115年中山大學電機碩士班-工程數學甲詳解

國立中山大學 115學年度碩士班考試入學招生考試試題

科目名稱:工程數學甲【電機系碩士班戊組選考、庚組、通訊所碩士班乙組選考、電波聯合碩士班選考】

下面1-10題為複選題,每題5分,總分50分。每題有五個選項,其中至少有一個是正確答案。答錯1個選項者,得3分,答錯2個選項者,得1分,答錯多於2個選項或未作答者,該題以零分計算。

解答:$$(A)\times: \text{This is not necessarily true.} \\(B)\times: Bx=d \text{ has no solution.}  \Rightarrow  d \not \in R(B) \Rightarrow d \not \in span\{a_1,a_2,a_3\} \\(C)\times: \text{This is not necessarily true} \\(D)\bigcirc: dim(R(C)) \gt dim(R(B)) \Rightarrow rank(C) \gt rank(B) \\(E)\bigcirc: \cases{d\not \in R(B) \Rightarrow rank([B,d])= rank(B)+1 \\ C\text{ has exactly one more column than B} \Rightarrow rank(C)\le rank(B)+1\\ rank(C) \gt rank(B)} \\\qquad \Rightarrow rank(C) =rank([B,d]) \\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(DE)}$$

解答:$$(A)\bigcirc: R(AA^T) \subseteq R(A) \Rightarrow y\in R(AA^T) \Rightarrow y\in R(A).R(A) \text{ and }N(A^T) \text{ are orthogonal} \\\qquad \text{complements. Therefore, }R(A) \cap N(A^T)= \vec 0. \text{ Now }y\ne0 \text{ and }y \in R(A) \Rightarrow y \not \in N(A^T) \Rightarrow A^T(y)\ne 0 \\(B)\times: R(A) \text{ and }N(A^T) \text{ are orthogonal complements. The only way this statement could} \\\qquad \text{ be true if }R(A)=\{\vec 0\}, \text{ meaning $A$ is the zero matrix. It dones not hold for general matrix.} \\(C)\times: A= \begin{bmatrix}0& 1\\0& 0 \end{bmatrix} \Rightarrow A^T = \begin{bmatrix}0&0\\1&0 \end{bmatrix} \Rightarrow \cases{R(A) \text{ is spanned by } \begin{bmatrix}1\\0 \end{bmatrix} \\R(A^T) \text{ is spanned by} \begin{bmatrix}0\\1 \end{bmatrix}} \Rightarrow R(A) \ne R(A^T) \\(D)\times:A= \begin{bmatrix}0& 1\\0& 0 \end{bmatrix} \Rightarrow A^T = \begin{bmatrix}0&0\\1&0 \end{bmatrix} \Rightarrow \cases{N(A)\text{ is spanned by} \begin{bmatrix}1\\0 \end{bmatrix} \\N(A^T) \text{ is spanned by} \begin{bmatrix}0\\ 1 \end{bmatrix}} \Rightarrow N(A)\ne N(A^T) \\(E)\times: A= \begin{bmatrix}1& 0\\0& 0 \end{bmatrix}\text{ and }x= \begin{bmatrix}1\\1 \end{bmatrix} \Rightarrow Ax= \begin{bmatrix}1\\ 0 \end{bmatrix} \ne 0. \text{ But }R(A^T) \text{ is spanned by} \begin{bmatrix}1\\0 \end{bmatrix},\\ \qquad \text{ that is, }A \not \in R(A^T) \\ \Rightarrow \text{The only true statement is }\bbox[red, 2pt]{(A)}$$

