高一數(shù)學(xué)下學(xué)期課后試題：反函數(shù)的概念

第7頁(yè)高一數(shù)學(xué)下學(xué)期課后試題：反函數(shù)的概念【】記得有一句話是這么說的：數(shù)學(xué)是一門描寫數(shù)字之間關(guān)系的科學(xué),是我們前進(jìn)的階梯。對(duì)于高中學(xué)生的我們,數(shù)學(xué)在生活中,考試科目里更是尤為重要,所以小編在此為您發(fā)布了文章：高一數(shù)學(xué)下學(xué)期課后試題：反函數(shù)的概念希望此文能給您帶來幫助。本文題目：高一數(shù)學(xué)下學(xué)期課后試題：反函數(shù)的概念1.函數(shù)y=的反函數(shù)是()A.y=(xR且x-4)B.y=(xR且x3)C.y=(xR且x)D.y=(xR且x-)答案：C解析：由y=,得x=.故所求反函數(shù)為y=(xR且x3).2.函數(shù)y=的反函數(shù)是()A.y=B.y=C.y=D.y=答案：A解析：當(dāng)x0時(shí),由y=x2,得x=-.故反函數(shù)為y=f-1(x)=-(x0).當(dāng)x0時(shí),由y=-x,得x=-2y.故反函數(shù)為y=f-1(x)=-2x(x0).y=f-1(x)=-x,x0,-2x,x0.3.假設(shè)函數(shù)f(x)的反函數(shù)f-1(x)=1+x2(x0),那么f(2)等于()A.1B.-1C.1和-1D.5答案：B解法一:由y=1+x2(x0),得x=-.故f(x)=-(x0),f(2)=-=-1.解法二:令1+x2=2(x0),那么x=-1,即f(2)=-1.4.假設(shè)函數(shù)y=f(x)的反函數(shù)是y=-(-10),那么原函數(shù)的定義域是()A.(-1,0)B.[-1,1]C.[-1,0]D.[0,1]答案：C解析：∵原函數(shù)的定義域?yàn)榉春瘮?shù)的值域,又-10,01,即y[-1,0].5.設(shè)y=+m和y=nx-9互為反函數(shù),那么m、n的值分別是()A.-6,3B.2,1C.2,3D.3,3答案：D解析：求出y=+m的反函數(shù)y=3x-3m,再與y=nx-9比照系數(shù)即得.6.f(x)=x2-1(x2),那么f-1(4)=______________.答案：解析：因?yàn)閒(x)=x2-1,x2,所以其反函數(shù)為f-1(x)=(x3).所以f-1(4)=.7.求以下函數(shù)的反函數(shù):(1)y=-(-1(2)y=-x2-2x+1(1(3)y=解：(1)由y=-,得y2=1-x2,即x2=1-y2.∵-10,x=-.又∵y=-,-10,-1所求反函數(shù)為y=-(-1(2)由y=-x2-2x+1=-(x+1)2+2,得(x+1)2=2-y.∵12,23.x+1=,即x=-1+.反函數(shù)為y=-1+(-7-2).(3)①由y=x2(x0),得x=-,即y=x2(x0)的反函數(shù)為y=-(x0).②由y=-x-1(x0),得x=-y-1,即y=-x-1(x0)的反函數(shù)為y=-x-1(x-1).由①②可知f(x)=的反函數(shù)為f-1(x)=能力提升踮起腳,抓得住!8.函數(shù)y=2|x|在下面的區(qū)間上,不存在反函數(shù)的是()A.[0,+])B.(-,0)]C.[-4,4]D.[2,4]答案：C解法一:函數(shù)假設(shè)在區(qū)間上單調(diào),那么存在反函數(shù),易知函數(shù)y=2|x|在[0,+),(-,0],[2,4]上單調(diào).解法二:當(dāng)x=4時(shí),y=8,知不是一一映射.9.函數(shù)f(x)是增函數(shù),它的反函數(shù)是f-1(x),假設(shè)a=f(2)+f-1(2),b=f(3)+f-1(3),那么下面結(jié)論中正確的選項(xiàng)是()A.abD.無(wú)法確定答案：A解析：∵f(x)是增函數(shù),故其反函數(shù)f-1(x)也是增函數(shù),f(3)f(2),f-1(3)f-1(2),即ba.10.f(x)=3x-2,那么f-1[f(x)]=__________________;f[f-1(x)]=__________________.答案：xx解析：∵f-1(x)=,f-1[f(x)]=[(3x-2)+2]=x,f[f-1(x)]=3-2=x.一般地,f[f-1(x)]與f-1[f(x)]的表達(dá)式總為x,但兩個(gè)函數(shù)定義域不一定相同,故不一定是同一個(gè)函數(shù).11.函數(shù)f(x)=ax2+(a+2)x-1在xR上存在反函數(shù),那么f-1(1)=_______________.答案：1解析：依題意a=0,f(x)=2x-1,令f-1(1)=b,那么f(b)=1,即2b-1=1b=1.12.函數(shù)f(x)=(x).(1)求它的反函數(shù);(2)求使f-1(x)=f(x)的實(shí)數(shù)a的值;(3)當(dāng)a=-1時(shí),求f-1(2).解：(1)設(shè)y=,∵x-a,反解得(y-3)x=2-ay.假設(shè)y=3,那么a=與a矛盾.y3.x=.f-1(x)=(x).(2)當(dāng)f-1(x)=f(x)時(shí),有,整理得(a+3)x2+(a2-9)x-2(a+3)=0.a+3=0,即a=-3.(3)當(dāng)a=-1時(shí),由(1)知f-1(x)=.f-1(2)=-4.13.f(x)=()2(x1),(1)求f(x)的反函數(shù)f-1(x),并求出反函數(shù)的定義域;(2)判斷并證明f-1(x)的單調(diào)性.解：(1)設(shè)y=()2x=,又x1,y1,即f-1(x)=,f-1(x)的定義域?yàn)閇0,1].(2)f-1(x)在[0,1)上單調(diào)遞增.證明如下:設(shè)0x1f-1(x1)-f-1(x2)=0.f-1(x)在[0,1]上單調(diào)遞增.拓展應(yīng)用跳一跳,夠得著!14.要使函數(shù)y=x2-2ax+1在區(qū)間[1,2]上存在反函數(shù),那么a的取值范圍是()A.a1B.a2C.a1或a2D.12答案：C解析：由得函數(shù)y=x2-2ax+1在區(qū)間[1,2]上單調(diào),那么a1或a2.15.函數(shù)y=f(x-1)的反函數(shù)為y=f-1(x-1),且f(1)=2,那么f(2)的值為______________.答案：1解析：y=f-1(x-1)x-1=f(y)x=f(y)+1,故y=f-1(x-1)的反函數(shù)為y=f(x)+1.故f(x-1)=f(x)+1,即f(x)=f(x-1)-1,那么f(2)=f(1)-1=1.16.(1)f(x)=(a、b、c是常數(shù))的反函數(shù)是f-1(x)=,求a+b+c的值.(2)設(shè)點(diǎn)P(-1,-2)既在函數(shù)f(x)=ax2+b(x0)的圖象上,又在f(x)的反函數(shù)的圖象上,求f-1(x).解：(1)設(shè)y=,解得x=,即f-1(x)=,因此,,由對(duì)應(yīng)項(xiàng)系數(shù)相等得a=3,b=5,c=-2,a+b+c=6.