五年高考3年模擬§32-導(dǎo)數(shù)的應(yīng)用課件

1、3.2導(dǎo)數(shù)的應(yīng)用高考理數(shù) (課標(biāo)專用)3.2導(dǎo)數(shù)的應(yīng)用高考理數(shù) (課標(biāo)專用)A組統(tǒng)一命題課標(biāo)卷題組考點(diǎn)一導(dǎo)數(shù)與函數(shù)的單調(diào)性1.(2015課標(biāo),12,5分,0.369)設(shè)函數(shù)f (x)是奇函數(shù)f(x)(xR)的導(dǎo)函數(shù), f(-1)=0,當(dāng)x0時(shí),xf (x)-f(x)0成立的x的取值范圍是()A.(-,-1)(0,1)B.(-1,0)(1,+)C.(-,-1)(-1,0)D.(0,1)(1,+)五年高考A組統(tǒng)一命題課標(biāo)卷題組五年高考答案A令g(x)=,則g(x)=,由題意知,當(dāng)x0時(shí),g(x)0,從而f(x)0;當(dāng)x(1,+)時(shí),g(x)0,從而f(x)0.又g(-x)=g(x),g(x)是偶函

2、數(shù),當(dāng)x(-,-1)時(shí),g(x)0;當(dāng)x(-1,0)時(shí),g(x)0,從而f(x)0(0(0)時(shí),考慮構(gòu)造函數(shù)g(x)=,再利用所構(gòu)造的函數(shù)的單調(diào)性求解.答案A令g(x)=,則g(x)=,由題意知,2.(2018課標(biāo)全國(guó),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,單調(diào)遞減,在單調(diào)遞增.(2)由(1)知, f(x)存在兩個(gè)極值點(diǎn)當(dāng)且僅當(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

3、+a=-2+a=-2+a,解析(1)f(x)的定義域?yàn)?0,+), f (x)=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,+)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.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)時(shí),可以構(gòu)造函數(shù)F(x)=f(x)-g(x).若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)時(shí),可以構(gòu)造函數(shù)F(x)=f(x)-g

4、(x),若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).所以a-2等價(jià)于-x2+2ln x20.方法總結(jié)利3.(2017課標(biāo)全國(guó),21,12分)已知函數(shù)f(x)=ae2x+(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.3.(2017課標(biāo)全國(guó),21,12分)已知函數(shù)f(x)=a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

5、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)單調(diào)遞減,在(-ln a,+)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)=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);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).解析本題

6、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.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).方法總結(jié)利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性:若f (x)0,xD恒成立,則在區(qū)間D上函數(shù)f(x)單調(diào)遞增;若f (x)ln,方法總結(jié)利用導(dǎo)數(shù)研究函數(shù)的單解析(1)f (x)=m(emx-1)+2x.若m0,則當(dāng)x(-,0)時(shí),emx-10, f (x)0.若m0, f (x)0;當(dāng)x(0,+)時(shí),emx-10.所以, f(x)在(-,0)單調(diào)遞減,在(0,+)單調(diào)遞增.(2)由(1)知,

7、對(duì)任意的m, f(x)在-1,0單調(diào)遞減,在0,1單調(diào)遞增,故f(x)在x=0處取得最小值.所以對(duì)于任意x1,x2-1,1,|f(x1)-f(x2)|e-1的充要條件是即設(shè)函數(shù)g(t)=et-t-e+1,則g(t)=et-1.當(dāng)t0時(shí),g(t)0時(shí),g(t)0.故g(t)在(-,0)單調(diào)遞減,在(0,+)單調(diào)遞增.又g(1)=0,g(-1)=e-1+2-e1時(shí),由g(t)的單調(diào)性知,g(m)0,即em-me-1;當(dāng)m0,即e-m+me-1.綜上,m的取值范圍是-1,1.思路分析(1)證明f(x)在(-,0)上單調(diào)遞減,在(0,+)上單調(diào)遞增,即證明無(wú)論m取何值,均有f (x)0在(-,0)上恒

8、成立,而f (x)=m(emx-1)+2x,故需對(duì)m的符號(hào)進(jìn)行分類討論.(2)x1,x2-1,1,均有|f(x1)-f(x2)|e-1,即f(x)max-f(x)mine-1,可由第(1)問(wèn)的單調(diào)性來(lái)確定f(x)max與f(x)min.當(dāng)m-1,1時(shí),g(m)0,g(-m)0,即式成5.(2016課標(biāo)全國(guó),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.5.(2016課標(biāo)全國(guó),21,12分)解析(1)f(x)的定義域?yàn)?-,-2)(-2,

9、+).(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, 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

10、g(xa)=.(8分)解析(1)f(x)的定義域?yàn)?-,-2)(-2,+)于是h(a)=,由=0,得y=單調(diào)遞增.所以,由xa(0,2,得=0,得y=單調(diào)遞增.解題關(guān)鍵考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極值(最值)1.(2017課標(biāo)全國(guó),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.-1B.-2e-3C.5e-3D.1答案A本題主要考查函數(shù)的極值與導(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)ex-1, f (x)=e

11、x-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.考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極值(最值)答案A本題主要考查2.已知函數(shù)極值點(diǎn)和極值求參數(shù)值的兩個(gè)要領(lǐng):(1)列式:根據(jù)極值點(diǎn)處導(dǎo)數(shù)為0和極值列方程組進(jìn)行求解.(2)驗(yàn)證:因?yàn)閷?dǎo)數(shù)為零的點(diǎn)不一定是函數(shù)的極值點(diǎn),所以求解后必須進(jìn)行驗(yàn)

