103年高考三級-工程數學詳解

103年公務人員高等考試三級考試

類 科：電力工程、電子工程、電信工程、醫學工程
科 目：工程數學
甲、申論題部分：（ 50 分）

解答：$$y'-6y+9\int_0^t y(\tau)\;d\tau = 4t^3e^{3t} \Rightarrow \mathcal{L}\{y'\}-6\mathcal{L}\{y\}+ 9\mathcal{L}\{ \int_0^t y(\tau)\;d\tau \} =4\mathcal{L}\{ t^3e^{3t} \}\\ \Rightarrow sY(s)-y(0)-6Y(s)+9{Y(s)\over s} =4\cdot   {3!\over (s-3)^4} \Rightarrow (s-6+{9\over s})Y(s) ={24\over (s-3)^4} \\ \Rightarrow {(s-3)^2\over s}Y(s) ={24\over (s-3)^4} \Rightarrow Y(s)={24s\over (s-3)^6} ={24\over (s-3)^5}+ {72\over (s-3)^6} \\ \Rightarrow y(t)= \mathcal{L}^{-1}\left\{4!\over (s-3)^5\right\}+{3\over 5}\mathcal{L}^{-1}\left\{5!\over (s-3)^6\right\} =t^4e^{3t}+{3\over 5}t^5e^{3t}\\ \Rightarrow \bbox[red,2pt]{y(t)= t^4e^{3t}+{3\over 5}t^5e^{3t}}$$

解答：$$A=\left[\begin{array} {ccccc|ccccc}1 & a_2 & a_3 & \cdots & a_n & 1 & 0 & 0 &\cdots & 0\\ \lambda_2& 1+\lambda_2 a_2 & \lambda_2 a_3 & \cdots & \lambda_2 a_n & 0 & 1 &0 &\cdots & 0 \\ \lambda_3 & \lambda_3a_2 & 1+ \lambda_3 a_3 & \cdots & \lambda_3 a_n & 0 & 0 & 1& \cdots &0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\\lambda_n & \lambda_n a_2 & \lambda_na_3 & \cdots & 1+\lambda_na_n & 0 & 0 & 0 & \cdots & 1\end{array}\right]\\ 第1列\times (-\lambda_k)加至第k列,k=2,\dots,n\\\Rightarrow \left[\begin{array} {ccccc|ccccc}1 & a_2 & a_3 & \cdots & a_n & 1 & 0 & 0 &\cdots & 0\\ 0& 1    & 0   & \cdots & 0   & -\lambda_2 & 1 &0 &\cdots & 0 \\ 0 & 0  & 1    & \cdots & 0   & -\lambda_3 & 0 & 1& \cdots &0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0  & 0  & \cdots & 1  & -\lambda_n & 0 & 0 & \cdots & 1\end{array}\right]\\第k列\times (-a_k)加至第1列,k=2,\dots,n\\ \Rightarrow  \left[\begin{array} {ccccc|ccccc}1 & 0 & 0 & \cdots & 0 & 1+\sum_{k=2}^n \lambda_k a_k & -a_2 & -a_3 &\cdots & -a_n\\ 0& 1    & 0   & \cdots & 0   & -\lambda_2 & 1 &0 &\cdots & 0 \\ 0 & 0  & 1    & \cdots & 0   & -\lambda_3 & 0 & 1& \cdots &0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0  & 0  & \cdots & 1  & -\lambda_n & 0 & 0 & \cdots & 1\end{array}\right] \\ \Rightarrow A^{-1} =\bbox[red,2pt]{\left[\begin{array} { }  1+\sum_{k=2}^n \lambda_k a_k & -a_2 & -a_3 &\cdots & -a_n\\   -\lambda_2 & 1 &0 &\cdots & 0 \\   -\lambda_3 & 0 & 1& \cdots &0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -\lambda_n & 0 & 0 & \cdots & 1\end{array}\right]}$$

