函數(shù)的奇偶性提升習(xí)題精編版

1、最新資料推薦函數(shù)的奇偶性提升習(xí)題姓名：譚學(xué)龍 學(xué)號(hào)：1443201000138專業(yè)：數(shù)學(xué)與應(yīng)用數(shù)學(xué)回顧：奇偶函數(shù)的性質(zhì)：(1 )函數(shù)具有奇偶性的必要條件是其定義域關(guān)于原點(diǎn)對(duì)稱；(2)f(x)是偶函數(shù)u f(x)的圖象關(guān)于y軸對(duì)稱；f(x)是奇函數(shù)。f(x)的圖象關(guān)于原點(diǎn)對(duì)稱；(3 )奇函數(shù)在對(duì)稱的單調(diào)區(qū)間內(nèi)有相同的單調(diào)性，偶函數(shù)在對(duì)稱的單調(diào)區(qū)間內(nèi)具 有相反的單調(diào)性.3 . f (x)為偶函數(shù) u f (x) = f (x) = f (| x |).4 .若奇函數(shù)f(x)的定義域包含0,則f (0) = 0.1 .已知定義在R上的奇函數(shù)f(x),當(dāng)x>0時(shí)，f(x) =x2+|x|1,那么

2、x<0時(shí)，f(x);x : a2 .若函數(shù)f(x)= ex a 在1-1,1 上是奇函數(shù)，則f(x)的解析式為x2 bx 13 .奇函數(shù)f (x)在區(qū)間3,7上是增函數(shù)，在區(qū)間3,6上的最大值為8,最小值為-1 ,貝U 2 f ( -6) f (-3)二 4 .已知函數(shù)f (x) =ax2+bx + 3a+b為偶函數(shù)，其定義域?yàn)?la -1,2a,則f (x)的值域.5 .判斷下列函數(shù)的奇偶性，并加以證明.x + 2,x < -1,1(1) f (x) =x*| x| ；(2) f (x) = -,-1 <x <1-x 2,x 16 .已知奇函數(shù)f (x)在(-1 ,

3、1)上是減函數(shù)，求滿足f (1-m) + f (1-m2) < 0的實(shí)數(shù)m的取值范圍.7 .已知f (x)是定義在R上的不恒為零的函數(shù)，且對(duì)任意的a,bw R都滿足f (a b) =af (b) +bf (a).(1)求 f(0), f(1)的值；(2)判斷f(x)的奇偶性，并證明你的結(jié)論.8 .設(shè)奇函數(shù)f (x)是定義在oci,)上的增函數(shù)，若不等式f (ax+6) + f (2 x2) c 0對(duì)于任意xw 12,4都成立，求實(shí)數(shù)a的取值范圍.1.【答案】x2|x+1+|x| -1, f(-x)-f (x)_f (x) = x2 +|x|-1 , f (x) = -x2 -|x| +1

4、x2.【答案】f(x)=F x2 1a【解析】: f(-x)=-f(x)f(_0) = _f (0), f(0) =0,- = 0,a = 01一x即 f(x)=x2 -bx 1"1)一,-1- ，b = 02-b 2 bf (x)在區(qū)間3,6上也為遞增函數(shù)，即 f (6) = 8, f (3) = -12f (-6) f(-3) =2f(6) - f(3)=15【解析】因?yàn)楹瘮?shù)f(x) = ax2 +bx + 3a + b為la 1,2a上的偶函數(shù)，所以a -1 2a =0, b =0,"3,即 f(x)=x2 +1,所以 f(x)b=0.35【解析】(1)定義域?yàn)?,1

5、，g(x) = x,1 |x| = g(x),所以g(x)是奇函數(shù).(2)函數(shù)的定義域?yàn)镽 ,當(dāng)x<1時(shí)，f (x)= x+2 ,此時(shí)x>1 ,【解析】 設(shè)xc0,則x>0, f (x) = x2f(x) =(x)+2=x + 2 = f(x) .當(dāng) x >1 時(shí)，f (x) = x +2 ,此時(shí)-x < -1 , f (x) = x + 2 = f (x). _1_當(dāng)iMxMl 時(shí)，f (x) = 5 = f (x).綜上可知對(duì)任意x W R都有f (-x) = f (x),所以f(x)為偶函數(shù).6.【解析】由已知f (1 m) <f (1 m2),由f

6、(x)為奇函數(shù)，所以f (1一m) < f (m2 1),又f (x)在(1,1)上是減函數(shù)，-1 <1 -m <1,0 <m <2,|4-1 < m?-1 <1,解得42 < m <0或0 < m< 四,1 -m>m2 -1.-2 <m<1.0 ： m ： 17【解析】(1) f(0)= f(0 0)=0 f(0)+0f (0)=0；f(1) = f (11)=1 ,f (1) + 1 f(1) = 2f(1), , f(1) = 0.(2丫 f (1)= f 1) ,(1)=(1)f(1)十(1)f (1) = 2f (1) = 0 ,二 f(-1) = 0.f( x) = f 1(-1) x ,1 - (-1) f (x) xf (-1)= -f (x) 0 = -f (x)故f (x)為奇函數(shù).22、8.【解析】由 f(ax+6) +f(2 x )<0得 f(ax + 6)< f(2x )v f (x)為奇函數(shù)，, f (ax + 6) < f(x2 -2).又f (x)在R上為增函數(shù)，原問題等價(jià)于ax + 6<x22對(duì)xe 12,4】都成立，即x2 -ax-8 >0對(duì) xe 12,4】都成立.令g(x)