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本帖最后由 elim 于 2020-7-23 13:05 编辑
题:设函数\(\,f:[0,1]\to\mathbb{R}\,\)满足:
\(\qquad(1)\quad f(x)\le f(y)\;(0\le x< y\le 1);\)
\(\qquad(2)\quad f(0)=0;\)
\(\qquad(3)\quad f(\frac{x}{3})=\frac{f(x)}{2};\)
\(\qquad(4)\quad f(1-x)=1-f(x),\)
\(\qquad\)求\(\,f\big(\frac{109}{2020}\big).\)
解:据\(\small(2),(4),\,f(1)=1-f(0)=1,\;f(\frac{1}{2})=1-f(\frac{1}{2})=\frac{1}{2}\)
\(\qquad\frac{1}{2}=\frac{f(1)}{2}\overset{(3)}{=}f(\frac{1}{3})=1-f(\frac{1}{3})\overset{(4)}{=}f(\frac{2}{3})\)
\(\qquad f(x)\overset{(1)}{=}\frac{1}{2}\;{\small(\frac{1}{3}\le x\le\frac{2}{3})}\overset{(3)}{\implies}f(x)=\frac{1}{2^3}\;\small(\frac{1}{3^2}\le x\le\frac{2}{3^3})\)
\(\because\quad{\small 2020< 27\times 109< 2\times 2020,}\;\therefore\;f(\frac{109}{2020})=\overset{\,}{\frac{1}{8}}.\quad\small\square\) |
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