泰勒展開(kāi)式與高考試題

1、中國(guó)高考數(shù)學(xué)母題一千題(第0001號(hào))愿與您共建真實(shí)的中國(guó)高考數(shù)學(xué)母題(楊培明泰勒展開(kāi)式與高考試題 在本講中,我們將專(zhuān)注于由泰勒展開(kāi)式命制高考試題的方法和途徑,由此,揭秘高考命題專(zhuān)家命制高考試題的手段和方向,以利于命制更優(yōu)秀的試題,為預(yù)測(cè)高考試題服務(wù).母題結(jié)構(gòu):(超越函數(shù)的泰勒展開(kāi)式):ex=1+ex(其中0<<1);ln(1+x)=x-+-+(-1)n-1+(-1)n()n+1.母題解析:略. 1.指數(shù)函數(shù) 子題類(lèi)型:(2010年課標(biāo)高考理科試題)設(shè)函數(shù)f(x)=ex-1-x-ax2.()若a=0,求f(x)的單調(diào)區(qū)間; ()若當(dāng)x0時(shí),f(x)0,求a

2、的取值范圍.解析:()當(dāng)a=0時(shí),由f(x)=ex-1-x(x)=ex-1,由此列表,由表知,f(x)在(-,0)上遞減.在(0,+)上遞增;()由ex=1+ex(其中0<<1)當(dāng)x0時(shí),ex1+=1+x+x2;由此可先證:ex-1-x-x20:令g(x)=ex-1-x-x2(x0),則(x)=ex-1-x,由()知,(x)在0,+)上單調(diào)遞增(x)(0)=0g(x)在0,+)上單調(diào)遞增g(x)g(0)=0當(dāng)a時(shí),f(x)=ex-1-x-ax2ex-1-x-x20成立;當(dāng)a>時(shí),(x)=ex-1-2ax(x)=ex-2a(x)在(0,ln2a)上單調(diào)遞減當(dāng)0<x<

3、ln2a時(shí),(x)<(0)=0f(x)在(0,ln2a)上單調(diào)遞減當(dāng)0<x<ln2a時(shí),f(x)<f(0)=0,不合題意.綜上,a的取值范圍是(-,.點(diǎn)評(píng):由ex=1+ex(其中0<<1)當(dāng)x0時(shí),ex1+x+x2;因此可直接的命制如上試題;由此,還可直接的命制:1.(2010年課標(biāo)高考文科試題)設(shè)函數(shù)f(x)=x(ex-1)-ax2.()若a=,求f(x)的單調(diào)區(qū)間; ()若當(dāng)x0時(shí),f(x)0,求a的取值范圍. 由ex=1+ex(其中0<<1)ex1+x(xR)e-1(x>-1)1-e-11-=(x>-1),由此可命制:2.(20

4、10年全國(guó)高考試題)設(shè)函數(shù)f(x)=1-e-x.()證明:當(dāng)x>-1時(shí),f(x); ()設(shè)當(dāng)x0時(shí),f(x),求a的取值范圍. 由ex=1+ex(其中0<<1)e-x=1-+(-1)n+(-1)n+1ex(其中0<<1);+得:ex+e-x=2+x2+當(dāng)x0時(shí),ex+e-x2+x2;-得:ex+e-x=2(x+)當(dāng)x0時(shí),ex+e-x2(x+),由此可命制:3.(2007年全國(guó)高考試題)設(shè)函數(shù)f(x)=ex-e-x.()證明:f(x)的導(dǎo)數(shù)(x)2; ()若對(duì)所有x0都有f(x)ax,求a的取值范圍.4.(原創(chuàng)試題)設(shè)函數(shù)f(x)=ex-e-x.()證明:當(dāng)x0時(shí)

