2014屆高考數(shù)學(xué)一輪 知識(shí)點(diǎn)各個(gè)擊破 第二章 課時(shí)跟蹤檢測(cè)(八)函數(shù)的圖象 文 新人教A版
課時(shí)跟蹤檢測(cè)(八)函數(shù)的圖象1函數(shù)f(x)2x3的圖象()A關(guān)于y軸對(duì)稱B關(guān)于x軸對(duì)稱C關(guān)于直線yx對(duì)稱 D關(guān)于原點(diǎn)對(duì)稱2函數(shù)y的圖象大致是()3(2012·北京海淀二模)為了得到函數(shù)ylog2(x1)的圖象,可將函數(shù)ylog2x的圖象上所有的點(diǎn)的()A縱坐標(biāo)縮短到原來的,橫坐標(biāo)不變,再向右平移1個(gè)單位長度B縱坐標(biāo)縮短到原來的,橫坐標(biāo)不變,再向左平移1個(gè)單位長度C橫坐標(biāo)伸長到原來的2倍,縱坐標(biāo)不變,再向右平移1個(gè)單位長度D橫坐標(biāo)伸長到原來的2倍,縱坐標(biāo)不變,再向左平移1個(gè)單位長度4(2011·陜西高考)設(shè)函數(shù)f(x)(xR)滿足f(x)f(x),f(x2)f(x),則yf(x)的圖象可能是()5(2012·濟(jì)南模擬)函數(shù)ylg的大致圖象為()6(2011·天津高考)對(duì)實(shí)數(shù)a和b,定義運(yùn)算“”:ab設(shè)函數(shù)f(x)(x22)(xx2),xR.若函數(shù)yf(x)c的圖象與x軸恰有兩個(gè)公共點(diǎn),則實(shí)數(shù)c的取值范圍是()A.B.C.D.7.已知函數(shù)f(x)的圖象如圖所示,則函數(shù)g(x)logf(x)的定義域是_8函數(shù)f(x)圖象的對(duì)稱中心為_9如圖,定義在1,)上的函數(shù)f(x)的圖象由一條線段及拋物線的一部分組成,則f(x)的解析式為_10已知函數(shù)f(x)(1)在如圖所示給定的直角坐標(biāo)系內(nèi)畫出f(x)的圖象;(2)寫出f(x)的單調(diào)遞增區(qū)間;(3)由圖象指出當(dāng)x取什么值時(shí)f(x)有最值11若直線y2a與函數(shù)y|ax1|(a0且a1)的圖象有兩個(gè)公共點(diǎn),求a的取值范圍12已知函數(shù)f(x)的圖象與函數(shù)h(x)x2的圖象關(guān)于點(diǎn)A(0,1)對(duì)稱(1)求函數(shù)f(x)的解析式;(2)若g(x)f(x),g(x)在區(qū)間(0,2上的值不小于6,求實(shí)數(shù)a的取值范圍1.(2013·威海質(zhì)檢)函數(shù)yf(x)(xR)的圖象如圖所示,下列說法正確的是()函數(shù)yf(x)滿足f(x)f(x);函數(shù)yf(x)滿足f(x2)f(x);函數(shù)yf(x)滿足f(x)f(x);函數(shù)yf(x)滿足f(x2)f(x)ABC D2若函數(shù)f(x)的圖象經(jīng)過變換T后所得圖象對(duì)應(yīng)函數(shù)的值域與函數(shù)f(x)的值域相同,則稱變換T是函數(shù)f(x)的同值變換下面給出四個(gè)函數(shù)及其對(duì)應(yīng)的變換T,其中變換T不屬于函數(shù)f(x)的同值變換的是()Af(x)(x1)2,變換T將函數(shù)f(x)的圖象關(guān)于y軸對(duì)稱Bf(x)2x11,變換T將函數(shù)f(x)的圖象關(guān)于x軸對(duì)稱Cf(x)2x3,變換T將函數(shù)f(x)的圖象關(guān)于點(diǎn)(1,1)對(duì)稱Df(x)sin,變換T將函數(shù)f(x)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱3已知函數(shù)yf(x)的定義域?yàn)镽,并對(duì)一切實(shí)數(shù)x,都滿足f(2x)f(2x)(1)證明:函數(shù)yf(x)的圖象關(guān)于直線x2對(duì)稱;(2)若f(x)是偶函數(shù),且x0,2時(shí),f(x)2x1,求x4,0時(shí)的f(x)的表達(dá)式答 題 欄A級(jí)1._ 2._ 3._ 4._ 5._ 6._ B級(jí)1._ 2._ 