12、證.方法總結(jié)1.利用導(dǎo)數(shù)研究函數(shù)極值問(wèn)題的兩個(gè)方向: 2.已知函數(shù)極值點(diǎn)和極值求參數(shù)值的兩個(gè)要領(lǐng):方法總結(jié)1.利2.(2014課標(biāo),12,5分)設(shè)函數(shù)f(x)=sin.若存在f(x)的極值點(diǎn)x0滿足+f(x0)2m2,則m的取值范圍是()A.(-,-6)(6,+)B.(-,-4)(4,+)C.(-,-2)(2,+)D.(-,-1)(1,+)2.(2014課標(biāo),12,5分)設(shè)函數(shù)f(x)=sin答案Cf (x)=cos,f(x)的極值點(diǎn)為x0,f (x0)=0,cos=0,x0=k+,kZ,x0=mk+,kZ,又+f(x0)2m2,+m2,kZ,即m2+3m2,kZ,m0,kZ,又存在x0滿足+

13、f(x0)2,m2-3,m24,m2或m-2,故選C.評(píng)析本題考查了函數(shù)的極值問(wèn)題,三角函數(shù)求值、恒成立等問(wèn)題.考查分析問(wèn)題、解決問(wèn)題的能力.答案Cf (x)=cos,評(píng)析本題考解析解法一: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=-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

14、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(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,g(t)為減函數(shù).當(dāng)t=時(shí),g(t)取得最大值,即f 2(x)的最大值為,得f(x)的最大值為,3.(2018課標(biāo)全國(guó),16,5分)已知函

15、數(shù)f(x)=2sin x+sin 2x,則f(x)的最小值是.答案- 解析解法一:由f(x)=2sin x+sin 2x,得f 又f(x)=2sin x+sin 2x為奇函數(shù),f(x)的最小值為-.解法三:f(x)=2sin x+sin 2x=2sin x(1+cos x)=8sincos3.f 2(x)=64sin2cos2cos2cos2=3sin2cos2cos2cos2=.當(dāng)且僅當(dāng)3sin2=cos2,即sin2=,cos2=時(shí)等號(hào)成立,所以f 2(x)的最大值為,則f(x)的最大值為,又f(x)=2sin x+sin 2x為奇函數(shù),f(x)的最小值為-.又f(x)=2sin x+si

16、n 2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)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 (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.(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.(

17、ii)若a0,設(shè)函數(shù)h(x)=ln(1+x)-.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).4.(2018課標(biāo)全國(guó),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.4.(2h(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a2x2+4ax+6a+1=0存在根x10,故

18、當(dāng)x(x1,0),且|x|min時(shí),h(x)0;當(dāng)x(0,1)時(shí),h(x)0.所以x=0是h(x)的極大值點(diǎn),從而x=0是f(x)的極大值點(diǎn).綜上,a=-.思路分析(1)a=0時(shí),寫(xiě)出f(x)的解析式,對(duì)f(x)求導(dǎo).易得f(0)=0,結(jié)合單調(diào)性可將問(wèn)題解決.(2)對(duì)a進(jìn)行分類討論,分析各類情況下的極大值點(diǎn),進(jìn)而得參數(shù)a的值.h(x)=-=.思路分析(1)a=0時(shí),寫(xiě)出f(x解后反思1.利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,大多數(shù)情況下歸結(jié)為對(duì)含有參數(shù)的一元二次不等式的解集的情況的討論,在能夠通過(guò)因式分解求出不等式對(duì)應(yīng)方程的根時(shí),依據(jù)根的大小進(jìn)行分類討論;在不能通過(guò)因式分解求出根的情況下,根據(jù)不等式對(duì)應(yīng)方

19、程的判別式進(jìn)行分類討論,討論函數(shù)的單調(diào)性是在函數(shù)的定義域內(nèi)進(jìn)行的.易錯(cuò)警示容易忽略函數(shù)定義域.函數(shù)解析式中含有對(duì)數(shù)型的式子,則其真數(shù)部分應(yīng)大于零.2.利用導(dǎo)數(shù)研究出函數(shù)的單調(diào)性和極值后,可以畫(huà)出草圖,進(jìn)行觀察分析,研究滿足條件的參數(shù)值或范圍.解后反思1.利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,大多數(shù)情況下歸結(jié)為對(duì)5.(2017課標(biāo)全國(guó),21,12分)已知函數(shù)f(x)=ax2-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)2-2.5.(2017課標(biāo)全國(guó),21,12分)已知函數(shù)f(x)=a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,

20、+).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.設(shè)h(x)=2x-2-ln x,則h(x)=2-.當(dāng)x時(shí),h(x)0.解析本題考查了導(dǎo)數(shù)的綜合應(yīng)用.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)

21、,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(0,1)得f(x0)f(e-1)=e-2,所以e-2f(x0)2-2.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.所以h(x)在單調(diào)遞減,在單調(diào)遞增.方法總結(jié)利用導(dǎo)數(shù)解考點(diǎn)三導(dǎo)數(shù)的綜合應(yīng)用1.(2015課標(biāo),12,5分,0.317)設(shè)函數(shù)f(x)=ex(2x-1)-ax+a,其中a1

