新課標(biāo)廣西2019高考數(shù)學(xué)二輪復(fù)習(xí)專題對(duì)點(diǎn)練92.1~2.4組合練.docx

專題對(duì)點(diǎn)練92.12.4組合練(限時(shí)90分鐘,滿分100分)一、選擇題(共9小題,滿分45分)1.設(shè)函數(shù)f(x)=x2+1,x1,lnx,x>1,則f(f(e)=()A.0B.1C.2D.ln(e2+1)2.設(shè)a=60.4,b=log0.40.5,c=log80.4,則a,b,c的大小關(guān)系是()A.a<b<cB.c<b<aC.c<a<bD.b<c<a3.已知函數(shù)y=loga(x+c)(a,c為常數(shù),其中a>0,a1)的圖象如圖所示,則下列結(jié)論成立的是()A.a>1,c>1B.a>1,0<c<1C.0<a<1,c>1D.0<a<1,0<c<14.(2018全國(guó),文9)函數(shù)y=-x4+x2+2的圖象大致為()5.函數(shù)y=1+log0.5(x-1)的圖象一定經(jīng)過(guò)點(diǎn)()A.(1,1)B.(1,0)C.(2,1)D.(2,0)6.若函數(shù)f(x)=cosx,xa,1x,x>a的值域?yàn)?1,1,則實(shí)數(shù)a的取值范圍是()A.1,+)B.(-,-1C.(0,1D.(-1,0)7.已知函數(shù)f(x)=x12,則()A.x0R,使得f(x)<0B.x(0,+),f(x)0C.x1,x20,+),使得f(x1)-f(x2)x1-x2<0D.x10,+),x20,+),使得f(x1)>f(x2)8.已知函數(shù)f(x)為偶函數(shù),當(dāng)x0時(shí),f(x)為增函數(shù),則“65<x<2”是“flog2(2x-2)>flog1223”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件9.已知f(x)=ax2+x,x>0,-x,x0,若不等式f(x-1)f(x)對(duì)一切xR恒成立,則實(shí)數(shù)a的最大值為()A.916B.-1C.-12D.1二、填空題(共3小題,滿分15分)10.已知x0,y0,且x+y=1,則x2+y2的取值范圍是.11.已知二次函數(shù)f(x)=ax2-2x+c的值域?yàn)?,+),則9a+1c的最小值為.12.(2018天津,文14)已知aR,函數(shù)f(x)=x2+2x+a-2,x0,-x2+2x-2a,x>0.若對(duì)任意x-3,+),f(x)|x|恒成立,則a的取值范圍是.三、解答題(共3個(gè)題,滿分分別為13分,13分,14分)13.(2018全國(guó),文21)已知函數(shù)f(x)=aex-ln x-1.(1)設(shè)x=2是f(x)的極值點(diǎn),求a,并求f(x)的單調(diào)區(qū)間;(2)證明:當(dāng)a1e時(shí),f(x)0.14.已知函數(shù)f(x)=ex-ax2-2x(aR).(1)當(dāng)a=0時(shí),求f(x)的最小值;(2)當(dāng)a<e2-1時(shí),證明不等式f(x)>e2-1在(0,+)上恒成立.15.(2018浙江,22)已知函數(shù)f(x)=x-ln x.(1)若f(x)在x=x1,x2(x1x2)處導(dǎo)數(shù)相等,證明:f(x1)+f(x2)>8-8ln 2;(2)若a3-4ln 2,證明:對(duì)于任意k>0,直線y=kx+a與曲線y=f(x)有唯一公共點(diǎn).專題對(duì)點(diǎn)練9答案1.C解析 f(e)=ln e=1,所以f(f(e)=f(1)=12+1=2.故選C.2.B解析 a=60.4>1,b=log0.40.5(0,1),c=log80.4<0,a>b>c.3.D解析 函數(shù)單調(diào)遞減,0<a<1,當(dāng)x=1時(shí),y=loga(1+c)<0,即1+c>1,即c>0,當(dāng)x=0時(shí),loga(x+c)=logac>0,即c<1,即0<c<1,故選D.4.D解析 當(dāng)x=0時(shí),y=2>0,排除A,B;當(dāng)x=12時(shí),y=-124+122+2>2.排除C.故選D.5.C解析 函數(shù)y=log0.5x恒過(guò)定點(diǎn)(1,0),而y=1+log0.5(x-1)的圖象是由y=log0.5x的圖象向右平移一個(gè)單位,向上平移一個(gè)單位得到,定點(diǎn)(1,0)平移以后即為定點(diǎn)(2,1),故選C.6.A解析 函數(shù)f(x)=cosx,xa,1x,x>a的值域?yàn)?1,1,當(dāng)xa時(shí),f(x)=cos x-1,1,滿足題意;當(dāng)x>a時(shí),f(x)=1x-1,1,應(yīng)滿足0<1x1,解得x1.a的取值范圍是1,+).7.B解析 由函數(shù)f(x)=x12,知在A中f(x)0恒成立,故A錯(cuò)誤,B正確;又f(x)=x12在0,+)上是遞增函數(shù),故C錯(cuò)誤;在D中,當(dāng)x1=0時(shí),不存在x20,+)使得f(x1)>f(x2),故D不成立.故選B.8.D解析 由f(x)是偶函數(shù)且當(dāng)x0時(shí),f(x)為增函數(shù),則x>0時(shí),f(x)是減函數(shù),故由flog2(2x-2)>flog1223,得|log2(2x-2)|<log1223=log232,故0<2x-2<32,解得1<x<74,故“65<x<2”是“1<x<74”的既不充分也不必要條件,故選D.9.B解析 作出函數(shù)f(x)和f(x-1)的圖象,當(dāng)a0時(shí),f(x-1)f(x)對(duì)一切xR不恒成立(如圖1).