解答：$$平面E:Ax+By+Cz=D 之法向量\vec n=(A,B,C)\\ 過原點且方向向量為\vec n之直線L:{x\over A} ={y\over B} ={z\over C}，L上的點P可表示成P(At,Bt,Ct),t\in \mathbb{R}\\ P\in E \Rightarrow A^2t+B^2t+C^2t = D \Rightarrow t={D\over A^2+B^2+C^2 } \\\Rightarrow P=\bbox[red,2pt]{\left({AD\over A^2+B^2+C^2 } ,{BD\over A^2+B^2+C^2 } ,{CD\over A^2+B^2+C^2 } \right)}$$

解答：

(一)$$藍色骰子出現6的機率為{1\over 6}，出現其它點數的機率為{5\over 6}；因此2個6、3個其它點數的機率為({1\over 6})^2({5\over 6})^3\\，再乘上其排列數C^5_2，機率為C^5_2({1\over 6})^2({5\over 6})^3；\\綠色骰子出現奇數或偶數的機率皆為{1\over 2}，三顆皆偶數且另兩顆皆奇數的機率為({1\over 2})^3({1\over 2})^2\\，再乘上其排列數C^5_3，機率為C^5_3({1\over 2})^3({1\over 2})^2；\\符合上述兩要求的機率為C^5_2({1\over 6})^2({5\over 6})^3\times C^5_3({1\over 2})^3({1\over 2})^2= \bbox[red,2pt]{5^5\over 2^8\cdot 3^5}$$(二)$$無論藍色或綠色骰子出現k個6的機率都是C^5_k({1\over 6})^k({5\over 6})^{5-k},k=0,..,5；\\因此兩骰子出現一樣多6的機率為 \bbox[red,2pt]{\sum_{k=0}^5 \left[ C^5_k({1\over 6})^k({5\over 6})^{5-k}\right]^2}$$

乙、測驗題部分：（50 分）

解答：$$\cases{x(t)=t\\ y(t)=t\\ z(t)=t^2} \Rightarrow \cases{dx=dt\\ dy=dt\\ dz=2tdt} \Rightarrow \int_C \varphi\cdot ds =\int_0^2 (t+t)\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}\;dt \\ =\int_0^2 2t\sqrt{2+4t^2}\;dt =\int_2^{18} {1\over 4}\sqrt u\;du (u=2+4t^2 \Rightarrow du=8t\;dt) ={1\over 6}(18^{3/2}-2^{3/2}) \\ ={1\over 6}(54\sqrt 2-2\sqrt 2) ={26\over 3}\sqrt 2，故選\bbox[red,2pt]{(D)}$$

解答：$$\nabla^2(fg)= f\nabla^2g+g\nabla^2f+2\nabla f\cdot\nabla g，故選\bbox[red,2pt]{(D)}$$

解答：$$\int_c F(r)dt = \int_0^\pi [4\sin^2(t),t,4\cos^2(t)]dt =\left. \left[2t-\sin(2t),{1\over 2}t^2,2\cos(2t)+2\right] \right|_0^\pi \\ =[2\pi,{1\over 2}\pi^2,2\pi]，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{u=(1,1,1)\\ v=(1,0,1)\\ w=(0,1,0)} \Rightarrow \cases{u\times (v\times w)=u\times (-1,0,1) = (1,-2,1)\\ (u\times v)\times w= (1,0,-1)\times w= (1,0,1)} \\ \Rightarrow u\times (v\times w) \ne (u\times v)\times w，故選\bbox[red,2pt]{(C)}$$

解答：$$A=\begin{bmatrix} -2 & 1 & 0\\ 1 &-2 & 0 \\0 & 0 & 1\end{bmatrix} \Rightarrow A^{-1}= \begin{bmatrix} -2/3 & -1/3 & 0\\ -1/3 &-2/3 & 0 \\0 & 0 & 1\end{bmatrix}  \Rightarrow \det(A^{-1})={4\over 9}-{1\over 9}={1\over 3}\\矩陣之行列式等於其特徵值之積，故選\bbox[red,2pt]{(D)}$$

