（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)檢測理.docx

專題突破練10專題二函數(shù)與導(dǎo)數(shù)過關(guān)檢測一、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|x1C.x|-1x1D.2.(2019全國卷1,理3)已知a=log20.2,b=20.2,c=0.20.3,則()A.abcB.acbC.cabD.bca3.(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)xb,則()A.ln(a-b)0B.3a0D.|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)x0時(shí),f(x)+f(x)x0,若a=12f12,b=-2f(-2),c=ln12fln12,則a,b,c的大小關(guān)系正確的是()A.acbB.bcaC.abcD.cab12.(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)x0時(shí),f(x)=-eax.若f(ln 2)=8,則a=.16.設(shè)邊長為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)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(x1x2)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)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)-1x0時(shí),f(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01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1f(x0)0,得M=x|x0,得N=x|x-1,MN=x|-1x1.2.B解析因?yàn)閍=log20.220=1,又00.20.30.20=1,即c(0,1),所以ac1,f()=-1+20,排除B,C.故選D.4.C解析當(dāng)x0,f(-x)=(-x)3+ln(1-x),f(x)是R上的奇函數(shù),當(dāng)xb,但ln(a-b)=0,排除A;取a=2,b=1,3a=9,3b=3,3a3b,排除B;y=x3是增函數(shù),ab,a3b3,故C正確;取a=1,b=-2,滿足ab,但|a|log220log216,4log220-1時(shí),f(x)0,函數(shù)f(x)遞增;當(dāng)x-1時(shí),f(x)1時(shí),g(x)0時(shí),f(x)+f(x)x0,當(dāng)x0時(shí),h(x)=f(x)+xf(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2ln212,bca.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)20,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)x0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(0x1),則DE=AD=xm,梯形的周長為x+2(1-x)+1=3-x(m),又SADE=34x2(m2),SABC=3412=34(m2),梯形的面積為34-34x2(m2),S=433x2-6x+91-x2(0x1),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),S0,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)30,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)a2時(shí),g1a-1=e1a-10,g(0)=2-a0,f(x)遞增;當(dāng)x(x1,0)時(shí),g(x)=f(x)0,g(0)=2-a0,x2(0,lna),g(x2)=0.當(dāng)x(0,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-1x10x20,1ax1+11,1x2+11+lna,ex2-ex1=x2-x1(x1+1)(x2+1),1(x1+1)(x2+1)-a0,1x2+1x1+1a(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+11),所以f(x)=2lnx+kxx+1(x0).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+20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+k20,x1x2=10,=(4+k)2-160,解得k0,有l(wèi)nxx-1.令g(x)=lnx-x+1(x0),由于g(1)=0,并且g(x)=1x-1,當(dāng)x1時(shí),g(x)0,則g(x)在(1,+)上為減函數(shù);當(dāng)0x0,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)-1x0時(shí),g(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)-1x0時(shí),f(x)0時(shí),f(x)0.(2)若a0,由(1)知,當(dāng)x0時(shí),f(x)(2+x)ln(1+x)-2x0=f(0),這與x=0是f(x)的極大值點(diǎn)矛盾.若a0,設(shè)函數(shù)h(x)=f(x)2+x+ax2=ln(1+x)-2x2+x+ax2.由于當(dāng)|x|0,故h(x)與f(x)符號相同.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+10,則當(dāng)0x-6a+14a,且|x|0,故x=0不是h(x)的極大值點(diǎn).若6a+10,則a2x2+4ax+6a+1=0存在根x10,故當(dāng)x(x1,0),且|x|min1,1|a|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a1,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0a1,則0x11,所以當(dāng)0xx1時(shí),u(x)0;當(dāng)x1x0;當(dāng)xx2時(shí),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0a0,g(x)=ex-10恒成立,所以g(x)=ex-(x+1)g(0)=0,即exx+10,則x2ex-m(x+1)lnxx2(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.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)00,h(x)=x2-mlnx0,所以x2ex-m(x+1)lnx0.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,x1時(shí),f(x)=x-(lnx)2,x1.f(x)=1-2(lnx)1x=x-2lnxx.令g(x)=x-2lnx,x1,則g(x)=1-2x=x-2x.當(dāng)x(1,2)時(shí),g(x)0,g(x)單調(diào)遞增,g(x)g(2)=2-2ln20,即f(x)0.f(x)在(1,+)上單調(diào)遞增.f(x)f(1)=1.故當(dāng)a=1,x1時(shí)f(x)1.(2)解f(x)=1-2alnxx=x-2alnxx(x0),令h(x)=x-2alnx(x0),則h(x)=1-2ax=x-2ax.當(dāng)a=0時(shí),f(x)=x無極大值.當(dāng)a0,h(x)在(0,+)上單調(diào)遞增,h(1)=10,h(e12a)=e12a-10,x1(e12a,1),使得h(x1)=0.當(dāng)x(0,x1)時(shí),f(x)0,f(x)單調(diào)遞增.f(x)在x=x1處有極小值,