115鳳山高中

填充第 6 題：

\(\displaystyle f\left(x\right) +2 f\left(\frac{x-3}{x+1}\right) = x\)   ..... 第一式



令 \(\displaystyle t=\frac{x-3}{x+1}\)，則 \(\displaystyle x=\frac{t+3}{1-t}\)，代入第一式，

得 \(\displaystyle f\left(\frac{t+3}{1-t}\right) + 2 f\left(t\right) = \frac{t+3}{1-t}\)

即  \(\displaystyle f\left(\frac{x+3}{1-x}\right) + 2 f\left(x\right) = \frac{x+3}{1-x}\)   ......第二式



將第一式中的自變數 \(x\) 以 \(\displaystyle \frac{x-3}{x+1}\) 代入，

得 \(\displaystyle f\left(\frac{x-3}{x+1}\right) + 2 f\left(\frac{3+x}{1-x}\right) = \frac{x-3}{x+1}\) ..... 第三式



由第一式\(\times1\) + 第二式\(\times4\) - 第三式\(\times2\)，得

\(\displaystyle 9 f\left(x\right) = x + 4 \cdot \frac{x+3}{1-x} - 2 \cdot \frac{x-3}{x+1}\)

\(\displaystyle \Rightarrow f\left(x\right) = \frac{1}{9}\left(x + 4 \cdot \frac{x+3}{1-x} - 2 \cdot \frac{x-3}{x+1}\right)\)

填充第4題：

設 \(\displaystyle A_k\left(x_k,y_k\right)\)，\(\displaystyle \overline{A_kF}=r_k\) 且 \(\displaystyle \angle A_kFA_0=\theta_k\)，則

\(\displaystyle \left\{\begin{matrix}x_k=3+r_k\cos\theta_k\\y_k=r_k\sin\theta_k\\\end{matrix}\right.\)，代入橢圓方程式：\(\displaystyle \frac{x^2}{36}+\frac{y^2}{27}=1\)，

得 \(\displaystyle \frac{\left(3+r_k\cos\theta_k\right)^2}{36}+\frac{\left(r_k\sin\theta_k\right)^2}{27}=1\)

\(\displaystyle \Rightarrow3\left(3+r_k\cos\theta_k\right)^2+4\left(r_k\sin\theta_k\right)^2=108\)

\(\displaystyle \Rightarrow\left(3\cos^2\theta_k+4\sin^2\theta_k\right){r_k}^2+\left(18\cos\theta_k\right)r_k-81=0\)

\(\displaystyle \Rightarrow\left(4-\cos^2\theta_k\right){r_k}^2+\left(18\cos\theta_k\right)r_k-81=0 \)



利用十字交乘法：  \(\displaystyle \underline{\begin{matrix}\left(2-\cos\theta_k\right)&9\\\left(2+\cos\theta_k\right)&-9\\\end{matrix}}\)

得 \(\displaystyle r_k=\frac{-9}{2-\cos\theta_k}\)（負不合）或 \(\displaystyle r_k=\frac{9}{2+\cos\theta_k}\)。




因此 \(\displaystyle x_k=3+r_k\cos\theta_k=3+\frac{9\cos\theta_k}{2+\cos\theta_k}\)，進而得

\(\displaystyle d_k=12-x_k=12-\left(3+\frac{9\cos\theta_k}{2+\cos\theta_k}\right)=\frac{18}{2+\cos\theta_k}\)。




故，

\(\displaystyle \sum_{k=1}^{11}{\frac{1}{d_k}}=\sum_{k=1}^{11}{\frac{2+\cos\theta_k}{18}}=\frac{2}{18}\times12+\frac{1}{18}\left(\cos0^{\circ}+\cos30^{\circ}+\cos60^{\circ}+\cdots+\cos330^{\circ}\right)=\frac{4}{3}\)。