高考專題1把不等式轉(zhuǎn)化為f >f 判斷函數(shù)f(x“f”等式(組)但要注意函數(shù)奇偶性的區(qū)別①f(xRf(x在(0,+∞單調(diào)遞增若解不等式f(x1f(x2②f(xRf(x在(0,+∞單調(diào)遞減若解不等式f(x1f(x22模型1.對于fr(xgr(x)構(gòu)造h(xf(x2.對于不等式fr(xk(k≠0g(x=f(xkx3fr(xf(x0g(xex拓展對于不等式fr(xkf(x0g(xekx4對于不等式fr5.xfr

-f

00

==xf拓展對于不等式xfr(xnf(x0g(xxn6.xfr

-f

0

=f(x(x≠0拓展xfr

-nf

0

=00(2)若f(x)<0則構(gòu)造h(x)=ln[-0)模型9.對于fr(x)lnx+

0)10.(1fr(xf(x)tanx(fr(xf(x)tanx)fr(x)cosxf(x)sinx00)，構(gòu)造h(xf(x)cosx．(2對于fr(x)cosxf(x)sinx00)h(xf(x)模型11.(1)fr(x)sinx+f(x)cosx(2)fr(x)sinx-

l題型 含一次函數(shù)結(jié)合(x)構(gòu) fr(xf(x(xRxR,fr(x1,f(32f(x>x-1的解集為 (-

(-

fr(xf(x(xRxR,fr(x2,f(23f(x>2x-1的解集為 (-

(-

(2425高三下·廣東深圳·月考f(xR上的奇函數(shù)，f(20x0時有xfr(x+f(x>0成立則不等式f(x)>0的解集是 (-2,0)∪(2, C.(-∞,-2)∪ D.(2,(24-25高三下·河北邯鄲·月考)已知函數(shù)fr(x)是定義域?yàn)镽的奇函數(shù)f(x)的導(dǎo)函數(shù)當(dāng)x<0時，xfr(x)-f(x)>0且f(2)=4則不等式f(x)≤2x的解集為( A.[- B.(-∞,-2]∪[2,C.[-2,0]∪[2, D.[-2,0)∪[2,(2024·廣東佛山·一模fr(xf(x的導(dǎo)數(shù)，f(1-xf(1x0，f(20當(dāng)x>1時，(x-1fr(x-f(x>0則使得f(x<0成立的x的取值范圍是 A.

B.

C.(-

D.(-

題型 含二次函數(shù)結(jié)合(x)構(gòu) 設(shè)函數(shù)f(x在R上存在導(dǎo)數(shù)fr(xxR有fr(xx若f(1k≥

-k則k的取值范圍是 (-

1(2324高三下·重慶·月考已知fr(x是函數(shù)f(x(xRxfr(x>2x，f(2=5則不等式f(x>x2+1的解集為 (-

(-

2(2024·山東聊城·三模f(xRfr(xx0時，fr12x+2的解集為 A.(- -1

1,1 -∞,-

已知函數(shù)f(x及其導(dǎo)數(shù)fr(xRx，f(x=f(-x

x10.不等式f(2x-

-f

<-

+3x的解集為 A.(- 2,+∞

2,2 -∞,2

題型 含冪函數(shù)結(jié)合(x)構(gòu) f(xRf(32x02f(xxfr(x則x2f(x<18的解集為 A.(-C.(-

B.(-D.(-∞,-

f(x(-∞,0)，f(-11fr(xxfr(x2f(x)>0則不等式f(x+2025)+(x+2025)2<0的解集為 A.(-2026, B.(-∞,- C.(-∞,- D.(-2026,-R上的奇函數(shù)f

的導(dǎo)函數(shù)為fr

x0f

+xfr

則不等式x2f(x<(3x-1)2f(3x-1的解集為 (-∞,

∪(1

(1,

1(2324高三上·江西·月考f(xR上的奇函數(shù)，fr(xx時，xfr(x+3f(x>0恒成立則不等式x3f(x+(2x-13f(1-2x<0的解集為 1C.(-

-∞,1

(2324高三上·江蘇蘇州·月考已知奇函數(shù)f(x(x≠0的導(dǎo)函數(shù)為fr(xf(-2=0．當(dāng)x>0時，3f(x>xfr(x則使得f(x>0成立的x的取值范圍為 A.(-∞,-C.(-∞,-

B.(-D.(-

(23-24高三上·湖南衡陽·月考)設(shè)函數(shù)fr(x)是函數(shù)f(x)(x∈R)的導(dǎo)函數(shù)若f(x)-f(-x)=2x3且當(dāng)x≥0時，fr(x)>3x2則不等式f(x)-f(x+1)+3x2+3x+1>0 -∞,-

B.(-∞,-

-1

D.(-2,題型 含指數(shù)型函數(shù)結(jié)合(x)構(gòu) (2425高三下·黑龍江雞西·月考Rf(xfr(x且f(x+fr(x<0，f(ln2=1則不等式f(xex>2的解集為 A.(-

B.(-

C.

D.(2425高三下·內(nèi)蒙古赤峰·月考f(xRfr(x足fr(x>f(x且f(x+5為偶函數(shù)，f(10=1則不等式f(x<ex的解集為 A.

B.(-

C.(-

D.(2324高三上·江西撫州·期中(-2,2f(xfr(xf(xe4xf(-x0,f(1e2x0時，fr(x2f(xe2xf(2-<e4的解集為 A.

B.(-

C.

D.2Rfr(x)等式f(x)<1+e-x的解集為( A.(- B.(0,C.(-∞,-1)∪(1, D.(-∞,-1)∪(2425高三下·重慶南岸·月考函數(shù)f(xR且f(xfr(xf(0=3則不等式f(x>2ex+1的解集為 A.(-

B.

C.(-

D.已知可導(dǎo)函數(shù)f

的導(dǎo)函數(shù)為fr

xR都有f

<fr +A.B.yf(x12024e2+x1f(xA.B. C.(- D.(-已知函數(shù)f

(0+∞

fr

>-f

ln2f(lnx<f(2的解集為

題型 含對數(shù)型函數(shù)結(jié)合(x)構(gòu) (2324高三上·江蘇揚(yáng)州·開學(xué)考試f(xRx0lnxfr(x

?f(x)<0則不等式(x-2)?f(x)>0的解集為 (-

(-

上的函數(shù)f

滿足xfr

x1f

=

f(ex>ex+x的解集為 A.

B.

C.

D.(2025·浙江·二模)已知函數(shù)f

為f

足fr

xlnx+f

x2

2a,bf

≥f(b-

a2 的最小值( b+A.22- B.

