112年身心障礙人員考試
考 試 別: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:y″+2y′+2y={10sin(2t),0<t<10,t≥1=10sin(2t)−10sin(2t)u(t−1)⇒L−1{y″+2y′+2y}=L−1{10sin(2t)−10sin(2t)u(t−1)}⇒s2Y(s)−sy(0)−y′(0)+2(sY(s)−y(0))+2Y(s)=20s2+4−10e−s(ssin2+2cos2)s2+4⇒(s2+2s+2)Y(s)−s+3=20s2+4−10e−s(ssin2+2cos2)s2+4⇒Y(s)=20(s2+4)(s2+2s+2)−10e−s(ssin2+2cos2)(s2+4)(s2+2s+2)+s−3s2+2s+2L−1{20(s2+4)(s2+2s+2)}=L−1{−2s−2(s2+4+2s+1(s+1)2+1+41(s+1)2+1}=−2cos(2t)−sin(2t)+2e−t(cost+2sint)L−1{s−3s2+2s+2}=L−1{s+1(s+1)2+1−4(s+1)2+1}=e−t(cost−4sint)L−1{10e−s(ssin2+2cos2)(s2+4)(s2+2s+2)}=L−1{10sin2⋅se−s(s2+4)(s2+2s+2)+20cos2⋅e−s(s2+4)(s2+2s+2)}=−sin2⋅u(t−1)(cos(2t−2)−2sin(2t−2)−e1−t(cos(t−1)−3sin(t−1)))−cos2⋅u(t−1)(2cos(2t−2)+sin(2t−2)−2e1−t(cos(t−1)+2sin(t−1))))因此L−1{Y(s)}=y(t)=−2cos(2t)−sin(2t)+3e−tcost+u(t−1)(sin(2t)+2cos(2t))−u(t−1)sin2⋅e1−t(cos(t−1)−3sin(t−1))−2u(t−1)cos2⋅e1−t(cos(t−1)+2sin(t−1))解答:(一)[00−2100121010103001]R3+R1,−R3+R2→[10110102−201−1103001]−R1+R3,R2/2→[10110101−101/2−1/2002−100]R3/2+R2,R3/2→[101101010−1/21/2−1/2001−1/200]−R3+R1→[1003/201010−1/21/2−1/2001−1/200]⇒A−1=[3/201−1/21/2−1/2−1/200](二)det
乙、測驗題部分:(50 分)
解答:(A) \bigcirc: 不平行的兩平面相交為一直線,有無限多組解\\ (B)\bigcirc: (2,1,0)均在兩平面上 \\ (C)\bigcirc: (-1,3,1)均在兩平面上 \\ (D)\times:直線的維度為1\\故選\bbox[red, 2pt]{(D)}解答:M=\left[\begin{matrix}-1 & 1 & -1 & 0\\0 & 1 & 0 & 0\\1 & 0 & 1 & 0\\0 & 0 & -1 & -1\end{matrix}\right] \Rightarrow rref(M)= \left[\begin{matrix}1 & 0 & 0 & -1\\0 & 1 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow rank(M)=3,故選\bbox[red, 2pt]{(C)}
解答:令X=\begin{bmatrix} 1\\ 1\end{bmatrix} \Rightarrow AX =\begin{bmatrix} 5 \\ 11 \end{bmatrix}, 不存在\lambda 滿足AX=\lambda X, 故選\bbox[red, 2pt]{(D)}
解答:T(x,y,z)=0 \Rightarrow \cases{2x=0\\ y-z=0} \Rightarrow \cases{x=0\\ y=z} \Rightarrow ker(T)=\{(0,a,a)\mid a\in \mathbb R\} 為一直線 \\ \Rightarrow 維度=1,故選\bbox[red, 2pt]{(A)}
解答:S=\{(a,0,0,b)\mid a,b\in \mathbb R\} \Rightarrow d(S,u) =\sqrt{(a-1)^2+1^2+3^2 +(b-5)^2}\\ 當a=1,b=5時,d(S,u)有最小值\sqrt{1+9}=\sqrt{10},故選\bbox[red, 2pt]{(B)}
解答:A,B相似\Rightarrow \det(B)=\det(A)=3 \Rightarrow \det(AB)=\det(A)\det(B)=9,故選\bbox[red, 2pt]{(D)}
