114年專科學力鑑定-工程數學詳解

教育部114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:$${dy\over dx} ={1\over 1+x^2} \Rightarrow \int 1\,dy= \int{1\over 1+x^2}\,dx \Rightarrow y=\tan^{-1}x+C，故選\bbox[red, 2pt]{(A)}$$

解答:$$L\{e^{-t}(t-1)^2\} =L\{e^{-t}(t^2-2t+1)\} =L\{t^2e^{-t}-2te^{-t}+e^{-t}\} ={2\over (s+1)^3}-{2\over (s+1)^2}+{1\over s+1}\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$(A)\times: f(x)=x^3+\sin x \Rightarrow f(-x)=-x^3-\sin x=-f(x) \Rightarrow 奇函數 \\(B) \times: f(x)=x^3 \cos(x) \Rightarrow f(-x) =-x^3 \cos x =-f(x) \Rightarrow 奇函數\\ (C)\times: f(x)=x^3+2 \Rightarrow f(-x) =-x^3+2 \ne -f(x)\Rightarrow 不是奇函數\\ (D) \bigcirc: f(x)=|x|+1 \Rightarrow f(-x)=|x|+1 =f(x) \Rightarrow 偶函數\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{\vec u=(2,0,-1) \\ \vec v=(-1,2,1) \\\vec w=(1,1,1)} \Rightarrow \vec u+2\vec v-\vec w =(2-2-1,0+4-1,-1+2-1) =(-1,3,0) =-\vec i+3\vec j\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{ A = \begin{bmatrix}2&-1\\3& 4 \end{bmatrix} \\[1ex] B= \begin{bmatrix}0&5\\ -2& 1 \end{bmatrix}} \Rightarrow 2A^T-B=2 \begin{bmatrix}2&3\\ -1&4 \end{bmatrix}-\begin{bmatrix}0&5\\ -2& 1 \end{bmatrix} = \begin{bmatrix}4& 1\\ 0& 7 \end{bmatrix}，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{M(x,y)=y^m-2\sin(2x^2) \\ N(x,y)= y^3+2xy+e^y} \Rightarrow \cases{M_y= my^{m-1}\\ N_x= 2y} \Rightarrow my^{m-1} =2y \Rightarrow m=2，故選\bbox[red, 2pt]{(B)}$$

解答:$$y''+2y'-3y=0 \Rightarrow \alpha^2+2\alpha-3=0 \Rightarrow (\alpha-1)(\alpha+3) =0 \Rightarrow \alpha=1,-3 \\ \Rightarrow \cases{r_1=1\\ r_2=-3} \Rightarrow r_1-r_2=4，故選\bbox[red, 2pt]{(C)}$$

解答:$$y''-3y'+2y=0 \Rightarrow \alpha^2-3\alpha+2=0 \Rightarrow (\alpha-1)(\alpha-2) =0 \Rightarrow \alpha=1,2 \\ \Rightarrow y_c=Ae^x+ Be^{2x} \Rightarrow y_p= Cx+D \Rightarrow y=y_c+y_p \Rightarrow y= Ae^x+ Be^{2x}+Cx+D，故選\bbox[red, 2pt]{(A)}$$

解答:$$L\{2*\sin(3t)\} =L\{2\} \cdot L\{ \sin(3t)\} ={2\over s} \cdot {3\over s^2+9}，故選\bbox[red, 2pt]{(D)}$$

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

解答:$$\vec u\bot \vec v \Rightarrow \vec u\cdot \vec v=0 \Rightarrow (1,-2m,m) \cdot(1,1,m)=1-2m+m^2=(m-1)^2=0 \Rightarrow m=1，故選\bbox[red, 2pt]{(D)}$$

解答:$$ \vec n= \vec u\times \vec v =(1,0,-1) \times (-1,2,2) =(2,-1,2) \Rightarrow \vec w \parallel \vec n \Rightarrow {a\over 2} ={b\over -1}={c\over 2} =k\\ \Rightarrow \cases{a=2k\\ b=-k\\ c=2k} \Rightarrow \sqrt{a^2+b^2+ c^2} =3 \Rightarrow 4k^2+k^2+4k^2=9 \Rightarrow k=\pm 1\\ \Rightarrow (a,b,c)=(2,-1,2)或(-2,1,-2) \Rightarrow |a+b+c|=3，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{\vec u=(1,0,-1) \\ \vec v=(-1,1,1) \\ \vec w =(1,1,1)} \Rightarrow \vec v\times \vec w=(0,2,-2) \Rightarrow \vec u\cdot (\vec v\times \vec w)=(1,0,-1) \cdot (0,2,-2) =2，故選\bbox[red, 2pt]{(C)}$$

