2026版《優(yōu)化設(shè)計(jì)大一輪》高考數(shù)學(xué)（優(yōu)化設(shè)計(jì)新高考版）微點(diǎn)突破2抽象函數(shù)

微點(diǎn)突破2抽象函數(shù)高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI2026常見的抽象函數(shù)對(duì)應(yīng)的基本初等函數(shù)模型如下:條件案例類型f(x+y)=f(x)+f(y)f(x)=x正比例函數(shù)f(xy)=f(x)f(y)f(x)=x3冪函數(shù)f(x+y)=f(x)f(y)f(x)=2x指數(shù)函數(shù)f(xy)=f(x)+f(y)f(x)=log2x對(duì)數(shù)函數(shù)f(x+y)+f(x-y)=2f(x)f(y)f(x)=cos

x余弦函數(shù)f(x+T)=f(x)(T≠0)f(x)=sin

x周期函數(shù)

A

[對(duì)點(diǎn)訓(xùn)練](1)(2024·新高考Ⅰ,8)已知函數(shù)f(x)的定義域?yàn)镽,f(x)>f(x-1)+f(x-2),且當(dāng)x<3時(shí),f(x)=x,則下列結(jié)論中一定正確的是(

)A.f(10)>100 B.f(20)>1000C.f(10)<1000 D.f(20)<10000解析

∵當(dāng)x<3時(shí),f(x)=x,∴f(1)=1,f(2)=2.∵f(x)的定義域?yàn)镽,且f(x)>f(x-1)+f(x-2),∴f(3)>f(2)+f(1)=3,f(4)>f(3)+f(2)>5,f(5)>f(4)+f(3)>8,f(6)>f(5)+f(4)>13,f(7)>f(6)+f(5)>21,f(8)>f(7)+f(6)>34,f(9)>f(8)+f(7)>55,f(10)>f(9)+f(8)>89,f(11)>f(10)+f(9)>144,f(12)>f(11)+f(10)>233,f(13)>f(12)+f(11)>377,f(14)>f(13)+f(12)>610,f(15)>f(14)+f(13)>987,f(16)>f(15)+f(14)>1

000.∴f(20)>1

000.結(jié)合各選項(xiàng)知,選項(xiàng)B一定正確.B(2)(多選題)(2023·新高考Ⅰ,11)已知函數(shù)f(x)的定義域?yàn)镽,f(xy)=y2f(x)+x2f(y),則(

)A.f(0)=0B.f(1)=0C.f(x)是偶函數(shù)D.x=0為f(x)的極小值點(diǎn)ABC解析

對(duì)于選項(xiàng)A,令x=0,y=0,f(0)=0,所以A正確;對(duì)于選項(xiàng)B,令x=1,y=1,f(1×1)=12×f(1)+12×f(1)=2f(1),解得f(1)=0,所以B正確;對(duì)于選項(xiàng)C,令x=-1,y=-1,f[(-1)×(-1)]=(-1)2×f(-1)+(-1)2×f(-1)=2f(-1),解得f(-1)=0;再令x=-1,y=x,f[(-1)×x]=x2×f(-1)+(-1)2×f(x),f(-x)=f(x),所以C正確;對(duì)于選項(xiàng)D,用特值法,函數(shù)f(x)=0,為常數(shù)函數(shù),且滿足f(xy)=y2f