解答:假設特徵值\cases{\lambda_1=-3\\ \lambda_2=1\\ \lambda_3=2}相對應的特徵向量為\cases{v_1\\ v_2\\ v_3}\\ 則(A^2+3A+2I)v_i= A^2v_i+ 3Av_i+2Iv_i =\lambda_i^2v_i +3\lambda_iv_i +2v_i =(\lambda_i^2 +3\lambda_i+2)v_i\\ \Rightarrow A^2+3A+2I的特徵值為\lambda_i^2 +3\lambda_i+2, 分別是\cases{9-9+2=2\\ 1+3+2=6\\ 4+6+2=12} \\ \Rightarrow \det(A^2+3A+2I) =2\times 6\times 12=144,故選\bbox[red, 2pt]{(A)}
解答:A=\begin{bmatrix}0.8 & 0.3\\ 0.2 & 0.7 \end{bmatrix} =\begin{bmatrix} -1 & 1.5 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 0.5 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} -0.4 & 0.6\\ 0.4& 0.4\end{bmatrix} \\ \Rightarrow A^{\infty} =\begin{bmatrix} -1 & 1.5 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} -0.4 & 0.6\\ 0.4& 0.4\end{bmatrix} =\begin{bmatrix} 0.6 & 0.6\\ 0.4 & 0.4 \end{bmatrix},故選\bbox[red, 2pt]{(B)}
解答:(A)\times: \cos(z)={1\over 2}(e^{iz}+e^{-iz}) \Rightarrow {d\over dz}\cos(z) ={1\over 2}(ie^{iz}-ie^{-iz}) =-{1\over 2i}(e^{iz}-e^{-iz}) =-\sin(z) \\ (B)\bigcirc: \cos(z) = \cos(x+iy) = \cos x\cos(iy) − \sin x\sin(iy) = \cos x\cosh y − i\sin x\sinh y \\ \qquad \Rightarrow \cases{u=\cos x\cosh y\\ v=-\sin x \sinh y} \Rightarrow \cases{{\partial v\over \partial x}=- \cos x\sinh y\\ {\partial u\over \partial y}=\cos x\sinh y} \Rightarrow {\partial v\over \partial x}=-{\partial u\over \partial y} \\(C)\times: u(x,y)=\cos x\cosh y=\cos x\cdot {e^y+e^{-y}\over 2 }\\(D)\times: v(x,y)=-\sin x\sinh y =-\sin x{e^y-e^{-y}\over 2}\\,故選\bbox[red, 2pt]{(B)}
解答:e^{2z+\ln(2)} =1+i =\sqrt 2(\cos{\pi\over 4}+i\sin{\pi\over 4}) =\sqrt 2e^{\pi i/4} =e^{{1\over 2}\ln(2)+{\pi\over 4}i} \\ \Rightarrow 2z+\ln(2) ={1\over 2}\ln(2)+{\pi\over 4}i \Rightarrow z=-{1\over 4}\ln(2)+{\pi \over 8}i,故選\bbox[red, 2pt]{(C)}
解答:z=t-it^2 \Rightarrow dz=dt-2itdt= (1-2it)dt \Rightarrow \int_C (1-z)dz =\int_0^1 (1-t+it^2)(1-2it)dt \\= \int_0^1 2t^3+3it^2-(2i+1)t+1 \,dt = \left. \left[ {1\over 2}t^4 +it^3-(i+{1\over 2})t^2+ t\right]\right|_0^1 \\={1\over 2}+i-(i+{1\over 2})+1=1,故選\bbox[red, 2pt]{(B)}
解答:z^2+i6z=z(z+6i)=0 \Rightarrow z=0,-6i; 由於0及-6i皆在路徑外,故選\bbox[red, 2pt]{(A)}
