113年身障三等-工程數學詳解

113年身心障礙人員考試

等 別： 三等考試
類 科： 電力工程
科 目： 工程數學

甲、 申論題部分：（ 50 分）

解答: $$\mathbf{(一)}\; 2x \cos(3y)-3x^2\sin(3y){dy\over dx}=0 \Rightarrow  2x \cos(3y)dx-3x^2\sin(3y) dy=0 \\\quad 令\cases{P(x,y)=2x\cos(3y)\\ Q(x,y)=-3x^2\sin(3y)} \Rightarrow \cases{P_y=-6x\sin(3y)\\ Q_x=-6x\sin(3y) } \Rightarrow P_x=Q_y \Rightarrow 正合,\bbox[red, 2pt]{故得證} \\\mathbf{(二)}\; \Phi(x,y)= \int 2x\cos(3y)\,dx = \int-3x^2\sin(3y)\,dy \\\quad \Rightarrow \Phi(x,y)=x^2\cos(3y)+ \phi(y)= x^2\cos(3y)+ \rho(x)+c_1 \\\quad \Rightarrow x^2\cos(3y)+c_1=0 \Rightarrow 一般解: \bbox[red, 2pt]{y={1\over 3} \cos^{-1}{-c_1\over x^2}, c_1為常數} \\\mathbf{(三)}\; y(1)=0 \Rightarrow 1\cdot 1+c_1=0 \Rightarrow c_1=-1 \Rightarrow x^2\cos(3y)=1 \Rightarrow \cos(3y)={1\over x^2}\\ \quad \Rightarrow \bbox[red, 2pt]{y={1\over 3} \cos^{-1}{1\over x^2}}$$

解答: $$\mathbf{(一)}\; \det(A)=14-20-12-4-15-56=-93\ne 0 \Rightarrow A可逆,\bbox[red, 2pt]{故得證} \\\mathbf{(二)}\; 由(一)知: \det(A)=\bbox[red, 2pt]{-93} \\\mathbf{(三)}\; [A\mid I]=\left[ \begin{array}{rrr|rrr}7 & -3 & -2 & 1 & 0 & 0\\5 & -2 & 4 & 0 & 1 & 0\\1 & 2 & -1 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_1/7 \to R_1} \left[ \begin{array}{rrr|rrr}1 & - \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} & 0 & 0\\5 & -2 & 4 & 0 & 1 & 0\\1 & 2 & -1 & 0 & 0 & 1\end{array} \right] \\\xrightarrow{R_2-5R_1 \to R_2, R_3-R_1\to R_3}\left[ \begin{array}{rrr|rrr}1 & - \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} & 0 & 0\\0 & \frac{1}{7} & \frac{38}{7} & - \frac{5}{7} & 1 & 0\\0 & \frac{17}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 1\end{array} \right] \xrightarrow{7R_2\to R_2} \left[ \begin{array}{rrr|rrr}1 & - \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} & 0 & 0\\0 & 1 & 38 & -5 & 7 & 0\\0 & \frac{17}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 1\end{array} \right] \\ \xrightarrow{R_1+(3/7)R_2\to R_1, R_3-(17/7)R_2 \to R_3}\left[ \begin{array}{rrr|rrr}1 & 0 & 16 & -2 & 3 & 0\\0 & 1 & 38 & -5 & 7 & 0\\0 & 0 & -93 & 12 & -17 & 1 \end{array} \right] \xrightarrow{-R_3/93 \to R_3} \left[ \begin{array}{rrr|rrr}1 & 0 & 16 & -2 & 3 & 0\\0 & 1 & 38 & -5 & 7 & 0\\0 &0 & 1 & - \frac{4}{31} & \frac{17}{93} & - \frac{1}{93}\end{array} \right] \\ \xrightarrow{R_1-16R_3\to R_1, R_2-38R_3 \to R_2} \left[ \begin{array}{rrr|rrr}1 & 0 & 0 & \frac{2}{31} & \frac{7}{93} & \frac{16}{93}\\0 & 1 & 0 & - \frac{3}{31} & \frac{5}{93} & \frac{38}{93}\\0 & 0 & 1 & - \frac{4}{31} & \frac{17}{93} & - \frac{1}{93}\end{array} \right] \Rightarrow A^{-1} = \bbox[red, 2pt]{\left[\begin{matrix}\frac{2}{31} & \frac{7}{93} & \frac{16}{93}\\- \frac{3}{31} & \frac{5}{93} & \frac{38}{93}\\- \frac{4}{31} & \frac{17}{93} & - \frac{1}{93}\end{matrix}\right]} \\\textbf{(四)}\; AB=I \Rightarrow \det(B)={1\over \det(A)} =-{1\over 93} \Rightarrow \det(B^4)= \bbox[red, 2pt]{1 \over 93^4}$$

