課時(shí)規(guī)范練12函數(shù)的圖象基礎(chǔ)鞏固組1.下列函數(shù)中,其圖象與函數(shù)y=lnx的圖象關(guān)于直線x=1對(duì)稱的是()A.y=ln(1-x) B.y=ln(2-x)C.y=ln(1+x) D.y=ln(2+x)2.(2024全國甲,7)函數(shù)f(x)=-x2+(ex-e-x)sinx在區(qū)間[-2.8,2.8]上的圖象大致為()A BC D3.已知定義在R上的奇函數(shù)f(x)在區(qū)間[0,+∞)上的圖象如圖所示,則不等式x2f(x)>2f(x)的解集為()A.(-2,0)∪(2,2)B.(-∞,-2)∪(2,+∞)C.(-∞,-2)∪(-2,0)∪(2,2)D.(-2,-2)∪(0,2)∪(2,+∞)4.函數(shù)f(x)=|x2-1A.函數(shù)f(x)是偶函數(shù)B.函數(shù)f(x)的最小值是0C.函數(shù)f(x)的單調(diào)遞增區(qū)間是[1,+∞)D.函數(shù)f(x)的圖象關(guān)于直線x=1對(duì)稱5.(2024廣東廣州一模)已知函數(shù)f(x)的部分圖象如圖所示,則f(x)的解析式可能是()A.f(x)=sin(tanx) B.f(x)=tan(sinx)C.f(x)=cos(tanx) D.f(x)=tan(cosx)6.若函數(shù)f(x)=(emx-n)2的大致圖象如圖所示,則()A.m>0,0<n<1 B.m>0,n>1C.m<0,0<n<1 D.m<0,n>1綜合提升組7.已知min{m,n}表示實(shí)數(shù)m,n中的較小數(shù),若函數(shù)f(x)=min{3+log14x,log2x},當(dāng)0<a<b時(shí),有f(a)=f(b),則ab的值為(A.6 B.8 C.9 D.168.(多選)(2024廣東深圳期中)已知f(x)=x2+2x+2,x≤0,1+lnx,x>0,若存在x1<x2<x3,使得f(x1)=f(A.實(shí)數(shù)m的取值范圍為(1,2]B.1≤x3≤eC.x1+x2=-2D.x1x2的最大值為19.(2024廣東深圳三模)函數(shù)y=|-x2+4x+5|的單調(diào)遞增區(qū)間是.

創(chuàng)新應(yīng)用組10.設(shè)f(x)=|log2x|,0<x<2,sin(πx4),2<x<10,若存在實(shí)數(shù)x1,x2,x3,x4滿足x1<x2<x3<x4,且f(x1)=f(A.(0,12) B.(4,16)C.(9,21) D.(15,25)11.(多選)若關(guān)于x的不等式lnx>(mx-2)x在區(qū)間(0,+∞)上有唯一的整數(shù)解,則實(shí)數(shù)m的取值可以是()A.1 B.76 C.54 D

課時(shí)規(guī)范練12函數(shù)的圖象1.B解析:設(shè)Q(x,y)是所求函數(shù)圖象上任一點(diǎn),則其關(guān)于直線x=1的對(duì)稱點(diǎn)P(2-x,y)在函數(shù)y=lnx的圖象上,所以y=ln(2-x).故選B.2.B解析:由題知函數(shù)f(x)的定義域?yàn)镽,關(guān)于原點(diǎn)對(duì)稱,f(-x)=-(-x)2+(e-x-ex)sin(-x)=-x2+(ex-e-x)sinx=f(x),所以函數(shù)f(x)為偶函數(shù),且函數(shù)的圖象關(guān)于y軸對(duì)稱,排除A,C;f(1)=-1+e-1esin1>-1+e-1esinπ6=-1+e2-12e>0,排除D.3.C解析:根據(jù)奇函數(shù)的圖象特征,作出f(x)在區(qū)間(-∞,0)上的圖象如圖所示,由x2f(x)>2f(x),得(x2-2)f(x)>0,等價(jià)于x解得x<-2或2<x<2或-2<x<0.