全國(guó)統(tǒng)考2025版高考數(shù)學(xué)大一輪復(fù)習(xí)第2章函數(shù)概念與基本初等函數(shù)Ⅰ第1講函數(shù)及其表示1備考試題文含解析

其次章函數(shù)概念與基本初等函數(shù)Ⅰ第一講函數(shù)及其表示練好題·考點(diǎn)自測(cè)1.下列說法中正確的個(gè)數(shù)是()(1)f(x)=1x-(2)已知f(x)=m(x∈R),則f(m3)=m3.(3)y=lnx2與y=2lnx表示同一函數(shù).(4)f(x)=x2+1,-1≤x≤1,A.0 B.1C.2 D.32.[2024江西模擬]已知函數(shù)f(x)的圖象如圖2-1-1所示,則函數(shù)f(x)的解析式可能是()圖2-1-1A.f(x)=(4x+4-x)|x|B.f(x)=(4x-4-x)log4|x|C.f(x)=(4x+4-x)log14|D.f(x)=(4x+4-x)log4|x|3.[2024全國(guó)卷Ⅱ,10,5分][文]下列函數(shù)中,其定義域和值域分別與函數(shù)y=10lgx的定義域和值域相同的是()A.y=x B.y=lgx C.y=2x D.y=14.[2024貴陽(yáng)市摸底測(cè)試]已知函數(shù)f(x)=sinπx6,x≤0,logA.12 B.-1C.32 D.-5.[2024北京,11,5分]函數(shù)f(x)=1x+1+lnx的定義域是6.[福建高考,4分]若函數(shù)f(x)=-x+6,x≤2,3+logax,x拓展變式1.已知函數(shù)f(x)=1x(x<0),x2(x≥0),g(x)=x+1,則(1)g(f(2.(1)已知函數(shù)f(x)=lg(2a·x-1)的定義域是(2,+∞),則實(shí)數(shù)a的取值集合是.

(2)已知函數(shù)f(x)=12(x-1)2+1的定義域與值域都是[1,b](b>1),則實(shí)數(shù)b的值為3.(1)已知函數(shù)f(x)=2x+1,x<1,x2+ax,x≥1且(2)[2024全國(guó)卷Ⅲ,16,5分][文]設(shè)函數(shù)f(x)=x+1,x≤0,2x,x>0,則滿意f(x)(3)[2024北京,14,5分]設(shè)函數(shù)f(x)=x①若a=0,則f(x)的最大值為;

②若f(x)無最大值,則實(shí)數(shù)a的取值范圍是.

