1. 解函数方程: $$f(x) + fleft({1over x} ight)lg x = b^2,$$ 其中 $xinmathbf{R^+}$, $b > 0$, $b e1$.
解答: $$egin{cases}f(x) + fleft({1over x} ight)lg x = b^2\ f({1over x}) + f(x)lg{1over x} = b^2end{cases}Rightarrow egin{cases} f(x) + fleft({1over x} ight)lg x = b^2\ f({1over x}) - f(x)lg x = b^2end{cases}$$ $$Rightarrow f(x) = {b^2(1 - lg x)over 1+lg^2 x}.$$ 注:
事实上已知条件中 $b > 0$ 且 $b e 1$ 乃冗余条件, 若依该已知则可将原题中 $b^2$ 改为 $b^x$ 较妥. 解决方法完全相同.
2. 解函数方程: $$f(x) + fleft({x - 1 over x} ight) = 1+x,$$ 其中 $x e0$, $x e1$.
解答: $$phi(x) = {x-1over x}Rightarrow phi^{(2)}(x) = {1over 1-x} Rightarrow phi^{(3)}(x) = x,$$ $$Rightarrow egin{cases} f(x) + fleft(phi(x) ight) = 1+x\ fleft(phi(x) ight) + fleft(phi^{(2)}(x) ight) = 1+phi(x)\ fleft(phi^{(2)}(x) ight) + f(x) = 1 + phi^{(2)}(x) end{cases}$$ $$Rightarrow egin{cases} f(x) + fleft({x-1 over x} ight) = 1+x\ fleft({x-1 over x} ight) + fleft({1over 1-x} ight) = {2x - 1 over x}\ fleft({1over 1-x} ight) + f(x) = {2-x over 1-x} end{cases}$$ $$Rightarrow f(x) = {1+x^2 - x^3 over 2x(1-x)}.$$
3. 函数 $f: mathbf{R} ightarrow mathbf{R}$满足 $$f(xy) = {f(x) + f(y) over x + y},$$ 其中 $x, yinmathbf{R}$, $x+y e0$. 求 $f(x)$.
解答:
令 $x = 0$, $y = 1$ 得 $$f(0) = f(0) + f(1) Rightarrow f(1) = 0.$$ 令 $y = 1$ 得 $$f(x) = {f(x) over x+1} Rightarrow xf(x) = 0 Rightarrow f(x) = 0.$$ 即当 $x e 0, -1$ 时, $f(x) = 0$.
令 $x = 2$, $y = 0$ 得 $$f(0) = {f(2) + f(0) over 2} Rightarrow f(0) = f(2) = 0,$$ 令 $x = -1$, $y = 0$ 得 $$f(0) = {f(-1) + f(0) over -1} Rightarrow f(-1) = 0.$$ 综上, $f(x) = 0$ ($xinmathbf{R}$).
4. 解函数方程: $$f(x+y) + f(x - y) = 2f(x)cos y.$$ 解答:
令 $x = {piover2} + t$, $y = {piover2}$ 得 $$f(t) + f(pi + t) = 0,$$ 令 $x = 0$, $y = t$ 得 $$f(t) + f(-t) = 2f(0)cos t,$$ 令 $x = {piover2}$, $y = {pi over 2} + t$ 得 $$f(t + pi) + f(-t) = -2fleft({piover2} ight)sin t.$$ 由此可得 $$f(t) = f(0)cos t + fleft({piover2} ight)sin t$$ $$Rightarrow f(x) = acos x + bsin x$$ 其中 $a = f(0)$, $b = fleft({piover2} ight)$.
5. 设 $f(x)$ 是定义在 $(0, +infty)$ 上的增函数, 且 $$fleft({xover y} ight) = f(x) - f(y),$$ 求证: $$fleft(x^n ight) = nf(x), ninmathbf{N^*}.$$ 解答:
用数学归纳法证明之.
当 $n = 1$ 时显然有 $f(x^1) = 1cdot f(x)$.
假设 $n = k$ 时 $fleft(x^k ight) = kcdot f(x)$ 成立, 则当 $n = k+1$ 时 $$fleft(x^{k+1} ight) = fleft({x^k over {1over x}} ight) = fleft(x^k ight) - fleft({1over x} ight)$$ $$= kf(x) - f(1) + f(x) = (k+1)cdot f(x),$$ 其中 $f(1) = 0$ 可令 $x = y$ 得出.
综上, $fleft(x^n ight) = ncdot f(x)$.
6. 设 $f(n)$ 是定义在正整数集上且取正整数值的函数, 对所有正整数 $m, n$, 都有 $$fleft(f(m) + f(n) ight) = m + n.$$ 求 $f(2016)$ 的值.
解答:
令 $f(m) = n$ 可得 $$f(2n) = fleft(f(m) + f(m) ight) = m + m = 2m,$$ 令上式中 $m = 1$, $n = r$, 即 $f(1) = r$ 可得 $$f(2r) = 2.$$ 若 $r = 1$, 则可顺次得出 $$f(1) = 1, f(2) = 2, f(3) = f(1 + 2) = fleft(f(1) + f(2) ight) = 3, cdotscdots, f(n) = n.$$ 若 $r > 1$, 不妨设 $r = a+1$, $f(a) = b$, (其中 $a, binmathbf{N^*}$) $$Rightarrow egin{cases}f(2b) = fleft(f(a) + f(a) ight) = 2a\ 2b + 2r = fleft(f(2b) + f(2r) ight) = f(2a + 2) = f(2r) = 2end{cases}$$ $$Rightarrow r + b = 1 Rightarrow r < 1.$$ 矛盾.
因此必有 $f(n) = n$ 成立.
由此可得 $f(2016) = 2016$.