乙、 測驗題部分：（ 50 分）

解答: $$rref(A)=\left[\begin{matrix}1 & 0 & 0 & -1\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow rank(A)=3,故選\bbox[red, 2pt]{(C)}$$

解答: $$y=x^m \Rightarrow \cases{y'=mx^{m-1}\\ y''=m(m-1)x^{m-2}} \Rightarrow x^2y''-xy'-3y=m(m-1)x^m-mx^m-3x^m \\=(m^2-2m-3)x^m=0 \Rightarrow m^2-2m-3=(m-3)(m+1)=0 \Rightarrow m=-1,3\\ \Rightarrow y=c_1x^{-1}+ c_2x^3,故選\bbox[red, 2pt]{(A)}$$

解答: $$T(x,y)=(2x+y,3x+4y) \Rightarrow  \begin{bmatrix}2 & 1 \\3 & 4 \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} =0 \equiv A\mathbf x=0\\ A= \begin{bmatrix}2 & 1 \\3 & 4 \end{bmatrix} \Rightarrow A^{-1}= \left[\begin{matrix}\frac{4}{5} & - \frac{1}{5}\\- \frac{3}{5} & \frac{2}{5}\end{matrix}\right] \Rightarrow rank(T)=2 \Rightarrow nullity(T)=0\\ 列空間維度=rank(T)=2\ne 0,故選\bbox[red, 2pt]{(D)}$$

解答: $$A為斜對稱 \Rightarrow A=-A^T \Rightarrow \begin{bmatrix}0 & -2& -3 \\a & 0 & 3 \\ b & c & 0\end{bmatrix} =\begin{bmatrix}0 & -a& -b \\ 2 & 0 & -c \\ 3 & -3 & 0\end{bmatrix} \Rightarrow \cases{a=2\\ b=3\\ c=-3} \\(D) \times: bc=3\times (-3)=-9 \ne 9  \\,故選\bbox[red, 2pt]{(D)}$$

解答: $$\begin{bmatrix}1+x & 2& 3& 4 \\1 & 2+x & 3 & 4\\ 1& 2& 3+x& 4\\ 1& 2& 3& 4+x \end{bmatrix} \xrightarrow{R_1-R_2\to R_1, R_3-R_2\to R_3, R_4-R_2 \to R_4} \begin{bmatrix}x & -x& 0& 0 \\1 & 2+x & 3 & 4\\ 0& -x& x& 0\\ 0& -x& 0& x \end{bmatrix} \\ \Rightarrow \det(A)=x\begin{vmatrix}2+x & 3& 4 \\-x & x& 0\\ -x & 0 & x \end{vmatrix} -\begin{vmatrix}-x & 0& 0 \\-x & x & 0\\ -x& 0& x\end{vmatrix} =x(x^3+9x^2)-(-x^3)=x^4+10x^3 =x^3(10+x)\\ ,故選\bbox[red, 2pt]{(A)}$$

