1.3.2奇偶性課后篇鞏固提升基礎(chǔ)鞏固1.下列函數(shù)是奇函數(shù)的是()A.y=x(x-1)x-1B.y=-3x2C.y=-|x|D.y=x3-35x解析先判斷函數(shù)的定義域是否關(guān)于原點(diǎn)對(duì)稱(chēng),再確定f(-x)與f(x)的關(guān)系.選項(xiàng)A中函數(shù)的定義域?yàn)?-,1)(1,+),不關(guān)于原點(diǎn)對(duì)稱(chēng),所以排除A;選項(xiàng)B,C中函數(shù)的定義域均是R,且函數(shù)均是偶函數(shù);選項(xiàng)D中函數(shù)的定義域是R,且f(-x)=-f(x),則此函數(shù)是奇函數(shù).答案D2.已知函數(shù)g(x)=f(x)-x,其中y=g(x)是偶函數(shù),且f(2)=1,則f(-2)=()A.-1B.1C.-3D.3解析g(x)=f(x)-x,f(2)=1,g(2)=f(2)-2=1-2=-1.y=g(x)是偶函數(shù),g(-2)=f(-2)+2=-1,f(-2)=-3.故選C.答案C3.已知f(x)為偶函數(shù),且當(dāng)x0時(shí),f(x)2,則當(dāng)x<0時(shí),有()A.f(x)2B.f(x)2C.f(x)-2D.f(x)R解析可畫(huà)出滿(mǎn)足題意的一個(gè)f(x)的大致圖象如圖所示,由圖易知當(dāng)x<0時(shí),有f(x)2.故選B.答案B4.已知函數(shù)f(x)=x|x|-2x,則下列結(jié)論正確的是()A.f(x)是偶函數(shù),遞增區(qū)間是(0,+)B.f(x)是偶函數(shù),遞增區(qū)間是(-,1)C.f(x)是奇函數(shù),遞減區(qū)間是(-1,1)D.f(x)是奇函數(shù),遞增區(qū)間是(-,0)解析由函數(shù)f(x)=x|x|-2x可得,函數(shù)的定義域?yàn)镽,且f(-x)=-x|-x|-2(-x)=-x|x|+2x=-f(x),故函數(shù)為奇函數(shù),函數(shù)f(x)=x|x|-2x=x2-2x,x0,-x2-2x,x<0,如圖所示,所以函數(shù)的遞減區(qū)間為(-1,1),故選C.答案C5.已知f(x)是奇函數(shù),當(dāng)x>0時(shí),f(x)=-x(1+x),當(dāng)x<0時(shí),f(x)等于()A.-x(1-x)B.x(1-x)C.-x(1+x)D.x(1+x)解析當(dāng)x<0時(shí),-x>0,則f(-x)=x(1-x).又f(x)是R上的奇函數(shù),所以當(dāng)x<0時(shí),f(x)=-f(-x)=-x(1-x).故選A.答案A6.設(shè)函數(shù)f(x)和g(x)分別是R上的偶函數(shù)和奇函數(shù),則下列結(jié)論恒成立的是()A.f(x)+|g(x)|是偶函數(shù)B.f(x)-|g(x)|是奇函數(shù)C.|f(x)|+g(x)是偶函數(shù)D.|f(x)|-g(x)是奇函數(shù)解析由f(x)是偶函數(shù),可得f(-x)=f(x),由g(x)是奇函數(shù)可得g(-x)=-g(x),故|g(x)|為偶函數(shù),f(x)+|g(x)|為偶函數(shù).答案A7.若函數(shù)f(x)=(x+1)(x+a)x為奇函數(shù),則a=.解析f(x)是奇函數(shù),f(-x)=-f(x),即(-x+1)(-x+a)-x=-(x+1)(x+a)x,顯然x0,整理得x2-(a+1)x+a=x2+(a+1)x+a,故a+1=0,解得a=-1.答案-18.已知f(x)=x5+ax3+bx-8,且f(-2)=10,則f(2)=.解析令h(x)=x5+ax3+bx,易知h(x)為奇函數(shù).因?yàn)閒(x)=h(x)-8,h(x)=f(x)+8,所以h(-2)=f(-2)+8=18.h(2)=-h(-2)=-18,所以f(2)=h(2)-8=-18-8=-26.答案-269.