解答:$$(A)\times: y\in R(A) \Rightarrow y=Ax \text{ for some }x\in \mathbb R^n \Rightarrow y=(BC)x= B(Cx) \Rightarrow y\in R(B) \\ \qquad \Rightarrow R(A) \subseteq R(B)  \Rightarrow R(B) \not \subseteq R(A) \\(B)\times: x\in N(C) \Rightarrow Cx=0 \Rightarrow Ax=(BC)x=B(Cx)=B(0)=0 \Rightarrow x\in N(A) \\ \qquad \Rightarrow N(C) \subseteq N(A) \Rightarrow N(A) \not \subseteq N(C) \\(C)\bigcirc: rank(A) =rank(BC) \le \min(rank(B), rank(C)), \text{ where rank(B) $\le k$} \Rightarrow rank(A)\le k \\(D)\times: \text{Let }\cases{A=C= \begin{bmatrix}0&0\\0& 0 \end{bmatrix} \\B= \begin{bmatrix}1&0\\0& 1 \end{bmatrix}} \Rightarrow \cases{rank(A)= rank(C)=0\\ rank(B)=1\\ k=2 \\A=BC} \Rightarrow rank(B)\ne rank(C) \\(E) \bigcirc: \cases{rank(A)\le rank(B)\\ rank(B)\le k\\ rank(A)=k} \Rightarrow rank(A)=rank(B) \\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(CE)}$$

解答:$$(A)\bigcirc: A^T=A \Rightarrow \text{$A$ is a real symmetric matrix} \Rightarrow \text{diagonalizable }\\(B) \bigcirc: \cases{\text{Row 1=Row 3=Row 5}=[3,0,3,0,3] \\ \text{Row 2= Row 4}=[0,2,0,2,0]} \Rightarrow \text{ two linearly independent rows} \Rightarrow rank(A)=2 \\ (C)\times: A \text{ is 5x5 and }rank(A)=2 \Rightarrow null(A)=5-2=3 \Rightarrow \lambda=0 \text{ is an eigenvalue with} \\\qquad \text{ multiplicity of 3. And } \cases{\mathbf v_1= [1\; 0 \;1 \;0 \;1]^T \Rightarrow A\mathbf v_1 =9\mathbf v_1 \\ \mathbf v_2=[0\; 1\; 0\; 1\; 0]^T \Rightarrow A\mathbf v_2= 4\mathbf v_2} \Rightarrow  \text{eigenvalues: 9,4,0,0,0} \\ \qquad \Rightarrow 2 \text{ is not an eigenvalue of }A \\(D)\times: 3 \text{ is not an eigenvalue of }A \\(E)\bigcirc: \text{Two positive eigenvalues: 9 and 4.} \\ \Rightarrow \text{The true statements:}\bbox[red, 2pt]{(ABE)}$$

解答:$$(A)\times:v_1\cdot v_2=0 \Rightarrow v_1 \bot v_2 \Rightarrow p= \text{proj}_{W}(x) ={x\cdot v_1\over |v_1|^2}v_1+{x\cdot v_2\over |v_2|^2}v_2= {3\over 2} \begin{bmatrix}1\\0\\1 \end{bmatrix}+{1\over 2} \begin{bmatrix}1\\0\\-1 \end{bmatrix}= \begin{bmatrix}2\\ 0\\1 \end{bmatrix} \\ \qquad \Rightarrow ||p|| = \sqrt{5} \ne 2 \\(B)\bigcirc: x-p= \begin{bmatrix}2\\3\\1 \end{bmatrix}- \begin{bmatrix}2\\0\\1 \end{bmatrix} = \begin{bmatrix}0\\3\\0 \end{bmatrix} \Rightarrow ||x-p||=3 \\(C)\bigcirc: (x-p) \bot w\in W \Rightarrow w^T(x-p) =0 \\(D)\times: x-p \text{ is orthogonal to $W$ and $x-p=[0,3,0]^T \ne 0 \Rightarrow x-p \not \in W$} \\(E) \times: p\cdot v_1=3\ne 0 \\ \Rightarrow \text{The true statements:}\bbox[red,2pt]{(BC)}$$