5、,f(x)2x+x3; ()若對(duì)所有x0都有f(x)ax+x3,求a的取值范圍. 2.對(duì)數(shù)函數(shù) 子題類(lèi)型:(2008年山東高考試題)已知函數(shù)f(x)=+aln(x-1),其中nN*,a為常數(shù).()當(dāng)n=2時(shí),求函數(shù)f(x)的極值; ()當(dāng)a=1時(shí),證明:對(duì)任意的正整數(shù)n,當(dāng)x2時(shí),有f(x)x-1.解析:()f(x)的定義域?yàn)?1,+),當(dāng)n=2時(shí),f(x)=+aln(x-1)(x)=;當(dāng)a0時(shí),f(x)無(wú)極值;當(dāng)a>0時(shí),f極小值(x)=f(1+)=(1+ln),無(wú)極大值;()當(dāng)a=1時(shí),f(x)=+ln(x-1),由ln(x-1)x-2,要證f(x)x-1,只需證1;由x21-x-1

6、|1-x|11.點(diǎn)評(píng):由ln(1+x)=x-+-+(-1)n-1+(-1)n()n+1ln(x+1)xln(x-1)x-2,由此,如上試題;由此,還可直接的命制:5.(2010年全國(guó)高考試題)已知函數(shù)f(x)=(x+1)lnx-x+1.()若x(x)x2+ax+1,求a的取值范圍; ()證明:(x-1)f(x)0. 由ln(x+1)x(x>-1)lnxx-1(x>0)ln-1(x>0)lnx1-=(x>0)ln(x+1)(x>-1),由此可命制:6.(2006年全國(guó)高考試題)設(shè)函數(shù)f(x)=(x+1)ln(x+1),若對(duì)所有的x0,都有f(x)ax成立,求實(shí)數(shù)a的

7、取值范圍.7.(2015年山東高考試題)設(shè)函數(shù)f(x)=ln(x+1)+a(x2-x),其中aR.()討論函數(shù)f(x)極值點(diǎn)的個(gè)數(shù),并說(shuō)明理由; ()若x>0,f(x)0成立,求a的取值范圍. 由ln(x+1)(x>-1)(x+1)ln(x+1)x,由此可命制:8.(2006年全國(guó)高考試題)設(shè)函數(shù)f(x)=(x+1)ln(x+1),若對(duì)所有的x0,都有f(x)ax成立,求實(shí)數(shù)a的取值范圍. 由ln(x+1)(x>-1)(x+1)ln(x+1)x(x+1)2ln(x+1)x+x2(x>-1),由此可命制:9.(原創(chuàng)試題)設(shè)函數(shù)f(x)=(x+1)2ln(x+1)-x2.(

8、)證明:當(dāng)x0時(shí),f(x)x; ()若對(duì)所有x0都有f(x)ax,求a的取值范圍. 由ln(1+x)=x-+-+(-1)n-1+(-1)n()n+1ln(1-x)=-x-()n+1;-得:ln=2(x+)當(dāng)x(0,1)時(shí),ln>2(x+),由此可命制:10.(2015年北京高考試題)已知函數(shù)f(x)=ln.()求曲線y=f(x)在點(diǎn)(0,f(0)處的切線方程;()求證:當(dāng)x(0,1)時(shí),f(x)>2(x+); ()設(shè)實(shí)數(shù)k使得f(x)>k(x+)對(duì)x(0,1)恒成立,求k的最大值. 3.指對(duì)函數(shù) 子題類(lèi)型:(2006年全國(guó)高考試題)已知函數(shù)f(x)=e-ax.()設(shè)a>

9、0,討論y=f(x)的單調(diào)性;()若對(duì)任意x(0,1)恒有f(x)>1,求a的取值范圍.解析:()由f(x)定義域?yàn)閤|x1,(x)=(x2-);當(dāng)0<a2時(shí),(x)0恒成立f(x)在(-,1)和(1,+)上單調(diào)遞增;當(dāng)a>2時(shí),(x)=(x+)(x-)f(x)在(-,-)、(,1)和(1,+)上單調(diào)遞增,在(-,)上單調(diào)遞減;()當(dāng)x(0,1)時(shí),f(x)>1e-ax>1ln-ax>0ln>ax;當(dāng)a2時(shí),ln>2xax;當(dāng)a>2時(shí),令g(x)=ln(1+x)-ln(1-x)-ax,則(x)=+-a=(x+)(x-)g(x)在(0,)上單

