（廣西課標(biāo)版）2020版高考數(shù)學(xué)二輪復(fù)習(xí) 專(zhuān)題能力訓(xùn)練8 利用導(dǎo)數(shù)解不等式及參數(shù)范圍 文

專(zhuān)題能力訓(xùn)練8利用導(dǎo)數(shù)解不等式及參數(shù)范圍一、能力突破訓(xùn)練1.已知函數(shù)f(x)=ax2+bx-ln x,a,bR.(1)若a<0,且b=2-a,試討論f(x)的單調(diào)性;(2)若對(duì)b-2,-1,x(1,e)使得f(x)<0成立,求實(shí)數(shù)a的取值范圍.2.(2018全國(guó),文21)已知函數(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)a1e時(shí),f(x)0.3.已知函數(shù)f(x)=ax+xln x的圖象在x=e(e為自然對(duì)數(shù)的底數(shù))處的切線(xiàn)的斜率為3.(1)求實(shí)數(shù)a的值;(2)若f(x)kx2對(duì)任意x>0成立,求實(shí)數(shù)k的取值范圍;(3)當(dāng)n>m>1(m,nN*)時(shí),證明:nmmn>mn.4.已知函數(shù)f(x)=ln x-ax,其中aR.(1)當(dāng)a=-1時(shí),判斷f(x)的單調(diào)性;(2)若g(x)=f(x)+ax在其定義域內(nèi)為減函數(shù),求實(shí)數(shù)a的取值范圍;(3)當(dāng)a=0時(shí),函數(shù)f(x)的圖象關(guān)于y=x對(duì)稱(chēng)得到函數(shù)h(x)的圖象,若直線(xiàn)y=kx與曲線(xiàn)y=2x+1h(x)沒(méi)有公共點(diǎn),求k的取值范圍.5.設(shè)函數(shù)f(x)=aln x,g(x)=12x2.(1)記g'(x)為g(x)的導(dǎo)函數(shù),若不等式f(x)+2g'(x)(a+3)x-g(x)在x1,e內(nèi)有解,求實(shí)數(shù)a的取值范圍;(2)若a=1,對(duì)任意的x1>x2>0,不等式mg(x1)-g(x2)>x1f(x1)-x2f(x2)恒成立,求m(mZ,m1)的值.6.已知函數(shù)f(x)=ln x-(x-1)22.(1)求函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)證明:當(dāng)x>1時(shí),f(x)<x-1;(3)確定實(shí)數(shù)k的取值范圍,使得存在x0>1,當(dāng)x(1,x0)時(shí),恒有f(x)>k(x-1).二、思維提升訓(xùn)練7.已知函數(shù)f(x)=x3+ax2+bx+1(a>0,bR)有極值,且導(dǎo)函數(shù)f'(x)的極值點(diǎn)是f(x)的零點(diǎn).(極值點(diǎn)是指函數(shù)取極值時(shí)對(duì)應(yīng)的自變量的值)(1)求b關(guān)于a的函數(shù)解析式,并寫(xiě)出定義域;(2)證明:b2>3a;(3)若f(x),f'(x)這兩個(gè)函數(shù)的所有極值之和不小于-72,求a的取值范圍.8.設(shè)函數(shù)f(x)=x3-ax-b,xR,其中a,bR.(1)求f(x)的單調(diào)區(qū)間;(2)若f(x)存在極值點(diǎn)x0,且f(x1)=f(x0),其中x1x0,求證:x1+2x0=0;(3)設(shè)a>0,函數(shù)g(x)=|f(x)|,求證:g(x)在區(qū)間-1,1上的最大值不小于14.專(zhuān)題能力訓(xùn)練8利用導(dǎo)數(shù)解不等式及參數(shù)范圍一、能力突破訓(xùn)練1.解(1)f'(x)=2ax+(2-a)-1x=2ax2+(2-a)x-1x=(ax+1)(2x-1)x.當(dāng)-1a<12,即a<-2時(shí),f(x)的單調(diào)遞增區(qū)間為-1a,12,單調(diào)遞減區(qū)間為0,-1a,12,+;當(dāng)-1a=12,即a=-2時(shí),f(x)在區(qū)間(0,+)上單調(diào)遞減;當(dāng)-1a>12,即0>a>-2時(shí),f(x)的單調(diào)遞增區(qū)間為12,-1a,單調(diào)遞減區(qū)間為0,12,-1a,+.