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2013届高考数学(理)一轮复习教案：第二篇函数与基本初等函数Ⅰ第3讲函数的奇偶性与周期性(人教A版).pdf

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《2013届高考数学(理)一轮复习教案：第二篇函数与基本初等函数Ⅰ第3讲函数的奇偶性与周期性(人教A版).pdf》由会员分享，可在线阅读，更多相关《2013届高考数学(理)一轮复习教案：第二篇函数与基本初等函数Ⅰ第3讲函数的奇偶性与周期性(人教A版).pdf（14页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、2013 届高考数学（理）一轮复习教案：第二篇函数与基本初等函数第 3 讲函数的奇偶性与周期性【2013年高考会这样考】1判断函数的奇偶性2利用函数奇偶性、周期性求函数值及求参数值3考查函数的单调性与奇偶性的综合应用【复习指导】本讲复习时应结合具体实例和函数的图象，理解函数的奇偶性、周期性的概念，明确它们在研究函数中的作用和功能重点解决综合利用函数的性质解决有关问题基础梳理1奇、偶函数的概念一般地，如果对于函数f(x)的定义域内任意一个x，都有 f(x)f(x)，那么函数f(x)就叫做偶函数一般地，如果对于函数f(x)的定义域内任意一个x，都有 f(x)f(x)，那么函数 f(x)就叫做奇函数

2、奇函数的图象关于原点对称；偶函数的图象关于y 轴对称2奇、偶函数的性质(1)奇函数在关于原点对称的区间上的单调性相同，偶函数在关于原点对称的区间上的单调性相反(2)在公共定义域内两个奇函数的和是奇函数，两个奇函数的积是偶函数；两个偶函数的和、积都是偶函数；一个奇函数，一个偶函数的积是奇函数3周期性(1)周期函数：对于函数 yf(x)，如果存在一个非零常数T，使得当 x 取定义域内的任何值时，都有 f(xT)f(x)，那么就称函数 yf(x)为周期函数，称 T 为这个函数的周期(2)最小正周期：如果在周期函数f(x)的所有周期中存在一个最小的正数，那么这个最小正数就叫做f(x)的最小正周期一条规

3、律奇、偶函数的定义域关于原点对称函数的定义域关于原点对称是函数具有奇偶性的必要不充分条件两个性质(1)若奇函数 f(x)在 x0 处有定义，则 f(0)0.(2)设 f(x)，g(x)的定义域分别是 D1，D2，那么在它们的公共定义域上：奇奇奇，奇奇偶，偶偶偶，偶偶偶，奇偶奇文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9

4、ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9

5、 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S

6、9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3

7、S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G

8、3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2

9、G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O

10、2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1三种方法判断函数的奇偶性，一般有三种方法：(1)定义法；(2)图象法；(3)性质法三条结论(1)若对于 R 上的任意的 x 都有 f(2ax)f(x)或 f(x)f(2ax)，则 yf(x)的图象关于直线 xa 对称(2)若对于 R 上的任意 x 都有 f(2ax)f(x)，且 f(2bx)f(x)(其中 ab)，则：yf(x)是以 2(ba)为周期的周期函数(3)若 f(xa)f(x)

11、或 f(xa)1f x或 f(xa)1f x，那么函数 f(x)是周期函数，其中一个周期为 T2a；(3)若 f(xa)f(xb)(ab)，那么函数 f(x)是周期函数，其中一个周期为T2|ab|.双基自测1(2011 全国)设 f(x)是周期为 2 的奇函数，当 0 x1 时，f(x)2x(1x)，则 f52()A.12B.14C.14D.12解析因为 f(x)是周期为 2 的奇函数，所以 f 52f52f1212.故选 A.答案A 2(2012 福州一中月考)f(x)1xx 的图象关于()Ay 轴对称B直线 yx 对称C坐标原点对称D直线 yx 对称解析f(x)的定义域为(，0)(0，)，

12、又 f(x)1x(x)1xx f(x)，则 f(x)为奇函数，图象关于原点对称答案C 文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M

