2022年高考數(shù)學(xué)一輪復(fù)習(xí)考點(diǎn)規(guī)范練16導(dǎo)數(shù)的綜合應(yīng)用含解析新人教A版

1、PAGE PAGE 7考點(diǎn)規(guī)范練16導(dǎo)數(shù)的綜合應(yīng)用基礎(chǔ)鞏固1.已知函數(shù)f(x)=x3+ax2+bx+c在x=-23與x=1處都取得極值.(1)求a,b的值及函數(shù)f(x)的單調(diào)區(qū)間;(2)若對(duì)于x-1,2,不等式f(x)c2恒成立,求c的取值范圍.解:(1)f(x)=x3+ax2+bx+c,f(x)=3x2+2ax+b.又f(x)在x=-23與x=1處都取得極值,f-23=129-43a+b=0,f(1)=3+2a+b=0,兩式聯(lián)立解得a=-12,b=-2,f(x)=x3-12x2-2x+c,f(x)=3x2-x-2=(3x+2)(x-1),令f(x)=0,得x1=-23,x2=1,當(dāng)x變化時(shí),

2、f(x),f(x)的變化情況如下表:x-,-23-23-23,11(1,+)f(x)+0-0+f(x)極大值極小值函數(shù)f(x)的遞增區(qū)間為-,-23與(1,+);遞減區(qū)間為-23,1.(2)f(x)=x3-12x2-2x+c,x-1,2,當(dāng)x=-23時(shí),f-23=2227+c為極大值,而f(2)=2+c,則f(2)=2+c為最大值,要使f(x)f(2)=2+c,解得c2.c的取值范圍為(-,-1)(2,+).2.設(shè)函數(shù)f(x)=ax2-a-ln x,g(x)=1x-eex,其中aR,e=2.718為自然對(duì)數(shù)的底數(shù).(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)x1時(shí),g(x)0;(3)確定a的所有

3、可能取值,使得f(x)g(x)在區(qū)間(1,+)內(nèi)恒成立.答案:(1)解f(x)=2ax-1x=2ax2-1x(x0).當(dāng)a0時(shí),f(x)0時(shí),由f(x)=0有x=12a.當(dāng)x0,12a時(shí),f(x)0,f(x)單調(diào)遞增.(2)證明令s(x)=ex-1-x,則s(x)=ex-1-1.當(dāng)x1時(shí),s(x)0,所以ex-1x,從而g(x)=1x-1ex-10.(3)解由(2),當(dāng)x1時(shí),g(x)0.當(dāng)a0,x1時(shí),f(x)=a(x2-1)-lnxg(x)在區(qū)間(1,+)內(nèi)恒成立時(shí),必有a0.當(dāng)0a1.由(1)有f12a0,所以此時(shí)f(x)g(x)在區(qū)間(1,+)內(nèi)不恒成立.當(dāng)a12時(shí),令h(x)=f(x

4、)-g(x)(x1).當(dāng)x1時(shí),h(x)=2ax-1x+1x2-e1-xx-1x+1x2-1x=x3-2x+1x2x2-2x+1x20.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,a12,+.3.已知函數(shù)f(x)=(x-k)ex+k,kZ.(1)當(dāng)k=0時(shí),求函數(shù)f(x)的單調(diào)區(qū)間;(2)若當(dāng)x(0,+)時(shí),不等式f(x)+50恒成立,求k的最大值.解:(1)當(dāng)k=0時(shí),f(x)=xex,f(x)=ex+xex=ex(x+1),當(dāng)x(-,-1)時(shí),f(x)0;f(x)在區(qū)間(-,-1)內(nèi)是減函數(shù),在

5、區(qū)間(-1,+)內(nèi)是增函數(shù).(2)不等式f(x)+50恒成立(x-k)ex+k+50在x(0,+)時(shí)恒成立,令F(x)=(x-k)ex+k+5,F(x)=ex(x-k+1)(xR),當(dāng)x(-,k-1)時(shí),f(x)0;f(x)在區(qū)間(-,k-1)內(nèi)是減函數(shù),在區(qū)間(k-1,+)內(nèi)是增函數(shù).若k-10,即k1,當(dāng)x(0,+)時(shí),F(x)F(0)0.而F(0)=50恒成立,k1符合題意.若k-10,即k1,當(dāng)x(0,+)時(shí),只需F(x)min=F(k-1)=-ek-1+5+k0即可.令h(k)=-ek-1+5+k,h(k)=1-ek-10,h(3)=-e2+80,h(4)=-e3+90,1k3,綜上

