《（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 文》由會(huì)員分享，可在線閱讀，更多相關(guān)《（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 文（18頁珍藏版）》請?jiān)谘b配圖網(wǎng)上搜索。
1、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,文3,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,文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)x0時(shí),f(x)=()A.-x3-ln(1-x)B.x3+ln(1-x)C.x3-ln(1-x)D.-x
2、3+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,文6)設(shè)f(x)為奇函數(shù),且當(dāng)x0時(shí),f(x)=ex-1,則當(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.caf(2-32)f(2-23)B.flog314f(2-23)f(2-32)C.f(2-32)f(2-23)flog314D.f(2-23)f(2-32)flog314二、填空題13.(2019全國卷1,文13)曲
3、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)1-x恒成立,求實(shí)數(shù)a的取值范圍.18.(2019山西運(yùn)城二模,文21)已知函數(shù)f(x)=xex-a(ln x+x),aR.(1)當(dāng)a=e時(shí),求f(x)的單調(diào)區(qū)間;(2)若f(x)有兩個(gè)零點(diǎn),求實(shí)數(shù)a的取值范圍.19.(2019全國卷1,文20)已知函數(shù)f(x)=2sin x-xcos x-x,f(x)為f(x)的導(dǎo)數(shù).(1)證明:f(x)在區(qū)間(0,)存在唯一零點(diǎn);(2)若x0,時(shí),f
4、(x)ax,求a的取值范圍.20.(2019山東泰安二模,文20)已知函數(shù)f(x)=(x-m)ln x(m0).(1)若函數(shù)f(x)存在極小值點(diǎn),求m的取值范圍;(2)當(dāng)m=0時(shí),證明:f(x)ex-1.21.(2019山東青島二模,文21)已知函數(shù)g(x)=lnxx-m(m0),h(x)=2x+m.(1)若g(x)在區(qū)間(0,e2上單調(diào)遞增,求實(shí)數(shù)m的取值范圍;(2)若m=-1,且f(x)=g(x)h(x),求證:對(duì)定義域內(nèi)的任意實(shí)數(shù)x,不等式f(x)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
5、),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)x0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.D解析f(x)是奇函數(shù),f(-x)=-f(x).當(dāng)x0,f(-x)=e-x-1=-f(x),即f(x)=-e-x+1.故選D.7.C解析由題意
6、可知,f(x)在0,+)內(nèi)單調(diào)遞增,在(-,0)內(nèi)單調(diào)遞增.因?yàn)閒(x)在區(qū)間(-,+)上是增函數(shù),所以a2-3a+20,解得1a2.8.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log232log220log216,4log220-1時(shí),f(x)0,函數(shù)f(x)遞增;當(dāng)x-1時(shí),f(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=hln1
7、2=h(-ln2)=h(ln2),且2ln212,bca.12.C解析f(x)是R上的偶函數(shù),flog314=f(-log34)=f(log34).又y=2x在R上單調(diào)遞增,log341=202-232-32.又f(x)在區(qū)間(0,+)內(nèi)單調(diào)遞減,f(log34)f(2-23)f(2-23)flog314.故選C.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x00,y=12x-3x,k=12x0-3x0=-12,x0=2
8、.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=-3.16.18因?yàn)楹瘮?shù)y=f(x+1)-2為奇函數(shù),所以函數(shù)f(x)的圖象關(guān)于點(diǎn)(1,2)對(duì)稱,g(x)=2x-1x-1=1x-1+2關(guān)于點(diǎn)(1,2)對(duì)稱,所以兩個(gè)函數(shù)圖象的交點(diǎn)也關(guān)于點(diǎn)(1,2)對(duì)稱,則(x1+x2+x6)+(y1+y2+y6)=23+43=18.故答案為18.17.(1)證明f(x)=ex-2x-a,令g(x)=ex-2x-a,則g(x)=ex-2.則當(dāng)x(-,l
9、n2)時(shí),g(x)0.所以函數(shù)g(x)在x=ln2時(shí)取最小值,即f(x)在x=ln2時(shí)取最小值,所以f(x)min=f(ln2)=2-2ln2-a.又a2-2ln2,所以f(x)min0.故當(dāng)a2-2ln2時(shí),導(dǎo)函數(shù)f(x)的最小值不小于0.(2)解當(dāng)x0時(shí),ex-x2-ax1-x,即aexx-x-1x+1.令h(x)=exx-x-1x+1(x0),則h(x)=ex(x-1)-x2+1x2=(x-1)(ex-x-1)x2.令(x)=ex-x-1(x0),則(x)=ex-10.當(dāng)x(0,+)時(shí),(x)單調(diào)遞增,(x)(0)=0.則當(dāng)x(0,1)時(shí),h(x)0,h(x)單調(diào)遞增.所以h(x)min
