2025年高考數(shù)學(xué)一輪復(fù)習(xí)-考點(diǎn)突破練15-函數(shù)的圖象與性質(zhì)-專(zhuān)項(xiàng)訓(xùn)練【含解析】

考點(diǎn)突破練15函數(shù)的圖象與性質(zhì)一、單項(xiàng)選擇題1.函數(shù)y=2x-3A.3B.(-∞,3)∪(3,+∞)C.32,3∪(3,D.(3,+∞)2.設(shè)函數(shù)f(x)=log2(6-x),x<1,2x-1A.2 B.6 C.8 D.103.函數(shù)f(x)在(-∞,+∞)上單調(diào)遞增,且為奇函數(shù),若f(2)=1,則滿足-1≤f(x-1)≤1的x的取值范圍是()A.[-2,2] B.[-1,3] C.[0,2] D.[1,3]4.已知函數(shù)f(x)=(4x-4-x)ln|x|的圖象大致為()5.已知函數(shù)f(x)對(duì)任意實(shí)數(shù)x都有f(2+x)=f(2-x),并且對(duì)任意x1,x2(x1≠x2)∈(-∞,2),都有f(x1)-f(A.f(0)<f(3)B.f(2)=f(-2)C.f(23)<f(-2)D.f(2-1)<f(2+1)6.已知函數(shù)f(x)=ex-x-1,x≤0,-f(-A.0,1e BC.(0,e) D.(e,+∞)7.已知函數(shù)f(x)是定義在R上的奇函數(shù),函數(shù)g(x)=|x-2|f(x)的圖象關(guān)于直線x=2對(duì)稱(chēng),若f(-1)=-1,則g(3)=()A.5 B.1 C.-1 D.-58.已知函數(shù)y=f(x-1)是定義在R上的偶函數(shù),且f(x)在(-∞,-1)上單調(diào)遞減,f(0)=0,則f(x)f(2x+1)<0的解集為()A.(-∞,-2)∪(0,+∞) B.(-2,0)C.-2D.-二、多項(xiàng)選擇題9.已知f(x)是定義在R上的奇函數(shù),若f(x+4)=f(x)且f(1)=2,則f(1)+f(2)+…+f(n)(n∈N*)的值可能為()A.-2 B.0 C.2 D.410.若函數(shù)f(x)同時(shí)具有性質(zhì):①對(duì)于任意的x,y∈R,f(x)+f(y)2≥fx+y2A.f(x)=|x|B.f(x)=ln(x+x2C.f(x)=2x+1D.f(x)=ln(|x|+1)11.已知函數(shù)f(x)對(duì)任意x∈R都有f(x+2)=-f(x),若函數(shù)y=f(x-1)的圖象關(guān)于直線x=1對(duì)稱(chēng),且對(duì)任意的x1,x2∈(0,2),且x1≠x2,都有f(x1)-f(x2)xA.f(x)是偶函數(shù)B.f(2022)=1C.f(x)的圖象關(guān)于點(diǎn)(1,0)中心對(duì)稱(chēng)D.f(-2)>f(-1)12.已知函數(shù)f(x)及其導(dǎo)函數(shù)f'(x)的定義域均為R,記g(x)=f'(x).若f32-2x,g(2A.f(0)=0B.g-12C.f(-1)=f(4)D.g(-1)=g(2)三、填空題13.已知函數(shù)f(x)=log4x,x>0,f14.寫(xiě)出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù)f(x)=.