C. D.(2324高三下·遼寧·月考已知函數(shù)f

,fr

，xfr(x)-f

lnx+f

0則不等式f

<0的解集是

題型 含三角函數(shù)結(jié)合(x)構(gòu)(2324高三上·黑龍江齊齊哈爾·期末已知函數(shù)f

是fr2fπ

.xsinx的解集為

有fr

sinx-f

cosx0則關(guān)于x的不等式f(x

π,πR上的函數(shù)f(x)滿足f(xf(-x2sinx0x1x2f(x1+sinx1-f(x2-x1-

0f(x+

f(x)+sinx-cosx -∞,π

π

-∞,π

-π定義域?yàn)?-π, 的函數(shù)f(x)滿足f(x)+f(-x)=0其導(dǎo)函數(shù)為fr(x)當(dāng)0≤ <πfr(x)cosx+f(x)sinx0x的不等式f(x

fπcosx為 (-π,-

(-π

∪(0,

(-π

(2324高三上·廣東·月考f(xfr(x(-π,π且f(x為偶函數(shù)，f( =-2，3f(xcosx+fr(xsinx>0則不等 fx+

cos3x-

0的解集為 (-π

(-2π,

(-2π 已知函數(shù)f(x及其導(dǎo)函數(shù)fr(x的定義域均為(ππf(x

≥f

tanx

2

< 的解集為 題型 其他類型構(gòu) (2324高三下·江蘇南通·月考f(xfr(x)f(1e0時，fr(x

ex則不等式f(x)-

<1的解集為 A.(0,1)∪(1,

C.

D.(2024·江蘇南通·模擬預(yù)測Ryf(xyfr(x若fr(x+(x+12也為偶函數(shù)且f(2a+4>f(a2+1則實(shí)數(shù)a的取值范圍是 A.(-∞,-C.(-

B.(-∞,-D.(-

已知函數(shù)f(x的定義域?yàn)?0,+∞fr(x(x+

2f

+xfr

xf

f

7f(x3

3x+12(x+3集為

(-

(-

(2324高三上·陜西安康·月考R上的連續(xù)函數(shù)f(x滿足f(x1x1時，(x-1fr(xf(x0其中fr(x是f(x的導(dǎo)數(shù).xe2xf(e2x+1≥(2x-af(2x+1-a恒成立則實(shí)數(shù)a的取值范圍為 (-∞,ln2-

ln2-

(-∞,-

-(2324高三上·重慶渝中·月考f(x(0,+∞fr

(x

2f

+xfr

xf

f

7f(x43x+(x+4

的解集為 (-

(-

(-檢 重難知識鞏 (2324高三上·湖北武漢·期中已知函數(shù)的定義域?yàn)镽，f(15對任意x(x)<2則f(x)>3+2x的解集為 A.(- B.(2, C. D.(-(2025·湖南·三模已知y=f(x是定義在(1,+∞=fr(x若xfr(x<f(x且f(3=6則不等式f(lnx>2lnx的解集為 A.

B.

C.

D.R上的函數(shù)f(x的導(dǎo)函數(shù)為fr(x)且滿足fr(xf(x),f(2024e2024不等式f(1lnx<3x的解集為 A.

B.

C.

D.(2024·寧夏銀川·三模)已知定義在R上的奇函數(shù)f(x)的圖象是一條連續(xù)不斷的曲線，fr(x是f(xx0時，3f(xxfr(x0且f(22則不等式(x1)3f(x+1)>16的解集為 A.(1, C.(- D.(-∞,-3∪(2425高三下·河北邢臺·月考f(xRf(11fr(x滿足fr(x-2f(x>0則不等式e2f(ln(x- <(x-12的解集為 A.

B.(1,e+

C.(e,e+

D.(e+fr

(2024·云南楚雄·一模f(xRxRf(x 1-

成立．若f(-1=-1則不等式f(x

的解集為 (-∞,-

(-

(-

(2324高三上·河北·月考已知函數(shù)f(x及其導(dǎo)函數(shù)fr(x+∞)且xfr(xx1)f(x恒成立，f(3e則不等式(x4)f(x43ex+2 A.(-4,- B.(- C.(- D.(-1,已知函數(shù)f(x)及其導(dǎo)函數(shù)fr(x)的定義域均為R且f(x)為偶函數(shù)f( =-3f(x)cosxfr(x)sinx0則不等式f(x+

cos3x-

0的解集為 A.(-B.(-A.(-π B.(-2π C.(-2πA.(-B.(- Rf(xfr(x，f(1exfr(xex C. D.設(shè)奇函數(shù)f(x的定義域?yàn)?ππf(xx -π

fr

cosx+f

sinx0若

f(m)<f

m)(-π,-

(0,

(-π,

(2324高三上·四川成都·月考Rf(xfr(x當(dāng)x≤0時滿足(x2+3fr(x+2xf(-x>0則不等式f(x≥f(2x+ 的解集為

x2+ 4(x2+x+

(-

(-∞,-

(2324高三下·上?！ぴ驴糵(xR且當(dāng)x<0時，2f(x+xfr(x)<0則不等式(x-20232f(x-2023)-f(-1)>0的解 (2425高三上·海南省直轄縣級單位·期中Rf(xf(220f(xfr(xfr(x6x22f(x2x32x(2025·山西·模擬預(yù)測f(xRfr(xfr(xf(x0e2x+2f(x2f(-x(2024·云南·模擬預(yù)測)已知fr(x是定義域?yàn)?0, 的函數(shù)f(x的導(dǎo)函數(shù)fr

sinx+f

cosx0則不等式f

sinx>1f

的解集 (2425高三下·廣東梅州·月考f(xRfr(x)xR有f(xf(-x2sinx且在[0上fr(xcosx．若fπ-t-f(tcostsint (2425高三上·廣東潮州·月考(0,+∞上的函數(shù)f(x的導(dǎo)函數(shù)為fr(x當(dāng)x>0時，xfr(x<2且f(e=5則不等式f(x2-4lnx<3的解集 (2425高三上·上?！て谥蠷f(xf(xfr(-xf

1f(x

1的解集 ex- (2425高三上·山東日照·月考f(xfr(x且F(x=ex+1f(x+1是偶函數(shù)其函數(shù)圖象為不間斷曲線且(x-1fr(x+f >0,不等式xf(lnx<e3f(3的解集 (2324高三下·北京·月考R上的函數(shù)f(x滿足f(2xf(-xx1xfr

fr

f

1則不等式f

的解集是x-(2024·河南·三模f(x(0,+∞，fr(xx(0,+∞，f(x>f(x-xfr lnx則不等式f(x(ex-1-1>0的解集 檢測 創(chuàng)新能力提 已知函數(shù)f

f

=e-1，fr

ex-2f(x>x2的解集為

(2223高二下·四川成都·月考f(x(0fr(x)fr2fx+2023的解集為

f

0(x+2023f(x+2023A.x|x>-C.x|-2023<x<0

B.x|x<-D.x|-2023<x<-(2024·四川德陽·三模f(xfr(xR(x-2fr(xf 0，f(4-xf(xe4-2xe3f(lnxxf(3 A.

