202X屆高考數(shù)學(xué)一輪復(fù)習(xí)第三章導(dǎo)數(shù)3.2導(dǎo)數(shù)的應(yīng)用課件

1、(2014天津文,19,14分)已知函數(shù)f(x)=x2-ax3(a0),xR.(1)求f(x)的單調(diào)區(qū)間和極值;(2)若對(duì)于任意的x1(2,+),都存在x2(1,+),使得f(x1)f(x2)=1.求a的取值范圍.23考點(diǎn)一導(dǎo)數(shù)與函數(shù)的單調(diào)性考點(diǎn)一導(dǎo)數(shù)與函數(shù)的單調(diào)性A A組自主命題組自主命題天津卷題組天津卷題組五年高考解析解析(1)由已知,有f (x)=2x-2ax2(a0).令f (x)=0,解得x=0或x=.當(dāng)x變化時(shí), f (x), f(x)的變化情況如下表:1ax(-,0)0f (x)-0+0-f(x)010,a1a1,a213a所以, f(x)的單調(diào)遞增區(qū)間是;單調(diào)遞減區(qū)間是(-,0

2、),.當(dāng)x=0時(shí), f(x)有極小值,且極小值f(0)=0;當(dāng)x=時(shí),f(x)有極大值,且極大值f=.(2)由f(0)=f=0及(1)知,當(dāng)x時(shí), f(x)0;當(dāng)x時(shí), f(x)2,即0a時(shí),由f=0可知,0A,而0 B,所以A不是B的子集.當(dāng)12,即a時(shí),有f(2)0,且此時(shí)f(x)在(2,+)上單調(diào)遞減,故A=(-, f(2),因而A(-,0);由f(1)0,有f(x)在(1,+)上的取值范圍包含(-,0),則(-,0)B.所以,AB.當(dāng)時(shí),有f(1)0,且此時(shí)f(x)在(1,+)上單調(diào)遞減,故B=,A=(-, f(2),所以A不是B的子集.綜上,a的取值范圍是.10,a1,a1a1a21

3、3a32a30,2a3,2a1|(1,), ( )0( )xf xf x32a3432a32a343232a321,0(1)f3 3,4 2評(píng)析評(píng)析 本題主要考查導(dǎo)數(shù)的運(yùn)算,利用導(dǎo)數(shù)研究函數(shù)的性質(zhì),考查化歸思想、分類討論思想、函數(shù)思想.考查綜合分析問(wèn)題和解決問(wèn)題的能力.1.(2019天津文,20,14分)設(shè)函數(shù)f(x)=ln x-a(x-1)ex,其中aR.(1)若a0,討論f(x)的單調(diào)性;(2)若0ax0,證明3x0-x12.1e考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極( (最最) )值值解析解析本小題主要考查導(dǎo)數(shù)的運(yùn)算、不等式證明、運(yùn)用導(dǎo)數(shù)研究函數(shù)的性質(zhì)等基礎(chǔ)知識(shí)和方法.考查函數(shù)思想、化

4、歸與轉(zhuǎn)化思想.考查綜合分析問(wèn)題和解決問(wèn)題的能力,體現(xiàn)了邏輯推理和數(shù)學(xué)運(yùn)算素養(yǎng).(1)由已知, f(x)的定義域?yàn)?0,+),且f (x)=-aex+a(x-1)ex=.因此當(dāng)a0時(shí),1-ax2ex0,從而f (x)0,所以f(x)在(0,+)內(nèi)單調(diào)遞增.(2)證明:(i)由(1)知, f (x)=.令g(x)=1-ax2ex,由0a0,且g=1-a=1-0,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1x0=0,所以f(x)在(0,x0)內(nèi)單調(diào)遞增;當(dāng)x(x0,+)時(shí), f (x)=1時(shí),h(x)=-11時(shí),h(x)h(1)=0,所以ln x

5、x-1.從而f=ln-a=ln-ln+1=hf(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).(ii)由題意,即從而ln x1=,即=.因?yàn)楫?dāng)x1時(shí),ln xx01,故=,兩邊取對(duì)數(shù),得ln ln ,于是x1-x02ln x02.10exx2011(1)1xxx20 x10exx20 x思路分析思路分析 (1)求出導(dǎo)函數(shù),結(jié)合a0判定f (x)0,從而得f(x)在(0,+)上遞增;(2)(i)利用導(dǎo)數(shù)的零點(diǎn)判斷f(x)的單調(diào)區(qū)間,結(jié)合零點(diǎn)存在性定理判斷f(x)的零點(diǎn)個(gè)數(shù);(ii)由極值點(diǎn)x0及零點(diǎn)x1建立方

6、程組從而得到=,又由x1時(shí),ln xx-1,得到=,兩邊取對(duì)數(shù)即可得結(jié)論.012011e1,ln(1) e ,xxaxxa x10exx2011ln1xxx10exx2011(1)1xxx20 x2.(2019天津理,20,14分)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時(shí),證明f(x)+g(x)0;(3)設(shè)xn為函數(shù)u(x)=f(x)-1在區(qū)間內(nèi)的零點(diǎn),其中nN,證明2n+-xncos x,得f (x)0,則f(x)單調(diào)遞減;當(dāng)x(kZ)時(shí),有sin x0,則f(x)單調(diào)遞增.所以, f(x)的單調(diào)遞增區(qū)間為(kZ), f(x)的單調(diào)

7、遞減區(qū)間為(kZ).(2)證明:記h(x)=f(x)+g(x).依題意及(1),有g(shù)(x)=ex(cos x-sin x),從而g(x)=-2exsin x.當(dāng)x時(shí),g(x)0,故h(x)=f (x)+g(x)+g(x)(-1)52,244kk32,244kk32,244kk52,244kk2x,4 2 2x=g(x)0,因此,h(x)在區(qū)間上單調(diào)遞減,進(jìn)而h(x)h=f=0.所以,當(dāng)x時(shí), f(x)+g(x)0.(3)證明:依題意,u(xn)=f(xn)-1=0,即cos xn=1.記yn=xn-2n,則yn,且f(yn)=cos yn=cos(xn-2n)=e-2n(nN).由f(yn)=

8、e-2n1=f(y0)及(1),得yny0.由(2)知,當(dāng)x時(shí),g(x)0,所以g(x)在上為減函數(shù),因此g(yn)g(y0)g=0.又由(2)知, f(yn)+g(yn)0,故-yn-=-=.所以,2n+-xn0和f (x)0時(shí)x的范圍.(2)記h(x)=f(x)+g(x),求h(x),從而得到函數(shù)h(x)在上的單調(diào)性,轉(zhuǎn)化為求h(x)的最小值,驗(yàn)證最小值非負(fù)即可.(3)記u(x)在區(qū)間內(nèi)的零點(diǎn)為xn,則xn,則有yn=xn-2n.與(2)聯(lián)系知f(yn)+g(yn)0,此時(shí)要先確定g(yn)的符號(hào),再將上式轉(zhuǎn)化為-yn-,然后進(jìn)一步證明-0,函數(shù)g(x)=|f(x)|,求證:g(x)在區(qū)間