7. _ 8. _ 9. _答 案課時(shí)跟蹤檢測(cè)(八)A級(jí)1D2.B3.A4.B5選D由題知該函數(shù)的圖象是由函數(shù)ylg|x|的圖象左移一個(gè)單位得到的,故其圖象為選項(xiàng)D中的圖象6選B由題意可知f(x)作出圖象,由圖象可知yf(x)與yc有兩個(gè)交點(diǎn)時(shí),c2或1<c<,即函數(shù)yf(x)c的圖象與x軸恰有兩個(gè)公共點(diǎn)時(shí)實(shí)數(shù)c的取值范圍是(,2.7解析:當(dāng)f(x)>0時(shí),函數(shù)g(x)logf(x)有意義,由函數(shù)f(x)的圖象知滿足f(x)>0的x(2,8答案:(2,88解析:f(x)1,把函數(shù)y的圖象向上平移1個(gè)單位,即得函數(shù)f(x)的圖象由y的對(duì)稱中心為(0,0),可得平移后的f(x)圖象的對(duì)稱中心為(0,1)答案:(0,1)9解析:當(dāng)1x0時(shí),設(shè)解析式為ykxb,則得yx1.當(dāng)x>0時(shí),設(shè)解析式為ya(x2)21,圖象過點(diǎn)(4,0),0a(42)21,得a.答案:f(x)10解:(1)函數(shù)f(x)的圖象如圖所示(2)由圖象可知,函數(shù)f(x)的單調(diào)遞增區(qū)間為1,0,2,5(3)由圖象知當(dāng)x2時(shí),f(x)minf(2)1,當(dāng)x0時(shí),f(x)maxf(0)3.11解:當(dāng)0a1時(shí),y|ax1|的圖象如圖1所示,由已知得02a1,即0a.當(dāng)a1時(shí),y|ax1|的圖象如圖2所示,由已知可得02a1,即0a,但a1,故a.綜上可知,a的取值范圍為.12解:(1)設(shè)f(x)圖象上任一點(diǎn)坐標(biāo)為(x,y),點(diǎn)(x,y)關(guān)于點(diǎn)A(0,1)的對(duì)稱點(diǎn)(x,2y)在h(x)的圖象上,2yx2,yx,即f(x)x.(2)由題意g(x)x,且g(x)x6,x(0,2x(0,2,a1x(6x),即ax26x1.令q(x)x26x1,x(0,2,q(x)x26x1(x3)28,x(0,2時(shí),q(x)maxq(2)7,故a的取值范圍為7,)B級(jí)1選C由圖象可知,函數(shù)f(x)為奇函數(shù)且關(guān)于直線x1對(duì)稱,所以f(1x)f(1x),所以f1(x1)f1(x1),即f(x2)f(x)故正確2選B對(duì)于A,與f(x)(x1)2的圖象關(guān)于y軸對(duì)稱的圖象對(duì)應(yīng)的函數(shù)解析式為g(x)(x1)2(x1)2,易知兩者的值域都為0,);對(duì)于B,函數(shù)f(x)2x11的值域?yàn)?1,),與函數(shù)f(x)的圖象關(guān)于x軸對(duì)稱的圖象對(duì)應(yīng)的函數(shù)解析式為g(x)2x11,其值域?yàn)?,1);對(duì)于C,與f(x)2x3的圖象關(guān)于點(diǎn)(1,1)對(duì)稱的圖象對(duì)應(yīng)的函數(shù)解析式為2g(x)2(2x)3,即g(x)2x3,易知值域相同;對(duì)于D,與f(x)sin的圖象關(guān)于點(diǎn)(1,0)對(duì)稱的圖象對(duì)應(yīng)的函數(shù)解析式為g(x)sin,其值域?yàn)?,1,易知兩函數(shù)的值域相同3解:(1)證明:設(shè)P(x0,y0)是函數(shù)yf(x)圖象上任一點(diǎn),則y0f(x0),點(diǎn)P關(guān)于直線x2的對(duì)稱點(diǎn)為P(4x0,y0)因?yàn)閒(4x0)f(2(2x0)f(2(2x0)f(x0)y0,所以P也在yf(x)的圖象上,所以函數(shù)yf(x)的圖象關(guān)于直線x2對(duì)稱(2)因?yàn)楫?dāng)x2,0時(shí),x0,2,所以f(x)2x1.又因?yàn)閒(x)為偶函數(shù),所以f(x)f(x)2x1,x2,0當(dāng)x4,2時(shí),4x0,2,所以f(4x)2(4x)12x7.而f(4x)f(x)f(x),所以f(x)2x7,x4,2所以f(x)