110年高考三級-工程數學詳解

110年公務人員高等考試三級考試試題

類 科： 電力工程、 電子工程、 電信工程
科 目： 工程數學

甲、 申論題部分： （ 50 分）

解答:

(先這樣..待續)

解答:$$\int_\varphi z^2\;dz = \int_0^{1+2i} z^2\;dz = \left. \left[ {1\over 3}z^3 \right] \right|_0^{1+2i} = \bbox[red,2pt]{-{11\over 3}-{2\over 3}i}$$

解答:$$撲克牌有四種花色，每一花色有13張版；因此抽3張為同花的次數為4C^{13}_4\\，52張牌任抽3張有C^{52}_3種可能；因此同花的機率為{4C^{13}_3 \over C^{52}_3} =\bbox[red,2pt]{22\over 425}$$

解答:

(一)$$A= \begin{bmatrix} 5 & -4 & 4\\ 12 &-11 &12 \\ 4 & -4 & 5 \end{bmatrix} \Rightarrow det(A)=-275-192-192+ 176+ 240+240= \bbox[red,2pt]{-3}$$(二)$$\det(A-\lambda I)=  \det\left(\begin{bmatrix}5-\lambda & -4 & 4 \\ 12 & -11-\lambda & 12\\ 4 & -4 & 5-\lambda \end{bmatrix}\right)=0 \Rightarrow (\lambda-1)^2(\lambda+3)=0\\ \Rightarrow 特徵值\lambda=1,-3;\\ \lambda_1=1 \Rightarrow (A-\lambda I)X = \begin{bmatrix}4 & -4 & 4 \\ 12 & -12 & 12\\ 4 & -4 & 4 \end{bmatrix}\begin{bmatrix}x_1 \\  x_2  \\ x_3 \end{bmatrix}=0 \Rightarrow x_1+x_3=x_2\\\quad \Rightarrow 取v_1=\begin{bmatrix}1 \\  1  \\ 0 \end{bmatrix}, v_2= \begin{bmatrix}-1 \\  0  \\ 1 \end{bmatrix}\\ \lambda_2=-3 \Rightarrow (A-\lambda I)X=\begin{bmatrix}8 & -4 & 4 \\ 12 & -8 & 12\\ 4 & -4 & 8 \end{bmatrix}\begin{bmatrix}x_1 \\  x_2  \\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ x_2=3x_1}\Rightarrow 取v_3=\begin{bmatrix}1 \\  3  \\ 1 \end{bmatrix}\\ \bbox[red,2pt]{特徵值為\lambda_1=1及\lambda_2=-3\\，\lambda_1 對應的特徵向量為(1,1,0)及(-1,0,1)；\lambda_2對應的特徵向量為(1,3,1)}$$(三)$$\\ 取P=[v_1 v_2 v_3]= \bbox[red,2pt]{\begin{bmatrix}1 & -1 & 1 \\ 1 & 0 & 3\\ 0 & 1 & 1 \end{bmatrix}} \Rightarrow P^{-1}=\begin{bmatrix}3 & -2 & 3 \\ 1 & -1 & 2\\ -1 & 1 & -1 \end{bmatrix} \\\Rightarrow P^{-1}AP= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & -3 \end{bmatrix}為一對角矩陣$$

解答：$$A,B為兩矩陣\Rightarrow \text{trace}(AB)=\text{trace}(BA)\\現在 Q=P\begin{bmatrix}-2 & 0 \\ 0 & -3 \end{bmatrix}P^{-1} \Rightarrow \text{trace}(Q)=\text{trace}\left( P\begin{bmatrix}-2 & 0 \\ 0 & -3 \end{bmatrix}P^{-1}\right) \\=\text{trace}\left( P^{-1}P\begin{bmatrix}-2 & 0 \\ 0 & -3 \end{bmatrix}\right) =\text{trace}\left(  \begin{bmatrix}-2 & 0 \\ 0 & -3 \end{bmatrix}\right) =-2-3=-5，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{T(x,y,z)= (x-y,z+y)\\ S(x,y,z)= (x+z,x+y)} \Rightarrow (T+S)(x,y,z)=(2x-y+z,x+2y+z)\\ (A)(T+S)(6,2,-10)=(0,0,0)\\(B)(T+S)(3,2,-5) =(-1,2)\ne (0,0)\\(C)(T+S)(3,-2,5) =(13,4)\ne (0,0)\\(D)(T+S)(-6,-2,-10) =(-20,-20)\ne (0,0)\\，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{\det(A)=21\\ \det(B)=-6 \\ \det(C)=168} \Rightarrow \det(ABC^{-1}) ={\det(A)\times \det(B)\over \det(C)} ={21\times (-6)\over 168} =-{3\over 4}=-0.75，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{x+y=3\\ x+2y=-1\\ x+3y=2\\ x+4y=7} \equiv \begin{bmatrix}1 & 1\\ 1& 2\\ 1& 3\\ 1& 4 \end{bmatrix}\begin{bmatrix}x\\ y \end{bmatrix} =\begin{bmatrix}3\\ -1\\ 2\\ 7 \end{bmatrix} \equiv U\cdot \begin{bmatrix}x\\ y \end{bmatrix}=d \\ \Rightarrow \begin{bmatrix}\hat x=\alpha\\ \hat y=\beta \end{bmatrix} =(U^TU)^{-1}U^Td =\begin{bmatrix}-1\\ 3/2 \end{bmatrix} \Rightarrow \cases{\alpha=-1 \\ \beta=3/2}，故選\bbox[red,2pt]{(C)}$$

