109年高考三級-工程數學詳解

109年公務人員高等考試三級考試

類科:醫學工程、 電力工程、 電子工程、 電信工程
科目:工程數學
甲、申論題： (50分)

解：

$$先求齊性解,即y''-{4\over x}y'+{4\over x^2}y=0 \Rightarrow x^2y''-4xy'+4y=0\\ 令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} 代回上 \Rightarrow m(m-1)x^m -4mx^m+4x^m=0 \\ \Rightarrow m(m-1)-4m+4=0 \Rightarrow m^2-5m+4=0 \Rightarrow (m-4)(m-1)=0 \\\Rightarrow m=4,1 \Rightarrow y_h=C_1x^4+C_2x\\ 接著用參數變異法求特解: 令\cases{y_1=x^4\\ y_2=x \\ r(x)=x^2 +x} \Rightarrow \cases{y_1'=4x^3\\ y_2'=1} \Rightarrow W=\left|\matrix{x^4 & x\\ 4x^3 & 1} \right|=-3x^4 \\ 令\cases{u'(x)=-y_2r(x)/W = 3(x+1)/x^2 \\v'(x)=y_1r(x)/W = -(x^2+x)/3} \Rightarrow \cases{u(x)=(\ln x-1/x)/3 \\ v(x)=-x^3/9-x^2/6} \\ \Rightarrow y_p= uy_1+vy_2= {1\over 3}x^4\ln x-{1\over 2}x^3-{1\over 9}x^4 \Rightarrow y=y_h+y_p =C_1x^4+C_2x +{1\over 3}x^4\ln x-{1\over 2}x^3-{1\over 9}x^4\\ \Rightarrow \bbox[red, 2pt]{y=C_3x^4+C_2x +{1\over 3}x^4\ln x-{1\over 2}x^3}$$

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解：

$$H(2-t)=1-H(t-2) \Rightarrow 3H(2-t)=3-3H(t-2) \Rightarrow L\{3H(2-t) \} =L\{3-3H(t-2) \}\\ =3L\{1\}-3L\{H(t-2) \} = {3\over s}-3\cdot {e^{-2s} \over s} = \bbox[red, 2pt]{3(1-e^{-2s})/s}$$

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解：

$$i=\cos{\pi\over 2}+i\sin{\pi\over 2} =e^{i\pi/2} \Rightarrow \log i= {\pi\over 2}i \\ \Rightarrow i^{1+i}= e^{(1+i)\log i} = e^{(1+i){\pi\over 2}i} =e^{-{\pi\over 2}+{\pi \over 2}i} =e^{-\pi/2}\cdot e^{\pi i/2}= ie^{-\pi/2} =a+bi \Rightarrow \bbox[red, 2pt]{\cases{a=0\\b=e^{-\pi/2}}}$$

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解：

$${\mathbf{u} \cdot \mathbf{v} \over |\mathbf{v}| } ={(3\hat{i}+ \hat{j}+\hat{k}) \cdot (\hat{i}+ \hat{j}+2\hat{k}) \over |(\hat{i}+ \hat{j}+2\hat{k})|} = {3+1+2 \over \sqrt{1+1+4}} ={6\over \sqrt 6} =\bbox[red, 2pt]{\sqrt 6}$$

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解：

$$\int_0^\infty ae^{-x}\;dx=1 \Rightarrow \left. \left[ -ae^{-x} \right]\right|_0^\infty = (0-(-a)) =1 \Rightarrow a=1 \\ \Rightarrow P(1\le x\le 2) = \int_1^2 e^{-x}\;dx \Rightarrow \left. \left[ -e^{-x}\right] \right|_1^2 =-e^{-2}+e^{-1} =\bbox[red, 2pt]{{1\over e}-{1\over e^2}}，故選\bbox[red, 2pt]{( )}$$