B.

C.

D.Rf(xf(2xf(2-xx2xfr(xf

2fr

f

1則不等式f

< 的解集是( x-A.

(-

C.

D.(-

(2324高三下·河南南陽·月考已知函數(shù)f(xRfr(x)，x滿足f(-xf(xcosx1x0時，fr(x1sinx0．若f(x+ +1sin(x+ ≥f(-x)則x的取值范圍 (2025·遼寧撫順·模擬預(yù)測f(xR，f(xfr(x3，f(14xxRxx時都有f(x1-f(x20lnf(x-3x

x1-的解集 (2425高三上·遼寧·期末Rf(xx0時，f(x且x>0時，e2x?fr(x-f <fr(x+f(x恒成立且f(1=e2-1則x≠0時不等f(x>e1+x-e1-x的解集 高考專題1把不等式轉(zhuǎn)化為f >f 判斷函數(shù)f(x“f”等式(組)但要注意函數(shù)奇偶性的區(qū)別①f

Rf

若解不等式f

f

②f

Rf

若解不等式f

f

2模型1.對于fr(xgr(x)構(gòu)造h(xf(x2對于不等式fr

=f

-kx+3對于不等式fr

0

=ex拓展對于不等式fr

+kf

0

=ekx4對于不等式fr5.xfr

-f

00

==xf拓展xfr

+nf

0

=xn6.xfr

-f

0

=f(x(x≠0拓展xfr

-nf

0

=7

00(2)若f(x)<0則構(gòu)造h(x)=ln[-0)模型9.對于fr(x)lnx+

0)10.(1fr(xf(x)tanx(fr(xf(x)tanx)fr(x)cosxf(x)sinx00)，構(gòu)造h(xf(x)cosx．(2對于fr(x)cosxf(x)sinx00)h(xf(x)模型11.(1)fr(x)sinx+f(x)cosx(2)fr(x)sinx-

l題型 含一次函數(shù)結(jié)合(x)構(gòu) 已知fr

是函數(shù)f

(x∈

xR,fr

1,f

2f(x>x-1的解集為 (-

(-

=f

x1

的單調(diào)性可得答案

=f

x1因?yàn)閒

2

=f

-3+1=

=fr

1因?yàn)閒r

1所以gr

=f

x1因此由f

x-1?f

-x+1>0?

?x>已知fr

是函數(shù)f

(x∈

xR,fr

2,f

3f(x>2x-1的解集為 (-

(-

2x1可化為f

-2x+1>f

221

=f

2x+1

2x1可化為f

-2x+1>f

-2×2+

=f

2x+1則原不等式可化為

g(2

=fr

-因?yàn)閒r

2

x

=f

故不等式f

2x1的解集為(2,+∞(2425高三下·廣東深圳·月考f(xR上的奇函數(shù)，f(20x0xfr

0成立則不等式f(x)>0的解集是 A.(-2,0)∪(2, C.(-∞,-2)∪ D.(2,解不等式

=xf

=xf

Fr

=xfr

x0xfr

x0時，F(xiàn)r(x0，F(xiàn)(x單調(diào)遞增；又因?yàn)閒(x是定義在R上的奇函數(shù)，F(xiàn)(-xF(x是偶函數(shù)，x0時，F(xiàn)(x是單調(diào)遞減函數(shù)

=-xf(-

=xf

=F(x又因?yàn)閒(20則f(-20,F(2F(-2F不等式f(x)>0等價 >0x0時，F(xiàn)(x0-2x所以不等式f(x0(-2,02.(24-25高三下·河北邯鄲·月考)已知函數(shù)fr(x)是定義域?yàn)镽的奇函數(shù)f(x)的導(dǎo)函數(shù)當(dāng)x<0時，xfr(x)-f(x)>0且f(2)=4則不等式f(x)≤2x的解集為( A.[- B.(-∞,-2]∪[2,C.[-2,0]∪[2, D.[-2,0)∪[2,F(x

F(x)在(-∞,0)上單調(diào)遞增.根據(jù)f(x)為R的奇函數(shù)得到F(x為RF(x的性質(zhì)可求f(x2x

Fr(x

xfr(x)- x0時，F(xiàn)r(x0F(x

在(-∞,0因?yàn)閒(x)F(xf

F(x在(0上單調(diào)遞減因?yàn)閒(2)=4所以F(2) =2故F(-2)=F(2)=x0f(x2xF(x

≥2=F(-2因?yàn)镕(x在(-∞,0-2xx0f(xRf(00滿足f(xx0f(x2xF(xF(x(0x2.綜上，f(x2x的解集為[-2,02.

≤2=F(25(2024·廣東佛山·一模fr

是函數(shù)f

的導(dǎo)數(shù)，f(1-

+f(1+

0當(dāng)x1時，(x-

fr

f

0

<0成立的x的取值范圍是

(-

(-

=f(xg(xx-1

0f(1-xf(1x0g(1-xg(1xg(xg(x(-∞,10000f(x0f(x0x的取值范圍f

fr

(x-1-f

x-1g(x

(x- 0g(x在(1,+∞g(2f(2f(1-xf(1x0(1-x-1g(1-x(1x-1g(1x-xg(1-xxg(1x0g(1-xg(1xg(xx1g(x在(-∞,1g(0g(2x0時，g(x0f(x1x2時，g(x0f(x所以使得f(x0x的取值范圍是(-∞,0(1,2題型 含二次函數(shù)結(jié)合(x)構(gòu) 設(shè)函數(shù)f(x在R上存在導(dǎo)數(shù)fr(xxR有fr(xx若f(1k≥

