).則g(x)在(0,+)上有最大值g(1)=0,即g(x)0,故原不等式成立.19.解(1)當(dāng)a=0時(shí),f(x)=(2+x)ln(1+x)-2x,f'(x)=ln(1+x)-x1+x,設(shè)函數(shù)g(x)=f'(x)=ln(1+x)-x1+x,則g'(x)=x(1+x)2,當(dāng)-1<x<0時(shí),g'(x)<0;當(dāng)x>0時(shí),g'(x)>0.故當(dāng)x>-1時(shí),g(x)g(0)=0,當(dāng)且僅當(dāng)x=0時(shí),g(x)=0,從而f'(x)0,當(dāng)且僅當(dāng)x=0時(shí),f'(x)=0.所以f(x)在(-1,+)單調(diào)遞增.又f(0)=0,故當(dāng)-1<x<0時(shí),f(x)<0;當(dāng)x>0時(shí),f(x)>0.(2)若a0,由(1)知,當(dāng)x>0時(shí),f(x)(2+x)ln(1+x)-2x>0=f(0),這與x=0是f(x)的極大值點(diǎn)矛盾.若a<0,設(shè)函數(shù)h(x)=f(x)2+x+ax2=ln(1+x)-2x2+x+ax2.由于當(dāng)|x|<min1,1|a|時(shí),2+x+ax2>0,故h(x)與f(x)符號(hào)相同.又h(0)=f(0)=0,故x=0是f(x)的極大值點(diǎn)當(dāng)且僅當(dāng)x=0是h(x)的極大值點(diǎn).h'(x)=11+x-2(2+x+ax2)-2x(1+2ax)(2+x+ax2)2=x2(a2x2+4ax+6a+1)(x+1)(ax2+x+2)2.若6a+1>0,則當(dāng)0<x<-6a+14a,且|x|<min1,1|a|時(shí),h'(x)>0,故x=0不是h(x)的極大值點(diǎn).若6a+1<0,則a2x2+4ax+6a+1=0存在根x1<0,故當(dāng)x(x1,0),且|x|<min1,1|a|時(shí),h'(x)<0,所以x=0不是h(x)的極大值點(diǎn).若6a+1=0,則h'(x)=x3(x-24)(x+1)(x2-6x-12)2.當(dāng)x(-1,0)時(shí),h'(x)>0;當(dāng)x(0,1)時(shí),h'(x)<0.所以x=0是h(x)的極大值點(diǎn),從而x=0是f(x)的極大值點(diǎn).綜上,a=-16.20.(1)解由已知k=-1,所以f(x)=(x2+a)e-x=x2+aex,所以f'(x)=x2+aex'=2xex-(x2+a)exe2x=-x2+2x-aex.若a1,在R上恒有u(x)=-(x-1)2+1-a0,所以f'(x)=-(x-1)2+1-aex0,所以f(x)在(0,+)上單調(diào)遞減.若a<1,u(x)=-(x-1)2+1-a圖象與x軸有兩個(gè)不同交點(diǎn).設(shè)u(x)=-(x-1)2+1-a=0的兩根分別為x1=1-1-a,x2=1+1-a.若0<a<1,則0<x1<1,x2>1,所以當(dāng)0<x<x1時(shí),u(x)<0;當(dāng)x1<x<x2時(shí),u(x)>0;當(dāng)x>x2時(shí),u(x)<0.所以,f(x)在(0,x1)上和(x2,+)上分別單調(diào)遞減;在(x1,x2)上單調(diào)遞增.若a0,則x1=1-1-a0,x2=1+1-a2,所以,當(dāng)x(0,x2)時(shí)總有u(x)>0;當(dāng)x(x2,+)時(shí),u(x)<0.所以f(x)在(0,x2)上單調(diào)遞增,在(x2,+)上單調(diào)遞減.綜上,若a1,則f(x)在(0,+)上為單調(diào)遞減;若0<a<1,則f(x)在(0,x1)上和(x2,+)上分別單調(diào)遞減,在(x1,x2)上單調(diào)遞增;若a0,則f(x)在(0,x2)上單調(diào)遞增,在(x2,+)上單調(diào)遞減.(2)證明由題知a=0,k=1,所以f(x)=x2ex,令g(x)=ex-(x+1),對(duì)任意實(shí)數(shù)x>0,g'(x)=ex-1>0恒成立,所以g(x)=ex-(x+1)>g(0)=0,即ex>x+1>0,則x2ex-m(x+1)lnx>x2(x+1)-m(x+1)lnx=(x+1)(x2-mlnx).令h(x)=x2-mlnx,所以h'(x)=(x2-mlnx)'=2x-mx=2x2-mx,因?yàn)?<m2e,所以h'(x)=2(x+m2)(x-m2)x所以當(dāng)x0,m2時(shí),h'(x)<0;當(dāng)xm2,+時(shí),h'(x)>0.所以h(x)=x2-mlnx在(0,+)上有最小值.所以hm2=m2-mlnm2=m21-lnm2.因?yàn)?<m2e,所以lnm21,所以1-lnm20.所以m21-lnm20,即當(dāng)0<m2e時(shí),對(duì)任意x>0,h(x)=x2-mlnx0,所以x2ex-m(x+1)lnx>0.所以關(guān)于x的方程f(x)-m(x+1)lnx=0無實(shí)根.21.(1)解當(dāng)a=1,x>1時(shí),f(x)=x-(lnx)2,x>1.f'(x)=1-2(lnx)×1x=x-2lnxx.令g(x)=x-2lnx,x>1,則g'(x)=1-2x=x-2x.當(dāng)x(1,2)時(shí),g'(x)<0,g(x)單調(diào)遞減,當(dāng)x(2,+)時(shí),g'(x)>0,g(x)單調(diào)遞增,g(x)g(2)=2-2ln2>0,即f'(x)>0.f(x)在(1,+)上單調(diào)遞增.f(x)>f(1)=1.故當(dāng)a=1,x>1時(shí)f(x)>1.(2)解f'(x)=1-2alnxx=x-2alnxx(x>0),令h(x)=x-2alnx(x>0),則h'(x)=1-2ax=x-2ax.當(dāng)a=0時(shí),f(x)=x無極大值.當(dāng)a<0時(shí),h'(x)>0,h(x)在(0,+)上單調(diào)遞增,h(1)=1>0,h(e12a)=e12a-1<0,x1(e12a,1),使得h(x1)=0.當(dāng)x(0,x1)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x(x1,+)時(shí),f'(x)>0,f(x)單調(diào)遞增.f(x)在x=x1處有極小值,f(x)無極大值.當(dāng)a>0時(shí),h(x)在(0,2a)上單調(diào)遞減,h(x)在(2a,+)上單調(diào)遞增,f(x)有極大值,h(2a)=2a-2aln(2a)=2a(1-ln2a)<0,即a>e2.又h(1)=1>0,h(e)=e-2a<0,x0(1,e),使得h(x0)=x0-2alnx0=0,即alnx0=x02;當(dāng)x(0,x0)時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)x(x0,e)時(shí),f'(x)<0,f(x)單調(diào)遞減.f(x)有極大值,綜上所述,a>e2.(3)證明由(2)可知,alnx0=x02,f(x0)=x0-a(lnx0)2=x0-x0lnx02(1<x0<e).設(shè)p(x)=x-xlnx2(1<x<e),則p'(x)=1-1+lnx2=1-lnx2>0.p(x)在(1,e)上單調(diào)遞增,p(1)<p(x)<p(e),即1<p(x)<e2,故1<f(x0)<e2.19