1、抽象函數專題幾類抽象函數模型抽象函數模型適用模型的初等函數f (xy)f (x)f (y)正比例函數f (x)kx(k0)f (xy)f (x)f (y)或f ()冪函數f (x)xnf (xy)f (x)f (y)或f (xy)指數函數f (x)ax(a>0且a1)f (xy)f (x)f (y)或f ()f (x)f (y)對數函數f (x)logax(a>0且a1)f (xT)f (x)正余弦函數f (x)sinx，f (x)cosxf (xy)正切函數f (x)tanx練習題1定義域為(0，¥)的函數f (x)滿足f (xy)f (x)f (y)，若f (4)2，

2、則f ()的值為_答案：解：因為f (4)f (2)f (2)，f (2)f ()f ()，所以f (4)4 f ()，f ()2函數f (x)滿足f (xy2)f (x)2f (y)2且f (1)0，則f (2018)的值為_答案：1009解：f (0)0，f (1)，f (x1)f (x)，f (2018)f (1)2017×10093(1)函數f (x)滿足f (xy)f (x)f (y)x y1，若f (1)1，則f (8)A1B1C19D43答案：D解：因為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)函數f (x)滿足f (xy)f (x)f (y)xy1，若f (1)1，則f (8)A1B1C19D43答案：C 解：因為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定義在R上的函數f (x)滿足f (x1x2)f (x1)f (x2)1，則下列說法正確的是Af (x)為奇函數Bf (x)為偶函數Cf (x)1為奇函數Df (x)1為偶函數答案：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)圖象關于點(0，1)對稱，所以f (x)1為奇函數5設f (x)是定義在(0，)上的單調增函數，滿足f (xy)f (x)f (y)，f (3)1，當f (x)f (x8)2時x的取值范圍是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)，因為f (x) 是定義在(0，)上的增函數，所以有解得8<x9.6定義在0，1上的函數f (x)滿足f (0)0，f (x)f (1x)2，f () f (x)，當0 x1< x21時，f (x1)f (x2)，則f ()的值為 答案： 7(1)已知函數f (x)滿足2xf (x)3f (x)x10，求f (x)的表達式解：因為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相減得4x2f (x)9f (x)2 x22 x3x30，所以f (x)(2)設函數f (x)滿足f (x)2f ()x(x0)，求證：|f (x)|證明：因為f (x)2f ()x ，所以f ()2f (x) ×2得2f ()4f (x)得f (x)， |f (x)|8(12分)定義在R上的單調函數f (x)滿足f (xy)f (x)f (y)，設f (3)log23(1)判斷函數的奇偶性；(2)若f (k3x)f (3x9x4)<0，求實數k的取值范圍解：(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)為奇函數(2)奇函數，所以， 是定義在上的單調函數，所以函數在上的單調遞增函數，奇函數，不等式等價于，因此，即，因為，當取等號，所以實數的取值范圍是9(12分)已知定義在R上的函數f (x)滿足f (x)f (y)f (xy)，當>0時，f (x)<0，且f (1)(1)判斷f (x)為奇偶性；(2)求證：f (x)在R上是減函數； (3)求f (x)在3，6上的最大值與最小值解：(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)為奇函數(2)設x1，x2R，xx2x1>0，那么yf (x2)f (x1)f (x2)f (x1)f (x2x1)f (x)因為x >0，所以y<0，所以f (x)在R上是減函數(3)因為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上的最大值為f (3)2，最小值為f (6)4 10(12分)已知定義在區間(0，)上的函數f (x)滿足f ()f (x2)f (x1)，且當x>1

9、時，f (x)<0(1)證明：f (x)為單調遞減函數(2)若f (3)1，求f (x)在2，9上的最小值解：(1)設x1，x2(0，)，xx2x1>0，那么yf (x2)f (x1)f ()因為當>1時，f (x)<0，>1，所以f ()<0，y>0，所以f (x)為單調遞減函數(2)因為f (x)在(0，)上是單調遞減函數，所以f (x)在2，9上的最小值為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上的最小值為211(12分)定義域為(，0)(0，)的函數f

