(完整word版)函数奇偶性练习题(内含答案),推荐文档.pdf

新希望培训学校资料MATHEMATICS 心在哪里，新的希望就在哪里函数奇偶性练习（内含答案）一、选择题1已知函数f（x）ax2bxc（a0）是偶函数，那么g（x）ax3bx2cx（）A奇函数B偶函数C既奇又偶函数D非奇非偶函数2已知函数f（x）ax2bx3ab是偶函数，且其定义域为a1，2a，则（）A31a，b0 Ba 1，b0 Ca1，b0 Da 3，b03已知f（x）是定义在 R上的奇函数，当x0 时，f（x）x22x，则f（x）在 R上的表达式是（）Ay x（x2）By x（x1）C y x（x 2）Dyx（x 2）4已知f（x）x5ax3bx8，且f（2）10，那么f（2）等于（）A 26 B 18 C 10 D 105函数1111)(22xxxxxf是（）A偶函数B奇函数C非奇非偶函数D既是奇函数又是偶函数6若)(x，g（x）都是奇函数，2)()(xbgaxf在（0，）上有最大值5，则f（x）在（，0）上有（）A最小值 5 B最大值 5 C最小值 1 D最大值 3二、填空题7函数2122)(xxxf的奇偶性为 _（填奇函数或偶函数）8若y（m1）x22mx3 是偶函数，则m_9已知f（x）是偶函数，g（x）是奇函数，若11)()(xxgxf，则f（x）的解析式为_10已知函数f（x）为偶函数，且其图象与x轴有四个交点，则方程f（x）0 的所有实根之和为_三、解答题11设定义在2，2上的偶函数f（x）在区间 0，2上单调递减，若f（1m）f（m），求实数m的取值范围精品资料-欢迎下载-欢迎下载 名师归纳-第 1 页，共 4 页 -新希望培训学校资料MATHEMATICS 心在哪里，新的希望就在哪里12已知函数f（x）满足f（xy）f（xy）2f（x）f（y）（xR，yR），且f（0）0，试证f（x）是偶函数13.已知函数f（x）是奇函数，且当x0 时，f（x）x32x21，求f（x）在 R上的表达式14.f（x）是定义在（，55，）上的奇函数，且f（x）在 5，）上单调递减，试判断f（x）在（，5上的单调性，并用定义给予证明15.设函数yf（x）（xR且x0）对任意非零实数x1、x2满足f（x1x2）f（x1）f（x2），求证f（x）是偶函数精品资料-欢迎下载-欢迎下载 名师归纳-第 2 页，共 4 页 -文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8新希望培训学校资料MATHEMATICS 心在哪里，新的希望就在哪里函数的奇偶性练习参考答案1解析：f（x）ax2bxc为偶函数，xx)(为奇函数，g（x）ax3bx2cxf（x）)(x满足奇函数的条件答案：A 2解析：由f（x）ax2bx3ab为偶函数，得b0又定义域为a1，2a，a12a，31a故选 A3解析：由x0 时，f（x）x22x，f（x）为奇函数，当x0 时，f（x）f（x）（x22x）x22xx（x2）,)0()0()2()2()(xxxxxxxf即f（x）x（|x|2）答案：D 4解析：f（x）8x5ax3bx为奇函数，f（2）818，f（2）8 18，f（2）26答案：A 5解析：此题直接证明较烦，可用等价形式f（x）f（x）0答案：B 6解析：)(x、g（x）为奇函数，)()(2)(xbgxaxf为奇函数又f（x）在（0，）上有最大值5，f（x）2 有最大值3f（x）2 在（，0）上有最小值3，f（x）在（，0）上有最小值1答案：C 7答案：奇函数8答案：0 解析：因为函数y（m1）x22mx3 为偶函数，f（x）f（x），即（m1）（x）22m（x）3（m1）x22mx3，整理，得m09解析：由f（x）是偶函数，g（x）是奇函数，可得11)()(xxgxf，联立11)()(xxgxf，11)1111(21)(2xxxxf答案：11)(2xxf10答案：0 11答案：21m12证明：令xy0，有f（0）f（0）2f（0）f（0），又f（0）0，可证f（0）1令x0，f（y）f（y）2f（0）f（y）f（y）f（y），故f（x）为偶函数13解析：本题主要是培养学生理解概念的能力f（x）x32x21因f（x）为奇函数，f（0）0当x0 时，x0，f（x）（x）32（x）21x32x2 1，精品资料-欢迎下载-欢迎下载 名师归纳-第 3 页，共 4 页 -文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 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ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8文档编码:CN4O2H3T10C9 HP7C2Q8W6W9 ZL9T2I1O6I8新希望培训学校资料MATHEMATICS 心在哪里，新的希望就在哪里f（x）x32x21因此，.)0()0()0(12012)(,2323xxxxxxxxf点评：本题主要考查学生对奇函数概念的理解及应用能力14解析：任取x1x25，则x1x25因f（x）在 5，上单调递减，所以f（x1）f（x2）f（x1）f（x2）f（x1）f（x2），即单调减函数点评：此题要注意灵活运用函数奇偶性和单调性，并及时转化15解析：由x1，x2R且不为 0 的任意性，令x1x21 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