22、,若存在唯一的整數(shù)x0使得f(x0)0,則a的取值范圍是()A.B.C.D. 考點(diǎn)三導(dǎo)數(shù)的綜合應(yīng)用答案D由f(x0)0,即(2x0-1)-a(x0-1)0得(2x0-1)a(x0-1).當(dāng)x0=1時(shí),得e1,則a.令g(x)=,則g(x)=.當(dāng)x時(shí),g(x)0,g(x)為增函數(shù),要滿足題意,則x0=2,此時(shí)需滿足g(2)ag(3),得3e2ae3,與a1矛盾,所以x01.因?yàn)閤01,所以a.答案D由f(x0)0,g(x)為增函數(shù),當(dāng)x(0,1)時(shí),g(x)0,g(x)為減函數(shù),要滿足題意,則x0=0,此時(shí)需滿足g(-1)ag(0),思路分析存在唯一的整數(shù)x0使得f(x0)0,即存在唯一整數(shù)x0

23、使(2x0-1)1,x01三種情況討論.評(píng)析本題主要考查導(dǎo)數(shù)的應(yīng)用及分類討論思想,分離參變量是解決本題的關(guān)鍵,本題綜合性較強(qiáng),屬難題.得a1(滿足a0,g(x)為增函數(shù)2.(2018課標(biāo)全國(guó),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.2.(2018課標(biāo)全國(guó),21,12分)已知函數(shù)f(x)=e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);(i

24、i)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(0,2)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.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);解析(1)當(dāng)a=1時(shí), f(x)1等價(jià)于(x2+1)e-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=

25、.方法總結(jié)利用導(dǎo)數(shù)研究不等式恒成立問(wèn)題,可以先構(gòu)造函數(shù),然后對(duì)構(gòu)造的新函數(shù)求導(dǎo),利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,求出最值,進(jìn)而得出相應(yīng)的含參不等式,從而求出參數(shù)的取值范圍;也可以先分離變量,再構(gòu)造函數(shù),直接把問(wèn)題轉(zhuǎn)化為函數(shù)的最值問(wèn)題.研究函數(shù)的零點(diǎn)個(gè)數(shù)問(wèn)題,可以通過(guò)導(dǎo)數(shù)研究函數(shù)的單調(diào)性、最值等.具體地,可畫(huà)出函數(shù)圖象,根據(jù)函數(shù)圖象的走勢(shì)規(guī)律,標(biāo)出函數(shù)極值點(diǎn)、最值點(diǎn)的位置求解.這種用數(shù)形結(jié)合思想分析問(wèn)題的方法,可以使問(wèn)題有一個(gè)清晰、直觀的整體展現(xiàn).若h(2),由于h(0)=1,方法總結(jié)利用解析本題考查導(dǎo)數(shù)的綜合應(yīng)用.(1)f(x)的定義域?yàn)?0,+).若a0,因?yàn)閒=-+aln 20,由f (x)=1

26、-=知,當(dāng)x(0,a)時(shí), f (x)0.所以f(x)在(0,a)單調(diào)遞減,在(a,+)單調(diào)遞增.故x=a是f(x)在(0,+)的唯一最小值點(diǎn).由于f(1)=0,所以當(dāng)且僅當(dāng)a=1時(shí), f(x)0.故a=1.(2)由(1)知當(dāng)x(1,+)時(shí),x-1-ln x0.令x=1+,得ln.從而ln+ln+ln+=1-1.故2,所以m的最小值為3.3.(2017課標(biāo)全國(guó),21,12分)已知函數(shù)f(x)=x-1-aln x.(1)若f(x)0,求a的值;(2)設(shè)m為整數(shù),且對(duì)于任意正整數(shù)n,1時(shí),x-1-ln x0.令x=1+,換元后可求出的范圍.一題多解(1)f (x)=1-=(x0).當(dāng)a0時(shí), f

27、(x)0,而f(1)=0,不合題意,a0,f(x)在(0,a)上單調(diào)遞減,在(a,+)上單調(diào)遞增.又f(x)0,f(a)0,即a-1-aln a0,記h(x)=x-1-xln x,則h(x)=1-ln x-1=-ln x.h(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減,h(x)h(1)=0,即當(dāng)且僅當(dāng)x=1時(shí),h(x)0,當(dāng)且僅當(dāng)a=1時(shí),式成立.a=1.思路分析(1)對(duì)a分類討論,并利用導(dǎo)數(shù)研究f(x)的單調(diào)性解析(1)f (x)=ex+e-x-20,等號(hào)僅當(dāng)x=0時(shí)成立.所以f(x)在(-,+)上單調(diào)遞增.(2)g(x)=f(2x)-4bf(x)=e2x-e-2x-4b(ex-e-

28、x)+(8b-4)x,g(x)=2e2x+e-2x-2b(ex+e-x)+(4b-2)=2(ex+e-x-2)(ex+e-x-2b+2).(i)當(dāng)b2時(shí),g(x)0,等號(hào)僅當(dāng)x=0時(shí)成立,所以g(x)在(-,+)上單調(diào)遞增.而g(0)=0,所以對(duì)任意x0,g(x)0.(ii)當(dāng)b2時(shí),若x滿足2ex+e-x2b-2,即0 xln(b-1+)時(shí),g(x)0.而g(0)=0,因此當(dāng)0 xln(b-1+)時(shí),g(x)0,ln 20.692 8;4.(2014課標(biāo),21,12分,0.151)已知函數(shù)f(x)=ex-e-x-2x.(1)討論f(x)的單調(diào)性;(2)設(shè)g(x)=f(2x)-4bf(x),當(dāng)

