2020版高考文科數(shù)學(北師大版)一輪復習試題：大題專項突破高考大題專項1含答案

232232高大項函與數(shù)合軸突利用求值值參圍.知函數(shù)f(x=x-k)e求f()的單調(diào)區(qū)間;求f()在區(qū)間[上最.(2018山東濰坊一,21)知函數(shù)()ln若a=-判斷fx)在(+∞)的單調(diào)性求函數(shù)f()在1,e]上的最小值..(2018山東師大附中一21)已知函數(shù)f)=(x-aaR)(1)當a=2時求數(shù)f()在x=的切線方;(2)求f)在區(qū)間[上最.(2018遼寧撫順3模,改已知函數(shù)fx=ax-2lnx(a∈).f(x)>任意∈+)恒成立,求取值范..函數(shù)f()=x+ax+b(x)=cx+d).若曲線(x和曲線()都過點P且點有相同的切線求a,,,值;若x≥-,f(x)≤(x),求的值范圍..(2018江西南昌一,21編)知函數(shù)f()=lnx-a∈R其中為然對數(shù)的底數(shù)若當x∈+∞時,fx)≥成立求的值范.突利用證問討零數(shù)32322323222.(2018全國文21)已知函數(shù)f(x=(1)求曲線y=f(x)點(-處的切線方;(2)證明當≥1時,fx)+≥.(2018河北保定一,21編)知函數(shù)f()設(shè)函數(shù)g(x)=x+1證:當x∈+)且a>0時,fx)>g(x).知函數(shù)f(x=ax-x+1,若(x)存在唯一的零點,且x>求取值范0.(2018安徽蕪湖期,21編)知函數(shù)f()ln(aR).函數(shù)x)在區(qū)間(1,e]上存在兩個不同零點求實數(shù)取值范..函數(shù)f()=lnx.(1)討論()的導函數(shù)f')點的個數(shù)(2)證明當a>f)≥2..(2018衡水中學押題,21)已知函數(shù)f(x)=+ax∈,曲線()的圖像在點0,f處的切線方程為y=bx.求函數(shù)()的解析式當x∈時,證:fx)-x+x;若f()>kx對意的x∈+)恒成立求數(shù)k的取值范圍k-k-2222202k-k-2222202高考大題專項一函與導數(shù)的綜合壓軸大題突破利用導數(shù)求極值、值、參數(shù)范圍.(1)f'()=x-k+1)e.(x)=0,x=k-1.x∈∞k-,f')<x∈k-1,+∞)f'()>0f)-k-1),(1,∞.≤≤,fx[f)[f(0);0<k-1<1,<k<fx)[1][,f(x[fk-=-e;1≥≥2f(x[0,1],f)[f(1)=-k)e.,≤f)[0,1]f(0)=-k;1<k<2fx)[fk-=-ek≥2,f)[f(1)=.

;.(1)a=-2,f'(x=2x∈+∞),x>∴()(+∞).

f')=x+

≥f'()≥f()[

∴(x)=f(1)=1.min0,)=0x=±

(),x=.0

≤1,a-2,-≤x∈[1,e],f'(>f(x),∴()=f=1.min1<<e,-

2

<a<-x∈x],)≤0,(xx∈[x,e],)≥f).00f()=f(x=-min

+aln

≥e,a≤,∈f'x)<0,(x)∴()=f=+a.min:≥-2,()1;-2ea-,fx)+a..(1)

<a<-2,f(

;f()=2)ef'(x)=(x-.f(0)=-=(0-1)=-.2,x+y+2=0f'x)=(x-a+1),f'x)=0,x=a-1①a-≤1,≤2,x∈f')≥f(x)[1,2]f)=f(1)=(1-a)emin②a-≥2,≥3,x∈f')≤f(x)[1,2]f)=f(2)=(2-a.min③11<2,2<a<f'(x(x)x:x1)11,2)f')-+fx)2a-132232222222-22-222a-132232222222-22-22f)[1,1],[a-.f)[(a-=-e:≤2,()=f=(1mina≥f)=f(2)=(2-a)e;min2<a<fx)=f=-.min

.f(x+x>0,a>-x+

x∈+,px)=-x+(x)=-2x+

+∞),qx)=-2x+2xq'x=-x-

x>1,(x)q()(1)=0,(x)<0(1,),()=-x+(+∞,p()(1)≥a[-1,).(1)f(0)=g=f'(0)=g'(0)=(x)=,g'x)=cx+d+c2,a=(1),fx)=x+x+2,g(x)=x+1)(x(x)-fx)=k(-x-x-2,()=k(x+2)-=2(x+e-.≥0,k≥()=x=-kx=-1①≤e-≤1x∈2,,F'()<1x∈x+()>01(x(x,(x+∞.11(x[-2,)F(x)1(x=x+--4x-+2)≥11x≥F()≥fx≤(x).②(x)=2e(2)(e-e)F'()>(x(-2,∞).(=0,≥-2,F)≥f()≤kg().③F(2)=-2e+2=-(e<0x≥-,f(x≤).,k[]..f)=

lnx-∈Rx)=

-,0f')=->fx)x∈+),fx=f(1)=min0f')=

-,∈+∞),

≥.①∈(0,e]x∈[1,∞≤e,f')=-≥f()[+∞f()=f(1)=0()min②∈(e,∞∈[1,+),f'(x=-=0f()x∈[1,x(x+,f(x<f=000x∈[1,∞)fx)≥0,(-∞2x+12x+x+12222232323322332x+12x+x+1222223232332233突破

利用導數(shù)證明問題及論零點個數(shù).解f'(x)=,f'(0)=(x)(0,-1)1=0(2)證明≥1f)+≥x+x-1+e.gx)=x+,g'()=2x+1+x<-1,(x)<0,g(x;x>-,g')>gx;g(≥g(=0f)+≥0.明令h)=f)(x)=x+-x-(x=-

px)=x-x-a=0,(x

∵p(1)=1-1px)=0∴>00,p)=0,-a=0,0(x)(0,x,(x,∞),0(x)=h=x+ln1=x+--=x-ln-.min000(x=2lnx>F'()=-F(x)(.∵(1)=-02=0,F(x)>0,(x)>0,min,∈(0,+∞)f))

>,.法1f(x)R0,(=-x+±原函數(shù)草圖∴a=;0,x-,fx)→-f=f)0x,;00f')=ax-x,f'x)=36x===<12∴

f'<f'()>0;(0,))<0∴()

,

,(0,+∞).∴f()x>0f)=f00min∵a<0,∴a<-2

>

+1>0

<>.解法y=ax3x1a≥;0,y=axy=-1,y),00

2,.解法a=-+=-t+3t,y=a(t=-t+3()=-t

+=-t

∈(1,1)g'(t>t>t<-1,(x<(t)(∞-,(1,1),(1,+∞t=-1,(t=-(t=-t+3t=g()=.minx+∞g(t→a<-,g()=-t+3t,.3322233222.f)=0,

(,y=agx)=

()=,(x)=0,x∈(1,),)<0,(1,)x∈(,g'()>0,(;gx)()=3e,g(==27>且g(e)=<27,ming(x)=(

].解fx)(+∞),f'()=-(x>0).a≤f')>0,f'(x0,-

,f')(0,).()><b<b<f'(b)<0,f').(2)證明(f'()(+∞xx∈(0,x,f')<x∈+),f')>000(x)(,(x,+∞),x=x,()f.000f=0

=0,+ax+aln0

≥2a+alnf)≥2ln.解,)=-x,==b.(0,0),(),1,(x)=-x1(2)證明()