113年專科學力鑑定-工程數學詳解

教育部113年自學進修專科學校學力鑑定考試試題本

專業科目(一)： 工程數學

解答:$$\cases{P(x,y)=2y^2+\cos x+1\\ Q(x,y)= 2y(x-2)} \Rightarrow \cases{P_y=4y\\ Q_x=2y} \Rightarrow u'={P_y-Q_x\over Q}u = {1\over x-2}u \\ \Rightarrow u'={1\over x-2}u \Rightarrow {1\over x-2}u'-{1\over (x-2)^2}u=0 \Rightarrow \left( {u\over x-2}\right)' =0 \\ \Rightarrow 積分因子u =x-2，故選\bbox[red, 2pt]{(B)}$$


解答:$$y'',y',y的係數都必須是常數，故選\bbox[red, 2pt]{(C)}$$

解答:$$y''-4y'+4y=0 \Rightarrow \lambda^2-4\lambda+4=0 \Rightarrow (\lambda-2)^2=0 \Rightarrow \lambda=2\\ \Rightarrow y_h=c_1e^{2x} +c_2xe^{2x} \Rightarrow y_p=Ax^2e^{2x} \Rightarrow y_p'=2Axe^{2x}+2Ax^2 e^{2x}\\ \Rightarrow y_p''= 2Ae^{2x}+ 8Axe^{2x}+ 4Ax^2e^{2x} \Rightarrow y_p''-4y_p'+4y_p= 2Ae^{2x}=2e^{2x} \Rightarrow A=1\\ \Rightarrow y_p=x^2e^{2x} \Rightarrow y=y_h+y_p =c_1e^{2x} +c_2xe^{2x} +x^2e^{2x},故選\bbox[red, 2pt]{(D)}$$

解答:$$y''+6y'+13y=0 \Rightarrow \lambda^2+6 \lambda+13=0 \Rightarrow \lambda=-3\pm 2i \Rightarrow y=e^{-3x}(A\cos 2x+B\sin2 x),故選\bbox[red, 2pt]{(A)}$$

解答:$$(A) \times:\cases{P(x,y)=2xy^2-5x\\ Q(x,y)=4y^4+x^3+3x+5y} \Rightarrow \cases{P_y=4xy\\ Q_x=3x^2+3} \Rightarrow P_y\ne Q_x \\(B)\bigcirc: \cases{P(x,y) =2\ln x+x^2y^3\\ Q(x,y)= y^2(1+x^3)} \Rightarrow P_y=3x^2y^2 =Q_x \Rightarrow \text{exact} \\(C)\times: \cases{P(x,y) =x^2y^3-3x-2y \\Q(x,y) =x^2y^2+x} \Rightarrow \cases{P_y=3x^2y^2-2\\ Q_x=2xy^2+1} \Rightarrow P_y\ne Q_x \\(D)\times: \cases{P(x,y)= \cos x^2-2y\sin x\\ Q(x,y)=x\sin y+x^2y} \Rightarrow \cases{P_y= -2\sin x\\ Q_x=\sin y+2xy} \Rightarrow P_y\ne Q_x\\ 故選\bbox[red, 2pt]{(B)}$$

解答:$$y''-{5\over x}y'+{8\over x^2}y=0 \Rightarrow x^2y''-5xy'+8y=0 \\令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \Rightarrow x^2y''-5xy'+8y=(m^2-6m+8)x^m=0 \Rightarrow m^2-6m+8=0 \Rightarrow (m-4)(m-2)=0\\ \Rightarrow m=2,4 \Rightarrow y=c_1x^2+c_2x^4, 故選\bbox[red, 2pt]{(B)}$$

解答:$$L^{-1}\left\{ {4s-2\over s^2-6s+13}\right\} =L^{-1}\left\{ {4(s-3)+10\over (s-3)^2+4} \right\} =4L^{-1}\left\{ {s-3\over (s-3)^2+4} \right\} +5L^{-1}\left\{ {2\over (s-3)^2+4} \right\} \\=4e^{3t} \cos 2t+5e^{3t}\sin 2t,故選\bbox[red, 2pt]{(C)}$$

解答:$$L \left\{ f(t)*g(t)\right\} =L \left\{ f(t) \right\} L \left\{ g(t) \right\} =L \left\{ t^2e^t\right\}L \left\{ \sin 3t\right\} =-\frac{\text{d^2}}{\text{d}s^2} L\{e^t\} \cdot {3\over s^2+3^2}\\=-\frac{\text{d^2}}{\text{d}s^2}\left({1\over s-1} \right) \cdot {3\over s^2+ 9} ={2\over (s-1)^3}\cdot {3\over s^2+ 9}={6 \over (s-1)^3(s^2+9)},故選\bbox[red, 2pt]{(D)}$$

解答:$$(A)\bigcirc: f(x)=x^2\sin{n\pi x\over \ell} 為奇函數 \Rightarrow \int_{-\ell}^\ell f(x)\,dx =0 \\ (B)\times: f(x)=x^2 \cos {n\pi x\over \ell}為偶函數 \Rightarrow \int_{-\ell}^\ell f(x)\,dx \ne0 \\(C)\times: f(x)=x^2 \cos {n\pi x\over \ell} \sin {n\pi x\over \ell}={1\over 2}x^2 \sin{2n\pi x\over \ell}為奇函數 \Rightarrow \int_{-\ell}^\ell f(x)\,dx =0 \\(D)\times: f(x)=x^2 \cos^2{n\pi x\over \ell} 為偶函數\Rightarrow \int_{-\ell}^\ell f(x)\,dx \ne0\\故選\bbox[red, 2pt]{(A)}$$