29、x0時(shí),g(x)0,求b的最大值;(3)已知1.414 21.414 3,估計(jì)ln 2的近似值(精確到0.001).解析(1)f (x)=ex+e-x-20,等號(hào)僅當(dāng)x=當(dāng)b=+1時(shí),ln(b-1+)=ln,g(ln)=-2+(3+2)ln 20,ln 21.5.(2014課標(biāo),21,12分,0.244)設(shè)函數(shù)f(x解析(1)函數(shù)f(x)的定義域?yàn)?0,+), f (x)=aexln x+ex-ex-1+ex-1.由題意可得f(1)=2, f (1)=e.故a=1,b=2.(2)證明:由(1)知, f(x)=exln x+ex-1,從而f(x)1等價(jià)于xln xxe-x-.設(shè)函數(shù)g(x)=xl

30、n x,則g(x)=1+ln x.所以當(dāng)x時(shí),g(x)0.故g(x)在上單調(diào)遞減,在上單調(diào)遞增,從而g(x)在(0,+)上的最小值為g=-.設(shè)函數(shù)h(x)=xe-x-,則h(x)=e-x(1-x).所以當(dāng)x(0,1)時(shí),h(x)0;當(dāng)x(1,+)時(shí),h(x)0時(shí),g(x)h(x),即f(x)1.解析(1)函數(shù)f(x)的定義域?yàn)?0,+), f (x思路分析(1)由可求得a,b.(2)f(x)1(x0),即exln x+ex-11(x0)exxln x+exx(x0)xln x-(x0),可分別研究函數(shù)g(x)=xln x與h(x)=-的最值.評(píng)析本題主要考查導(dǎo)數(shù)的幾何意義、利用導(dǎo)數(shù)研究函數(shù)的單

31、調(diào)性及最值問(wèn)題,考查等價(jià)轉(zhuǎn)化思想及邏輯推理能力.思路分析(1)由可求得a,b.評(píng)析本題主要考查導(dǎo)數(shù)的幾6.(2015課標(biāo),21,12分,0.192)已知函數(shù)f(x)=x3+ax+,g(x)=-ln x.(1)當(dāng)a為何值時(shí),x軸為曲線y=f(x)的切線?(2)用minm,n表示m,n中的最小值,設(shè)函數(shù)h(x)=minf(x),g(x)(x0),討論h(x)零點(diǎn)的個(gè)數(shù).6.(2015課標(biāo),21,12分,0.192)已知函數(shù)f(解析(1)設(shè)曲線y=f(x)與x軸相切于點(diǎn)(x0,0),則f(x0)=0, f (x0)=0,即解得x0=,a=-.因此,當(dāng)a=-時(shí),x軸為曲線y=f(x)的切線.(5分)(

32、2)當(dāng)x(1,+)時(shí),g(x)=-ln x0,從而h(x)=minf(x),g(x)g(x)0,故h(x)在(1,+)無(wú)零點(diǎn).當(dāng)x=1時(shí),若a-,則f(1)=a+0,h(1)=minf(1),g(1)=g(1)=0,故x=1是h(x)的零點(diǎn);若a-,則f(1)0,h(1)=minf(1),g(1)=f(1)0,所以只需考慮f(x)在(0,1)的零點(diǎn)個(gè)數(shù).(i)若a-3或a0,則f (x)=3x2+a在(0,1)無(wú)零點(diǎn),故f(x)在(0,1)單調(diào).而f(0)=, f(1)=a+,所以當(dāng)a-3時(shí), f(x)在(0,1)有一個(gè)零點(diǎn);當(dāng)a0時(shí), f(x)在(0,1)沒(méi)有零點(diǎn).(ii)若-3a0,即-a

33、0,則f(x)在(0,1)無(wú)零點(diǎn);若f =0,即a=-,則f(x)在(0,1)有唯一零點(diǎn);若f 0,即-3a-,由于f(0)=, f(1)=a+,所以當(dāng)-a-時(shí), f(x)在(0,1)有兩個(gè)零點(diǎn);當(dāng)-3-或a-時(shí),h(x)有一個(gè)零點(diǎn);當(dāng)a=-或a=- 時(shí),h(x)有兩個(gè)零點(diǎn);當(dāng)- a- 時(shí),h(x)有三個(gè)零點(diǎn).(12分)得最小值,最小值為f =+.考點(diǎn)一導(dǎo)數(shù)與函數(shù)的單調(diào)性1.(2014遼寧,11,5分)當(dāng)x-2,1時(shí),不等式ax3-x2+4x+30恒成立,則實(shí)數(shù)a的取值范圍是()A.-5,-3B.C.-6,-2D.-4,-3B組自主命題省(區(qū)、市)卷題組答案C由題意知x-2,1都有ax3-x2

34、+4x+30,即ax3x2-4x-3在x-2,1上恒成立.當(dāng)x=0時(shí),aR.當(dāng)0 x1時(shí),a=-+.令t=(t1),g(t)=-3t3-4t2+t,因?yàn)間(t)=-9t2-8t+10(t1),所以g(t)在1,+)上單調(diào)遞減,g(t)max=g(1)=-6(t1),所以a-6.當(dāng)-2xk1,則下列結(jié)論中一定錯(cuò)誤的是()A.fC.f 答案C構(gòu)造函數(shù)g(x)=f(x)-kx+1,則g(x)=f (x)-k0,g(x)在R上為增函數(shù).k1,0,則gg(0).而g(0)=f(0)+1=0,g=f-+10,即f-1=,所以選項(xiàng)C錯(cuò)誤,故選C.2.(2015福建,10,5分)若定義在R上的函數(shù)f(x)滿3

35、.(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.答案 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,解得-1a.方法小結(jié)函數(shù)不等式的求解思路