9、-1,1上的最大值.不小于14解析解析(1)由f(x)=x3-ax-b,可得f (x)=3x2-a.下面分兩種情況討論:當(dāng)a0時(shí),有f (x)=3x2-a0恒成立,所以f(x)的單調(diào)遞增區(qū)間為(-,+).當(dāng)a0時(shí),令f (x)=0,解得x=或x=-.當(dāng)x變化時(shí), f (x), f(x)的變化情況如下表:33a33ax-f (x)+0-0+f(x)單調(diào)遞增極大值單調(diào)遞減極小值單調(diào)遞增3a,3 3a33a3a,333a33a,3所以f(x)的單調(diào)遞減區(qū)間為,單調(diào)遞增區(qū)間為,.(2)證明:因?yàn)閒(x)存在極值點(diǎn),所以由(1)知a0,且x00.由題意,得f (x0)=3-a=0,即=,進(jìn)而f(x0)

10、=-ax0-b=-x0-b.又f(-2x0)=-8+2ax0-b=-x0+2ax0-b=-x0-b=f(x0),且-2x0 x0,由題意及(1)知,存在唯一實(shí)數(shù)x1滿足 f(x1)=f(x0),且x1x0,因此x1=-2x0.所以x1+2x0=0.(3)證明:設(shè)g(x)在區(qū)間-1,1上的最大值為M,maxx,y表示x,y兩數(shù)的最大值.下面分三種情況討論:當(dāng)a3時(shí),-11,由(1)知, f(x)在區(qū)間-1,1上單調(diào)遞減,所以f(x)在區(qū)間-1,1上的取值范圍為f(1), f(-1),因此M=max|f(1)|,|f(-1)|=max|1-a-b|,|-1+a-b|=max|a-1+b|,|a-1

11、-b|=33,33aa3,3a 3,3a20 x20 x3a30 x23a30 x83a23a33a33a1,0,1,0.ab bab b 所以M=a-1+|b|2.當(dāng)a3時(shí),-1-1,由(1)和(2)知f(-1)f =f , f(1)f =f ,所以f(x)在區(qū)間-1,1上的取值范圍為f , f ,因此M=max,=max=max=+|b|=.當(dāng)0a時(shí),-1-1,由(1)和(2)知f(-1)f =f ,所以f(x)在區(qū)間-1,1上的取值范圍為f(-1), f(1),因此M=max|f(-1)|,|f(1)|=max|-1+a-b|,|1-a-b|=max|1-a+b|,|1-a-b|=1-a

12、+|b|.綜上所述,當(dāng)a0時(shí),g(x)在區(qū)間-1,1上的最大值不小于.2 33a33a1414思路分析思路分析(1)求含參數(shù)的函數(shù)f(x)的單調(diào)區(qū)間,需要進(jìn)行分類討論;(2)由第(1)問(wèn)可知a0,要證x1+2x0=0,只需證出f(-2x0)=f(x0),其中x1=-2x0,即可得結(jié)論;(3)求g(x)在-1,1上的最大值,對(duì)a分情況討論即可.評(píng)析評(píng)析 本題主要考查導(dǎo)數(shù)的運(yùn)算、利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)、證明不等式等基礎(chǔ)知識(shí)和方法.考查分類討論思想和化歸思想.考查綜合分析問(wèn)題和解決問(wèn)題的能力.考點(diǎn)三導(dǎo)數(shù)的綜合應(yīng)用1.(2018天津,20,14分)已知函數(shù)f(x)=ax,g(x)=logax,其中a1

13、.(1)求函數(shù)h(x)=f(x)-xln a的單調(diào)區(qū)間;(2)若曲線y=f(x)在點(diǎn)(x1, f(x1)處的切線與曲線y=g(x)在點(diǎn)(x2,g(x2)處的切線平行,證明x1+g(x2)=-;(3)證明當(dāng)a時(shí),存在直線l,使l是曲線y=f(x)的切線,也是曲線y=g(x)的切線.2lnlnlnaa1ee解析解析本小題主要考查導(dǎo)數(shù)的運(yùn)算、導(dǎo)數(shù)的幾何意義、運(yùn)用導(dǎo)數(shù)研究指數(shù)函數(shù)與對(duì)數(shù)函數(shù)的性質(zhì)等基礎(chǔ)知識(shí)和方法.考查函數(shù)與方程思想、化歸思想.考查抽象概括能力、綜合分析問(wèn)題和解決問(wèn)題的能力.(1)由已知,h(x)=ax-xln a,有h(x)=axln a-ln a.令h(x)=0,解得x=0.由a1,

14、可知當(dāng)x變化時(shí),h(x),h(x)的變化情況如下表:x(-,0)0(0,+)h(x)-0+h(x)極小值所以函數(shù)h(x)的單調(diào)遞減區(qū)間為(-,0),單調(diào)遞增區(qū)間為(0,+).(2)證明:由f (x)=axln a,可得曲線y=f(x)在點(diǎn)(x1, f(x1)處的切線斜率為ln a.由g(x)=,可得曲線y=g(x)在點(diǎn)(x2,g(x2)處的切線斜率為.因?yàn)檫@兩條切線平行,故有l(wèi)n a=,即x2(ln a)2=1.兩邊取以a為底的對(duì)數(shù),得logax2+x1+2logaln a=0,所以x1+g(x2)=-.(3)證明:曲線y=f(x)在點(diǎn)(x1,)處的切線l1:y-=ln a(x-x1).曲線y

15、=g(x)在點(diǎn)(x2,logax2)處的切線l2:y-logax2=(x-x2).要證明當(dāng)a時(shí),存在直線l,使l是曲線y=f(x)的切線,也是曲線y=g(x)的切線,只需證明當(dāng)a時(shí),存在x1(-,+),x2(0,+),使得l1與l2重合.即只需證明當(dāng)a時(shí),1xa1lnxa21lnxa1xa21lnxa1xa2lnlnlnaa1xa1xa1xa21lnxa1ee1ee1ee方程組有解.由得x2=,代入,得-x1ln a+x1+=0.因此,只需證明當(dāng)a時(shí),關(guān)于x1的方程存在實(shí)數(shù)解.設(shè)函數(shù)u(x)=ax-xaxln a+x+,即要證明當(dāng)a時(shí),函數(shù)y=u(x)存在零點(diǎn).u(x)=1-(ln a)2xa

