台灣聯合大學系統114學年度碩士班招生考試
類組：電機類
科目：工程數學A
單一選擇題，共20題，每題5分
解答:$$A= \left[ \begin{matrix}1 & 1 & 0 & 2 & 2\\0 & 2 & -1 & 1 & 3\\1 & 0 & -4 & -3 & 5\\0 & 3 & 1 & 4 & 2\end{matrix} \right] \Rightarrow \text{rref}(A)= \left[\begin{matrix}1 & 0 & 0 & 1 & 1\\0 & 1 & 0 & 1 & 1\\0 & 0 & 1 & 1 & -1\\0 & 0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow \text{rank}(A)=3，故選\bbox[red, 2pt]{(C)}$$
解答:$$\text{Dim}(T)=6 \Rightarrow \text{Rank}(T) = \text{Dim}(T)-\text{Null}(T) =6-2=4，故選\bbox[red, 2pt]{(D)}$$
解答:$$\text{If }A \text{ is invertible and }B=0, \text{ then }\cases{\det(A) \ne 0\\ \det(B)=0} \Rightarrow \det(AB)= \det(A) \det(B)=0，故選\bbox[red, 2pt]{(E)}$$
解答:$$A = \begin{bmatrix} 1& 1\\ 0& 2\end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow \lambda=1,2 \Rightarrow \cases{\lambda_1=1 \Rightarrow \text{eigenvector }v_1= \begin{pmatrix} 1\\0\end{pmatrix} \\ \lambda_2=2 \Rightarrow \text{eigenvector }v_2= \begin{pmatrix} 1\\1\end{pmatrix} }\\ \Rightarrow A^T= \begin{bmatrix} 1&0\\ 1& 2\end{bmatrix} \Rightarrow \det(A^T-\lambda I)=0 \Rightarrow \lambda=1,2 \Rightarrow \cases{\lambda_1=1 \Rightarrow \text{eigenvector }w_1= \begin{pmatrix} 1\\ -1\end{pmatrix} \\ \lambda_2=2 \Rightarrow \text{eigenvector }w_2= \begin{pmatrix} 0\\1\end{pmatrix} } \\ \Rightarrow A \text{ and }A^T \text{ have different eigenvectors.}，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{u=(2,6) \\w=(k,4k)} \Rightarrow \text{proj}_w(u) ={u\cdot w\over w\cdot w}w ={26k\over 17k^2} (k,4k) = \left({26\over 17},{104\over 17} \right)，故選\bbox[red, 2pt]{(B)}$$
解答:$$a= {\langle f,1\rangle \over \langle 1,1\rangle} = { \int_{-\pi}^\pi t\,dt =0 \over \langle 1,1\rangle} =0\\ b={\langle f,\sin(t)\rangle \over \langle \sin(t),\sin(t)\rangle} ={\int_{-\pi}^\pi t\sin(t)\,dt \over \int_{-\pi}^\pi \sin^2(t)} ={2\pi\over \pi} =2\\ c= {\langle f,\cos(t)\rangle \over \langle \cos(t),\cos(t)\rangle} ={\int_{-\pi}^\pi t\cos(t)\,dt =0\over \langle \cos(t),\cos(t)\rangle} =0 \\ \Rightarrow v=0+2\sin(t)+0\cos(t)=2\sin(t)，故選\bbox[red, 2pt]{(C)}$$
解答:$$\begin{vmatrix}7&1&-2& 0 & 0\\ -1& 2& 1& 0 & 0\\ 4&4& 2& 3& 4\\ 2& 2& 2\sin t& 2& 2\\ 2& 1& 0& 1&2 \end{vmatrix} \xrightarrow{R_1+7R_2\to R_1, R_3+4R_2\to R_3, R_4+2R_2 \to R_4, R_5+2R_2\to R_5} \begin{vmatrix}0&15 &5& 0 & 0\\ -1& 2& 1& 0 & 0\\ 0&11 & 6& 3& 4\\ 0& 6& 2+2\sin t& 2& 2\\ 0& 5& 2& 1&2 \end{vmatrix} \\=\begin{vmatrix}15 &5& 0 & 0 \\ 11 & 6& 3& 4\\ 6& 2+2\sin t& 2& 2\\ 5& 2& 1&2 \end{vmatrix}=15 \begin{vmatrix} 6& 3& 4\\ 2+2\sin t& 2& 2\\ 2& 1&2 \end{vmatrix} -5\begin{vmatrix} 11 & 3& 4\\ 6 & 2& 2\\ 5& 1&2 \end{vmatrix}=15(4-4\sin t)-5\cdot 0 \\ =60-60 \sin(t) \Rightarrow \det(A(t))=-60\sin(t)+60 \Rightarrow {d\over dt}\det(A(t))=-60\cos(t) \\\Rightarrow \det(A(\pi/2))= {d\over dt}\det(A(\pi/2)) =0，故選\bbox[red, 2pt]{(C)}$$