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解：

(一)$$A=\left[ \matrix{1 & 0 & -1 \\ 0 & 1 & 2\\ 2 & 1 & 0}\right] \Rightarrow det(A)=0+0+0+2-2+0=\bbox[red, 2pt]{0}$$(二)$$A=\left[ \matrix{1 & 0 & -1 \\ 0 & 1 & 2\\ 2 & 1 & 0}\right] \Rightarrow det(A-\lambda I)=0 \Rightarrow \left| \matrix{1 -\lambda& 0 & -1 \\ 0 & 1-\lambda & 2\\ 2 & 1 & -\lambda}\right|=0 \Rightarrow -\lambda(1-\lambda)^2=0\\ \Rightarrow 特徵值為0及1\\ \lambda_1=0 \Rightarrow (A-\lambda_1)X=0 \Rightarrow \left[ \matrix{1 & 0 & -1 \\ 0 & 1 & 2\\ 2 & 1 & 0}\right] \left[ \matrix{x_1 \\ x_2\\ x_3}\right]=0\Rightarrow \cases{x_3=x_1 \\ x_2=-2x_1},取v_1=\left[ \matrix{1 \\ -2\\ 1}\right]\\ \lambda_2=1 \Rightarrow (A-\lambda_2)X=0 \Rightarrow \left[ \matrix{0 & 0 & -1 \\ 0 & 0 & 2\\ 2 & 1 & -1}\right] \left[ \matrix{x_1 \\ x_2\\ x_3}\right]=0 \Rightarrow \cases{x_3=0\\ 2x_1+x_2=x_3} ,取v_2=\left[ \matrix{1 \\ -2\\ 0}\right]\\答: 特徵值為\bbox[red, 2pt]{0及1},相對的持徵向量為\bbox[red, 2pt]{\left[ \matrix{1 \\ -2\\ 1}\right]及\left[ \matrix{1 \\ -2\\ 0}\right]}$$(三)$$A=\left[ \matrix{1 & 0 & -1 \\ 0 & 1 & 2\\ 2 & 1 & 0}\right] \xrightarrow{-2r_1+r_3}\left[ \matrix{1 & 0 & -1 \\ 0 & 1 & 2\\ 0 & 1 & 2}\right] \xrightarrow{-r_2+r_3}\left[ \matrix{1 & 0 & -1 \\ 0 & 1 & 2\\ 0 & 0 & 0}\right] \\ AX=0 \Rightarrow \left[ \matrix{1 & 0 & -1 \\ 0 & 1 & 2\\ 0 & 0 & 0}\right]\left[ \matrix{x_1 \\ x_2\\ x_3}\right]=0 \Rightarrow \cases{x_1=x_3\\ x_2+2x_3=0} \Rightarrow \left[ \matrix{x_1 \\ x_2\\ x_3}\right]=t\left[ \matrix{1 \\ -2\\ 1}\right],t\in R\\ Null(A)=\bbox[red, 2pt]{\text{span}\left(\left[ \matrix{1 \\ -2\\ 1}\right]\right)}$$

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解：

$$\cases{A(1,2,2)\\ B(0,1,-2) \\C(1,4,1) \\ D(2,5,5)} \Rightarrow \cases{\overrightarrow{AB}= (-1,-1,-4)\\ \overrightarrow{AC}= (0,2,-1)\\ \overrightarrow{AD}= (1,3,3) } \Rightarrow \overrightarrow{AC}=\overrightarrow{AB} +\overrightarrow{AD} \Rightarrow 由\overrightarrow{AB}及\overrightarrow{AD}求面積\\ \Rightarrow 面積= \sqrt{ |\overrightarrow{AB}|^2|\overrightarrow{AD}|^2-(\overrightarrow{AB} \cdot \overrightarrow{AD})^2} = \sqrt{18\times 19-16^2} =\sqrt{86} ，故選\bbox[red, 2pt]{( C)}$$

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解：

$$\cases{S_1:x+2y-2z=3 \\ S_2:2x+4y-4z=7} \Rightarrow \cases{S_1:x+2y-2z=3 \\ S_2:x+2y-2z=7/2} \\\Rightarrow \text{dist}(S_1,S_2)= {|3-7/2|\over \sqrt{1^2+2^2+2^2}} ={1/2 \over 3}=1/6，故選\bbox[red, 2pt]{(A )}$$

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解：

$$\det(A)=-2 \Rightarrow \det(-2A) =(-2)^4\times \det(A) =16\times (-2)=-32，故選\bbox[red, 2pt]{(D )}$$

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解：

$$\left[\begin{array}{rrr|rrr} 1& 0 & 2 & 1 & 0 & 0\\ 0 & -1 & 1 & 0 & 1 & 0\\ 3& 2 & 0 & 0 & 0 &1\end{array} \right] \xrightarrow{-3r_1+r_3} \left[\begin{array}{rrr|rrr} 1& 0 & 2 & 1 & 0 & 0\\ 0 & -1 & 1 & 0 & 1 & 0\\ 0& 2 & -6 & -3 & 0 &1\end{array} \right] \xrightarrow{r_3/2}\\\left[\begin{array}{rrr|rrr} 1& 0 & 2 & 1 & 0 & 0\\ 0 & -1 & 1 & 0 & 1 & 0\\ 0& 1 & -3 & -3/2 & 0 &1/2\end{array} \right]  \xrightarrow{r_2+r_3} \left[\begin{array}{rrr|rrr} 1& 0 & 2 & 1 & 0 & 0\\ 0 & -1 & 1 & 0 & 1 & 0\\ 0& 0 & -2 & -3/2 & 1 &1/2\end{array} \right]  \xrightarrow{-r_2,-r_3/2}\\\left[\begin{array}{rrr|rrr} 1& 0 & 2 & 1 & 0 & 0\\ 0 & 1 & -1 & 0 & -1 & 0\\ 0& 0 & 1 & 3/4 & -1/2 &-1/4\end{array} \right]   \xrightarrow{-2r_3+r_1, r_3+r_1} \left[\begin{array}{rrr|rrr} 1& 0 & 0 & -1/2 & 1 & 1/2\\ 0 & 1 & 0 & 3/4 & -3/2 & -1/4\\ 0& 0 & 1 & 3/4 & -1/2 &-1/4\end{array} \right] \\\Rightarrow A^{-1}=\left[\begin{array}{rrr}  -1/2 & 1 & 1/2\\   3/4 & -3/2 & -1/4\\  3/4 & -1/2 &-1/4\end{array} \right] =\left[\begin{array}{rrr}  a & b & c\\   d & e & f\\  g & h &i\end{array} \right] \Rightarrow a+e+i=-1/2-3/2-1/4=-9/4\\，故選\bbox[red, 2pt]{(C )}$$