k則k的取值范圍是 (-

1

=f

1x2g(1-kg(k

=f

x2

=fr

-x>g(xR由f(1-

f

≥

k得g(1-

0即g(1-

≥g(k

R1kkk1(xRx(xRx(2324高三下·重慶·月考fr(xffr(x2x，f(25則不等式f(xx21的解集為(-

(-

2

=f

x2

1不等式f

x2

=f

x2

=fr

-2x>

Rf

5

=f

-22=不等式f

x21即f

x21

x即不等式f

x21(2,+∞(2024·山東聊城·三模)設(shè)函數(shù)f

的定義域?yàn)镽導(dǎo)數(shù)為fr

2x-1且對于任意的實(shí)數(shù)x,f(-+2的解集為

=f

2xf(2x-

f

<3x2-A.(- -1

1,1 -∞,-

=f

x2xg(xRg(x在(0,+∞(-

f(2x-

f

3x25x2轉(zhuǎn)化為f(2x-1(2x-12+(2x-

<f=f

-x2+x即g(2x-

<

=f

-x2+則g(-

=f(-

x2-x=f

+2x-x2-x=即g(x為Rx0時，fr(x2xgr

=fr

2x10g(x在

(-

因?yàn)閒(2x-

f

<3x2-5x+所以f(2x-

-(2x-12+(2x-

<f

-x2+即g(2x-

<

2x-

x(2x-12

<x<從而求解不式

=f

x2x研究函數(shù)g(x已知函數(shù)f

及其導(dǎo)數(shù)fr

Rx，f

=f(-

-

x10.不等式f(2x-

f

<-

+3x的解集為

(-

22

-∞,2

=f

1x2x

f(xg(x即可得解

=f

1x2xgr

=fr

+x+x0時，fr(xx10g(x在(0,+∞由f

=f(-

2x

-1x2-x=g(-

-1x2+x-即g(xg(-xg(xg(x在(-∞,0則不等式f(2x-

f

<-

3x可化為：g(2x-

(2x-22-(2x-

+1x2+x<-3x2+ g(2x-2g(x2x-2x(2x-22(2x-2x(2x-2-x0(3x-2(x-2x(2,2

=f

1x2x

的單調(diào)性與對稱性題型 含冪函數(shù)結(jié)合(x)構(gòu) f(xRf(32x02f(xxfr(x則x2f(x<18的解集為 A.(-C.(-

B.(-D.(-∞,-

f(x)函數(shù)f(xg(xx2f(x在x0時，2f(xxfr(x0，2xf(xx2fr(x∴gr(x=2xf(x+x2fr(x>g(xx2f(x(0∵f(3)=x2f(x18x2f(x1832g(|x|)x0|x|x0(-3,3).已知函數(shù)f(x的定義域?yàn)?-∞,0)，f(-11其導(dǎo)函數(shù)fr2f(x)>0則不等式f(x+2025)+(x+2025)2<0的解集為

滿足xfr(x)-A.(-2026, B.(-∞,- C.(-∞,- D.(-2026,-F(x

F(x2025F(-

0)

x(xfr(x)- 結(jié)合題干可知Fr(x)<0即F(x)在(-∞,0)0(x2025)20(x2025)2可得f(x(x+

<-1=f(-1)(-即F(x+2025)<F(-F(x(-∞,0上單調(diào)遞減和定義域可得：-1x20250，即-2026x<-2025，f(x2025(x2025)20(-2026-R上的奇函數(shù)f

的導(dǎo)函數(shù)為fr

x0f

+xfr

則不等式x2f(x<(3x-1)2f(3x-1的解集為 (-∞,

∪(1

(1,

1

x2f x2f

x2

g

=xf

+2f(x

+xfr x

x2f由題可知f

+2f

0g(x

y

x2f x2f(x3x1)2f(3x-1g(xg(3x-1x3x1x1(2324高三上·江西·月考f(xR上的奇函數(shù)，fr(xx時，xfr(x+3f(x>0恒成立則不等式x3f(x+(2x-13f(1-2x<0的解集為 1C.(-

-∞,1

=x3f

【解析】令g(x=x3f(x則gr(x=3x2f(x+x3fr(x=x2xfr(x+3f x0時，gr(x0g(x(0,+∞因?yàn)閒(x為奇函數(shù)所以g(-x=(-x3f(-x=-x3?-f =x3f(x=g(x即g(x為偶函數(shù)所以原不等式變?yōu)間(x<g(2x-1所以g( <g(2x-1

2x-1x

x故原不等式的解集為(-∞, ∪(1,+∞(2324高三上·江蘇蘇州·月考已知奇函數(shù)f(x(x≠0的導(dǎo)函數(shù)為fr(xf(-2=0．當(dāng)x>0時，3f(x>xfr(x則使得f(x>0成立的x的取值范圍為 A.(-∞,-C.(-∞,-

B.(-D.(-

f f(-0F(-2F(20x0x0f

x3fr

-3x2f

xfr

-3f【解析】F(x

F(x

x0時，F(xiàn)rf

=xfr

-3f 0故F(x 在(0,+∞上單調(diào)遞減又f(x(x≠0為奇函數(shù)，f(-20故f(2且F

f 定義域?yàn)?-

∪(0,+∞F(-

=f(-x=-f(x=f(x=F(x(-xf

-

fF(x

F(x

在(-

F(-

=F

=x0fx0f

0F0F

00x0故x<-故使得f

0成立的x的取值范圍為(-∞,-

∪(0,2(23-24高三上·湖南衡陽·月考)設(shè)函數(shù)fr(x)是函數(shù)f(x)(x∈R)的導(dǎo)函數(shù)若f(x)-f(-x)=2x3且當(dāng)x≥0時，fr(x)>3x2則不等式f(x)-f(x+1)+3x2+3x+1>0 -∞,-

(-∞,-

-1

D.(-2,F(xf(xx3由題意判斷出F(x1)31)1|【解析】令F(x)=f(x可知F(x的定義域?yàn)镽且Fr(x)=fr(xf(xf(-x2x3f(xx3f(-x(-x3，F(xiàn)(x)3x2可知F(x在

對于不等式f(xf(x13x23x11)31|F(xf(xx3題型 含指數(shù)型函數(shù)結(jié)合(x)構(gòu) (2425高三下·黑龍江雞西·月考Rf(xfr(x且f(x+fr(x<0，f(ln2=1則不等式f(xex>2的解集為 (-

(-

F(xf(xexF(xF(xRF(ln22F(x的單調(diào)性可解函數(shù)不等式【解析】f(xfr(x0F(xf(x則Fr(x=fr(xex+f(xex=exf(x+fr <0故F(x在R上單調(diào)遞減F(ln2f(ln2eln22所以不等式f(xex2F(xF(ln2的解集為(-∞,ln2(2425高三下·內(nèi)蒙古赤峰·月考f(xRfr(x足fr(x>f(x且f(x+5為偶函數(shù)，f(10=1則不等式f(x<ex的解集為

