2022版高考數(shù)學(xué)一輪復(fù)習(xí) 高考大題專項(xiàng)導(dǎo)數(shù)的綜合應(yīng)用新人教A版

1、2022版高考數(shù)學(xué)一輪復(fù)習(xí) 高考大題專項(xiàng)導(dǎo)數(shù)的綜合應(yīng)用新人教a版2022版高考數(shù)學(xué)一輪復(fù)習(xí) 高考大題專項(xiàng)導(dǎo)數(shù)的綜合應(yīng)用新人教a版年級(jí)：姓名：高考大題專項(xiàng)(一)導(dǎo)數(shù)的綜合應(yīng)用突破1利用導(dǎo)數(shù)研究與不等式有關(guān)的問題1.(2020全國1,理21)已知函數(shù)f(x)=ex+ax2-x.(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)當(dāng)x0時(shí),f(x)12x3+1,求a的取值范圍.2.(2020山東濰坊二模,20)已知函數(shù)f(x)=1x+aln x,g(x)=exx.(1)討論函數(shù)f(x)的單調(diào)性;(2)證明:當(dāng)a=1時(shí),f(x)+g(x)-1+ex2ln xe.3.已知函數(shù)f(x)=ln x+ax(ar)的

2、圖象在點(diǎn)1e,f1e處的切線斜率為-e,其中e為自然對(duì)數(shù)的底數(shù).(1)求實(shí)數(shù)a的值,并求f(x)的單調(diào)區(qū)間;(2)證明:xf(x)xex.4.(2020廣東湛江一模,文21)已知函數(shù)f(x)=ln ax-bx+1,g(x)=ax-ln x,a1.(1)求函數(shù)f(x)的極值;(2)直線y=2x+1為函數(shù)f(x)圖象的一條切線,若對(duì)任意的x1(0,1),x21,2都有g(shù)(x1)f(x2)成立,求實(shí)數(shù)a的取值范圍.5.(2020山東濟(jì)寧5月模擬,21)已知兩個(gè)函數(shù)f(x)=exx,g(x)=lnxx+1x-1.(1)當(dāng)t0時(shí),求f(x)在區(qū)間t,t+1上的最大值;(2)求證:對(duì)任意x(0,+),不等

3、式f(x)g(x)都成立.6.(2020湖北武漢二月調(diào)考,理21)已知函數(shù)f(x)=(x-1)ex-kx2+2.(1)略;(2)若x0,+),都有f(x)1成立,求實(shí)數(shù)k的取值范圍.7.(2020山東濟(jì)南一模,22)已知函數(shù)f(x)=a(ex-x-1)x2,且曲線y=f(x)在(2,f(2)處的切線斜率為1.(1)求實(shí)數(shù)a的值;(2)證明:當(dāng)x0時(shí),f(x)1;(3)若數(shù)列xn滿足exn+1=f(xn),且x1=13,證明:2n|exn-1|0),試研究函數(shù)g(x)的極值情況;(2)記函數(shù)f(x)=f(x)-xex在區(qū)間(1,2)上的零點(diǎn)為x0,記m(x)=minf(x),xex,若m(x)=

4、n(nr)在區(qū)間(1,+)上有兩個(gè)不等實(shí)數(shù)解x1,x2(x12x0.突破2利用導(dǎo)數(shù)研究與函數(shù)零點(diǎn)有關(guān)的問題1.(2020山東煙臺(tái)一模,21)已知函數(shù)f(x)=1+lnxx-a(ar).(1)若f(x)0在(0,+)上恒成立,求a的取值范圍,并證明:對(duì)任意的nn*,都有1+12+13+1nln(n+1);(2)設(shè)g(x)=(x-1)2ex,討論方程f(x)=g(x)的實(shí)數(shù)根的個(gè)數(shù).2.(2019全國2,理20)已知函數(shù)f(x)=ln x-x+1x-1.(1)討論f(x)的單調(diào)性,并證明f(x)有且僅有兩個(gè)零點(diǎn);(2)設(shè)x0是f(x)的一個(gè)零點(diǎn),證明曲線y=ln x在點(diǎn)a(x0,ln x0)處的切

