高考數(shù)學(xué)專題復(fù)習(xí)導(dǎo)練測(cè) 第二章 函數(shù)與基本初等函數(shù)(I)階段測(cè)試(三)課件 理 新人教A版.ppt
數(shù)學(xué) A(理),45分鐘階段測(cè)試(三),第二章 函數(shù)概念與基本初等函數(shù),2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,一、選擇題 1.若函數(shù)f(x)x2axb的圖象與x軸的交點(diǎn)為(1,0)和(3,0),則函數(shù)f(x)( ) A.在(,2上單調(diào)遞減,在2,)上單調(diào)遞增 B.在(,3)上單調(diào)遞增 C.在1,3上單調(diào)遞增 D.單調(diào)性不能確定,2,3,4,5,6,7,8,9,10,1,解析 畫出函數(shù)f(x)的草圖如圖.,易知f(x)的對(duì)稱軸為x 2,,故f(x)在(,2上單調(diào)遞減,在2,)上單調(diào)遞增.,答案 A,3,4,5,6,7,8,9,10,1,2,2.設(shè)f(x)為定義在R上的奇函數(shù),當(dāng)x0時(shí),f(x)log3(1x),則f(2)等于( ) A.1 B.3 C.1 D.3,解析 由題意得,f(2)f(2)log3(12)1.,A,2,4,5,6,7,8,9,10,1,3,3.(2014遼寧)已知a ,blog2 ,c ,則( ) A.abc B.acb C.cab D.cba,解析 0 1,,即01,所以cab.,C,2,3,5,6,7,8,9,10,1,4,4.(2014浙江)在同一直角坐標(biāo)系中,函數(shù)f(x)xa(x0),g(x)logax的圖象可能是( ),2,3,5,6,7,8,9,10,1,4,解析 方法一 當(dāng)a1時(shí),yxa與ylogax均為增函數(shù),但yxa遞增較快,排除C; 當(dāng)0a1時(shí),yxa為增函數(shù),ylogax為減函數(shù),排除A. 由于yxa遞增較慢,所以選D. 方法二 冪函數(shù)f(x)xa的圖象不過(0,1)點(diǎn),排除A;,2,3,5,6,7,8,9,10,1,4,B項(xiàng)中由對(duì)數(shù)函數(shù)f(x)logax的圖象知01,而此時(shí)冪函數(shù)f(x)xa的圖象應(yīng)是增長(zhǎng)越來(lái)越快的變化趨勢(shì),故C錯(cuò). 答案 D,2,3,4,6,7,8,9,10,1,5,5.已知定義在R上的函數(shù)yf(x)對(duì)于任意的x都滿足f(x1)f(x),當(dāng)1x1時(shí),f(x)x3,若函數(shù)g(x)f(x)loga|x|至少有6個(gè)零點(diǎn),則a的取值范圍是( ),2,3,4,6,7,8,9,10,1,5,解析 由f(x1)f(x)得f(x1)f(x2), 因此f(x)f(x2),即函數(shù)f(x)是周期為2的周期函數(shù). 函數(shù)g(x)f(x)loga|x|至少有6個(gè)零點(diǎn)可轉(zhuǎn)化成yf(x)與h(x)loga|x|兩函數(shù)圖象交點(diǎn)至少有6個(gè),需對(duì)底數(shù)a進(jìn)行分類討論.,2,3,4,6,7,8,9,10,1,5,若a1,則h(5)loga55.,2,3,4,6,7,8,9,10,1,5,若0a1,則h(5)loga51,即0a .,所以a的取值范圍是(0, (5,).,答案 A,2,3,4,5,7,8,9,10,1,6,二、填空題 6.客車從甲地以60 km/h的速度勻速行駛1小時(shí)到達(dá)乙地,在乙地停留了半小時(shí),然后以80 km/h的速度勻速行駛1小時(shí)到達(dá)丙地.客車從甲地出發(fā),經(jīng)過乙地,最后到達(dá)丙地所 經(jīng)過的路程s與時(shí)間t的函數(shù)解析式是_.,2,3,4,5,6,8,9,10,1,7,7.方程4x|12x|5的實(shí)數(shù)解x_.,解析 當(dāng)x0時(shí),方程4x|12x|5可化為:4x2x60, 解得2x3(舍)或2x2,故x1; 當(dāng)x0時(shí),方程4x|12x|5可化為:4x2x40.,綜上可知:x1.,1,2,3,4,5,6,9,10,1,7,8,8.關(guān)于函數(shù)f(x)lg (x0),有下列命題: 其圖象關(guān)于y軸對(duì)稱; 當(dāng)x0時(shí),f(x)是增函數(shù);當(dāng)x0時(shí),f(x)是減函數(shù); f(x)的最小值是lg 2; f(x)在區(qū)間(1,0),(2,)上是增函數(shù); f(x)無(wú)最大值,也無(wú)最小值. 其中所有正確結(jié)論的序號(hào)是_.,2,3,4,5,6,9,10,1,7,8,解析 根據(jù)已知條件可知f(x)lg (x0)為偶函數(shù),顯然利用偶函數(shù)的性質(zhì)可知命題正確;,對(duì)真數(shù)部分分析可知最小值為2,因此命題成立; 利用復(fù)合函數(shù)的性質(zhì)可知命題成立; 命題,單調(diào)性不符合復(fù)合函數(shù)的性質(zhì),因此錯(cuò)誤; 命題中,函數(shù)有最小值,因此錯(cuò)誤,故填寫. 答案 ,2,3,4,5,6,7,8,10,1,9,三、解答題 9.已知yf(x)是定義域?yàn)镽的奇函數(shù),當(dāng)x0,)時(shí),f(x)x22x.,2,3,4,5,6,7,8,10,1,9,(1)寫出函數(shù)yf(x)的解析式;,解 當(dāng)x(,0)時(shí),x(0,). yf(x)是奇函數(shù), f(x)f(x)(x)22(x)x22x,,2,3,4,5,6,7,8,10,1,9,(2)若方程f(x)a恰有3個(gè)不同的解,求a的取值范圍.,解 當(dāng)x0,)時(shí),f(x)x22x(x1)21,最小值為1; 當(dāng)x(,0)時(shí),f(x)x22x1(x1)2,最大值為1.,據(jù)此可作出函數(shù)yf(x)的圖象(如圖所示), 根據(jù)圖象得,若方程f(x)a恰有3個(gè)不同的解, 則a的取值范圍是(1,1).,2,3,4,5,6,7,8,9,1,10,10.已知函數(shù)f(x)log4(ax22x3). (1)若f(1)1,求f(x)的單調(diào)區(qū)間;,解 f(1)1,log4(a5)1, 因此a54,a1,這時(shí)f(x)log4(x22x3). 由x22x30得1x3, 函數(shù)f(x)的定義域?yàn)?1,3).,2,3,4,5,6,7,8,9,1,10,令g(x)x22x3, 則g(x)在(1,1)上遞增,在(1,3)上遞減. 又ylog4x在(0,)上遞增, 所以f(x)的單調(diào)遞增區(qū)間是(1,1), 遞減區(qū)間是(1,3).,2,3,4,5,6,7,8,9,1,10,(2)是否存在實(shí)數(shù)a,使f(x)的最小值為0?若存在,求出a的值;若不存在,說明理由.,解 假設(shè)存在實(shí)數(shù)a使f(x)的最小值為0, 則h(x)ax22x3應(yīng)有最小值1,,