2019版高考數(shù)學(xué)總復(fù)習(xí) 第二章 函數(shù)、導(dǎo)數(shù)及其應(yīng)用 8 指數(shù)與指數(shù)函數(shù)課時(shí)作業(yè) 文.doc
課時(shí)作業(yè)8指數(shù)與指數(shù)函數(shù)一、選擇題1(2018河北八所重點(diǎn)中學(xué)一模)設(shè)a>0,將表示成分?jǐn)?shù)指數(shù)冪的形式,其結(jié)果是()Aa BaCa Da解析:a,故選C.答案:C2若函數(shù)f(x)a|2x4|(a>0,a1)滿足f(1),則f(x)的單調(diào)遞減區(qū)間是()A(,2 B2,)C2,) D(,2解析:由f(1)得a2.又a>0,所以a,因此f(x)|2x4|因?yàn)間(x)|2x4|在2,)上單調(diào)遞增,所以f(x)的單調(diào)遞減區(qū)間是2,)答案:B3(2018河南南陽、信陽等六市一模)已知a、b(0,1)(1,),當(dāng)x>0時(shí),1<bx<ax,則()A0<b<a<1 B0<a<b<1C1<b<a D1<a<b解析:x>0時(shí),1<bx,b>1.x>0時(shí),bx<ax,x>0時(shí),x>1.>1,a>b.1<b<a.答案:C4已知f(x)3xb(2x4,b為常數(shù))的圖象經(jīng)過定點(diǎn)(2,1),則f(x)的值域?yàn)?)A9,81 B3,9C1,9 D1,)解析:由f(x)過定點(diǎn)(2,1)可知b2,因?yàn)閒(x)3x2在2,4上是增函數(shù),所以f(x)minf(2)1,f(x)maxf(4)9.故f(x)的值域?yàn)?,9答案:C5(2018貴州適應(yīng)性考試)函數(shù)yax21(a>0且a1)的圖象恒過的點(diǎn)是()A(0,0) B(0,1)C(2,0) D(2,1)解析:法一:因?yàn)楹瘮?shù)yax(a>0,a1)的圖象恒過點(diǎn)(0,1),將該圖象向左平移2個(gè)單位,再向下平移1個(gè)單位得到y(tǒng)ax21(a>0,a1)的圖象,所以yax21(a>0,a1)的圖象恒過點(diǎn)(2,0),選項(xiàng)C正確法二:令x20,x2,得f(2)a010,所以yax21(a>0,a1)的圖象恒過點(diǎn)(2,0),選項(xiàng)C正確答案:C6已知函數(shù)f(x)則函數(shù)f(x)是()A偶函數(shù),在0,)單調(diào)遞增B偶函數(shù),在0,)單調(diào)遞減C奇函數(shù),且單調(diào)遞增D奇函數(shù),且單調(diào)遞減解析:易知f(0)0,當(dāng)x>0時(shí),f(x)12x,f(x)2x1,而x<0,則f(x)2x1f(x);當(dāng)x<0時(shí),f(x)2x1,f(x)12x,而x>0,則f(x)12(x)12xf(x)即函數(shù)f(x)是奇函數(shù),且單調(diào)遞增,故選C.答案:C7(2018安徽省高三階段檢測)函數(shù)y4cosxe|x|(e為自然對數(shù)的底數(shù))的圖象可能是()解析:因?yàn)楹瘮?shù)y4cosxe|x|,所以f(x)4cos(x)e|x|f(x),所以函數(shù)f(x)是偶函數(shù),其圖象關(guān)于y軸對稱,排除選項(xiàng)B,D.又f(0)4cos0e03,所以選項(xiàng)A滿足條件故選A.答案:A8(2018湖北四市聯(lián)考)已知函數(shù)f(x)2x2,則函數(shù)y|f(x)|的圖象可能是()解析:y|f(x)|2x2|易知函數(shù)y|f(x)|的圖象的分段點(diǎn)是x1,且過點(diǎn)(1,0),(0,1),|f(x)|0.