2020版高考數(shù)學(xué)理科(人教B版)一輪復(fù)習(xí)高考大題專項1函數(shù)與導(dǎo)數(shù)的綜合壓軸大題

22-22-高大項函數(shù)數(shù)綜軸題突利用求值值參圍.知函數(shù)f(x=x-k)e(1)求f)的單調(diào)區(qū)間;(2)求f)在區(qū)間[上最小..(2018福建龍巖4質(zhì)檢21改)知函數(shù)f=x-2)e-a求函數(shù)()=f(x+的極值點.(2018山東師大附中一,已知函數(shù)f(x)=(x-a∈R)(1)當(dāng)a=2時求數(shù)f()在0的切線方;(2)求f)在區(qū)間[上最小..(2018陜西咸陽一,21編)知fx=-aln(aR)當(dāng)1時,不等式f)>(x-對任意∈+恒立求實數(shù)的值范..函數(shù)f()=x

+ax+bgx)=

(cx+d)若曲線(x)曲線()都過點P(0,2),且在點有相同的切線(1)求a,,值;(2)若x≥-,f(x)≤(x求的值范圍..(2018河北江西南昌一,已知函數(shù)f(x)=+bx在點(1,處的切線是0(1)求函數(shù)f(x)極值(2)當(dāng)

f)+(m<0)成立時,求實數(shù)m的值范圍為然對數(shù)的底數(shù)).

3232232322突利用證問討零數(shù).(2018全國文21)已知函數(shù)fx)=(1)求曲線y=f(x)點(-處的切線方;(2)證明當(dāng)≥1時,fx)+≥0

-.(2018河北保定一,21編)知函數(shù)f)=(aR)若fx)有兩個極值點,x證1明f.知函數(shù)f(x=ax-x+若f(存在唯一的零點且x>0,求a的值范圍.0.(2018安徽蕪湖期,21編)知函數(shù)f)=xlnx(a∈R).函數(shù)(在(1,e]上存在兩個不同零點求實數(shù)取值范..(2018河南鄭州一,21)知函數(shù)f(x)=(1)討論函數(shù)f()的單調(diào);

aR且a.(2)當(dāng)x

時,試判斷函數(shù))=x-1)e

+x-m零點個數(shù)..(2018河北衡水中學(xué)押題已知函數(shù)f(x)=

-x

,∈,線x)的圖象在點(f處的切線方程為(1)求函數(shù)y=f(x)解析式(2)當(dāng)x∈時,證:fx)-x+x;(3)若f)>kx對意的x∈+恒成立求數(shù)k的值范

k-k-202a-12a-k-k-202a-12a-高考大題專項一函與導(dǎo)數(shù)的綜合突破利用導(dǎo)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')>f)-1),(+).≤≤,(x[,f)[f;0<k-1<1,<k<f[1],[k-1,1]f)[fk-1)=-;1≥≥2f)[0,1]f(x)[f(1)=(1)e,≤f)[(0)=-k1<k<2fx)[fk-1)=-e;k≥2,f)[f(1)=(1-k)e.()=x+1)e-a(x)=(2)e

-2(x+=(x+2)(e

-a(ⅰ)≤0,(∞,,(x)<0,(+)(x)>(ⅱ)0,g'x)=x=-ln(2a①,ln(2)=-2,()≥0;②,ln(2)>-2,(2,ln(2()<0;(--(ln(2a+)g'(x)>0③,ln(2)<-2,(a),-2),g'x)<(-))(2,+),(x)>,≤0,(x)-2,0<a<

,(x)-2,a);-2..(1)

,();a>

,()),f()=2)e,f'()=(.f(0)=-=(0=-1y=-x-2,x+y+=f'x)=(x-a+1),f'()=0,x=a-①a-≤1,≤2,x∈f'()≥0,(x)[1,2].f)=f(1)=(1.min②a-≥2,≥3,x∈f'()≤0,(x)[1,2].f)=f(2)=(2min③11<2,2<a<3,f'(x(x)x:x(1,a-1a-1,2)f')-0+()f)[a-1],[f)[fa-=-e.:≤2,()=f=(1-a)e;mina≥f)=f(2)=(2-a;min2<a<fx)=f1)=-min.f)=-a+x--m1)>(x=