解答:若未給定初始值,有無窮多解,故選\bbox[red, 2pt]{(A)}
解答:\cases{u(x,y)={2y\over x}-9\\ v(x,y)= 3-{6x\over y}} \Rightarrow \cases{x^2y^2 u(x,y)=2xy^3-9x^2y^2 \\ x^2y^2v(x,y)= 3x^2y^2-6x^3y} \\ \Rightarrow \cases{{\partial \over \partial y}x^2y^2 u(x,y) =6xy^2-18x^2y \\{\partial \over \partial x}x^2y^2v(x,y) =6xy^2-18x^2y} \Rightarrow {\partial \over \partial y}x^2y^2 u(x,y)={\partial \over \partial x}x^2y^2v(x,y)\\ \Rightarrow x^2y^2 為積分因子,故選\bbox[red, 2pt]{(D)}
解答:y=a_0+ a_1x+a_2x^2 +\cdots+a_nx^n+\cdots\\ \Rightarrow \cases{xy= a_0x+ a_1x^2+ a_2x^3+\cdots + a_nx^{n+1}+ \cdots\\ y'=a_1 +2a_2x + \cdots +na_nx^{n-1}+ \cdots} \\ \Rightarrow y'+xy = a_1+(a_0+2x_2)x +\cdots +(a_{n-1}+ (n+1)a_{n+1})x^n+\cdots =0\\ \Rightarrow a_{n-1}+ (n+1)a_{n+1}=0 \Rightarrow a_{n+1}=-{1\over n+1}a_{n-1},故選\bbox[red, 2pt]{(C)}
解答:g(t)=\begin{cases}0, & t\lt 5\\ t-5, &t\ge 5 \end{cases} \Rightarrow \mathcal L\{ g(t)\} =\int_5^\infty (t-5)e^{-st}\,dt =\left. \left[-{e^{-st} \over s^2}(s(t-5)+1) \right] \right|_5^\infty \\ =0+{e^{-5t}\over s^2}={e^{-5t}\over s^2},故選\bbox[red, 2pt]{(B)}
解答:(A)\times: f為偶函數 \Rightarrow b_n=0\\ (B)\times: f為奇函數\Rightarrow a_n=0\\ (C)\times: A={n\pi \over L} \ne {2n\pi \over L} \\ (D)\bigcirc: f為奇函數\Rightarrow a_n=0 \Rightarrow f(x)= \sum_{n=1}^\infty b_n \sin{n\pi\over L}x \\ \qquad \Rightarrow \lim_{x\to L^-}f(x) =\lim_{x\to L^-}\sum_{n=1}^\infty b_n \sin{n\pi\over L}x =\sum_{n=1}^\infty b_n\lim_{x\to L^-} \left( \sin{n\pi\over L}x\right) =0\\,故選\bbox[red, 2pt]{(D)}
解答:{沒肺癌且不抽菸\over 沒肺癌且抽菸+沒肺癌且不抽菸} ={70\% \times 90\% \over 30\%\times 60\%+ 70\% \times 90\%} \\ ={63\over 18+63}={7\over 9},故選\bbox[red, 2pt]{(A)}
解答:P(2\le X\le 4)=0.4 \Rightarrow {4-2\over A}=0.4 \Rightarrow A=2\times {10\over 4}=5,故選\bbox[red, 2pt]{(C)}
解答:Y=-4X+1 \Rightarrow Var(Y)=Var(-4X+1)= 16Var(X)=16\\ 又4X+Y=1 \Rightarrow Var(4X+Y)=Var(1)=0 \Rightarrow 16Var(X)+Var(Y)+8Cov(X,Y)=0\\ \Rightarrow 32+8Cov(X,Y)=0 \Rightarrow Cov(X,Y)=-4,故選\bbox[red, 2pt]{(D)}
解答:S=\{(a,0,0,b)\mid a,b\in \mathbb R\} \Rightarrow d(S,u) =\sqrt{(a-1)^2+1^2+3^2 +(b-5)^2}\\ 當a=1,b=5時,d(S,u)有最小值\sqrt{1+9}=\sqrt{10},故選\bbox[red, 2pt]{(B)}
