111年地方特考-工程數學詳解

 111 年特種考試地方政府公務人員考試試題

等 別：三等考試
類 科：電力工程、電子工程
科 目：工程數學

甲、申論題部分：（50分）

解答: $$\mathbf{(一)}\;f(x,y)=(x^2+2y)e^{-(x^2+y^2)} \Rightarrow \nabla f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}) \\\qquad =((2x-2x^3-4xy)e^{-(x^2+y^2)} ,(2-2x^2y-4y^2)e^{-(x^2+y^2)} \Rightarrow \nabla f(P_0) =\nabla f(1,-1)=\bbox[red, 2pt]{(4e^{-2},0)}\\ \mathbf{(二)}\; \vec u=(1,0) \Rightarrow 方向導數 \nabla f\cdot \vec u =(4e^{-2},0)\cdot (1,0)= \bbox[red, 2pt]{4e^{-2}}$$

解答: $$\mathbf{(一)}\;積分因子I(x)=e^{\int {3\over x}dx} =\bbox[red, 2pt]{x^3}\\\mathbf{(二)}\;  I(x) y'+I(x){3\over x}y=I(x)x \Rightarrow x^3y'+ 3x^2y =x^4 \Rightarrow (x^3y)'=x^4 \Rightarrow x^3y =\int x^4\,dx \\ \Rightarrow x^3y ={1\over 5}x^5+c_1 \Rightarrow y={1\over 5}x^2+{c_1\over x^3},又y(1)=3 \Rightarrow {1\over 5}+c_1=3 \Rightarrow c_1={14\over 5} \\ \Rightarrow \bbox[red, 2pt]{y={1\over 5}x^2+ {14\over 5x^3}}$$

解答: $$\mathbf{(一)}\;\lambda_1=1 \Rightarrow (A-\lambda_1 I)x=0 \Rightarrow \begin{bmatrix}-1& 0 & -2\\ 1 & 1& 1\\ 1& 0 & 2 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+2x_3=0\\ x_2-x_3=0 }\\\qquad  \Rightarrow 基本解: \{x_3\begin{pmatrix} -2\\ 1\\ 1\end{pmatrix}\}\\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)x=0 \Rightarrow \begin{bmatrix}-2& 0 & -2\\ 1 & 0& 1\\ 1& 0 & 1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1+x_3=0 \\\qquad \Rightarrow 基本解: \{x_2\begin{pmatrix} 0\\ 1\\ 0\end{pmatrix}+x_3 \begin{pmatrix} -1\\ 0\\ 1\end{pmatrix}\}\\ 因此特徵向量的一般形式為\;\bbox[red, 2pt]{a\begin{pmatrix} -2\\ 1\\ 1\end{pmatrix},b\begin{pmatrix} 0\\ 1\\ 0\end{pmatrix},c\begin{pmatrix} -1\\ 0\\ 1\end{pmatrix},其中a,b,c\in \mathbb R} \\\mathbf{(二)}\; 若v為A的特徵向量,則Av=\lambda v \Rightarrow A^2v =\lambda Av=\lambda^2 v \Rightarrow A^8 v=\lambda^8 v\\ \qquad \Rightarrow v也是A^8的特徵向量 \Rightarrow A與A^8有相同的特徵向量\\ \qquad 即\bbox[red, 2pt]{a\begin{pmatrix} -2\\ 1\\ 1\end{pmatrix},b\begin{pmatrix} 0\\ 1\\ 0\end{pmatrix},c\begin{pmatrix} -1\\ 0\\ 1\end{pmatrix},其中a,b,c\in \mathbb R}$$

