高三理科數(shù)學(xué)新課標(biāo)二輪復(fù)習(xí)專題整合高頻突破習(xí)題：專題二 函數(shù)與導(dǎo)數(shù) 專題能力訓(xùn)練5 Word版含答案

高考數(shù)學(xué)精品復(fù)習(xí)資料 2019.5專題能力訓(xùn)練5基本初等函數(shù)、函數(shù)的圖象和性質(zhì)能力突破訓(xùn)練1.(20xx湖北六校聯(lián)考)下列函數(shù)在其定義域上既是奇函數(shù)又是減函數(shù)的是()A.f(x)=-x|x|B.f(x)=xsin xC.f(x)=1xD.f(x)=x122.已知a=21.2,b=12-0.8,c=2log52,則a,b,c的大小關(guān)系為()A.c<b<aB.c<a<bC.b<a<cD.b<c<a3.函數(shù)y=ex+e-xex-e-x的圖象大致為()4.(20xx全國(guó),理5)函數(shù)f(x)在區(qū)間(-,+)單調(diào)遞減,且為奇函數(shù),若f(1)=-1,則滿足-1f(x-2)1的x的取值范圍是()A.-2,2B.-1,1C.0,4D.1,35.已知函數(shù)f(x)=2x-1-2,x1,-log2(x+1),x>1,且f(a)=-3,則f(6-a)=()A.-74B.-54C.-34D.-146.(20xx安徽池州模擬)已知函數(shù)的定義域?yàn)镽,且滿足下列三個(gè)條件:對(duì)任意的x1,x24,8,當(dāng)x1<x2時(shí),都有f(x1)-f(x2)x1-x2>0;f(x+4)=-f(x);y=f(x+4)是偶函數(shù).若a=f(6),b=f(11),c=f(2 017),則a,b,c的大小關(guān)系正確的是()A.a<b<cB.b<a<cC.a<c<bD.c<b<a7.已知a>b>1,若logab+logba=52,ab=ba,則a=,b=.8.若函數(shù)f(x)=xln(x+a+x2)為偶函數(shù),則a=.9.已知函數(shù)f(x)是定義在R上的偶函數(shù),且在區(qū)間0,+)上單調(diào)遞增.若實(shí)數(shù)a滿足f(log2a)+f(log12a)2f(1),則a的取值范圍是.10.設(shè)奇函數(shù)y=f(x)(xR),滿足對(duì)任意tR都有f(t)=f(1-t),且當(dāng)x0,12時(shí),f(x)=-x2,則f(3)+f-32的值等于.11.設(shè)函數(shù)f(x)=(x+1)2+sinxx2+1的最大值為M,最小值為m,則M+m=.12.若不等式3x2-logax<0在x0,13內(nèi)恒成立,求實(shí)數(shù)a的取值范圍.思維提升訓(xùn)練13.函數(shù)y=cos6x2x-2-x的圖象大致為()14.(20xx江西百校聯(lián)盟聯(lián)考)已知f(x)是定義在R上的偶函數(shù),當(dāng)x>0時(shí),f(x)=ax+log5x,x>4,x2+2x+3,0<x4,若f(-5)<f(2),則a的取值范圍為()A.(-,1)B.(-,2)C.(-2,+)D.(2,+)15.已知函數(shù)f(x)(xR)滿足f(-x)=2-f(x),若函數(shù)y=x+1x與y=f(x)圖象的交點(diǎn)為(x1,y1),(x2,y2),(xm,ym),則i=1m(xi+yi)=()A.0B.mC.2mD.4m16.已知f(x)是定義在R上的偶函數(shù),且在區(qū)間(-,0)上單調(diào)遞增.若實(shí)數(shù)a滿足f(2|a-1|)>f(-2),則a的取值范圍是.17.設(shè)f(x)是定義在R上且周期為2的函數(shù),在區(qū)間-1,1上,f(x)=ax+1,-1x<0,bx+2x+1,0x1,其中a,bR.若f12=f32,則a+3b的值為.18.(20xx山東,理15)若函數(shù)exf(x)(e=2.718 28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)中所有具有M性質(zhì)的函數(shù)的序號(hào)為.f(x)=2-xf(x)=3-xf(x)=x3f(x)=x2+219.