故不等式解集為(-∞,-2)∪(-2,0)∪(2,2).4.B解析:畫出函數(shù)f(x)的圖象,如圖.可知函數(shù)f(x)是非奇非偶函數(shù),A錯(cuò)誤;函數(shù)f(x)的最小值是0,B正確;函數(shù)f(x)的單調(diào)遞增區(qū)間是(1,+∞)和(-1,0),C錯(cuò)誤;f(0)=1,f(2)=ln2,f(0)≠f(2),所以函數(shù)圖象不關(guān)于直線x=1對(duì)稱,D錯(cuò)誤,故選B.5.D解析:觀察圖象可知函數(shù)為偶函數(shù),對(duì)于A,f(-x)=sin(tan(-x))=sin(-tanx)=-sin(tanx)=-f(x),為奇函數(shù);對(duì)于B,f(-x)=tan(sin(-x))=tan(-sinx)=-tan(sinx)=-f(x),為奇函數(shù);同理,C,D選項(xiàng)為偶函數(shù),而對(duì)于C,其定義域?yàn)?π2+kπ,π2+kπ,不是R,舍去,故D正確.6.B解析:令f(x)=0,得emx=n,即mx=lnn,解得x=1mlnn,由圖象知x=1mlnn>0,當(dāng)m>0時(shí),n>1,當(dāng)m<0時(shí),0<n<1,故排除A,D,當(dāng)m<0時(shí),易知y=emx是減函數(shù),當(dāng)x→+∞時(shí),y→0,f(x)→n2,故排除C,故選7.B解析:作出函數(shù)f(x)的圖象,如圖中實(shí)線所示,由f(a)=f(b)可知,log2a=log14b+3,所以log2a+log4b=3,即log2a+log2b=log2(ab)=3,所以ab=8.AC解析:畫出f(x)的圖象如圖所示,要想y=m與y=f(x)有三個(gè)不同的交點(diǎn),需要m∈(1,2],A正確;由題意可知x1<x2≤0,x3∈(1,e]且x1,x2關(guān)于直線x=-1對(duì)稱,故x1+x2=-2,B錯(cuò)誤,C正確;則-x1-x2=2≥2(-x1)·(-x2)=2x1x2,解得x1x2≤1,當(dāng)且僅當(dāng)x1=x2=-1時(shí),等號(hào)成立,但x1≠-1,x2≠-1,故選AC.9.(-1,2),(5,+∞)解析:函數(shù)y=|-x2+4x+5|=-由|-x2+4x+5|=0,解得x=-1或x=5,函數(shù)y=|-x2+4x+5|的圖象如圖所示,由圖可知,函數(shù)y=|-x2+4x+5|的單調(diào)遞增區(qū)間為(-1,2),(5,+∞).10.A解析:函數(shù)的圖象如圖所示.因?yàn)閒(x1)=f(x2),所以-log2x1=log2x2,所以x1x2=1.因?yàn)閒(x3)=f(x4),所以x3+x4=12,2<x3<4,8<x4<10,所以(x3-2)(x4-2)x1x2=x3x4-2(x3+x4)+4=x3x4-20=x3(12-x又因?yàn)?<x3<4,所以0<-(x3-6)2+16<12,故選A.11.CD解析:依題意lnx>(mx-2)x等價(jià)于lnxx>mx-2.設(shè)g(x)=lnxx,h(x)=mx-2,x∈(0,+∞),則g'(x)=1-lnxx2,令g'(x)=0,得x=e,當(dāng)x∈(0,e)時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x∈(e,+∞)時(shí),g'(x)<0,g(x)單調(diào)遞減,所以g(x)max=g(e)=1e,g(1)=0,當(dāng)x>1時(shí),恒有g(shù)(x)>0.又函數(shù)y=h(x)的圖象是恒過點(diǎn)(0,-2)的直線,在同一平面直角坐標(biāo)系內(nèi)作出函數(shù)g因?yàn)殛P(guān)于x的不等式lnx>