2022年高考數(shù)學(xué)一輪復(fù)習(xí)考點(diǎn)規(guī)范練8指數(shù)與指數(shù)函數(shù)含解析新人教A版2

1、PAGE PAGE 6考點(diǎn)規(guī)范練8指數(shù)與指數(shù)函數(shù)基礎(chǔ)鞏固1.化簡(jiǎn)664x6y4(x0,y0)得()A.2xy23B.2xy32C.-2xy32D.-2xy23答案:D2.已知f(x)=3x-b(2x4,b為常數(shù))的圖象經(jīng)過點(diǎn)(2,1),則f(x)的值域?yàn)?)A.9,81B.3,9C.1,9D.1,+)答案:C解析:由f(x)過點(diǎn)(2,1)可知b=2,因?yàn)閒(x)=3x-2在區(qū)間2,4上是增函數(shù),所以f(x)min=f(2)=32-2=1,f(x)max=f(4)=34-2=9.故選C.3.已知f(x)是定義在R上的奇函數(shù),當(dāng)x0時(shí),f(x)=3x+m(m為常數(shù)),則f(-log35)的值為()

2、A.4B.-4C.6D.-6答案:B解析:由題意知,f(0)=30+m=0,解得m=-1,故x0時(shí),f(x)=3x-1.所以f(-log35)=-f(log35)=-(3log35-1)=-4,故選B.4.設(shè)a=30.7,b=13-0.8,c=log0.70.8,則a,b,c的大小關(guān)系為()A.abcB.bacC.bcaD.ca30.7=a30=1,c=log0.70.8log0.70.7=1,cab.故選D.5.函數(shù)y=xax|x|(0a0,-ax,x0時(shí),函數(shù)y是一個(gè)指數(shù)函數(shù),其底數(shù)0a1,所以函數(shù)y在區(qū)間(0,+)內(nèi)單調(diào)遞減;當(dāng)x0時(shí),函數(shù)y的圖象與指數(shù)函數(shù)y=ax(x0,且1bxax,

3、則()A.0ba1B.0ab1C.1baD.1a0,1bx1,a1.bx1,ab1,即ab,故選C.7.若函數(shù)f(x)=a|2x-4|(a0,a1)滿足f(1)=19,則f(x)的單調(diào)遞減區(qū)間是()A.(-,2B.2,+)C.-2,+)D.(-,-2答案:B解析:由f(1)=19得a2=19,故a=13a=-13舍去,即f(x)=13|2x-4|.由于y=|2x-4|在區(qū)間(-,2上單調(diào)遞減,在區(qū)間2,+)內(nèi)單調(diào)遞增,故f(x)在區(qū)間(-,2上單調(diào)遞增,在區(qū)間2,+)內(nèi)單調(diào)遞減.故選B.8.若2x-2y0B.ln(y-x+1)0D.ln|x-y|0答案:A解析:2x-2y3-x-3-y,2x-

4、3-x2y-3-y.f(t)=2t-3-t在R上為增函數(shù),且f(x)f(y),x0,y-x+11,ln(y-x+1)ln1=0.故選A.9.已知偶函數(shù)f(x)滿足f(x)=2x-4(x0),則x|f(x-2)0=()A.x|x4B.x|x4C.x|x6D.x|x2答案:B解析:因?yàn)閒(x)為偶函數(shù),所以當(dāng)x0時(shí),f(x)=f(-x)=2-x-4.所以f(x)=2x-4,x0,2-x-4,x0時(shí),有x-20,2x-2-40或x-20,解得x4或x0,a1)過定點(diǎn).答案:(1,1)解析:由|x-1|=0,即x=1,此時(shí)y=1,故函數(shù)恒過定點(diǎn)(1,1).11.已知函數(shù)y=ax-b(a0,且a1)的圖

5、象經(jīng)過第二、三、四象限,則ab的取值范圍為.答案:(0,1)解析:因?yàn)閥=ax-b的圖象經(jīng)過第二、三、四象限,所以函數(shù)y=ax-b單調(diào)遞減且其圖象與y軸的交點(diǎn)在y軸的負(fù)半軸上.令x=0,得y=a0-b=1-b,則需0a1,1-b0,即0a1.故ab(0,1).12.函數(shù)y=14x-12x+1在區(qū)間-3,2上的值域是.答案:34,57解析:令t=12x,由x-3,2,得t14,8.則y=t2-t+1=t-122+34t14,8.當(dāng)t=12時(shí),ymin=34;當(dāng)t=8時(shí),ymax=57.故所求函數(shù)的值域?yàn)?4,57.13.已知函數(shù)f(x)=ax+b(a0,a1)的定義域和值域都是-1,0,則a+b

6、=.答案:-32解析:當(dāng)a1時(shí),f(x)在區(qū)間-1,0上單調(diào)遞增,則a-1+b=-1,a0+b=0,無解.當(dāng)0a1時(shí),f(x)在區(qū)間-1,0上單調(diào)遞減,則a-1+b=0,a0+b=-1,解得a=12,b=-2,a+b=-32.能力提升14.當(dāng)x(-,-1時(shí),不等式(m2-m)4x-2x0恒成立,則實(shí)數(shù)m的取值范圍是()A.(-2,1)B.(-4,3)C.(-1,2)D.(-3,4)答案:C解析:原不等式可變形為m2-m12x.函數(shù)y=12x在區(qū)間(-,-1上是減函數(shù),12x12-1=2.當(dāng)x(-,-1時(shí),m2-m12x恒成立等價(jià)于m2-m2,解得-1m0,且a1,若函數(shù)y=|ax-2|與y=3

7、a的圖象有兩個(gè)交點(diǎn),則實(shí)數(shù)a的取值范圍是.答案:0,23解析:當(dāng)0a1時(shí),作出函數(shù)y=|ax-2|的圖象,如圖1.圖1若直線y=3a與函數(shù)y=|ax-2|(0a1)的圖象有兩個(gè)交點(diǎn),則由圖象可知03a2,所以0a1時(shí),作出函數(shù)y=|ax-2|的圖象,如圖2.圖2若直線y=3a與函數(shù)y=|ax-2|(a1)的圖象有兩個(gè)交點(diǎn),則由圖象可知03a0,且a1)有兩個(gè)零點(diǎn),則實(shí)數(shù)a的取值范圍是.答案:(1,+)解析:令ax-x-a=0,即ax=x+a.若0a1,則y=ax與y=x+a的圖象有如圖所示的兩個(gè)公共點(diǎn).故a的取值范圍是(1,+).17.記x2-x1為區(qū)間x1,x2的長(zhǎng)度,已知函數(shù)y=2|x|,x-2,a(a0),其值域?yàn)閙,n,則區(qū)間m,n的長(zhǎng)度的最小值是.答案:3解析:令f(x)=y=2|x|,則f(x)=2x,0 xa,2-x,-2x0時(shí),f(x)在區(qū)間-2,0)上為減函數(shù),在區(qū)間0,a上為增函數(shù),當(dāng)02時(shí),f(x)max=f(a)=2a4,值域?yàn)?,2a.綜上(1)(2),可知m,n的長(zhǎng)度的最小值