111年公務人員高等考試三級考試
類 科: 核子工程
科 目: 微積分與微分方程
解答:(一)f(x)=x2−6x+10=(x−3)2+1⇒{最小值=f(3)=1最大值=f(0)=10⇒{最小值:1最大值:10(二)∞∑n=0(x−2)nn+1⇒{an+1=(x−2)n+1n+2an=(x−2)nn+1⇒limn→∞|an+1an|=limn→∞|(x−2)⋅n+1n+2|=|x−2|<1⇒1<x<3;檢查收斂區間兩端點:{x=3⇒∑∞n=0(x−2)nn+1=∑∞n=01n+1調和級數⇒發散x=1⇒∑∞n=0(x−2)nn+1=∑∞n=0(−1)nn+1交錯調和級數⇒收斂因此,收斂半徑=1,收斂區間=[1,3)


解答:球坐標{x=ρsinϕcosθy=ρsinϕsinθz=ρcosϕ⇒∭

解答:
(一)f(x,y)=x^2+ y^3 \Rightarrow \nabla f=(f_x,f_y) = (2x, 3y^2) \Rightarrow \nabla f(1,-1)=\bbox[red,2pt]{(2,3)}(二)方向導數最大值=||(2,3)|| = \sqrt{2^2+3^2} = \bbox[red, 2pt]{\sqrt{13}}

f(t)=\begin{cases} 1,& t\in [0,2)\\ 0, &t\ge 2\end{cases} \Rightarrow \mathcal{L}\{f(t)\}= F(s)= \int_0^\infty f(t)e^{-st}\,dt =\int_0^2 e^{-st}\,dt =\bbox[red, 2pt]{{1\over s}(1-e^{-2s})}(二)f(t)=\begin{cases} 1,& t\in [0,2)\\ 0, &t\ge 2\end{cases} \Rightarrow \mathcal{L}\{f(t)\}= F(s)= \int_0^\infty f(t)e^{-st}\,dt =\int_0^2 e^{-st}\,dt ={1\over s}(1-e^{-2s})\\ y''(t)-y(t)=f(t) \Rightarrow \mathcal{L}\{y''(t)\}-\mathcal{L}\{y(t)\} =\mathcal{L}\{f(t)\} \\\Rightarrow s^2Y(s)-sy(0)-y'(0) -(sY(s)-y(0))={1\over s}(1-e^{-2s}) \Rightarrow (s^2-s)Y(s) ={1\over s}(1-e^{-2s})+s-1\\ \Rightarrow Y(s)={1\over s^2(s-1)}(1-e^{-2s})+{1\over s} =\left( {1\over s-1}-{1\over s}-{1\over s^2}\right)(1-e^{-2s})+{1\over s} \\={1\over s-1}-{1\over s^2}-{1\over s-1}e^{-2s}+ {1\over s}e^{-2s} +{1\over s^2}e^{-2s} \\\Rightarrow y(t)=\mathcal{L^{-1}}\{Y(s)\} =\mathcal{L^{-1}}\{{1\over s-1}\} -\mathcal{L^{-1}}\{{1\over s^2}\} -\mathcal{L^{-1}}\{{1\over s-1}e^{-2s}\} +\mathcal{L^{-1}}\{{1\over s }e^{-2s}\} +\mathcal{L^{-1}}\{{1\over s^2 }e^{-2s}\} \\=e^t-t-u(t-2)e^{t-2}+u(t-2) +(t-2)u(t-2) \\ \Rightarrow \bbox[red,2pt]{y(t)=e^t-t+u(t-2)(t-1-e^{t-2})}
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