(-

(-

exf(x

f(101f(01g(01g(xg(0)利用單

fr(x)ex-

fr(x)-

g(x

0g(xRf(101f(x5f(-x5f(x5f(0f(101g(0

=于是f

1g(0)即不等式f(xex的解集為(-∞,0(2324高三上·江西撫州·期中(-2,2f(xfr(xf(xe4xf(-x0,f(1e2x0時，fr(x2f(xe2xf(2-<e4的解集為

(-

構(gòu)造函數(shù)

f e2xg(2-

<

g(x在(-

f(xf

+e4xf(-

+g(-

=f(x+f(-xf

+e4xf(-

=

fr(x-2f 又g(x

在(-

f(11e2xf(2-

<e4?f(2-x<1?g(2-e2(2-

<g(12-2-2Rfr(x)等式f(x)<1+e-x的解集為( (- B.(0,C.(-∞,-1)∪(1, D.(-∞,-1)∪g(xexf(xex1利用導(dǎo)數(shù)確定單調(diào)性并求解不等式【解析】g(xexf(xex1由f(02g(0f(02f(x1fr(x)gr(xexf(xfr(xexexf(xfr(x1g(xRf(x1e-xexf(x1exg(x0g(0)，x0所以不等式f(x1e-x(-∞,0).(2425高三下·重慶南岸·月考函數(shù)f

R且f

<fr

f

3

2ex+1的解集為 (-

(-

f(x)-

fr(x)ex-(f(x)-

fr(x)-f(x)+對g(xg(x

已知f(xfr(x1即fr(xf(x10而ex0gr(x0恒成立這說明函數(shù)g(x在這說明函數(shù)g(x在R上單調(diào)遞增已知f(03g(0)=f(0)-1

=不等式f(x2ex1可變形為f(x12ex即f(x)-

2g(xg(xRx0.f(x2ex1(0+∞).已知可導(dǎo)函數(shù)f

的導(dǎo)函數(shù)為fr

xR都有f

<fr +yf(x12024e2+x1f(x2024e2x1

(-

(-g(x

f(x- f(1- 2024數(shù)不等式即可

xR都有f

<fr + 2fx-

-fr

<f(x-

fr

e2x-2e2xf(x-

fr

-2f(x-g(x

g(x

g(xRyf(x12024e2+x1所以f

2024e210f(1-

=-而不等式f

2024e2x1等價于f(x-

<-2014=f(1-1g(xg(1g(xRxf(x(0+∞fr(x>-f(xln2f(lnx<f(2的解集為

fr(x>-f(xln2和f(lnxf(2F(x2xf(x(x>0

【解析】F(x2xf(x(x>0因?yàn)閒r(x>-f(xln2即f(xln2fr(x則Fr(x=2xf(xln2+fr >0(x>0F(x在(0,+∞F(lnxF(22lnxf(lnx22f(2f(lnxf(20lnx2lne2，1x

f(lnxf(2(1,e2

題型 含對數(shù)型函數(shù)結(jié)合(x)構(gòu) (2324高三上·江蘇揚(yáng)州·開學(xué)考試f(xRx0lnxfr(x

?f(x)<0則不等式(x-2)?f(x)>0的解集為 (-

(-

F(xlnxf(x，x(0,+∞F(xlnxf(xxf(xR上的奇函數(shù)得到f(x0則Fr

=

f

+lnx?fr(xx0時，lnxfr(x

?f(x)<F(xlnxf(xx(0,+∞上單調(diào)遞減，F(xiàn)(1ln1f(1x1時，F(xiàn)(x0,lnx0，f(x0x1時，F(xiàn)(x0,lnx0，f(x又函數(shù)f(xx0時，f(x因?yàn)榭蓪?dǎo)函數(shù)f(xRf(0(x2f(x0x20xx(0,+∞(x2f(x0的解集為(0,2(0,+∞f(xxfr(xx1f(eln(ee+1f(ex>ex+x的解集為

g(x=f(xlnxxgr

=fr

-1-1

xfr

-x- xfr(x-x- 因?yàn)閤f(x>x+1所 >0即g

上恒成立

上單調(diào)遞增

=f

1e0

又f

ex+x?f

lnex-ex>0?

(2025·浙江·二模)已知函數(shù)f

為f

足fr

xlnx+f

x2

2a,bf

≥f(b-

a2 的最小值( b+A.22- B.

C. D.

x2

f(xx(lnxf(x

rxlnxf

=

+ln2f

=

Cln2·2C0f

所以fr

=4xlnx-(2lnx

=2x(2lnx-10在(2,+∞(2lnx2

f

又因?yàn)閒(a2

≥f(b-

a2b1即a2+ ≥b-1+2 =b+1+2 -2，b+

b+

b+tb1t4yt

2,t4所以t+2-2≥4+2-2=5所以a2+ ≥5當(dāng)且僅當(dāng)b=3時等號成立.

b+ (2324高三下·遼寧·月考已知函數(shù)f

,fr

，xfr(x)-f

lnx+f

0則不等式f

<0的解集是

=f(x lnx

而得到f(xf ? xfr

f

f

xfr

f lnx+fg(x

?lnx

?x

<g(x在(0,+∞f ?ln1=x1時，g(x00x1時，g(xx1時，g(x0x1時，g(x

=f=f

?lnx?lnx

0且0且

00所以f

<<當(dāng)x=1時因?yàn)閤fr(x-f lnx+f(x<0令x=1得f(1<0所以在(0,+∞f(x0題型 含三角函數(shù)結(jié)合(x)構(gòu)(2324高三上·黑龍江齊齊哈爾·期末已知函數(shù)f

是fr2fπ

.xsinx的解集為

有fr

sinx-f

cosx0則關(guān)于x的不等式f(x

π,πg(shù)(xsinx求解不等式即得

(0,π)

fr(x)sinx-

<

f(g(x(0,πf(x2f

sinx > ， sing(xg

0x<π(0π28R上的函數(shù)f(x)滿足f(xf(-x2sinx0x1x2f(x1+sinx1-f(x2-x1-

0f(x+

f(x)+sinx-cosx -∞,π

π

-∞,π

-πg(shù)(xR0,+∞fx+

f(xsinxcosxg(x+

【解析】設(shè)g(xf(xsinx則g(-xf(-xsin(-由f(x)=f(-x2sinx得f(xsinx=f(-xsin(-x)所以g(xx00

<

f(x1+sinx1-f(x2-x1-

,+∞又g(xg(x在(-∞,0f(x+

+cosx=fx+

+sinx+

f(xsinxg(x

所以x+

xxπ定義域?yàn)?-π, 的函數(shù)f(x)滿足f(x)+f(-x)=0其導(dǎo)函數(shù)為fr(x)當(dāng)0≤ <πfr(x)cosx+f(x)sinx0x的不等式f(x