16、x,可知x(-,0)時(shí),u(x)0;x(0,+)時(shí),u(x)單調(diào)遞減,又u(0)=10,u=1-0,使得u(x0)=0,即1-(ln a)2x0=0.由此可得u(x)在(-,x0)上單調(diào)遞增,在(x0,+)上單調(diào)遞減.u(x)在x=x0處取得極大值u(x0).因?yàn)閍,故ln ln a-1,1112121ln,ln1lnlog,lnxxxaaaxaax aaxa121(ln )xaa1xa1xa1lna2lnlnlnaa1ee1lna2lnlnlnaa1ee21(ln )a21(ln )aa0 xa1ee所以u(píng)(x0)=-x0ln a+x0+=+x0+0.下面證明存在實(shí)數(shù)t,使得u(t)時(shí),有u

17、(x)(1+xln a)(1-xln a)+x+=-(ln a)2x2+x+1+,所以存在實(shí)數(shù)t,使得u(t)0.因此,當(dāng)a時(shí),存在x1(-,+),使得u(x1)=0.所以,當(dāng)a時(shí),存在直線l,使l是曲線y=f(x)的切線,也是曲線y=g(x)的切線.0 xa0 xa1lna2lnlnlnaa201(ln )xa2lnlnlnaa22lnlnlnaa1lna1lna2lnlnlnaa1lna2lnlnlnaa1ee1ee2.(2015天津,20,14分)已知函數(shù)f(x)=nx-xn,xR,其中nN*,且n2.(1)討論f(x)的單調(diào)性;(2)設(shè)曲線y=f(x)與x軸正半軸的交點(diǎn)為P,曲線在點(diǎn)P

18、處的切線方程為y=g(x),求證:對(duì)于任意的正實(shí)數(shù)x,都有f(x)g(x);(3)若關(guān)于x的方程f(x)=a(a為實(shí)數(shù))有兩個(gè)正實(shí)數(shù)根x1,x2,求證:|x2-x1|0,即x1時(shí),函數(shù)f(x)單調(diào)遞增;當(dāng)f (x)1時(shí),函數(shù)f(x)單調(diào)遞減.所以, f(x)在(-,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減.(2)證明:設(shè)點(diǎn)P的坐標(biāo)為(x0,0),則x0=, f (x0)=n-n2.曲線y=f(x)在點(diǎn)P處的切線方程為y=f (x0)(x-x0),即g(x)=f (x0)(x-x0).令F(x)=f(x)-g(x),即F(x)=f(x)-f (x0)(x-x0),則F(x)=f (x)-f (x0

19、).由于f (x)=-nxn-1+n在(0,+)上單調(diào)遞減,故F(x)在(0,+)上單調(diào)遞減.又因?yàn)镕(x0)=0,所以當(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)內(nèi)單調(diào)遞增,在(x0,+)上單調(diào)遞減,所以對(duì)于任意的正實(shí)數(shù)x,都有F(x)F(x0)=0,即對(duì)于任意的正實(shí)數(shù)x,都有f(x)g(x).(3)證明:不妨設(shè)x1x2.由(2)知g(x)=(n-n2)(x-x0).設(shè)方程g(x)=a的根為x2,可得x2=+x0.當(dāng)n2時(shí),g(x)在(-,+)上單調(diào)遞減.11nn2ann又由(2)知g(x2)f(x2)=a=g(x2),可得x2x2.類似地,設(shè)曲

20、線y=f(x)在原點(diǎn)處的切線方程為y=h(x),可得h(x)=nx.當(dāng)x(0,+)時(shí), f(x)-h(x)=-xn0,即對(duì)于任意的x(0,+), f(x)h(x).設(shè)方程h(x)=a的根為x1,可得x1=.因?yàn)閔(x)=nx在(-,+)上單調(diào)遞增,且h(x1)=a=f(x1)h(x1),因此x1x1.由此可得x2-x1x2-x1=+x0.因?yàn)閚2,所以2n-1=(1+1)n-11+=1+n-1=n,故2=x0.所以,|x2-x1|+2.an1an11Cn11nn1an評(píng)析評(píng)析 本題主要考查導(dǎo)數(shù)的運(yùn)算、導(dǎo)數(shù)的幾何意義、利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)、證明不等式等基礎(chǔ)知識(shí)和方法.考查分類討論思想、函數(shù)思想

21、和化歸思想.考查綜合分析問(wèn)題和解決問(wèn)題的能力.3.(2014天津,20,14分)設(shè)f(x)=x-aex(aR),xR.已知函數(shù)y=f(x)有兩個(gè)零點(diǎn)x1,x2,且x10在R上恒成立,可得f(x)在R上單調(diào)遞增,不合題意.a0時(shí),由f (x)=0,得x=-ln a.當(dāng)x變化時(shí), f (x), f(x)的變化情況如下表:x(-,-ln a)-ln a(-ln a,+)f (x)+0-f(x)-ln a-1這時(shí), f(x)的單調(diào)遞增區(qū)間是(-,-ln a);單調(diào)遞減區(qū)間是(-ln a,+).于是,“函數(shù)y=f(x)有兩個(gè)零點(diǎn)”等價(jià)于如下條件同時(shí)成立:(i)f(-ln a)0;(ii)存在s1(-,-

22、ln a),滿足f(s1)0;(iii)存在s2(-ln a,+),滿足f(s2)0,即-ln a-10,解得0ae-1.而此時(shí),取s1=0,滿足s1(-,-ln a),且f(s1)=-a0;取s2=+ln,滿足s2(-ln a,+),且f(s2)=+0.由已知,x1,x2滿足a=g(x1),a=g(x2).由a(0,e-1),及g(x)的單調(diào)性,可得x1(0,1),x2(1,+).對(duì)于任意的a1,a2(0,e-1),設(shè)a1a2,g(1)=g(2)=a1,其中0112;g(1)=g(2)=a2,其中011a2,即g(1)g(1),可得11;2a2a22eaa22lneaaexxexx1exx類

23、似可得20,得1,且解得x1=,x2=.所以x1+x2=.(*)令h(x)=,x(1,+),則h(x)=.令u(x)=-2ln x+x-,得u(x)=.當(dāng)x(1,+)時(shí),u(x)0.因此,u(x)在(1,+)上單調(diào)遞增,故對(duì)于任意的x(1,+),u(x)u(1)=0,由此可得h(x)0,故h(x)在(1,+)上單調(diào)遞增.21212121xx1ex2ex21xx21xx2121,ln ,xtxxxtln1tt ln1ttt (1)ln1ttt(1)ln1xxx212ln(1)xxxx1x21xx因此,由(*)可得x1+x2隨著t的增大而增大.而由(2)知t隨著a的減小而增大,所以x1+x2隨著a

