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1、專(zhuān)題突破練9 利用導(dǎo)數(shù)證明問(wèn)題及討論零點(diǎn)個(gè)數(shù)
1.設(shè)函數(shù)f(x)=e2x-aln x.
(1)討論f(x)的導(dǎo)函數(shù)f'(x)零點(diǎn)的個(gè)數(shù);
(2)證明:當(dāng)a>0時(shí),f(x)≥2a+aln2a.
2.(2019陜西咸陽(yáng)一模,文21)設(shè)函數(shù)f(x)=x+1-mex,m∈R.
(1)當(dāng)m=1時(shí),求f(x)的單調(diào)區(qū)間;
(2)求證:當(dāng)x∈(0,+∞)時(shí),lnex-1x>x2.
3.(2019河南洛陽(yáng)三模,理21)已知函數(shù)f(x)=ln x-kx,其中k∈R為常數(shù).
(1)討論函數(shù)f(x)的單調(diào)性;
(2)若f(
2、x)有兩個(gè)相異零點(diǎn)x1,x2(x12-ln x1.
4.已知函數(shù)f(x)=ln x+ax2+(2a+1)x.
(1)討論f(x)的單調(diào)性;
(2)當(dāng)a<0時(shí),證明f(x)≤-34a-2.
5.設(shè)函數(shù)f(x)=ln x-x+1.
(1)討論f(x)的單調(diào)性;
(2)證明當(dāng)x∈(1,+∞)時(shí),11,證明當(dāng)x∈(0,1)時(shí),1+(c-1)x>cx.
3、
6.(2019湖南六校聯(lián)考,文21)已知函數(shù)f(x)=ex,g(x)=ax2+x+1(a>0).
(1)設(shè)F(x)=g(x)f(x),討論函數(shù)F(x)的單調(diào)性;
(2)若0g(x)在(0,+∞)上恒成立.
7.(2019天津卷,文20)設(shè)函數(shù)f(x)=ln x-a(x-1)ex,其中a∈R.
(1)若a≤0,討論f(x)的單調(diào)性;
(2)若0x0,證明3x0-x1>2.
4、參考答案
專(zhuān)題突破練9 利用導(dǎo)數(shù)證明問(wèn)
題及討論零點(diǎn)個(gè)數(shù)
1.解(1)f(x)的定義域?yàn)?0,+∞),f'(x)=2e2x-ax(x>0).
當(dāng)a≤0時(shí),f'(x)>0,f'(x)沒(méi)有零點(diǎn),
當(dāng)a>0時(shí),因?yàn)閑2x單調(diào)遞增,-ax單調(diào)遞增,
所以f'(x)在(0,+∞)單調(diào)遞增.
又f'(a)>0,當(dāng)b滿(mǎn)足00時(shí),f'(x)存在唯一零點(diǎn).
(2)由(1),可設(shè)f'(x)在(0,+∞)的唯一零點(diǎn)為x0,當(dāng)x∈(0,x0)時(shí),f'(x)<0;當(dāng)x∈(x0,+∞)時(shí),f'(x)>0.
故f(x)在(0,x0)單調(diào)遞減,在(x0,+
5、∞)單調(diào)遞增,所以當(dāng)x=x0時(shí),f(x)取得最小值,最小值為f(x0).
因?yàn)?e2x0-ax0=0,
所以f(x0)=a2x0+2ax0+aln2a≥2a+aln2a.
故當(dāng)a>0時(shí),f(x)≥2a+aln2a.
2.(1)解當(dāng)m=1時(shí),f(x)=x+1-ex,f'(x)=1-ex.令f'(x)=0,則x=0.
當(dāng)x<0時(shí),f'(x)>0;當(dāng)x>0時(shí),f'(x)<0,
故函數(shù)f(x)的單調(diào)遞增區(qū)間是(-∞,0);單調(diào)遞減區(qū)間是(0,+∞).
(2)證明由(1)知,當(dāng)m=1時(shí),f(x)max=f(0)=0,
∴當(dāng)x∈(0,+∞)時(shí),x+1-ex<0,即ex>x+1.
當(dāng)x∈(
6、0,+∞)時(shí),要證lnex-1x>x2,
只需證ex-1>xex2.
令F(x)=ex-1-xex2=ex-x(e)x-1,
F'(x)=ex-(e)x-x(e)xlne=(e)x(e)x-1-x2=ex2ex2-1-x2.
由ex>x+1可得,ex2>1+x2,
則x∈(0,+∞)時(shí),F'(x)>0恒成立,即F(x)在(0,+∞)內(nèi)單調(diào)遞增,
∴F(x)>F(0)=0.
即ex-1>xex2,∴l(xiāng)nex-1x>x2.
3.(1)解f'(x)=1x-k=1-kxx(x>0),
①當(dāng)k≤0時(shí),f'(x)>0,f(x)在區(qū)間(0,+∞)內(nèi)遞增,
②當(dāng)k>0時(shí),由f'(x)>0,
7、得0x2>0,
∵f(x1)=0,f(x2)=0,
∴l(xiāng)nx1-kx1=0,lnx2-kx2=0,
∴l(xiāng)nx1-lnx2=k(x1-x2),lnx1+lnx2=k(x1+x2),
要證明lnx2>2-lnx1,即證明lnx1+lnx2>2,故k(x1+x2)>2,
即lnx1-lnx2x1-x2>2x1+x2,即lnx1x2>2(x1-x2)x1+x2,
設(shè)t=x1x2>1,上式轉(zhuǎn)化為lnt>2(t-1)t+1(t>1).
