1、强化训练函数的性质1下列函数中，既是偶函数又在区间(1,2)内单调递减的是()Af (x) Bf (x)Cf (x)2x2x Df (x)cos x答案B解析函数f (x)是偶函数，且在(1,2)内单调递减，符合题意2函数f (x)x(x0)是()A奇函数，且在(0,3)上是增函数B奇函数，且在(0,3)上是减函数C偶函数，且在(0,3)上是增函数D偶函数，且在(0,3)上是减函数答案B解析因为f (x)xf (x)，所以函数f (x)x为奇函数又f(x)1，在(0,3)上f(x)0恒成立，所以f (x)在(0,3)上是减函数3若函数f (x)ax2bx8(a0)是偶函数，则g(x)2ax3b
2、x29x是()A奇函数 B偶函数C非奇非偶函数 D既奇又偶函数答案A解析由f (x)是偶函数可得b0，g(x)2ax39x，g(x)是奇函数4(2019湖北武汉重点中学联考)已知偶函数f (x)在0，)上单调递减，f (1)1，若f (2x1)1，则x的取值范围为()A(，1 B1，)C0,1 D(，01，)答案C解析由题意，得f (x)在(，0上单调递增，且f (1)1，所以f (2x1)f (1)，则|2x1|1，解得0x1.故选C.5若定义在R上的奇函数f (x)满足对任意的xR，都有f (x2)f (x)成立，且f (1)8，则f (2 019)，f (2 020)，f (2 021)
3、的大小关系是()Af (2 019)f (2 020)f (2 020)f (2 021)Cf (2 020)f (2 019)f (2 021)Df (2 020)f (2 021)f (2 019)答案A解析因为定义在R上的奇函数f (x)满足对任意的xR，都有f (x2)f (x)成立，所以f (x4)f (x)，即函数f (x)的周期为4，且f (0)0，f (2)f (0)0，f (3)f (1)8，所以f (2 019)f (45043)f (3)8，f (2 020)f (4505)f (0)0，f (2 021)f (45051)f (1)8，即f (2 019)f (2 02
4、0)0时，f (x)f (x2)1，则f (2 019)f (2 017)1f (2 015)2f (1)1 009f (1)1 010, 而f (1)0，故f (2 019)1 010.15已知定义在R上的奇函数f (x)满足f (x4)f (x)，且在区间0,2上是增函数若方程f (x)m(m0)在区间8,8上有四个不同的根x1，x2，x3，x4，则x1x2x3x4_.答案8解析因为定义在R上的奇函数满足f (x4)f (x)，所以f (x4)f (x)由f (x)为奇函数，所以函数图象关于直线x2对称，且f (0)0.由f (x4)f (x)知f (x8)f (x)，所以函数的周期为8.
5、又因为f (x)在区间0,2上是增函数，所以函数在区间2,0上也是增函数，作出函数f (x)的大致图象如图所示，那么方程f (x)m(m0)在区间8,8上有四个不同的根x1，x2，x3，x4，不妨设x1x2x3x4，由对称性可知x1x212，x3x44，所以x1x2x3x48.16函数f (x)的定义域为Dx|x0，且满足对于任意x1，x2D，有f (x1x2)f (x1)f (x2)(1)求f (1)的值；(2)判断f (x)的奇偶性并证明你的结论；(3)如果f (4)1，f (x1)2，且f (x)在(0，)上是增函数，求x的取值范围解(1)因为对于任意x1，x2D，有f (x1x2)f (x1)f (x2)，所以令x1x21，得f (1)2f (1)，所以f (1)0.(2)f (x)为偶函数，证明如下：令x1x21，有f (1)f (1)f (1)，所以f (1)f (1)0.令x11，x2x，有f (x)f (1)f (x)，所以f (x)f (x)，又f (x)的定义域关于原点对称，所以f (x)为偶函数(3)依题设有f (44)f (4)f (4)2，由(2)知，f (x)是偶函数，所以f (x1)2，等价于f (|x1|)f (16)又f (x)在(0，)上是增函数所以0|x1|16，解得15x17且x1.所以x的取值范围是x|15x17且x1
