1、抽象函数专题 第 1 页(共 9 页)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(k 0)f (xy)f ( x)f (y)或 f ( )xy f（x）f（y）幂函数 f (x)x nf (xy)f (x )f (y)或 f (xy )f（x）f（y） 指数函数 f (x)a x(a0 且 a1)f (xy)f (x) f (y) 或 f ( )f (x)f (y)xy对数函数 f (x)log ax(a0 且 a1)f (xT)f (x) 正余弦函数 f (x)sinx ，f (x)cos xf (xy) f（
2、x） f（y）1 f（x）f（y） 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，则 f ( )的值为2_答案： 12解：因为 f (4)f (2)f (2) ，f (2)f ( )f ( )，2 2所以 f (4)4 f ( )，f ( ) 2 2122函数 f (x)满足 f (xy 2)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 100912 12 123(1)函数 f
3、 (x)满足 f (xy )f (x)f (y)x y1，若 f (1)1，则 f (8)A1 B1 C19 D43答案：D解：因为 f (1)1，y 1 代入 f (xy)f (x)f (y)x y1，得f (x 1)f (x) x 2，因此：f (2)f (1) 3f (3)f (2) 4f (8) f (7)9抽象函数专题 第 2 页(共 9 页)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)A1 B1 C19 D43答案：C 解：因为 f (1)1，y 1 代入 f (xy)f (x)f (y)xy 1，
4、得f (x 1)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 (x x)f (x) f (x )x 2 1f (0)f (x) f (x )x 21f (x)f (x) x 224定义在 R 上的函数 f (x)满足 f (x1x 2)f (x 1)f (x 2) 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：
5、C 解：x1x 20 代入 f (x1x 2)f ( x1)f (x 2)1，得 f (0)1x1x，x 2x 代入 f (x1x 2)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，) B(8，9 C8，9 D(0 ，8)抽象函数专题 第 3 页(共 9 页)答案：B解：211f (3)f (3)f (9)，由 f (x)f (x8)2，可得 fx(x8)f
6、 (9)，因为 f (x) 是定义在(0 ，)上的增函数，所以有Error!解得 80 时 ， f (x)0，那么yf (x 2)f (x 1)f (x 2)f (x 1)f (x2x 1)f (x )因为x 0，所以y 1 时，f (x)x2x10，那么yf (x 2)f (x 1)f ( )x2x1抽象函数专题 第 5 页(共 9 页)因为当 1 时 ， f (x)1，所以 f ( )0，所以 f (x)为单调递减函数x2x1 x2x1(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 (x 2)得，f ( )
7、f (9)f (3)，而 f (3)1，所以 f (9)2x1x2 93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 (x)满足 f (x) f (y) f (xy)(1)求证：f ( )f (x)；1x(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)1x 1x 1x(2)取 y1 代入 f
8、(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，x 2 (0，)，xx 2x 10，那么yf (x 2)f (x 1)f (x 2) f ( )f ( )1x1 x2x1因为 1，所以 f ( )0，y 0，所以 f (x)在(0，)上单调递增x2x1 x2x1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，x 2 (，0)，xx 2x 10
9、，那么yf (x 2)f (x 1)f (x 2) f ( )f ( )f ( )1x1 x2x1 x1x2因为 1，所以 f ( )0，y 0 时，f (x)1，且 f (0)0(1)求证：f (0)1；(2)求证：f (x)0；(3)求证：f (x) 是 R 上的增函数；抽象函数专题 第 6 页(共 9 页)(4)若 f (x)f (2xx 2)1，求 x 的取值范围解：(1)取 ab0 代入 f (ab) f (a)f (b)，得 f (0)2 f (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
