高中數(shù)學(xué)第二章基本初等函數(shù)（Ⅰ）2.1指數(shù)函數(shù)2.1.2指數(shù)函數(shù)及其性質(zhì)第1課時(shí)指數(shù)函數(shù)的圖象和性質(zhì)練習(xí).docx

2.1.2指數(shù)函數(shù)及其性質(zhì)第1課時(shí)指數(shù)函數(shù)的圖象和性質(zhì)課時(shí)過(guò)關(guān)能力提升基礎(chǔ)鞏固1.函數(shù)y=2-x的圖象是()解析:y=2-x=12x,y=2-x的圖象即為y=12x的圖象.故選B.答案:B2.函數(shù)f(x)=45x-1的值域是()A.(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(duì)稱(chēng),則f(3)等于()A.18B.14C.8D.4解析:由已知得f(x)=12x,故f(3)=123=18.答案:A4.函數(shù)y=742-x的定義域是()A.RB.(-,2C.2,+)D.(0,+)解析:由2-x0,得x2.答案:B5.已知函數(shù)f(x)=2x,x0,x+1,x0.若f(a)+f(1)=0,則實(shí)數(shù)a的值等于()A.-3B.-1C.1D.3解析:依題意,f(a)=-f(1)=-21=-2,2x0,a0.f(a)=a+1=-2,故a=-3.答案:A6.若y=(a-3)(a-2)x是指數(shù)函數(shù),則a=.解析:由題意,得a-3=1,a-20,a-21,解得a=4.答案:47.函數(shù)f(x)=a3-x-1(a0,且a1)的圖象恒過(guò)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(guò)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,t0.y=t2+t-3=t+122-134在(0,+)上為增函數(shù),y14-134=-3,函數(shù)的值域?yàn)?-3,+).答案:(-3,+)9.求下列函數(shù)的值域:(1)y=2-1x;(2)y=51-x.解:(1)-1x0,y=2-1x1.y0,且y1,所求函數(shù)的值域是(0,1)(1,+).(2)1-x0,y=51-x50=1.所求函數(shù)的值域是1,+).10.已知函數(shù)f(x)=a-22x+1(xR),a為實(shí)數(shù).(1)試證明對(duì)任意實(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是任意兩個(gè)實(shí)數(shù),且x1x2,則f(x1)-f(x2)=a-22x1+1-a-22x2+1=2(2x1-2x2)(2x1+1)(2x2+1).x1x2,2x1-2x20,2x1+10,2x2+10.f(x1)-f(x2)0,即f(x1)0,且a1解析:由指數(shù)函數(shù)的定義,得a2-3a+3=1,a0,a1,解得a=2.答案:C2.函數(shù)y=ax-a(a0,且a1)的圖象可能是()解析:當(dāng)x=1時(shí),y=a1-a=0,即圖象過(guò)點(diǎn)(1,0),只有C中圖象符合.答案:C3.當(dāng)x0時(shí),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|2B.|a|1D.|a|2解析:依題意得a2-11,a22,|a|2.答案:D4.函數(shù)y=8-23-x(x0)的值域是()A.0,8)B.(0,8)C.0,8D.(0,8答案:A5.已知函數(shù)f(x)=32x3+32x,則f1101+f2101+f100101=.解析:f(x)+f(1-x)=9x3+9x+91-x3+91-x=9x3+9x+39x+3=1,所以原式=f1101+f100101+f2101+f99101+f50101+f51101=1+1+1=50.答案:506.已知集合A=x|y=x+4-2x,xR,B=y|y=4x+12-2x-3,xA.(1)求集合A;(2)求集合B.解:(1)要使函數(shù)y=x+4-2x有意義,自變量x的取值需滿足x0,4-2x0,解得0x2,則A=x|0x2.(2)設(shè)2x=t,當(dāng)xA,即x0,2時(shí),