4.[2024山東,10,5分][文]若函數(shù)exf(x)(e=2.71828…是自然對(duì)數(shù)的底數(shù))在f(x)的定義域上單調(diào)遞增,則稱函數(shù)f(x)具有M性質(zhì).下列函數(shù)中具有M性質(zhì)的是()A.f(x)=2-x B.f(x)=x2C.f(x)=3-x D.f(x)=cosx5.函數(shù)y=f(x)的圖象是如圖2-1-3所示的折線段OAB,其中A(1,2),B(3,0),函數(shù)g(x)=x·f(x),那么函數(shù)g(x)的值域?yàn)?)A.[0,2] B.[0,94C.[0,32]圖2-1-3答案其次章函數(shù)概念與基本初等函數(shù)Ⅰ第一講函數(shù)及其表示1.B對(duì)于(1),定義域是空集,不滿意函數(shù)的概念,故(1)錯(cuò)誤;對(duì)于(2),f(x)是常數(shù)函數(shù),所以f(m3)=m,故(2)錯(cuò)誤;對(duì)于(3),兩個(gè)函數(shù)的定義域不同,故不是同一函數(shù),(3)錯(cuò)誤;對(duì)于(4),結(jié)合分段函數(shù)可知(4)正確.所以正確命題的個(gè)數(shù)為1,故選B.2.D對(duì)于A,f(x)大于等于0恒成立,與圖象不符,解除;對(duì)于B,當(dāng)x<-1時(shí),f(x)<0,與圖象不符,解除;對(duì)于C,當(dāng)x>1時(shí),f(x)<0,與圖象不符,解除.選D.3.D解法一函數(shù)y=10lgx的定義域?yàn)?0,+∞),當(dāng)x>0時(shí),y=10lgx=x,故函數(shù)的值域?yàn)?0,+∞).只有選項(xiàng)D符合.解法二易知函數(shù)y=10lgx中x>0,解除選項(xiàng)A,C;因?yàn)?0lgx必為正值,所以解除選項(xiàng)B.選D.4.D∵f(x)=sinπ6x,x≤0,log13x,x>05.(0,+∞)函數(shù)f(x)=1x+1+lnx的自變量滿意x+1≠0,6.(1,2]因?yàn)閒(x)=-x+6,x≤2,3+logax,x>2,所以當(dāng)x≤2時(shí),f(1.(1)1x+1(x<0),x2+1(x≥0)當(dāng)x<0時(shí),f(x)=1x,則g(f(x))=1x+1;當(dāng)x≥0時(shí),f∴g(f(x))=1(2)1x+1(x<-1),(x+1)2(x≥-1)令g(x)=x+1<0,得x<-1,則此時(shí)f(g(x))=1x+1.令g(x)=x2.(1){-1}由題意得,不等式2a·x-1>0的解集為(2,+∞),由2a·x-1>0可得x>12a,∴12a(2)3f(x)=12(x-1)2+1,x∈[1,b]且b>1,f(1)=1,f(b)=12(b-1)2+1,函數(shù)圖象的對(duì)稱軸為直線x=1,且f(x)在[1,b]上單調(diào)遞增.∴函數(shù)的值域?yàn)閇1,12(b-1)2+1].由已知得12(b-1)2+1=b,解得b3.(1)542依題意知f(-2)=2-2+1=54.因?yàn)閒(0)=20+1=2,所以f(f(0))=f(2)=22+2a=4a,解得a(2)(-14,+∞)當(dāng)x≤0時(shí),f(x)+f(x-12)=x+1+x-12+1=2x+32>1,即-14<x≤0;當(dāng)0<x≤12時(shí),f(x)+f(x-12)=2x+x+1>1恒成立;當(dāng)x>12時(shí),f(x)+f(x(3)①2若a=0,則f(x)=x3-3x,x≤0時(shí),f'(x)=3x2-3=3(x+1)(x-1),令f'(x)>0,得x<-1,令f'(x)<0,得-1<x≤0,所以函數(shù)f(x)在(-∞,-1]上單調(diào)遞增,在(-1,0]上單調(diào)遞減,所以函數(shù)f(x)在(-∞,0]上的最大值為f(-1)=2.綜上可得,函數(shù)f(x)的最大值為2.②(-∞,-1)函數(shù)y=x3-3x與y=-2x的大致圖象如圖D2-1-1所示,若函數(shù)f(x)=x3-3x,x≤a,-圖D2-1-14.A對(duì)于選項(xiàng)A,f(x)=2-x=(12)x,則exf(x)=ex·(12)x=(e2)x,∵e2>1,∴exf(x)在R上單調(diào)遞增,∴f(x)=2-x具有M性質(zhì).對(duì)于選項(xiàng)B,f(x)=x2,exf(x)=exx2,[exf(x)]'=ex(x2+2x),令ex(x2+2x)>0,得x>0或x<-2;令ex(x2+2x)<0,得-2<x<0,∴函數(shù)exf(x)在(-∞,-2)和(0,+∞)上單調(diào)遞增,在(-2,0)上單調(diào)遞減,∴f(x)=x2不具有M性質(zhì).對(duì)于選項(xiàng)C,f(x)=3-x=(13)x,則exf(x)=ex·(13)x=(e3)x,∵0<e3<1,∴y=(e3)x在R上單調(diào)遞減,∴f(x)=3-x不具有M性質(zhì).對(duì)于選項(xiàng)D,f(x)=cosx,exf(x)=excosx,則[exf(x)]'=ex(cosx-sinx)≥0在R上不恒成立,故exf(x)=excosx在R上不是單調(diào)遞增的,所以5.B由題圖可知,直線OA的方程是y=2x;因?yàn)閗AB=0-23-1=-1,所以直線AB的方