2025高考數(shù)學(xué)一輪復(fù)習(xí)-2.8-函數(shù)的圖象-專項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-2.8-函數(shù)的圖象-專項(xiàng)訓(xùn)練【A級(jí)基礎(chǔ)鞏固】1.要得到函數(shù)y=log2(2x+4)的圖象,只需將函數(shù)y=log2(x+2)的圖象()A.向左平移2個(gè)單位長(zhǎng)度B.向右平移2個(gè)單位長(zhǎng)度C.向上平移1個(gè)單位長(zhǎng)度D.向下平移1個(gè)單位長(zhǎng)度2.函數(shù)f(x)=(x-1x)sinx的圖象可能為(3.已知圖①對(duì)應(yīng)的函數(shù)為y=f(x),則圖②對(duì)應(yīng)的函數(shù)是()A.y=f(-|x|) B.y=f(-x) C.y=f(|x|) D.y=-f(-x)4.若函數(shù)y=f(x)的大致圖象如圖所示,則f(x)的解析式可能是()A.f(x)=x|x|-C.f(x)=xx2-5.已知函數(shù)f(x)=1-x1+A.f(x)在(-1,+∞)上單調(diào)遞增B.f(x)的圖象關(guān)于點(diǎn)(-1,1)對(duì)稱C.f(x)為奇函數(shù)D.f(x)的圖象關(guān)于直線y=x對(duì)稱6.若函數(shù)f(x)=ax-2x7.如圖,已知函數(shù)f(x)的圖象為折線ACB(含端點(diǎn)A,B),其中A(-4,0),B(4,0),C(0,4),則不等式f(x)>log2(x+2)的解集是.

8.設(shè)函數(shù)f(x)=x+1x9.已知函數(shù)f(x)=|log2x|,0<x<2,-f(x2)=f(x3),則x1·x2·x3的取值范圍是.

INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.如圖為某無(wú)人機(jī)飛行時(shí),從某時(shí)刻開始15min內(nèi)的速度V(x)(單位:m/min)與時(shí)間x(單位:min)的關(guān)系.若定義“速度差函數(shù)”v(x)為無(wú)人機(jī)在時(shí)間段[0,x]內(nèi)的最大速度與最小速度的差,則v(x)的圖象為()11.已知定義在區(qū)間(0,2)上的函數(shù)y=f(x)的圖象如圖所示,則y=-f(2-x)的圖象為()12.若平面直角坐標(biāo)系內(nèi)A,B兩點(diǎn)滿足:(1)點(diǎn)A,B都在f(x)的圖象上;(2)點(diǎn)A,B關(guān)于原點(diǎn)對(duì)稱,則對(duì)稱點(diǎn)對(duì)(A,B)是函數(shù)f(x)的一個(gè)“和諧點(diǎn)對(duì)”,(A,B)與(B,A)可看作一個(gè)“和諧點(diǎn)對(duì)”.已知函數(shù)f(x)=x2+2xA.1個(gè) B.2個(gè) C.3個(gè) D.4個(gè)13.已知函數(shù)f(x)=log2(x+1),x>3,13|f(x2),f(x1)+f(x3)=4,則x3x1已知函數(shù)f(x)=|3-2x-x2|的圖象和直線2x+ay+7=0有三個(gè)交點(diǎn),求a的值.參考答案【A級(jí)基礎(chǔ)鞏固】1.解析:因?yàn)閥=log2(2x+4)=log2[2(x+2)]=log22+log2(x+2)=1+log2(x+2),故只需將函數(shù)y=log2(x+2)的圖象向上平移1個(gè)單位長(zhǎng)度,即可得到y(tǒng)=log2(2x+4)的圖象.故選C.2.解析:f(x)的定義域?yàn)閧x|x≠0},因?yàn)閒(-x)=(-x+1x)sin(-x)=(x-1當(dāng)x∈(0,1)時(shí),x-1x=x2-3.解析:根據(jù)函數(shù)圖象知,當(dāng)x≤0時(shí),所求函數(shù)圖象與已知函數(shù)相同,當(dāng)x>0時(shí),所求函數(shù)圖象與x<0時(shí)圖象關(guān)于y軸對(duì)稱,即所求函數(shù)為偶函數(shù)且x≤0時(shí)與y=f(x)相同,故B,D不符合要求,當(dāng)x≤0時(shí),y=f(-|x|)=f(x),y=f(|x|)=f(-x),故A正確,C錯(cuò)誤.