2024-2025學(xué)年高中數(shù)學(xué)第二章基本初等函數(shù)Ⅰ2.1.2指數(shù)函數(shù)及其性質(zhì)第1課時指數(shù)函數(shù)的圖象和性質(zhì)練習(xí)含解析新人教A版必修1

PAGEPAGE4第1課時指數(shù)函數(shù)的圖象和性質(zhì)課時過關(guān)·實(shí)力提升基礎(chǔ)鞏固1.若函數(shù)y=(a2-3a+3)·ax是指數(shù)函數(shù),則有()A.a=1或a=2 B.a=1C.a=2 D.a>0,且a≠1解析:由指數(shù)函數(shù)的定義,得a2-答案:C2.函數(shù)f(x)=4A.(0,+∞) B.(-1,+∞)C.(-∞,0) D.(-∞,-1)解析:f(x)的定義域是R,y=45x的值域是(0,+∞),則f(x)的值域是(-1,答案:B3.若函數(shù)y=f(x)的圖象與y=2x的圖象關(guān)于y軸對稱,則f(3)等于()A.解析:由已知得f(x)=12x答案:A4.函數(shù)y=7A.R B.(-∞,2] C.[2,+∞) D.(0,+∞)解析:由2-x≥0,得x≤2.答案:B5.若函數(shù)f(x)=ax-b的圖象如圖所示,其中a,b為常數(shù),則下列結(jié)論正確的是()A.a>1,b<0 B.a>1,b>0C.0<a<1,b>0 D.0<a<1,b<0解析:由圖象知,f(x)=ax-b是由y=ax向左平移得到的,所以b<0.由題中圖象遞減,得0<a<1.答案:D6.已知函數(shù)f(x)=ax+b(a>0,且a≠1),經(jīng)過點(diǎn)(-1,5),(0,4),則f(-2)的值為.

解析:由已知得所以f(x)=12x+3,所以答案:77.函數(shù)f(x)=a3-x-1(a>0,且a≠1)的圖象恒過定點(diǎn)的坐標(biāo)是.

解析:令3-x=0,解得x=3,則f(3)=a0-1=0,即函數(shù)f(x)的圖象恒過定點(diǎn)(3,0).答案:(3,0)8.函數(shù)y=4x+2x-3的值域?yàn)?

解析:函數(shù)的定義域是R.設(shè)2x=t,則y=t2+t-3,t>0.∵y=t2+t-3=t+122-134在(0,+∞)上為增函數(shù),∴答案:(-3,+∞)9.求下列函數(shù)的值域:(1)y=解:(1)∵-1x≠0,∴y=2-1x≠1.∴所求函數(shù)的值域是(0,1)∪(1,+∞).(2)∵1-x≥0,∴y=51∴所求函數(shù)的值域是[1,+∞).10.已知函數(shù)f(x)=ax-1(x≥0)的圖象經(jīng)過點(diǎn)2,12(1)求a的值;(2)求函數(shù)y=f(x)(x≥0)的值域.解:(1)因?yàn)楹瘮?shù)f(x)的圖象經(jīng)過點(diǎn)所以a2-1=12(2)由(1)知函數(shù)為f(x)=12x-1(x≥0),由x≥0,得x-1≥實(shí)力提升1.已知函數(shù)y=1A.1 B.2 C.4 D.8解析:由兩個函數(shù)的圖象關(guān)于y軸對稱,可知12a-4與答案:C2.函數(shù)y=ax-a(a>0,且a≠1)的圖象可能是()解析:∵當(dāng)x=1時,y=a1-a=0,即圖象過點(diǎn)(1,0),∴只有C中圖象符合.答案:C3.當(dāng)x>0時,函數(shù)f(x)=(a2-1)x的值總大于1,則實(shí)數(shù)a的取值范圍是()A.1<|a|<BC.|a|>1 D.|a|>解析:依題意得a2-1>1,a2>2,∴|a|>答案:D4.函數(shù)y=8-23-x(x≥0)的值域是()A.[0,8) B.(0,8) C.[0,8] D.(0,8]答案:A5.★已知函數(shù)f(x)=3解析:f(x)+f(1-x)=所以原式=f1101+答案:506.★已知函數(shù)f(x)=a-22x+1(x∈(1)試證明對隨意實(shí)數(shù)a,f(x)為增函數(shù);(2)試確定a的值,使f(x)為奇函數(shù).(1)證明設(shè)x1,x2是隨意兩個實(shí)數(shù),且x1<x2,則f(x1)-f(x2)=∵x1<x2,∴又2x>0,∴∴f(x1)-f(x2)<0,即f(x1)<f(x2)