高考抽象函数专题

1、抽象函数专题几类抽象函数模型抽象函数模型适用模型的初等函数f (xy)f (x)f (y)正比例函数f (x)kx(k0)f (xy)f (x)f (y)或f ()幂函数f (x)xnf (xy)f (x)f (y)或f (xy)指数函数f (x)ax(a>0且a1)f (xy)f (x)f (y)或f ()f (x)f (y)对数函数f (x)logax(a>0且a1)f (xT)f (x)正余弦函数f (x)sinx，f (x)cosxf (xy)正切函数f (x)tanx练习题1定义域为(0，¥)的函数f (x)满足f (xy)f (x)f (y)，若f (4)2，

2、则f ()的值为_答案：解：因为f (4)f (2)f (2)，f (2)f ()f ()，所以f (4)4 f ()，f ()2函数f (x)满足f (xy2)f (x)2f (y)2且f (1)0，则f (2018)的值为_答案：1009解：f (0)0，f (1)，f (x1)f (x)，f (2018)f (1)2017×10093(1)函数f (x)满足f (xy)f (x)f (y)x y1，若f (1)1，则f (8)A1B1C19D43答案：D解：因为f (1)1，y1代入f (xy)f (x)f (y)x y1，得f (x1)f (x)x 2，因此：f (2)f (

3、1)3f (3)f (2)4f (8) f (7)9累加，得f (8)43(2)函数f (x)满足f (xy)f (x)f (y)xy1，若f (1)1，则f (8)A1B1C19D43答案：C 解：因为f (1)1，y1代入f (xy)f (x)f (y)xy1，得f (x1)f (x)x 2，因此：f (1)f (0)2f (0)f (1)1f (1)f (2)0f (2)f (3)1f (3)f (4)2f (4)f (5)3f (5)f (6)4f (6)f (7)5f (7)f (8)6累加，得f (8)19另外：f (xx)f (x)f (x)x 21f (0)f (x)f (x)

4、x 21f (x)f (x)x 224定义在R上的函数f (x)满足f (x1x2)f (x1)f (x2)1，则下列说法正确的是Af (x)为奇函数Bf (x)为偶函数Cf (x)1为奇函数Df (x)1为偶函数答案：C 解：x1x20代入f (x1x2)f (x1)f (x2)1，得f (0)1x1x，x2x 代入f (x1x2)f (x1)f (x2)1，得f (x)f (x)2，f (x)图象关于点(0，1)对称，所以f (x)1为奇函数5设f (x)是定义在(0，)上的单调增函数，满足f (xy)f (x)f (y)，f (3)1，当f (x)f (x8)2时x的取值范围是A(8，)

5、 B(8，9 C8，9 D(0，8)答案：B解：211f (3)f (3)f (9)，由f (x)f (x8)2，可得fx(x8)f (9)，因为f (x) 是定义在(0，)上的增函数，所以有解得8<x9.6定义在0，1上的函数f (x)满足f (0)0，f (x)f (1x)2，f () f (x)，当0 x1< x21时，f (x1)f (x2)，则f ()的值为 答案： 7(1)已知函数f (x)满足2xf (x)3f (x)x10，求f (x)的表达式解：因为2xf (x)3f (x)x10，所以2xf (x)3f (x)x10×2x得4x2f (x)6 x f

6、(x)2 x22 x 0；×3得6xf (x)9f (x)3x30相减得4x2f (x)9f (x)2 x22 x3x30，所以f (x)(2)设函数f (x)满足f (x)2f ()x(x0)，求证：|f (x)|证明：因为f (x)2f ()x ，所以f ()2f (x) ×2得2f ()4f (x)得f (x)， |f (x)|8(12分)定义在R上的单调函数f (x)满足f (xy)f (x)f (y)，设f (3)log23(1)判断函数的奇偶性；(2)若f (k3x)f (3x9x4)<0，求实数k的取值范围解：(1)取xy0代入f (x)f (y)f (