解答：$$(A) M(2,-4,6)^t= (8,-16,24)^t=4(2,-4,6)^t \Rightarrow (2,-4,6)為特徵向量\\(B)M(2,0,6)^t= (4,-16,12)^t \ne \lambda(2,0,6)\\ (C) M(2,2,0)^t= (-4,-4,0)^t =-2(2,2,0)^t\\ (D) M(-1,0,1)^t= (2,0,-2)^t =-2(-1,0,1)^t\\，故選\bbox[red,2pt]{(B)}$$

解答：$$複數也可以為特徵值，故選\bbox[red,2pt]{(D)}$$

解答：$$e^z=1+2i=\sqrt 5({1\over \sqrt 5}+{2\over \sqrt 5}i) =\sqrt 5(\cos \theta +i\sin\theta) =e^{\ln \sqrt 5+i\theta}  \\ \Rightarrow z=\ln \sqrt 5+i\theta,其中\cases{\cos \theta=1/\sqrt 5\\ \sin \theta=2/\sqrt 5} \Rightarrow \tan \theta=2 \Rightarrow \theta =\tan^{-1} 2\\ \Rightarrow z={1\over 2}\ln(5)+ i\tan^{-1}(2)，故選\bbox[red,2pt]{(B)}$$

解答：$$\det(A-\lambda I)=\begin{vmatrix}2-\lambda & 1 & 1-i\\ 1 &-\lambda & -i\\ 1+i & i & 2-\lambda \end{vmatrix} = \lambda^3-4\lambda^2+2=0 \Rightarrow \cases{\lambda_1+\lambda_2 +\lambda_3= 4\\ \lambda_1\lambda_2+ \lambda_2\lambda_3 +\lambda_3\lambda_1 = 0 \\ \lambda_1\lambda_2\lambda_3= -2}\\，故選\bbox[red,2pt]{(D)}$$

解答：$$c_n={(3n)!\over (n!)^3} \Rightarrow {c_{n+1}\over c_n}={(3n+3)!\over ((n+1)!)^3} \cdot{ (n!)^3\over (3n)!} ={(3n+3)(3n+2)(3n+1)\over (n+1)^3} ={3(3n+2)(3n+1)\over (n+1)^2} \\ \Rightarrow \lim_{n\to \infty}\left|{c_{n+1}\over c_n} \right| =\lim_{n\to \infty}{3(3n+2)(3n+1)\over (n+1)^2}=27 \Rightarrow 收斂半徑={1\over 27}\\ \Rightarrow 符合條件|z-2i|\lt {1\over 27}收斂，故選\bbox[red,2pt]{(B)}$$

解答：$$\int_C ze^{1/z}dz = \int_C z(1+{1\over z}+{1\over 2!z^2} +{1\over 3!z^3} +\cdots)\;dz = \int_C  (z+1+{1\over 2!z } +{1\over 3!z^2} +\cdots)\;dz \\={1\over 2}\int_C {1\over z}\;dz = {1\over 2}\times 2\pi i=\pi i，故選\bbox[red,2pt]{(B)}$$

解答：$$(C)x^5y''+ 4x^4y'+2y=0 \Rightarrow y''+{4\over x}y'+ {2\over x^5}=0\\ x=0在{2\over x^5}是一個\text{pole of order 5}，超過\text{order 2}，為一個\text{irregular singular point}，故選\bbox[red,2pt]{(C)}$$

解答：$$Y(s)={-s+2 \over s^2+6s+8}={2\over s+2}-{3\over s+4} \\\Rightarrow y(x)= \mathcal{L}^{-1}\{Y(s)\}= 2\mathcal{L}^{-1}\{ {1\over s+2}\}-3\mathcal{L}^{-1}\{ {1\over s+4}\} =2e^{-2x}-3e^{-4x} \\ \Rightarrow y(x)=2e^{-2x}-3e^{-4x} \Rightarrow \cases{y_0=  y(0)=2-3=-1\\ y'+ay= (2a-4)e^{-2x}+(-3a+12)e^{-4x}=4e^{-2x} \Rightarrow a=4}\\ \Rightarrow a+y_0=4-1=3，故選\bbox[red,2pt]{(B)}$$