故選A.4.解析:由圖可知,當(dāng)x∈(0,1)時(shí),f(x)<0,取x=12,則對(duì)于B,f(12)=對(duì)于D,f(12)=121當(dāng)x>0時(shí),對(duì)于A,f(x)=xx-1=1+15.解析:化簡(jiǎn)得f(x)=1-x1+x=f(x)可以看作是函數(shù)g(x)=2x先畫出g(x)=2x的圖象,再進(jìn)行平移畫出f(x)=2明顯可見,對(duì)于原函數(shù)g(x)=2x(-∞,0)和(0,+∞)上分別單調(diào)遞減,所以g(x)經(jīng)過(guò)平移后變成的f(x)在(-1,+∞)上單調(diào)遞減,關(guān)于(-1,-1)對(duì)稱,為非奇非偶函數(shù),圖象關(guān)于直線y=x對(duì)稱,所以D正確,A,B,C錯(cuò)誤.故選D.6.解析:f(x)=ax-a+當(dāng)a=2時(shí),f(x)=2,不合題意;當(dāng)a≠2時(shí),f(x)關(guān)于點(diǎn)(1,a)對(duì)稱,故a=1.答案:17.解析:在同一坐標(biāo)系中作出函數(shù)y=f(x)和y=log2(x+2)的圖象,易知當(dāng)x=2時(shí),f(x)=log2(x+2)=2,所以不等式f(x)>log2(x+2)的解集是(-2,2).答案:(-2,2)8.解析:當(dāng)x>0時(shí),f(x)=x+1x函數(shù)y=x-1x作出函數(shù)f(x)的圖象如圖所示,由圖象可知,要使方程f(x)=a有三個(gè)實(shí)數(shù)解,則需a>2,則符合題意的一個(gè)a的值為3.答案:3(答案不唯一,只要滿足a>2均可)9.解析:作出函數(shù)f(x)的圖象.不妨設(shè)x1<x2<x3,由圖可得,|log2x1|=|log2x2|=-x3+3∈(0,1),所以log2x1=-log2x2,即x1x2=1,由f(x1)=f(x2)=f(x3)得,x3∈(2,3),所以x1·x2·x3的取值范圍是(2,3).答案:(2,3)INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.解析:由題意可得,當(dāng)x∈[0,6]時(shí),無(wú)人機(jī)做勻加速運(yùn)動(dòng),V(x)=60+403“速度差函數(shù)”v(x)=403當(dāng)x∈(6,10]時(shí),無(wú)人機(jī)做勻速運(yùn)動(dòng),V(x)=140,“速度差函數(shù)”v(x)=80;當(dāng)x∈(10,12]時(shí),無(wú)人機(jī)做勻加速運(yùn)動(dòng),V(x)=40+10x,“速度差函數(shù)”v(x)=-20+10x;當(dāng)x∈(12,15]時(shí),無(wú)人機(jī)做勻減速運(yùn)動(dòng),“速度差函數(shù)”v(x)=100,選項(xiàng)C滿足“速度差函數(shù)”解析式.故選C.11.解析:法一由定義在(0,2)上的函數(shù)y=f(x)的圖象可知f(x)=x當(dāng)0<2-x<1,即1<x<2時(shí),f(2-x)=2-x,當(dāng)1≤2-x<2,即0<x≤1時(shí),f(2-x)=1,所以y=-f(2-x)=-1法二y=f(x)y=f(-x)y=f(2-x)y=-f(2-x).故選B.12.解析:作出函數(shù)y=x2+2x(x<0)的圖象關(guān)于原點(diǎn)對(duì)稱的圖象(如圖中的虛線部分),看它與函數(shù)y=2e13.解析:由已知得,當(dāng)x>3時(shí),f(x)>2,當(dāng)x∈[-9,3]時(shí),0≤f(x)≤2,并且圖象關(guān)于x=-3對(duì)稱,函數(shù)圖象如圖所示,如果x1>-3,則x1<x2,且f(x1)=f(x2)不成立,所以x1∈[-9,-3),x2∈(-3,3],并且有x1+x2=-6;由f(x1)+f(x3)=4可知,必有4>f(x3)=4-f(x1)≥2,所以4>log2(x3+1)≥2,3≤x3<15,所以x3x1+x2=-16答案:(-52,-114.解:畫出函數(shù)f(x)的圖象,如圖所示,因?yàn)橹本€2x+ay+7=0過(guò)定點(diǎn)(-723-2x-x2(-3<x<-1)