24、的減小而增大.評(píng)析評(píng)析 本題主要考查函數(shù)的零點(diǎn)、導(dǎo)數(shù)的運(yùn)算、利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)等基礎(chǔ)知識(shí)和方法.考查函數(shù)思想、化歸思想.考查抽象概括能力、綜合分析問(wèn)題和解決問(wèn)題的能力.4.(2013天津,20,14分)已知函數(shù)f(x)=x2ln x.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)證明:對(duì)任意的t0,存在唯一的s,使t=f(s);(3)設(shè)(2)中所確定的s關(guān)于t的函數(shù)為s=g(t),證明:當(dāng)te2時(shí),有.25ln ( )lng tt12解析解析(1)函數(shù)f(x)的定義域?yàn)?0,+).f (x)=2xln x+x=x(2ln x+1),令f (x)=0,得x=.當(dāng)x變化時(shí), f (x), f(x)的變

25、化情況如下表:1exf (x)-0+f(x)極小值10,e1e1,e所以函數(shù)f(x)的單調(diào)遞減區(qū)間是,單調(diào)遞增區(qū)間是.(2)證明:當(dāng)0 x1時(shí), f(x)0.令h(x)=f(x)-t,x1,+).由(1)知,h(x)在區(qū)間(1,+)內(nèi)單調(diào)遞增.h(1)=-t0.故存在唯一的s(1,+),使得t=f(s)成立.(3)證明:因?yàn)閟=g(t),由(2)知,t=f(s),且s1,所以=,其中u=ln s.要使成立,只需0ln ue2時(shí),若s=g(t)e,則由f(s)的單調(diào)性,有t=f(s)f(e)=e2,矛盾.所以se,即u1,從而ln u0成立.另一方面,令F(u)=ln u-,u1.f (u)=-

26、,令f (u)=0,得u=2.當(dāng)1u0;當(dāng)u2時(shí), f (u)1, f(u)F(2)0.因此ln ue2時(shí),有0,2x0,ln 20.當(dāng)f(x)=2-x時(shí),exf(x)在f(x)的定義域上單調(diào)遞增,故具有M性質(zhì),易知B、C、D不具有M性質(zhì),故選A.e2xxe2xxe2xx2e 2e 2 ln2(2 )xxxxxe2xx1.求f(x)的導(dǎo)函數(shù)f (x).方法小結(jié)方法小結(jié)利用導(dǎo)數(shù)研究函數(shù)f(x)的單調(diào)性的一般步驟:2.令f (x)0(或f (x)0),得區(qū)間I.3.判斷單調(diào)性,當(dāng)xI時(shí), f(x)單調(diào)增(或減).2.(2016課標(biāo),12,5分)若函數(shù)f(x)=x-sin 2x+asin x在(-,

27、+)單調(diào)遞增,則a的取值范圍是()A.-1,1 B.C. D. 1311,31 1,3 311,3 答案答案 C f (x)=1-cos 2x+acos x=1-(2cos2x-1)+acos x=-cos2x+acos x+, f(x)在R上單調(diào)遞增,則f (x)0在R上恒成立,令cos x=t,t-1,1,則-t2+at+0在-1,1上恒成立,即4t2-3at-50在-1,1上恒成立,令g(t)=4t2-3at-5,則解得-a,故選C.232343534353(1)4350,( 1)4350,gaga1313疑難突破疑難突破由函數(shù)的單調(diào)性求參數(shù)范圍,利用導(dǎo)數(shù)將問(wèn)題轉(zhuǎn)化為恒成立問(wèn)題,再利用二

28、次函數(shù)來(lái)解決.3.(2017江蘇,11,5分)已知函數(shù)f(x)=x3-2x+ex-,其中e是自然對(duì)數(shù)的底數(shù).若f(a-1)+f(2a2)0,則實(shí)數(shù)a的取值范圍是 .1ex答案答案 11,2解析解析本題考查用導(dǎo)數(shù)研究函數(shù)單調(diào)性、函數(shù)單調(diào)性的應(yīng)用.易知函數(shù)f(x)的定義域關(guān)于原點(diǎn)對(duì)稱.f(x)=x3-2x+ex-,f(-x)=(-x)3-2(-x)+e-x-=-x3+2x+-ex=-f(x),f(x)為奇函數(shù),又f (x)=3x2-2+ex+3x2-2+2=3x20(當(dāng)且僅當(dāng)x=0時(shí),取“=”),從而f(x)在R上單調(diào)遞增,所以f(a-1)+f(2a2)0f(a-1)f(-2a2)-2a2a-1,

29、解得-1a.1ex1ex1ex1ex12方法小結(jié)方法小結(jié)函數(shù)不等式的求解思路:(1)轉(zhuǎn)化為f(x)f(g(x);(2)結(jié)合單調(diào)性轉(zhuǎn)化為(x)g(x)或(x)g(x).4.(2018課標(biāo),21,12分)已知函數(shù)f(x)=-x+aln x.(1)討論f(x)的單調(diào)性;(2)若f(x)存在兩個(gè)極值點(diǎn)x1,x2,證明:2,令f (x)=0,得x=或x=.當(dāng)x時(shí), f (x)0.所以f(x)在,上單調(diào)遞減,在上單調(diào)遞增.21xax221xaxx242aa242aa240,2aa24,2aa2244,22aaaa240,2aa24,2aa2244,22aaaa(2)由(1)知, f(x)存在兩個(gè)極值點(diǎn)當(dāng)且

30、僅當(dāng)a2.由于f(x)的兩個(gè)極值點(diǎn)x1,x2滿足x2-ax+1=0,所以x1x2=1,不妨設(shè)x11,由于=-1+a=-2+a=-2+a,1212( )()f xf xxx121x x1212lnlnxxxx1212lnlnxxxx2222ln1xxx所以a-2等價(jià)于-x2+2ln x20.設(shè)函數(shù)g(x)=-x+2ln x,由(1)知,g(x)在(0,+)上單調(diào)遞減,又g(1)=0,從而當(dāng)x(1,+)時(shí),g(x)0,所以-x2+2ln x20,即a-2.1212( )()f xf xxx21x1x21x1212( )()f xf xxx5.(2017課標(biāo),21,12分)已知函數(shù)f(x)=ae2x