5、線也是曲線y=ex的切線.3.(2020北京通州區(qū)一模,19)已知函數(shù)f(x)=xex,g(x)=a(ex-1),ar.(1)當(dāng)a=1時(shí),求證:f(x)g(x);(2)當(dāng)a1時(shí),求關(guān)于x的方程f(x)=g(x)的實(shí)數(shù)根的個(gè)數(shù).4.(2020湖南長郡中學(xué)四模,文21)已知函數(shù)f(x)=2ae2x+2(a+1)ex.(1)略;(2)當(dāng)a(0,+)時(shí),函數(shù)f(x)的圖象與函數(shù)y=4ex+x的圖象有唯一的交點(diǎn),求a的取值集合.5.(2020天津和平區(qū)一模,20)已知函數(shù)f(x)=ax+bxex,a,br,且a0.(1)若函數(shù)f(x)在x=-1處取得極值1e,求函數(shù)f(x)的解析式;(2)在(1)的條件

6、下,求函數(shù)f(x)的單調(diào)區(qū)間;(3)設(shè)g(x)=a(x-1)ex-f(x),g(x)為g(x)的導(dǎo)函數(shù),若存在x0(1,+),使g(x0)+g(x0)=0成立,求ba的取值范圍.6.已知函數(shù)f(x)=ln x,g(x)=2a3x3+2(1-a)x2-8x+8a+7.(1)若曲線y=g(x)在點(diǎn)(2,g(2)處的切線方程是y=ax-1,求函數(shù)g(x)在0,3上的值域;(2)當(dāng)x0時(shí),記函數(shù)h(x)=f(x),f(x)g(x),g(x),f(x)g(x),若函數(shù)y=h(x)有三個(gè)零點(diǎn),求實(shí)數(shù)a的取值集合.參考答案高考大題專項(xiàng)(一)導(dǎo)數(shù)的綜合應(yīng)用突破1利用導(dǎo)數(shù)研究與不等式有關(guān)的問題1.解(1)當(dāng)a=

7、1時(shí),f(x)=ex+x2-x,f(x)=ex+2x-1.故當(dāng)x(-,0)時(shí),f(x)0.所以f(x)在(-,0)上單調(diào)遞減,在(0,+)上單調(diào)遞增.(2)f(x)12x3+1等價(jià)于12x3-ax2+x+1e-x1.設(shè)函數(shù)g(x)=12x3-ax2+x+1e-x(x0),則g(x)=-12x3-ax2+x+1-32x2+2ax-1e-x=-12xx2-(2a+3)x+4a+2e-x=-12x(x-2a-1)(x-2)e-x.若2a+10,即a-12,則當(dāng)x(0,2)時(shí),g(x)0.所以g(x)在(0,2)上單調(diào)遞增,而g(0)=1,故當(dāng)x(0,2)時(shí),g(x)1,不合題意.若02a+12,即-

8、12a12,則當(dāng)x(0,2a+1)(2,+)時(shí),g(x)0.所以g(x)在(0,2a+1),(2,+)上單調(diào)遞減,在(2a+1,2)上單調(diào)遞增.由于g(0)=1,所以g(x)1當(dāng)且僅當(dāng)g(2)=(7-4a)e-21,即a7-e24.所以當(dāng)7-e24a12時(shí),g(x)1.若2a+12,即a12,則g(x)12x3+x+1e-x.由于07-e24,12,故由可得12x3+x+1e-x1.故當(dāng)a12時(shí),g(x)1.綜上,a的取值范圍是7-e24,+.2.(1)解函數(shù)的定義域?yàn)?0,+),f(x)=-1x2+ax=ax-1x2,當(dāng)a0時(shí),f(x)0時(shí),由f(x)0,得x1a,由f(x)0,得0x0時(shí),

