108年專技高考_電機工程技師-工程數學詳解

108年專門職業及技術人員高等考試

等        別：高等考試
類        科：電機工程技師
科        目：工程數學

解：

$$令u=a+x \Rightarrow du=dx \Rightarrow u^2y''-2y=3u^2+1\\ 先求齊次解，即u^2y''-2y=0 \Rightarrow y_h=C_1u^2+C_2{1\over u}\\ 接下來利用\text{ variant of parameters}來求y_p，即y''-{2\over u^2}y=3+{1\over u^2} 的解\\ 令 \begin{cases}y_1 = u^2\\ y_2=1/u\end{cases}  \Rightarrow \begin{cases}y'_1 = 2u\\ y'_2=-1/u^2\end{cases}  \Rightarrow W= \begin{vmatrix}y_1 & y'_1 \\y_2 & y'_2 \end{vmatrix} = \begin{vmatrix}u^2 & 2u \\1/u & -1/u^2 \end{vmatrix} =-3\\  \Rightarrow y_p= -y_1 \int {y_2r(u)\over W}du+y_2 \int {y_1r(u)\over W}du,其中r(u)=3+{1\over u^2}\\ =-u^2\int {(1/u)( 3+1/u^2)\over -3}du +{1\over u}\int {u^2(3+1/u^2) \over -3}du \\ =u^2 \int \left({1\over u} + {1\over 3u^3}\right)du -{1\over u}\int \left( u^2+{1\over 3} \right) du = u^2\left( \ln |u|- {1\over 6u^2}\right)- {1\over u} \left( {1\over 3}u^3 +{1\over 3}u\right) \\ =u^2\ln |u|- {1\over 6}-{1\over 3}u^2-{1\over 3}= u^2\ln |u|-{1\over 3}u^2-{1\over 2} \\ \Rightarrow y=y_h+y_p = C_1u^2+C_2{1\over u}+ u^2\ln |u|-{1\over 3}u^2-{1\over 2}\\ \Rightarrow \bbox[red, 2pt]{y=C_1(a+x)^2+C_2{1\over a+x}+ (a+x)^2\ln |a+x|-{1\over 3}(a+x)^2-{1\over 2}}$$

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解：
$$det(A-\lambda I)=0 \Rightarrow  \begin{vmatrix}2-\lambda & 0 & 0 \\1 & -\lambda & 2 \\ 0 & 0 & 3-\lambda \end{vmatrix} =0 \Rightarrow -\lambda(\lambda-3)(\lambda-2)=0  \Rightarrow \bbox[red, 2pt]{特徵值\lambda = 0,2,3};\\ \lambda=0  \Rightarrow (A-\lambda I)X=0  \Rightarrow \begin{bmatrix}2 & 0 & 0 \\1 & 0 & 2 \\ 0 & 0 & 3 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow  \begin{cases}x_1 = 0\\x_1+2x_3=0\\ x_3=0\end{cases}  \Rightarrow 取u_1=\begin {bmatrix} 0\\1\\0\end{bmatrix}\\ \lambda=2  \Rightarrow (A-\lambda I)X=0  \Rightarrow \begin{bmatrix}0 & 0 & 0 \\1 & -2 & 2 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow  \begin{cases}x_1=2x_2\\ x_3=0\end{cases}  \Rightarrow 取u_2=\begin {bmatrix} 2\\1\\0\end{bmatrix}\\ \lambda=3  \Rightarrow (A-\lambda I)X=0  \Rightarrow \begin{bmatrix}-1 & 0 & 0 \\1 & -3 & 2 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow  \begin{cases}x_1 = 0\\3x_2=2x_3\end{cases}  \Rightarrow 取u_3=\begin {bmatrix} 0\\2\\3\end{bmatrix}\\ \Rightarrow \bbox[red,, 2pt]{特徵向量為\begin {bmatrix} 0\\1\\0\end{bmatrix}, \begin {bmatrix} 2\\1\\0\end{bmatrix}, \begin {bmatrix} 0\\2\\3\end{bmatrix}};$$

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解：
(一) $$\nabla T=( \frac{\partial f}{\partial x} ,\frac{\partial f}{\partial y} ,\frac{\partial f}{\partial z} ) =(y+z,x+z,y+x) \Rightarrow \nabla T(1,1,1)=\bbox[red, 2pt]{(2,2,2)}=2\hat i+2\hat j+2\hat k$$ (二) $$\vec{v}=3\hat i-4\hat k \Rightarrow \vec{u}={\vec{v}\over |\vec{v}|} ={3\over 5}\hat i-{4\over 5}\hat k \Rightarrow \nabla T(1,1,1)\cdot \vec{u}=(2,2,2)\cdot  (3/5,0,-4/5)\\={6\over 5}+0-{8\over 5} =\bbox[red, 2pt]{-{2\over 5}}$$

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解：
$$\unicode{x222F}_S F\cdot dA =\iiint_T \nabla\cdot (x \hat i+y\hat j+z\hat k)dV = \iiint_T ({\partial \over \partial x}x + {\partial \over \partial y} y+ {\partial \over \partial z} z)dV = \iiint_T 3\;dV\\ 由於S:x^2+y^2+z^2=1為半徑為1的球體，體積為{4\over 3}\pi \Rightarrow \iiint_T 3\;dV={4\over 3}\pi \times 3=\bbox[red, 2pt]{4\pi}\\ 其中T代表表面為S的封閉區域$$

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解：
$$令 \begin{cases}f(z) = 1/(z+1)\\ g(z)=1/(z-1)\end{cases}  \Rightarrow \oint_c {dz \over z^2-1}= \oint_c {dz \over (z+1)(z-1)}=\oint_c {f(z)dz \over z-1}=\oint_c {g(z)dz \over z+1} \\ (一)2\pi i\times f(1)= 2\pi i \times {1\over 2}= \bbox[red, 2pt]{\pi i}\\ (二)2\pi i\times g(-1) =2\pi i\times {1\over -2} = \bbox[red, 2pt]{-\pi i}$$

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解：
(一) $$E[X(X-4)]=5 \Rightarrow E[X^2-4X]=5\\ \Rightarrow E[X^2]-4E[X]=5  \Rightarrow E[X^2] =5+4E[X]= 5+4\times 2=\bbox[red, 2pt]{13}$$ (二) $$E(-4X+10)= -4E(x)+10=-4\times 2+10 = \bbox[red, 2pt]{2}$$ (三) $$Var(X)=E(X^2)-(E(X))^2 = 13-2^2=9  \Rightarrow Var(-4X+10)= 16\cdot Var(X)=16\times 9=144\\ \Rightarrow \sqrt{Var(-4X+10)} =\sqrt{144}= \bbox[red, 2pt]{12}$$

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