31、+(a-2)ex-x.(1)討論f(x)的單調(diào)性;(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.解析解析本題考查了利用導(dǎo)數(shù)討論函數(shù)的單調(diào)性和函數(shù)的零點(diǎn)問(wèn)題.(1)f(x)的定義域?yàn)?-,+), f (x)=2ae2x+(a-2)ex-1=(aex-1)(2ex+1).(i)若a0,則f (x)0,則由f (x)=0得x=-ln a.當(dāng)x(-,-ln a)時(shí), f (x)0.所以f(x)在(-,-ln a)上單調(diào)遞減,在(-ln a,+)上單調(diào)遞增.(2)(i)若a0,由(1)知, f(x)至多有一個(gè)零點(diǎn).(ii)若a0,由(1)知,當(dāng)x=-ln a時(shí), f(x)取得最小值,最小值為f(-ln a

32、)=1-+ln a.當(dāng)a=1時(shí),由于f(-ln a)=0,故f(x)只有一個(gè)零點(diǎn);當(dāng)a(1,+)時(shí),由于1-+ln a0,即f(-ln a)0,故f(x)沒(méi)有零點(diǎn);1a1a當(dāng)a(0,1)時(shí),1-+ln a0,即f(-ln a)-2e-2+20,故f(x)在(-,-ln a)有一個(gè)零點(diǎn).1a設(shè)正整數(shù)n0滿足n0ln,則f(n0)=(a+a-2)-n0-n0-n00.由于ln-ln a,因此f(x)在(-ln a,+)有一個(gè)零點(diǎn).綜上,a的取值范圍為(0,1).31a0en0en0en02n31a6.(2015重慶,19,12分)已知函數(shù)f(x)=ax3+x2(aR)在x=-處取得極值.(1)確定

33、a的值;(2)若g(x)=f(x)ex,討論g(x)的單調(diào)性.43解析解析(1)對(duì)f(x)求導(dǎo)得f (x)=3ax2+2x,因?yàn)閒(x)在x=-處取得極值,所以f =0,即3a+2=-=0,解得a=.(2)由(1)得g(x)=ex,故g(x)=ex+ex=ex=x(x+1)(x+4)ex.令g(x)=0,解得x=0或x=-1或x=-4.當(dāng)x-4時(shí),g(x)0,故g(x)為減函數(shù);當(dāng)-4x0,故g(x)為增函數(shù);434316943163a83123212xx2322xx3212xx3215222xxx12當(dāng)-1x0時(shí),g(x)0時(shí),g(x)0,故g(x)為增函數(shù).綜上知g(x)在(-,-4)和(

34、-1,0)內(nèi)為減函數(shù),在(-4,-1)和(0,+)內(nèi)為增函數(shù).評(píng)析評(píng)析 本題考查了導(dǎo)數(shù)的運(yùn)算,利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值的方法;考查了運(yùn)算求解能力及分析論證能力.考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極( (最最) )值值1.(2017課標(biāo),11,5分)若x=-2是函數(shù)f(x)=(x2+ax-1)ex-1的極值點(diǎn),則f(x)的極小值為()A.-1 B.-2e-3 C.5e-3 D.1答案答案 A本題主要考查導(dǎo)數(shù)的應(yīng)用.由題意可得f (x)=ex-1x2+(a+2)x+a-1.x=-2是函數(shù)f(x)=(x2+ax-1)ex-1的極值點(diǎn),f (-2)=0,a=-1,f(x)=(x2-x-1)e

35、x-1, f (x)=ex-1(x2+x-2)=ex-1(x-1)(x+2),x(-,-2),(1,+)時(shí), f (x)0, f(x)單調(diào)遞增;x(-2,1)時(shí), f (x)0, f(x)單調(diào)遞減.f(x)極小值=f(1)=-1.故選A.思路分析思路分析由x=-2是函數(shù)f(x)的極值點(diǎn)可知f (-2)=0,從而求出a的值,將a的值代入導(dǎo)函數(shù)f (x),求出f(x)的單調(diào)區(qū)間,判斷極小值點(diǎn),從而求出函數(shù)的極小值.2.(2016四川,6,5分)已知a為函數(shù)f(x)=x3-12x的極小值點(diǎn),則a=()A.-4 B.-2 C.4 D.2答案答案 D由題意可得f (x)=3x2-12=3(x-2)(x+

36、2),令f (x)=0,得x=-2或x=2,則f (x), f(x)隨x的變化情況如下表:x(-,-2)-2(-2,2)2(2,+)f (x)+0-0+f(x)極大值極小值函數(shù)f(x)在x=2處取得極小值,則a=2.故選D.評(píng)析評(píng)析 本題考查了函數(shù)的極值問(wèn)題.正確理解函數(shù)的極值點(diǎn)的概念是解題的關(guān)鍵.3.(2018課標(biāo),16,5分)已知函數(shù)f(x)=2sin x+sin 2x,則f(x)的最小值是 .答案答案- 3 32解析解析解法一:由f(x)=2sin x+sin 2x,得f (x)=2cos x+2cos 2x=4cos2x+2cos x-2,令f (x)=0,得cos x=或cos x=

37、-1,可得當(dāng)cos x時(shí), f (x)0, f(x)為增函數(shù),所以當(dāng)cos x=時(shí), f(x)取最小值,此時(shí)sin x=.又因?yàn)閒(x)=2sin x+2sin xcos x=2sin x(1+cos x),1+cos x0恒成立,f(x)取最小值時(shí),sin x=-,f(x)min=2=-.解法二:f(x)=2sin x+sin 2x=2sin x+2sin xcos x=2sin x(1+cos x),f 2(x)=4sin2x(1+cos x)2=4(1-cos x)(1+cos x)3.令cos x=t,t-1,1,設(shè)g(t)=4(1-t)(1+t)3,g(t)=-4(1+t)3+12(

38、1+t)2(1-t)=4(1+t)2(2-4t).當(dāng)t時(shí),g(t)0,g(t)為增函數(shù);當(dāng)t時(shí),g(t)0時(shí), f (x)0,f(x)在(0,+)上為增函數(shù),又f(0)=1,f(x)在(0,+)上沒(méi)有零點(diǎn),a0.當(dāng)0 x時(shí), f (x)時(shí), f (x)0, f(x)為增函數(shù),x0時(shí), f(x)有極小值,為f=-+1.f(x)在(0,+)內(nèi)有且只有一個(gè)零點(diǎn),f=0,a=3.f(x)=2x3-3x2+1,則f (x)=6x(x-1).令f (x)=0,得x=0或x=1.3a3a3a327a3ax-1(-1,0)0(0,1)1f (x) + - f(x)-4增1減0f(x)在-1,1上的最大值為1,

