回復 22# beaglewu 的帖子
第1題
設四次多項式函數\(f(x)=(x^2+2x+3)(x^2+2x-2)+5x^2+10x+4\),當\(x=\alpha\)時,\(f(x)\)有最小值\(m\),則數對\((\alpha,m)=\) 。
[解答]
\(x\)應為實數
\(\begin{align}
& f\left( x \right)={{\left( {{x}^{2}}+2x \right)}^{2}}+\left( {{x}^{2}}+2x \right)-6+5\left( {{x}^{2}}+2x \right)+4 \\
& ={{\left( {{x}^{2}}+2x \right)}^{2}}+6\left( {{x}^{2}}+2x \right)-2 \\
& ={{\left[ \left( {{x}^{2}}+2x \right)+3 \right]}^{2}}-11 \\
& ={{\left[ {{\left( x+1 \right)}^{2}}+2 \right]}^{2}}-11 \\
& ...... \\
\end{align}\)
第4題
設二次多程式\((m^2+1)x^2-4mx+2=0\)有兩正根\(\alpha\)與\(\beta\),且\(2\alpha \beta=\alpha-3\beta\),則\(m=\) 。
[解答]
\(\begin{align}
& \alpha \beta =\frac{2}{{{m}^{2}}+1} \\
& \alpha =\frac{2m+\sqrt{2{{m}^{2}}-2}}{{{m}^{2}}+1},\beta =\frac{2m-\sqrt{2{{m}^{2}}-2}}{{{m}^{2}}+1} \\
& 2\alpha \beta =\alpha -3\beta \\
& \frac{4}{{{m}^{2}}+1}=\frac{4\sqrt{2{{m}^{2}}-2}-4m}{{{m}^{2}}+1} \\
& ...... \\
\end{align}\)
第5題
\((1+2x)^n\)展開式中\(x^3\)的係數為\(a_n(n\ge 3)\),則\(\displaystyle \sum_{n=3}^{100}\frac{1}{a_n}=\) 。
[解答]
\(\begin{align}
& {{a}_{n}}=C_{3}^{n}\times {{2}^{3}}=\frac{8n\left( n-1 \right)\left( n-2 \right)}{6} \\
& \frac{1}{{{a}_{n}}}=\frac{3}{4}\times \frac{1}{n\left( n-1 \right)\left( n-2 \right)}=\frac{3}{8}\times \left[ \frac{1}{\left( n-2 \right)\left( n-1 \right)}-\frac{1}{\left( n-1 \right)n} \right] \\
& ...... \\
\end{align}\)