课程链接：目标2017高中数学联赛基础班-2（赵胤授课）

1. 解函数方程: $$f(x) + f\left({1\over x}\right)\lg x = b^2,$$ 其中 $x\in\mathbf{R^+}$, $b > 0$, $b\ne1$.

解答: $$\begin{cases}f(x) + f\left({1\over x}\right)\lg x = b^2\\ f({1\over x}) + f(x)\lg{1\over x} = b^2\end{cases}\Rightarrow \begin{cases} f(x) + f\left({1\over x}\right)\lg x = b^2\\ f({1\over x}) - f(x)\lg x = b^2\end{cases}$$ $$\Rightarrow f(x) = {b^2(1 - \lg x)\over 1+\lg^2 x}.$$ 注:

事实上已知条件中 $b > 0$ 且 $b \ne 1$ 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: $$f(x) + f\left({x - 1 \over x}\right) = 1+x,$$ 其中 $x\ne0$, $x\ne1$.

解答: $$\phi(x) = {x-1\over x}\Rightarrow \phi^{(2)}(x) = {1\over 1-x} \Rightarrow \phi^{(3)}(x) = x,$$ $$\Rightarrow \begin{cases} f(x) + f\left(\phi(x)\right) = 1+x\\ f\left(\phi(x)\right) + f\left(\phi^{(2)}(x)\right) = 1+\phi(x)\\ f\left(\phi^{(2)}(x)\right) + f(x) = 1 + \phi^{(2)}(x) \end{cases}$$ $$\Rightarrow \begin{cases} f(x) + f\left({x-1 \over x}\right) = 1+x\\ f\left({x-1 \over x}\right) + f\left({1\over 1-x}\right) = {2x - 1 \over x}\\ f\left({1\over 1-x}\right) + f(x) = {2-x \over 1-x} \end{cases}$$ $$\Rightarrow f(x) = {1+x^2 - x^3 \over 2x(1-x)}.$$

3. 函数 $f: \mathbf{R}\rightarrow \mathbf{R}$满足 $$f(xy) = {f(x) + f(y) \over x + y},$$ 其中 $x, y\in\mathbf{R}$, $x+y\ne0$. 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 $x\ne 0, -1$ 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, $f(x) = 0$ ($x\in\mathbf{R}$).

4. 解函数方程: $$f(x+y) + f(x - y) = 2f(x)\cos y.$$ 解答:

令 $x = {\pi\over2} + t$, $y = {\pi\over2}$ 得 $$f(t) + f(\pi + t) = 0,$$ 令 $x = 0$, $y = t$ 得 $$f(t) + f(-t) = 2f(0)\cos t,$$ 令 $x = {\pi\over2}$, $y = {\pi \over 2} + t$ 得 $$f(t + \pi) + f(-t) = -2f\left({\pi\over2}\right)\sin t.$$ 由此可得 $$f(t) = f(0)\cos t + f\left({\pi\over2}\right)\sin t$$ $$\Rightarrow f(x) = a\cos x + b\sin x$$ 其中 $a = f(0)$, $b = f\left({\pi\over2}\right)$.

5. 设 $f(x)$ 是定义在 $(0, +\infty)$ 上的增函数, 且 $$f\left({x\over y}\right) = f(x) - f(y),$$ 求证: $$f\left(x^n\right) = nf(x),\ n\in\mathbf{N^*}.$$ 解答:

用数学归纳法证明之.

当 $n = 1$ 时显然有 $f(x^1) = 1\cdot f(x)$.

假设 $n = k$ 时 $f\left(x^k\right) = k\cdot f(x)$ 成立, 则当 $n = k+1$ 时 $$f\left(x^{k+1}\right) = f\left({x^k \over {1\over x}}\right) = f\left(x^k\right) - f\left({1\over x}\right)$$ $$= kf(x) - f(1) + f(x) = (k+1)\cdot f(x),$$ 其中 $f(1) = 0$ 可令 $x = y$ 得出.

综上, $f\left(x^n\right) = n\cdot f(x)$.

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 $$f\left(f(m) + f(n)\right) = m + n.$$ 求 $f(2016)$ 的值.

解答:

令 $f(m) = n$ 可得 $$f(2n) = f\left(f(m) + f(m)\right) = m + m = 2m,$$ 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 $$f(1) = 1,\ f(2) = 2,\ f(3) = f(1 + 2) = f\left(f(1) + f(2)\right) = 3,\ \cdots\cdots,\ f(n) = n.$$ 若 $r > 1$, 不妨设 $r = a+1$, $f(a) = b$, (其中 $a, b\in\mathbf{N^*}$) $$\Rightarrow \begin{cases}f(2b) = f\left(f(a) + f(a)\right) = 2a\\ 2b + 2r = f\left(f(2b) + f(2r)\right) = f(2a + 2) = f(2r) = 2\end{cases}$$ $$\Rightarrow r + b = 1 \Rightarrow r < 1.$$ 矛盾.

因此必有 $f(n) = n$ 成立.

由此可得 $f(2016) = 2016$.

腾讯课堂目标2017高中数学联赛基础班-2作业题解答-7