fπcosx為 (-π,-

(-π

∪(0,

(-π

g(xcosxg(xg(xgπ【解析】f(xf(-x0x(ππf(x

fr(x)cosx+ g(xcosx0x

<0則g(x)在 cos顯然函數(shù)g(x

g(x在(π

是遞減從而g(x)在(-π, 上 不等式f(x

fπcosx化為f(x)

f(cos

g(xg

π<x<π 所以不等式f(x) f(πcosx的解集為(π,π (2324高三上·廣東·月考f(xfr(x(-π,π且f(x為偶函數(shù)，f( =-2，3f(xcosx+fr(xsinx>0則不等 fx+

cos3x-

0的解集為 (-π

(-2π,

(-2π

g(xf

sin3x,x(ππ 性解不等式【解析】g(xf(xsin3x,x(ππg(shù)r

=3f

sin2xcosx+fr

sin3x sin2x3f(xcosx+fr(xsinxx(ππsin2x03f(xcosxfr(xsinx 可知gr(x≥0且僅當(dāng)x=0時gr(x=0則g(x在(-π, 上單調(diào)遞增又因?yàn)閒

=fπ

=-g(

=f-

sin3-

=

=g-

-

<x<πg(shù)(x

=fx+

sin3x+

=fx+

f(x+

cos3x-

0g(x+

g-ππx+π<π2πx f(x+

cos3x-

0(2π,0g(xf(xsin3x,x(ππg(shù)(x 1利用誘導(dǎo)公式可得f(x+πcos3x-1>0等價于g(x+ >g(-π即可得結(jié)果 已知函數(shù)f(x及其導(dǎo)函數(shù)fr(x的定義域均為(ππf(x

≥f

tanx

2

< 的解集為 的單調(diào)性且是偶函數(shù)將問題轉(zhuǎn)化為F(x F 即可依據(jù)函數(shù)的單調(diào)性和奇偶性求解

≥f

tanx=f(xsinxfr(xcosxf(xsinxfr(xcosxf(x(cosx)r因此f(xcosxr≥0從而F(x=f(xcosx在(0, 上單調(diào)遞增又f(x是(-π, 上的偶函數(shù)且y=cosx是偶函數(shù) F(-xf(-xcos(-xf(xcos(xF(x即F(x是(-π, 上的偶函數(shù)故F(x在(-π,0上單調(diào)遞減

2

1又f

即f(x

cosx1即F

<所以F

<Fπ

F

<x<πx(ππ 題型 其他類型構(gòu) (2324高三下·江蘇南通·月考f(xfr(x)f(1e0時，fr(x

ex則不等式f(x)-

<1的解集為 A.(0,1)∪(1,

C.

D.ex0f(1e調(diào)性即可得解

0又f(x)-

1f(xlnxex0g(x

=f

ex

=fr

x0時，fr(x

ex

=fr

-ex<g(x在(0,+∞f(1eg(1f(1ln1e1eef(x)-

1f(xlnxex0

<

x(2024·江蘇南通·模擬預(yù)測Ryf(xyfr(x若fr(x+(x+12也為偶函數(shù)且f(2a+4>f(a2+1則實(shí)數(shù)a的取值范圍是 A.(-∞,-C.(-

B.(-∞,-D.(-

1)2f(-xf(x)-fr(-xfr(x)，g(xfr(xx1)2，fr(x(x12則g(-xg(x)即fr(-xx1)2=fr(xx1)2，-fr(xx1)2fr(xx1)2，所以fr(xx0時，fr(x2x0f(x(0,+∞f(x(-∞,0由f(2a+4)>f(a2+1)即f(2a+ >f(a2+1所以2a4a21-(a212a4a21a<-1aa的取值范圍是(-∞,-1(3,+∞g(xfr(xx1)2從而推導(dǎo)出fr(x2x，已知函數(shù)f(x的定義域?yàn)?0,+∞fr(x(x+

2f

+xfr

xf

f

7f(x3

3x+12(x+3集為

(-

(-

x2f=x 為g(x

<

0xf(xx+1，

2f

+xfr

xf

?2f

+xfr(x

fr

x2fx+1

f

r=2xf

fr

f

r>x2f(x ?x+1x+1

f(xrx2f(x

x2f

x2f

r(x+

-x2fg(x

x1gr(x

(x+

>

x2f=x

3x+

(x+32f(x+因?yàn)閒

的定義域可知，x30f(x

<(x+32

x+ <又因?yàn)閒

7g(6

=62f =

x2f即上面不等式可轉(zhuǎn)化為g(x

<

g(x

x

0x36-3x(2324高三上·陜西安康·月考定義在R上的連續(xù)函數(shù)f

滿足f(x

x1時，(x-1fr(xf(x0其中fr(x是f(x的導(dǎo)數(shù).xe2xf(e2x+1≥(2x-af(2x+1-a恒成立則實(shí)數(shù)a的取值范圍為 (-∞,ln2-

ln2-

(-∞,-

-F(xxf(x1F(xa2xe2xxRg(x2xe2x的最大值可得【解析】F(xxf(x1Fr(xxfr(x1f(x1x0時，xfr(x1f(x10Fr(xF(x在(0,+∞F(xF(0因?yàn)閒(x1F(xF(xR上單調(diào)遞增e2xf(e2x1(2x-af(2x1-aF(e2xF(2x-ae2x2xaa2xe2xxR恒成立.令g(x2xgr(x22e2xgr(x0x0x0時，gr(x0，g(xx0時，gr(x0，g(xg(xx0g(xmaxg(0=-1a≥-1a-1,+∞(2324高三上·重慶渝中·月考f(x(0,+∞fr

(x

2f

+xfr

xf

f

7f(x43x+(x+4

的解集為 (-

(-

(-

x2f

【分析由已知有(xf >x+1構(gòu)造g(x=xf(x則x+1-g(x<0構(gòu)造G(x=g(x

(x+1-x+1則G(x (x+1 >0即G(x在(0,+∞上是單調(diào)遞增把目標(biāo)G(x4G(6

2f

+xfr

xf

f

r=2xf

fr

x2fx+1

-gr(x(x+

xf(xx1g(x

x+ <0，x∈(0,+∞x10g(xgr(x(x1

(x+1-

x1G

(x+1 >G(x在(0,+∞不等式f(x3x+不等式f(x3x+(x+42f(x+等價 x+ <3即G(x+g(x+4x+5<而G(6 =36f3G(x4G(6G(x在(0,+∞(x4<6g(xx2f(x