9、f(x)在0,1a上單調(diào)遞減,在1a,+上單調(diào)遞增.(2)證明因?yàn)閤0,所以不等式等價(jià)于ex-ex+1elnxx,設(shè)f(x)=ex-ex+1,f(x)=ex-e,所以當(dāng)x(1,+)時(shí),f(x)0,f(x)單調(diào)遞增;當(dāng)x(0,1)時(shí),f(x)0,g(x)單調(diào)遞增,當(dāng)x(e,+)時(shí),g(x)g(x),即ex-ex+1elnxx,故原不等式成立.3.(1)解因?yàn)楹瘮?shù)f(x)的定義域?yàn)?0,+),f(x)=1x-ax2,所以f1e=e-ae2=-e,所以a=2e,所以f(x)=1x-2ex2.令f(x)=0,得x=2e,當(dāng)x0,2e時(shí),f(x)0,所以f(x)在0,2e上單調(diào)遞減,在2e,+上單調(diào)遞增

10、.(2)證明設(shè)h(x)=xf(x)=xlnx+2e,由h(x)=lnx+1=0,得x=1e,所以當(dāng)x0,1e時(shí),h(x)0,所以h(x)在0,1e上單調(diào)遞減,在1e,+上單調(diào)遞增,所以h(x)min=h1e=1e.設(shè)t(x)=xex(x0),則t(x)=1-xex,所以當(dāng)x(0,1)時(shí),t(x)0,t(x)單調(diào)遞增,當(dāng)x(1,+)時(shí),t(x)t(x),即xf(x)xex.4.解(1)a1,函數(shù)f(x)的定義域?yàn)?0,+).f(x)=lnax-bx+1=lna+lnx-bx+1,f(x)=1x-b=1-bxx.當(dāng)b0時(shí),f(x)0,f(x)在(0,+)上為增函數(shù),無極值;當(dāng)b0時(shí),由f(x)=0

11、,得x=1b.當(dāng)x0,1b時(shí),f(x)0,f(x)單調(diào)遞增;當(dāng)x1b,+時(shí),f(x)f(x2)成立,只需g(x1)minf(x2)max.g(x)=a-1x=ax-1x,由g(x)=0,得x=1a.a1,01a1.當(dāng)x0,1a時(shí),g(x)0,g(x)單調(diào)遞增.g(x)g1a=1+lna,即g(x1)min=1+lna.f(x2)=1x2-b在x21,2上單調(diào)遞減,f(x2)max=f(1)=1-b=3-ae.1+lna3-ae.即lna+ae-20.設(shè)h(a)=lna+ae-2,易知h(a)在(1,+)上單調(diào)遞增.又h(e)=0,實(shí)數(shù)a的取值范圍為(e,+).5.(1)解由f(x)=exx得,

12、f(x)=xex-exx2=ex(x-1)x2,當(dāng)x1時(shí),f(x)1時(shí),f(x)0,f(x)在區(qū)間(-,1)上單調(diào)遞減,在區(qū)間(1,+)上單調(diào)遞增.當(dāng)t1時(shí),f(x)在區(qū)間t,t+1上單調(diào)遞增,f(x)的最大值為f(t+1)=et+1t+1.當(dāng)0t1,f(x)在區(qū)間(t,1)上單調(diào)遞減,在區(qū)間(1,t+1)上單調(diào)遞增,f(x)的最大值為f(x)max=maxf(t),f(t+1).下面比較f(t)與f(t+1)的大小.f(t)-f(t+1)=ett-et+1t+1=(1-e)t+1ett(t+1).t0,1-e0,當(dāng)0t1e-1時(shí),f(t)-f(t+1)0,故f(x)在區(qū)間t,t+1上的最大值

13、為f(t)=ett,當(dāng)1e-1t1時(shí),f(t)-f(t+1)0,f(x)在區(qū)間t,t+1上的最大值為f(t+1)=et+1t+1.綜上可知,當(dāng)01e-1時(shí),f(x)在區(qū)間t,t+1上的最大值為f(t+1)=et+1t+1.(2)證明不等式f(x)g(x)即為exxlnxx+1x-1.x0,不等式等價(jià)于exlnx-x+1,令h(x)=ex-(x+1)(x0),則h(x)=ex-10,h(x)在(0,+)上為增函數(shù),h(x)h(0)=0,即exx+1,所以,要證exlnx-x+1成立,只需證x+1lnx-x+1成立即可.即證2xlnx在(0,+)上成立.設(shè)(x)=2x-lnx,則(x)=2-1x=

