第四節(jié)函數(shù)的圖象課時(shí)作業(yè)練1.已知函數(shù)f(x)的圖象如圖所示,則函數(shù)g(x)=log2f(x)的定義域是.答案(2,8解析要使g(x)有意義,需有f(x)>0,由f(x)的圖象可知,當(dāng)x(2,8時(shí), f(x)>0,故g(x)的定義域?yàn)?2,8.2.若函數(shù)f(x)=log2|ax-1|的圖象關(guān)于直線(xiàn)x=2對(duì)稱(chēng),則非零實(shí)數(shù)a=.答案12解析由題意得 f(4)=f(0),即log2|4a-1|=0,則4a-1=1或4a-1=-1,解得a=12或a=0.因?yàn)閍為非零實(shí)數(shù),所以a=12.3.已知直線(xiàn)y=a與函數(shù)f(x)=2x、g(x)=32x的圖象分別相交于A、B兩點(diǎn),則A、B兩點(diǎn)之間的距離為.答案log23解析由題意知A(log2a,a),Blog2a3,a,所以A、B兩點(diǎn)之間的距離=|xA-xB|=log23.4.(2019江蘇南京模擬)若函數(shù)y=12|1-x|+m的圖象與x軸有公共點(diǎn),則m的取值范圍是.答案-1m<0解析作出函數(shù)y=12|1-x|的圖象(如圖所示),欲使y=12|1-x|+m的圖象與x軸有交點(diǎn),則-1m<0.5.已知最小正周期為2的函數(shù)f(x)在區(qū)間-1,1上的解析式為f(x)=x2,則函數(shù)f(x)在實(shí)數(shù)集R上的圖象與函數(shù)g(x)=|log5x|的圖象的交點(diǎn)的個(gè)數(shù)是.答案5解析作出兩函數(shù)圖象如圖,由圖象可得交點(diǎn)個(gè)數(shù)是5.6.若函數(shù)f(x)是定義在R上的偶函數(shù),且在(-,0上是減函數(shù), f(2)=0,則使得f(x)<0的x的取值范圍是.答案(-2,2)解析由題意知f(-2)=f(2)=0,則f(x)的大致圖象如圖所示.所以所求的x的取值范圍是(-2,2).7.已知f(x)是定義在R上的偶函數(shù),且對(duì)于任意的x0,+),滿(mǎn)足f(x+2)=f(x).若當(dāng)x0,2)時(shí), f(x)=|x2-x-1|,則函數(shù)y=f(x)-1在區(qū)間-2,4上的零點(diǎn)個(gè)數(shù)為.答案7解析作出函數(shù)y=f(x),x-2,4及y=1的圖象如圖所示,可得兩圖象有7個(gè)交點(diǎn),即函數(shù)y=f(x)-1,x-2,4有7個(gè)零點(diǎn).8.(2017江蘇邗江中學(xué)上學(xué)期高三月考)函數(shù)f(x)=log2|x-1|的單調(diào)遞減區(qū)間是.答案(-,1)解析函數(shù)f(x)=log2|x-1|=log2(x-1),x>1,log2(1-x),x<1,作出圖象如圖,由圖象可知, f(x)的單調(diào)遞減區(qū)間是(-,1).9.(2019鹽城時(shí)楊中學(xué)高三模擬)已知函數(shù)f(x)=logax,x>0,|x+3|,-4x<0,其中a>0且a1,y=f(x)的圖象上有且只有兩對(duì)點(diǎn)關(guān)于y軸對(duì)稱(chēng),則實(shí)數(shù)a的取值范圍是.答案a4解析由題意知y=loga(-x),x<0與y=|x+3|,-4x<0的圖象有2個(gè)交點(diǎn),則a>1,且loga41=logaa,解得a4.10.已知函數(shù)f(x)=x1+|x|.(1)畫(huà)出函數(shù)f(x)的圖象;(2)指出函數(shù)f(x)的單調(diào)區(qū)間.解析(1)f(x)=x1+|x|=x1+x,x0,x1-x,x<0.