已知奇函數(shù)f(x)的定義域?yàn)?5,5,且在區(qū)間0,5上的圖象如圖所示.(1)畫(huà)出在區(qū)間-5,0上的圖象.(2)寫(xiě)出使f(x)<0的x的取值集合.解(1)因?yàn)楹瘮?shù)f(x)是奇函數(shù),所以y=f(x)在-5,5上的圖象關(guān)于原點(diǎn)對(duì)稱(chēng).由y=f(x)在0,5上的圖象,可知它在-5,0上的圖象,如圖所示.(2)由圖象知,使函數(shù)值f(x)<0的x的取值集合為(-2,0)(2,5).10.已知f(x)為奇函數(shù),且當(dāng)x<0時(shí),f(x)=x2+3x+2.若當(dāng)x1,3時(shí),f(x)的最大值為m,最小值為n,求m-n的值.解當(dāng)x<0時(shí),f(x)=x2+3x+2,且f(x)是奇函數(shù),當(dāng)x>0時(shí),-x<0,則f(-x)=x2-3x+2.故當(dāng)x>0時(shí),f(x)=-f(-x)=3x-x2-2.當(dāng)x1,32時(shí),f(x)是增函數(shù);當(dāng)x32,3時(shí),f(x)是減函數(shù).因此當(dāng)x1,3時(shí),f(x)max=f32=14,f(x)min=f(3)=-2.m=14,n=-2,從而m-n=94.能力提升1.若函數(shù)f(x)=(m-1)x2+(m-2)x+(m2-7m+12)為偶函數(shù),則m的值是()A.1B.2C.3D.4解析f(-x)=(m-1)x2-(m-2)x+(m2-7m+12),f(x)=(m-1)x2+(m-2)x+(m2-7m+12),由f(-x)=f(x),得m-2=0,即m=2.答案B2.設(shè)f(x)是奇函數(shù),對(duì)任意的實(shí)數(shù)x,y,有f(x+y)=f(x)+f(y),當(dāng)x>0時(shí),f(x)<0,則f(x)在區(qū)間a,b上()A.有最大值f(a)B.有最小值f(a)C.有最大值fa+b2D.有最小值fa+b2解析任取x1<x2,則x2-x1>0.當(dāng)x>0時(shí),f(x)<0,f(x2-x1)<0,即f(x2)+f(-x1)<0.f(x)是奇函數(shù),f(x2)-f(x1)<0,f(x2)<f(x1),f(x)在R上是減函數(shù).f(x)在區(qū)間a,b上有最大值f(a),最小值f(b).故選A.答案A3.已知定義在R上的函數(shù)f(x)在(-,-2)上是減函數(shù),若g(x)=f(x-2)是奇函數(shù),且g(2)=0,則不等式xf(x)0的解集是()A.(-,-4-2,+)B.-4,-20,+)C.(-,-22,+)D.(-,-40,+)解析g(x)=f(x-2)的圖象是將函數(shù)f(x)的圖象向右平移2個(gè)單位得到的,又g(x)=f(x-2)的圖象關(guān)于原點(diǎn)對(duì)稱(chēng),所以函數(shù)f(x)的圖象關(guān)于點(diǎn)(-2,0)對(duì)稱(chēng),大致圖象如圖所示,且f(0)=g(2)=0,f(-4)=g(-2)=-g(2)=0,f(-2)=g(0)=0,結(jié)合函數(shù)的圖象,由xf(x)0可知x0,f(x)0或x<0,f(x)0.結(jié)合圖象可知x0或-2x<0或x-4.故不等式xf(x)0的解集是(-,-4-2,+),故選A.答案A4.已知f(x)=(k-2)x2+(k-3)x+3是偶函數(shù),則f(x)的遞減區(qū)間為.解析由偶函數(shù)的定義知k=3,f(x)=x2+3,其圖象開(kāi)口向上,所以f(x)的遞減區(qū)間是(-,0.答案(-,05.已知函數(shù)f(x)為R上的奇函數(shù),且當(dāng)x0時(shí),f(x)=-x2+1x+1+t,則f(-2)=.解析因?yàn)楹瘮?shù)f(x)為R上的奇函數(shù),所以f(0)=0,即-02+10+1+t=0,解得t=-1.所以f(x)=-x2+1x+1-1.所以f(2)=-22+12+1-1=-143.