36、:(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).3.(2017江蘇,11,5分)已知函數(shù)f(x)=x3-2x4.(2015北京,18,13分)已知函數(shù)f(x)=ln.(1)求曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程;(2)求證:當(dāng)x(0,1)時(shí), f(x)2;(3)設(shè)實(shí)數(shù)k使得f(x)k對(duì)x(0,1)恒成立,求k的最大值.4.(2015北京,18,13分)已知函數(shù)f(x)=ln.解析(1)因?yàn)閒(x)=ln(1+x)-ln(1-x),所以f (x)=+, f (0)=2.又因?yàn)閒(0)=0,所以曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程為

37、y=2x.(2)證明:令g(x)=f(x)-2,則g(x)=f (x)-2(1+x2)=.因?yàn)間(x)0(0 xg(0)=0,x(0,1),即當(dāng)x(0,1)時(shí), f(x)2.(3)由(2)知,當(dāng)k2時(shí), f(x)k對(duì)x(0,1)恒成立.當(dāng)k2時(shí),令h(x)=f(x)-k,則h(x)=f (x)-k(1+x2)=.解析(1)因?yàn)閒(x)=ln(1+x)-ln(1-x),所以當(dāng)0 x時(shí),h(x)0,因此h(x)在區(qū)間上單調(diào)遞減.當(dāng)0 x時(shí),h(x)h(0)=0,即f(x)2時(shí), f(x)k并非對(duì)x(0,1)恒成立.綜上可知,k的最大值為2.所以當(dāng)0 x時(shí),h(x)-e1-x在區(qū)間(1,+)內(nèi)恒成立

38、(e=2.718為自然對(duì)數(shù)的底數(shù)).5.(2016四川,21,14分)設(shè)函數(shù)f(x)=ax2-a解析(1)f (x)=2ax-=(x0).當(dāng)a0時(shí), f (x)0時(shí),由f (x)=0,有x=.此時(shí),當(dāng)x時(shí), f (x)0, f(x)單調(diào)遞增.(2)令g(x)=-,s(x)=ex-1-x.則s(x)=ex-1-1.而當(dāng)x1時(shí),s(x)0,所以s(x)在區(qū)間(1,+)內(nèi)單調(diào)遞增.又由s(1)=0,有s(x)0,從而當(dāng)x1時(shí),g(x)0.當(dāng)a0,x1時(shí), f(x)=a(x2-1)-ln xg(x)在區(qū)間(1,+)內(nèi)恒成立時(shí),必有a0.當(dāng)0a1.解析(1)f (x)=2ax-=(x0).由(1)有f0

39、,所以此時(shí)f(x)g(x)在區(qū)間(1,+)內(nèi)不恒成立.當(dāng)a時(shí),令h(x)=f(x)-g(x)(x1).當(dāng)x1時(shí),h(x)=2ax-+-e1-xx-+-=0.因此,h(x)在區(qū)間(1,+)內(nèi)單調(diào)遞增.又因?yàn)閔(1)=0,所以當(dāng)x1時(shí),h(x)=f(x)-g(x)0,即f(x)g(x)恒成立.綜上,a.思路分析對(duì)于(1),先求導(dǎo),然后分a0和a0兩種情況判斷f (x)的符號(hào),從而確定f(x)的單調(diào)性;對(duì)于(2),令g(x)=-,s(x)=ex-1-x,則s(x)與g(x)在(1,+)上正負(fù)一致,易證x1時(shí)s(x)0,從而g(x)0,再對(duì)a進(jìn)行分類:a0;0ag(x)是否恒成立,最后得出結(jié)論.評(píng)析本

40、題主要考查導(dǎo)數(shù)的應(yīng)用,利用導(dǎo)數(shù)判斷函數(shù)的單調(diào)性,并由此確定函數(shù)的最值,也考查了分類討論思想和轉(zhuǎn)化與化歸思想,將疑難問(wèn)題進(jìn)行轉(zhuǎn)化,化繁為簡(jiǎn).由(1)有f0,思路分析對(duì)于(1考點(diǎn)二導(dǎo)數(shù)與函數(shù)的極值(最值)1.(2014陜西,10,5分)如圖,某飛行器在4千米高空水平飛行,從距著陸點(diǎn)A的水平距離10千米處開(kāi)始下降,已知下降飛行軌跡為某三次函數(shù)圖象的一部分,則該函數(shù)的解析式為()A.y=x3-xB.y=x3-xC.y=x3-xD.y=-x3+x答案A根據(jù)題意知,所求函數(shù)在(-5,5)上單調(diào)遞減.對(duì)于A,y=x3-x,y=x2-=(x2-25),x(-5,5),y0時(shí), f (x)0,f(x)在(0,+

41、)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=3.2.(2018江蘇,11,5分)若函數(shù)f(x)=2x3-axx-1(-1,0)0(0,1)1f (x) + - f(x)-4增1減0f(x)=2x3-3x2+1,則f (x)=6x(x-1).f(x)在-1,1上的最大值為1,最小值為-4.最大值與最小值的和為-3.x-1(-1,0)0(0,1)1f (x) + - f(x3.(2016北京,14,5分)設(shè)函數(shù)f(x)

