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

qq1.png (11.66 KB)

2017-6-23 11:22

可得解為 \(\displaystyle \frac{2}{3}<x<1\) 或 \(x>4\)

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 \(f\left(1\right)=f\left(2\right)=f\left(3\right)=\cdots=f(2007)=f(2008)=f(2009)=2007\)