當(dāng)x0時(shí), f(x)=x1+x=1-1x+1,其圖象可由函數(shù)y=-1x(x>0)的圖象向左平移1個(gè)單位,向上平移1個(gè)單位后得到.又易知f(x)為奇函數(shù),所以其圖象關(guān)于原點(diǎn)對(duì)稱(chēng).函數(shù)f(x)的圖象如圖.(2)由圖象知,函數(shù)f(x)在(-,+)上是單調(diào)增函數(shù),故其單調(diào)增區(qū)間為R.11.(2017江蘇泰興中學(xué)等四校高三檢測(cè))已知f(x)為定義在R上的奇函數(shù),當(dāng)x>0時(shí),f(x)為二次函數(shù),且滿(mǎn)足f(2)=1,f(x)在(0,+)上的兩個(gè)零點(diǎn)為1和3.(1)求函數(shù)f(x)在R上的解析式;(2)作出f(x)的圖象,并根據(jù)圖象討論關(guān)于x的方程f(x)-c=0(cR)根的個(gè)數(shù).解析(1)由題意知,當(dāng)x>0時(shí),可設(shè)f(x)=a(x-1)(x-3),a0,f(2)=1,a=-1,f(x)=-x2+4x-3.當(dāng)x<0時(shí),-x>0,f(x)為R上的奇函數(shù),f(-x)=-f(x),f(x)=-f(-x)=-(-x)2+4(-x)-3=x2+4x+3,當(dāng)x=0時(shí), f(0)=0,所以f(x)=-x2+4x-3,x>0,0,x=0,x2+4x+3,x<0.(2)作出f(x)的圖象(如圖所示).由f(x)-c=0得c=f(x),則問(wèn)題轉(zhuǎn)化為求y=c與y=f(x)圖象的交點(diǎn)個(gè)數(shù),根據(jù)交點(diǎn)討論方程的根:當(dāng)c3或c-3時(shí),兩圖象有1個(gè)交點(diǎn),則方程有1個(gè)根;當(dāng)1<c<3或-3<c<-1時(shí),兩圖象有2個(gè)交點(diǎn),則方程有2個(gè)根;當(dāng)c=-1或c=1時(shí),兩圖象有3個(gè)交點(diǎn),則方程有3個(gè)根;當(dāng)0<c<1或-1<c<0時(shí),兩圖象有4個(gè)交點(diǎn),則方程有4個(gè)根;當(dāng)c=0時(shí),兩圖象有5個(gè)交點(diǎn),則方程有5個(gè)根.12.已知函數(shù)f(x)的圖象與函數(shù)h(x)=x+1x+2的圖象關(guān)于點(diǎn)A(0,1)對(duì)稱(chēng).(1)求函數(shù)f(x)的解析式;(2)若g(x)=f(x)+ax,且g(x)在區(qū)間(0,2上為減函數(shù),求實(shí)數(shù)a的取值范圍.解析(1)設(shè)函數(shù)f(x)圖象上任一點(diǎn)的坐標(biāo)為(x,y),則點(diǎn)(x,y)關(guān)于點(diǎn)A(0,1)的對(duì)稱(chēng)點(diǎn)(-x,2-y)在函數(shù)h(x)的圖象上,2-y=-x+1-x+2,即y=x+1x,f(x)=x+1x.(2)g(x)=f(x)+ax=x+a+1x,則g(x)=1-a+1x2.g(x)在(0,2上遞減,g(x)0在(0,2上恒成立,即ax2-1在(0,2上恒成立,則a(x2-1)max,x(0,2,實(shí)數(shù)a的取值范圍是3,+).13.已知函數(shù)f(x)=ax2+1bx+c(a>0,bN*,cR)是奇函數(shù),當(dāng)x>0時(shí), f(x)有最小值2,且f(1)<52.(1)求函數(shù)f(x)的解析式;(2)函數(shù)f(x)的圖象上是否存在關(guān)于點(diǎn)(1,0)對(duì)稱(chēng)的兩點(diǎn)?