14、2x-1x,當(dāng)0x12時(shí),(x)12時(shí),(x)0,(x)單調(diào)遞增,(x)min=12=1-ln12=1+ln20,(x)0在(0,+)上成立,對(duì)任意x(0,+),不等式f(x)g(x)都成立.6.解(1)略(2)f(x)=xex-2kx=x(ex-2k),當(dāng)k0時(shí),ex-2k0,所以,當(dāng)x0時(shí),f(x)0時(shí),f(x)0,則f(x)在區(qū)間(-,0)上單調(diào)遞減,在區(qū)間(0,+)上單調(diào)遞增,所以f(x)在區(qū)間0,+)上的最小值為f(0),且f(0)=1,符合題意;當(dāng)k0時(shí),令f(x)=0,得x=0或x=ln2k,所以當(dāng)00,f(x)單調(diào)遞增,所以f(x)在區(qū)間0,+)上的最小值為f(0),且f(0)

15、=1,符合題意;當(dāng)k12時(shí),ln2k0,當(dāng)x(0,ln2k)時(shí),f(x)0,f(x)在區(qū)間(0,ln2k)上單調(diào)遞減,所以f(ln2k)1,只需證h(x)=ex-12x2-x-10.h(x)=ex-x-1,令c(x)=ex-x-1,則c(x)=ex-1.因?yàn)楫?dāng)x0時(shí),c(x)0,所以h(x)=ex-x-1在(0,+)上單調(diào)遞增,所以h(x)=ex-x-1h(0)=0.所以h(x)=ex-12x2-x-1在(0,+)上單調(diào)遞增,所以h(x)=ex-12x2-x-1h(0)=0成立.所以當(dāng)x0時(shí),f(x)1.(3)證明(方法1)由(2)知當(dāng)x0時(shí),f(x)1.因?yàn)閑xn+1=f(xn),所以xn+

16、1=lnf(xn).設(shè)g(xn)=lnf(xn),則xn+1=g(xn),所以xn=g(xn-1)=g(g(xn-2)=g(g(x1)0.要證2n|exn-1|1,只需證|exn-1|12n.因?yàn)閤1=13,所以|ex1-1|=e13-1.因?yàn)閑-323=e-2780,所以e1332,所以|ex1-1|=e13-112.故只需證|exn+1-1|12|exn-1|.因?yàn)閤n(0,+),故只需證exn+1-112exn-12,即證f(xn)-10,(x)=12x2+x-2ex+x+2,令(x)=12x2+x-2ex+x+2,則(x)=12x2+2x-1ex+1,令(x)=12x2+2x-1ex+

17、1,則(x)=12x2+3x+1ex0,所以(x)在區(qū)間(0,+)上單調(diào)遞增,故(x)=12x2+2x-1ex+1(0)=0.所以(x)在區(qū)間(0,+)上單調(diào)遞增,故(x)=12x2+x-2ex+x+2(0)=0.所以(x)在區(qū)間(0,+)上單調(diào)遞增,所以(x)=12x2-2ex+12x2+2x+2(0)=0,所以原不等式成立.(方法2)由(2)知當(dāng)x0時(shí),f(x)1.因?yàn)閑xn+1=f(xn),所以xn+1=lnf(xn).設(shè)g(xn)=lnf(xn),則xn+1=g(xn),所以xn=g(xn-1)=g(g(xn-2)=g(g(x1)0.要證2n|exn-1|1,只需證|exn-1|12n

18、.因?yàn)閤1=13,所以|ex1-1|=e13-1.因?yàn)閑-323=e-2780,所以e1332,所以|ex1-1|=e13-112.故只需證|exn+1-1|12|exn-1|.因?yàn)閤n(0,+),故只需證exn+1-112exn-12,即證f(xn)-10.因?yàn)?x)=12(x2-4)ex+12(x2+4x+4)=12(x+2)(x-2)ex+(x+2),設(shè)u(x)=(x-2)ex+(x+2),故只需證u(x)0.u(x)=(x-1)ex+1,令v(x)=(x-1)ex+1,則v(x)=xex0,所以v(x)在區(qū)間(0,+)上單調(diào)遞增,故v(x)=(x-1)ex+1v(0)=0,所以u(píng)(x)