解答：$$令T=\begin{bmatrix} a& b\\ c& d\end{bmatrix}，則\cases{T(1,-1)=(3,2)\\ T(1,1)= (1,-5)} \Rightarrow \cases{a-b=3\\ c-d=2\\ a+b=1\\ c+d=-5} \Rightarrow \cases{a=2\\ b=-1\\ c=-3/2\\ d=-7/2} \\ \Rightarrow T=\begin{bmatrix} 2& -1\\ -3/2& -7/2 \end{bmatrix} \Rightarrow S=T^{-1}=\begin{bmatrix} 7/17& -2/17\\ -3/17& -4/17 \end{bmatrix} \\ \Rightarrow S(2,-7) =\begin{bmatrix} 7/17& -2/17\\ -3/17& -4/17 \end{bmatrix} \begin{bmatrix} 2\\ -7\end{bmatrix} = \begin{bmatrix} 28/17\\ 22/17 \end{bmatrix} =\begin{bmatrix} p\\ q \end{bmatrix} \\ \Rightarrow \cases{p=28/17\\ q=22/17 } \Rightarrow p+q={50\over 17} =2.94，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{2x_1 -3x_4=1\\ 5x_2+ \alpha x_3=-7\\ x_1+6x_2 +2x_3=13\\ x_2-2x_3+3x_4=-2} \Rightarrow A=\begin{bmatrix} 2 & 0 & 0 & -3\\ 0 & 5 & \alpha & 0\\ 1 & 6 & 2& 0\\ 0 & 1 &-2 &3\end{bmatrix}\\ 聯立方程式無解\Rightarrow \det(A)=0 \Rightarrow -33\alpha +90=0 \Rightarrow \alpha={30\over 11} \approx 2.73，故選\bbox[red,2pt]{(C)}$$

解答：$$\det\left(\begin{bmatrix} -1& 3 & 0\\ -2 & x& -1  \\ 0 & 2&1 \end{bmatrix}\right)= -x+4=0 \Rightarrow x=4，故選\bbox[red,2pt]{(D)}$$

解答：$$\int_0^{\pi/3} \sqrt{x'(t)^2+ y'(t)^2+ z'(t)^2}\;dt = \int_0^{\pi/3} \sqrt{4\cos^2(t)+ 4\sin^2(t)+ 25}\;dt \\= \int_0^{\pi/3} \sqrt{29}\;dt = {\sqrt {29}\over 3}\pi \approx 5.64，故選\bbox[red,2pt]{(C)}$$

解答：$$3e^{i\pi/3}+ 5e^{-i\pi/4} +2 e^{i\pi} =3(\cos{\pi\over 3} +i\sin{\pi\over 3}) +5(\cos{-\pi\over 4} +i\sin{-\pi\over 4}) +2(\cos{\pi } +i\sin{\pi })\\ =3({1\over 2}+ {\sqrt 3\over 2}i) +5({\sqrt 2\over 2}-{\sqrt 2\over 2}i)-2 =({3+5\sqrt 2\over 2}-2)+i({3\sqrt 3-5\sqrt 2\over 2}) \\\Rightarrow \cases{x={3+5\sqrt 2\over 2}-2 \gt 0\\ y={3\sqrt 3-5\sqrt 2\over 2}\lt 0} \Rightarrow x\cdot y\lt 0，故選\bbox[red,2pt]{(A)}$$

解答：$$\sin(\pi+i)=-\sin(i)=-i\sinh(1)= -i\left({e-e^{-1}\over 2}\right),沒有實部，故選\bbox[red,2pt]{(D)}$$

解答：$$f(z)=3\cdot {4z^2+z+2 \over z(4z^2-17z+4)} =3\cdot {4z^2+z+2 \over 4z(z-1/4)(z-4)} \Rightarrow \cases{Res(f,0)=3\times {1\over 2}=1.5\\ Res(f,1/4)=3\times (-2/3)=-2} \\ \Rightarrow \oint_C f(z)\;dz= 2\pi i(Res(f,0)+Res(f,1/4))=2\pi i\times (-0.5)=-\pi i，故選\bbox[red,2pt]{(B)} $$

解答：$$\int_C f(z)\;dz = \int_0^{1+i} z^2+1\;dz = \left. \left[{1\over 3}z^3+z \right] \right|_0^{1+i} ={1\over 3}+{5\over 3}i，故選\bbox[red,2pt]{(D)}$$