已知函數(shù)f(x)=ex-e-x(xR,且e為自然對(duì)數(shù)的底數(shù)).(1)判斷函數(shù)f(x)的奇偶性與單調(diào)性.(2)是否存在實(shí)數(shù)t,使不等式f(x-t)+f(x2-t2)0對(duì)一切x都成立?若存在,求出t;若不存在,請(qǐng)說明理由.參考答案專題能力訓(xùn)練5基本初等函數(shù)、函數(shù)的圖象和性質(zhì)能力突破訓(xùn)練1.A解析函數(shù)f(x)=-x2,x0,x2,x<0在其定義域上既是奇函數(shù)又是減函數(shù),故選A.2.A解析b=12-0.8=20.8<21.2=a,且b>1,又c=2log52=log54<1,c<b<a.3.A解析函數(shù)有意義,需使ex-e-x0,其定義域?yàn)閤|x0,排除C,D.因?yàn)閥=ex+e-xex-e-x=e2x+1e2x-1=1+2e2x-1,所以當(dāng)x>0時(shí)函數(shù)為減函數(shù).故選A.4.D解析因?yàn)閒(x)為奇函數(shù),所以f(-1)=-f(1)=1,于是-1f(x-2)1等價(jià)于f(1)f(x-2)f(-1).又f(x)在區(qū)間(-,+)單調(diào)遞減,所以-1x-21,即1x3.所以x的取值范圍是1,3.5.A解析f(a)=-3,當(dāng)a1時(shí),f(a)=2a-1-2=-3,即2a-1=-1,此等式顯然不成立.當(dāng)a>1時(shí),f(a)=-log2(a+1)=-3,即a+1=23,解得a=7.f(6-a)=f(-1)=2-1-1-2=14-2=-74.6.B解析由得f(x)在區(qū)間4,8上單調(diào)遞增;由得f(x+8)=-f(x+4)=f(x),故f(x)是周期為8的周期函數(shù),所以c=f(20xx)=f(2528+1)=f(1),b=f(11)=f(3);再由可知f(x)的圖象關(guān)于直線x=4對(duì)稱,所以b=f(11)=f(3)=f(5),c=f(1)=f(7).結(jié)合f(x)在區(qū)間4,8上單調(diào)遞增可知,f(5)<f(6)<f(7),即b<a<c.故選B.7.42解析設(shè)logba=t,由a>b>1,知t>1.由題意,得t+1t=52,解得t=2,則a=b2.由ab=ba,得b2b=bb2,即得2b=b2,即b=2,a=4.8.1解析f(x)是偶函數(shù),f(-1)=f(1).又f(-1)=-ln(-1+a+1)=lna+1+1a,f(1)=ln(1+a+1),因此ln(a+1+1)-lna=ln(a+1+1),于是lna=0,a=1.9.12,2解析由題意知a>0,又log12a=log2a-1=-log2a.f(x)是R上的偶函數(shù),f(log2a)=f(-log2a)=f(log12a).f(log2a)+f(log12a)2f(1),2f(log2a)2f(1),即f(log2a)f(1).又f(x)在0,+)上單調(diào)遞增,|log2a|1,-1log2a1,a12,2.10.-14解析根據(jù)對(duì)任意tR都有f(t)=f(1-t)可得f(-t)=f(1+t),即f(t+1)=-f(t),進(jìn)而得到f(t+2)=-f(t+1)=-f(t)=f(t),得函數(shù)y=f(x)的一個(gè)周期為2,則f(3)=f(1)=f(0+1)=-f(0)=0,f-32=f12=-14,所以f(3)+f-32=0+-14=-14.11.2解析f(x)=(x+1)2+sinxx2+1=1+2x+sinxx2+1,設(shè)g(x)=2x+sinxx2+1,則g(-x)=-g(x),故g(x)是奇函數(shù).由奇函數(shù)圖象的對(duì)稱性知g(x)max+g(x)min=0,則M+m=g(x)+1max+g(x)+1min=2+g(x)max+g(x)min=2.12.解由題意知3x2<logax在x0,13內(nèi)恒成立.