解答: $$\mathbf{(一)}\;半徑為1的圓形標靶面積=1^2\pi= \pi,Ｘ代表到靶心距離,則f_X(x)={kx\over \pi}\\並滿足\int f_Ｘ\,dx=1 \Rightarrow \int_0^1 {kx\over \pi}dx=1 \Rightarrow {k\over 2\pi}=1 \Rightarrow k=2\pi \Rightarrow f_X(X)={2\pi X\over \pi}=\bbox[red,2pt]{2X}　\\\mathbf{(二)}\; F_Y(y) =P(Y\lt y) =P(X^2\lt y) = P(X\lt \sqrt y) =\int_0^{\sqrt y} 2x\,dx=y \\ \qquad \Rightarrow F_Y(y)=y \Rightarrow f_Y(y)={d\over dy}F_Y(y)= 1 \Rightarrow f_Y(y)=\bbox[red,2pt]1$$

解答: $$\mathbf{(一)}\; f(z)={1\over z^2+4} ={1\over (z+2i)(z-2i)} \Rightarrow f(z)\text{ has simple poles at }z=\pm 2i \\ \Rightarrow z=\pm 2i皆在C_1內 \Rightarrow \cases{Res(f,2i)= \left. {1\over z+2i} \right|_{z=2i} ={1\over 4i}\\ Res(f,-2i)= \left. {1\over z-2i} \right|_{z=-2i} ={1\over -4i}} \\ \Rightarrow \oint_{C_1} f(z)\,dz = 2\pi i(Res(f,2i)+Res(f,-2i)) =2\pi i\times 0=\bbox[red, 2pt] 0 \\ \mathbf{(二)}\;\cases{z=2i在C_2內\\ z=-2i不在C_2內}\Rightarrow \oint_{C_1} f(z)\,dz = 2\pi i\times Res(f,2i)=\bbox[red, 2pt]{\pi \over 2}$$

乙、測驗題部分：（50分）

解答: $$\begin{vmatrix}\vec i & \vec j & \vec k \\1 & -1 & 3 \\2 & 0 & -4\end{vmatrix} =4\vec i+10\vec j+2\vec k=(4,10,2) =(a,b,c)\Rightarrow a\times b\times c=80 ,故選\bbox[red, 2pt]{(B)}$$

解答: $$\begin{bmatrix}1 & 1  \\1 & 1\\ 1& 1\end{bmatrix}\begin{bmatrix}x_1 \\x_2\end{bmatrix} = \begin{bmatrix} 1 \\1\end{bmatrix} \Rightarrow x_1+x_2=1有無窮多解 ,故選\bbox[red, 2pt]{(C)}$$

解答: $$\cos \phi ={(1,3,2)\cdot (0,-1,4) \over \Vert(1,3,2) \Vert \Vert(0,-1,4) \Vert } ={5\over \sqrt{16}\cdot \sqrt{17}} \approx {5\over 16} \Rightarrow \sin \phi =\sqrt{1-({5\over 16})^2} ={\sqrt{231} \over 16} \\ 又 15\lt \sqrt{231}\lt 16 \Rightarrow  \Rightarrow {15\over 16}\lt {\sqrt{231}\over 16}\lt 1 \Rightarrow 0.94\lt {\sqrt{231}\over 16}\lt 1 ,故選\bbox[red, 2pt]{(A)}$$

解答: $$無法構成基底代表三向量為線性相依,即a(1,2,3)+b(1,0,-1)=(3,1,\alpha) \\ \Rightarrow \cases{a+b=3\\2a=1\\ 3a-b=\alpha} \Rightarrow \cases{a=1/2\\ b=5/2} \Rightarrow \alpha={3\over 2}-{5\over 2}=-1 ,故選\bbox[red, 2pt]{(C)}$$

解答: $$\det(A)=7 \ne 0 \Rightarrow A為可逆,故選\bbox[red, 2pt]{(A)}$$

解答: $$\begin{bmatrix} 2& -3 & 1\\ 1& -2& 1\\ 1& -3 & 2\end{bmatrix} \begin{bmatrix}-1\\ 0 \\ 1 \end{bmatrix} =\begin{bmatrix} -1\\ 0 \\ 1\end{bmatrix} 符合Av=v\Rightarrow v是特徵向量 ,故選\bbox[red, 2pt]{(D)}$$