解答：$$\mathcal{L}[\cos(\omega t+\phi)]= \mathcal{L}[ \sin(\omega t+\phi+\pi/2) ] ={\omega\cos( \phi+\pi/2)+ s\sin(\phi+\pi/2)\over s^2+\omega^2}\\ ={-\omega\sin( \phi )+ s\cos(\phi )\over s^2+\omega^2}，故選\bbox[red,2pt]{(D)}$$

解答：$$y''+\lambda y=0 \Rightarrow 特徵方程式r^2+\lambda=0 \Rightarrow r= \pm \sqrt{-\lambda} \\ \text{Cases 1:}\lambda \lt 0 \Rightarrow y=c_1e^{\sqrt{-\lambda}x} +c_2e^{-\sqrt{-\lambda}x}，將y(0)=y(L)=0代入\Rightarrow c_1=c_2=0\\\text{Cases 2:}\lambda = 0 \Rightarrow y''=0 \Rightarrow y=c_1x+c_2，將y(0)=y(L)=0代入\Rightarrow c_1=c_2=0\\\text{Cases 3:}\lambda = \sigma^2\gt  0\Rightarrow r=\pm \sigma i \Rightarrow y= c_1\cos (\sigma x) +c_2\sin(\sigma x)\\ \qquad\quad將y(0)=y(L)=0代入\Rightarrow \cases{c_1=0\\ c_2\sin(\sigma L)=0} \\ \qquad\quad \Rightarrow \sigma ={m\pi\over L} \Rightarrow  \lambda_m= \sigma^2={m^2 \pi^2\over L^2},m=1,2,\dots，故選\bbox[red,2pt]{(B)}$$

解答：$$y=x^2  \Rightarrow y''+Ay'+By = 2+2Ax+Bx^2，\\無法找到常數A,B使得對任意x皆滿足2+2Ax+Bx^2=0，故選\bbox[red,2pt]{(A)}$$

解答：$$b_n={2\over L}\int_0^L x^2\sin{2n\pi\over L}x\;dx \\={2\over L} \left.\left[ -{L\over 2n\pi}x^2\cos\left({2n\pi\over L}x \right)+{L^2\over 2n^2\pi^2}x\sin \left({2n\pi\over L}x \right) +{L^3\over 4n^3\pi^3} \cos\left({2n\pi\over L}x \right) \right] \right|_0^L\\ ={2\over L}\left(-{L^3\over 2n\pi} \right) =-{L^2\over n\pi}，故選\bbox[red,2pt]{(C)}$$

解答：$$p(1,1)+p(1,2)+p(2,1)=1 \Rightarrow 2c+3c+5c=1 \Rightarrow c=1/10\\ \Rightarrow \cases{p(y=1)=p(1,1)+p(2,1)=7c=7/10\\ p(y=2)=p(1,2)=3c=3/10} \\ \Rightarrow \cases{E(Y)=\sum yp(y)=1\cdot p(y=1)+2\cdot p(y=2) =7/10+6/10=13/10\\ E(Y^2)=\sum y^2p(y) =1^2p(y=1)+2^2p(y=2)=7/10+12/10=19/10} \\ \Rightarrow Var(Y)=E(Y^2)-(E(Y))^2={19\over 10}-{169\over 100} ={21\over 100}，故選\bbox[red,2pt]{(C)}$$

解答：$$令\cases{A=(正,正)\\ B=(反,正)\\ C=(正,反)\\ D=(反,反)} \Rightarrow \cases{P(A)=1/6\\ P(B)=1/3\\ P(C)=1/6\\ P(D)=1/3}\\丟三次且符合題意的情形:AAB,AAC,AAD及其排列；\\因此\cases{P(AAB)=C^3_1P(A)P(A)P(B)=1/36\\ P(AAC)=C^3_1P(A)P(A)P(C)=1/72\\ P(AAD)=C^3_1P(A) P(A)P(D)= 1/36} \\ \Rightarrow P(AAB)+ P(AAC)+ P(AAD)=5/72，故選\bbox[red,2pt]{(B)}$$

解答：$$正確篩檢+錯誤篩檢= 0.05\times 0.78+0.95\times 0.06=0.096，故選\bbox[red,2pt]{(B)}$$

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