在同一平面直角坐標(biāo)系內(nèi),分別作出函數(shù)y=3x2和y=logax的圖象.觀察兩函數(shù)圖象,當(dāng)x0,13時(shí),若a>1,函數(shù)y=logax的圖象顯然在函數(shù)y=3x2圖象的下方,所以不成立;當(dāng)0<a<1時(shí),由圖可知,y=logax的圖象必須過點(diǎn)13,13或在這個(gè)點(diǎn)的上方,則loga1313,所以a127,所以127a<1.綜上,實(shí)數(shù)a的取值范圍為127a<1.思維提升訓(xùn)練13.D解析y=cos6x2x-2-x為奇函數(shù),排除A項(xiàng);y=cos6x有無窮多個(gè)零點(diǎn),排除C項(xiàng);當(dāng)x在原點(diǎn)右側(cè)附近時(shí),可保證2x-2-x>0,cos6x>0,則此時(shí)y>0,故選D.14.B解析因?yàn)閒(x)是定義在R上的偶函數(shù),所以f(-5)=f(5)=5a+log55=1+5a,則不等式f(-5)<f(2)可化為f(5)<f(2).又f(2)=4+4+3=11,所以由5a+1<11可得a<2,故選B.15.B解析由f(-x)=2-f(x),得f(x)的圖象關(guān)于點(diǎn)(0,1)對(duì)稱.而y=x+1x=1+1x的圖象是由y=1x的圖象向上平移一個(gè)單位長(zhǎng)度得到的,故y=x+1x的圖象關(guān)于點(diǎn)(0,1)對(duì)稱.則函數(shù)y=x+1x與y=f(x)圖象的交點(diǎn)也關(guān)于點(diǎn)(0,1)對(duì)稱,且每一組對(duì)稱點(diǎn)(xi,yi),(xi,yi)(i=1,2,m)滿足xi+xi=0,yi+yi=2,所以i=1m(xi+yi)=i=1mxi+i=1myi=m20+m22=m.16.12,32解析由題意知函數(shù)f(x)在區(qū)間(0,+)上單調(diào)遞減,又f(x)是偶函數(shù),則不等式f(2|a-1|)>f(-2)可化為f(2|a-1|)>f(2),則2|a-1|<2,|a-1|<12,解得12<a<32.故答案為12,32.17.-10解析f32=f12,f12=f-12,12b+232=-12a+1,易求得3a+2b=-2.又f(1)=f(-1),-a+1=b+22,即2a+b=0,a=2,b=-4,a+3b=-10.18.解析對(duì),設(shè)g(x)=ex2-x,則g(x)=ex2-x+2-xln12=ex2-x1+ln12>0,g(x)在R上單調(diào)遞增,具有M性質(zhì);對(duì),設(shè)g(x)=ex3-x,則g(x)=ex3-x+3-xln13=ex3-x1+ln13<0,g(x)在R上單調(diào)遞減,不具有M性質(zhì);對(duì),設(shè)g(x)=exx3,則g(x)=exx2(x+3),令g(x)=0,得x1=-3,x2=0,g(x)在區(qū)間(-,-3)上單調(diào)遞減,在區(qū)間(-3,+)上單調(diào)遞增,不具有M性質(zhì);對(duì),設(shè)g(x)=ex(x2+2),則g(x)=ex(x2+2x+2),x2+2x+2=(x+1)2+1>0,g(x)>0,g(x)在R上單調(diào)遞增,具有M性質(zhì).故填.19.解(1)f(x)=ex-1ex,且y=ex是增函數(shù),y=-1ex是增函數(shù),f(x)是增函數(shù).f(x)的定義域?yàn)镽,且f(-x)=e-x-ex=-f(x),f(x)是奇函數(shù).(2)由(1)知f(x)是增函數(shù)且為奇函數(shù).f(x-t)+f(x2-t2)0對(duì)xR恒成立,f(x-t)f(t2-x2),t2-x2x-t,x2+xt2+t對(duì)xR恒成立.又t+122x+12min2對(duì)一切xR恒成立,t+1220,t=-12.即存在實(shí)數(shù)t=-12,使不等式f(x-t)+f(x2-t2)0對(duì)一切x都成立.