台灣聯合大學系統113學年度碩士班招生考試
類組：電機類 科目：工程數學A（3003）
多種選擇題，共20題，每題5分
解答：$$y'-y=1 \Rightarrow y'e^{-x}-ye^{-x} =e^{-x} \Rightarrow (ye^{-x}) '=e^{-x} \Rightarrow ye^{-x} =\int e^{-x}\,dx =-e^{-x}+C\\ \Rightarrow y=Ce^{x}-1,故選\bbox[red, 2pt]{(C)}$$
解答：$$(1-x^2)y- xy'=0 \Rightarrow (1-x^2)y=xy' \Rightarrow {1\over y}y'={1-x^2 \over x} \Rightarrow \int {1\over y}dy =\int{1-x^2\over x}\,dx \\ \Rightarrow \ln y=\ln x-{1\over 2}x^2+c_1 \Rightarrow y=e^{\ln x-{1\over 2}x^2+c_1} =x\cdot e^{-x^2/2}\cdot e^{c_1} =c_2xe^{-x^2/2}, 故選\bbox[red, 2pt]{(D)}$$
解答：$$y''= \sin (-x) \Rightarrow y'= \int \sin(-x)\,dx =\cos(-x)+c_1 \Rightarrow y= \int \left(\cos(-x)+c_1 \right)\,dx \\=-\sin(-x)+c_1x+c_2,故選\bbox[red, 2pt]{(C)}$$
解答：$$xy'+ y=3 \Rightarrow (xy)'=3 \Rightarrow xy =\int 3\,dx =3x+c \Rightarrow y=3+{c\over x},故選\bbox[red, 2pt]{(A)}$$
解答：$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \Rightarrow x^2y''-5xy'+8y=m(m-1)x^m-5mx^m+8x^m = (m^2-6m+8)x^m=0 \\ \Rightarrow m^2-6m+8=0 \Rightarrow (m-4)(m-2)=0 \Rightarrow m=2,4, 故選\bbox[red, 2pt]{(B)}$$
解答：$$f(x)=e^{2x} \Rightarrow f^{[n]}(x) =2^ne^{2x} \Rightarrow f(x) = \sum_{n=0}^\infty {f^{[n]}(3)\over n!}(x-3)^n = \sum_{n=0}^\infty {2^ne^6\over n!}(x-3)^n \\ \Rightarrow \text{ the coefficient of }(x-3)^5={2^5e^6\over 5!} ={32\times {e^6\over 5!}}, 故選\bbox[red, 2pt]{(D)}$$
解答：$$(A)\times: f(x)= \sin(2x+{\pi\over 4}) \Rightarrow f'(\pi/2) =-\sqrt 2 \Rightarrow \text{the coefficient of }(x-\pi/2) =-\sqrt 2\ne 0 \\(B)\times: f(x)=\cos(2(x-\pi/2)) =c_0+ c_1(x-\pi/2)^2+ c_2(x-\pi/2)^4+ \cdots \\\qquad \Rightarrow \text{the coefficient of }x \ne 0 \\(C)\times: f(x)=e^{x^2-\pi} \Rightarrow f'(\pi/2)=\pi e^{\pi^2/4-\pi} \ne 0 \\(D)\times: f(x)={1\over 1+x^2-\pi} \Rightarrow f'(\pi/2) =-{\pi\over (1+\pi^2/4-\pi)^2} \ne 0\\ (E)\times: f(x)={x\over 1+x^3-\pi} \Rightarrow f'(\pi/2) = {1-\pi^3/4-\pi\over ((1+\pi^3/6)-\pi)^2} \ne 0 \\ 故\bbox[cyan, 2pt]{None}$$
解答：$$I = \int_{-\pi/2}^{\pi/2} \frac{1}{1+ \sin^2 \theta} d\theta = 2 \int_{0}^{\pi/2} \frac{1}{1+ \sin^2 \theta} d\theta = 2 \int_{0}^{\pi/2} \frac{\sec^2 \theta}{\sec^2 \theta + \tan^2 \theta} d\theta= 2 \int_{0}^{\pi/2} \frac{\sec^2 \theta}{1 + 2\tan^2 \theta} d\theta \\ u=\tan \theta \Rightarrow du=\sec^2 \theta \,d\theta \Rightarrow I = 2 \int_{0}^{\infty} \frac{1}{1 + 2u^2} du =2 \cdot \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{1}{1+v^2} dv \; (v=\sqrt 2u) \\= \sqrt 2 \left. \left[ \tan^{-1}v \right] \right|_0^\infty =\sqrt 2\cdot {\pi\over 2} ={\pi\over \sqrt 2}, 故選\bbox[red, 2pt]{(D)}$$
