PAGEPAGE1第6講函數(shù)的奇偶性與周期性1.[2024·衡水調(diào)研]下列函數(shù)中,在[-1,1]上與函數(shù)y=cos2x2的單調(diào)性和奇偶性都相同的是 (A.y=2x-2-x B.y=|x|+1C.y=x2(x+2) D.y=-x2+22.函數(shù)f(x+1)是偶函數(shù),則函數(shù)f(x)的圖像關(guān)于 ()A.直線x=1對(duì)稱 B.直線x=-1對(duì)稱C.點(diǎn)(1,0)對(duì)稱 D.點(diǎn)(-1,0)對(duì)稱3.[2024·岳陽(yáng)一中月考]已知函數(shù)f(x)=mxmx+1+2024tanx+x2(m>0,m≠1),若f(1)=3,則f(-1)A.-3 B.-1 C.3 D.04.[2024·南昌模擬]已知定義在R上的函數(shù)f(x)滿意:對(duì)隨意實(shí)數(shù)x都有f(x+3)=f(x-3),f(-x)=f(x),且當(dāng)x∈[-3,0]時(shí),f(x)=log12(6+x),則f(2024)的值為5.[2024·廣州一調(diào)]已知函數(shù)f(x)=2x2x-1+a為奇函數(shù)6.已知f(x)=asinx+b3x+4,若f(lg3)=3,則flg13=A.13 B.-1C.5 D.87.[2024·石家莊一模]設(shè)f(x)是定義在[-2b,3+b]上的偶函數(shù),且在[-2b,0]上為增函數(shù),則f(x-1)≥f(3)的解集為 ()A.[-3,3] B.[-2,4]C.[-1,5] D.[0,6]8.已知函數(shù)f(x)滿意條件:?x∈R,f(x)+f(-x)=0且f(x+t)-f(x)<0(t>0),則函數(shù)f(x)的解析式可能是 ()A.f(x)=xsinx+3 B.f(x)=x3C.f(x)=-sinx D.f(x)=-3x9.[2024·漳州二檢]已知函數(shù)f(x)是定義在R上的周期為6的奇函數(shù),且滿意f(1)=1,f(2)=3,則f(8)-f(5)= ()A.-4 B.-2 C.2 D.410.已知f(x)是偶函數(shù),且在[0,+∞)上函數(shù)f(x)=34x,0≤x<1,3-A.f-32B.f-32C.f-32≥D.f-32≤11.[2024·淄博模擬]已知定義在R上的函數(shù)f(x)滿意條件:①對(duì)隨意x∈R,都有f(x+2)+f(x)=1;②對(duì)隨意不同的x1,x2∈[0,2],都有(x1-x2)[f(x1)-f(x2)]>0;③函數(shù)f(x+2)的圖像關(guān)于y軸對(duì)稱.若a=f(4.5),b=f(6.5),c=f(7),則a,b,c的大小關(guān)系為.

12.[2024·上饒模擬]已知f(x)是R上的偶函數(shù),且在[0,+∞)上單調(diào)遞增,若f(a-3)<f(4),則a的取值范圍為.

13.設(shè)f(x)是定義在R上的奇函數(shù),且對(duì)隨意實(shí)數(shù)x,恒有f(x+2)=-f(x),當(dāng)x∈[0,2]時(shí),f(x)=2x-x2.(1)求證:f(x)是周期函數(shù);(2)當(dāng)x∈[2,4]時(shí),求f(x)的解析式;(3)求f(0)+f(1)+f(2)+…+f(2024)的值.14.已知f(x)=px2+23x+q是奇函數(shù)(1)求實(shí)數(shù)p,q的值;(2)推斷函數(shù)f(x)在(-∞,-1)上的單調(diào)性,并加以證明.15.已知定義在R上的奇函數(shù)f(x)在[0,+∞)上單調(diào)遞減,若當(dāng)x∈[-1,2]時(shí),f(x3-2x+a)<f(x+1)恒成立,則a的取值范圍為 ()A.(-3,+∞) B.(-∞,-3)C.(3,+∞) D.(-∞,3)16.[2024·濰坊模擬]設(shè)函數(shù)f(x)(x∈R)滿意f(x-π)=f(x)-sinx,當(dāng)-π<x≤0時(shí),f(x)=0,則f2018π3=課時(shí)作業(yè)(六)1.D[解析]函數(shù)y=cos2x2在[-1,0]上單調(diào)遞增,在[0,1]上單調(diào)遞減,且函數(shù)y=cos2x2為偶函數(shù),而函數(shù)y=-x2+2也在[-1,0]上單調(diào)遞增,在[0,1]上單調(diào)遞減,且函數(shù)y=-x2+2也為偶函數(shù),故選2.A[解析]因?yàn)閒(x+1)是偶函數(shù),所以f(x+1)的圖像關(guān)于y軸對(duì)稱,而把f(x+1)的圖像向右平移1個(gè)單位長(zhǎng)度可得f(x)的圖像,故f(x)的圖像關(guān)于直線x=1對(duì)稱,故選A.3.D[解析]由題設(shè)有f(-x)=m-xm-x+1-2025tanx+x2=1mx+1-2025tanx+x2,故有f(x)+f(-x)=1+2x2,所以f(1)+f(-1)=3,從而4.-2[解析]由f(x+3)=f(x-3),可得f(x+6)=f(x),所以函數(shù)f(x)的周期是6.由f(-x)=f(x),得函數(shù)f(x)為偶函數(shù),則f(2024)=f(2)=f(-2)=-2.5.-12[解析]∵f(x)=2x2∴f(1)+f(-1)=0,即2+a-1+a=0,∴a=-126.C[解析]∵f(x)=asinx+b3x+4,∴f(x)+f(-x)=8,又∵lg13=-lg3,∴f(lg3)+flg13=∴flg13=5,故選C.7.B[解析]∴f(x)是定義在[-2b,3+b]上的偶函數(shù),∴2b=3+b,∴b=3.∵函數(shù)f(x)在[-6,0]上為增函數(shù),∴f(x)在[0,6]上為減函數(shù),∴f(x-1)≥f(3),即|x-1|≤3,故-2≤x≤4.故選B.8.D[解析]∵f(x)+f(-x)=0,∴f(-x)=-f(x),∴f(x)為奇函數(shù).∵f(x+t)-f(x)<0,∴f(x+t)<f(x),又∵t>0,∴f(x)在R上為減函數(shù),∴f(x)是奇函數(shù)且在R上是減函數(shù).對(duì)于A,f(x)=xsinx+3為偶函數(shù),∴該選項(xiàng)不合題意;對(duì)于B,f(x)=x3在R上為增函數(shù),∴該選項(xiàng)不合題意;對(duì)于C,f(x)=-sinx在R上不單調(diào),∴該選項(xiàng)不合題意;對(duì)于D,f(x)=-3x為奇函數(shù),且在R上為減函數(shù),∴該選項(xiàng)符合題意.故選D.9.D[解析]由題意,得f(8)=f(2+6)=f(2)=3,f(5)=f(-1+6)=f(-1)=-f(1)=-1,則f(8)-f(5)=4.10.C[解析]依據(jù)題意,在[0,+∞)上函數(shù)f(x)=(34)