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解：

$$\cases{2x+y=u \\3x+4y=v} \Rightarrow \cases{x=4u/5-v/5 \\y=-3u/5+2v/5} \Rightarrow T^{-1}(u,v)=(4u/5-v/5,-3u/5+2v/5)\\ \Rightarrow T^{-1}(5,6)=(4-6/5,-3+12/5)=(14/5,-3/5)，故選\bbox[red, 2pt]{(A )}$$

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解：

$$\det(A-\lambda I)=0 \Rightarrow \left\vert\matrix{1-\lambda & 3\\2& -\lambda} \right \vert=0 \Rightarrow (\lambda-3)(\lambda+2)=0 \Rightarrow \lambda_1=-2,\lambda_2=3\\ (A-\lambda_1I)X=0 \Rightarrow \left[\matrix{3 & 3\\2& 2} \right]\left[\matrix{x_1 \\x_2} \right] \Rightarrow x_1+x_2=0,取v_1=\left[\matrix{-1 \\1} \right] \\ (A-\lambda_2I)X=0 \Rightarrow \left[\matrix{-2 & 3\\2& -3} \right]\left[\matrix{x_1 \\x_2} \right] \Rightarrow 2x_1=3x_2,取v_2 = \left[\matrix{3/2 \\1} \right] \\ 令P=[av_1 \;bv_2] = \left[\matrix{-a & 3b/2\\a& b} \right]=X \xrightarrow{a=-1,b=2} X=\left[\matrix{1 & 3\\ -1& 2} \right] \Rightarrow 答案選\bbox[red,2pt]{(A)}\\若令P=[av_2 \;bv_1] = \left[\matrix{3a/2 & -b\\a& b} \right]=X \xrightarrow{a=2,b=-1} X=\left[\matrix{3 & 1\\2& -1} \right]\Rightarrow 答案選\bbox[red,2pt]{(C)}$$

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解：

$$1+i=\sqrt 2(\cos {\pi\over 4}+i\sin {\pi\over 4})=\sqrt 2e^{i\pi/4} \Rightarrow (1+i)^{1/4} =(\sqrt 2e^{i\pi/4} )^{1/4} =2^{1/8}e^{i\pi/16}\\=2^{1/8}(\cos {\pi\over 16}+i\sin {\pi\over 16})，故選\bbox[red, 2pt]{(A )}$$

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解：

$$本題題意:f(z)={6-2i \over 1-i}，求\overline{f(z)}。\\ =f(z)={(6-2i)(1+i) \over (1-i)(1+i)} = {8+4i \over 2} =4+2i \Rightarrow \overline{f(z)}=4-2i，故選\bbox[red, 2pt]{(C )}$$

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解：

$$f(z)={ie^z \over (z-1+i)^2} \Rightarrow \text{Res}f(1-i) = \lim_{z\to 1-i} {d\over dz}\left[(z-(1-i))^2f(z) \right]  = \lim_{z\to 1-i} {d\over dz}\left[ie^z \right] \\ =\lim_{z\to 1-i} ie^z =ie^{1-i} =ie^1\cdot e^{i(-1)} =ie(\cos(-1)+i\sin(-1))=ie(\cos 1-i\sin 1)\\ \Rightarrow \oint_c f(z)\;dz = 2\pi \times\text{Res}f(1-i) =2\pi i\times ie(\cos 1-i\sin 1)=-2\pi e(\cos 1-\sin 1)，故選\bbox[red, 2pt]{(B )}$$

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解：

$$(C)\times: z= {1+i\over \sqrt 2} \Rightarrow |z|\not \lt 1 \Rightarrow 發散，故選\bbox[red, 2pt]{(C )}$$