設(shè)g(t)=lnt-2
8、(t-1)t+1,∴g'(t)=(t-1)2t(t+1)2>0,
∴g(t)在(1,+∞)上單調(diào)遞增,
∴g(t)>g(1)=0,∴l(xiāng)nt>2(t-1)t+1,∴l(xiāng)nx1+lnx2>2,即lnx2>2-lnx1.
4.(1)解f(x)的定義域?yàn)?0,+∞),f'(x)=1x+2ax+2a+1=(x+1)(2ax+1)x.
若a≥0,則當(dāng)x∈(0,+∞)時(shí),f'(x)>0,故f(x)在(0,+∞)單調(diào)遞增.
若a<0,則當(dāng)x∈0,-12a時(shí),f'(x)>0;
當(dāng)x∈-12a,+∞時(shí),f'(x)<0.
故f(x)在0,-12a單調(diào)遞增,在-12a,+∞單調(diào)遞減.
(2)證明由(1)
9、知,當(dāng)a<0時(shí),f(x)在x=-12a取得最大值,最大值為f-12a=ln-12a-1-14a.
所以f(x)≤-34a-2等價(jià)于ln-12a-1-14a≤-34a-2,
即ln-12a+12a+1≤0.
設(shè)g(x)=lnx-x+1,則g'(x)=1x-1.
當(dāng)x∈(0,1)時(shí),g'(x)>0;
當(dāng)x∈(1,+∞)時(shí),g'(x)<0.
所以g(x)在(0,1)單調(diào)遞增,在(1,+∞)單調(diào)遞減.
故當(dāng)x=1時(shí),g(x)取得最大值,最大值為g(1)=0.
所以當(dāng)x>0時(shí),g(x)≤0.
從而當(dāng)a<0時(shí),ln-12a+12a+1≤0,
即f(x)≤-34a-2.
5.(1)解由
10、題設(shè),f(x)的定義域?yàn)?0,+∞),f'(x)=1x-1,
令f'(x)=0解得x=1.
當(dāng)00,f(x)單調(diào)遞增;
當(dāng)x>1時(shí),f'(x)<0,f(x)單調(diào)遞減.
(2)證明由(1)知f(x)在x=1處取得最大值,最大值為f(1)=0.
所以當(dāng)x≠1時(shí),lnx1,
設(shè)g(x)=1+(c-1)x-cx,
則g'(x)=c-1-cxlnc,
令g'(x)=0,解得x0=lnc-1lnclnc.
當(dāng)x0,g(
11、x)單調(diào)遞增;
當(dāng)x>x0時(shí),g'(x)<0,g(x)單調(diào)遞減.
由(2)知10.
所以當(dāng)x∈(0,1)時(shí),1+(c-1)x>cx.
6.(1)解F(x)=g(x)f(x)=ax2+x+1ex,F'(x)=-ax2+(2a-1)xex=-ax(x-2a-1a)ex,
①若a=12,則F'(x)=-ax2ex≤0,
∴F(x)在R上單調(diào)遞減.
②若a>12,則2a-1a>0,
當(dāng)x<0或x>2a-1a時(shí),F'(x)<0,當(dāng)00,
∴F(x)在(-∞,0)
12、,2a-1a,+∞上單調(diào)遞減,在0,2a-1a上單調(diào)遞增.
③若00時(shí),F'(x)<0,當(dāng)2a-1a0.
∴F(x)在-∞,2a-1a,(0,+∞)上單調(diào)遞減,在2a-1a,0上單調(diào)遞增.
(2)證明∵00恒成立,
∴h'(x)在(0,+∞)上單調(diào)遞增.
又h'(0)=0,∴當(dāng)x∈(0,+∞)
13、時(shí),h'(x)>0,
∴h(x)在(0,+∞)上單調(diào)遞增,
∴h(x)>h(0)=0,∴ex-12x2-x-1>0,ex>12x2+x+1,
∴ex>12x2+x+1≥ax2+x+1,
綜上,f(x)>g(x)在(0,+∞)上恒成立.
7.(1)解由已知,f(x)的定義域?yàn)?0,+∞),且f'(x)=1x-[aex+a(x-1)ex]=1-ax2exx.
因此當(dāng)a≤0時(shí),1-ax2ex>0,從而f'(x)>0,
所以f(x)在(0,+∞)內(nèi)單調(diào)遞增.
(2)證明①由(1)知,f'(x)=1-ax2exx.令g(x)=1-ax2ex,由0
14、調(diào)遞減,又g(1)=1-ae>0,且gln1a=1-aln1a21a=1-ln1a2<0,
故g(x)=0在(0,+∞)內(nèi)有唯一解,從而f'(x)=0在(0,+∞)內(nèi)有唯一解,不妨設(shè)為x0,則1g(x0)x=0,
所以f(x)在(0,x0)內(nèi)單調(diào)遞增;
當(dāng)x∈(x0,+∞)時(shí),f'(x)=g(x)x1時(shí),h'(x)=1x-1<0,故h(x)在(1,+∞)內(nèi)單調(diào)遞減,從而當(dāng)x>1時(shí),h(x)<
15、h(1)=0,所以lnxf(1)=0,所以f(x)在(x0,+∞)內(nèi)有唯一零點(diǎn).又f(x)在(0,x0)內(nèi)有唯一零點(diǎn)1,從而,f(x)在(0,+∞)內(nèi)恰有兩個(gè)零點(diǎn).
②由題意,f'(x0)=0,f(x1)=0,即ax02ex0=1,lnx1=a(x1-1)ex1,從而lnx1=x1-1x02ex1-x0,即ex1-x0=x02lnx1x1-1.因?yàn)楫?dāng)x>1時(shí),lnxx0>1,故ex1-x02.
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（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 專(zhuān)題突破練9 利用導(dǎo)數(shù)證明問(wèn)題及討論零點(diǎn)個(gè)數(shù) 文
