（適用于新高考新教材）高考數(shù)學(xué)一輪總復(fù)習(xí)第三章函數(shù)與基本初等函數(shù)課時(shí)規(guī)范練12函數(shù)的圖象

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(1x) B.y=ln(2x)C.y=ln(1+x) D.y=ln(2+x)2.(2022河北石家莊一模)函數(shù)f(x)=x32x3.(2022河南商丘三模)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-A.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.已知函數(shù)f(x)=x2+14,g(x)=sinx,則圖象為下圖的函數(shù)可能是(A.y=f(x)+g(x)1B.y=f(x)g(x)1C.y=f(x)g(x)D.y=g6.若函數(shù)f(x)=(emxn)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.(2022福建福州一中三模)已知函數(shù)f(x)=ln(x+x2+1)·cos(3x+φ),則當(dāng)φ∈[0,π]時(shí),f(x)的圖象不可能是(8.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.169.已知函數(shù)f(x)=logax,x>0,|x+3|,-4≤x<A.0,14 B.0,14∪(1,+∞)C.14,1∪(1,+∞) D.(0,1)∪(1,4)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fA.(0,12) B.(4,16)C.(9,21) D.(15,25)11.(多選)(2022重慶巴蜀中學(xué)高三檢測(cè))若關(guān)于x的不等式lnx>(mx2)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(2x,y)在函數(shù)y=lnx的圖象上,所以y=ln(2x).故選B.2.A解析:因?yàn)楹瘮?shù)f(x)=x32x+2-x的定義域?yàn)镽,且f(x)=f(x),所以f(x)為奇函數(shù),其圖象關(guān)于原點(diǎn)對(duì)稱,據(jù)此排除B,D選項(xiàng);易知當(dāng)x→+∞時(shí),f(x)=x32x+2-x>0,2x→+∞,2x→0,x3→+∞,因?yàn)橹笖?shù)函數(shù)y=2x比冪函數(shù)y=x3的增長(zhǎng)速度快,所以f(x)→0,即當(dāng)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(x22)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:由題圖可知,圖象關(guān)于原點(diǎn)對(duì)稱,則所求函數(shù)為奇函數(shù).由已知得f(x)=x2+14為偶函數(shù),g(x)=sinx函數(shù)y=f(x)+g(x)14=x2+sinx為非奇非偶函數(shù),故選項(xiàng)A不符合函數(shù)y=f(x)g(x)14=x2sinx為非奇非偶函數(shù),故選項(xiàng)B不符合函數(shù)y=f(x)g(x)=x2+14sinx,由題圖可知,所求函數(shù)在區(qū)間0,π4上不單調(diào),而函數(shù)y=f(x)g(x)=x2+14sin6.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)→n27.D解析:設(shè)g(x)=ln(x+x2+1),則它的定義域?yàn)橐驗(yàn)間(x)+g(x)=ln(x+x2+1)+ln(x+x2+1)=ln(x2+1x所以g(x)=g(x),即g(x)為奇函數(shù).當(dāng)φ=0時(shí),y=cos3x為偶函數(shù),則f(x)=ln(x+x2+1)·cos3x為奇函數(shù),且f(0)=fπ6=fπ2=0,fπ18>0,所以選項(xiàng)B可能;當(dāng)φ=π時(shí),y=cos(3x+π)=cos3x為偶函數(shù),則f(x)=ln(x+x2+1)·cos3x為奇函數(shù),且f(0)=fπ6=fπ2=0,fπ18<0,所以選項(xiàng)A可能;當(dāng)φ=π2時(shí),y=cos3x+π2=sin3x為奇函數(shù),則f(x)=ln(x+x2+1)·sin3x為偶函數(shù),且f(0)=fπ3=f2π3=0,fπ18<0,所以選項(xiàng)C可能.故選D.8.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=9.C解析:當(dāng)4≤x<0時(shí),函數(shù)y=|x+3|圖象關(guān)于原點(diǎn)對(duì)稱的圖象對(duì)應(yīng)的函數(shù)為y=|x+3|,即y=|x+3|(0<x≤4),若函數(shù)f(x)的圖象上有且只有一對(duì)點(diǎn)關(guān)于原點(diǎn)對(duì)稱,則等價(jià)于函數(shù)f(x)=logax與y=|x+3|(0<x≤4)圖象只有一個(gè)交點(diǎn),分別作出兩個(gè)函數(shù)的圖象,如圖.若a>1,函數(shù)f(x)=logax與y=|x+3|(0<x≤4)圖象有唯一的交點(diǎn),滿足條件;若0<a<1,當(dāng)x=4時(shí),y=|4+3|=1,所以要使函數(shù)f(x)=logax與y=|x+3|(0<x≤4)圖象有唯一的交點(diǎn),則要滿足f(4)<1,即loga4<1=logaa1,解得14<a<1.綜上實(shí)數(shù)a的取值范圍是14,1∪(1,+∞),故選C.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=x3x42(x3+x4)+4=x3x420=x又因?yàn)?<x3<4,所以0<(x36)2+16<12,故選A.11.CD解析:依題意lnx>(mx2)x等價(jià)于lnxx>mx2.設(shè)g(x)=lnxx,h(x)=mx2,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)因?yàn)?/p>