2025高考數(shù)學(xué)一輪復(fù)習(xí)-2.6-指數(shù)函數(shù)-專項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-2.6-指數(shù)函數(shù)-專項(xiàng)訓(xùn)練【A級基礎(chǔ)鞏固】1.下列函數(shù)中,值域是(0,+∞)的為()A.y=3x-1 B.y=(C.y=1-(12.設(shè)函數(shù)f(x)=ex-e-x,則f(x)()A.是奇函數(shù),且在(-∞,+∞)單調(diào)遞增B.是奇函數(shù),且在(-∞,+∞)單調(diào)遞減C.是偶函數(shù),且在(-∞,+∞)單調(diào)遞增D.是偶函數(shù),且在(-∞,+∞)單調(diào)遞減3.函數(shù)f(x)=ax-2+1(其中a>0,a≠1)的圖象恒過的定點(diǎn)是()A.(2,1) B.(2,2) C.(1,1) D.(1,2)4.已知a=0.40.3,b=0.30.3,c=0.30.4,則()A.a>c>b B.a>b>cC.c>a>b D.b>c>a5.(多選題)已知函數(shù)f(x)=|2x-1|,實(shí)數(shù)a,b滿足f(a)=f(b)(a<b),則()A.2a+2b>2B.?a,b∈R,使得0<a+b<1C.2a+2b=2D.a+b<06.設(shè)f(x)是定義在R上的偶函數(shù),且當(dāng)x≤0時,f(x)=2-x.若對任意的x∈[m,m+1],不等式f(x)≥f2(x-m)恒成立,則正數(shù)m的取值范圍為()A.[1,+∞) B.(1,+∞)C.(0,1) D.(0,1]7.函數(shù)y=(12)-x8.已知函數(shù)f(x)=4x-2x+2-1,x∈[0,3],則其值域?yàn)?

9.已知定義域?yàn)镽的函數(shù)f(x)=ax-(k-1)a-x(a>0,且a≠1)是奇函數(shù).(1)求實(shí)數(shù)k的值;(2)若f(1)<0,判斷函數(shù)f(x)的單調(diào)性,若f(m2-2)+f(m)>0,求實(shí)數(shù)m的取值范圍.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級能力提升】10.(多選題)函數(shù)f(x)=11+e-A.函數(shù)f(x)的圖象關(guān)于原點(diǎn)對稱B.函數(shù)f(x)的值域?yàn)?0,1)C.不等式f(x)>12D.f(x)是增函數(shù)11.函數(shù)f(x)=2x2-4x12.函數(shù)f(x)=a2x+ax+1(a>0,且a≠1)在[-1,1]上的最大值為13,求實(shí)數(shù)a的值.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級應(yīng)用創(chuàng)新練】13.(多選題)已知a+3a=b+5b=3,則下列不等關(guān)系正確的是()A.0<a<b<1 B.0<b<a<1C.b+3a<a+5b D.blna>alnb參考答案【A級基礎(chǔ)鞏固】1.解析:函數(shù)y=3x函數(shù)y=(13)x函數(shù)y=1-函數(shù)y=312.解析:因?yàn)閒(-x)=e-x-e-(-x)=-(ex-e-x)=-f(x),且x∈R,所以f(x)為奇函數(shù),因?yàn)閥=e-x在(-∞,+∞)上單調(diào)遞減,則y=-e-x單調(diào)遞增,又y=ex在(-∞,+∞)上單調(diào)遞增,所以f(x)=ex-e-x在(-∞,+∞)上單調(diào)遞增.故選A.3.解析:令x-2=0,得x=2,代入f(x)得f(2)=2,因此函數(shù)f(x)=ax-2+1(其中a>0,a≠1)的圖象恒過的定點(diǎn)是(2,2).故選B.4.解析:因?yàn)閥=0.3x是減函數(shù),所以0.30.3>0.30.4,所以b>c>0.而ab=(0.40.3)即a>b,所以a>b>c.故選B.5.解析:畫出函數(shù)f(x)=|2x-1|的圖象,如圖所示.