10、(x)滿足f (x)f (y)f (xy)(1)求證：f ()f (x)；(2)求證：f (x)為偶函數；(3)當>1時，f (x)>0，求證：f (x)在(，0)上單調遞減解：(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)為偶函數(3)解法1：設x1，x2(0，)，

11、xx2x1>0，那么yf (x2)f (x1)f (x2)f ()f ()因為>1，所以f ()>0，y>0，所以f (x)在(0，)上單調遞增由(2)知f (x)為偶函數，所以f (x)在(，0)上單調遞減解法2：設x1，x2(，0)，xx2x1>0，那么yf (x2)f (x1)f (x2)f ()f ()f ()因為>1，所以f ()>0，y<0，所以f (x)在(，0)上單調遞減12(12分)設定義在R上的函數yf (x)滿足f (ab)f (a)·f (b)當x>0時，f (x)>1，且f (0)0(1)求證：f

12、(0)1；(2)求證：f (x)>0；(3)求證：f (x)是R上的增函數；(4)若f (x)·f (2xx2)>1，求x的取值范圍解：(1)取ab0代入f (ab)f (a)·f (b)，得f (0)2f (0)，因為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)當x>0時，f (x)>1；x0時，f (x)1；當x<0時，x>0，f (x)>1，所以f (x)(0，1)綜上，f (x)>0(3)設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 因為 xx2x1>0，所以f (x)>1，故y>0，f (x)是R上的增函數(4)f (x)·f (2xx2)f (x2xx2)f (3xx2)，1f (0)，所以不等式f (x)·f (2xx2)>1可化為f (3xx2)> f (0)由(2)知3xx2>0，得x的取值范圍為(0，3)13(12分)已知定義在R上的不恒為零的函數f (x)滿足 f (xy)y f (x)x f (y)(1)判斷f (x)的奇

14、偶性；(2)若f (2)2，設an ，bn ，求證數列an 為等差數列，數列bn 為等比數列解：(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)為奇函數(2)因為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，所以數列an 為等差數列a1 1，所以 an a1(n1)×1n，所以f (2n)2n因為2，所

15、以數列bn 為等比數列14(12分)定義在(0，¥)上的函數f (x)滿足：對任意實數m，f (xm)mf (x)；f (2)1(1)求證：f (xy)f (x)f (y)；(2)求證：f (x)是(0，¥)上的單調增函數；(3)若f (x)f (x3)2，求x的取值范圍解：(1)因為x，y均為正數，根據指數函數性質可知，總有實數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 ()設x1，x2(0，¥)，xx2x1>0，那么yf (x2)f (x1)f (x2)f ()f ()因為>1，根據指數函數性質可知，總有正實數r，使得 2r，所以yf (2r)r>0因此f (x)是(0，¥)上的單調增函數(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的取值范圍為(3，4 15(12分)定義在0，1上的函數f (x)滿足f

17、 (x) 0，f (1)1當x1 0，x2 0，x1x2 1時，f (x1x2) f (x1)f (x2) (1)求f (0)；(2)求f (x)最大值；(3)當x0，1時，4f (x)24(2a)f (x)54a，求實數a 的取值范圍解：(1)因為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)設x1，x20，1，xx2x1>0，則x0，1，所以f (x) 0yf (x2)f (x1)f (x1x) f (x1) f (x1 )f (x) f (x1)f (x) 0所以函數f (x)在0，1上不是減函數

18、，f (x)最大值是f (1)1(3)當x0，1時，f (x) 0，1若f (x)1，則44(2a)54a1，不等式4f (x)24(2a)f (x)54a成立若f (x) 0，1)，分離參數a1f (x) 因為1f (x) 21，當f (x)時等號成立所以實數a的取值范圍是(¥，1備選：1(12分，重慶)已知定義域為R的函數f (x)滿足f (f (x)x2x)f (x)x2x(1)若f (2)3，求f (1)；(2)求f (0)；(3)設有且僅有一個實數x0，使得f (x0)x0，求函數f (x)的解析表達式2(12分)已知函數f (x)滿足f (xy)f (y)(x2y1)x，且f (1)0(1)求f (0)的值；(2)當x1，x2(0，)時， f (x1)2<logax2，求a的取值范圍3(12分)已知偶函數f (x)滿足f (xy)f (x)f (