高考第一輪文科數(shù)學(xué)（人教A版）課時(shí)規(guī)范練9　指數(shù)與指數(shù)函數(shù)

課時(shí)規(guī)范練9指數(shù)與指數(shù)函數(shù)基礎(chǔ)鞏固組1.若函數(shù)f(x)=12a-1·ax是指數(shù)函數(shù),則fA.-2 B.2 C.-22 D.222.函數(shù)f(x)=x|x|·3.設(shè)a=1234,b=1534A.a<b<c B.c<a<bC.b<c<a D.b<a<c4.函數(shù)y=2x2-2x+2,xA.R B.[4,32]C.[2,32] D.[2,+∞)5.計(jì)算:12-1+(3-22)0-96.函數(shù)y=a2x-2+3(a>0,且a≠1)的圖象恒過定點(diǎn).

7.(2022廣東茂名三模)函數(shù)f(x)=9x+31-2x的最小值是.

8.若a>1,則不等式a2x+1<ax2+29.函數(shù)y=ax(a>0,且a≠1)在區(qū)間[1,2]上的最大值比最小值大a2,則實(shí)數(shù)a的值是綜合提升組10.(2022河南鄭州二模)f(x)=ex-x在區(qū)間[-1,1]上的最小值是()A.1+1e C.e+1 D.e-111.(2023陜西西安聯(lián)考)若曲線f(x)=mxex-n在點(diǎn)(1,f(1))處的切線為y=ex,則mn=.

12.已知函數(shù)f(x)=2x-4x.(1)解不等式f(x)>16-9×2x;(2)若關(guān)于x的方程f(x)=m在[-1,1]上有解,求m的取值范圍.創(chuàng)新應(yīng)用組13.(2022陜西商洛一模)若對(duì)任意的x∈(1,+∞),恒有eax+1eax≥elnx+1elnxA.(-∞,e]B.-∞,-1e∪1e,+∞C.-∞,1eD.1e,+∞14.設(shè)f(x)=|2x-1-1|,a<c且f(a)>f(c),則2a+2c4.(填“>”“<”或“=”)

參考答案課時(shí)規(guī)范練9指數(shù)與指數(shù)函數(shù)1.B因?yàn)楹瘮?shù)f(x)=12a-1·ax是指數(shù)函數(shù),所以12a-1=1,即a=4,所以f(x)=4x,所以2.D函數(shù)f(x)=x|x|·3x的定義域?yàn)閧x|x≠0},且f(x)=x|x|·3x=133.D由于y=12x在R上為減函數(shù),所以1234<1212?a<c,由于y=x34在4.C函數(shù)y=2x2-2x+2,是由y=2t和t=x2-2x+2,x∈[-1,2]復(fù)合而成,因?yàn)閠=x2-2x+2=(x-1)2+1的對(duì)稱軸為x=1,開口向上,所以t=x2-2x+2在[-1,1)上單調(diào)遞減,在[1,2]上單調(diào)遞增,所以當(dāng)x=-1時(shí),tmax=(-1)2-2×(-1)+2=5,當(dāng)x=1時(shí),tmin=1-2×1+2=1,所以1≤t≤5.因?yàn)閥=2t在R上單調(diào)遞增,所以2=21≤y=2t≤25=32,所以函數(shù)y=2x2-5.43+π原式=2+1+1-2320.5+|2-π|=2+1+1-236.(1,4)根據(jù)題意,在函數(shù)y=a2x-2+3中,令2x-2=0,解得x=1,此時(shí)f(1)=a2-2+3=4,即函數(shù)的圖象恒過定點(diǎn)(1,4).7.23f(x)=9x+31-2x=9x+39x≥29x·39x=23,當(dāng)且僅當(dāng)9x=39x,即x=14時(shí)取等號(hào)8.(-∞,-2)∪(2,+∞)∵a>1,則由不等式a2x+1<ax2+2x-3可得2x+1<x2+2x-3,即(x+2)(x-2)>0,解得9.12或32若0<a<1,則函數(shù)y=ax根據(jù)題意有a-a2=a2,解得a=12或0(舍去),所以a=若a>1,則函數(shù)y=ax在區(qū)間[1,2]上單調(diào)遞增,根據(jù)題意有a2-a=a2,解得a=32或0(舍去),所以a=綜上所述,a=1210.B∵f(x)=ex-x,∴f'(x)=ex-1,令f'(x)=0,解得x=0,∴當(dāng)x<0時(shí),f'(x)<0,函數(shù)f(x)=ex-x單調(diào)遞減,當(dāng)x>0時(shí),f'(x)>0,函數(shù)f(x)=ex-x單調(diào)遞增,∴函數(shù)f(x)=ex-x在[-1,1]上的最小值為f(0)=e0-0=1,故選B.11.-e4將x=1代入y=ex,得切點(diǎn)為(1,e),將切點(diǎn)坐標(biāo)代入f(x)=mxex-n,得e=me-n,又f'(x)=mex(x+1),f'(1)=2me=e,則m=12,所以e=12e-n,n=-e2,12.解(1)∵f(x)>16-9×2x,∴(2x)2-10×2x+16<0,∴(2x-2)(2x-8)<0,∴2<2x<8,∴1<x<3.∴不等式f(x)>16-9×2x的解集為{x|1<x<3}.(2)令t=2x,∵x∈[-1,1],∴t∈12,2,∴關(guān)于x的方程f(x)=m在[-1,1]上有解轉(zhuǎn)化為t-t2=m在t∈又y=t-t2=-t-122+14在t∈12,2上為減函數(shù),∴ymax=14,故m的取值范圍是-213.B令g(x)=ex+e-x,則g(x)為偶函數(shù),當(dāng)x∈(0,+∞)時(shí),g(x)=ex+1ex在(0,+∞)上單調(diào)遞增,由eax+1eax≥elnx+1elnx,得g(ax)≥g(lnx),即g(|ax|)≥g(|lnx|),∵x>1,∴|a|x≥lnx恒成立,即|a|≥lnxx對(duì)任意的x∈(1,+∞)恒成立,設(shè)h(x)=lnxx,則h'(x)=1-lnxx2,在(1,e)上,h'(x)>0,h(x)單調(diào)遞增,在(e,+∞)上,h'(x)<0,h(x)單調(diào)遞減,所以當(dāng)x=e時(shí),h(x)有最大值h(e)14.<f(x)在(-∞,1]上是減函數(shù),在(1,