-grx+

(x+

0

=g(x判斷單調(diào)性為關(guān)鍵x+檢 重難知識鞏 (2324高三上·湖北武漢·期中已知函數(shù)的定義域?yàn)镽，f(15對任意x(x)<2則f(x)>3+2x的解集為 A.(- B.(2, C. D.(-構(gòu)造函數(shù)g(x=f(x2x3利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性即可求解【解析】設(shè)g(x=f(x2x3則gr(x=fr(xxR，gr(x0，g(xR∵f(1)=5，∴g(1=f(1)-2-3=g(xg(10x∴f(x32x(-∞,1).(2025·湖南·三模已知y=f

=fr(x若xfr(x<f(x且f(3=6則不等式f(lnx>2lnx的解集為

f

1lnx3即可求解f

xfr

-f

g(x

xfr(xf(xxfr(xf(x0gr(xf則g(x (x>1在區(qū)間(1,+∞上單調(diào)遞減f(lnx>2=f又f

6

2lnx

lnx>1

1lnxexR上的函數(shù)f(x的導(dǎo)函數(shù)為fr(x)且滿足fr(xf(x),f(2024e2024不等式f(1lnx<3x的解集為

D.D.F(x

【解析】F(xf(x)fr(xf(x則Fr(x

fr(x)ex-f(x)ex

fr(x)-

所以函數(shù)F(x)在R

f(1 可轉(zhuǎn)化 F(2024可轉(zhuǎn)化

1不等式f

<3 e1 e

1lnx20240xe6072f(1lnx3x(0,e6072 (2024·寧夏銀川·三模)已知定義在R上的奇函數(shù)f(x)的圖象是一條連續(xù)不斷的曲線，fr(x是f(xx0時，3f(xxfr(x0且f(22則不等式(x1)3f(x+1)>16的解集為 A.(1, C.(- D.(-∞,-3∪(x1)3f(x116g(xx3f(xx0g(xx3f【解析】令g(x=x3f(x則gr(x=3x2f(x+x3fr(x=x23f(x+xfr x0時，3f(xxfr(x0g(x(0,+∞又f(xg(x(x13f(x123f(2x12x<-3x1.(2425高三下·河北邢臺·月考f(xRf(11fr(x滿足fr(x-2f(x>0則不等式e2f(ln(x- <(x-12的解集為 A.

B.(1,e+

C.(e,e+

D.(e+

f(xg(xe2f(ln(x-1<(x-12g(x的單調(diào)性即可求解f

fr

-2f

e2xg(x

0

在R又不等式e2f(ln(x- <(x-12等價于f(ln(x- <1，(x-1 即g(ln(x- =f(ln(x- =f(ln(x-1 =f(ln(x-1 <1=f(1=g(1，e2ln(x-

eln(x-1

(x-1

ln(x-11(x-1>0x(1,efr

(2024·云南楚雄·一模f(xRxRf(x 1-

成立．若f(-1=-1則不等式f(x

的解集為 (-∞,-

(-

(-

g(x3xf(xg(xg(xg(1fr >0得f(x+f(xln3>令g(x=3xf(x則gr(x=3x(ln3?f(x+fr >g(xR又f(-1=-1，f(xf(11，g(13f(1f(x31-x3xf(x3g(xg(1x(2324高三上·河北·月考已知函數(shù)f(x及其導(dǎo)函數(shù)fr(x+∞)且xfr(xx1)f(x恒成立，f(3e則不等式(x4)f(x43ex+2 A.(-4,- B.(- C.(- D.(-1,

,x01)f(x)g(x

,x0gr(x

xfr(x)+f(x)-

0g(x(0又(x4)f(x43ex+2得(x4)f(x0x43-4x<-1．

<3f(3)g(x4g(x

,x0(x4)f(x43ex+2g(x4g(3)利用單調(diào)性解即可已知函數(shù)f(x)及其導(dǎo)函數(shù)fr(x)的定義域均為R且f(x)為偶函數(shù)f( =-3f(x)cosxfr(x)sinx0則不等式f(x+

cos3x-

0的解集為 A.(-π B.(-2π

C.(-2π, D.(π

結(jié)合函數(shù)的單調(diào)性及一元一次不等式的解法即可求解則gr(x)=3f(x)sin2xcosx+fr(x)sin3x=sin2x3f(x)cosx所以g(x在R上單調(diào)遞增

0gr(x又因?yàn)閒(xf(-

=fπ

=-g(

=-1

3f-

=1g(x

=fx+

sin3x+

=fx+

f(x+

cos3x-

0g(x+

g-πxππx2π f(x+

cos3x-

0(2π,+∞Rf(xfr(x，f(1exfr(xex (-

(-

=f(x xF(1

=f 10x1時，F(xiàn)(x0從而求出解集f【解析】F(x

-因?yàn)閒r

f

<

fr

f(x-F(xfF(xfr(x-f -1故F

=f xRf又f(1=e故F(1 -1=0f

xex?f(x-x>0?F

f(xxex(-∞,1設(shè)奇函數(shù)f(x的定義域?yàn)?ππf(xx -π

fr

cosx+f

sinx0若

f(m)<f

m)(-π,-

(0,

(-π, D.(π,

=f(x

yg(x間(π,0g(x

<gπ

y

的單調(diào)性可解出所求不等式【解析】g(xf(x(ππ

f(-

f因?yàn)楹瘮?shù)y=f(x為奇函數(shù)所以g(-x=cos(- =-cosx=-g(x則函數(shù)g(x=f(x是定義在(-π, 上的奇函數(shù)

=fr

cosx+f

x(π,0fr(xcosxf(xsinxx(π,0時，gr(x0g(xf(xx(π,0 則函數(shù)g(x=f(x是(-π, 上的奇函數(shù)并且單調(diào)遞增

f(m)<f

cos(-m)=f

m<πcosmf

f(cos

<gπm<ππm<ππm<π (2324高三上·四川成都·月考Rf(xfr(x當(dāng)x≤0時滿足(x2+3fr(x+2xf(-x>0則不等式f(x≥f(2x+ 的解集為

x2+ 4(x2+x+

(-

(-∞,-

=f(x g(xx23

R在(-∞,0Rf

x2+因?yàn)閒(x是定義在Rf(-x=-f(xf(- fg(-x=x23x23=-g(xg(xRx0時，(x23fr(x2xf(-x0(x23fr(x2xf(x (x2+3fr(x-2xfg(x