解答:$$  f(x) 為偶函數 \Rightarrow \cases{b_n=0\\ a_n\ne 0\\ f(-x)= f(x)}，故選\bbox[red, 2pt]{(D)}$$

解答:$$  \cases{\vec u=(a,-5,7) \\ \vec v=(6,b,-3)\\ \vec w-(8, 6,c)} \Rightarrow \vec u+\vec v-2\vec w=(a+6-16,-5+b-12,7-3-2c)\\ =(a-10,b-17,4-2c)=(2,-10,-4) \Rightarrow \cases{a=12\\ b=7\\ c=4} \Rightarrow a-b-c=12-7-4=1，故選\bbox[red, 2pt]{(C)}\\ 公布的答案是\bbox[cyan,2pt]{(A)}$$


解答:$$\cases{\vec a=(4,5,3)\\ \vec b=(t,3,4)} \Rightarrow \cos \theta= {\vec a\cdot \vec b\over |\vec a||\vec b|} ={4t+15+12\over \sqrt{4^2+5^2+3^2}\cdot \sqrt{t^2+3^2+4^2}} ={4t+27\over \sqrt{50} \cdot \sqrt{t^2+25}} ={7\over 50} \\ \Rightarrow 7\sqrt{t^2+25} =\sqrt{50}(4t+27) \Rightarrow 49(t^2+25) =50(4t+27)^2 \Rightarrow 751t^2+10800t+35225=0\\ \Rightarrow (t+5)(751t+7045)=0 \Rightarrow t=-5,-{7045\over 751},故選\bbox[red, 2pt]{(A)}$$

解答:$$(6,8,9)\times (1,-2,4)=(8\cdot 4+2\cdot 9, 9\cdot 1-4\cdot 6, (-2)\cdot 6-1\cdot 8) =(50, -15,-20) ,故選\bbox[red, 2pt]{(D)}$$

解答:$$$$
解答:$$(\vec u\times \vec v)\cdot \vec w =(19,3,-7) \cdot (6,2,9)=57,故選\bbox[red, 2pt]{(C)}$$

解答:$$2A+B=\begin{bmatrix}2a & 2 \\6 & 4 \end{bmatrix} +\begin{bmatrix}3 & 1 \\2 & 6 \end{bmatrix} =\begin{bmatrix}2a+3 & 3 \\8 & 10 \end{bmatrix} \\ \Rightarrow (2A+B)C=  \begin{bmatrix}2a+3 & 3 \\8 & 10 \end{bmatrix}  \begin{bmatrix}1 & 2 \\3 &b \end{bmatrix} =\begin{bmatrix}2a+12 & 4a+6+3b \\38 & 10b+16 \end{bmatrix} =\begin{bmatrix}14 & 19 \\38 & 46 \end{bmatrix} \\ \Rightarrow \cases{a=1\\b=3} \Rightarrow b-a=2,故選\bbox[red, 2pt]{(D)}$$


解答:$$(A)\times: \cases{A是2\times 3\\ B是3\times 2},兩者不能相加 \\(B)\times: \cases{A是2\times 3 \Rightarrow A^T是3\times 2\\ B是3\times 2} ,兩者不能相乘 \\(C)\times: \cases{A是2\times 3\\ B是3\times 2 \Rightarrow B^T是2\times 3},兩者不能相乘  \\(D) \bigcirc: \cases{A是2\times 3 \Rightarrow AA^T是2\times 2\\ B是3\times 2 \Rightarrow B^TB是2\times 2},兩者可以相加 \\故選\bbox[red, 2pt]{(D)}$$

解答:$$A=\begin{bmatrix}3 & 1& 4 \\2 & 1& 3\\ 1& 0 & 1 \end{bmatrix} \xrightarrow{R_1-3R_1\to R_1, R_2-2R_3\to R_2} \begin{bmatrix}0 & 1& 1 \\0 & 1& 1\\ 1& 0 & 1 \end{bmatrix} \xrightarrow{R_2-R_1\to R_2} \begin{bmatrix}0 & 1& 1 \\0 & 0& 0\\ 1& 0 & 1 \end{bmatrix} \\ \xrightarrow{R_1 \leftrightarrow R_3} \begin{bmatrix}1& 0 & 1\\0 & 0& 0\\0 & 1& 1   \end{bmatrix} \xrightarrow{R_2\leftrightarrow{R_3}} \begin{bmatrix}1& 0 & 1\\0 & 1& 1\\0 & 0& 0   \end{bmatrix} \Rightarrow rank(A)=2,故選\bbox[red, 2pt]{(B)}  $$

解答:$$\begin{vmatrix}1& 3& 4\\ 2& -1 & 1\\ 1& 2& t \end{vmatrix} =0 \Rightarrow 21-7t=0 \Rightarrow t=3,故選\bbox[red, 2pt]{(D)}$$

解答:$$\det(A-\lambda I)=\begin{vmatrix}3-\lambda & b\\ a& 5-\lambda \end{vmatrix} =\lambda^2-8\lambda+15-ab=0有二重根 \\ \Rightarrow 判別式64-4(15-ab)=0 \Rightarrow ab=-1,故選\bbox[red, 2pt]{(A)}$$


解答:$$AB=I \Rightarrow B=A^{-1} \Rightarrow B=\begin{bmatrix}4 & 7 \\5 & 9 \end{bmatrix}^{-1}= \begin{bmatrix}9 & -7 \\-5 & 4 \end{bmatrix} =\begin{bmatrix}a & b \\c & d \end{bmatrix} \\ \Rightarrow \cases{a=9\\ d=4} \Rightarrow a-d=5,故選\bbox[red, 2pt]{(C)}$$

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解題僅供參考，學力鑑定歷年試題及詳解