回復 1# thankyou 的帖子
第1題:
當 \(x<9\) 時,\(\displaystyle y=\sum_{k=1}^{20} \left|x-k\right|+3x+2\) 的各折線段斜率皆為負,
當 \(x>9\) 時,\(\displaystyle y=\sum_{k=1}^{20} \left|x-k\right|+3x+2\) 的各折線段斜率皆為正,
且因為 \(\displaystyle y=\sum_{k=1}^{20} \left|x-k\right|+3x+2\) 的圖形為連續的折線段,
所以當 \(x=9\) 時,圖形有最低點,即 \(x=9\) 時,\(y\) 有最小值為 \(131\) 。
第 2 題:
因為 \(\displaystyle \lim_{x\to1}\frac{\sqrt{2x^2+a}-x+b}{\left(x-1\right)^2}\) 為定值,
所以 \(\displaystyle \lim_{x\to1}\left(\sqrt{2x^2+a}-x+b\right)=0\)
\(\displaystyle \Rightarrow \sqrt{2+a}-1+b=0\Rightarrow a=b^2-2b-1\)
\(\displaystyle \Rightarrow \lim_{x\to1}\frac{\sqrt{2x^2+a}-x+b}{\left(x-1\right)^2}=\lim_{x\to1}\frac{\sqrt{2x^2+b^2-2b-1}-x+b}{\left(x-1\right)^2}\)
\(\displaystyle =\lim_{x\to1}\frac{2x^2+b^2-2b-1-\left(x-b\right)^2}{\left(x-1\right)^2\left(\sqrt{2x^2+b^2-2b-1}+x-b\right)}\)
\(\displaystyle =\lim_{x\to1}\frac{x^2-1+2b\left(x-1\right)}{\left(x-1\right)^2\left(\sqrt{2x^2-b-1}+x-b\right)}\)
\(\displaystyle =\lim_{x\to1}\frac{\left(x+1+2b\right)\left(x-1\right)}{\left(x-1\right)^2\left(\sqrt{2x^2-b-1}+x-b\right)}\)
\(\displaystyle =\lim_{x\to1}\frac{x+1+2b}{\left(x-1\right)\left(\sqrt{2x^2-b-1}+x-b\right)}\)
因為上式為定值,所以 \(\displaystyle \lim_{x\to1}\left(x+1+2b\right)=0\)
\(\Rightarrow 1+1+2b=0\Rightarrow b=-1, a=b^2-2b-1=2\)
第 3 題:
當 \(-1<x<1\) 時,\(\displaystyle f(x)=\lim_{n\to\infty}\frac{x^{2n-1}+ax+b}{x^{2n}+1}=ax+b\)
\(\displaystyle \Rightarrow \lim_{x\to1^{-}}f(x)=a+b\) 且 \(\displaystyle \lim_{x\to-1^{+}}f(x)=-a+b\)
當 \(x>1\) 或 \(x<-1\) 時,\(\displaystyle f(x)=\lim_{n\to\infty}\frac{x^{2n-1}+ax+b}{x^{2n}+1}=\lim_{n\to\infty}\frac{\displaystyle x^{-1}+\frac{a}{x^{2n-1}}+\frac{b}{x^{2n}}}{\displaystyle 1+\frac{1}{x^{2n}}}=\frac{1}{x}\)
\(\displaystyle \Rightarrow \lim_{x\to1^{+}}f(x)=1\) 且 \(\displaystyle \lim_{x\to-1^{-}}f(x)=-1\)
當 \(x=1\) 時,\(\displaystyle f(1)=\lim_{n\to\infty}\frac{1^{2n-1}+a+b}{1^{2n}+1}=\frac{1+a+b}{2}\)
當 \(x=-1\) 時,\(\displaystyle f(-1)=\lim_{n\to\infty}\frac{\left(-1\right)^{2n-1}-a+b}{\left(-1\right)^{2n}+1}=\frac{-1-a+b}{2}\)
因為 \(f(x)\) 為連續函數,
所以 \(\displaystyle \lim_{x\to1^{+}}f(x)=\lim_{x\to1^{-}}f(x)=f(1)\) 且 \(\displaystyle \lim_{x\to-1^{+}}f(x)=\lim_{x\to-1^{-}}f(x)=f(-1)\)
\(\displaystyle \Rightarrow 1=a+b=\frac{1+a+b}{2}\) 且 \(\displaystyle -a+b=-1=\frac{-1-a+b}{2}\)
可解得 \(a=1, b=0\)