度高一數(shù)學(xué)人教A版必修一練習(xí)第一章集合與函數(shù)的概念1.3.2第一課時(shí)函數(shù)奇偶性的定義與判定

奇偶性第一課時(shí)函數(shù)奇偶性的定義與判定【選題明細(xì)表】知識(shí)點(diǎn)、方法題號(hào)奇偶函數(shù)的圖象特征2,4,6,11奇偶性的概念與判定1,3,10,11奇偶性的應(yīng)用5,7,8,9,121.函數(shù)f(x)=x4+2x2是(B)(A)奇函數(shù)(B)偶函數(shù)(C)既是奇函數(shù)又是偶函數(shù)(D)非奇非偶函數(shù)解析:因?yàn)閒(x)=(x)4+2(x)2=x4+2x2=f(x),所以函數(shù)f(x)=x4+2x2是偶函數(shù).故選B.2.已知函數(shù)f(x)=x3+1x(A)原點(diǎn)對(duì)稱 (B)y軸對(duì)稱(C)y=x對(duì)稱 (D)y=x對(duì)稱解析:函數(shù)的定義域?yàn)?∞,0)∪(0,+∞),因?yàn)閒(x)=(x)3+1-x=(x3+1所以函數(shù)為奇函數(shù).所以函數(shù)f(x)=x3+1x3.如果f(x)是定義在R上的奇函數(shù),那么下列函數(shù)中,一定為偶函數(shù)的是(B)(A)y=x+f(x) (B)y=xf(x)(C)y=x2+f(x) (D)y=x2f解析:因?yàn)閒(x)是奇函數(shù),所以f(x)=f(x).對(duì)于A,g(x)=x+f(x)=xf(x)=g(x),所以y=x+f(x)是奇函數(shù).對(duì)于B,g(x)=xf(x)=xf(x)=g(x),所以y=xf(x)是偶函數(shù).對(duì)于C,g(x)=(x)2+f(x)=x2f(x),所以y=x2+f(x)為非奇非偶函數(shù),對(duì)于D,g(x)=(x)2f(x)=x2所以y=x2f故選B.4.下列結(jié)論中正確的是(B)(A)偶函數(shù)的圖象一定與y軸相交(B)奇函數(shù)y=f(x)在x=0處有定義,則f(0)=0(C)奇函數(shù)y=f(x)的圖象一定過(guò)原點(diǎn)(D)圖象過(guò)原點(diǎn)的奇函數(shù)必是單調(diào)函數(shù)解析:A項(xiàng)中若定義域不含0,則圖象與y軸不相交,C項(xiàng)中若定義域不含0,則圖象不過(guò)原點(diǎn),D項(xiàng)中奇函數(shù)不一定單調(diào),故選B.5.已知f(x)=ax3+bx+1(ab≠0),若f(2018)=k,則f(2018)等于(D)(A)k (B)k (C)1k (D)2k解析:設(shè)g(x)=ax3+bx,易知g(x)為奇函數(shù),則f(x)=g(x)+1.因?yàn)閒(2018)=k,則g(2018)=f(2018)1=k1,所以g(2018)=g(2018)=1k.所以f(2018)=g(2018)+1=1k+1=2k.故選D.6.如圖,給出奇函數(shù)y=f(x)的局部圖象,則f(2)+f(1)的值為(A)(A)2 (B)2(C)1 (D)0解析:由圖知f(1)=12,f(2)=3又f(x)為奇函數(shù),所以f(2)+f(1)=f(2)f(1)=321故選A.7.若函數(shù)f(x)=kx2+(k1)x+3是偶函數(shù),則k等于.

解析:由于函數(shù)f(x)=kx2+(k1)x+3是偶函數(shù),因此k1=0,k=1.答案:18.設(shè)f(x)是(∞,+∞)上的奇函數(shù),且f(x+2)=f(x),當(dāng)0≤x≤1時(shí),f(x)=x,則f(7.5)=.

解析:由f(x+2)=f(x),得f(7.5)=f(5.5+2)=f(5.5)=f(3.5+2)=f(3.5)=f(1.5+2)=f(1.5)=f(0.5+2)=f(0.5)=f(0.5)=0.5.答案:0.59.已知函數(shù)f(x)=12x(1)若g(x)=f(x)a為奇函數(shù),求a的值;(2)試判斷f(x)在(0,+∞)內(nèi)的單調(diào)性,并用定義證明.解:(1)由已知g(x)=f(x)a得,g(x)=1a2x因?yàn)間(x)是奇函數(shù),所以g(x)=g(x),即1a2-x=(1a2(2)函數(shù)f(x)在(0,+∞)內(nèi)為增函數(shù).證明:設(shè)0<x1<x2,則f(x1)f(x2)=12x1（12x2因?yàn)?<x1<x2,所以x1x2<0,x1x2>0,從而2(x1-x所以函數(shù)f(x)在(0,+∞)內(nèi)是單調(diào)增函數(shù).10.下列函數(shù)中,既是奇函數(shù)又是增函數(shù)的為(D)(A)y=x+1 (B)y=x2 (C)y=1x 解析:A.y=x+1為非奇非偶函數(shù),不滿足條件.B.y=x2是偶函數(shù),不滿足條件.C.y=1xD.設(shè)f(x)=x|x|,則f(x)=x|x|=f(x),則函數(shù)為奇函數(shù),當(dāng)x>0時(shí),y=x|x|=x2,此時(shí)為增函數(shù),當(dāng)x≤0時(shí),y=x|x|=x2,此時(shí)為增函數(shù).綜上在R上函數(shù)為增函數(shù).故選D.11.已知函數(shù)f(x)是定義域?yàn)镽的奇函數(shù),當(dāng)x>0時(shí),f(x)=x22x.(1)求出函數(shù)f(x)在R上的解析式;(2)畫(huà)出函數(shù)f(x)的圖象.解:(1)①由于函數(shù)f(x)是定義域?yàn)镽的奇函數(shù),則f(0)=0;②當(dāng)x<0時(shí),x>0,因?yàn)閒(x)是奇函數(shù),所以f(x)=f(x),所以f(x)=f(x)=[(x)22(x)]=x22x,綜上,f(x)=x(2)圖象如圖.12.設(shè)函數(shù)f(x)在R上是偶函數(shù),且在區(qū)間(∞,0)上遞增,且f(2a2+a+1)<f(2a2-2a+3),求a的取值范圍.解:由f(x)在R上是偶函數(shù),且在區(qū)間(∞,0)上遞增,可知f(x)在(0,+∞)上遞減.因