解答: $$\det(A-\lambda I)=0 \Rightarrow -(\lambda+2)(\lambda-2)(\lambda-4)=0 \Rightarrow \lambda=-2,2,4\\ \lambda_1=-2 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow v_1= k\begin{pmatrix}1 \\-1 \\1\end{pmatrix} \\\lambda_2=2 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow v_2= s\begin{pmatrix}-1 \\-1 \\1\end{pmatrix} \\\lambda_3= 4 \Rightarrow (A-\lambda_3 I)v =0 \Rightarrow v_3= t\begin{pmatrix}-1 \\1 \\1\end{pmatrix}\\ \Rightarrow \cases{(A) I=v_3\\ (B) J=v_2\\ (C)\times \\(D) L= v_1},故選\bbox[red, 2pt]{(C)}$$

解答: $$D=\begin{bmatrix}3 & 4 \\-1 & 7 \end{bmatrix} \Rightarrow \det(D-\lambda I)=0 \Rightarrow (\lambda-5)^2=0 \Rightarrow \lambda=5, 二重根\\ (A-\lambda I)v=0 \Rightarrow v=k\begin{bmatrix}2 \\1 \end{bmatrix},僅有一獨立向量 \Rightarrow D無法對角化,故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{A(1,0,1)\\ B(2,x,4)\\ C(5,5,7)\\ D(8,8,10)} \Rightarrow \cases{\overrightarrow{AC} =(4,5,6)\\ \overrightarrow{AD}=(7,8,9)} \Rightarrow \vec n= \overrightarrow{AC} \times \overrightarrow{AD} =(-3,6,-3) \\ 平面E:-3(x-1)+6y-3(z-1)=0 \Rightarrow x-2y+z=2 \\ B在E上\Rightarrow 2-2x+4=2 \Rightarrow x=2,故選\bbox[red, 2pt]{(B)}$$

解答: $$(y+x^3)dx=xdy \Rightarrow y'={dy\over dx}={y\over x}+x^2\\ y=ax^3+bx^c \Rightarrow \cases{y(1)=a+b=2 \\ y'=3ax^2+bcx^{c-1}={y\over x}+x^2 =ax^2+bx^{c-1}+x^2=(a+1)x^2 bx^{c-1}}\\\cases{3a=a+1\\ bc=b} \Rightarrow \cases{a=1/2\\ b=3/2\\ c=1},故選\bbox[red, 2pt]{(D)}$$

解答: $$f(x)為奇函數 \Rightarrow a_n=0\\ b_n= \int_{-1}^0 -\sin(n\pi x)\,dx +\int_0^1 \sin(n\pi x)\,dx =\left.\left[ {1\over n\pi} \cos(n\pi x)\right] \right|_{-1}^0 +\left.\left[ -{1\over n\pi} \cos(n\pi x)\right] \right|_{0}^1 \\={2 \over n\pi}(1-(-1)^n) \Rightarrow f(x) =\sum_{n=1}^\infty {2\over n\pi}(1-(-1)^n) \sin(n\pi x) \\ ={4\over \pi}\sin(\pi x)+0\sin(2\pi x)+{4\over 3\pi} \sin(3\pi x) + 0 \sin(4\pi x) +\cdots \\ ={4\over \pi}(\sin(\pi x)+0 \sin(2\pi x)+{1\over 3} \sin(3\pi x)+ 0\sin(4\pi x)+ \cdots ) \\ \Rightarrow \cases{a=1\\ b=0\\ c={1\over 3}\\d=0} \Rightarrow a+b+c+d = {4\over 3},故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{P(x,y)= 4x^2y-3xy^2\\ Q(x,y)=x^3-2x^2y} \Rightarrow \cases{P_y=4x^2-6xy\\ Q_x=3x^2-4xy} \Rightarrow {P_y-Q_x\over Q} ={1\over x} \\ \Rightarrow u'={1\over x}u \Rightarrow 積分因子u(x)=x =x^my^n \Rightarrow \cases{m=1\\ n=0 } \Rightarrow m+n=1,故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\ln x^2=2\ln x\\ \ln x^3=3\ln x} \Rightarrow \ln x,\ln x^2, \ln x^3不為線性獨立,故選\bbox[red, 2pt]{(B)}$$