解答:$$8^2-4\cdot 1\cdot 7=36 \gt 0 \Rightarrow \text{The curve is a hyperbola(雙曲線) not an ellipse.}，故選\bbox[red, 2pt]{(A)}$$
解答:$$T(a,b) =(3a+(2+i)b, (2-i)a+7b) \Rightarrow A= \begin{bmatrix} 3& 2+i\\ 2-i& 7\end{bmatrix} \Rightarrow A^* =\begin{bmatrix} 3& 2+i\\ 2-i& 7\end{bmatrix} \\ \Rightarrow A=A^* \Rightarrow T \text{ is self-adjoint} \\ \Rightarrow AA^* =AA=A^*A \Rightarrow T \text{ is normal} \\ AA^*=A^2 \ne I \Rightarrow T \text{ is NOT unitary. And }AA^T \ne I \Rightarrow T \text{ is NOT orthogonal.} \\ \\ ，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{x+2y+z=4\\ x-y+2z=-11 \\ x+5y=19} \Rightarrow \begin{bmatrix} 1& 2& 1\\ 1&-1& 2\\ 1& 5& 0\end{bmatrix} \begin{bmatrix} x\\y \\z\end{bmatrix} = \begin{bmatrix} 4\\ -11\\19\end{bmatrix} \\ A= \left[ \begin{array}{rrr|r}1 & 2 & 1 & 4\\1 & -1 & 2 & -11\\1 & 5 & 0 & 19 \end{array} \right] \Rightarrow \text{rref}(A) = \left[ \begin{array}{rrr|r}1 & 0 & \frac{5}{3} & -6\\0 & 1 & - \frac{1}{3} & 5\\0 & 0 & 0 & 0 \end{array} \right] \Rightarrow \cases{x+5z/3=-6\\ y-z/3=5} \\ \Rightarrow (x,y,z) =(-6-{5\over 3}k, 5+{1\over 3}k, k),k\in \mathbb R \\ (A) \bigcirc: \text{There are more than one solutions.} \\(B) \bigcirc: k=0 \Rightarrow (x,y,z)=(-6,5,0) \\(C)\bigcirc: \cases{x+5z/3=0\\ y-z/3=0} \Rightarrow (x,y,z)=(-5t/3, t/3,t), t\in \mathbb R \Rightarrow \text{ more than one solution} \\(D) \bigcirc: t=6 \Rightarrow (x,y,z) =(-10,2,6) \\(E)\times: f(k) =(-6-{5\over 3}k)^2+ (5+{1\over 3}k)^2+ k^2) \Rightarrow f(k=-3) \text{ has the minimal value} \\ \quad \Rightarrow (x,y,z) =(-1,4,-3) \text{ is the minimal solution}\\，故選\bbox[red, 2pt]{(E)}$$
解答:$$z= x+i y, x,y\in \mathbb R \Rightarrow e^z= e^{x+iy} =e^x\cdot e^{iy} =e^x (\cos y+ i\sin y) \\ \Rightarrow \cases{u(x,y)=e^x\cos y\\ v(x,y)=e^x\sin y} \Rightarrow \cases{u_x=e^x\cos y\\ v_y = e^x \cos y\\ u_y=-e^x\sin y\\ v_x =e^x \sin y} \Rightarrow \cases{u_x= v_y\\ u_y=-v_x} \Rightarrow e^z \text{ is analytic}，故選\bbox[red, 2pt]{(D)}$$
解答:$$f(z)=z^2 \Rightarrow f'(z)/f(z)=2/z \text{ is NOT analytic at }z=0，故選\bbox[red, 2pt]{(D)}$$
解答:$$1+2z+(2z)^2+ \cdots \text{converges only if } |2z|\lt 1 \Rightarrow |z|\lt {1\over 2} \\ \Rightarrow \text{For }{1\over 2}\le z\lt 1, \text{ the series diverges. Then it is NOT analytic}，故選\bbox[red, 2pt]{(E)}$$
解答:$$(A)\bigcirc:\sin(x) ={e^{ix}-e^{-ix} \over 2i} \Rightarrow \sin(i z) ={e^{i(iz)}-e^{-i(iz)} \over 2i} ={e^{-z} -e^{z} \over 2i} ={i(e^z-e^{-z})\over 2} =i\sinh(z)\\(B)\times: z=x+i y \Rightarrow f(x,y)=\sin(i(x+iy)) = i\sinh(x+iy) =i (\sinh(x) \cosh(iy)+ \cosh(x)\sinh(iy)) \\\qquad = i\sinh(x) \cos(y)- \cosh(x)\sin(y) ={i\over 2} \cos(y)(e^x-e^{-x})-{1\over 2}\sin(y) (e^x+e^{-x}) \text{ NOT periodic in }x\\(C) \bigcirc: \text{By (D)}, \sin(iz) \text{ is an entire function }\Rightarrow \sin(i z) \text{ is analytic}\\ (D) \bigcirc: e^z \text{ and }e^{-z} \text{ are both entire functions} \Rightarrow \sin(iz) =i\cdot {e^z-e^{-z} \over 2} \text{ is an entire function}\\(E)\bigcirc: \sin(x) ={e^{ix}-e^{-ix} \over 2i} \Rightarrow \sin(i z) ={e^{i(iz)}-e^{-i(iz)} \over 2i} ={e^{-z} -e^{z} \over 2i}\\，故選\bbox[red, 2pt]{(B)}$$