39、最小值為-4.最大值與最小值的和為-3.5.(2018課標(biāo),21,12分)已知函數(shù)f(x)=(2+x+ax2)ln(1+x)-2x.(1)若a=0,證明:當(dāng)-1x0時(shí), f(x)0時(shí), f(x)0;(2)若x=0是f(x)的極大值點(diǎn),求a.解析解析本題考查導(dǎo)數(shù)與函數(shù)的單調(diào)性、導(dǎo)數(shù)與函數(shù)的極值.(1)證明:當(dāng)a=0時(shí), f(x)=(2+x)ln(1+x)-2x, f (x)=ln(1+x)-.設(shè)函數(shù)g(x)=f (x)=ln(1+x)-,則g(x)=.當(dāng)-1x0時(shí),g(x)0時(shí),g(x)0.故當(dāng)x-1時(shí),g(x)g(0)=0,且僅當(dāng)x=0時(shí),g(x)=0,從而f (x)0,且僅當(dāng)x=0時(shí), f

40、(x)=0.所以f(x)在(-1,+)單調(diào)遞增.又f(0)=0,故當(dāng)-1x0時(shí), f(x)0時(shí), f(x)0.1xx1xx2(1)xx(2)(i)若a0,由(1)知,當(dāng)x0時(shí), f(x)(2+x)ln(1+x)-2x0=f(0),這與x=0是f(x)的極大值點(diǎn)矛盾.(ii)若a0,設(shè)函數(shù)h(x)=ln(1+x)-.2( )2f xxax222xxax由于當(dāng)|x|0,故h(x)與f(x)符號(hào)相同.又h(0)=f(0)=0,故x=0是f(x)的極大值點(diǎn)當(dāng)且僅當(dāng)x=0是h(x)的極大值點(diǎn).h(x)=-=.如果6a+10,則當(dāng)0 x-,且|x|0,故x=0不是h(x)的極大值點(diǎn).如果6a+10,則a2

41、x2+4ax+6a+1=0存在根x10,故當(dāng)x(x1,0),且|x|min時(shí),h(x)0;當(dāng)x(0,1)時(shí),h(x)0,函數(shù)g(x)單調(diào)遞增;當(dāng)a0時(shí),x時(shí),g(x)0,函數(shù)g(x)單調(diào)遞增,x時(shí),函數(shù)g(x)單調(diào)遞減.所以當(dāng)a0時(shí),g(x)的單調(diào)增區(qū)間為(0,+);當(dāng)a0時(shí),g(x)的單調(diào)增區(qū)間為,單調(diào)減區(qū)間為.(2)由(1)知, f (1)=0.當(dāng)a0時(shí), f (x)單調(diào)遞增,1x12axx10,2a1,2a10,2a1,2a所以當(dāng)x(0,1)時(shí), f (x)0, f(x)單調(diào)遞增.所以f(x)在x=1處取得極小值,不合題意.當(dāng)0a1,由(1)知f (x)在內(nèi)單調(diào)遞增,可得當(dāng)x(0,1)時(shí)

42、, f (x)0.所以f(x)在(0,1)內(nèi)單調(diào)遞減,在內(nèi)單調(diào)遞增,所以f(x)在x=1處取得極小值,不合題意.當(dāng)a=時(shí),=1, f (x)在(0,1)內(nèi)單調(diào)遞增,在(1,+)內(nèi)單調(diào)遞減,所以當(dāng)x(0,+)時(shí), f (x)0, f(x)單調(diào)遞減,不合題意.當(dāng)a時(shí),00, f(x)單調(diào)遞增,當(dāng)x(1,+)時(shí), f (x).12思路分析思路分析 (1)求出函數(shù)的導(dǎo)數(shù),對(duì)a進(jìn)行分類討論;(2)由第(1)問(wèn)知f (1)=0,對(duì)a進(jìn)行分類討論,然后利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性和極值來(lái)驗(yàn)證是否滿足條件,從而求出a的取值范圍.評(píng)析評(píng)析 本題考查導(dǎo)數(shù)的應(yīng)用,利用導(dǎo)數(shù)求函數(shù)的單調(diào)性和極值要注意“定義域優(yōu)先”原則,注

43、意對(duì)a分類討論.7.(2016課標(biāo),21,12分)(1)討論函數(shù)f(x)=ex的單調(diào)性,并證明當(dāng)x0時(shí),(x-2)ex+x+20;(2)證明:當(dāng)a0,1)時(shí),函數(shù)g(x)=(x0)有最小值.設(shè)g(x)的最小值為h(a),求函數(shù)h(a)的值域.22xx2exaxax解析解析(1)f(x)的定義域?yàn)?-,-2)(-2,+).(2分)f (x)=0,且僅當(dāng)x=0時(shí), f (x)=0,所以f(x)在(-,-2),(-2,+)單調(diào)遞增.因此當(dāng)x(0,+)時(shí), f(x)f(0)=-1.所以(x-2)ex-(x+2),(x-2)ex+x+20.(4分)(2)g(x)=(f(x)+a).(5分)由(1)知,y

44、=f(x)+a單調(diào)遞增.對(duì)任意a0,1), f(0)+a=a-10, f(2)+a=a0.因此,存在唯一xa(0,2,使得f(xa)+a=0,即g(xa)=0.(6分)當(dāng)0 xxa時(shí), f(x)+a0,g(x)xa時(shí), f(x)+a0,g(x)0,g(x)單調(diào)遞增.(7分)因此g(x)在x=xa處取得最小值,最小值為g(xa)=.(8分)2(1)(2)e(2)e(2)xxxxxx22e(2)xxx3(2)e(2)xxa xx32xx2e(1)axaaa xx2e()(1)axaaaf xxxe2axax 于是h(a)=,由=0,得y=單調(diào)遞增.所以,由xa(0,2,得=0;當(dāng)x時(shí),g(x)0,

45、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)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(-,0.0,2,20,2,222.(2019課標(biāo)文,21,12分)已知函數(shù)f(x)=(x-1)ln x-x-1.證明:(1)f(x)存在唯一的極值點(diǎn);(2)f(x

46、)=0有且僅有兩個(gè)實(shí)根,且兩個(gè)實(shí)根互為倒數(shù).解析解析本題主要考查利用導(dǎo)數(shù)研究函數(shù)的極值點(diǎn)及方程根的問(wèn)題,考查推理論證能力、運(yùn)算求解能力,考查函數(shù)與方程思想、轉(zhuǎn)化與化歸思想,考查的核心素養(yǎng)是邏輯推理及數(shù)學(xué)運(yùn)算.(1)f(x)的定義域?yàn)?0,+).f (x)=+ln x-1=ln x-.因?yàn)閥=ln x單調(diào)遞增,y=單調(diào)遞減,所以f (x)單調(diào)遞增.又f (1)=-10,故存在唯一x0(1,2),使得f (x0)=0.又當(dāng)xx0時(shí), f (x)x0時(shí), f (x)0, f(x)單調(diào)遞增.因此, f(x)存在唯一的極值點(diǎn).(2)由(1)知f(x0)0,1xx1x1x12ln4 12所以f(x)=0在