42、=若a=0,則f(x)的最大值為;若f(x)無(wú)最大值,則實(shí)數(shù)a的取值范圍是.答案2;(-,-1)3.(2016北京,14,5分)設(shè)函數(shù)f(x)=答案2解析若a=0,則f(x)=當(dāng)x0時(shí), f(x)=-2x0;當(dāng)x0時(shí), f (x)=3x2-3=3(x-1)(x+1),當(dāng)x0, f(x)是增函數(shù),當(dāng)-1x0時(shí), f (x)0, f(x)是減函數(shù),f(x)f(-1)=2.f(x)的最大值為2.在同一平面直角坐標(biāo)系中畫(huà)出y=-2x和y=x3-3x的圖象,如圖所示,當(dāng)a2時(shí), f(x)max=a3-3a.綜上,當(dāng)a(-,-1)時(shí), f(x)無(wú)最大值. 疑難突破分段函數(shù)需要分段討論.對(duì)于,由于x0時(shí),

43、f(x)0時(shí), f(x)=-24.(2017北京,19,13分)已知函數(shù)f(x)=excos x-x.(1)求曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程;(2)求函數(shù)f(x)在區(qū)間上的最大值和最小值.4.(2017北京,19,13分)已知函數(shù)f(x)=exco解析本題考查導(dǎo)數(shù)的幾何意義,考查利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、最值.(1)因?yàn)閒(x)=excos x-x,所以f (x)=ex(cos x-sin x)-1, f (0)=0.又因?yàn)閒(0)=1,所以曲線y=f(x)在點(diǎn)(0, f(0)處的切線方程為y=1.(2)設(shè)h(x)=ex(cos x-sin x)-1,則h(x)=ex(cos

44、 x-sin x-sin x-cos x)=-2exsin x.當(dāng)x時(shí),h(x)0,所以h(x)在區(qū)間上單調(diào)遞減.所以對(duì)任意x有h(x)h(0)=0,即f (x)0,得f(x)在定義域內(nèi)的單調(diào)遞增區(qū)間,令f (x)0時(shí),m(x)0;當(dāng)x0時(shí),m(x)0,當(dāng)x0時(shí),h(x)0時(shí),h(x)0,h(x)單調(diào)遞增,解析本題考查導(dǎo)數(shù)的幾何意義和極值.所以當(dāng)x=0時(shí)h(x)取到極小值,極小值是h(0)=-2a-1;(ii)當(dāng)a0時(shí),h(x)=2(ex-eln a)(x-sin x),由h(x)=0得x1=ln a,x2=0.當(dāng)0a1時(shí),ln a0,當(dāng)x(-,ln a)時(shí),ex-eln a0,h(x)單調(diào)遞

45、增;當(dāng)x(ln a,0)時(shí),ex-eln a0,h(x)0,h(x)0,h(x)單調(diào)遞增.所以當(dāng)x=ln a時(shí)h(x)取到極大值,極大值為h(ln a)=-a(ln a)2-2ln a+sin(ln a)+cos(ln a)+2,當(dāng)x=0時(shí)h(x)取到極小值,極小值是h(0)=-2a-1;當(dāng)a=1時(shí),ln a=0,所以當(dāng)x(-,+)時(shí),h(x)0,函數(shù)h(x)在(-,+)上單調(diào)遞增,無(wú)極值;當(dāng)a1時(shí),ln a0,所以當(dāng)x(-,0)時(shí),ex-eln a0,h(x)單調(diào)遞增;所以當(dāng)x=0時(shí)h(x)取到極小值,極小值是h(0)=-2a-當(dāng)x(0,ln a)時(shí),ex-eln a0,h(x)0,h(x)

46、0,h(x)單調(diào)遞增.所以當(dāng)x=0時(shí)h(x)取到極大值,極大值是h(0)=-2a-1;當(dāng)x=ln a時(shí)h(x)取到極小值,極小值是h(ln a)=-a(ln a)2-2ln a+sin(ln a)+cos(ln a)+2.綜上所述:當(dāng)a0時(shí),h(x)在(-,0)上單調(diào)遞減,在(0,+)上單調(diào)遞增,函數(shù)h(x)有極小值,極小值是h(0)=-2a-1;當(dāng)0a1時(shí),函數(shù)h(x)在(-,0)和(ln a,+)上單調(diào)遞增,在(0,ln a)上單調(diào)遞減,函數(shù)h(x)有極大值,也有極小值,極大值是h(0)=-2a-1,極小值是h(ln a)=-a(ln a)2-2ln a+sin(ln a)+cos(ln

47、a)+2.當(dāng)x(0,ln a)時(shí),ex-eln a0,h(x)0.(1)討論f(x)在其定義域上的單調(diào)性;(2)當(dāng)x0,1時(shí),求f(x)取得最大值和最小值時(shí)的x的值.解析(1)f(x)的定義域?yàn)?-,+), f (x)=1+a-2x-3x2.令f (x)=0,得x1=,x2=,x1x2,所以f (x)=-3(x-x1)(x-x2).當(dāng)xx2時(shí), f (x)0;當(dāng)x1x0.故f(x)在(-,x1)和(x2,+)內(nèi)單調(diào)遞減,在(x1,x2)內(nèi)單調(diào)遞增.(2)因?yàn)閍0,所以x10.當(dāng)a4時(shí),x21.由(1)知, f(x)在0,1上單調(diào)遞增.所以f(x)在x=0和x=1處分別取得最小值和最大值.當(dāng)0a

48、4時(shí),x21.由(1)知, f(x)在0,x2上單調(diào)遞增,在x2,1上單調(diào)遞減.所以f(x)在x=x2=處取得最大值.6.(2014安徽,18,12分)設(shè)函數(shù)f(x)=1+(1+又f(0)=1, f(1)=a,所以當(dāng)0a1時(shí), f(x)在x=1處取得最小值;當(dāng)a=1時(shí), f(x)在x=0處和x=1處同時(shí)取得最小值;當(dāng)1a,則當(dāng)x時(shí), f (x)0.所以f(x)在x=2處取得極小值0.若a,則當(dāng)x(0,2)時(shí),x-20,ax-1x-10,所以2不是f(x)的極小值點(diǎn).綜上可知,a的取值范圍是.考點(diǎn)三導(dǎo)數(shù)的綜合應(yīng)用解析(1)因?yàn)閒(x)=ax2-(方法總結(jié)函數(shù)極值問(wèn)題的常見(jiàn)類型及解題策略(1)已知