解答:$$(A) \times: g(z) \text{ has a pole of order 2 at }z=0. \\(B) \times: g(z)={f(z)\over z^2} ={1\over z^2} \left( f(0)+ f'(0)z+ {f''(0)\over 2!}z^2+ \cdots\right) ={f(0)\over z^2}+ {f'(0) \over z}+ {f''(0) \over 2!}+ \cdots \\\qquad \Rightarrow g(z) \text{ has a residue given by }f'(0) \ne f(0) \\(C) \bigcirc: R=|\text{center-singuarity}|= |2i-0|=2 \\(D) \times: z=1 \text{ is not a pole }\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$y'+3y=0 \Rightarrow y_h= c_1e^{-3x} \\ y_p= e^{-x}(Ax^2+Bx+C) \Rightarrow y_p'= e^{-x}(-Ax^2+(2A-B)x+ B-C) \\ \Rightarrow y_p'+3y_p = e^{-x}(2Ax^2+(2A+2B)x +B+ 2C) =e^{-x}x^2 \Rightarrow \cases{2A=1\\ 2A+2B=0\\ B+2C=0} \\ \Rightarrow \cases{A=1/2\\ B=-1/2\\ C=1/4} \Rightarrow y_p= e^{-x}({1\over 2}x^2-{1\over 2}x+{1\over 4}) ={1\over 4}(2x^2-2x+ 1)e^{-x} \Rightarrow y=y_h+y_p \\ \Rightarrow y=c_1e^{-3x} + {1\over 4}(2x^2-2x+ 1)e^{-x}，故選\bbox[red, 2pt]{(D)}$$
解答:$$y=x^m \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''+y= m(m-1)x^m+x^m =(m^2-m+1)x^m=0\\ \Rightarrow m^2-m+1=0 \Rightarrow m={1\pm \sqrt 3i\over 2} \Rightarrow y_h= \sqrt x \left(c_1 \cos \left({\sqrt 3\over 2}\ln(x) \right) +c_2 \sin \left({\sqrt 3\over 2}\ln(x) \right)\right) \\ y_p =Ax^2+Bx+C \Rightarrow y_p''=2A \Rightarrow x^2y_p''+y_p =3Ax^2+Bx+C=3x^2 \Rightarrow \cases{A=1\\ B=C=0} \\ \Rightarrow y_p= x^2 \Rightarrow y=y_p +y_h \Rightarrow y= x^2+ \sqrt x \left(c_1 \cos \left({\sqrt 3\over 2}\ln(x) \right) +c_2 \sin \left({\sqrt 3\over 2}\ln(x) \right)\right)，故選\bbox[red, 2pt]{(C)}$$
解答:$$L\{[t]\} =\int_0^\infty [t]e^{-st}\,dt = \int_1^2 e^{-st}\,dt + \int_2^32e^{-st}\,dt + \int_3^4 3e^{-st}\,dt +\cdots \\= -{1\over s} \left(e^{-2s}-e^{-s} \right)-{2\over s} \left(e^{-3s}-e^{-2s} \right) -{3\over s} \left(e^{-4s}-e^{-3s} \right) ={1\over s} \left( e^{-s} +e^{-2s}+e^{-3s}+ \cdots\right) \\={1\over s} \cdot {e^{-s} \over 1-e^{-s}} \Rightarrow L\{f(t)\} =L\{t\} -L\{[t]\} ={1\over s^2}-{e^{-s} \over s(1-e^{-s})} ={ 1-e^{-s}(1+ s)\over s^2(1-e^{-s})}，故選\bbox[red, 2pt]{(A)}$$
解答:$$L^{-1} \left\{{1\over s^3+1} \right\} =L^{-1} \left\{{1\over 3(s+1)}+{-s+2\over 3(s^2-s+1)} \right\} \\=L^{-1} \left\{{1\over 3(s+1)}-{1\over 3}\cdot {s-1/2\over (s-1/2)^2+3/4}+ {1\over 2}\cdot {1\over (s-1/2)^2+3/4} \right\}\\ = {1\over 3}e^{-t}-{1\over 3}e^{t/2} \cos(\sqrt 3t/2) +{1\over 2}\cdot {2\over \sqrt 3}e^{t/2} \sin \left({\sqrt 3 t\over 2} \right)\\ ={1\over 3}e^{-t}-{1\over 3}e^{t/2}\cos(\sqrt 3 t/2)+{\sqrt 3\over 3} e^{t/2}\sin(\sqrt 3t/2)，故選\bbox[red, 2pt]{(B)}$$
解題僅供參考，其他碩士班試題及詳解
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