解答:$$B= \begin{bmatrix}3\\ 4 \end{bmatrix} \Rightarrow BB^T =\begin{bmatrix}3\\ 4 \end{bmatrix} \begin{bmatrix} 3& 4\end{bmatrix} = \begin{bmatrix}9& 12\\ 12& 16 \end{bmatrix} \Rightarrow BB^T+A= \begin{bmatrix}9& 12\\ 12& 16 \end{bmatrix}+ \begin{bmatrix}1&-1\\0& 2 \end{bmatrix} = \begin{bmatrix}10&11\\ 12& 18 \end{bmatrix}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$\begin{bmatrix}1&-2 &1 \\ 0& 2&-8\\ 5& 0& -5 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix}0\\8\\ 10 \end{bmatrix} \Rightarrow \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}= \begin{bmatrix}1&-2 &1 \\ 0& 2&-8\\ 5& 0& -5 \end{bmatrix}^{-1}  \begin{bmatrix}0\\8\\ 10 \end{bmatrix}  \\=\left[\begin{matrix}- \frac{1}{6} & - \frac{1}{6} & \frac{7}{30}\\- \frac{2}{3} & - \frac{1}{6} & \frac{2}{15}\\- \frac{1}{6} & - \frac{1}{6} & \frac{1}{30}\end{matrix}\right] \begin{bmatrix}0\\8\\ 10 \end{bmatrix} =\begin{bmatrix}1\\0\\ -1 \end{bmatrix}  \Rightarrow x_1+x_2+ x_3= 1+0-1=0，故選\bbox[red, 2pt]{(B)}$$

解答:$$[A\mid I] = \left[ \begin{array}{rr|rr}3&-2 &1&0\\ 1& 4& 0& 1 \end{array} \right] \xrightarrow{R_1-3R_2\to R_1}  \left[ \begin{array}{rr|rr}0 & -14 & 1 & -3\\1 & 4 & 0 & 1 \end{array} \right] \xrightarrow{R_1 \leftrightarrow R_2} \left[ \begin{array}{rr|rr}1 & 4 & 0 & 1 \\0 & -14 & 1 & -3  \end{array} \right] \\ \xrightarrow{ R_2/(-14) \to R_2} \left[ \begin{array}{rr|rr}1 & 4 & 0 & 1\\0 & 1 & - \frac{1}{14} & \frac{3}{14} \end{array} \right] \xrightarrow{R_1-4R_2 \to R_1} \left[ \begin{array}{rr|rr}1 & 0 & \frac{2}{7} & \frac{1}{7}\\0 & 1 & - \frac{1}{14} & \frac{3}{14} \end{array} \right] \\ \Rightarrow A^{-1} = \begin{bmatrix}   \frac{2}{7} & \frac{1}{7}\\  - \frac{1}{14} & \frac{3}{14}\end{bmatrix}，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{P(x,y)=3x^2+y^2 \\ Q(x,y) =-xy} \Rightarrow \cases{P_y=2y\\ Q_x=-y} \Rightarrow u'={P_y-Q_x\over Q}u ={3y\over -xy}u=-{3\over x}u \\ \Rightarrow u'=-{3\over x}u \Rightarrow \int {1\over u}\,du=\int-{3\over x}\,dx \Rightarrow \ln u=-3\ln x \Rightarrow 積分因子u=x^{-3}，故選\bbox[red, 2pt]{(C)}$$

解答:$${dy\over dx}-y=2xe^x \Rightarrow e^{-x} {dy\over dx}-ye^{-x}=2x  \Rightarrow \left( ye^{-x} \right)' =2x \Rightarrow ye^{-x} = \int 2x\,dx =x^2+C \\ \Rightarrow y=x^2e^{x} +Ce^{x}\Rightarrow y(0)=C \Rightarrow y=x^2 =1 \Rightarrow y=x^2e^x+e^x \Rightarrow y(1)=2e，故選\bbox[red, 2pt]{(D)}$$

解答:$$L^{-1} \left\{ {s+4\over (s-2)(s+1)}\right\} =L^{-1} \left\{ {2\over s-2}-{1\over s+1}\right\} =2e^{2t}-e^{-t}，故選\bbox[red, 2pt]{(A)}$$

解答:$$A\vec v= \lambda \vec v \Rightarrow  \begin{bmatrix}2& a\\ 3& 4 \end{bmatrix} \begin{bmatrix}-1\\2 \end{bmatrix}  ={5\over 2} \begin{bmatrix}-1\\ 2 \end{bmatrix} \Rightarrow \begin{bmatrix}-2+2a\\ 5 \end{bmatrix} = \begin{bmatrix}-5/2\\ 5 \end{bmatrix} \Rightarrow -2+2a=-{5\over 2} \Rightarrow a=-{1\over 4}\\，故選\bbox[red, 2pt]{(B)}$$

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