6、,k的最大值為3.4.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)a0,故f(x)在區(qū)間(0,+)單調(diào)遞增.若a0;當(dāng)x-12a,+時(shí),f(x)0.故f(x)在區(qū)間0,-12a內(nèi)單調(diào)遞增,在區(qū)間-12a,+內(nèi)單調(diào)遞減.(2)證明由(1)知,當(dāng)a0;當(dāng)x(1,+)時(shí),g(x)0時(shí),g(x)0.從而當(dāng)a0時(shí),ln-12a+12a+10,即f(x)-34a-2.能力提升5.設(shè)函數(shù)f(x)=(1-x2)ex.(1)討論f(x)的單調(diào)性;(2)當(dāng)x0時(shí),f(x)ax+1,求a的取值范圍.解:(1)f(x)=(1-2x-x2)ex.令f(x)=0得x=-1-2,

7、x=-1+2.當(dāng)x(-,-1-2)時(shí),f(x)0;當(dāng)x(-1+2,+)時(shí),f(x)0.所以f(x)在區(qū)間(-,-1-2),(-1+2,+)內(nèi)單調(diào)遞減,在區(qū)間(-1-2,-1+2)內(nèi)單調(diào)遞增.(2)f(x)=(1+x)(1-x)ex.當(dāng)a1時(shí),設(shè)函數(shù)h(x)=(1-x)ex,h(x)=-xex0),因此h(x)在區(qū)間0,+)內(nèi)單調(diào)遞減,而h(0)=1,故h(x)1,所以f(x)=(x+1)h(x)x+1ax+1.當(dāng)0a0(x0),所以g(x)在區(qū)間0,+)內(nèi)單調(diào)遞增,而g(0)=0,故exx+1.當(dāng)0 x(1-x)(1+x)2,(1-x)(1+x)2-ax-1=x(1-a-x-x2),取x0=5

8、-4a-12,則x0(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0)ax0+1.當(dāng)a0時(shí),取x0=5-12,則x0(0,1),f(x0)(1-x0)(1+x0)2=1ax0+1.綜上,a的取值范圍是1,+).6.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)=0有且僅有兩個(gè)實(shí)根,且兩個(gè)實(shí)根互為倒數(shù).答案:證明(1)f(x)的定義域?yàn)?0,+).f(x)=x-1x+lnx-1=lnx-1x.因?yàn)閥=lnx單調(diào)遞增,y=1x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)x

9、x0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,所以f(x)=0在區(qū)間(x0,+)內(nèi)存在唯一根x=.由x01得11x0.又f1=1-1ln1-1-1=f()=0,故1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).7.已知函數(shù)f(x)=x3-kx+k2.(1)討論f(x)的單調(diào)性;(2)若f(x)有三個(gè)零點(diǎn),求k的取值范圍.解:(1)f(x)=3x2-k.當(dāng)k=0時(shí),f(x)=x3,故f(x)在(-,+)單調(diào)遞增;當(dāng)k0,故f(x)在(-,+)單調(diào)遞增.當(dāng)k0時(shí),令f(x)=0,得

10、x=3k3.當(dāng)x-,-3k3時(shí),f(x)0;當(dāng)x-3k3,3k3時(shí),f(x)0.故f(x)在-,3k3,3k3,+單調(diào)遞增,在3k3,3k3單調(diào)遞減.(2)由(1)知,當(dāng)k0時(shí),f(x)在(-,+)單調(diào)遞增,f(x)不可能有三個(gè)零點(diǎn).當(dāng)k0時(shí),x=-3k3為f(x)的極大值點(diǎn),x=3k3為f(x)的極小值點(diǎn).此時(shí),-k-1-3k33k3k+1且f(-k-1)0,f-3k30.根據(jù)f(x)的單調(diào)性,當(dāng)且僅當(dāng)f3k30,即k2-2k3k90時(shí),f(x)有三個(gè)零點(diǎn),解得k427.因此k的取值范圍為0,427.高考預(yù)測(cè)8.已知函數(shù)f(x)=x(ln x-ax)(aR).(1)若a=1,求函數(shù)f(x)的圖象在點(diǎn)(1,f(1)處的切線方程;(2)若函數(shù)f(x)有兩個(gè)極值點(diǎn)x1,x2,且x1-12.答案:(1)解由已知,f(x)=x(lnx-x),當(dāng)x=1時(shí),f(x)=-1,f(x)=lnx+1-2x,當(dāng)x=1時(shí),f(x)=-1,所以所求切線方程為x+y=0.(2)證明由已知可得f(x)=lnx+1-2ax=0有兩個(gè)相異實(shí)根x1,x2,令h(x)=f(x),則h(