(x2+3 >g(x在(-∞,0g(xRg(xRf f(2x+

f(2x+則不等 x2+ 4(x2+x+

=(2x+12+3g(xg(2x1x2x1x≤-f f(2x+所以不等 x2+ 4(x2+x+

的解集為(-∞,-1(2324高三下·上?！ぴ驴糵(xR(x)

xfr(x0(x-20232f(x2023f(-10F(xx2f(xF(x(-∞,0(0,+∞(x-20232f(x2023f(-10F(x-2023F(-【解析】令F(xx2f(xFr(x2xf(xx2fr(xx2f(xxfr(x)x0時，2f(xxfr(xx0時，F(xiàn)r(xx2f(xxfr(x)F(x在(-∞,0f(xR所以f(-xf(xF(-xx)2f(-xx2f(xF(xF(x在(0,+∞所以F(x-2023)=(x-2023)2f(x-2023),F(-1)=(-1)2f(-1)=f(-F(x-2023F(-1所以x-202312022x(2425高三上·海南省直轄縣級單位·期中Rf(xf(220f(xfr(xfr(x6x22f(x2x32xg(xf(x2x32xg(xRg(20g(xg(22xg(xRg(2f(22232220164g(xg(2x所以不等式f(x2x32x的解集為(2,+∞(2025·山西·模擬預(yù)測f(xRfr(xfr(xf(x<0則不等式e2x+2f(x+2<f(-x的解集 g(xexf(x利用函數(shù)的單調(diào)性解不等式即可【解析】設(shè)g(x=exf(x,則gr(x=exfr(x+f <g(xRe2x+2f(x2f(-xex+2f(x2e-xf(-xg(x2g(-x故x2>-xx>-e2x+2f(x2f(-x(-1,+∞故答案為：(-(2024·云南·模擬預(yù)測)已知fr(x是定義域?yàn)?0, 的函數(shù)f(x的導(dǎo)函數(shù)fr

sinx+f

cosx0則不等式f

sinx>1f

的解集

=f

sinx

gπ

x<πf

sinx>1f

的解集

=f

sinx,x∈0,

=fr

sinx+f

cosx<

f

sinx>1f

?fx<

sinx>f

sinπ即g(x>g( 得 π0<x<0x<π(0π (2425高三下·廣東梅州·月考f(xRfr(x)xR有f(xf(-x2sinx且在[0上fr(xcosx．若fπ-t-f(tcostsint g(xf(xsinx【解析】令g(xf(xsinx由f(xf(-x得g(x)-g(-x)=f(x)-sinxf(-x)+sinx]=0函數(shù)g(x)cosx0fπ-

sint

+sinπ-

[g(t)+sint]>cost-故答案為：g(t)g(|t|)|t|故答案為：πt(π(2425高三上·廣東潮州·月考(0,+∞上的函數(shù)f(x的導(dǎo)函數(shù)為fr(x當(dāng)x>0時，xfr(x<2且f(e=5則不等式f(x2-4lnx<3的解集 g(xf(x24lnx3

=f

4lnx-

x2fr(x2-則 =f(e-4lne-3=5-2-3=0，g(x=2xf -x= x0時，xfr(xgr(x0g(x在(0,+∞g(xf(x24lnx3則g(x< 所以x ef(x24lnx3(e,+∞故答案為：(e,+∞(2425高三上·上?！て谥蠷f(xf(xfr(-xf

1f(x

1的解集 ex- 【解析】因?yàn)閒(x是定義在Rf(-xfr(-xfr(x).已知f(xfr(-x0可得f(xfr(x令g(xex-2f(x)，gr(xex-2f(xex-2fr(xex-2f(x由于f(xfr(x0又ex-20所以gr(x0這表明g(x在R上單調(diào)遞增不等式f(x11ex-1f(x1ex-f(2)g(xx12x1.故不等式f(x11(1ex-(2425高三上·山東日照·月考f(xfr(x且F(x=ex+1f(x+1是偶函數(shù)其函數(shù)圖象為不間斷曲線且(x-1fr(x+f >0,不等式xf(lnx<e3f(3的解集 F(x所以Fr(x=ex+1f(x+1+ex+1fr(x+1=ex+1f(x+1+fr(x+ 又因?yàn)?x-1f(x+fr >0用x+1代替x得：xf(x+1+fr(x+ >所以當(dāng)x>0時，f(x+1+fr(x+1>0所以ex+1f(x+1+fr(x+ >F(x在(0,+∞上單調(diào)遞增F(xy(-∞,0上單調(diào)遞減lnxt1xet+1xf(lnxet+1f(t1F(te3f(3e2+1f(21F(2xf(lnxe3f(3F(tF(2(-∞,0(0,+∞t2?-2t-1lnxt13

<x<xf(lnxe3f(3的解集為(1,e3函數(shù)不等式轉(zhuǎn)化為代數(shù)不等式求解(2324高三下·北京·月考R上的函數(shù)f(x滿足f(2xf(-xx1xfr

fr

f

1則不等式f

的解集是x-【分析依據(jù)題意判斷函數(shù)f(x關(guān)于直線x=1對稱結(jié)合(x-1f =(x-1fr(xf(x0構(gòu)造函數(shù)g(xx1)f(x)易得其關(guān)于(1,0【解析】R上的函數(shù)f(x滿足f(2xf(-所以函數(shù)f(xx1f(x1f(1x),x因?yàn)楫?dāng)x>1時,有xfr(x)+f(x)>fr(x),即[(x-1)f(x)]=(x-1)fr(x)+f(x)>故令g(xx1)f(x則g(xx1)f(x)在(1上單調(diào)遞增,g(1xg(1xxf(1xxf(1x0，所以g(xx1)f(x關(guān)于點(diǎn)(1,0所以g(xx1)f(x在Rf(2g(221)f(21x1時,f(xxx1)f(x1g(x11x2x1時,f(xxx1)f(x1g(x1所以x<1且x>2,即無解.所以不等式f(x)< 的解集是(1,2.x-[(x1)f(xx1)fr(xf(x0g(xx1)f(x)(2024·河南·三模f(x(0,+∞，fr(xx(0,+∞，f(x>f(x-xfr lnx則不等式f(x(ex-1-1>0的解集

=f

?lnx

0與

0f

?lnx

=f

?lnx

=fr

?lnx

?1-lnx

=f

f

-xfr(x

g(x在(0,+∞ex-110x1ex-110x而

0?x>

<0?0<x<

f

(ex-1-

0f

?lnx

g(x0g(10g(xg(1x(1,+∞檢測 創(chuàng)新能力提 已知函數(shù)f

f

=e-1，fr

ex-2f(x>x2的解集為

ex1x2g