《（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 專題突破練9 利用導(dǎo)數(shù)證明問(wèn)題及討論零點(diǎn)個(gè)數(shù) 理》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9 利用導(dǎo)數(shù)證明問(wèn)題及討論零點(diǎn)個(gè)數(shù) 理（13頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、專題突破練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福建漳州質(zhì)檢二,理21)已知函數(shù)f(x)=xln x. (1)若函數(shù)g(x)=f(x)x2-1x,求g(x)的極值; (2)證明:f(x)+1

2、數(shù). (1)討論函數(shù)f(x)的單調(diào)性; (2)若f(x)有兩個(gè)相異零點(diǎn)x1,x2(x12-ln x1. 4.(2019四川成都二模,理21)已知函數(shù)f(x)=ln x+a1x-1,a∈R. (1)若f(x)≥0,求實(shí)數(shù)a取值的集合; (2)證明:ex+1x≥2-ln x+x2+(e-2)x. 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. 6.已

3、知函數(shù)f(x)=(x-2)ex+a(x-1)2有兩個(gè)零點(diǎn). (1)求a的取值范圍; (2)設(shè)x1,x2是f(x)的兩個(gè)零點(diǎn),證明:x1+x2<2. 7.(2019天津卷,理20)設(shè)函數(shù)f(x)=excos x,g(x)為f(x)的導(dǎo)函數(shù). (1)求f(x)的單調(diào)區(qū)間; (2)當(dāng)x∈π4,π2時(shí),證明f(x)+g(x)π2-x≥0; (3)設(shè)xn為函數(shù)u(x)=f(x)-1在區(qū)間2nπ+π4,2nπ+π2內(nèi)的零點(diǎn),其中n∈N,證明2nπ+π2-xn

4、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,+∞)單調(diào)遞增,所以當(dāng)x=x0時(shí),f(x)取得最小值,最

5、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∵g(x)=lnx-1x(x>0), ∴g'(x)=2-lnxx2. 令g'(x)>0,解得0e2, 故g(x)在(0,e2)內(nèi)遞增,在(e2,+∞)內(nèi)遞減, 故g(x)的極大值為g(e2)=1e2,沒(méi)有極小值. (2)證明要證f(x)+10. 先證明lnx≤x-1,取h(x)=lnx-x+1, 則h'(x)=1-x

6、x, 易知h(x)在(0,1)內(nèi)遞增,在(1,+∞)內(nèi)遞減, 所以h(x)≤h(1)=0,即lnx≤x-1,當(dāng)且僅當(dāng)x=1時(shí),取“=”, 故xlnx≤x(x-1),ex-x2-xlnx-1≥ex-2x2+x-1, 故只需證明當(dāng)x>0時(shí),ex-2x2+x-1>0恒成立. 令k(x)=ex-2x2+x-1(x≥0),則k'(x)=ex-4x+1. 令F(x)=k'(x),則F'(x)=ex-4,令F'(x)=0,解得x=2ln2. ∵F'(x)遞增,∴當(dāng)x∈(0,2ln2]時(shí),F'(x)≤0,F(x)遞減,即k'(x)遞減, 當(dāng)x∈(2ln2,+∞)時(shí),F'(x)>0,F(x)遞增

7、,即k'(x)遞增, 且k'(2ln2)=5-8ln2<0,k'(0)=2>0,k'(2)=e2-8+1>0, 由零點(diǎn)存在定理,可知?x1∈(0,2ln2),?x2∈(2ln2,2),使得k'(x1)=k'(x2)=0, 故0x2時(shí),k'(x)>0,k(x)遞增, 當(dāng)x10, 故當(dāng)x>0時(shí),k(x)>0,原不等式成

8、立. 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,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-lnx2

9、x1-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(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)=1x-ax2=x-ax2(x>0). 當(dāng)a≤0時(shí),f'(x)>0,函數(shù)f(x)在(0,+∞)上單調(diào)遞增.又f(1)=0, 因此當(dāng)00時(shí),可

10、得函數(shù)f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增, 故當(dāng)x=a時(shí),函數(shù)f(x)取得最小值,則f(a)=lna+1-a≥0. 令g(a)=lna+1-a,g(1)=0. 由g'(a)=1a-1=1-aa,可知當(dāng)a=1時(shí),函數(shù)g(a)取得最大值,而g(1)=0, 因此只有當(dāng)a=1時(shí)滿足f(a)=lna+1-a≥0.故a=1. 故實(shí)數(shù)a取值的集合是{1}. (2)證明由(1)可知,當(dāng)a=1時(shí),f(x)≥0,即lnx≥1-1x在(0,+∞)內(nèi)恒成立. 要證明ex+1x≥2-lnx+x2+(e-2)x,即證明ex≥1+x2+(e-2)x,即ex-1-x2-(e-2)x≥0.

