1、课时质量评价(八)(建议用时：45分钟)A组全考点巩固练1(多选题)已知yf (x)是定义在R上的奇函数，则下列函数中为奇函数的是()Ayf (|x|)Byf (x)Cyxf (x)Dyf (x)xBD解析：由奇函数的定义f (x)f (x)验证对于选项A，f (|x|)f (|x|)，为偶函数；对于选项B，f (x)f (x)f (x)，为奇函数；对于选项C，xf (x)xf (x)xf (x)，为偶函数；对于选项D，f (x)(x)f (x)x，为奇函数故选BD2(多选题)设函数f (x)x3，则f (x)()A是奇函数B是偶函数C在(0，)单调递增D在(，0)单调递减AC解析：因为f (
2、x)x3，则f (x)x3f (x)，即f (x)为奇函数根据幂函数的性质可知，yx3在(0，)上单调递增，在(，0)上单调递增，故y1在(0，)上单调递减，在(，0)上单调递减，所以函数f (x)x3在(，0)上单调递增，在(0，)上单调递增3已知函数f (x)x1，f (a)3，则f (a)的值为()A3 B1 C1 D2B解析：由题意得f (a)f (a)a1(a)12, 所以f (a)2f (a)231.故选B4设函数f (x)是定义在R上的奇函数，且f (x)则f (7)()A3 B3 C2 D2B解析：因为函数f (x)是定义在R上的奇函数，且f (x)所以f (7)f (7)lo
3、g2(71)3.5若定义在R上的偶函数f (x)和奇函数g(x)满足f (x)g(x)ex，则g(x)()Aexex B(exex) C(exex) D(exex)D解析：因为f (x)g(x)ex，所以f (x)g(x)f (x)g(x)ex，所以g(x)(exex)6已知函数f (x)为奇函数，当x0时，f (x)x2x，则当x0时，函数f (x)的最大值为_解析：设x0，所以f (x)x2x.又函数f (x)为奇函数，所以f (x)f (x)x2x，所以当x0，f (x2)对任意xR恒成立，则f (2 023)_.1解析：因为f (x)0，f (x2)，所以f (x4)f (x2)2f
4、(x)，即函数f (x)的周期是4，所以f (2 023)f (50641)f (1)因为函数f (x)为偶函数，所以f (2 023)f (1)f (1)当x1时，f (12)，得f (1). 由f (x)0，得f (1)1，所以f (2 023)f (1)1.13定义在实数集R上的函数f (x)满足f (x)f (x2)0，且f (4x)f (x)现有以下三种叙述：8是函数f (x)的一个周期；f (x)的图象关于直线x2对称；f (x)是偶函数其中正确的序号是_解析：由f (x)f (x2)0，得f (x2)f (x)，则f (x4)f (x2)f (x)，即4是f (x)的一个周期，8
5、也是f (x)的一个周期，故正确；由f (4x)f (x)，得f (x)的图象关于直线x2对称，故正确；由f (4x)f (x)与f (x4)f (x)，得f (4x)f (x)，f (x)f (x)，即函数f (x)为偶函数，故正确14已知函数f (x)对任意x，yR，总有f (x)f (y)f (xy)，且当x0时，f (x)0，f (1).(1)求证：f (x)是R上的单调增函数；(2)求f (x)在3,3上的最大值解：(1)令xy0，得f (0)f (0)f (00)所以f (0)0.令yx，得f (x)f (x)f (0)0.所以f (x)是奇函数任取x1x2，有f (x1)f (x
6、2)f (x1)f (x2)f (x1x2)f (x2x1)因为x2x10，所以f (x2x1)0.所以f (x1)f (x2)0，f (x1)f (x2)所以f (x)在R上为增函数(2)由(1)得f (x)在R上单调递增，所以函数f (x)的最大值为f (3)，且f (3)f (21)f (2)f (1)f (11)f (1)3f (1)32.15设f (x)是(，)上的奇函数，f (x2)f (x)当0x1时，f (x)x.(1)求f ()的值；(2)当4x4时，求f (x)的图象与x轴所围成图形的面积解：(1)由f (x2)f (x)，得f (x4)f (x2)2)f (x2)f (x)，所以f (x)是以4为周期的周期函数，所以f ()f (14)f (4)f (4)(4)4.(2)由f (x)是奇函数且f (x2)f (x)，得f (x1)2f (x1)f (x1)，即f (1x)f (1x)故函数yf (x)的图象关于直线x1对称又当0x1时，f (x)x，且f (x)的图象关于原点成中心对称，则f (x)的图象如图所示当4x4时，设f (x)的图象与x轴围成的图形面积为S，则S4SOAB4214.