114年中山大學資工碩士班-工程數學詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：工程數學【資工系碩士班乙組】

             

解答: $$\cases{2x_2+3x_3+ 4x_4=1\\ x_1-3x_2+4x_3+5x_4=2 \\ -3x_1+10x_2-6x_3-7x_4=-4} \Rightarrow \begin{bmatrix} 0& 2& 3& 4\\ 1& -3& 4& 5\\ -3& 10& -6& -7\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix}=\begin{bmatrix} 1\\ 2\\ -4\end{bmatrix} \\ \Rightarrow \left[ \begin{array} {rrrr| r} 0& 2& 3& 4& 1\\ 1& -3& 4& 5& 2\\ -3& 10& -6& -7& -4\end{array} \right] \xrightarrow{R_3+3R_2 \to R_3}  \left[ \begin{array} {rrrr| r} 0 & 2 & 3 & 4 & 1\\1 & -3 & 4 & 5 & 2\\0 & 1 & 6 & 8 & 2 \end{array} \right] \xrightarrow{R_1\leftrightarrow R_2 } \left[ \begin{array} {rrrr| r} 1 & -3 & 4 & 5 & 2\\0 & 2 & 3 & 4 & 1\\0 & 1 & 6 & 8 & 2 \end{array} \right] \\ \xrightarrow{R_1+3R_3\to R_1, R_2-2R_3\to R_2}  \left[ \begin{array} {rrrr| r} 1 & 0 & 22 & 29 & 8\\0 & 0 & -9 & -12 & -3\\0 & 1 & 6 & 8 & 2 \end{array} \right] \xrightarrow{R_2 \leftrightarrow R_3} \left[ \begin{array} {rrrr| r} 1 & 0 & 22 & 29 & 8\\0 & 1 & 6 & 8 & 2\\0 & 0 & -9 & -12 & -3 \end{array} \right] \\ \xrightarrow{(-1/9)R_3 \to R_3} \left[ \begin{array} {rrrr| r} 1 & 0 & 22 & 29 & 8\\0 & 1 & 6 & 8 & 2\\0 & 0 & 1 & \frac{4}{3} & \frac{1}{3} \end{array} \right]  \xrightarrow{ R_1-22R_3\to R_2, R_2-6R_3\to R_2} \left[ \begin{array} {rrrr| r} 1 & 0 & 0 & - \frac{1}{3} & \frac{2}{3}\\0 & 1 & 0 & 0 & 0\\0 & 0 & 1 & \frac{4}{3} & \frac{1}{3} \end{array} \right] \\ \Rightarrow \cases{x_1-{1\over 3}x_4={2\over 3}\\ x_2=0\\ x_3+ {4\over 3}x_4 ={1\over 3}} \Rightarrow \text{ solutions of the system: } \bbox[red, 2pt]{\left \{\left({2\over 3}+{1\over 3}t, 0, {1\over 3}-{4\over 3}t, t \right), t\in \mathbb R \right\}}$$

解答: $$\textbf{(2.1) } \det\left( \begin{bmatrix} 2& 0 & 0 & -3& 1 \\ 0 & 0& 0& 0 &7\\ -3& 2& 0 & -1& -6\\ 2& -2 & -1 & 1& 4\\ 0 &0& 0 &4& 3\end{bmatrix} \right) = -\det\left( \begin{bmatrix}  0 & 0& 0& 0 &7\\2& 0 & 0 & -3& 1\\ -3& 2& 0 & -1& -6\\ 2& -2 & -1 & 1& 4\\ 0 &0& 0 &4& 3\end{bmatrix} \right) \\\quad =-7\det \left( \begin{bmatrix}   2& 0 & 0 & -3 \\ -3& 2& 0 & -1& \\2 & -2 & -1 & 1 \\ 0 &0& 0 &4 \end{bmatrix} \right) = 7\det \left(\begin{bmatrix}0 &0& 0 &4   \\ -3& 2& 0 & -1& \\2 & -2 & -1 & 1 \\  2& 0 & 0 & -3 \end{bmatrix} \right) \\\quad = -28 \det \left( \begin{bmatrix}  -3& 2& 0    \\2 & -2 & -1   \\  2& 0 & 0   \end{bmatrix} \right) = 28 \det \left( \begin{bmatrix}  2& 0 & 0\\2 & -2 & -1   \\ -3& 2 & 0 \end{bmatrix} \right) = 56 \det  \left( \begin{bmatrix}   -2 & -1   \\  2 & 0 \end{bmatrix} \right) = \bbox[red, 2pt]{112} \\\textbf{(2.2) }   \begin{bmatrix}  1& b& b^2\\ b& b^2 & b^3\\ b^2& b^3 &b^4\end{bmatrix}  \xrightarrow{R_2-bR_1 \to R_2} \begin{bmatrix}  1& b& b^2\\ 0& 0 & 0\\ b^2& b^3 &b^4\end{bmatrix}  \\\quad \Rightarrow \det\left( \begin{bmatrix}  1& b& b^2\\ b& b^2 & b^3\\ b^2& b^3 &b^4\end{bmatrix} \right) =\det\left( \begin{bmatrix}  1& b& b^2\\ 0& 0 & 0\\ b^2& b^3 &b^4\end{bmatrix} \right) =\bbox[red, 2pt] 0$$