x,0≤x<1,3-94x,x≥1,則函數(shù)f(x)在[0,+∞)上為減函數(shù).又f(x)是偶函數(shù),∴f-32=f32,由a2+2a+52=(a+1)2+32≥32,得f32≥fa2+2a+52,即11.a<c<b[解析]因?yàn)閷?duì)隨意不同的x1,x2∈[0,2],都有(x1-x2)[f(x1)-f(x2)]>0,所以f(x)在[0,2]上為增函數(shù);因?yàn)楹瘮?shù)f(x+2)的圖像關(guān)于y軸對(duì)稱,所以f(x)的圖像關(guān)于直線x=2對(duì)稱;因?yàn)閒(x+2)+f(x)=1,所以f(x+4)+f(x+2)=1,即f(x+4)=f(x),所以f(x)的周期為4.因此a=f(4.5)=f(0.5),b=f(6.5)=f(2.5)=f(1.5),c=f(7)=f(3)=f(1),因?yàn)閒(0.5)<f(1)<f(1.5),所以a<c<b.12.(-1,7)[解析]∵f(x)是R上的偶函數(shù),且在[0,+∞)上單調(diào)遞增,∴不等式f(a-3)<f(4)等價(jià)于|a-3|<4,即-4<a-3<4,得-1<a<7,即a的取值范圍是(-1,7).13.解:(1)證明:∵f(x+2)=-f(x),∴f(x+4)=-f(x+2)=f(x),∴f(x)是周期為4的周期函數(shù).(2)當(dāng)x∈[-2,0]時(shí),-x∈[0,2],由已知得f(-x)=2(-x)-(-x)2=-2x-x2,又f(x)是奇函數(shù),∴-f(x)=f(-x)=-2x-x2,∴當(dāng)x∈[-2,0]時(shí),f(x)=x2+2x.又當(dāng)x∈[2,4]時(shí),x-4∈[-2,0],∴f(x-4)=(x-4)2+2(x-4).又f(x)是周期為4的周期函數(shù),∴當(dāng)x∈[2,4]時(shí),f(x)=f(x-4)=(x-4)2+2(x-4)=x2-6x+8,∴當(dāng)x∈[2,4]時(shí),f(x)=x2-6x+8.(3)易知f(0)=0,f(1)=1,f(2)=0,f(3)=-1.∵f(x)是周期為4的周期函數(shù),∴f(0)+f(1)+f(2)+f(3)=f(4)+f(5)+f(6)+f(7)=…=f(2008)+f(2009)+f(2010)+f(2011)=f(2012)+f(2013)+f(2014)+f(2015)=0,∴f(0)+f(1)+f(2)+…+f(2024)=f(2024)+f(2024)+f(2024)=f(0)+f(1)+f(2)=1.14.解:(1)因?yàn)閒(x)=px2所以定義域關(guān)于原點(diǎn)對(duì)稱,所以q=0,所以f(x)=px又f(2)=53,所以4p+26=53(2)由(1)知f(x)=2x2+23x,則f(x)在(-∞,-任取x1<x2<-1,則f(x1)-f(x2)=2x12+23因?yàn)閤1<x2<-1,所以x2-x1>0,1-x1x2<0,x1x2>0,所以f(x1)-f(x2)<0,即f(x1)<f(x2),所以函數(shù)f(x)在(-∞,-1)上是增函數(shù).15.C[解析]∵定義在R上的奇函數(shù)f(x)在[0,+∞)上單調(diào)遞減,∴f(x)在R上是減函數(shù).若當(dāng)x∈[-1,2]時(shí),f(x3-2x+a)<f(x+1)恒成立,則當(dāng)x∈[-1,2]時(shí),x3-2x+a>x+1恒成立,即a>-x3+3x+1恒成立.設(shè)g(x)=-x3+3x+1(x∈[-1,2]),令g'(x)=-3x2+3=0,得x=±1.∴在[-1,1)上,g'(x)≥0,g(x)是增函數(shù);在[1,2]上,g'(x)≤0,g(x)是減函數(shù).故g(x)的最大值為g(1)=3,∴a>3.故選C.16.