+x-(x-1),=F(x)>0∈+),

22222-2-2-22222222-2-2-222F'()=

+-m,F'=

+-m″(x=

-,1F″()>0,()(+),F'(x)>F'=

+me+()>0,x(+)(x)>F(1)=0,;+<(ln)=>0,x∈(1,lnm),F'(x=11F'()<0,F()(x),()<F(1)=0,m11m≤e+..(1)f(0)=g(0)=f'(0)=(0)=.(x)=,g'x)=

(cx+d+c),2,a=(1),fx)=x+x+(x)=2e(x+1).F(x)=kg(x)-f)=2e(x+1)-x-2,()=k(x+2)-=2(x+e-≥0,k≥1.()=x=-,x=-212①≤e-≤1x∈2,,F'()<1x∈x+),F'()>0.1(x(x,(x,+.(x)[+F(x.1(x=x+---x+≥x≥-2,(x≥0,(x(x).111②(x)=2e(2)(e-e)F'()>()(+(=0,≥-2,(≥0,(x≤).③F(2)=-2k+2=-2e(k-e<0x≥-,f(x≤x,k[].(1)f()=ln(ax∴)=+b=+b,∵(f∴=+b=f(1)=∴a=1,()=x-x+∴)=-=f)f=e-1=0,.

-

∴f)(0,1)(1,+).(1)(x)=

f)+

-

x(m<0),

ln1+

-

x(m<∈(0,+)

2+gx)=h()=-(x)=

-

(x=-m<0,0<x<()<0,()>0;,(x>h'(x)<∴g()(0,1),(+,)(1)=hx)((+)minhx)=h(1)=-∴g(()x=,gx≥h(),()≥h(x),minmax-1,≥-m<∴m[-突破

利用導(dǎo)數(shù)證明問題及論零點個數(shù)

--2x+-x2x+x+21----2222323233223--2x+-x2x+x+21----2222323233223.解f'(x)=

(0)=(x)(0,-1)x-y-1=0(2)證明≥1f)+≥(x+x-+.(x)+x-1+(x)=2x++x<-1,(x)<0,g(x;x>-1,g'()>);(x)≥g(1)=f)+≥0.明f'(x)=

-

-

(x>px)=x+)(x)(+),x,p)=0(+)1--x-

4,∴(x+fx=-121

+x-2

=x-12

f=f=

--

-

=

-

-(a-∴ha)=ln

-

=-a-+=++4),(a)=<--∴h()(+),h(4)=(a<∴.法1f(x)R0,fx=-x+1,原函數(shù)草圖∴a=;0,x-,fx)-f=fx)0;00f')=ax-x,f'x)=axx===<∴f')<12

)>0;(0,+)f'()<0∴()

,(+)∴f()x>f)=f>000min.解法y=ax3x-a≥;0,y=axy=3x1(x0-2,.解法a=-+3=-t+3tg()=-t+t,

+>0

<1a>4∵a<()=-t

+=-t

t((t>t>1t<-,g')<.gt(∞,1),(-(+)t=-1,(t=-t)=-t+3t=g(t=min

33322----33322----x+g()→+a<-y=ag()=-t.

+t,.f)=0,

(y=a的(x=

g'(=

-

(x)=0x∈(1,),(x<0,(1,)x∈(,g'()>(;gx)()=()==>且g(e)=<27,ming(x=∴a(].

,.(1))=

-

(f'(x>f(x(0,+;0,)>f'x<<x<

,fx(+),(x)∵x

,g=(lnx-1)e+x-m,(lnx-.hx)=(ln1)e()=-e

+(a=1,f)=x+-∴()≥(1)=0,

[,+x-1≥0

∴()=-

+1≥+1>∴h(x=x-

+x,x∴h()=h=-2h(x)=.minmax.解,)=-2,

∴m<-2

m>e-2me(0)==b.((xa=-(x)=-x1(2)證明()=f(x+x-x=e-