又函數(shù)f(x)為R上的奇函數(shù),所以f(-2)=-f(2)=143.答案1436.定義在(-8,a)上的奇函數(shù)f(x)在區(qū)間2,7上是增函數(shù),在區(qū)間3,6上的最大值為a,最小值為-1,則2f(-6)+f(-3)=.解析根據(jù)題意,f(x)是定義在(-8,a)上的奇函數(shù),則a=8.又由f(x)在區(qū)間2,7上是增函數(shù),且在區(qū)間3,6上的最大值為a=8,最小值為-1,則f(6)=a=8,f(3)=-1.函數(shù)是奇函數(shù),則f(-6)=-8,f(-3)=1.則2f(-6)+f(-3)=2(-8)+1=-15.答案-157.判斷函數(shù)f(x)=x2-2x+3,x>0,0,x=0,-x2-2x-3,x<0的奇偶性.解方法一:(圖象法)畫(huà)出函數(shù)f(x)的圖象,如圖所示,定義域關(guān)于原點(diǎn)對(duì)稱(chēng),函數(shù)f(x)是奇函數(shù).方法二:(定義法)f(x)的定義域?yàn)镽(關(guān)于原點(diǎn)對(duì)稱(chēng)).當(dāng)x=0時(shí),-x=0,f(-x)=f(0)=0,f(x)=f(0)=0,f(-x)=-f(x);當(dāng)x>0時(shí),-x<0,f(-x)=-(-x)2-2(-x)-3=-(x2-2x+3)=-f(x);當(dāng)x<0時(shí),-x>0,f(-x)=(-x)2-2(-x)+3=-(-x2-2x-3)=-f(x).由可知,當(dāng)xR時(shí),都有f(-x)=-f(x),f(x)為奇函數(shù).8.已知f(x)為定義在R上的偶函數(shù),當(dāng)x-1時(shí),f(x)=x+b,且f(x)的圖象經(jīng)過(guò)點(diǎn)(-2,0),在y=f(x)的圖象中有一部分是頂點(diǎn)為(0,2),過(guò)點(diǎn)(-1,1)的一段拋物線.(1)試求出f(x)的表達(dá)式;(2)求出f(x)的值域.解(1)f(x)的圖象經(jīng)過(guò)點(diǎn)(-2,0),0=-2+b,即b=2.當(dāng)x-1時(shí),f(x)=x+2.f(x)為偶函數(shù),當(dāng)x1時(shí),f(x)=f(-x)=-x+2.當(dāng)-1x1時(shí),依題意設(shè)f(x)=ax2+2(a0),則1=a(-1)2+2,a=-1.當(dāng)-1x1時(shí),f(x)=-x2+2.綜上,f(x)=x+2,x-1,-x2+2,-1<x<1,-x+2,x1.(2)當(dāng)x-1時(shí),f(x)=x+2(-,1;當(dāng)-1<x<1時(shí),f(x)=-x2+2(1,2;當(dāng)x1時(shí),f(x)=-x+2(-,1.綜上所述,f(x)(-,2.9.已知函數(shù)f(x)=x+ab+x2是定義在(-1,1)上的奇函數(shù),且f12=25.(1)用定義證明:f(x)在(-1,1)上是增函數(shù);(2)若實(shí)數(shù)m滿(mǎn)足f(m-1)+f(1-2m)<0,求m的取值范圍.(1)證明函數(shù)f(x)=x+ab+x2是定義在(-1,1)上的奇函數(shù),f(0)=0,ab=0,a=0.f12=25,b=1,f(x)=x1+x2.設(shè)x1,x2是(-1,1)上任意兩個(gè)實(shí)數(shù),且-1<x1<x2<1,f(x2)-f(x1)=x21+x22-x11+x12=(x2-x1)(1-x1x2)(1+x22)(1+x12).x1,x2(-1,1),且x2-x1>0,1-x1x2>0,(x2-x1)(1-x1x2)(1+x22)(1+x12)>0,f(x2)>f(x1),f(x)在(-1,1)上單調(diào)遞增.(2)解f(x)=x1+x2是(-1,1)上的奇函數(shù)且單調(diào)遞增,f(m-1)+f(1-2m)<0,f(m-1)<f(2m-1).-1<m-1<1,-1<2m-1<1,m-1<2m-1,綜上得0<m<1.