10、調(diào)遞減g(x)<g(0)=0ln<ax,不合題意.綜上,a的取值范圍是(-,2.點(diǎn)評(píng):指數(shù)與對(duì)數(shù)函數(shù)的結(jié)合有兩個(gè)方面:指數(shù)函數(shù)與對(duì)數(shù)函數(shù)的相互轉(zhuǎn)化;指數(shù)與對(duì)數(shù)同出現(xiàn)在一個(gè)函數(shù)中.11.(2014年課標(biāo)高考試題)設(shè)函數(shù)f(x)=aexlnx+,曲線y=f(x)在點(diǎn)(1,f(1)處的切線為y=e(x-1)+2.()求a,b; ()證明:f(x)>1.12.(2013年課標(biāo)高考試題)已知函數(shù)f(x)=ex-ln(x+m).()設(shè)x=0是f(x)的極值點(diǎn),求m,并討論f(x)的單調(diào)性; ()當(dāng)m2時(shí),證明f(x)>0. 4.子題詳解:1.解:()由(x)=(x+1)(ex-1)

11、f(x)在(-,-1)和(0,+)上單調(diào)遞增,在(-1,0)上單調(diào)遞減;()當(dāng)x0時(shí),f(x)0當(dāng)x0時(shí),x(ex-1-ax)0當(dāng)x0時(shí),ex1+ax;當(dāng)x0時(shí),ex1+x當(dāng)a1時(shí),ex1+x1+ax;當(dāng)a>1時(shí),令g(x)=ex-ax-1,則(x)=ex-ag(x)在(0,lna)上遞減g(x)<g(0)=0,不合題意.綜上,a(-,1.4.解:()當(dāng)x>-1時(shí),由f(x)(x+1)(1-e-x)xexx+1;()當(dāng)x0時(shí),f(x)=1-e-x0f(x)0ax+1>0a0;由f(x)(ax+1)(1-e-x)x(e-x-1)(ax+1)+x0;令g(x)=(e-x-1

12、)(ax+1)+x,則g(0)=0,(x)=e-x(a-1-ax)+1-a;當(dāng)0a時(shí),由exx+1-x1-ex(x)e-xa-1+a(1-ex)+1-a=(2a-1)(e-x-1)0g(x)g(0)=0;當(dāng)a>時(shí),由e-x1-x-xe-x-1(x)=e-x(a-1-ax)+(1-a+ax)-ax(e-x-1)(a-1-ax)+a(e-x-1)=(e-x-1)(2a-1-ax)當(dāng)x(0,)時(shí),(x)<0g(x)<g(0)=0.故a0,.3.解:()由(x)=ex+e-x2=2,當(dāng)且僅當(dāng)x=0時(shí)取得等號(hào);()由ex1+x,且e-x1-xf(x)=ex-e-x2x;當(dāng)a2時(shí),f(x

13、)=ex-e-x2xax;當(dāng)a>2時(shí),令g(x)=f(x)-ax,則(x)=ex+e-x-a=e-x(e2x-aex+1)g(x)在(0,ln)上單調(diào)遞減g(x)<g(0)=0,不合題意.綜上,a的取值范圍是(-,2.4.解:()令g(x)=f(x)-(2x+x3)(x0),則(x)=ex+e-x-(2+x2)(x)=ex-e-x-2x(x)=ex+e-x-20(x)在0,+)上單調(diào)遞增(x)(0)=0(x)在0,+)上單調(diào)遞增(x)(0)=0g(x)在0,+)上單調(diào)遞增g(x)g(0)=0f(x)2x+x3;()由()知,當(dāng)a2時(shí),對(duì)所有x0都有f(x)ax+x3;當(dāng)a>2