7、xy)，得f (0)0取yx代入f (x)f (x)f (0)，得f (x)f (x)所以f (x)为奇函数(2)奇函数，所以， 是定义在上的单调函数，所以函数在上的单调递增函数，奇函数，不等式等价于，因此，即，因为，当取等号，所以实数的取值范围是9(12分)已知定义在R上的函数f (x)满足f (x)f (y)f (xy)，当>0时，f (x)<0，且f (1)(1)判断f (x)为奇偶性；(2)求证：f (x)在R上是减函数； (3)求f (x)在3，6上的最大值与最小值解：(1)取xy0代入f (x)f (y)f (xy)，得f (0)0取yx代入f (x)f (x)f (0

8、)，得f (x)f (x)所以f (x)为奇函数(2)设x1，x2R，xx2x1>0，那么yf (x2)f (x1)f (x2)f (x1)f (x2x1)f (x)因为x >0，所以y<0，所以f (x)在R上是减函数(3)因为f (1)，所以f (2)f (1)f (1)；f (3)f (1)f (2)2；f (3) f (3)2；f (6)f (3)f (3)4由(2)知f (x)在3，6上，所以求f (x)在3，6上的最大值为f (3)2，最小值为f (6)4 10(12分)已知定义在区间(0，)上的函数f (x)满足f ()f (x2)f (x1)，且当x>1

9、时，f (x)<0(1)证明：f (x)为单调递减函数(2)若f (3)1，求f (x)在2，9上的最小值解：(1)设x1，x2(0，)，xx2x1>0，那么yf (x2)f (x1)f ()因为当>1时，f (x)<0，>1，所以f ()<0，y>0，所以f (x)为单调递减函数(2)因为f (x)在(0，)上是单调递减函数，所以f (x)在2，9上的最小值为f (9)由f ()f (x1)f (x2)得，f ()f (9)f (3)，而f (3)1，所以f (9)2所以f (x)在2，9上的最小值为211(12分)定义域为(，0)(0，)的函数f

10、(x)满足f (x)f (y)f (xy)(1)求证：f ()f (x)；(2)求证：f (x)为偶函数；(3)当>1时，f (x)>0，求证：f (x)在(，0)上单调递减解：(1)取xy1代入f (x)f (y)f (xy)，得f (1)0取y 代入f (x)f (y)f (xy)，得f (x)f ()0，故f ()f (x)(2)取y1代入f (x)f (y)f (xy)，得f (x)f (1)f (x)取xy1代入f (x)f (y)f (xy)，f (1)f (1)f (1)，所以f (1)0所以f (x)f (x)，f (x)为偶函数(3)解法1：设x1，x2(0，)，

11、xx2x1>0，那么yf (x2)f (x1)f (x2)f ()f ()因为>1，所以f ()>0，y>0，所以f (x)在(0，)上单调递增由(2)知f (x)为偶函数，所以f (x)在(，0)上单调递减解法2：设x1，x2(，0)，xx2x1>0，那么yf (x2)f (x1)f (x2)f ()f ()f ()因为>1，所以f ()>0，y<0，所以f (x)在(，0)上单调递减12(12分)设定义在R上的函数yf (x)满足f (ab)f (a)·f (b)当x>0时，f (x)>1，且f (0)0(1)求证：f

12、(0)1；(2)求证：f (x)>0；(3)求证：f (x)是R上的增函数；(4)若f (x)·f (2xx2)>1，求x的取值范围解：(1)取ab0代入f (ab)f (a)·f (b)，得f (0)2f (0)，因为f (0)0，所以f (0)1(2)ax，bx代入f (ab)f (a)·f (b)，得f (0)f (x)·f (x)，即f (x)当x>0时，f (x)>1；x0时，f (x)1；当x<0时，x>0，f (x)>1，所以f (x)(0，1)综上，f (x)>0(3)设x1，x2R，xx2