47、(x0,+)內(nèi)存在唯一根x=.由x01得1x0.又f=ln-1=0,111111( )f 故是f(x)=0在(0,x0)的唯一根.綜上, f(x)=0有且僅有兩個(gè)實(shí)根,且兩個(gè)實(shí)根互為倒數(shù).1思路分析思路分析(1)求函數(shù)極值點(diǎn)的個(gè)數(shù),實(shí)質(zhì)是求導(dǎo)函數(shù)零點(diǎn)的個(gè)數(shù),注意應(yīng)用零點(diǎn)存在性定理;(2)由第(1)問(wèn)易知方程f(x)=0在(x0,+)上存在唯一根,根據(jù)所要證明的結(jié)論,只需求出f=0即可.13.(2019課標(biāo)文,20,12分)已知函數(shù)f(x)=2x3-ax2+2.(1)討論f(x)的單調(diào)性;(2)當(dāng)0a0,則當(dāng)x(-,0)時(shí), f (x)0;當(dāng)x時(shí), f (x)0.故f(x)在(-,0),單調(diào)遞增

48、,在單調(diào)遞減;若a=0, f(x)在(-,+)單調(diào)遞增;若a0;當(dāng)x時(shí), f (x)0.3a,3a0,3a,3a0,3a,3a,03a故f(x)在,(0,+)單調(diào)遞增,在單調(diào)遞減.(2)當(dāng)0a3時(shí),由(1)知, f(x)在單調(diào)遞減,在單調(diào)遞增,所以f(x)在0,1的最小值為f=-+2,最大值為f(0)=2或f(1)=4-a.于是m=-+2,M=所以M-m=當(dāng)0a2時(shí),可知2-a+單調(diào)遞減,所以M-m的取值范圍是.當(dāng)2a0時(shí), f(x)在(-,0),單調(diào)遞增,將這兩個(gè)區(qū)間合并表示為f(x)在(-,0)單調(diào)遞增導(dǎo)致錯(cuò)誤,從而失分.,3a,3a4.(2019課標(biāo)理,20,12分)已知函數(shù)f(x)=2

49、x3-ax2+b.(1)討論f(x)的單調(diào)性;(2)是否存在a,b,使得f(x)在區(qū)間0,1的最小值為-1且最大值為1?若存在,求出a,b的所有值;若不存在,說(shuō)明理由.解析解析本題主要考查利用導(dǎo)數(shù)求函數(shù)的單調(diào)性以及求函數(shù)的最值問(wèn)題,通過(guò)導(dǎo)數(shù)的應(yīng)用考查學(xué)生的運(yùn)算求解能力以及分類討論思想,考查了數(shù)學(xué)運(yùn)算的核心素養(yǎng).(1)f (x)=6x2-2ax=2x(3x-a).令f (x)=0,得x=0或x=.若a0,則當(dāng)x(-,0)時(shí), f (x)0;當(dāng)x時(shí), f (x)0.故f(x)在(-,0),單調(diào)遞增,在單調(diào)遞減.若a=0, f(x)在(-,+)單調(diào)遞增.若a0;當(dāng)x時(shí), f (x)0.故f(x)在,

50、(0,+)單調(diào)遞增,在單調(diào)遞減.3a,3a0,3a,3a0,3a,3a,03a,3a,03a(2)滿足題設(shè)條件的a,b存在.(i)當(dāng)a0時(shí),由(1)知, f(x)在0,1單調(diào)遞增,所以f(x)在區(qū)間0,1的最小值為f(0)=b,最大值為f(1)=2-a+b.此時(shí)a,b滿足題設(shè)條件當(dāng)且僅當(dāng)b=-1,2-a+b=1,即a=0,b=-1.(ii)當(dāng)a3時(shí),由(1)知, f(x)在0,1單調(diào)遞減,所以f(x)在區(qū)間0,1的最大值為f(0)=b,最小值為f(1)=2-a+b.此時(shí)a,b滿足題設(shè)條件當(dāng)且僅當(dāng)2-a+b=-1,b=1,即a=4,b=1.(iii)當(dāng)0a3時(shí),由(1)知, f(x)在0,1的最

51、小值為f=-+b,最大值為b或2-a+b.若-+b=-1,b=1,則a=3,與0a3矛盾.若-+b=-1,2-a+b=1,則a=3或a=-3或a=0,與0a0,g0;當(dāng)x時(shí),g(x)0.所以g(x)在(-1,)單調(diào)遞增,在單調(diào)遞減,故g(x)在存在唯一極大值點(diǎn),即f(x)在存在唯一極大值點(diǎn).11x21(1) x1,221,2,2,21,21,2(2)f(x)的定義域?yàn)?-1,+).(i)當(dāng)x(-1,0時(shí),由(1)知,f(x)在(-1,0)單調(diào)遞增,而f(0)=0,所以當(dāng)x(-1,0)時(shí),f(x)0,故f(x)在(-1,0)單調(diào)遞減.又f(0)=0,從而x=0是f(x)在(-1,0的唯一零點(diǎn).(

52、ii)當(dāng)x時(shí),由(1)知,f(x)在(0,)單調(diào)遞增,在單調(diào)遞減,而f(0)=0,f0;當(dāng)x時(shí),f(x)0,所以當(dāng)x時(shí),f(x)0.從而,f(x)在沒(méi)有零點(diǎn).(iii)當(dāng)x時(shí),f(x)0,f()1,所以f(x)0,從而f(x)在(,+)沒(méi)有零點(diǎn).綜上, f(x)有且僅有2個(gè)零點(diǎn)點(diǎn).0,2,22,2,2,22120,20,2,2,22,2思路分析思路分析(1)寫(xiě)出函數(shù)f(x)的表達(dá)式,利用其導(dǎo)函數(shù)研究單調(diào)性及極值點(diǎn).(2)以x為主元進(jìn)行分類討論,分別在各個(gè)區(qū)間上,由導(dǎo)函數(shù)的單調(diào)性判斷f(x)與0的關(guān)系,得到f(x)的單調(diào)性,從而求得在各個(gè)區(qū)間的零點(diǎn)個(gè)數(shù).6.(2018課標(biāo)文,21,12分)已知函

53、數(shù)f(x)=aex-ln x-1.(1)設(shè)x=2是f(x)的極值點(diǎn),求a,并求f(x)的單調(diào)區(qū)間;(2)證明:當(dāng)a時(shí), f(x)0.1e解析解析(1)f(x)的定義域?yàn)?0,+),f (x)=aex-.由題設(shè)知, f (2)=0,所以a=.從而f(x)=ex-ln x-1, f (x)=ex-.當(dāng)0 x2時(shí), f (x)2時(shí), f (x)0.所以f(x)在(0,2)單調(diào)遞減,在(2,+)單調(diào)遞增.(2)當(dāng)a時(shí), f(x)-ln x-1.設(shè)g(x)=-ln x-1,則g(x)=-.當(dāng)0 x1時(shí),g(x)1時(shí),g(x)0.所以x=1是g(x)的最小值點(diǎn).1x212e212e212e1x1eeexe