解答: $$y'(t) = \int_0^t y(\tau) \cos(t-\tau)d\tau \Rightarrow L\{y'(t)\} =L\left\{ \int_0^t y(\tau) \cos(t-\tau)d\tau \right\} \\ \Rightarrow sY(s)-1=L\{y(t)\} L\{\cos t\} =Y(s)\cdot {s\over s^2+1} \Rightarrow (s-{s\over s^2+1})Y(s)=1 \\ \Rightarrow Y(s)={s^2+1\over s^3} ={1\over s}+{1\over s^3} \Rightarrow y(t)= L^{-1}\{{1\over s}\} +L^{-1}\{{1\over s^3}\}=1+{1\over 2}t^2,故選\bbox[red, 2pt]{(D)}$$

解答: $$Res(f,z=0) =\left. {z+1\over (z-2)^2(z+4)}\right|_{z=0} ={1\over 4\cdot 4} ={1\over 16} \\ Res(f,z=2)= \left. {d\over dz} {z+1\over z(z+4)}\right|_{z=2} = \left.  {1\over z(z+4)} -{(z+1)(2z+4) \over (z^2+4z)^2}\right|_{z=2} =-{1\over 12} \\ \oint_C f(z)\,dz = 2\pi i ({1\over 16}-{1\over 12})=-{1\over 24}\pi i,故選\bbox[red, 2pt]{(C)}$$

解答: $$f(z)={1\over (z-1)(z-2)} ={1\over z-2}-{1\over z-1} =-{1\over 2}\cdot {1\over 1-{z\over 2}}-{1\over z}\cdot {1\over 1-{1\over z}} \\=-{1\over 2}(1+{z\over 2}+{z^2\over 4}+{z^3\over 8}+\cdots )-{1\over z}(1+{1\over z}+{1\over z^2} ++{1\over z^3} +\cdots) \\ =\cdots -{1\over z^2}-{1\over z}-{1\over 2}-{z\over 4}+ \cdots \Rightarrow \cases{a=-1\\ b=-1\\ c=-1/2\\ d=-1/4},故選\bbox[red, 2pt]{(B)}$$

解答: $$z=x+i y \Rightarrow z^2=(x^2-y^2)+i(2xy) \Rightarrow f(z)=Re(z^2)-i Im(z^2)=(x^2-y^2)-i(2xy) \\ \Rightarrow \cases{u(x,y)=x^2-y^2\\ v(x,y)=-2xy} \Rightarrow \cases{u_x=2x\\ v_y=-2x} \Rightarrow u_x\ne v_y \Rightarrow f(z)不可解析,故選\bbox[red, 2pt]{(B)}$$

解答: $$z=2+2i \Rightarrow z^2=8i \Rightarrow z^4=-64 \Rightarrow z^4-6i z^2+16=-64-6i\times (8i)+16\\=-64+48+16=0 \Rightarrow 2+2i為其一根,故選\bbox[red, 2pt]{(D)}$$

解答: $$Var(Z)= Var(3X-2Y+3)=9Var(x)+4Var(Y)-12Cov(X,Y) =54 \\ \Rightarrow 9\cdot 2+4Var(Y)-12 \cdot (-2)=54 \Rightarrow 4Var(Y)=12 \Rightarrow Var(Y)= \sigma_Y^2=3,故選\bbox[red, 2pt]{(C)}$$

解答: $$f_X(x)=\int_x^1 xy\,dy={1\over 2}x-{1\over 2}x^3 \Rightarrow E[X]= \int_0^1 \left({1\over 2}x-{1\over 2}x^3 \right)\,dx = {1\over 15},故選\bbox[red, 2pt]{(C)}$$

解答: $$a,b,c均不通的機率=(1-0.4)(1-0.5)(1-0.4)=0.6\cdot 0.5\cdot 0.6= 0.12 \\ \Rightarrow 流通機率=1-0.12 =0.88,故選\bbox[red, 2pt]{(A)}$$

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解題僅供參考,其他歷年試題及詳解