解答:$$\cases{L(b_1)=2b_1 \\ L(b_2)=b_2 \\ L(b_3)=-b_3 } \Rightarrow A= \begin{bmatrix}2&0&0\\0&1&0\\0&0& -1 \end{bmatrix} \Rightarrow \cases{(D)\bigcirc: \det(A)=-2\\ (E)\times: rank(A)=3 \ne 2} \\ \text{Let }x=c_1b_1+ c_2b_2+ c_3 b_3 \Rightarrow \begin{bmatrix}c_1\\c_1\\0 \end{bmatrix}+ \begin{bmatrix}c_2\\-c_2\\ 0 \end{bmatrix}+ \begin{bmatrix}0\\0\\c_3 \end{bmatrix} = \begin{bmatrix}2\\0\\2 \end{bmatrix} \Rightarrow \cases{c_1=1\\c_2=1\\c_3=3} \\ \Rightarrow x= 1b_1+ 1b_2+3b_3 \Rightarrow L^5(x)= L^5(b_1+b_2+ 3b_3) =L^5(b_1) +L^5(b_2) +3L^5(b_3)\\ =2^5b_1+ 1^5b_2+ 3(-1)^5b_3=32b_1+b_2-3b_3 =32 \begin{bmatrix}1\\1\\0 \end{bmatrix} + \begin{bmatrix}1 \\-1\\0 \end{bmatrix}-3 \begin{bmatrix}0\\0\\1 \end{bmatrix} = \begin{bmatrix}33\\ 31\\-3  \end{bmatrix} = \begin{bmatrix}y_1\\ y_2\\ y_3 \end{bmatrix} \\ \Rightarrow \cases{(A)\bigcirc: y_1 =33\\ (B)\bigcirc: y_2=31\\ (C) \bigcirc: y_3=-3} \\ \Rightarrow \text{The true statements are: }\bbox[red, 2pt]{(ABCD)}$$

解答:$$(A)\bigcirc: F(x,y,z)=x^2+y^2-z=0 \Rightarrow \nabla F=(2x,2y,-1) =2x \mathbf i+2y\mathbf j-\mathbf k \\(B)\bigcirc: \nabla F(P)=\nabla F(2,-2,8) =(4,-4,-1) \Rightarrow \text{ tangent plane: }4(x-2)-4(y+2)-(z-8)=0 \\ \qquad \Rightarrow 4x-4y-z=8 \\(C)\times: \nabla F(P)=\nabla F(2,-2,8) =(4,-4,-1) \ne (4,-4,-8)\\(D)\times: |\nabla F(2,-2,8)|= \sqrt{4^2+(-4)^2+(-1)^2} = \sqrt{33} \ne \sqrt{32} \\(E)\bigcirc: \text{normal line: }{x-2\over 4}={y+2\over -4}={z-8\over -1}=t \Rightarrow \cases{x=4t+2\\ y=-4t-2\\ z=-t+8} \\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(ABE)}$$

解答:$$x^2+y^2=1 \Rightarrow \cases{x=\cos t\\ y=\sin t} \Rightarrow z=\cot^{-1} {\cos t\over \sin t} =\cot^{-1} \cot t=t \Rightarrow r(t)=( \cos t,\sin t, t) \text{ for }t\in (0,\pi) \\ \Rightarrow r'(t)=(-\sin t, \cos t, 1) \Rightarrow ||r'(t)||= \sqrt{(-\sin t)^2+ \cos^2t+1^2} =\sqrt 2 \Rightarrow r''(t)=(-\cos t, -\sin t,0) \\(A)\times: \bbox[cyan, 2pt]{題目有誤}, (1,0,\pi/2) \text{ does not lie on the curve.} \\(B)\times: \kappa ={||r'(t) \times r''(t)|| \over ||r'(t)||^3} ={\sqrt 2\over (\sqrt 2)^3} ={1\over 2} \ne 2 \\(C)\bigcirc: \rho={1\over \kappa}={1/(1/2)} =2 \\(D) \bigcirc: {r'(t) \over ||r'(t)||}= {(-\sin t, \cos t,1) \over \sqrt 2} \Rightarrow \left. {(-\sin t, \cos t,1) \over \sqrt 2}\right|_{t=\pi/2} = {1\over \sqrt 2}(-1,0,1) ={-\mathbf  i+\mathbf k\over \sqrt 2} \\(E)\times: T'(t)= {d\over dt} \left( {(-\sin t, \cos t,1) \over \sqrt 2} \right) ={1\over \sqrt 2}(-\cos t, -\sin t,0) \Rightarrow ||T'(t)||={1\over \sqrt 2} \\ \qquad \Rightarrow {T'(t) \over ||T'(t)||}= (-\cos t, -\sin t,0) \Rightarrow \left. (-\cos t, -\sin t,0)\right|_{t=\pi/2} =(0,-1,0) = -\mathbf j \ne \mathbf j \\ \Rightarrow \text{The correct statements: }\bbox[red, 2pt]{(CD)}$$

