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1、專題突破練9　利用導(dǎo)數(shù)證明問題及討論零點(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 專題突破練9　利用導(dǎo)數(shù)證明問 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. 13