11、 令h(x)=ex-1-x2-(e-2)x,x>0. h'(x)=ex-2x-(e-2),令u(x)=ex-2x-(e-2), u'(x)=ex-2,令u'(x)=ex-2=0,解得x=ln2. 則函數(shù)u(x)在(0,ln2)內(nèi)單調(diào)遞減,在(ln2,+∞)內(nèi)單調(diào)遞增. 即函數(shù)h'(x)在(0,ln2)內(nèi)單調(diào)遞減,在(ln2,+∞)內(nèi)單調(diào)遞增. 而h'(0)=1-(e-2)=3-e>0, h'(ln2)0,h(x)單調(diào)遞增; 當(dāng)x∈(x0,1)時(shí),h'(x)<0,h(x)單

12、調(diào)遞減. 當(dāng)x∈(1,+∞)時(shí),h'(x)>0,h(x)單調(diào)遞增. 又h(0)=1-1=0,h(1)=e-1-1-(e-2)=0, 故對(duì)?x>0,h(x)≥0恒成立,即ex-1-x2-(e-2)x≥0. 綜上可知,ex+1x≥2-lnx+x2+(e-2)x成立. 5.(1)解由題設(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),lnx

13、故當(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(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)=(x-1)ex+2a(x-1)=(x-1)(ex+2

14、a). ①設(shè)a=0,則f(x)=(x-2)ex,f(x)只有一個(gè)零點(diǎn). ②設(shè)a>0,則當(dāng)x∈(-∞,1)時(shí),f'(x)<0; 當(dāng)x∈(1,+∞)時(shí),f'(x)>0, 所以f(x)在(-∞,1)單調(diào)遞減,在(1,+∞)單調(diào)遞增. 又f(1)=-e,f(2)=a,取b滿足b<0且ba2(b-2)+a(b-1)2=ab2-32b>0, 故f(x)存在兩個(gè)零點(diǎn). ③設(shè)a<0,由f'(x)=0得x=1或x=ln(-2a). 若a≥-e2,則ln(-2a)≤1, 故當(dāng)x∈(1,+∞)時(shí),f'(x)>0, 因此f(x)在(1,+∞)單調(diào)遞增. 又當(dāng)x≤1時(shí),

15、f(x)<0,所以f(x)不存在兩個(gè)零點(diǎn). 若a<-e2,則ln(-2a)>1, 故當(dāng)x∈(1,ln(-2a))時(shí),f'(x)<0; 當(dāng)x∈(ln(-2a),+∞)時(shí),f'(x)>0. 因此f(x)在(1,ln(-2a))單調(diào)遞減, 在(ln(-2a),+∞)單調(diào)遞增. 又當(dāng)x≤1時(shí)f(x)<0,所以f(x)不存在兩個(gè)零點(diǎn). 綜上,a的取值范圍為(0,+∞). (2)證明不妨設(shè)x1f(2-x2), 即f(2-x2)<0.

16、 由于f(2-x2)=-x2e2-x2+a(x2-1)2, 而f(x2)=(x2-2)ex2+a(x2-1)2=0, 所以f(2-x2)=-x2e2-x2-(x2-2)ex2. 設(shè)g(x)=-xe2-x-(x-2)ex, 則g'(x)=(x-1)(e2-x-ex). 所以當(dāng)x>1時(shí),g'(x)<0, 而g(1)=0, 故當(dāng)x>1時(shí),g(x)<0. 從而g(x2)=f(2-x2)<0, 故x1+x2<2. 7.(1)解由已知,有f'(x)=ex(cosx-sinx).因此,當(dāng)x∈2kπ+π4,2kπ+5π4(k∈Z)時(shí),有sinx>cosx,得f'(x)<0,則f(x)單調(diào)遞

17、減; 當(dāng)x∈2kπ-3π4,2kπ+π4(k∈Z)時(shí),有sinx0,則f(x)單調(diào)遞增. 所以,f(x)的單調(diào)遞增區(qū)間為2kπ-3π4,2kπ+π4(k∈Z),f(x)的單調(diào)遞減區(qū)間為2kπ+π4,2kπ+5π4(k∈Z). (2)證明記h(x)=f(x)+g(x)π2-x. 依題意及(1),有 g(x)=ex(cosx-sinx), 從而g'(x)=-2exsinx. 當(dāng)x∈π4,π2時(shí),g'(x)<0, 故h'(x)=f'(x)+g'(x)π2-x+g(x)(-1)=g'(x)π2-x<0. 因此,h(x)在區(qū)間π4,π2上單調(diào)遞減,進(jìn)而h(x)

18、≥hπ2=fπ2=0. 所以,當(dāng)x∈π4,π2時(shí),f(x)+g(x)π2-x≥0. (3)證明依題意,u(xn)=f(xn)-1=0,即exncosxn=1.記yn=xn-2nπ,則yn∈π4,π2,且f(yn)=eyncosyn=exn-2nπcos(xn-2nπ)=e-2nπ(n∈N). 由f(yn)=e-2nπ≤1=f(y0)及(1),得yn≥y0. 由(2)知,當(dāng)x∈π4,π2時(shí),g'(x)<0,所以g(x)在π4,π2上為減函數(shù), 因此g(yn)≤g(y0)