解答：$$y=\sum_{a=0}^\infty a_nx^n =a_0+ a_1x+ a_2x^2+\cdots +a_nx^n+\cdots \\ \Rightarrow 3xy =3a_0x +3a_1x^2 +3a_2x^3+ \cdots+ 3a_nx^{n+1}+\cdots \\\Rightarrow y'=a_1+2a_2x + 3a_3x^2+\cdots +na_nx^{n-1} +\cdots\\\Rightarrow x^2y'= a_1x^2+ 2a_2x^3+ 3a_3x^4+\cdots + na_nx^{n+1}+\cdots\\ \Rightarrow y''=2a_2+ 6a_3x+12a_4x^2+\cdots +n(n-1)a_nx^{n-2}+\cdots\\ \Rightarrow y''-x^2y'-3xy = 2a_2+(6a_3-3a_0)x + (12a_4-4a_1)x^2+\cdots\\ 初始值\cases{y(0)=1\\ y'(0)=-2} \Rightarrow \cases{a_0=1\\ a_1=-2};又y''-x^2y'-3xy=0 \Rightarrow \cases{a_2=0\\ 6a_3-3a_0=0 \Rightarrow a_3=1/2}\\ 即\cases{a_0=1\\ a_1=-2\\ a_2=0 \\ a_3=1/2}，故選\bbox[red,2pt]{(D)}$$

解答：$${dy\over dx} ={3x^2-1\over 2y+5} \Rightarrow \int 2y+5\;dy=\int 3x^2-1\;dx \Rightarrow y^2+5y=x^3-x+C\\ \Rightarrow y^2+5y+{25\over 4}=x^3-x+{25\over 4} +C \Rightarrow (y+{5\over 2})^2=x^3-x+{25\over 4} +C \\ \Rightarrow y=-{5\over 2}\pm \sqrt{x^3-x+{25\over 4} +C}\\又 y(1)=-1 \Rightarrow \sqrt{{25\over 4} +C}= {3\over 2} \Rightarrow C=-4 \Rightarrow y(0)=-{5\over 2}\pm {3\over 2}=-4或-1，故選\bbox[red,2pt]{(C)}$$

解答：$$(A)\times: 因為有(y(t))^2項\\ (C)\times: 因為有y(t)\cdot y''(t)項\\ (D)\times: 因為有y(t)^{1/2}項\\，故選\bbox[red,2pt]{(B)}$$

解答：$${dy\over dt}+2y = t\delta(t-2) \Rightarrow \mathcal{L}\{{dy\over dt}\}+2\mathcal{L}\{y\} =\mathcal{L}\{t\delta(t-2)\} \\\Rightarrow sY(s)-y(0)+2Y(s)= (-e^{-2s})'=2e^{-2s}  \Rightarrow Y(s)={2e^{-2s}\over s+2}，故選\bbox[red,2pt]{(A)}$$

解答：$$y={1\over 2}e^{-2x}-{1\over 3}e^{-3x}+ {1\over 6}x+C\Rightarrow y'=-e^{-2x}+e^{-3x}+ {1\over 6} \Rightarrow y''= 2e^{-2x} -3e^{-3x} \\\Rightarrow y''+5y'+6y =x+ 6C+{5\over 6}=x \Rightarrow C=-{5\over 36}\\ 又\cases{y(0)={1\over 6}+C=A\\ y'(0)={1\over 6} =B} \Rightarrow \cases{A={1\over 36}\\ B={1\over 6}\\ C=-{5\over 36}} \Rightarrow A+B+C={1\over 18}，故選\bbox[red,2pt]{(C)}$$

解答：$$假設\cases{甲勝率:p=3/5\\ 乙勝率:q=1-p=2/5}，則第4局打完結束的情形:\cases{第4局打完甲勝:前3局甲2勝1敗\\ 第4局打完乙勝:前3局乙2勝1敗} \\ \Rightarrow \cases{C^3_2p^2q\times p=3\cdot 54/5^4\\ C^3_2q^2p\times q=3\cdot 24/5^4} \Rightarrow 機率為{3(54+24)\over 5^4} = {234\over 625}，故選\bbox[red,2pt]{(C)}$$

解答：$$P(Y=3\mid X=2)={P_{XY}(2,3)\over P(X=2)} ={P_{XY}(2,3)\over P_{XY}(2,1)+ P_{XY}(2,3)+P_{XY}(2,3)}\\={0.18\over 0.25+0.07+0.18}={0.18\over 0.5} =0.36，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{X\sim N(45,15^2)\\ Y\sim N(65,10^2)} \Rightarrow \cases{(A) Var(Y)=Var(X/3+50)=Var(X)/9=25 \ne 10^2\\ (B)Var(Y)= Var(2X/3+35)=4Var(X)/9=100=10^2\\ (C)Var(Y)= Var(2X/5+47)= 4Var(X)/25= 36\ne 10^2\\ (D)Var(Y)= Var(X+20) = Var(X)=225\ne 10^2}\\，故選\bbox[red,2pt]{(B)}$$

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