解答：$$\int_0^\infty {1\over 1+x^n}\,dx = {\pi \over n \sin(\pi/n)} \; \href{https://math.stackexchange.com/questions/247866/show-that-int-0-infty-frac11xn-dx-frac-pi-n-sin-pi-n-wh}{公式來源} \\ \Rightarrow \int_0^\infty {dx\over 1+x^{56}} ={\pi\over 56 \sin(\pi/56)}, 故選\bbox[red, 2pt]{(A)}$$
解答：$$z^2+1=0 \Rightarrow \cases{z=i \in C\\ z=-i \not \in C} \Rightarrow \text{Res}(f,i) ={Ln(1+i) \over 2i} ={\ln \sqrt 2+\pi i/4\over 2i} ={\pi\over 8}-i{\ln 2\over 4} \\ \Rightarrow I=\oint_C f(z)\,dz =2\pi i \left( {\pi\over 8}-i{\ln 2\over 4} \right) ={\pi \ln 2\over 2}+i {\pi^2 \over 4} \Rightarrow Im(I)={\pi^2 \over 4},故選\bbox[red, 2pt]{(B)}$$
解答：$$(A) \times: \det(A)=0 \Rightarrow A \text{ is NOT invertible} \\(B) \bigcirc: A= \begin{bmatrix} -i& i\\ 1& 1\end{bmatrix} \begin{bmatrix}0& 0\\0& 2 \end{bmatrix} \begin{bmatrix}i/2& 1/2\\ -i/2& 1/2 \end{bmatrix} \\(C)\bigcirc: A= \begin{bmatrix} 1& i\\-i& 1 \end{bmatrix} \Rightarrow A^* =\begin{bmatrix} 1& i\\-i& 1 \end{bmatrix} \Rightarrow A=A^* \Rightarrow A^*A= AA^* \Rightarrow A \text{ is normal} \\(D)\bigcirc: A^*=A\\ (E)\times: A^*A = \begin{bmatrix}2& 2i\\ -2i& 2\end{bmatrix} \ne I_2 \\ 故選\bbox[red, 2pt]{(BCD)}$$
解答：$$(D)\times: \cases{\text{elementary row operations} \sim O(n^3) \\ \text{cofactor expansion }\sim O(n!)} \Rightarrow \cases{ O(n!) >>O(n^3), \text{ for large }n\\ O(n!) \approx O(n^3), \text{for small }n}\\ 故選\bbox[red, 2pt]{(D)}$$
解答：$$(A) \times: W=\{ \vec 0\} \Rightarrow W \text{ is a subspace and contain only one vector} \\(D)\times: V=W \Rightarrow V \subseteq W, \text{ but dim}(V) =\text{dim}(W)\\ 故選\bbox[red, 2pt]{(BCE)}$$
解答：$$(A) \times: ker(T)=\{(0,0,0)\} =\{\vec 0\} \Rightarrow \text{ the basis of ker}(T) = \varnothing \\(B)\bigcirc: T \text{ is 1-1, iff ker}(T)= \varnothing \\(C)\bigcirc:T \text{ is onto}\\ (D)\bigcirc: A= \begin{bmatrix}3& 2& 0\\ -2& 3& 0\\ 0& 0& 5 \end{bmatrix} \Rightarrow \det(A)=65 \ne 0 \\(E)\times: \det(A-\lambda I) =0 \Rightarrow \lambda =5, 3\pm 2i \not \in \mathbb R\\ 故選\bbox[red, 2pt]{(BCD)}$$
解答：$$\begin{vmatrix} 6& 1&-2& 0& 0\\ 2& 2& 3& 2& 2\\ 4& 3& 1& 3& 4\\-1 &3& 2& 0& 0\\ 2& 1& -1+\cos t& 1& 2 \end{vmatrix} \xrightarrow{R_1+6R_4 \to R_1, R_2+2R_4 \to R_2, R_3+4R_4 \to R_3, R_5+2R_4 \to R_5} \\\begin{vmatrix} 0& 19& 10& 0& 0\\ 0& 8& 7& 2& 2\\ 0& 15 & 9& 3& 4\\-1 &3& 2& 0& 0\\ 0& 7& 3+\cos t& 1& 2 \end{vmatrix} = \begin{vmatrix} 19& 10& 0& 0\\ 8& 7& 2& 2\\ 15 & 9& 3& 4 \\ 7& 3+\cos t& 1& 2 \end{vmatrix} =19 \begin{vmatrix} 7& 2& 2\\ 9& 3& 4 \\ 3+\cos t& 1& 2 \end{vmatrix} -10 \begin{vmatrix} 8& 2& 2\\ 15 & 3& 4 \\ 7& 1& 2 \end{vmatrix} \\=19(2+2\cos t)-10\cdot 0= 38(1+\cos t) =0 \Rightarrow t=\pm \pi , 故選\bbox[red, 2pt]{(AE)}$$