(2)對(duì)b-2,-1,x(1,e)使得ax2+bx-lnx<0成立,即ax2-x-lnx<0在區(qū)間(1,e)內(nèi)有解,即a<lnx+xx2在(1,e)內(nèi)有解,即a<lnx+xx2max.令g(x)=lnx+xx2,則g'(x)=-x(x-1+2lnx)x4.x(1,e),g'(x)<0,即在區(qū)間(1,e)內(nèi)g(x)單調(diào)遞減.a<g(1)=1.故實(shí)數(shù)a的取值范圍為a<1.2.(1)解f(x)的定義域?yàn)?0,+),f'(x)=aex-1x.由題設(shè)知,f'(2)=0,所以a=12e2.從而f(x)=12e2ex-lnx-1,f'(x)=12e2ex-1x.當(dāng)0<x<2時(shí),f'(x)<0;當(dāng)x>2時(shí),f'(x)>0.所以f(x)在區(qū)間(0,2)內(nèi)單調(diào)遞減,在區(qū)間(2,+)內(nèi)單調(diào)遞增.(2)證明當(dāng)a1e時(shí),f(x)exe-lnx-1.設(shè)g(x)=exe-lnx-1,則g'(x)=exe-1x.當(dāng)0<x<1時(shí),g'(x)<0;當(dāng)x>1時(shí),g'(x)>0.所以x=1是g(x)的最小值點(diǎn).故當(dāng)x>0時(shí),g(x)g(1)=0.因此,當(dāng)a1e時(shí),f(x)0.3.(1)解f(x)=ax+xlnx,f'(x)=a+lnx+1.又f(x)的圖象在x=e處的切線(xiàn)的斜率為3,f'(e)=3,即a+lne+1=3,a=1.(2)解由(1)知,f(x)=x+xlnx,若f(x)kx2對(duì)任意x>0成立,則k1+lnxx對(duì)任意x>0成立.令g(x)=1+lnxx,則問(wèn)題轉(zhuǎn)化為求g(x)的最大值,g'(x)=1x·x-(1+lnx)x2=-lnxx2.令g'(x)=0,解得x=1.當(dāng)0<x<1時(shí),g'(x)>0,g(x)在區(qū)間(0,1)內(nèi)是增函數(shù);當(dāng)x>1時(shí),g'(x)<0,g(x)在區(qū)間(1,+)內(nèi)是減函數(shù).故g(x)在x=1處取得最大值g(1)=1,k1即為所求.(3)證明令h(x)=xlnxx-1,則h'(x)=x-1-lnx(x-1)2.由(2)知,x1+lnx(x>0),h'(x)0,h(x)是區(qū)間(1,+)內(nèi)的增函數(shù).n>m>1,h(n)>h(m),即nlnnn-1>mlnmm-1,mnlnn-nlnn>mnlnm-mlnm,即mnlnn+mlnm>mnlnm+nlnn,lnnmn+lnmm>lnmmn+lnnn.整理,得ln(mnn)m>ln(nmm)n.(mnn)m>(nmm)n,nmmn>mn.4.解(1)函數(shù)f(x)的定義域?yàn)?0,+),且f'(x)=x-1x2,當(dāng)0<x<1時(shí),f'(x)<0,當(dāng)x>1時(shí),f'(x)>0,f(x)在區(qū)間(0,1)內(nèi)為減函數(shù),在區(qū)間(1,+)內(nèi)為增函數(shù).(2)由g(x)=f(x)+ax=lnx-ax+ax,可知函數(shù)g(x)的定義域?yàn)?0,+),g'(x)=ax2+x+ax2.g(x)在其定義域內(nèi)為減函數(shù),x(0,+),g'(x)0.ax2+x+a0a(x2+1)-xa-xx2+1a-xx2+1min.又xx2+1=1x+1x12,-xx2+1-12,當(dāng)且僅當(dāng)x=1時(shí)取等號(hào).a-12.(3)當(dāng)a=0時(shí),f(x)=lnx,h(x)=ex.直線(xiàn)l:y=kx與曲線(xiàn)y=2x+1h(x)=2x+1ex沒(méi)有公共點(diǎn),等價(jià)于關(guān)于x的方程(k-2)x=1ex(*)在R上沒(méi)有實(shí)數(shù)解,當(dāng)k=2時(shí),方程(*)可化為1ex=0,其在R上沒(méi)有實(shí)數(shù)解.