10.
已知\( F \)為拋物線\( \Gamma \)：\( x^2=4y \)的焦點，\(A\)、\(B\)為\( \Gamma \)上焦弦，滿足\( AF=\lambda FB \)，過\(A\)、\(B\)分別做切線交於\(M\)點，
(1)試證：\( FM⊥AB \)
(2)請問\( \lambda=\)？時，\( \Delta ABM \)面積有最小值。
[解答]
\(\displaystyle A\left( a,\frac{{{a}^{2}}}{4} \right),B\left( b,\frac{{{b}^{2}}}{4} \right),a>0>b\)
易知\(ab=-4\)
作BC垂直準線於C，作AD垂直準線於D
則\(\begin{align}
  & \Delta ABM=\frac{1}{2}ABCD=\frac{1}{2}\times \frac{1}{2}\times \left( \frac{{{a}^{2}}}{4}+1+\frac{{{b}^{2}}}{4}+1 \right)\left( a-b \right) \\
& =\frac{1}{2}\times \frac{1}{2}\times \left( \frac{{{a}^{2}}}{4}+1+\frac{{{\left( -\frac{4}{a} \right)}^{2}}}{4}+1 \right)\left( a+\frac{4}{a} \right) \\
& =\frac{{{\left( a+\frac{4}{a} \right)}^{3}}}{16} \\
& \ge 4 \\
\end{align}\)
等號成立於\(a=2,b=-2,\lambda =1\)時

\(M\)是直線\(MA\)和\(MB\)的交點
用點斜式把兩條直線方程式找出來，解聯立即知\(M\)在準線上

2.
設數列\( \langle\; a_n \rangle\; \)的前\(n\)項和\( \displaystyle S_n=\frac{4}{3}a_n-\frac{1}{3}\cdot 2^{n+1}+\frac{2}{3}\)，\(n=1,2,3,\ldots\)
(1)求首項\(a_1\)
(2)求一般項\(a_n\)
(3)設\( \displaystyle T_n=\frac{2^n}{S_n} \)，\(n=1,2,3,\ldots\)，證明：\( \displaystyle \sum_{i=1}^n T_i<\frac{3}{2} \)


2-3
\(\begin{align}
  & {{a}_{n}}={{4}^{n}}-{{2}^{n}} \\
&  \\
& {{T}_{n}}=\frac{{{2}^{n}}}{{{S}_{n}}}=\frac{{{2}^{n}}}{\sum\limits_{k=1}^{n}{\left( {{4}^{k}}-{{2}^{k}} \right)}}=\frac{{{2}^{n}}}{\frac{4}{3}\left( {{4}^{n}}-1 \right)-2\left( {{2}^{n}}-1 \right)} \\
& =\frac{3\times {{2}^{n}}}{{{4}^{n+1}}-4-3\times {{2}^{n+1}}+6} \\
& =\frac{3}{2}\times \frac{{{2}^{n}}}{2\times {{2}^{2n}}-3\times {{2}^{n}}+1} \\
& =\frac{3}{2}\times \frac{{{2}^{n}}}{\left( {{2}^{n}}-1 \right)\left( {{2}^{n+1}}-1 \right)} \\
& =\frac{3}{2}\left( \frac{1}{{{2}^{n}}-1}-\frac{1}{{{2}^{n+1}}-1} \right) \\
&  \\
& \sum\limits_{i=1}^{n}{{{T}_{i}}}=\frac{3}{2}\left( 1-\frac{1}{{{2}^{n+1}}-1} \right)<\frac{3}{2} \\
\end{align}\)

2-2
\(\begin{align}
  & {{a}_{n}}={{S}_{n}}-{{S}_{n-1}} \\
& =\frac{4}{3}{{a}_{n}}-\frac{{{2}^{n+1}}}{3}+\frac{2}{3}-\left( \frac{4}{3}{{a}_{n-1}}-\frac{{{2}^{n}}}{3}+\frac{2}{3} \right) \\
& =\frac{4}{3}{{a}_{n}}-\frac{4}{3}{{a}_{n-1}}-\frac{{{2}^{n}}}{3} \\
& {{a}_{n}}=4{{a}_{n-1}}+{{2}^{n}} \\
& {{a}_{n}}+{{2}^{n}}=4\left( {{a}_{n-1}}+{{2}^{n-1}} \right) \\
&  \\
& {{a}_{2}}+{{2}^{2}}=4\left( {{a}_{1}}+2 \right) \\
& {{a}_{3}}+{{2}^{3}}=4\left( {{a}_{2}}+{{2}^{2}} \right) \\
& : \\
& : \\
& {{a}_{n}}+{{2}^{n}}=4\left( {{a}_{n-1}}+{{2}^{n-1}} \right) \\
&  \\
& {{a}_{n}}+{{2}^{n}}={{4}^{n-1}}\times \left( 2+2 \right) \\
& {{a}_{n}}={{4}^{n}}-{{2}^{n}} \\
\end{align}\)

第8題
\(\begin{align}
  & f\left( x \right)={{x}^{2}}+ax+1 \\
& \left( 1 \right)\ 0<-\frac{a}{2}<\frac{1}{2}\quad ,\quad -\frac{{{a}^{2}}-4}{4}\ge 0 \\
& \left( 2 \right)\ -\frac{a}{2}\ge \frac{1}{2}\quad ,\quad f\left( \frac{1}{2} \right)=\frac{1}{4}+\frac{a}{2}+1\ge 0 \\
& \left( 3 \right)\ -\frac{a}{2}\le 0\quad ,\quad f\left( 0 \right)=1\ge 0 \\
\end{align}\)

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