49、導(dǎo)函數(shù)圖象判斷函數(shù)極值的情況.先找導(dǎo)數(shù)為0的點(diǎn),再判斷導(dǎo)數(shù)為0的點(diǎn)的左、右兩側(cè)導(dǎo)數(shù)的符號(hào).(2)已知函數(shù)求極值.求f (x)求方程f (x)=0的根列表檢驗(yàn)f (x)在f (x)=0的根的附近兩側(cè)的符號(hào)下結(jié)論.(3)已知極值求參數(shù).若函數(shù)f(x)在點(diǎn)(x0,y0)處取得極值,則f (x0)=0,且在該點(diǎn)左、右兩側(cè)導(dǎo)數(shù)值的符號(hào)相反.方法總結(jié)函數(shù)極值問(wèn)題的常見(jiàn)類型及解題策略2.(2018江蘇,17,14分)某農(nóng)場(chǎng)有一塊農(nóng)田,如圖所示,它的邊界由圓O的一段圓弧MPN(P為此圓弧的中點(diǎn))和線段MN構(gòu)成.已知圓O的半徑為40米,點(diǎn)P到MN的距離為50米.現(xiàn)規(guī)劃在此農(nóng)田上修建兩個(gè)溫室大棚,大棚內(nèi)的地塊形狀

50、為矩形ABCD,大棚內(nèi)的地塊形狀為CDP,要求A,B均在線段MN上,C,D均在圓弧上.設(shè)OC與MN所成的角為.(1)用分別表示矩形ABCD和CDP的面積,并確定sin 的取值范圍;(2)若大棚內(nèi)種植甲種蔬菜,大棚內(nèi)種植乙種蔬菜,且甲、乙兩種蔬菜的單位面積年產(chǎn)值之比為43.求當(dāng)為何值時(shí),能使甲、乙兩種蔬菜的年總產(chǎn)值最大. 2.(2018江蘇,17,14分)某農(nóng)場(chǎng)有一塊農(nóng)田,如圖所示解析本題主要考查三角函數(shù)的應(yīng)用、用導(dǎo)數(shù)求最值等基礎(chǔ)知識(shí),考查直觀想象和數(shù)學(xué)建模及運(yùn)用數(shù)學(xué)知識(shí)分析和解決實(shí)際問(wèn)題的能力.(1)設(shè)PO的延長(zhǎng)線交MN于H,則PHMN,所以O(shè)H=10米.過(guò)O作OEBC于E,則OEMN,所以C

51、OE=,故OE=40cos 米,EC=40sin 米,則矩形ABCD的面積為240cos (40sin +10)=800(4sin cos +cos )平方米,CDP的面積為240cos (40-40sin )=1 600(cos -sin cos )平方米.解析本題主要考查三角函數(shù)的應(yīng)用、用導(dǎo)數(shù)求最值等基礎(chǔ)知識(shí),考過(guò)N作GNMN,分別交圓弧和OE的延長(zhǎng)線于G和K,則GK=KN=10米.令GOK=0,則sin 0=,0.當(dāng)時(shí),才能作出滿足條件的矩形ABCD,所以sin 的取值范圍是.答:矩形ABCD的面積為800(4sin cos +cos )平方米,CDP的面積為1 600(cos -sin

52、 cos )平方米,sin 的取值范圍是.(2)因?yàn)榧?、乙兩種蔬菜的單位面積年產(chǎn)值之比為43,所以設(shè)甲的單位面積的年產(chǎn)值為4k,乙的單位面積的年產(chǎn)值為3k(k0).則年總產(chǎn)值為4k800(4sin cos +cos )+3k1 600(cos -sin cos )=8 000k(sin cos +cos ),.設(shè)f()=sin cos +cos ,.過(guò)N作GNMN,分別交圓弧和OE的延長(zhǎng)線于G和K,則GK=則f ()=cos2-sin2-sin =-(2sin2+sin -1)=-(2sin -1)(sin +1),令f ()=0,得=,當(dāng)時(shí), f ()0,所以f()為增函數(shù);當(dāng)時(shí), f ()

53、0,判斷是否存在b0,使函數(shù)f(x)與g(x)在區(qū)間(0,+)內(nèi)存在“S點(diǎn)”,并說(shuō)明理由.3.(2018江蘇,19,16分)記f (x),g(x)解析本題主要考查利用導(dǎo)數(shù)研究初等函數(shù)的性質(zhì),考查綜合運(yùn)用數(shù)學(xué)思想方法分析與解決問(wèn)題的能力以及邏輯推理能力.(1)證明:函數(shù)f(x)=x,g(x)=x2+2x-2,則f (x)=1,g(x)=2x+2,由f(x)=g(x)且f (x)=g(x),得此方程組無(wú)解.因此, f(x)=x與g(x)=x2+2x-2不存在“S點(diǎn)”.(2)函數(shù)f(x)=ax2-1,g(x)=ln x,則f (x)=2ax,g(x)=,設(shè)x0為f(x)與g(x)的“S點(diǎn)”,由f(x

