請教兩題
第一題 不等式\(\displaystyle {\rm{1 + }}\lg (3x^2 + x - 2) - \lg (x + 1) < |x - 2| \)ANS:\(\displaystyle (\frac{2}{3},1) \cup (4,\infty ) \)
第二題
函數\(f\)定義在整數集上，且滿足
\(
\begin{array}{l}
\displaystyle f(n) = \left\{ \begin{array}{l}
\displaystyle n - 2,n \ge 2009 \\
\displaystyle f[f(n + 3)],n < 2009 \\
\end{array} \right. \\
\end{array}
\)
則\(\displaystyle f(9)=？\)
ANS:2007
回復 1# eyeready 的帖子
第一題：由 \(\log\) 的定義有效，可知 「\(3x^2+x-2>0\) 且 \(x+1>0\)」，得 \(\displaystyle x>\frac{2}{3}\)
再來處理題目，
\(1+\log\left(3x^2+x-2\right)-\log\left(x+1\right)<\left|x-2\right|\)
\(\displaystyle \Rightarrow \log\frac{3x^2+x-2}{x+1}<\left|x-2\right|-1\)
\(\displaystyle \Rightarrow \log\left(3x-2\right)<\left|x-2\right|-1\)
當 \(x\leq2\) 時， \(y=\log\left(3x-2\right)\) 與 \(y=1-x\) 交於 \(\left(1,0\right)\)
當 \(x\geq2\) 時， \(y=\log\left(3x-2\right)\) 與 \(y=x-3\) 交於 \(\left(4,1\right)\)
可以畫出如下圖
[attach]4197[/attach]
可得解為 \(\displaystyle \frac{2}{3}<x<1\) 或 \(x>4\)
回復 1# eyeready 的帖子
第二題：為了方便起見，定義 \(f^1\left(x\right)=f\left(x\right)\)，\(f^n\left(x\right)=f\left(f^{n-1}\left(x\right)\right), \forall n=2,3,\cdots\)
則 \(f\left(2009\right) = 2007\) 且
\(f\left(f\left(2009\right)\right) = f\left(2007\right) = f\left(f\left(2010\right)\right) = f\left(2008\right) = f\left(f\left(2011\right)\right) = f\left(2009\right)\)
合併以上二者，可得 \(f^n\left(2009\right) = f\left(2009\right)=2007, \forall n\in\mathbb{N}\)
\(f\left(9\right) = f\left(f\left(12\right)\right)=f\left(f\left(f\left(15\right)\right)\right)=\cdots= f^{667}\left(2010\right)\)
\( = f^{666}\left(2010-2\right) = f^{666}\left(2008\right) = f^{667}\left(2011\right) =f^{666}\left(2009\right) = 2007\)
註：寫完回首，發現結論是 \(f\left(1\right)=f\left(2\right)=f\left(3\right)=\cdots=f(2007)=f(2008)=f(2009)=2007\)
回復 3# weiye 的帖子
謝謝學長的幫忙！PS:祝學長有個愉快的假期^_^
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