14、時(shí),令h(x)=f(x)-(ax+x3)(x0),則(x)=ex+e-x-(a+x2)(x)=ex-e-x-2x=(x)(0)=0(x)在0,+)上單調(diào)遞增;由(0)=2-a<0(x)存在零點(diǎn)x0>0h(x)在(0,x0)上單調(diào)遞減當(dāng)x(0,x0)時(shí),h(x)<h(0)=0當(dāng)x(0,x0)時(shí),f(x)<ax+x3.綜上,a(-,2.5.解:()由x(x)x2+ax+1lnxx+aa-1;()當(dāng)0<x<1時(shí),f(x)=(x+1)lnx-x+1=xlnx+lnx-x+1<lnx-x+1<0(x-1)f(x)>0;當(dāng)x1時(shí),f(x)=(x+1)l

15、nx-x+1=lnx+x(lnx+-1)(由lnx1-)0(x-1)f(x)0.6.解:當(dāng)a1時(shí),由ln(x+1)(x+1)ln(x+1)xf(x)ax對(duì)于x0恒成立;當(dāng)a>1時(shí),令g(x)=f(x)-ax,則(x)=ln(x+1)+1-a,由(x)=0x=ea-1-1g(x)在(0,ea-1-1)內(nèi)遞減g(x)<g(0)=0f(x)<ax,不合.7.解:()由(x)=(2ax2+ax-a+1);當(dāng)0a時(shí),(x)0f(x)在(-1,+)上單調(diào)遞增,無(wú)極值點(diǎn);當(dāng)a<0時(shí),f(x)有一個(gè)極值點(diǎn);當(dāng)a>時(shí),f(x)有兩個(gè)極值點(diǎn);()由ln(x+1)知,a(x-x2)a(

16、1-x)a(1-x2)10a1時(shí),ln(x+1)a(x-x2)f(x)0成立;當(dāng)a>1時(shí),f(x)在(0,(-1)內(nèi)遞減f(x)<f(0)=0,不符合,舍去;當(dāng)a<0時(shí),f(x)在(-1),+)內(nèi)遞減f(x)<f(0)=0,不符合.綜上,a0,1.8.解:由(x)=ln(x+1)+1在區(qū)間0,+)內(nèi)單調(diào)遞增,且f(x)在x=0處的切線為y=xf(x)x.故a1.9.解:()當(dāng)x0時(shí),f(x)x(x+1)2ln(x+1)-x2xln(x+1)x(x>-1)成立;()由()知,當(dāng)a1時(shí),對(duì)所有x0都有當(dāng)x0時(shí),f(x)ax;當(dāng)a>1時(shí),令g(x)=f(x)-ax

17、,則(x)=2(x+1)ln(x+1)-x+1-a(x)=2ln(x+1)+1>0(x)在0,+)上單調(diào)遞增;由(0)=1-a<0(x)存在零點(diǎn)x0>0g(x)在(0,x0)上單調(diào)遞減當(dāng)x(0,x0)時(shí),g(x)<g(0)=0當(dāng)x(0,x0)時(shí),f(x)<ax.綜上,a(-,1.10.解:()由(0)=2切線:y=2x;()略;()由()知,當(dāng)k2時(shí),f(x)>2(x+)k(x+)對(duì)x(0,1)恒成立;當(dāng)k>2時(shí),令g(x)=f(x)-k(x+),則(x)=(x)-k(1+x2)=(x2-)g(x)在(0,)上單調(diào)遞減當(dāng)0<x<時(shí),g(x)<g(0)=0f(x)<k(x+),不合題意.綜上,k的最大值為2.11.解:()由f(1)=2,(1)=eb=2,a=1;()因f(x)=exlnx+,所以,f(x)>1exlnx+>1ex(lnx+)>1;由ex=1+ex(其中0<<1)ex1+xex-1xexex(當(dāng)且僅當(dāng)x=1時(shí),等號(hào)成立)e-lnxe(-lnx)lnx-(當(dāng)且僅當(dāng)x=時(shí),等號(hào)成立)當(dāng)x(0,+)時(shí),ex(lnx+)>ex(-+)=1.12.解:()由(0)=0m=1(x)=ex-(x)=