當(dāng)k2時(shí),方程(*)可化為1k-2=xex.令g(x)=xex,則有g(shù)'(x)=(1+x)ex.令g'(x)=0,得x=-1,當(dāng)x在區(qū)間(-,+)內(nèi)變化時(shí),g'(x),g(x)的變化情況如下表:x(-,-1)-1(-1,+)g'(x)-0+g(x)-1e當(dāng)x=-1時(shí),g(x)min=-1e,同時(shí)當(dāng)x趨于+時(shí),g(x)趨于+,故g(x)的取值范圍為-1e,+.因此當(dāng)1k-2-,-1e時(shí),方程(*)無(wú)實(shí)數(shù)解,解得k的取值范圍是(2-e,2).綜合,可知k的取值范圍是(2-e,2.5.解(1)不等式f(x)+2g'(x)(a+3)x-g(x),即alnx+2x(a+3)x-12x2,化簡(jiǎn),得a(x-lnx)12x2-x.由x1,e知x-lnx>0,因而a12x2-xx-lnx.設(shè)y=12x2-xx-lnx,則y'=(x-1)(x-lnx)-1-1x12x2-x(x-lnx)2=(x-1)12x+1-lnx(x-lnx)2.當(dāng)x(1,e)時(shí),x-1>0,12x+1-lnx>0,y'>0在x1,e時(shí)成立.由不等式有解,可得aymin=-12,即實(shí)數(shù)a的取值范圍是-12,+.(2)當(dāng)a=1時(shí),f(x)=lnx.由mg(x1)-g(x2)>x1f(x1)-x2f(x2)恒成立,得mg(x1)-x1f(x1)>mg(x2)-x2f(x2)恒成立,設(shè)t(x)=m2x2-xlnx(x>0).由題意知x1>x2>0,則當(dāng)x(0,+)時(shí)函數(shù)t(x)單調(diào)遞增,t'(x)=mx-lnx-10恒成立,即mlnx+1x恒成立.因此,記h(x)=lnx+1x,得h'(x)=-lnxx2.函數(shù)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在區(qū)間(1,+)內(nèi)單調(diào)遞減,函數(shù)h(x)在x=1處取得極大值,并且這個(gè)極大值就是函數(shù)h(x)的最大值.由此可得h(x)max=h(1)=1,故m1,結(jié)合已知條件mZ,m1,可得m=1.6.(1)解f'(x)=1x-x+1=-x2+x+1x,x(0,+).由f'(x)>0,得x>0,-x2+x+1>0,解得0<x<1+52.故f(x)的單調(diào)遞增區(qū)間是0,1+52.(2)證明令F(x)=f(x)-(x-1),x(0,+),則有F'(x)=1-x2x.當(dāng)x(1,+)時(shí),F'(x)<0,所以F(x)在區(qū)間1,+)內(nèi)單調(diào)遞減,故當(dāng)x>1時(shí),F(x)<F(1)=0,即當(dāng)x>1時(shí),f(x)<x-1.(3)解由(2)知,當(dāng)k=1時(shí),不存在x0>1滿(mǎn)足題意.當(dāng)k>1時(shí),對(duì)于x>1,有f(x)<x-1<k(x-1),則f(x)<k(x-1),從而不存在x0>1滿(mǎn)足題意.當(dāng)k<1時(shí),令G(x)=f(x)-k(x-1),x(0,+),則有G'(x)=1x-x+1-k=-x2+(1-k)x+1x.由G'(x)=0得,-x2+(1-k)x+1=0.解得x1=1-k-(1-k)2+42<0,x2=1-k+(1-k)2+42>1.當(dāng)x(1,x2)時(shí),G'(x)>0,故G(x)在區(qū)間1,x2)內(nèi)單調(diào)遞增.從而當(dāng)x(1,x2)時(shí),G(x)>G(1)=0,即f(x)>k(x-1),綜上,k的取值范圍是(-,1).二、思維提升訓(xùn)練7.(1)解由f(x)=x3+ax2+bx+1,得f'(x)=3x2+2ax+b=3x+a32+b-a23.當(dāng)x=-a3時(shí),f'(x)有極小值b-a23.因?yàn)閒'(x)的極值點(diǎn)是f(x)的零點(diǎn),所以f-a3=-a327+a39-ab3+1=0,又a>0,故b=2a29+3a.