解答：$$\det(A-\lambda I)=-(\lambda-1)(\lambda+1)^2=0 \Rightarrow \lambda=1,-1\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}0 & -2 & 8 \\0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_2=0\\ x_3=0} \\\qquad \Rightarrow \mathbf v= x_1 \begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix}, \text{ choose }v_1=\begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix} \\ \lambda_2=-1 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}2 & -2 & 8 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1-x_2+4x_3=0 \\\qquad \Rightarrow \mathbf v= x_2 \begin{bmatrix}1 \\1\\ 0 \end{bmatrix}+ x_3 \begin{bmatrix}-4 \\0 \\ 1 \end{bmatrix}, \text{ choose }v_2=\begin{bmatrix}1 \\1\\ 0 \end{bmatrix}, v_3= \begin{bmatrix}-4 \\0 \\ 1 \end{bmatrix}, 故選\bbox[red, 2pt]{(C)}$$
解答：$$A= \begin{bmatrix}1 & 1 & -4 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\0 & -1 & 0 \\0 & 0 & -1 \end{bmatrix} \begin{bmatrix}1 & -1 & 4 \\0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ \Rightarrow A^{2301}=\begin{bmatrix}1 & 1 & -4 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1^{2301} & 0 & 0 \\0 & (-1)^{2301} & 0 \\0 & 0 & (-1)^{2301} \end{bmatrix} \begin{bmatrix}1 & -1 & 4 \\0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = A\\ \Rightarrow A^{2301}=\begin{bmatrix}1 & -2 & 8 \\0 & -1 & 0 \\0 & 0 & -1 \end{bmatrix}, 故選\bbox[red, 2pt]{(D)}$$
解答：$$A=MM^T= \begin{bmatrix}333& 81\\81&117 \end{bmatrix} \Rightarrow \det(A-\lambda I)= \lambda^2-450\lambda+32400 =(\lambda-360) (\lambda-90)=0 \\ \Rightarrow \text{eigenvalues: }\cases{\lambda_1 =360\\ \lambda_2=90} \Rightarrow \text{singular values:} \cases{\sqrt{\lambda_1} =6\sqrt{10} \\ \sqrt{\lambda_2} =3\sqrt{10}},故選\bbox[red, 2pt]{(E)}$$
解答：$$\text{the largest singular value: }6\sqrt{10}, 故選\bbox[red, 2pt]{(D)}$$
解答：$$(A) \times: \beta=-1 \Rightarrow \det(B-\lambda I) =-(\lambda-1)(\lambda+2)^2 \Rightarrow \cases{\lambda=1\gt 0\\ \lambda=-2\lt 0} \\(B) \times:\beta=0 \Rightarrow \det(B-\lambda I) =-(\lambda+1)^2(\lambda -2) \Rightarrow \cases{\lambda=-1\lt 0\\ \lambda=2 \gt 0} \\(C)\times: \beta=1 \Rightarrow \det(B-\lambda I) =-\lambda^2(\lambda-3) \Rightarrow \lambda \ge 0 \Rightarrow \text{positive semi-definite} \\(D)\bigcirc: \beta=2 \Rightarrow \det(B-\lambda I) =-(\lambda-1)^2 (\lambda-4) \Rightarrow \lambda\gt 0 \Rightarrow \text{positive definite} \\(E) \times:\beta=i \Rightarrow \det(B-\lambda I) =-(\lambda-(2+i)) (\lambda+(1-i))^2 \Rightarrow \lambda \in \mathbb C \\故選 \bbox[red, 2pt]{(D)}$$
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解題僅供參考，其他碩士班歷年考題