解答: $$A=\left(\begin{matrix}\frac{4}{5} & \frac{3}{10} \\\frac{1}{5} & \frac{7}{10}\end{matrix}\right) =\left(\begin{matrix}-1 & \frac{3}{2} \\1 & 1\end{matrix}\right) \left(\begin{matrix}\frac{1}{2} & 0 \\0 & 1\end{matrix}\right)\left(\begin{matrix}\frac{-2}{5} & \frac{3}{5} \\\frac{2}{5} & \frac{2}{5} \end{matrix}\right) \Rightarrow A^\infty=\left(\begin{matrix}-1 & \frac{3}{2} \\1 & 1 \end{matrix} \right) \left(\begin{matrix}\frac{1}{2^\infty} & 0 \\0 & 1^\infty\end{matrix}\right) \left( \begin{matrix}\frac{-2}{5} & \frac{3}{5} \\\frac{2}{5} & \frac{2}{5}\end{matrix} \right) \\=\left(\begin{matrix}-1 & \frac{3}{2} \\1 & 1 \end{matrix} \right) \left(\begin{matrix}0 & 0 \\0 & 1 \end{matrix}\right) \left( \begin{matrix}\frac{-2}{5} & \frac{3}{5} \\\frac{2}{5} & \frac{2}{5}\end{matrix} \right)=  \left( \begin{matrix}\frac{3}{5} & \frac{3}{5} \\\frac{2}{5} & \frac{2}{5}\end{matrix} \right) \Rightarrow A^\infty \begin{pmatrix} 100\\ 0\end{pmatrix}=\begin{pmatrix} 60 \\ 40\end{pmatrix} ,故選\bbox[red, 2pt]{(C)}$$

解答: $$f(z)=\bar z \Rightarrow \lim_{z\to 0}{f(z)-f(0)\over z-0} =\lim_{z\to 0}{\bar z\over z} =\lim_{(x,y)\to (0,0)}{x-iy\over x+iy}=\begin{cases} 1,& x\to 0,y=0\\ -1,& x=0, y\to 0\end{cases} \\ \Rightarrow \bar z在z=0不可微 ,故選\bbox[red, 2pt]{(D)}$$

解答: $$留數={1\over z}的係數=5-2i=a+bi \Rightarrow a+b=3 ,故選\bbox[red, 2pt]{(C)}$$

解答: $$\Gamma可以表示成r(t)=(t,t^2),0\le t\le 1,因此z=t+it^2 \Rightarrow dz=dt+2itdt =(1+2it)dt\\ \Rightarrow \int_\Gamma z^2\,dz = \int_0^1 (t+it^2)^2(1+2it)dt = \int_0^1 ((t^2-5t^4)+(4t^3-2t^5)i)dt\\ ={1\over 3}-1+(1-{1\over 3})i=-{2\over 3}+{2\over 3}i \Rightarrow a\cdot b=-{2\over 3}\times{2\over 3}=-{4\over 9}=-0.44,故選\bbox[red, 2pt]{(B)}$$

解答: $$g(z)=\int_C {s^2-s+2\over (s-z)^2} \,ds\Rightarrow g(1)=\int_C {s^2-s+2\over (s-1)^2} \,ds\\ 令f(s)=s^2-s+2 \Rightarrow f'(s)=2s-1 \Rightarrow f'(1)=1 \Rightarrow \int_C {s^2-s+2\over (s-1)^2} \,ds =2\pi i f'(1)=2\pi i\\ ,故選\bbox[red, 2pt]{(A)}$$

解答: $$f=x^2\sin(xy) \Rightarrow \nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) =(2x \sin(xy)+x^2y \cos(xy),x^3\cos(xy)) \\ \Rightarrow \nabla f(1,\pi)=(2\sin \pi+\pi \cos(\pi), \cos(\pi)) =(-\pi,-1),故選\bbox[red, 2pt]{(C)}$$