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解：

$$\lambda^2+1.25\lambda -0.875=0 \Rightarrow (\lambda+1.75)(\lambda-0.5)=0 \Rightarrow \lambda_1=0.5,\lambda_2=-1.75 \\ \Rightarrow y=C_1e^{0.5x} +C_2e^{-1.75x}，故選\bbox[red, 2pt]{(B )}$$

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解：

$$令\cases{P(x,y)=3x^2+xy^\alpha \\ Q(x,y)=-x^2y^{\alpha-1}y'} \Rightarrow 3x^2+xy^\alpha -x^2y^{\alpha-1}y'=0 \equiv P(x,y)dx+Q(x,y)dy=0 \\ 正合\Rightarrow {\partial P\over \partial y} =  {\partial Q\over \partial x} \Rightarrow \alpha xy^{\alpha-1} =-2xy^{\alpha-1}\Rightarrow \alpha=-2，故選\bbox[red, 2pt]{(A)}$$

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解：

$$\lambda^4+4\lambda^3 +7\lambda^2+6\lambda+2=0 \Rightarrow (\lambda+1)^2(\lambda^2+2\lambda+2) \Rightarrow \lambda=-1(重根), -1\pm i \\ \Rightarrow y=(C_1+C_2x)e^{-x} +e^{-x}(C_3\cos x+C_4 \sin x)，故選\bbox[red, 2pt]{(C)}$$

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解：

$$由於v是常數,不是x的函數,因此將原式同除以x\\,即x^2y''+xy'+(k^2x^2-v^2)y=0 \Rightarrow xy''+y'+(k^2x-{v^2\over x})y=0\\ 因此(p(x)y')'= xy''+y'= (xy')' \Rightarrow p(x)=x，故選\bbox[red, 2pt]{( C)}$$

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解：

$$F(s)=L\{f(t)\} = \int_0^\infty f(t)e^{-st}\;dt \Rightarrow F(2)=L\{t\cos (t)\} = \int_0^\infty t\cos(t) e^{-2t}\;dt\\ 由於 F(s)=L\{t\cos (t)\} ={s^2-1\over (s^2+1)^2} \Rightarrow F(2)= {3\over 25}，故選\bbox[red, 2pt]{(B)}$$

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解：

$${e^{-2s} \over s(s^2+4)} = e^{-2s}\left({1\over 4}({1\over s}-{s\over s^2+4)} \right) \Rightarrow L^{-1}\{{e^{-2s} \over s(s^2+4)}\} ={1 \over 4}\left(L^{-1}\{ e^{-2s}\cdot {1\over s}\} -L^{-1}\{ e^{-2s}\cdot {s\over s^2+4}\}\right)\\ = {1 \over 4}\left(u(t-2) -u(t-2)\cos2(t-2)\}\right) = \left({1\over 4}-{1\over 4}\cos 2(t-2) \right)u(t-2)，故選\bbox[red, 2pt]{(C )}$$

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解：

$$f(x)=1+\sin^2 (2x) \Rightarrow \cases{f(x)為偶函數\Rightarrow b_n=0(只有(C),(D)可能正確)\\週期為{\pi \over 2}}\\ a_0={2\over \pi}\int_0^{\pi/2} (1+\sin^2(2x))\;dx ={2\over \pi} \left.\left[{3\over 2}x-{1\over 8}\sin(4x) \right] \right|_0^{\pi/2} ={3\over 2} \Rightarrow 只有(C)正確，故選\bbox[red, 2pt]{(C)}$$

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解：

$$(C)\bigcirc: A、B獨立\Rightarrow \cases{P(A\cap B)= P(A)P(B)\\ P(A\cap \bar B) = P(A)P(\bar B)} \Rightarrow \cases{P(A\mid B)={P(A\cap B) \over P(B)}={P(A)P(B) \over P(B)}=P(A)\\ P(A\mid \bar B)= {P(A)P(\bar B) \over P(\bar B)}=P(A)} \\\qquad \Rightarrow P(A\mid B)=P(A\mid \bar B)\\(D)\times:P(A\cup B)=P(A)+P(B)- P(A \cap B)\ne P(A)+P(B)，故選\bbox[red, 2pt]{(D )}$$

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解：

$$\int_{-\infty}^\infty f(x)\;dx =1  \Rightarrow \int_0^2 C(4x-2x^2)\;dx=1 \Rightarrow C\left. \left[2x^2-{2\over 3}x^3 \right] \right|_0^2=1 \Rightarrow C(8-{16\over 3})=1 \\ \Rightarrow C= {3\over 8}，故選\bbox[red, 2pt]{(C )}$$

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解：

$$E(X)={d\over dt}M_X(t)|_{t=0} ={d\over dt}e^{t+2t^2}|_{t=0} =(1+4t)e^{t+2t^2}|_{t=0} = (1+0)e^0=1，故選\bbox[red, 2pt]{(C )}$$

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