解答:$$Au_{xx}+ Bu_{xy}+ Cu_{yy}+ Du_x +Eu_y+Fu=G \Rightarrow \Delta =B^2-4AC=x^2-4\cdot 0\cdot 0 =x^2 \gt 0 \Rightarrow \text{ hyperbolic} \\ \Rightarrow \cases{(C) \bigcirc: \text{parabolic is INCORRECT} \\ (D)\bigcirc: \text{elliptic is INCORRECT}} \Rightarrow G=0 \Rightarrow \text{ the equation is homogeneous} \Rightarrow (E)\times: \text{CORRECT}\\ u(x,y)= X(x)Y(y) \Rightarrow xX'Y'+2yXY=0 \Rightarrow {xX'\over X}=-{2yY\over Y'}=\lambda \\\text{Solving for }X: {X'\over X}={\lambda\over x} \Rightarrow X=c_1x^\lambda \\ \text{Solving for }Y: {Y'\over Y}=-{2y\over \lambda} \Rightarrow \ln|Y|=-{y^2\over \lambda}+C \Rightarrow Y=c_2e^{-y^2/\lambda} \\ \Rightarrow u(x,y)= c_3 x^\lambda e^{-y^2/\lambda} =cx^{Ak }e^{By^E/Fk} \Rightarrow \lambda=Ak \Rightarrow u=cx^{Ak}e^{-y^2/Ak} \Rightarrow {By^E\over Fk}=-{y^2\over Ak} \\ \Rightarrow \cases{E=2\\ B/F=-1/A \Rightarrow AB=-F} \\\Rightarrow  (A)\times: A+B=0 \Rightarrow B=-A \Rightarrow F=A^2 \Rightarrow \cases{A=1\\ B=-1\\ E=2\\ F=1} \Rightarrow u=cx^ke^{-y^2/k} \Rightarrow \text{CORRECT} \\\Rightarrow (B)\times:E+F=3 \Rightarrow F=1 \Rightarrow AB=-1 \Rightarrow A=\pm 1, B=\mp1 \Rightarrow \text{Both pairs yield valid solutions} \\\qquad \Rightarrow \text{CORRECT} \\ \Rightarrow \text{The INCORRECT statements: }\bbox[red, 2pt]{CD}$$