解答: $$y''+y'+y=0的解為e^{-t/2}(c_1\cos(\sqrt 3t/2)+ c_2\sin (\sqrt 3t/2)) \\ \Rightarrow y''+y'+y=\sin (\omega t)的解為週期函數,但頻率不一定是\omega ,故選\bbox[red, 2pt]{(D)}$$

解答: $$y=\sum_{n=0}^\infty a_n x^n\Rightarrow y'=\sum_{n=1}^\infty na_nx^{n-1} \Rightarrow  y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}  \Rightarrow y''(0)=2a_2\\ 將x=0代入原式y''(x)+xy'(x)+ e^xy(x)=x^2+1 \Rightarrow y''(0)+0+2=1 \Rightarrow 2a_2+2=1 \\ \Rightarrow a_2=-{1\over 2},故選\bbox[red, 2pt]{(B)}$$

解答: $$L與L^{-1}皆為線性轉換,因此(A)及(B)皆正確\\　又\cases{u=f(t)\\ dv=e^{-st}dt} \Rightarrow \cases{du= f'(t)dt\\ v=-{1\over s}e^{-st}} \Rightarrow sL\{f\}= s\int_0^\infty f(t)e^{-st}\,dt \\= s \left( \left.-{1\over s}f(t)e^{-st}\right|_0^{\infty} +{1\over s}\int_0^\infty f'(t)e^{-st}\,dt\right) =f(0)+L\{f'(t)\} =L\{f'(t)\}\\ \Rightarrow sL\{f\}= L\{f'\},因此(D)正確 ,故選\bbox[red, 2pt]{(C)}$$

解答: $$若g(x)=\begin{cases} 1,& |x|\lt 1\\ 0,& 其它\end{cases} \Rightarrow G(\omega)={2\sin(\omega)\over \omega} \Rightarrow f(t)=g(t-1) \\\Rightarrow F(\omega)=G(\omega)e^{-j\omega} ={2\sin(\omega)\over \omega} e^{-j\omega} ,故選\bbox[red, 2pt]{(A)}$$

解答: $$f(x)=\begin{cases} 1-|x|,& |x|\le 1\\ 0,& |x|\gt 1\end{cases} =\begin{cases} 1+x,& -1\le x\le 0\\ 1-x,& 0\le x\le 1\\0,& 其它\end{cases} \Rightarrow E(X)=0 (圖形對稱y軸)\\ \Rightarrow E(X^2)=\int_{-1}^0 x^2(1+x)\,dx + \int_0^1 x^2(1-x)\,dx = {1\over 6} \Rightarrow Var(X)=E(X^2)-(E(X))^2 ={1\over 6}\\ ,故選\bbox[red, 2pt]{(C)}$$

解答: $$\iint f(x,y)\,dxdy=1 \Rightarrow \int_0^1 \int_0^1 Axy^2\,dx dy= \int_0^1{1\over 2}Ay^2 \,dy= {1\over 6}A=1 \Rightarrow A=6\\ f_X(x) =\int_0^1 6xy^2\,dy=2x \Rightarrow E(X) =\int_0^1 x f_X(x) \,dx = \int_0^1 2x^2\,dx = {2\over 3},故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{P(n=5)=0.4^5\\ P(n=4)=C^{5}_4\cdot 0.4^4\cdot 0.6\\ P(n=3)= C^5_3 \cdot 0.4^3\cdot 0.6^2} \Rightarrow P(n=5)+ P(n=4)+P(n=3) \\=0.4^3(0.4^2+3\cdot 0.4+10\cdot 0.6^2) =0.064\cdot 4.96 =0.31744,故選\bbox[red, 2pt]{(C)}$$

解答: $$e^{i\theta} =\cos \theta+i \sin \theta \Rightarrow e^{\pi i/2} =i \Rightarrow \ln i={\pi \over 2}i\\因此 i^{2i} =e^{2i\ln i} = e^{2i \cdot  \pi i/2} =e^{-\pi},故選\bbox[red, 2pt]{(D)}$$

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解題僅供參考,其他試題及詳解