54、0)=g(x0)且f (x0)=g(x0),得即(*)得ln x0=-,即x0=,則a=.當(dāng)a=時(shí),x0=滿足方程組(*),即x0為f(x)與g(x)的“S點(diǎn)”,因此,a的值為.解析本題主要考查利用導(dǎo)數(shù)研究初等函數(shù)的性質(zhì),考查綜合(3)f (x)=-2x,g(x)=,x0, f (x0)=g(x0)b=-0 x0(0,1),f(x0)=g(x0)-+a=-a=-,令h(x)=x2-a=,x(0,1),a0,設(shè)m(x)=-x3+3x2+ax-a,x(0,1),a0,則m(0)=-a0m(0)m(1)0,存在b0,使函數(shù)f(x)與g(x)在區(qū)間(0,+)內(nèi)存在“S點(diǎn)”.思路分析本題是新定義情境下運(yùn)

55、用導(dǎo)數(shù)研究函數(shù)零點(diǎn)問(wèn)題,前兩問(wèn)只需按新定義就能解決問(wèn)題,第三問(wèn)中先利用f (x0)=g(x0)對(duì)x0加以限制,然后將f(x0)=g(x0)轉(zhuǎn)化成a=-,從而轉(zhuǎn)化為研究h(x)=,x(0,1),a0的零點(diǎn)存在性問(wèn)題,再研究函數(shù)m(x)=-x3+3x2+ax-a,x(0,1),a0,由m(0)0,可判斷出m(x)在(0,1)上存在零點(diǎn),進(jìn)而解決問(wèn)題.(3)f (x)=-2x,g(x)=,x0, f 4.(2018浙江,22,15分)已知函數(shù)f(x)=-ln x.(1)若f(x)在x=x1,x2(x1x2)處導(dǎo)數(shù)相等,證明: f(x1)+f(x2)8-8ln 2;(2)若a3-4ln 2,證明:對(duì)于

56、任意k0,直線y=kx+a與曲線y=f(x)有唯一公共點(diǎn).解析本題主要考查函數(shù)的單調(diào)性,導(dǎo)數(shù)的運(yùn)算及其應(yīng)用,同時(shí)考查邏輯思維能力和綜合應(yīng)用能力.(1)函數(shù)f(x)的導(dǎo)函數(shù)f (x)=-,由f (x1)=f (x2)得-=-,因?yàn)閤1x2,所以+=.由基本不等式得=+2,因?yàn)閤1x2,所以x1x2256.由題意得f(x1)+f(x2)=-ln x1+-ln x2=-ln(x1x2).設(shè)g(x)= -ln x,則g(x)=(-4),所以4.(2018浙江,22,15分)已知函數(shù)f(x)=-ln所以g(x)在256,+)上單調(diào)遞增,故g(x1x2)g(256)=8-8ln 2,即f(x1)+f(x2

57、)8-8ln 2.(2)令m=e-(|a|+k),n=+1,則f(m)-km-a|a|+k-k-a0,f(n)-kn-ann0,直線y=kx+a與曲線y=f(x)有唯一公共點(diǎn).由(1)可知g(x)g(16),又a3-4ln 2,一題多解(1)f (x)=-,且f (x1)=f (x2)(x1x2).設(shè)f (x1)=t,則-=t的兩根為x1,x2.即2t()2-+2=0有兩個(gè)不同的正根x1,x2.即f(x1)+f(x2)=+-ln(x1x2)=+2ln t.設(shè)g(t)=+2ln t,則g(t)=-+=g=8-8ln 2,f(x1)+f(x2)8-8ln 2.一題多解(1)f (x)=-,(2)設(shè)

58、h(x)=f(x)-kx-a=-ln x-kx-a,只需證明:當(dāng)a3-4ln 2時(shí),對(duì)于任意的k0,函數(shù)h(x)在(0,+)上只有唯一的零點(diǎn).取m=e-|a|-k,則h(m)=+|a|+k-ke-|a|-k-a+k(1-e-|a|-k)k(1-e-|a|-k)0.又x0時(shí),-kx-k=.即h(x)-a-ln x,取n=,則h(n)-a-ln n=0,而-|a|-k-a-km0.由于h(m)0,h(n)0,h(x)在(m,n)上至少有一個(gè)零點(diǎn),即h(x)在(0,+)上至少有一個(gè)零點(diǎn).h(x)=-k-k=-k,當(dāng)k時(shí),h(x)在(0,+)上單調(diào)遞減,即當(dāng)k時(shí),h(x)在(0,+)上只有一個(gè)零點(diǎn).(

59、2)設(shè)h(x)=f(x)-kx-a=-ln x-kx-a當(dāng)0k時(shí),h(x)=0有兩個(gè)不同的正根,(其中).此時(shí)h(x)在(0,)上為減函數(shù),在(,)上為增函數(shù),在(,+)上為減函數(shù).h(x)=0,k=-,則h()=-ln -k-a=-ln +1-a,h()=-=,h()在(0,16)上為減函數(shù),在(16,+)上為增函數(shù),h()h(16)=3-ln 16-a=3-4ln 2-a0.又當(dāng)=16時(shí),k=,又0k0,x(0,時(shí),h(x)0.即h(x)在(0,上沒(méi)有零點(diǎn),但h(x)在(,+)上有一個(gè)零點(diǎn).當(dāng)0k0,直線y=kx+a與曲線y=f(x)有唯一的公共點(diǎn).當(dāng)0k1.(1)求函數(shù)h(x)=f(x)

60、-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.解析本題主要考查導(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,可知當(dāng)x變化時(shí),h(x),h(x)的變化情況如下表:x(-,0)