解答:$$(A)\bigcirc: y'=y^3-3y+2 =(y-2)(y-1) \Rightarrow \cases{y=0(y\lt 1)\Rightarrow y'=2\gt 0\\ y=3/2(1\lt y\lt 2)\Rightarrow y'=-1/4\lt 0 \\ y=3(y\gt 2) \Rightarrow y'=2 \gt 0} \\ \qquad \Rightarrow \text{Because solutions below $y=1$ increase towards it and solutions above $y=1$ decrease}\\\text{ towards it, $y=1$ is an asymptotically stable equilibrium, which is known as an attractor.} \\(B) \times: {dy\over (y-1)(y-2)} =dx \Rightarrow \int \left( {1\over y-1}-{1\over y-1} \right)\,dy= \int 1\,dx \Rightarrow \ln \left|{y-2\over y-1} \right| =x+ c_1\\ \qquad \Rightarrow {y-2\over y-1}=c_2e^x \Rightarrow y=2 \text{ is a particular solution, if we set }c_2=0.\\ (C)\times: {y-2\over y-1}=c_2e^x \Rightarrow y={2-c_2e^x\over 1-c_2e^x} \ne {ce^x-2\over ce^x+1} \\(D)\times: y(0)=-1 \Rightarrow {-1-2\over -1-1}=c_2e^0 \Rightarrow c_2={3\over 2} \Rightarrow y={2-(3/2)e^x\over 1-(3/2)e^x} ={4-3e^x\over 2-3e^x} \\ \qquad 2-3e^x =0 \Rightarrow x=\ln(2/3) \lt 0 \Rightarrow \text{ the largest interval: }(\ln(2/3), \infty) \ne [0, \infty) \\(E)\bigcirc:  \text{By the fundamental existence and uniqueness theorem for 1-st ODEs,} \\\qquad \text{ it quarantees that there is some  open interval centered at $x=0$ where }\\\qquad \text{ a unique local soluton exists.} \\ \Rightarrow \text{The correct statements are }\bbox[red, 2pt]{(AE)}$$

下面11-15題為單選題,總分20分。每題答對4分,答錯或未作答者以0分計。

$$\text{For the following problems, let $j=\sqrt{-1}$ denote the imaginary unit.}$$

解答:$$z=3-4j \Rightarrow |z|= \sqrt{3^2+ (-4)^2} =5 \Rightarrow \bbox[red, 2pt]{(B)}$$

解答:$$z=x+iy \Rightarrow \bar z=x-iy \Rightarrow \cases{u(x,y)=x\\ v(x,y)=-y} \Rightarrow \cases{u_x=1\\ v_y=-1} \Rightarrow u_x\ne v_y \\ \Rightarrow \text{The correct answer is }\bbox[red, 2pt]{(C)}$$

解答:$$f(z)=z^2 \Rightarrow f'(z)=2z \Rightarrow f'(1+j) =2(1+j)=2+2j \Rightarrow \text{The correct answer is }\bbox[red, 2pt]{(B)}$$

解答:$$(A)\times: \cases{u(x,y)=x^2+y^2 \\ v(x,y)=0} \Rightarrow u_x\ne v_y \\(B) \bigcirc: \cases{u(x,y)=x^2-y^2\\ v(x,y)=2xy} \Rightarrow \cases{u_x=2x\\ u_y=-2y\\ v_x=2y\\ v_y=2x} \Rightarrow \cases{u_x=v_y\\ u_y=-v_x} \\(C)\times \cases{u(x,y) =x^2\\ v(x,y)=y^2} \Rightarrow u_x \ne v_y \\ (D)\times: z=x+iy \Rightarrow |z|^2=x^2+y^2  \Rightarrow \text{this is the exact same function as Option (A).} \\ \Rightarrow \text{The correct answer is }\bbox[red, 2pt]{(B)}$$

解答:$$f(z)={1\over z} \Rightarrow \text{Res}(f,0)=1 \Rightarrow \oint_{|z|=2} f(z)\,dz =2\pi j\cdot \text{Res}(f,0)=2\pi j \Rightarrow \text{Correct statement: }\bbox[red, 2pt]{(D)}$$

下面16-21題為單選題,總分30分。每題答對5分,答錯或未作答者以0分計。

解答:$$L\{t \sin(3t)\} =-{d\over dt} L\{\sin(3t)\}=-{d\over dt} \left({3\over s^2+9} \right) ={ 6s\over (s^2 +9)^2} \\ \Rightarrow L\{f(t)\} =L\{ \cos(3t)\} +L\{3t\sin(3t)\}={s\over s^2+9}+ 3\cdot {6s\over (s^2+9)^2} ={s(s^2+9)+ 18s\over (s^2+9)^2} \\ ={s(s^2+27) \over (s^2+9)^2} \Rightarrow \bbox[red, 2pt]{(C)}$$