因?yàn)閒(x)有極值,故f'(x)=0有實(shí)根,從而b-a23=19a(27-a3)0,即a3.當(dāng)a=3時(shí),f'(x)>0(x-1),故f(x)在R上是增函數(shù),f(x)沒(méi)有極值;當(dāng)a>3時(shí),f'(x)=0有兩個(gè)相異的實(shí)根x1=-a-a2-3b3,x2=-a+a2-3b3.列表如下:x(-,x1)x1(x1,x2)x2(x2,+)f'(x)+0-0+f(x)極大值極小值故f(x)的極值點(diǎn)是x1,x2.從而a>3.因此b=2a29+3a,定義域?yàn)?3,+).(2)證明由(1)知,ba=2aa9+3aa.設(shè)g(t)=2t9+3t,則g'(t)=29-3t2=2t2-279t2.當(dāng)t362,+時(shí),g'(t)>0,從而g(t)在區(qū)間362,+內(nèi)單調(diào)遞增.因?yàn)閍>3,所以aa>33,故g(aa)>g(33)=3,即ba>3.因此b2>3a.(3)解由(1)知,f(x)的極值點(diǎn)是x1,x2,且x1+x2=-23a,x12+x22=4a2-6b9.從而f(x1)+f(x2)=x13+ax12+bx1+1+x23+ax22+bx2+1=x13(3x12+2ax1+b)+x23(3x22+2ax2+b)+13a(x12+x22)+23b(x1+x2)+2=4a3-6ab27-4ab9+2=0.記f(x),f'(x)所有極值之和為h(a),因?yàn)閒'(x)的極值為b-a23=-19a2+3a,所以h(a)=-19a2+3a,a>3.因?yàn)閔'(a)=-29a-3a2<0,于是h(a)在區(qū)間(3,+)內(nèi)單調(diào)遞減.因?yàn)閔(6)=-72,于是h(a)h(6),故a6.因此a的取值范圍為(3,6.8.(1)解由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)a>0時(shí),令f'(x)=0,解得x=3a3,或x=-3a3.當(dāng)x變化時(shí),f'(x),f(x)的變化情況如下表:x-,-3a3-3a3-3a3,3a33a33a3,+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)遞增所以f(x)的單調(diào)遞減區(qū)間為-3a3,3a3,單調(diào)遞增區(qū)間為-,-3a3,3a3,+.(2)證明因?yàn)閒(x)存在極值點(diǎn),所以由(1)知a>0,且x00.由題意,得f'(x0)=3x02-a=0,即x02=a3,進(jìn)而f(x0)=x03-ax0-b=-2a3x0-b.又f(-2x0)=-8x03+2ax0-b=-8a3x0+2ax0-b=-2a3x0-b=f(x0),且-2x0x0,由題意及(1)知,存在唯一實(shí)數(shù)x1滿(mǎn)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),-3a3-1<13a3,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-b|=a-1+b,b0,a-1-b,b<0.所以M=a-1+|b|2.當(dāng)34a<3時(shí),-23a3-1<-3a3<3a3<123a3,由(1)和(2)知f(-1)f-23a3=f3a3,f(1)f23a3=f-3a3,所以f(x)在區(qū)間-1,1上的取值范圍為f3a3,f-3a3,因此M=maxf3a3,f-3a3=max-2a93a-b,2a93a-b=max2a93a+b,2a93a-b=2a93a+|b|29×34×3×34=14.當(dāng)0<a<34時(shí),-1<-23a3<23a3<1,由(1)和(2)知f(-1)<f-23a3=f3a3,f(1)>f23a3=f-3a3,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+|b|>14.綜上所述,當(dāng)a>0時(shí),g(x)在區(qū)間-1,1上的最大值不小于14.13