13、x1>0，那么yf (x2)f (x1)f (x1x)f (x1)f (x1)f (x)f (x1)f (x1)f (x)1 因为 xx2x1>0，所以f (x)>1，故y>0，f (x)是R上的增函数(4)f (x)·f (2xx2)f (x2xx2)f (3xx2)，1f (0)，所以不等式f (x)·f (2xx2)>1可化为f (3xx2)> f (0)由(2)知3xx2>0，得x的取值范围为(0，3)13(12分)已知定义在R上的不恒为零的函数f (x)满足 f (xy)y f (x)x f (y)(1)判断f (x)的奇

14、偶性；(2)若f (2)2，设an ，bn ，求证数列an 为等差数列，数列bn 为等比数列解：(1)取xy1代入f (xy)y f (x)x f (y)，得f (1)0取xy1代入f (xy)y f (x)x f (y)，得f (1)0取y1代入f (x)f (x)x f (1)，得f (x)f (x) ，所以f (x)为奇函数(2)因为f (2n1)f (2·2n)2 f (2n)2n f (2)，所以f (2n1)2 f (2n)2n1同除以2n1，得 1，即an1an1，所以数列an 为等差数列a1 1，所以 an a1(n1)×1n，所以f (2n)2n因为2，所

15、以数列bn 为等比数列14(12分)定义在(0，¥)上的函数f (x)满足：对任意实数m，f (xm)mf (x)；f (2)1(1)求证：f (xy)f (x)f (y)；(2)求证：f (x)是(0，¥)上的单调增函数；(3)若f (x)f (x3)2，求x的取值范围解：(1)因为x，y均为正数，根据指数函数性质可知，总有实数m，n使得x2m，y2n于是f (xy)f (2m2n)f (2m+n)(m+n)f (2)m+n而mm f (2) f (2m) f (x)， nn f (2) f (2n) f (y)，所以f (xy)f (x)+f (y)(2)取xy1代入f

16、 (xy)f (x)f (y)，得f (1)0取y 代入f (1)f (x)f ()，得f (x)f ()设x1，x2(0，¥)，xx2x1>0，那么yf (x2)f (x1)f (x2)f ()f ()因为>1，根据指数函数性质可知，总有正实数r，使得 2r，所以yf (2r)r>0因此f (x)是(0，¥)上的单调增函数(3)由(1)知若f (x)f (x3)f (x23 x)，2 f (2)f (2)f (4)所以不等式f (x)f (x3)2即f (x23 x)f (4)由得x的取值范围为(3，4 15(12分)定义在0，1上的函数f (x)满足f

17、 (x) 0，f (1)1当x1 0，x2 0，x1x2 1时，f (x1x2) f (x1)f (x2) (1)求f (0)；(2)求f (x)最大值；(3)当x0，1时，4f (x)24(2a)f (x)54a，求实数a 的取值范围解：(1)因为f (x) 0，所以f (0) 0取x1x20代入f (x1x2) f (x1)f (x2)得f (0) 0，因此f (0)0(2)设x1，x20，1，xx2x1>0，则x0，1，所以f (x) 0yf (x2)f (x1)f (x1x) f (x1) f (x1 )f (x) f (x1)f (x) 0所以函数f (x)在0，1上不是减函数

18、，f (x)最大值是f (1)1(3)当x0，1时，f (x) 0，1若f (x)1，则44(2a)54a1，不等式4f (x)24(2a)f (x)54a成立若f (x) 0，1)，分离参数a1f (x) 因为1f (x) 21，当f (x)时等号成立所以实数a的取值范围是(¥，1备选：1(12分，重庆)已知定义域为R的函数f (x)满足f (f (x)x2x)f (x)x2x(1)若f (2)3，求f (1)；(2)求f (0)；(3)设有且仅有一个实数x0，使得f (x0)x0，求函数f (x)的解析表达式2(12分)已知函数f (x)满足f (xy)f (y)(x2y1)x，且f (1)0(1)求f (0)的值；(2)当x1，x2(0，)时， f (x1)2<logax2，求a的取值范围3(12分)已知偶函数f (x)满足f (xy)f (x)f (