10、 )，即 f (x) 1f（ x）当 x0 时，f ( x)1；x0 时，f (x) 1；当 x0，f (x )1，所以 f (x) (0，1)1f（ x）综上，f (x)0(3)设 x1，x 2 R，xx 2x 10，那么yf (x 2)f (x 1)f (x 1x) f (x1)f (x 1)f (x) f (x 1)f (x 1)f (x)1 因为 xx 2x 10，所以 f (x)1，故y0，f (x)是 R 上的增函数(4)f (x)f (2xx 2)f (x2xx 2)f (3xx 2)，1f (0)，所以不等式 f (x)f (2xx 2)1 可化为 f (3xx 2) f (0
11、)由(2)知 3xx 20，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)的奇偶性；(2)若 f (2)2， ，设 an ，b n ，求证数列 an 为等 差 数 列 ，*Nf（ 2n）2n f（ 2n）n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)
12、 ，所以 f (x)为奇函数(2)因为 f (2n1 )f (22 n)2 f (2n) 2n f (2)，所以 f (2n1 ) 2 f (2n) 2n1 同除以2n 1，得 1，即 an1 a n1，所以数列 an 为等 差 数 列 f（2n+1）2n+1 f（2n）2na1 1，所以 an a 1 (n1) 1n ，所以 f (2n)2 nf（2）2因为 2，所以数列b n 为等比数列bn+1bn14(12 分)定义在(0，) 上的函数 f (x)满足：对任意实数 m，f (x m)mf (x )；f (2)1抽象函数专题 第 7 页(共 9 页)(1)求证：f (xy)f (x )f
13、(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 x2 m，y2 n于是 f (xy)f (2 m2n)f (2 m+n)(m +n)f (2)m+n而 mm f (2) f (2 m) f (x) ， nn f (2) f (2 n) f (y)，所以 f (xy)f (x)+f (y) (2)取 xy1 代入 f (xy)f (x)f (y)，得 f (1) 0取 y 代入 f (1)f (x)f ( )，得f (x) f ( )1x 1x 1x
14、设 x1，x 2 (0，)，xx 2x 10，那么yf (x 2)f (x 1)f (x 2)f ( )f ( )1x1 x2x1因为 1，根据指数函数性质可知，总有正实数 r，使得 2 r，所以yf (2r)x2x1 x2x1r 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 (x 3)2 即 f (x23 x)f (4)由Error! 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 (x) 0，f (1)1当 x1 0，x 2 0，
15、x 1x 2 1 时，f (x1x 2) f (x1)f (x 2) (1)求 f (0)；(2)求 f (x)最大值；(3)当 x 0，1时，4f (x )24(2a) f (x)54a ，求实数 a 的取值范 围 0解：(1)因为 f (x) 0，所以 f (0) 0取 x1x 20 代入 f (x1x 2) f (x1)f (x 2)得 f (0) 0，因此 f (0)0(2)设 x1，x 2 0，1，x x2x 10，则x0，1，所以 f (x) 0yf (x 2)f (x 1)f (x 1x) f (x1) f (x1 )f (x ) f (x 1)f (x) 0所以函数 f (x)
16、在0，1上不是减函数，f (x)最大值是 f (1)1(3)当 x 0，1时，f (x ) 0，1若 f (x)1，则 44(2 a)54a 1 ，不等式 4f (x)24(2 a)f (x)54a0抽象函数专题 第 8 页(共 9 页)成立0若 f (x) 0，1)，分离参数 a1f (x) 141 f (x)因为 1f (x) 2 1，当 f (x) 时等号成立141 f (x) 12所以实数 a 的取值范围是( ，1备选：抽象函数专题 第 9 页(共 9 页)1(12 分，重庆)已知定义域为 R 的函数 f (x)满足 f (f (x)x 2x)f (x)x 2x (1)若 f (2)3
17、，求 f (1)；(2)求 f (0)；(3)设有且仅有一个实数 x0，使得 f (x0)x 0，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，x 2(0， )时， f (x1)21 时，f (x )0，f (2) 1(1)求证：f (x) 在 (0，)上是增函数；(2)解不等式 f (2x21) 时，f (x)0求12证：f (x)是单调递增函数5(12 分)已知函数 f (x)满足 f (xy)f (x)f (y)，且 f (x)0，当 x1 时，f (x)1试判断 f (x)在(0， )上的单调性6(12 分)已知函数 f (x)的定义域关于原点对称，且满足 f (xy) ，存在正常数 a，f (x)f (y) 1f (x) f (y)使 f (a)1求证： f (x)是奇函数