解答:A,B相似\Rightarrow \det(B)=\det(A)=3 \Rightarrow \det(AB)=\det(A)\det(B)=9,故選\bbox[red, 2pt]{(D)}
解答:假設特徵值\cases{\lambda_1=-3\\ \lambda_2=1\\ \lambda_3=2}相對應的特徵向量為\cases{v_1\\ v_2\\ v_3}\\ 則(A^2+3A+2I)v_i= A^2v_i+ 3Av_i+2Iv_i =\lambda_i^2v_i +3\lambda_iv_i +2v_i =(\lambda_i^2 +3\lambda_i+2)v_i\\ \Rightarrow A^2+3A+2I的特徵值為\lambda_i^2 +3\lambda_i+2, 分別是\cases{9-9+2=2\\ 1+3+2=6\\ 4+6+2=12} \\ \Rightarrow \det(A^2+3A+2I) =2\times 6\times 12=144,故選\bbox[red, 2pt]{(A)}
解答:A=\begin{bmatrix}0.8 & 0.3\\ 0.2 & 0.7 \end{bmatrix} =\begin{bmatrix} -1 & 1.5 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 0.5 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} -0.4 & 0.6\\ 0.4& 0.4\end{bmatrix} \\ \Rightarrow A^{\infty} =\begin{bmatrix} -1 & 1.5 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} -0.4 & 0.6\\ 0.4& 0.4\end{bmatrix} =\begin{bmatrix} 0.6 & 0.6\\ 0.4 & 0.4 \end{bmatrix},故選\bbox[red, 2pt]{(B)}
解答:(A)\times: \cos(z)={1\over 2}(e^{iz}+e^{-iz}) \Rightarrow {d\over dz}\cos(z) ={1\over 2}(ie^{iz}-ie^{-iz}) =-{1\over 2i}(e^{iz}-e^{-iz}) =-\sin(z) \\ (B)\bigcirc: \cos(z) = \cos(x+iy) = \cos x\cos(iy) − \sin x\sin(iy) = \cos x\cosh y − i\sin x\sinh y \\ \qquad \Rightarrow \cases{u=\cos x\cosh y\\ v=-\sin x \sinh y} \Rightarrow \cases{{\partial v\over \partial x}=- \cos x\sinh y\\ {\partial u\over \partial y}=\cos x\sinh y} \Rightarrow {\partial v\over \partial x}=-{\partial u\over \partial y} \\(C)\times: u(x,y)=\cos x\cosh y=\cos x\cdot {e^y+e^{-y}\over 2 }\\(D)\times: v(x,y)=-\sin x\sinh y =-\sin x{e^y-e^{-y}\over 2}\\,故選\bbox[red, 2pt]{(B)}
解答:e^{2z+\ln(2)} =1+i =\sqrt 2(\cos{\pi\over 4}+i\sin{\pi\over 4}) =\sqrt 2e^{\pi i/4} =e^{{1\over 2}\ln(2)+{\pi\over 4}i} \\ \Rightarrow 2z+\ln(2) ={1\over 2}\ln(2)+{\pi\over 4}i \Rightarrow z=-{1\over 4}\ln(2)+{\pi \over 8}i,故選\bbox[red, 2pt]{(C)}
解答:z=t-it^2 \Rightarrow dz=dt-2itdt= (1-2it)dt \Rightarrow \int_C (1-z)dz =\int_0^1 (1-t+it^2)(1-2it)dt \\= \int_0^1 2t^3+3it^2-(2i+1)t+1 \,dt = \left. \left[ {1\over 2}t^4 +it^3-(i+{1\over 2})t^2+ t\right]\right|_0^1 \\={1\over 2}+i-(i+{1\over 2})+1=1,故選\bbox[red, 2pt]{(B)}