解答:$$x_1(t) = \sum_{k=0}^{100} ({1\over 2})^k e^{jk{2\pi\over 50}t} \Rightarrow a_k=({1\over 2})^k, 0\le k\le 100 \text{ and }a_k=0, \text{for all negative }k \\ \Rightarrow \cases{a_1=1/2\\ a_{-1}=0} \Rightarrow a_1 \ne a_{-1} \Rightarrow x_1(t) \text{ is NOT even} \\ x_2(t)= \sum_{k=-100}^{100} \cos(k\pi) e^{jk{2 \pi \over 50}t} \Rightarrow a_k=\cos(k\pi) \Rightarrow a_{-k} =\cos(-k\pi) =\cos (k\pi) =a_k \\ \qquad \Rightarrow x_2(t) \text{ is even} \\ x_3(t) =\sum_{k=-100}^{100} j\sin{k\pi\over 2} e^{jk{2\pi\over 50}t} \Rightarrow a_k=j\sin{k\pi\over 2} \Rightarrow a_{-k} =-j\sin{k\pi\over 2} =-a_k \Rightarrow x_3(t) \text{ is NOT even} \\ \Rightarrow \text{The correct answer is }\bbox[red, 2pt]{(B)}$$

解答:$$(A)\times:\cases{a_1=j/2\\ a_{-1} =j/2 \Rightarrow a_{-1}^* =-j/2} \Rightarrow a_1 \ne a_{-1}^* \Rightarrow x(t) \text{ is NOT real} \\(B) \bigcirc: a_k=j({1\over 2})^{|k|} \Rightarrow a_{-k}=a_k \Rightarrow x(t) \text{ is even} \\(C)\times: \text{The derivative of an even function is an odd function} \\ \Rightarrow \text{The correct answer is }\bbox[red, 2pt]{(B)}$$

解答:$$\mathcal F\{x(at)\} = {1\over |a|}X({\omega\over a}), \text{ now }x(t)=e^{-|t|} \text{ and }\mathcal F\{x(t)\}=X(\omega) ={2\over 1+\omega^2} \\ \Rightarrow \mathcal F\{x(2t)\} ={1\over |2|} X({\omega\over 2}) ={1\over 2}\cdot {2\over 1+(\omega/2)^2} ={4\over 4+ \omega^2} \Rightarrow \bbox[red, 2pt]{(A)}$$

解答:$$(A) \times: X(j\omega) =u(\omega)-u(\omega-2) \Rightarrow \cases{X(j1)=1\\ X(-j1)=0} \Rightarrow X(-j1) \ne -X(j1) \Rightarrow \text{Not odd} \\(C)\times: X(j\omega)= {\sin(2\omega) \over \omega} e^{j(2\omega+\pi/2)} =j{\sin(2\omega) \over \omega} e^{j(2\omega)} \Rightarrow \text{Not pure imaginary} \\ (D)\times: X(j\omega)= \sum_{k=-\infty}^\infty ({1\over 2})^{|k|} \delta(\omega-{k\pi\over 4}) \Rightarrow X(-j\omega)=X(j\omega) \Rightarrow \text{it is even not odd} \\ \Rightarrow  \bbox[red, 2pt]{(B)}$$

解答:$$x(t) =L^{-1} \left\{{s-1\over s^2+3s+2} \right\} =L^{-1} \left\{-{2\over s+1} +{3\over s+2}\right\} = -2e^{-t}+3e^{-2t} \text{ for }t\ge 0 \\ \Rightarrow x(t)= (-2e^{-t}+3e^{-2t})u(t) \Rightarrow \bbox[red, 2pt]{(A)}$$

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解題僅供參考，其他碩士班題及詳解