54、exeex1x故當(dāng)x0時(shí),g(x)g(1)=0.因此,當(dāng)a時(shí), f(x)0.1e方法總結(jié)方法總結(jié)利用導(dǎo)數(shù)證明不等式的常用方法:(1)證明f(x)g(x),x(a,b),可以構(gòu)造函數(shù)F(x)=f(x)-g(x).如果F(x)0,則F(x)在(a,b)上是減函數(shù),同時(shí)若F(a)0,由減函數(shù)的定義可知,x(a,b)時(shí),有F(x)0,即證明了f(x)g(x),x(a,b),可以構(gòu)造函數(shù)F(x)=f(x)-g(x),如果F(x)0,則F(x)在(a,b)上是增函數(shù),同時(shí)若F(a)0,由增函數(shù)的定義可知,x(a,b)時(shí),有F(x)0,即證明了f(x)g(x).7.(2018課標(biāo)文,21,12分)已知函數(shù)f

55、(x)=x3-a(x2+x+1).(1)若a=3,求f(x)的單調(diào)區(qū)間;(2)證明:f(x)只有一個(gè)零點(diǎn).13解析解析(1)當(dāng)a=3時(shí), f(x)=x3-3x2-3x-3, f (x)=x2-6x-3.令f (x)=0,解得x=3-2或x=3+2.當(dāng)x(-,3-2)(3+2,+)時(shí), f (x)0;當(dāng)x(3-2,3+2)時(shí), f (x)0,所以f(x)=0等價(jià)于-3a=0.設(shè)g(x)=-3a,則g(x)=0,當(dāng)且僅當(dāng)x=0時(shí)g(x)=0,所以g(x)在(-,+)單調(diào)遞增.故g(x)至多有一個(gè)零點(diǎn),從而f(x)至多有一個(gè)零點(diǎn).又f(3a-1)=-6a2+2a-=-6-0,故f(x)有一個(gè)零點(diǎn).綜

56、上, f(x)只有一個(gè)零點(diǎn).133333333333321xxx321xxx2222(23)(1)xxxxx13216a16138.(2018課標(biāo),21,12分)已知函數(shù)f(x)=ex-ax2.(1)若a=1,證明:當(dāng)x0時(shí), f(x)1;(2)若f(x)在(0,+)只有一個(gè)零點(diǎn),求a.解析解析(1)當(dāng)a=1時(shí), f(x)1等價(jià)于(x2+1)e-x-10.設(shè)函數(shù)g(x)=(x2+1)e-x-1,則g(x)=-(x2-2x+1)e-x=-(x-1)2e-x.當(dāng)x1時(shí),g(x)0,h(x)沒(méi)有零點(diǎn);(ii)當(dāng)a0時(shí),h(x)=ax(x-2)e-x.當(dāng)x(0,2)時(shí),h(x)0.所以h(x)在(0,

57、2)單調(diào)遞減,在(2,+)單調(diào)遞增.故h(2)=1-是h(x)在0,+)的最小值.24ea若h(2)0,即a,h(x)在(0,+)沒(méi)有零點(diǎn);若h(2)=0,即a=,h(x)在(0,+)只有一個(gè)零點(diǎn);2e42e4若h(2),由于h(0)=1,所以h(x)在(0,2)有一個(gè)零點(diǎn).由(1)知,當(dāng)x0時(shí),exx2,所以h(4a)=1-=1-1-=1-0.故h(x)在(2,4a)有一個(gè)零點(diǎn).因此h(x)在(0,+)有兩個(gè)零點(diǎn).綜上, f(x)在(0,+)只有一個(gè)零點(diǎn)時(shí),a=.2e43416eaa32216(e)aa3416(2 )aa1a2e49.(2017課標(biāo),21,12分)已知函數(shù)f(x)=ax2-

58、ax-xln x,且f(x)0.(1)求a;(2)證明:f(x)存在唯一的極大值點(diǎn)x0,且e-2 f(x0)2-2.解析解析本題考查了導(dǎo)數(shù)的綜合應(yīng)用.(1)f(x)的定義域?yàn)?0,+).設(shè)g(x)=ax-a-ln x,則f(x)=xg(x), f(x)0等價(jià)于g(x)0.因?yàn)間(1)=0,g(x)0,故g(1)=0,而g(x)=a-,g(1)=a-1,得a=1.若a=1,則g(x)=1-.當(dāng)0 x1時(shí),g(x)1時(shí),g(x)0,g(x)單調(diào)遞增.所以x=1是g(x)的極小值點(diǎn),故g(x)g(1)=0.綜上,a=1.(2)由(1)知f(x)=x2-x-xln x, f (x)=2x-2-ln x

59、.設(shè)h(x)=2x-2-ln x,則h(x)=2-.1x1x1x當(dāng)x時(shí),h(x)0.10,21,2所以h(x)在單調(diào)遞減,在單調(diào)遞增.又h(e-2)0,h0;當(dāng)x(x0,1)時(shí),h(x)0.因?yàn)閒 (x)=h(x),所以x=x0是f(x)的唯一極大值點(diǎn).由f (x0)=0得ln x0=2(x0-1),故f(x0)=x0(1-x0).由x0得f(x0)f(e-1)=e-2,所以e-2f(x0)2-2.10,21,21210,21,210,214方法總結(jié)方法總結(jié)利用導(dǎo)數(shù)解決不等式問(wèn)題的一般思路:(1)恒成立問(wèn)題常利用分離參數(shù)法轉(zhuǎn)化為最值問(wèn)題求解.若不能分離參數(shù),可以對(duì)參數(shù)進(jìn)行分類討論.(2)證明不

60、等式問(wèn)題可通過(guò)構(gòu)造函數(shù)轉(zhuǎn)化為函數(shù)的最值問(wèn)題求解.10.(2016課標(biāo)文,21,12分)已知函數(shù)f(x)=(x-2)ex+a(x-1)2.(1)討論f(x)的單調(diào)性;(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.解析(1)f (x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).(i)設(shè)a0,則當(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.(2分)(ii)設(shè)a-,則ln(-2a)0;當(dāng)x(ln(-2a),1)時(shí), f (x)0.所以f(x)在(-,ln(-2a),(1,+)單調(diào)遞增,在(ln(-2a),1)單調(diào)遞減.(4分)若a1,故當(dāng)x(-