圖1圖2當(dāng)a<0時(shí),f(x-1)過(guò)定點(diǎn)(1,0)(如圖2),當(dāng)x>0時(shí),f(x)=ax2+x的兩個(gè)零點(diǎn)為x=0和x=-1a,要使不等式f(x-1)f(x)對(duì)一切xR恒成立,則只需要-1a1,得a-1,即a的最大值為-1.10.12,1解析 x2+y2=x2+(1-x)2=2x2-2x+1,x0,1,所以當(dāng)x=0或1時(shí),x2+y2取最大值1;當(dāng)x=12時(shí),x2+y2取最小值12.因此x2+y2的取值范圍為12,1.11.6解析 二次函數(shù)f(x)=ax2-2x+c的值域?yàn)?,+),可得判別式=4-4ac=0,即有ac=1,且a>0,c>0,可得9a+1c29ac=23=6,當(dāng)且僅當(dāng)9a=1c,即有c=13,a=3時(shí),取得最小值6.12.18,2解析 當(dāng)x>0時(shí),f(x)|x|可化為-x2+2x-2ax,即x-122+2a-140,所以a18;當(dāng)-3x0時(shí),f(x)|x|可化為x2+2x+a-2-x,即x2+3x+a-20.對(duì)于函數(shù)y=x2+3x+a-2,其圖象的對(duì)稱軸方程為x=-32.因?yàn)楫?dāng)-3x0時(shí),y0,所以當(dāng)x=0時(shí),y0,即a-20,所以a2.綜上所述,a的取值范圍為18,2.13.解 (1)f(x)的定義域?yàn)?0,+),f(x)=aex-1x.由題設(shè)知,f(2)=0,所以a=12e2.從而f(x)=12e2ex-ln x-1,f(x)=12e2ex-1x.當(dāng)0<x<2時(shí),f(x)<0;當(dāng)x>2時(shí),f(x)>0.所以f(x)在(0,2)單調(diào)遞減,在(2,+)單調(diào)遞增.(2)當(dāng)a1e時(shí),f(x)exe-ln x-1.設(shè)g(x)=exe-ln x-1,則g(x)=exe-1x.當(dāng)0<x<1時(shí),g(x)<0;當(dāng)x>1時(shí),g(x)>0.所以x=1是g(x)的最小值點(diǎn).故當(dāng)x>0時(shí),g(x)g(1)=0.因此,當(dāng)a1e時(shí),f(x)0.14.(1)解 a=0時(shí),f(x)=ex-2x,f(x)=ex-2,令f(x)>0,解得x>ln 2,令f(x)<0,解得x<ln 2,故f(x)在(-,ln 2)遞減,在(ln 2,+)遞增,故f(x)min=f(ln 2)=2-2ln 2.(2)證明 f(x)=ex-2ax-2,f(1)=e-2-2a>e-2-2e2-1=0,f(0)=-1<0,故存在x0(0,1),使得f(x0)=0,令h(x)=ex-2ax-2,則x(0,+)時(shí),h(x)=ex-2a>ex+2-e>0,故h(x)在(0,+)遞增且h(x0)=0,故x=x0是h(x)的唯一零點(diǎn),且在x=x0處f(x)取最小值f(x0)=ex0-x0(ax0+2),又h(x0)=0,即ex0-2ax0-2=0,得ax0+1=ex02,故f(x0)=ex01-x02-x0,構(gòu)造函數(shù)g(t)=et1-t2-t,則g(t)=et12-t2-1,g(t)=et-t2,故t(0,1)時(shí),g(t)<0,g(t)在(0,1)遞減,故t(0,1)時(shí),g(t)<g(0)<0,故g(t)在(0,1)遞減,故f(x0)在(0,1)遞減,故f(x)min=f(x0)>e11-12-1=e2-1,原結(jié)論成立.15.證明 (1)函數(shù)f(x)的導(dǎo)函數(shù)f(x)=12x-1x,由f(x1)=f(x2),得12x1-1x1=12x2-1x2,因?yàn)閤1x2,所以1x1+1x2=12.由基本不等式,得12x1x2=x1+x224x1x2,因?yàn)閤1x2,所以x1x2>256.由題意得f(x1)+f(x2)=x1-ln x1+x2-ln x2=12x1x2-ln(x1x2).設(shè)g(x)=12x-ln x,則g(x)=14x(x-4),所以x(0,16)16(16,+)g(x)-0+g(x)2-4ln 2所以g(x)在256,+)上單調(diào)遞增,故g(x1x2)>g(256)=8-8ln 2,即f(x1)+f(x2)>8-8ln 2.(2)令m=e-(|a|+k),n=|a|+1k2+1,則f(m)-km-a>|a|+k-k-a0,f(n)-kn-a<n1n-an-kn|a|+1n-k<0,所以,存在x0(m,n),使f(x0)=kx0+a.所以,對(duì)于任意的aR及k(0,+),直線y=kx+a與曲線y=f(x)有公共點(diǎn).由f(x)=kx+a,得k=x-lnx-ax.設(shè)h(x)=x-lnx-ax,則h(x)=lnx-x2-1+ax2=-g(x)-1+ax2.其中g(shù)(x)=x2-ln x.由(1)可知g(x)g(16).又a3-4ln 2,故-g(x)-1+a-g(16)-1+a=-3+4ln 2+a0,所以h(x)0,即函數(shù)h(x)在(0,+)上單調(diào)遞減.因此方程f(x)-kx-a=0至多1個(gè)實(shí)根.綜上,當(dāng)a3-4ln 2時(shí),對(duì)于任意k>0,直線y=kx+a與曲線y=f(x)有唯一公共點(diǎn).