解答: $$\textbf{(3.1) }\lambda^2+5\lambda+ 6=0 \Rightarrow (\lambda+3)(\lambda+ 2)=0 \Rightarrow \lambda=-2,-3 \Rightarrow x(t)= c_1e^{-2t}+ c_2e^{-3t} \\\quad \Rightarrow x'(t)=-2c_1e^{-2t}  -3c_2e^{-3t} \Rightarrow \cases{x(0)= c_1+c_2 = 2\\ x'(0)=-2c_1-3c_2 =3} \Rightarrow \cases{c_1=9\\ c_2=-7} \\\quad \Rightarrow \bbox[red, 2pt]{x(t)= 9e^{-2t} -7e^{-3t}}\\ \textbf{(3.2) }x'(t)=-18e^{-2t}+21 e^{-3t}=0 \Rightarrow 6e^{-2t}= 7e^{-3t} \Rightarrow e^t={7\over 6} \Rightarrow t= \ln{7\over 6} \\\quad \Rightarrow x(\ln{7\over 6})= {9\over (7/6)^2}-{7 \over (7/6)^3} ={3\cdot 6^2\over 7^2} = \bbox[red, 2pt]{108\over 49}$$

解答: $$x_p= A\cos t+ B\sin t \Rightarrow x_p'=-A\sin t+B\cos t \Rightarrow x_p''=-A\cos t-B\sin t \\ \Rightarrow x_p''-3x_p'-4x_p = (-5A-3B) \cos t+(3A-5B)\sin t=2\sin t \\ \Rightarrow \cases{-5A-3B=0\\ 3A-5B=2} \Rightarrow \cases{A=3/17\\ B=-5/17} \Rightarrow \bbox[red, 2pt] {x_p={3\over 17} \cos t-{5\over 17}\sin t}$$

解答: $$2L\{x''\}+ L\{x'\}+ 2L\{x\}=L\{\delta(t-5)\} \Rightarrow 2s^2X(s)+ sX(s) +2X(s) =e^{-5s} \\ \Rightarrow X(s)={e^{-5s} \over 2s^2+ s+2} \Rightarrow x(t) =L^{-1}\{X(s)\} = L^{-1}\{ {e^{-5s} \over 2s^2+ s+2}\} \\ L^{-1}\left\{{1\over 2s^2+s+2}\right\} = L^{-1}\left\{{1\over 2 }\cdot {1\over (s+{1\over 4})^2+ {15 \over 16}}\right\} = {1\over 2}e^{-t/4} \cdot {4\over \sqrt{15}} \sin{\sqrt{15} t \over 4} ={2\over \sqrt{15}}e^{-t/4} \sin{\sqrt{15}t\over 4} \\ \Rightarrow L^{-1} \left\{{e^{-5s}\over 2s^2+s+2}\right\} =\bbox[red, 2pt]{x(t) = {2\over \sqrt{15}} e^{-(t-5)/4} \sin{\sqrt{15}(t-5)\over 4}u(t-5)}$$

解答: $$\textbf{(6.1)$$

$$\textbf{(6.2) }f(-x)=-f(x) \Rightarrow f(x) \text{ is odd }\Rightarrow a_n=0\\\qquad b_n= {4\over \pi}\int_0^{\pi/2} 2x^2 \sin(2nx)\,dx = {8\over \pi } \cdot  \left(({1\over 4n^3}-{\pi^2 \over 8n})(-1)^n -{1\over 4n^3}\right)\\ = \left({2\over n^3\pi}-{\pi\over n} \right)(-1)^n-{2\over n^3\pi}  \Rightarrow \bbox[red, 2pt]{f(x) =\sum_{n=1}^\infty \left( \left({2\over n^3\pi}-{\pi\over n} \right)(-1)^n-{2\over n^3\pi} \right) \sin(2n x)}$$

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解題僅供參考，碩士班歷年試題及詳解