①定義域?yàn)镽;②值域?yàn)?-∞,1);③對(duì)任意x1,x2∈(0,+∞)且x1≠x2,均有f(x115.已知函數(shù)f(x)=1x2+1-ln|x|,則使不等式f(2t+1)>f(t+3)成立的實(shí)數(shù)t的取值范圍是16.若f(x)=lna+11-x+b是奇函數(shù),則a=

參考答案與解析1.C解析要使函數(shù)y=2x-3+1x-3有意義,則2x-3≥0,x-3≠0,解得2.B解析因?yàn)閒(x)=log2(6-x),x<1,2x-1,x≥1,所以f(-2)=log28=3,f(log3.B解析f(x)是奇函數(shù),故f(-2)=-f(2)=-1.又f(x)是增函數(shù),-1≤f(x-1)≤1,所以f(-2)≤f(x-1)≤f(2),則-2≤x-1≤2,解得-1≤x≤3.4.A解析由題意,函數(shù)f(x)=(4x-4-x)ln|x|,定義域?yàn)?-∞,0)∪(0,+∞),關(guān)于原點(diǎn)對(duì)稱(chēng),f(-x)=(4-x-4x)ln|-x|=-(4x-4-x)ln|x|=-f(x),所以函數(shù)f(x)為奇函數(shù),所以排除C,D選項(xiàng),當(dāng)x→+∞時(shí),f(x)>0,可排除B.5.C解析由函數(shù)f(x)對(duì)任意實(shí)數(shù)x都有f(2+x)=f(2-x),可得函數(shù)f(x)的圖象關(guān)于直線x=2對(duì)稱(chēng),又由對(duì)任意x1,x2∈(-∞,2),都有f(x可得函數(shù)f(x)在區(qū)間(-∞,2)上單調(diào)遞減,則在區(qū)間(2,+∞)上單調(diào)遞增,由f(0)=f(4)>f(3),所以A不正確;由f(2)<f(-2),所以B不正確;由f(23)<f(6)=f(-2),所以C正確;由|2-1-2|>|2+1-2|,所以f(2-1)>f(2+1),所以D不正確.6.C解析因?yàn)閒(0)=0,x>0時(shí),f(x)=-f(-x),分析易得x<0時(shí)也有f(x)=-f(-x),即函數(shù)f(x)是奇函數(shù),x≤0時(shí),f(x)=ex-x-1,f'(x)=ex-1≤0,所以f(x)是減函數(shù),所以奇函數(shù)f(x)在R上是減函數(shù),又f(-1)=1e,所以f(1)=-f(-1)=-1不等式f(lnx)>-1e為f(lnx)>f(1),所以lnx<1,0<x<e所以x的取值范圍是(0,e).7.B解析因?yàn)間(x)的圖象關(guān)于直線x=2對(duì)稱(chēng),則g(x+2)=|x|f(x+2)是偶函數(shù),g(2-x)=|-x|f(2-x)=|x|f(2-x),且g(x+2)=|x|f(x+2),所以|x|f(2-x)=|x|f(2+x)對(duì)任意的x∈R恒成立,所以f(2-x)=f(2+x),因?yàn)閒(-1)=-1且f(x)為奇函數(shù),所以f(3)=f(2+1)=f(2-1)=-f(-1)=1,因此g(3)=|3-2|f(3)=f(1)=1.8.C解析因?yàn)楹瘮?shù)y=f(x-1)是定義在R上的偶函數(shù),所以y=f(x)的圖象關(guān)于直線x=-1對(duì)稱(chēng).因?yàn)閒(x)在(-∞,-1)上單調(diào)遞減,所以在(-1,+∞)上單調(diào)遞增.因?yàn)閒(0)=0,所以f(-2)=f(0)=0.所以當(dāng)x∈(-∞,-2)∪(0,+∞)時(shí),f(x)>0;當(dāng)x∈(-2,0)時(shí),f(x)<0.