解答：$$(A) \times: \det(A)=0 \Rightarrow A \text{ is NOT invertible} \\(B) \bigcirc: A= \begin{bmatrix} -i& i\\ 1& 1\end{bmatrix} \begin{bmatrix}0& 0\\0& 2 \end{bmatrix} \begin{bmatrix}i/2& 1/2\\ -i/2& 1/2 \end{bmatrix} \\(C)\bigcirc: A= \begin{bmatrix} 1& i\\-i& 1 \end{bmatrix} \Rightarrow A^* =\begin{bmatrix} 1& i\\-i& 1 \end{bmatrix} \Rightarrow A=A^* \Rightarrow A^*A= AA^* \Rightarrow A \text{ is normal} \\(D)\bigcirc: A^*=A\\ (E)\times: A^*A = \begin{bmatrix}2& 2i\\ -2i& 2\end{bmatrix} \ne I_2 \\ 故選\bbox[red, 2pt]{(BCD)}$$
解答：$$(D)\times: \cases{\text{elementary row operations} \sim O(n^3) \\ \text{cofactor expansion }\sim O(n!)} \Rightarrow \cases{ O(n!) >>O(n^3), \text{ for large }n\\ O(n!) \approx O(n^3), \text{for small }n}\\ 故選\bbox[red, 2pt]{(D)}$$
解答：$$(A) \times: W=\{ \vec 0\} \Rightarrow W \text{ is a subspace and contain only one vector} \\(D)\times: V=W \Rightarrow V \subseteq W, \text{ but dim}(V) =\text{dim}(W)\\ 故選\bbox[red, 2pt]{(BCE)}$$
解答：$$(A) \times: ker(T)=\{(0,0,0)\} =\{\vec 0\} \Rightarrow \text{ the basis of ker}(T) = \varnothing \\(B)\bigcirc: T \text{ is 1-1, iff ker}(T)= \varnothing \\(C)\bigcirc:T \text{ is onto}\\ (D)\bigcirc: A= \begin{bmatrix}3& 2& 0\\ -2& 3& 0\\ 0& 0& 5 \end{bmatrix} \Rightarrow \det(A)=65 \ne 0 \\(E)\times: \det(A-\lambda I) =0 \Rightarrow \lambda =5, 3\pm 2i \not \in \mathbb R\\ 故選\bbox[red, 2pt]{(BCD)}$$
解答：$$\begin{vmatrix} 6& 1&-2& 0& 0\\ 2& 2& 3& 2& 2\\ 4& 3& 1& 3& 4\\-1 &3& 2& 0& 0\\ 2& 1& -1+\cos t& 1& 2 \end{vmatrix} \xrightarrow{R_1+6R_4 \to R_1, R_2+2R_4 \to R_2, R_3+4R_4 \to R_3, R_5+2R_4 \to R_5} \\\begin{vmatrix} 0& 19& 10& 0& 0\\ 0& 8& 7& 2& 2\\ 0& 15 & 9& 3& 4\\-1 &3& 2& 0& 0\\ 0& 7& 3+\cos t& 1& 2 \end{vmatrix} = \begin{vmatrix} 19& 10& 0& 0\\ 8& 7& 2& 2\\ 15 & 9& 3& 4 \\ 7& 3+\cos t& 1& 2 \end{vmatrix} =19 \begin{vmatrix} 7& 2& 2\\ 9& 3& 4 \\ 3+\cos t& 1& 2 \end{vmatrix} -10 \begin{vmatrix} 8& 2& 2\\ 15 & 3& 4 \\ 7& 1& 2 \end{vmatrix} \\=19(2+2\cos t)-10\cdot 0= 38(1+\cos t) =0 \Rightarrow t=\pm \pi , 故選\bbox[red, 2pt]{(AE)}$$
解答：$$\det(A-\lambda I)=-(\lambda-1)(\lambda+1)^2=0 \Rightarrow \lambda=1,-1\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}0 & -2 & 8 \\0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_2=0\\ x_3=0} \\\qquad \Rightarrow \mathbf v= x_1 \begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix}, \text{ choose }v_1=\begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix} \\ \lambda_2=-1 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}2 & -2 & 8 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1-x_2+4x_3=0 \\\qquad \Rightarrow \mathbf v= x_2 \begin{bmatrix}1 \\1\\ 0 \end{bmatrix}+ x_3 \begin{bmatrix}-4 \\0 \\ 1 \end{bmatrix}, \text{ choose }v_2=\begin{bmatrix}1 \\1\\ 0 \end{bmatrix}, v_3= \begin{bmatrix}-4 \\0 \\ 1 \end{bmatrix}, 故選\bbox[red, 2pt]{(C)}$$
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