13、1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5

14、M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M

15、5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1

16、M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D

17、1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10

18、D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W1

19、0D1M5M13(2011 广东)设函数 f(x)和 g(x)分别是 R 上的偶函数和奇函数，则下列结论恒成立的是()Af(x)|g(x)|是偶函数Bf(x)|g(x)|是奇函数C|f(x)|g(x)是偶函数D|f(x)|g(x)是奇函数解析由题意知 f(x)与|g(x)|均为偶函数，A 项：偶偶偶；B 项：偶偶偶，B 错；C 项与 D 项：分别为偶奇偶，偶奇奇均不恒成立，故选A.答案A 4(2011 福建)对于函数 f(x)asin xbxc(其中，a，bR，cZ)，选取 a，b，c 的一组值计算 f(1)和 f(1)，所得出的正确结果一定不可能是()A4 和 6 B3 和 1 C2和 4

20、D1 和 2 解析 f(1)asin 1bc，f(1)asin 1bc 且 cZ，f(1)f(1)2c是偶数，只有 D 项中两数和为奇数，故不可能是D.答案D 5(2011 浙江)若函数 f(x)x2|xa|为偶函数，则实数a_.解析法一 f(x)f(x)对于 xR 恒成立，|xa|xa|对于 xR 恒成立，两边平方整理得ax0 对于 xR 恒成立，故 a0.法二由 f(1)f(1)，得|a1|a1|，得 a0.答案 0文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档

21、编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文

22、档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1

23、文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M

24、1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5

25、M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M

26、5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1

27、M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1考向一判断函数的奇偶性【例 1】?下列函数：f(x)1x2x21；f(x)x3x；f(x)ln(xx21)；f(x)3x3x2；f(x)lg1x1x.其中奇函数的个数是()A2 B3 C4 D5 审题视点利用函数奇偶性的定义判断解析f(x)1x2

28、x21的定义域为 1,1，又 f(x)f(x)0，则 f(x)1x2x21是奇函数，也是偶函数；f(x)x3x 的定义域为 R，又 f(x)(x)3(x)(x3x)f(x)，则 f(x)x3x 是奇函数；由 xx21x|x|0知 f(x)ln(xx21)的定义域为 R，又 f(x)ln(xx21)ln1xx21ln(xx21)f(x)，则 f(x)为奇函数；文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10

29、W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR1

30、0W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR

31、10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 Z

32、R10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9

33、ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9

34、 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S

35、9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1f(x)3x3x2的定义域为 R，又 f(x)3x3x23x3x2f(x)，则 f(x)为奇函数；由1x1x0 得1x1，f(x)ln1x1x的定义域为(1,1)，又 f(x)ln1x1xln1x1x1ln1x1xf(x)，则 f(x)为奇函数答案D 判断函数的奇偶性的一般方法是：(1)求函数的定义域；(2)证

36、明 f(x)f(x)或 f(x)f(x)成立；或者通过举反例证明以上两式不成立如果二者皆未做到是不能下任何结论的，切忌主观臆断【训练 1】判断下列函数的奇偶性：(1)f(x)4x2|x3|3；(2)f(x)x2|xa|2.文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6

37、 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C

38、6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1

39、C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O

40、1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4

41、O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X

42、4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10

43、X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1解(1)解不等式组4x20，|x3|30，得2x0，或 0 x2，因此函数 f(x)的定义域是 2,0)(0,2，则 f(x)4x2x.f(x)4 x2x4x2xf(x)，所以 f(x)是奇函数(2)f(x)的定义域是(，)当 a0 时，f(x)x2|x|2，f(x)x2|x|2x2|x|2f(x)因此 f(x)是偶函数；当 a0 时，f(a)a22，f(a)a2|2a|2，f(a)f(a)，且 f(a)f(a)因此 f(x)既不是偶函数也不是奇

44、函数考向二函数奇偶性的应用【例 2】?已知 f(x)x12x112(x0)(1)判断 f(x)的奇偶性；(2)证明：f(x)0.审题视点 (1)用定义判断或用特值法否定；(2)由奇偶性知只须求对称区间上的函数值大于 0.(1)解法一f(x)的定义域是(，0)(0，)f(x)x12x112x22x12x1.f(x)x22x12x1x22x12x1f(x)故 f(x)是偶函数法二f(x)的定义域是(，0)(0，)，文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV

45、2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:C

46、V2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:

47、CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码

48、:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编

49、码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档

50、编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文档编码:CV2K10X4O1C6 HE2D2O2G3S9 ZR10W10D1M5M1文

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关键词：: 2013 高考数学一轮复习教案第二函数基本初等奇偶性周期性人教

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本文标题：2013届高考数学(理)一轮复习教案：第二篇函数与基本初等函数Ⅰ第3讲函数的奇偶性与周期性(人教A版).pdf
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