19、在區(qū)間(0,+)上單調(diào)遞增,故u(x)=(x-2)ex+(x+2)u(0)=0,所以原不等式成立.8.(1)解由題意,得f(x)=lnx+1,故g(x)=ax2-(a+2)x+lnx+1,故g(x)=2ax-(a+2)+1x=(2x-1)(ax-1)x,x0,a0.令g(x)=0,得x1=12,x2=1a.當(dāng)0a12,由g(x)0,得0x1a;由g(x)0,得12x2時(shí),1a0,得0x12;由g(x)0,得1ax12.所以g(x)在x=1a處取極大值g1a=-1a-lna,在x=12處取極小值g12=-a4-ln2.綜上,當(dāng)0a2時(shí),g(x)在x=1a處取極大值-1a-lna,在x=12處取極

20、小值-a4-ln2.(2)證明f(x)=xlnx-xex,定義域?yàn)閤(0,+),f(x)=1+lnx+x-1ex.當(dāng)x(1,2)時(shí),f(x)0,即f(x)在區(qū)間(1,2)上單調(diào)遞增.又因?yàn)閒(1)=-1e0,且f(x)在區(qū)間(1,2)上的圖象連續(xù)不斷,故根據(jù)函數(shù)零點(diǎn)存在定理,f(x)在區(qū)間(1,2)上有且僅有一個(gè)零點(diǎn).所以存在x0(1,2),使得f(x0)=f(x0)-x0ex0=0.且當(dāng)1xx0時(shí),f(x)x0時(shí),f(x)xex.所以m(x)=minf(x),xex=xlnx,1xx0.當(dāng)1x0,得m(x)單調(diào)遞增;當(dāng)xx0時(shí),m(x)=xex,由m(x)=1-xex0,得m(x)單調(diào)遞減.

21、若m(x)=n在區(qū)間(1,+)上有兩個(gè)不等實(shí)數(shù)解x1,x2(x12x0,即證x22x0-x1.又因?yàn)?x0-x1x0,而m(x)在區(qū)間(x0,+)上單調(diào)遞減,所以可證m(x2)m(2x0-x1).由m(x1)=m(x2),即證m(x1)m(2x0-x1),即x1lnx12x0-x1e2x0-x1.記h(x)=xlnx-2x0-xe2x0-x,1x0;當(dāng)t(1,+)時(shí),(t)0,故0(t)1,所以-1e-2x0-xe2x0-x1-1e0,即h(x)單調(diào)遞增,故當(dāng)1xx0時(shí),h(x)h(x0)=0,即x1lnx12x0,得證.突破2利用導(dǎo)數(shù)研究與函數(shù)零點(diǎn)有關(guān)的問題1.(1)證明由f(x)0可得,a

22、1+lnxx(x0),令h(x)=1+lnxx,則h(x)=1xx-(1+lnx)x2=-lnxx2.當(dāng)x(0,1)時(shí),h(x)0,h(x)單調(diào)遞增;當(dāng)x(1,+)時(shí),h(x)0,h(x)單調(diào)遞減,故h(x)在x=1處取得最大值,要使a1+lnxx,只需ah(1)=1,故a的取值范圍為1,+).顯然,當(dāng)a=1時(shí),有1+lnxx1,即不等式lnx1(nn*),則有l(wèi)nn+1nn+1n-1=1n,所以ln21+ln32+lnn+1nln(n+1).(2)解由f(x)=g(x),可得1+lnxx-a=(x-1)2ex,即a=1+lnxx-(x-1)2ex,令t(x)=1+lnxx-(x-1)2ex,