10、=h(1)=e-1,所以ae-1.所以實(shí)數(shù)a的取值范圍為(-,e-1.18.解(1)f(x)定義域?yàn)?0,+),當(dāng)a=e時(shí),f(x)=(1+x)(xex-e)x.當(dāng)0x1時(shí),f(x)1時(shí),f(x)0,f(x)在(0,1)上為減函數(shù);在(1,+)上為增函數(shù).(2)記t=lnx+x,則t=lnx+x在(0,+)上單調(diào)遞增,且tR.f(x)=xex-a(lnx+x)=et-at=g(t).f(x)在(0,+)上有兩個(gè)零點(diǎn)等價(jià)于g(t)=et-at在tR上有兩個(gè)零點(diǎn).當(dāng)a=0時(shí),g(t)=et在R上單調(diào)遞增,且g(t)0,故g(t)無零點(diǎn);當(dāng)a0恒成立,g(t)在R上單調(diào)遞增,又g(0)=10,g1a
11、=e1a-10時(shí),由g(t)=et-a=0可知g(t)在t=lna時(shí)有唯一的一個(gè)極小值g(lna)=a(1-lna),若0a0,g(t)無零點(diǎn);若a=e,g(t)極小值=0,g(t)只有一個(gè)零點(diǎn);若ae,g(t)極小值=a(1-lna)0,由y=lnxx在(e,+)上為減函數(shù),可知當(dāng)ae時(shí),eaaea2,從而g(a)=ea-a20,g(t)在(0,lna)和(lna,+)上各有一個(gè)零點(diǎn).綜上可知,當(dāng)ae時(shí),f(x)有兩個(gè)點(diǎn),故所求a的取值范圍是(e,+).19.(1)證明設(shè)g(x)=f(x),則g(x)=cosx+xsinx-1,g(x)=xcosx.當(dāng)x0,2時(shí),g(x)0;當(dāng)x2,時(shí),g(
12、x)0,g()=-2,故g(x)在(0,)存在唯一零點(diǎn).所以f(x)在(0,)存在唯一零點(diǎn).(2)解由題設(shè)知f()a,f()=0,可得a0.由(1)知,f(x)在(0,)只有一個(gè)零點(diǎn),設(shè)為x0,且當(dāng)x(0,x0)時(shí),f(x)0;當(dāng)x(x0,)時(shí),f(x)0,所以f(x)在(0,x0)單調(diào)遞增,在(x0,)單調(diào)遞減.又f(0)=0,f()=0,所以,當(dāng)x0,時(shí),f(x)0.又當(dāng)a0,x0,時(shí),ax0,故f(x)ax.因此,a的取值范圍是(-,0.20.(1)解函數(shù)的定義域?yàn)?0,+),f(x)=x-mx+lnx=1-mx+lnx.當(dāng)m=0時(shí),f(x)=0得x=1e,當(dāng)x0,1e時(shí),f(x)0,x
13、=1e是函數(shù)f(x)的極小值點(diǎn),滿足題意.當(dāng)m0時(shí),令g(x)=f(x),g(x)=mx2+1x=x+mx2.令g(x)=0,解得x=-m.當(dāng)x(0,-m)時(shí),g(x)0.g(x)min=g(-m)=2+ln(-m),若g(-m)0,即m-e-2,則f(x)=g(x)0恒成立,f(x)在(0,+)上單調(diào)遞增,無極值點(diǎn),不滿足題意.若g(-m)=2+ln(-m)0,即-e-2m0,g(-m)g(1-m)0.又g(x)在(-m,+)上單調(diào)遞增,g(x)在(-m,+)上恰有一個(gè)零點(diǎn)x1.當(dāng)x(-m,x1)時(shí),f(x)=g(x)0,x1是f(x)的極小值點(diǎn),滿足題意,綜上,-e-20,xlnx0,f(
14、x)(1)=e-10,(x)在(1,+)上單調(diào)遞增,h(x)=(x)(1)=e-10,h(x)在(1,+)上單調(diào)遞增,h(x)h(1)=e-10,當(dāng)x1時(shí),xlnxex-1成立,綜上,f(x)ex-1.21.(1)解由已知g(x)=lnxx-m(m0)的定義域?yàn)?0,+),所以g(x)=(lnx)(x-m)-lnx(x-m)(x-m)2=1-mx-lnx(x-m)2.因?yàn)間(x)在(0,e2上單調(diào)遞增,所以對(duì)任意x(0,e2,都有g(shù)(x)=1-mx-lnx(x-m)20.所以1-mx-lnx0,所以mx1-lnx,即mx(1-lnx).令h(x)=x(1-lnx),h(x)=-lnx,所以當(dāng)0
15、x0.當(dāng)x=1時(shí),h(1)=0,當(dāng)x1時(shí),h(x)0,所以函數(shù)h(x)=x(1-lnx)在(0,1)上單調(diào)遞增,在(1,e2上單調(diào)遞減.當(dāng)0x0,所以h(x)min=h(e2)=e2(1-lne2)=-e2,所以m-e2,故實(shí)數(shù)m的取值范圍是(-,-e2.(2)證明當(dāng)m=-1時(shí),f(x)=g(x)h(x)=lnxx+12x-1=2lnxx2-1,對(duì)定義域內(nèi)的任意正數(shù)x,不等式f(x)1x恒成立,即對(duì)定義域內(nèi)的任意正數(shù)x,2lnxx2-11時(shí),x2-10;當(dāng)0x1時(shí),x2-11時(shí),2xlnxx2-1;當(dāng)0xx2-1.令G(x)=x2-1-2xlnx,所以G(x)=(x2-1-2xlnx)=2x-(2xlnx)=2x-2lnx-2=2(x-lnx-1).令m(x)=x-lnx-1,則m(x)=(x-lnx-1)=1-1x=x-1x.所以x=1是m(x)的極值點(diǎn),從而m(x)有極小值m(1)=0,所以G(x)=2(x-lnx-1)0恒成立.所以G(x)=x2-1-2xlnx在(0,+)上單調(diào)遞增.又因?yàn)镚(1)=0,所以當(dāng)x1時(shí),G(x)=x2-1-2xlnx0,即2xlnxx2-1恒成立;當(dāng)0x1時(shí),G(x)=x2-1-2xlnxx2-1恒成立.所以,對(duì)定義域內(nèi)的任意實(shí)數(shù)x,不等式2lnxx2-11x恒成立.18
（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 文