由f(x)f(2x+1)<0,得x<-2或x9.BC解析由題設(shè),f(x)是周期為4的奇函數(shù),且f(1)=2,則f(-2)=f(-2+4)=f(2)=-f(2),即f(2)=0.f(-1)=f(-1+4)=f(3)=-f(1)=-2,f(0)=f(0+4)=f(4)=0,所以f(1)=f(1)+f(2)=2,f(1)+f(2)+f(3)=f(1)+f(2)+f(3)+f(4)=0,當(dāng)n=4k或n=4k+3,k∈N*時(shí),f(1)+f(2)+…+f(n)=0;當(dāng)n=4k+1或n=4k+2,k∈N時(shí),f(1)+f(2)+…+f(n)=2.10.AC解析對(duì)于B,易知,此函數(shù)定義域關(guān)于原點(diǎn)對(duì)稱(chēng),f(-x)=ln(-x+x2+1)=ln(x+x2+1)-1=-ln(x+x2+1)=-f(x),故f(x)=ln(x+x2+1)為奇函數(shù),對(duì)于A,f(x)+f(y對(duì)于C,f(x)+f(y對(duì)于D,x=3,y=1時(shí),f(x)+f(y)2=ln22<ln11.ACD解析對(duì)于選項(xiàng)A,由函數(shù)f(x-1)的圖象關(guān)于直線x=1對(duì)稱(chēng),根據(jù)函數(shù)的圖象變換,可得函數(shù)f(x)的圖象關(guān)于直線x=0對(duì)稱(chēng),所以函數(shù)f(x)為偶函數(shù),所以A正確;對(duì)于選項(xiàng)B,由函數(shù)f(x)對(duì)任意x∈R都有f(x+2)=-f(x),可得f(x+4)=-f(x+2)=f(x),所以函數(shù)f(x)是周期為4的周期函數(shù).因?yàn)閒(-2)=0,可得f(2)=0,則f(2022)=f(505×4+2)=f(2)=0,所以B錯(cuò)誤;又因?yàn)楹瘮?shù)f(x)為偶函數(shù),即f(-x)=f(x),所以f(x+2)=-f(x)=-f(-x),可得f(x+2)+f(-x)=0,所以函數(shù)f(x)的圖象關(guān)于點(diǎn)(1,0)中心對(duì)稱(chēng),所以C正確;由對(duì)任意的x1,x2∈(0,2),且x1≠x2,都有f(x可得函數(shù)f(x)在區(qū)間(0,2)上是單調(diào)遞增的,又因?yàn)楹瘮?shù)為偶函數(shù),故函數(shù)f(x)在區(qū)間(-2,0)上是單調(diào)遞減的,故f(-2)>f(-1),所以D正確.12.BC解析∵f32-2x是偶函數(shù),∴f32+2x=f32-2x,∴函數(shù)f(x)的圖象關(guān)于直線x=32對(duì)稱(chēng),∴f(-1)=f(4).故C正確;∵g(2+x)為偶函數(shù),∴g(2-x)=g(2+x),∴g(x)的圖象關(guān)于直線x=2對(duì)稱(chēng).∵g(x)=f'(x),g(x)的圖象關(guān)于直線x=2對(duì)稱(chēng),∴f(x)的圖象關(guān)于點(diǎn)(2,t)(t∈R)對(duì)稱(chēng).∵f(x)的圖象關(guān)于直線x=32對(duì)稱(chēng)∴g(x)的圖象關(guān)于點(diǎn)32,∴f(x)與g(x)均是周期為2的函數(shù).∴f(0)=f(2)=t(不恒等于0),故A錯(cuò)誤;g-12=g32=0,∴構(gòu)造函數(shù)f(x)=sin(πx)符合題目要求,g(x)=πcos(πx),而g(-1)=πcos(-π)=-π,g(2)=πcos2π=π,故D錯(cuò)誤.故選BC.13.12解析因?yàn)閒(x)=log4x,x>0,f(x+3),x≤0,則f(-14.1-12x(答案不唯一)解析f(x)=1-12x,定義域?yàn)镽;12x>0,f(x)=1-12此函數(shù)是增函數(shù),滿足對(duì)任意x1,x2∈(0,+∞)且x1≠x2,均有f(x115.-43,-12∪-12,2解析函數(shù)f(x)的定義域?yàn)閧x|x≠0},f(-x)=1(-x)2+1-ln|-x|=1x2+1-ln|x|=f(x),故函數(shù)f(x因?yàn)楹瘮?shù)y=1x2+1,y