解答:z^2+i6z=z(z+6i)=0 \Rightarrow z=0,-6i; 由於0及-6i皆在路徑外,故選\bbox[red, 2pt]{(A)}
解答:若未給定初始值,有無窮多解,故選\bbox[red, 2pt]{(A)}
解答:\cases{u(x,y)={2y\over x}-9\\ v(x,y)= 3-{6x\over y}} \Rightarrow \cases{x^2y^2 u(x,y)=2xy^3-9x^2y^2 \\ x^2y^2v(x,y)= 3x^2y^2-6x^3y} \\ \Rightarrow \cases{{\partial \over \partial y}x^2y^2 u(x,y) =6xy^2-18x^2y \\{\partial \over \partial x}x^2y^2v(x,y) =6xy^2-18x^2y} \Rightarrow {\partial \over \partial y}x^2y^2 u(x,y)={\partial \over \partial x}x^2y^2v(x,y)\\ \Rightarrow x^2y^2 為積分因子,故選\bbox[red, 2pt]{(D)}
解答:y=a_0+ a_1x+a_2x^2 +\cdots+a_nx^n+\cdots\\ \Rightarrow \cases{xy= a_0x+ a_1x^2+ a_2x^3+\cdots + a_nx^{n+1}+ \cdots\\ y'=a_1 +2a_2x + \cdots +na_nx^{n-1}+ \cdots} \\ \Rightarrow y'+xy = a_1+(a_0+2x_2)x +\cdots +(a_{n-1}+ (n+1)a_{n+1})x^n+\cdots =0\\ \Rightarrow a_{n-1}+ (n+1)a_{n+1}=0 \Rightarrow a_{n+1}=-{1\over n+1}a_{n-1},故選\bbox[red, 2pt]{(C)}
解答:g(t)=\begin{cases}0, & t\lt 5\\ t-5, &t\ge 5 \end{cases} \Rightarrow \mathcal L\{ g(t)\} =\int_5^\infty (t-5)e^{-st}\,dt =\left. \left[-{e^{-st} \over s^2}(s(t-5)+1) \right] \right|_5^\infty \\ =0+{e^{-5t}\over s^2}={e^{-5t}\over s^2},故選\bbox[red, 2pt]{(B)}
解答:(A)\times: f為偶函數 \Rightarrow b_n=0\\ (B)\times: f為奇函數\Rightarrow a_n=0\\ (C)\times: A={n\pi \over L} \ne {2n\pi \over L} \\ (D)\bigcirc: f為奇函數\Rightarrow a_n=0 \Rightarrow f(x)= \sum_{n=1}^\infty b_n \sin{n\pi\over L}x \\ \qquad \Rightarrow \lim_{x\to L^-}f(x) =\lim_{x\to L^-}\sum_{n=1}^\infty b_n \sin{n\pi\over L}x =\sum_{n=1}^\infty b_n\lim_{x\to L^-} \left( \sin{n\pi\over L}x\right) =0\\,故選\bbox[red, 2pt]{(D)}
解答:{沒肺癌且不抽菸\over 沒肺癌且抽菸+沒肺癌且不抽菸} ={70\% \times 90\% \over 30\%\times 60\%+ 70\% \times 90\%} \\ ={63\over 18+63}={7\over 9},故選\bbox[red, 2pt]{(A)}
解答:P(2\le X\le 4)=0.4 \Rightarrow {4-2\over A}=0.4 \Rightarrow A=2\times {10\over 4}=5,故選\bbox[red, 2pt]{(C)}
解答:Y=-4X+1 \Rightarrow Var(Y)=Var(-4X+1)= 16Var(X)=16\\ 又4X+Y=1 \Rightarrow Var(4X+Y)=Var(1)=0 \Rightarrow 16Var(X)+Var(Y)+8Cov(X,Y)=0\\ \Rightarrow 32+8Cov(X,Y)=0 \Rightarrow Cov(X,Y)=-4,故選\bbox[red, 2pt]{(D)}
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