23、則t(x)=-lnxx2-(x2-1)ex,當(dāng)x(0,1)時(shí),t(x)0,t(x)單調(diào)遞增;當(dāng)x(1,+)時(shí),t(x)0,t(x)單調(diào)遞減,故t(x)在x=1處取得最大值t(1)=1,又當(dāng)x0時(shí),t(x)-,當(dāng)x+時(shí),t(x)-,所以,當(dāng)a=1時(shí),方程f(x)=g(x)有一個(gè)實(shí)數(shù)根;當(dāng)a1時(shí),方程f(x)=g(x)沒有實(shí)數(shù)根.2.(1)解f(x)的定義域?yàn)?0,1)(1,+).因?yàn)閒(x)=1x+2(x-1)20,所以f(x)在區(qū)間(0,1),(1,+)上單調(diào)遞增.因?yàn)閒(e)=1-e+1e-10,所以f(x)在區(qū)間(1,+)上有唯一零點(diǎn)x1,即f(x1)=0.又01x11,f1x1=-lnx

24、1+x1+1x1-1=-f(x1)=0,故f(x)在區(qū)間(0,1)上有唯一零點(diǎn)1x1.綜上,f(x)有且僅有兩個(gè)零點(diǎn).(2)證明因?yàn)?x0=e-lnx0,故點(diǎn)b-lnx0,1x0在曲線y=ex上.由題設(shè)知f(x0)=0,即lnx0=x0+1x0-1,故直線ab的斜率k=1x0-lnx0-lnx0-x0=1x0-x0+1x0-1-x0+1x0-1-x0=1x0.曲線y=ex在點(diǎn)b-lnx0,1x0處切線的斜率是1x0,曲線y=lnx在點(diǎn)a(x0,lnx0)處切線的斜率也是1x0,所以曲線y=lnx在點(diǎn)a(x0,lnx0)處的切線也是曲線y=ex的切線.3.(1)證明設(shè)函數(shù)f(x)=f(x)-g(

25、x)=xex-aex+a.當(dāng)a=1時(shí),f(x)=xex-ex+1,所以f(x)=xex.所以當(dāng)x(-,0)時(shí),f(x)0.所以f(x)在(-,0)上單調(diào)遞減,在(0,+)上單調(diào)遞增.所以當(dāng)x=0時(shí),f(x)取得最小值f(0)=0.所以f(x)0,即f(x)g(x).(2)解設(shè)函數(shù)f(x)=f(x)-g(x)=xex-aex+a.當(dāng)a1時(shí),f(x)=(x-a+1)ex,令f(x)0,即(x-a+1)ex0,解得xa-1;令f(x)0,即(x-a+1)ex0,解得x1,所以h(a)0.所以h(a)在(1,+)上單調(diào)遞減.所以h(a)h(1)=0,所以f(a-1)0,所以f(x)在區(qū)間(a-1,a)

26、上存在一個(gè)零點(diǎn).所以在a-1,+)上存在唯一的零點(diǎn).又因?yàn)閒(x)在區(qū)間(-,a-1)上單調(diào)遞減,且f(0)=0,所以f(x)在區(qū)間(-,a-1)上存在唯一的零點(diǎn)0.所以函數(shù)f(x)有且僅有兩個(gè)零點(diǎn),即方程f(x)=g(x)有兩個(gè)實(shí)數(shù)根.4.解(1)略.(2)設(shè)t=ex,則f(t)=2at2+2(a+1)t的圖象與y=4t+lnt的圖象只有一個(gè)交點(diǎn),其中t0,則2at2+2(a+1)t=4t+lnt只有一個(gè)實(shí)數(shù)解,即2a=2t+lntt2+t只有一個(gè)實(shí)數(shù)解.設(shè)g(t)=2t+lntt2+t,則g(t)=-2t2+t-2tlnt+1-lnt(t2+t)2,g(1)=0.令h(t)=-2t2+t-2tlnt+1-lnt,則h(t)=-4t-1t-2lnt-1.設(shè)y=1t+2lnt,令y=-1t2+2t=2t-1t2=0,解得t=12,則y,y隨t的變化如表所示t0,121212,+y-0+y2-2ln2則當(dāng)t=12時(shí),y=1t+2lnt取最小值為2-2ln2=2(1-ln2)0.所以-1t-2lnt0,即h(t)=-4t-1t-2lnt-10,g(t)單調(diào)遞增;當(dāng)t(1,+)時(shí),g(t)0得x12,令f(x)0得-1x0或0x0),所以g(x)=bx2+ax-bx-aex.由g(x)+g(x)=0,得ax-bx-2aex+bx2+a