由圖知1-2a=2b-1,則2a+2b=2,故A錯誤,C正確;由基本不等式可得2=2a+2b>22a·2b=26.解析:因?yàn)楹瘮?shù)f(x)是定義在R上的偶函數(shù),且當(dāng)x≤0時,f(x)=2-x,則當(dāng)x>0時,-x<0,f(x)=f(-x)=2x,所以f(x)在(0,+∞)上單調(diào)遞增,故f(x)在[m,m+1]上單調(diào)遞增,f(x)=2|x|.又f(x)≥f2(x-m)且m>0,x∈[m,m+1],所以x-m≥0,所以f(x-m)=2x-m,故原不等式等價于2x≥(2x-m)2,化簡得22m≥2x,即2m≥x,所以m+1≤2m,可得m≥1.故選A.7.解析:令t=-x2+2x,則y=(12)t,t∈[-∞,1],因?yàn)閠=-x2+2x在(-∞,1)上單調(diào)遞增,在[1,+∞)上單調(diào)遞減,y=(12)t在定義域內(nèi)為減函數(shù),所以y=(12答案:[1,+∞)8.解析:令t=2x,因?yàn)閤∈[0,3],所以1≤t≤8,所以g(t)=t2-4t-1=(t-2)2-5,t∈[1,8].又y=g(t)的圖象關(guān)于直線t=2對稱,開口向上,所以g(t)在[1,2)上單調(diào)遞減,在[2,8]上單調(diào)遞增,且|8-2|>|2-1|,所以t=2時,函數(shù)取得最小值,即g(t)min=-5,t=8時,函數(shù)取得最大值,即g(t)max=31,所以f(x)∈[-5,31].答案:[-5,31]9.解:(1)因?yàn)閒(x)是定義域?yàn)镽的奇函數(shù),所以f(0)=a0-(k-1)a0=1-(k-1)=0,所以k=2,經(jīng)檢驗(yàn)k=2符合題意,所以k=2.(2)f(x)=ax-a-x(a>0,且a≠1),因?yàn)閒(1)<0,所以a-1a從而y=ax在R上單調(diào)遞減,y=a-x在R上單調(diào)遞增,故由單調(diào)性的性質(zhì)可判斷f(x)=ax-a-x在R上單調(diào)遞減,不等式f(m2-2)+f(m)>0,可化為f(m2-2)>f(-m),所以m2-2<-m,即m2+m-2<0,解得-2<m<1,所以實(shí)數(shù)m的取值范圍是(-2,1).INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級能力提升】10.解析:函數(shù)f(x)的定義域?yàn)镽,且f(0)=12≠0,所以函數(shù)f(x)的圖象不關(guān)于原點(diǎn)對稱,A錯誤;因?yàn)閑-x+1>1,所以f(x)=11+e-x∈(0,1),B正確;由f(x)=11+e11.解析:函數(shù)f(x)的定義域?yàn)?-∞,0]∪[2,+∞).由復(fù)合函數(shù)的單調(diào)性可知,f(x)在(-∞,0]上單調(diào)遞減,在[2,+∞)上單調(diào)遞增.而f(0)=4,f(2)=1.所以函數(shù)f(x)的最小值為1.答案:112.解:由f(x)=a2x+ax+1,令ax=t,則t>0,則y=t2+t+1=(t+12)2+3其對稱軸為t=-12該二次函數(shù)在[-12①若a>1,由x∈[-1,1],得t=ax∈[1a故當(dāng)t=a,即x=1時,ymax=a2+a+1=13,解得a=3或a=-4(舍去).②若0<a<1,由x∈[-1,1],可得t=ax∈[a,1a],故當(dāng)t=1ymax=(1a)2+1解得a=13或a=-1綜上可得,a=3或13INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級應(yīng)用創(chuàng)新練】13.解析:由a+3a=b+5b=3可知,若a≤0,b≤0,則3a≤1,5b≤1,則a+3a=b+5b=3不成立,又a≥1,b≥1時,3a≥3,5b≥5,故0<a<1,0<b<1,又3a=3-a,5b=3-b,則a,b可看作y=3x,y=5x的圖象與直線y=3-x交點(diǎn)的橫坐標(biāo),作出y=3x,y=5x與y=3-x的圖象如圖,結(jié)合圖象可知0<b<a<1,故A錯誤,B正確;由0<b<a<1,a+3a=b+5b=