又|f(x)|在(,1)上單調(diào)遞減答案:B9關(guān)于x的方程2xa2a在(,1上有解,則實(shí)數(shù)a的取值范圍是()A2,1)(0,1 B2,1(0,1C2,1)(0,2 D2,1(0,2解析:方程2xa2a在(,1上有解,又y2x(0,2,0<a2a2,即解得2a<1或0<a1.答案:A10(2018河南三門峽一模,6)設(shè)函數(shù)f(x)若f(a1)f(2a1),則實(shí)數(shù)a的取值范圍是()A(,1 B(,2C2,6 D2,)解析:易知f(x)是定義域R上的增函數(shù)f(a1)f(2a1),a12a1,解得a2.故實(shí)數(shù)a的取值范圍是(,2故選B.答案:B二、填空題11化簡:022(0.01)0.5_.解析:原式111.答案:12不等式2>x4的解集為_解析:不等式2>x4可化為x4,等價(jià)于x22x<x4,即x23x4<0,解得1<x<4.答案:x|1<x<413函數(shù)yxx1在區(qū)間3,2上的值域是_解析:因?yàn)閤3,2,若令tx,則t,故yt2t12.當(dāng)t時(shí),ymin;當(dāng)t8時(shí),ymax57.故所求函數(shù)的值域?yàn)?答案:14已知a>0,且a1,若函數(shù)y|ax2|與y3a的圖象有兩個(gè)交點(diǎn),則實(shí)數(shù)a的取值范圍是_解析:當(dāng)0<a<1時(shí),作出函數(shù)y|ax2|的圖象,如圖a.若直線y3a與函數(shù)y|ax2|(0<a<1)的圖象有兩個(gè)交點(diǎn),則由圖象可知0<3a<2,所以0<a<.當(dāng)a>1時(shí),作出函數(shù)y|ax2|的圖象,如圖b,若直線y3a與函數(shù)y|ax2|(a>1)的圖象有兩個(gè)交點(diǎn),則由圖象可知0<3a<2,此時(shí)無解所以a的取值范圍是.答案:能力挑戰(zhàn)15(2018北京模擬)已知函數(shù)f(x)ax,其中a>0,且a1,如果以P(x1,f(x1),Q(x2,f(x2)為端點(diǎn)的線段的中點(diǎn)在y軸上,那么f(x1)f(x2)等于()A1 BaC2 Da2解析:以P(x1,f(x1),Q(x2,f(x2)為端點(diǎn)的線段的中點(diǎn)在y軸上,x1x20,又f(x)ax,f(x1)f(x2)ax1ax2ax1x2a01,故選A.答案:A16已知函數(shù)f(x)a|x1|(a>0,a1)的值域?yàn)?,),則f(4)與f(1)的大小關(guān)系是_解析:因?yàn)閨x1|0,函數(shù)f(x)a|x1|(a>0,a1)的值域?yàn)?,),所以a>1.由于函數(shù)f(x)a|x1|在(1,)上是增函數(shù),且它的圖象關(guān)于直線x1對稱,則函數(shù)在(,1)上是減函數(shù),故f(1)f(3),f(4)>f(1)答案:f(4)>f(1)17記x2x1為區(qū)間x1,x2的長度,已知函數(shù)y2|x|,x2,a(a0),其值域?yàn)閙,n,則區(qū)間m,n的長度的最小值是_解析:令f(x)y2|x|,則f(x)(1)當(dāng)a0時(shí),f(x)2x在2,0上為減函數(shù),值域?yàn)?,4(2)當(dāng)a>0時(shí),f(x)在2,0)上遞減,在0,a上遞增,當(dāng)0<a2時(shí),f(x)maxf(2)4,值域?yàn)?,4;當(dāng)a>2時(shí),f(x)maxf(a)2a>4,值域?yàn)?,2a綜合(1)(2),可知m,n的長度的最小值為3.答案:3