若存在,求出點(diǎn)的坐標(biāo);若不存在,說(shuō)明理由.解析(1)f(x)是奇函數(shù),f(-1)=-f(1),即a+1-b+c=-a+1b+c,c=0.當(dāng)x>0時(shí), f(x)=ax2+1bx=axb+1bx2ab2,當(dāng)且僅當(dāng)x=1a時(shí)取等號(hào),則2ab2=2,a=b2.又f(1)=a+1b<52,即2b2-5b+2<0,12<b<2,又bN*,b=1,a=1,f(x)=x2+1x.(2)設(shè)存在一點(diǎn)(x0,y0)在函數(shù)f(x)的圖象上,且關(guān)于點(diǎn)(1,0)的對(duì)稱(chēng)點(diǎn)(2-x0,-y0)也在函數(shù)f(x)的圖象上,則y0=x02+1x0,-y0=(2-x0)2+12-x0,消去y0化簡(jiǎn)得x02-2x0-1=0,則x0=12,函數(shù)f(x)的圖象上存在關(guān)于點(diǎn)(1,0)對(duì)稱(chēng)的兩點(diǎn)(1+2,22)和(1-2,-22).基礎(chǔ)滾動(dòng)練(滾動(dòng)循環(huán)夯實(shí)基礎(chǔ))1.(2018江蘇南京多校高三段考)已知集合A=-1,2,2m-1,集合B=2,m2,若BA,則實(shí)數(shù)m=.答案12.(2018江蘇,5,5分)函數(shù)f(x)=log2x-1的定義域?yàn)?答案2,+)解析由題意可得log2x-10,即log2x1,x2.函數(shù)的定義域?yàn)?,+).3.函數(shù)f(x)的圖象如圖所示,點(diǎn)O,A,B的坐標(biāo)分別為(0,0),(1,2),(3,1),則f1f(3)的值等于.答案24.(2018江蘇南京多校高三模擬)設(shè)f(x)是定義在R上的周期為2的函數(shù),當(dāng)x-1,1)時(shí),f(x)=-4x2+2,-1x<0,x,0x<1,則f32=.答案1解析f32=f-12=-1+2=1.5.(2017興化第一中學(xué)高三月考)已知偶函數(shù)f(x)在0,+)上單調(diào)遞減, f52=0,若f(2x-1)0,則x的取值范圍是.答案-34,74解析由題意知f(2x-1)0-522x-152,解得-34x74.6.已知A=y|y=x2-4x+3,xR,B=y|y=-x2-2x+2,xR,則AB=.答案-1,3解析由題意得A=-1,+),B=(-,3,所以AB=-1,3.7.已知函數(shù)f(x)=xx+1+x+1x+2+x+2x+3+x+3x+4,則f-52+2+f-52-2=.答案8解析因?yàn)閒(x)=4-1x+1+1x+2+1x+3+1x+4,所以f(x)+f(-5-x)=8-1x+1+1x+2+1x+3+1x+4-1-4-x+1-3-x+1-2-x+1-1-x=8,所以f-52+2+f-52-2=8.8.(2017江蘇如東高級(jí)中學(xué)高三期中)定義域?yàn)镽的函數(shù)f(x)=1-2x2x+1.若對(duì)于任意的tR,不等式f(t2-2t)<-f(2t2-k)恒成立,求k的取值范圍.解析任取x1,x2R,且令x1<x2,則f(x1)-f(x2)=1-2x12x1+1-1-2x22x2+1=2(2x2-2x1)(2x2+1)(2x1+1),因?yàn)閤2>x1,所以2(2x2-2x1)>0,又(2x2+1)(2x1+1)>0,所以f(x1)-f(x2)>0,則函數(shù)f(x)為R上的減函數(shù),易知f(x)為R上的奇函數(shù),故不等式f(t2-2t)<-f(2t2-k)可化為f(t2-2t)< f(k-2t2),即t2-2t>-2t2+k,即k<3t2-2t在tR上恒成立,而3t2-2t的最小值為-13,所以k<-13.